#### Read RC09_Shear2.ppt text version

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EXAMPLE 6-1 Design of Vertical Stirrups WSD v.s. SDM

Reinforced Concrete Design

Home work: Select the stirrup spacing for the beam shown below.

Shear & Diagonal Tension #2

Shear Design Summary WSD v.s. SDM Design Examples Location of Max. Shear for the Design of Beams Shear Span Deep Beam

Mongkol JIRAVACHARADET

f c' = 280 ksc, and fy = 4,000 ksc Use DB10 stirrups.

Show your results on a scaled sketch. PL = 5 tons PD = 2 tons A A 2.5 m 4.0 m 2.5 m Section A-A PL = 5 tons PD = 2 tons wL = 3 t/m wD = 2 t/m 40 cm

d = 53 cm

SURANAREE UNIVERSITY OF TECHNOLOGY

INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING

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Shear Design Summary

WSD

Shear: V = VDL + VLL

Shear Diagram : (ton) 29.5 w = 5 t/m 17 10 4m 2.5 m

WSD

SDM

Shear: Vu = 1.4 VDL + 1.7 VLL Vn = Vu /

Concrete: Vc = 0.29 fc b d Steel: Vs = V - Vc Spacing: s = Av fs d / Vs Min. Stirrup: smax = Av fy / 3.5 b Chk. light shear: V 0.795 fc b d smax d/2 60 cm Chk. heavy shear: V 1.32 fc b d smax d/4 30 cm

Concrete: Vc = 0.53 fc b d Steel: Vs = Vn - Vc Spacing: s = Av fy d / Vs Min. Stirrup: smax = Av fy / 3.5 b Chk. light shear: Vs 1.1 fc b d smax d/2 60 cm Chk. heavy shear: Vs 2.1 fc b d smax d/4 30 cm

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2.5 m

-10 -17

Assume column width = 0.3 m @ critical section V = 29.5 5(0.15+0.53) Shear strength of concrete Vc = 0.29(280)1/2(40)(53)/1,000 Required shear strength of steel Vs = 26.10 10.29 Check lightly shear rein. 0.795(280)1/2(40)(53)/1,000

-29.5 = 26.10 ton = 10.29 ton = 15.81 ton = 28.20 ton > V OK

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WSD Select RB9 : Av = 2(0.636) = 1.27 cm2, fs = 1,200 ksc Stirrup spacing : s = 1.27(1.2)(53)/15.81 = 5.11 cm < [53/2=26.5 cm] < 60 cm USE Stirrup RB9 @ 0.05 m Shear @ x = 2.5 m, V = 10 ton Use min. stirrup : s = 1.27(2,400)/3.5(40) = 21.8 cm < [53/2=26.5 cm] < 60 cm USE Stirrup RB9 @ 0.20 m Select RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2,400 ksc Stirrup spacing : s = 1.27(2.4)(53)/30.00 = 5.38 cm < [53/2=26.5 cm] < 60 cm USE Stirrup RB9 @ 0.05 m Shear @ x = 2.5 m, Vu/ = 15.8/0.85 = 18.6 ton Use min. stirrup : s = 1.27(2,400)/3.5(40) = 21.8 cm < [53/2=26.5 cm] < 60 cm USE Stirrup RB9 @ 0.20 m

SDM

2.5 m

3.7 m

2.5 m

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2.5 m

3.7 m

2.5 m

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Shear Diagram : (ton) 46.85

wu = 1.4(2)+1.7(3) = 7.9 t/m Pu = 1.4(2)+1.7(5) = 11.3 ton wu = 7.9 t/m 27.1 15.8 4m 2.5 m

SDM

Location of Maximum Shear for the Design of Beams

ACI 11.1.3.1 For nonprestressed members, sections located less than a distance d from face of support shall be permitted to be designed for Vu computed at a distance d.

2.5 m

-15.8 -27.1 d -46.85 = 41.48 ton = 18.80 ton = 30.00 ton = 39.02 ton > Vs OK

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d d d

Assume column width = 0.3 m @ critical section Vu = 46.85 7.9(0.15+0.53) Shear strength of concrete Vc = 0.53(280)1/2(40)(53)/1,000 Required shear strength of steel Vs = 41.48/0.85 18.80 Check lightly shear rein. 1.1(280)1/2(40)(53)/1,000

Shear force diagram for design

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Typical support conditions for locating Vu

d Critical section Vu Vu Vu d

EXAMPLE 6-2 More Detailed Design of Vertical Stirrups SDM The simple beam supports a uniformly distributed service dead load of 2 t/m, including its own weight, and a uniformly distributed service live load of 2.5 t/m. Design vertical stirrups for this beam. The concrete strength is 250 ksc, the yield strength of the flexural reinforcement is 4,000 ksc.

DL = 2 t/m LL = 2.5 t/m

wu = 1.4(2) + 1.7(2.5) = 7.05 t/m

d = 64 cm

Beam loaded near bottom Critical section Vu Vu Critical section Beam supported by tension force d

Beam column joint

L = 10 m

wLu = 1.7(2.5) = 4.25 t/m wuL/2 = 7.05(10)/2 = 32.25 t/m

30 cm

wLuL/8 = 4.25(10)/8 = 5.31 t/m

32.25/0.85 = 37.94 ton Vu/ Diagram : 5.31/0.85 = 6.25 ton

Beam with concentrated load close to support

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Shear at Midspan of Uniformly Loaded Beams

In normal building, the dead load is always present over the full span, the live load may act over the full span, or over part of the span. LL full span DL full span w L Vu = u 2

assume column width = 0.40 cm Vu / at d = 37.94 (0.84/5)(37.94 6.25) = 32.62 ton Shear strength of concrete Vc = 0.53 fc b d = 0.53 250 (30)(64) /1,000 = 16.09 ton 37.94 t

84 cm

Critical section 32.62 t Required Vs 16.09 t Vc 8.05 t 6.25 t Midspan

Max. shear @ ends LL half span DL full span Vu =

Vu/

w Lu L 8 Support

0.5Vc

Max. shear @ midspan w L Vu = u 2 w L Vu = Lu 8

Is the cross section large enough? Vn,max = Vc + 2.1 fc b d = 16.09 + 2.1 250 (30)(64) /1,000 = 79.84 > 32.62 ton Vc + 1.1 fc b d = 16.09 + 1.1 250 (30)(64) /1,000 = 55.6 > 32.62 ton

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Shear force envelop :

OK

smax d / 2 60 cm

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Minimum stirrup : (ACI 11.5.6.3) USE RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2400 ksc A v,min = 0.2 fc Rearranging gives smax = smax = A v fy 0.2 fc b A v fy 3.5b = bs fy = 1.27(2,400) = 32 cm 0.2 250 (30) (ACI Eq. 11-13) s=15 cm @ x = 140 cm [email protected] [email protected] s=29 cm @ x = 239 cm [email protected]

but not less than

1.27(2,400) = 29 cm 3.5(30)

20 cm

[email protected] cm 1 cm

Use smax = 29 cm < [d/2 = 64/2 = 32 cm] < 60 cm Compute stirrup resuired at d from support s= A v fy d Vu / - Vc

Support

500 cm OK OK

Midspan

RB9 @ 0.11 m : [email protected] = 142 cm > 140 cm = 1.27(2.4)(64) = 11.8 cm 32.62 - 16.09 RB9 @ 0.15 m : [email protected] = 247 cm > 239 cm RB9 @ 0.29 m : 247 + [email protected] = 479 cm

Use [email protected] Change spacing to s = 15 cm where this is acceptable, and then to the maximum spacing of 29 cm. Compute Vu/ where s can be increased to 15 cm. Vu A v fyd 1.27(2.4)(64) = + Vc = + 16.09 = 29.1 ton s 15

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37.94 t

84 cm

Critical section 32.62 t 29.1 t Vc = 16.09 t 8.05 t 6.25 t

Shear Span (a = M /V )

Distance a over which the shear is constant P P

Vu/

a

a

0.5Vc 500 cm x 37.94 - 29.1 × 500 = 140 cm from support x= 37.94 - 6.25 Change s to 29 cm, compute Vu/ Vu = 1.27(2.4)(64) + Vc = + 16.09 = 22.82 ton s 29 37.94 - 22.82 x= × 500 = 239 cm from support 37.94 - 6.25 A v fy d

Support

Midspan

Shear Diagram

V = +P + V = -P M = Va

Moment Diagram

+

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Crack Pattern in Several Lengths of Beam Mark 1 2 3 4 5 6 7/1 8/1 10/1 9/1

Span (m) 0.90 1.15 1.45 1.70 1.95 2.35 3.10 3.60 4.70 5.80

a/d 1.0 1.5 2.0 2.5 3.0 4.0 5.0 6.0 8.0 7.0

DEEP BEAM

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Brunswick Building. Note the deep concrete beams at the top of the ground columns. These 168-ft beams, supported on four columns and loaded by closely spaced fascia columns above, are 2 floors deep. Shear stresses and failure mechanisms were studied on a small concrete model. (Chicago, Illinois)

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Variation in Shear Strength with a/d for rectangular beams

Flexural moment strength

Inclined cracking strength, Vc

Shear-compression strength Failure moment = Va

Deep beams

Shear-tension and shear-compression failures

Flexural failures

Diagonal tension failures

0

1

2

3 a/d

4

5

6

7

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Deep Beams

Deep beams are structural elements loaded as beams in which a significant amount of the load is transferred to the supports by a compression thrust joining the load and the reaction.

Design Criteria for Shear in Deep Beams

Basic Shear Strength: where

Vn Vu Vn = Vc + Vs

When shear span a = M /V to depth ratio < 2 Mechanism:

Location for Computing Factored Shear:

(a) Simply Supported Beams

Use both horizontal and vertical may prevent cracks

(Critical section located at distance z from face of support) Compressive struts - z = 0.15Ln d for uniform loading - z = 0.50a d for concentrated loading (b) Continuous Beams Critical section located at face of support

If unreinforced, large cracks may open at lower midspan.

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Limitation on Nominal Shear Strength

Vn,max = 2.7 fc b d

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Definition of Deep Beam

ACI 10.7.1 Deep beams are members loaded on one face and supported on the opposite face so that compression struts can develop between the loads and the supports, and have either: (a) clear spans, Ln, equal to or less than four times the overall member depth; or

Shear Strength of Concrete, Vc

M V d Vc = 3.5 - 2.5 u 0.50 fc + 176 u b d 1.6 fc b d Vu d Mu where 1.0 3.5 - 2.5 Mu 2.5 Vud

If some minor unsightly cracking is not tolerated, the designer can use h Ln / h 4 Ln (b) regions with concentrated loads within twice the member depth from the face of the support. P x h x < 2h Simplified method: Vc = 0.53 fc b d

Shear Reinforcement, Vs

A 1 + Ln / d Vs = v + sv 12 Av = sv =

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A vh 11 - Ln / d fy d sh 12 Avh = sh =

( .2), F ( .),

( .2) F ( .)

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Minimum Shear Reinforcement

maximum sv maximum sh and minimum A vh = 0.0015 b sh d 30 cm 5 d 30 cm 5

(b)

F

0.50a = 0.5(1.20) = 0.60

FF

< [d = 0.90

a = 1.20 ]

.

F (c)

F 0.60 . F F

Mu 102(60) = = 0.67 Vu d 102(90) 3.5 - 2.5 Mu = 3.5 - 2.5(0.67) = 1.83 < 2.5 Vu d OK

F

1.7 LL = 1.7(60) = 102

F

F

minimum A v = 0.0025 b sv

V d v c = 1.83 0.50 fc + 176 u Mu

w =

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4(10.18) = 0.0129 35(90)

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5.6 60 F d = 90 . F f c= 280 F

1.20 m

F

F F ./ .2

60 t

F

3.6 fy = 4,000

60 t

F 35 ./ .2

.

176 ( 0.0129 ) v c = 1.83 0.50 280 + 0.67 = 1.83[8.37 + 3.39] = 21.5

./ .2

1.20 m

Upper limit: v c = 1.6 fc = 1.6 240 = 24.8 kg/cm2

90 cm

Vc = vc bw d = 21.5(35)(90)/1,000 = 67.8

4DB36

(d)

F

Required Vn = Vu

5 cm 40 cm

[email protected] = 3.5 m 3.6 m

5 cm 40 cm

35 cm

=

102 = 120 ton 0.85

(a)

F

F

Ln/d = 360/90 = 4 Vn

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Vn,max = 2.7 fc b d = 2.7 280(35)(90) /1,000 = 142

> 120 F

OK

F

> Vc (120 > 67.8)

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(e)

Av 1 + Ln / d Avh 11 - Ln / d Vs + = fd s 12 s2 12 y Ln/d = 4 : Vs = 120 67.8 = 52.2 b = 35

.

fy = 4,000

./ .2

A v 5 A vh 7 52.2 + = = 0.145 sv 12 sh 12 4.0(90) min Av = 0.0015 b sv min Avh = 0.0025 b sh max sv = d/5 =18 max sh = d/5 = 18

. .

sh = 18

2 2

F

DB12

F

F

F

.

min Avh = 0.0025(35)(18) = 1.58

F

F

Avh = 2(1.13) = 2.26

2

> 2.2

OK

Shear2_29

F

Avh

A v 5 2.26 7 + = 0.145 s 12 18 12 Av 12 = [0.145 - 0.0732] = 0.172 s 5 DB12: Av = 2(1.13) = 2.26

2

F F

DB12

s = 2.26/0.172 = 13.1 12

. < [d/5 = 18 F

.]

OK

.

[email protected] 90 cm

[email protected] 4DB36

[email protected] = 3.6 m 40 cm 3.6 m 40 cm

35 cm Shear2_30

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