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Strength of materials

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Strength of Materials
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
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Prof. M. S. Sivakumar
Strength of Materials
Problem 1: Derivation of Shear stress in rectangular crosssection
Derive an expression for the shear stress distribution in a beam of solid rectangular crosssection 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,
VQ
It
V
y.dA
=
It ∫
τ=
A1
=
=
V
Ib
d/2
∫
by dy
y1
V⎡
( d / 2 )2 − ( y1 )2 ⎤⎥⎦ ---------------------- (1)
⎢
⎣
2I
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.
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Strength of Materials
Vd 2
8I
3 V
=
2 bh
3V
=
2A
τmax =
Top
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Strength of Materials
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
y NA =
20 × 100 × 10 + 20 × 100 × 70
= 40mm
20 × 100 + 20 × 100
Q = ∫ ydA of shaded area about neutral axis.
Q = 20 x 100 x 30 = 6 x 104
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Strength of Materials
V = 1KN
3
20 × 1003
2 100 × 20
I=
+ 20 × 100 × 30 +
+ 100 × 20 × 302
12
12
= 5.33 × 106
q=
VQ 103 × 6 × 104 × φ(10−3 )3
N
=
= 1.125 × 104
4
I
m
5.33 × 106 × 10−3
(
)
= 11.25
KN
m
Top
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Strength of Materials
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:
τmax =
3V
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,
τmax =
3 × 15 × 103
2 × 50 × 50 × 10−6
= 9 Mpa
Top
Indian Institute of Technology Madras
Prof. M. S. Sivakumar
Strength of Materials
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:
τ=
Formula used:
VQ
It
V = 2KN
⎞
10 × 1003 ⎛ 100 × 103
I=
+⎜
+ 100 × 10 × 552 ⎟ × 2 = 6.9 × 106 mm 4
⎜ 12
⎟
12
⎝
⎠
Q is maximum at the midpoint as shown below
Q = 50 × 10 × 55
τmax =
2 × 103 × 50 × 10 × 55 × 10−3
(
)
( 6.9 × 10 )(10 )
−3
6
−3 4
Indian Institute of Technology Madras
× 10 × 10
3
= 0.79 MPa
Strength of Materials
Prof. M. S. Sivakumar
Top
Indian Institute of Technology Madras
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