$\Sigma M_D = 0$
$20R_1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5)$
$R_1 = 5000 \, \text{lb}$
$\Sigma M_B = 0$
$20R_2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15)$
$R_2 = 2000 \, \text{lb}$
To draw the Shear Diagram
- VA = -1000 lb
- VB = VA + Area in load diagram; VB = -1000 - 400(5) = -3000 lb; VB2 = -3000 + R1 = 2000 lb
- VC = VB2 + Area in load diagram; VC = 2000 + 0 = 2000 lb; VC2 = 2000 - 2000 = 0
- VD = VC2 + Area in load diagram; VD = 0 + 200(10) = 2000 lb
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 - ½ (1000 + 3000)(5)
MB = -10000 lb·ft
- MC = MB + Area in shear diagram
MC = -10000 + 2000(10) = 10000 lb·ft
- MD = MC + Area in shear diagram
MD = 10000 - ½ (10)(2000) = 0
- For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B.
- The moment curve AB is a downward parabola with vertex at A'. A' is the location of zero shear for segment AB at point outside the beam.