# Solution to Problem 430 | Relationship Between Load, Shear, and Moment

**Problem 430**

Beam loaded as shown in Fig. P-430.

Without writing shear and moment equations, draw the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear.

**Solution 430**

$\Sigma M_D = 0$

$20R_1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5)$

$R_1 = 5000 \, \text{lb}$

$\Sigma M_B = 0$

$20R_2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15)$

$R_2 = 2000 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= -1000 lb - V
_{B}= V_{A}+ Area in load diagram; V_{B}= -1000 - 400(5) = -3000 lb; V_{B2}= -3000 + R1 = 2000 lb - V
_{C}= V_{B2}+ Area in load diagram; V_{C}= 2000 + 0 = 2000 lb; V_{C2}= 2000 - 2000 = 0 - V
_{D}= V_{C2}+ Area in load diagram; V_{D}= 0 + 200(10) = 2000 lb

**To draw the Moment Diagram**

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

M_{B}= 0 - ½ (1000 + 3000)(5)

M_{B}= -10000 lb·ft - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= -10000 + 2000(10) = 10000 lb·ft - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 10000 - ½ (10)(2000) = 0 - For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B.
- The moment curve AB is a downward parabola with vertex at A'. A' is the location of zero shear for segment AB at point outside the beam.

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