# Solution to Problem 448 | Relationship Between Load, Shear, and Moment

**Problem 448**

Shear diagram as shown in Fig. P-448.

In the following problem, draw moment and load diagrams corresponding to the given shear diagram. Specify values at all change of load positions and at all points of zero shear.

**Solution 448**

**To draw the Load Diagram**

- A uniformly distributed load in AB is acting downward at a magnitude of 40/2 = 20 kN/m.
- Upward concentrated force of 40 + 36 = 76 kN acts at B. No load in segment BC.
- A downward point force acts at C at a magnitude of 36 - 16 = 20 kN.
- Downward uniformly distributed load in CD has a magnitude of (16 + 24)/4 = 10 kN/m & causes zero shear at point F, 1.6 m from C.
- Another upward concentrated force acts at D at a magnitude of 20 + 24 = 44 kN.
- The load in segment DE is uniform and downward at 20/2 = 10 kN/m.

**To draw the Moment Diagram**

- M
_{A}= 0 - M
_{B}= M_{A}+ Area in shear diagram

M_{B}= 0 - ½ (40)(2) = -40 kN·m

M_{AB}is downward parabola with vertex at A. - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= -40 + 36(1) = -4 kN·m

M_{BC}is linear and upward - M
_{F}= M_{C}+ Area in shear diagram

M_{F}= -4 + ½ (16)(1.6) = 8.8 kN·m - M
_{D}= M_{F}+ Area in shear diagram

M_{D}= 8.8 - ½ (24)(2.4) = -20 kN·m

M_{CD}is downward parabola with vertex at F. - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= -20 + ½ (20)(2) = 0

M_{DE}is downward parabola with vertex at E.

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