Question 9.5: A simple beam AB with an overhang BC supports a concentrated...
A simple beam AB with an overhang BC supports a concentrated load P at the end of the overhang (Fig. 9-15a). The main span of the beam has length L and the overhang has length L/2.
Determine the equations of the deflection curve and the deflection δ_{C} at the end of the overhang (Fig. 9-15b). Use the third-order differential equation of the deflection curve (the shear-force equation). Note: The beam has constant flexural rigidity EI.
Use a four-step problem-solving approach.
1. Conceptualize:
Differential equations of the deflection curve: Because reactive forces act at supports A and B, write separate differential equations for parts AB and BC of the beam. Begin by finding the shear forces in each part of the beam.
The downward reaction at support A is equal to P/2, and the upward reaction at support B is equal to 3P/2 (see Fig. 9-15a). It follows that the shear forces in parts AB and BC are
V=-\frac{P}{2} \quad(0<x<L) (9-52a)
V=P \quad\left(L<x<\frac{3 L}{2}\right) (9-52b)
in which x is measured from end A of the beam (Fig. 9-15b).
2. Categorize: The third-order differential equations for the beam now become [see Eq. (9-16b)]:
E I v^{\prime \prime}=M \quad E I v^{\prime \prime \prime}=V \quad E I v^{\prime \prime \prime \prime}=-q (9-16a,b,c)
E I v^{\prime \prime \prime}=-\frac{P}{2} \quad(0<x<L) (a)
E I v^{\prime \prime \prime}=P \quad\left(L<x<\frac{3 L}{2}\right) (b)
3. Analyze:
Bending moments in the beam: Integration of the preceding two equations yields the bending-moment equations:
M=E I v^{\prime \prime}=-\frac{P x}{2}+C_{1} \quad(0 \leq x \leq L) (c)
M=E I v^{\prime \prime}=P x+C_{2} \quad\left(L \leq x \leq \frac{3 L}{2}\right) (d)
The bending moments at points A and C are zero; hence, the following boundary conditions apply to this beam:
v^{\prime \prime}(0)=0 \quad v^{\prime \prime}\left(\frac{3 L}{2}\right)=0Use these conditions with Eqs. (c) and (d) to get
C_{1}=0 \quad C_{2}=-\frac{3 P L}{2}Therefore, the bending moments are
M=E I v^{\prime \prime}=-\frac{P x}{2} \quad(0 \leq x \leq L) (9-53a)
M=E I v^{\prime \prime}=-\frac{P(3 L-2 x)}{2} \quad\left(L \leq x \leq \frac{3 L}{2}\right) (9-53b)
These equations can be verified by determining the bending moments from free-body diagrams and equations of equilibrium.
Slopes and deflections of the beam: The next integrations yield the slopes:
E I v^{\prime}=-\frac{P x^{2}}{4}+C_{3} \quad(0 \leq x \leq L)E I v^{\prime}=-\frac{P x(3 L-x)}{2}+C_{4} \quad\left(L \leq x \leq \frac{3 L}{2}\right)
The only condition on the slopes is the continuity condition at support B. According to this condition, the slope at point B as found for part AB of the beam is equal to the slope at the same point as found for part BC of the beam. Therefore, substitute x = L into each of the two preceding equations for the slopes and obtain
-\frac{P L^{2}}{4}+C_{3}=-P L^{2}+C_{4}This equation eliminates one constant of integration because C_{4} can be expressed in terms of C_{3}:
C_{4}=C_{3}+\frac{3 P L^{2}}{4} (e)
The third and last integrations give
E I v=-\frac{P x^{3}}{12}+C_{3} x+C_{5} \quad(0 \leq x \leq L) (f)
E I v=-\frac{P x^{2}(9 L-2 x)}{12}+C_{4} x+C_{6} \quad\left(L \leq x \leq \frac{3 L}{2}\right) (g)
For part AB of the beam (Fig. 9-15a), there are two boundary conditions on the deflections, namely, the deflection is zero at points A and B:
v(0)=0 \quad \text { and } \quad v(L)=0Apply these conditions to Eq. (f) to obtain
C_{5}=0 \quad C_{3}=\frac{P L^{2}}{12} (h,i)
Substitute the preceding expression for C_{3} in Eq. (e) to get
C_{4}=\frac{5 P L^{2}}{6} (j)
For part BC of the beam, the deflection is zero at point B. Therefore, the boundary condition is
v(L) = 0
Apply this condition to Eq. (g), and also substitute Eq. (j) for C_{4} to get
C_{6}=-\frac{P L^{3}}{4} (k)
All constants of integration have now been evaluated.
The deflection equations are obtained by substituting the constants of integration (Eqs. h, i, j, and k) into Eqs. (f) and (g). The results are
v=\frac{P x}{12 E I}\left(L^{2}-x^{2}\right) (0 \leq x \leq L) (9-54a)
v=-\frac{P}{12 E I}\left(3 L^{3}-10 L^{2} x+9 L x^{2}-2 x^{3}\right) \quad\left(L \leq x \leq \frac{3 L}{2}\right) (9-54b)
Note that the deflection is always positive (upward) in part AB of the beam [Eq. (9-54a)] and always negative (downward) in the overhang BC [Eq. (9-54b)].
Deflection at the end of the overhang: Find the deflection \delta_{C} at the end of the overhang (Fig. 9-15b) by substituting x = 3L / 2 in Eq. (9-54b):
\delta_{C}=-v\left(\frac{3 L}{2}\right)=\frac{P L^{3}}{8 E I} (9-55)
4. Finalize: The required deflections of the overhanging beam [Eqs. (9-54) and (9-55)] were obtained by solving the third-order differential equation of the deflection curve. The deflection curve is shown in Fig. 9-15b.
