Any help on this statics problem?
This is the exact problem I'm having trouble on for my Building Tech II Class. Any help on where to begin would be great.
- WhomeLv 77 years agoFavorite Answer
Let tension in each section be labeled with a T followed by the end points involved.
The horizontal forces about D must equal zero, assume right is positive direction
Tdesin75 - Tcd(4/5) = 0
Tde = 4Tcd/5sin75
Summing vertical forces about D to zero, up is assumed positive direction.
Tcd(3/5) + Tdecos75 - 200 = 0
substitute the result from the first equation
Tcd(3/5) + (4Tcd/5sin75)cos75 - 200 = 0
Tcd[(3/5) + (4/5sin75)cos75] = 200
Tcd = 245.6 lb < === ANS a
Tde = 4(245.6)/5sin75
Tde = 203.4 lb < === ANS a
Summing horizontal forces about C to zero
Tcd(4/5) - Taccos60 = 0
Tac = 245.6(4/5) / cos60
Tac = 393 lb < === ANS a
Summing vertical forces about C to zero, assume BC is in compression
Tbc - Tacsin60 - Tcd(3/5) = 0
Tbc = 393sin60 + 245.6(3/5)
Tbc = 488 lb < === ANS a
as this result is positive, our assumption about the pole being in compression is valid.
The highest tension exists in section AC. If we reduce its load to a maximum of 320 lb. the other forces will decrease in the same proportion.
320 / 393 = 0.814
Tac = 393(0.814) = 320 lb T
Tbc = 488(0.814) = 397 lb C
Tcd = 245.6(0.814) = 200 lb T
Tde = 203.4(0.814) = 165.5 lb T
W = 200(0.814) = 162.8 lb
As the pole stress is below its maximum compression stress of 450 lb, it holds as well