Mechanical Engineering Principles

4 MECHANICAL ENGINEERING PRINCIPLES

x

L

Reaction

force

γ

Applied

force

Figure 1.

For a shear force, strain is denoted by the sym-
bolγ(Greek letter gamma) and, with reference to
Figure 1.5, is given by:

γ=

x

L

Problem 4. A bar 1.60 m long contracts

axially by 0.1 mm when a compressive load

is applied to it. Determine the strain and the

percentage strain.

Strainε=

contraction

original length

=

0 .1mm

1. 60 × 103 mm

=

0. 1

1600

= 0. 0000625

Percentage strain= 0. 0000625 × 100 = 0 .00625%

Problem 5. A wire of length 2.50 m has a

percentage strain of 0.012% when loaded

with a tensile force. Determine the extension

of the wire.

Original length of wire= 2 .50 m=2500 mm

and strain=

0. 012

100

= 0. 00012

Strainε=

extensionx

original lengthL

hence,extension x=εL=( 0. 00012 )( 2500 )

= 0 .30 mm

Problem 6. (a) A rectangular metal bar has

a width of 10 mm and can support a

maximum compressive stress of 20 MPa;

determine the minimum breadth of the bar

when loaded with a force of 3 kN.

(b) If the bar in (a) is 2 m long and

decreasesinlengthby0.25mmwhenthe

force is applied, determine the strain and the

percentage strain.

(a) Since stress,σ=

forceF

areaA

then, area,A=

F

σ

=

3000 N

20 × 106 Pa

= 150 × 10 −^6 m^2

=150 mm^2

Cross-sectional area=width×breadth, hence

breadth=

area

width

=

150

10

=15 mm

(b) Strain, ε =

contraction

original length

=

0. 25

2000

= 0. 000125

Percentage strain= 0. 000125 × 100

= 0 .0125%

Problem 7. A pipe has an outside diameter

of 25 mm, an inside diameter of 15 mm and

length 0.40 m and it supports a compressive

load of 40 kN. The pipe shortens by 0.5 mm

when the load is applied. Determine (a) the

compressive stress, (b) the compressive

strain in the pipe when supporting this load.

Compressive forceF=40 kN=40000 N,

and cross-sectional areaA=

π

4

(D^2 −d^2 ),

whereD=outside diameter=25 mm and

d=inside diameter=15 mm. Hence

A=

π

4

( 252 − 152 )mm^2

=

π

4

( 252 − 152 )× 10 −^6 m^2

= 3. 142 × 10 −^4 m^2