For the beam and loading shown, determine the slope and deflection at point B.

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Accepted Answer

Principle of Superposition. The given loading can be obtained by superposing the loadings shown in the following “picture equation.” The beam AB is, of course, the same in each part of the figure.

For each of the loadings I and II, we now determine the slope and deflection at B by using the table of Beam Deflections and Slopes in Appendix D.

Beam Deflections and Slopes
Beam and Loading	Elastic Curve	Maximum Deflection	Slope at End	Equation of Elastic Curve
		[latex]-\frac{P L^{3}}{3 E I}[/latex]	[latex]-\frac{P L^{2}}{2 E I}[/latex]	[latex]y=\frac{P}{6 EI}\left(x^{3}-3 L x^{2} ight)[/latex]
		[latex]-\frac{w L^{4}}{8 E I}[/latex]	[latex]-\frac{w L^{3}}{6 E I}[/latex]	[latex]y=-\frac{w}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2} ight)[/latex]
		[latex]-\frac{M L^{2}}{2 E I}[/latex]	[latex]-\frac{M L}{E I}[/latex]	[latex]y=-\frac{M}{2 E I} x^{2}[/latex]
		[latex]-\frac{P L^{3}}{48 E I}[/latex]	[latex]\pm \frac{P L^{2}}{16 E I}[/latex]	For [latex]x \leq \frac{1}{2} L[/latex] : [latex]y=\frac{P}{48 E I}\left(4 x^{3}-3 L^{2} x ight)[/latex]
		For a > b: [latex]-\frac{P b\left(L^{2}-b^{2} ight)^{3 / 2}}{9 \sqrt{3} E I L}[/latex] at [latex]x_{m}=\sqrt{\frac{L^{2}-b^{2}}{3}}[/latex]	[latex]u _{A}=-\frac{P b\left(L^{2}-b^{2} ight)}{6 E I L}[/latex] [latex]u_{B}=+\frac{P a\left(L^{2}-a^{2} ight)}{6 E I L}[/latex]	For x < a: [latex]y=\frac{P b}{6 E I L}\left[x^{3}-\left(L^{2}-b^{2} ight) x ight][/latex] For x = a: [latex]y=-\frac{P a^{2} b^{2}}{3 E I L}[/latex]
		[latex]-\frac{5 w L^{4}}{384 E I}[/latex]	[latex]\pm \frac{w L^{3}}{24 E I}[/latex]	[latex]y=-\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x ight)[/latex]
		[latex]\frac{M L^{2}}{9 \sqrt{3} E I}[/latex]	[latex]u _{A}=+\frac{M L}{6 E I}[/latex] [latex]u _{B}=-\frac{M L}{3 E I}[/latex]	[latex]y=-\frac{M}{6 E I L}\left(x^{3}-L^{2} x ight)[/latex]

Loading I

[latex]\left( u _{B} ight)_{ I }=-\frac{w L^{3}}{6 E I}[/latex] [latex]\left(y_{B} ight)_{ I }=-\frac{w L^{4}}{8 E I}[/latex]

Loading II

[latex]\left( u _{C} ight)_{ II }=+\frac{w(L / 2)^{3}}{6 E I}=+\frac{w L^{3}}{48 E I}[/latex] [latex]\left(y_{C} ight)_{ II }=+\frac{w(L / 2)^{4}}{8 E I}=+\frac{w L^{4}}{128 E I}[/latex]

In portion CB, the bending moment for loading II is zero and thus the elastic curve is a straight line.

[latex]\left( u _{B} ight)_{ II }=\left( u _{C} ight)_{ II }=+\frac{w L^{3}}{48 E I}[/latex]

[latex]\begin{aligned}\left(y_{B} ight)_{ II } &=\left(y_{C} ight)_{ II }+\left( u _{C} ight)_{ II }\left(\frac{L}{2} ight) \&=\frac{w L^{4}}{128 E I}+\frac{w L^{3}}{48 E I}\left(\frac{L}{2} ight)=+\frac{7 w L^{4}}{384 E I}\end{aligned}[/latex]

Slope at Point B

[latex]u _{B}=\left( u _{B} ight)_{ I }+\left( u _{B} ight)_{ II }=-\frac{w L^{3}}{6 E I}+\frac{w L^{3}}{48 E I}=-\frac{7 w L^{3}}{48 E I}[/latex] [latex]u _{B}=\frac{7 w L^{3}}{48 E I} c[/latex]

Deflection at B

[latex]y_{B}=\left(y_{B} ight)_{ I }+\left(y_{B} ight)_{ II }=-\frac{w L^{4}}{8 E I}+\frac{7 w L^{4}}{384 E I}=-\frac{41 w L^{4}}{384 E I}[/latex] [latex]y_{B}=\frac{41 w L^{4}}{384 E I} \downarrow[/latex]

Question 9.S-P.7: For the beam and loading shown, determine the slope and defl...

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