Truss Analysis by hand calculation

[jsu0512]

I'm trying to do hand calculation to check the moment diagram i'm getting from SAP 2000 for the attached diagram.

The horizontal beam (top & bottom ) is HSS 6"x3"x1/8" and continues between the support that are 6m distance to each other.

The vertical frames are pin connected between the top and bottom HSS beams.

There is linearly distributed load of 3kN/m along the bottom HSS beam.

Can anyone shed me some light on how to do this with hand calculation?

It would be solvable by hand if there is bracing in each opening but this diagram has no opening.

I want to confirm the moment from SAP result by hand calculation.

 

[jayrod12]

Well SAP will give you an answer based on stiffness, i.e. deflection, so you'd need to do the same.

the bottom beam will impart load on the tension hanger at the middle which will put a point load on the upper beam. The overall diagram will have the deflection at mid-span of both beams be equal. It's a bit tough to do by hand, but with enough iteration I think you can make a go of it.

So what you need to do in my mind (keeping in mind this may not be the most efficient way of doing it),
1)Assume the tension member acts like a rigid support to determine the amount of tension going through it.
2)Use that tension load as a point load on your upper beam, and determine that beams deflection.
3)Go back to the lower beam and recalculate as a simple span beam spanning 6m with a udl on it going down, and a point load at the middle going up, solve for the point load for the expected deflection calculated for the top beam.
4)Using the tension load determined in step 3, recalculate expected deflection in top beam like point 2.
5) repeat steps 2-4 until the solution converges.

 

[rb1957]

well that is not a classical truss ... no diagonals. The lower chord is carrying the applied load in bending to the supports, the other members are (for all intents and purposes) unloaded.



another day in paradise, or is paradise one day closer ?

 

[dik]

That's a Vierendeel truss... tricky to do by hand, but can be done... no pinned connections allowed.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[rb1957]

what, the upper chord is loaded by rotation of the side verticals ?

yes, the fixity coefficients of the joints is quite hard to hand calc.

still not a classical truss (axial members, no transverse loads)

another day in paradise, or is paradise one day closer ?

 

[ProgrammingPE]

Since you are checking this using pinned connections (typically a vierendeel truss like this uses fixed connections), you can actually determine the forces directly fairly easily. Assuming the center vertical member is rigid, the deflection in the top chord and bottom chord will be identical at the midspan.

The deflection for a simply supported beam with a point load at midspan is Δtop = P*l³/(48*E*I). This will be the midspan deflection for the top chord. For the bottom chord, you will have a vertical point load at the midspan and the uniform load, so its midspan deflection will be Δbottom = 5*w*l^4/(384*E*I) - P*l³/(48*E*I). You can then determine the tension in the center vertical member, P, by setting Δtop = Δbottom which will give you P = 5.625kN.

The moment in the top chord will then be Mmid = P*l/4 = 8.4kN-m. The moment in the bottom chord will be Mx = w*x/2*(l-x) - P/2*x which reaches its max around 2.1m at 6.4kN-m. I'm guessing the discrepancy is because the SAP model is also including the self weights.

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[BAretired]

It is very straightforward to do that calculation by hand. Neglecting the strain in the middle vertical, the deflection of the two beams are equal. The top beam has load T acting down. The bottom beam has uniform load w acting down and concentrated load T acting up. The midspan deflection of top and bottom beam is equal.

so 5wL^4/384EI - TL^3/48EI = TL^3/48EI
5wL^4/384 = TL^3/24;
T = 5wL*24/384 = 0.3125wL

If the strain in the vertical is worth considering, it can easily be included.



BA

 

[dik]

The horizontal shear is taken by the web verticals and this force is transferred to the to and bottom chords as a moment... I've done a lot of pedestrian bridges using this type of framing. It's quite elegant, IMHO...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[dik]

Without fixity of some sort at the connections think of it as a long slender beam with a depth of it equal to the depth of the bottom chord...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[MotorCity]

Two suggestions: If you want a quick, accurate, easy check.....use moment distribution method of analysis. If you want a quick, conservative check, just check the bottom beam to carry the load (neglect frame action). Its a pretty small span and rather inefficient compared to a traditional truss configuration.

 

[BAretired]

Without fixity of some sort at the connections think of it as a long slender beam with a depth of it equal to the depth of the bottom chord...

It is more like two beams acting together, dik. Fixity of the exterior verticals would stiffen the two beams. Fixity of the middle vertical would have no effect under uniform load because there is no rotation at midspan for either top or bottom beam.

BA

 

[BAretired]

Two suggestions: If you want a quick, accurate, easy check.....use moment distribution method of analysis. If you want a quick, conservative check, just check the bottom beam to carry the load (neglect frame action). Its a pretty small span and rather inefficient compared to a traditional truss configuration.

Moment distribution is not required. The joints are all pinned. This is not a truss. There is no frame action. It is simply two beams sharing a load.

BA

 

[dik]

maybe one beam if there is only a pinned connection from the vertical to the top chord...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

Both beams are continuous, dik.

BA

 

[MotorCity]

If all joints are pinned, it will be unstable.

 

[jsu0512]

ProgrammingPE and BAretired.

Thank you for the simple solution. It works and matches with the moment and shear diagrams i'm getting from SAP program.

I really appreciate it.

 

[dik]

Thanks, MC... if the top chord is loaded, the load is transferred to the bottom (albeit as point loads) and the top chord has no contribution... if the bottom chord is loaded, the load is not transferred to the top and the top chord has no contribution.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

If all joints are pinned, it will be unstable.

That is true. The upper beam needs to be laterally supported.

BA

 

[dik]

...and if we're being picky (which I think we are a little) the upper beam needs to be vertically supported... I think I've had enough fun (I think we both know what we are talking about)...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

...and if we're being picky (which I think we are a little) the upper beam needs to be vertically supported...

The upper beam is vertically supported.

BA