# Solution to Problem 429 | Relationship Between Load, Shear, and Moment

**Problem 429**

Beam loaded as shown in Fig. P-429.

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 429**

$\Sigma M_C = 0$

$4R_1 + 120(2)(1) = 100(2) + 120(2)(3)$

$R_1 = 170 \, \text{lb}$

$\Sigma M_A = 0$

$4R_2 = 120(2)(1) + 100(2) + 120(2)(5)$

$R_2 = 410 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= R_{1}= 170 lb - V
_{B}= V_{A}+ Area in load diagram

V_{B}= 170 - 120(2) = -70 lb

V_{B2}= -70 - 100 = -170 lb - V
_{C}= V_{B2}+ Area in load diagram

V_{C}= -170 + 0 = -170 lb

V_{C2}= -170 + R_{2}

V_{C2}= -170 + 410 = 240 lb - V
_{D}= V_{C2}+ Area in load diagram

V_{D}= 240 - 120(2) = 0 **Solving for x:**

x / 170 = (2 - x) / 70

70x = 340 - 170x

x = 17 / 12 ft = 1.42 ft

**To draw the Moment Diagram**

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

M_{x}= 0 + (17/12)(170)

M_{x}= 1445/12 = 120.42 lb·ft - M
_{B}= M_{x}+ Area in shear diagram

M_{B}= 1445/12 - ½ (2 - 17/12)(70)

M_{B}= 100 lb·ft - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 100 - 170(2) = -240 lb·ft - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= -240 + ½ (2)(240) = 0

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