1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Design Example
– Shear
and Torsion
Massachusetts
Institute
of Technology
1.054/1.541 Mechanics and Design of Concrete Structures (3-0-9)
Design Example
Shear and Torsion
Objective: To examine the adequacy of given cross section based on shear and torsion capacities.
Problem: At a section of a beam, the internal forces are Vu = 45 kips, Mu = 300 kips-ft, and Tu =
120 kips-ft. The material strengths are f’c = 4 ksi and fy = 60 ksi. Assume that the
distance from beam faces to the center of stirrups is 2 in and d is 21 in.
a
a
Figure 1. Plan view of overpass
Tu
Mu
24‘’
Vu
16‘’
Figure 2. Cross section a-a
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Design Example – Shear and Torsion
Task: Answer or accomplish the following questions and tasks based on given assumption.
–
Will this given cross section (16 in wide and 24 in deep) be adequate for shear and
torsion requirements? If not, what width is required?
Assume that the depth is to be held at 24 in.
Select the required torsion, shear, and bending reinforcement for the minimum
required width (to nearest inch).
Summarize reinforcement on a sketch of the cross section.
–
–
–
[Design Procedures]
1. Check maximum torsion capacity
(ACI)
For the given section:
∑x
2
CT =
Tc =
Section/(Equation)
y = (16 ) ( 24 ) = 6144 in3
2
(16 )( 21) = 0.05469 in-1
bw d
=
2
∑ x y 6144
0.8 f c' ⋅ ∑ x 2 y
⎛ 0.4Vu ⎞
1+ ⎜
⎟
⎝ CT Tu ⎠
2
=
0.8 4000 ⋅ ( 6144 )
0.4 × 45
⎛
⎞
1+ ⎜
⎟
0.05469
120
12
×
×
⎝
⎠
2
= 318.88 × 103 lbs-in
Since (Ts )max = 4Tc
11.6.1 &
⇒ (Tu )max = φ (Tc + Ts ) = φ (Tc + 4Tc ) = 4.25Tc ( φ = 0.85 for torsion)
11.6.2.2
⇒ (Tu )max = 1355 ×103 lbs-in
However, (Tu )actual = 120 kips-ft = 1440 ×103 lbs-in
(Tu )actual > (Tu )max ⇒ Section is NOT adequate.
2. Selection of cross section dimensions
Let h = 24 in = constant.
(Tu )actual
0.8 4000 ⋅ x 2 ( 24 )
< 4.25Tc ⇒ 1440 ×10 lbs-in < 4.25
1.02579
3
Assume CT to be about the same ⇒ same dimension.
⇒ x > 16.92 in
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Design Example – Shear and Torsion
Try x =17 in,
∑x
Tc =
2
y = (17 ) ( 24 ) = 6936 in3, CT = 0.05147 in-1.
2
0.8 4000 ⋅ ( 6936 )
0.4 × 45
⎛
⎞
1+ ⎜
⎟
⎝ 0.05147 ×120 × 12 ⎠
2
= 341× 103 lbs-in
⇒ (Tu )max = 4.25Tc = 1449 ×103 lbs-in > Tu
(O.K.)
∴ Section 17′′ × 24′′ is OK. (Although predicting heavy reinforcement.)
3. Selection of stirrups
Ts =
⎛ 1440
⎞
− Tc = ⎜
− 341⎟ × 103 = 1353 × 103 lbs-in
φ
⎝ 0.85
⎠
Tu
x1 = 17 − 2 ( 2 ) = 13 in
y1 = 24 − 2 ( 2 ) = 20 in
α T = 0.66 + 0.33
y1
= 1.168
x1
At
Ts
1353 × 103
=
=
= 0.0743 in2/in
s αT x1 y1 f y 1.168 (13)( 20 ) ( 60 ×103 )
25bw ( 25 )(17 )
⎛ At ⎞
=
= 0.00708 in2/in
⎜ ⎟ =
f yv
60000
⎝ s ⎠ min
Vc =
Vs =
2 f c' bw d
⎛
T ⎞
1 + ⎜ 2.5CT u ⎟
Vu ⎠
⎝
2
=
(O.K.)
2 4000 (17 )( 21)
120 ×12 ⎞
⎛
1 + ⎜ 2.5 ( 0.05147 )
⎟
45 ⎠
⎝
2
11.6.5.3
= 10657 lbs
⎛ 45
⎞
− Vc = ⎜
− 10.66 ⎟ × 103 = 42.28 × 103 lbs
φ
⎝ 0.85
⎠
Vu
Av
V
42.28 × 103
= s =
= 0.0336 in 2 /in
s
f y d ( 60 ×103 ) 21
(11-15)
A 1 Av
0.0336
⎛ A⎞
Since ⎜ ⎟
= t+
= 0.0743 +
= 0.0911 in 2 /in
2
⎝ s ⎠stirrup s 2 s
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
smax =
Spring 2004
Design Example – Shear and Torsion
x1 + y1 13 + 20
=
= 8.25 in
4
4
4 f c' bw d = 4 4000 (17 )( 21) = 90314 lbs > Vs (Not applicable)
smax =
d 21
=
= 10.5 in
2 2
11.5.4.3
11.5.4.1
)
(
φ 0.5 f c' ∑ x 2 y = 0.85 ( 0.5 ) 4000 ( 6936 ) = 186 × 103 lbs-in < Tu (Applicable)
( Av + 2 A t )min =
50bw s 50 (17 ) s
=
= 0.0142 s
fy
60 × 103
⇒ ( Astirrup )min =
0.0142 s
= 0.0071s in2
2
(11-23)
Thus, choose Astirrup and s to satisfy:
(i) s >
A
in;
0.0911
(ii) s < 8.25 in;
(iii) A > 0.0071s in2.
Try #5 stirrups @ 3.5in, A = 0.31 in2.
smin =
A
= 3.4 in
0.0911
smax = 8.25 in
(O.K.)
(O.K.)
Amin = 0.0071s = 0.0071( 3.5) = 0.0249 in2
(O.K.)
⇒ #5 stirrups @ 3.5in are O.K.
4. Selection of longitudinal reinforcement
A
⎛x +y ⎞
Al = 2 At ⎜ 1 1 ⎟ = 2 t ( x1 + y1 ) = 2 ( 0.0743)(13 + 20 ) = 4.9 in2
s
⎝ s ⎠
Al ,min =
5 f c' Acp
f yl
−
f yv
At
ph
s
f yl
(11-24)
Acp is the area enclosed by outside perimeter of concrete cross section, in2, and ph is the
perimeter of centerline of outermost closed transverse torsional reinforcement, in.
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Design Example – Shear and Torsion
Assuming f yv = f yl ( f yv : yield strength of stirrups/ f yl : yield strength of longitudinal steels)
and substituting all known numbers ( Acp = 17 × 24 = 408 in2, ph = 2 (17 − 4 + 24 − 4 ) = 66 in)
provides Al ,min < 0, therefore (11-24) is disregarded.
Use
⇒
50bw s
50bw At
< 2 At or
<
to find 2 At .
2 fy
fy
s
(11-23)
50 (17 )
50bw
A
=
= 0.0071 < t = 0.0743
3
2 f y 2 ( 60 ×10 )
s
Substitute for 2 At in (11-23).
⎡
⎛
⎢ 400 x ⎜ T
u
⎜
Al = ⎢
⎢ f y ⎜ T + Vu
⎜ u 3C
⎢
T
⎝
⎣
⎤
⎞
⎟
A⎥
⎟ − 2 t ⎥ ( x1 + y1 )
s⎥
⎟
⎟
⎥
⎠
⎦
⎡
⎤
⎛
⎞
⎢ 400 (17 ) ⎜
⎥
⎟
1440 ×103
=⎢
⎟ − 2 ( 0.0743) ⎥ ( 33) < 0
3
3 ⎜
⎢ 60 × 10 ⎜ 1440 ×103 + 45 ×10 ⎟
⎥
⎜
⎟
⎢⎣
⎥⎦
3 × 0.05147 ⎠
⎝
This leads to a negative value → disregard it!
⇒ Al = 4.9 in2 for torsion
a⎞
⎛
M u = φ ( 0.85 f c' ab ) ⎜ d − ⎟
2⎠
⎝
(1)
0.85 f c' ab
As =
fy
(2)
Eq.(1) provides
a⎞
⎛
300 ×12 = 0.9 ( 0.85 )( 4 ) a (17 ) ⎜ 21 − ⎟ ⇒ 26.01a 2 − 1092.42a + 3600 = 0
2⎠
⎝
⇒ a = 3.605 in
Eq.(2) provides As = 3.47 in2 → ρ =
Amin =
3.47
= 0.0097
(17 )( 21)
200
= 0.0033 in 2 < As = 3.47 in 2
fy
(O.K.)
In balanced state,
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Cbalanced =
b ( 0.003)
= 12.6 in ⇒ abalanced = 0.85Cbalanced = 10.71 in
0.003 + ε y
( As )balanced = ( As )b =
⇒ ρb =
Spring 2004
Design Example – Shear and Torsion
0.85 ( 4 )(10.71)(17 )
= 10.32 in2
60
10.32
= 0.0289
(17 )( 21)
ρ max = 0.75 ρb = 0.0217
(O.K.)
⇒ As = 3.47 in2 for flexure
5. Summary
Use
4 #10 @ the bottom (3.47 in2 for M, 1.61 in2 for T)
2 #6 @ intermediate level (1.50 in2 for T)
3 #6 @ the top (1.79 in2 for T, 0.46 in2 excess)
We have an excess of steel. In the worse case, we have a moment Mu without Tu. The 5.08in2
of steel at the bottom can all be considered for flexural tensile reinforcement purpose. In that
case,
ρ=
5.08
= 0.0142 < 0.75 ρb (Check for flexural capacity as singly-reinforced section)
(17 )( 21)
which is still O.K.
The results are illustrated in Fig. 3:
3 #6
24‘’
2 #6
#5 @ 3.5 ‘’
4 #10
17‘’
Figure 3. Configuration of the cross section
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Design Example – Shear and Torsion
Comments on the final design:
1. Different design configurations are possible, in general. Various combinations of
different sizes of steel bars can achieve same reinforcement ratio. However, relevant
designs are made typically considering the convenience of construction and the spacing
between any two steel bars (the concrete between two steel bars will crash undesirably if
the spacing between them is not enough).
2. Considering the constructability, four corner positions are usually required to deploy
longitudinal bars to fix stirrups.
3. It is preferable to use same size of steel bars on each cross section for economic reason
unless it is not possible to achieve the requirement of reinforcement.
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