Calculating & Using Bending Moments: Cantilevers vs. Euler's Strut

[thepopasmurf]

Would anyone be able to explain how to calculate and use bending moments? I have been studying cantilevers and Euler's strut and I don't understand the following:

Why for a cantilever the bending moment is positive, and for the Euler strut the bending moment is negative. As far as I can tell two are bending in the same way.

 

[RedX]

Would anyone be able to explain how to calculate and use bending moments? I have been studying cantilevers and Euler's strut and I don't understand the following:

Why for a cantilever the bending moment is positive, and for the Euler strut the bending moment is negative. As far as I can tell two are bending in the same way.

Is an Euler strut like a mcphereson strut? If so I don't see how they can be different either. A cantilever is just a bar that sticks out of a wall and a force is applied to the free end. A strut sticks out of a wall and into another wall, but you should be able to replace one of the walls with the force that the wall exerts, thereby reducing it to the cantilever case. For example for the mcphereson strut you can replace the end where it attaches to the steering knuckle with an applied force equal to the force that the knuckle exerts on the strut.

 

[thepopasmurf]

I'm not familiar with McPhereson struts. They don't appear to be the same. I'm referring to buckling of a thin rod rod under a compressive force (for some reason it's called the Euler strut).

 

[Studiot]

I have been studying cantilevers and Euler's strut and I don't understand the following:

Why for a cantilever the bending moment is positive, and for the Euler strut the bending moment is negative. As far as I can tell two are bending in the same way.

Cantilevers, like other forms of beam, are usually loaded in a direction transverse to the main axis of the beam.
Struts are compression members that should not be subject to transverse loading. the compression loading is along the main axis of the strut.

I suggest you re-read you study material in the light of this information and come back if things are still not clear.

@redx
I am not an automotive engineer but I know that a Mcpherson strut is a complex component that is also subject to torsion and not applicable here.

go well

 

[Studiot]

See post#4 of this thread for a graphical build up of how a moment can arise in an axially loaded strut.
Euler's method applies and solves the mathematics resulting from this mechanism.

https://www.physicsforums.com/showthread.php?t=484310

 

[thepopasmurf]

Hi,

So I have been going over my notes (I also looked at your link). I think I understand it but there is still one thing.

So I am trying to show that the bending moment M= -Fy (See Diagram Attached)

From my diagram you can see that I can get this answer by splitting the strut into two halves, but I would like to get this result by just considering the beam as a whole. My problem with that is that then the moments from both of the end forces cancel out. How do I reconcile this without treating the middle of the beam as some special boundary?

Edit: I should clarify that in my diagram, the moments, M are the moments needed to stop the piece from rotating

 

[Studiot]

You have given little away about the level at which you are approaching this and elastic instability (buckling) is not usually studied till well after cantilevers as it requires more detailed mathematical manipulation.

Essentially Euler's method works as follows

Write the differential equation for the moment.

$$M = EI\frac{{{d^2}y}}{{d{x^2}}}$$

This cannot be solved directly since M is not a function of x

But noting as in the linked thread that M = axial force x eccentricity and that y is the eccentricity.

$$M = EI\frac{{{d^2}y}}{{d{x^2}}} = - Fy$$


There are now several ways to solve this differential equation.

One is to expand and use the integrating factor 2dy

Expand

$$EI\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = - Fy$$

Multiply through by 2dy

$$EI{\left( {\frac{{dy}}{{dx}}} \right)^2} = - F{y^2} + C$$

This is now a standard differential equation in x and y.
The constant of integration can be eliminated by noting that at maximum deflection the slope of the curve is zero ie

$$\frac{{dy}}{{dx}} = 0$$

hope this helps.

 

[thepopasmurf]

But noting as in the linked thread that M = axial force x eccentricity and that y is the eccentricity.

$$M = EI\frac{{{d^2}y}}{{d{x^2}}} = - Fy$$

I accept this. Can I clarify something: for a point on the rod, what do you consider to be the axial force? It seems that it must change when you go past the centre point.


RE my level: I am a physics student and I am comfortable with the physical and mathematical concepts involved except for this one bit

 

[Studiot]

Can I clarify something: for a point on the rod, what do you consider to be the axial force? It seems that it must change when you go past the centre point.

Why do you think this?

The differential equation is derived by considering a free body diagram extending from one end of the member. As you progress towards the other end the forces acting across (at) successive sections are considered. Such a scheme will only ever include the axial load at one end. In general there is a resultant force and moment at any section which ends the free body, balanced by an equal and opposite force and moment supplied by the other part of the member across that section. But we don't include the balancing forces and moments in our reckoning.