# Solution to Problem 403 | Shear and Moment Diagrams

**Problem 403**

Beam loaded as shown in Fig. P-403.

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 403**

**From the load diagram:**

$\Sigma M_B = 0$

$5R_D + 1(30) = 3(50)$

$R_D = 24 \, \text{kN}$

$\Sigma M_D = 0$

$5R_B = 2(50) + 6(30)$

$R_B = 56 \, \text{kN}$

**Segment AB:**

$V_{AB} = -30 \, \text{kN}$

$M_{AB} = -30x \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = -30 + 56$

$V_{BC} = 26 \, \text{kN}$

$M_{BC} = -30x + 56(x - 1)$

$M_{BC} = 26x - 56 \, \text{kN}\cdot\text{m}$

**Segment CD:**

$V_{CD} = -30 + 56 - 50$

$V_{CD} = -24 \, \text{kN}$

$M_{CD} = -30x + 56(x - 1) - 50(x - 4)$

$M_{CD} = -30x + 56x - 56 - 50x + 200$

$M_{CD} = -24x + 144 \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram:**

- In segment AB, the shear is uniformly distributed over the segment at a magnitude of -30 kN.
- In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.
- In segment CD, the shear is uniformly distributed at a magnitude of -24 kN.

**To draw the Moment Diagram:**

- The equation M
_{AB}= -30x is linear, at x = 0, M_{AB}= 0 and at x = 1 m, M_{AB}= -30 kN·m. - M
_{BC}= 26x - 56 is also linear. At x = 1 m, M_{BC}= -30 kN·m; at x = 4 m, M_{BC}= 48 kN·m. When M_{BC}= 0, x = 2.154 m, thus the moment is zero at 1.154 m from B. - M
_{CD}= -24x + 144 is again linear. At x = 4 m, M_{CD}= 48 kN·m; at x = 6 m, M_{CD}= 0.

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