Fixed-end moments for Trapezoidal load on part of span?
Fixed-end moments for Trapezoidal load on part of span?
(OP)
Hi,
Trying to solve an indeterminate beam using moment distribution displacement method, I used the general fixed-end moment equations, but one span has a trapezoidal load on part of the span which does not have a fixed-end moment equation,
here is the problem: https://goo.gl/photos/dnY2sz6vZ4BxESA8A
How to calculate the fixed-end moments for that span?
Many Thanks!
Trying to solve an indeterminate beam using moment distribution displacement method, I used the general fixed-end moment equations, but one span has a trapezoidal load on part of the span which does not have a fixed-end moment equation,
here is the problem: https://goo.gl/photos/dnY2sz6vZ4BxESA8A
How to calculate the fixed-end moments for that span?
Many Thanks!






RE: Fixed-end moments for Trapezoidal load on part of span?
EDIT: Whoops, I now see you wrote that it has a trapezoidal load on only part of a span so the above does not apply.
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
Kindest regards!
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
I'd solve the span as a doubly cantilevered beam with a partial span loading. Not That hard ... 2 unknowns, unit force method would probably be my choice.
another day in paradise, or is paradise one day closer ?
RE: Fixed-end moments for Trapezoidal load on part of span?
Kindest regards!
RE: Fixed-end moments for Trapezoidal load on part of span?
if doing mdm, is there a way to derive the influence and carry-over coefficients ? ie mdm initally locks each support (fixed) and then releases each in turn, redistributes, and repeats
another day in paradise, or is paradise one day closer ?
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
The equations below calculate the fixed end moments and shears for the left and right ends of the member under consideration. s1, s2, and s3 are the lengths of the beam in front of the load, under the load, and behind the load respectively. w1 and w2 are the load intensities at the beginning and end of the trapezoidal loading. I've also included some write up for the triangular decomposition method. There a,b,c = s1,s2,s3. I think you'll get the idea. I developed equations for outputting shear, moment, and deflection too if you need those. The deflection equations are wild. I had to use MathCAD to keep the algebra straight.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Fixed-end moments for Trapezoidal load on part of span?
Yes! For fixed fixed, you will have two unknowns that you will need to solve for (do this by solving two equilibrium equations on the conjugate beam).
Once you solve for the moments, you can continue on to find the displacements.
RE: Fixed-end moments for Trapezoidal load on part of span?
By the way I've vetted these two equations against ETABS and they work great,
UDL on part of the span: http://www.engineersedge.com/beam_bending/beam_ben...
IDL on part of the span: http://www.engineersedge.com/beam_bending/beam_ben...
If you had to solve this without using these equations, what method would you suggest?
Kindest regards!
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
It would be if both ends are considered fixed horizontally, but one end can be free to move horizontally while fixed against rotation and vertical displacement.
To solve the problem without KootK's equations, I would calculate the rotations at each end due to the trapezoidal loading on a simple span, then calculate the moments required to reduce both rotations to zero.
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
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RE: Fixed-end moments for Trapezoidal load on part of span?
Thanks. Intended for automation of course. I've often wondered how commercial packages handle this. Crazy equations but they cover uniform, triangular, trapezoidal, increasing, decreasing, and negative loads all in one go. Basically, any kind of standard distributed load. Very efficient in a programmatic sense.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Fixed-end moments for Trapezoidal load on part of span?
Not quite. That only works because the load is uniform and symmetric. In your case, the load is variable.
In the general case, using the Conjugate Beam Method, do the following:
1. Draw the Bending Moment diagram for the real beam assuming a simple span and the actual loading.
2. Load the Conjugate Beam with a fictitious load corresponding to M/EI of the real beam.
3. Calculate the shear at each end of the conjugate beam (which is also the rotation of the real beam).
4. Calculate the end moments required to bring the end rotations to zero.
If that sounds like too much work, you could use a reasonable approximation of the actual load.
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
Download ConBeamU
Screenshot attached
On the FEA sheet (Fixed End Actions, not Finite Element Analysis) there are some examples compared with standard formulas, including Example 6 (starting row 131) which is a partial triangular load, using a formula from the Reinforced Concrete Designers Handbook by Reynolds and Steedman. The formulas are a bit simpler than Kootk's, but of course you would need to combine two to get trapezoidal loading.
The FEAU function will return fixed end moments and forces, and REAU will work with spring restraints. They both have versions without the U, that assume consistent units. The functions don't use standard formulas, they use Macaulay's method to calculate slopes and deflections for a cantilever, then solve to find end actions for the specified end conditions.
The functions work for beams with any number of segments with different section properties, and the spreadsheet also has functions for continuous beams, with any number of supports and segments, using the same approach.
Search the blog for Macaulay and/or ConbeamU for more details.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Fixed-end moments for Trapezoidal load on part of span?
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Fixed-end moments for Trapezoidal load on part of span?
If more precision is required, M1 = ∫aa+b wx(L-x)2dx/L2 where:
a = start of trapezoidal loading
b = length of load
w = load per unit length
For trapezoidal loading, w = w1 + (w2-w1)(x-a)/b
Similarly,
M2 = ∫aa+b wx2(L-x)dx/L2
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
just to make sure, if I want to calculate M1 with the integration, are the ( a , b , w1 , w2 ) in this image correct
or do I need to switch w1 and w2? https://goo.gl/photos/BJhcfstsicLBqoUWA
I entered the values of a=1.5 b=4 w1=50 w2=70 L=7 like this : http://s19.postimg.org/3o3cljuv7/screenshot_46.jpg
and it gave incorrect answer. what am I doing wrong in the integration?
Kindest regards!
RE: Fixed-end moments for Trapezoidal load on part of span?
Your diagram is the reverse of mine (I move from left to right), but it should produce the correct answer as your numbering system is consistent with a right to left system. Your labeling appears correct; do not switch w1 and w2.
Your limits of integration are (0, L) which is incorrect. The limits of integration are (a, a+b) or (1.5, 5.5). You should be integrating only the loaded portion of span, not the whole span because integration is a summation of the effects of the load.
I cannot follow your expression to be integrated. There should be some terms with higher powers of x.
For example, in the expression:
M1 = ∫aa+b wx(L-x)2dx/L2
there are terms with x, x2 and x3 even when w is constant.
When w varies as a function of x, the expression for M1 contains a term with x4 as well as the lower powers of x.
In the case of trapezoidal loading,
w = w1 + (w2-w1)(x-a)/b
which can be written w = w1 - (w2-w1)a/b + (w2-w1)x/b
In the present case, this would be w = 42.5 + 5x
and when you do the multiplication, you should arrive at the correct expression to be integrated.
Edit: shown in red.
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
is the expression correct now? http://s19.postimg.org/424ol5eyr/screenshot_47.jpg
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
It should be:
M1 = ∫1.55.5 (42.5+5x)x(7-x)2dx/49
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
RE: Fixed-end moments for Trapezoidal load on part of span?
If you divide your result (1690.64938) by (1.5)5.5, the corrected result is 181.78.
BA
RE: Fixed-end moments for Trapezoidal load on part of span?
BA you were very helpful and I hope someone else benefits from this..
Thanks!