Problem 704 | Solution of Propped Beam

$EI \, y'' = M$

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

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

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

 

Apply boundary conditions to solve for integration constants C₁ and C₂:

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

Deflection at A is zero. Thus, the deflection due to R_A denoted by δ₁ is equal to the sum of deflection at B denoted as δ₂ 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

The reaction at the simple support R_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

1. The shear at A is R_A
2. There is no load between segment BC, thus, the shear over BC is uniform and equal to R_A.
3. 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.
4. 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

1. The moment at A is zero.
2. The shear between AB is uniform and positive, thus the moment between AB is linearly increasing (straight line) from zero to aR_A.
3. The moment at B is aR_A which is equal to the area of the shear diagram of segment BC.
4. 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.
5. The moment at D is equal to the sum of the moment at B and the area of the shear diagram of segment BD.
6. 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.