Mechanical Engineering Principles

20 MECHANICAL ENGINEERING PRINCIPLES

Load (kN) 0 8 19 29 36

Extension (mm) 0 0.015 0.038 0.060 0.072

The maximum load carried by the specimen

is 50 kN and its length after fracture is

52 mm. Determine (a) the modulus of

elasticity, (b) the ultimate tensile strength,

(c) the percentage elongation of the mild

steel.

The load/extension graph is shown in Figure 2.4.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

10

20

25

30

40

A

Load / kN

Extension / mm

B

C

Figure 2.4

(a) Gradient of straight line is given by:

BC

AC

=

25000

0. 05 × 10 −^3

= 500 × 106 N/m

Young’s modulus of elasticity

=(gradient of graph)

(

L

A

)

,where

L=40 mm (gauge length)

= 0 .040 m and area,

A=100 mm^2 = 100 × 10 −^6 m^2.

Young’s modulus of elasticity

=( 500 × 106 )

(

0. 040

100 × 10 −^6

)

= 200 × 109 Pa=200 GPa

(b) Ultimate tensile strength

=

maximum load

original cross-sectional area

=

50000 N

100 × 10 −^6 m^2

= 500 × 106 Pa

=500 MPa

(c) Percentage elongation

=

increase in length

original length

× 100

=

52 − 40

40

× 100 =

12

40

× 100 =30%

Problem 3. The results of a tensile test are:

Diameter of specimen 15 mm; gauge length

40 mm; load at limit of proportionality

85 kN; extension at limit of proportionality

0.075 mm; maximum load 120 kN; final

length at point of fracture 55 mm.

Determine (a) Young’s modulus of elasticity,

(b) the ultimate tensile strength, (c) the stress

at the limit of proportionality, (d) the

percentage elongation.

(a) Young’s modulus of elasticity is given by:

E=

stress

strain

=

F

A

x

L

=

FL

Ax

where the load at the limit of proportionality,

F=85 kN=85000 N,

L=gauge length=40 mm= 0 .040 m,

A=cross-sectional area=

πd^2

4

=

π( 0. 015 )^2

4

= 0 .0001767 m^2 , and

x=extension= 0 .075 mm= 0 .000075 m.

Hence, Young’s modulus of elasticity

E=

FL

Ax

=

( 85000 )( 0. 040 )

( 0. 0001767 )( 0. 000075 )

= 256. 6 × 109 Pa= 256 .6GPa