Question 15.6: For the beam and loading shown, determine the reaction at th...
For the beam and loading shown, determine the reaction at the fixed support C.
STRATEGY: The beam is statically indeterminate to the second degree.
Strategically selecting the reactions at C as redundants, you can use the method of superposition and model the given problem by using a summation of load cases for which deflection formulae are readily available.
MODELING: Assuming the axial force in the beam to be zero, the beam ABC is indeterminate to the second degree, and we choose two reaction components as redundants: the vertical force \pmb{R}_C and the couple \pmb{M}_C. The deformations caused by the given load P, the force \pmb{R}_C , and the couple \pmb{M}_C are considered separately, as shown in Fig. 1.
ANALYSIS: For each load, the slope and deflection at point C is found by using the table of Beam Deflections and Slopes in Appendix C.
Appendix C Beam Deflections and Slopes
| Equation of Elastic Curve | Slope at End | Maximum Deflection | Elastic Curve | Beam and Loading |
| y=\frac{P}{6EI}(x^3-3Lx^2) | -\frac{PL^2}{2EI} | -\frac{PL^3}{3EI} |
|
1
|
| y=-\frac{w}{24EI}(x^4-4Lx^3+6L^2x^2) | -\frac{wL^3}{6EI} | -\frac{wL^4}{8EI} |
|
2
|
| y=-\frac{M}{2EI}x^2 | -\frac{ML}{EI} | -\frac{ML^2}{2EI} |
|
3
|
| \text{For }x\leq \frac{1}{2}L: \\ y=\frac{P}{48EI}(4x^3-3L^2x) | \pm \frac{PL^2}{16EI} | -\frac{PL^3}{48EI} |
|
4
|
| \text{For }x\lt a: \\ y=\frac{Pb}{6EIL}[x^3-(L^2-b^2)x] \\ \text{For }x=a: \ y=-\frac{Pa^2b^2}{3EIL} | \theta_A=-\frac{Pb(L^2-b^2)}{6EIL} \\ \theta_B=+\frac{Pa(L^2-a^2)}{6EIL} | \text{For }a\gt b: \\ -\frac{Pb(L^2-b^2)^{3/2}}{9\sqrt{3}EIL} \\ \text{at }x_m=\sqrt{\frac{L^2-b^2}{3} } |
|
5
|
| y=-\frac{w}{24EI}(x^4-2Lx^3+L^3x) | \pm \frac{wL^3}{24EI} | -\frac{5wL^4}{384EI} |
|
6
|
| y=-\frac{M}{6EIL}(x^3-L^2x) | \theta_A=+\frac{ML}{6EI} \\ \theta_B=-\frac{ML}{3EI} | \frac{ML^2}{9\sqrt{3}EI} |
|
7
|
Load P. For this load, portion BC of the beam is straight.
(\theta_C)_P=(\theta_B)_P=-\frac{Pa^2}{2EI} \quad \quad (y_C)_P=(y_B)_P+(\theta_B)_Pb \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac{Pa^3}{3EI}-\frac{Pa^2}{2EI}b=-\frac{Pa^2}{6EI}(2a+3b)
Force \pmb{R_C} (\theta_C)_R=+\frac{R_CL^2}{2EI} \quad \quad \quad (y_C)_R=+\frac{R_CL^2}{3EI}
Couple \pmb{M_C} (\theta_C)_M=+\frac{M_CL}{EI} \quad \quad \quad (y_C)_M=+\frac{M_CL^2}{2EI}
Boundary Conditions. At end C, the slope and deflection must be zero:
\begin{matrix} \left[x=L,\theta_C=0\right]: && \theta_C=(\theta_C)_P+(\theta_C)_R+(\theta_C)_M \\ &&0=-\frac{Pa^2}{2EI}+\frac{R_CL^2}{2EI}+\frac{M_CL}{EI} && \pmb{(1)} \\ \left[x=L,y_C=0\right]: && y_C=(y_C)_P+(y_C)_R+(y_C)_M \\ && 0=-\frac{Pa^2}{6EI}(2a+3b)+\frac{R_CL^3}{3EI}+\frac{M_CL^2}{2EI} && \pmb{(2)} \end{matrix}
Reaction Components at C. Solve Eqs. (1) and (2) simultaneously:
R_C=+\frac{Pa^2}{L^3}(a+3b) \quad \quad \pmb{R}_C=\frac{Pa^2}{L^3}(a+3b)\uparrow \\ M_C=-\frac{Pa^2b}{L^2} \quad \quad \quad \quad \quad \pmb{M}_C=\frac{Pa^2b}{L^2}\circlearrowright
The methods of statics are used to determine the reaction at A, shown in Fig. 2.
REFLECT and THINK: Note that an alternate strategy that could have been used in this particular problem is to treat the couple reactions at the ends as redundant. The application of superposition would then have involved a simply-supported beam, for which deflection formulae are also readily available.
