Solid mechanics (shear stress) question

[milan666]

Hi, if the shear stress on an element in a solid is SAy/bI, where S is the shear force and y is the distance from the neutral axis, what value of y do i use if the point I'm calculating is on the neutral axis?

 

[Mapes]

On first glance, you would use zero. But it would help to know the shape of the solid, the type of loading, how you got SAy/bI, and what the other variables mean for me to be confident with this recommendation.

 

[milan666]

Its a cantilever beam, with a rectangular cross-section, and i have to find the shear and normal stresses on a point which is at the neutral axis. If y is zero, then the shear force would be zero. Does that mean that there are no shear stresses at a point on the neutral axis? I can't post the actual problem itself cause I am at uni and that would be plagarism.

 

[Mapes]

OK, so what is the loading, why are you using SAy/bI, and what do the other variables mean? If the results of an equation don't match your intuition, it could be sign that you're applying the wrong equation.

 

[milan666]

Thats what we were taught to use. S is the shear force at that point on the beam, A is the area of the cross section, y is the distance form the neutral axis, b is the width of the cross section, I is the second moment of area.

 

[Mapes]

Thats what we were taught to use. S is the shear force at that point on the beam, A is the area of the cross section, y is the distance form the neutral axis, b is the width of the cross section, I is the second moment of area.

Are you sure A is the area of the entire cross section and y the distance from the neutral axis, or is A the area of some region above or below the neutral axis, and y the distance from the centroid of that region to the neutral axis?

 

[Cvan]

At the neutral axis, normal stress will be zero (strain is zero at the neutral axis, this is part of how 'neutral axis' is defined).

Also, at the neutral axis, since you have a compressive force on one side and a tensile force on the other, SHEAR stress is a maximum.

"An Introduction to the Mechanics of Solids (second edition with SI units)" by Crandall, dahl, and Lardner lists as Equation 7.27 the following for shear stress in the case of beam described as you do

[tex]
\tau_{xy}=\frac{V}{2 I_{zz}} \left[\left(\frac{h}{2}\right)^2-y_1^2\right]
[/tex]

where h is the height, y1 is the distance from the neutral axis, V is the shear force, and Izz is the moment of inertia--or second moment of area, whichever terminology you're used to (at least I think these are right from when I took mech of matl's last semester).

The important take-away is that shear stress will have a maximum at the neutral axis (y1=0), i.e. you'll need more glue to hold it together there than anywhere else, all other things being simple in an isotropic material.

 

[milan666]

Oh ok i got it now, y is the distance from the neutral axis to the centroid of the area where you make the cut. Thanks!