Problem 1: Derivation of Shear stress in rectangular crosssection Problem 2: Computation of Shear stresses
Problem 3: Computation of Shear stresses Problem 4: Computation of Shear stresses
Problem 1: Derivation of Shear stress in rectangular crosssection
Derive an expression for the shear stress distribution in a beam of solid rectangular cross- section transmitting a vertical shear V.
The cross sectional area of the beam is shown in the figure. A longitudinal cut through the beam at a distance y1, from the neutral axis, isolates area klmn. (A1).
Shear stress,
1
1
A d / 2
y
VQ It
V y.dA It
V by dy Ib
τ =
=
=
∫
∫
( ) ( )
2 1V d / 2 y 2I
= ⎡⎢⎣ − 2⎤
⎥⎦ --- (1)
The Shear Stress distribution is as shown below
Max Shear Stress occurs at the neutral axis and this can be found by putting y = 0 in the equation 1.
2 max
Vd 8I 3 V 2 bh 3 V 2 A
τ =
=
=
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Problem 2: Computation of Shear stresses
A vertical shear force of 1KN acts on the cross section shown below. Find the shear at the interface (per unit length)
Solution:
Formula used: q = VQ/I
We first find the distance of the neutral axis from the top fiber.
All dimensions in mm
NA
20 100 10 20 100 70
y 4
20 100 20 100
× × + × ×
= =
× + × 0mm
A
Q=
∫
yd of shaded area about neutral axis.V = 1KN
( )
3 3
2 2
6
3 4 3 3
4
6 3 4
20 100 100 20
I 20 100 30 100 20 30
12 12
5.33 10
VQ 10 6 10 (10 ) N
q 1.125 10
I m
5.33 10 10
11.25 KN m
−
−
× ×
= + × × + + × ×
= ×
× × × φ
= = = ×
× ×
=
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Problem 3: Computation of Shear stresses
A 6m long beam with a 50 mm × 50 mm cross section is subjected to uniform loading of 5KN/m. Find the max shear stress in the beam
Solution:
max
3V τ = 2A
We first find the section of maximum shear force. We know this is at the supports and is equal to
5 6 15KN 2
× =
We also know that max.shear stress occurs at the centre (for a rectangular cross section) and is 1.5 times the average stress.
So,
3
max 6
3 15 10
9 Mpa 2 50 50 10−
τ = × × =
× × ×
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Problem 4: Computation of Shear stresses
The cross section of an I beam is shown below. Find the max.shear stress in the flange if it transmits a vertical shear of 2KN.
Solution:
Formula used: VQ τ = It V=2KN
3 3
2 6
10 100 100 10
I 100 10 55 2 6.9 10 mm
12 12
⎛ ⎞
× ×
= +⎜⎜⎝ + × × ⎟⎟⎠× = ×
4
Q is maximum at the midpoint as shown below
Q = 50 × 10 × 55
( ) ( )( )
3 3 3
max 6 3 4 3
2 10 50 10 55 10
0.79 MPa
6.9 10 10 10 10
−
− −
× × × × ×
τ = =
× × ×
