# Solution to Problem 449 | Relationship Between Load, Shear, and Moment

**Problem 449**

Shear diagram as shown in Fig. P-449.

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 449**

**To draw the Load Diagram**

- Downward 4000 lb force is concentrated at A and no load in segment AB.
- The shear in BC is uniformly increasing, thus a uniform upward force is acting at a magnitude of (3700 + 4000)/2 = 3850 lb/ft. No load in segment CD.
- Another point force acting downward with 3700 - 1700 = 1200 lb at D and no load in segment DE.
- The shear in EF is uniformly decreasing, thus a uniform downward force is acting with magnitude of (1700 + 3100)/8 = 600 lb/ft.
- Upward force of 3100 lb is concentrated at end of span F.

**To draw the Moment Diagram**

- The locations of zero shear (points G and H) can be easily determined by ratio and proportion of triangle.
- M
_{A}= 0; M_{B}= M_{A}+ Area in shear diagram

M_{B}= 0 - 4000(3) = -12,000 lb·ft - M
_{G}= M_{B}+ Area in shear diagram

M_{G}= -12,000 - ½ (80/77)(4000)

M_{G}= -14,077.92 lb·ft - M
_{C}= M_{G}+ Area in shear diagram

M_{C}= -14,077.92 + ½ (74/77)(3700)

M_{C}= -12,300 lb·ft - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= -12,300 + 3700(3) = -1200 lb·ft - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= -1200 + 1700(4) = 5600 lb·ft - M
_{H}= M_{E}+ Area in shear diagram

M_{H}= 5600 + ½ (17/6)(1700)

M_{H}= 8,008.33 lb·ft - M
_{F}= M_{H}+ Area in shear diagram

M_{F}= 8,008.33 - ½ (31/6)(3100) = 0

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