A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kn. The modulus of elasticity for steel may be taken as 200 kn/mm2 . Find stress, strain and elongation of the bar due to applied load.
Answers
The Cross sectional area is 200.96 mm2 , Stress is 223. 92 N/mm2 , Strain is 0.0011198 and Elongation is 0.558 mm.
Explanation:
Load p =45 KN = 45 x 10^3 N
Young’s modulus E = 200 KN /mm^2
Length of the rod, l = 500 mm
Diameter of the rod = 16 mm
Solution:
Cross sectional area; A = πd^2/4 =π x (16)^2/4 = 200.96 mm^2
Stress σ =load / area =45 x 10^3/200.96 = 223. 92 N/mm^2
Strain, e = stress / young’s modulus =223.96 /200 x 10^3 = 0.0011198
Elongation δl = Pl/AE = (45 x 10^3 x 500) / (200.96 x 200 x 10^3)
= 22,500,000/ 40,192,000
= 0.558 mm
Thus the Cross sectional area is 200.96 mm^2 , Stress is 223. 92 N/mm^2 , Strain is 0.0011198 and Elongation is 0.558 mm.
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Answer:
elongation is 0.497
Explanation:
40kN is there you have taken 45