Trapezoidal loads shear - moment diagram 2

[JoeH78]

Hi all,

I'm experiencing a difficulty understanding how the trapezoidal loads are distributed and how to shear moment diagrams are drawn for structural members subjected to trapezoidal loading.

For example below example is frame member subject to trapeze load, due to the fact that this memeber is extracted from 3D frame building it has initial shear and moment at ends resulted from fixed - end moments(equivalent nodal loads) loading.

So my hand-calculations is as below(x distance as variabel) and want to compare them with software calulated values.
Vshear = 150 - q*x2/(2*A);
Mmoment=150*x -q*x3/(2*3*A);

A = triangular horizontal leg-length of trapezoidal load. (A=100cm , span = 400cm; q=1kg/cm )

the above formulas produces the consistent results for boundary conditions and same with software results but in-between values are wrong.

For example if x = 30cm then Vshear = 145.5 kg where as software finds 142.5 kg.

http://i55.tinypic.com/1zdcfop.jpg

Any comments will be appreciated.
Regards,

 

[ishvaaag]

Your value is OK for a fully symmetrical trapecial load, see attachment. However you say the beam is extracted from frame analysis, is it entirely symmetrical? If not a term algebraic (M1+M2)·x/L will be modifying the shear at station x.

If not, it might be -to be proven- a device of the frame element taking shear deformation into account, that is by default the situation in SAP2000.

 

[JoshPlumSE]

I've seen plenty of cases where shear deformation can make small differences between the analysis results and the hand calc results.

 

[rb1957]

if "q*x2" is q squared (usually q^2) then yes i agree with your expression.

ishvaaag makes a valid point, 'cept the results show equal moments at both ends.

i wonder if your FE program is dividing your beam into elements and giving constant results for each element ... try changing the position by 1mm, or 1cm ...

what ugly units your program uses ! (IMHO)

 

[JoeH78]

thanks in advance,

However you say the beam is extracted from frame analysis, is it entirely symmetrical? If not a term algebraic (M1+M2)·x/L will be modifying the shear at station x.

Structure itself is fully symetrical and this is beam, connected to columns.
I wonder why (M1+M2)·x/L should contribute to the shear force as long as we write vertical force equilibrium ?

i wonder if your FE program is dividing your beam into elements and giving constant results for each element ... try changing the position by 1mm, or 1cm ...

Even for small fraction of length e.g. 31cm gives 142.25 kg so that means consistent results in my idea .

As a result, do I have to deduce that as floating point deviation of computer programs (which doesn't sound conceivable) or am I really making something wrong?

what ugly units your program uses ! (IMHO)

same goes for inches, feet, pounds as well (IMHO)

 

[ishvaaag]

joehigashi, sorry, (M1+M2)/L, it is the moment hanging line what is linear, no the shear from the difference of moment at ends.

 

[ishvaaag]

... which adds (if there's any) a constant shear all over the length.

 

[rb1957]

"Even for small fraction of length e.g. 31cm gives 142.25 kg so that means consistent results in my idea." actually this supports the idea that the program is using small constant elements ... if this is right, do a bunch x's 1cm apart, i think there'll be a step change in the shear.

 

[ishvaaag]

The issue under control either through

-Automatic Frame Mesh
-Placing a Point where a value is wanted

See pdf attachment.

 

[JoeH78]

THANK YOU SO MUCH,

EXCELLENT OVERVIEW.

 

[rb1957]

"I try to overcome the issue by setting the number of inner points in the beam but It seems unable to set a number proper for a 30/400 point be directly readable."

pretty much exactly what i thought it was doing ... instead of a continuous change of shear, it's results are a staircase of very small steps, defined by the number of interior points. Quite possibly it is plotting the element value (ie between two interior points) at the mid point of the two interior values, which is about as correct as you're going to get, and looks nice and smooth.