Question 15.CA.6: Determine the slope and deflection at D for the beam and loa...
Determine the slope and deflection at D for the beam and loading shown (Fig. 15.18a), knowing that the flexural rigidity of the beam is EI = 100 MN⋅m².
The slope and deflection at any point of the beam can be obtained by superposing the slopes and deflections caused by the concentrated load and by the distributed load (Fig. 15.18b).
Since the concentrated load in Fig. 15.18c is applied at quarter span, the results for the beam and loading of Concept Application 15.3 can be used to write
\begin{aligned}\left(\theta_D\right)_p &=-\frac{P L^2}{32 E I}=-\frac{\left(150 \times 10^3\right)(8)^2}{32\left(100 \times 10^6\right)}=-3 \times 10^{-3} \text{ rad} \\\left(y_D\right)_P &=-\frac{3 P L^3}{256 E I}=-\frac{3\left(150 \times 10^3\right)(8)^3}{256\left(100 \times 10^6\right)}=-9 \times 10^{-3} m \\&=-9 mm\end{aligned}On the other hand, recalling the equation of the elastic curve obtained for a uniformly distributed load in Concept Application 15.2, the deflection in Fig. 15.18d is
y=\frac{w}{24 E I}\left(-x^4+2 L x^3-L^3 x\right) (1)
Differentiating with respect to x gives
\theta=\frac{d y}{d x}=\frac{w}{24 E I}\left(-4 x^3+6 L x^2-L^3\right) (2)
Making w = 20 kN/m, x = 2 m, and L = 8 m in Eqs. (1) and (2), we obtain
\begin{aligned}\left(\theta_D\right)_w &=\frac{20 \times 10^3}{24\left(100 \times 10^6\right)}(-352)=-2.93 \times 10^{-3} \text{ rad} \\\left(y_D\right)_w &=\frac{20 \times 10^3}{24\left(100 \times 10^6\right)}(-912)=-7.60 \times 10^{-3} m \\&=-7.60 mm\end{aligned}Combining the slopes and deflections produced by the concentrated and the distributed loads,
\begin{aligned}\theta_D &=\left(\theta_D\right)_p+\left(\theta_D\right)_w=-3 \times 10^{-3}-2.93 \times 10^{-3} \\&=-5.93 \times 10^{-3} \text{ rad} \\y_D &=\left(y_D\right)_P+\left(y_D\right)_W=-9 mm-7.60 mm=-16.60 mm\end{aligned}