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STATICS - CASE STUDY SOLUTION


Joint A
  Joint A

 

The line of action of the force is 30 cm from joint A. Therefore, the magnitude of the bending moment is

     MA = F dA

           = (80 N) (30 cm) = 2,400 N-cm

   

Joint B
  Joint B

 

The distance from joint B to the line of action of F is found from the Pythagorean theorem:

     dB = (302 + 452)0.5 = 54.08 cm

Substitute and solve for the magnitude of the moment:

     MB = (80 N) (54.08 cm) = 4,327 N-cm

 

     

Joint C
  Joint C

 

Calculate the moment arm and the moment at joint C in the same manner as for joint B:

     dC = (152 + 452)0.5 = 47.43 cm

     MC = (80 N) (47.43 cm) = 3,795 N-cm

 

     

Joint D
  Joint D and E

 

Calculate the moment of F at joints D and E in the same manner as for joints B and C:

     dD = (152 + 152)0.5 = 21.21 cm

     MD = (80 N) (21.21 cm) = 1,697 N-cm

     dE = (302 + 152)0.5 = 33.54 cm

     ME = (80 N) (33.54 cm) = 2,683 N-cm

 

     

Joint E
  Maximum Moment

 

The joint with the largest moment is joint B, which has a moment of 4,327 N cm, or in more common units 43.27 N-m.

 

     
   
 
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