# Problem 706 | Solution of Propped Beam with Decreasing Load

**Example 03**

The propped beam shown in Fig. P -706 is loaded by decreasing triangular load varying from w_{o} from the simple end to zero at the fixed end. Find the support reactions and sketch the shear and moment diagrams

**Solution by Double Integration Method**

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

$y = \dfrac{w_o}{L}(L - x)$

Solving for moment equation

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

$M = R_Ax - \frac{1}{3}w_ox^2 - \frac{1}{6}x^2y$

$M = R_Ax - \frac{1}{3}w_ox^2 - \frac{1}{6}x^2 \left[ \dfrac{w_o}{L}(L - x) \right]$

$M = R_Ax - \dfrac{w_o}{3}x^2 - \dfrac{w_o}{6L}(Lx^2 - x^3)$

Doing the double integration

$EI \, y'' = R_Ax - \dfrac{w_o}{3}x^2 - \dfrac{w_o}{6L}(Lx^2 - x^3)$

$EI \, y' = \dfrac{R_A}{2}x^2 - \dfrac{w_o}{9}x^3 - \dfrac{w_o}{6L} \left( \dfrac{Lx^3}{3} - \dfrac{x^4}{4} \right) + C_1$

$EI \, y = \dfrac{R_A}{6}x^3 - \dfrac{w_o}{36}x^4 - \dfrac{w_o}{6L} \left( \dfrac{Lx^4}{12} - \dfrac{x^5}{20} \right) + C_1x + C_2$

**Boundary conditions**

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

At x = L, y = 0

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^4}{36} - \dfrac{w_o}{6L} \left( \dfrac{L^4}{12} - \dfrac{L^4}{20} \right) + C_1L + 0$

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^4}{36} - \dfrac{w_oL^4}{180} + C_1L$

$C_1 = \dfrac{w_oL^3}{30} - \dfrac{R_AL^2}{6}$

At x = L, y' = 0

$0 = \dfrac{R_AL^2}{2} - \dfrac{w_oL^3}{9} - \dfrac{w_o}{6L} \left( \dfrac{L^4}{3} - \dfrac{L^4}{4} \right) + \left( \dfrac{w_oL^3}{30} - \dfrac{R_AL^2}{6} \right)$

$0 = \dfrac{R_AL^2}{2} - \dfrac{w_oL^3}{9} - \dfrac{w_oL^3}{72} + \dfrac{w_oL^3}{30} - \dfrac{R_AL^2}{6}$

$\dfrac{R_AL^2}{3} = \dfrac{11w_oL^3}{120}$

$R_A = \dfrac{11w_oL}{40}$ *answer*

**Solution by Superposition Method**

From the figure, it is clear that

$\delta_1 + \delta_3 = \delta_2$

$\dfrac{R_AL^3}{3EI} + \dfrac{w_oL^4}{30EI} = \dfrac{w_oL^4}{8EI}$

$\dfrac{R_A}{3} + \dfrac{w_oL}{30} = \dfrac{w_oL}{8}$

$\dfrac{R_A}{3} = \dfrac{11w_oL}{120}$

$R_A = \dfrac{11w_oL}{40}$ *answer*

**Superposition Method Using Point Load and Integration**

For this problem $P = y \, dx$, $a = L - x$, and $\delta = d \delta_2$.

$d\delta_2 = \dfrac{y\,dx(L-x)^2}{6EI}[ \, 3L - (L - x) \, ]$

From the figure,

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

$y = \dfrac{w_o}{L}(L - x)$

Thus,

$d\delta_2 = \dfrac{\left[ \dfrac{w_o}{L}(L - x) \right](L - x)^2}{6EI}[ \, 3L - (L - x) \, ] \, dx$

$d\delta_2 = \dfrac{w_o(L - x)^3}{6L \, EI}[ \, 3L - (L - x) \, ] \, dx$

$d\delta_2 = \dfrac{w_o}{2EI}(L - x)^3 \, dx - \dfrac{w_o}{6L \, EI}(L - x)^4 \, dx$

$\displaystyle \delta_2 = \dfrac{w_o}{2EI} \int_0^L (L - x)^3 \, dx - \dfrac{w_o}{6L \, EI} \int_0^L (L - x)^4 \, dx$

$\delta_1$ = end deflection due to R_{A}

$\delta_1 = \dfrac{R_AL^3}{3EI}$

$\delta_1 = \delta_2$

$\displaystyle \dfrac{R_AL^3}{3EI} = \dfrac{w_o}{2EI} \int_0^L (L - x)^3 \, dx - \dfrac{w_o}{6L \, EI} \int_0^L (L - x)^4 \, dx$

$\displaystyle R_A = \dfrac{3w_o}{2L^3} \int_0^L (L - x)^3 \, dx - \dfrac{w_o}{2L^4} \int_0^L (L - x)^4 \, dx$

$R_A = \dfrac{3w_o}{2L^3} \left[ -\dfrac{(L - x)^4}{4} \right] - \dfrac{w_o}{2L^4} \left[ -\dfrac{(L - x)^5}{5} \right]$

$R_A = -\dfrac{3w_o}{8L^3} \left[ 0^4 - L^4 \right] + \dfrac{w_o}{10L^4} \left[ 0^5 - L^5 \right]$

$R_A = \dfrac{3w_oL}{8} - \dfrac{w_oL}{10}$

$R_A = \dfrac{11w_oL}{40}$ *answer*

**Shear and Moment Diagrams**

_{A}was solved using two different methods above.

$R_A = \dfrac{11w_oL}{40}$

*answer*

Solving for vertical reaction at B

$\Sigma F_V = 0$

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

$\dfrac{11w_oL}{40} + R_B = \dfrac{w_oL}{2}$

$R_B = \dfrac{9w_oL}{40}$ *answer*

Solving for moment reaction at B

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

$M_B = \left( \dfrac{9w_oL}{40} \right)L – \dfrac{w_oL^2}{3}$

$M_B = \dfrac{9w_oL^2}{40} – \dfrac{w_oL^2}{3}$

$M_B = –\dfrac{7w_oL^2}{120}$ *answer*

**To Draw the Shear Diagram**

- The shear at A is equal to R
_{A} - The load at AB is increasing from -w
_{o}at A to zero at B, thus, the slope of shear diagram from A to B is also increasing from -w_{o}at A to zero at B. - The load between AB is 1st degree (linear), thus, the shear diagram between AB is 2nd degree (parabolic) with vertex at B and open upward.
- The magnitude of shear at B is equal to -R
_{B}. It is equal to the shear at A minus the triangular load between AB. See the magnitude of R_{B}above. - The shear diagram from A to B will become zero somewhere along AB, the point is denoted by C in the figure. To locate point C, two solutions are presented below.

**Location of C, the point of zero shear**

Point C is the location of zero shear which may also be the location of maximum moment.**By squared property of parabola**

$\dfrac{(L - x_C)^2}{R_B} = \dfrac{L^2}{R_A + R_B}$$(L - x_C)^2 = \dfrac{R_BL^2}{R_A + R_B}$

$(L - x_C)^2 = \dfrac{(\frac{9}{40}w_oL)L^2}{\frac{11}{40}w_oL + \frac{9}{40}w_oL}$

$(L - x_C)^2 = \dfrac{\frac{9}{40}w_oL^3}{\frac{1}{2}w_oL}$

$(L - x_C)^2 = \frac{9}{20}L^2$

$L - x_C = \pm \frac{3}{\sqrt{20}}L$

$x_C = \left(1 \pm \dfrac{3}{\sqrt{20}} \right)L$

$x_C = 1.6708L$ (

*absurd*)

$x_C = 0.3292L$*answer***By shear equation**

Another method of solving for x_{C}is to pass an exploratory section anywhere on AB and sum up all the vertical forces to the left of the exploratory section. The location of x_{C}is where the sum of all vertical forces equate to zero. Consider the figure shown to the right. Note that this figure is the same figure we used to find the reaction R_{A}by double integration method shown above. The double integration method shows the relationship of x and y which is $y = \dfrac{w_o}{L}(L - x)$.

Sum of all vertical forces

$\Sigma F_V = R_A - \frac{1}{2}w_ox - \frac{1}{2}xy$$\Sigma F_V = \dfrac{11w_oL}{40} - \dfrac{w_ox}{2} - \dfrac{w_ox}{2L}(L - x)$

$\Sigma F_V = \dfrac{11w_oL}{40} - \dfrac{w_ox}{2} - \dfrac{w_ox}{2} + \dfrac{w_ox^2}{2L}$

$\Sigma F_V = \dfrac{11w_oL}{40} - w_ox + \dfrac{w_ox^2}{2L}$

At point C, ΣF

_{V}= 0 and x = x_{C}

$0 = \dfrac{11w_oL}{40} - w_ox_C + \dfrac{w_o{x_C}^2}{2L}$$0 = \dfrac{11L}{40} - x_C + \dfrac{{x_C}^2}{2L}$

$0 = \frac{11}{20}L^2 - 2Lx_C + {x_C}^2$

${x_C}^2 - 2Lx_C = -\frac{11}{20}L^2$

${x_C}^2 - 2Lx_C + L^2= -\frac{11}{20}L^2 + L^2$

$(x_C - L)^2= \frac{9}{20}L^2$

$x_C - L= \pm \frac{3}{\sqrt{20}}L$

$x_C = \left(1 \pm \dfrac{3}{\sqrt{20}} \right)L$

$x_C = 1.6708L$ (

*absurd*)

$x_C = 0.3292L$*answer*

**To Draw the Moment Diagram**

- The moment at simple support A is zero.
- The shear diagram of AB is 2
^{nd}degree curve, thus, the moment diagram between AB is 3^{rd}degree curve. - The moment at C can be computed in two ways; (a) by solving the area of shear diagram between A and C, and (b) by using the moment equation. For method (a), see the following links for similar situation of solving a partial area of parabolic spandrel.
The solution below is using the approach mentioned in (b). From double integration method of solving R

_{A}, the moment equation is given by$M = R_Ax - \dfrac{w_o}{3}x^2 - \dfrac{w_o}{6L}(Lx^2 - x^3)$

$M = \dfrac{11w_oL}{40}x - \dfrac{w_o}{3}x^2 - \dfrac{w_o}{6L}(Lx^2 - x^3)$

For x = x

_{C}= 0.3292L, M = M_{C}$M_C = \dfrac{11w_oL}{40}(0.3292L) - \dfrac{w_o}{3}(0.3292L)^2 - \dfrac{w_o}{6L}[ \, L(0.3292L)^2 - (0.3292L)^3 \, ]$

$M_C = 0.0423w_oL^2$

- In the same manner of solving for M
_{C}, M_{B}can be found by using x = L. Thus,$M_B = \dfrac{11w_oL^2}{40} - \dfrac{w_oL^2}{3} - \dfrac{w_o}{6L}(L^3 - L^3)$

$M_B = -\dfrac{7w_oL^2}{120}$ which confirms the solution above for M

_{B}. - To locate the point of zero moment denoted by D in the figure, we will again use the moment equation; now with M = 0 and x = x
_{D}.$0 = \dfrac{11w_oL}{40}x_D - \dfrac{w_o}{3}{x_D}^2 - \dfrac{w_o}{6L}(L{x_D}^2 - {x_D}^3)$

$0 = \frac{11}{40}L - \dfrac{1}{3}x_D - \dfrac{x_D}{6L}(L - x_D)$

$0 = \frac{33}{20}L^2 - 2Lx_D - x_D(L - x_D)$

$0 = \frac{33}{20}L^2 - 3Lx_D + {x_D}^2$

${x_D}^2 - 3Lx_D = -\frac{33}{20}L^2$

${x_D}^2 - 3Lx_D + \frac{9}{4}L^2 = -\frac{33}{20}L^2 + \frac{9}{4}L^2$

$(x_D - \frac{3}{2}L)^2 = \frac{3}{5}L^2$

$x_D - \frac{3}{2}L = \pm \frac{3}{5}L$

$x_D = \left( \frac{3}{2} \pm \frac{3}{5} \right)L$

$x_D = 2.2746L$ (

*absurd*)

$x_D = 0.7254L$*answer*