# Solution to Problem 428 | Relationship Between Load, Shear, and Moment

**Problem 428**

Beam loaded as shown in Fig. P-428.

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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 428**

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$\Sigma M_D = 0$

$5R_1 = 50(0.5) + 25$

$R_1 = 10 \, \text{kN}$

$\Sigma M_A = 0$

$5R_2 + 25 = 50(4.5)$

$R_2 = 40 \, \text{kN}$

**To draw the Shear Diagram**

- V
_{A}= R_{1}= 10 kN - V
_{B}= V_{A}+ Area in load diagram

V_{B}= 10 + 0 = 10 kN - V
_{C}= V_{B}+ Area in load diagram

V_{C}= 10 + 0 = 10 kN - V
_{D}= V_{C}+ Area in load diagram

V_{D}= 10 - 10(3) = -20 kN

V_{D2}= -20 + R_{2}= 20 kN - V
_{E}= V_{D2}+ Area in load diagram

V_{E}= 20 - 10(2) = 0 **Solving for x:**

x / 10 = (3 - x) / 20

20x = 30 - 10x

x = 1 m

**To draw the Moment Diagram**

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

M_{B}= 0 + 1(10) = 10 kN·m

M_{B2}= 10 - 25 = -15 kN·m - M
_{C}= M_{B2}+ Area in shear diagram

M_{C}= -15 + 1(10) = -5 kN·m - M
_{x}= M_{C}+ Area in shear diagram

M_{x}= -5 + ½(1)(10) = 0 - M
_{D}= M_{x}+ Area in shear diagram

M_{D}= 0 - ½(2)(20) = -20 kN·m - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= -20 + ½ (2)(20) = 0

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