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.

Problem 1-2

SOLUTION:

Part a

The free-body diagram of section a-a is shown below.1-2a.JPG

 

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.

1-2b

 

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

 

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