Mechanical Engineering Principles

10 MECHANICAL ENGINEERING PRINCIPLES

40

A

80

120

160

200

240

280

320

360

C

B

D

Extension /mm

Load / N

0 246810121416 18

F

Figure 1.

Length of specimen,L= 8 .0mand

cross-sectional areaA=

πd^2

4

=

π( 0. 0013 )^2

4

= 1. 327 × 10 −^6 m^2

Hencemodulus of elasticity,E

=( 33. 33 × 103 )

(

8. 0

1. 327 × 10 −^6

)

=201 GPa

The limit of proportionality is at point D in
Figure 1.9 where the graph no longer follows a
straight line. This point corresponds to a load of
250 N as shown.

Stress at the limit of proportionality

=

force

area

=

250

1. 327 × 10 −^6

= 188. 4 × 106 Pa=188.4 MPa

Note that for structural materials the stress at the
elastic limit is only fractionally larger than the stress
at the limit of proportionality, thus it is reasonable
to assume that the stress at the elastic limit is the
same as the stress at the limit of proportionality; this
assumption is made in the remaining exercises. In
Figure 1.9, the elastic limit is shown as pointF.

Now try the following exercise

Exercise 3 Further problems on Hooke’s

law

1. A wire is stretched 1.5 mm by a force of
   300 N. Determine the force that would
   stretch the wire 4 mm, assuming the
   elastic limit of the wire is not exceeded.
   [800 N]


2. A rubber band extends 50 mm when a
   force of 300 N is applied to it. Assum-
   ing the band is within the elastic limit,
   determine the extension produced by a
   force of 60 N. [10 mm]


3. A force of 25 kN applied to a piece
   of steel produces an extension of
   2 mm. Assuming the elastic limit is
   not exceeded, determine (a) the force
   required to produce an extension of
   3.5 mm, (b) the extension when the
   applied force is 15 kN.

[(a) 43.75 kN (b) 1.2 mm]

4. A test to determine the load/extension
   graph for a specimen of copper gave the
   following results:

Load (kN) 8.5 15.0 23.5 30.

Extension (mm) 0.04 0.07 0.11 0.

Plot the load/extension graph, and from

the graph determine (a) the load at an

extension of 0.09 mm, and (b) the exten-

sion corresponding to a load of 12.0 N.

[(a) 19 kN (b) 0.057 mm]

5. A circular bar is 2.5 m long and has a
   diameter of 60 mm. When subjected to a
   compressive load of 30 kN it shortens by
   0.20 mm. Determine Young’s modulus
   of elasticity for the material of the bar.

[132.6 GPa]

6. A bar of thickness 20 mm and hav-
   ing a rectangular cross-section carries a
   load of 82.5 kN. Determine (a) the min-
   imum width of the bar to limit the max-
   imum stress to 150 MPa, (b) the mod-
   ulus of elasticity of the material of the
   bar if the 150 mm long bar extends