# Bar with varying cross-section

## Description

A flat bar of thickness $t$, is loaded by a point force at B.

Determine the free-end displacement (meaning at point C).

Given data:

• $F=40 \kN$
• $E=200 \GPa$
• $t=25 \ \rm{mm}$

The displacement at C is $u_{\rm{C}} = 0.097$ mm.

## Solution

The displacement at C , $u_{\rm{C}}=u(x=2L)$, is determined from the relationship between strain and displacement (definition of normal strain):

$$\epsilon = \od{u}{x} \qgives u_{\rm{C}} = \int_{0}^{2L} \epsilon(x) \dx$$

Thus, to calculate the displacement, we need to determine an expression for how the strain varies along the bar.

### Strain

Hooke’s law gives
$$\sigma(x) = E \ \epsilon(x) \qgives \epsilon(x) =\frac{\sigma(x)}{E} =\frac{N(x)}{EA(x)}$$

So, we need the distribution of normal force.

### Normal forces

Consider sections through the bar, to the left and right of point B, and make free-body diagrams. Equilibrium gives:

Part AB

$$\eqleft N_{\rm{AB}} – F = 0 \qgives N_{\rm{AB}} = F$$

Part BC

$$\eqleft N_{\rm{BC}} = 0$$

Since the normal force is zero in part BC, also the strain will be zero ($\epsilon_{\rm{BC}}=0$). Consequently, part BC doesn’t deform.

### Cross-sectional area

The area varies linearly from $0.15t$ in A to $0.05t$ in C., this is expressed as

$$A(x) = 0.15t – 0.10t \frac{x}{2L} = t(0.15 – 0.10 \frac{x}{2L})$$

There are several ways to arrive at this expression, for example one can use equation for a straight line:

Determine the inclination $k$ in $(A=kx+m)$ as  $\frac{0.05t-0.15t}{2L-0}$ and the point $m$ from the condition $A(0)=0.15t = k\cdot 0 + m \gives m=0.15t$

The strain for part AB can now be written as $\epsilon_{\rm{AB}}(x) = \frac{F}{Et(0.15 – 0.10 \frac{x}{2L})}$

### Displacement

Finally, the displacement is determined by integrating the strain over the bar

\begin{align} u_{\rm{C}} &= \int_{0}^{2L} \epsilon(x) \dx = \int_{0}^{L} \frac{F}{Et(0.15 – 0.10 \frac{x}{2L})} \dx \\ &= \frac{F}{Et} \int_{0}^{L} \frac{1}{(0.15 – 0.10 \frac{x}{2L})} \dx = \frac{F}{Et} \left. \frac{\ln(0.15 – 0.10 \frac{x}{2L}) }{-\frac{0.10}{2L}} \right|_0^L\\ &=\ldots \approx 0.097 \ \rm{mm} \end{align}