Question 9.8: 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 R_B is now considered as an unknown load (Fig. 9.37a) 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 \left(y_{B}\right)_{w} caused at B by the uniformly distributed load w (Fig. 9.37b) and the deflection \left(y_{B}\right)_{R} produced at the same point by the redundant reaction R_B (Fig. 9.37c).

From the table of Appendix D (cases 2 and 1), we find that

\left(y_{B}\right)_{w}=-\frac{w L^{4}}{8 E I}                  \left(y_{B}\right)_{R}=+\frac{R_{B} L^{3}}{3 E I}

Beam Deflections and Slopes
Beam and Loading	Elastic Curve	Maximum Deflection	Slope at End	Equation of Elastic Curve
		-\frac{P L^{3}}{3 E I}	-\frac{P L^{2}}{2 E I}	y=\frac{P}{6 EI}\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}}	u _{A}=-\frac{P b\left(L^{2}-b^{2}\right)}{6 E I L}   u_{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}	u _{A}=+\frac{M L}{6 E I}   u _{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, we have

and, solving for R_B,                      R_{B}=\frac{3}{8} w L                  R _{B}=\frac{3}{8} w L \uparrow

Drawing the free-body diagram of the beam (Fig. 9.38) and writing the corresponding equilibrium equations, we have

+\uparrow \sum F_{y}=0:                    R_{A}+R_{B}-w L=0                    (9.52)

+\circlearrowleft ∑M_A =0:                          M_{A}+R_{B} L-(w L)\left(\frac{1}{2} L\right)=0                (9.53)

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 M_A is now considered as an unknown load (Fig. 9.39a) 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 \left( u _{A}\right)_{w} caused at A by the uniformity distributed load w (Fig. 9.39b) and the slope \left( u _{A}\right)_{M} produced at the same point by the unknown couple M_A (Fig 9.39c).

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

\left( u _{A}\right)_{w}=-\frac{w L^{3}}{24 E I}                    \left( u _{A}\right)_{M}=\frac{M_{A} L}{3 E I}

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

and, solving for M_A,

M_{A}=\frac{1}{8} w L^{2}                        M_{A}=\frac{1}{8} w L^{2}\circlearrowleft

The values of R_A and R_B may then be found from the equilibrium equations (9.52) and (9.53).