Figure P.15.24 shows two cantilevers, the end of one being vertically above the end of the other and connected to it by a spring AB. Initially the system is unstrained. A weight, W, placed at A causes a vertical deflection at A of δ1 and a vertical deflection at B of δ2. When the spring is removed

Question

Figure P.15.24 shows two cantilevers, the end of one being vertically above the end of the other and connected to it by a spring AB. Initially the system is unstrained. A weight, W, placed at A causes a vertical deflection at A of $\delta_{1}$ and a vertical deflection at B of $\delta_{2}$ . When the spring is removed the weight W at A causes a deflection at A of $\delta_{3}$ . Find the extension of the spring when it is replaced and the weight, W, is transferred to B.

Ans. $\delta_{2}\left(\delta_{1}-\delta_{2}\right) /\left(\delta_{3}-\delta_{1}\right)$

Ans. [latex]\delta_{2}\left(\delta_{1}-\delta_{2}\right) /\left(\delta_{3}-\delta_{1}\right)[/latex]

Accepted Answer

From Maxwell’s reciprocal theorem the deflection at A due to W at B is equal to the deflection at B due to W at A, i.e. [latex]\delta_{2}[/latex]. What is now required is the deflection at B due to W at B.
Since the deflection at A with W at A and the spring removed is [latex]\delta_{3}[/latex] the load in the spring at A with W at B is [latex]\left(\delta_{2} / \delta_{3}\right) W[/latex] which must equal the load in the spring at B with W at B. Therefore the resultant load at B with W at B is

[latex]W-\left(\frac{\delta_{2}}{\delta_{3}}\right) W=W\left(1-\frac{\delta_{2}}{\delta_{3}}\right)[/latex] (i)

The load W at A with the spring in place produces a deflection of [latex]\delta_{1}[/latex] at A. Therefore the resultant load at A is [latex]\left(\delta_{1} / \delta_{3}\right) W[/latex] so that, if the load in the spring at A with W at A is F, then [latex]W-F=\left(\delta_{1} / \delta_{3}\right) W[/latex], i.e.

[latex]F=W\left(1-\frac{\delta_{1}}{\delta_{3}}\right)[/latex] (ii)

This then is the load at B with W at A and it produces a deflection [latex]\delta_{2}[/latex]. Therefore, from Eqs (i) and (ii) the deflection at B due to W at B is

[latex]\left[\frac{W\left(1-\frac{\delta_{2}}{\delta_{3}}\right)}{W\left(1-\frac{\delta_{1}}{\delta_{3}}\right)}\right] \delta_{2}[/latex]

and the extension of the spring with W at B is

[latex]\left[\frac{\left(1-\frac{\delta_{2}}{\delta_{3}}\right)}{\left(1-\frac{\delta_{1}}{\delta_{3}}\right)}\right] \delta_{2}-\delta_{2}[/latex]

i.e.

[latex]\delta_{2}\left[\frac{\left(\delta_{1}-\delta_{2}\right)}{\left(\delta_{3}-\delta_{1}\right)}\right][/latex] .

Question 15.P.24: Figure P.15.24 shows two cantilevers, the end of one being v...

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