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.