Combined loading problem and failure

AI Thread Summary
The discussion focuses on analyzing the combined loading problem of a rod under a force P of 120N, examining whether it yields at two points using Tresca and von Mises criteria. Calculations reveal shear stress, bending moments, and torque at both points, with specific stress values determined using Mohr's circle method. At point 1, the shear stress from torque is calculated, while at point 2, the absence of normal stress due to its position on the neutral axis is noted. Participants clarify the contributions of various stress components, leading to a better understanding of stress distribution, particularly the parabolic nature of shear stress in beams. The conversation concludes with an acknowledgment of the need for further research on shear stress variation in circular cross-sections.
Righthanded123
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Homework Statement


upload_2018-2-28_13-2-45.png

The problem asks to find out whether the rod yields at points in section A according to the tresca and von mises criteria
P is 120N

Homework Equations


Shear stress= VQ/It
Stress= Mx/I
shear stress= Tp/J

The Attempt at a Solution


I picked 2 points at section A.
The first point is on the
upload_2018-2-28_13-6-52.png

I calculated that P will cause a shear stress Vy along the Y axis which is -120N
P will cause a bending moment Mz about the z axis 120N x 250mm
and P will cause a torque Tx about the x-axis 120N x 125mm

At point 1
Vy will not cause a shear stress because Q=0
Tx will cause a shear stress using Tp/J
and Mz will cause a compressive stress which will be at a maximum
Then i found the principal stresses using the mohr circle method
Then using the von mises and tresca criteria I compared the stresses to the yield stress of 200MN/mm to determine if the rod will yield.

At point 2
Vy will case a shear stress found by VQ/It
Tx will cause a shear stress using Tp/J
and Mz will not cause a normal stress because the distance from neutral axis for point 2 is 0 ?

This where I am stuck I am not sure point 2 only has shear stresses as I have calculated
and If i have picked enough points, as i imagine it will be the same values.
 

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As I understand it, you've identified the reaction at A as the linear superposition of (1) A shear force in the y direction (2) a bending moment about the z axis and (3) a twisting moment about the x axis. The components of the stress tensor involved are xx, xy, and xz. What values to you get for these three stress tensor components at points 1 and 2, and how did you obtain them?
 
Looking at the base of the road, my 2 points are
upload_2018-2-28_16-28-39.png


so I got that..
Vy= -120N
Torque(Tx) = 120 x 0.125=15Nm(clockwise)
Bending moment (Mz)= 120 x 0.25= 30Nm
J = 2.03x10^-9
I =1.02x10^-9
Q=4.20x10^-7

At point 1, I think there will be no contribution from Vy to shear force acting on point 1
The torque will cause shear force which is found by (Tx * 0.006)/ J = 44.2 MPa (in the positive y direction)
The bending will cause a normal stress which is found by (Mz * 0.006)/I= 176.8MPa (in compression?)
upload_2018-2-28_16-26-43.png


The principal stresses I found using the
upload_2018-2-28_16-11-17.png


σ1=10.43
σ2=-187.23

For point B
I did the same thing basically but I found that there will be no contribution from Mz
This is because in the equation , Mx/I , x would be 0 because point 2 is on the neutral axis
so the contribution will be from Vy and Tz, which didn't make sense ..?
 

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Please bear with me while I spend a little time working this problem independently, so that I can better compare with what you did. Hope to get back in a few hours.

Chet
 
I confirm your reaction forces and torques, and your magnitudes of the shear and normal stresses.

Now, at point 1, I get the following state of stress:
$$\sigma_{xx}=-177.0\ MPa$$
$$\sigma_{xy}=-1.06\ MPa$$
$$\sigma_{xz}=44.25\ MPa$$
$$\sigma_{yy}=\sigma_{zz}=\sigma_{yz}=0$$

At point 2, I get the following state of stress:
$$\sigma_{xx}=0\ MPa$$
$$\sigma_{xy}=-1.06-44.25=-45.31\ MPa$$
$$\sigma_{xz}=0$$
$$\sigma_{yy}=\sigma_{zz}=\sigma_{yz}=0$$

I think our results are comparable, except for the xy shear stress at point 1 (which is not equal to zero). OK so far?
 
Thank you very much, this has really helped raise my confidence when approaching problems like this.
I had another go of it and I still don't understand the xy shear stress at point 1.
If it comes from the shear force Vy, wouldn't shear stress be 0 at point 1
The shear stress distribution is parabolic, so the maximum shear stress would be at the neutral axis and 0 at the ends where point 1 is, like in this image
shear-stress-rectangle.png


there's something I am missing but I don't know what
 

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Oh! It think you're right that in beam bending loading, the shear stress is distributed parabolically. Sorry. My mistake.
 
really?! I got something right
I think I messed up my calculation for Q though
Thank you very much for your help!
 
Righthanded123 said:
really?! I got something right
I think I messed up my calculation for Q though
Thank you very much for your help!
To get the type (a) loading shear stress distribution, one probably needs to research how that shear stress varies with position over the cross section for a beam of circular cross section. I'm too lazy to do that, but I don't think it is going to contribute very much to the shear stress at point 2. So, now you are OK in calculating the principal stresses (or at least approximating them) at point 2, correct?

EDIT: http://www.nptel.ac.in/courses/112107146/lects%20&%20picts/image/lect28%20and%2029/lecture%2028%20and%2029.htm
 
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