# Solution to Problem 405 | Shear and Moment Diagrams

**Problem 405**

Beam loaded as shown in Fig. P-405.

Write shear and moment equations for the beams in the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.

**Solution 405**

$\Sigma M_A = 0$

$10R_C = 2(80) + 5[10(10)]$

$R_C = 66 \, \text{kN}$

$\Sigma M_C = 0$

$10R_A = 8(80) + 5[10(10)]$

$R_A = 114 \, \text{kN}$

**Segment AB:**

$V_{AB} = 114 - 10x \, \text{kN}$

$M_{AB} = 114x - 10x(x/2)$

$M_{AB} = 114x - 5x^2 \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = 114 - 80 - 10x$

$V_{BC} = 34 - 10x \, \text{kN}$

$M_{BC} = 114x - 80(x - 2) - 10x(x/2)$

$M_{BC} = 160 + 34x - 5x^2 \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram:**

- For segment AB, V
_{AB}= 114 - 10x is linear; at x = 0, V_{AB}= 14 kN; at x = 2 m, V_{AB}= 94 kN. - V
_{BC}= 34 - 10x for segment BC is linear; at x = 2 m, V_{BC}= 14 kN; at x = 10 m, V_{BC}= -66 kN. When V_{BC}= 0, x = 3.4 m thus V_{BC}= 0 at 1.4 m from B.

**To draw the Moment Diagram:**

- M
_{AB}= 114x - 5x^{2}is a second degree curve for segment AB; at x = 0, M_{AB}= 0; at x = 2 m, M_{AB}= 208 kN·m. - The moment diagram is also a second degree curve for segment BC given by M
_{BC}= 160 + 34x - 5x^{2}; at x = 2 m, M_{BC}= 208 kN·m; at x = 10 m, M_{BC}= 0. - Note that the maximum moment occurs at point of zero shear. Thus, at x = 3.4 m, M
_{BC}= 217.8 kN·m.

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