Derivation of Elongation in a Tapered Bar with Rectangular Cross-Section Under an Applied Force

 

Derivation of Elongation in a Tapered Bar with Rectangular Cross-Section Under an Applied Force

In this blog, we will derive the elongation of a tapered bar with a rectangular cross-section when subjected to an axial force $P$. A tapered bar is one whose cross-sectional dimensions change along its length. This derivation is essential for understanding how such a bar deforms under load, which is crucial for various engineering applications such as beams, columns, and structural supports.

Assumptions

1. The material of the bar is homogeneous and isotropic.
2. The deformation is within the elastic limit of the material (Hooke's law is applicable).
3. The taper is linear, meaning the cross-sectional dimensions change linearly from one end to the other.
4. The axial force $P$ is applied uniformly along the length of the bar.

Geometry of the Tapered Bar

Consider a tapered bar with the following characteristics:

• Length of the bar: $L$
• Width at the larger end: $b_1$
• Width at the smaller end: $b_2$
• Constant thickness: $t$

Since the bar is tapered linearly, the width $b(x)$ at a distance $x$ from the larger end is given by: $b(x) = b_1 - \left( \frac{b_1 - b_2}{L} \right) x$

Differential Element Analysis

To derive the elongation, consider an infinitesimally small element $dx$ at a distance $x$ from the larger end. The width of this element is $b(x)$. The cross-sectional area $A(x)$ of the element is: $A(x) = b(x) \cdot t$ Substituting $b(x)$: $A(x) = t \left( b_1 - \left( \frac{b_1 - b_2}{L} \right) x \right)$

Elongation of the Differential Element

The elongation $d\delta$ of the differential element $dx$ under the axial force $P$ is given by Hooke's law: $d\delta = \frac{P \, dx}{A(x) \, E}$ where $E$ is the modulus of elasticity of the material.

Substituting $A(x)$ into the elongation equation: $d\delta = \frac{P \, dx}{t \left( b_1 - \left( \frac{b_1 - b_2}{L} \right) x \right) E}$

Total Elongation of the Bar

To find the total elongation $\delta$, integrate the differential elongation over the length of the bar from $x = 0$ to $x = L$: $\delta = \int_0^L \frac{P \, dx}{t \left( b_1 - \left( \frac{b_1 - b_2}{L} \right) x \right) E}$

Let $k = \frac{b_1 - b_2}{L}$. Then the integral becomes: $\delta = \frac{P}{t E} \int_0^L \frac{dx}{b_1 - kx}$

To solve this integral, use the substitution $u = b_1 - kx$, hence $du = -k \, dx$ and $dx = -\frac{du}{k}$. The limits of integration change accordingly:

• When $x = 0$, $u = b_1$
• When $x = L$, $u = b_2$

The integral becomes: $\delta = \frac{P}{t E} \int_{b_1}^{b_2} \frac{-\frac{du}{k}}{u}$ $\delta = \frac{P}{t E k} \int_{b_1}^{b_2} \frac{du}{u}$ $\delta = \frac{P}{t E k} \left[ \ln u \right]_{b_1}^{b_2}$ $\delta = \frac{P}{t E k} \left( \ln b_2 - \ln b_1 \right)$ $\delta = \frac{P}{t E k} \ln \left( \frac{b_2}{b_1} \right)$

Substitute $k = \frac{b_1 - b_2}{L}$: $\delta = \frac{P L}{t E (b_1 - b_2)} \ln \left( \frac{b_2}{b_1} \right)$

Simplification and Final Formula

Simplify the expression to get the final formula for the elongation $\delta$ of the tapered bar: $\delta = \frac{P L}{t E (b_1 - b_2)} \ln \left( \frac{b_2}{b_1} \right)$

Thus, the elongation of a tapered bar with a rectangular cross-section under an axial force $P$ is given by: $\delta = \frac{P L}{t E (b_1 - b_2)} \ln \left( \frac{b_2}{b_1} \right)$

Conclusion

In this blog, we derived the elongation of a tapered bar with a rectangular cross-section when subjected to an axial force. The key steps involved analyzing a differential element of the bar, applying Hooke's law, and integrating over the length of the bar to find the total elongation. This derivation is crucial for understanding how tapered structural elements deform under load, which is essential for designing safe and efficient engineering components.