Solution to Problem 408 | Shear and Moment Diagrams

$6R_D = 1[2(50)] + 5[2(20)]$

$R_D = 50 \, \text{kN}$
 

$\Sigma M_D = 0$

$6R_A = 5[2(50)] + 1[2(20)]$

$R_A = 90 \, \text{kN}$
 

Segment AB:
$V_{AB} = 90 - 50x \, \text{kN}$

$M_{AB} = 90x - 50x(x/2)$

$M_{AB} = 90x - 25x^2 \, \text{kN}\cdot\text{m}$
 
 

Segment BC:
$V_{BC} = 90 - 50(2)$

$V_{BC} = -10 \, \text{kN}$

$M_{BC} = 90x - 2(50)(x - 1)$

$M_{BC} = -10x + 100 \, \text{kN}\cdot\text{m}$
 
 

Segment CD:
$V_{CD} = 90 - 2(50) - 20(x - 4)$

$V_{CD} = -20x + 70 \, \text{kN}$
 

$M_{CD} = 90x - 2(50)(x - 1) - 20(x - 4)(x - 4)/2$

$M_{CD} = 90x - 100(x - 1) - 10(x - 4)^2$

$M_{CD} = -10x^2 + 70x - 60 \, \text{kN}\cdot\text{m}$
 

To draw the Shear Diagram:

1. V_AB = 90 - 50x is linear; at x = 0, V_BC = 90 kN; at x = 2 m, V_BC = -10 kN. When V_AB = 0, x = 1.8 m.
2. V_BC = -10 kN along segment BC.
3. V_CD = -20x + 70 is linear; at x = 4 m, V_CD = -10 kN; at x = 6 m, V_CD = -50 kN.

To draw the Moment Diagram:

1. M_AB = 90x - 25x² is second degree; at x = 0, M_AB = 0; at x = 1.8 m, M_AB = 81 kN·m; at x = 2 m, MAB = 80 kN·m.
2. M_BC = -10x + 100 is linear; at x = 2 m, M_BC = 80 kN·m; at x = 4 m, M_BC = 60 kN·m.
3. M_CD = -10x² + 70x - 60; at x = 4 m, M_CD = 60 kN·m; at x = 6 m, M_CD = 0.