1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Massachusetts Institute of Technology
Outline 10
1.054/1.541 Mechanics and Design of Concrete Structures (3-0-9)
Outline 10
Torsion, Shear, and Flexure
‰
Torsion
o Stress distribution on a cross section subject to torsion
T
T
x
x: narrow side
τ max
y
y: wide side
o Maximum shear stress, τ max
τ max = η
T
x2 y
where η = shape factor, T = torque, x , y = dimensions of the cross
section. The shape factor is different for linear and nonlinear cases.
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
‰
Spring 2004
Outline 10
Failure mode
o Torsion failure of plain concrete occurs suddenly with an inclined
tension crack in one of the wider faces, then extending into the
narrow faces. Concrete crushing occurs in the opposite wider face.
‰
Torsional strength, Tup, of plain concrete
o Several theories have been presented for computing torsional strength
of plain concrete including elastic, plastic, and skew bending theories.
o Skew bending:
x
T
M π
Ttw
y
4
b
Æ T is the applied torque and M, Ttw, are the bending and twisting
moments, respectively, on the
Æ M=
(
S=
π
4
plane.
T
, b = 2⋅y,
2
)
2 y x2
6
=
x2 y
3 2
Tup
3T
M
σt =
= 2 2 = 2up ,
S x y
x y
3 2
where Tup = ultimate torsion for plain concrete when σ reaches σ t .
Æ Tup =
x2 y
σt
3
(
)
σ t = 0.85 f r ≈ 0.85 7.5 f c' = 6 f c'
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Outline 10
x2 y
⋅ 6 f c' = 2 x 2 y f c' = Tup
Æ Tcr =
3
‰
Torsional strength contributed by steel
o Consider the system consisting of longitudinal and transverse
(stirrups) steel:
( x1 , y1 are the dimensions of steel frame as shown.)
s
s
x1
π
4
y1
y1
o Torsional moment with respect to axis of the vertical stirrups
Ts1 = ( Atα1 f s )
y1
x1
s
where At = area of one stirrup leg,
f s = stirrup stress, and
s = stirrup spacing.
o Torsional moment with respect to axis of the horizontal stirrups
Ts 2 = ( Atα 2 f s )
x1
y1
s
o Total torsional moment
Ts =
At f s
x1 y1 (α1 + α 2 )
s
Ts =
At f s
x1 y1α t
s
( α t is determined from experiment.)
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
‰
Spring 2004
Outline 10
Design concept
o Total ultimate torsion capacity, Tu
Tu = Tc + Ts
where Tc = torsional capacity contributed by concrete, and
Ts = torsional capacity contributed by reinforcement.
Tc = β Tup ( β ≈ 0.4 )
Thus,
Tc = 0.8 f c' x 2 y
The coefficient β represents reduction in torsional strength
provided by concrete after cracking. Upon cracking of concrete
stress and strain are partially transferred to steel. Stiffness and
strength of the system will depend on the amount of transverse
and longitudinal reinforcements.
o The final failure may be in one of the following ways:
1. Under reinforced Æ Both transverse and longitudinal steel yield
before failure.
2. Over reinforced Æ Concrete crushes before yielding of steel.
3. Partially over (under) reinforced
o For under reinforced elements, α t is independent of the steel ratio.
Tu
αt
Tc
At f s
4/8
x1 y1
s
1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Spring 2004
Outline 10
o Code suggestion:
α t = 0.66 + 0.33
‰
y1
≤ 1.50
x1
Role of longitudinal steel
1. It anchors the stirrups, particularly at corners.
2. It provides dowel resistance.
3. It controls crack widening.
‰
Condition of under reinforcement
Al ≤ 2 At
x1 + y1
s
where Al = volume per length of longitudinal steel.
Æ Steel yields first.
‰
Torsion combined with flexure
‰
Torsion combined with shear
o Generally shear exists simultaneously with bending.
Æ The existence of shear will reduce the resisting ability in torsion.
Thus, it is necessary to consider the case of torsion combined with
shear.
o For RC beams with transverse reinforcement:
ƒ
Pure torsion: Tu = Tc + Ts
ƒ
Pure shear: Vu = Vc + Vs
Tc
T
T0
Ts
+
=
Vc
Vs
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T 2V
+
=1
T0 3 V0
2T V
+ =1
3 T0 V0
V
V0
1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
‰
Spring 2004
Outline 10
ACI Code
o Design for torsion
Æ Same interaction as in members without transverse reinforcement.
o Excess torque
Æ Over and above that resisted by concrete, the same amount of
reinforcement is provided in members subject to torsion plus shear
as would be required for purely torsional members.
Æ This torsional reinforcement is added to that required for carrying
bending moments and flexural shears.
o Tu ≤ φ Tn = φ (Tc + Ts )
where Tu = factored torque,
φ = capacity reduction factor for torsion = 0.75,
Tn = nominal strength for torsion,
Tc = torsional moment carried by concrete, and
Ts = torsional moment carried by steel.
o Tc =
T0
2
⎛ T ⎞ ⎛V ⎞
1+ ⎜ 0 ⎟ ⎜ c ⎟
⎝ V0 ⎠ ⎝ Tc ⎠
2
where T0 = 0.8 f c' x 2 y = pure torsion and
shear.
T0
x 2 y 0.4
bd
Æ
= 0.4
=
and CT = 2
V0
bd CT
x y
o Assume
Vc Vu
= , such that
Tc Tu
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V0 = 2 f c' bd = pure
1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
Æ Tc =
Ts =
0.8 f c' x 2 y
⎛ 0.4 Vu ⎞
1+ ⎜
⎟
⎝ CT Tu ⎠
α t At f y
s
,
2
x1 y1 ,
Vc =
Vs =
Spring 2004
Outline 10
2 f c' bd
⎛
T ⎞
1 + ⎜ 2.5CT u ⎟
Vu ⎠
⎝
2
Av f y d
s
Æ Tu = φ (Tc + Ts ) = φTn
Ts =
Æ At =
Tu − φTc
φ
(T − φTc ) s
sTs
= u
α t f y x1 y1 α tφ f y x1 y1
o Ts ≤ 4Tc is required to assure yielding of steel first.
o Minimum spacing of torsional stirrups Æ 4 ( x1 + y1 ) or 12 in.
‰
Condition of neglecting torsional effects
o Torsional effects may be neglected if
n
(
Tu < 0.5φ f c' ∑ xi yi
i =1
where
∑(x y )
n
i =1
‰
2
2
i
i
)
= sum of the small rectangles for irregular shapes.
Hollow sections
h
y
x
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1.054/1.541 Mechanics and Design of Concrete Structures
Prof. Oral Buyukozturk
o When h >
o When
x
, consider the cross section as solid.
4
x
x
≤ h ≤ , assume it as solid but multiply
10
4
o When h <
Spring 2004
Outline 10
h
∑ ( x y ) by ⎛⎜⎝ 4 x ⎞⎟⎠ .
2
x
, consider it as a thin-walled section. Æ Check for
10
instability (local buckling).
‰
General formulation of post-cracking behavior of flexure, shear,
and tension interaction in R/C beams
‰
Discussion of applications: Concrete guideway systems from
monorail and maglev transportation infrastructure.
‰
Design Example – Shear and torsion
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