Question 15.S-P.3: For the uniform beam AB, (a) determine the reaction at A, (b...
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.)
Bending Moment. Using the free body shown, we write
+\curvearrowright \sum{M_{D}} = 0: R_{A}x – \frac{1}{2} \left( \frac{w_{0} x^{2}}{L} \right) \frac{x}{3} – M = 0 M = R_{A}x – \frac{w_{0} x^{3}}{6L}Differential Equation of the Elastic Curve. We use Eq. (15.4) and write
EI \frac{d^{2}y}{dx^{2}} =\frac{M(x)}{EI} (15.4)
EI \frac{d^{2}y}{dx^{2}} = R_{A}x – \frac{w_{0} x^{3}}{6L}
Noting that the flexural rigidity EI is constant, we integrate twice and find
EI \frac{dy}{dx} = EI θ = \frac{1}{2} R_{A}x^{2} -\frac{w_{0}x^{4}}{ 24L} + C_{1} (1)
EI y = \frac{1}{6} R_{A}x^{3} -\frac{w_{0}x^{5}}{120L} + C_{1}x + C_{2} (2)
Boundary Conditions. The three boundary conditions that must be satisfied are shown on the sketch
[ x = 0, y = 0 ] : C_{2} = 0 (3)
[ x = L, θ = 0 ] : \frac{1}{2} R_{A}L^{2} – \frac{w_{0}L^{3}}{24} + C_{1} = 0 (4)
[ x = L, y = 0 ] : \frac{1}{6} R_{A}L^{3} – \frac{w_{0}L^{4}}{120} + C_{1}L + C_{2} = 0 (5)
a. Reaction at A. Multiplying Eq. (4) by L, subtracting Eq. (5) member by member from the equation obtained, and noting that C_{2} = 0, we have
\frac{1}{3} R_{A}L^{3} – \frac{1}{30} w_{0}L^{4} = 0 R_{A} = \frac{1}{10} w_{0}L↑
We note that the reaction is independent of E and I. Substituting R_{A} = \frac{1}{10} w_{0}L into Eq. (4), we have
\frac{1}{2} ( \frac{1}{10} w_{0}L)L^{2} – \frac{1}{24} w_{0}L^{3} + C_{1} = 0 C_{1} = – \frac{1}{120} w_{0}L^{3}
b. Equation of the Elastic Curve. Substituting for R_{A}, C_{1}, and C_{2} into Eq. (2), we have
EI y = \frac{1}{6} \left( \frac{1}{10} w_{0}L\right) x^{3} -\frac{w_{0}x^{5}}{120L} – \left( \frac{1}{120} w_{0}L^{3}\right) x
y =\frac{w_{0}}{120EIL} (-x^{5} + 2L^{2}x^{3} – L^{4}x)
c. Slope at A. We differentiate the above equation with respect to x:
θ = \frac{dy}{dx} = \frac{w_{0}}{120EIL} (-5x^{4} + 6L^{2}x^{2} – L^{4})
Making x = 0, we have θ_{A} = – \frac{w_{0}L^{3}}{120EI} θ_{A} = – \frac{w_{0}L^{3}}{120EI}⦪
