Reinforced Concrete Design Analysis and Design for Torsion 1

13
Reinforced Concrete Design
Analysis and Design for Torsion 1
Torsional effects in reinforced concrete
Torsion in plain concrete members
Torsion in reinforced concrete members
Combined shear and torsion
ACI code provisions for torsion design
Mongkol JIRAVACHARADET
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Torsional effects in reinforced concrete
T
m t
T
Torsion at a cantilever slab

T
A
m t
T
A B
Stiff edge beam
B
Torsion at an edge beam
A B
Flexible edge beam

Torsion in plain concrete members
T
Rectangular section: max 2
y/x
α
τ
=
T
αx
y
1.0 1.5 2.0 3.0 5.0 ∞
0.208 0.219 0.246 0.267 0.290 1/3
y
x
T
τ max

Cracking Strength
Plain concrete rectangular sections in torsion
Torsion cracks
T
T
T t
T b 45 o
45 o
T
τ
T
τ
f t max at 45 o
Bending: T b = T cos45 o
Torsion: T t = T cos45 o
τ
τ

a
T t
T
T b
45 o
a
Sectional Modulus:
Concrete crack occurs when
f t max at 45 o
ft,max = 0.80fr = 0.80 × 2.0 f ′ c = 1.6 f ′ c
f r = Modulus of rupture
2
1 ⎛ y ⎞ 3 ⎛ 2 ⎞ x y
a−a = a−a = =
o o
S I /( x / 2)
12
⎜ x
cos45
⎟ ⎜
x
⎟
⎝ ⎠ ⎝ ⎠ 6cos45
T 6cos45 3T
f T f ′
o
t,max =
b, cr
=
Sa−a o
cr cos45 2
x y
cr = = 1.6
2
x y
c
2
x y
cr =
c
( 1.6 )
T f ′
3
τ
τ

Strength of Reinforced Concrete
Rectangular sections in torsion
ACI 1995: Solid section is analysed as a hollow-box section.
T n
T cr solid section
T cr hollow section
0
0 1 2 3
Percent of torsional reinforcement
Solid Hollow

Shear Stress in Thin-walled Tube
A 0
Cracking Torque ( T cr ):
t
Shear flow:
Shear stress:
T
τ = = 1.1 f ′
cr
cr
2A0t
c
( )
T = 1.1 f ′ 2A
t
cr c
0
q
=
τ =
=
T
2A
0
0
kg/cm
q
kg/cm
t
T
2A
t
2

2A 3A
ACI318-95: A0 = , t =
3 4 p
cp cp
p cp : outside perimeter of concrete cross-section
A cp : area enclosed by p cp
T
cr
1.1
=
2
c cp
Neglect torsion when: T ≤ φT
/ 4
p
f ′ A
cp
u cr
cp
kg-cm
0.85 for torsion

7.1 กก
3 กก
15 .
กกก ก f c = 240 กก./. 2
60 cm
15 cm
30 cm
3 ton
p cp = 2(60+30) = 180 .
Acp = (60)(30) = 1,800 . 2
ก T cr = 1.1 (1,800) 2 /(180 1,000)
240
= 307 -. = 3.07 -.
ก φT cr /4 = 0.85(3.07)/4 = 0.65 -
ก
Tu = (3)(0.15) = 0.45 - < 0.65 - OK

Torsion in Reinforced Concrete Members
x 0
b
y 0
h
Gross area A oh = x 0 y 0
Shear perimeter p h = 2(x 0 + y 0)
x 0 , y 0 = Distance from center to
center of stirrup

Space Truss Analogy
Torsional resistance =
sum of contribution from
shear in each four walls
V 2
V 1
V 3
V 4
Crack
Stirrup
θ
Torsional crack
T
x o
45 o
y o
Longitudinal bar
Concrete compression struts
T

V4x T 4 =
2
o
x o
V4 = n At fyv
V 4
o
n = y cot 45 / s = y / s
V
4
=
o o
At fyv yo
s
A f x y
T = = T = T =
T
2s
t yv o o
4 1 2 3
y o
V 4
s
A t f yv
A t f yv
y o cot θ
A t f yv
A t = Area of one leg stirrup
f yv = Yield strength of transverse
reinforcement
n = Number of cracks intercept
stirrup

T = T + T + T + T
n
y o
V 4
1 2 3 4
ACI318-95:
Diagonal compression struts
θ
Axial force due to torsion:
=
∆N 4 = V 4 cot θ
2 At fyv xo yo
s
∆N 4 /2
∆N 4 /2
=
=
At fyv yo
s
2 At fyv Aoh
2A
f A
T = where A = 0.85A
s
t yv 0
n 0
oh
θ
s
V 4 /sinθ
V 4 cot θ
x oy o
V 4

Summing over all four sides,
Longitudinal
reinforcement
∆ N = ∆ N1 + ∆ N2 + ∆ N3 + ∆N4
A l f yl
At fyv
= 2( xo + yo
)
s
=
At fyv ph
s
A f t Al = ph
s f
yv
yl
perimeter of stirrup
A l = Total area of longitudinal reinforcement to resist torsion
f yl = Yield strength of longitudinal reinforcement

Combined Shear and Torsion
Shear stress: τ v = V / b w d
Torsional stress: τ t = T / (2A 0 t)
for cracked section: A 0 = 0.85A oh , t = A oh / p h
Hollow section
Torsional
stresses
Shear
stresses
V T p
τ = τ v + τ t = +
b d A
w 1.7
h
2
oh
Torsional
stresses
Solid section
Shear
stresses
2 2
⎛ V ⎞ ⎛ T p ⎞ h
τ = ⎜ ⎟ + ⎜ 2 ⎟
⎝ bwd ⎠ ⎝1.7 Aoh
⎠

ACI Provisions for Torsion Design
Strength requirement:
T ≤ φT
u n
where T u = required torsional strength at factored load
T n = nominal torsional strength of member
φ = 0.85 strength reduction factor for torsion
ACI318-95:
A oh = shaded area
2A
f A
T = where A = 0.85A
s
t yv 0
n 0
oh

Minimal Torsion
Neglect torsion when
where
T
cr
=
1.1
p
f ′ A
2
c cp
cp
T ≤ φT
u cr
kg-cm
⎛ V ⎞
c
Limitation on shear stress τ max ≤ φ ⎜ + 2.1 f ′ c ⎟
⎝ bwd ⎠
Hollow section:
Solid section:
/ 4
V ⎛ ⎞
u Tu ph Vc
+ ≤ φ + 2.1 ′
2 ⎜ fc
⎟
bwd 1.7Aoh
⎝ bwd ⎠
2 2
⎛ V ⎞ ⎛ ⎞ ⎛ ⎞
u Tu ph Vc
⎜ ⎟ + ⎜ ≤ φ + 2.1 ′
2 ⎟ ⎜ fc
⎟
⎝ bwd ⎠ ⎝1.7 Aoh ⎠ ⎝ bwd ⎠

Reinforcement for torsion
2Ao
At f yv
Tu / φ = ,
s
Ao = 0.85Aoh
Tu s
At =
2φ
A f
, f yv ≤ 4,000 kg/cm
o yv
Combined shear and torsion stirrup
(for closed loop stirrup)
Av + t Av 2At
= +
s s s
v+ t v t
A v
A t
2

Max. spacing to control spiral cracking smax ≤ p / 8 ≤ 30 cm
Min. area of close stirrup ( 2 ) 3.5 w b s
Av + At
≥
f
f ′ A t
Min. area of longitudinal steel
⎛ A ⎞ f
Al ,min = − ⎜ ⎟ ph
f ⎝ s ⎠ f
where
A 1.8b
≥
s f
t w
- Spacing of A l ≤ 30 cm
- Distributed around perimeter
yv
- Bar size ≥ 10 mm ≥ 1/24 stirrup spacing
- At least one bar on each corner
h
yv
1.3 c cp yv
yv yl