Max principal stress equation

Reminder of max principal stress equation:

Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Patran Nastran Shear Stress in Beam Example

A beam under bending is subject to both normal and shear stresses. The shear stress in a beam is defined by the following equation.

Where,
V = Vertical shear force
Q = Static moment of area
I = Moment of inertia of the beam cross section
t = thickness of material

This post will show how the above equation can be used to calculate shear stress in a rectangular. The results will be validated using a simple patran/nastran model.

Assumed a beam with pinned-pinned boundary and a point load applied at the mid point as shown below:

Let,

P=1000lbs
A=25in
B=25in

Hence using any beam software or simple hand calcs, it can be shown that the vertical shear load is 500lbs between x=0in and application point P and 500lbs (but opposite direction) between application point P and x=50in.

Assuming the beam is a rectangular plate with a width of 0.5in and a height of 10in. The moment of inertia can be calculated.

I = (width x height^3)/12 = 41.67in^4

Calculating the Q (Static moment of area) at different height on the beam. the following table can be constructed.

y	A	Q	VQ/it
4.5	0.25	1.125	27
4.25	0.75	3.1875	76.5
3.75	1.25	4.6875	112.5
3.25	1.75	5.6875	136.5
2.75	2.25	6.1875	148.5

Where y is the area/segment of interest to the neutral axis.

A = area of interest.
VQ/it = shear stress of beam.

The next step is to show how to validate this shear stress in patran/nastran.

First create a beam in Patran using Quads elements as shown below. The thickness of each quad should be 0.5in and the material properties are based on aluminum which has a youngs modulus of 10300000psi and poison ratio of 0.33. Using an applied load of 1000lbs as per the example problem above, the patran model is shown below.

note that there is 50 elements in width and 10 element in height.

next we can verify the shear stress using RESULTS->CREATE->MARKER->TENSOR. Select tensor and select component xy. in target entities select element. In the figure below a few elements near the neutral axis is selected. The shear stress should be around 148.5 psi.

The shear values are consistent with "hand calculations". 

Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Tangent Modulus

Tangent modulus is the slope of the stress strain curve. Below the proportional limit the tangent modulus is equivalent to the young's modulus. Hence tangent modulus is most useful above proportional limit (i.e: in the plastic range). Refer to the stress strain curve below:

Note that the tangent modulus is very useful when it comes to calculation of allowables in the plastic strain region

Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Stress Strain Curves Explained

Most stress strain curves are determined by testing coupon material in axial loading, hence stress strain curves are sometimes describe as "axial stress strain curves" because this better represents what is tested. Understanding the relation between stress and strain is one of the foundation for every stress engineer. one example of stress strain curve is shown below. the source of this figure is from Wikipedia.

the "flat top" (i.e: strain beyond point 2) is a plastic zone in which the strain is not linearly proportional to the stress.

Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Patran Nastran buckling analysis for a flat plate

This post shows how to find the buckling index using Patran/Nastran for a flat aluminum plate under shear loading.

1. Create a flat plate geometry in Patran using Geometry create-> surface -> XYZ. The size of the plate used in this example is 8x10.

2. Create an isotropic material: under properties -> Create -> isotropic -> manual input. and in input properties enter the following values.

3. Create 2d shell properties: Create -> 2d -> shell. In property set name put plate. in input properties select isotropic material craeted in step 2 and enter thickness as 0.063.

4. Create mesh seeds as shown:

you should see the following.

5. Mesh the surface:

6. Apply the following boundary conditions to the plate.

7. Apply the following applied loads. 100lbs at each node on each edge as shown in opposite direction. this gives a total of a 1000lbs in opposite direction at each edge giving a shear stress of 1587psi. 

8. Run the model in Nastran using solution 105 (Buckling) and you should get a factor of 1.1383. The first mode of buckling deformation is shown. 

Note that the major advantage is that one can compare buckling index of different configuration. plate with hole, elliptical cutouts etc. which is not always available using standard formulas in standard textbooks. 


Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Buckling Failure of Metallic Structure

Buckling failure is another form of compression failure. When structural engineers talk about buckling they are normally referring to Euler Johnson column buckling failure or plate buckling failure. I will cover both of these in this post.

Plate buckling refer to failure of thin sheets, which can occur under shear or compression loading. The allowable buckling stress is calculated using the following equation:



Where

v = poisson ratio

b = width of plate

t = thickness of plate

E = Youngs modulus

k = constant function of plate width and length.


The constant k above can be found in many text books such as bruhn, flabel, niu ... Note that the derivation of this constant is quite extensive.

Note that the derivation of this constant is quite extensive and not the intent of this website. The figure below shows some of the k values for a flat plate under compression.



Column buckling is very similar to plate buckling in terms of mathematical equations. The critical column buckling stress is calculated as follow:




Where L is the length of the column.

Note that the above equation is commonly referred to as the Euler Johnson equation.



Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions. 

Crippling Failure Part 2

One of the most common equation for crippling is shown below:

Fcc = (Ce*(Fcy*Ec)^0.5)/(b'/t)^3/4

Where
Fcc = Allowable crippling stress
Fcy = yield compression stress allowable
Ec = elastic modulus of material in compression
Ce = material boundary coefficient
b'/t = equivalent width over thickness ratio

Reference: Practical Stress Analysis for Design Engineers, Jean-Claude Flabel

As mentioned in the part 1 post of crippling failure, you can observed from the crippling allowable equation that it is not a function of length but only a function of material properties and width/thickness ratio whereas buckling is a function of length. This is an important distinction between crippling and buckling. For clarity the Euler column buckling equation is shown below:

Pcr-buckling = (Pi^2*E*I)/L^2

Where
Pcr = Critical buckling load
E = Elastic modulus of material in compression
I = moment of inertia about the failure axis of the column
L = column length

The above buckling equation is based on a pinned-pinned type boundary column. As shown the buckling is a function of column length.

Example Problem:

Assuming a flange with a width of 1.2 in and a thickness of 0.1in thick. The crippling allowable stress can be calculated:

Ce = 0.317 (1 edge free aluminum - Ref Flabel)
Fcy = 34 ksi (2024-T3 ref MMPDS)
Ec = 11.0 x 10^6

Fcc = 0.295*SQRT(34000*11*10^6)/(1.2/0.1)^(3/4) = 27.9ksi

Disclaimer: The content on this site is provided as general information only and should not be taken as engineering advice. All site content, including advertisements, shall not be construed as a recommendation to specifically analyze a stress problem. The ideas expressed on this site are solely the opinions of the author(s) and do not necessarily represent the opinions of sponsors or firms affiliated with the author(s). Any action that you take as a result of information, analysis, or advertisement on this site is ultimately your responsibility. Consult professional adviser before making any engineering decisions.