Like others, I'm getting CD = 2404.01 # (Tension), and CI = 3334.62 # (Compression). My calculation was as follows: ΣFy = 0 = 1000 - (3606)Sin(56.3°) + (CD)Sin(56.3°) => (CD)Sin(56.3°) = (3606)Sin(56.3°) - 1000 => (CD)Sin(56.3°) = 2000.03 => CD = 2000.03/Sin(56.3°) = 2404.01 # For evaluating forces in x-axis, CD sign will remain positive as it is tension. ΣFx = 0 = (CD)Cos(56.3°) + CI + (3606)Cos(56.3°) => CI = -(CD)Cos(56.3°) - (3606)Cos(56.3°) => CI = -(2404.01)Cos(56.3°) - (3606)Cos(56.3°) => CI = -3334.62 #
for those confused about point E only having a horizontal reaction force component. Point E is a fixed support. The only other force acting on point E is ED (either in tension or compression, regardless it is a horizontal force).Therefore, the only Reaction force possible from the fixed support at E is the Ex force that balances out the ED force.
@@ricardoramosiii8796 remember this important fact about trusses: each member can only be axially loaded (meaning the bar is either stretched or compressed). So looking at point e, we can see that the horizontal member, ED, is the only member touching that support point. Since it is horizontal, and can only be axially loaded, we can conclude that there is no reaction force in the y direction. We know that there is no moment reaction because the force’s line of action passes through the point in question.
yes, that essentially what he did. notice that the only other force at E is ED (either in tension or compression, regardless it is a horizontal force). Therefore the only Reaction force at E is Ex.
My answers I got: FCJ = 3605.6 T FCB = -1000 C FIJ = -6000 C FCI = -3333.3 C What do others think I double checked my work and couldnt find a single thing wrong.
Jeff is right Robert. You need to follow the equation equal to zero. -4000-4CB=0; -4000=4CB; -1000=CB (Now -ve changes sign on the diagram). Hope that helps.
I am at problem no 10 right now and I will comment when I done all trusses problem than I will watch ur next machine and frame videos then I solve rest problems let's see how much time I take 26 dec 2021
Like others, I'm getting CD = 2404.01 # (Tension), and CI = 3334.62 # (Compression). My calculation was as follows:
ΣFy = 0 = 1000 - (3606)Sin(56.3°) + (CD)Sin(56.3°)
=> (CD)Sin(56.3°) = (3606)Sin(56.3°) - 1000
=> (CD)Sin(56.3°) = 2000.03
=> CD = 2000.03/Sin(56.3°) = 2404.01 #
For evaluating forces in x-axis, CD sign will remain positive as it is tension.
ΣFx = 0 = (CD)Cos(56.3°) + CI + (3606)Cos(56.3°)
=> CI = -(CD)Cos(56.3°) - (3606)Cos(56.3°)
=> CI = -(2404.01)Cos(56.3°) - (3606)Cos(56.3°)
=> CI = -3334.62 #
Me too..
me too my dudes, Prof Jeff mustve made an oopsie
me too my dood, Prof Hanson mustve made an oopsie xDDD
@@dimitriladas2888 a little oopsie is acceptable 🙊
@@dimitriladas2888 even professors can do an oppsie
for those confused about point E only having a horizontal reaction force component. Point E is a fixed support. The only other force acting on point E is ED (either in tension or compression, regardless it is a horizontal force).Therefore, the only Reaction force possible from the fixed support at E is the Ex force that balances out the ED force.
doesn't a fixed support have a reaction in the y-axis and also have a moment since its fixed and can't move?
@@ricardoramosiii8796 remember this important fact about trusses: each member can only be axially loaded (meaning the bar is either stretched or compressed). So looking at point e, we can see that the horizontal member, ED, is the only member touching that support point. Since it is horizontal, and can only be axially loaded, we can conclude that there is no reaction force in the y direction. We know that there is no moment reaction because the force’s line of action passes through the point in question.
Thank you for that info.
@@kylejohnson8447 so how do we differentiate when it’s a 2 force member or just a fixed support with reaction forces?
CD= 2403# T
CI= 3332.6 # C
Member CD is 2404.01
Yea, this is what I got as well
and it's in tension
I concur
Thanks, thought i was going crazy
me too so CI 3334C
for the joint C calculation, the CD should equal 2404 in tension and the IJ should equal 3333 in compression.
actually IJ is correct and it is 6000, but as for CD and CI they are incorrect. CD=2404 and CI=4334.67.
@@mohammadhamza3937 the IJ I mentioned is actually CI, which is a mistake. but the value of CI should be 3333 I think.
@@gavinliu9710 agreed. CI is around 3333
yeap
IJ is 6000 compresion
DR. Hanson, thank you for another awesome Truss and Combo Problem and their solution.
Wonderful Statics lessons!! really taking the time to explain the problems along with a good sense of humor :))
Ahh he almost got his test question right! Second to last step! A+ on the explanation all the same.
Hello from Turkey Mr. Jeff :)
CD=2404# T and CI=3335#C Please Jeff can you check
I got confused about how is that possible to eliminate the vertical force at point E
its considered as a cable note there is no memberCE so it can bw considered as a cable ie in tension !
member CD is 2404 tension and CI is 3334 tension also.
CI gives negative value so it will be compression.
Can you come up with an ultimate problem that includes friction or a machine?
Could you also solve for the moment around A to find IJ?
At E, can we solve it as a fixed support with 2 reaction forces and 1 moment?
yes, that essentially what he did. notice that the only other force at E is ED (either in tension or compression, regardless it is a horizontal force). Therefore the only Reaction force at E is Ex.
it looked like E was just a support with some kind of rope or line attached to the truss at D, so I was right to assume only a horizontal force there
Isn’t the connection at E a fixed support? I thought fixed connections couldn’t be two-force members because they apply a couple?
My answers I got:
FCJ = 3605.6 T
FCB = -1000 C
FIJ = -6000 C
FCI = -3333.3 C
What do others think I double checked my work and couldnt find a single thing wrong.
5:37 CJ 🖤
Mesa CC should just give my tuition to Dr Jeff Hanson
I might be mistaking but I think cb has the wrong sign.... otherwise I'm confused
Jeff is right Robert. You need to follow the equation equal to zero. -4000-4CB=0; -4000=4CB; -1000=CB (Now -ve changes sign on the diagram). Hope that helps.
What difference would it make if i cut thru the members vertically?
when getting reactions at support, do i always have to get the moment in that support?
please can you do a video on solving space truss problems?
I tried using the method of joints for the entire truss, but I got bored after about 20 minutes, haha!
it's good practice though
I am at problem no 10 right now and I will comment when I done all trusses problem than I will watch ur next machine and frame videos then I solve rest problems let's see how much time I take 26 dec 2021
Its been 2 and a half years bro
5:16 🥲
We can start with the above half and take moment at c we get the force in the IJ member by one step in the beginning of solution
This is the trick you used in the last example 😁😁
Nope! you have to find Ex first using global equ. after that the easiest way is what Dr.Hanson did
can I do moment 3 times? and solve the problem with just moment equation.
i guess you could try
theoretically it's possible
but why would you torture your self with moments
their calculation takes more time
Something is wrong. Your CD is supposed to be 2404.01 KN
JOHNNY WEAKSAUCEEEEE
member IJ is 0 since 4000-1000+3606sin56.3 +IJ =0
where 3606sin56.3=0.02
sorry, you are right