Strength of Materials Problems Part 2

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14 Οκτ 2011 (πριν από 6 χρόνια και 7 μήνες)

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Indian Institute of Technology Madras Prof. M. S. Sivakumar

Strength of Materials Prof. M. S. Sivakumar

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

Strength of Materials Prof. M. S. Sivakumar

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 y
1,
from the neutral axis, isolates area klmn. (A
1
).
Shear stress,

1
1
A
d/2
y
VQ
It
V
y.dA
It
V
b
ydy
Ib
τ =
=
=

( ) ( )
2
1
V
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.

Strength of Materials Prof. M. S. Sivakumar

2
max
Vd
8I
3 V
2 bh
3 V
2 A
τ =
=
=

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Strength of Materials Prof. M. S. Sivakumar

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=

Q = 20 x 100 x 30 = 6 x 10
4
Strength of Materials Prof. M. S. Sivakumar

V = 1KN

( )
3 3
2 2
6
3 4 3 3
4
4
6 3
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
KN
11.25
m

× ×
= + × × + + × ×
= ×
× × ×φ
= = = ×
× ×
=

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Strength of Materials Prof. M. S. Sivakumar

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
2A
τ =

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
9Mpa
2 50 50 10

× ×
τ = =
× × ×

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Strength of Materials Prof. M. S. Sivakumar

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
It
τ =

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
4
6 3 3
2 10 50 10 55 10
0.79 MPa
6.9 10 10 10 10

− −
× × × × ×
τ = =
× × ×

Strength of Materials Prof. M. S. Sivakumar