Method of Pins- Trusses (conceptual)

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giacomh
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Homework Statement



I am designing a bridge that looks like this:

http://imgur.com/zFsW7

and using the method of joints to find the tension in each pin. The bridge has to hold 10,000 lbs on each pin, and we have to find the maximum tension in each member (so tension of each member when the weight is on A, then 10,000 B, etc.)

I'm checking my answer by placing the 10,000 lbs on pin I and checking symmetry. (I can't really show work because its in an excel spreadsheet and there are quite a few joints). Every answer is tension is symmetrical except GH/HK and FH/HJ.


Attempt at problem

So, based on summations in the x direction for the pins G and K, GH=DG and HK=KL. So the entire top truss should have equivalent tensions (besides CA and OP)? I don't understand how that works. Will GH=HK no matter where the weight is placed on the bridge?
 
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giacomh said:

Homework Statement



I am designing a bridge that looks like this:

http://imgur.com/zFsW7

and using the method of joints to find the tension in each pin. The bridge has to hold 10,000 lbs on each pin, and we have to find the maximum tension in each member (so tension of each member when the weight is on A, then 10,000 B, etc.)

I'm checking my answer by placing the 10,000 lbs on pin I and checking symmetry. (I can't really show work because its in an excel spreadsheet and there are quite a few joints). Every answer is tension is symmetrical except GH/HK and FH/HJ.


Attempt at problem

So, based on summations in the x direction for the pins G and K, GH=DG and HK=KL. So the entire top truss should have equivalent tensions (besides CA and OP)? I don't understand how that works. Will GH=HK no matter where the weight is placed on the bridge?

Just checking - but you mention the tensions in HK for example.
Did you have a "negative Tension" value - indicating it is actually compression in that member?
 
Yeah, GH, HK, DG, and KL are all compressive