$\Sigma F_V = 0$
$4w_o + 2 \,[ \, \frac{1}{2}w_o(1) \, ] = 20(4) + 2(50)$
$5w_o = 180$
$w_o = 36 \, \text{kN/m}$
To draw the Shear Diagram
- VA = 0
- VB = VA + Area in load diagram
VB = 0 + ½ (36)(1) = 18 kN
VB2 = VB - 50 = 18 - 50
VB2 = -32 kN
- The net uniformly distributed load in segment BC is 36 - 20 = 16 kN/m upward.
VC = VB2 + Area in load diagram
VC = -32 + 16(4) = 32 kN
VC2 = VC - 50 = 32 - 50
VC2 = -18 kN
- VD = VC2 + Area in load diagram
VD = -18 + ½ (36)(1) = 0
- The shape of shear at AB and CD are parabolic spandrel with vertex at A and D, respectively.
- The location of zero shear is obviously at the midspan or 2 m from B.
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 + 1/3 (1)(18)
MB = 6 kN·m
- Mmidspan = MB + Area in shear diagram
Mmidspan = 6 - ½ (32)(2)
Mmidspan = -26 kN·m
- MC = Mmidspan + Area in shear diagram
MC = -26 + ½ (32)(2)
MC = 6 kN·m
- MD = MC + Area in shear diagram
MD = 6 - 1/3 (1)(18) = 0
- The moment diagram at AB and CD are 3rd degree curve while at BC is 2nd degree curve.
