# Solution to Problem 434 | Relationship Between Load, Shear, and Moment

**Problem 434**

Beam loaded as shown in Fig. P-434.

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 434**

$\Sigma M_E = 0$

$6R_1 + 120 = 20(4)(6) + 60(4)$

$R_1 = 100 \, \text{ kN}$

$\Sigma M_B = 0$

$6R_2 = 20(4)(0) + 60(2) + 120$

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

**To draw the Shear Diagram**

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

V_{B}= 0 - 20(2) = -40 kN

V_{B2}= V_{B}+ R_{1}= -40 + 100 = 60 kN] - V
_{C}= V_{B2}+ Area in load diagram

V_{C}= 60 - 20(2) = 20 kN

V_{C2}= V_{C}- 60 = 20 - 60 = -40 kN - V
_{D}= V_{C2}+ Area in load diagram

V_{D}= -40 + 0 = -40 kN - V
_{E}= V_{D}+ Area in load diagram

V_{E}= -40 + 0 = -40 kN

V_{E2}= V_{E}+ R_{2}= -40 + 40 = 0

**To draw the Moment Diagram**

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

M_{B}= 0 - ½ (40)(2) = -40 kN·m - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= -40 + ½ (60 + 20)(2) = 40 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 40 - 40(2) = -40 kN·m

M_{D2}= M_{D}+ M = -40 + 120 = 80 kN·m - M
_{E}= M_{D2}+ Area in shear diagram

M_{E}= 80 - 40(2) = 0 - Moment curve BC is a downward parabola with vertex at C'. C' is the location of zero shear for segment BC.
**Location of zero moment at segment BC:**

By squared property of parabola:

3 - x)2 / 50 = 32 / (50 + 40)

3 - x = 2.236

x = 0.764 m from B

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