**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*