Problem 705 | Solution of Propped Beam with Increasing Load

$\dfrac{y}{x} = \dfrac{w_o}{L}$

$y = \dfrac{w_ox}{L}$
 

 

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 C₂ = 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

Resolve the propped beam into two cantilever beams, one with uniformly varying load and the other with concentrated load as shown below. The concentrated load is the reaction at A.
 

 

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

From the solutions above,
$R_A = \dfrac{w_oL}{10}$

 
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

1. The shear at A is equal to R_A
2. 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.
3. 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.
4. 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.
5. 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

1. 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.
2. The moment at A is zero.
3. Moment at C is equal to the moment A plus the area of shear diagram between A and C.

   $M_C = \frac{2}{3}x_CR_A$

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

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

4. 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.
5. 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