Question 11.SP.6: For the beam and loading shown, determine the deflection at ......
For the beam and loading shown, determine the deflection at point D. Use E = 29 × 10^6 psi.
STRATEGY: Add a dummy load associated with the desired vertical deflection at joint D. Use a free-body diagram to determine the reactions due to both the dummy load and the distributed load. The moments in each segment are then written as a function of the coordinate along the beam. Equation (11.57) is used to determine the deflection.
x_j=\frac{\partial U}{\partial P_j}=\int_0^L \frac{M}{E I} \frac{\partial M}{\partial P_j} d x (11.57)
MODELING and ANALYSIS:
Castigliano’s Theorem. We introduce a dummy load Q as shown in Fig. 1. Using Castigliano’s theorem and noting that EI is constant, write
y_D=\int \frac{M}{E I}\left(\frac{\partial M}{\partial Q}\right) d x=\frac{1}{E I} \int M\left(\frac{\partial M}{\partial Q}\right) d x (1)
The integration will be performed separately for segments AD and DB.
Reactions. Using the free-body diagram of the entire beam (Fig. 2) gives
R _A=\frac{w b^2}{2 L}+Q \frac{b}{L} \uparrow \quad R _B=\frac{w b\left(a+\frac{1}{2} b\right)}{L}+Q \frac{a}{L} \uparrow
Portion AD of Beam. Using the free-body diagram shown in Fig. 3,
M_1=R_A x=\left(\frac{w b^2}{2 L}+Q \frac{b}{L}\right) x \quad \frac{\partial M_1}{\partial Q}=+\frac{b x}{L}
Substituting into Eq. (1) and integrating from A to D gives
\frac{1}{E I} \int M_1 \frac{\partial M_1}{\partial Q} d x = \frac{1}{E I} \int_{0}^{a}R_A x \left(\frac{bx}{L}\right)dx=\frac{R_Aa^3b}{3EIL}
Then substitute for R_A and set the dummy load Q equal to zero
\frac{1}{E I} \int M_1 \frac{\partial M_1}{\partial Q} d x=\frac{w a^3 b^3}{6 E I L^2} (2)
Portion DB of Beam. Using the free-body diagram shown in Fig. 4, the bending moment at a distance ν from end B is
M_2=R_B ν-\frac{w ν^2}{2}=\left[\frac{w b\left(a+\frac{1}{2} b\right)}{L}+Q \frac{a}{L}\right] ν-\frac{w ν^2}{2} \quad \frac{\partial M_2}{\partial Q}=+\frac{a ν}{L}
Substitute into Eq. (1) and integrate from point B (where ν = 0) to point D (where ν = b) for
\frac{1}{E I} \int M_2 \frac{\partial M_2}{\partial Q} d ν=\frac{1}{E I} \int_0^b\left(R_B ν-\frac{w ν^2}{2}\right)\left(\frac{a ν}{L}\right) d ν=\frac{R_B a b^3}{3 E I L}-\frac{w a b^4}{8 E I L}
Substituting for R_B and setting Q = 0,
\frac{1}{E I} \int M_2 \frac{\partial M_2}{\partial Q} d ν=\left[\frac{w b\left(a+\frac{1}{2} b\right)}{L}\right] \frac{a b^3}{3 E I L}-\frac{w a b^4}{8 E I L}=\frac{5 a^2 b^4+a b^5}{24 E I L^2} w (3)
Deflection at Point D. Recalling Eqs. (1), (2), and (3),
y_D=\frac{w a b^3}{24 E I L^2}\left(4 a^2+5 a b+b^2\right)=\frac{w a b^3}{24 E I L^2}(4 a+b)(a+b)=\frac{w a b^3}{24 E I L}(4 a+b)
From Appendix E, I = 68.9 in^4 for a W10 × 15 beam. Substituting the numerical values for I, w, a, b, and L,
y_D=0.262 \text { in. } \downarrow
