# Problem 1-2|Stress | Mechanics of Materials| Ninth Edition| R.C. Hibbeler|

#### Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.

SOLUTION:

Part a

The free-body diagram of section a-a is shown below.

Solve for normal force at a-a (Na) by summing forces in the normal direction

$\sum F_x=0$

$N_a-500=0$

$N_a=500\:lb$

Solve for the shear force at a-a (Va) by summing forces in the shear direction

$\sum F_y=0$

$V_a=0$

Part b

The free-body diagram of section b-b is shown below.

Solve for normal force at b-b (Nb) by summing forces in the normal direction

$\sum F_x=0$

$N_b-500cos(30)=0$

$N_b=433\:lb$

Solve for the shear force at b-b(Vb) by summing forces in the shear direction

$\sum F_y=0$

$V_b-500sin(30)=0$

$V_b=250\:lb$