An easier way to solve this is by taking a moment at A and solving for Fy, then using the right side of the truss. Make a cut at the same spot you did in the problem and take a moment at C. Then you don't need to solve for the reactions at A and you don't need to use any of the loads at I or C in solving for HB.
Is there a reason you chose to section between CD vs BC? When Solving thru BC you remove the 20k from your equations, simplifying them and return an answer of approx. 28.1 which would still be the correct answer. Do you gain a more accurate number by including more? M(C)= -20k(60')+10k(30')-F(IH)cos14(33') = ~28.1k
I was wondering the same thing, but then it clicked that if you cut at BC you will have to include force at IC in addition to IH taking M(B), still possible but won't be able to solve it with one equation. Doing the cut at CD lets you take M(C) eliminating forces at IC and CH and the 20k on that joint, leaving IH to solve more simply.
Ncees pe hb 1.1 Felt like a statics 101 homework problem. Lengthy, messy problem, with multiple ways to mess up. Good pe practice but not a pe-grade problem. Thanks Cody. Ans = 28.52 k =
Ncees pe hb 1.1 Felt like a statics 101 homework problem. Lengthy, messy problem, with multiple ways to mess up. Good pe practice but not a pe-grade problem. Thanks Cody. Ans = 28.52 k =
An easier way to solve this is by taking a moment at A and solving for Fy, then using the right side of the truss. Make a cut at the same spot you did in the problem and take a moment at C. Then you don't need to solve for the reactions at A and you don't need to use any of the loads at I or C in solving for HB.
Is there a reason you chose to section between CD vs BC? When Solving thru BC you remove the 20k from your equations, simplifying them and return an answer of approx. 28.1 which would still be the correct answer. Do you gain a more accurate number by including more?
M(C)= -20k(60')+10k(30')-F(IH)cos14(33') = ~28.1k
I was wondering the same thing, but then it clicked that if you cut at BC you will have to include force at IC in addition to IH taking M(B), still possible but won't be able to solve it with one equation. Doing the cut at CD lets you take M(C) eliminating forces at IC and CH and the 20k on that joint, leaving IH to solve more simply.
Ncees pe hb 1.1
Felt like a statics 101 homework problem. Lengthy, messy problem, with multiple ways to mess up. Good pe practice but not a pe-grade problem.
Thanks Cody.
Ans = 28.52 k =
Ncees pe hb 1.1
Felt like a statics 101 homework problem. Lengthy, messy problem, with multiple ways to mess up. Good pe practice but not a pe-grade problem.
Thanks Cody.
Ans = 28.52 k =