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

 

[rb1957]

no (again) ron ... you Cannot draw a FBD of a beam with no moments at the ends and only two transverse forces (at the ends) ... a typical SS beam works 'cause there is a transverse load applied in the span, reacted at the ends (in equilibrium).

 

[rb1957]

john,

it's a Long time since i've done mechanisms ! structures are easy 'cause they're in static equilibrium. it relatively easy to relate the movement of the input shaft to the movement of the output shaft. the two cranks are connected by a fixed link. The question is how much load is created in the rod ?

If it's in static equilibrium, then why does it rotate ? if it's less why does it stop rotating ? my "guess" is that the rod in the rod is the static equilirium load, and that the rotation is controlled by the movement of the input shaft.

 

[msquared48]

OK. I think that I see what could be the problem .

The OP stated initially that the rod is bending - fact.

The problem is that the diagram we are looking at is a two dimensional diagram, but the problem is three dimensional. I think what is happening is that the crank shaft axes and the rod axis are not in the same plane, inducing moments into the pinned joints at the ends of the rod. A moment has to exist or the rod would not be bending, unless it was bent prior to installation.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[Ron]

Mike...exactly!

 

[rb1957]

good catch ... i'd missed that ... the next thing is which plane is it bending about (as mike mentions, bending out-of-plane is possible). it could also be an onset of buckling ... an enforced displacement on crank 1, a rigid crank 2 would mean the rod would have to change it's length (by bending) or possibly the ends are not truely pinned ... they might have a single fastener, but they might be clamped up tight so a sizeable friction moment could develop.

 

[desertfox]

I don't think the OP is talking about out of plane bending, looking at his diagrams the forces he as resolved are at right angles to the cranks, which then gives him forces on the connecting link which don't run parallel with its axis.
All the OP needs to do is draw a line parallel to the axis of the connecting rod at each crank end, then from that line draw two perpendicular lines one at each end which pass through the crank centre,knowing the torque on the crank and dividing this by the new lengths measured he can find the resultant force along the axis of the connecting rod.

desertfox

 

[msquared48]

He has to have out of plane bending. Otherwise, how are you going to connect the rod to the cranks?

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[desertfox]

I didn't say he didn't have out of plane bending but I believe that isn't forming part of his analysis or question, what he is doing is purely 2D just like the text book 4 bar linkages.
However you can connect two cranks with a connecting rod without out of plane bending by machining each crank with a clevis and fitting the link in between with pins.

desertfox

 

[desertfox]

Hi DiscipleofScience

I have uploaded a file that might help, I think your resolving forces in the X and Y directions which is fine, however your resultant force on the connecting link with a simple pin joint at each end is down the central axis between the two pins, one way of getting this resultant is shown in the file.

desertfox

 

[DiscipleofScience]

Hi All

The front crank is the drive end and only ever moves the bell cranks say half a turn either way from the mid point, most of the time it is static, holding the rear crank in position.

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.

Because the rod is not square to the bell crank (in order to maximize the mechanical advantage it is futher out on the arm of the seccond crank) this means that the applied axial load is not all in the horizontal plane, relative to the rod, there is a small vertical component of applied force FB. Is this vertical component lost?

I suppose maximum efficieny occurs when the bell crank is at 90 degrees. Is this too small to be worried about?

Also - why do I have 2 equations for FD?

http://files.engineering.com/getfil...1-1c9d-483f-b90c-fd95b3e67b5c&file=img001.jpg

regards

John

 

[desertfox]

Hi DiscipleofScience

Our posts crossed see my uploaded file.

Thanks for clarifying the mechanism.

Regards

desertfox

 

[desertfox]

hi

Just looked at your calculations, firstly once you have Fb you need to divide it by Cos Alpha that gives you the force in the connecting rod that then transfers its force to Lc, then if you take the force in the connecting rod and draw its line of action perpendicular to the pivot point at Lc you can workout the torque acting on it.
Lastly the moment where Lc is you have:- Fd*(Lc+Ld) I think that is completly wrong.

desertfox

 

[DiscipleofScience]

desertfox,

thankyou for the sketch, yeah I think I am getting there with this, we are on the same lines

the fact is the small horizontal component has reduced the force from end to end, maximum efficieny is when the angle between F2 and FR (on your sketch) is zero. This horizontal component doesn't travel down the rod but is reacted againts the primary force (the vertical force)

Plugging in some numbers:

Force 2kN
L1 = 0.3m
L2 = 0.4m
angle of crank relative to the rod (angle between F2 and FR) = 10 degrees

LR = 0.4 x Cos 10 = 0.39

so FR = 2x103 x (0.3 + 0.39) = 1.38kN

a loss of (2x103 x (0.3 + 0.4)) - 1.38x103 = 20N (1%)

 

[DiscipleofScience]

Sorry - had a rethink there! Just to make a correction about FB


It should be (2x103 x 0.3)/0.4 = 1.5kN

take away the horizontal component

The horizontal component (to the rod) is 1.5x103 x Cos 10 = 1.477kN

A reduction of 22.7N

 

[DiscipleofScience]

desertfox - I think you are right about the moment about C

FC = FD x LD / LC and it acts directly on the pivot

so FB x LR x Sin 40 = 0, which I think it means its trying to say that there are no moments at all on C

regards

John

 

[desertfox]

Hi John

Fr in my sketch should be greater than F2 because the right angled distance from the line of action of FR is smaller than the distance than L2, therefore to find Fr divide F2 by the Cos alpha angle which in your case is Cos of 10 degrees.

desertfox

 

[rb1957]

like desertfox said, a couple posts back, the force in the rod is Fb/cos(alpha).

looking at your sketch i think there's a problem in your understanding of torque the torque of Fa is Fa*La Only if La is perpendicular to Fa (which it doesn't look to be in the sketch)

the torque arm at crank 2 of the rod force is less than Lc ('cause the rod is not normal to the arm of crank 2).

similarly Fd.

but none of this explains why the rod is bending (in real life).

 

[rb1957]

your solution of sum moments @ C is completely messed up (i'm way less dipolmatic than desert fox) ...

Fb creates no moment about C since it passed thru C.

it'd be Way more rational to take moments about the pivot of crank2.

 

[rb1957]

i'm not sure that alpha, as shown in the sketch, means anything at all. you need the perpendicular distance between the pivot point and the force. this is not the length of the crank arm (unless the force is at 90deg to the crank arm). it is the crank arm*cos(angle) ...

 

[DiscipleofScience]

thanks rb1957, you make a good point that the input and output forces are also not perpendicular to the crank arm, again it will be a case of resolving horizontal and vertical components relative to the crank arm, with the horizontal force being reacted back (lost) at the crank stud.

I am with you also on the Force*Cos(angle)

working with your desertfox's sketch

LR = Sin (90-alpha) x L2 = Cos (alpha) x L2

So basically FR = (F1(v) x L1) / Cos (alpha) x L2

(v) denotes vertical component

[Interesting to note that as alpha decreases FR increases)

please tell me I'm right ;-)

cheers

John

I agree my attempts at taking moments are lame - thanks for