Mechanical Engineering Principles

THE EFFECTS OF FORCES ON MATERIALS 5

(a) Compressive stress,

σ=

F

A

=

40000 N

3. 142 × 10 −^4 m^2

= 12. 73 × 107 Pa=127.3 MPa

(b) Contraction of pipe when loaded,
x= 0 .5mm= 0 .0005 m, and original length
L= 0 .40 m. Hence, compressive strain,

ε=

x

L

=

0. 0005

0. 4

= 0. 00125 (or 0 .125%)

Problem 8. A circular hole of diameter

50 mm is to be punched out of a 2 mm thick

metal plate. The shear stress needed to cause

fracture is 500 MPa. Determine (a) the

minimum force to be applied to the punch,

and (b) the compressive stress in the punch

at this value.

(a) The area of metal to be sheared,A=perimeter

of hole×thickness of plate.

Perimeter of hole=πd=π( 50 × 10 −^3 )

= 0 .1571 m.

Hence, shear area,A= 0. 1571 × 2 × 10 −^3

= 3. 142 × 10 −^4 m^2

Since shear stress=

force

area

,

shear force =shear stress×area

=( 500 × 106 × 3. 142 × 10 −^4 )N

= 157 .1kN,

which is the minimum force to be applied to

the punch.

(b) Area of punch=

πd^2

4

=

π( 0. 050 )^2

4

= 0 .001963 m^2

Compressive stress=

force

area

=

157. 1 × 103 N

0 .001963 m^2

= 8. 003 × 107 Pa

=80.03 MPa,

which is the compressive stress in the punch.

Problem 9. A rectangular block of plastic

material 500 mm long by 20 mm wide by

300 mm high has its lower face glued to a

bench and a force of 200 N is applied to the

upper face and in line with it. The upper face

moves 15 mm relative to the lower face.

Determine (a) the shear stress, and (b) the

shear strain in the upper face, assuming the

deformation is uniform.

(a) Shear stress,τ=

force

area parallel to the force

Area of any face parallel to the force

=500 mm×20 mm

=( 0. 5 × 0. 02 )m^2 = 0 .01 m^2

Hence,shear stress,

τ=

200 N

0 .01 m^2

=20000 Paor20 kPa

(b) Shear strain,γ=

x

L

(see side view

in Figure 1.6)

=

15

300

= 0. 05 (or5%)

x= 15 mm

L= 300 mm

Reaction

force

γ

Applied

force

500 mm

Figure 1.