5- Shear Stress Distribution
5.1 Shear stress distribution due to bending
M+dM
Longitudinal stress at A
𝜎A =
Longitudinal stress at B
𝜎B =
Longitudinal force at A
𝜎𝐴 𝐴𝑟𝑒𝑎 =
Longitudinal force at B
𝜎𝐵 𝐴𝑟𝑒𝑎 =
F = FA − FB =
ℎ dM
F = ∫𝑦
𝜏=
ℎ dM
∫𝑦
I
I
M+dM
I
I
M
I
𝑦
𝑦
𝑦 (bdy) −
M
I
M+dM
I
M
I
𝑦 (bdy) = FA
𝑦 (bdy) = FB
𝑦 (bdy) =
dM
I
𝑦 (bdy)
𝑦 (bdy)
Shear force
Area
=
𝑄
𝐴𝑠
𝑦 (bdy) = 𝜏 𝑏𝑑𝑥 ,
, 𝑄 = 𝜏 𝐴𝑠 = 𝜏 𝑏𝑑𝑥
ℎ
∫𝑦 𝑦 (bdy) = 𝐴𝑦̅
(( First moment of area of shaded portion ))
dM
I
𝐴𝑦̅ = 𝜏 𝑏𝑑𝑥
𝜏=
𝐴𝑦̅ 𝑑𝑀
𝐼𝑏 𝑑𝑥
𝜏=
𝑑𝑀
𝑑𝑥
𝑄𝐴𝑦̅
I𝑏
: rate of change in B.M. and equal to S.F or 𝑄 at the section
where:
𝜏 : shear stress
𝑄 : shear force
𝐴 : area of shaded portion
𝑦̅ : distance from N.A. to the centroid of the shaded area
I : moment of inertia
b : Area width
5.2 Applications
5.2.1 Rectangular Section
𝑄𝐴𝑦̅ 𝑄 𝐷/2
𝜏=
= ∫ 𝑦𝑑𝐴
I𝑏
I𝑏 𝑦
𝐷
𝐴 = ( − 𝑦) 𝑏
2
1 𝐷
1
𝐷
𝑦̅ = [𝑦 + ( − 𝑦)] = (𝑦 + )
2 2
2
2
𝜏=
𝑄 𝐷
1
𝐷
[( − 𝑦) 𝑏 (𝑦 + )]
I𝑏 2
2
2
𝑄 𝐷2
𝜏 = ( − 𝑦2)
2I 4
at 𝑦 =
𝑄 𝐷2 𝐷2
𝜏= ( − )=0
2I 4
4
𝐷
2
at 𝑦 = 0
𝑄 𝐷2
𝜏 = ( − 0)
2I 4
(N.A.)
𝑏𝐷3
IN.A =
12
𝑄
𝜏avg =
𝐷𝑏
𝜏max = 𝜏N.A. =
𝜏N.A. = 𝜏max
3
𝜏
2 avg
𝜏N.A. = 𝜏max
𝑄𝐷2
3𝑄
=
=
𝑏𝐷3 2𝑏𝐷
8
12
𝑄𝐷2
=
8I
Example: A simply supported beam having a rectangular cross-section with 10 cm wide and 15 cm
deep carries a uniformly distributed load on a span 180 cm. If the permissible stress is 2.75 kN/cm2 in
bending and 0.21 kN/cm2 in shear, calculate the maximum load the beam can carry.
Solution:
𝜎𝑏𝑚𝑎𝑥
15 mm
Bending
𝑤𝐿2
𝑦𝑚𝑎𝑥
𝑀𝑚𝑎𝑥 𝑦𝑚𝑎𝑥
=
= 8 3
𝑏ℎ
I
12
𝑤 (180)2 15
8
2
2.75 ∗ 103 =
10 (15)3
12
10 mm
𝑤 = 254.6 N/cm
Shear
𝜏N.A. = 𝜏max
𝑄𝐷2
=
8I
𝑤 (180)
(15)2
2
0.21 ∗ 10 =
10 (15)3
8 12
3
𝑤 = 233.3 N/cm
Permissible load
𝐴
5.2.2 Triangular Section
𝑥
2
ℎ
3
x : distance from A (top)
𝑄𝐴𝑦̅
𝜏=
I𝑏
1
𝐴𝐴𝐷𝐸 = 𝐷𝐸 𝑥
2
𝐷𝐸 𝐵𝐶
=
𝑥
ℎ
𝐴𝐴𝐷𝐸
2
𝑥
3
𝐷
𝑥
+
ℎ
𝑦
𝐸
𝑦̅
+
1
ℎ
3
𝐵
𝐶
𝑏
𝐷𝐸 =
1𝑏
𝑏𝑥 2
=
𝑥𝑥=
,
2ℎ
2ℎ
𝑏𝑥 2 2
(ℎ − 𝑥)
𝑄
𝜏 = [ 2ℎ 3
]
𝑏
I
𝑥
ℎ
𝑏
𝑥
ℎ
2
𝑦̅ = (ℎ − 𝑥)
3
𝜏=
𝑄
(ℎ𝑥 − 𝑥 2 )
3I
at
𝑥 = 0 (at top A )
at 𝑥 = ℎ
(at base)
2
at 𝑥 = ℎ
3
To find 𝜏max ,
𝜏max
(at N.A. )
𝜏=0
𝜏N.A.
𝑄
2
2
=
(ℎ ℎ − ( ℎ)2 )
3I
3
3
d𝜏
𝑄
=0=
(ℎ − 2𝑥)
d𝑥
3I
𝑄
ℎ
ℎ
=
(ℎ ( ) − ( )2 )
3I
2
2
𝜏avg =
IN.A
𝜏=0
𝜏max
𝑥=
𝑄ℎ2
=
12I
𝑄
𝑄
2𝑄
=
=
Area
1
𝑏ℎ
𝑏ℎ
2
𝑏ℎ3
=
36
𝜏N.A.
2𝑄ℎ2
8𝑄
=
=
𝑏ℎ3 3𝑏ℎ
27
36
4
𝜏N.A. = 𝜏avg
3
𝜏max
𝑄ℎ2
3𝑄
=
=
𝑏ℎ3
𝑏ℎ
12
36
3
𝜏max. = 𝜏avg
2
ℎ
+
𝑏
ℎ
2
𝜏N.A.
2𝑄ℎ2
=
27 I
5.2.3 Circular Section
𝜏=
𝑏
𝑄𝐴𝑦̅
I𝑏
𝑟
𝑟
𝑄𝐴𝑦̅ 𝑄
𝜏=
= ∫ 𝑦𝑑𝐴
I𝑏
I𝑏 𝑦
𝑑𝐴 = 𝑏 𝑑𝑦 = 2√𝑟 2 − 𝑦 2 𝑑𝑦
𝑟
𝐴𝑦̅ = ∫ 𝑦 2√𝑟 2 − 𝑦 2 𝑑𝑦
𝑦
3
2 2
2
𝑏3
2 2
3
𝐴𝑦̅ = (𝑟 − 𝑦 ) =
𝑏 =
3
3×8
12
𝑄 𝑏3
𝑄𝑏 2
𝜏= ( )=
,
I𝑏 12
12 I
𝑄(2√𝑟 2 − 𝑦 2 )2
𝜏=
12 I
𝑄(𝑟 2 − 𝑦 2 )
𝜏=
3I
at
at
𝐵
𝐴
𝑦=𝑟
𝜏=0
𝑦 = 0 (at N.A. )
𝜏avg
𝑄
𝑄
=
=
,
Area
𝜋𝑟 2
𝜏N.A.
𝑄𝑟 2
4𝑄
= 𝜋
=
2
3 𝑟4 3 𝜋 𝑟
4
𝜏N.A.
IN.A =
𝑄𝑟 2
=
= 𝜏max
3I
𝜋 4
𝑟
4
𝜏N.A. = 𝜏max =
4
𝜏
3 avg
𝑑𝑦
𝑦
5.2.4 I-Section
a- if (y) greater than
𝑑
2
𝐷
− 𝑦)
2
1 𝐷
1 𝐷
𝑦̅ = 𝑦 + ( − 𝑦) = ( + 𝑦)
2 2
2 2
𝑦
Shaded area = 𝐵 (
𝜏=
𝐷
𝑑
𝑄
1 𝐷
𝐷
[𝐵 ( − 𝑦) × ( + 𝑦)]
I𝑏
2
2 2
𝜏=
𝑄𝐵 𝐷2
( − 𝑦2)
2I𝑏 4
at
𝑦=
at
𝑑
𝑦=
2
𝐷
2
𝜏=0
𝑄𝐵 𝐷2 𝑑 2
𝑄
𝜏1 =
( − ) = (𝐷2 − 𝑑 2 )
2I𝐵 4
4
8I
𝑄𝐵 𝐷2 𝑑 2
𝑄𝐵 2
(𝐷 − 𝑑 2 )
𝜏2 =
( − )=
2I𝑏′ 4
4
8I𝑏′
b- if (y) smaller than
𝑑
2
𝐷 𝑑 𝑑 1 𝐷 𝑑
(𝐴𝑦̅)𝐹𝑙𝑎𝑛𝑔𝑒 = 𝐵 ( − ) [ + ( − )]
2 2 2 2 2 2
(𝐴𝑦̅)𝐹𝑙𝑎𝑛𝑔𝑒
(𝐴𝑦̅)𝑊𝑒𝑏
𝐵 𝐷2 𝑑 2
𝐵
= ( − ) = (𝐷2 − 𝑑 2 )
2 4
4
8
𝑑
1 𝑑
𝑏′ 𝑑2
= 𝑏 ( − 𝑦) [ 𝑦 + ( − 𝑦)] =
( − 𝑦2)
2
2 2
2 4
′
Web
𝑏′
Flange
𝐵
𝑦̅
(𝐴𝑦̅) 𝑇𝑜𝑡𝑎𝑙 = (𝐴𝑦̅)𝐹𝑙𝑎𝑛𝑔𝑒 + (𝐴𝑦̅)𝑊𝑒𝑏
(𝐴𝑦̅)𝐹𝑙𝑎𝑛𝑔𝑒
𝐵
𝑏′ 𝑑2
2
2
= (𝐷 − 𝑑 ) + ( − 𝑦 2 )
8
2 4
𝑄 𝐵
𝑏′ 𝑑2
2
2
𝜏=
[ (𝐷 − 𝑑 ) + ( − 𝑦 2 )]
I𝑏′ 8
2 4
at
𝑦=0
(at N.A. )
𝜏N.A. = 𝜏max
𝑄 𝐵
𝑏′ 𝑑2
2
2
=
[ (𝐷 − 𝑑 ) +
]
I𝑏′ 8
8
Example: Draw shear stress distribution for figure shown below when the shear force 𝑄 =10 kN.
2 cm
3 cm
6 cm
2 cm
8 cm
Solution:
𝜏=
𝑄𝐴𝑦̅
I𝑏
IN.A
𝑏ℎ3 𝜋 4
=
−
𝑑
12 64
IN.A
8 × 103 𝜋 4
=
−
6 = 630.1 cm4
12
64
𝑄 𝐴𝑦̅
10 × 103 𝐴𝑦̅
𝐴𝑦̅
𝜏= ( )=
( ) = 16.85
I 𝑏
630.1
𝑏
𝑏
𝑝𝑜𝑖𝑛𝑡 (1)
𝐴𝑟𝑒𝑎 = 0,
𝑝𝑜𝑖𝑛𝑡 (2)
𝐴𝑟𝑒𝑎 = 8 × 1 = 8,
𝝉𝟐 = 16.85
36
= 𝟕𝟒. 𝟔𝟏𝟓 𝐤𝐍/𝐜𝐦𝟐
8
𝑝𝑜𝑖𝑛𝑡 (3)
𝐴𝑟𝑒𝑎 = 8 × 2 = 16,
𝝉𝟑 = 16.85
64
= 𝟏𝟑𝟐. 𝟔𝟒 𝐤𝐍/𝐜𝐦𝟐
8
𝑝𝑜𝑖𝑛𝑡 (4)
𝑦̅ = 5,
𝐴𝑦̅ = 0 ,
𝑏=8,
∴ 𝝉𝟏 = 𝟎
𝑦̅ = 4.5, 𝐴𝑦̅ = 36 , 𝑏 = 8
𝑦̅ = 4, 𝐴𝑦̅ = 64 , 𝑏 = 8
3
3.5
2
(𝐴𝑦̅)4 = 8 × 3.5 [
+ 1.5] − [ (32 − 1.52 )2 ] = 79.3 cm3
2
3
𝝉𝟒 = 16.85
𝑝𝑜𝑖𝑛𝑡 (5)
𝝉𝟓 = 16.85
79.3
= 𝟒𝟕𝟎 𝐤𝐍/𝐜𝐦𝟐
2.8
𝜋
4𝑟
1
(𝐴𝑦̅)5 = 8 × 5 × 2.5 − [ 𝑑 2 × ] = 100 − 62 × 3 = 64 cm3
4
3𝜋
3
64
= 𝟓𝟑𝟎 𝐤𝐍/𝐜𝐦𝟐
2
1
2
3
4
5