Statically indeterminate beam problem

[sethmcnear]

I have a beam 85.4 inches long that is fixed at one end. There is a load of 600 lbs at the other end. It is supported by a pin support (0 defelection) at 6.8 inches from the fixed end. I need the reaction forces at the fixed end, at the support and the moment at the fixed end. I have done the calculations and gotten the following:

Reaction at fixed end: -10,404 lb
Reaction at support: 11,004 lb
Moment at fixed end: 23,587.2 inlbs counterclockwise

It's been a long time since I've done this and want to check my work.

Thanks,
Seth

 

[JStephen]

The moment of inertia and modulus of elasticity are also required for a solution.

Actual loading would be highly dependent upon the vertical placement of that support at 6.8".

Assuming your numbers are correct, the maximum moment occurs at that support, not at the fixed end.

 

[UcfSE]

Look for some software to check your work. I assume you are checking hand calculations.

 

[rb1957]

i guess your reaction set is balanced, equilibrium is one check, but will be satisfied with many solutions.

i'd try two different methods. i prefer unit force method and 3 moment equation.

i don't know how you calc'd the problem, but unit force method is easy to apply. count the pin support as redundant; solve the cantilever, determine the deflection at the pin support position, then apply a unit force on the cantilever at the pin support position, calculate the deflection (this is simply end deflection of a cantilever 6.8" long !); then the pin support reaction is the ratio of these deflections (how many unit forces do you have to apply to restore the statically determinate deflection to zero).

 

[sethmcnear]

I apologize. The support at 6.8 inches should actually be considered as a roller support on the underside of the beam, thereby preventing deflection in the y-direction but not slope.
As far as needing the modulus of elasticity and the moment of inertia, I didn't think those were required to solve for the reaction forces and reaction moment at the fixed end. I would need those if I were to calculate the moment at some other point on the beam, the beam's deflection, or it's slope.
The method I used was to create the moment equation for the system using singularity functions. I did this by writing the load equation and integrating twice. I then integrated the moment equation twice and used my boundary conditions (zero deflection at x=0 and x=6.8, zero slope at x=0) to solve my constants of integration and one of the unknowns in terms of another. In all cases, EI ended up being multiplied by zero so did not factor into the results. I then used the equilibrium equations for force and moment combined with the deflection equation to solve the three unknowns. Was this the incorrect approach or did I not apply it correctly?

 

[sethmcnear]

Yes, my reaction set is balanced but you're right, there are several results that would satisfy the equilibrium equations.
The support force is actually being used to represent the resultant of a reactionary load being applied on the beam by a triangular support gusset. I assumed the load from the gusset would be linearly applied to the beam with a triangular area (not sure how accurate that is, but didn't know how else to do it) and calculated it's position (6.8" from the fixed end) that way.
The methodolgy I used is listed in the post above. I'll take a look at your recommended technique.

 

[rb1957]

you're ok with that appraoch, as you don't calculate deflection, but rather deflection*EI. since EI is constant it washes out. becareful in case you have a case where I is not constant.

i'm not 100% sure about your method (it's been too long since i heard "singularity equations", though maybe i'd recognise one if i saw it !) ... i can see the moment equation of the beam is two spans, the connection between them is slope continuity and a zero deflection, but i don't see immediately how you can express the moment due to the simple support. try the unit force approach i outlined above.

 

[rb1957]

you're ok ...
i quickly solved for the simply support load; i got 11,003,
i take it that you applied the load in the -ve direction

 

[RPstress]

Actually, I think this is one of those odd occasions (well, not that odd) when you have indeterminacy which is not dependent on stiffness. For the simple case where the constraint at 6.8" is perfectly rigid and fixity at the moment reaction is 100%, I get the same numbers as sethmcnear (well, 10043 lb, 11003 lb and 23580 lb-in). Just to double check I went OTT and did a beam FE with two different stiffnesses (I's of 20 in^4 and 100 in^4) and got identical reactions. The moment goes through zero about 3" from the end and at the moment reaction the top surface (assuming load direction is downwards) of the beam is in compression, unlike most of the length. Worst moment is 47160 lb-in at the prop, bottom surface in compression.

 

[sethmcnear]

I figured EI washed out. And it won't be changing, so I can consider it a constant. I will need the moments in the beam eventually, but my main concern is to verify I've done the initial calculations correctly first.
In developing my load equation using singularity functions, I treated the moment at the fixed end as a moment acting at zero: -M*<x-0>^-2. Integration brought me to -M*<x-0>^0 as the term representing that moment in my moment equation. Since x will always be greater than 0, the singularity goes to 1, leaving me with -M. I'm trying your method shortly.
It's been over 4 years since I graduated and I didn't have to use any of this stuff in my previous job. I'm the only one at this job that has any experience with it, so I don't really have any back-up. I'm trying to rmemeber and relearn enough from textbooks to get the results.

 

[sethmcnear]

Great!! Thank you to all of you for your help and input.

 

[sethmcnear]

And yes, the load at the end is in the -ve direction. Just because I don't have a lot of experience with it, when would your indeterminancy be dependend on the stiffness? I'm guessing you mean the stiffness of your supports, i.e. whether or not they would deflect/stretch before the object they were supporting?

 

[vonlueke]

sethmcnear: Using classical beam theory, I get the following results. Vertical reaction at fixed end, -10402.9 lbf. Vertical reaction at roller support, 11002.9 lbf. Moment reaction at fixed end, 23580.0 in-lbf clockwise (assuming fixed-end support is at the left-hand end of beam). Maximum beam moment, 47160.0 in-lbf at the roller support, concave downward.

However, the span between your fixed end and roller support is short. So the answer to your problem appears to be highly sensitive to the cross section dimensions and elastic modulus of your beam, such that the above results might not even be in the correct ball park. Assuming your supports are truly rigid, you would need to specify cross section dimensions and tensile modulus of elasticity.

 

[JStephen]

My approach would be to pull out either the AISC-ASD or Formulas for Stress and Strain, as the case may be. Look up cantilever beam loading. Calculate deflection at the support for the end load only, then calculate deflection for the support from the support load and equate the two.

As to whether you need the E & I- I would probably go ahead and calculate a number for that deflection, so I would use it. Depends on what you need, I guess. As the formulas become more complex, it becomes easier to work with a specific deflection rather than a term. And then, I'm thinking in terms of checking allowable stresses along the beam, which assumes you have beam properties anyway.

Anyway, using your method, you should be able to check quite easily by looking up those same formuals, calculate deflection due to both loads, and see if it's zero.

 

[sethmcnear]

vonlueke: Thank you for the response. I mentioned in one of my previous posts, the 'roller' support I'm using is actually representing the resultant reaction force of the load from a triangular support gusset (two gussets really).

The beam is actually a manipulator arm with 3 segments which I am treating as a single 'system'. The first two segments are 4X4 inch tube with a wall thickness of 3/16ths (hot rolled steel). My goal is to calculate the stress experienced by the first segment. I know there are more accurate methods to use in solving this problem but my experience with this sort of problem is limited.

 

[vonlueke]

sethmcnear: It turns out the classical beam theory results are conservative for this problem (for forces and moments). The results are listed in my previous post. The critical cross section is at the roller support. Also check the cross section there for crippling or local buckling (which I didn't attempt). Good luck.

 

[sethmcnear]

vonlueke: Am I right in assuming that by 'conservative' you mean the classic beam theory results are on the high side and that those provided by an analysis using a more accurate representation would be lower?

 

[vonlueke]

sethmcnear: Yes, that's what I meant by the term conservative.

 

[rb1957]

simplistically, 600lb*86" = 51600in.lbs
4"x4"x0.19" square tube,
I = 4*0.19*1.905^2*2+0.19*3.625^3/12*2
= 7in4
bending stress = 51600*2/7 = 15ksi ("squat", and that's the technical term !)

but i think the issue is going to be the welds. at the end of the beam, assume the moment is carried by 24" welds, 4" apart. this means that each weld carries 51600/4 = 12900 lbs, or 3175 lbs/in (in tension). not being a "weld" guy, this looks like a lot

 

[rb1957]

btw, at 4"x4"x0.19", by 86" long, your beam is going to weigh in at between 80-100 lbs ... fairly sizable compared to your 600 lb load