# Solution to Problem 439 | Relationship Between Load, Shear, and Moment

**Problem 439**

A beam supported on three reactions as shown in Fig. P-439 consists of two segments joined by 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 439**

$\Sigma M_H = 0$

$8R_1 = 4000(4)$

$R_1 = 2000 \, \text{lb}$

$8R_1 = 4000(4)$

$R_1 = 2000 \, \text{lb}$

$\Sigma M_A = 0$

$8V_H = 4000(4)$

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

$\Sigma M_D = 0$

$10R_2 = 2000(14) + 400(10)(5)$

$R_2 = 4800 \, \text{lb}$

$\Sigma M_H = 0$

$14R_3 + 4(4800) = 400(10)(9)$

$R_3 = 1200 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= 0 - V
_{B}= 2000 lb

V_{B2}= 2000 - 4000 = -2000 lb - V
_{H}= -2000 lb - V
_{C}= -2000 lb

V_{C}= -2000 + 4800 = 2800 lb - V
_{D}= 2800 - 400(10) = -1200 lb - Location of zero shear:

x / 2800 = (10 - x) / 1200

1200x = 28000 - 2800x

x = 7 ft

**To draw the Moment Diagram**

- M
_{A}= 0 - M
_{B}= 2000(4) = 8000 lb·ft - M
_{H}= 8000 - 4000(2) = 0 - M
_{C}= -400(2)

M_{C}= -8000 lb·ft - M
_{x}= -800 + ½ (2800)(7)

M_{x}= 1800 lb·ft - M
_{D}= 1800 - ½(1200)(3)

M_{D}= 0 - Zero M is 4 ft from R
_{2}

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