Mechanical Engineering Principles

THE EFFECTS OF FORCES ON MATERIALS 15

Rearranging gives: σ 2 =

PE 2

(A 1 E 1 +A 2 E 2 )

( 1. 8 )

and σ 1 =

PE 1

(A 1 E 1 +A 2 E 2 )

( 1. 9 )

N.B. IfPis a compressive force, then bothσ 1 and
σ 2 will be compressive stresses (i.e. negative), and
vice-versa ifPwere tensile.

Problem 21. A concrete pillar, which is

reinforced with steel rods, supports a

compressive axial load of 2 MN.

(a) Determine stressesσ 1 andσ 2 given the

following:

For the steel,A 1 = 4 × 10 −^3 m^2 and

E 1 = 2 × 1011 N/m^2

For the concrete,A 2 = 0 .2m^2 and

E 2 = 2 × 1010 N/m^2

(b) What percentage of the total load does

the steel reinforcement take?

(a) From equation (1.9),

σ 1 =−

PE 1

(A 1 E 1 +A 2 E 2 )

=−

2 × 106 × 2 × 1011

(

4 × 10 −^3 × 2 × 1011 + 0. 2 × 2 × 1010

)

=−

4 × 1017

(

8 × 108 + 40 × 108

)=

4 × 1017

48 × 108

=

109

12

=− 83. 3 × 106

i.e.the stress in the steel,

σ 1 =− 83 .3MPa ( 1. 10 )

From equation (1.8),

σ 2 =−

PE 2

(A 1 E 1 +A 2 E 2 )

=−

2 × 106 × 2 × 1010

( 4 × 10 −^3 × 2 × 1011 + 0. 2 × 2 × 1010 )

=−

4 × 1016

( 8 × 108 + 40 × 108 )

=

4 × 1016

48 × 108

=

108

12

=− 8. 3 × 106

i.e.the stress in the concrete,

σ 2 =− 8 .3MPa ( 1. 11 )

(b) Force in the steel,

F 1 =σ 1 A 1

=− 83. 3 × 106 × 4 × 10 −^3

= 3. 33 × 105 N

Therefore,the percentage total load taken by

the steel reinforcement

=

F 1

total axial load

×100%

=

3. 33 × 105

2 × 106

×100%= 16 .65%

Problem 22. If the pillar of problem 21

were subjected to a temperature rise of

25 °C, what would be the values of stresses

σ 1 andσ 2?

Assume the coefficients of linear expansion

are, for steel,α 1 = 14 × 10 −^6 /°C, and for

concrete,α 2 = 12 × 10 −^6 /°C.

Asα 1 is larger thanα 2 , the effect of a temperature

rise will cause the ‘thermal stresses’ in the steel to be

compressive and those in the concrete to be tensile.

From equation (1.5),the thermal stress in the steel,

σ 1 =−

(α 1 −α 2 )E 1 E 2 A 2 T

(A 1 E 1 +A 2 E 2 )

=−

( 14 × 10 −^6 − 12 × 10 −^6 )× 2 × 1011

× 2 × 1010 × 0. 2 × 25

48 × 108

=−

40 × 1015

48 × 108

= 0. 833 × 107

=− 8 .33 MPa ( 1. 12 )

From equation (1.3), the thermal stress in the

concrete,

σ 2 =

σ 1 A 1

A 2

=−

(− 8. 33 × 106 )× 4 × 10 −^3

0. 2

= 0 .167 MPa ( 1. 13 )

From equations (1.10) to (1.13):

σ 1 =− 83. 3 − 8. 33 =− 91 .63 MPa

and σ 2 =− 8. 3 + 0. 167 =− 8 .13 MPa