# Solution to Problem 427 | Relationship Between Load, Shear, and Moment

**Problem 427**

Beam loaded as shown in Fig. P-427.

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. (Note to instructor: Problems 403 to 420 may also be assigned for solution by semi-graphical method describes in this article.)

**Solution 427**

$\Sigma M_C = 0$

$12R_1 = 100(12)(6) + 800(3)$

$R_1 = 800 \, \text{lb}$

$\Sigma M_A = 0$

$12R_2 = 100(12)(6) + 800(9)$

$R_2 = 1200 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= R_{1}= 800 lb - V
_{B}= VA + Area in load diagram

V_{B}= 800 - 100(9)

V_{B}= -100 lb

V_{B2}= -100 - 800 = -900 lb - V
_{C}= V_{B2}+ Area in load diagram

V_{C}= -900 - 100(3)

V_{C}= -1200 lb - Solving for x:

x / 800 = (9 - x) / 100

100x = 7200 - 800x

x = 8 ft

**To draw the Moment Diagram**

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

M_{x}= 0 + ½(8)(800) = 3200 lb·ft; - M
_{B}= M_{x}+ Area in shear diagram

M_{B}= 3200 - ½(1)(100) = 3150 lb·ft - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 3150 - ½(900 + 1200)(3) = 0 - The moment curve BC is downward parabola with vertex at A'. A' is the location of zero shear for segment BC.

Tags: