Solution to Problem 558 | Unsymmetrical Beams

From Solution 557, tension governs at both positive and negative maximum moments.
 

At M = -0.5w_o x² N·m:
$20 = \dfrac{0.5w_o x^2(180)(1000)}{36 \times 10^6}$

$w_o = 8000 / x^2$
 

At M = -0.5w_o x² + 0.5w_o(5 - x)² N·m:
$20 = \dfrac{[ \, 0.5w_o x^2 + 0.5w_o(5 - x)^2 \, ] \, (50)(1000)}{36 \times 10^6}$

$14\,400 = -0.5w_o x^2 + 0.5w_o(5 - x)^2$

$28\,800 = -w_o x^2 + w_o(5 - x)^2$

$28\,800 = -w_o x^2 + w_o(25 - 10x + x^2)$

$28\,800 = -w_o x^2 + (25 - 10x)w_o + w_o x^2$

$28\,800 = (25 - 10x)w_o$

$28\,800 = (25 - 10x)(8\,000/x^2)$

$(28\,800x^2 / 8000) = 25 - 10x$

$\frac{18}{5}x^2 = 25 - 10x$

$18x^2 = 125 - 50x$

$18x^2 + 50x - 125 = 0$

$x = 1.59 \, \text{ m and } \, -4.37 \, \text{(absurd)}$
 

use x = 1.59 m answer
 

$w_o = 8000 / 1.59^2$

$w_o = 3164.43 \, \text{N/m}$

$w_o = 3.16 \, \text{kN/m} \,\, $ answer