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.

http://www.chegg.com/homework-help/questions-and-a...

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  • Whome
    Lv 7
    7 years ago
    Favorite 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

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