Reinforced Concrete Design

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

9

Reinforced Concrete Design

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

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

Shear & Diagonal Tension #2

Show your results on a scaled sketch.

Shear Design Summary WSD v.s. SDM

PL = 5 tons PD = 2 tons

Design Examples

40 cm

PL = 5 tons PD = 2 tons wL = 3 t/m wD = 2 t/m

A

Location of Max. Shear for the Design of Beams

d = 53 cm

Shear Span

A 2.5 m

Deep Beam

4.0 m

2.5 m Section A-A

Mongkol JIRAVACHARADET

SURANAREE

INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY

SCHOOL OF CIVIL ENGINEERING

Shear Design Summary

Shear2_03

29.5

WSD Shear: V = VDL + VLL

w = 5 t/m

SDM

17

Shear: Vu = 1.4 VDL + 1.7 VLL

10

Vn = Vu / φ

4m

Concrete: Vc = 0.29 fc′ b d

Concrete: Vc = 0.53 fc′ b d

Steel: Vs = V - Vc

Steel: Vs = Vn - Vc

Spacing: s = Av fs d / Vs

Spacing: s = Av fy d / Vs

Min. Stirrup: smax = Av fy / 3.5 b

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

WSD

Shear Diagram : (ton)

Chk. light shear:

2.5 m

2.5 m

-10 -17

Assume column width = 0.3 m

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 Shear2_02

@ critical section V = 29.5 – 5(0.15+0.53)

-29.5 = 26.10 ton

Shear strength of concrete Vc = 0.29(280)1/2(40)(53)/1,000

= 10.29 ton

Required shear strength of steel Vs = 26.10 – 10.29

= 15.81 ton

Check lightly shear rein. 0.795(280)1/2(40)(53)/1,000

= 28.20 ton > V OK Shear2_04

WSD

SDM

Select RB9 : Av = 2(0.636) = 1.27 cm2, fs = 1,200 ksc

Select RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2,400 ksc

Stirrup spacing : s = 1.27(1.2)(53)/15.81 = 5.11 cm < [53/2=26.5 cm] < 60 cm

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, V = 10 ton

Shear @ x = 2.5 m, Vu/φ φ = 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 min. stirrup : s = 1.27(2,400)/3.5(40) = 21.8 cm < [53/2=26.5 cm] < 60 cm



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2.5 m

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2.5 m

2.5 m

3.7 m

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3.7 m

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2.5 m

Shear2_05

Shear Diagram : (ton) 46.85

SDM

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

Shear2_07

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.

wu = 7.9 t/m 27.1 15.8 4m

2.5 m

2.5 m

-15.8

Assume column width = 0.3 m

d

d

-27.1 -46.85

@ critical section Vu = 46.85 – 7.9(0.15+0.53)

= 41.48 ton

Shear strength of concrete Vc = 0.53(280)1/2(40)(53)/1,000

= 18.80 ton

Required shear strength of steel Vs = 41.48/0.85 – 18.80

= 30.00 ton

Check lightly shear rein. 1.1(280)1/2(40)(53)/1,000

= 39.02 ton > Vs OK Shear2_06

d

d

Shear force diagram for design Shear2_08

Typical support conditions for locating Vu

EXAMPLE 6-2 More Detailed Design of Vertical Stirrups SDM

d

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.

d

Critical section Vu

Vu

Vu

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

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

Beam loaded near bottom

d = 64 cm

Beam column joint

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

L = 10 m

Critical section Vu Vu

30 cm

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

32.25/0.85 = 37.94 ton 5.31/0.85 = 6.25 ton

Vu/φ Diagram :

d

Critical section Beam supported by tension force

Beam with concentrated load close to support Shear2_09

Shear2_11

assume column width = 0.40 cm

Shear at Midspan of Uniformly Loaded Beams

Vu / φ at d = 37.94 – (0.84/5)(37.94 – 6.25) = 32.62 ton 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.

Shear strength of concrete Vc = 0.53 fc′ b d = 0.53 250 (30)(64) /1,000 = 16.09 ton

LL full span DL full span

Max. shear @ ends LL half span DL full span

Vu =

w L Vu = u 2

Required Vs

84 cm

16.09 t

Vu/φ

w Lu L 8

Max. shear @ midspan w L Vu = u 2

Shear force envelop :

Critical section 32.62 t

37.94 t

Vc

8.05 t 6.25 t

0.5Vc

Midspan

Support Is the cross section large enough?

w L Vu = Lu 8

Vn,max = Vc + 2.1 fc′ b d = 16.09 + 2.1 250 (30)(64) /1,000 = 79.84 > 32.62 ton

OK

Vc + 1.1 fc′ b d = 16.09 + 1.1 250 (30)(64) /1,000 = 55.6 > 32.62 ton Shear2_10

⇒ smax ≤ d / 2 ≤ 60 cm

Shear2_12

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 =

but not less than

smax =

bs fy

A v fy 0.2 fc′ b A v fy 3.5b

=

=

s=15 cm @ x = 140 cm

(ACI Eq. 11-13)

s=29 cm @ x = 239 cm

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11@11 cm 1 cm

7@15 cm

8@29 cm

1.27(2,400) = 32 cm 0.2 250 (30)

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

20 cm

Use smax = 29 cm < [d/2 = 64/2 = 32 cm] < 60 cm

500 cm

Support

Midspan

Compute stirrup resuired at d from support s=

A v fy d Vu / φ − Vc

=

1.27(2.4)(64) = 11.8 cm 32.62 − 16.09

Use [email protected]. Change spacing to s = 15 cm where this is acceptable, and then to the maximum spacing of 29 cm.

RB9 @ 0.11 m : 20+1+11@11 = 142 cm > 140 cm

OK

RB9 @ 0.15 m : 142+7@15 = 247 cm > 239 cm

OK

RB9 @ 0.29 m : 247 + 8@29 = 479 cm

Compute Vu/φ φ where s can be increased to 15 cm. Vu A v fyd 1.27(2.4)(64) = + Vc = + 16.09 = 29.1 ton φ s 15

Shear2_13

Critical section 32.62 t

37.94 t

Shear Span (a = M /V )

29.1 t

84 cm

Shear2_15

Distance a over which the shear is constant Vc = 16.09 t

Vu/φ

0.5Vc 500 cm

Support

a

8.05 t 6.25 t

Midspan

x

Shear Diagram

P

P

V = +P + -

37.94 − 29.1 × 500 = 140 cm from support x= 37.94 − 6.25

V = -P

Change s to 29 cm, compute Vu/φ Vu = φ

M = Va Moment Diagram

A v fy d

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

Shear2_14

+

Shear2_16

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

Shear2_17

DEEP BEAM

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)

Shear2_19

Variation in Shear Strength with a/d for rectangular beams Flexural moment strength

Shear-compression strength Failure moment = Va

Inclined cracking strength, Vc

Deep beams

Shear-tension and shear-compression failures

Flexural failures Diagonal tension failures

0

1

2

3

4

5

6

7

a/d Shear2_18

Shear2_20

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

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

Basic Shear Strength:

φVn ≥ Vu

where

Vn = Vc + Vs

Location for Computing Factored Shear:

Mechanism:

(a) Simply Supported Beams (Critical section located at distance z from face of support)

Use both horizontal and vertical may prevent cracks

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

Limitation on Nominal Shear Strength

If unreinforced, large cracks may open at lower midspan.

Vn,max = 2.7 fc′ b d Shear2_21

Definition of Deep Beam

Shear2_23

Shear Strength of Concrete, Vc  M  V d Vc =  3.5 − 2.5 u   0.50 fc′ + 176 ρ u  b d ≤ 1.6 fc′ b d V d Mu  u  

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

where 1.0 ≤ 3.5 − 2.5

Mu ≤ 2.5 Vud

If some minor unsightly cracking is not tolerated, the designer can use Ln / h ≤ 4

h

Simplified method:

Ln

Shear Reinforcement, Vs

(b) regions with concentrated loads within twice the member depth from the face of the support. P x

Vc = 0.53 fc′ b d

h

x < 2h

 A  1 + Ln / d  Vs =  v   +  sv  12 

Shear2_22

A vh  11 − Ln / d     fy d sh  12 

Av =

(.2),

sv =

! (.),

Avh = sh =

(.2) " (.) Shear2_24

Minimum Shear Reinforcement

(b) , ('2>#$ $ $ !< )*( ?(+&,&'%

maximum sv ≤

d ≤ 30 cm 5

maximum sh ≤

d ≤ 30 cm 5

a = 1.20 .

0.50a = 0.5(1.20) = 0.60 < [d = 0.90 ]

, ('2>@ 0.60 . ) 42'")*( (c) $ "% ?(= , '2> $ $ !< )*(! %

and

1.7 LL = 1.7(60) = 102 minimum A vh = 0.0015 b sh

$ % 1 , $ $ !< )*( ?(+&,'23 ( , ('2>

minimum A v = 0.0025 b sv

Mu 102(60) = = 0.67 Vu d 102(90)

'%@;#$ % 1%

3.5 − 2.5

Mu = 3.5 − 2.5(0.67) = 1.83 < 2.5 Vu d

OK

 V d v c = 1.83 0.50 fc′ + 176 ρω u  Mu  

ρw = Shear2_25

5.6 #$ % &'(' $ $ #)*(" $ +&, ) 60 + )*(+ &'', % 3.6 % ', 35 . %' 1!#2 324 d = 90 . +&, fc= 280 ./.2 fy = 4,000 ./.2 60 t

Shear2_27

 176 ( 0.0129 )  v c = 1.83 0.50 280 +  0.67   = 1.83[8.37 + 3.39] = 21.5 ./.2

60 t

1.20 m

4(10.18) = 0.0129 35(90)

$%&

1.20 m

Upper limit: v c = 1.6 fc′ = 1.6 240 = 24.8 kg/cm2 90 cm

$ "% (d) !"#

4DB36 5 cm 40 cm

35@10 = 3.5 m 3.6 m

5 cm

Required Vn =

35 cm

Vu

φ

=

102 = 120 ton 0.85

40 cm

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

(a) 2) ; ' !< % 1 =#$ % Ln/d = 360/90 = 4

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

= 142

) Shear2_26

Vn

> 120

OK

, > Vc (120 > 67.8) ( , #2 Shear2_28

(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

max sv = d/5 =18 .

min Avh = 0.0025 b sh

max sh = d/5 = 18

+&,

DB12

' + ' + (,

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

% + ,

.

Avh = 2(1.13) = 2.26

2

sh = 18

.

2

> 2.2

2

OK Shear2_29

%

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

, ( +&,

DB12

s = 2.26/0.172 = 13.1 . < [d/5 = 18

.]

OK

!< @ * 12 . ( &'%

90 cm

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30@12 = 3.6 m 40 cm

3.6 m

35 cm 40 cm Shear2_30