Question 15.3: For the uniform beam AB (a) determine the reaction at A, (b)...
For the uniform beam AB (a) determine the reaction at A, (b) derive the equation of the elastic curve, (c) determine the slope at A. (Note that the beam is statically indeterminate to the first degree.)
STRATEGY: The beam is statically indeterminate to the first degree.
Treating the reaction at A as the redundant, write the bending-moment equation as a function of this redundant reaction and the existing load. After substituting the bending-moment equation into the differential equation of the elastic curve, integrating twice, and applying the boundary conditions, the reaction can be determined. Use the equation for the elastic curve to find the desired slope.
MODELING: Using the free body shown in Fig. 1, obtain the bending moment diagram:
+\circlearrowright \sum{M_D=0:} \quad \quad R_Ax-\frac{1}{2}\left(\frac{w_0x^2}{L} \right) \frac{x}{3}-M=0 \quad \quad M=R_Ax-\frac{w_0x^3}{6L}
ANALYSIS:
Differential Equation of the Elastic Curve. Use Eq. (15.4) for
\frac{d^2y}{dx^2}=\frac{M(x)}{EI} \quad \quad \quad \pmb{(15.4)} \\ EI\frac{d^2y}{dx^2}=R_Ax-\frac{w_0x^3}{6L}
Noting that the flexural rigidity EI is constant, integrate twice and find
EI\frac{dy}{dx} =EI \ \theta=\frac{1}{2}R_Ax^2-\frac{w_0x^4}{24L}+C_1 \quad \quad \quad \pmb{(1)} \\ EI \ y=\frac{1}{6}R_Ax^3-\frac{w_0x^5}{120L}+C_1x +C_2 \quad \quad \quad \pmb{(2)}
Boundary Conditions. The three boundary conditions that must be satisfied are shown in Fig. 2.
[x=0,y=0]: \quad \quad C_2=0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \pmb{(3)} \\ [x=L, \theta=0]: \quad \quad \frac{1}{2}R_AL^2-\frac{w_0L^3}{24}+C_1=0 \quad \quad \quad \quad \quad \quad \pmb{(4)} \\ [x=L,y=0]:\quad \quad \frac{1}{6}R_AL^3 -\frac{w_0L^4}{120}+C_1L+C_2=0 \quad \quad \quad \pmb{(5)}a. Reaction at A. Multiplying Eq. (4) by L, subtracting Eq. (5) member by member from the equation obtained, and noting that C_2 = 0, give
\frac{1}{3}R_AL^3-\frac{1}{30}w_0L^4=0 \quad \quad \quad \pmb{R}_A=\frac{1}{10}w_0L\uparrow
The reaction is independent of E and I. Substituting R_A=\frac{1}{10}w_0L into Eq. (4),
\frac{1}{2}\left(\frac{1}{10}w_0L \right) L^2-\frac{1}{24}w_0L^3+C_1=0 \quad \quad \quad C_1=-\frac{1}{120}w_0L^3
b. Equation of the Elastic Curve. Substituting for R_A, C_1, \text{ and } C_2 into Eq. (2),
EI \ y=\frac{1}{6}(\frac{1}{10}w_0L )x^3-\frac{w_0x^5}{120L}-\left(\frac{1}{120}w_0L^3 \right)x
y=\frac{w_0}{120EIL}(-x^5+2L^2x^3-L^4x)
c. Slope at A (Fig. 3). Differentiate the equation of the elastic curve with respect to x :
\theta=\frac{dy}{dx}=\frac{w_0}{120EIL}(-5x^4+6L^2x^2-L^4)
Making x = 0, \theta_A=-\frac{w_0L^3}{120EI} \quad \quad \quad \theta_A=\frac{w_0L^3}{120EI}</span></span>\measuredangle
