For the uniform beam and loading shown, determine (a) the reaction at each support, (b) the slope at end A.

Question

Accepted Answer

Principle of Superposition. The reaction [latex]R_B[/latex] is designated as redundant and considered as an unknown load. The deflections due to the distributed load and to the reaction [latex]R_B[/latex] are considered separately as shown below.

For each loading the deflection at point B is found 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}\right)[/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}\right)[/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\right)[/latex]
		For a > b: [latex]-\frac{P b\left(L^{2}-b^{2}\right)^{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}\right)}{6 E I L}[/latex] [latex]u_{B}=+\frac{P a\left(L^{2}-a^{2}\right)}{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}\right) x\right][/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\right)[/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\right)[/latex]

Distributed Loading. We use case 6, Appendix D

[latex]y=-\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right)[/latex]

At point B, [latex]x=\frac{2}{3} L[/latex]:

[latex]\left(y_{B}\right)_{w}=-\frac{w}{24 E I}\left[\left(\frac{2}{3} L\right)^{4}-2 L\left(\frac{2}{3} L\right)^{3}+L^{3}\left(\frac{2}{3} L\right)\right]=-0.01132 \frac{w L^{4}}{E I}[/latex]

Redundant Reaction Loading. From case 5, Appendix D, with [latex]a=\frac{2}{3} L[/latex] and [latex]b=\frac{1}{3} L[/latex], we have

[latex]\left(y_{B}\right)_{R}=-\frac{P a^{2} b^{2}}{3 E I L}=+\frac{R_{B}}{3 E I L}\left(\frac{2}{3} L\right)^{2}\left(\frac{L}{3}\right)^{2}=0.01646 \frac{R_{B} L^{3}}{E I}[/latex]

a. Reactions at Supports. Recalling that [latex]y_B = 0[/latex], we write

[latex]y_{B}=\left(y_{B}\right)_{w}+\left(y_{B}\right)_{R}[/latex]

[latex]0=-0.01132 \frac{w L^{4}}{E I}+0.01646 \frac{R_{B} L^{3}}{E I}[/latex] [latex]R _{B}=0.688 w L \uparrow[/latex]

Since the reaction [latex]R_B[/latex] is now known, we may use the methods of statics to determine the other reactions:

[latex]R _{A}=0.271 w L \uparrow[/latex] [latex]R _{C}=0.0413 w L \uparrow[/latex]

b. Slope at End A. Referring again to Appendix D, we have

Distributed Loading. [latex]\left( u _{A}\right)_{w}=-\frac{w L^{3}}{24 E I}=-0.04167 \frac{w L^{3}}{E I}[/latex]

Redundant Reaction Loading. For [latex]P=-R_{B}=-0.688 w L[/latex] and [latex]b=\frac{1}{3} L[/latex]

[latex]\left( u _{A}\right)_{R}=-\frac{P b\left(L^{2}-b^{2}\right)}{6 E I L}=+\frac{0.688 w L}{6 E I L}\left(\frac{L}{3}\right)\left[L^{2}-\left(\frac{L}{3}\right)^{2}\right][/latex] [latex]\left( u _{A}\right)_{R}=0.03398 \frac{w L^{3}}{E I}[/latex]

Finally, [latex]u _{A}=\left( u _{A}\right)_{w}+\left( u _{A}\right)_{R}[/latex]

[latex]u _{A}=-0.04167 \frac{w L^{3}}{E I}+0.03398 \frac{w L^{3}}{E I}=-0.00769 \frac{w L^{3}}{E I}[/latex] [latex]u_{A}=0.00769 \frac{w L^{3}}{E I} c[/latex]

Question 9.S-P.8: For the uniform beam and loading shown, determine (a) the re...

Related Answered Questions

The rigid bar DEF is welded at point D to the uniform steel beam AB. For the loading shown, determine (a) the equation of the elastic curve of the beam, (b) the deflection at the midpoint C of the beam. Use E = 29 × 10^6 psi. ...

Verified Answer:

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

Verified Answer:

Determine the maximum deflection of the beam of Example 9.12. ...

Verified Answer:

For the beam and loading shown, (a) determine the deflection at end A, (b) evaluate yA for the following data: W10 × 33: I = 170 in^4 E = 29 × 10^6 psi a = 3 ft = 36 in. L = 5.5 ft = 66 in. w = 13.5 kips/ft = 1125 lb/in. ...

Verified Answer:

Determine the reactions at the supports for the prismatic beam and loading shown in Fig. 9.36. (This is the same beam and loading as in Example 9.05 of Sec. 9.5.) ...

Verified Answer:

For the beam and loading shown, determine the reaction at the fixed support C. ...

Verified Answer:

For the uniform beam AB, (a) determine the reaction at A, (b) derive the equation of the elastic curve, (c) determine the slope at A. (Note that the beam is statically indeterminate to the first degree.) ...

Verified Answer:

Determine the reaction at the supports for the prismatic beam and loading shown (Fig. 9.71). ...

Verified Answer:

Determine the slope and deflection at end B of the prismatic beam of Example 9.09, drawing the bending-moment diagram by parts. ...

Verified Answer:

Determine the slope and deflection at end B of the prismatic cantilever beam AB when it is loaded as shown (Fig. 9.48), knowing that the flexural rigidity of the beam is EI = 10 MN · m². ...

Verified Answer: