Shear sharing distribution

[designrider]

If 2 rectangular beam sections of the same material share load effects, moment sharing is distributed by the ratio of respective member stiffness. That is each individual I (moment of inertia) divided by the total or summed moment of inertia. How is shear sharing distributed? Is it by a ratio of Q (first moment of area), A (area), I (moment of inertia), or something else?

 

[FishTheStructure]

If you refer to basic beam mechanics, shear and bending moment from a single load are not separate items which can take different load paths through independent members. In fact, shear is the derivative of the bending moment: M(x) = EI d2w/dx2, V(x) = EI d3w/dx3 where w(x) is the deflection at point x along the length. From the above equations, it can be seen that both shear and moment are dependent on the stiffness (EI) of the member. Thus if one load is being shared by two members, it will be distributed by the ratio of EI for both shear and bending moment.

Great spirits have always encountered violent opposition from mediocre minds - Albert Einstein

 

[designrider]

I think I solved my own question:

Similar to moment effect distribution described above, the LOADING will be distributed by the ratio of respective member stiffness. That is each individual I (moment of inertia) divided by the total or summed moment of inertia (the well known "stiffness attracts load"). Therefore, the all the effects of such loading including moment and SHEAR will also distribute on the basis of member stiffness. Simply said, define LOAD sharing first, then calculate load EFFECTS on each member.

 

[designrider]

Thanks, your comments also resolve the question. I didn't see your post before typing my own.

 

[slickdeals]

Assume you have two wide-flange beams sitting on top of each other. Both of them have same web thickness. One has a narrow flange but is deeper (Beam A), while the other has a wider flange but is shallower (Beam B). Both have the same moment of inertia.

Shear area of beam A is greater (due to being deeper). Will shear distribution still remain proportional to stiffness?

 

[dhengr]

Designrider:
You have to define and explain your problem in a little more detail than you have so far. Some sketches, sizes, dimensions, loads would be helpful in doing this. Normally we look at this problem and assume compatibility btwn. the two beams, in which case they take their share of the load in proportion the their relative stiffnesses, EI1 and EI2. But, Slickdeals suggests they may be stacked on top of each other. And then the question is, are they connected in any way at their faying surface, the flg. to flg. surface btwn them? If they are not properly connected together, there will be a shearing type slippage btwn. them at this faying surface; their load share will be in proportion to their relative stiffness, and the stresses (bending and shear) are figured independently for each beam. When they are properly connected together at their faying surface, through shear flow, they become one combined member with a new moment of inertia, and the stresses are figured accordingly, on that combined member.

Alternatively, you could have the two members side by side and properly connected together, or not connected, except through the load. They will still tend to carry the load in proportion to their relative stiffnesses, but now the stiffness of the load, its C.G., its ability to distribute the load this way without rolling, etc., its ability to force compatibility btwn. the two beams must be brought into play. If they are properly connected together, you might still have a torsional problem, with one side of the new combined beam being weaker than the other, and this has to be accounted for in some way. It relates back to the load rolling problem above.

 

[JAE]

slickdeals,

Good set-up and question.

The "typical" way that we analyze beams is to account for flexural stiffness only....i.e. we use EI and equations for deflection like 5wl^4 / 384EI to determine "stiffness".

In reality there is also shear deformation in members that we typically do not account for (unless you are running software with shear deflections used).
If we were to account for shear stiffness/flexibility then the two beams you describe would have different total stiffnesses.

Your Beam A with deeper depth and narrower flange would be slightly stiffer because while both beams have the same EI value, they differ in their EA values.

I believe that for most steel wide flanges, flexural stiffness is more prominent than shear stiffness except for very short spans.