Question 15.4: For the beam and loading shown, determine the slope and defl...
For the beam and loading shown, determine the slope and deflection at point B.
STRATEGY: Using the method of superposition, you can model the given problem using a summation of beam load cases for which deflection formulae are readily available.
MODELING: Through the principle of superposition, the given loading can be obtained by superposing the loadings shown in the following picture equation of Fig. 1. The beam AB is the same in each part of the figure.
ANALYSIS: For each of the loadings I and II (detailed further in Fig. 2), determine the slope and deflection at B by using the table of Beam Deflections and Slopes in Appendix C.
Appendix C Beam Deflections and Slopes
| Equation of Elastic Curve | Slope at End | Maximum Deflection | Elastic Curve | Beam and Loading |
| y=\frac{P}{6EI}(x^3-3Lx^2) | -\frac{PL^2}{2EI} | -\frac{PL^3}{3EI} |
|
1
|
| y=-\frac{w}{24EI}(x^4-4Lx^3+6L^2x^2) | -\frac{wL^3}{6EI} | -\frac{wL^4}{8EI} |
|
2
|
| y=-\frac{M}{2EI}x^2 | -\frac{ML}{EI} | -\frac{ML^2}{2EI} |
|
3
|
| \text{For }x\leq \frac{1}{2}L: \\ y=\frac{P}{48EI}(4x^3-3L^2x) | \pm \frac{PL^2}{16EI} | -\frac{PL^3}{48EI} |
|
4
|
| \text{For }x\lt a: \\ y=\frac{Pb}{6EIL}[x^3-(L^2-b^2)x] \\ \text{For }x=a: \ y=-\frac{Pa^2b^2}{3EIL} | \theta_A=-\frac{Pb(L^2-b^2)}{6EIL} \\ \theta_B=+\frac{Pa(L^2-a^2)}{6EIL} | \text{For }a\gt b: \\ -\frac{Pb(L^2-b^2)^{3/2}}{9\sqrt{3}EIL} \\ \text{at }x_m=\sqrt{\frac{L^2-b^2}{3} } |
|
5
|
| y=-\frac{w}{24EI}(x^4-2Lx^3+L^3x) | \pm \frac{wL^3}{24EI} | -\frac{5wL^4}{384EI} |
|
6
|
| y=-\frac{M}{6EIL}(x^3-L^2x) | \theta_A=+\frac{ML}{6EI} \\ \theta_B=-\frac{ML}{3EI} | \frac{ML^2}{9\sqrt{3}EI} |
|
7
|
Loading I
(\theta_B)_I=-\frac{wL^3}{6EI} \quad \quad \quad (y_B)_1=-\frac{wL^4}{8EI}
Loading II
(\theta_C)_{II}=+\frac{w(L/2)^3}{6EI}=+\frac{wL^3}{48EI} \quad \quad (y_C)_{II}=+\frac{w(L/2)^4}{8EI}=+\frac{wL^4}{128EI}
In portion CB, the bending moment for loading II is zero. Thus, the elastic curve is a straight line.
\begin{matrix} (\theta_B)_{II}=(\theta_C)_{II}=+\frac{wL^3}{48EI} && (y_B)_{II}=(y_C)_{II}+(\theta_C)_{II}\left(\frac{L}{2}\right) \\ && \quad =\frac{wL^4}{128EI}+\frac{wL^3}{48EI}\left(\frac{L}{2} \right) =+\frac{7wL^4}{384EI} \end{matrix}
Slope at Point B
\theta_B=(\theta_B)_I+(\theta_B)_{II}=-\frac{wL^3}{6EI}+\frac{wL^3}{48EI}=-\frac{7wL^3}{48EI} \quad \quad \theta_B=\frac{7wL^3}{48EI}\measuredangle
Deflection at B
y_B=(y_B)_I+(y_B)_{II}=-\frac{wL^4}{8EI}+\frac{7wL^4}{384EI}=-\frac{41wL^4}{384EI} \quad \quad \quad y_B=\frac{41wL^4}{384EI}\downarrow
REFLECT and THINK: Note that the formulae for one beam case can sometimes be extended to obtain the desired deflection of another case, as you saw here for loading II.
