# Problem 704 | Solution of Propped Beam

**Problem 704**

Find the reactions at the supports and draw the shear and moment diagrams of the propped beam shown in Fig. P-704.

**Solution by Double Integration Method**

$EI \, y'' = R_Ax – \frac{1}{2}w_o \langle x – a \rangle^2$

$EI \, y' = \frac{1}{2}R_Ax^2 – \frac{1}{6}w_o \langle x – a \rangle^3 + C_1$

$EI \, y = \frac{1}{6}R_Ax^3 – \frac{1}{24}w_o \langle x – a \rangle^4 + C_1x + C_2$

Apply boundary conditions to solve for integration constants C_{1} and C_{2}:

At $x = 0$, $y = 0$, hence $C_2 = 0$

At $x = L$, $y = 0$, hence

$0 = \frac{1}{6}R_AL^3 – \frac{1}{24}w_o (L – a)^4 + C_1L + 0$

$C_1L = \frac{1}{24}w_o (L – a)^4 - \frac{1}{6}R_AL^3$

$C_1 = \dfrac{w_o}{24L} (L – a)^4 - \dfrac{R_AL^2}{6}$

At $x = L$, $y' = 0$, hence

$0 = \frac{1}{2}R_AL^2 – \frac{1}{6}w_o (L – a)^3 + \dfrac{w_o}{24L} (L – a)^4 - \dfrac{R_AL^2}{6}$

$\dfrac{R_AL^2}{2} – \dfrac{w_o}{6}(L – a)^3 + \dfrac{w_o}{24L}(L – a)^4 - \dfrac{R_AL^2}{6} = 0$

$\dfrac{R_AL^2}{3} = \dfrac{w_o}{6}(L – a)^3 - \dfrac{w_o}{24L}(L – a)^4$

$R_A = \dfrac{w_o}{2L^2}(L – a)^3 - \dfrac{w_o}{8L^3}(L – a)^4$

$R_A = \dfrac{w_ob^3}{2L^2} - \dfrac{w_ob^4}{8L^3}$

$R_A = \dfrac{w_ob^3}{8L^3}(4L - b)$ *answer*

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**Solution by Superposition Method**

_{A}denoted by δ

_{1}is equal to the sum of deflection at B denoted as δ

_{2}and the vertical deflection at A due to rotation of B which is denoted by θ.

$\delta_1 = \delta_2 + a\theta$

$\dfrac{R_AL^3}{3EI} = \dfrac{w_ob^4}{8EI} + a\left( \dfrac{w_ob^3}{6EI} \right)$

$\dfrac{R_AL^3}{3} = \dfrac{w_ob^4}{8} + \dfrac{w_oab^3}{6}$

$R_A = \dfrac{3w_ob^4}{8L^3} + \dfrac{w_oab^3}{2L^3}$

$R_A = \dfrac{w_ob^3}{L^3}\left( \dfrac{3b}{8} + \dfrac{a}{2} \right)$

$R_A = \dfrac{w_ob^3}{L^3}\left( \dfrac{3b}{8} + \dfrac{L - b}{2} \right)$

$R_A = \dfrac{w_ob^3}{L^3}\left( \dfrac{3b + 4L - 4b}{8} \right)$

$R_A = \dfrac{w_ob^3}{8L^3}(4L - b)$ *answer*

**Shear and Moment Diagrams**

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

$R_A = \dfrac{w_ob^3}{8L^3}(4L – b)$

*answer*

With value of R_{A} known, it is now easy to solve for M_{C} and R_{C}.

**Solving for fixed-end moment**

$M_C = R_AL – 0.5w_ob^2$

$M_C = \dfrac{w_ob^3}{8L^2}(4L – b) – \dfrac{w_ob^2}{2}$

$M_C = \dfrac{w_ob^2}{8L^2}[ \, b(4L – b) - 4L^2 \, ]$

$M_C = \dfrac{w_ob^2}{8L^2}[ \, 4Lb – b^2 - 4L^2 \, ]$

$M_C = -\dfrac{w_ob^2}{8L^2}(4L^2 - 4Lb + b^2)$ *answer*

**Solving for fixed-end shear**

$\Sigma F_V = 0$

$R_C + R_A = w_ob$

$R_C + \dfrac{w_ob^3}{8L^3}(4L – b) = w_ob$

$R_C = - \dfrac{w_ob^3}{8L^3}(4L – b) + w_ob$

$R_C = \dfrac{w_ob}{8L^3} [ \,-b^2(4L – b) + 8L^3 \, ]$

$R_C = \dfrac{w_ob}{8L^3} [ \,-4Lb^2 + b^3 + 8L^3 \, ]$

$R_C = \dfrac{w_ob}{8L^3} (8L^3 - 4Lb^2 + b^3)$ *answer*

**Shape of Shear Diagram**

- The shear at A is R
_{A} - There is no load between segment BC, thus, the shear over BC is uniform and equal to R
_{A}. - The load over segment CD is uniform and downward, thus, the shear on this segment is linear and decreasing from R
_{A}to -R_{C}. - The shear diagram between BC is zero at D. The location of D can be found by ratio and proportion of the shear triangles of segments BD and DC.

**Shape of Moment Diagram**

- The moment at A is zero.
- The shear between AB is uniform and positive, thus the moment between AB is linearly increasing (straight line) from zero to aR
_{A}. - The moment at B is aR
_{A}which is equal to the area of the shear diagram of segment BC. - The shear between BC is linear with zero at point D, thus the moment diagram of segment BC is a second degree curve (parabola) with vertex at D. The parabola is open downward because the shear from B to C is decreasing.
- The moment at D is equal to the sum of the moment at B and the area of the shear diagram of segment BD.
- Finally, the moment at C is equal to the moment at D minus the area of the shear diagram of segment DC. You can check the accuracy of the moment diagram by finding the moment at C in the load diagram.

In the event that you need to determine the location of zero moment (for construction joint most probably), compute for the area of the moment diagram from D to C and use the squared property of parabola to locate the zero moment. An easier way to find the point of zero moment is to write the moment equation (similar to what we did in double integration method above), equate the equation to zero and solve for x.