Question 9.SP.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:
\begin{array}{ll} \text { W10 } \times 33: I=171 \text{in} ^4 & E=29 \times 10^6 \text{psi}\\ a=3 \text{ft} =36 \text{in} . & L=5.5 \text{ft} =66 \text{in} . \\ w=13.5 \text{kips} /\text{ft} =1125 \text{lb}/\text{in}. & \end{array}
STRATEGY: To apply the moment-area theorems, you should first obtain the M/EI diagram for the beam. Then, by placing the reference tangent at a support, you can evaluate the tangential deviations at other strategic points that, through simple geometry, will enable the determination of the desired deflection.
MODELING and ANALYSIS:
M∕EI Diagram. Referring to Fig. 1, draw the bending-moment diagram.
Since the flexural rigidity EI is constant, the M/EI diagram is as shown, which consists of a parabolic spandrel of area A_1 and a triangle of area A_2 .
\begin{aligned} & 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} \end{aligned}
Reference Tangent at B. The reference tangent is drawn at point B in Fig. 2. Using the second moment-area theorem, the tangential deviation of C with respect to B is
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″A′B and CC′B,
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,
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
\begin{aligned} & 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 \end{aligned}
b. Evaluation of y_A . Substituting the data,
\begin{array}{r} y_A=\frac{(1125 lb / \text {in.})(36 in .)^4}{8\left(29 \times 10^6 lb / in ^2\right)\left(171 in ^4\right)}\left(1+\frac{4}{3} \frac{66 \text { in. }}{36 \text { in. }}\right) \\ y_A=0.1641 \text { in. } \downarrow \end{array}
REFLECT and THINK: Note that an equally effective alternate strategy would be to draw a reference tangent at point C.
