Question 9.18: A simple beam with an overhang supports a uniform load of in...
A simple beam with an overhang supports a uniform load of intensity q on span AB and a concentrated load P at end C of the overhang (Fig. 9-42).
Determine the deflection \delta_{C} and angle of rotation \theta_{C} at point C. (Use the modified form of Castigliano’s theorem.)
Deflection \delta_{C} at the end of the overhang (Fig. 9-42b). Since the load P corresponds to this deflection, we do not need to supply a fictitious load. Instead, we can begin immediately to find the bending moments throughout the length of the beam. The reaction at support A is
R_{A}=\frac{qL}{2}-\frac{P}{2}as shown in Fig. 9-43. Therefore, the bending moment in span AB is
M_{AB}=R_{A}x_{1}-\frac{qx^{2}_{1}}{2}=\frac{qLx_{1}}{2}-\frac{Px_{1}}{2}-\frac{qx^{2}_{1}}{2} (0\leq x_{1}\leq L)where x_{1} is measured from support A (Fig. 9-43). The bending moment in the overhang is
M_{BC}=-Px_{2} (0\leq x_{2}\leq \frac{L}{2})where x_{2} is measured from point C (Fig. 9-43).
Next, we determine the partial derivatives with respect to the load P:
\frac{\partial M_{BC} }{\partial P}=-x_{2} (0\leq x_{2}\leq \frac{L}{2})
Now we are ready to use the modified form of Castigliano’s theorem (Eq. 9-88) to obtain the deflection at point C:
\delta_{i}=\frac{\partial}{\partial P _{i} }\int{\frac{M^{2}dx }{2EI} } =\int{\left(\frac{M}{EI}\right)}\left(\frac{\partial M}{\partial P_{i}} \right)dx (9-88)
\delta_{C}=\int{\left(\frac{M}{EI}\right)}\left(\frac{\partial M}{\partial P} \right)dx
=\frac{1}{EI}\int_{0}^{L}{M_{AB}\left(\frac{\partial M_{AB} }{\partial P} \right)dx}+\frac{1}{EI}\int_{0}^{L/2}{M_{BC}\left(\frac{\partial M_{BC} }{\partial P} \right)dx}
Substituting the expressions for the bending moments and partial derivatives, we get
\delta_{C}=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_{1}}{2}-\frac{Px_{1}}{2}-\frac{qx^{2}_{1}}{2} \right)\left(-\frac{x_{1} }{2} \right) dx_{1}}+\frac{1}{EI}\int_{0}^{L/2}{(-Px_{2} )(-x_{2} )dx_{2} }By performing the integrations and combining terms, we obtain the deflection:
\delta_{C}=\frac{PL^{3}}{8EI}-\frac{qL^{4}}{48EI} (9-89)
Since the load P acts downward, the deflection \delta_{C} is also positive downward. In other words, if the preceding equation produces a positive result, the deflection is downward. If the result is negative, the deflection is upward.
Comparing the two terms in Eq. (9-89), we see that the deflection at the end of the overhang is downward when P > qL/6 and upward when P < qL/6.
Angle of rotation \theta_{C} at the end of the overhang (Fig. 9-42b). Since there is no load on the original beam (Fig. 9-42a) corresponding to this angle of rotation, we must supply a fictitious load. Therefore, we place a couple of moment M_{C} at point C (Fig. 9-44). Note that the couple M_{C} acts at the point on the beam where the angle of rotation is to be determined. Furthermore, it has the same clockwise direction as the angle of rotation (Fig. 9-42).
We now follow the same steps as when determining the deflection at C. First, we note that the reaction at support A (Fig. 9-44) is
Consequently, the bending moment in span AB becomes
M_{AB}=R_{A}x_{1}-\frac{qx^{2}_{1}}{2}=\frac{qLx_{1} }{2}- \frac{Px_{1}}{2}-\frac{M_{C}x_{1} }{L} -\frac{qx^{2}_{1}}{2} (0\leq x_{1}\leq L)Also, the bending moment in the overhang becomes
M_{BC}=-Px_{2}-M_{C} (0\leq x_{2}\leq \frac{L}{2})The partial derivatives are taken with respect to the moment M_{C}, which is the load corresponding to the angle of rotation. Therefore,
\frac{\partial M_{AB} }{\partial M_{C} }=-\frac{x_{1} }{L} (0\leq x_{1}\leq L)\frac{\partial M_{BC} }{\partial M_{C} }=-1 \left(0\leq x_{2}\leq \frac{L}{2}\right)
Now we use the modified form of Castigliano’s theorem (Eq. 9-88) to obtain the angle of rotation at point C:
\theta_{C}=\int{\left(\frac{M}{EI}\right)}\left(\frac{\partial M}{\partial M_{C}} \right)dx=\frac{1}{EI}\int_{0}^{L}{M_{AB}\left(\frac{\partial M_{AB} }{\partial M_{C}} \right)dx}+\frac{1}{EI}\int_{0}^{L/2}{M_{BC}\left(\frac{\partial M_{BC} }{\partial M_{C}} \right)dx}
Substituting the expressions for the bending moments and partial derivatives, we obtain
\theta_{C}=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_{1}}{2}-\frac{Px_{1}}{2}-\frac{M_{C}x_{1}}{L}-\frac{qx^{2}_{1}}{2} \right)\left(-\frac{x_{1} }{L} \right) dx_{1}}+\frac{1}{EI}\int_{0}^{L/2}{(-Px_{2}-M_{C} )(-1)dx_{2} }Since M_{C} is a fictitious load, and since we have already taken the partial derivatives, we can set M_{C} equal to zero at this stage of the calculations and simplify the integrations:
\theta_{C}=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_{1}}{2}-\frac{Px_{1}}{2}-\frac{qx^{2}_{1}}{2} \right)\left(-\frac{x_{1} }{2} \right) dx_{1}}+\frac{1}{EI}\int_{0}^{L/2}{(-Px_{2} )(-1 )dx_{2} }After carrying out the integrations and combining terms, we obtain
\theta_{C}=\frac{7PL^{2}}{24EI}-\frac{qL^{3}}{24EI} (9-90)
If this equation produces a positive result, the angle of rotation is clockwise. If the result is negative, the angle is counterclockwise.
Comparing the two terms in Eq. (9-90), we see that the angle of rotation is clockwise when P > qL/7 and counterclockwise when P < qL/7.
If numerical data are available, it is now a routine matter to substitute numerical values into Eqs. (9-89) and (9-90) and calculate the deflection and angle of rotation at the end of the overhang.
