PCA%20-%208_Connection_Design

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PCI 6 Edition Connection Design

Presentation Outline • Structural Steel Design • Limit State Weld Analysis • Strut – Tie Analysis for Concrete Corbels • Anchor Bolts • Connection Examples

Changes • New method to design headed studs (Headed Concrete Anchors - HCA) • Revised welding section – Stainless Materials – Limit State procedure presented

• Revised Design Aids (moved to Chapter 11) • Structural Steel Design Section – Flexure, Shear, Torsion, Combined Loading – Stiffened Beam seats

• Strut – Tie methodology is introduced • Complete Connection Examples

Structural Steel Design • Focus on AISC LRFD 3rd Edition – Flexural Strength – Shear Strength – Torsional Strength – Combined Interaction

• Limit State Methods are carried through examples

Structural Steel Details

• Built-up Members • Torsional Strength • Beam Seats

Steel Strength Design • Flexure

φMp = φ·Fy·Zs Where: φMp = Flexural Design Strength Fy = Yield Strength of Material Zs = Plastic Section Modulus

Steel Strength Design • Shear

φVn = φ(0.6·Fy)·Aw Where: φVp = Shear Design Strength Aw = Area subject to shear

Steel Strength Design • Torsion (Solid Sections)

φTn = φ(0.6·Fy)·α·h·t2 Where: φTp = α = h = t =

Torsional Design Strength Torsional constant Height of section Thickness

Torsional Properties

• Torsional Constant, α • Rectangular Sections

Steel Strength Design • Torsion (Hollow Sections)

φTn = 2·φ(0.6·Fy)·Ᾱ·t Where: φTp = Torsional Design Strength Ᾱ = Area enclosed by centerline of walls t

= Wall thickness

Torsional Properties

• Hollow Sections

Ᾱ = w·d

Combined Loading Stress • Normal Stress

P Mc M fn = , , A I S

• Bending Shear Stress fv

bending

• Torsion Shear Stress fv

torsion

VQ V = , It A

Tc T T = , , 2 J αht 2At

Combined Loading • Stresses are added based on direction • Stress Limits based on Mohr’s circle analysis – Normal Stress Limits φfun = φ ⋅ fy φ = 0.90

– Shear Stress Limits φfuv = φ ⋅ 0.60fy

(

φ = 0.90

)

Built-Up Section Example

Example

∑F

x

=0

T−C =0 A tF y −A cFy = 0

At = Ac

Determine Neutral Axis Location, y

Tension Area

Compression Area

A t = 4in ⋅ y Tension = Compression

A c = 2 ⋅ 3 in ⋅1in +  3 in − y  ⋅ 4in  8  8 A c = 2.25 − 4 ⋅ y 4 ⋅ y = 2.25 − 4 ⋅ y 2.25 y= = 0.281 in 8

Define Plastic Section Modulus, Zp

Either Tension or Compression Area x Distance between the Tension / Compression Areas Centroids

(

Zp = A t H − y t − y c

)

Determine Centroid Locations • Tension

y 0.281 yt = = = 0.14 in 2 2

• Compression yc

__

Ay ∑ = ∑A

= 0.683 in

Calculate Zp

( ) Z = (4 ⋅ y )(H − y − y ) = (4 ⋅ 0.281)(1.375 − 0.14 − 0.683) Zp = A t H − y t − y c

p

Zp

t

Zp = 0.62 in3

c

Beam Seats • Stiffened Bearing – Triangular – Non-Triangular

Triangular Stiffeners • Design Strength

φVn=φ·Fy·z·b·t Where: φVn = Stiffener design strength φ = Strength reduction factor = 0.9 b = Stiffener projection t = Stiffener thickness z = Stiffener shape factor

Stiffener Shape Factor

b 0.75 ≤ ≤ 2.0 a 2

 b  b  b z = 1.39 − 2.2   + 1.27   + 0.25    a  a  a

3

Thickness Limitation

b 250 ≤ t Fy

Triangular Stiffener Example Given: A stiffened seat connection shown at right. Stiffener thickness, ts = 3/8 in.

Fy = 36 ksi

Problem: Determine the design shear resistance of the stiffener.

Shape Factor b 8 = + 0.8 > 0.75 and < 1.0 a 10 2

 b  b  b z = 1.39 − 2.2   + 1.27   + 0.25    a  a  a

( )

3

( ) + 0.25 (0.8 )

z = 1.39 − 2.2 0.8 + 1.27 0.8 z = 0.315

2

3

Thickness Limitation b 250 ≤ t Fy 8 = 21.3 0.375

250 36

21.3 ≤ 41.7

= 41.7

Design Strength φVn = φ ⋅ Fy ⋅ z ⋅ b ⋅ t

(

)(

)( )(

φVn = 0.9 36 ksi 0.315 8 in 0.375 in φVn = 28.9 kips

)

Weld Analysis • Elastic Procedure • Limit State (LRFD) Design introduced • Comparison of in-plane “C” shape – Elastic Vector Method - EVM – Instantaneous Center Method – ICM

Elastic Vector Method – (EVM) • Stress at each point calculated by mechanics of materials principals

fx = fy = fz = fr =

Px Aw Py Aw Pz Aw

+ + +

Mz y Ip Mz x Ip Mx y I xx

+

My x I yy

fx 2 + fy 2 + fz2

Elastic Vector Method – (EVM) • Weld Area ( Aw ) based on effective throat • For a fillet weld:

Aw =

a 2

lw

Where: a = Weld Size lw = Total length of weld

Instantaneous Center Method (ICM) â&#x20AC;˘ Deformation Compatibility Solution â&#x20AC;˘ Rotation about an Instantaneous Center

Instantaneous Center Method (ICM) • Increased capacity – More weld regions achieve ultimate strength – Utilizes element vs. load orientation

• General solution form is a nonlinear integral • Solution techniques – Discrete Element Method – Tabular Method

ICM Nominal Strength • An elements capacity within the weld group is based on the product of 3 functions.

– Strength – Angular Orientation – Deformation Compatibility

Rn = j

f ⋅ g ⋅h

Strength, f

f = 0.6 ⋅ FEXX ⋅ A w Aw - Weld area based on effective throat

Angular Orientation, g Weld capacity increases as the angle of the force and weld axis approach 90o Rj = R â&#x2039;&#x2026; g

()

g = 1.0 + 0.5 sin θ

3

2

Deformation Compatibility, h   r ∆u ∆u   rcritical rcritical 1.9 − 0.9  h= −0.32 −0.32  0.209 θ + 2 a 0.209 θ + 2 a     r

(

)

(

0.3

)

Where the ultimate element deformation ∆u is:

(

∆u = 1.087 θ + 6

)

−0.64

a ≤ 0.17a

Element Force

{

R n = 0.6FEXX A w j

}

()

 ⋅ 1.0 + 0.5 sin θ 

3

2

 ⋅ 

0.3     r r   ∆u ∆ u     rcritical rcritical 1.9 − 0.9    −0.32 −0.32   0.209 θ + 2 a 0.209 θ + 2 a         

(

)

(

)

Where: r and θ are functions of the unknown location of the instantaneous center, x and y

Equations of Statics ∑F

y

∑M

IC

=0

Number of Elements

∑

R n − Pn = 0 yj

j=1

=0

Number of Elements

∑ j=1

(

)

R n rj − Pn e + r0 = 0 j

Tabulated Solution • AISC LRFD 3rd Edition, Tables 8-5 to 8-12 φVn = C·C1· D·l Where: D = C = C1 = l =

number of 16ths of weld size tabulated value, includes φ electrode strength factor weld length

Comparison of Methods â&#x20AC;˘ Page 6-47:

Corbel Design • Cantilever Beam Method • Strut – Tie Design Method • Design comparison – Results comparison of Cantilever Method to Strut – Tie Method

• Embedded Steel Sections

Cantilever Beam Method Steps Step 1 – Determine maximum allowable shear Step 2 – Determine tension steel by cantilever Step 3 – Calculate effective shear friction coeff.

Step 4 – Determine tension steel by shear friction Step 5 – Compare results against minimum Step 6 – Calculate shear steel requirements

Cantilever Beam Method • Primary Tension Reinforcement • Greater of Equation A or B   a  h  Eq. A  Vu   + Nu     d     d   1   2Vu   Eq. B A s =  + Nu  φfy   3µ e   1 As = φfy

• Tension steel development is critical both in the column and in the corbel

Cantilever Beam Method • Shear Steel

A h ≥ 0.5  A s + A n  • Steel distribution is within 2/3 of d

Cantilever Beam Method Steps Step 1 – Determine bearing area of plate Step 2 – Select statically determinate truss Step 3 – Calculate truss forces Step 4 – Design tension ties Step 5 – Design Critical nodes Step 6 – Design compression struts Step 7 – Detail Accordingly

Strut – Tie Analysis Steps Step 1 – Determine of bearing area of plate A pl ≥

Vu φ ⋅ 0.85 ⋅ f`c

φ = 0.75

Strut â&#x20AC;&#x201C; Tie Analysis Steps Step 2 â&#x20AC;&#x201C; Select statically determinate truss AC I provides guidelines for truss angles, struts, etc.

Strut â&#x20AC;&#x201C; Tie Analysis Steps Step 3 â&#x20AC;&#x201C; Determine of forces in the truss members Method of Joints or Method of Sections

Strut – Tie Analysis Steps Step 4 – Design of tension ties As =

Fnt φfy

φ = 0.75

Strut – Tie Analysis Steps Step 5 – Design of critical nodal zone fcu = 0.85 ⋅ β n ⋅f`c where: βn = 1.0 in nodal zones bounded by structure or bearing areas = 0.8 in nodal zones anchoring one tie = 0.6 in nodal zones anchoring two or more ties

Strut – Tie Analysis Steps Step 6 – Check compressive strut limits based on Strut Shape The design compressive strength of a strut without compressive reinforcement φFns = φ·fcu·Ac where: φ = 0.75 Ac = width of corbel × width of strut

Strut – Tie Analysis Steps Compression Strut Strength • From ACI 318-02, Section A.3.2:

fcu = 0.85 ⋅ βs ⋅ f`c Where: βs – function of strut shape / location = 0.60λ, bottle shaped strut = 0.75, when reinforcement is provided = 1.0, uniform cross section = 0.4, in tension regions of members = 0.6, for all other cases

Strut â&#x20AC;&#x201C; Tie Analysis Steps

Step 7 â&#x20AC;&#x201C; Consider detailing to ensure design technique

Corbel Example

Given: Vu = 80 kips Nu = 15 kips fy = Grade 60 f′c = 5000 psi Bearing area – 12 x 6 in.

Problem: Find corbel depth and reinforcement based on Cantilever Beam and Strut – Tie methods

Step 1CBM – Cantilever Beam Method (CBM)

h = 14 in d = 13 in. a = ¾ lp = 6 in.

From Table 4.3.6.1 2

Vumax = 1000 ⋅ λ A

cr

=

( )( )( ) = 196 kips ≥ 80 kips

1000 12 14 14 1000

Step 2CBM – Tension Steel

• Cantilever Action 1 As = φfy

  a  h  1  Vu   + Nu    =  d   .75 60   d 

= 1.18 in2

 6  14   80   + 15     13     13 

( )

Step 3CBM – Effective Shear Friction Coefficient

( )( )( )( )

1000 ⋅ λ ⋅ b ⋅ h ⋅ µ 1000 1 14 14 1.4 µe = = Vu 80 = 3.43 ≥ 3.4 Use µ e = 3.4

Step 4CBM – Tension Steel

• Shear Friction

( ) ( ) ( )

 1   2Vu  1   As = + N =  u φfy   3µ e   0.75 60 = 0.68 in2

  2 80     + 15    3 3.4    

Step 5CBM – As minimum

A s,min

f`c

( )( )

5 = 0.4 ⋅ b ⋅ d = 0.4 14 13 φfy 60

= 0.61 in2

As based on cantilever action governs As = 1.18 in2

Step 6CBM – Shear Steel  15    A h = 0.5  A s − A n  = 0.5 1.18 −  0.75 60  = 0.42 in

   

( )

Use (2) #3 ties = (4) (0.11 in2) = 0.44 in2 Spaced in top 2/3 (13) = 8 ½ in

Step 1ST â&#x20AC;&#x201C; Strut - Tie Solution (ST) Determination of bearing plate size and protection for the corner against spalling Required plate area:

A bearing =

(

Vu

Ď&#x2020; 0.85f`c

)

=

80

(

0.75 0.85f`c

)

= 25.1 in2 Use 12 by 6 in. plate, area = 72 in2 > 25.1 in2

Step 2ST – Truss Geometry tan θR=Nu / Vu = (15)/(80) = 0.19 l1 = (h - d) tanθR + aw + (hc - cc) = (14 - 13)(0.19) + 6 + (14 - 2.25) = 17.94 in. l2 = (hc - cc) – ws/2 = (14 - 2.25) - ws/2 = 11.75 - ws/2

Step 2ST – Truss Geometry Find ws Determine compressive force, Nc, at Node ‘p’: ∑Mm = 0 Vu·l 1+Nu·d – Nc·l 2=0 [Eq. 1]

(80)(17.94) + (15)(13) – Nc(11.75 – 0.5ws) = 0 [Eq. 2]

Step 2ST – Truss Geometry • Maximum compressive stress at the nodal zone p (anchors one tie, βn = 0.8) fcu = 0.85·βn·f`c = 0.85(0.8)(5)= 3.4 ksi An = area of the nodal zone = b·ws = 14ws

Step 2ST – Determine ws , l2 • From Eq. 2 and 3 0.014Nc2 - 11.75Nc - 1630 = 0 Nc = 175 kips ws = 0.28Nc = (0.28)(175) = 4.9in l2 = 11.75 - 0.5 ws = 11.75 - 0.5(4.9) = 9.3

Step 3ST – Solve for Strut and Tie Forces • Solving the truss ‘mnop’ by statics, the member forces are: Strut op Tie no Strut np Tie mp Tie mn

= = = = =

96.0 kips (c) 68.2 kips (t) 116.8 kips (c) 14.9 kips (t) 95.0 kips (t)

Step 4ST – Critical Tension Requirements

• For top tension tie ‘no’ Tie no = 68.2 kips (t)

As =

Fnt φfy

=

62

( )

0.75 60

= 1.52in2 Provide 2 – #8 = 1.58 in2 at the top

Step 5ST – Nodal Zones • The width `ws’’ of the nodal zone ‘p ’ has been chosen in Step 2 to satisfy the stress limit on this zone • The stress at nodal zone ‘o ’ must be checked against the compressive force in strut ‘op ’ and the applied reaction, Vu • From the compressive stress flow in struts of the corbel, Figure 6.8.2.1, it is obvious that the nodal zone ‘p ’ is under the maximum compressive stress due to force Nc. • Nc is within the acceptable limit so all nodal zones are acceptable.

Step 6ST – Critical Compression Requirements

• Strut ‘np’ is the most critical strut at node ‘p’.

The nominal compressive strength of a strut without compressive reinforcement Fns = fcu·Ac Where: Ac = width of corbel × width of strut

Step 6ST – Strut Width

• Width of strut ‘np’ Strut Width =

ws o

sin(54.4 )

= 6.03 in

=

4.9 sin(54.4 o )

Step 6ST – Compression Strut Strength • From ACI 318-02, Section A.3.2:

fcu = 0.85 ⋅ β s ⋅ f`c Where - bottle shaped strut, βs = 0.60λ

()

fcu = 0.85 ⋅ 0.6 1 ⋅ 5 = 2.55 ksi

(

) ( )(

)

φFns = φ ⋅ f cu ⋅ Ac = 0.75 2.55  14 6.03  = 161.5 kips  

161 kips ≥ 116.8 kips

OK

Step 7ST – Surface Reinforcement

• Since the lowest value of βs was used, surface reinforcement is not required based on ACI 318 Appendix A

Example Conclusion

Cantilever Beam Method

Strut-and-Tie Method

Embedded Steel Sections

Concrete and Rebar Nominal Design Strengths

• Additional Tension Compression Reinforcement Capacity 2 ⋅ A s ⋅ fy

Vr =

6e 1+

4.8s

le

le

−1

Corbel Capacity

• Reinforced Concrete

(

φVn = φ Vc + VR φ = 0.75

)

Steel Section Nominal Design Strengths • Flexure - Based on maximum moment in section; occurs when shear in steel section = 0.0 φVn =

φ ⋅ Zs ⋅ fy a+

0.5 ⋅ Vu

0.85 ⋅ f`c ⋅b

Where: b = effective width on embed, 250 % x Actual φ = 0.9

Steel Section Nominal Design Strengths

• Shear

φVs = φ ⋅ 0.6fy ⋅ h ⋅ t where: h, t = depth and thickness of steel web φ = 0.9

Anchor Bolt Design

â&#x20AC;˘ ACI 318-2002, Appendix D, procedures for the strength of anchorages are applicable for anchor bolts in tension.

Strength Reduction Factor

Function of supplied confinement reinforcement Ď&#x2020; = 0.75 with reinforcement Ď&#x2020; = 0.70 with out reinforcement

Headed Anchor Bolts No = Cbs·AN·Ccrb·Ψed,N Where: Ccrb = Cracked concrete factor, 1 uncracked, 0.8 Cracked AN = Projected surface area for a stud or group Ψed,N =Modification for edge distance Cbs = Breakout strength coefficient Cbs = 2.22 ⋅ λ ⋅

f 'c

3

hef

Hooked Anchor Bolts No = 126·f`c·eh·do·Ccrp Where: eh = hook projection ≥ 3do do = bolt diameter Ccrp = cracking factor (Section 6.5.4.1)

Column Base Plate Design â&#x20AC;˘ Column Structural Integrity requirements 200Ag

Completed Connection Examples • Examples Based – Applied Loads – Component Capacity

• Design of all components – Embeds – Erection Material – Welds

• Design for specific load paths

Completed Connection Examples • Cladding “Push / Pull” • Wall to Wall Shear

• Wall Tension • Diaphragm to Wall Shear

Questions?