Chapter 2, Problem 124RP

The uniform wire ABC, of unstretched length 2l, is attached to the supports shown and a vertical load P is applied at the midpoint B. Denoting by A the cross-sectional area of the wire and by E the modulus of elasticity, show that, for δ << l, the deflection at the midpoint B is

$\delta =l\sqrt[3]{\frac{P}{AE}}$

Fig. P2.124

To determine

To prove the deflection of uniform wire ABC at midpoint B is $\delta =l\sqrt[3]{\frac{P}{AE}}$.

Explanation of Solution

Given information:

The unstretched length is $2l$.

The vertical load is P.

Calculation:

Sketch the uniform wire ABC as shown in Figure 1.

Refer to Figure 1.

The value of $\sin \theta \approx \tan \theta \approx \frac{\delta }{l}$.

Find the value of vertical load $\left(P_{AB}\right)$ using equilibrium Equation.

$\sum F_Y=0$

$2P_{AB}\sin \theta -P=0$ (1)

Substitute for $\sin \theta$ in Equation (1).

$\begin{array}{l}2P_{AB}\sin \theta -P=0\\ P_{AB}=\frac{P}{2\sin \theta }\\ P_{AB}=\frac{P}{2\sin \left(\frac{\delta }{l}\right)}\\ P_{AB}=\frac{Pl}{2\delta }\end{array}$

Find the elongation at point AB using the relation:

${\delta }_{AB}=\frac{P_{AB}l}{AE}$ (2)

Substitute $\frac{Pl}{2\delta }$ for ${\delta }_{AB}$ in Equation (2).

$\begin{array}{c}{\delta }_{AB}=\frac{\left(\frac{Pl}{2\delta }\right)l}{AE}\\ =\frac{P{l}^2}{2AE\delta }\end{array}$

Sketch the right angle triangle as shown in Figure 2.

Refer to Figure 2.

$\begin{array}{l}{\left(l+{\delta }_{AB}\right)}^2=l^2+{\delta }^2\\ {\delta }^2=l^2+2l{\delta }_{AB}+{\delta }_{AB}^2-l^2\\ {\delta }^2=2l{\delta }_{AB}\left(1+\frac{1}{2}\frac{{\delta }_{AB}}{l}\right)\end{array}$

$1+\frac{1}{2}\frac{{\delta }_{AB}}{l}$ Very small value compared to $2l{\delta }_{AB}$.

${\delta }^2=2l{\delta }_{AB}$ (3)

Substitute $\frac{P{l}^2}{2AE\delta }$ for ${\delta }_{AB}$.

$\begin{array}{l}{\delta }^2=2l\left(\frac{P{l}^2}{2AE\delta }\right)\\ {\delta }^2=\frac{P{l}^3}{AE\delta }\\ {\delta }^3\approx \frac{P{l}^3}{AE}\\ \delta =l\sqrt[3]{\frac{P}{AE}}\end{array}$

Hence, the deflection of uniform wire ABC at midpoint B is $\underset{\_}{\delta =l\sqrt[3]{\frac{P}{AE}}}$ is proved.