Problem 736 | Shear and moment diagrams of fully restrained beam under triangular load

$EI \, \theta_{AC} = (Area_{AC}) = 0$

$\frac{1}{2}(12)(12R_A) + 12M_A - \frac{1}{4}(8)(5760) = 0$

$72R_A + 12M_A - 11\,520 = 0$

$6R_A + M_A = 960$   →   Equation (1)
 

 

$EI\,t_{C/A} = (Area_{AC}) \cdot \bar{X}_C = 0$

$\frac{1}{2}(12)(12R_A)[ \, \frac{1}{3}(12) \, ] + 12M_A[ \, \frac{1}{2}(12) \, ] - \frac{1}{4}(8)(5760)[ \, \frac{1}{5}(8) \, ] = 0$

$288R_A + 72M_A - 18\,432 = 0$

$4R_A + M_A = 256$   →   Equation (2)
 

Solving Equations (1) and (2)
$R_A = 352 \, \text{ lb}$           answer

$M_A = -1152 \, \text{ lb}\cdot\text{ft}$           answer
 

Checking
$EI\,t_{A/C} = (Area_{AC}) \cdot \bar{X}_A = 0$

$\frac{1}{2}(12)(12R_A)[ \, \frac{2}{3}(12) \, ] + 12M_A[ \, \frac{1}{2}(12) \, ] - \frac{1}{4}(8)(5760)[ \, 4 + \frac{4}{5}(8) \, ] = 0$

$576R_A + 72M_A - 119\,808 = 0$

$576(352) + 72(-1152) - 119 808 = 0$

$0 = 0$       okay!
 

$M_C = 12R_A + M_A - 5760$

$M_C = 12(352) - 1152 - 5760$

$M_C = -2688 \, \text{ lb}\cdot\text{ft}$           answer
 

$\Sigma F_V = 0$

$R_A + R_C = \frac{1}{2}(8)(540)$

$352 + R_C = 2160$

$R_C = 1\,808 \, \text{ lb}$           answer
 

To draw the shear diagram

1. V_A = 352 lb
2. V_B = V_A + Load_AB
   V_B = 352 + 0
   V_B = 352 lb
3. There is no load between AB, thus, shear in AB is constant.
4. V_C = V_B + Load_BC
   V_C = 352 - (1/2)(8)(540)
   V_C = -1808 lb
5. Load between B and C is linearly decreasing from zero to -540 lb/ft, thus, shear in segment BC is a concave downward second degree curve (parabola) with vertex at B.
6. Location of point D by squared property of parabola:
   $\dfrac{{x_D}^2}{352} = \dfrac{8^2}{352 1808}$

   $x_D = 3.23 \, \text{ ft}$   to the right of B

 

 

To draw the moment diagram

1. M_A = -1152 lb·ft
2. M_B = M_A + V_AB
   M_B = -1152 + 352(4)
   M_B = 256 lblb·ftft
3. The shear between A and B is uniform and positive, thus, the moment in AB is linear and increasing.
4. M_D = M_B + V_BD
   M_D = 256 + (2/3)(x_D)(352)
   M_D = 256 + (2/3)(3.23)(352)
   M_D = 1013.97 lb·ft
5. M_C = M_D + V_DC
    

   Solving for VDC
   V_DC = (-1/3)(8)(352 + 1808) + (1/3)(x_D)(352) + 352(8 - x_D)
   V_DC = -5760 + (1/3)(3.23)(352) + 352(8 - 3.23)
   V_DC = -3701.97 lb
    

   Thus,
   M_C = 1013.97 - 3701.97
   M_C = -2688 lb·ft

6. The shear diagram from B to C is a parabola, thus, the moment diagram of segment BC is a third degree curve. The value of shear from B to C decreases, thus, the slope of moment diagram between B and C also decreases making the cubic curve concave downward.