rod that is pin jointed at both ends 1

[DiscipleofScience]

Hi

I am using a rod in compression and am fairly comfortable with using Eulers theorem.

However how could I resolve the situation where a compressive force is at an angle - see picture

I believe that this would actually force a torque on the rodding and perhaps should be more of a concern than Euler.

Would I resolve the force into the x and y components relative to the centre line of the rod and the force that is not resolved to the centreline is basically the torque that is acting on the rodding itself rather than transmitting force.

http://files.engineering.com/getfil...4-a853-4515-89fa-c2fa4d7cb705&file=cranks.jpg

I am trying to understand the forces in this rodding as presently it is bending.

regards

John

 

[desertfox]

Hi DiscipleofScience

Using your diagram earlier with the enlarged view of the crank and Fb, if Fb as drawn is perpendicular to the crank and the rod on the otherside is at an angle alpha, then the force in the rod is given by Fb/Cos Alpha, now imagine just altering the rod angle alpha (everything else staying the same) so that instead of being 10 degrees it is now 30 degrees, therefore as Alpha increases then the force in the rod FR increases, if Alpha decreases then the force FR decreases, I think your statement in brackets is the wrong way round.

Your statement in your original post:-

"I am trying to understand the forces in this rodding as presently it is bending".

Is it actually bending on the actual mechanism or are you thinking its bending because of your forces in your diagrams.
If its bending for real then what might be happening is that its buckling under the compression forces, we could determine that but would need the forces in the mechanism and the connecting rod sizes.

desertfox

 

[DiscipleofScience]

Yes it is bending and its due to Euler type compression forces, I have based calculations for sizing the new rod on the output force so say it delivers a maximum of 6kN to the back then that is what I have based my calcs on with a Factor of Safety of 4.

I just need to develop the free body diagram.

 

[desertfox]

Hi John

Look at the example on the link I left you earlier and the sketch I uploaded recently.
From what you are saying I believe the links just buckling, have you considered using a circular connecting link ie uniform 'I' value in all directions.

desertfox

 

[rb1957]

how readily does crank2 turn ? you've said that crank 1 rotates. if crank2 is seized then that'll cause problems for the rod.

have your calcs shown that the original rod buckled ?

alpha is defined two ways in your sketch, as the angle between the rod and the normal to the crank1 arm, and as the angle between the rod and the crank2 arm ?? are the cranks parallel ?

if alpha (=a) is the angle between the rod and the crank1 arm, note this is not your definition, for you have alpha as the angle between the rod and the normal of the arm; i think you need three other angles defined ...
the angle (=b) betwwen the input for and crank1 arm,
the angle (=c) between the rod and the crank2 arm, and
the angle (=d) between the output force and crank2 arm.

then using the arm lengths A for crank1 at input force,
B crank1 at rod attmt,
C at crank2 at rod attmt, and
D at crank2 at output force.

then Prod*B*sin(a) = Pinput*A*sin(b)
and Prod*C*sin(c) = Poutput*D*sin(d)

 

[DiscipleofScience]

hi desertfox - yes thankyou I have studied the example and the sketch very carefully and tried to use the same principles to produce the free body diagram for the pivot of the first crank - which gives me the force in the connecting rod.


hi rb1957
The Euler equation bears what we are seeing. I have attached a free body diagram which hopefully explains it better, this is a difficult to get across with text!

I can use the same principle for the rear crank, there's really no reliable dimensionable relationship between the position of the front and rear crank except the force travelling down the rod. So I almost have to treat them as seperate entities.

At least now I have a formula for the force in the rod/link thanks to you guys.

http://files.engineering.com/getfil...f-4f91-4772-a416-ba34d5a69d1e&file=img002.jpg

 

[DiscipleofScience]

just to be a pedant, I need to point out in the summation of moments bit: one should be negative so that it becomes positive value when moves to the other side of the equation.

 

[rb1957]

if the arrows are taken as +ve directions then it works, just that the math is a little sloppy.
Sum Mb = 0 ...Pav*0.305 = Pbh*x (why is x a variable ?)

the reactions at the pivot are wrong ... there are contributions from both forces (Fa and Frod) in both directions.

 

[DiscipleofScience]

hi rb1957 -

x is variable to suit the installation (a sleeve on a crank arm)

I have the reactions for FA as FA Vertical and FA Horizontal similarly for FC (the rod) FCV and FCH - all relative to the crank. The moments (FCH and FCV) don't act on the centre pin.

I have drawn the pin reaction arrows the wrong way around.

 

[rb1957]

the pivot reacts the forces applied to the crank.

Fa has h- and v-components, both these are reacted at the pivot.

can you draw a free body diagram ? sum forces in both directions ??

i'd've thought x was known (since you've already got this thing built, no?)

 

[DiscipleofScience]

rb1957, is this okay for the FBD?

http://files.engineering.com/getfil...3-a87f-4850-81da-5a9e9f275582&file=img003.jpg

 

[TERIO]

Your summation of moments is fine. For summations of forces you need to sum all forces in the chosen direction, e.g.,

[Σ]Fx = 0 = FAx-RBH-FCy

 

[DiscipleofScience]

Terio - would I include FCy since it is not acting on B directly (B is a pivot), but is a moment around B?

 

[TERIO]

FCy acts on B because a shear force equal to FCy acts on all cross-sections of the arm between B and C. To see this cut you free body diagram somewhere between B and C and look at the forces (and moments) that would have to act on the cut section to keep the part with C from flying away.

 

[rb1957]

No ...

sumFx = 0 ... Fbh = Fax-Fcy

your diagram is not incorrect, but it is very sloppy ... you're using "datums of convenience" ... it's "better" to have a consistent +ve direction and to measure angles to the same datum. 'cause then the math works out better. with your approach, you need to be looking at thepicture to understnad why you're combining x- and y- components, +ve and -ve signs.

 

[rb1957]

btw, as a nit-pick, the rod looks as though it's attached to the cranks by some brkts. the line of action of the rod force is thru the pins that attach the rod to the brkts (and not where the brkts attach to the cranks).

 

[DiscipleofScience]

rb1957 - not sure what you mean by 'datums of convenience', which datum should I use? Do you mean instead of referring to say FCy I should in fact state 'Fc Sin thetaB'?

I have swapped the labels around for FCy and FCx - schoolboy error!

Okay Terio and rb1957 I'll take it as read that FCy and FAy should be included so therefore amended diagram is attached.

rb1957 - I'm probably overegging this anyway by taking account of 'theta A' angle since is it is only a few degrees at most, for this purpose I'll keep it simple since I am seem to be struggling with this as it is

http://files.engineering.com/getfil...d-20f9-42fe-a234-f11f2ef26b36&file=img004.jpg

 

[zekeman]

"This is a simple rod connected by 2 bell cranks, bending is strictly in 2D due to the fixings as desertfox alludes to, it is a a fork type connection."

Yes,if you "see" the rod bending in that 2 dimensional structure it is due to the dynamics of the problem , i.e. acceleration in the mechanism and friction.

If the mechanism is dead slow or almost stopped, then the static case is in play and there can be virtually NO moment carried in the rod, even with friction present

 

[desertfox]

hi DiscipleofScience

It appears to me that your summation of moments in your last post is incorrect Fcy doesn't have a moment it is a vertical force going through the pivot at B and the force you have actually found in equation 1 is not Fcy but Fcx.
I will try to help you some more on probably Friday.

desertfox

 

[rb1957]

he's relabelled Fcy as Fcx ... corrected one equation, forgot the other.

"datums" of convenience" ... your +ve direction for each force suits the direction you think the force will be acting ... Fax is +ve in one direction, Fcx the other. "Normally" you'd start by defining a +ve co-ordinates system x +ve to the right, and Fc will be a -ve number. The advantage of this system is you can say Fax+Fbx+Fcx = 0 (instead of Fax-Fbx-Fcx = 0); i suspect Fbx is negative.

not sure how this helps answer your original problem ... why is the rod bending ? If the calc'd load exceeds the euler buckling load, you then have to ask yourself, what's controlling the displacement of A ? ('cause if the rod buckles, it's stiffness goes to zero and the crank'll rotate, unless A has soe displacement control).

 

[zekeman]

I wrote
"Yes,if you "see" the rod bending in that 2 dimensional structure it is due to the dynamics of the problem , i.e. acceleration in the mechanism and friction.

If the mechanism is dead slow or almost stopped, then the static case is in play and there can be virtually NO moment carried in the rod, even with friction present"
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Firthwer comment:
While the dynamic situation would cause some bending in the rod I do not believe it to be of the magnitude to cause the "observed " bending.

So, I basically agree with the conclusion of others that since there is NO moment acting at either end of the rod there cannot be any bending within the rod. As a consequence, if the OP observes bending then it's cause is Euler buckling due to a single compressive force in the rod, period.

I think the OP should post the forces and rod dimensions so we can put this to bed. If the numbers suggest Euler buckling then case closed. If not, then we have to look for something in the pin connections that could "clamp" one or both ends and thus cause the "observed" bending.