# Solution to Problem 432 | Relationship Between Load, Shear, and Moment

**Problem 432**

Beam loaded as shown in Fig. P-432.

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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 432**

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$\Sigma M_E = 0$

$5R_1 + 120 = 6(60) + 40(3)(3.5)$

$R_1 = 132 \, \text{kN}$

$\Sigma M_B = 0$

$5R_2 + 60(1) = 40(3)(1.5) + 120$

$R_2 = 48 \, \text{kN}$

**To draw the Shear Diagram**

- V
_{A}= -60 kN - V
_{B}= V_{A}+ Area in load diagram

V_{B}= -60 + 0 = -60 kN

V_{B2}= V_{B}+ R_{1}= -60 + 132 = 72 kN - V
_{C}= V_{B2}+ Area in load diagram

V_{C}= 72 - 3(40) = -48 kN - V
_{D}= V_{C}+ Area in load diagram

V_{D}= -48 + 0 = -48 kN - V
_{E}= V_{D}+ Area in load diagram

V_{E}= -48 + 0 = -48 kN

V_{E2}= V_{E}+ R_{2}= -48 + 48 = 0 - Solving for x:

x / 72 = (3 - x) / 48

48x = 216 - 72x

x = 1.8 m

**To draw the Moment Diagram**

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

M_{B}= 0 - 60(1) = -60 kN·m - M
_{x}= M_{B}+ Area in shear diagram

M_{X}= -60 + ½ (1.8)(72) = 4.8 kN·m - M
_{C}= M_{X}+ Area in shear diagram

M_{C}= 4.8 - ½ (3 - 1.8)(48) = -24 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= -24 - ½ (24 + 72)(1) = -72 kN·m

M_{D2}= -72 + 120 = 48 kN·m - M
_{E}= M_{D2}+ Area in shear diagram

M_{E}= 48 - 48(1) = 0 - The location of zero moment on segment BC can be determined using the squared property of parabola. See the solution of Problem 434.

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