$\Sigma M_B = 0$
$2w_o (5) = 10(4)(0) + 20(2) + 40(3)$
$w_o = 16 \, \text{kN/m}$
$\Sigma M_{midpoint \,\, of \,\, EF} = 0$
$5R_1 = 10(4)(5) + 20(3) + 40(2)$
$R_1 = 68 \, \text{kN}$
To draw the Shear Diagram
- MA = 0
- MB = MA + Area in load diagram
MB = 0 - 10(2) = -20 kN
MB2 + MB + R1 = -20 + 68 = 48 kN
- MC = MB2 + Area in load diagram
MC = 48 - 10(2) = 28 kN
MC2 = MC - 20 = 28 - 20 = 8 kN
- MD = MC2 + Area in load diagram
MD = 8 + 0 = 8 kN
MD2 = MD - 40 = 8 - 40 = -32 kN
- ME = MD2 + Area in load diagram
ME = -32 + 0 = -32 kN
- MF = ME + Area in load diagram
MF = -32 + wo(2)
MF = -32 + 16(2) = 0
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 - ½ (20)(2) = -20 kN·m
- MC = MB + Area in shear diagram
MC = -20 + ½ (48 + 28)(2)
MC = 56 kN·m
- MD = MC + Area in shear diagram
MD = 56 + 8(1) = 64 kN·m
- ME = MD + Area in shear diagram
ME = 64 - 32(1) = 32 kN·m
- MF = ME + Area in shear diagram
MF = 32 - ½(32)(2) = 0
- The location and magnitude of moment at C' are determined from shear diagram. By squared property of parabola, x = 0.44 m from B.
