Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

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

Indian Institute of Technology Madras

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

Indian Institute of Technology Madras

2

max

Vd

8I

3 V

2 bh

3 V

2 A

τ =

=

=

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

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.

Q = 20 x 100 x 30 = 6 x 10

4

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

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

Indian Institute of Technology Madras

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

Indian Institute of Technology Madras

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

Indian Institute of Technology Madras

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