Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

When do we use Von mises stress and max principal stress in aerostructures stress analysis 5

Status
Not open for further replies.

MCA1983

Aerospace
Sep 30, 2023
17
When do we use von mises stress and max principal stress in analyzing aerostructures? I got to know based on ductile and brittle materials we use.. but is there any other criteria to use. when subjected to tension compression or bending loads ? Which stress do we use or consider as critical ..

Thanks !
 
Replies continue below

Recommended for you

VM is a yield strength criteria, not ultimate strength, for ductile metals.
Max prin is often used for fatigue checks (max tensile stress) for metals.
Max prin is also used for brittle materials, such as ceramics, cast iron, etc.
Neither are used for composite materials.

Bending stress. Sigh. Is just tension or compression at the extreme fibers/surfaces.
 
The basic problem is: how do you use simple 1D test data (e.g. uniaxal tension, pure shear) to predict failure of real parts that are exposed to combined states of stress, possibly in 3D. Von Mises and Principal Stresses are one way to do that.

Since the VM stress is by definition a positive number, engineers often prefer to use principal stresses, because failure modes and strength values can be very different in tension and compression and the principal stresses can be positive or negative.

AS SWC mentioned, the VM criterion is used to predict yielding in ductile metals. Since it is easy to check a single number, the VM stress is often used to filter results of large finite element models to locate critical areas. However, in a linear FEM all stresses above yield (Fty) are by definition inaccurate. If you wanted accurate post-yield stresses from a FEM, you would have to perform a nonlinear material (plastic) analysis. If (and only if) you performed a plastic analysis, then you could check for failure by comparing VM to the ultimate strength (Ftu). I'm not saying you have to do this, I'm just saying you could do this if you wanted.

Instead of using either VM or principal stresses, another approach to combined loads has been to use "interaction equations". Interaction equations use "stress ratios" which consist of the ratio of applied load to allowable load for each load acting separately. For example, consider a beam with combined axial tension and bending. The stress ratio in tension is Rt = Papplied/Pallowable, and for bending it is Rb = Mapplied/Mallowable. The "applied" loads might come from a FEM, and the "allowable" loads are from theory or test for the loads acting separately. Assuming a linear interaction is relevant, failure is predicted when Rt + Rb = 1. Any classic aircraft stress book, such as Bruhn or Niu, gives interaction equations for many different cases.

Also as SWC mentioned, VM and principal stresses are really only applicable to isotropic materials, not composites. Failure theories for composites is a whole separate topic.

Bending of beams made of ductile metals is a special case. Aerospace criteria usually states: no yielding at limit load and no failure at ultimate load. But that means that you may be (and often are) in the plastic range at ultimate load. The stress formula Mc/I is not applicable then. However, traditional aerospace defines a fictitious stress, called "modulus of rupture" that accounts for the material nonlinearity. It takes the failure moment and multiplies that by (c/I). This stress will be higher than Ftu, which is why I call it "fictitious". The modulus of rupture is a function of the shape of the beam section as well as the stress-strain diagram (is it a gradual yield or an abrupt yield). This can also become a whole separate topic, but don't be surprised if you see a bending allowable stress greater than the ultimate tensile strength!
 
I prefer to use principal for 2D and vM for 3D structures

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
 
At the end of the day these are just failure theories we attempt to correlate to test data. Reading a machine design book chapter on static failure theories will tell you a lot. I recommend Machine Design: An Integrated Approach by Robert Norton.

The most "accurate" failure theory, or which is conservative, depends on the actual state of stress.

2023-10-11_15-18-32_aqlg1j.jpg

2023-10-11_15-17-55_xz1ts7.jpg

2023-10-11_15-17-16_mmo2a5.jpg


As I've said before on these forums (you can probably find a lot of posts about this exact topic if you search) - von Mises equivalent stress is just an equivalent uniaxial stress magnitude that causes the same shear distortion strain energy as the actual applied stress state, meant to be used with a typical uniaxial stress strain curve. The area below a typical stress strain curve represents the total strain energy absorbed. But we may recognize that the hydrostatic component of the strain energy contributes basically nothing to failure.

Actually the theory comes from Maxwell, Huber, Hencky, and von Mises independently.

The whole crux of what the various authors contribute is to take advantage of the fact that the second main invariant of the deviatoric stress tensor is always positive and can be assessed as a scalar value.

The theory is that the shear distortion strain energy (deviatoric strain energy) associated with change in shape of a stress element is key to material failure. The hydrostatic component contributes essentially nothing to failure. (And this is borne out by test data)

As such, for a complex stress state, we want a method of comparing deviatoric strain energy per unit volume to an imposed yield value. Von Mises “stress” does indeed have units of stress (intentionally) but is perhaps more rightly thought of as distortion strain energy per unit volume.

We can obtain this by starting with the stress deviator (deviatoric stress tensor) and imposing a condition relating to the distortion yield (pure shear yield stress) as follows:

J_2 is the second invariant of the stress deviator (simply the difference of the applied and hydrostatic stresses)
2023-10-11_15-27-33_ee0wal.jpg
resulting in J_2=1/2*(S1^2+S2^2+S3^2 )


The criteria would then be f(J_2 )= √(J_2 )-k=0, where k is the pure shear yield strength such that when the complex stress deviator reaches this value, failure occurs. Herein lie the two biggest limitations of the theory

We will further on generally define k=σ_y/√3, so materials must approximate this. Also, F_ty and F_cy must be equal, otherwise we’d need to impose multiple values of k.

Huber pointed out that defining the effective stress as σ_eff=√(3*J_2 ) leads to the simple criteria 0= σ_eff - σ_y

Writing σ_eff with J_2 in terms of the principal or applied stresses and rearrangement leads to the well-known formulae for von Mises effective stress.

There are more "accurate" theories for uneven materials based on J2 and J3, but they are kind of "in the noise" of the test data anyway.

SWComposites said:
Max prin is often used for fatigue checks (max tensile stress) for metals

This is not inherently incorrect, but if I can add a little nuance... linear elastic fracture mechanics is specifically where you might see insistence on principal stress, because it is normally thought that cracks will grow perpendicular to this vector. Actually, cracks grow in he direction that maximizes strain energy release rate, so you can have slant cracks in mixed mode scenarios. But point taken. However:
1. We are usually making conservative envelopes that combine worst-case geometric correction and worst stress orientation anyway, and
2. For stress-life or strain-life, we are often dealing with multiaxial stress states, in which case most of the prevailing theories would have you work with an effective stress, NOT a principal stress.

Keep em' Flying
//Fight Corrosion!
 
Thanks everyone for your replies.. definitely it will help my knowledge.. always thankfully to all..
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor