Forces in a Howe truss (2D) pinned at BOTH ends

Summary:

...sections method seems not to work when both ends are pinned... it only does when one end is pinned and the other has a roller. I'm not an expert, would you help me?

Main Question or Discussion Point

Dear experts,

How would I find the forces for the truss with both ends pinned? (Fig.1, bottom)

I'm trying to determine the forces in all members of some trusses. I've successfully applied the method of sections to a Howe truss (top truss in Fig.1) and I' ve verified it by FEA with MecWay (free for less than 1000 nodes). However, when I try to do the same to another truss but with both ends pinned (bottom truss in Fig.1), sections method seems not to work. I think I can't find some reactions at the pinned ends in order to make the sections method work.

As a workaround, I obtained the shear force V(x) and momentum M(x) of an "equivalent" beam, i.e. with both ends clamped, and used these V(x) and M(x) in the Free Body Diagrams... Unfortunately, again, this only worked for the truss pinned at one end with a roller in the other.

Figure 1. Static 3D analysis of two Howe trusses performed in MecWay v12 with “line2” truss elements (one per beam) and equal cross-section. Both trusses were loaded with 1 kN/m line load on top chord. The top truss is pinned on the left end and has a roller on the other end. The bottom truss is pinned at both ends.

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PhanthomJay
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There are 8 unknown reaction loads, 2 at each of the 4 pins. Because of symmetry and other factors, this can be reduced to just 2 unknowns, the horizontal top pin reactions, equal at each, and the bottom horizontal pin reactions, equal at each. They cannot be solved without doing an indeterminate analysis, which is a bit tedious using hand calcs.

Thank you very much PhanthomJay!

You're right, I forgot performing the basic indeterminacy check for trusses... sorry.

To solve my problem, I've been studying a bit some info about indeterminate analysis for trusses. I think it is right leaving some of them here for future reference. You can find some examples in PDF format here. In addition, I've found some YouTube videos about this topic.
It seems that it is possible to obtain all member forces for my externally indeterminate truss once all the reactions are known. I've checked this by obtaining all forces from the reactions found in the automated Static analysis (Mecway) and the statics equations for the truss with both ends pinned (in Excel).

From the info studied, the recipe to solve my indeterminate truss would be as follows, would you check if I'm right?
1. Compute all member forces in the real system (using real loads) with 3 redundant reactions removed, e.g. the x-axis displacement in both bottom chord ends and in the left end of top chord.
2. Compute the displacement for each released reaction separately in the released structure (no redundant reactions but with real loads).
3. In the next step apply a unit virtual load, along each redundant reaction and calculate the corresponding displacement using the unit load method.
4. Obtain the 3 compatibility equations that simultaneously vanish the 3 displacements of the released reactions.
5. Solve the 3 unknown reactions from such compatibility equations.
6. Plug the reactions into the real truss equations to obtain all forces.

By the way, please, correct me if I'm wrong but from the indeterminacy check formula (m+r >= 2j) I've obtained a Degree of Indeterminacy (DoI) of 3 instead of the 2 DoIs as you said.

Top truss (pin+roller):
37 + 3 = 2·20 --> 40 = 40 --> Statically determinate
r, reactions: 3
j, joints: 20 (2N)
m, members: 37 (4N-3)
struts: 9 (N-1)
diagonals: 10 (N)
top chord: 10 (N)
bottom chord: 8 (N-2)

Bottom truss (pinned both sides):
39 + 8 = 2·22 --> 47 > 44 --> Statically indeterminate (DoI 3)
r, reactions: 8
j, joints: 22 (2N)
m, members: 39 (4N-3)
struts: 9 (N-1)
diagonals: 10 (N)
top chord: 10 (N)
bottom chord: 10 (N)

Perhaps there is some way to reduce the 3 DoIs to just 2 by using some symmetry. If so, would you explain it or tell me where I would get more info about it?

PhanthomJay
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Your method seems ok except when calculating deflections under Unit load at a released joint, you must calc the deflections at the other released joints as well.

I never trusted those formulas for degrees of indeterminacy. I reckon from symmetry that the 2 top horizontal reactions at left and right must be equal and opposite.also bot two horizontal reactions left and right must be equal and opposite.

also note you should take those given loads in between joints and split the half to each adjacent joint.

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You're right, I calculated all possible displacements between the 3 unit forces (Q1, Q2, and Q3, one for each released reaction), i.e. 9 cross-terms in total. And, yes, I did apply half of each member force to the corresponding node, thanks!

Regarding to the symmetry and indeterminacy issue, I forgot to say that my loading case should be general, i.e. not just a uniform distributed load (it was used in the figure for simplicity). Thus, if any vertical load distribution is possible the total number of external DoIs should be 3, shouldn't it?

Now, I've checked in Mecway that my spreadsheet for pinned-ends Howe trusses works fine with any vertical load distribution. Thank you!

However, the truss must withstand some additional horizontal loading (applied at both ends). Do you think that I would simply use superposition principle and the unit loads (Qx) to calculate all forces when the truss is both vertically and horizontally loaded? If not, how should I do this?

By the way, what should be the boundary conditions to cross-check this new loading case in Mecway? I'm not sure, but I think I should use, for example, pinned ends on the right side and horizontal sliders (only vertical motion constrained) on the left.

PhanthomJay
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You're right, I calculated all possible displacements between the 3 unit forces (Q1, Q2, and Q3, one for each released reaction), i.e. 9 cross-terms in total. And, yes, I did apply half of each member force to the corresponding node, thanks!

Regarding to the symmetry and indeterminacy issue, I forgot to say that my loading case should be general, i.e. not just a uniform distributed load (it was used in the figure for simplicity). Thus, if any vertical load distribution is possible the total number of external DoIs should be 3, shouldn't it?
yes, correct, for the general case, there are 6 unknown reactions (vertical and horizontal reaction at each of the 2 bottom pins, and a horizontal reaction at each of the two top pins), and since you have the 3 static equilibrium equations with 6 unknowns, the truss is externally statically indeterminate to the 3rd degree.
Now, I've checked in Mecway that my spreadsheet for pinned-ends Howe trusses works fine with any vertical load distribution. Thank you!

However, the truss must withstand some additional horizontal loading (applied at both ends). Do you think that I would simply use superposition principle and the unit loads (Qx) to calculate all forces when the truss is both vertically and horizontally loaded? If not, how should I do this?
yes, superposition principle is fine, that is good, but calculate not only the unit loads but also the deformations.
By the way, what should be the boundary conditions to cross-check this new loading case in Mecway? I'm not sure, but I think I should use, for example, pinned ends on the right side and horizontal sliders (only vertical motion constrained) on the left.
Well you can't change a pin support to a roller support for the actual loads, you've got to keep the supports as given, and use the same released supports as the ones you used for the vertical loads, but be careful when choosing which support reaction to release, so as to keep the truss stable and make calculations easier.

Hi, thanks for the rapid response, it is highly appreciated!

Would you check the following procedure to consider an additional horizontal load?

To include an horizontal load $P_H$, I first obtained all the member forces for the released system $F_{Vi}$ (only vertically loaded) following the post #3 recipe. Then, and following the superposition principle, I added the member forces $F_{Hi}$ originated by the horizontal unitary loads $Q_x$ (where sub-index $x$ references the 1, 2, and 3 released reactions) after being multiplied by the half of the additional horizontal force $P_H/2$ (to consider that the horizontal force is equally distributed between both chords). Thus, the force in the $i^{th}$ member simply becomes:
$$F_i = F_{Vi} + F_{Hi}$$
Next, I used the following formula to obtain the 3x3 matrix of displacements ($\delta$) that determines the truss response to each $Q_x$:
$$(\delta_{xy}) = \frac{1}{AE} \sum^N_i Q_{xi} \cdot Q_{yi} \cdot L_i$$
Where $(\delta_{xy})$ are the individual displacements at $x^{th}$ released reaction caused by $y^{th}$ unit force ($Q_y$). Accordingly, $Q_{xi}$ (or $Q_{yi}$) are the forces in the $i^{th}$ members due to the $x^{th}$ (or $y^{th}$) unit loads applied at $x$ (or $y$) released reactions. $N$, $A$, and $E$ are the number of beams, the constant beam cross-section, and the Young's modulus, respectively. The displacements consequence of the internal forces ($F_i$) are computed similarly:
$$(\Delta_x) = \frac{1}{AE} \sum^N_i Q_{xi} \cdot F_i \cdot L_i$$
Finally, I solved the unknown reactions ($R_x$) from the linear system of compatibility equations:
$$(\delta_{xy}) · (R_x) = -(\Delta_x)$$
This way, the reactions obtained without $P_H$ were 7 kN (pointing out of the truss) in top chord reactions and 9.5 kN (pointing towards the truss) in bottom. If a compression $P_H$ of 10 kN is included, they become 12 kN and 4.5 kN, respectively.

Is this right? Am I missing something?

How would I check it in Finite Element Analysis? I mean, if I directly apply three 5 kN loads (one at each released reaction) plus the computed reaction forces (including the horizontal loading) to the released Finite Elements model shown at Figure 1 (bottom), the program will simply compute the resultant force at the reactions and obtain the same forces as in the truss without any $P_H$... which should not be correct. Would you help me?

(Sorry for the long and complex post I was not able to simplify it any further)

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Hi again, I think I'm trying to do something wrong...

The all-ends pinned truss can't respond to a compression horizontal load. The results I obtained in last post indicate so. I think that the equations are telling us that it is not possible to maintain the horizontal displacements 0 unless the reaction forces exactly balance the horizontal load, which is trivial.

One of the ways to overcome this would be to consider only the right end pinned (both chords) and to keep the left side as rollers. This way the structure should be able to respond to the applied loads and there will be just one reaction to compute (the bottom chord horizontal reaction). However, I foresee that the deformations will not be symmetrical (for symmetrical loading cases).

As another workarround, do you think that I would remove all horizontal reactions and still be able to solve the equations? Horizontal loads would be applied symmetrically at both ends. The vertical loads should not produce any whole-rigid-body lateral displacement. Furthermore, since the 4 horizontal reactions would be removed, the system should be statically determinate in some sense, shouldn't it?

PhanthomJay
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Gold Member
Hi. Holiday delays, sorry.

You certainly have picked a tedious problem. It’s bad enough with just one degree of indeterminancy. With 3 degrees of indeterminancy, it is extremely difficult to do because it is so easy to lose track of your proper forces , deflections, unit loads and their deflections, compatibility equations, etc. You are very organized with your work, and that is fantastic, because even if you can’t get a solution, you are well on your way to excel in real world engineering. I have never had to solve a 3 degrees indeterminate truss by hand in 50 years of structural engineering practice. The computers handle it fine though, but the computer program is useless unless you understand the basics and be 100 percent sure you have correctly input the loads and correctly identified the boundary conditions at the supports, and then you must do a watered down hand calc to be sure the computer is giving correct results. I’m not sure why you are entering the released statically determinate truss into the computer model. Enter the actual loaded truss and actual support conditions and it will do all the math for you. Maybe I misunderstand your approach.
While superposition is valid, it’s best here to include the horizontal load up front with the vertical loads, why not? You release the 3 reactions as before and continue with the same approach. You can’t release 4 horizontal reaction components because you will have instability in truss.

it is a bear of a problem.

Hi PhantomJay, thank you very much for your support and huge experience!!!

First, let me tell you more about what I want to do. I want to optimise a Howe truss girder under combined vertical and horizontal loading (see next figures in this post). I'm just using the computer model to validate the (partial) results. That's why I used the 3-reactions released system in Mecway or other dummy trusses and loading, i.e. just to cross-check that I was doing things right.

My ultimate goal is defining the optimal dimensions (e.g. member thicknesses, truss height, number of sections, etc...) for a given span and loading conditions using just basic formulas (e.g. joints and sections methods). I know that there would exist many automated software tools to rapidly optimise such trusses, but unfortunately I don't have any valid license. Perhaps you can recommend me some tool for this task, any free?

The additional motivation to accomplish this very tedious task is, as you said, to "understand the basics and be 100 percent sure you have correctly input the loads and correctly identified the boundary conditions at the support". Furthermore, I think that a "simple" spreadsheet able to rapidly test different truss configurations will be more practical than almost any software I would afford.

Back to the problem, I have to say that it was possible to solve the truss with just 4 rollers as boundary conditions. First, as you recommended, I included the horizontal loads up front with the vertical loads. Then, I enforced that all horizontal loads are self-cancelled in the problem definition (this is what I meant when I said "Horizontal loads would be applied symmetrically at both ends"). This way, the system becomes statically determinate, even without pins, since I was able to obtain all forces using just the joints and sections methods. I will not trust the general indeterminacy formulas anymore :-). I've successfully validated this approach in Mecway.

I want to share with you and the community the formulas to obtain all the forces in a Howe truss supported by 4 rollers and under arbitrary vertical loads at top chord plus balanced horizontal loads at the ends, they are summarised in the next two figures (Figures 2 and 3):

In addition, I used the unit force method to predict the displacements at both ends (top and bottom chords) and at any arbitrary top chord joint. The formulas to obtain the members forces caused by such unitary loads are shown here (Figures 4 and 5):

Using these formulas with the displacements of the compatibility equations (now a 2x2 matrix), it is also possible to obtain the forces required to convert the truss ends from rollers to pins, i.e. to compute the Ptx and Pbx forces that lead to zero displacement upon loading. This will help me to determine the optimal boundary conditions, i.e. those that lead to the lowest weight.

Now, as part of the optimisation process, I want to use the minimal cross-section area for each member. Is it as simple as using the material strength ($\sigma$) and each member forces ($F_i$) to obtain the corresponding areas ($A_i$) as well as performing the Euler buckling checks just in compression members?
$$A_i=F_i/\sigma$$
Please, let me know if you see mistakes in my formulas or drawings, and thanks again for your kind help!

PhanthomJay
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Oh sure I missed that symmetry of horizontal loading so it’s ok to release all 4 horizontal reactions in this case, but bear in mind that this is still an unstable truss because any differential horizontal load will send this truss rolling to neverland. But it’s fine in this case since your actual truss will be all pinned at the supports.
I’m old school so I don’t know much about optimization programs, maybe Mathlab? Or check out those spaghetti trusses students create to hold a certain weight. I mostly rely on our forefathers 100 years ago and more who were geniuses without any computer aid and I follow their designs. It’s a trial and error process where the idea is to come up with the best truss height and number of bays and truss type to keep the bottom chord at a minimal stress with the least midpoint deflection and the lightest overall truss weight, but you’ve got to consider fabrication cost and the labor cost of building the truss, you might have lighter members but more of them in a rather labor intensive construction installation. Watch for buckling also bad compression members may have to be braced laterally in both directions to minimize l/r ratios.
Well, have a Happy New Year!

Hi, thanks for your useful comments! I wish you happy new year too!

At this point, I think I should open a new thread about truss optimisation. But, just one question more here:

Since the truss is now statically determinate, all member loads are totally defined for current geometry. However, I intuitively suppose that when the member cross-sections are different, the loads distribution would change significatively.

Is this change in loads distribution only caused by the geometry change or there is another reason?

If just geometry changes cause load redistribution, to obtain the true forces in de members when different cross-sections are used, I would update the joints geometry by computing all displacements produced by initial forces, and then repeat this process iteratively until convergence. What do you think?

PhanthomJay
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I am not sure what you mean when you say the truss is now statically determinate. The truss with the 4 pins is statically indeterminate. The truss with the 4 rollers is statically determinate, but that’s not the actual truss, it’s just the first step in determining actual member forces and support reactions in the indeterminate truss. When a truss is statically determinate, for a given geometry, it doesn’t matter what is the cross section area of each member. The member forces and support reactions will always be the same. The loading distribution is a function of geometry only. Once you have those forces, you can size the member as you noted, the area required is just the force divided by the allowable stress, but with a buckling check. But rather than size each individual member, it might be best to pick just a couple or so sizes to ease fabrication and avoid confusion during installation.
Now for the indeterminate truss both the geometry and the member sizes will affect the load distribution, because forces depend on the member stiffnesses (AE/L). You solve as per your noted method.

Hi, sorry for the delay, I've been searching optimisation software for trusses these days. You can check the new thread about this topic that I recently opened here.

When I said that the truss is now statically determinate I meant that I was able to obtain all forces in all members and the reactions using just the equations from statics (method of joints and sections), i.e. without any energy based methods (releasing reactions and applying unit forces to compute the displacements that restore the constraints).

Thank you very much for your kind clarifications, I'm learning a lot!

PhanthomJay
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Or if possible, use a 'standard' configuration based on prior history, and size your members accordingly.

You have done real good work thus far. I do however get lost in your subscripts for member designations, I am a numbers man ...plug in the numbers and chug it out.....

Hi PantomJay,

You're right, all pinned truss is statically indeterminate, all rollers truss with self-compensating loads is statically determinate. And don't worry for my indices, they are quite hard to follow and not properly explained. I will explain if somebody asks some day.

I've abandoned the idea of optimising the truss by hand. My spreadsheet results will serve just to cross-check future computational optimisation results. By manually trying a few tube diameters (the same for all) with the minimum thicknesses to withstand the loads while not buckling is sufficient to obtain the minimal achiveable mass. However, global buckling must be assesed later and surely a few more tubes must be added to prevent lateral buckling. In reality, I will use two of such trusses in parallel to prevent global buckling.

Since I didn't obtained much responses for truss optimisation software here, I posted a new thread in Grasshopper forum (an architectural parametric design and analysis software). It has some plugins, i.e. Karamba3D and Peregrine, that integrates FEA with cross-section and topology optimisation tools. Definitively, this seems the way to go.

Thanks a lot for your support and help!

PhanthomJay
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Gold Member
Excellent work. Yes, I agree ,global buckling would not be a problem in the plane of the truss, but it might be a problem out of plane unless it was restrained in that direction. I thought at first that you were a student, but you know more about trusses than most. Good luck!!

I'm a structural biology researcher at work, but a self taught student in engineering with rare hobbies :-)

Thank you very much for your dedication and educational compromise here, it is highly appreciated.

PhanthomJay