For wood:

$P_w = \sigma_w \, A_w$

$P_w = 1800 \, [ \, \frac{1}{4} \pi (8^2) \,]$

$P_w = 90\,477.9 \, \text{lb}$

From FBD of Wood:

$P = P_w = 90\,477.9 \, \text{lb}$

For concrete:

$P_c = \sigma_c \, A_c$

$P_c = 650(12^2)$

$P_c = 93\,600 \, \text{lb}$

From FBD of Concrete:

$P = P_c = 93\,600 \, \text{lb}$

For safe load P,

$P = 90\,478 \, \text{lb}$ *answer*

## Comments

## Question on both the FBD of

Question on both the FBD of wood and steel. How come it is presented in such a way that there is a load P on top of the object and each object has a corresponding P going up. I thought we are computing for the P on top of the objects? Kindly explain and thank you.