Question 15.CA.7: Determine the reactions at the supports for the prismatic be...

We consider the reaction at B as redundant and release the beam from the support. The reaction \pmb{R}_B is now considered as an unknown load (Fig. 15.19b) and will be determined from the condition that the deflection of the beam at B must be zero. The solution is carried out by considering separately the deflection (y_B)_w caused at B by the uniformly distributed load w (Fig. 15.19c) and the deflection (y_B)_R produced at the same point by the redundant reaction \pmb{R}_B (Fig. 15.19d).

From the table of Appendix E (cases 2 and 1),

Beam and Loading	Elastic Curve	Maximum Deflection	Slop at End	Equation of Elastic Curve
		-\frac{P L^3}{3 E I}	-\frac{P L^2}{2 E I}	y=\frac{P}{6 E I}\left(x^3-3 L x^2\right)
		-\frac{w L^4}{8 E I}	-\frac{w L^3}{6 E I}	y=-\frac{w}{24 E I}\left(x^4-4 L x^3+6 L^2 x^2\right)
		-\frac{M L^2}{2 E I}	-\frac{M L}{E I}	y=-\frac{M}{2 E I} x^2
		-\frac{P L^3}{48 E I}	\pm \frac{P L^2}{16 E I}	For x \leq \frac{1}{2} L: y=\frac{P}{48 E I}\left(4 x^3-3 L^2 x\right)
		For a > b: -\frac{P b\left(L^2-b^2\right)^{3 / 2}}{9 \sqrt{3} E I L} at x_m=\sqrt{\frac{L^2-b^2}{3}}	\theta_A=-\frac{P b\left(L^2-b^2\right)}{6 E I L} \theta_B=+\frac{P a\left(L^2-a^2\right)}{6 E I L}	For x < a: y=\frac{P b}{6 E I L}\left[x^3-\left(L^2-b^2\right) x\right] For x = a:    y=-\frac{P a^2 b^2}{3 E I L}
		-\frac{5 w L^4}{384 E I}	\pm \frac{w L^3}{24 E I}	y=-\frac{w}{24 E I}\left(x^4-2 L x^3+L^3 x\right)
		\frac{M L^2}{9 \sqrt{3} E I}	\theta_A=+\frac{M L}{6 E I} \theta_B=-\frac{M L}{3 E I}	y=-\frac{M}{6 E I L}\left(x^3-L^2 x\right)

Writing that the deflection at B is the sum of these two quantities and that it must be zero,

and, solving for R_B, \quad R_B=\frac{3}{8} w L \quad \pmb{R}_B=\frac{3}{8} w L \uparrow

Drawing the free-body diagram of the beam (Fig. 15.19e) and writing the corresponding equilibrium equations,

\begin{aligned}+\uparrow \Sigma F_y=0: \quad &R_A+R_B-w L=0 \\&R_A=w L-R_B=w L-\frac{3}{8} w L=\frac{5}{8} w L \\&\pmb{R}_A=\frac{5}{8} w L \uparrow\end{aligned}     (1)

\begin{aligned}+\circlearrowleft \Sigma M_A=0 \quad M_A+R_B L-(w L)\left(\frac{1}{2} L\right)&=0 \\M_A=\frac{1}{2} w L^2-R_B L=\frac{1}{2} w L^2-\frac{3}{8} w L^2&=\frac{1}{8} w L^2 \\\pmb{M}_A&=\frac{1}{8} w L^2 \circlearrowleft\end{aligned}     (2)

Alternative Solution. We may consider the couple exerted at the fixed end A as redundant and replace the fixed end by a pin-and-bracket support. The couple \pmb{M}_A is now considered as an unknown load (Fig. 15.19f) and will be determined from the condition that the slope of the beam at A must be zero. The solution is carried out by considering separately the slope (θ_A)_w caused at A by the uniformly distributed load w (Fig. 15.19g) and the slope (θ_A)_M produced at the same point by the unknown couple \pmb{M}_A (Fig. 15.19h).

Using the table of Appendix E (cases 6 and 7) and noting that A and B must be interchanged in case 7,

Writing that the slope at A is the sum of these two quantities and that it must be zero gives

where M_A is

The values of R_A and R_B are found by using the equilibrium equations (1) and (2).