Mechanical Engineering Principles

THE EFFECTS OF FORCES ON MATERIALS 3

whereF is the force in Newton’s and Ais the
cross-sectional area in square metres. For tensile
and compressive forces, the cross-sectional area is
that which is at right angles to the direction of the
force. For a shear force the shear stress is equal
toF/A, where the cross-sectional areaA is that
which is parallel to the direction of the force. The
symbol used for shear stress is the Greek letter
tau,τ.

Problem 2. A rectangular bar having a

cross-sectional area of 75 mm^2 has a tensile

force of 15 kN applied to it. Determine the

stress in the bar.

Cross-sectional areaA=75 mm^2 = 75 × 10 −^6 m^2
and forceF=15 kN= 15 × 103 N

Stress in bar,σ=

F

A

=

15 × 103 N

75 × 10 −^6 m^2

= 0. 2 × 109 Pa=200 MPa

Problem 3. A circular wire has a tensile

force of 60.0 N applied to it and this force

produces a stress of 3.06 MPa in the wire.

Determine the diameter of the wire.

ForceF= 60 .0Nand
stressσ= 3 .06 MPa= 3. 06 × 106 Pa

Since σ=

F

A

then area, A=

F

σ

=

60 .0N

3. 06 × 106 Pa

= 19. 61 × 10 −^6 m^2 = 19 .61 mm^2

Cross-sectional areaA=πd

2

4 ;

hence 19. 61 =

πd^2

4

, from which,

d^2 =

4 × 19. 61

π

from which,d=

√(

4 × 19. 61

π

)

i.e.diameter of wire = 5.0 mm

Now try the following exercise

Exercise 1 Further problems on stress

1. A rectangular bar having a cross-sectional
   area of 80 mm^2 has a tensile force of
   20 kN applied to it. Determine the stress
   in the bar. [250 MPa]


2. A circular cable has a tensile force of
   1 kN applied to it and the force produces
   a stress of 7.8 MPa in the cable. Calculate
   the diameter of the cable. [12.78 mm]


3. A square-sectioned support of side 12 mm
   is loaded with a compressive force of
   10 kN. Determine the compressive stress
   in the support. [69.44 MPa]


4. A bolt having a diameter of 5 mm is
   loaded so that the shear stress in it is
   120 MPa. Determine the value of the
   shear force on the bolt. [2.356 kN]


5. A split pin requires a force of 400 N to
   shear it. The maximum shear stress before
   shear occurs is 120 MPa. Determine the
   minimum diameter of the pin.
   [2.06 mm]


6. A tube of outside diameter 60 mm and
   inside diameter 40 mm is subjected to a
   load of 60 kN. Determine the stress in the
   tube. [38.2 MPa]

1.6 Strain

The fractional change in a dimension of a material

produced by a force is called thestrain. For a tensile

or compressive force, strain is the ratio of the change

of length to the original length. The symbol used for

strain isε(Greek epsilon). For a material of length

Lmetres which changes in length by an amountx

metres when subjected to stress,

ε=

x

L

Strain is dimension-less and is often expressed as a

percentage, i.e.

percentage strain=

x

L

× 100