# Solution to Problem 441 | Relationship Between Load, Shear, and Moment

**Problem 441**

A beam ABCD is supported by a roller at A and a hinge at D. It is subjected to the loads shown in Fig. P-441, which act at the ends of the vertical members BE and CF. These vertical members are rigidly attached to the beam at B and C. (Draw shear and moment diagrams for the beam ABCD only.)

**Solution 441**

$M_B = 14(2)$

$M_B = 28 \, \text{kN}\cdot\text{m}$ counterclockwise

$F_{CH} = \frac{3}{5} (10)$

$F_{CH} = 6 \, \text{ kN}$ to the right

$F_{CV} = 4/5 (10)$

$F_{CV} = 8 \, \text{ kN}$ upward

$M_C = F_{CH} (2) = 6(2)$

$M_C = 12 \, \text{ kN}\cdot\text{m}$ clockwise

$\Sigma M_D = 0$

$6R_A + 12 + 8(2) = 28$

$R_A = 0$

$\Sigma M_A = 0$

$6R_{DV} + 12 = 28 + 8(4)$

$R_{DV} = 8 \, \text{kN}$

$\Sigma F_H = 0$

$R_{DH} = 14 + 6$

$R_{DH} = 20 \, \text{kN}$

**To draw the Shear Diagram**

- Shear in segments AB and BC is zero.
- V
_{C}= 8 - V
_{D}= V_{C}+ Area in load diagram

V_{D}= 8 + 0 = 8 kN

V_{D2}= V_{D}- R_{DV}

V_{D2}= 8 - 8 = 0

**To draw the Moment Diagram**

- Moment in segment AB is zero
- M
_{B}= -28 kN·m - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= -28 + 0 = -28 kN·m

M_{C2}= M_{C}+ 12 = -28 + 12

M_{C2}= -16 kN·m - M
_{D}= M_{C2}+ Area in shear diagram

MD = -16 + 8(2)

MD = 0