# Maximum moment

1. Aug 14, 2006

### teng125

http://www.sendspace.com/file/r2469z

attached file

find the stresses at points A;B and C when the maximum moment is applied.

should i consider the total three beams or only one beam if i want to calculate for the moment of inertia in order to use the formula (bending stress = moment / moment of inertia) to answer the ques above at point A,B and C ??

pls help

thanx

2. Aug 14, 2006

### Staff: Mentor

I think you might want to put your file on another site.

3. Aug 15, 2006

### teng125

pls try again because i have just try it and it can be downloaded

4. Aug 15, 2006

### Meson

Given the maximum bending stress has been found, then the bending stress distribution is given by

$$\sigma(y)=\frac{\sigma_{max}}{C}y$$

where C is the distance between the neutral axis (where the layer of atoms experience zero resultant stress) to the outermost fiber.

The bending stress distribution varies with the position y. So by substituting the distance y into $$\sigma(y)$$ gives the corresponding bending stress at the designated location.

5. Aug 15, 2006

### teng125

what is the difference between allowable,limit and max stress??

i think that allowable and limit stress is the same while how can i calculate max stress if allowable and limit stress is given??

6. Aug 15, 2006

### teng125

so here,C is the same for all points but y is the distance from neutral axis to the respective points??

am i right??

7. Aug 15, 2006

### Meson

Good question.

The limit stress, by definition is the stress a particular material can withstand before it
a) exceeds the yield strength (yield point) OR
b) exceeds the tensile strength (also known as breaking point, ultimate tensile strength: UTS)
However, for most of the machines, we do not want the stress experienced by its construction material (steel etc) to exceeds the yield strength for they will fail to work as desired or designed within the range of plastic deformation. Thus b) is out of interest. Hence, the limit stress $$\sigma_{lim}$$ is practically the yield stress, $$R_{p\ 0.2}$$, that is

$$\sigma_{lim}=R_{p\ 0.2}$$

according to DIN-EN-10002-1, 1991

The allowable stress, is the stress that is permitted to be loaded on a given machine. It is always less than the limit stress. The reason is that while a particular material is theoritically able to withstand its limit stress, practically this is not the case. Most material experience plastic deformation under a stress of not even one third of the limit stress. It depends on many factors, such as the quality of the material, operation environment and control, risk and cost, just to name a few. We call this the safety factor. There should be a complete list stating these factor in your study material (you are a mechanical engineering student, aren't you hmm?). Hence, the allowable stress $$\sigma_{all}$$ is also known as the actual stress $$\sigma_{act}$$. This applies to all three types of stress: tension, compression and shear.

$$\sigma_{all}=\frac{\sigma_{lim}}{n_{s}}$$

where $$n_{s}$$ is the safety factor.

When a bending moment is applied to a material, each layer of atom in the material experiences different stress. There exists a location where the stress is maximum and we call this the maximum stress. It has nothing to do with limit stress or allowable stress. It depends on the force applied on the material, which yields a bending moment and hence the stress.

$$\sigma_{max}=\frac{M_{b}}{W_{by}}$$

where $$M_{b}$$ is the bending moment applied and $$W_{by}$$ is the Widerstandsmoment.

Bingo.

Last edited: Aug 15, 2006
8. Aug 15, 2006

### Meson

Also worth mentioned that the bending stress distribution is best to be written in the form

$$\sigma=\frac{y}{C}\sigma_{max}$$ where the ratio $$\frac{y}{C}$$ is emphasized. Thus y can takes the following range of values:

-C < y < C

with + indicating tensile, - indicating compression stress

EDIT:
I think the explanation about limit stress, allowable stress and maximum stress given above might somehow gives you a hint in finding the maximum stress given limit stress and/or allowable stress.

$$\sigma_{max}=\sigma_{all}=\frac{\sigma_{lim}}{n_{s}}$$

Yes, there isn't much useful information we can obtain from this relation as the question isn't meaningful. However, usually we are interested in the maximum MOMENT given the limit stress and/or allowable stress for we will know the maximum load the material of interest can withstand.

Last edited: Aug 15, 2006