# Solution to Problem 447 | Relationship Between Load, Shear, and Moment

**Problem 447**

Shear diagram as shown in Fig. P-447.

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 447**

**To draw the Load Diagram**

- A 2400 lb upward force is acting at point A. No load in segment AB.
- A point force of 2400 - 400 = 2000 lb is acting downward at point B. No load in segment BC.
- Another downward force of magnitude 400 + 4000 = 4400 lb at point C. No load in segment CD.
- Upward point force of 4000 + 1000 = 5000 lb is acting at D. No load in segment DE.
- A downward force of 1000 lb is concentrated at point E.

**To draw the Moment Diagram**

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

M_{B}= 0 + 2400(2) = 4800 lb·ft

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

M_{C}= 4800 + 400(3) = 6000 lb·ft

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

M_{D}= 6000 - 4000(2) = -2000 lb·ft

M_{CD}is linear and downward - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= -2000 + 1000(2) = 0

M_{DE}is linear and upward

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