# Solution to Problem 445 | Relationship Between Load, Shear, and Moment

**Problem 445**

Beam carrying the loads shown in Fig. P-445.

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 445**

$\Sigma M_{R2} = 0$

$5R_1 = 80(3) + 90(2)$

$R_1 = 84 \, \text{kN}$

$5R_1 = 80(3) + 90(2)$

$R_1 = 84 \, \text{kN}$

$\Sigma M_{R1} = 0$

$5R_2 = 80(2) + 90(3)$

$R_2 = 86 \, \text{kN}$

Checking

$R_1 + R_2 = F_1 + F_2$ (*okay!*)

**To draw the Shear Diagram**

- V
_{A}= R_{1}= 84 kN - V
_{B}= V_{A}+ Area in load diagram

V_{B}= 84 - 20(1) = 64 kN - V
_{C}= V_{B}+ Area in load diagram

V_{C}= 64 - ½ (20 + 80)(3) = -86 kN - V
_{D}= V_{C}+ Area in load diagram

V_{D}= -86 + 0 = -86 kN

V_{D2}= V_{D}+ R_{2}= -86 + 86 = 0 **Location of zero shear:**

From the load diagram:

y / (x + 1) = 80 / 4

y = 20(x + 1)

V_{E}= V_{B}+ Area in load diagram

0 = 64 - ½ (20 + y)x

(20 + y)x = 128

[20 + 20(x + 1)]x = 128

20x^{2}+ 40x - 128 = 0

5x^{2}+ 10x - 32 = 0

x = 1.72 and -3.72

use x = 1.72 m from B- By squared property of parabola:

z / (1 + x)^{2}= (z + 86) / 4^{2}

16z = 7.3984z + 636.2624

8.6016z = 254.4224

z = 73.97 kN

**To draw the Moment Diagram**

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

M_{B}= 0 + ½ (84 + 64)(1) = 74 kN·m - M
_{E}= M_{B}+ Area in shear diagram

M_{E}= 74 + A_{1}(see figure for A_{1}and A_{2})**For A**_{1}:

A_{1}= 2/3 (1 + 1.72)(73.97) - 64(1) - 2/3 (1)(9.97)

A_{1}= 63.5

M_{E}= 74 + 63.5 = 137.5 kN·m - M
_{C}= M_{E}+ Area in shear diagram

M_{C}= M_{E}- A_{2}**For A2:**

A_{2}= 1/3 (4)(73.97 + 86) - 1/3 (1 + 1.72)(73.97) - 1.28(73.97)

A_{2}= 51.5

M_{C}= 137.5 - 51.5 = 86 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 86 - 86(1) = 0

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