From steelfiber to industrial floor

Page 1

From steelfiber to industrial floor Technical design guidelines

Hendrik Thooft Bekaert Building Products R&D Mgr. Date City

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Table of content • Steelfiber • Steelfiber concrete – EN 14651 Beamtest – Design values

• Basic design principles • ULS – Resisting forces – Acting forces • SLS • Conclusions

Chap 1 ; steelfiber ; 3 types

3

4 5

Chap 1 ; steelfiber ; wire properties Tensile curves different wire qualities 2500

Stress (MPa)

2000

1500 3D 65/60BG 4D 65/60BG 5D 65/60BG

1000

PP yarn

PP Yarn

500

E mod Steel E mod PP

0

0,00

1,00

2,00

3,00

4,00

5,00

Strain (%)

6,00

7,00

8,00

9,00

Chap 1 ; steelfiber ; Pull out

300 N

500 N

1200 N +140 %

+67 % +300 %

Chap 2 ; steelfiber concrete ; material characterisation

EN 14651 hsp = 125 mm b = 150 mm

Chap 2 ; steelfiber concrete ; Performance

Chap 2 ; steelfiber concrete ; minimal reinforcement ď łN fLK fR1k

0 0.5 Jointed slabs on ground: Jointless slabs on ground: Structures

fR3k

đ?’‡đ?‘šđ?&#x;?đ?’Ž đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł đ?’‡đ?‘šđ?&#x;?đ?’Ž đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł

2.5 đ?’‡đ?‘šđ?&#x;‘đ?’Ž ≼ đ?&#x;Ž, đ?&#x;’ and đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł ≼ đ?&#x;Ž, đ?&#x;‘ đ?’‡

đ?‘šđ?&#x;‘đ?’Ž ≼ đ?&#x;Ž, đ?&#x;” and đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł ≼ đ?&#x;Ž, đ?&#x;’

fR1k/fLk > 0.4 and fR3k/fR1k > 0.5 *

*(modelcode 2010 - 5.6) Fibre reinforcement can substitute (also partially) conventional reinforcement at ultimate limit state if the following relationships are fulfilled:

Chap 3 ; Basic design principles 2. SLS ; Design of operational entegrity and durability

1. ULS ; Design of structural interity and safety

• •

Fu F*cr

gmaterial

Fcr Fk

gload

dcr

d*cr

du

Crack opening Deflections

Chap 3 ; Basic design principles 1. ULS ; Design of structural interity and safety

đ??¸đ?‘‘ đ?‘…đ?‘‘

đ?‘Źđ?’… < đ?‘šđ?’…

denotes the design action-effect denotes the design resistance

��� ≤ ��� ��� ≤ ���

Chap 3 ; Basic design principles

Partial Safety Factor – floors on grade permanent loads gG

1,20

variable loads gQ

1,20

concrete in compression and tension gc

1,50

steel fibre reinforced concrete in tension gSF

1,20

steel bars / steel fabric gS

1,15

đ??¸đ?‘‘

the design actions

Chap 4 ; �� – ULS – load cases ��� : is the occuring moment from factored loads ��� • • • •

: Is the occuring shear stresses from factored loads

Point loads (Single or back to back racking systems,‌) Wheel loads (Forcklift trucks, Trucks, ‌) Line loads (Non-bearing walls,‌) UDL loads (Bulk storage,‌)

Chap 4 ; đ?‘Źđ?’… : ULS đ?‘´đ?‘Źđ?’… design principle elastic-elastic section based: mel

elastic-plastic section based: mel

plastic-plastic system based: mpl

Resistance

section based: fflď‚ W or feqď‚ W

system based: (ffl + feq)ď‚ W

system based: (ffl + feq)ď‚ W

f fl f eq mel mpl m m’ W

first crack flexural strength Losberg design post crack flexural strength guideline bending moment according linear-elastic design ÎŁ (m, m’) according plastic calculation (~85% á 90% max mel) positive bending moments according plastic calculation negative bending moments according plastic calculation section modulus

bending moment

Chap 4 ; đ?‘Źđ?’… : ULS −đ?‘Źđ?’?đ?’‚đ?’”đ?’•đ?’Šđ?’„ đ?’?đ?’†đ?’?đ?’ˆđ?’‰đ?’• đ?‘™đ?‘’đ?‘™ =

4

đ??¸đ?‘? ∙ â„Ž3 12 ∙ 1 − đ?œˆ 2 ∙ đ?‘˜ Defining k compaction modulus EV1 and EV2

Multi layer model steel fibre concrete slab

subbase layer 3

h3, E3

k2

subbase layer 2

h2, E2

k1

subbase layer 1

h1, E1

k0

subsoil

E k h

k3

dynamic modulus of elasticity modulus of subgrade reaction layer thickness

k0 [N/mm²] or [MN/m²] [N/mm³] [mm]

đ??¸

đ??¸

đ?‘‰1 đ?‘‰2 đ?‘˜ ≈ 550 = 550∙đ?›ź ,

with

đ??¸

đ?›ź = đ??¸đ?‘‰2 ≤ 2,4 (2,0) and k in N/mmÂł đ?‘‰1

EV1: compaction modulus determined in 1st load cycle in N/mm² or MN/m² EV2: compaction modulus determined in 2nd load cycle in N/mm² or MN/m² Îą: a value ≤ 2,0 is recommended, the maximum value is 2,4

Chap 4 ; đ?‘Źđ?’… : ULS −đ?‘Źđ?’?đ?’‚đ?’”đ?’•đ?’Šđ?’„ đ?’?đ?’†đ?’?đ?’ˆđ?’‰đ?’• Defining k Direct plate test

CBR to k-value •

•

CBR California Bearing Ratio E dynamic modulus of elasticity k modulus of subgrade reaction

[%] [N/mm²] or [MN/m²] [N/mm³]

standard plate diameter for the determination of the k-value – 762 mm (DIN) = 30 in (ASTM) – 750 mm (BS) correct the k-value if a smaller slab diameter is used – the use of the standard plate (750 / 762 mm or 30 in) is strongly recommended – k750 has to be used for all calculations 750đ?‘šđ?‘š ; D ≤ 750 đ?‘šđ?‘š đ??ˇ đ??ˇ ; đ??ˇ ≼ 300 đ?‘šđ?‘š 750đ?‘šđ?‘š

–

đ?‘˜đ??ˇ = đ?‘˜750đ?‘šđ?‘š ∙

–

đ?‘˜750 = đ?‘˜đ??ˇ ∙

Chap 4 ; 𝑬𝒅 – ULS – 𝑴𝑬𝒅 - 1 center pointload t rc m’Ed

VEd

m’Ed

actual pressure distribution

r0

assumed pressure distribution p0d

pd,max r0 t

∙ p0d

𝑚𝐸𝑑 + 𝑚′

𝐸𝑑

=

𝑉𝐸𝑑 ∙ 1−𝛾∙𝜋∙ 2∙𝜋

𝑟0 2 𝑙𝑒𝑙

𝑠𝑥 ∙ 1 − 23 ∙ 𝑟𝑡0+2∙𝜋∙𝑟

0

𝑟 ∙ 1− 0 2∙𝑡

𝑖𝑓 𝑠𝑥 = 0: 𝑟0 𝑙𝑒𝑙

rc m’Ed

=

3

𝑟𝑐 Τ𝑙𝑒𝑙

𝛾∙𝜋∙

mEd r0

3𝑟 1− ∙ 𝑜 4 𝑡

=

3

3Τ ∙𝜋∙𝑠 Τ𝑙 𝑦 𝑒𝑙 4 3𝑟 𝛾∙𝜋∙ 1− ∙ 𝑜 4 𝑡

𝑖𝑓 𝑠𝑥 > 0: 𝑟0 𝑙𝑒𝑙

=

4𝑟 1+ ∙ 𝑐

2∙𝛾∙𝜋∙

3 𝑠𝑥 2 𝑟𝑜 1 𝑠𝑥 1𝑟 1− ∙ + ∙ ∙ 1− ∙ 𝑜 3 𝑡 𝜋 𝑟0 2 𝑡

2𝑟 + ∙ 0∙ 3 𝑠𝑥

3𝑟 1− ∙ 0 4 𝑡

=

1+ 2∙𝛾∙𝜋∙

2𝑟 1− ∙ 𝑜 3 𝑡

1 𝑠 + ∙ 𝑥∙ 𝜋 𝑟0

𝜋∙𝑠𝑦 𝑠𝑥

1 𝑟𝑜 2 𝑡

1− ∙

2𝑟 3𝑟 + ∙ 0 ∙ 1− ∙ 0 3 𝑠𝑥

4 𝑡

Chap 4 ; �� – ULS – ��� - 1 pointload at a joint tk

t rc

rc m’Ed

VEd

VEd

m’Ed

m’Ed mEd actual

rk

presumed p0d

pd,max

p0d r0 t

∙ p0d r0

a actual presumed m’Ed

rc

mEd

r0

đ?‘šđ??¸đ?‘‘ + đ?‘šâ€˛

rk = r0∙tana đ?‘&#x;0 đ?‘™đ?‘’đ?‘™

=

3

đ??¸đ?‘‘

=

đ?‘‰đ??¸đ?‘‘ ∙ 4

1+

đ?‘ đ?‘Ś ∙ tan đ?›ź − 43∙đ?›žđ?‘˜ ∙ đ?‘&#x;0

đ?‘ đ?‘Ś đ?‘ đ?‘Ľ + ∙cot đ?›ź đ?‘™đ?‘’đ?‘™ đ?‘™đ?‘’đ?‘™ 3đ?‘&#x; đ?‘Ą đ?›žđ?‘˜ ∙ 1− ∙ 0 ∙ 1+ ∙tan đ?›ź 8 đ?‘Ą đ?‘Ąđ?‘˜

1,5∙cot đ?›źâˆ™

đ?‘&#x;0 2 3đ?‘&#x; 5 đ?‘&#x; ∙ 1− ∙ 0 − ∙ 0 ∙tan đ?›ź đ?‘™đ?‘’đ?‘™ 8 đ?‘Ą 16 đ?‘Ąđ?‘˜

∙ tan2 đ?›ź

Chap 4 ; 𝑬𝒅 ∶ 𝐔𝐋𝐒 − 𝐏𝐮𝐧𝐜𝐡𝐢𝐧𝐠 − 𝐆𝐞𝐧𝐞𝐫𝐚𝐥 𝐩𝐫𝐢𝐧𝐜𝐢𝐩𝐥𝐞 𝑉𝐸𝑑 − 𝑉𝑠𝑜𝑖𝑙 = 𝑉𝐸𝑑,𝑟𝑒𝑑 ≤ 𝑉𝑅𝑑,𝑐 + 𝑉𝑅𝑑,𝑓

Chap 4 ; 𝑬𝒅 ∶ 𝐔𝐋𝐒 − 𝐕𝐬𝐨𝐢𝐥 𝐮𝐧𝐝𝐞𝐫 𝐞𝐝𝐠𝐞 𝐥𝐨𝐚𝐝 tedge,eq rc VEd

𝐑𝐞𝐬𝐢𝐬𝐭𝐢𝐧𝐠 𝐬𝐨𝐢𝐥 𝐟𝐨𝐫𝐜𝐞 − tedge,eq rc

acrit

acrit

if 𝒕𝒆𝒅𝒈𝒆,𝒆𝒒 − 𝒓𝒄 ≥ 𝒂𝒄𝒓𝒊𝒕 ⇒ VEd

𝑝𝑚,𝑎 = pm,a

actual

Psoil p0d

pd,max

𝑡𝑒𝑑𝑔𝑒,𝑒𝑞 − 𝑎𝑐𝑟𝑖𝑡 + 𝑟𝑐 ∙ 𝑝𝑂𝑑 𝑡𝑒𝑑𝑔𝑒,𝑒𝑞

𝑝𝑚,𝑏 = 13 ∙ 𝑝0𝑑 − 𝑝𝑚,𝑎

presumed

pm,b

p0d

𝑝𝑚 = 𝑝𝑚,𝑎 + 𝑝𝑚,𝑏 ⇒ 𝑃𝑠𝑜𝑖𝑙 = 𝑝𝑚 ∙ 𝜋/2 ∙ 𝑎𝑐𝑟𝑖𝑡 + 𝑟𝑐

if 𝒕𝒆𝒅𝒈𝒆,𝒆𝒒 − 𝒓𝒄 < 𝒂𝒄𝒓𝒊𝒕 ⇒ rc

𝑝𝑚,𝑎 = 0 acrit

𝑝𝑚,𝑏 = 13 ∙ 𝑝0𝑑 𝑝𝑚 = 𝑝𝑚,𝑏 ⇒ 𝑃𝑠𝑜𝑖𝑙 = 𝑝𝑚 ∙ 𝜋/2 ∙ 𝑡𝑒𝑑𝑔𝑒,𝑒𝑞 2

2

Chap 4 ; 𝑬𝒅 ∶ 𝐔𝐋𝐒 − 𝐕𝐬𝐨𝐢𝐥 𝐮𝐧𝐝𝐞𝐫 𝐜𝐞𝐧𝐭𝐞𝐫 𝐥𝐨𝐚𝐝

𝐑𝐞𝐬𝐢𝐬𝐭𝐢𝐧𝐠 𝐬𝐨𝐢𝐥 𝐟𝐨𝐫𝐜𝐞 − if 𝒕 − 𝒓𝒄 ≥ 𝒂𝒄𝒓𝒊𝒕 ⇒

t rc VEd

actual pressure distribution

pm,a

assumed pressure distribution

Psoil p0d pd,max

𝑝𝑚,𝑎

acrit

pm,b

𝑡 − 𝑎𝑐𝑟𝑖𝑡 + 𝑟𝑐 = ∙ 𝑝𝑂𝑑 𝑡

𝑝𝑚,𝑏 = 13 ∙ 𝑝0𝑑 − 𝑝𝑚,𝑎 𝑝𝑚 = 𝑝𝑚,𝑎 + 𝑝𝑚,𝑏

Average pressure under punching cone

if 𝒕 − 𝒓𝒄 < 𝒂𝒄𝒓𝒊𝒕 ⇒ 𝑝𝑚,𝑎 = 0 rc

acrit

𝑝𝑚,𝑏 = 13 ∙ 𝑝0𝑑 𝑝𝑚 = 𝑝𝑚,𝑏 ⇒ 𝑃𝑠𝑜𝑖𝑙 = 𝑝𝑚 ∙ 𝜋 ∙ 𝑡 2

Chap 4 ; đ?‘Źđ?’… âˆś đ??”đ??‹đ??’ − đ??•đ??„đ??? − đ??’đ??Ąđ??žđ??šđ??Ť đ??Źđ??­đ??Ťđ??žđ??Źđ??Ź đ?‘Łđ??¸đ?‘‘

đ?›˝ ∙ ( đ?‘‰đ??¸đ?‘‘ − đ?‘ƒđ?‘ đ?‘œđ?‘–đ?‘™ ) = đ?‘˘đ?‘– ∙ đ?‘‘

Location

DIN EC2 + NAD + DAfStb*

Internal

ďƒž b = 1,10

Edge

ďƒž b = 1,40

Corner

ďƒž b = 1,50

VEd : acting vertical load d : slab thickness ui : control perimeter

đ?‘Žđ?‘?đ?‘&#x;đ?‘–đ?‘Ą

đ?‘&#x;w đ?‘˘0

�1

Chap 4 ; �� – ULS – Load transfer at saw cut joints Load Transfer at Saw Cuts Scenario

indoor

outdoor

temperatur final e difference shrinkage ΔT [K] đ?œ€đ?‘ ∞ [‰]

temperatur final sum of e difference shrinkage strains [‰] ΔT [K] đ?œ€đ?‘ ∞ [‰]

sum of strains [‰]

1

1,00

0,8

0,80

0,6

0,60

very cold

30

0,25

0,55

30

0,25

0,55

tempered

20

0,25

0,45

5

0,40

0,45

0,4

0,40

humid

5

0,20

0,25

0

0,20

0,20

0,2

0,20

hot

20

0,40

0,60

0

0,60

0,60

∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą = ∆ đ?‘‡ ∙ đ?›ź đ?‘‡ + đ?œ€đ?‘ ∞ ∙ đ?‘™đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą đ?‘¤đ?‘–đ?‘Ąâ„Ž đ?›ź đ?‘‡ = 10 ∙ 10−6 Τđ??ž

joint efficiency

load multiplicator

0

0,00 0,0

1,0

2,0

3,0

4,0

5,0

Joint Opening [mm]

đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą đ?‘œđ?‘?đ?‘’đ?‘›đ?‘–đ?‘›đ?‘” ∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą ≤ 1.524 đ?‘šđ?‘š: ′ ÎŚđ?‘’đ?‘“đ?‘“ = −0.29 ∙ ∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą + 1.0636 Load reduction factor

đ?œ’đ?‘ đ?‘Žđ?‘¤ đ?‘?đ?‘˘đ?‘Ą = 1 − ÎŚđ?‘’đ?‘“đ?‘“ Τ2

đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą đ?‘œđ?‘?đ?‘’đ?‘›đ?‘–đ?‘›đ?‘”∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą > 1.524 đ?‘šđ?‘š: ′ ÎŚđ?‘’đ?‘“đ?‘“ = −0.0679 ∙ ∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą 3 + 0.5675 ∙ ∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą 2 − 1.5681 ∙ ∆đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą + 1.9282 ′ ÎŚđ?‘’đ?‘“đ?‘“ = đ?‘šđ?‘Žđ?‘Ľ đ?‘šđ?‘–đ?‘› 1; ÎŚđ?‘’đ?‘“đ?‘“ ;0

Ref. : PCA Bulletin D124 - Aggregate Interlock at Joints in Concrete Pavements

Chap 4 ; �� –ULS– Load transfer at joint profiles Load reduction factor

đ?œ’đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą đ?‘?đ?‘&#x;đ?‘œđ?‘“đ?‘–đ?‘™đ?‘’ = đ?‘“đ?‘Ą(đ?‘Ąđ?‘Śđ?‘?đ?‘’, đ?‘‘đ?‘œđ?‘¤đ?‘’đ?‘™ đ?‘’đ?‘“đ?‘“đ?‘’đ?‘?đ?‘Ą)

* For đ??Œ values contact the supplier

đ?œ’đ??¸đ?‘?đ?‘™đ?‘–đ?‘?đ?‘ đ?‘’ = 100%

đ?‘Ąđ?‘–đ?‘™đ?‘™ 20 đ?‘šđ?‘š đ?‘—đ?‘œđ?‘–đ?‘›đ?‘Ą đ?‘œđ?‘?đ?‘’đ?‘›đ?‘–đ?‘›đ?‘” Slab breaks before joint

Ref. : TR34 3rd edition Queensland University of Technology “jointing systems in concrete structures� - 2002

đ?‘…đ?‘‘ The design resistance

Chap 4 ; �� – ULS ��� : is the resisting moment capacity of the cross section, derived from factored material properties

��� : is the resisting shear capacity of the cross section , derived from the factored material properties.

• Ft(fibre type) • Ft (fibre dosage) • Ft (concrete quality)

Chap 4 ; đ?‘šđ?’… âˆś đ??”đ??‹đ??’ − đ??Œđ??‘đ??? Material Characterization (EN 14645)

đ??›đ??šđ??Źđ??˘đ??œ đ??Šđ??Ťđ??˘đ??§đ??œđ??˘đ??Šđ??Ľđ??ž −

Bending stresses

Constitutive law (Model Code 2010)

Tension stresses

compression zone

Section moment capacity

tension zone (steel fibers)

Moment resitance đ?‘´đ?‘šđ?’…

Chap 4 ; 𝑹𝒅 ∶ 𝐔𝐋𝐒 − 𝐌𝐑𝐝 𝑴𝑹𝒅

η∙fcd

𝐟𝐨𝐫𝐦𝐮𝐥𝐚𝐞 −

ec ≤ ecu x

λ∙x η ∙ x ∙ fcd

fctd,s

df h

h-x

d Ff

es ≤ esu fctd,ε ≤ fctd,u

as

ef ≤ efu

Fs

𝑓𝑐𝑡𝑑,𝑠 + 𝑓𝑐𝑡𝑑,𝜀 𝐹𝑐𝑑 = 𝐹𝑠𝑑 + 𝐹𝑓𝑑 ⇒ 𝑏 ∙ 𝜆 ∙ 𝑥 ∙ 𝜂 ∙ 𝑓𝑐𝑑 = 𝐴𝑠 ∙ 𝜎𝑠𝑑 + ∙ ℎ−𝑥 ∙𝑏 2 𝜆∙𝑥 𝑀𝑅𝑑 = −𝑀𝑐𝑑 + 𝑀𝑠𝑑 + 𝑀𝑓𝑑 = −𝐹𝑐𝑑 ∙ + 𝐹𝑠𝑑 ∙ 𝑑 + 𝐹𝑓𝑑 ∙ 𝑑𝑓 2  The higher fr1 and fr3, the higher Ffd, the higher Mrd

𝑴𝑹𝒅

Chap 4 ; 𝑹𝒅 ∶ 𝐔𝐋𝐒 − 𝐯𝐑𝐝

𝐟𝐨𝐫𝐦𝐮𝐥𝐚𝐞 −

DIN EC2 + NAD + DAfStb Only conventional

DIN EC2 + NAD + DAfStb

𝑣𝑅𝑑,𝑐,𝑚𝑎𝑥 = max 𝑣𝑅𝑑,𝑐 , 𝑣𝑚𝑖𝑛

deff

𝑑𝑒𝑓𝑓 = h − c − 2∙d2 s (conv,combi) 𝑑𝑒𝑓𝑓 = ℎ(fiber only) (Reinforcement is assumed in both directions)

Characteristic factor approach

𝑓𝑐𝑡𝑅,𝑢 = 𝜅𝐹 ∙ 𝜅𝐺 ∙ 0,51 ∙ 𝛽𝑢 ∙ 𝑓𝑅,4𝑚

𝑓

Combined

𝑣𝑅𝑑,𝑐 = max 𝑣𝑅𝑑,𝑐 , 𝑣𝑚𝑖𝑛 + 𝑣𝑅𝑑,𝑐𝑓 ≤ 1,4 ∙ max 𝑣𝑅𝑑,𝑐 , 𝑣𝑚𝑖𝑛

Fiber only

𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 + 𝑣𝑅𝑑,𝑐𝑓 ≤ 1,4 ∙ 𝑣𝑚𝑖𝑛

𝑓

𝑓

Fiber contribution

𝑣𝑅𝑑,𝑐𝑓 = 0,85 ∙ 𝑓

𝑓

𝛼𝑐 ∙ 𝑓𝑐𝑡𝑅,𝑢 𝑓

𝑓

𝑓

𝛾𝑐𝑡 𝑓

𝑓

𝑓

𝑓

𝑓

𝑓

𝑓

𝛼𝑐ℎ𝑎𝑟 ≤ 𝜅𝐹 ∙ 𝜅𝐺 ∙ 0,51 𝑓 𝜅𝐹 = 0,5 𝑓 𝑓 𝜅𝐺 = 1,0 + 𝐴𝑐𝑡 ∙ 0,5 ≤ 1,70 𝑓 𝐴𝑐𝑡 =∙ 𝑢1 ∙ 𝑑𝑒𝑓𝑓

𝛼𝑐 = 0,85 | 𝑓𝑐𝑡𝑅,𝑢 = 𝜅𝐹 ∙ 𝜅𝐺 ∙ 𝑓𝑐𝑡0,𝑢

𝑓 𝜅𝐹 𝑓 𝑓𝑐𝑡0,𝑢

𝑓 | 𝜅𝐺

𝑓 𝐴𝑐𝑡

= 0,5 = 1,0 + ∙ 0,5 ≤ 1,70 𝑓 𝑓 = 𝑓𝑐𝑓𝑙𝑘,𝐿2 ∙ 𝛽𝑢 | 𝑓𝑐𝑓𝑙𝑘,𝐿2 ≤ 0,51 ∙ 𝑓𝑅,4𝑚

𝜌𝑙 𝜌𝑙 = min

𝑓

𝛽𝑢 = 0,37 | 𝛾𝑐𝑡 = 1,25

𝑣𝑅𝑑,𝑐

𝑣𝑅𝑑,𝑐 = 𝐶𝑅𝑑,𝑐 ∙ 𝑘 ∙ 100 ∙ 𝜌𝑙 ∙ 𝑓𝑐𝑘

1Τ3

1 𝐶𝑅𝑑,𝑐 = 0,18Τ𝛾𝑐

൞ 2 𝐶𝑅𝑑,𝑐 = 0,18Τ𝛾𝑐 ∙ 0,1 𝑢0 Τ𝑑𝑒𝑓𝑓 + 0,6 3 𝐶𝑅𝑑,𝑐 = 0,15Τ𝛾𝑐 (1) elevated slab, foundation slab, (2) inner columns of elevated slab with u0/deff <4, (3) pad foundations,

𝛾𝑐 = 1,5

𝑘 = 1 + 200Τ𝑑𝑒𝑓𝑓 ≤ 2,0 (𝑑𝑒𝑓𝑓 𝑚𝑚 )

𝑣𝑚𝑖𝑛

𝑑𝑒𝑓𝑓 ≤ 600𝑚𝑚 → vmin = 0,0525Τγc ∙ k 3Τ2 ∙ fck 𝑑𝑒𝑓𝑓 > 800𝑚𝑚 → vmin = 0,0375Τγc ∙ k 3Τ2 ∙ fck 600𝑚𝑚 < 𝑑𝑒𝑓𝑓 ≤ 800𝑚𝑚 → interpolate

Remarks

ds 2 fcd 2 ; 0,02; 0,5 ∙ s ∙ deff γyd s = spacing

π∙

Chap 4 ; đ?‘šđ?’… âˆś đ??”đ??‹đ??’ đ??œđ??¨đ??§đ??œđ??Ľđ??Žđ??Źđ??˘đ??¨đ??§ The ULS requirement is met when

��� ≤ ��� ��� ≤ ��� For all load cases Using the required load safety factors material safety factors

Chap 5 ; SLS – Concrete cracks uncracked

cracked

fctm = 3 MPa Ec = 30,000 MPa

1m

Δec / Fs

1m

differential deformation Δec: Δec = fctm / Ec = 3 MPa / 30,000 MPa Δec = 0.1 ‰ This corresponds to 10K temperature difference or 0.1 ‰ shrinkage deformation.

cracking force Fcr: Fcr = 3 MPa ∙1 m ∙ 1m = 3 MN

Chap 5 ; SLS – cracking Shrinkage deformation Δes = 0.4 ‰ Maximum strain in concrete prior to cracking Δec = 0.1 ‰

Fully restraint concrete element 5m

fctm

3 MPa Deformation cracked concrete

100 % Restraint concrete elements will crack by

Deformation restraint shrinkage

  

E = 30,000 N/mm²

0.1 ‰

0.4 ‰

ec

Shrinkage Temperature variations Acting forces

Chap 5 ; SLS – cracking leff = 50 mm

Hooke’s law 60 mm

- Same size Hooked end steelfiber = working on end anchorage - No shape outlines - Low resolution pictures leff = 30 mm

50 mm

Assumption ! : Plastic fibre = working on bond anchorage

 E e l e l

Chap 5 ; SLS – 0,2 mm crack opening l e leff

Δl (mm)

Leff (mm)

ε (%)

Steelfiber

0,2

50

0,4

Plastic fibre

0,2

30

0,67

Tensile curves different wire qualities 2500

2000

Stress (MPa)

Small crack openings 1500 3D 65/60BG 4D 65/60BG 1000

5D 65/60BG

800 Mpa

PP yarn

580 Mpa500

45 Mpa 0 0,00

1,00

2,00

3,00

4,00

5,00

Strain (%)

6,00

7,00

8,00

9,00

Steel : High stresses = effective Plastic : Low stresses = not effective

Chap 5 ; SLS – cracking To avoid cracking • Δec < 0.1 ‰ • Reduce friction between the floor and the subbase • Plastic sheet between floor and subbase • Flat subbase , avoid points of achorage • Load the floor only after 28 days (or later) When Δec > 0.1 ‰ cracks will appear and the reinforcement must limit the crack opening

• At 0,2 mm crack opening, steel works at high stresses (580 – 800 N/mm²) • At 0,2 mm crack opening macro synthetic fibers have low effectiveness ( appr. 45 N/mm²)

Chap 5 ; SLS – Detailing Mesh

Rebar

Saw cuts joints

A-A

B

B

Chap 5 ; SLS – 3 possible floor systems Jointed floors • Saw cuts at appr. 6 x 6 m • Single Plastic sheet •

đ?’‡đ?‘šđ?&#x;?đ?’Ž đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł

đ?’‡

đ?‘šđ?&#x;‘đ?’Ž ≼ đ?&#x;Ž, đ?&#x;’ , đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł ≼ đ?&#x;Ž, đ?&#x;‘

• For lightly traficed floors • Maintance costs at joints

Jointless floors • Joint profiles at appr. 35 x 35 m • Double plastic sheet •

đ?’‡đ?‘šđ?&#x;?đ?’Ž đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł

đ?’‡

đ?‘šđ?&#x;‘đ?’Ž ≼ đ?&#x;Ž, đ?&#x;” , đ?’‡đ?’„đ?’•đ?’Ž,đ?‘ł ≼ đ?&#x;Ž, đ?&#x;’

• For distribution centres • Low maintance cost at the expansion joints. (in function of the type of joint)

Seamless floors • No joints. (all day joints are tight together • Single plastic sheet • Dramix 4D + mesh according special SLS design



w k  sr ,max ďƒ— e f sm  e cm



• For high output distribution centres • No movement joints so no maintance costs

CONCLUSIONS • A state of the art floor design is complicated – ULS to define slab thickness, and reinforcement (steel fibre type and dosage or combined reinforcement Steelfibers + Mesh) – SLS check if specific serviceability requirements are met.

• A correct definition of the design parameters is essential – Validaton of the subbase by onsite checking – Obtaining the correct load case from the client – Quality control of the steelfibre concrete performance

• A correct design is only a part of the succes of a project in addition to … – – – –

An experienced flooring contractor Quality materials (concrete, steelfibers, additives,…) Quality joints Quality control on the jobsite

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