• Righthanded123
In summary: 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
Righthanded123

## Homework Statement

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

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

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?)

The principal stresses I found using the

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

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

Combined loading is a mechanical engineering term that refers to the simultaneous action of two or more types of loads on a structure or component. This can include forces, moments, and/or thermal loads. In real-world applications, structures are often subjected to multiple types of loads, and understanding how they interact is crucial for designing safe and efficient systems.

The combined loading problem is the task of analyzing and predicting the behavior of a structure or component under the influence of multiple types of loads. This involves determining the stresses, strains, and deflections that will result from the applied loads, as well as identifying any potential failure modes.

There are several factors that can contribute to failure in combined loading, including material properties, loading conditions, and geometric constraints. In some cases, a structure or component may be able to withstand a single type of load, but when multiple loads are applied simultaneously, the resulting stresses may exceed the material's strength and cause failure.

The analysis of a combined loading problem involves using mathematical equations and principles of mechanics to determine the stresses, strains, and deflections in a structure or component. This can be done through hand calculations or with the help of computer-aided design (CAD) software. It is important to consider all loading conditions and their interactions to accurately predict the behavior of the system.

To prevent failure in combined loading, engineers must carefully consider the design of a structure or component. This includes selecting appropriate materials, considering the loading conditions that the system will be subjected to, and incorporating factors of safety to account for uncertainties. Additionally, regular testing and maintenance can help identify potential issues and prevent catastrophic failure.

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