Combined Stresses

Design a left crank arm for a bicycle.

Tutorial Notes: combined-stresses-worked problems.pdf empty.one

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Combined Stresses Theory

1. Adding Stresses
2. Combining Bending and Tension Stresses
3. Eccentric Loading
4. Combined Shear and Tensile Stresses
5. Stresses on a Cutting Plane
6. Multiple Tensile Stresses
7. Combined Stresses General Case
8. Von Mises Stress
9. Combined Stresses in Shafts
OBJECTIVES
When you have completed this unit you will be able to:
- Calculate the resultant of tension and compression loads acting on the same axis.
- Calculate the resultant of tension (compression) and bending stresses.
- Calculate the principal tensile stresses.
- Calculate the maximum shear stress.
- Use combined stresses for shaft design.

1. Adding Stresses

How do we add up or combine stresses (to get the resultant stress)?

Just like colinear forces, we stresses are in the same direction we can simply ad them algebraicly.

For example: Tension and compression stresses acting along the same axis. Wesimply take the tensile stress as positive and the compressive stress as negative - then add them up.

A tensile stress of 1000N is combined with a compressive stress of 2000N. The resultant is a compressive stress of 1000N.

2. Combining Bending and Tension Stresses

When a beam is bent it produces tension and compressive stresses along the axis of the beam. These axial stresses can be algebraically added to other tension and compression stresses acting along the axis of the beam. (Below)

The following example involves a combination of bending and tensile stresses caused by a single force.

3. Eccentric Loading

This is sometimes referred to as eccentric loading.
The load F (below) does not act on the centroid of the section. Consequently the load F; produces,
(a) Compressive stresses (axial stress),
(b) Bending stresses (i.e. axial tension and compression).

4. Combined Shear and Tensile Stresses

Information about material strength is obtained from the tensile test. In these tests, only pure tension in one direction exists in the specimen. However, in practice you do not have just pure one direction tension in components, but a combination of stresses acting simultaneously. Consequently it is necessary to combine all these simultaneously acting stresses to get an equivalent stress to compare with the results of a tensile test. Figure 5 shows a typical belt driven shaft. The shaft is subjected to torsion due to the power transmitted and to bending by the belt loads.

To design a shaft you need to be able to find the equivalent maximum stress resulting from the bending stress and the shear stress. However, you cannot simply add tensile (or compressive) and shear stresses together directly. The procedure for finding the maximum equ1valent stress is outlined in the following sections.

5. Stresses on a Cutting Plane

Figure 6 shows a specimen under tension. If you place an imaginary cut across this specimen at angle 8 you will see that you require a normal stress f n and a shear stress f s to maintain equilibrium. As the angle 8 varies, so will the magnitude of the normal and shear stresses.

As a matter of interest the normal stress is a maximum at θ = 0 and this is where the shear stress is zero.

The normal stress (to the chosen plane) that corresponds to a zero shear stress is called a principal stress.

The shear stress reaches a maximum value for θ = 45°. This is called the maximum shear stress.

Extra: Here's a lecture about the derivation of principal stress and how it relates to shear stress. It looks complicated but still gives the overall idea.

https://www.youtube.com/watch?v=20oIAoZyb0s

6. Multiple Tensile Stresses

In a cylindrical pressure vessel, the walls are simultaneously subjected to:
(a) hoop stress,
(b) longitudinal stress.

In the situation of the pressure vessel it is not clear as to what the resultant equivalent stress is. A part of the cylindrical wall of the pressure vessel is shown in Figure 8 with the two stresses acting at right angles.

As with the tensile specimen, the stresses on a surface at angle θ can be evaluated. In this manner the maximum or principal stresses can be evaluated (i.e. the maximum equivalent stress resulting from hoop stress fh and and axial stress fL).

7. Combined Stresses General Case

The pressure vessel wall in Figure 7 had two stress acting at right angles.

A more general case is when a component may be subjected to two normal stresses acting at right angles plus shear stresses. This is shown in Figure 9.

Analysis of the stresses in Figure 9 produces the following results.

8. Von Mises Stress

Unsigned comment added by 75.162.56.142 (talk) 23:43, 9 March 2008 (UTC)

What is Von Mises Stress exactly?

(BTW: Von Mises Stress is actually a misnomer. This criterion is also referred to as the Maxwell–Huber–Hencky–von Mises theory, and Mr Von Mises wasn't even the first person here! Oh, and he is not to be confused with the other Von Mises either, of Austrian school of economics fame, who said that when central banks interfere with the economy they turn an inevitable correction into an unnecessary disaster.)

Meanwhile, back to unnecessary disasters of the engineering sort...

In an elastic body that is subject to a system of loads in 3 dimensions, a complex 3 dimensional system of stresses is developed (as you might imagine). That is, at any point within the body there are stresses acting in different directions, and the direction and magnitude of stresses changes from point to point. The Von Mises criterion is a formula for calculating whether the stress combination at a given point will cause failure. There are three "Principal Stresses" that can be calculated at any point, acting in the x, y and z directions (often labelled as 1,2 and 3 axes - to not get mixed up with the model axes).

(The 1,2, and 3 directions are the "principal axes" for the point and their orientation changes from point to point, but that is a technical issue.) Von Mises found that, even though none of the principal stresses exceeds the yield stress of the material, it is possible for yielding to result from the combination of stresses. The Von Mises criteria is a formula for combining these 3 stresses into an equivalent stress, which is then compared to the yield stress of the material.

The equivalent stress is often called the "Von Mises Stress" as a shorthand description. It is not really a stress, but a number that is used as an index. If the "Von Mises Stress" exceeds the yield stress, then the material is considered to be at the failure condition. The formula is actually pretty simple, if you want to know it:

Where , and are the principal stresses and is the equivalent stress, or "Von Mises Stress".

Pretty simple, but finding the principal stresses at any point in the body is the tricky part.

The following table summarizes von Mises yield criterion for the different stress conditions.

Load scenario	Restrictions	Simplified von Mises equation
General	No restrictions	$\sigma_v = \sqrt{\tfrac{1}{2}[(\sigma_{11} - \sigma_{22})^2 + (\sigma_{22} - \sigma_{33})^2 + (\sigma_{33} - \sigma_{11})^2 + 6(\sigma_{12}^2 + \sigma_{23}^2 + \sigma_{31}^2)]}$
Principal stresses	No restrictions	$\sigma_v = \sqrt{\tfrac{1}{2}[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2]}$
General plane stress	$\sigma_3=0\!$ $\sigma_{31}=\sigma_{23}=0\!$	$\sigma_v = \sqrt{\sigma_{11}^2- \sigma_{11}\sigma_{22}+ \sigma_{22}^2+3\sigma_{12}^2}\!$
Principal plane stress	$\sigma_3=0\!$ $\sigma_{12}=\sigma_{31}=\sigma_{23}=0\!$	$\sigma_v = \sqrt{\sigma_1^2- \sigma_1\sigma_2+ \sigma_2^2}\!$
Pure shear	$\sigma_1=\sigma_2=\sigma_3=0\!$ $\sigma_{31}=\sigma_{23}=0\!$	$\sigma_v = \sqrt{3}\|\sigma_{12}\|\!$
Uniaxial	$\sigma_2=\sigma_3=0\!$ $\sigma_{12}=\sigma_{31}=\sigma_{23}=0\!$	$\sigma_v = \sigma_1\!$

Notes:

Subscripts 1,2,3 can be replaced with x,y,z, or any other orthogonal coordinate system labelling method
Shear stress is denoted here as $\sigma_{ij}$ ; in practice it is also denoted as $\tau_{ij}$

9. Combined Stresses in Shafts

Shafts are subjected to bending moments and torsional moments, which produce tensile and compressive stresses, and shear stresses respectively. Figure 10 shows a shaft with belt drives. The belt tensions cause bending in the shaft.

To design a shaft you have to deal with the combined bending and torsional stress. Figure 11 shows a shaft cross section and the stresses in the shaft.

You will note that the maximum bending stress occurs on the surface of the shaft.
The maximum torsional stress also occurs on the surface of the shaft. The resultant of these two stresses is found
using the method of combined stresses as discussed earlier in this unit.

Combining the theory with the basic equations of torsion in a cylindrical shaft, we can derive a simplifieed equation;

Note: The equivalent moment Me does not actually exist. It is simply used in the bending stress equation to produce the value of the maximum tensile stress in the shaft.

The following examples apply these equations to shafts under bending and torsion.

FEA practice

Torsion

First things first. Let's look at torsion.

A simple shaft, Diam 20 x 200 long, steel, with an applied moment (torque) of 100 Nm, and a fixed constaint on the other end.

The FEA results:

Where element size of the mesh was refined as below;

Using the "probe" we can determine the stress at exact locations o the part.

On the top toolbar, click on Probe button in the Result group, then click a spot on the model. This will paste a label over that spot. Every spot you click will get a new label. Once you have finished clicking to make labels, get out of probe (escape) and then you can drag the labels out the way to reveal an arrow to the original spot you clicked.

The label will display whichever measurement you are set to, Von Mises, 1st Principle, etc (in the left Browser Bar)