# Solution to Problem 420 | Shear and Moment Diagrams

**Problem 420**

A total distributed load of 30 kips supported by a uniformly distributed reaction as shown in Fig. P-420.

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 420**

$w = 30(1000)/12$

$w = 2500 \, \text{lb/ft}$

$\Sigma F_V = 0$

$R = W$

$20r = 30(1000)$

$r = 1500 \, \text{lb/ft}$

**First segment (from 0 to 4 ft from left):**

$V_1 = 1500x$

$M_1 = 1500x(x/2)$

$M_1 = 750x^2$

**Second segment (from 4 ft to mid-span):**

$V_2 = 1500x - 2500(x - 4)$

$V_2 = 10000 - 1000x$

$M_2 = 1500x(x/2) - 2500(x - 4)(x - 4)/2$

$M_2 = 750x^2 - 1250(x - 4)^2$

**To draw the Shear Diagram:**

- For the first segment, V
_{1}= 1500x is linear; at x = 0, V_{1}= 0; at x = 4 ft, V_{1}= 6000 lb. - For the second segment, V
_{2}= 10000 - 1000x is also linear; at x = 4 ft, V_{1}= 6000 lb; at mid-span, x = 10 ft, V_{1}= 0. - For the next half of the beam, the shear diagram can be accomplished by the concept of symmetry.

**To draw the Moment Diagram:**

- For the first segment, M
_{1}= 750x^{2}is a second degree curve, an open upward parabola; at x = 0, M_{1}= 0; at x = 4 ft, M_{1}= 12000 lb·ft. - For the second segment, M
_{2}= 750x^{2}- 1250(x - 4)^{2}is a second degree curve, an downward parabola; at x = 4 ft, M_{2}= 12000 lb·ft; at mid-span, x = 10 ft, M_{2}= 30000 lb·ft. - The next half of the diagram, from x = 10 ft to x = 20 ft, can be drawn by using the concept of symmetry.