Question 9.S-P.12: For the beam and loading shown, (a) determine the deflection...
For the beam and loading shown, (a) determine the deflection at end A, (b) evaluate y_A 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.
(M/EI) Diagram. We first draw the bending-moment diagram. Since the flexural rigidity EI is constant, we obtain the (M/EI) diagram shown, which consists of a parabolic spandrel of area A_1 and a triangle of area A_2.
A_{1}=\frac{1}{3}\left(-\frac{w a^{2}}{2 E I}\right) a=-\frac{w a^{3}}{6 E I}A_{2}=\frac{1}{2}\left(-\frac{w a^{2}}{2 E I}\right) L=-\frac{w a^{2} L}{4 E I}
Reference Tangent at B. The reference tangent is drawn at point B as shown. Using the second moment-area theorem, we determine the tangential deviation of C with respect to B:
t_{C / B}=A_{2} \frac{2 L}{3}=\left(-\frac{w a^{2} L}{4 E I}\right) \frac{2 L}{3}=-\frac{w a^{2} L^{2}}{6 E I}
From the similar triangles A^{\prime \prime} A^{\prime} B and C C^{\prime} B, we find
A^{\prime \prime} A^{\prime}=t_{C / B}\left(\frac{a}{L}\right)=-\frac{w a^{2} L^{2}}{6 E I}\left(\frac{a}{L}\right)=-\frac{w a^{3} L}{6 E I}
Again using the second moment-area theorem, we write
t_{A / B}=A_{1} \frac{3 a}{4}=\left(-\frac{w a^{3}}{6 E I}\right) \frac{3 a}{4}=-\frac{w a^{4}}{8 E I}
a. Deflection at End A
y_{A}=A^{\prime \prime} A^{\prime}+t_{A / B}=-\frac{w a^{3} L}{6 E I}-\frac{w a^{4}}{8 E I}=-\frac{w a^{4}}{8 E I}\left(\frac{4}{3} \frac{L}{a}+1\right)y_{A}=\frac{w a^{4}}{8 E I}\left(1+\frac{4}{3} \frac{L}{a}\right) \downarrow
b. Evaluation of y_{A}. Substituting the data given, we write
y_{A}=\frac{(1125 lb / in .)(36 in .)^{4}}{8\left(29 \times 10^{6} lb / in ^{2}\right)\left(170 in ^{4}\right)}\left(1+\frac{4}{3} \frac{66 in .}{36 in .}\right)y_{A}=0.1650 in . \downarrow
