Why Compression force in Tension members?

The explanation to this issue can be found here: https://communities.bentley.com/products/ram-staad/w/structural_analysis_and_design__wiki/26838/compression-force-developed-in-tension-members

I am still waiting for you answer.

If a tension-only member is vertical or sloping, the selfweight of the member will induce a varying axial force which will be zero at the top node and compressive at the bottom node. In combination cases where the selfweight load case is a component, this will cause the axial force to be tensile at the top and compressive at the bottom. In such cases, the nature of the force at the 2 ends cannot be used as a criteria to decide if the member is in tension or in compression. So, the program uses the sign of force at the mid-point of the member.

Member 294 in your model seems to belong to this category. For load case 145, find the force in the middle by taking the average of the forces at the start and at the end. If it is a negative value, then, the member is considered to be in tension.

However, during steel design, the member has to be designed for the forces at the start node as well as the end node. At the end which is in compression, the member is checked against the AISC clauses for compression.

Thanks for your reply. But, as per your comments i toggle the self weight of the tension members, but it fails again in compression. I have also taken the average of the forces at the start and at the end. And it is a negative value. So, the member should be considered as tension members. But why it fails again in compression? When i increase the size of the members then it passes. But why in given size its behaving such like. What is the solution of this? How can i pass this members without increasing size? Staad model is attached again for your kind review.Tyre Curing Press Bldg..std

Can you please mention the version of STAAD Pro you are using?

Sure. STAAD.Pro Advanced CONNECT Edition - Version 21.00.02.43

If you consider member 294, the axial forces at the start and end nodes are as mentioned next

start node 28 = 0.076 kN ( Axial compression )

end node 657 = 0.081 kN (Axial tension) 

So the axial force at the center of the member is -0.081+0.076/2 = -0.0025 which is tensile and that why the member has not been deactivated from the analysis.

Using MAIN and TMAIN as 1 would allow you to bypass the slenderness checks and that is the reason why you are not getting a slenderness failure in spite of the fact that the actual slenderness for the member is way above the allowable limits. However the member would still be designed at all 13 sections. At the first section the axial compression comes out to be 0.076 KN which is more than the capacity in compression and that is why the section fails the code check.  

If you do not have to design these sections, you may just leave those out from your CHECK CODE. If you would like to do a CHECK CODE then you would have to use a bigger section.

If you consider member 294, the axial forces at the start and end nodes are as mentioned next

start node 28 = 0.076 kN ( Axial compression )

end node 657 = 0.081 kN (Axial tension) 

So the axial force at the center of the member is -0.081+0.076/2 = -0.0025 which is tensile and that why the member has not been deactivated from the analysis.

Using MAIN and TMAIN as 1 would allow you to bypass the slenderness checks and that is the reason why you are not getting a slenderness failure in spite of the fact that the actual slenderness for the member is way above the allowable limits. However the member would still be designed at all 13 sections. At the first section the axial compression comes out to be 0.076 KN which is more than the capacity in compression and that is why the section fails the code check.  

If you do not have to design these sections, you may just leave those out from your CHECK CODE. If you would like to do a CHECK CODE then you would have to use a bigger section.

Thanks a lot for your information's. But, please let me clear about the words, "member would still be designed at all 13 sections". I am little bit confused about that. 

STAAD.Pro considers 13 equally spaced sections along the member length (including the start and the end locations) and carries out the design at each section considering all loads. The one with highest utilization ratio is then reported. 

Thanks a lot for your information. Now it is clear to me.