Question 9.18: A simple beam with an overhang supports a uniform load of an......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Deflection \delta_C at the end of overhang.
1, 2. Conceptualize, Categorize: Since the load P corresponds to this deflection (Fig. 9-42), a fictitious load is not needed. Instead, begin immediately to find the bending moments throughout the length of the beam. The reaction at support A is
\quad\quad R_{A}={\frac{q L}{2}}-{\frac{P}{2}}
as shown in Fig. 9-43. Therefore, the bending moment in span AB is
\quad\quad\quad\quad M_{A B}=R_{A}x_{1}-{\frac{q x_{1}^{2}}{2}}={\frac{q L x_{1}}{2}}-{\frac{P x_{1}}{2}}-{\frac{q x_{1}^{2}}{2}}~~(0≤ x_{1}≤ L)
where x_1 is measured from support A (Fig. 9-43). The bending moment in the overhang is
\quad\quad\quad\quad M_{B C}=-P x_{2} \quad {\Bigg\lgroup}0\leq x_{2}\leq{\frac{L}{2}}{\Bigg\rgroup}
where x_2 is measured from point C (Fig. 9-43).
Next, determine the partial derivatives with respect to the load P:
\quad\quad\quad\quad {\frac{\partial M_{A B}}{\partial P}}=-{\frac{x_{1}}{2}}\;\;\;(0\leq x_{1}\leq L) \\ {\frac{\partial M_{BC}}{\partial P}}=-{{x_{2}}}\;\;\;{\Bigg\lgroup}0\leq x_{1}\leq \frac{L}{2}{\Bigg\rgroup}
3. Analyze: Now use the modified form of Castigliano’s theorem [Eq. (9-118)] to obtain the deflection at point C:

\quad\quad\quad\quad \delta_{C}\ =\int {\Bigg\lgroup}\frac{M}{E{I}}{\Bigg\rgroup}{\Bigg\lgroup}\frac{\partial M}{\partial P}{\Bigg\rgroup}d x \\  \quad\quad\quad\quad =\frac{1}{E I}{\int_{0}^{L}}\,M_{A B}{\Bigg\lgroup}\frac{{\partial}M_{A B}}{\partial P}{\Bigg\rgroup}d x+\frac{1}{E I}\int_{0}^{L /{2}}M_{B C}\left(\frac{{\partial}M_{B C}}{\partial P}\right)d\alpha
Substitute the expressions for the bending moments and partial derivatives to get
\quad\quad\quad\quad \delta_{C}=\frac{1}{E I}\int_{0}^{L}\!{\Bigg\lgroup}\frac{q L\,x_{1}}{2}-\,\frac{P\,x_{1}}{2}\,-\,\frac{q\,x_{1}^2}{2}{\Bigg\rgroup}\!{\Bigg\lgroup}-\frac{x_{1}}{2}{\Bigg\rgroup}d x_{1}\,+\,\frac{1}{E I}\int_{0}^{L/2}(-P x_{2})(-x_{2})\,d x_{2}
Perform the integrations and combining terms to obtain the deflection:
\quad\quad\quad\quad \delta_{C}={\frac{P{{L}}^{3}}{8E I}}-{\frac{q{{L}}^{4}}{48E I}}\quad\quad\quad(9-120)
4. Finalize: 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.
Compare the two terms in Eq. (9-120) to see that the deflection at the end of the overhang is downward when P > qL / 6 and upward when P < qL / 6.
Part (a): Angle of rotation \theta_C at the end of overhang.
1, 2. Conceptualize, Categorize: Since there is no load on the original beam (Fig. 9-42a) corresponding to this angle of rotation (Fig. 9-42b), supply a fictitious load. 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).
Follow the same s teps as when determining the deflection at C. First, note that the reaction at support A (Fig. 9-44) is
\quad\quad\quad\quad {R_{A}}=\frac{q L}{2}-\frac{P}{2}-\frac{M_{C}}{L}
Consequently, the bending moment in span AB becomes
\quad\quad\quad\quad M_{A B}=R_{A}x_{1}\,-\,{\frac{q x_{1}^{2}}{2}}\,=\,{\frac{q L x_{1}}{2}}\,-\,{\frac{P x_{1}}{2}}\,-\,{\frac{M_{C}x_{1}}{L}}\,-\,{\frac{q x_{1}^{2}}{2}}\;\;\;(\,0\leq\,x_{1}\leq L)
Also, the bending moment in the overhang becomes
\quad\quad\quad\quad M_{B C}=-P x_{2}-M_{C}\quad{\Bigg\lgroup}0\leq x_{2}\leq{\frac{L}{2}}{\Bigg\rgroup}
The partial derivatives are taken with respect to the moment M_C, which is the load corresponding to the angle of rotation. Therefore,
\quad\quad\quad\quad \frac{\partial M_{A B}}{\partial M_{C}}=-\frac{x_{1}}L\ \ (0\leq x_{1}\leq L) \\ \quad\quad\quad\quad \frac{\partial M_{BC}}{\partial M_{C}}=-1 \quad\quad \ \ {\Bigg\lgroup}0\leq x_{2}\leq \frac{L}{2}{\Bigg\rgroup}
3. Analyze: Now use the modified form of Castigliano’s theorem [Eq. (9-118)] to obtain the angle of rotation at point C:
\quad\quad\quad\quad \theta_{C}\;\;=\;\;\int{\Bigg\lgroup}{\frac{M}{E{I}}}{\Bigg\rgroup}{\Bigg\lgroup}{\frac{\partial M}{\partial M_{C}}}{\Bigg\rgroup}d x \\ \quad\quad\quad\quad =\frac{1}{E I}\int_{0}^{L}M_{A B}{\Bigg\lgroup}\frac{\partial M_{A B}}{\partial M_{C}}{\Bigg\rgroup}d x\,+\,\frac{1}{E I}\,\int_{0}^{{L}/{2}} M_{B C}{\Bigg\lgroup}\frac{\partial M_{B C}}{\partial M_{C}}{\Bigg\rgroup}d x
Substitute the expressions for the bending moments and partial derivatives to obtain
\quad\quad\quad\quad \theta_{C}=\frac{1}{E I}{\int}\left(\frac{q L x_{1}}{2}-\frac{P x_{1}}{2}-\frac{M_{c}x_{1}}{L}-\frac{q x_{1}^{~2}}{2}\right)\left(-\frac{x_{1}}{L}\right)d x_{1} \\ \quad\quad\quad\quad + \frac{1}{EI}\int_{0}^{L/2}(-Px_2 – M_C)(-1)dx_2
Since M_C is a fictitious load, and since the partial derivatives have already been taken, set M_C equal to zero at this stage of the calculations and simplify the integrations:
\quad\quad\quad\quad \theta_{C}={\frac{1}{E I}}{\int_{0}^{L}}{\Bigg\lgroup}{\frac{q L x_{1}}{2}}-{\frac{P x_{1}}{2}}-{\frac{q x_{1}^{2}}{2}}{\Bigg\rgroup}{\Bigg\lgroup}-{\frac{x_{1}}{L}}{\Bigg\rgroup}d x_{1}+{\frac{1}{E I}}\int_{0}^{L/2}(-P x_{2})(-1)\,d x_{2}
After carrying out the integrations and combining terms, the rotation at C is
\quad\quad\quad\quad \theta_{C}={\frac{7PL^{2}}{24E I}}-{\frac{q L^{3}}{24E I}}\quad\quad (9-121)
4. Finalize: 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. (9-121), note 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. (9-120) and (9-121) and calculate the deflection and angle of rotation at the end of the overhang.