Question 15.S-P.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_{max} for the following data:
W14 × 68 I = 723 in^{4} E = 29 × 10^{6} psi
P = 50 kips L = 15 ft = 180 in. a = 4 ft = 48 in.
Free-Body Diagrams. Reactions: R_{A} = Pa/L↓ R_{B} = P(1 + a/L)↑ Using the free-body diagram of the portion of beam AD of length x, we find
M = -P\frac{a}{L} x (0 < x < L)
Differential Equation of the Elastic Curve. We use Eq. (15.4) and write
EI \frac{d^{2}y}{dx^{2}} =\frac{M(x)}{EI}
(15.4)
EI \frac{d^{2}y}{dx^{2}} = -P \frac{a}{L} x
Noting that the flexural rigidity EI is constant, we integrate twice and find
EI \frac{dy}{dx} = – \frac{1}{2} P \frac{a}{L} x^{2} + C_{1} (1)
EI y = – \frac{1}{6} P \frac{a}{L} x^{3} + C_{1}x + C_{2} (2)
Determination of Constants. For the boundary conditions shown, we have
[x = 0, y = 0]: From Eq. (2), we find C_{2} = 0
[x = L, y = 0]: Again using Eq. (2), we write
EI(0) = – \frac{1}{6} P \frac{a}{L} L^{3} + C_{1}L C_{1} = + \frac{1}{6} PaL
a. Equation of the Elastic Curve. Substituting for C_{1} and C_{2} into Eqs. (1) and (2), we have
EI \frac{dy}{dx} = – \frac{1}{2} P \frac{a}{L} x^{2}+\frac{1}{6} PaL \frac{dy}{dx}= \frac{PaL}{6EI} \left[ 1 – 3\left( \frac{x}{L}\right)^{2} \right] (3)
EI y = – \frac{1}{6} P \frac{a}{L} x^{3} + \frac{1}{6} PaLx y = \frac{PaL^{2}}{6EI} \left[ \frac{x}{L} – \left( \frac{x}{L}\right)^{3}\right ] (4)
b. Maximum Deflection in Portion AB. The maximum deflection y_{max} occurs at point E where the slope of the elastic curve is zero. Setting dy/dx = 0 in Eq. (3), we determine the abscissa x_{m} of point E:
0 = \frac{PaL}{6EI} \left[ 1 – 3\left( \frac{x_{m}}{L} \right)^{2} \right] x_{m} = \frac{L}{\sqrt{3}} = 0.577L
We substitute x_{m}/L = 0.577 into Eq. (4) and have
y_{max} = \frac{PaL^{2}}{6EI} [ (0.577) – (0.577)^{3} ] y_{max} = 0.0642 \frac{PaL^{2}}{EI}
c. Evaluation of y_{max}. For the data given, the value of y_{max} is
y_{max} = 0.0642 \frac{(50 kips) (48 in.) (180 in.)^{2}}{(29 × 10^{6} psi) (723 in^{4})} y_{max} = 0.238 in.
