A steel bar having cross sectional area 201.06 mm^2 & 2.0 m long is subjected to tensile force of 35 kN. If Modulus of elasticity is 2 x 10^5 N/mm^2, then change in length of the bar is
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Step-by-step explanation:
ANSWER
σ=Stress and ε=strain
σ=
A
F
=
3×10
−5
m
2
(550kg)×(9.81m/s
2
)
=0.18㎬
ε=
l
0
Δl
=
Υ
σ
=
200×10
9
0.18×10
9
=9×10
−4
Δl=εl
0
=(9×10
−4
)(2m)=0.0018m=1.8mm
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