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 2

q2 Two wires each of diameter 0.25 cm are shown in Fig 2.1. One is made of steel and 1.5 m long and the other 1.0 m long and made of brass. If the extensions of both wires are the same, find the (i) the extension of each wire, and (ii) the Young's modulus for brass.
Young's Modulus for steel is 2.0 X 1011 Nm-2.
Fig 2.1
 

What do we know?
Radius of Steel Wire = 1.25 X 10-3m
Radius of Brass Wire = 1.25 X 10-3m
Length of Steel Wire = 1.5 m
Length of Brass Wire = 1.0 m
Force on Steel Wire = 60N + 40N = 100N
Force on Brass Wire = 60N
Young's Modulus for steel is 2.0 X 1011 Nm-2
Extension of brass wire = Extension of Steel wire

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Young's Modulus = [[Stress] / [Strain]]
Strain = [[ΔL] / [L]]
So we can get ΔL for the steel wire if we calculate Stress of Steel Wire and use the Young's modulus to find strain.
Strain = [[ΔL] / [L]]
Then we calculate the Stress and Strain of the Brass wire and from this we can find the Young's Modulus of the Brass wire.

Answer:
Steel Wire
Stress of Steel Wire= [(Force) / (Cross Sectional Area)]
Stress of Steel Wire= [(100N) / (πr2)]
Stress of Steel Wire= [(100N) / [(π) {1.25 X 10-3 m}2)]
Stress of Steel Wire= 2.04 X 107 Nm-2
Modulus of Steel Wire= [[Stress] / [Strain]]
Strain of Steel Wire= [[Stress] / [Modulus]]
Strain of Steel Wire= 2.04 X 107 Nm-2 / 2.0 X 1011 Nm-2
Strain of Steel Wire= 1.02 X 10-4 Nm-2
[[ΔL] / [L]] = 1.02 X 10-4 Nm-2
[ΔL] of Steel Wire = 1.02 X 10-4 Nm-2 / [1.5m]
[ΔL] of Steel Wire = 1.53 X 10-4 m
[ΔL] of Brass Wire = 1.53 X 10-4 m

Brass

Stress of Brass Wire = [(Force) / (Cross Sectional Area)]
Stress of Brass Wire = [(60N) / (πr2)]
Stress of Brass Wire = [(60N) / [(π) {1.25 X 10-2 m}2)]
Stress of Brass Wire = 1.22 X 105 Nm-2
Strain of Brass Wire = [[ΔL] / [L]]
Strain of Brass Wire = (1.53 X 10-4 m) / 1.0 m
Strain of Brass Wire = (1.53 X 10-4)
Modulus of Brass Wire = [[Stress] / [Strain]]
Modulus of Brass Wire = [(1.22 X 107 Nm-2) / (1.53 X 10-4)]
Modulus of Brass Wire = 8.02 X 1010 Nm-2

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