A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/Sq(mm) ; determine : the elongation in rod?
elongation of rod = original length x strain,
we know that original length is equal to 150cm
strain = stress/modulus of elasticity = force/ Area x modulus of elasticity
= 20000/(0.785 x 20 x 20 x 200000) = 0.000318
by putting value of strain in elongation formula we get
elongation of rod = strain x original length
= 0.000318 x 150
= 0.0477 cm
Question 72
A rod of 150cm long and of diameter 2.0 cm is subjected to an axial pullof 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/sq(mm), determine strain?
we know that,
Modulus of elasticity is stress/strain,
stress is equal to force/area, hence by putting value of Area = 0.785 x diameter x diameter, and force is equal to 20000 N, we get stress equal to 63.662 N/sq(mm),
now, from the equation of hookes law,
strain = stress/modulus of elasticity = 63.662/200000 = 0.000318
Question 73
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/Sq(mm) ; determine : the stress?
Area of Rod:- (Ï€/4)(radius^2) :- 100Ï€ sq(mm),
now,
Stress= Force/Area = 20000/100Ï€ = 63.662 N/sq(mm)
Question 74
what is lateral strain?
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Question 75
what is longitudinal strain?
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Question 76
what is poisson ratio?
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Question 77
what is Factor of safety?
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Question 78
what is modulus of rigidity or shear modulus?
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Question 79
what is modulus of elasticity or young modulus?
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Question 80
what is hookes law?
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