# Solution to Problem 451 | Relationship Between Load, Shear, and Moment

**Problem 451**

Shear diagram as shown in Fig. P-451.

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 451**

**To draw the Load Diagram**

- Upward concentrated load at A is 10 kN.
- The shear in AB is a 2nd-degree curve, thus the load in AB is uniformly varying. In this case, it is zero at A to 2(10 + 2)/3 = 8 kN at B. No load in segment BC.
- A downward point force is acting at C in a magnitude of 8 - 2 = 6 kN.
- The shear in DE is uniformly increasing, thus the load in DE is uniformly distributed and upward. This load is spread over DE at a magnitude of 8/2 = 4 kN/m.

**To draw the Moment Diagram**

- To find the location of zero shear, F:

x^{2}/10 = 3^{2}/(10 + 2)

x = 2.74 m - M
_{A}= 0 - M
_{F}= M_{A}+ Area in shear diagram

M_{F}= 0 + 2/3 (2.74)(10) = 18.26 kN·m - M
_{B}= M_{F}+ Area in shear diagram

M_{B}= 18.26 - [1/3 (10 + 2)(3) - 1/3 (2.74)(10) - 10(3 - 2.74)]

M_{B}= 18 kN·m - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 18 - 2(1) = 16 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 16 - 8(1) = 8 kN·m - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= 8 - ½ (2)(8) = 0 - The moment diagram in AB is a second degree curve, at BC and CD are linear and downward. For segment DE, the moment diagram is parabola open upward with vertex at E.

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