Question

A metal bar, which is part of a frame, is 50mm in diameter and 300mm long. It has a tensile force acting on it of 40 kN which tends to stretch it. The Young's modulus is 205 GPa. Calculate the stress and strain in the bar and the amount it stretches.

Answer 1

Under a tensile force of 40 kN, a metal bar with a diameter of 50mm and a length of 300mm, made of a material with Young's modulus of 205 GPa, experiences a stress of 2.04 x 10^7 Pa, a strain of 9.95 x 10^-5, and stretches by about 0.02985 mm.

Step-by-step explanation:

To calculate the stress, strain, and the amount the metal bar stretches, we'll need to use formulas from mechanics of materials and elasticity.

The stress (\(\sigma\)) in the bar is defined as the force (F) divided by the cross-sectional area (A) of the bar:

• Stress (\(\sigma\)) = Force (F) / Area (A)

Where the area for a circular cross-section is:

• Area (A) = \(\pi d^2/4\)

Substituting the given values:

• \(A = \pi (0.05 m)^2 / 4 = 1.96 \times 10^{-3} m^2\)
• \(\sigma = (40 \times 10^3 N) / (1.96 \times 10^{-3} m^2) = 2.04 \times 10^7 N/m^2\) or Pa

The strain (\(\epsilon\)) is defined as the change in length (\(\Delta L\)) divided by the original length (\(L_0\)):

• Strain (\(\epsilon\)) = Change in Length (\(\Delta L\)) / Original Length (\(L_0\))

From Hooke's Law, we know that Stress = Young's Modulus (E) \(\times\) Strain (\(\epsilon\)), therefore:

• \(\epsilon = \sigma / E\)

Using the given Young's modulus:

• \(\epsilon = (2.04 \times 10^7 Pa) / (205 \times 10^9 Pa) = 9.95 \times 10^{-5}\)

The amount the bar stretches (\(\Delta L\)) can be found by:

• \(\Delta L = \epsilon \times L_0\)
• \(\Delta L = 9.95 \times 10^{-5} \times 0.3 m = 2.985 \times 10^{-5} m = 0.02985 mm\)

Therefore, under the tensile force of 40 kN, the metal bar experiences a stress of \(2.04 \times 10^7 Pa\), strain of \(9.95 \times 10^{-5}\), and it stretches by approximately \(0.02985 mm\).