Physics, asked by Atharva5926, 10 months ago

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 alenjohns111
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 mariospartan
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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