Mechanical Engineering Principles

THE EFFECTS OF FORCES ON MATERIALS 9

When subjected to a compressive force the

length of the rod is 199.6 mm. Determine

(a) the stress in the rod when loaded, and

(b) the magnitude of the force.

Take the modulus of elasticity for aluminium

alloyas70GPa.

(a) Original length of rod,L = 200 mm, final

length of rod= 199 .6 mm, hence contraction,

x= 0 .4 mm. Thus, strain,

ε=

x

L

=

0. 4

200

= 0. 002

Modulus of elasticity,E=

stressσ

strainε

, hence

stress,σ=Eε= 70 × 109 × 0. 002

= 140 × 106 Pa=140 MPa

(b) Since stressσ=

forceF

areaA

, then force,F=σA

Cross-sectional area,A=

πd^2

4

=

π( 0. 010 )^2

4

= 7. 854 × 10 −^5 m^2.

Hence,compressive force,

F=σA= 140 × 106 × 7. 854 × 10 −^5

=11.0 kN

Problem 15. A brass tube has an internal

diameter of 120 mm and an outside diameter

of 150 mm and is used to support a load of

5 kN. The tube is 500 mm long before the

load is applied. Determine by how much the

tube contracts when loaded, taking the

modulus of elasticity for brass as 90 GPa.

Force in tube,F = 5kN=5000 N, and cross-

sectional area of tube,

A=

π

4

(D^2 −d^2 )=

π

4

( 0. 1502 − 0. 1202 )

= 0 .006362 m^2.

Stress in tube, σ =

F

A

=

5000 N

0 .006362 m^2

= 0. 7859 × 106 Pa.

Since the modulus of elasticity,

E=

stressσ

strainε

,

then strain, ε=

σ

E

=

0. 7859 × 106 Pa

90 × 109 Pa

= 8. 732 × 10 −^6.

Strain,

ε=

contractionx

original lengthL

thus, contraction,x =εL= 8. 732 × 10 −^6 × 0. 500

= 4. 37 × 10 −^6 m.

Thus, when loaded, the tube contracts by

4.37μm.

Problem 16. In an experiment to determine

the modulus of elasticity of a sample of mild

steel, a wire is loaded and the corresponding

extension noted. The results of the experi-

ment are as shown.

Load (N) 0 40 110 160 200 250 290 340

Extension (mm) 0 1.2 3.3 4.8 6.0 7.5 10.0 16.

Draw the load/extension graph.

The mean diameter of the wire is 1.3 mm

and its length is 8.0 m. Determine the

modulus of elasticityEof the sample, and

the stress at the limit of proportionality.

A graph of load/extension is shown in Figure 1.

E=

σ

ε

=

F

A

x

L

=

(

F

x

)(

L

A

)

F

x

is the gradient of the straight line part of the

load/extension graph.

Gradient,

F

x

=

BC

AC

=

200 N

6 × 10 −^3 m

= 33. 33 × 103 N/m

Modulus of elasticity=(gradient of graph)

(

L

A

)