# Solving an indeterminate truss with static redundancy of 2 confusion

I have the following truss and I want to calculate the reactions using the flexibility method

Using this method means that I need to break this indeterminate system into a determinate system and two redundant systems (with the reintroduced member force) in this case. In other to obtain these subsystems, I need to release members (or reactions).

Say I want to release 2 of the members from the truss, how do I know which members to release?

Do the members that I remove have to be the members that are connected the supports? (ie to use the compatibility equation).

So a follow up question is where does the compatibility equations apply? (Only at the supports?)

I have tried to remove 2 members that are connected to the supports but using a software it tells me that the structure is unstable. (with full hinges applied etc)

I then proceed to remove other 2 members, they are stable, but they are not connected to supports. Does the compatibility equation apply here?

• Remove the diagonal redundant truss member as it carries zero load on calculations Oct 30, 2022 at 9:20

Hint - On top of pin supported ends (one more than the equations for global equilibrium conditions), the two diagonal members in the middle panel possess one degree of redundancy. So you should release the horizontal restraint at one of the pin supports, and release one diagonal member addressed above.

For compatibility, apply a unit load at the released support, and a pair of unit loads at the joints connecting the released diagonal member, the unit loads should be in the opposite directions.

The compatibility should be carried out at the joint/member with the red unit load, to close out the horizontal gap at the released support, caused by the releases.

Note:

• Hi r13 thanks for the hints yet again. Can you elaborate more in terms of how you know that I should remove one of the two diagonal members in the middle panel? How did you get one degree of redundancy? I just can't see it. Also why the horizontal restraint at one of the pin supports? At the moment, choosing to release the members seems kinda arbitrary to me. Also when does the compatibility equations applies? Only at the supports? Many thanks again r13 May 14, 2022 at 15:20
• You have a total of 2 degrees of freedom, one at the support (4 unknowns for 3 equations), and one at the cross diagonal members. The released structure must be both stable and structurally determinate. You can't release the vertical restraint at the support, as the structure will become unstable. It goes without saying, that one of the cross diagonal members must be released in order for the remaining structure to be a determinate system that can be solved by usual methods.
– r13
May 14, 2022 at 15:30
• Thanks. Can you explain the compatibility equations as well? Like where/when can I apply the equation? Like why are you allowed to apply the compatibility equation at the diagonals? I hope this question make sense. I hate to say this, but I am still confused about the diagonal part. I don't understand how you are just able to look at a truss structure and know which members to remove? How did you figure out the there is one df at the cross diagonal members? May 14, 2022 at 15:53
• I hope you've realized that the additional diagonal member causes internal indeterminacy. A quick check on the simply supported truss system (after release one unknown at one of the supports) - (10 j)oints, 18 (m)embers, 3 unknown (r)eactions . i = (m+r)-2j = 18+3-2*10 = 1 degree of redundancy. Now we know one of the truss members must be removed to get rid of the redundancy, yet, the truss remains stable. So removing one of the diagonal from the cross diagonal members is the most reasonable/logical choice. So far so good?
– r13
May 14, 2022 at 16:52
• yes I think I can see that. Sorry for the late reply, I was unwell. May 20, 2022 at 16:34