Understanding Truss Problems: Solving for Reaction Forces and Member Forces

[smashbrohamme]

ok this is a basic truss problem and I have to solve every member in the truss(see picture)

Basically I started off solving my reaction forces...For the bottom left Pin, i got 8kips goin left, and 9kips acting down...for the bottom right roller I got 9kips acting up.

So after working my way through the reaction forces I started making FBD's

I started with E..and ended up making Fec = 9kips going upwards, and Fef = 8kips going right.

needless to say I worked my way up the truss and found out when I got to FBD of B (which is my last FBD) I did something wrong, because wouldn't BD and BA be zero force members?

here is my current answer, I won't use compression and tension due to confusion, I will say the directed location from left, right, up and down.

Fec= 9kips up
Fef = 8kips right
Fcf = 10 kips, up
Fdf = 15kips, down
fcd = 4kips, right
Fac = 15 kips, down.
fad = 5kips, up
fbd = 12kips, up. *this is where I stoped* because obiviously I messed something up.

 

[PhanthomJay]

ok this is a basic truss problem and I have to solve every member in the truss(see picture)

Basically I started off solving my reaction forces...For the bottom left Pin, i got 8kips goin left, and 9kips acting down...for the bottom right roller I got 9kips acting up.

good start!

So after working my way through the reaction forces I started making FBD's

I started with E..and ended up making Fec = 9kips going upwards, and Fef = 8kips going right.

Yes, at joint E

needless to say I worked my way up the truss and found out when I got to FBD of B (which is my last FBD) I did something wrong, because wouldn't BD and BA be zero force members?

yes correct, they are both zero force members.

here is my current answer, I won't use compression and tension due to confusion, I will say the directed location from left, right, up and down.

left or right or up or down depends on what joint you are looking at. If Fec is 9 kips up at joint E (tension member), then Fec (or Fce, if you like) is 9 kips down at joint C (must still be a tension member).

Fec= 9kips up
Fef = 8kips right
Fcf = 10 kips, up

OK to here

Fdf = 15kips, down

Stop! You have a signage error. See my tip in red above.

 

[smashbrohamme]

Ok, here is how I got the 15 kips at Fdf going into tension. which is acting down on the actual drawing.

FBD of E shows that Fef is 8kips into tension..

so taking that into FBD of F you can make a equilibrium equation into the X direction making 8kips-Fcf(9/11.25) giving me Fcf= 10 Compression.

So you got Fcf going 10 up breaking it into a vector of 6, and you got the reaction force of 9kips going up from point F, that would have to make Fdf going into tension of 15kips right?

 

[PhanthomJay]

Ok, here is how I got the 15 kips at Fdf going into tension. which is acting down on the actual drawing.

FBD of E shows that Fef is 8kips into tension..

so taking that into FBD of F you can make a equilibrium equation into the X direction making 8kips-Fcf(9/11.25) giving me Fcf= 10 Compression.

Yes, correct

So you got Fcf going 10 up breaking it into a vector of 6, and you got the reaction force of 9kips going up from point F, that would have to make Fdf going into tension of 15kips right?

No. This is where you are making an error. The words 'up' or 'down' when looking at internal forces in members or joints have no meaning until you reference those words to a particular joint or member, in consideration of Newton's third law. When looking at a free body of a joint, member tension forces ALWAYS pull AWAY from the joint, and member compression forces ALWAYS push TOWARD the joint. So when looking at joint F, since Fcf is in compression, it points TOWARD the joint, so its vertical component of 6 kips must also point TOWARD the joint (acting DOWN). Then when you look at joint F in the vertical direction, you have 9 kips UP from the reaction force at F, and 6 kips DOWN from the vertical component of the member force Fcf, acting TOWARD the joint, for a total vertical contribution from those forces equal to 3 kips UP. Therefore, for equilibrium, the force in member DF, Fdf, must be equal to 3 kips acting down on the joint, which is a compression force of 3 kips in member Fdf. Now continue, and please watch those nasty plus and minus signs! . It is helpful also to try to visualize the problem...those 4 kip applied forces tend to squeeze (compress) the right side member DF and the diagonals, while stretching (tensioning) the left side vertical members.

 

[smashbrohamme]

wow that explains everything...but let me make sure I still got this right

A member is labeled wether its a tension member or a compression member.

So if you take Member EC or CE...saying wether it goes up or down is a dumb way to label it...because for instance, From EC its in tension, which makes it point up...but when I make my FBD of Point C...the member between E and C will no longer point up, because that is a tension member, and once you make a FBD of Point C...force CE will point down!

wow sadly I just took a test on this crap, and if this is right then I am screwed lol.

 

[PhanthomJay]

wow that explains everything...but let me make sure I still got this right

A member is labeled wether its a tension member or a compression member.

So if you take Member EC or CE...saying wether it goes up or down is a dumb way to label it...because for instance, From EC its in tension, which makes it point up...but when I make my FBD of Point C...the member between E and C will no longer point up, because that is a tension member, and once you make a FBD of Point C...force CE will point down!

Yes, this is correct.

 

[smashbrohamme]

thanks buddy, I am going to redo the problem when I get home tonight and see if I come up with a logical reason for the zero force members.

you must really love this stuff to always help kids out like me!