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Biaxial bending & Sign convention

Biaxial bending & Sign convention

Biaxial bending & Sign convention

(OP)
Hi all.

I did a couple of quick calcs to better understand the phenomenon of biaxial bending. One thing that is currently tripping me up is the sign convention. For example, if we have a rectangular beam as follows

with b = 15" and h = 30" and a coordinate system established as show with origin at exactly the middle. Ix = 33750 in2 and Iy = 8437.5 in2. Furthermore we have a moment Mx of 100 lb.in occuring about the +x axis and a moment My of 100 lb.in occuring about the +y axis. Intuitively, we can tell Mx is causing compression at the bottom and tension at the top of this section. Likewise we can also say My is causing compression at the right and tension at the left of the section. However, when we run the numbers:

sigma_top = Mx*y/Ix = (100 lb.in)*(15 in)/33750 in4 = 0.044 psi
sigma_bot = Mx*y/Ix = (100 lb.in)*(-15 in)/33750 in4 = -0.044 psi
sigma_right = My*x/Iy = (100 lb.in)*(7.5 in)/8437.5 in4 = 0.088 psi
sigma_left = My*x/Iy = (100 lb.in)*(-7.5 in)/8437.5 in4 = -0.088 psi

It is highly confusing to me that sigma_bot and sigma_right, which are both in compression, have opposing signs. Equally confusing is the fact that sigma_top and sigma_left, which are both in tension, have opposing signs. Can anyone bring light to this basic mechanics of materials issue? thanks

RE: Biaxial bending & Sign convention

RH screw rule ... (+ve with your thumb, no, the right one!, along the axis)

+ve Mx creates tension for +ve y,
+ve My creates tension for -ve x

another day in paradise, or is paradise one day closer ?

RE: Biaxial bending & Sign convention

(OP)
Right. I agree with that.

+ve Mx creates tension for +ve y

sigma_top = Mx*y/Ix = (100 lb.in)*(15 in)/33750 in4 = 0.044 psi

+ve My creates tension for -ve x

sigma_left = My*x/Iy = (100 lb.in)*(-7.5 in)/8437.5 in4 = -0.088 psi


The question is, if they are both in tension, why do they have opposite signs?


RE: Biaxial bending & Sign convention

because y and x in those calculations don't have a sign convention technically. It's arbitrary. They are strictly distances away from your centroid.

RE: Biaxial bending & Sign convention

no, x and y have a positive direction, they are not scalar. the error is in stress = My*x/I, since +ve My produces +ve stress for -ve x it is -My*x/I

another day in paradise, or is paradise one day closer ?

RE: Biaxial bending & Sign convention

(OP)
rb1957:

My is positive as given in the problem. x is negative because we are looking at the left. The general equation as far as I know is My/I, not -My/I. What determines whether we need to put a negative sign in front or not?

RE: Biaxial bending & Sign convention

Quote (awa5114)

What determines whether we need to put a negative sign in front or not?

Your selection of coordinate system. Change the direction of your positive x vector and this 'problem' goes away.

Coordinate systems are not a 'real' feature. They are imaginary, and can be selected such that they make problems easy or hard, intuitive or not, etc etc etc.

RE: Biaxial bending & Sign convention

the sign convention !

+ve My is creating +ve stress (tension) for -ve x ... different to +ve Mx creates tension for +ve y
stress = Mx*y/I but My*-x/I ... and you need to use signed distance 'cause +Mx produces compression for -ve y.

another day in paradise, or is paradise one day closer ?

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