truss tie members eccentric 3

[biheja43]

I have got case where the tie members in the truss eccentric from where they should connect to the top and bottom chord so moment will appear on the top and bottom chord because of this eccentricity but how to calculate the moment diagram according to this eccentricity according to static

 

[BAretired]

I am not quite clear on your question. What do you mean by tie members? Are you referring to web members?

If web members of a truss do not intersect the neutral axis of the top or bottom chord at a point, the truss can be analyzed as a rigid frame. Unless the eccentricity is substantial, this is usually not done, however.

BA

 

[biheja43]

well what I've got is triangulated mild steel truss framework from 100x100x10RHS chords with 80x80x6shs internal struts and 100x20 flat bar internal ties and all the internal tie members are eccentricby 30mm at chord connections i need to find out the local bending moment in the chords using statics.
thanks

 

[BAretired]

Maybe it would be best if you provide a sketch of your truss. The force in the tie exerts a moment to the joint of 30*F where F is the force in the tie. This can be distributed to all of the members meeting at the joint in accordance with their stiffness.

Personally, with such a small eccentricity, I would be inclined to ignore it.

BA

 

[biheja43]

thanks for the reply and this is the diagram in all cases i need to calculate it and not to ignore it.
thanks

 

[hokie66]

The chords and verticals of your truss are lightly loaded, so can well handle the moment due to the eccentricity. Just distribute the moment as described by BA in proportion to the stiffness of the three SHS members forming the joint.

The more important issue is how you connect the flat bar diagonal tension members. A gusset through the SHS members would be one solution. Using another SHS for these webs might be a better detail.

 

[rscassar]

It is eccentric but the moments in the SHS tube will not be that great and you should be able to transfer the shear through the panel to the strut. How have you detailed the flat bar connecting to the 100x6SHS tube. I would have gone a thicker thickness SHS and called up a full strength butt weld between the 20plate and the 100SHS.

 

[biheja43]

So what you saying is calculate the moment on the top and bottom chords by taking the value of the tie internal force as if its connect to the chord without eccentricity and then multiply it by the eccentric distance and that moment will act on joint where they should connect then distribute the force generate from this moment according to the stiffness of each member connect to the joint (exclude the tie member) and to draw the moment distributed diagram will draw the top chord and the bottom chord moment diagram as a beam with the original load and the moment from the eccentricity and the beam will be simply support beam and the connection points on the chord to take as fixed hope this is the case or please correct this information if I was wrong .
thanks

 

[hokie66]

You are making this too complicated. If there is eccentricity (and there need not be), just take the force in the diagonal, multiply it by the eccentricity, and distribute it into the other three members at the joint. In fact, in your case, the stresses are low and you can take it all into one member. But again, the connection of the flat bar is the issue.

 

[csd72]

The way I treat this is as foloows:

Sketch the centrelines of the chords and the diagonals.

Measure the vertical distance between where the chords intersect and the centreline of the chord.

Now multiply the horizontal components of both diagonals by this eccentricty to get the total eccentric moment.

If your diagonals are evenly spaced then half of this will be distributed each way and you will get a sawtooth type configuration for your moment diagram.

As the chords are much less stiff then I usually ignore the distribution into these.

It is worth noting that the maximum moment from this eccentricity does not coincide with the maximum compression in the chord.

 

[BAretired]

A 100 x 20 flat bar has a radius of gyration of 5.77mm. The l/r ratio for the diagonal member is 1470. This is too flexible even for a tension member. You should use a member with minimum r of L/300 = 28.3 which would require HSS 75x75.

So, I would suggest you use HSS 75x75 for all web members, both vertical and diagonal.

BA