# Solution to Problem 438 | Relationship Between Load, Shear, and Moment

**Problem 438**

The beam loaded as shown in Fig. P-438 consists of two segments joined by a frictionless hinge at which the bending moment is zero.

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 438**

From the FBD of the section to the left of hinge

$\Sigma M_H = 0$

$4R_1 = 200(6)(3)$

$R_1 = 900 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= 0 - V
_{B}= V_{A}+ Area in load diagram

V_{B}= 0 - 200(2) = -400 lb

V_{B2}= V_{B}+ R_{1}= -400 + 900 = 500 lb - V
_{H}= V_{B2}+ Area in load diagram

V_{H}= 500 - 200(4) = -300 lb - V
_{C}= V_{H}+ Area in load diagram

V_{C}= -300 - 200(2) = -700 lb - Location of zero shear:

x / 500 = (4 - x) / 300

300x = 2000 - 500x

x = 2.5 ft

**To draw the Moment Diagram**

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

M_{B}= 0 - ½ (400)(2) = -400 lb·ft - M
_{x}= M_{B}+ Area in load diagram

M_{x}= -400 + ½ (500)(2.5)

M_{x}= 225 lb·ft - M
_{H}= M_{x}+ Area in load diagram

M_{H}= 225 - ½(300)(4 - 2.5) = 0 ok! - M
_{C}= M_{H}+ Area in load diagram

M_{C}= 0 - ½ (300 + 700)(2)

M_{C}= -1000 lb·ft - The location of zero moment in segment BH can easily be found by symmetry.