# Solution to Problem 435 | Relationship Between Load, Shear, and Moment

**Problem 435**

Beam loaded and supported as shown in Fig. P-435.

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 435**

$\Sigma M_B = 0$

$2w_o (5) = 10(4)(0) + 20(2) + 40(3)$

$w_o = 16 \, \text{kN/m}$

$\Sigma M_{midpoint \,\, of \,\, EF} = 0$

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

$R_1 = 68 \, \text{kN}$

**To draw the Shear Diagram**

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

M_{B}= 0 - 10(2) = -20 kN

M_{B2}+ M_{B}+ R_{1}= -20 + 68 = 48 kN - M
_{C}= M_{B2}+ Area in load diagram

M_{C}= 48 - 10(2) = 28 kN

M_{C2}= M_{C}- 20 = 28 - 20 = 8 kN - M
_{D}= M_{C2}+ Area in load diagram

M_{D}= 8 + 0 = 8 kN

M_{D2}= M_{D}- 40 = 8 - 40 = -32 kN - M
_{E}= M_{D2}+ Area in load diagram

M_{E}= -32 + 0 = -32 kN - M
_{F}= M_{E}+ Area in load diagram

M_{F}= -32 + w_{o}(2)

M_{F}= -32 + 16(2) = 0

**To draw the Moment Diagram**

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

M_{B}= 0 - ½ (20)(2) = -20 kN·m - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= -20 + ½ (48 + 28)(2)

M_{C}= 56 kN·m - M
_{D}= M_{C}+ Area in shear diagram

M_{D}= 56 + 8(1) = 64 kN·m - M
_{E}= M_{D}+ Area in shear diagram

M_{E}= 64 - 32(1) = 32 kN·m - M
_{F}= M_{E}+ Area in shear diagram

M_{F}= 32 - ½(32)(2) = 0 - The location and magnitude of moment at C' are determined from shear diagram. By squared property of parabola, x = 0.44 m from B.

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