Mechanical Engineering Principles

14 MECHANICAL ENGINEERING PRINCIPLES

these two positions (i.e. at the position A-A in
Figure 1.13). To achieve this, it will be necessary
for bar (2) to be pulled out by a distanceε 2 Land
for bar (1) to be pushed in by a distance ε 1 L,
where

ε 1 =compressive strain in (1)

and ε 2 =tensile strain in (2)

From considerations of compatibility (‘deflection’)
in Figure 1.13,

α 1 LT−ε 1 L=α 2 LT+ε 2 L

i.e. ε 1 =(α 1 −α 2 )T−ε 2

Now,σ 1 =E 1 ε 1 andσ 2 =E 2 ε 2

Hence, σ 1 =(α 1 −α 2 )E 1 T−σ 2

E 1

E 2

( 1. 2 )

To obtain the second simultaneous equation, it will
be necessary to consider equilibrium of the com-
pound bar.

Let F 1 =unknown compressive force in bar (1)

and F 2 =unknown tensile force in bar (2)

Now, from equilibrium considerations,

F 1 =F 2

butσ 1 =FA^1
1

andσ 2 =

F 2

A 2

Therefore, σ 1 A 1 =σ 2 A 2

or σ 1 =

σ 2 A 2

A 1

( 1. 3 )

Equating equations (1.2) and (1.3) gives

σ 2 A 2

A 1

=(α 1 −α 2 )E 1 T−σ 2

E 1

E 2

from which,

σ 2 =

(α 1 −α 2 )E 1 T

(

E 1

E 2

+

A 2

A 1

) =

(α 1 −α 2 )E 1 T

(

A 1 E 1 +A 2 E 2

E 2 A 1

)

i.e. σ 2 =

(α 1 −α 2 )E 1 E 2 A 1 T

(A 1 E 1 +A 2 E 2 )

(tensile) ( 1. 4 )

and σ 1 =

(α 1 −α 2 )E 1 E 2 A 2 T

(A 1 E 1 +A 2 E 2 )

(compressive)

( 1. 5 )

Problem 20. If the solid bar of Problem 19

did not suffer temperature change, but instead

was subjected to a tensile axial forceP,as

shown in Figure 1.14, determineσ 1 andσ 2.

L

P P

Bar 2 Bar 1

Figure 1.14 Compound bar under axial tension

There are two unknown forces in this bar, namely,

F 1 andF 2 ; therefore, two simultaneous equations

will be required.

The first of these simultaneous equations can be

obtained by considering compatibility, i.e.

deflection of bar (1)=deflection of bar (2)

or δ 1 =δ 2

Butδ 1 =ε 1 Land δ 2 =ε 2 L

Therefore, ε 1 L=ε 2 L

or ε 1 =ε 2

Now,ε 1 =

σ 1

E 1

andε 2 =

σ 2

E 2

Hence,

σ 1

E 1

=

σ 2

E 2

or σ 1 =

σ 2 E 1

E 2

( 1. 6 )

The second simultaneous equation can be obtained

by considering the equilibrium of the compound bar.

Let F 1 =tensile force in bar (1)

and F 2 =tensile force in bar (2)

Now, from equilibrium conditions

P=F 1 +F 2

i.e. P=σ 1 A 1 +σ 2 A 2 ( 1. 7 )

Substituting equation (1.6) into equation (1.7) gives:

P=

σ 2 E 1

E 2

A 1 +σ 2 A 2 =σ 2

(

E 1 A 1

E 2

+A 2

)

=σ 2

(

A 1 E 1 +A 2 E 2

E 2

)