Physics, asked by karthikeyankarthik23, 6 months ago

A mild steel rod of 20 mm diameter and 300 mm long is enclosed centrally inside a hollow copper tube of external diameter 30 mm and internal diameter of 25 mm. The ends of the tube and the rods are brazed together and the composite bar is subjected to an axial pull of 40 kN. if E for steel and copper is 200 GN/m and 100 GN/m respectively. Find the stresses developed in the rod and the tube. Also find the extension of the rod

Answers

Answered by shreyasrout66000
29

Explanation:

6. A mild steel rod of 20 mm diameter and 300 mm long is enclosed centrally inside a hollow upper tube of external diameter 30 mm and internal diameter of 25 mm. The ends of the tube and rods are brazed together and the composite bar is subjected to an axial pull of 40 kN. If E for steel and copper is 200 GN/m² and 100 GN/m² respectively, find the stresses developed in the rod and tube. Also, find the extension of the rod.

Answered by adventureisland
21

The extension of the rod is the 94.8MPa.

Explanation:

Diameter of steel rod=20mm

External diameter of copper tube=30mm

Internal dimeter of copper tube=25mm

Total load(p)=40kN=40*103N

Modulus of elasticity of steel(E_{s})=200Gpa

Modulus of elasticity of copper(E_{c})=100GPa

A_{s}=\frac{\pi }{4} *(20)^{2}

=314.2mm^{2}

A_{c}=\frac{\pi }{4}[(30)^{2}-(25)^{2}]

=216mm^{2}

σs=\frac{E_{s}}{E_{c}}*σc

=\frac{200}{100} *σc

=2σc

40*10^{3}=(σsA_{s})+(σcA_{c})

=(2σc*314.2)+(σc*216)

=844.4σc

σc=\frac{40*10^{3}}{844.4} =47.4\frac{N}{mm^{2}}

σs=2σc

=2*47.4

=94.8MP_{a}.

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