# Solution to Problem 450 | Relationship Between Load, Shear, and Moment

**Problem 450**

Shear diagram as shown in Fig. P-450.

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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 450**

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**To draw the Load Diagram**

- The shear diagram in AB is uniformly upward, thus the load is uniformly distributed upward at a magnitude of 900/4 = 225 lb/ft. No load in segment BC.
- A downward point force acts at point C with magnitude of 900 lb. No load in segment CD.
- Another concentrated force is acting downward at D with a magnitude of 900 lb.
- The load in DE is uniformly distributed downward at a magnitude of (1380 - 900)/4 = 120 lb/ft.
- An upward load is concentrated at E with magnitude of 480 + 1380 = 1860 lb.
- 480/4 = 120 lb/ft is distributed uniformly over the span EF.

**To draw the Moment Diagram**

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

M_{B}= 0 + ½ (4)(900) = 1800 lb·ft - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 1800 + 900(2) = 3600 lb·ft - M
_{D}= MC + Area in shear diagram

M_{D}= 3600 + 0 = 3600 lb·ft - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= 3600 - ½ (900 + 1380)(4)

M_{E}= -960 lb·ft - M
_{F}= M_{E}+ Area in shear diagram

M_{F}= -960 + ½ (480)(4) = 0 - The shape of moment diagram in AB is upward parabola with vertex at A, while linear in BC and horizontal in CD. For segment DE, the diagram is downward parabola with vertex at G. G is the point where the extended shear in DE intersects the line of zero shear.
- The moment diagram in EF is a downward parabola with vertex at F.