Question 15.1: The overhanging steel beam ABC carries a concentrated load P...
The overhanging steel beam ABC carries a concentrated load P at end C. For portion AB of the beam, (a) derive the equation of the elastic curve, (b) determine the maximum deflection, (c) evaluate y_{\text{max}} for the following data:
\begin{matrix} W14 \times 68 & I=722 \text{ in}^4 & E=29 \times 10^6 \text{ psi} \\ P=50 \text{ kips} & L=15 \text{ ft}=180 \text{ in.} & a=4 \text{ ft}=48 \text{ in.} \end{matrix}
STRATEGY: You should begin by determining the bending-moment equation for the portion of interest. Substituting this into the differential equation of the elastic curve, integrating twice, and applying the boundary conditions, you can then obtain the equation of the elastic curve. Use this equation to find the desired deflections.
MODELING: Using the free-body diagram of the entire beam (Fig. 1) gives the reactions: \pmb{\text{R}}_A=Pa/L\downarrow \ \pmb{\text{R}}_B=P(1+a/L)\uparrow . The free-body diagram of the portion of beam AD of length x (Fig. 1) gives
M=-P\frac{a}{L}x \quad \quad \quad (0\lt x\lt L)
ANALYSIS:
Differential Equation of the Elastic Curve. Using Eq. (15.4) gives
EI\frac{d^2y}{dx^2}=-P\frac{a}{L}x
Noting that the flexural rigidity EI is constant, integrate twice and find
EI\frac{dy}{dx}=-\frac{1}{2}P\frac{a}{L}x^2+C_1 \quad \quad \quad \quad \quad \pmb{(1)} \\ EI \ y=-\frac{1}{6}P\frac{a}{L}x^3+C_1x+C_2 \quad \quad \quad \quad \quad \pmb{(2)}
Determination of Constants. For the boundary conditions shown (Fig. 2),
[x=0,y=0]: From Eq.(2), C_2=0
[x=L,y=0]: Again using Eq. (2),
EI(0)=-\frac{1}{6} P\frac{a}{L}L^3+C_1L \quad \quad C_1=+\frac{1}{6}PaL
a. Equation of the Elastic Curve. Substituting for C_1 \text{ and }C_2 into Eqs. (1) and (2),
EI\frac{dy}{dx} =-\frac{1}{2} P\frac{a}{L}x^2+\frac{1}{6}PaL \quad \quad \quad \frac{dy}{dx}=\frac{PaL}{6EI}\left[1-3\left(\frac{x}{L} \right)^2 \right] \quad \quad \quad \pmb{(3)} \\ EI \ y=-\frac{1}{6} P\frac{a}{L}x^3+\frac{1}{6}PaLx \quad \quad \quad y=\frac{PaL^2}{6EI}\left[\frac{x}{L} -\left(\frac{x}{L} \right)^3 \right] \quad \quad \quad \pmb{(4)}
b. Maximum Deflection in Portion AB. The maximum deflection y_{\text{max}} occurs at point E where the slope of the elastic curve is zero (Fig. 3). Setting dy/dx = 0 in Eq. (3), the abscissa x_m of point E is
0=\frac{PaL}{6EI}\left[1-3\left(\frac{x_m}{L} \right)^2 \right] \quad \quad x_m=\frac{L}{\sqrt{3}}=0.577 \text{L}
Substitute x_m/L into Eq. (4):
y_{\text{max}}=\frac{PaL^2}{6EI}[(0.577)-(0.577)^3] \quad \quad y_{\text{max}}=0.0642\frac{PaL^2}{EI}
c. Evaluation of \pmb{y_{\text{max}}}. For the data given, the value of y_{\text{max}} is
y_{\text{max}}=0.0642\frac{(50 \text{ kips})(48 \text{ in.})(180 \text{ in.})^2}{(29 \times 10^6 \text{ psi})(722 \text{ in}^4)} \quad \quad y_{\text{max}}=0.238 \text{ in.}
REFLECT and THINK: Because the maximum deflection is positive, it is upward. As a check, we see that this is consistent with the deflected shape anticipated for this loading (Fig. 3).
