Having a brain fart, need help please

[sbozy25]

Ok, so I am having a monumental brain fart right now and it is really bothering me.

Here is the situation:

I have an axle with an upper and 2 lower torque rods that connect and all run longitudinally. The upper torque rod normally sits on the CL of the axle, but in this case, the customer wants to move it off center so they can use a common bracket. They are concerned that this offset will cause an issue with the end torque rods and their bushings.

I'm trying to find what the moment will be at each end as a result of this uneven load. To simplify this problem, I'm assuming this is an supported beam, with 3 loads. I pulled out my Shigly to make sure I did this right, but I'm getting some astronomical values of moments to the 8th and 9th power. That can't be right, can it??? What's frustrating about this, is that any college freshman could probably do this... If only I could remember back that far, but it has been quite a few years, and I only have to do calcs like this once every 5 to 10 years...

I've attached a picture with the loads and dimensions.


Looking at Shigly, I am using the formulas for fixed supports with an intermediate load.

Which would mean the moment to the left side would be [(Fab^2)(3a+b)]/l^2 and the moment on the right side would be [(Fba^2)]/l^2

Is that right, or am I missing something with the FL values?

Definition of irony: A Ford Focus driver with ADD...

 

[sbozy25]

Sorry, I wrote the wrong formula down..

I meant, the left side is (Fab^2)/l^2 and the right is (Fba^2)/l^2

Definition of irony: A Ford Focus driver with ADD...

 

[structSU10]

Those formulas are correct - What units are you using? Consistency with the units may be the issue. If applied load is 40142# - moment at the left should be about 19000 #*ft, moment at the right should be about 12300 #*ft.

 

[desertfox]

The way your calculating this as a beam all you will get is forces not moments, using your sketch I get one reaction as 24336.854lbf and 15805lbf.

desertfox

 

[sbozy25]

I might have had a unit issue. I broke the equations out and did one section at a time and then added everything up.

Here is what I have...

Lets call the left side 1 and the right side 1.

M1 = (Fu*a*b^2)/l^2 = 227,960 inLb = 18,996.7 ftLB
M2 = (Fu*a^2*b)/l^2 = 148,045 inLb = 12,337 ftLb

R1 = [(Fu*b^2)(3a+b)/l^3] - FL = -4,386.8Lb, meaning the reaction force is in the same direction as FL
R2 = [(Fu*a^2)(3b+a)/l^3] - FL = -16,991.6LB, this reaction force is also in the same direction as FL

This is all assuming the ends are fixed. I think these numbers make sense. The book equation for R1 and R2 don't have the subtraction of FL, but it is acting at the same spot, so if I'm thinking of this right, they would factor into the equation for reaction forces. however, because they are at the end, they will not factor into the moments at the end.

Does this make sense, or am I way in left field picking daisies?

Definition of irony: A Ford Focus driver with ADD...

 

[desertfox]

Hi
My apologies I just looked at the sketch and thought you were assuming a simple supported beam, I see now the ends are fixed and I agree with the formula's in your last post for the moment reactions.

 

[flash3780]

I'm a visual-type-of-person, so were it me... I'd make a beam force/moment diagram. Something like this:

That's one way to express your bending loads in a familiar-looking format.

 

[BrianPetersen]

If this is a driven rear axle for a motor vehicle, it's quite possible to install an upper link off center in this manner. If done correctly, the torque reactions into the chassis can be used to offset the drive torque from the propeller shaft so that the bodyshell doesn't try to twist itself against the springs.

The forces on the left and right trailing arms will most certainly be unequal - and potentially quite large - when you do this. But it can be done, and it has been done.

 

[chicopee]

sbozy25, your free body diagram does not have the correct value for the left reactive force. Based on the FU value and the physical dimensions, the left reaction s/b 24,336.8 lbs and the right reaction s/b 15,805.2 lbs, both of which add up to FU of 40,142 lbs. There is no end moments and the maximum moment is at FU and that value is 376,005 in-lbs.
So if you do have the correct value for FU and the left reaction, then you are missing another load which could be the approximate uniform load of the axle housing. If that is so, then the values you gave us would make more sense,however, we can not guess as to what is possibly missing.

 

[sbozy25]

Update...

My brain fart must have passed last night. I was sitting in my office enjoying a nice glass of Crown Royal Black, when it hit me.


Due to the location of the FL value, I don't need to look at them at first.

I had to write two equations and solve for my unknown values.

FU = R1+R2
Sum of moment at point 1 = 0 = FU*a - R2*L

Solving for R2, I get 18,551 lbs, then when I put it back into equation 1 and solve for R1, I get 28,565 lbs.

Then I know that if the upper torque rod were on center, R1 and R2 would have to be 1/2 Fu, which is 23,558 lbs.

Next, to find my load bias as a result of the offset torque rod, I simply find the delta between the calculated R1 and Fu, it would be the same for R2... I get a 5,007 lbs difference. I know my bushing rate is 230,000 lbs/in. So, to find how much the bushing will move at either end, I divide the load difference by the bushing rate, and I get 0.02 in, and when that is trigged out, I find the angular deflection to be 0.001°. That gives me how much extra the bushings will move at location 1 and location 2 as a result of the offset rod.



Definition of irony: A Ford Focus driver with ADD...

 

[desertfox]

hi sbozy25

I'm a little confused, in your original post Fu= 40,142lbs now if summing moments:- 0=Fu*a-R2*L

0= 40,142*15.45 - R2*39.24

so R2= 40,142*15.45/39.24
R2 = 15805lbf
So I can't see where you got R2=18551 lbs

desertfox