Question

a wire is tied straight between two rigid poles 10m a part has initial tensile stress 10N/mm^2 at 32°c. calculate stress in wire if temperature reduces to minus 8°c take E=75×10^5N/mm^2α=20×10^-6/°c.​

Answer 1

Answer:

Here, l=4.0m;Δl=2×10

−3

m;a=2.0×10

−6

m

2

,Y=2.0×10

11

N/m

2

(i) The energy density of stretched wire

u=

2

1

× stress × strain

=

2

1

×Y×(strain)

2

=

2

1

×2.0×10

11

×(2×10

−3

)/4)

2

=0.25×10

5

=2.5×10

4

J/m

3

.

(ii) Elastic potential energy = energy density × volume

=2.5×10

4

×(2.0×10

−6

)×4.0J=20×10

−2

=0.20J.