# Solution to Problem 431 | Relationship Between Load, Shear, and Moment

**Problem 431**

Beam loaded as shown in Fig. P-431.

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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 431**

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$\Sigma M_D = 0$

$7R_1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2)$

$R_1 = 70 \, \text{kN}$

$\Sigma M_A = 0$

$7R_2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10)$

$R_2 = 200 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A}= R_{1}= 70 kN - V
_{B}= V_{A}+ Area in load diagram

V_{B}= 70 - 10(2) = 50 kN

V_{B2}= 50 - 50 = 0 - V
_{C}= V_{B2}+ Area in load diagram

V_{C}= 0 - 10(1) = -10 kN - V
_{D}= VC + Area in load diagram

V_{D}= -10 - 30(4) = -130 kN

V_{D2}= -130 + R_{2}

V_{D2}= -130 + 200 = 70 kN - V
_{E}= V_{D2}+ Area in load diagram; V_{E}= 70 - 10(3) = 40 kN

V_{E2}= 40 - 40 = 0

**To draw the Moment Diagram**

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

M_{B}= 0 + ½ (70 + 50)(2) = 120 kN·m - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 120 - ½ (1)(10) = 115 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 115 - ½ (10 + 130)(4)

M_{D}= -165 kN·m - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= -165 + ½ (70 + 40)(3) = 0 - Moment curves AB, CD and DE are downward parabolas with vertices at A', B' and C', respectively. A', B' and C' are corresponding zero shear points of segments AB, CD and DE.
- Locating the point of zero moment:

a / 10 = (a + 4) / 130

130a = 10a + 40

a = 1/3 m

y / (x + a) = 130 / (4 + a)

y = 130(x + 1/3) / (4 + 1/3)

y = 30x + 10

M_{C}= 115 kN·m

M_{zero}= M_{C}+ Area in shear

0 = 115 - ½ (10 + y)x

(10 + y)x = 230

(10 + 30x + 10)x = 230

30x^{2}+ 20x - 230 = 0

3x^{2}+ 2x - 23 = 0

x = 2.46 m

Zero moment is at 2.46 m from C

Another way to solve the location of zero moment is by the squared property of parabola (see Problem 434). The point of zero moment is an ideal location for the construction joint.

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