A steel rod of 20 mm diameter is enclosed centrally in a hollow copper tube of external diameter 30mm and internal diameter 25 mm. The composite bar is then subjected to an axial pull of 40 kn. Find the stresses in the rod and tube. Take es = 200 kn/mm2 and ec = 100 kn/mm2.
Answers
Answered by
11
Explanation:
stress on steel = 256.64N/mm2.
strain in copper =strain in steel
Ps/Es = Pc/Ec
PC = 127.32
total load= load of copper and steel
= Ps×Pi/4 ×(20)sq + Pc×Pi/4 ×(30sq -20sq) =P
P = 1.29 ×10 raise to 5 N/mmsq.
Answered by
3
Given:
Diameter =20mm
External diameter = 30mm
internal diameter = 25 mm
es = 200kn/mm²
ec = 100 kn/mm²
To Find:
The stresses in the rod and tube.
Explanation:
Steel stress = 256.64N/mm²
strain in copper = is equal to the strain in steel
Therefore Ps/Es = Pc/Ec
That is Pc = 127.32
Now the total load is equal to the copper and steel load
= Ps×(Pi/4) ×(20) + Pc×(Pi/4) ×(30-20) =P
P = 1.29 ×10⁵N/mm²
Answer =1.29 ×10⁵N/mm²
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