Internal force, members of truss

[jhox08]

please see the attached image, I'm supposed to find the internal force in each member of the truss.





I used moments to find My=40 and Gy=120, I then moved on to FBD of joint G where I found Fgh=30kN in tension, and Fga=50kN in compression, next I did joint A to find Fab=30kN in tension and Fah=60kN in compression...I'm stuck on joint H, I have too many unknowns for my equations...I also get confused with the signs in this problems, any help is greatly appreciated...

also, please let me know if I'm wrong on any of my values...thanks

 

[PhanthomJay]

I'm struggling to read your loads...looks like 30 and 30 at the top, and 40 and 20 at the bottom? If so, loads add up to 120, so reactions at G and M must add to 120...I get My = 40 up and Gy = 80 up...but maybe I am misreading the figure. Please confirm loading. When isolating joints, you should remember that if the force in a member points toward the joint, the member is in compression...if the force in the member points away from the joint , it is in tension.

 

[jhox08]

yes you're correct on the loads and I agree Gy is 80, mistake on my end...at joint G, I get Tgh=30 in tension, Tga=50 in compression, and this is where I am pretty shaky, since Tga is in compression at joint G it must point into joint A as well, so the FBD for joint A is 30kN down, Tab to the right, Tah down and Tab=50 into the joint, would my equilibrium equations be Fx=0=Tab +(3/5)Tag and since Tag is in compression do I keep the negative sign in the equilibrium equation so its Fx=0=Tab+(3/5)(-50), so Tab=30 in tension?...same confusion with Fy...Fy=0=-30-Tah+(3/5)Tab...Fy=0=-30-Tah+(3/5)(-50)...Tah=-60

 

[p21bass]

Sorry - never mind...

 

[PhanthomJay]

yes you're correct on the loads and I agree Gy is 80, mistake on my end...at joint G, I get Tgh=30 in tension, Tga=50 in compression, and this is where I am pretty shaky, since Tga is in compression at joint G it must point into joint A as well, so the FBD for joint A is 30kN down, Tab to the right, Tah down and Tab=50 into the joint, would my equilibrium equations be Fx=0=Tab +(3/5)Tag and since Tag is in compression do I keep the negative sign in the equilibrium equation so its Fx=0=Tab+(3/5)(-50), so Tab=30 in tension?...

It's so easy to let that plus and minus sign bite you, isn't it? Tab is 30 in compression. And you are messing up your subscripts, another easy thing to do. Tga is 50 into the joint..it's vector components are then 30 to the right and 40 up..that means Tab must point to the left for equilibrium, therefore , it is in compression

...same confusion with Fy...Fy=0=-30-Tah+(3/5)Tab...Fy=0=-30-Tah+(3/5)(-50)...Tah=-60

You've got 30 applied down, and 40 up from vert component of Tga, that's a net of 10 up, so for equilibrium, Tah must be 10 down, or in tension. ...continue...

 

[jhox08]

first of all thank you so very much for your help...how am I messing up my subscripts?...so continuing on, I've redone all of my calculations...Tga=50 C, Tgh=30 T, Tah=10 T, Tab=30 C, Thb=37.5 T, Thj=7.5 T, Tjb=20 T, Tjk=7.5 T, I really hope I'm right with these...

I don't feel very confident with the following, I get twisted about when to use sin and cos of the angle

Tbk=25 C from equation Fy=0=-20-(4/5)Tbk
Tbc=45 C from Fx=0=Tbc+30+(3/5)Tbk
Tkc=15 C from Fy=0=Tkc+(3/5)Tkb
Tkl=12.5 C from Fx=0=Tkl-7.5+(4/5)Tkb
Tcl=56.25 C from Fy=0=-30-15-(4/5)Tcl
Tcd=11.25 C from Fx=0=45+Tcd-(3/5)Tcl