# Problem 705 | Solution of Propped Beam with Increasing Load

**Problem 705**

Find the reaction at the simple support of the propped beam shown in Fig. P-705 and sketch the shear and moment diagrams.

**Solution of Propped Reaction by Double Integration Method**

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$y = \dfrac{w_ox}{L}$

Moment at x:

$M = R_Ax - \frac{1}{2}xy(\frac{1}{3}x)$

$M = R_Ax - \frac{1}{6}x^2y$

$M = R_Ax - \dfrac{x^2}{6}\left( \dfrac{w_ox}{L} \right)$

$M = R_Ax - \dfrac{w_ox^3}{6L}$

Thus,

$EI \, y'' = R_Ax - \dfrac{w_ox^3}{6L}$

$EI \, y' = \dfrac{R_Ax^2}{2} - \dfrac{w_ox^4}{24L} + C_1$

$EI \, y = \dfrac{R_Ax^3}{6} - \dfrac{w_ox^5}{120L} + C_1x + C_2$

At x = 0, y = 0, thus C_{2} = 0

At x = L, y’ = 0

$0 = \dfrac{R_AL^2}{2} - \dfrac{w_oL^4}{24L} + C_1$

$C_1 = \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2}$

Thus, the deflection equation is

$EI \, y = \dfrac{R_Ax^3}{6} - \dfrac{w_ox^5}{120L} + \left( \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2} \right)x$

At x = L, y = 0

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^5}{120L} + \left( \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2} \right)L$

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^4}{120} + \dfrac{w_oL^4}{24} - \dfrac{R_AL^3}{2}$

$0 = -\dfrac{R_AL^3}{3} + \dfrac{w_oL^4}{30}$

$\dfrac{R_AL^3}{3} = \dfrac{w_oL^4}{30}$

$R_A = \dfrac{w_oL}{10}$ *answer*

**Solution of Propped Reaction by the Method of Superposition**

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The deflection at A is zero. Thus, by superposition method, the deflection due to triangular load is equal to the deflection due to concentrated load.

$\delta_1 = \dfrac{w_o L^4}{30EI}$ → deflection due to triangular load

$\delta_2 = \dfrac{R_A L^3}{3EI}$ → deflection due to concentrated load

$\delta_1 = \delta_1$

$\dfrac{w_o L^4}{30EI} = \dfrac{R_A L^3}{3EI}$

$R_A = \dfrac{w_o L}{10}$ *answer*

**Sketching the Shear and Moments Diagrams**

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$R_A = \dfrac{w_oL}{10}$

Solving for reaction at B, R_{B}

$\Sigma F_V = 0$

$R_A + R_B = \frac{1}{2}Lw_o$

$\dfrac{w_oL}{10} + R_B = \dfrac{w_oL}{2}$

$R_B = \dfrac{2w_oL}{5}$ *answer*

Solving for moment at B, M_{B}

$M_B = R_AL – \frac{1}{2}Lw_o(\frac{1}{3}L)$

$M_B = \left( \dfrac{w_oL}{10} \right) - \dfrac{w_oL^2}{6}$

$M_B = \dfrac{w_oL^2}{10} - \dfrac{w_oL^2}{6}$

$M_B = -\dfrac{w_oL^2}{15}$ *answer*

**To Draw the Shear Diagram**

- The shear at A is equal to R
_{A} - The load between AB is negatively increasing from zero at A to w
_{o}at B, thus, the slope of the shear diagram between A and B is decreasing from zero at A to -w_{o}at B. - The load between AB is linear, thus, the shear diagram between AB is a parabola (2
^{nd}degree curve) with vertex at A and open downward as stated with the decreasing slope in number 2. - The shear at B is equal to -R
_{B}. See the magnitude of R_{B}in the solution above. To compute; shear at B = shear at A - load between AB. - The shear diagram between AB is zero at C as shown. Location of C, denoted by x
_{C}can be found by squared property of parabola as follows.$\dfrac{{x_C}^2}{R_A} = \dfrac{L^2}{R_A + R_B}$

${x_C}^2 = \dfrac{R_AL^2}{R_A + R_B}$

${x_C}^2 = \dfrac{(\frac{1}{10}w_oL)L^2}{\frac{1}{10}w_oL + \frac{2}{5}w_oL}$

${x_C}^2 = \dfrac{\frac{1}{10}w_oL^3}{\frac{1}{2}w_oL}$

${x_C}^2 = \frac{1}{5}L^2$

$x_C = \frac{1}{\sqrt{5}}L$ from left support

**To Draw the Moment Diagram**

- The shear diagram from A to B is a 2nd degree curve, thus, the moment diagram between A and B is a third degree curve.
- The moment at A is zero.
- Moment at C is equal to the moment A plus the area of shear diagram between A and C.
$M_C = \frac{2}{3}x_CR_A$

$M_C = \frac{2}{3}(\frac{1}{\sqrt{5}}L)(\frac{1}{10}w_oL)$

$M_C = \dfrac{w_oL^2}{15\sqrt{5}}$

- The moment at B is equal to -M
_{B}, see the magnitude of M_{B}from the solution above. It is more easy to compute the moment at B by using the load diagram instead of shear diagram. In case, you need to solve the moment at B by the use of shear diagram; M_{B}= M_{C}- Area of shear between CB. You can follow the link for an example of finding the area of shear diagram of similar shape. - The moment is zero at point D. To locate this point, equate the moment equation developed in double integration method to zero.
$M = R_Ax – \dfrac{w_ox^3}{6L}$ See double integration method above for finding M.

$M = \dfrac{w_oLx}{10} – \dfrac{w_ox^3}{6L}$

At D, x = x

_{D}and M = 0$0 = \dfrac{w_oLx_D}{10} – \dfrac{w_o{x_D}^3}{6L}$

$\dfrac{w_o{x_D}^3}{6L} = \dfrac{w_oLx_D}{10}$

$\dfrac{{x_D}^2}{6L} = \dfrac{L}{10}$

${x_D}^2 = \dfrac{3L^2}{5}$

$x_D = \dfrac{\sqrt{3}L}{\sqrt{5}} = \dfrac{\sqrt{3}L}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}$

$x_D = \frac{\sqrt{15}}{5}L$ from left support