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 6

The stress at which plastic deformation begins is 3.45X108 Nm-2, and the modulus of elasticity is 1.03X1011 Nm-2, for a brass alloy.
(a) What is the maximum load that may be applied to a wire of radius 6.43mm without plastic deformation?
(b) If the original length of the specimen is 5.00 cm, what is the maximum length to which it may be stretched without causing plastic deformation?

We know
Maximum Stress without plastic deformation = 3.45X108 Nm-2
Young's Modulus of Elasticity = 1.03X1011 Nm-2
Radius of wire = 6.43X10-3 m
Cross Sectional Area = πr2 m2
Cross Sectional Area = 1.30X10-4 m2
Original Specimen Length = 5.00X10-2 m

Stress = [[Force] / [Cross Sectional Area]]
Young's Modulus = [[Stress] / [Strain]]
Strain = [[ΔL] / [L]]

(a) Stress = { [Force] / [Cross Sectional Area]}
(3.45X108 Nm-2) = Force / (1.30X10-4 m2)
Force = (3.45X108 Nm-2) X (1.30X10-4 m2)
Maximum Force without plastic deformation= 6.49X104 N

(b) Young's Modulus = [[Stress] / [Strain]]
(1.03X1011 Nm-2) = (3.45X1011 Nm-2) / Strain
Strain = (3.45X108 Nm-2) / (1.03X1011 Nm-2)
Strain = 3.35X10-3

Strain = [[ΔL] / [L]]
3.35X10-3 = ΔL / 5.00X10-2 m
ΔL = 3.35 X10-3 X 5.00X10-2 m
ΔL = 1.68 X10-4 m.

L + ΔL = 5.00X10-2 m + 1.68 X10-4 m.
Maximum Length without Plastic Deformation = 5.02X10-2 m.
Maximum Length without Plastic Deformation = 5.02 cm.

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