Question

A bar of metal 100 mm × 50 mm in cross section is 250 mm long. It carries a tensile load of 400 kN in the direction of its length, a compressive load of 4000 kN on its 100 mm × 250 mm faces and a tensile load of 2000 kN on its 50 mm × 250 mm face. If E = 2 × 105 N/mm2 and Poisson’s ratio is 0.25, find the change in volume of the bar. What change must be made in the 4000 kN load in order that there shall be no change in the volume of the bar.

Answer 1

Explanation:

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Answer 2

The change in volume of the bar is 250 mm³ and the load should be increased to 6000 kN.

Step-by-step Explanation:

Given: Dimensions of the bar 100 mm × 50 mm × 250 mm

The tensile load on the 100 mm × 50 mm face (P₁) = 400kN

The tensile load on the 100 mm × 50 mm face (P₂) = 400kN

The tensile load on the 100 mm × 50 mm face (P₃) = 400kN

Young's Modulus (E) = 2 × 10⁵ N/mm²

Poisson's Ratio (σ) = 0.25

To Find: Change in the volume and the compressive load so that the volume of the bar is unchanged

Solution:

• Finding tensile stress across the dimensions of the metal

Given the dimensions of the bar and tensile loads P₁, P₂, and P₃ such that the tensile stress along P₁ is,

$P_{x} = \frac{P_1}{A} = \frac{400 \times 1000}{100 \times 50} = 80 \ N/mm^2$

Similarly, tensile stress along P₂ and compressive stress along P₃ are -

$P_{y} = \frac{P_2}{A} = \frac{2000 \times 1000}{250 \times 50} = 160 \ N/mm^2$

$P_{z} = \frac{P_3}{A} = - \frac{4000 \times 1000}{100 \times 250} = - 160 \ N/mm^2$

• Finding change in the volume

The volumetric strain of the bar is given by-

$e_v = \frac{(p_x + p_y + p_z) (1-2\sigma)}{E}$

Putting the values of $P_x$, $P_y$, $P_z$ and the given values in the above expression, we get-

$\Rightarrow e_v = \frac{(80 + 160 - 160) (1-2(0.25))}{2 \times 10^{5} } = 20 \times 10^{-5}$

Since, $\text{change in the volume} = e_v \times \text{original volume}$, therefore,

$\Rightarrow 20 \times 10^{-5} \times 250 \times 100 \times 50 = 250 \ mm^3$

$\Rightarrow \text{Change in the volume} = 250 \ mm^3$

• Finding the compressive load so that the volume of the bar will remain unchanged

For no change in the volume, the value of $P_z$ should satisfy the following equation;

$P_x + P_y + P_z = 0$

$\Rightarrow 80 + 160 + P_z = 0$

$\Rightarrow P_z = - 240 \ N/mm^{2}$

Therefore, the compressive load is $P_3 = P_z A$

$\Rightarrow P_3 = 240 \times 100 \times 250 = 6000 \ kN$

Hence, the change in volume of the bar is 250 mm³ while the load should be increased to 6000 kN so that the volume of the bar will remain unchanged.