A long cylinder is subjected to end shear Q0. Plot the value of Mx as a function of Q0 from βx = 0 to βx = 4.0 and derive the location of the maximum value of Mx.

Question

A long cylinder is subjected to end shear [latex]Q_0[/latex]. Plot the value of Mx as a function of [latex]Q_0[/latex] from βx = 0 to βx = 4.0 and derive the location of the maximum value of [latex]M_x[/latex].

Accepted Answer

From Eq. 5.25,

[latex]M_{x}=-D\left(\frac{d^{2} w}{d x^{2}}\right)=\frac{1}{2 \beta}\left(2 \beta M_{0} A_{\beta x}+2 Q_{0} D_{\beta x}\right)[/latex]

[latex]Q_{x}=\left(2 \beta M_{0} D_{\beta x}-Q_{0} B_{\beta x}\right)[/latex] (5.25)

[latex]M_{x}=\frac{Q_{0}}{\beta} D_{\beta x}[/latex]

Referring to Table 5.1 for values of [latex]D_{\beta x}[/latex], a plot of [latex]M_{x} /\left(Q_{0} / \beta\right)[/latex] can be constructed as shown in Fig. 5.11.

Table 5.1 Values of Functions [latex]A_{\beta x}, B_{\beta x}, C_{\beta x}, D_{\beta x}[/latex]
[latex]\beta _{x}[/latex]	[latex]A_{\beta x}[/latex]	[latex]B_{\beta x}[/latex]	[latex]C_{\beta x}[/latex]	[latex]D_{\beta x}[/latex]
0	1.0000	1.0000	1.0000	1.0000
0.05	0.9976	0.9025	0.9500	0.0475
0.10	0.9907	0.8100	0.9003	0.0903
0.15	0.9797	0.7224	0.8510	0.1286
0.20	0.9651	0.6398	0.8024	0.1627
0.30	0.9267	0.4888	0.7077	0.2189
0.40	0.8784	0.3564	0.6174	0.2610
0.50	0.8231	0.2415	0.5323	0.2908
0.55	0.7934	0.1903	0.4919	0.3016
0.60	0.7628	0.1431	0.4530	0.3099
0.80	0.6354	-0.0093	0.3131	0.3223
1.00	0.5083	-0.1108	0.1988	0.3096
1.20	0.3899	-0.1716	0.1091	0.2807
1.40	0.2849	-0.2011	0.0419	0.2430
1.60	0.1959	-0.2077	-0.0059	0.2018
1.80	0.1234	-0.1985	-0.0376	0.1610
2.00	0.0667	-0.1794	-0.0563	0.1231
2.50	-0.0166	-0.1149	-0.0658	0.0491
3.00	-0.0423	-0.0563	-0.0493	0.0070
3.5	-0.0389	-0.0177	-0.0283	-0.0106
4.0	-0.0258	0.0019	-0.0120	-0.0139
5.0	-0.0045	0.0084	0.0019	-0.0065
6.0	0.0017	0.0031	0.0024	-0.0007
7.0	0.0013	0.0001	0.0007	0.0006

To find the maximum moment [latex]M_{x}[/latex], the above equation can be differentiated with respect to x and equated to zero. Hence,

[latex]M_{x}=\frac{Q_{0}}{\beta} \cdot e^{-\beta x} \sin ^{\beta x}[/latex]

[latex]\frac{d M_{x}}{d x}=0=\frac{Q_{0}}{\beta}\left(-\beta e^{-\beta x} \sin \beta x+\beta e^{-\beta x} \cos \beta x\right)[/latex]

or

[latex]\beta x=\frac{\pi}{4}[/latex]

and maximum moment is given by

[latex]M=\frac{0.322 Q_{0}}{\beta}[/latex]

Question 5.4: A long cylinder is subjected to end shear Q0. Plot the value...

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