Ad validation guide 2013 en parte2

ADVANCE DESIGN VALIDATION GUIDE

5.52 EC2 Test 40: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1) Test ID: 5153 Test status: Passed

5.52.1 Description Verifies a square cross section column made of concrete C30/37 subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1) Nominal rigidity method - The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a symmetrically reinforced section.

The column is considered connected to the ground by a fixed connection and free to the top part.

5.52.2 Background Nominal rigidity method.

Verifies the adequacy of a rectangular cross section made from concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

5.52.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

Loadings from the structure: 15kN axial force ► 150kMm rotation moment applied to the column top ► The self-weight is neglected Exploitation loadings: ► 7kN axial force ► 100kNm rotation moment applied to the column top ►

■

■ ■ ■ ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Concrete cover 5cm Transversal reinforcement spacing a = 40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced

501

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part.

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*15+150*7=30.75kN=0.03075MN MEd=1.35*150+1.50*100=352.50kNm=0.352MNm

■

e0 =

MEd 0.352 = = 11.45m NEd 0.03075

5.52.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 11.60m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

502

2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50

ADVANCE DESIGN VALIDATION GUIDE

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 150 + 0.30 *100 + ei = + 0.30 = 10.56m N Eqp 0 15 + 0.30 * 7

N Eqp1 = 15 + 0.30 * 7 = 17.10kN M Eqp1 = N Eqp1 * e1 = 17.10 *10.56 = 180.58kNm = 0.181MNm The first order ULS moment is defined latter in this example:

The creep coefficient

ϕ (∞,t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

ϕ RH

16.8 16.8 = = 2.72 30 + 8 f cm

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

f cm > 35MPa ⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ f ⎝ cm ⎠

0.2

therefore:

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ f ⎝ cm ⎠

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

= 0.944

and

0.2

= 0.984

503

ADVANCE DESIGN VALIDATION GUIDE

h0 =

2 * Ac 2 * 500 * 500 = = 250 mm ⇒ ϕ RH u 2 * (500 + 500 )

50 ⎛ ⎞ 1− ⎜ ⎟ 100 ⎟ * 0.984 = 1.72 = ⎜1 + * 0 .

944 0.1 * 3 250 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.28 *

0.181 = 1.17 0.352 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

The second order effects; The buckling calculation:

For an isolated column, the slenderness limit check is done using the next formula:

20 * A * B * C n

λlim =

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1) Where:

n=

N Ed 0.031 = = 0.0062 Ac * f cd 0.50² * 20

A=

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.17 = 0.81

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.81 * 1.1 * 0.7 = 158 .42 0.0062

λ = 80.37 < λlim = 158.42 Therefore, the second order effects can be neglected.

Calculation of the eccentricities and solicitations corrected for ULS:

The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525MNm Therefore, we must calculate: ■ ■

504

The eccentricity of the first order ULS moment, due to the stresses applied The additional eccentricity considered for the geometrical imperfections

ADVANCE DESIGN VALIDATION GUIDE

Initial eccentricity:

e0 =

M Ed 0.3525 = = 11 .46 m N Ed 0.03075

Additional eccentricity:

ei =

l0 11.6 = = 0.03m 400 400

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

N Ed = 0.03075MN

e1 = e0 + ei = 11.49m M Ed = e1 * N Ed = 11.49 * 0.03075 = 0.353MNm M = N Ed * e0 ⎧⎪ 20mm ⎧⎪20mm ⎧ 20mm = 20mm e0 = max ⎨ h = max ⎨ 500mm = max ⎨ ⎩16.7 mm ⎪⎩ 30 ⎪⎩ 30 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

The theoretical reinforcement will be determined by the following diagram

The input parameters of the diagram are:

μ=

0.353 M Ed = = 0.141 2 b * h * f cd 0.50 * 0.502 * 20

ν=

0.03075 N Ed = = 0.00615 b * h * f cd 0.5 * 0.5 * 20 505

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

ω = 0.35 The reinforcement area will be:

∑ As =

ω * 0.502 * 20 434.78

= 40.25cm2

The total area will be 40.25cm2.

Finite elements modeling

■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

Theoretical reinforcement area(cm2)

(reference value: 19.32cm2)

506

which means 20.13cm2 per face.

ADVANCE DESIGN VALIDATION GUIDE

Theoretical value (cm2)

(reference value: 38.64 cm2)

5.52.2.3 Reference results Result name

Az R

Result description

Reference value 2

19.32 cm2

Reinforcement area [cm ] 2

38.64 cm2

Theoretical reinforcement area [cm ]

5.52.3 Calculated results

Result name Az

Result description Az

Value -19.32 cm²

Error 0.0000 %

507

ADVANCE DESIGN VALIDATION GUIDE

5.53 EC2 Test 41: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1) Test ID: 5195 Test status: Passed

5.53.1 Description Verifies a square cross section concrete column made of concrete C30/37 subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1) Nominal rigidity method.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity and then calculate the frames by considering a symmetrically reinforced section.

The column is considered connected to the ground by a fixed connection and free to the top part.

5.53.2 Background Nominal rigidity method.

Verifies the adequacy of a rectangular cross section made from concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

5.53.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

Loadings from the structure: 150kN axial force ► 15kMm rotation moment applied to the column top ► The self-weight is neglected Exploitation loadings: ► 100kN axial force ► 7kNm rotation moment applied to the column top ►

■

508

■

ψ 2 = 0,3

■ ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Concrete cover 3cm and 5cm Transversal reinforcement spacing a=40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part.

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd = 1.35*150+1.50*100 = 352.50kN = 0.035MN MEd = 1.35*15+1.50*7 = 30.75kNm = 0.03075MNm

■

e0 =

MEd 0.03705 = = 0.087m NEd 0.353

5.53.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 11.60m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50

509

ADVANCE DESIGN VALIDATION GUIDE

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 15 + 0.30 * 7 + ei = + 0.30 = 0.125m N Eqp 0 150 + 0.30 *100

N Eqp1 = 150 + 0.30 *100 = 180kN M Eqp1 = N Eqp1 * e1 = 180 * 0.125 = 22.50kNm = 0.0225MNm The first order ULS moment is defined latter in this example:

M Ed 1 = 0.041MNm The creep coefficient

ϕ (∞,t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

16.8 16.8 = = 2.72 30 + 8 f cm

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

f cm

⎛ 35 ⎞ ⎟⎟ > 35MPa therefore: α1 = ⎜⎜ f ⎝ cm ⎠

0.7

(for t0= 28 days concrete age).

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ ⎟⎟ = 0.944 and α 2 = ⎜⎜ f ⎝ cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.2

= 0.984

50 ⎞ ⎛ 1− ⎟ ⎜ 2 * Ac 2 * 500 * 500 100 * 0.944 ⎟ * 0.984 = 1.72 h0 = = = 250 mm ⇒ ϕ RH = ⎜1 + 2 * (500 + 500 ) u 0.1 * 3 250 ⎟ ⎜ ⎟ ⎜ ⎠ ⎝ 510

ADVANCE DESIGN VALIDATION GUIDE

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.28 *

0.0225 = 1.25 0.041 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

The second order effects; The buckling calculation:

For an isolated column, the slenderness limit verification is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

n=

N Ed 0.353 = = 0.071 Ac * f cd 0.50² * 20

A=

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.25 = 0.80

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.80 * 1.1 * 0.7 = 46 .24 0.071

λ = 80.37 > λlim = 46.24 Therefore, the second order effects must be taken into account.

Calculation of the eccentricities and solicitations corrected for ULS:

The stresses for the ULS load combination are: ■ NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN ■ MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm Therefore, we must calculate: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

e0 =

M Ed 0.03075 = = 0.087 m N Ed 0.353

Additional eccentricity:

ei =

l0 11.6 = = 0.03m 400 400

511

ADVANCE DESIGN VALIDATION GUIDE

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

N Ed = 0.353MN e1 = e0 + ei = 0.117m M Ed = e1 * N Ed = 0.117 * 0.353 = 0.041MNm M = N Ed * e0 ⎧⎪ 20mm ⎧⎪20mm ⎧ 20mm = 20mm e0 = max ⎨ h = max ⎨ 500mm = max ⎨ ⎩16.7 mm ⎪⎩ 30 ⎪⎩ 30 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concrete section will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum percentage area.

The reinforcement will be determined using a compound bending with compressive stress. The determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:

h 0.50 ⎞ ⎛ M ua = M G 0 + N * (d − ) = 0.041 + 0.353 * ⎜ 0.45 − ⎟ = 0.112 MNm 2 2 ⎠ ⎝ Verification about the partially compressed section:

μ BC = 0,8 *

μ cu =

h h 0,50 0,50 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,494 d d 0,45 0,45

M ua 0.112 = = 0.055 bw * d ² * f cd 0.50 * 0.45² * 20

μcu = 0.055 < μBC = 0.494 therefore the section is partially compressed

512

ADVANCE DESIGN VALIDATION GUIDE

The calculation for the tensioned steel in pure bending:

μcu = 0.055

[

]

αu = 1,25 * 1 − (1 − 2 * 0,055) = 0,071 zc = d * (1 − 0,4 * α u ) = 0,45 * (1 − 0,4 * 0,071) = 0,437m A =

M ua 0,112 = = 5,89cm ² zc * f yd 0,437 * 434,78

The calculation for the compressed steel in bending:

For the compound bending:

A = A−

N 0.353 = 5.89 * 10 − 4 − = −2.23cm 2 Fyd 434.78

The minimum column percentage reinforcement must be considered:

As , min =

0,10 * N Ed 0.10 * 0.353 = = 0.81cm ² Fyd 434 .78

Therefore, a 5cm2 reinforcement area will be considered.

5.53.2.3 Calculation of the second order effects: Estimation of the nominal rigidity:

It is estimated the nominal rigidity of a post or frame member from the following formula:

EI = K c * Ecd * I c + K s * Es * I s Where:

Ecd =

Ecm 1.2

f cm = f ck + 8Mpa = 38Mpa ⎛f ⎞ Ecm = 22000 * ⎜ cm ⎟ ⎝ 10 ⎠

Ecd = Ic =

0.3

⎛ 38 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 32836.57 Mpa

Ecm 32837 = = 27364Mpa 1.2 1.2

b * h 3 0.50 4 = = 5,208 .10 − 3 m 4 12 12 inertia of the concrete section only

Es = 200000Mpa Is

: Inertia

ρ=

As 5.10−4 = = 0.002 Ac 0.50 * 0.50

513

ADVANCE DESIGN VALIDATION GUIDE

0.002 ≤ ρ =

f ck = 20

k1 = n=

As < 0.01 Ac

30 = 1.22 20

N Ed 0.353 = = 0.071 Ac * f cd 0.50² * 20

k2 = n *

λ 170

= 0.071*

80.37 = 0.0336 ≤ 0.20 170

2

2

A ⎛h 5 * 10 −4 ⎛ 0.50 ⎞ ⎞ Is = 2 * s * ⎜ − c ⎟ = 2 * *⎜ − 0.05 ⎟ = 2.10 − 5 m 4 2 ⎝2 2 ⎠ ⎝ 2 ⎠

Ks = 1

and

Kc =

k1 * k 2 1.22 * 0.0336 = = 0.018 1 + ϕ ef 1 + 1.25

Therefore:

EI = 0.018 * 27364 * 5,208 *10−3 + 1 * 200000 * 2 *10−5 = 6.56MNm² Stress correction:

The total moment, including second order effects, is defined as a value plus the moment of the first order:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

M 0 Ed = 0.041MNm

(moment of first order (ULS) taking into account geometric imperfections)

N Ed = 0.353MN (normal force acting at ULS).

In addition:

β=

β=

π² c0

π² 8

and

because the moment is constant (no horizontal force at the top of post).

= 1.234

NB = π ² *

514

c0 = 8

EI 6.56 =π²* = 0.48 MN 2 l0 11.60²

ADVANCE DESIGN VALIDATION GUIDE

It was therefore a moment of 2nd order which is:

M Ed

⎡ ⎤ ⎢ 1.234 ⎥ = 0.041 * ⎢1 + ⎥ = 0.182MNm 0.48 ⎢ − 1⎥ ⎣ 0.353 ⎦

There is thus a second order moment of 0.182MNm Calculation of the flexural combined reinforcement

The theoretical reinforcement will be determined by the following diagram:

M Ed = 0.182MNm N Ed = 0.353MN

The input parameters of the diagram are:

μ=

M Ed 0.182 = = 0.073 2 b * h * f cd 0.50 * 0.502 * 20

ν=

N Ed 0.353 = = 0.071 b * h * f cd 0.5 * 0.5 * 20

Therefore:

ω = 0.11 The reinforcement area will be:

515

ADVANCE DESIGN VALIDATION GUIDE

∑A

s

=

ω * b * h * f cd

0.11* 0.502 * 20 = = 12.65cm2 434.78

f yd

This means a total of 12.65cm2 The initial calculations must be repeated by increasing the section; a 6cm2 reinforcement section will be considered.

Additional iteration:

One more iteration by considering an initial section of 6.5cm ² Estimation of the nominal rigidity:

EI = K c * Ecd * I c + K s * Es * I s Where:

Ecd =

Ecm 1.2

f cm = f ck + 8Mpa = 38Mpa ⎛f ⎞ Ecm = 22000 * ⎜ cm ⎟ ⎝ 10 ⎠

Ecd =

0.3

⎛ 38 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 32836.57 Mpa

Ecm 32837 = = 27364Mpa 1.2 1.2

b * h 3 0.50 4 Ic = = = 5,208 .10 − 3 m 4 12 12

considering only the concrete section only

Es = 200000Mpa Is

: Inertia

6.5 * 10−4 As ρ= = = 0.0026 Ac 0.50 * 0.50 0.002 ≤ ρ =

f ck = 20

k1 = n=

As < 0.01 Ac

30 = 1.22 20

N Ed 0.353 = = 0.071 Ac * f cd 0.50 ² * 20

k2 = n *

λ 170

= 0.071*

80.37 = 0.0336 ≤ 0.20 170

2

2

A ⎛h 6.5 * 10 −4 ⎛ 0.50 ⎞ ⎞ Is = 2 * s * ⎜ − c ⎟ = 2 * − 0.05 ⎟ = 2.6 × 10 − 5 m 4 ⎜ 2 ⎝2 2 ⎠ ⎝ 2 ⎠

Ks = 1

516

and

Kc =

k1 * k 2 1.22 * 0.0336 = = 0.018 1 + ϕ ef 1 + 1.25

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

EI = 0.018* 27364* 5,208*10−3 + 1* 200000* 2.6 *10−5 = 7.78MNm² Stress correction:

The total moment, including second order effects, is defined as a value plus the moment of the first order:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

M 0 Ed = 0.041MNm

(moment of first order (ULS) taking into account geometric imperfections)

N Ed = 0.353MN (normal force acting at ULS).

In addition:

β=

β=

π² c0

π² 8

and

c0 = 8

because the moment is constant (no horizontal force at the top of post).

= 1.234

NB = π ² *

EI 7.78 =π²* = 0.57 MN 2 l0 11 .60²

It was therefore a moment of 2nd order which is:

M Ed

⎡ ⎤ ⎢ 1.234 ⎥ = 0.041 * ⎢1 + ⎥ = 0.123MNm 0.57 ⎢ − 1⎥ ⎣ 0.353 ⎦

There is thus a second order moment of 0.123MNm

517

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the flexural compound reinforcement

The theoretical reinforcement will be determined, from the following diagram:

M Ed = 0.123MNm N Ed = 0.353MN

The input parameters of the diagram are:

μ=

0.123 M Ed = = 0.049 2 b * h * f cd 0.50 * 0.502 * 20

ν=

0.353 N Ed = = 0.071 b * h * f cd 0.5 * 0.5 * 20

Therefore:

ω = 0.05 The reinforcement area will be:

∑ As =

ω * b * h * f cd f yd

=

0.05 * 0.502 * 20 = 5.75cm2 434.78

This means a total of 5.75cm2

518

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

Theoretical reinforcement area(cm2)

(reference value: 5.75cm2=2*2.88cm2)

Theoretical value (cm2)

(reference value: 6.02 cm2)

5.53.2.4 Reference results Result name

Result description

Reference value

Az

Reinforcement area [cm2]

3.01 cm2

R

Theoretical reinforcement area [cm2]

6.02 cm2

5.53.3 Calculated results Result name Az

Result description Az

Value -3.01 cm²

Error -0.0000 %

519

ADVANCE DESIGN VALIDATION GUIDE

5.54 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinear stress-strain diagram (Class X0) Test ID: 5213 Test status: Passed

5.54.1 Description Verifies a rectangular cross section beam made of concrete C30/37 subjected to eccentric loading - Bilinear stressstrain diagram (Class X0).

The verification of the bending stresses at ultimate limit state is performed.

Simple Bending Design for Ultimate Limit State During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.

- Support at start point (x = 0) fixed connection - Support at end point (x = 5.00) fixed connection

520

ADVANCE DESIGN VALIDATION GUIDE

5.55 EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stressstrain diagram (Class XC1) Test ID: 5225 Test status: Passed

5.55.1 Description Verifies the adequacy of a rectangular cross section beam made of concrete C25/30 supporting a balcony - Bilinear stress-strain diagram (Class XC1).

Simple Bending Design for Ultimate Limit State Verifies the column resistance to rotation moment along its length. During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.

5.55.2 Background Simple Bending Design for Ultimate Limit State Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist rotation moment along its length. During this test, the calculation of stresses is performed, along with the calculation of the longitudinal and transversal reinforcement.

5.55.2.1 Model description ■ ■ ■ ■

Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; Analysis type: static linear (plane problem); Element type: linear.

The geometric dimension of the beam are:

The following load cases and load combination are used: ■ ■ ■ ■ ■ ■ ■ ■ ■

Concrete type: C25/30 Reinforcement type: S500B Exposure class: XC1 Balcony load: 1kN/m2 The weight of the beam will be considered in calculation Concrete density: 25kN/m3 The beam is considered fixed at both ends Concrete cover: 40mm Beam length: 4.00m

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■

Height: h = 0.75 m, Width: b = 0.25 m, Length: L = 4.00 m, Section area: A = 0.1875 m2 , Concrete cover: c=4cm

Boundary conditions

The boundary conditions are described below: ■

Outer: Support at start point (x=0) fixed connection, Support at end point (x = 4.00) fixed connection Inner: None.

► ►

■

5.55.2.2 Reference results in calculating the concrete beam The ULS load calculation:

The first step of the calculation is to determine charges transmitted to the beam: ■ Vertical loads applied to the beam (kN/ml) from the load distribution over the balcony ■ Rotation moment applied to the beam (kN/ml) from the load distribution over the balcony Each action (self-weight and distributed load) is determined by summing the resulting vertical loads along the eaves and the torsional moment by multiplying the resultant by the corresponding lever arm.

CAUTION, different lever arm must be considered from the center of the beam (by adding therefore the half-width).

The results are displayed in the table below:

Load calculation:

From previously calculated results, the following stresses can be determined:

522

VEd

■

Shear:

■

Bending moment:

■

Torque:

TEd

M Ed

ADVANCE DESIGN VALIDATION GUIDE

Shear and bending moment:

One can determine the load at ULS taken over by the beam:

Pu = 1,35 * 13.44 + 1,5 * 2 = 21.14 KN / m For a beam fixed on both ends, the following values will be obtained: Maximum shear (ULS):

VEd =

Pu * l 21,14 * 4 = = 42,3KN = 0,042MN 2 2

Bending moment at the supports:

M Ed =

Pu * l ² 21,14 * 4² = = 28,2 KNm = 0,028MNm 12 12

Maximum Moment at middle of span:

M Ed =

Pu * l ² 21,14 * 4² = = 14,1KNm = 0,014MNm 24 24

Torsion moment:

For a beam subjected to a torque constant:

mtu = 1,35 * 8,59 + 1,5 * 2,25 = 14,97KNm/ m = 0,015MNm/ m l 4 TEd = mtu * = 0,015 * = 0,03MNm 2 2 5.55.2.3 Bending rebar Span reinforcement (at bottom fiber)

μcu =

0,014 = 0,007 0,25 * 0.70² * 16.67

(

)

αu = 1,25 * 1 − 1 − 2 * 0,007 = 0,0088 zc = 0.70 * (1 − 0,4 * 0,0088 ) = 0,697 m

Au =

0,014 = 4.62 * 10 − 4 m ² = 0.46cm ² 0,697 * 434,78

523

ADVANCE DESIGN VALIDATION GUIDE

Minimum reinforcement percentage verification:

As , min

Cracking matrix required (calculation hypothesis):

As , min

f ct , eff ⎧ * bw * d ⎪0.26 * = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

f ct , eff = f ctm = 2.56 Mpa

f ct , eff ⎧ 2.56 * bw * d = 0.26 * * 0.25 * 0.70 = 2.27cm² ⎪0.26 * = Max⎨ = 2.27cm² f yk 500 ⎪ 0.0013 * bw * d = 0.0013 * 0.25 * 0.70 = 2.27cm² ⎩

Therefore, it retains 2.27 cm ².

Reinforcement on supports (at top fiber)

μcu =

0,028 = 0,014 0,25 * 0.70² * 16.67

(

)

αu = 1,25 * 1 − 1 − 2 * 0,014 = 0,018 zc = 0.70 * (1 − 0,4 * 0,018 ) = 0,695m

Au =

0,028 = 9.27 * 10 − 4 m ² = 0.93cm ² 0,695 * 434,78

It also retains 2.27 cm ² (minimum percentage).

Shear reinforcement

VEd = 0,042MN The transmission to the support is not direct; it is considered a connecting rod inclined by 45˚, therefore

cot θ = 1

Concrete rod verification:

VRd , max = bw * z * v * f cd *

cot θ + cot α 1 + cot ²θ According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.2.3(4)

z = zc = 0.695m (from the design in simple bending of support) Vertical frames:

25 ⎤ ⎡ v = 0,6 * ⎢1 − = 0.54 ⎣ 250 ⎥⎦ VRd , max = bw * zc * v * f cd *

VRd , max =

cot θ + cot α bw * zc * v * f cd 0.25 * 0.695 * 0.54 * 16 .67 = = = 0.78 MN 1 + cot ²θ tgθ + cot θ 2

v1 * f cd * zu * bw tgθ + cot θ According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.2.3(3)

VEd = 0.042MN < VRd , max = 0.78MN

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ADVANCE DESIGN VALIDATION GUIDE

Calculation of transverse reinforcement:

Asw VEd . * tgθ 0.042 = = = 1.39cm² / ml s zu * f yd 0.695 * 434.78

(over shear)

From the minimum reinforcement percentage:

Asw ≥ ρ w,min .bw . sin α s According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 9.2.2(5) With:

ρ w, min =

0,08 * f ck 0,08 * 25 = = 0.0008 500 f yk

Asw ≥ 0.0008 * 0.25 = 2cm² / ml s Therefore:

Asw ≥ 2cm ² / ml s Torsion calculation: Torsion moment was calculated before:

TEd = 0,03MNm

Torsional shear stress:

τ t ,i =

TEd 2 * tef ,i * Ak According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.3.2(1)

2 * c = 8cm ⎧⎪ tef ,i = max ⎨ A = (25 * 75) = 9.375cm = 9.375cm ⎪⎩ u 2( 25 + 75)

Ak = (25 − 9.375) * (75 − 9.375) = 1025cm² = 0.1025m² τ t,i =

TEd 0.03 = = 1.56Mpa 2 * t ef ,i * A k 2 * 0.09375 * 0.1025

Concrete verification:

Calculate the maximum allowable stress in the rods:

TRd ,max = 2 * v * α cw * f cd * Ak * tef ,i * sin θ * cos θ According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.3.2(4)

f ⎞ ⎛ v = 0,6 * ⎜1 − ck ⎟ = 0.54 ⎝ 250 ⎠ TRd , max = 2 * 0.54 *16.67 * 0.1025 * 0.09375 * 0.70 * 0.70 = 0.085MN Because of the combined share/moment effect, we must calculate:

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ADVANCE DESIGN VALIDATION GUIDE

TEd TRd , max

+

VEd VRd , max

≤ 1,0 According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.3.2(4)

0.03 0.042 + = 0.399 ≤ 1,0 0.085 0.78 Torsion longitudinal reinforcement

ΣAl =

TEd * u k * cot θ 2 * Ak * f yd According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.3.2(3)

uk = 2 * [(b − tef ) + (h − tef )] = 2 * [(25 − 9.375) + (75 − 9.375)] = 162.5cm = 1.625m

ΣAl =

0.03 *1.625 = 5.47cm² 2 * 0.1025 * 434.78

Torsion transversal reinforcement

AswT TEd 0.03 = = = 3.36cm ² / ml sT 2 * Ak * f yd * cot θ 2 * 0.1025 * 434 .78 Therefore

AswT ≥ 3,36cm² / ml for each face. sT

Finite elements modeling

■ ■ ■

526

Linear element: S beam, 11 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

ULS load combinations(kNm)

Torsional moment (Ted=29.96kNm)

Longitudinal reinforcement (5.46cm2)

Transversal reinforcement (3.36cm2/ml)

5.55.2.4 Reference results Result name

Result description

Reference value

Mx

Torsional moment [kNm]

29.96 cm2

Al

Longitudinal reinforcement [cm2]

5.46 cm2

2

Ator,y

3.36 cm2/ml

Transversal reinforcement [cm /ml]

5.55.3 Calculated results

Result name Mx

Result description Mx

Value 29.9565 kN*m

Error 0.0000 %

Al

Al

5.4595 cm²

0.0000 %

Ator,y / face

Ator,y/face

3.35969 cm²

0.0001 %

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ADVANCE DESIGN VALIDATION GUIDE

5.56 EC2 Test 47: Verifying a rectangular concrete beam subjected to a tension distributed load Bilinear stress-strain diagram (Class XD2) Test ID: 5333 Test status: Passed

5.56.1 Description Verifies a rectangular cross section beam made of concrete C25/30 subjected to a tension distributed load - Bilinear stress-strain diagram (Class XD2).

The verification of the bending stresses at ultimate stress limit and serviceability limit state is performed.

Simple Bending Design for Ultimate and Service State Limit During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.57 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - Inclined stress strain behavior law Test ID: 4522 Test status: Passed

5.57.1 Description Verifies the longitudinal reinforcement area of a beam under a linear load - inclined stress strain behavior law.

Verification is done according to Eurocodes 2 norm with French Annex.

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ADVANCE DESIGN VALIDATION GUIDE

5.58 EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram Test ID: 4976 Test status: Passed

5.58.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The objective is to verify: - The stresses results - The longitudinal reinforcement - The verification of the minimum reinforcement percentage

5.58.2 Background Simple Bending Design for Ultimate Limit State Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.58.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combinations are used: ■ ■

Loadings from the structure: G = 70 kN/m, Exploitation loadings (category A): Q = 80kN/m,

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q The objective is to verify: ■ ■ ■

The stresses results The longitudinal reinforcement The verification of the minimum reinforcement percentage

Simply supported beam

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 1.25 m, Width: b = 0.65 m, Length: L = 14 m, Section area: A = 0.8125 m2 , Concrete cover: c = 4.50 cm Effective height: d = h-(0.6*h+ebz) = 1.130 m; d’ = ebz = 0.045m

Materials properties

Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XD1 Concrete density: 25kN/m3 Stress-strain law for reinforcement: Bilinear stress-strain diagram Cracking calculation required

■

Concrete C25/30: fcd =

fck 25 = = 16,67MPa γ c 1,5

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■

2/3 fctm = 0.30 * fck = 0.30 * 25 2 / 3 = 2.56MPa

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1 ■

Steel S500 : f yd =

f yk γs

=

500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

■

⎛ f +8⎞ ⎟⎟ E cm = 22000 * ⎜⎜ ck ⎝ 10 ⎠

0.3

⎛ 25 + 8 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 31476MPa

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1 Boundary conditions

The boundary conditions are described below: ■

Outer: Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 14) restrained in translation along Z. Inner: None.

► ►

■

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*70+1.5*80=214.5*103 N/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=70+80=150*103 N/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q=70+0.3*80=94*103 N/ml

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ADVANCE DESIGN VALIDATION GUIDE

■

Load calculations:

MEd =

214.5 * 14² = 5.255 * 10 6 Nm 8

MEcq = 150

150 * 14² = 3.675 * 10 6 Nm 8

94 * 14² = 2.303 * 10 6 Nm 8

MEqp =

5.58.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity: ■

(∞, t 0 ) =

RH

* β( fcm ) * β( t 0 ) According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2

■

β( fcm ) =

16.8 fcm

16.8

=

25 + 8

= 2.925 According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.4

■

β( t 0 ) =

1

=

0.1 + t 00.20

1 0.1 + 28 0.20

= 0.488 at t0=28 days According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.5

The value of the φRH coefficient depends of the concrete quality:

■

RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 = ⎜1 + * α1 ⎟ * α 2 ⎜ 0.1 * 3 h ⎟ 0 ⎜ ⎟ ⎝ ⎠ According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

■

α1 = α 2 = 1

■

⎛ 35 If not: α1 = ⎜⎜ ⎝ fcm

fcm ≤ 35MPa

if

⎞ ⎟ ⎟ ⎠

0.7

⎛ 35 and α 2 = ⎜⎜ ⎝ fcm

⎞ ⎟ ⎟ ⎠

0.2

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c

f cm = f ck + 8Mpa = 33Mpa ⇒ α1 = α 2 = 1

■

In our case we have:

■

2 * Ac 2 * 650 * 1250 h0 = = = 427.63mm ⇒ 2 * (650 + 1250) u

50 100 = 1+ = 1.66 0.1 * 3 427.63 1−

RH

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.6 ■

(∞, t 0 ) =

RH

* β( fcm ) * β( t 0 ) = 1.66 * 2.925 * 0.488 = 2.375

The coefficient of equivalence is determined by the following formula: ■

αe =

Es E cm 1 + (∞, t 0 ) *

MEqp MEcar

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ADVANCE DESIGN VALIDATION GUIDE

Defined earlier:

M Ecar = 3675 *103 Nm M Eqp = 2303 * 103 Nm This gives:

αe =

200000 31476 1 + 2.70 *

= 15.82

2303 * 10 3 3975 * 10 3

5.58.2.3 Reference results in calculating the concrete beam reduced moment limit For the calculation of steel ULS, we consider the moment reduced limit of

μ lu = 0.372

for a steel grade 500Mpa.

Therefore, make sure to enable the option "limit μ (0.372 / S 500) " in Advance Design.

Reference reinforcement calculation at SLU:

The calculation of the reinforcement is detailed below: ■ ■

Effective height: d=0.9*h=1.125 m Calculation of reduced moment:

μcu =

M Ed 5255.25 *103 Nm = = 0,383 bw * d ² * f cd 0,65m *1.125² m 2 *16,67 MPa

μcu = 0.383 < μlu = 0.372

therefore the compressed reinforced must be resized and then the concrete

section must be adjusted.

Calculation of the tension steel section (Section A1):

The calculation of tensioned steel section must be conducted with the corresponding moment of

M Ed 1 = μlu * bw * d ² * f cd = 0.372 * 0.65 *1.125² *16.67 = 5.10 *106 Nm

[

]

[

The α value: α lu = 1.25 * 1 − (1 − 2 * μ lu ) = 1.25 * 1 − (1 − 2 * 0.372) = 0.618

■

Calculation of the lever arm zc: z lu = d * (1 − 0.4 * α lu ) = 1.130 * (1 − 0.4 * 0.618) = 0.851m

■

Calculation of the reinforcement area:

M Ed 1 5.10 * 10 −6 Nm A1 = = = 137.35cm² zlu . f yd 0.851m * 434.78MPa

532

]

■

:

ADVANCE DESIGN VALIDATION GUIDE

Compressed steel reinforcement reduction (Section As2):

Reduction coefficient:

ε sc =

3,5 3 .5 (0.618 * 1.130 − 0.045 ) = 0.00327 (α lu * d − d ' ) = 1000 * α lu * d 1000 * 0.618 * 1.130

ε sc = 0.00327 > ε yd = 0.00217 ⇒ σ sc = f yd = 434.78MPa Compressed reinforcement calculation:

As 2 =

M Ed − M Ed 1 5.255 − 5.15 = = 2.32cm ² ( d − d ' )σ sc (1.130 − 0.045) × 434 .78

The steel reinforcement condition:

A2 = As 2 .

σ sc f yd

= 2.32cm ²

Total area to be implemented:

In the lower part: As1=A1+A2=142.42 cm2 In the top part: As2=2.32 cm2 Reference reinforcement calculation at SLS:

The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinder strength of concrete at 28 days, at 0.6*fck The assumptions are:

150 * 14² = 3.675 * 10 6 Nm 8

■

The SLS moment: MEcq = 150

■

The equivalence coefficient: α e = 15.82

■ The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:

x1 = Fc =

αe * σc αe * σc + σs

.d =

15.82 * 15 × 1.125 = 0.419m 15.82 * 15 + 400

1 1 * b w * x1 * σ c = * 0.65 * 0.419 × 15 = 2.04 * 10 6 N 2 2

zc = d −

x1 0.419 = 1.125 − = 0.985m 3 3

Mrb = Fc * z c = 2.04 * 10 6 * 0.985 = 2.01 * 10 6 Nm Therefore the compressed reinforced established earlier was correct.

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ADVANCE DESIGN VALIDATION GUIDE

Theoretical section 1 (tensioned reinforcement only)

M 1 = M rb = 2.01*106 Nm

α1 =

x1 0.419 = = 0.372 d 1.125

zc = d − A1 =

x1 0.419 = 1.125 − = 0.985m 3 3

M1 2.01 = = 51 .01cm ² zc × σ s 0.985 × 400

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

M 2 = M ser , cq − M rb = (3.675 − 2.01) * 106 = 1.665 * 106 Nm Compressed reinforcement stresses:

σ sc = α e * σ c *

α1 * d − d ' α1 * d

σ sc = 15.82 *15 *

0.372 *1.125 − 0.045 = 211.78MPa 0.372 *1.125

Compressed reinforcement area:

A' =

M2 1.645 = = 71.92cm ² (d − d ' ) * σ sc (1.125 − 0.045 ) * 211 .78

Complementary tensioned reinforcement area:

As = A' *

211 .78 σ sc = 71.92 * = 38.08cm ² 400 σs

Section area:

Tensioned reinforcement: 51.01+38.08=89.09 cm2 Compressed reinforcement: 71.92 cm2 Considering an envelope calculation of ULS and SLS, it will be obtained: Tensioned reinforcement ULS: A=141.42cm2 Compressed reinforcement SLS: A=71.92cm2 To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount of tensioned reinforcement (after ULS: Au=141.44cm2) Reference reinforcement additional iteration for calculation at SLS:

For this iteration the calculation will be started from the section of the tensioned reinforcement found when calculating the SLS: Au=141.44cm2.

For this particular value, it will be calculated the resistance obtained for tensioned reinforcement:

σs =

AELS 89.18 ×σs = × 400 = 252 MPa AELU 141 .42

This is a SLS calculation, considering this limitation: Calculating the moment resistance MRb for detecting the presence of compressed steel reinforcement:

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ADVANCE DESIGN VALIDATION GUIDE

x1 =

αe *σ c 15.82 *15 *d = *1.130 = 0.55m 15.82 *15 + 252 αe *σ c + σ s

1 1 Fc = * bw * x1 *σ c = * 0.65 * 0.55 *15 = 2.67 *106 N 2 2 zc = d −

x1 0.55 = 1.125 − = 0.94m 3 3

M rb = Fc * zc =

⎛ 1 x ⎞ 1 0.55 ⎞ ⎛ 6 * bw * x1 * σ c * ⎜⎜ d − 1 ⎟⎟ = * 0.65 * 0.55 * 15 * ⎜1.125 − ⎟ = 2.52 * 10 Nm 2 3⎠ 2 3 ⎠ ⎝ ⎝

According to the calculation above Mser,cq is grater then Mrb the compressed steel reinforcement is set.

Theoretical section 1 (tensioned reinforcement only)

M 1 = M rb = 2.52 *106 Nm

α1 =

x1 0.55 = = 0.488 d 1.125

zc = d − A1 =

x1 0.548 = 1.125 − = 0.94m 3 3

2.52 M1 = = 106 .38cm ² zc × σ s 0.94 × 252

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

M 2 = M ser , cq − M rb = (3.675 − 2.52 ) * 10 6 = 1.155 * 10 6 Nm Compressed reinforcement stresses:

σ sc = α e * σ c *

α1 *d − d ' α1 *d

σ sc = 15.82 *15 *

0.485 *1.125 − 0.045 = 217.73MPa 0.485 *1.125

Compressed reinforcement area:

A' =

M2 1.155 = = 49.12cm ² ( d − d ' ) * σ sc (1.125 − 0.045 ) * 217 .73

Complementary tensioned reinforcement area:

As = A' *

217 .73 σ sc = 49.12 * = 42.44cm ² 252 σs

Section area:

Tensioned reinforcement: 106.38+42.44=148.82 cm2 Compressed reinforcement: 49.12 cm2

535

ADVANCE DESIGN VALIDATION GUIDE

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

f ct , eff ⎧ * bw * d ⎪0.26 * = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

As , min

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where:

f ct ,eff = f ctm = 2.56 MPa

from cracking conditions

Therefore:

As , min

2.56 ⎧ * 0.65 *1.125 = 9.73 * 10− 4 m² ⎪0.26 * = max ⎨ = 9.73cm² 500 ⎪⎩ 0.0013 * 0.65 *1.125 = 9.51 * 10− 4 m²

Finite elements modeling

■ ■ ■

Linear element: S beam, 15 nodes, 1 linear element.

ULS and SLS load combinations(kNm)

Simply supported beam subjected to bending ULS (reference value: 5255kNm)

SLS –Characteristic (reference value: 3675kNm)

SLS –Quasi-permanent (reference value: 162.94kNm)

536

ADVANCE DESIGN VALIDATION GUIDE

Theoretical reinforcement area(cm2)

(reference value: A’=148.82cm2 A=49.12cm2)

Minimum reinforcement area(cm2)

(reference value: 9.73cm2)

5.58.2.4 Reference results Result name

Result description

Reference value

My,ULS

My corresponding to the 101 combination (ULS) [kNm]

5255 kNm

My,SLS,c

My corresponding to the 102 combination (SLS) [kNm]

3675 kNm

My,SLS,q

My corresponding to the 103 combination (SLS) [kNm]

2303 kNm

Az (A’)

Theoretical reinforcement area [cm2]

148.82 cm2

Az (A)

Theoretical reinforcement area [cm2]

49.12 cm2

2

9.73 cm2

Minimum reinforcement area [cm ]

Amin

5.58.3 Calculated results

Result name My

Result description

Value

Error

My USL

-5255.25 kN*m

0.0000 %

My

My SLS cq

-3675 kN*m

0.0000 %

My

My SLS pq

-2303 kN*m

0.0000 %

Az

Az

-147.617 cm²

-0.0003 %

Amin

Amin

9.79662 cm²

0.0000 %

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ADVANCE DESIGN VALIDATION GUIDE

5.59 EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram Test ID: 4970 Test status: Passed

5.59.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

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ADVANCE DESIGN VALIDATION GUIDE

5.60 EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stressstrain diagram Test ID: 4979 Test status: Passed

5.60.1 Description The purpose of this test is to verify the software results for the My resulted stresses for the USL load combination and for the results of the theoretical reinforcement area Az.

5.60.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.60.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■ ■ ■

Loadings from the structure: G = 45 kN/m Exploitation loadings (category A): Q = 37.4kN/m, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q

■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to verify: ■

The theoretical reinforcement area results

Simply supported beam

539

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics:

■ ■ ■

Beam length: 8m Concrete cover: c=3.50 cm Effective height: d=h-(0.6*h+ebz)=0.435 m; d’=ebz=0.035m

Materials properties

Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XC1 Concrete density: 25kN/m3 Reinforcement steel ductility: Class B The calculation is made considering bilinear stress-strain diagram

■

Concrete C25/30:

f cd =

f ck

γc

=

25 = 10.67MPa 1,5

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■

f ctm = 0.30 * f ck2 / 3 = 0.30 * 25 2 / 3 = 2.56 MPa According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

■

Steel S500B :

f yd =

f yk

γs

=

500 = 434.78MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ► ►

540

ADVANCE DESIGN VALIDATION GUIDE

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*45+1.5*37.4=116.85kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=45+34.7=79.7kN/ml

■

Load calculations:

M Ed =

116.85 * 8² = 934.8kNm 8

5.60.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:

hf ⎛ M btu = beff * h f * ⎜⎜ d − 2 ⎝

⎞ 0.15 ⎞ ⎛ ⎟⎟ * f cd == 1.00 * 0.15 * ⎜ 0.435 − ⎟ * 16.67 = 9kNm 2 ⎠ ⎝ ⎠

Comparing Mbtu with MEd:

M btu = 900kNm < M Ed = 934.8kNm Therefore, the concrete section is not entirely compressed; Therefore, the calculations considering the T section are required.

5.60.2.3 Reference reinforcement calculation: Theoretical section 2:

The moment corresponding to this section is:

hf ⎛ M Ed 2 = (beff − bw ) * h f * f cd * ⎜⎜ d − 2 ⎝ = 720kNm

⎞ 0.15 ⎞ ⎛ ⎟⎟ = (1 − 0.20) * 0.15 *16.67 * ⎜ 0.435 − ⎟= 2 ⎝ ⎠ ⎠

According to this value, the steel section is:

A2 =

M Ed 2 0.720 = = 46.00cm ² hf ⎞ 0.15 ⎞ ⎛ ⎛ ⎜⎜ d − ⎟⎟ * f yd ⎜ 0.435 − ⎟ * 347.8 2 ⎠ ⎝ 2 ⎠ ⎝

541

ADVANCE DESIGN VALIDATION GUIDE

Theoretical section 1:

The theoretical section 1 corresponds to a calculation for a rectangular shape beam section.

M Ed 1 = M Ed − M Ed 2 = 935 − 720 = 215kNm M Ed 1 0.215 = = 0.341 bw * d ² * Fcd 0.20 * 0.435 ² *16.67

μ cu =

For a S500B reinforcement and for a XC1 exposure class, there will be a: μ cu > μlu = 0.372 , therefore there will be no compressed reinforcement (the compression concrete limit is not exceeded because of the exposure class) There will be a calculation without considering compressed reinforcement:

[

]

[

]

α u = 1.25 * 1 − (1 − 2 * μcu ) = 1.25 * 1 − (1 − 2 * 0.341) = 0.544 z c1 = d * (1 − 0.4 * α u ) = 0.435 * (1 − 0.40 * 0.544) = 0.340m A1 =

M Ed 1 0.215 = = 14.52cm ² z c1 * f yd 0.340 * 347.78

Theoretical section 1:

In conclusion, the entire reinforcement steel area is A=A1+A2=46+14.52=60.52cm2 Finite elements modeling

■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

ULS load combinations(kNm)

Simply supported beam subjected to bending

Theoretical reinforcement area(cm2)

For Class B reinforcement steel ductility (reference value: A=60.52cm2)

5.60.2.4 Reference results Result name

Az (Class B)

Result description

Reference value 2

60.52 cm2

Theoretical reinforcement area [cm ]

5.60.3 Calculated results Result name Az

542

Result description Az

Value -60.5167 cm²

Error 0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.61 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stressstrain diagram Test ID: 4980 Test status: Passed

5.61.1 Description The purpose of this test is to verify the My resulted stresses for the USL load combination and the results of the theoretical reinforcement area Az. This test performs verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.61.2 Background Verifies the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.61.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■ ■ ■

Loadings from the structure: G = 0 kN/m (the dead load is not taken into account) Exploitation loadings (category A): Q = 18.2kN/m, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q

■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to test: ■

The theoretical reinforcement area

Simply supported beam

543

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics:

■ ■ ■

Beam length: 8m Concrete cover: c=3.50 cm Effective height: d=h-(0.6*h+ebz)=0.482 m; d’=ebz=0.035m

Materials properties

Rectangular solid concrete C25/30 and S400B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XC1 Concrete density: 25kN/m3 Reinforcement steel ductility: Class B The calculation is made considering bilinear stress-strain diagram

■

Concrete C25/30:

f cd =

f ck

γc

=

25 = 10.67 MPa 1,5

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■

f ctm = 0.30 * f ck2 / 3 = 0.30 * 25 2 / 3 = 2.56 MPa According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

■

Steel S400B :

f yd =

f yk

γs

=

400 = 347.83MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x = 0) restrained in translation along X, Y and Z, Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ► ►

544

ADVANCE DESIGN VALIDATION GUIDE

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*0+1.5*18.2=27.3kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=0+18.2=18.2kN/ml

■

Load calculations:

M Ed =

27.3 * 8² = 218.40kNm 8

M Ecq =

18.20 * 8² = 145.60kNm 8

5.61.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:

hf ⎛ M btu = beff * h f * ⎜⎜ d − 2 ⎝

⎞ 0.12 ⎞ ⎛ ⎟⎟ * f cd == 1.10 * 0.12 * ⎜ 0.482 − ⎟ *16.67 = 928kNm 2 ⎠ ⎝ ⎠

Comparing Mbtu with MEd:

M Ed = 218.40kNm < M btu = 928kNm Therefore, the concrete section is not entirely compressed; This requires a calculation considering a rectangular section of b=110cm and d=48.2cm.

Reference longitudinal reinforcement calculation:

μ cu =

M Ed 1 0.218 = = 0.051 bw * d ² * Fcd 1.10 * 0.482 ² *16.67

[

]

[

]

α u = 1.25 * 1 − (1 − 2 * μcu ) = 1.25 * 1 − (1 − 2 * 0.051) = 0.066 z c = d * (1 − 0.4 *α u ) = 0.482 * (1 − 0.40 * 0.066) = 0.469m A=

M Ed 0.218 = = 13.38cm² z c * f yd 0.469 * 347.83

Finite elements modeling

■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

545

ADVANCE DESIGN VALIDATION GUIDE

ULS load combinations(kNm)

Simply supported beam subjected to bending

Theoretical reinforcement area(cm2)

For Class B reinforcement steel ductility (reference value: A=13.38cm2)

5.61.2.3 Reference results Result name

Result description

Reference value

Az (Class B)

Theoretical reinforcement area [cm2]

13.38 cm2

5.61.3 Calculated results

Result name Az

546

Result description Az

Value -13.3793 cm²

Error 0.0002 %

ADVANCE DESIGN VALIDATION GUIDE

5.62 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stress-strain diagram Test ID: 4981 Test status: Passed

5.62.1 Description Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The purpose of the test is to verify the results using a constitutive law for reinforcement steel, on the inclined stressstrain diagram.

The objective is to verify: - The stresses results - The longitudinal reinforcement corresponding to Class A reinforcement steel ductility - The minimum reinforcement percentage

5.62.2 Background Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The purpose of this test is to verify the software results for using a constitutive law for reinforcement steel, on the inclined stress-strain diagram.

5.62.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Loadings from the structure: G = 25 kN/m (including the dead load) Exploitation loadings (category A): Q = 30kN/m,

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q ■ Reinforcement steel: Class A ■ The calculation is performed considering inclined stress-strain diagram ■ Cracking calculation required The objective is to verify: ■ ■ ■

The stresses results The longitudinal reinforcement corresponding to doth Class A reinforcement steel ductility The minimum reinforcement percentage

547

ADVANCE DESIGN VALIDATION GUIDE

Simply supported beam

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.65 m, Width: b = 0.28 m, Length: L = 6.40 m, Section area: A = 0.182 m2 , Concrete cover: c=4.50 cm Effective height: d=h-(0.6*h+ebz)=0.806 m; d’=ebz=0.045m

Materials properties

Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■

Exposure class XD3 Concrete density: 25kN/m3 Reinforcement steel ductility: Class A The calculation is performed considering inclined stress-strain diagram Cracking calculation required

■

Concrete C25/30:

f cd =

f ck

γc

=

30 = 20MPa 1,5

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■

f ctm = 0.30 * f ck2 / 3 = 0.30 * 30 2 / 3 = 2.90 MPa According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

■

⎛ f +8⎞ ⎟⎟ E cm = 22000 * ⎜⎜ ck ⎝ 10 ⎠

■

Steel S500 : f yd =

f yk γs

0.3

=

⎛ 30 + 8 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 32837MPa

500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

548

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 6.40) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ►

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*(25)+1.5*30=78.75kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=25+30=55kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q=25+0.6*30=43kNm

■

Load calculations:

M Ed =

78.75 * 6.40² = 403.20kNm 8

M Ecq =

55 * 6.40² = 281.60kNm 8

M Eqp =

43 * 6.40² = 220.16kNm 8

5.62.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.4

at t0=28 days According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.5

549

ADVANCE DESIGN VALIDATION GUIDE

The value of the φRH coefficient depends of the concrete quality:

RH

RH ⎞ ⎛ 1− ⎟ ⎜ 100 * α1 ⎟ * α 2 = ⎜1 + ⎟ ⎜ 0.1 * 3 h 0 ⎟ ⎜ ⎠ ⎝ According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

α1 = α 2 = 1 if fCM = 35MPa ⎛ 35 If not: α1 = ⎜⎜ ⎝ fcm

⎞ ⎟ ⎟ ⎠

0.7

⎛ 35 and α 2 = ⎜⎜ ⎝ fcm

⎞ ⎟ ⎟ ⎠

0.2

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c In this case, fcm = fck + 8Mpa = 38Mpa 0.7

⎛ 35 α1 = ⎜⎜ ⎝ fcm

⎞ ⎟ ⎟ ⎠

⎛ 35 α 2 = ⎜⎜ ⎝ fcm

⎞ ⎟ ⎟ ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

= 0.944 0.2

= 0.984

2 * Ac 2 * 280 * 650 h0 = = = 195.7mm ⇒ u 2 * ( 280 + 650)

RH

50 ⎞ ⎛ 1− ⎟ ⎜ 100 ⎟ * 0.984 = 1.78 ⎜ = 1+ ⎜ 0.1 * 3 195.7 ⎟ ⎟ ⎜ ⎠ ⎝

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.6

(∞, t 0 ) =

RH

* β( fcm ) * β( t 0 ) = 1.78 * 2.73 * 0.488 = 2.37

The coefficient of equivalence is determined using the following formula:

αe =

Es E cm

= MEqp

1 + (∞, t 0 ) *

MEcar

200000 = 17.40 32837 220.16 1 + 2.37 * 281.60

5.62.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD3, we will verify the section having non-compressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”: This value can be determined by the next formula if

μ luc = K (α e ) *

f ck < 50MPa valid for a constitutive law to horizontal plateau:

f ck ( 4.62 − 1.66 * γ ) * f ck + (165.69 − 79.62 * γ )

(

K (α e ) = 10−4 * a + b * α e + c * α e

2

)

The values of the coefficients "a", "b" and "c" are defined in the following table:

550

Diagram for inclined tier

Diagram for horizontal plateau

a

75,3 * f ck − 189,8

71,2 * fck + 108

b

− 5,6 * fck + 874,5

− 5,2 * fck + 847,4

c

0,04 * fck − 13

0,03 * f ck − 12,5

ADVANCE DESIGN VALIDATION GUIDE

This gives us: a = 75.3 * 25 + 189 .8 = 2069 .2

b = −5,2 * 30 + 874 .5 = 706 .5 c = 0,04 * 30 − 13 = −11.8

K (α e ) = 10 −4 (2069.2 + 706.5 * 17.60 − 11.80 * 17.60² ) = 1.087 Then:

γ=

437.76 = 1,425 307.2

μ luc = K ( α e ).

f ck = 0 . 272 ( 4 . 62 − 1 . 66 * γ ) * f ck + (165 . 69 − 79 . 62 * γ )

Calculation of reduced moment:

μ cu =

M Ed 0.403 *103 Nm = = 0,225 bw * d ² * f cd 0,50m * 0.566² m 2 * 20MPa

μ cu = 0.225 < μ luc = 0.272

therefore, there is no compressed reinforcement

Reference reinforcement calculation at SLS:

The calculation of the reinforcement is detailed below: ■ ■

Effective height: d = 0.566m Calculation of reduced moment:

μ cu = 0.225

■

Calculation of the lever arm zc:

z c = d * (1 − 0,4 *α u ) = 0,566 * (1 − 0,4 * 0,323) = 0,493m ■

Calculation of the reinforcement area:

Au = ■

M Ed 0,233 = = 11,68 *10− 4 m² = 11,68cm² zc * f yd 0,458 * 434,78

Calculation of the stresses from the tensioned reinforcement: Reinforcement steel elongation:

ε su =

1−αu

αu

* ε cu 2 =

1 − 0,323 * 3.5 = 7.35 ‰ 0,323

Tensioned reinforcement efforts:

σ su = 432,71 + 952,38 * ε su = 432,71 + 952,38 * 0.00735 = 439.71MPa ≤ 454MPa ■

Reinforcement section calculation:

Au =

M Ed 0.403 = = 18 .60 *10 − 4 m ² = 18 .60 cm ² z c * f yd 0.493 * 439 .71

551

ADVANCE DESIGN VALIDATION GUIDE

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is: f ct,eff ⎧ * bw * d ⎪0.26 * A s,min = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where: fct, eff = fctm = 2.90MPa from cracking conditions

Therefore:

As ,min

2.90 ⎧ * 0.28 * 0.566 = 2.39 * 10 −4 m² ⎪0.26 * = max ⎨ = 2.39cm² 500 ⎪⎩ 0.0013 * 0.28 * 0.566 = 2.06 * 10 −4 m²

Finite elements modeling

■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

ULS and SLS load combinations (kNm)

Simply supported beam subjected to bending

ULS (reference value: 403.20kNm)

SLS (reference value: 281.60kNm)

Theoretical reinforcement area(cm2)

For Class A reinforcement steel ductility (reference value: A=18.60cm2)

552

ADVANCE DESIGN VALIDATION GUIDE

Minimum reinforcement area(cm2)

(reference value: 52.39cm2)

5.62.2.4 Reference results Result name

Result description

Reference value

My,ULS

My corresponding to the 101 combination (ULS) [kNm]

403.20 kNm

My,SLS

My corresponding to the 102 combination (SLS) [kNm]

281.60 kNm

Az (Class A)

Theoretical reinforcement area [cm2]

18.60 cm2

Amin

Minimum reinforcement area [cm2]

2.39 cm2

5.62.3 Calculated results

Result name My

Result description My USL

Value -394.8 kN*m

Error 2.0833 %

My

My SLS

-275.733 kN*m

2.0833 %

Az

Az

-18.4844 cm²

0.6248 %

Amin

Amin

-2.38697 cm²

0.0001 %

553

ADVANCE DESIGN VALIDATION GUIDE

5.63 EC2 Test 4: Verifying a rectangular concrete beam subjected to Pivot A efforts â&#x20AC;&#x201C; Inclined stress-strain diagram Test ID: 4977 Test status: Passed

5.63.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The purpose of this test is to verify the software results for Pivot A efforts. For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied.

The objective is to verify: - The stresses results - The longitudinal reinforcement corresponding to both Class A and Class B reinforcement steel ductility - The minimum reinforcement percentage

554

ADVANCE DESIGN VALIDATION GUIDE

5.64 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1) Test ID: 5146 Test status: Passed

5.64.1 Description Verifies the adequacy of a concrete (C25/30) column with circular cross section using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1).

Simplified Method The column is considered connected to the ground by an articulated connection (all the translations are blocked). To the top part the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.

5.64.2 Background Simplified Method Verifies the adequacy of a circular cross section made from concrete C25/30.

5.64.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

■ ■ ■ ■ ■

Loadings from the structure: ► 5000kN axial force ► The self-weight is neglected Concrete cover 5cm Concrete C25/30 Steel reinforcement S500B Relative humidity RH=50% Buckling length L0=5.00m

555

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■

Radius: r = 0.40 m, Length: L = 5.00 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a articulated connection (all the translations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

Loading

The beam is subjected to the following load combinations: ■

Load combinations:

N ED = 1.35 * 5000 = 6750kN 5.64.2.2 Reference results in calculating the concrete column Scope of the method:

According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(1) L0 :buckling length L0=L=5m According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(2) i :the radius of giration of uncraked concrete section

I : second moment of inertia for circular cross sections

A : cross section area , therefore:

556

ADVANCE DESIGN VALIDATION GUIDE

The method of professional rules can be applied as: ■ ■ ■ Reinforcement calculation:

λ = 25 ≤ 60 , therefore: α=

0.84 ⎛λ ⎞ 1+ ⎜ ⎟ ⎝ 62 ⎠

2

=

0.84 ⎛ 25 ⎞ 1+ ⎜ ⎟ ⎝ 62 ⎠

2

= 0.722

k h = (0.70 + 0.5 * D ) * (1 − 8 * ρ * δ ) If the ρ and δ values are unknown, the term

will be considered:

(1 − 8 * ρ * δ ) = 0.95 , therefore: k h = (0.70 + 0.5 * D) * (1 − 6 * ρ * δ ) = (0.70 + 0.5 * 0.8) * 0.95 = 1.045 k s = 1.6 − 0.65 *

f cd = f yd =

As =

f yk 500

= 1.6 − 0.65 *

500 = 0.95 500

f ck 25 = = 16 .67 MPa 1.5 1 .5 f yk 1 .5

1 f yd

=

500 = 333 .33 MPa 1 .5

⎛ N ed ⎞ 1 6.750 ⎛ ⎞ * ⎜⎜ − π * r 2 * f cd ⎟⎟ = *⎜ − π * 0.4 2 *16.67 ⎟ = 31.14 cm ² ⎠ ⎝ kh * ks *α ⎠ 333 .33 ⎝ 1.045 * 0.95 * 0.722

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

Theoretical reinforcement area(cm2)

(reference value: 17.43cm2)

557

ADVANCE DESIGN VALIDATION GUIDE

5.64.2.3 Reference results Result name

Result description

Reference value

Ay

Reinforcement area [cm2]

17.43cm2

R

Theoretical reinforcement area [cm2]

34.86 cm2

5.64.3 Calculated results

Result name Az

558

Result description Az

Value -17.4283 cm²

Error 0.0002 %

ADVANCE DESIGN VALIDATION GUIDE

5.65 EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method - Bilinear stressstrain diagram (Class XC1) Test ID: 5211 Test status: Passed

5.65.1 Description Verifies the adequacy of a square cross section column made of concrete C30/37 subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.

Nominal curvature method.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

The column is considered connected to the ground by a fixed connection and free to the top part.

5.65.2 Background Nominal curvature method.

Verify the adequacy of a rectangular cross section made from concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

5.65.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

■

■ ■ ■ ■ ■ ■ ■

Loadings from the structure: ► 150 kN axial force ► 15 kMm rotation moment applied to the column top ► the self-weight is neglected Exploitation loadings: ► 100 kN axial force ► 7 kNm rotation moment applied to the column top

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Concrete cover 3cm and 5cm Transversal reinforcement spacing a=40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5 cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part.

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*150+1.50*100=352.50kN=0.353MN MEd=1.35*15+1.50*7=30.75kNm=0.03075MNm

■

560

ADVANCE DESIGN VALIDATION GUIDE

5.65.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 11.60m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 N Eqp 0

+ ei =

15 + 0.30 * 7 + 0.30 = 0.125m 150 + 0.30 *100

N Eqp1 = 150 + 0.30 *100 = 180kN M Eqp1 = N Eqp1 * e1 = 180 * 0.125 = 22.50kNm = 0.0225MNm The first order ULS moment is defined latter in this example:

M Ed 1 = 0.041MNm The creep coefficient

ϕ (∞,t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) β ( f cm ) = β (t0 ) =

16.8 16.8 = = 2.72 30 + 8 f cm

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

561

ADVANCE DESIGN VALIDATION GUIDE

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

f cm

⎛ 35 ⎞ ⎟⎟ > 35MPa therefore: α1 = ⎜⎜ f ⎝ cm ⎠

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ ⎟⎟ = 0.944 and α 2 = ⎜⎜ f ⎝ cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.2

= 0.984

50 ⎛ ⎞ 1− ⎜ ⎟ 2 * Ac 2 * 500 * 500 100 * 0.944 ⎟ * 0.984 = 1.72 = = 250 mm ⇒ ϕ RH = ⎜1 + h0 = u 2 * (500 + 500 ) 0.1 * 3 250 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.28 *

0.0225 = 1.25 0.041

The second order effects; The buckling calculation:

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

n=

0.353 N Ed = = 0.071 Ac * f cd 0.50² * 20

A=

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.25 = 0.80

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.80 * 1.1 * 0.7 = 46 .24 0.071

λ = 80.37 > λlim = 46.24 Therefore, the second order effects cannot be neglected.

562

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the eccentricities and solicitations corrected for ULS :

The stresses for the ULS load combination are: NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm Therefore it must be determined: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

e0 =

M Ed 0.03075 = = 0.087 m N Ed 0.353

Additional eccentricity:

ei =

l0 11.6 = = 0.03m 400 400

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

N Ed = 0.353MN e1 = e0 + ei = 0.117m M Ed = e1 * N Ed = 0.117 * 0.353 = 0.041MNm M = N Ed * e0 ⎧⎪20mm ⎧⎪ 20mm ⎧ 20mm h e0 = max ⎨ = max ⎨ 500mm = max ⎨ = 20mm ⎩16.7 mm ⎪⎩ 30 ⎪⎩ 30 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concrete section will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum percentage area.

The reinforcement is determined using a compound bending with compressive stress. The determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:

563

ADVANCE DESIGN VALIDATION GUIDE

h 0.50 ⎞ ⎛ M ua = M G 0 + N * ( d − ) = 0.041 + 0.353 * ⎜ 0.45 − ⎟ = 0.112 MNm 2 2 ⎠ ⎝ Verification about the partially compressed section:

μ BC = 0,8 *

μ cu =

h h 0,50 0,50 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,494 d d 0,45 0,45

0.112 M ua = = 0.055 bw * d ² * f cd 0.50 * 0.45² * 20

μcu = 0.055 < μBC = 0.494 therefore the section is partially compressed The calculation for the tensioned steel in pure bending:

μcu = 0.055

[

]

αu = 1,25 * 1 − (1 − 2 * 0,055) = 0,071 zc = d * (1 − 0,4 * α u ) = 0,45 * (1 − 0,4 * 0,071) = 0,437m A =

M ua 0,112 = = 5,89cm ² zc * f yd 0,437 * 434,78

The calculation for the compressed steel in bending:

For the compound bending:

The minimum column percentage reinforcement must be considered:

Therefore a 5cm2 reinforcement area will be considered

5.65.2.3 The nominal curvature method (second order effect): The curvature calculation:

Considering a reinforcement of 5cm ² (considered symmetric), one can determine the curvature from the following formula:

1 1 = K r * Kϕ * r r0 According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(1)

f yd

434.78 ε yd 1 Es = = = 200000 = 0.0107m −1 r0 (0,45 * d ) (0,45 * d ) 0,45 * 0.45

564

ADVANCE DESIGN VALIDATION GUIDE

Kr is the correction coefficient depending of the normal force:

nu − n ≤1 nu − n bal

Kr =

According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(3)

n=

ω=

N Ed 0.353 = = 0.0706 Ac * f cd 0.50² * 20

As * f yd Ac * f cd

=

5 *10−4 * 434.78 = 0.0435 0.50² * 20

nu = 1 + ω = 1 + 0.0435 = 1.0435 nbal = 0,4

Kr =

1.0435 − 0.0706 = 1.51 1.0435 − 0.40

Condition: K r

Kϕ

≤ 1 , therefore we consider: K r = 1

creep coefficient:

K ϕ = 1 + β * ϕ ef ≥ 1 According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(4)

, Therefore:

1 1 = K r * Kϕ * = 1 * 1 * 0.0107 = 0.0107 m −1 r r0 Calculation moment:

The moment of calculation is defined by the formula:

M Ed = M 0 Ed + M 2 Where:

M 0 Ed

M2

is moment of the first order including geometric imperfections.

is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

M 2 = N Ed * e2 e2 =

1 l02 11 .60 ² * = 0.0107 * = 0.144 m r c 10

M 2 = N Ed * e2 = 0.353 * 0.144 = 0.051MNm M Ed = M 0 Ed + M 2 = 0.041 + 0.051 = 0.09MNm 565

ADVANCE DESIGN VALIDATION GUIDE

The reinforcement must be sized considering the demands of the second degree effects, as follows:

Reinforcement calculation according the second order effects:

The interaction diagram will be used.

The input parameters in the diagram are:

According to the diagram, it will be obtained a minimal percentage reinforcement:

As , min =

0,10 * N Ed 0.10 * 0.353 = = 0.81cm ² ≥ 0.002 * Ac = 5cm 2 f yd 434 .78

Finite elements modeling

■ ■ ■

566

Linear element: S beam, 7 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Theoretical reinforcement area(cm2)

(reference value: Au=5cm2)

Theoretical value (cm2)

(reference value: 5 cm2)

5.65.2.4 Reference results Result name

Result description

Reference value 2

Amin

Reinforcement area [cm ]

5 cm2

R

Theoretical reinforcement area [cm2]

5 cm2

5.65.3 Calculated results

Result name Amin

Result description

Amin

Value

5 cm²

Error

0.0000 %

567

ADVANCE DESIGN VALIDATION GUIDE

5.66 EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction Inclined stress-strain diagram (Class X0) Test ID: 5232 Test status: Passed

5.66.1 Description Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Inclined stress-strain diagram (Class X0).

Tie sizing Inclined stress-strain diagram Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following rotation efforts: NEd=1.35*233.3+1.5+56.67=400kNm The boundary conditions are described below: - Support at start point (x=0) fixed connection - Support at end point (x = 5.00) translation along the Z axis is blocked

5.66.2 Background Tie sizing Bilinear stress-strain diagram / Inclined stress-strain diagram Verifies the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.

5.66.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

Loadings from the structure: Fx,G = 233.33 kN The dead load is neglected

■ ■

Exploitation loadings (category A): Fx,Q = 56.67kN The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q

Units

Metric System

568

ADVANCE DESIGN VALIDATION GUIDE

Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■ ■

Height: h = 0.15 m, Width: b = 0.15 m, Length: L = 5.00 m, Section area: A = 0.00225 m2 , Concrete cover: c=3cm Reinforcement S400, Class: B Fck=20MPa The load combinations will produce the following rotation efforts: NEd=1.35*233.3+1.5+56.67=400kNm

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) fixed connection, Support at end point (x = 5.00) translation along the Z axis is blocked Inner: None. ► ►

5.66.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then a inclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

As ,U ≥

N Ed

σ s ,U

=

0.400 = 11,50 * 10− 4 m² = 11,50cm² 400 1.15

It will be used a 4HA20=A=12.57cm2 Calculations according a inclined stress-strain diagram

S 400 − ClassB ⇒ σ s ,U = 373MPa As ,U ≥

N Ed

σ s ,U

=

0.400 = 10,72 * 10 − 4 m ² = 10,70cm ² 373

It can be seen that the gain is not negligible (about 7%).

Checking the condition of non-fragility:

As ≥ Ac *

f ctm f yk

f ctm = 0.30 * f ck2 / 3 = 0.30 * 202 / 3 = 2,21MPa

Ac = 0.15 * 0.15 = 0.0225m² As ≥ Ac *

f ctm 2.21 = 0.0225 * = 1.24cm ² f yk 400

569

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75 cm2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)

5.66.2.3 Reference results Result name

Result description

Reference value

Az,1

Longitudinal reinforcement obtained using the bilinear stress-strain diagram [cm2] longitudinal reinforcement obtained using the inclined stress-strain diagram [cm2]

5.75 cm2

Az,2

5.37 cm2

5.66.3 Calculated results

Result name Az

570

Result description

Az-ii

Value

-5.36526 cm²

Error

0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.67 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1) Test ID: 5205 Test status: Passed

5.67.1 Description Verifies a square cross section column made of concrete C30/37 subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.

Nominal curvature method.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

The column is considered connected to the ground by a fixed connection and free to the top part.

5.67.2 Background Nominal curvature method.

Verifies the adequacy of a square cross section made from concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

5.67.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■

■

Loadings from the structure: ► 15kN axial force ► 150kMm rotation moment applied to the column top ► The self-weight is neglected Exploitation loadings: ► 7kN axial force ► 100kNm rotation moment applied to the column top

■

ψ 2 = 0,3

■ ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Concrete cover 3cm and 5cm Transversal reinforcement spacing a=40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced

571

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part.

Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: ► ►

■

e0 =

NEd =1.35*15+1.50*7=30.75kN=0.03075MN MEd=1.35*150+1.50*100=352.50kNm=0.352MNm MEd 0.352 = = 11.45m NEd 0.03075

5.67.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 11.60m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

572

2 3 * l0 2 3 * 11.60 = = 80.37 0.50 a

ADVANCE DESIGN VALIDATION GUIDE

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 N Eqp 0

+ ei =

150 + 0.30 *100 + 0.30 = 10.56m 15 + 0.30 * 7

N Eqp1 = 15 + 0.30 * 7 = 17.10kN M Eqp1 = N Eqp1 * e1 = 17.10 *10.56 = 180.58kNm = 0.181MNm The first order ULS moment is defined latter in this example:

M Ed 1 = 0.3523 MNm The creep coefficient

ϕ (∞,t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( fcm ) = β (t0 ) =

16.8 16.8 = = 2.72 f cm 30 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

f cm

⎛ 35 ⎞ ⎟⎟ > 35MPa therefore: α1 = ⎜⎜ f ⎝ cm ⎠

0.7

(for t0= 28 days concrete age).

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ ⎟⎟ = 0.944 and α 2 = ⎜⎜ f ⎝ cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.2

= 0.984

573

ADVANCE DESIGN VALIDATION GUIDE

h0 =

2 * Ac 2 * 500 * 500 = = 250 mm ⇒ ϕ RH u 2 * (500 + 500 )

50 ⎛ ⎞ 1− ⎜ ⎟ 100 ⎟ * 0.984 = 1.72 = ⎜1 + * 0 .

944 0.1 * 3 250 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.28 *

0.181 = 1.17 0.352 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

The second order effects; The buckling calculation:

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

n=

0.031 N Ed = = 0.0062 Ac * f cd 0.50² * 20

A=

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.17 = 0.81

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.81 * 1.1 * 0.7 = 158 .42 0.0062

λ = 80.37 < λlim = 158.42 Therefore, the second order effects can be neglected.

574

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the eccentricities and solicitations corrected for ULS:

The stresses for the ULS load combination are: ■ NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN ■ MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525 MNm Therefore it must be determined: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

e0 =

0.3525 M Ed = = 11 .46 m 0.03075 N Ed

Additional eccentricity:

ei =

l0 11.6 = = 0.03m 400 400

The first order eccentricity: stresses correction: The forces correction, used for the combined flexural calculations:

N Ed = 0.03075MN e1 = e0 + ei = 11.49m M Ed = e1 * N Ed = 11.49 * 0.03075 = 0.353MNm M = N Ed * e0 500 mm ⎞ h ⎞ ⎛ ⎛ e0 = max ⎜ 20 mm; ⎟ = max ⎜ 20 mm; ⎟ = max( 20 mm; 16.7 mm ) = 20 mm 30 ⎠ 30 ⎠ ⎝ ⎝ According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

The theoretical reinforcement will be determined by the following diagram:

575

ADVANCE DESIGN VALIDATION GUIDE

The input parameters of the diagram are:

μ=

0.353 M Ed = = 0.141 2 b * h * f cd 0.50 * 0.502 * 20

ν=

0.03075 N Ed = = 0.00615 b * h * f cd 0.5 * 0.5 * 20

Therefore:

ω = 0.35 The reinforcement area will be:

∑ As =

ω * 0.502 * 20 434.78

= 40.25cm2

which means 20.13cm2 per face.

The total area will be 40.25cm2.

The nominal curvature methods (second order effect):

The curvature calculation: Considering a reinforcement of 40.25cm² (considered symmetric), one can determine the curvature from the following formula:

1 1 = K r * Kϕ * r r0 According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(1)

f yd

434.78 ε yd 1 Es = = = 1000 = 0.0107m −1 r0 (0,45 * d ) (0,45 * d ) 0,45 * 0.45 Kr is the correction coefficient depending of the normal force:

nu − n ≤1 nu − n bal

Kr =

According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(3)

n=

ω=

0.03075 N Ed = = 0.00615 Ac * f cd 0.50² * 20

As * f yd Ac * f cd

=

40.25 *10−4 * 434.78 = 0.350 0.50² * 20

nu = 1 + ω = 1 + 0.350 = 1.350 nbal = 0,4

Kr =

1.350 − 0.00615 = 1.41 1.350 − 0.40

Condition: K r

Kϕ

≤ 1 , therefore it will be considered: K r = 1

creep coefficient:

K ϕ = 1 + β * ϕ ef ≥ 1 According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(4)

576

ADVANCE DESIGN VALIDATION GUIDE

therefore

Calculation moment:

The moment of calculation is estimated from the formula:

M Ed = M 0 Ed + M 2 Where:

M 0 Ed

M2

is moment of the first order including geometric imperfections.

is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

M 2 = N Ed * e2 e2 =

1 l02 11 .60 ² * = 0.0107 * = 0.144 m r c 10

M 2 = N Ed * e2 = 0.03075 * 0.144 = 0.00443MNm M Ed = M 0 Ed + M 2 = 0.353 + 0.00443 = 0.357 MNm The reinforcement must be sized considering the demands of the second degree effects, as follows:

N Ed = 0.03075MN M Ed = 0.357 MNm

577

ADVANCE DESIGN VALIDATION GUIDE

Reinforcement calculation according the second order effects:

The interaction diagram will be used.

The input parameters in the diagram are:

ω = 0 .35

will be obtained, which gives:

∑A

s

=

ω * b * h * f cd f yd

Meaning a 20.13 cm2 per side (which confirms the initial section) Finite elements modeling

■ ■ ■

578

Linear element: S beam, 7 nodes, 1 linear element.

=

0.35 * 0.50² * 20 = 0 = 40.25cm² 434.78

ADVANCE DESIGN VALIDATION GUIDE

Theoretical reinforcement area(cm2)

(reference value: 40.25cm2=2*20.13cm2)

Theoretical value (cm2)

(reference value: 38.64 cm2)

5.67.2.3 Reference results Result name

Az R

Result description

Reference value 2

19.32 cm2

Reinforcement area [cm ] 2

Theoretical reinforcement area [cm ]

38.64 cm2

5.67.3 Calculated results Result name Az

Result description Az

Value -19.32 cm²

Error 0.0000 %

579

ADVANCE DESIGN VALIDATION GUIDE

5.68 EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction Bilinear stress-strain diagram (Class X0) Test ID: 5231 Test status: Passed

5.68.1 Description Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Bilinear stress-strain diagram (Class X0).

Tie sizing Bilinear stress-strain diagram Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following rotation efforts: NEd=1.35*233.3+1.5+56.67=400kNm The boundary conditions are described below: - Support at start point (x=0) fixed connection - Support at end point (x = 5.00) translation along the Z axis is blocked

5.68.2 Background Tie sizing Bilinear stress-strain diagram / Inclined stress-strain diagram Determine the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.

5.68.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used:

580

■

Loadings from the structure: Fx,G = 233.33 kN The dead load is neglected

■ ■

Exploitation loadings (category A): Fx,Q = 56.67kN The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■ ■

Height: h = 0.15 m, Width: b = 0.15 m, Length: L = 5.00 m, Section area: A = 0.00225 m2 , Concrete cover: c=3cm Reinforcement S400, Class: B Fck=20MPa The load combinations will produce the following rotation efforts: NEd=1.35*233.3+1.5+56.67=400kNm

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x = 0) fixed connection, Support at end point (x = 5.00) translation along the Z axis is blocked Inner: None. ► ►

5.68.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then a inclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

As ,U ≥

N Ed

σ s ,U

=

0.400 = 11,50 * 10− 4 m² = 11,50cm² 400 1.15

It will be used a 4HA20=A=12.57cm2 Calculations according a inclined stress-strain diagram

S 400 − ClassB ⇒ σ s ,U = 373MPa As ,U ≥

N Ed

σ s ,U

=

0.400 = 10,72 * 10 − 4 m ² = 10,70cm ² 373

It can be seen that the gain is not negligible (about 7%).

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Checking the condition of non-fragility:

As ≥ Ac *

f ctm f yk

f ctm = 0.30 * f ck2 / 3 = 0.30 * 202 / 3 = 2,21MPa

Ac = 0.15 * 0.15 = 0.0225m² As ≥ Ac *

f ctm 2.21 = 0.0225 * = 1.24cm ² f yk 400

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75 cm2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)

5.68.2.3 Reference results Result name

Result description

Reference value

Az,1

Longitudinal reinforcement obtained using the bilinear stress-strain diagram [cm2] Longitudinal reinforcement obtained using the inclined stress-strain diagram [cm2]

5.75 cm2

Az,2

5.37 cm2

5.68.3 Calculated results Result name Az

582

Result description Az-i

Value -5.75001 cm²

Error 0.0000 %

6 General Application

ADVANCE DESIGN VALIDATION GUIDE

6.1

Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238) Test ID: 4480 Test status: Passed

6.1.1 Description Performs the finite elements calculation on a model with 2 joined vertical elements with the clipping option enabled.

6.2

Verifying the precision of linear and planar concrete covers (TTAD #12525) Test ID: 4547 Test status: Passed

6.2.1 Description Verifies the linear and planar concrete covers precision.

6.3

Defining the reinforced concrete design assumptions (TTAD #12354) Test ID: 4528 Test status: Passed

6.3.1 Description Verifies the definition of the reinforced concrete design assumptions.

6.4

Verifying the synthetic table by type of connection (TTAD #11422) Test ID: 3646 Test status: Passed

6.4.1 Description Performs the finite elements calculation and the steel calculation on a structure with four types of connections.

Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. The model connections: columns base plates, beam - column fixed connections, beam - beam fixed connections and gusset plate. Live loads, snow loads and wind loads are applied.

6.5

Importing a cross section from the Advance Steel profiles library (TTAD #11487) Test ID: 3719 Test status: Passed

6.5.1 Description Imports the Canam Z 203x10.0 section from the Advance Steel Profiles library in the Advance Design list of available cross sections.

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6.6

Creating and updating model views and post-processing views (TTAD #11552) Test ID: 3820 Test status: Passed

6.6.1 Description Creates and updates the model view and the post-processing views for a simple steel frame.

6.7

Creating system trees using the copy/paste commands (DEV2012 #1.5) Test ID: 4164 Test status: Passed

6.7.1 Description Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is in the source system.

6.8

Creating system trees using the copy/paste commands (DEV2012 #1.5) Test ID: 4162 Test status: Passed

6.8.1 Description Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is on the same level as the source system.

6.9

Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102) Test ID: 4095 Test status: Passed

6.9.1 Description Creates a new .fto file using the "New" command from the "Standard" toolbar. Creates a rigid punctual support and tests the CAD coordinates.

6.10 Verifying mesh, CAD and climatic forces - LPM meeting Test ID: 4079 Test status: Passed

6.10.1 Description Generates the "Description of climatic loads" report for a model consisting of a concrete structure with dead loads, wind loads, snow loads and seism loads. Performs the model verification, meshing and finite elements calculation.

Generates the "Displacements of linear elements by element" report.

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6.11 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724) Test ID: 4361 Test status: Passed

6.11.1 Description Generates liquid pressure on a concrete structure with horizontal and vertical surfaces. Generates the loadings description report.

6.12 Changing the default material (TTAD #11870) Test ID: 4436 Test status: Passed

6.12.1 Description Selects a different default material for concrete elements.

6.13 Verifying the objects rename function (TTAD #12162) Test ID: 4229 Test status: Passed

6.13.1 Description Verifies the renaming function from the properties list of a linear element. The name contains the "_" character.

6.14 Launching the verification of a model containing steel connections (TTAD #12100) Test ID: 4096 Test status: Passed

6.14.1 Description Launches the verification function for a model containing a beam-column steel connection.

6.15 Verifying the appearance of the local x orientation legend (TTAD #11737) Test ID: 4109 Test status: Passed

6.15.1 Description Verifies the color legend display. On a model with a planar element, the color legend is enabled; it displays the elements by the local x orientation color.

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6.16 Verifying geometry properties of elements with compound cross sections (TTAD #11601) Test ID: 3546 Test status: Passed

6.16.1 Description Verifies the geometry properties of a steel structure with elements which have compound cross sections (CS4 IPE330 UPN240). Generates the geometrical data report.

6.17 Verifying material properties for C25/30 (TTAD #11617) Test ID: 3571 Test status: Passed

6.17.1 Description Verifies the material properties on a model with a concrete (C25/30) bar. Generates the material description report.

6.18 Verifying element creation using commas for coordinates (TTAD #11141) Test ID: 4554 Test status: Passed

6.18.1 Description Verifies element creation using commas for coordinates in the command line.

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7 Import / Export

ADVANCE DESIGN VALIDATION GUIDE

7.1

Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) Test ID: 3628 Test status: Passed

7.1.1 Description Exports a linear element to GTC.

7.2

Exporting an Advance Design model to DO4 format (DEV2012 #1.10) Test ID: 4101 Test status: Passed

7.2.1 Description Launches the "Export > Text file" command and saves the current project as a .do4 archive file.

The model contains all types of structural elements, loads and geometric objects.

7.3

Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) Test ID: 4193 Test status: Passed

7.3.1 Description Exports the analysis model to ADA (through GTC) with: - Export results: enabled - Export meshed model: disabled

7.4

Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172) Test ID: 4297 Test status: Passed

7.4.1 Description Imports a GTC file from SuperSTRESS. The file contains steel linear elements with haunch sections.

7.5

Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) Test ID: 4195 Test status: Passed

7.5.1 Description Exports the analysis model to ADA (through GTC) with: - Export results: disabled - Export meshed model: enabled

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7.6

Exporting linear elements to IFC format (TTAD #10561) Test ID: 4530 Test status: Passed

7.6.1 Description Exports to IFC format a model containing linear elements having sections of type "I symmetric" and "I asymmetric".

7.7

Importing IFC files containing continuous foundations (TTAD #12410) Test ID: 4531 Test status: Passed

7.7.1 Description Imports an IFC file containing a continuous foundation (linear support) and verifies the element display.

7.8

Importing GTC files containing "PH.RDC" system (TTAD #12055) Test ID: 4548 Test status: Passed

7.8.1 Description Imports a GTC file exported from Advance Design. The file contains the automatically created system "PH.RDC".

7.9

Exporting a meshed model to GTC (TTAD #12550) Test ID: 4552 Test status: Passed

7.9.1 Description Exports a meshed model to GTC. The meshed planar element from the model contains a triangular mesh.

7.10 Verifying the load case properties from models imported as GTC files (TTAD #12306) Test ID: 4515 Test status: Passed

7.10.1 Description Performs the finite elements calculation on a model with dead load cases and exports the model to GTC. Imports the GTC file to verify the load case properties.

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7.11 Verifying the releases option of the planar elements edges after the model was exported and imported via GTC format (TTAD #12137) Test ID: 4506 Test status: Passed

7.11.1 Description Exports to GTC a model with planar elements on which the edges releases were defined. Imports the GTC file to verify the planar elements releases option.

7.12 System stability when importing AE files with invalid geometry (TTAD #12232) Test ID: 4479 Test status: Passed

7.12.1 Description Imports a complex model containing elements with invalid geometry.

7.13 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9) Test ID: 4445 Test status: Passed

7.13.1 Description Verifies the GTC files exchange (import/export) between Advance Design and SuperSTRESS.

7.14 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) Test ID: 4388 Test status: Passed

7.14.1 Description Imports a GTC file from SuperSTRESS. The file contains elements with circular hollow section. Verifies if the cross sections are imported from the attached "UK Steel Sections" database.

7.15 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) Test ID: 4389 Test status: Passed

7.15.1 Description Imports a GTC file from SuperSTRESS when the "UK Steel Sections" database is not attached. The file contains elements with circular hollow section. Verifies the cross sections definition.

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8 Connection Design

ADVANCE DESIGN VALIDATION GUIDE

8.1

Deleting a welded tube connection - 1 gusset bar (TTAD #12630) Test ID: 4561 Test status: Passed

8.1.1 Description Deletes a welded tube connection - 1 gusset bar after the joint was exported to ADSC.

8.2

Creating connections groups (TTAD #11797) Test ID: 4250 Test status: Passed

8.2.1 Description Verifies the connections groups function.

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9 Mesh

ADVANCE DESIGN VALIDATION GUIDE

9.1

Verifying the mesh for a model with generalized buckling (TTAD #11519) Test ID: 3649 Test status: Passed

9.1.1 Description Performs the finite elements calculation and verifies the mesh for a model with generalized buckling.

9.2

Verifying mesh points (TTAD #11748) Test ID: 3458 Test status: Passed

9.2.1 Description Performs the finite elements calculation and verifies the mesh nodes of a concrete structure.

The structure consists of concrete linear elements (R20*20 cross section) and rigid supports; the loads applied on the structure: dead loads, live loads, wind loads and snow loads, according to Eurocodes.

9.3

Creating triangular mesh for planar elements (TTAD #11727) Test ID: 3423 Test status: Passed

9.3.1 Description Creates a triangular mesh on a planar element with rigid supports and self weight.

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10 Reports Generator

ADVANCE DESIGN VALIDATION GUIDE

10.1 Verifying the modal analysis report (TTAD #12718) Test ID: 4576 Test status: Passed

10.1.1 Description Generates and verifies the modal analysis report.

10.2 Verifying the shape sheet strings display (TTAD #12622) Test ID: 4559 Test status: Passed

10.2.1 Description Verifies the shape sheet strings display for a steel beam with circular hollow cross-section.

10.3 Verifying the shape sheet for a steel beam (TTAD #12455) Test ID: 4535 Test status: Passed

10.3.1 Description Verifies the shape sheet for a steel beam.

10.4 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230) Test ID: 4502 Test status: Passed

10.4.1 Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the start point of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.5 Verifying the Max row on the user table report (TTAD #12512) Test ID: 4558 Test status: Passed

10.5.1 Description Verifies the Max row on the user table report.

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10.6 Verifying the steel shape sheet display (TTAD #12657) Test ID: 4562 Test status: Passed

10.6.1 Description Verifies the steel shape sheet display when the fire calculation is disabled.

10.7 Verifying the EC2 calculation assumptions report (TTAD #11838) Test ID: 4544 Test status: Passed

10.7.1 Description Verifies the EC2 calculation assumptions report.

10.8 Verifying the shape sheet report (TTAD #12353) Test ID: 4545 Test status: Passed

10.8.1 Description Generates and verifies the shape sheet report.

10.9 Verifying the model geometry report (TTAD #12201) Test ID: 4467 Test status: Passed

10.9.1 Description Generates the "Model geometry" report to verify the model properties: total weight, largest structure dimensions, center of gravity.

10.10 Verifying the global envelope of linear elements forces result (on end points and middle of super element) (TTAD #12230) Test ID: 4488 Test status: Passed

10.10.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on end points and middle of super element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

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10.11 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230) Test ID: 4489 Test status: Passed

10.11.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start and end of super element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

10.12 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD #12230, TTAD #12261) Test ID: 4503 Test status: Passed

10.12.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the end point of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.13 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4492 Test status: Passed

10.13.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on each 1/4 of mesh element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

10.14 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230) Test ID: 4493 Test status: Passed

10.14.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on all quarters of super element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

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10.15 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230) Test ID: 4487 Test status: Passed

10.15.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on all quarters of super element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

10.16 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230) Test ID: 4496 Test status: Passed

10.16.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the start point of super element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

10.17 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230, TTAD #12261) Test ID: 4497 Test status: Passed

10.17.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the end point of super element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

10.18 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD #12230, #12261) Test ID: 4491 Test status: Passed

10.18.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the end point of super element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

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10.19 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230) Test ID: 4495 Test status: Passed

10.19.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on start and end of super element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

10.20 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4486 Test status: Passed

10.20.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on each 1/4 of mesh element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

10.21 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230) Test ID: 4490 Test status: Passed

10.21.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start point of super element by generating the "Global envelope of linear elements forces result report".

The model consists of a concrete portal frame with rigid fixed supports.

10.22 Verifying the global envelope of linear elements displacements (on end points and middle of super element) (TTAD #12230) Test ID: 4494 Test status: Passed

10.22.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on end points and middle of super element by generating the "Global envelope of linear elements displacements report".

The model consists of a concrete portal frame with rigid fixed supports.

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10.23 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4498 Test status: Passed

10.23.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on each 1/4 of mesh element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.24 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230) Test ID: 4501 Test status: Passed

10.24.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on start and end of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.25 Verifying the global envelope of linear elements stresses (on end points and middle of super element) (TTAD #12230) Test ID: 4500 Test status: Passed

10.25.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on end points and middle of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.26 Verifying the Min/Max values from the user reports (TTAD# 12231) Test ID: 4505 Test status: Passed

10.26.1Description Performs the finite elements calculation and generates a user report containing the results of Min/Max values.

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10.27 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230) Test ID: 4499 Test status: Passed

10.27.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on all quarters of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.28 Creating the rules table (TTAD #11802) Test ID: 4099 Test status: Passed

10.28.1Description Generates the "Rules description" report as .rtf and .txt file.

The model consists of a steel structure with supports and a base plate connection. Two rules were defined for the steel calculation.

10.29 Creating the steel materials description report (TTAD #11954) Test ID: 4100 Test status: Passed

10.29.1Description Generates the "Steel materials" report as a .txt file.

The model consists of a steel structure with supports and a base plate connection.

10.30 System stability when the column releases interfere with support restraints (TTAD #10557) Test ID: 3717 Test status: Passed

10.30.1Description Performs the finite elements calculation and generates the systems description report for a structure which has column releases that interfere with the supports restraints.

The structure consists of steel beams and steel columns (S235 material, HEA550 cross section) with rigid fixed supports.

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10.31 Generating the critical magnification factors report (TTAD #11379) Test ID: 3647 Test status: Passed

10.31.1Description Performs the finite elements calculation and the steel calculation on a structure with four types of connections.

Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. The model connections: columns base plates, beam - column fixed connections, beam - beam fixed connections and gusset plate. Live loads, snow loads and wind loads are applied.

10.32 Modal analysis: eigen modes results for a structure with one level Test ID: 3668 Test status: Passed

10.32.1Description Performs the finite elements calculation and generates the "Characteristic values of eigen modes" report.

The one-level structure consists of linear and planar concrete elements with rigid supports. A modal analysis is defined.

10.33 Generating a report with modal analysis results (TTAD #10849) Test ID: 3734 Test status: Passed

10.33.1Description Generates a report with modal results for a model with seismic actions.

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11 Seismic Analysis

ADVANCE DESIGN VALIDATION GUIDE

11.1 Verifying the spectrum results for EC8 seism (TTAD #11478) Test ID: 3703 Test status: Passed

11.1.1 Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 French standards.

11.2 Verifying the spectrum results for EC8 seism (TTAD #12472) Test ID: 4537 Test status: Passed

11.2.1 Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 Romanian standards (SR EN 1998-1/NA).

11.3 Verifying the combinations description report (TTAD #11632) Test ID: 3544 Test status: Passed

11.3.1 Description Generates a dead load case, two live load cases and a snow load case, defines the concomitance between the generated load cases and generates the combinations description report.

11.4 EC8 : Verifying the displacements results of a linear element according to Czech seismic standards (CSN EN 1998-1) (DEV2012 #3.18) Test ID: 3626 Test status: Passed

11.4.1 Description Verifies the displacements results of an inclined linear element according to Eurocodes 8 Czech standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports.

The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads (CSN EN 1998-1).

11.5 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) Test ID: 3576 Test status: Passed

11.5.1 Description Performs the finite elements calculation on a concrete structure. Generates the "Signed concomitant linear elements envelopes on Fx report".

The structure has concrete beams and columns, two concrete walls and a windwall. Loads applied on the structure: self weight and a planar live load of -40.00 kN.

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11.6 EC8 French Annex: verifying torsors on walls Test ID: 4803 Test status: Passed

11.6.1 Description Verifies the torsors on walls. Eurocode 8 with French Annex is used.

11.7 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure Test ID: 4810 Test status: Passed

11.7.1 Description Verifies torsors on grouped walls from a multi-storey concrete structure. EC8 with French Annex is used.

11.8 Verifying the damping correction influence over the efforts in supports (TTAD #13011).

Test ID: 4853 Test status: Passed

11.8.1 Description Verifies the damping correction influence over the efforts in supports. The model has 2 seismic cases. Only one case uses the damping correction. The seismic spectrum is generated according to the Eurocodes 8 - French standard (NF EN 1993-1-8/NA).

11.9 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778) Test ID: 5425 Test status: Passed

11.9.1 Description Verifies the seismic results according to EC8 French Annex for a single bay single story structure made of concrete.

11.10 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706) Test ID: 4859 Test status: Passed

11.10.1Description Verifies the sum of actions on supports and nodes restraints for a simple structure subjected to seismic action according to EC8 French annex.

11.11 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974) Test ID: 4858 Test status: Passed

11.11.1Description Verifies efforts and torsors on several planar elements of a concrete structure subjected to horizontal seismic action (according to PS92 norm).

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11.12 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD #13797) Test ID: 5455 Test status: Passed

11.12.1Description Generates reports with forces results per modes on a selection of elements (linear and planar elements) from a concrete structure subjected to seismic action (EC8 French Annex).

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12 Steel Design

ADVANCE DESIGN VALIDATION GUIDE

12.1 EC3 Test 2: Class section classification and share verification of an IPE300 beam subjected to linear uniform loading Test ID: 5410 Test status: Passed

12.1.1 Description Classification and verification of an IPE 300 beam made of S235 steel.

The beam is subjected to a 50 kN/m linear uniform load applied gravitationally.

The force is considered to be a live load and the dead load is neglected.

12.1.2 Background Classification and verification of sections for an IPE 300 beam made from S235 steel. The beam is subjected to a 50 kN/m linear uniform load applied gravitationally. The force is considered to be a live load and the dead load is neglected.

12.1.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Exploitation loadings (category A): Q = -50kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System Materials properties

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Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y and Z axis and rotation restrained on X axis Inner: None. ► ►

12.1.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 In this case the stresses distribution along the section is like in the picture below: ■ ■ ■

compression for the top flange compression and tension for the web tension for the bottom flange

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To determine the web class it will be used the Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

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Therefore:

This means that the column web is Class 1.

To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

615

ADVANCE DESIGN VALIDATION GUIDE

c 56.45mm = = 5.276 t 10.7 mm

ε = 0.92 Therefore:

c 56.45mm = = 5.276 ≤ 9 * ε = 9 * 0.92 = 8.28 t 10.7 mm

this means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the class section for the entire beam section will be considered Class 1.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

12.1.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd shall be determined as follows:

V pl , Rd =

Av *

fy

γM0

3 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)

Where: Av: section shear area for rolled profiles

Av = A − 2 * b * t f + (t w + 2 * r ) * t f

2

A: cross-section area A=53.81cm b: overall breadth b=150mm h: overall depth h=300mm

hw: depth of the web hw=248.6mm r: root radius r=15mm tf: flange thickness tf=10.7mm tw: web thickness tw=7.1mm

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) fy: nominal yielding strength for S275 fy=275MPa

γ M 0 : partial safety coefficient γ M 0 = 1 Therefore:

V pl , Rd =

616

Av *

fy

γM0

3 =

25.68 * 10 4 * 1

275 3 = 0.4077 MN = 407.7kN

ADVANCE DESIGN VALIDATION GUIDE

For more: Verification of the shear buckling resistance for webs without stiffeners:

hw η ≤ 72 * tw ε According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(7)

η = 1.20 * ε=

1 γ M1 = 1.20 * = 1.20 1 γM0

235 235 = = 0.92 275 fy

1.20 hw 248.6 η = = 35.01 ≤ 72 * = 72 * = 93.91 7.1 0.92 tw ε There is no need for shear buckling resistance verification According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2) Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

617

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

12.1.2.4 Reference results Result name

Result description

Reference value

Work ratio

%

31 %

12.1.3 Calculated results

Result name Fz

Work ratio - Fz

618

Result description Fz

Work ratio Fz

Value -125 kN

0%

Error -0.0000 %

3.2258 %

ADVANCE DESIGN VALIDATION GUIDE

12.2 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 column Test ID: 5424 Test status: Passed

12.2.1 Description Classification and verification on combined bending of an IPE 300 beam made of S235 steel.

The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis.

The beam is subjected to a -10 kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis.

Both forces are considered as live loads.

The dead load will be neglected.

12.2.2 Background Classification and verification on combined bending of sections for an IPE 300 beam made from S235 steel. The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis. The beam is subjected to a -10kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis. Both forces are considered live loads. The dead load will be neglected.

12.2.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -10kN/m, Q2 = 10kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

619

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Materials properties

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotation blocked along X axis.

Inner: None. ►

12.2.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 In this case, the beam is subjected to linear uniform equal loads, one vertical and one horizontal, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below:

620

ADVANCE DESIGN VALIDATION GUIDE

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

621

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the beam web is Class 1.

Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

The section geometrical properties are described in the picture below:

According to Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

622

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the beam left top flanges are Class 1.

Overall the beam top flange cross-section class is Class 1.

In the same way will be determined that the beam bottom flange cross-section class is also Class 1 A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

12.2.2.3 Reference results for calculating the combined biaxial bending α

β

⎡ M Y , Ed ⎤ ⎡ M z , Ed ⎤ ⎢ ⎥ +⎢ ⎥ ≤1 M ⎢⎣ M Ny , Rd ⎥⎦ Nz , Ed ⎣ ⎦ According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(5) In which α and β are constants, which may conservatively be taken as unity, otherwise as follows: For I and H sections:

α =2

β = max(n;1) n=

0 N Ed = = 0 therefore β = 1 N pl , Rd N pl , Rd

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

M Ny , Rd =

M pl , y , Rd * (1 − n) 1 − 0 .5 * a

but

M Ny , Rd ≤ M ply , Rd According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)

n=

0 N Ed = =0 N pl , Rd N pl , Rd

a=

( A − 2 * b * t f ) (53.81 *10−4 − 2 * 0.15 * 0.0107) = = 0.403 ≤ 0.5 53.81 *10− 4 A

M N , y , Rd =

M pl , y , Rd (1 − 0.5 * 0.403)

=

M pl , y , Rd 0.8

0.8 * M N , y , Rd = M pl , y , Rd ⇒ M N , y , Rd > M pl , y , Rd

but

M Ny , Rd ≤ M ply , Rd

623

ADVANCE DESIGN VALIDATION GUIDE

Therefore, it will be considered:

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

M Nz , Rd =

M pl , z , Rd * (1 − n) 1 − 0.5 * a

but

M Nz , Rd ≤ M plz , Rd

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)

0.8 * M N , z , Rd = M pl , z , Rd ⇒ M N , z , Rd > M pl , z , Rd M N , z , Rd = M pl , z , Rd =

wpl , z * f y

γM0

but

M Nz , Rd ≤ M plz , Rd therefore it will be considered:

125.20 *10−6 * 235 = = 0.030MNm 1

In conclusion: α

β

2 1 ⎡ M Y , Ed ⎤ ⎡ M z , Ed ⎤ ⎡ 0.03125 ⎤ ⎡ 0.03125 ⎤ + = + ⎥ ⎢ ⎢ ⎥ ⎢⎣ 0.148 ⎥⎦ ⎢⎣ 0.029375 ⎥⎦ = 1.09 > 1 M ⎢⎣ M Ny , Rd ⎥⎦ Nz Ed , ⎦ ⎣

Finite elements modeling

■ ■ ■

624

Linear element: S beam, 6 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

12.2.2.4 Reference results Result name

Combined bending

oblique

Result description

Reference value

%

111 %

12.2.3 Calculated results

Result name Work ratio - Oblique

Result description Work ratio-Oblique

Value 0%

Error 0.9009 %

625

ADVANCE DESIGN VALIDATION GUIDE

12.3 EC3 Test 3: Class section classification and share and bending moment verification of an IPE300 column Test ID: 5411 Test status: Passed

12.3.1 Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

In the middle, the column is subjected to a 200 kN force applied on the web direction, defined as a live load.

The dead load will be neglected.

12.3.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 200kN force applied on the web direction, defined as a live load. The dead load will be neglected.

12.3.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

626

Exploitation loadings (category A): Q = -200kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Materials properties

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (z = 5.00) free.

Inner: None. ►

12.3.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the column base) is like in the picture below:

627

ADVANCE DESIGN VALIDATION GUIDE

To determine the web class, we will use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

628

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the column web is Class 1.

To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.

The section geometrical properties are described in the picture below:

629

ADVANCE DESIGN VALIDATION GUIDE

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

Therefore:

This means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6).

12.3.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd , is determined as follows:

V pl , Rd =

Av *

fy

γM0

3 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)

Where: Av section shear area for rolled profiles A

cross-section area A=53.81cm2

b

overall breadth b=150mm

h

overall depth h=300mm

Av = A − 2 * b * t f + (t w + 2 * r ) * t f

hw depth of the web hw=248.6mm r

root radius r=15mm

tf

flange thickness tf=10.7mm

tw

web thickness tw=7.1mm

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) nominal yielding strength for S235 fy=235MPa

fy

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

V pl , Rd =

630

Av *

fy

γM0

3 =

25.68 * 10 4 * 1

235 3 = 0.3484 MN = 348.42kN

ADVANCE DESIGN VALIDATION GUIDE

12.3.2.4 Reference results in calculating the bending moment resistance

200 VEd = = 57.4% > 50% V pl , Rd 348.42 The shear force is greater than half of the plastic shear resistance. Its effect on the moment resistance must be taken into account.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(1)(2)

Where:

2

⎛ 2VEd ⎞ ⎛ 2 * 0.200 ⎞ 2 ⎜ ρ =⎜ − 1⎟ = ⎜ − 1⎟ = 0.0223 ⎟ ⎝ 0.348 V ⎠ ⎝ pl , Rd ⎠ According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(3)

Av = A − 2 * b * t f + (t w + 2 * r ) * t f Av section shear area for rolled profiles A cross-section area A=53.81cm2 b

overall breadth b=150mm

h

overall depth h=300mm

hw depth of the web hw=248.6mm r

root radius r=15mm

tf

flange thickness tf=10.7mm

tw web thickness tw=7.1mm

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

M y ,V , Rd

⎛ ⎛ ρ * Av ² ⎞ 0.0223 * (25.68 * 10 −4 )² ⎞ ⎜⎜ wpl − ⎟⎟ * f y ⎜⎜ 628.40 * 10 − 6 − ⎟⎟ * 235 t 4 4 * 0 .

0071 w ⎠ ⎠ =⎝ =⎝ = 0.146MNm γM0 1

M Ed = 2.5m * 200kN = 500kNm = 0.5MNm M Ed 0.500 = = 342% M y ,V , Rd 0.146

631

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

12.3.2.5 Reference results Result name

Result description

Reference value

Shear z direction work ratio

%

57 %

Combined oblique bending

%

341 %

12.3.3 Calculated results

Result name Work ratio - Fz

Work ratio - Oblique

632

Result description Work ratio Fz

Work ratio - Oblique

Value 0%

Error 1.7544 %

0%

0.2933 %

ADVANCE DESIGN VALIDATION GUIDE

12.4 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column Test ID: 5421 Test status: Passed

12.4.1 Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied on all the length of the column, on the web direction, both defined as live loads.

The dead load will be neglected.

12.4.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. The column is subjected to a 500kN compressive force applied on top and a 5kN/m uniform linear load applied for all the length of the column, on the web direction, both defined as live loads. The dead load will be neglected.

12.4.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = 500kN, Q2 = 5kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System

633

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (z = 5.00) free.

Inner: None. ►

12.4.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.

634

ADVANCE DESIGN VALIDATION GUIDE

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

635

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

Therefore:

This means that the column web is Class 3.

Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

636

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

Therefore:

This means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 3 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 3.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6) Cross sections for class 3, the maximum longitudinal stress should check:

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.4(3) In this case:

In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.2(1) This means:

637

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

12.4.2.3 Reference results Result name

Result description

Reference value

Combined oblique bending

%

87 %

12.4.3 Calculated results

Result name Work ratio - Oblique

638

Result description Work ratio- Oblique

Value 0%

Error 1.1494 %

ADVANCE DESIGN VALIDATION GUIDE

12.5 EC3 test 4: Class section classification and bending moment verification of an IPE300 column Test ID: 5412 Test status: Passed

12.5.1 Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

In the middle, the column is subjected to a 50 kN force applied on the web direction, defined as a live load.

The dead load will be neglected.

12.5.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 50kN force applied on the web direction, defined as a live load. The dead load will be neglected.

12.5.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Exploitation loadings (category A): Q = 50kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

639

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Materials properties

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (z = 5.00) free.

Inner: None. ►

12.5.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 In this case, the column is subjected to a lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.

640

ADVANCE DESIGN VALIDATION GUIDE

The Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

641

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the column web is Class 1.

Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

The section geometrical properties are described in the picture below:

642

ADVANCE DESIGN VALIDATION GUIDE

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

Therefore:

This means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

12.5.2.3 Reference results in calculating the bending moment resistance

M Ed ≤1 M c , Rd According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(1)

M c , Rd =

wpl * f y

γM0

for Class1 cross sections According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(2)

Where:

w pl = 628.40cm 3 fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

M y ,V , Rd =

wpl * f y

γM0

=

628.40 * 10−6 * 235 = 0.147674MNm 1

M Ed = 2.5m * 50kN = 125kNm = 0.125MNm M Ed 0.125 = = 84% M y ,V , Rd 0.148 Finite elements modeling

■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

643

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

12.5.2.4 Reference results Result name

Result description

Reference value

Combined oblique bending

%

85 %

12.5.3 Calculated results

Result name Work ratio - Oblique

644

Result description Work ratio - Oblique

Value 0%

Error 1.1765 %

ADVANCE DESIGN VALIDATION GUIDE

12.6 EC3 Test 1: Class section classification and compression verification of an IPE300 column Test ID: 5383 Test status: Passed

12.6.1 Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load.

The dead load will be neglected.

12.6.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. On top, the column is subjected to a 100kN force applied gravitationally, defined as a live load. The dead load will be neglected.

12.6.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■

Exploitation loadings (category A): Q = -100kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

645

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Materials properties

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis, Support at end point (z = 5.00) free.

Inner: None. ► ►

12.6.2.2 Reference results for calculating the cross section class In this case, the column is subjected only to compression, therefore the distribution of stresses along the section is like in the picture below:

646

ADVANCE DESIGN VALIDATION GUIDE

To determine the web class, we use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

Îľ =1 Therefore:

This means that the column web is Class 2.

647

ADVANCE DESIGN VALIDATION GUIDE

To determine the flanges class, we will use Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

c 56.45mm = = 5.276 t 10.7 mm

ε =1 Therefore:

c 56.45mm = = 5.276 ≤ 9 * ε = 9 10.7 mm t

this means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 2 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 2.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

648

ADVANCE DESIGN VALIDATION GUIDE

12.6.2.3 Reference results in calculating the compressive resistance Nc,Rd The design resistance of the cross-section force Nc,Rd shall be determined as follows: For Class 1, 2 or 3 cross-section

N c , Rd =

A* fy

γM0

Where: A

section area A=53.81cm2

Fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

N c , Rd =

A* fy

γM0

=

53.81 * 10 −4 * 235 = 1.264535 MN = 1264.54kN 1 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.4(2)

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:

649

ADVANCE DESIGN VALIDATION GUIDE

12.6.2.4 Reference results Result name

Result description

Reference value

Work ratio

%

8%

12.6.3 Calculated results Result name Work ratio - Fx

650

Result description Work ratio Fx

Value 8%

Error 0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

12.7 Generating the shape sheet by system (TTAD #11471) Test ID: 3575 Test status: Passed

12.7.1 Description Generates shape sheets by system, on a model with 2 systems.

12.8 Verifying the calculation results for steel cables (TTAD #11623) Test ID: 3560 Test status: Passed

12.8.1 Description Performs the finite elements calculation and the steel calculation for a steel model with cables (D4) and a static nonlinear case. Generates the steel analysis report: data and results.

12.9 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) Test ID: 3641 Test status: Passed

12.9.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a horizontal steel element, verifies the cross section class.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support at one end and a rigid support with translation restraints on Y and Z and rotation restraint on X at the other end. A linear live load of -10.00 kN on FX and a punctual live load of 1000 kN on FX are applied.

12.10 Verifying the shape sheet results for a column (TTAD #11550) Test ID: 3640 Test status: Passed

12.10.1Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a vertical steel element.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support. A vertical live load of -200.00 kN is applied.

12.11 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) Test ID: 3612 Test status: Passed

12.11.1Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet results report.

The structure is a simple vault consisting of three steel elements (S235 material, IPEA240 cross section) with two rigid fixed supports. The loads applied on the structure: self weight and a linear live load of -10.00kN on FZ.

651

ADVANCE DESIGN VALIDATION GUIDE

12.12 Verifying the cross section optimization according to EC3 (TTAD #11516) Test ID: 3620 Test status: Passed

12.12.1Description Verifies the cross section optimization of a steel element, according to Eurocodes 3.

Performs the finite elements calculation and the steel calculation. Generates the "Envelopes and shapes optimization" report.

The steel bar has a IPE360 cross section, a rigid hinge support at one end and a rigid support with translation restraints on X, Y and Z and rotation restraint on X. Two loads are applied: a punctual dead load of -1.00 kN on FZ and a punctual live load of -40.00 kN on FZ.

12.13 Verifying shape sheet on S275 beam (TTAD #11731) Test ID: 3434 Test status: Passed

12.13.1Description Performs the steel calculation for two horizontal bars and generates the shape sheets report.

The bars have cross sections from different catalogues (1016x305x487 UKB and UKB1016x305x487). They are made of the same material (S275); each is subjected to a -500.00 kN linear horizontal dead load and has two rigid supports at both ends.

12.14 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) Test ID: 3406 Test status: Passed

12.14.1Description Performs the steel calculation for two vertical bars with different thicknesses and generates the shape sheets report.

The bars are made of the same material (S275 H - EN 10210-1), have rectangular hollow cross sections, but with different thicknesses (R80*40/4.1 and R80*40/3.9). Each is subjected to a -150.00 kN vertical live load and has a rigid support.

12.15 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) Test ID: 4549 Test status: Passed

12.15.1Description Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computed and when it is not.

12.16 Changing the steel design template for a linear element (TTAD #12491) Test ID: 4540 Test status: Passed

12.16.1Description Selects a different design template for steel linear elements.

652

ADVANCE DESIGN VALIDATION GUIDE

12.17 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) Test ID: 4529 Test status: Passed

12.17.1Description Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from the specialized calculation (chords).

12.18 EC3: Verifying the buckling length results (TTAD #11550) Test ID: 4481 Test status: Passed

12.18.1Description Performs the steel calculation and verifies the buckling length results according to Eurocodes 3 - French standards.

The shape sheet report is generated.

The model consists of a vertical linear element (IPE300 cross section, S275 material) with a rigid fixed support at the base. A punctual live load of 200.00 kN is applied.

12.19 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975) Test ID: 4484 Test status: Passed

12.19.1Description Runs the Steel elements verification and generates the "Envelopes and optimizing profiles" report in order to verify the work ratios. The verification is performed using the EC3 norm with Romanian annex.

12.20 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method Test ID: 3814 Test status: Passed

12.20.1Description Verifies the buckling length for a steel portal frame with one level, using the roA roB method, according to CM66.

Generates the "Buckling and lateral-torsional buckling lengths" report.

12.21 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method Test ID: 3813 Test status: Passed

12.21.1Description Verifies the buckling length for a steel portal frame with one level, using the kA kB method, according to CM66.

Generates the "Buckling and lateral-torsional buckling lengths" report.

653

ADVANCE DESIGN VALIDATION GUIDE

12.22 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method Test ID: 3819 Test status: Passed

12.22.1Description Verifies the buckling length for a steel portal frame, using the kA kB method, according to Eurocodes 3.

Generates the "Buckling and lateral-torsional buckling lengths" report.

12.23 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) Test ID: 4289 Test status: Passed

12.23.1Description Verifies the steel shape optimization when using sections from Advance Steel Profiles database. Performs the finite elements calculation and the steel elements calculation and generates the steel shapes report.

The structure consists of columns with UKC152x152x23 cross section, beams with UKB254x102x22 cross section and roof beams with UKB127x76x13 cross section. Dead loads and live loads are applied on the structure.

12.24 Verifying the buckling coefficient Xy on a class 2 section Test ID: 4443 Test status: Passed

12.24.1Description Performs the finite elements calculation and the steel elements calculation. Verifies the buckling coefficient Xy on a class 2 section and generates the shape sheets report.

The model consists of a vertical linear element (I26*0.71+15*1.07 cross section and S275 material) with a rigid hinge support at the base and a rigid support with translation restraints on X and Y and rotation restraint on Z, at the top. A punctual live load is applied.

654

13 Timber Design

ADVANCE DESIGN VALIDATION GUIDE

13.1 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) Test ID: 4509 Test status: Passed

13.1.1 Description Defines the "Design experts" properties for a timber linear element, in a model created with a previous version of the program.

13.2 Verifying the timber elements shape sheet (TTAD #12337) Test ID: 4538 Test status: Passed

13.2.1 Description Verifies the timber elements shape sheet.

13.3 Verifying the units display in the timber shape sheet (TTAD #12445) Test ID: 4539 Test status: Passed

13.3.1 Description Verifies the Afi units display in the timber shape sheet.

13.4 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending Test ID: 4901 Test status: Passed

13.4.1 Description Verifies the fire resistance of a rectangular cross section purlin made from solid timber C24 to resist simple bending.

The purlin is exposed to fire on 3 faces for 30 minutes. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.

656

ADVANCE DESIGN VALIDATION GUIDE

13.5 EC5: Verifying a C24 timber beam subjected to shear force Test ID: 5036 Test status: Passed

13.5.1 Description Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

13.5.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

13.5.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test D; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■ ■

Loadings from the structure: G = 0.5 kN/m2, Exploitation loadings (category A): Q = 1.5 kN/m2, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2

Simply supported beam

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■

Height: h = 0.225 m, Width: b = 0.075 m, Length: L = 5.00 m, Distance between adjacent beams (span): d = 0.5 m, Section area: A = 16.875 x 10-3 m2 ,

Materials properties

Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■

Characteristic shear strength: fv,k = 2.5 x 106 Pa, Service class 1.

657

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (z=0) restrained in translation along X, Y and Z, Support at end point (z = 5.00) restrained in translation along X, Y, Z and restrained in rotation along X. Inner: None. ► ►

Loading

The beam is subjected to the following loadings: ■ ■

External:

Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m, Internal: None.

13.5.2.2 Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam shear stresses at ultimate limit state, the formula (6.13) from EN 1995-1-1 norm is used. Before using it, some parameters involved in calculations, like kmod, kcr, γM, kf, beff, heff, have to be determined.

After this the reference solution, which includes the design shear stress about the principal y axis, the design shear strength and the corresponding work ratios, is calculated.

Reference solution for ultimate limit state verification

Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to determine some parameters involved in calculations (kmod, γM, kcr, kf, beff, heff).

■

Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■

Partial factor for material properties: γM = 1.3

■

Cracking factor, kcr : kcr = 0.67 (for solid timber)

■

Factor depending on the shape of the cross section, kf: kf = 3/2 (for a rectangular cross section)

■

Effective width, beff: beff = kcr x b = 0.67 x 0.075m = 0.05025m

■

Effective height, heff: heff = h = 0.225m

■

Design shear stress (induced by the applied forces):

τd = ■

f v ,k ×

k mod

γM

= 2.5 × 10 6 Pa ×

0.8 = 1.538 × 10 6 Pa 1.3

Work ratio according to formulae 6.13 from EN 1995-1-1 norm:

τd f v ,d

658

beff × heff

3 × 3656.25 N 2 = = 0.485075 × 10 6 Pa 0.05025m × 0.225m

Design shear strength: fv,d =

■

k f × Fv ,d

≤ 1 .0

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

Shear force, Fz, diagram

Simply supported beam subjected to bending Shear force diagram [N]

Design shear stress diagram

Simply supported beam subjected to bending Design shear stress [Pa]

Shear strength work ratio diagram

Simply supported beam subjected to bending Work ratio S_d [%]

13.5.2.3 Reference results Result name

Result description

Reference value

Fz

Shear force [N]

3656.25 N

Stress S_d

Design shear stress [Pa]

485074.63 Pa

Work ratio S_d

Shear work ratio (6.13) [%]

32 %

659

ADVANCE DESIGN VALIDATION GUIDE

13.5.3 Calculated results

Result name Fz

660

Result description Shear force

Value -3.65625 kN

Error -0.0000 %

Design shear stress

485075 Pa

-0.0001 %

Shear strength work ratio

0.315299 %

-0.0002 %

ADVANCE DESIGN VALIDATION GUIDE

13.6 EC5: Verifying a timber column subjected to compression forces Test ID: 4823 Test status: Passed

13.6.1 Description Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timber C18.

13.6.2 Background Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C18. The verification is made according to formula (6.35) from EN 1995-1-1 norm.

13.6.2.1 Model description ■ ■ ■

Reference: Guide de validation Eurocode 5, test B; Analysis type: static linear (plane problem); Element type: linear.

Simply supported column

Units

Metric System Geometry

Column cross section characteristics: ■ ■ ■

Height: h = 0.15 m, Width: b = 0.10 m, Section area: A = 15.0 x 10-3 m2

661

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Rectangular solid timber C18 is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Longitudinal elastic modulus: E = 0.9 x 1010 Pa, Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.6 x 1010 Pa, Characteristic compressive strength along the grain: fc,0,k = 18 x 106 Pa, Service class 1.

Boundary conditions

The boundary conditions: ■

Outer:

■

Support at base (z=0) restrained in translation along X, Y and Z, Support at top (z = 3.2) restrained in translation along X, Y and restrained in rotation along Z. Inner: None. ► ►

Loading

The column is subjected to the following loadings: ■ ■

External: Point load at z = 3.2: Fz = N = -20000 N, Internal: None.

13.6.2.2 Reference results in calculating the timber column subjected to compression force The formula (6.35) from EN 1995-1-1 is used in order to verify a timber column subjected to compression force.

Before applying this formula we need to determine some parameters involved in calculations, such as: slenderness ratios, relative slenderness ratios, instability factors. After this, the reference solution is calculated. This includes: the design compressive stress, the design compressive strength and the corresponding work ratio.

Slenderness ratios

The most important slenderness is calculated relative to the z axis, as it will be the axis of rotation if the column buckles.

■

Slenderness ratio corresponding to bending about the z axis:

λ z = 12 ■

m × lg lc 1× 3.2m = 12 = 12 = 110.85 0.1m b b

Slenderness ratio corresponding to bending about the y axis (informative):

λ y = 12

m × lg lc 1× 3.2m = 12 = 12 = 73.9 0.15m h h

Relative slenderness ratios

The relative slenderness ratios are: ■

Relative slenderness ratio corresponding to bending about the z axis:

λrel , z = ■

λz π

f c ,0,k E0,05

=

m × l g × 12

f c , 0,k

b ×π

E0,05

1× 3.2m × 12 0.1m × π

18 × 10 6 Pa = 1.933 0.6 × 1010 Pa

Relative slenderness ratio corresponding to bending about the y axis (informative):

λrel , y =

λy π

f c , 0,k E0,05

=

m × l g × 12

f c ,0,k

h ×π

E0,05

So there is a risk of buckling, because λrel,max ≥ 0.3.

662

=

=

1× 3.2m × 12 0.15m × π

18 × 10 6 Pa = 1.288 0.6 × 1010 Pa

ADVANCE DESIGN VALIDATION GUIDE

Instability factors

In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1: βc = 0.2 (according to relation 6.29 from EN 1995-1-1) The instability factors are: ■

kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z2] (according to relation 6.28 from EN 1995-1-1)

■

ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (informative)

■

k c, z =

■

k c, y =

1 k z + k z2 − λ2rel , z

(according to relation 6.26 from EN 1995-1-1)

1

(informative)

k y + k y2 − λ2rel , y

Reference solution

Before calculating the reference solution (design compressive stress, design compressive strength and work ratio) we need to determine some parameters involved in calculations (kmod, γM).

■

Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■ ■

Partial factor for material properties: γM = 1.3 Design compressive stress (induced by the applied forces): σc,0,d =

■

Design compressive strength: fc,0,d =

■

N A f c , 0,k

k mod

γM

Work ratio: Work ratio =

σ c ,0,d k c , z × f c , 0,d

≤ 1.0

(according to relation 6.35 from EN 1995-1-1)

Finite elements modeling

■ ■ ■

Linear element: S beam, 4 nodes, 1 linear element.

663

ADVANCE DESIGN VALIDATION GUIDE

Work ratio diagram

Simply supported column subjected to compression force Work ratio

13.6.2.3 Reference results Result name

Result description

Reference value

kc,z

Instability factor

0.2400246

kc,y

Instability factor (informative)

0.488869

ﾏツ,0,d Work ratio

Design compressive stress [Pa]

1333333 Pa

Work ratio [%]

50 %

13.6.3 Calculated results Result name Kc,z

Value 0.240107 adim

Error 0.0001 %

Instability factor

0.488612 adim

0.0000 %

Stress SFx

Design compressive stress

1.33333e+006 Pa

0.0003 %

Work ratio

Work ratio

50 %

0.0000 %

Kc,y

664

Result description Instability factor

ADVANCE DESIGN VALIDATION GUIDE

13.7 EC5: Verifying a timber beam subjected to combined bending and axial tension Test ID: 4872 Test status: Passed

13.7.1 Description Verifies a rectangular cross section rafter made from solid timber C24 to resist combined bending and axial tension.

The verification of the cross-section subjected to combined stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

13.7.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending and axial tension. The verification of the deflections at serviceability limit state is also performed.

13.7.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m.

The following load cases and load combination are used: ■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S; ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S. All loads will be projected on the rafter direction since its slope is 50% (26.6°).

Simply supported rafter subjected to projected loadings

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.05 m, Length: L = 5.00 m, Section area: A = 10 x 10-3 m2 ,

■

b × h 2 0.05 ⋅ 0.20 2 = = 0.000333m 3 .

Elastic section modulus about the strong axis y: W y = 6 6

665

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■ ■

Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, Characteristic bending strength: fm,k = 24 x 106 Pa, Service class 2.

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (z=0) restrained in translation along Y, Z and restrained in rotation along X. ► Support at end point (z = 5.00) restrained in translation along X, Y, Z. Inner: None. ►

Loading

The rafter is subjected to the following projected loadings (at ultimate limit state): ■

■

External: ► Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, ► Tensile load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m = = 2191.22 N Internal: None.

13.7.2.2 Reference results in calculating the timber beam subjected to combined stresses In order to verify the timber beam subjected to combined stresses at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be determined. After this the reference solution, which consists of the design tensile stress, the design tensile strength and the corresponding work ratio and also the work ratios of the combined stresses, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.

Reference solution for ultimate limit state verification

Before calculating the reference solution (the design tensile stress, the design tensile strength and the corresponding work ratio, and also the work ratios of the combined stresses) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ ■

Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1) Partial factor for material properties: γM = 1.3

■

Depth factor (“h” represents the width in millimeters because the element is tensioned):

⎧⎛ 150 ⎞ 0.2 ⎧⎛ 150 ⎞ 0.2 ⎧1.25 ⎪⎜ ⎪⎜ ⎟ ⎟ kh = min ⎨⎝ h ⎠ = min ⎨⎝ 50 ⎠ = min ⎨ = 1.25 1 .

3 ⎩ ⎪1.3 ⎪1.3 ⎩ ⎩ ■

System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■

Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7

■

Design tensile stress (induced by the ultimate limit state force, N): σt,0,d =

666

N 2191.22 N = = 219122 Pa A 10 ×10 −3 m 2

ADVANCE DESIGN VALIDATION GUIDE

■

Design tensile strength:

■

σ t , 0,d f t , 0,d

(according to relation 6.1 from EN 1995-1-1)

My

6 × q × L2 = = Wy 8 × b × h 2

N × 5.00 2 m 2 m = 8.2045 × 10 6 Pa 2 2 8 × 0.05m × 0.2 m

6 × 875.15

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0.9 × 1.0 × 1.0 = 16.615 × 10 6 Pa 1.3

Work ratio according to formulae 6.17 from EN 1995-1-1 norm:

σ t , 0 ,d f t , 0 ,d ■

≤ 1.0

Design bending strength: fm,y,d =

■

0.9 × 1.25 = 12.115 × 10 6 Pa 1.3

Design bending stress (induced by the applied forces):

σm,y,d = ■

γM

× k h = 14 × 10 6 ×

Work ratio: SFx =

■

k mod

f t , 0 ,k ×

ft,0,d =

+

σ m , y ,d f m , y ,d

+ km

σ m , z ,d f m , z ,d

≤1

Work ratio according to formulae 6.18 from EN 1995-1-1 norm:

σ t , 0 ,d f t , 0 ,d

+ km

σ m , y ,d f m , y ,d

+

σ m , z ,d f m , z ,d

≤1

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

winst (Q) ≤ w fin ≤

L 300

L 125

wnet , fin ≤

L 200

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are: ■

Instantaneous deflection (for a base variable action):

winst (Q) = 0.00914m ⇒ winst (Q) = ■

L 547.05

Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

winst = d CQ = 0.01371m ⇒ winst =

L 364.7

667

ADVANCE DESIGN VALIDATION GUIDE

■

In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the deformation factor (kdef) has to be chosen:

k def = 0.8

(value determined for service class 2, according to table 3.2 from EN 1995-1-1)

wcreep = 0.8 × d QP = 0.8 × 0.0064m = 0.00512m ⇒ wcreep = ■

L 976.6

Final deflection:

w fin = winst + wcreep = 0.01371m + 0.00512m = 0.01883m ⇒ w fin = ■

Net deflection:

wnet , fin = w fin + wc = 0.01883m + 0m = 0.01883m ⇒ wnet , fin = Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

SFx work ratio diagram

Simply supported beam subjected to tensile forces Work ratio SFx

Strength work ratio diagram

Simply supported beam subjected to combined stresses Strength work ratio

Instantaneous deflection winst(Q)

Simply supported beam subjected to snow loads Instantaneous deflection winst(Q) [m]

668

L 265.5

L 265.5

ADVANCE DESIGN VALIDATION GUIDE

Instantaneous deflection winst(CQ)

Simply supported beam subjected to a characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

Final deflection wfin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

Net deflection wnet,fin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

13.7.2.3 Reference results Result name

Result description

Reference value

SFx

SFx work ratio [%]

1.808 %

Strength work ratio

Work ratio (6.17) [%]

51.19 %

winst (Q)

Deflection for a base variable action [m]

0.00914 m

dCQ

Deflection for a characteristic combination of actions [m]

0.01371 m

winst

Instantaneous deflection [m]

0.01371 m

kdef

Deformation coefficient

0.8

dQP

Deflection for a quasi-permanent combination of actions [m]

0.0064 m

wfin

Final deflection [m]

0.01883 m

wnet,fin

Net deflection [m]

0.01883 m

669

ADVANCE DESIGN VALIDATION GUIDE

13.7.3 Calculated results

Result name Work ratio SFx

670

Result description SFx work ratio

Value 0.0181483 %

Error -0.0003 %

Work ratio

Strength work ratio

0.511941 %

0.0001 %

D

w_inst(Q)

0.00871459 m

-4.6583 %

D

deflection for a characteristic combination

0.0130719 m

-4.6584 %

Winst

instantaneous deflection

0.0137105 m

0.0003 %

Kdef

deformation coefficient

0.8 adim

0.0000 %

D

deformation for a quasi-permanent combination

0.00610023 m

-4.6584 %

Wfin

final deflection

0.0188291 m

0.0002 %

Wnet,fin

net final deflection

0.0188291 m

0.0002 %

ADVANCE DESIGN VALIDATION GUIDE

13.8 EC5: Verifying a timber column subjected to tensile forces Test ID: 4693 Test status: Passed

13.8.1 Description Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.

13.8.2 Background Verifies the adequacy of the tension resistance for a rectangular cross section made from solid timber C24. The verification is made according to formula (6.1) from EN 1995-1-1 norm.

13.8.2.1 Model description ■ ■ ■

Reference: Guide de validation Eurocode 5, test A; Analysis type: static linear (plane problem); Element type: linear.

Column with fixed base

Units

Metric System Geometry

Cross section characteristics: ■ ■ ■ ■

Height: h = 0.122 m, Width: b = 0.036m, Section area: A = 43.92 x 10-4 m2 I = 5.4475 x 10-6 m4.

671

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■ ■

Longitudinal elastic modulus: E = 1.1 x 1010 Pa, Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, Service class 2.

Boundary conditions

The boundary conditions: ■

Outer: Fixed at base (z = 0), Free at top (z = 5),

■

Inner: None.

Loading

The column is subjected to the following loadings: ■ ■

External: Point load at z = 5: Fz = N = 10000 N, Internal: None.

13.8.2.2 Reference results in calculating the timber column subjected to tension force Reference solution

The reference solution is determined by formula (6.1) from EN 1995-1-1. Before applying this formula we need to determine some parameters involved in calculations (kmod, γM, kh). After this, the design tensile stress, the design tensile strength and the corresponding work ratio are calculated.

■

Modification factor for duration of load and moisture content: kmod = 0.9

■ ■

Partial factor for material properties: γM = 1.3 Depth factor (“h” represents the width, because the element is tensioned):

⎧⎛ 150 ⎞ 0.2 ⎪⎜ ⎟ kh = min ⎨⎝ h ⎠ ⎪1.3 ⎩ ■

Design tensile stress (induced by the ultimate limit state force, N): σt,0,d =

■

Design tensile strength: ft,0,d =

■

N A f t , 0,k ×

k mod

γM

× kh

Work ratio: SFx =

σ t ,0,d f t , 0,d

≤ 1.0

(according to relation 6.1 from EN 1995-1-1)

Finite elements modeling

■ ■ ■

672

Linear element: S beam, 6 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Work ratio SFx diagram

Column with fixed base, subjected to tension force Work ratio SFx

13.8.2.3 Reference results Result name

Result description

Reference value

ﾏフ,0,d SFx

Design tensile stress [Pa]

2276867.03 Pa

Work ratio [%]

18 %

13.8.3 Calculated results Result name Stress SFx

Work ratio

Result description Design tensile stress

Work ratio

Value 2.27687e+006 Pa

18 %

Error -0.0001 %

0.0000 %

673

ADVANCE DESIGN VALIDATION GUIDE

13.9 EC5: Shear verification for a simply supported timber beam Test ID: 4822 Test status: Passed

13.9.1 Description Verifies a rectangular cross section beam made from solid timber C24 to shear efforts. The verification of the shear stresses at ultimate limit state is performed.

674

ADVANCE DESIGN VALIDATION GUIDE

13.10 EC5: Verifying a timber purlin subjected to oblique bending Test ID: 4878 Test status: Passed

13.10.1Description Verifies a rectangular timber purlin made from solid timber C24 to resist oblique bending. The verification is made following the rules from Eurocode 5 French annex.

13.10.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to oblique bending. The verification of the bending stresses at ultimate limit state is performed.

13.10.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.3; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m.

The following load cases and load combination are used: ■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2; All loads will be projected on the purlin direction since the roof slope is 17°.

Simply supported purlin subjected to loadings

Units

Metric System Geometry

Purlin cross section characteristics: ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.10 m, Length: L = 3.5 m, Section area: A = 0.02 m2 ,

■

Elastic section modulus about the strong axis, y:

Wy =

b × h 2 0 .1 ⋅ 0 .20 2 = = 0 .000666 m 3 , 6 6

■

Elastic section modulus about the strong axis, z:

Wz =

b 2 × h 0 .12 ⋅ 0 .20 = = 0 .000333 m 3 .

6 6

675

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■

Characteristic bending strength: fm,k = 24 x 106 Pa, Service class 2.

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (z=0) restrained in translation along X, Y, Z; Support at end point (z = 3.5) restrained in translation along X, Y, Z and restrained in rotation along X. Inner: None. ► ►

Loading

The purlin is subjected to the following projected loadings (at ultimate limit state): ■

■

External: ► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x sin17° = 1101.22 N/m, ► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x cos17° = 3601.92 N/m, Internal: None.

13.10.2.2Reference results in calculating the timber purlin subjected to oblique bending In order to verify the timber purlin subjected to oblique bending at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, have to be determined. After this, the reference solution (which includes the design bending stress about y axis, the design bending stress about z axis and the maximum work ratio for strength verification) is calculated.

Reference solution for ultimate limit state verification

Before calculating the reference solution (design bending stress about y axis, design bending stress about z axis and maximum work ratio for strength verification) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■

Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)

■

Partial factor for material properties: γM = 1.3

■

Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0

■

System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■

Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7

■

Design bending stress about y axis (induced by uniformly distributed load, qz):

σm,y,d = ■

676

My Wy

=

qz × L = 8 ×Wy 2

N × 3.5 2 m 2 m = 8.2814 × 10 6 Pa 3 8 × 0.000666m

3601.92

Design bending stress about z axis (induced by uniformly distributed load, qy):

ADVANCE DESIGN VALIDATION GUIDE

σm,z,d = ■

Mz = = Wz 8 × Wz

N × 3.5 2 m 2 m = 5.0638 × 10 6 Pa 3 8 × 0.000333m

1101.22

Design bending strength: fm,y,d = fm,z,d =

■

q y × L2

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0.9 × 1.0 × 1.0 = 16.615 × 10 6 Pa 1.3

Maximum work ratio for strength verification; it represents the maximum value between the work ratios obtained with formulae 6.17 and 6.18 from EN 1995-1-1 norm:

σ ⎧σ m , y , d ⎫ + k m m , z ,d ⎪ ⎪ f m , z ,d ⎪ ⎪ f m , y ,d max ⎨ ⎬ ≤1 σ σ ⎪k m , y , d + m , z , d ⎪ ⎪ ⎪ m f m , y ,d f m , z ,d ⎭ ⎩ Finite elements modeling

■ ■ ■

Linear element: S beam, 5 nodes, 1 linear element.

677

ADVANCE DESIGN VALIDATION GUIDE

Stress SMy diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMy [Pa]

Stress SMz diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMz [Pa]

Maximum work ratio for strength verification

Strength of a simply supported purlin subjected to oblique bending Work ratio [%]

13.10.2.3Reference results Result name

Result description

Reference value

Smy

Design bending stress about y axis [Pa]

8281441 Pa

SMz

Design bending stress about z axis [Pa]

5063793 Pa

Work ratio

Maximum work ratio for strength verification [%]

71.2 %

13.10.3Calculated results

678

Result name Stress SMy

Result description Design bending stress about y axis

Value 8.47672e+006 Pa

Error 0.0000 %

Stress SMz

Design bending stress about z axis

5.18319e+006 Pa

-0.0000 %

Work ratio

Maximum work ratio for strength verification

0.71153 %

-0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

13.11 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression Test ID: 4877 Test status: Passed

13.11.1Description Verifies the lateral torsional stability for a rectangular timber beam subjected to combined bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.

679

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13.12 EC5: Verifying a timber beam subjected to simple bending Test ID: 4682 Test status: Passed

13.12.1Description Verifies a rectangular cross section beam made from solid timber C24 to resist simple bending. Verifies the bending stresses at ultimate limit state, as well as the deflections at serviceability limit state.

13.12.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending. Verification of the bending stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

13.12.2.1Model description ■ Reference: Guide de validation Eurocode 5, test C; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

The following load cases and load combination are used: ■ ■ ■ ■ ■

Loadings from the structure: G = 0.5 kN/m2, Exploitation loadings (category A): Q = 1.5 kN/m2, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2 Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q

Simply supported beam

Units

Metric System Geometry

Below are described the beam cross section characteristics:

680

■ ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.075 m, Length: L = 4.50 m, Distance between adjacent beams (span): d = 0.5 m, Section area: A = 15.0 x 10-3 m2 ,

■

Elastic section modulus about the strong axis y:

Wy =

b × h 2 0.075 ⋅ 0.20 2 = = 0.0005m 3 6 6

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■ ■

Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, Characteristic bending strength: fm,k = 24 x 106 Pa, Service class 1.

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (z=0) restrained in translation along X, Y and Z, ► Support at end point (z = 4.5) restrained in translation along X, Y, Z and restrained in rotation along X. Inner: None. ►

Loading

The beam is subjected to the following loadings: ■

External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■

Internal: None.

13.12.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam bending stresses at ultimate limit state, the formulae (6.11) and (6.12) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be calculated. After this, the reference solution, which includes the design bending stress about the principal y axis, the design bending strength and the corresponding work ratios, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.

Reference solution for ultimate limit state verification

Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■

Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■

Partial factor for material properties: γM = 1.3

■

Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0

■

System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■

Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7

■

Design bending stress (induced by the applied forces): σm,d =

■

My Wy

=

6 × q × L2 6 × 1.4625 × 4.5 2 = = 7.4039 × 10 6 Pa 2 2 8× b × h 8 × 0.075 × 0.2

Design bending strength: fm,d =

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0.8 × 1.0 × 1.0 = 14.769 × 10 6 Pa 1.3

681

ADVANCE DESIGN VALIDATION GUIDE

■

Work ratio according to formulae 6.11 from EN 1995-1-1 norm:

σ m ,d f m ,d ■

≤ 1.0

Work ratio according to formulae 6.12 from EN 1995-1-1 norm:

km ×

σ m,d f m,d

≤ 1.0

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

winst (Q) ≤ w fin ≤

L 300

L 125

wnet , fin ≤

L 200

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are the followings: ■

Instantaneous deflection (for a base variable action):

winst (Q) = 0.00749m ⇒ winst (Q) = ■

Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

winst = d CQ = 0.00999m ⇒ winst = ■

L 600.8 L 450.45

In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the deformation factor (kdef) has to be chosen:

k def = 0.6

(calculated value for service class 1, according to table 3.2 from EN 1995-1-1)

wcreep = 0.6 × d QP = 0.6 × 0.00475m = 0.00285m ⇒ wcreep = ■

L 1578.95

Final deflection:

w fin = winst + wcreep = 0.00999m + 0.00285m = 0.01284m ⇒ w fin = ■

Net deflection:

wnet , fin = w fin + wc = 0.01284m + 0m = 0.01284m ⇒ wnet , fin = Finite elements modeling

■ ■ ■

682

L 350.47

Linear element: S beam, 6 nodes, 1 linear element.

L 350.47

ADVANCE DESIGN VALIDATION GUIDE

Work ratio diagram

Simply supported beam subjected to bending Strength work ratio

13.12.2.3Reference results Result name

Result description

Reference value

Ď&#x192;m,d Strength work ratio

Design bending stress [Pa]

7403906.25 Pa

Work ratio (6.11) [%]

50 %

winst (Q)

Deflection for a base variable action [m]

0.00749 m

dCQ

Deflection for a characteristic combination of actions [m]

0.00999 m

winst

Instantaneous deflection [m]

0.00999 m

kdef

Deformation coefficient

0.6

dQP

Deflection for a quasi-permanent combination of actions [m]

0.00475 m

wfin

Final deflection [m]

0.01284 m

wnet,fin

Net deflection [m]

0.01284 m

13.12.3Calculated results Result name Stress

Result description Design bending stress

Value 7.40391e+006 Pa

50 %

Error -0.0001 %

Work ratio

Work ratio (6.11)

0.0000 %

D

Deflection for a base variable action

0.00714509 m

-4.6335 %

D

Deflection for a characteristic combination of actions

0.00952678 m

-4.6335 %

Winst

Instantaneous deflection

0.00998966 adim

-0.0000 %

Kdef

Deformation coefficient

0.6 adim

0.0000 %

D

Deflection for a quasi-permanent combination of actions

0.00452522 m

-4.6336 %

Wfin

Final deflection

0.0128367 adim

0.0001 %

Wnet,fin

Net deflection

0.0128367 adim

0.0001 %

683

ADVANCE DESIGN VALIDATION GUIDE

13.13 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression Test ID: 4879 Test status: Passed

13.13.1Description Verifies the stability of a rectangular timber purlin made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.

13.13.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made according to formulae (6.23) and (6.24) from EN 1995-1-1 norm.

13.13.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.4; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m.

The following load cases and load combination are used: ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W; Loadings from the structure: G = 550 N/m2; Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; Axial compression force due to wind effect on the supporting elements: W = 15000 N; Uniformly distributed load corresponding to the ultimate limit state combination: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.

All loads will be projected on the purlin direction since its slope is 30% (17°).

Simply supported purlin subjected to loadings

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

684

Height: h = 0.20 m, Width: b = 0.10 m, Length: L = 3.50 m, Section area: A = 0.02 m2 ,

ADVANCE DESIGN VALIDATION GUIDE

■

Elastic section modulus about the strong axis, y:

Wy =

b × h 2 0 .1 ⋅ 0 .20 2 = = 0 .000666 m 3 , 6 6

■

Elastic section modulus about the strong axis, z:

Wz =

b 2 × h 0 .12 ⋅ 0 .20 = = 0 .000333 m 3 .

6 6

Materials properties

Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material: ■ ■ ■ ■ ■

Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, Characteristic bending strength: fm,k = 24 x 106 Pa, Longitudinal elastic modulus: E = 1.1 x 1010 Pa, Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, Service class 2.

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Support at start point (z = 0) restrained in translation along X, Y, Z; Support at end point (z = 3.50) restrained in translation along Y, Z and restrained in rotation along X. Inner: None. ► ►

Loading

The purlin is subjected to the following projected loadings (corresponding to the ultimate limit state combination): ■

■

External: ► Axial compressive load: N =0.9 x W = 13500 N; ► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x sin17° = 1101.22 N/m, ► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x cos17° = 3601.92 N/m, Internal: None.

13.13.2.2Reference results in calculating the timber purlin subjected to combined stresses In order to verify the stability of a timber purlin subjected to biaxial bending and axial compression at ultimate limit state, the formulae (6.23) and (6.24) from EN 1995-1-1 norm are used. Before applying these formulae we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we’ll calculate the maximum work ratio for stability verification, which represents in fact the reference solution.

Slenderness ratios

The slenderness ratios corresponding to bending about y and z axes are determined as follows: ■

Slenderness ratio corresponding to bending about the z axis:

λ z = 12 ■

m × lg lc 1× 3.5m = 12 = 12 = 121.24 b b 0.1m

Slenderness ratio corresponding to bending about the y axis:

λ y = 12

m × lg lc 1× 3.5m = 12 = 12 = 60.62 h h 0.2m

685

ADVANCE DESIGN VALIDATION GUIDE

Relative slenderness ratios

The relative slenderness ratios are: ■

Relative slenderness ratio corresponding to bending about the z axis:

λz π

λrel , z = ■

E0, 05

=

121.24

π

21× 10 6 Pa = 2.056 0.74 × 1010 Pa

Relative slenderness ratio corresponding to bending about the y axis:

λrel , y = ■

f c , 0,k

λy π

f c , 0,k E0,05

=

60.62

π

21× 10 6 Pa = 1.028 0.74 × 1010 Pa

Maximum relative slenderness ratio:

λrel ,max = max(λrel , z , λrel , y ) = 2.056

So there is a risk of buckling because λrel,max ≥ 0.3.

Instability factors

In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1: βc = 0.2 (according to relation 6.29 from EN 1995-1-1) The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z2] (according to relation 6.28 from EN 1995-1-1) ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)

k c, z = k c, y =

1 k z + k z2 − λ2rel , z 1 k y + k y2 − λ2rel , y

(according to relation 6.26 from EN 1995-1-1)

(according to relation 6.25 from EN 1995-1-1)

Reference solution for ultimate limit state verification

Before calculating the reference solution (the maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1) it is necessary to determine the design compressive stress, the design compressive strength, the design bending stress, the design bending strength and some parameters involved in calculations (kmod, γM, kh, ksys, km).

■

Design compressive stress (induced by the axial compressive load from the corresponding ULS combination, N): σc,0,d =

■

Design bending stress about the y axis (induced by uniformly distributed load, qz):

σm,y,d = ■

686

My

q × L2 = z = Wy 8 ×Wy

N × 3.50 2 m 2 m = 8.2814 × 10 6 Pa 8 × 0.000666m 3

3601.92

Design bending stress about the z axis (induced by uniformly distributed load, qy):

σm,z,d = ■

N 13500 N = = 675000 Pa A 0.02m 2

N 2 1101.22 × 3.50 2 m 2 M z qy × L m = = 5.0638 × 10 6 Pa = 3 Wz 8 × Wz 8 × 0.000333m

Modification factor for duration of load (instantaneous action) and moisture content (service class 2):

ADVANCE DESIGN VALIDATION GUIDE

kmod = 1.1 (according to table 3.1 from EN 1995-1-1) ■

Partial factor for material properties: γM = 1.3

■

Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0

■

System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■

Design compressive strength: fc,0,d =

■

f c , 0,k ×

γM

= 21× 10 6 ×

1.1 = 17.769 × 10 6 Pa 1.3

Design bending strength: fm,y,d = fm,z,d =

■

k mod

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

1.1 × 1.0 × 1.0 = 20.308 × 10 6 Pa 1.3

Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:

σ m , y ,d σ ⎧ σ c ,0,d ⎫ + + k m ⋅ m, z ,d ⎪ ⎪ f m, y ,d f m , z ,d ⎪ ⎪ k c , y ⋅ f c , 0,d max ⎨ ⎬ ≤1 σ σ σ m y d , , c d m z d , 0 , , , ⎪ ⎪ + km ⋅ + ⎪ k c , z ⋅ f c , 0,d ⎪ f f m , y ,d m , z ,d ⎭ ⎩ Finite elements modeling

■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element.

687

ADVANCE DESIGN VALIDATION GUIDE

Instability factor, kcy

Simply supported purlin subjected to biaxial bending and axial compression Kc,y

Instability factor, kcz

Simply supported purlin subjected to biaxial bending and axial compression Kc,z

Maximum work ratio for stability verification

Simply supported purlin subjected to biaxial bending and axial compression Work ratio [%]

13.13.2.3Reference results Result name

Result description

Reference value

Kc,y

Instability factor, kc,y

0.67

Kc,z

Instability factor, kc,z

0.21

Work ratio

Maximum work ratio for stability verification [%]

71.1 %

13.13.3Calculated results

Result name Kc,y

688

Result description Instability factor, kc,y

Value 0.665025 adim

Error -0.0001 %

Kc,z

Instability factor, kc,z

0.212166 adim

0.0002 %

Work ratio

Maximum work ratio for stability verification

0.706586 %

-0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

13.14 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes Test ID: 4896 Test status: Passed

13.14.1Description Verifies the residual cross section of a column exposed to fire for 60 minutes. The column is made from glued laminated timber GL24 and it has only 3 faces exposed to fire. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.

13.14.2Background Verifies the adequacy of the cross sectional resistance for a rectangular cross section, which is made from glued laminated timber GL24, exposed to fire for 60 minutes on 3 faces. The verification is made according to chapter 4.2.2 from EN 1995-1-2 norm.

13.14.2.1Model description ■ ■ ■

Reference: Guide de validation Eurocode 5, test F.1; Analysis type: static linear (plane problem); Element type: linear.

Timber column with fixed base

Units

Metric System Geometry

Below are described the column cross section characteristics: ■ ■ ■ ■

Depth: h = 0.60 m, Width: b = 0.20 m, Section area: A = 0.12 m2 Height: H = 5.00 m

689

ADVANCE DESIGN VALIDATION GUIDE

Materials properties

Glued laminated timber GL24 is used. The following characteristics are used in relation to this material: ■

Density: ρ = 380 kg/m3,

■

Design charring rate: βn = 0.7 x 10-3 m/min,

Boundary conditions

The boundary conditions are described below: ■

Outer:

■

Fixed at base (Z = 0), ► Support at top (Z = 5.00) restrained in translation along X and Y, Inner: None. ►

Loading

The column is subjected to the following loadings: ■ ■

External: Point load at Z = 5.00: Fz = N = - 100000 N, Internal: None.

13.14.2.2Reference results in calculating the cross sectional resistance of a timber column exposed to fire Reference solution

The reference solution (residual cross section) is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).

■ ■

Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; Coefficient depending of fire resistance time and also depending if the members are protected or not: k0 = 1.0 (according to table 4.1 from EN 1995-1-2)

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Notional design charring depth: dchar,n = β n ⋅ t = 0.7 ⋅ 10 − 3

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m ⋅ 60 min = 0.042m (according to relation 3.2 from EN 1995-1-2) min

Effective charring depth: def = dchar,n + k 0 ⋅ d0

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Residual cross section: Afi = (h − def ) × (b − 2 ⋅ def )

Finite elements modeling

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690

Linear element: S beam, 6 nodes, 1 linear element.

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Residual cross section

Column with fixed base exposed to fire for 60 minutes Afi

13.14.2.3Reference results Result name

Result description

Reference value

Afi

Residual cross section [m2]

0.056202 m2

13.14.3Calculated results

Result name Afi

Result description Residual cross section

Value 0.056202 adim

Error -0.0000 %

691

14 XML Template Files

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14.1 Loading a template with the properties of a planar element (DEV2012 #1.4) Test ID: 4344 Test status: Passed

14.1.1 Description Loads an XML file as a properties template and applies it on a planar element. Saves the properties of the element on which the template was applied, in order to compare the files.

14.2 Loading a template with the properties of a linear element (DEV2012 #1.4) Test ID: 4207 Test status: Passed

14.2.1 Description Loads an XML file as a properties template and applies it on a linear element. Saves the properties of the element on which the template was applied, in order to compare the files.

14.3 Saving the properties of a planar element as a template (DEV2012 #1.4) Test ID: 4176 Test status: Passed

14.3.1 Description Saves the properties of a planar element as an XML file which can be later used as a template for other planar elements.

14.4 Saving the properties of a linear element as a template (DEV2012 #1.4) Test ID: 4169 Test status: Passed

14.4.1 Description Saves the properties of a linear element as an XML file which can be later used as a template for other linear elements.

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