4 She

 

Cha

TWWhenwhenimpoFailutrunca).

The fthe tonatur CritCriticfrom

is bo concecapita TW TW spanelcriticthe co

ar Strength T

apter 4:

W Shear on TW slabs an slabs carry rtance.

ure may occuated cone or

failure surfacop surface. Tre and amoun

tical sectiocal section fothe peripher

. The perime

entrated loadals, drop pan

Shear Forshear force Vl centerlines al shear periolumn.

TW Slabs         

: Shear S

of Slabs are supporteconcentrate

ur by punchir pyramid ar

ce extends frThe angle of nt of reinforc

on for TW or shear is tary of the sup

eter bo is loc

ds, or reactionels, or shear

rce Vu: Vu can be cal

around the cimeter, unles

                Prof

Strength

ed directly byd loads, as in

ing shear, wound the col

rom the bottinclination wcement. It m

shear: aken perpendpport, as show

cated at dista

ons, or from r caps (ACI

lculated as thcolumn less ss significan

f Dr Bayan Sa

h of TW

y columns, an footings, s

with the diagolumn, capita

om of slab, awith the hori

may range be

dicular to thewn in fig. b.

ance d/2 from

changes in s22.6.4); see

he total factothe load app

nt moments m

alim

W Slabs w

as in flat slabshear near co

onal crack foal, or drop pa

at the suppoizontal θ, fig

etween 20 an

e plane of th The perime

m edges or c

slab thicknesfigure;

ored load onplied within must be trans

 

w/o Bea

bs and flat polumns is of

ollowing theanel as show

ort, diagonallg. b, dependsnd 45o.

he slab and aeter of the cri

corners of co

ss such as ed

n the area bouthe area defsferred from

ams

lates, or f critical

e surface of awn below (fig

ly upward tos upon the

a distance d/2itical section

olumns,

dges of

unded by fined by the

m the slab to

1 

a g.

o

2 n

4 She

 

Fig.

ar Strength T

: Critical

TW Slabs         

section for

                Prof

r TW shea

f Dr Bayan Sa

ar

alim  2 

4 She

 

TW The Tequat

The Tthree Vc = Vc = Wher

Vc = Whe λs = s

NoteFor β

ar Strength T

Shear strTW shear strtions:

TW shear strequations:

= (1/3) [λs

= (1/12) [

re β = long λ = lightw

= (1/12) [(

re, αs = 40 = 30 = 20

size effect fa

es: β ≤ 2.0, eq. (a

TW Slabs         

ength, Vc rength (stres

rength (force

s λ √f′c bo

(2 + 4 / β

side / short sweight concr

(2 + αs d

0 for interior for edge co for corner cctor

a) governs. F

                Prof

ss in psi, fc’ i

e in N, fc’ in

d ]

β) λs λ √f′

side of the corete factor = =

/ bo) λs λ

column, lumn, and

column.

For an interi

f Dr Bayan Sa

in psi) shall b

MPa) shall

′c bo d ]

olumn 1.0 for norm

= 0.75 for lig

√f′c bo d ]

ior column w

alim

be the small

be the small

mal weight cghtweight co

]

(ACI 22.6

with bo / d ≤

 

lest of the fo

lest of the fo

(ACI 22

(ACI 2

concrete oncrete.

(ACI 2

6.5.3)

≤ 20, eq. (a)

ollowing thre

ollowing

2.6.5.2a)

22.6.5.2b)

22.6.5.2c)

governs

3 

ee

)

)

4 She

 

Shea φVn

TW The upermSuggleg, obotto

a)Sin

ar Strength T

ar Strengt

n = Vu = φ

When V

When V

Shear strei. increasiii. increasiii. provid If shear In this c and max φ

Shear rei

use of bars, wmitted providegested rebar sor closed stirm of the slab

ngle-leg

TW Slabs         

th Require

φ (Vc + Vs

Vu ≤ φ Vc

Vu > φ Vc

ength may being concrete sing slab thicding shear re

r reinforce

case, reduc

Vc

x nominal

(Vc + Vs)

nforcemen

wires, or sined that d ≥ 1shear reinforrrups that areb (ACI 8.7.6

b) mult

                Prof

ements

Vs) , φ

OK, no s

N.G.

e increased bstrength fc′

ckness at coleinforcement

ement is re

ce Vc to:

c = (1/6) [

l strength:

) ≤ (1/2)

nt in slabs

gle or multip150 mm, andrcement cone engaging lo6.2); see Fig

iple-leg stirr

f Dr Bayan Sa

φ = 0.75

shear rein

by: ′ lumn support

equired:

[ λ √f′c bo

:

[ λ √f′c bo

s

ple-leg stirrud 16 db. sist of propeongitudinal r.

rup or bar

alim

nforcement

rt, i.e., using

d ]

o d ]

ups as shear

erly anchoredreinforceme

c)

 

(A

t is requir

g a drop pane

reinforceme

d single-leg,ent at both th

closed stirru

ACI 22.6)

red.

el

ent in slabs i

, multiple-he top and

ups

4 

s

4 She

 

Arraa

Calc Av =Av = SpaciThe s ExamConscolumand ifOverfy = 4Live Solu

ar Strength T

angement a) Interior

culation of

= Vs s / fy cross sec areing limits arshear reinfor

mple 4-1: ider an inter

mn. Panel sizf not adequaall slab thick

420 MPa. load = 5 kN

ution:

TW Slabs         

of stirrupr column

f the area

d ea of all the e shown in frcement can

rior panel of ze l1 = l2 = 6ate, increase kness h = 19

N/m2

                Prof

p shear rei

of shear r

legs of shearfigs above. be terminate

f a flat plate s6.5 m. Determ

the shear str90 mm (d = 1

f Dr Bayan Sa

inforceme

reinforcem

r reinforcem

ed when Vu ≤

slab system mine shear srength by co150 mm). fc

alim

ent: b) edge

ment:

ment

≤ φ (1/6) λ √

supported bystrength of slonsidering cl′ = 28 MPa,

 

column

√fc′ bo d (AC

y a 300 mmlab at columlosed stirrup, normal wei

CI 22.6.6-7)

m square mn support,

s option. ight concrete

5 

.

e,

4 She

 

qu = c + d Vu = bo = 4β = 3bo / dφ Vc Deter φ Vc

Vu ≤ For s567.6SolviStirruTryinOne- (Use bars a

ar Strength T

1.2×(0.19×2d = 300 + 150

qu [tributar4(450) = 18000 / 300 = 1

d = 1800 / 15=φ(1/3)[λ√f

rmine distanc = φ (1/6)[λ

φVc = φ(1/6quare colum

64 ×103 = 0.7ing, a = 799.ups are terming s = 75 mm-leg Av = (1/8

= (567.10 No.10 stat corners).

TW Slabs         

24) + 1.6×5 =0 = 450 mm ry area] = 100 mm < 2.0

50 = 12 < 20f′cbo d] = 0.7

nce from side√f′cbo d] = 1

6)[λ√f′cbo d]mn: bo = 4 (co75 (1/6)×√2.3 mm, say ainated at d/2

m; 8)Vs s / fy d =.64 – 178.6)irrups @ 75m

                Prof

= 5.5 + 8 = 1

3.5[6.5×6.5

, equation c5(1/3)[1√28

es of column78.6 kN

|← ol side + a √8× 4 (300 + a = 800 mm2 = 75 mm in

= (Vu − φVc)103(75) / [8×mm along ea

f Dr Bayan Sa

13.5 kN/m2

– 0.45×0.45

governs. 8×1800×150]

n where stirr

← a →|

√2 ) a √2 ) × 150

nside the crit

) s / 8 fy d ×420×150] =ach arm/colu

alim

5] = 567.64 k

]= 357.18 kN

rups may be

|300mm| ↔ Col

0

tical perimet

= 58 mm2; Uumn line, wi

 

kN

N<Vu =567.

terminated (

side

ter bo.

Use No.10 (Aith 4 No. 10

64 kN N.G

(distance a)

Av = 71 mm2)longitudinal

6 

G.

) l

4 She

 

EffeUnbabeamspandtransfFig.

T

fl

ɣf Mu

oppo

T

Mu = Mu =

ar Strength T

ects of Momalanced mom

ms between sudrel beam) isferred to the

The fraction olexure γf is:

u is transferrsite faces of

The fraction o

0.30 Mo (if computed f

TW Slabs         

ment Tranment Mu can upports, shes especially column, wh

of unbalance

ed within anf the column

of unbalance

f DDM is useframe mome

                Prof

nsfer occur at the

ear strength acritical, beca

hich is in add

ed moment t

ACI Eq

n effective wor capital. (

ed moment t

ed) ent (if EFM

f Dr Bayan Sa

slab-columnat an exteriorause the totadition to the

transferred b

q(8.4.2.2.2)

width = 1.5 hh = slab thic

transferred b

M is used)

alim

n connectionr slab-colum

al exterior nedirect shear

by

)

on each sidckness includ

by eccentricit

ACI E

Mu

tranby:

Mu

 

ns. For slabsmn connectioegative momr due to grav

de of columnding drop pa

ty of shear i

Eq. (8.4.4.2.2

at connectionnsferred to co

1. Flexure;2. Shear; ɣ

i,e,

= ɣf Mu + ɣv

s without on (without ment must bevity loads; se

n, outside anel if any):

s:

2)

n is olumn

; ɣf Mu ɣv Mu

v Mu

7 

e ee

4 She

 

wherto the

Shea

wherJ = parea Ac andsectio

ar Strength T

e b1 and b2 ae direction o

Critial se

ar Stress a

e: Ac = area roperty of crAc.

d c′ = distancon in the dire

TW Slabs         

are the dimenf analysis; s

ection for int

and Streng

of concrete ritical sectio

ces from cenection of ana

                Prof

nsions of theee Fig.

terior column

gth Consid

section resison analogous

ntroidal axis alysis.

f Dr Bayan Sa

e perimeter o

n

dering Eff

sting shear trs to polar mo

of critical se

alim

of the critica

Critial

fects of Mo

ransfer, Ac =oment of ine

ection to the

 

al section, w

section for e

oment Tra

= bod ertia of segm

perimeter o

ith b1 paralle

edge column

ansfer

ments forming

of the critical

8 

el

n

g

l

4 She

 

ExpFor

ar Strength T

ressions foRectangul

TW Slabs         

or Ac, c, c′lar column

                Prof

′, J / c, andns;

f Dr Bayan Sa

d J / c′ ,

alim  9 

4 She

 

ExamConssquarslab ah = 1of theVu = SoluEdgeb1 = b2 = bo = 2c = bAc = (J/c =ɣf = 1ɣv = 1Unba

= =φvc =Shear CAssumd = 1 Cvu1 ≤ be pr HW1

HW240 kN

ar Strength T

mple 4-2: ider an exterre column. Dand column 180 mm. (d ≈e slab. Consi260 kN, Mo

ution: column benc1 + d/2 = 4c2 + d = 4002 (475) + 55

b12 / (2b1 + b

(2b1 + b2) d [2b1

2d(b1 +1/[1 + (2/3)√1 - ɣf = 0.365alanced Mu =

= 260×103/22= 1.156 + 0.6= φ(1/3)√f′c =r reinforcem

Check effectiming No. 1050 mm OK

Check maximφ(1/2) √f′c =

rovided. (Oth

1: Provide cl

2: Rework ENm

TW Slabs         

Shear Strrior (edge) p

Determine shsupport. Ove≈ 150 mm). ider

o = 230 kNm

nding perpen00 + 75 = 47

0 + 150 = 5550 = 1500 mmb2) = 150.4 m= 225,000 m2b2) + d3(2b

√(b1/b2)] = 0.5 = 0.3 Mo = 69

25000 + 0.3643 = 1.799 = 0.75(1/3) √

ment must be

ive depth, d 0 stirrups (db

(min 150 mmmum shear st= 0.75(1/2) √herwise, incr

losed stirrup

Ex.1 consider

                Prof

rength of Spanel of a flahear strengtherall slab thiAssume tha

m, fc′ = 28 M

ndicular to ed75 mm

50 mm m

mm mm2 b1 + b2)] / 6b.635

9 kNm

65(69×106)/MPa √28 = 1.323 provided

b = 9.5 mm),m and 16 db

tress permitt√28 = 1.984rease slab th

ps as slab she

ring the effe

f Dr Bayan Sa

Slab with Tat plate slab sh for transferickness

at the Direct

MPa, fy = 42

dge (Case C

b1= 3.918×1

/3.918×107

MPa < 1.79

=150 mm) Oed with bar r MPa >1.799

hickness)

ear reinforce

cts of mome

alim

Transfer osystem supp

r of direct sh

Design Meth

20 MPa.

C),

07 mm3

99 MPa N.G

OK reinforceme9 MPa OK S

ement at edg

ent transfer.

 

of Momenported by a 4hear and mom

hod is used

G.

ent. Shear reinfor

ge column.

Take unbala

1

nt 400-mm ment betwee

for analysis

rcement can

anced Mu =

10 

en

n

4 She

 

EffeThe eopeniarea. Slab lengtcentrFor sof tha

ar Strength T

ect of openeffect of opeing is locate

opening effeh equal to thoid of the colabs with shat without sh

TW Slabs         

ning on sheenings in slabd within 4 h

ect is evaluathe projectionolumn and taear reinforce

hear reinforc

                Prof

ear strengbs on concre

h from a colu

ted by reducn of the openangent to theement, the incement; see F

f Dr Bayan Sa

gth ete shear streumn peripher

cing the perining enclosede opening; seneffective poFig. (b).

alim

ength shall bry, a concen

imeter of thed by two-linee Fig (a). ortion of the

 

be consideredntrated load o

e critical secnes extending

perimeter b

1

d when the or reaction

ction bo by a g from the

bo is one-half

11 

f