Question 15.CA.3: For the prismatic beam and load shown (Fig. 15.11a), determi...
For the prismatic beam and load shown (Fig. 15.11a), determine the slope and deflection at point D.
Divide the beam into two portions, AD and DB, and determine the function y(x) that defines the elastic curve for each of these portions.
- From A to D (x < L/4). Draw the free-body diagram of a portion of beam AE of length x < L/4 (Fig. 15.11b). Take moments about E to obtain
M_1=\frac{3 P}{4} x (1)
and recalling Eq. (15.4), we write
\frac{d^2 y}{d x^2}=\frac{M(x)}{E I} (15.4)
E I \frac{d^2 y_1}{d x^2}=\frac{3}{4} P x (2)
where y_1(x) is the function that defines the elastic curve for portion AD of the beam. Integrating in x,
EI \theta_1=E I \frac{d y_1}{d x}=\frac{3}{8} P x^2+C_1 (3)
EI y_1=\frac{1}{8} P x^3+C_1 x+C_2 (4)
- From D to B (x > L/4). Now draw the free-body diagram of a portion of beam AE of the length x > L/4 (Fig. 15.11c) and write
M_2=\frac{3 P}{4} x-P\left(x-\frac{L}{4}\right) (5)
and recalling Eq. (15.4) and rearranging terms, we have
E I \frac{d^2 y_2}{d x^2}=-\frac{1}{4} P x+\frac{1}{4} P L (6)
where y_2(x) is the function that defines the elastic curve for portion DB of the beam. Integrating in x,
EI \theta_2=E I \frac{d y_2}{d x}=-\frac{1}{8} P x^2+\frac{1}{4} P L x+C_3 (7)
EI y_2=-\frac{1}{24} P x^3+\frac{1}{8} P L x^2+C_3 x+C_4 (8)
Determination of the Constants of Integration. The conditions satisfied by the constants of integration are summarized in Fig. 15.11d. At the support A, where the deflection is defined by Eq. (4), x = 0 and y_1 = 0. At the support B, where the deflection is defined by Eq. (8), x = L and y_2 = 0. Also, the fact that there can be no sudden change in deflection or in slope at point D requires that y_1 = y_2 and θ_1 = θ_2 when x = L/4. Therefore,
\left[x=0, y_1=0\right], \text {Eq. (4)}: \quad 0=C_2 (9)
\left[x=L, y_2=0\right], \text {Eq. (8)}: \quad 0=\frac{1}{12} P L^3+C_3 L+C_4 (10)
\left[x=L / 4, \theta_1=\theta_2\right], Eqs. (3) and (7):
\frac{3}{128} P L^2+C_1=\frac{7}{128} P L^2+C_3 (11)
\left[x=L / 4, y_1=y_2\right], Eqs. (4) and (8):
\frac{P L^3}{512}+C_1 \frac{L}{4}=\frac{11 P L^3}{1536}+C_3 \frac{L}{4}+C_4 (12)
Solving these equations simultaneously,
C_1=-\frac{7 P L^2}{128}, \quad C_2=0, \quad C_3=-\frac{11 P L^2}{128}, \quad C_4=\frac{P L^3}{384}Substituting for C_1 and C_2 into Eqs. (3) and (4), x ≤ L/4 is
EI \theta_1=\frac{3}{8} P x^2-\frac{7 P L^2}{128} (13)
EI y_1=\frac{1}{8} P x^3-\frac{7 P L^2}{128} x (14)
Letting x = L/4 in each of these equations, the slope and deflection at point D are
\theta_D=-\frac{P L^2}{32 E I} \quad \text { and } \quad y_D=-\frac{3 P L^3}{256 E I}Note that because \theta_D \neq 0, the deflection at D is not the maximum deflection of the beam.