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 δ_C and angle of rotation θ_C at point C. (Use the modified form of Castigliano’s theorem.)
Deflection δ_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_Ax_1\ -\ \frac{qx_1^2}{2}=\frac{qLx_1}{2}\ -\ \frac{Px_1}{2}\ -\ \frac{qx_1^2}{2} \quad \quad (0≤x_1≤L)
where x_1 is measured to the right from support A (Fig. 9-43). The bending moment in the overhang is
M_{BC}=-Px_2\quad \quad \left(0≤x_2≤\frac{L}{2}\right)
where x_2 is measured to the left from point C (Fig. 9-43).
Next, we determine the partial derivatives with respect to the load P:
\frac{∂M_{AB}}{∂P}=-\frac{x_1}{2}\quad \quad (0≤x_1≤L) \\ \frac{∂M_{BC}}{∂P}=-x_2\quad \quad \left(0≤x_2≤\frac{L}{2}\right)
Now that we have obtained expressions for the bending moments and partial derivatives, we can use the modified form of Castigliano’s theorem (Eq. 9-88) to obtain the deflection at point C:
δ_i=\frac{∂}{∂P_i}\int{\frac{M^2dx}{2EI}=\int{\left(\frac{M}{EI}\right)\left(\frac{∂M}{∂P_i}\right)dx}} (9-88)
δ_C=\int{\left(\frac{M}{EI}\right)\left(\frac{∂M}{∂P}\right)dx}\\ \ \quad = \frac{1}{EI}\int_{0}^{L}{M_{AB}\left(\frac{∂M_{AB}}{∂P}dx\right)}+ \frac{1}{EI}\int_{0}^{L/2}{M_{BC}\left(\frac{∂M_{BC}}{∂P}\right)dx}
Substituting the expressions for the bending moments and partial derivatives, we get
δ_C=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_1}{2}\ -\ \frac{Px_1}{2}\ -\ \frac{qx_1^2}{2}\right)dx}+\frac{1}{EI}\int_{0}^{L/2}{(-Px_2)(-x_2)dx_2}
By performing the integrations and combining terms, we get
δ_C=\frac{PL^3}{8EI}\ -\ \frac{qL^4}{48EI} (9-89)
Since the load P acts downward, the deflection δ_C is 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 qL > 6P.
Angle of rotation dc 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.
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
R_A=\frac{qL}{2}\ -\ \frac{P}{2}\ -\ \frac{M_C}{L}
Consequently, the bending moment in span AB becomes
M_{AB}=R_Ax_1\ -\ \frac{qx_1^2}{2}=\frac{qLx_1}{2}\ -\ \frac{Px_1}{2}\ -\ \frac{M_Cx_1}{L}\ -\ \frac{qx_1^2}{2} \quad \quad (0≤x_1≤L)
and the bending moment in the overhang becomes
M_{BC}=-Px_2\ -\ M_C\quad \quad \left(0≤x_2≤\frac{L}{2}\right)
The partial derivatives are taken with respect to the moment M_C, which is the load corresponding to the angle of rotation. Therefore,
\frac{∂M_{AB}}{∂M_{C}}=-\frac{x_1}{L}\quad \quad (0≤x_1≤L) \\ \frac{∂M_{BC}}{∂M_{C}}=-1\quad \quad \left(0≤x_2≤\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:
θ_C=\int{\left(\frac{M}{EI}\right)\left(\frac{∂M}{∂M_C}\right)dx}\\ \ \quad = \frac{1}{EI}\int_{0}^{L}{M_{AB}\left(\frac{∂M_{AB}}{∂M_C}\right)dx}+ \frac{1}{EI}\int_{0}^{L/2}{M_{BC}\left(\frac{∂M_{BC}}{∂M_C}\right)dx}Substituting the expressions for the bending moments and partial derivatives, we obtain
θ_C=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_1}{2}\ -\ \frac{Px_1}{2}\ -\ \frac{M_Cx_1}{L}\ -\ \frac{qx_1^2}{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:
θ_C=\frac{1}{EI}\int_{0}^{L}{\left(\frac{qLx_1}{2}\ -\ \frac{Px_1}{2}\ -\ \frac{qx_1^2}{2}\right)\left(-\frac{x_1}{L}\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
θ_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 qL > 7P.
If numerical data are available, it is now a routine matter to substitute numbers into Eqs. (9-89) and (9-90) and calculate the deflection and angle of rotation at the end of the overhang.
