# Solution to Problem 446 | Relationship Between Load, Shear, and Moment

**Problem 446**

Beam loaded and supported as shown in Fig. P-446.

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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 446**

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$\Sigma F_V = 0$

$4w_o + 2 \,[ \, \frac{1}{2}w_o(1) \, ] = 20(4) + 2(50)$

$5w_o = 180$

$w_o = 36 \, \text{kN/m}$

**To draw the Shear Diagram**

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

V_{B}= 0 + ½ (36)(1) = 18 kN

V_{B2}= V_{B}- 50 = 18 - 50

V_{B2}= -32 kN - The net uniformly distributed load in segment BC is 36 - 20 = 16 kN/m upward.

V_{C}= V_{B2}+ Area in load diagram

V_{C}= -32 + 16(4) = 32 kN

V_{C2}= V_{C}- 50 = 32 - 50

V_{C2}= -18 kN - V
_{D}= V_{C2}+ Area in load diagram

V_{D}= -18 + ½ (36)(1) = 0 - The shape of shear at AB and CD are parabolic spandrel with vertex at A and D, respectively.
- The location of zero shear is obviously at the midspan or 2 m from B.

**To draw the Moment Diagram**

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

M_{B}= 0 + 1/3 (1)(18)

M_{B}= 6 kN·m - M
_{midspan}= M_{B}+ Area in shear diagram

M_{midspan}= 6 - ½ (32)(2)

M_{midspan}= -26 kN·m - M
_{C}= Mmidspan + Area in shear diagram

M_{C}= -26 + ½ (32)(2)

M_{C}= 6 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 6 - 1/3 (1)(18) = 0 - The moment diagram at AB and CD are 3rd degree curve while at BC is 2nd degree curve.

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