Mechanical Engineering Principles

12 MECHANICAL ENGINEERING PRINCIPLES

1.10 Modulus of rigidity

Experiments have shown that under pure torsion (see
Chapter 10), up to the limit of proportionality, we
have Hooke’s law in shear, where

shear stress

shear strain

=rigidity (or shear) modulus

or

τ

γ

=G ( 1. 1 )

whereτ =shear stress,
γ =shear strain (see Figures 1.5 and 1.6) and
G=rigidity (or shear) modulus

1.11 Thermal strain

If a bar of lengthLand coefficient of linear expan-
sion α were subjected to a temperature rise of
T, its length will increase by a distanceαLT,as
described in Chapter 20. Thus the new length of the
bar will be:

L+αLT=L( 1 +αT)

Now, as the original length of the bar wasL,then
thethermal straindue to a temperature rise will be:

ε=

extension

original length

=

αLT

L

i.e. ε=αT

However, if the bar were not constrained, so that
it can expand freely, there will be no thermal
stress.
If, however, the bar were prevented from expand-
ing then there would be a compressive stress in
the bar.

Now ε=

original length−new length

original length

=

L−L( 1 +αT)

L

=

L−L−LαT

L

i.e. ε=−αT

and, since stress =strain×E,then

σ =−αTE

Problem 18. A steel prop is used to

stabilise a building, as shown in Figure 1.11.

(a) If the compressive stress in the bar at

20 °C is 30 MPa, what will be the stress in

the prop if the temperature is raised to

35 °C? (b) At what temperature will the prop

cease to be effective?

TakeE= 2 × 1011 N/m^2 and

α= 14 × 10 −^6 /°C.

Prop

Rigid floor

Building

Figure 1.11

(a) Additional thermal strain,

εT =−αT

=−( 14 × 10 −^6 /°C)×( 35 − 20 )°C

i.e.εT =− 14 × 10 −^6 × 15 =− 210 × 10 −^6

Additional thermal stress,

σT =EεT

= 2 × 1011 N/m^2 ×(− 210 × 10 −^6 )

i.e.σT =−42 MPa

Hence, the stress at 35°C=initial stress+σT

=(− 30 − 42 )MPa

i.e.σ =−72 MPa

(b) For the prop to be ineffective, it will be nec-

essary for the temperature to fall so that there

is no stress in the prop, that is, from 20°C

the temperature must fall so that the initial

stress of 30 MPa is nullified. Hence, drop in

stress=−30 MPa