Shear sharing distribution
Shear sharing distribution
(OP)
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?






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