Combined loading problem and failure

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Righthanded123
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Homework Statement


Diagram of a 12 mm diameter bent rod fixed to a base at section A, with a 250 mm vertical segment and a 125 mm horizontal segment ending in a force point P, with x, y and z axes shown.

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
Diagram of an L-shaped rod fixed in section A, with a 250 mm vertical leg along x, a 125 mm horizontal leg, a 12 mm diameter cross-section, load point P and axes x, y, z. Red labels 1 and 2 mark points on the section.

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.
 
Attachments
  • Diagram of a 12 mm diameter bent rod fixed to a base at section A, with a 250 mm vertical segment and a 125 mm horizontal segment ending in a force point P, with x, y and z axes shown.
    upload_2018-2-28_13-2-45.png
    7.8 KB · Views: 1,652
  • Diagram of an L-shaped rod fixed in section A, with a 250 mm vertical leg along x, a 125 mm horizontal leg, a 12 mm diameter cross-section, load point P and axes x, y, z. Red labels 1 and 2 mark points on the section.
    upload_2018-2-28_13-6-52.png
    9.5 KB · Views: 953
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
Hand-drawn diagram of a circular cross-section in the y-z plane with a dashed axis through its center, showing labelled arrows T2 and M2, and points 1 and 2 marked in red on the circle.


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?)
Hand-drawn sketch of a square cross-section with a thick black outline, a yellow arrow pointing down at its top with the label 176, and a red arrow pointing right above the top edge labelled 44.2.


The principal stresses I found using the
Equation for principal normal stresses: σ1,2 = (σx + σy)/2 ± sqrt(((σx − σy)/2)² + τxy²).


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 ..?
 
Attachments
  • Equation for principal normal stresses: σ1,2 = (σx + σy)/2 ± sqrt(((σx − σy)/2)² + τxy²).
    upload_2018-2-28_16-11-17.png
    487 bytes · Views: 743
  • Hand-drawn sketch of a square cross-section with a thick black outline, a yellow arrow pointing down labeled -176, and a red arrow pointing right labeled 44.2 with a red arrow pointing up on the right side.
    upload_2018-2-28_16-21-35.png
    1.6 KB · Views: 626
  • Hand-drawn cross-section of a circular shaft with x and z axes, a dashed line along a diagonal labeled y, and points 1 and 2 marked in red on the circle's edge.
    upload_2018-2-28_16-24-55.png
    2.9 KB · Views: 598
  • Hand-drawn sketch of a square cross-section with a thick black outline, a yellow arrow pointing down at its top with the label 176, and a red arrow pointing right above the top edge labelled 44.2.
    upload_2018-2-28_16-26-43.png
    1.6 KB · Views: 771
  • Hand-drawn diagram of a circular cross-section in the y-z plane with a dashed axis through its center, showing labelled arrows T2 and M2, and points 1 and 2 marked in red on the circle.
    upload_2018-2-28_16-28-39.png
    4.7 KB · Views: 827
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
Rectangular cross-section with width b and height h, a shaded strip near the top labelled y1 measured from the neutral axis z, and a parabolic shear stress distribution labelled tau max peaking at the centre.


there's something I am missing but I don't know what
 
Attachments
  • Rectangular cross-section with width b and height h, a shaded strip near the top labelled y1 measured from the neutral axis z, and a parabolic shear stress distribution labelled tau max peaking at the centre.
    shear-stress-rectangle.png
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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