At M = +1.0P lb·ft the upper fiber is in compression while the lower fiber is in tension.
$M = M_r$
$M = \dfrac{f_b I}{y}$
For fibers in compression (upper fiber):
$M_c = \dfrac{10(192)(1000)}{2.5}$
$1.0P = 768\,000 \, \text{lb}\cdot\text{in}$
$1.0P = 64\,000 \, \text{lb}\cdot\text{ft}$
$P = 64\,000 \, \text{lb}$
For fibers in tension (lower fiber):
$M_t = \dfrac{4(192)(1000)}{4}$
$1.0P = 192\,000 \, \text{lb}\cdot\text{in}$
$1.0P = 16\,000 \, \text{lb}\cdot\text{ft}$
$P = 16\,000 \, \text{lb}$
At M = -1.5P lb·ft, the upper fiber is in tension while the lower fiber is in compression.
$M = M_r$
$M = \dfrac{f_b I}{y}$
For fibers in compression (lower fiber):
$M_c = \dfrac{10(192)(1000)}{4}$
$1.5P = 480\,000 \, \text{lb}\cdot\text{in}$
$1.5P = 40\,000 \, \text{lb}\cdot\text{ft}$
$P = 26\,666.67 \, \text{lb}$
For fibers in tension (upper fiber):
$M_t = \dfrac{4(192)(1000)}{2.5}$
$1.5P = 307\,200 \, \text{lb}\cdot\text{in}$
$1.5P = 25\,600 \, \text{lb}\cdot\text{ft}$
$P = 17\,066.67 \, \text{lb}$
The safe load P = 16 000 lb answer
