Understanding Elongation Due to Self-Weight

When a prismatic bar is suspended from one end, its own weight acts as a distributed load along its length. This self-weight causes stress within the bar, which varies from zero at the free end to maximum at the point of suspension (the ceiling). This varying stress leads to a total elongation of the bar. We need to find out how this total elongation changes if all the dimensions of the bar are doubled.

The formula for the total elongation ($\delta$) of a prismatic bar of length \(L\) due to its own weight is given by: $$ \delta = \frac{W_{total} L}{2 A E} $$ where:

• \(W_{total}\) is the total weight of the bar
• \(L\) is the length of the bar
• \(A\) is the cross-sectional area of the bar
• \(E\) is the Young's modulus of the material

The total weight \(W_{total}\) can also be expressed using the density \(\rho\) and volume \(V\) of the bar: \(W_{total} = \rho g V\), where \(g\) is the acceleration due to gravity. The volume of a prismatic bar is \(V = A L\). So, \(W_{total} = \rho g A L\). Substituting this into the elongation formula: $$ \delta = \frac{(\rho g A L) L}{2 A E} $$ $$ \delta = \frac{\rho g A L^2}{2 A E} $$ We can cancel the area \(A\) from the numerator and denominator: $$ \delta = \frac{\rho g L^2}{2 E} $$ This formula shows that the total elongation due to self-weight depends on the material properties (\(\rho\) and \(E\)) and the length \(L\) of the bar. It is independent of the cross-sectional area \(A\) (width and thickness), assuming the cross-section is uniform (prismatic).

Analyzing the Effect of Doubling Dimensions

Let's consider the original dimensions and the new dimensions after they are doubled.

• Original Length = \(L_1\)
• Original Width = \(w_1\)
• Original Thickness = \(t_1\)
• Original Cross-sectional Area = \(A_1 = w_1 t_1\)

The original total elongation due to self-weight is: $$ \delta_1 = \frac{\rho g L_1^2}{2 E} $$

Now, all dimensions are doubled:

• New Length = \(L_2 = 2 L_1\)
• New Width = \(w_2 = 2 w_1\)
• New Thickness = \(t_2 = 2 t_1\)
• New Cross-sectional Area = \(A_2 = w_2 t_2 = (2 w_1)(2 t_1) = 4 w_1 t_1 = 4 A_1\)

The material properties, density (\(\rho\)) and Young's modulus (\(E\)), remain the same because the bar is made of the same material.

The new total elongation due to self-weight is calculated using the new length \(L_2\): $$ \delta_2 = \frac{\rho g L_2^2}{2 E} $$ Substitute \(L_2 = 2 L_1\) into the equation: $$ \delta_2 = \frac{\rho g (2 L_1)^2}{2 E} $$ $$ \delta_2 = \frac{\rho g (4 L_1^2)}{2 E} $$ $$ \delta_2 = 4 \left( \frac{\rho g L_1^2}{2 E} \right) $$

Comparing Original and New Elongation

We can see that the expression inside the parenthesis is the original elongation, \(\delta_1\). $$ \delta_2 = 4 \delta_1 $$

This shows that the new total elongation (\(\delta_2\)) is 4 times the original total elongation (\(\delta_1\)). Even though the cross-sectional area increased by 4 times, this increase in area was cancelled out in the formula derivation because the total weight also increased proportionally (Volume \(V_2 = L_2 A_2 = (2L_1)(4A_1) = 8 L_1 A_1 = 8 V_1\), so \(W_{total, 2} = 8 W_{total, 1}\)). However, the final formula for elongation due to self-weight only depends on the square of the length. Since the length was doubled, \(L^2\) becomes \((2L)^2 = 4L^2\), causing the elongation to increase by 4 times.

The total elongation produced by its own weight will increase by 4 times.

Revision Table: Prismatic Bar Elongation

Additional Information on Stress and Strain

For a prismatic bar under axial stress, strain is directly proportional to stress, according to Hooke's Law, within the elastic limit.

• Stress (\(\sigma\)): Force per unit area. In the case of self-weight, the force is the weight of the part of the bar below a certain point. This force varies along the length.
• Strain (\(\epsilon\)): Change in length per unit original length. Strain is related to stress by the material's Young's Modulus \(E\): \(\sigma = E \epsilon\), or \(\epsilon = \frac{\sigma}{E}\).
• Young's Modulus (E): A measure of the stiffness of an elastic material, defined as the ratio of stress to strain. It is a material property.
• Density (\(\rho\)): Mass per unit volume. This is also a material property.

When calculating elongation due to self-weight, we consider a small differential element of length \(dy\) at a distance \(y\) from the free end. The weight of the bar below this element is the force acting on it. The elongation of this small element is \(d\delta\). The total elongation is found by integrating \(d\delta\) over the entire length of the bar.

The force acting on the element at distance \(y\) from the free end is the weight of the bar below it: \(W_y = (\rho A y) g\) The stress at this location is \(\sigma_y = \frac{W_y}{A} = \frac{(\rho A y) g}{A} = \rho g y\). The strain at this location is \(\epsilon_y = \frac{\sigma_y}{E} = \frac{\rho g y}{E}\). The elongation of the differential element \(dy\) is \(d\delta = \epsilon_y dy = \frac{\rho g y}{E} dy\). To find the total elongation, integrate from \(y=0\) to \(y=L\): $$ \delta = \int_{0}^{L} \frac{\rho g y}{E} dy = \frac{\rho g}{E} \int_{0}^{L} y dy = \frac{\rho g}{E} \left[ \frac{y^2}{2} \right]_{0}^{L} = \frac{\rho g}{E} \left( \frac{L^2}{2} - 0 \right) = \frac{\rho g L^2}{2 E} $$ This confirms the formula used, highlighting why the cross-sectional area cancels out and the elongation is proportional to the square of the length.