Question

Determine the minimum diameter of cross
Sectional area of alluminium
length im Subjected to
an axial pull of
30 kN which will limit the maximam elongtion to 1mm . Take F = 70GPa for aluminium.
​

Answer 1

Answer:

Given that,

Radius r=10mm

Length l=1.0m

Force F=100kN

Young modulus Y=2.0×10

11

Nm

2

Now, the stress is

Stress =

3.14×10

2

m

2

100×10

3

N

=3.18×10

8

Nm

−2

Now, the elongation is

ΔL=

AY

FL

ΔL=

2.0×10

11

3.18×10

8

×1

ΔL=1.59×10

−3

m

ΔL=1.59mm

Now, the strain is

Strain

=

L

ΔL

=

1.0

1.59×10

−3

=1.59×10

−3

Hence, this is the required solution.