Solution to Problem 405 | Shear and Moment Diagrams

$10R_C = 2(80) + 5[10(10)]$

$R_C = 66 \, \text{kN}$
 

$\Sigma M_C = 0$

$10R_A = 8(80) + 5[10(10)]$

$R_A = 114 \, \text{kN}$
 

Segment AB:
$V_{AB} = 114 - 10x \, \text{kN}$

$M_{AB} = 114x - 10x(x/2)$

$M_{AB} = 114x - 5x^2 \, \text{kN}\cdot\text{m}$
 

Segment BC:
$V_{BC} = 114 - 80 - 10x$

$V_{BC} = 34 - 10x \, \text{kN}$

$M_{BC} = 114x - 80(x - 2) - 10x(x/2)$

$M_{BC} = 160 + 34x - 5x^2 \, \text{kN}\cdot\text{m}$
 

To draw the Shear Diagram:

1. For segment AB, V_AB = 114 - 10x is linear; at x = 0, V_AB = 14 kN; at x = 2 m, V_AB = 94 kN.
2. V_BC = 34 - 10x for segment BC is linear; at x = 2 m, V_BC = 14 kN; at x = 10 m, V_BC = -66 kN. When V_BC = 0, x = 3.4 m thus V_BC = 0 at 1.4 m from B.

To draw the Moment Diagram:

1. M_AB = 114x - 5x² is a second degree curve for segment AB; at x = 0, M_AB = 0; at x = 2 m, M_AB = 208 kN·m.
2. The moment diagram is also a second degree curve for segment BC given by M_BC = 160 + 34x - 5x²; at x = 2 m, M_BC = 208 kN·m; at x = 10 m, M_BC = 0.
3. Note that the maximum moment occurs at point of zero shear. Thus, at x = 3.4 m, M_BC = 217.8 kN·m.