Mechanical Engineering Principles

8 MECHANICAL ENGINEERING PRINCIPLES

Whenx=2 mm,F=250 N, thus 2=k( 250 ),

from which, constantk=

2

250

=

1

125

Whenx=5 mm, then 5=kF

i.e. 5 =

(

1

125

)

F

from which, force F= 5 ( 125 )=625 N

Thus to stretch the wire 5 mm a force of 625 N

is required.

Problem 11. A force of 10 kN applied to a

component produces an extension of

0.1 mm. Determine (a) the force needed to

produce an extension of 0.12 mm, and

(b) the extension when the applied force is

6 kN, assuming in each case that the limit of

proportionality is not exceeded.

From Hooke’s law, extensionxis proportional to

force F within the limit of proportionality, i.e.

x∝Forx=kF,wherekis a constant. If a force

of 10 kN produces an extension of 0.1 mm, then

0. 1 =k( 10 ), from which, constantk=

0. 1

10

= 0. 01

(a) When an extensionx= 0 .12 mm, then

0. 12 =k(F), i.e. 0. 12 = 0. 01 F, from which,

forceF=

0. 12

0. 01

=12 kN

(b) When forceF=6kN,then
extensionx=k( 6 )=( 0. 01 )( 6 )=0.06 mm

Problem 12. A copper rod of diameter

20 mm and length 2.0 m has a tensile force

of 5 kN applied to it. Determine (a) the

stress in the rod, (b) by how much the rod

extends when the load is applied. Take the

modulus of elasticity for copper as 96 GPa.

(a) ForceF=5kN=5000 N and

cross-sectional areaA=

πd^2

4

=

π( 0. 020 )^2

4

= 0 .000314 m^2

Stress,σ=

F

A

=

5000 N

0 .000314 m^2

= 15. 92 × 106 Pa=15.92 MPa

(b) SinceE=

σ

ε

then

strainε=

σ

E

=

15. 92 × 106 Pa

96 × 109 Pa

= 0. 000166

Strainε=

x

L

, hence extension,

x=εL=( 0. 000166 )( 2. 0 )= 0 .000332 m

i.e.extension of rod is 0.332 mm.

Problem 13. A bar of thickness 15 mm and

having a rectangular cross-section carries a

load of 120 kN. Determine the minimum

width of the bar to limit the maximum stress

to 200 MPa. The bar, which is 1.0 m long,

extends by 2.5 mm when carrying a load of

120 kN. Determine the modulus of elasticity

of the material of the bar.

Force, F = 120 kN = 120000 N and cross-

sectional areaA=( 15 x) 10 −^6 m^2 ,wherexis the

width of the rectangular bar in millimetres

Stressσ=

F

A

,from which,

A=

F

σ

=

120000 N

200 × 106 Pa

= 6 × 10 −^4 m^2

= 6 × 102 mm^2 =600 mm^2

Hence 600= 15 x, from which,

width of bar,x=

600

15

=40 mm

and extension of bar= 2 .5mm= 0 .0025 m

Strainε=

x

L

=

0. 0025

1. 0

= 0. 0025

Modulus of elasticity,E =

stress

strain

=

200 × 106

0. 0025

= 80 × 109 =80 GPa

Problem 14. An aluminium alloy rod has a

length of 200 mm and a diameter of 10 mm.