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
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
USE Stirrup RB9 @ 0.20 m
USE Stirrup RB9 @ 0.20 m
[email protected]
[email protected]
2.5 m
[email protected]
[email protected]
2.5 m
2.5 m
3.7 m
[email protected]
3.7 m
[email protected]
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
[email protected]
[email protected]
[email protected]
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
[email protected]
[email protected] 4DB36
30@12 = 3.6 m 40 cm
3.6 m
35 cm 40 cm Shear2_30
