UNIT 4
Conditions for Static Equilibrium
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Mechanical Properties of Solids
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6

 

 

Mechanical Properties of Solids

Example 1

ex1 A horizontal aluminum rod 4.8 cm in diameter projects 5.3 cm from a wall. A 120 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0 X 1010 Nm-2. Neglecting the rod’s mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.
Fig 1.1
 

What do we know?
F = mg, where m is the mass of the object.
If r is the radius of the rod then A = πr2.
Length of rod = 4.8 X 10-2 m
Radius of Rod = 5.3 X 10-2 m
Sheer Force = 1.2 X 103 N
Shear Modulus of Aluminum is 3.0 X 1010 Nm-2

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Shear Modulus = [[Stress] / [Strain]]
Strain = [[ΔX] / [L]]

So we can get ΔX if we calculate Strain

Thus, the shear stress is:

 

(a) F/A = (mg) / (πr2)

F/A = [(1200kg) X (9.8ms-2)] / [ π(0.0242)]

F/A = 6.5 X 106 Nm-2

(b) The shear modulus G is given by

G = [(F / A) / (ΔX / L)]

G = [(F L) / (A ΔX)]

ΔX= [(F/A) L / G]

ΔX= [(6.5 X 106 Nm-2) (0.053m) / (3.0 X 1010 Nm-2)]

ΔX= 1.1 X 10-5m.

 
   
<
>
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.