Interesting question: design of a truck-based crane

[eitan77]

Homework Statement

suggest a design of a truck-based crane. The boom of the crane is 50 m long, and it shall be able to lift a weight of 10 tons.
what is the optimal way to attach the boom to the crane platform to reduce the risk of the crane turning over and what is the range of the angles phi for which the crane is still stable?

I would love to know what you think about my solution.
Are the assumptions correct?
Is there a better way to attach the boom to the crane platform (which is not in the center of the platform)?

Relevant Equations

$\sum F =0$
$\sum M =0$

 

[berkeman]

Is there a requitement that the boom attach to the vehicle body in the center of the body? That is not how most mobile cranes are designed/built. And is the intention for this crane to be able to move with a load hoisted, or can you assume you will be stationary when hoisting (so you can add outriggers)...?

https://www.warrenforensics.com/2013/04/09/crane_balancing_act/

 

[eitan77]

Is there a requitement that the boom attach to the vehicle body in the center of the body? That is not how most mobile cranes are designed/built. And is the intention for this crane to be able to move with a load hoisted, or can you assume you will be stationary when hoisting (so you can add outriggers)...?

View attachment 335956
https://www.warrenforensics.com/2013/04/09/crane_balancing_act/

It can be assumed that the movement is quasi-static.
there is no requirement that the boom attach to the vehicle body in the center, the only requirement is to find the optimal way to attach the boom to the crane platform to reduce the risk of the crane turning over.

 

[berkeman]

Do you understand why the boom is attached the way it is in my figure?

And that article seems pretty on-point for this assignment. Have you had a chance to read it over yet? Are you allowed to have outriggers?

https://e-one.com/product/cr-137/

 

[eitan77]

Do you understand why the boom is attached the way it is in my figure?

And that article seems pretty on-point for this assignment. Have you had a chance to read it over yet? Are you allowed to have outriggers?

View attachment 335957
https://e-one.com/product/cr-137/

To be honest I'm not sure why it's attached this way

 

[berkeman]

To be honest I'm not sure why it's attached this way

Try drawing free body diagrams (FBDs) for the two cases of center-attached and rear-attached booms. Leave the counterweight off for simplicity at first. Then calculate how much weight can be lifted for a typical lift angle for each case.

 

[jbriggs444]

To be honest I'm not sure why it's attached this way

Think about stowage. You want the unextended boom to fit mostly within the footprint of the truck. Otherwise driving will be clumsy and you will be an impediment to traffic.

That is not a physics concern as such. It is a practical concern. Good design is about more than just physics.

 

[eitan77]

one more question, how can I find the maximal angles $\theta and \phi$ in which the crane is still stable?
it seems I'm missing equations:

f_11 +f_22 +f_21 +f_12 =mg +w
-h(f_22 +f_12 ) +mg*h/2 -$w[50cos( \theta )sin(\phi )-h/2 ]=0$
-mg* l/2 + l(f_11) +f_12 ) +$w[50cos( \theta )cos(\phi ) -l ]=0$

 

[haruspex]

one more question, how can I find the maximal angles $\theta and \phi$ in which the crane is still stable?
it seems I'm missing equations:
View attachment 335968
View attachment 335969

f_11 +f_22 +f_21 +f_12 =mg +w
-h(f_22 +f_12 ) +mg*h/2 -$w[50cos( \theta )sin(\phi )-h/2 ]=0$
-mg* l/2 + l(f_11) +f_12 ) +$w[50cos( \theta )cos(\phi ) -l ]=0$

Consider the different ways in which the system may become unstable and what those imply for $f_{mn}$ forces in each case.
What about the case where the boom extends away from the truck's mass centre? Or are you assuming it would not be used that way?

 

[eitan77]

Consider the different ways in which the system may become unstable and what those imply for $f_{mn}$ forces in each case.
What about the case where the boom extends away from the truck's mass centre? Or are you assuming it would not be used that way?

Thanks, regarding the second question I assume not.

I'm not sure, the only thing I can think of is to separate the cases where the body rotates only around the
x-axis and then only around the y-axis, and in both cases check what happens when the forces on the two legs farthest from the axis of rotation are zeroed out.

But I'm not sure that assuming that the forces on both legs cancel out together is a reasonable assumption.

 

[haruspex]

separate the cases where the body rotates only around the
x-axis and then only around the y-axis, and in both cases check what happens when the forces on the two legs farthest from the axis of rotation are zeroed out.

That is what I would do. Moment balance around an X axis is independent from that around a Y axis.

 

[eitan77]

That is what I would do. Moment balance around an X axis is independent from that around a Y axis.

Thank you again,

Does that make sense to you?
For some reason I feel that something is wrong.

 

[kuruman]

Does that make sense to you?
For some reason I feel that something is wrong.

I share your feeling.

Consider the projection of the load weight on the ground at point G. Must G be contained within the footprint ABCD as your torque balance equation seems to imply? Would the truck tip over if it extended beyond that? Don't forget that the weight of the truck is 5 times the load weight and is $\frac{l}{2}$ to the left of line BC. This means that if the boom were horizontal at $\varphi=0$, the distance from BC to the projection point P can be $5\times \frac{l}{2}$ before the truck tips over.

While you rethink this, you might consider the following, not in order of importance.

• Since only the boom length is given numerically, you gain nothing by substituting "50" for it. Use a symbol like $R$ to make the equations easier to read and test for dimensional consistency. It's also esthetically more pleasing to use "double" width and height for the rectangle to avoid division by 2.
• As you know, some reference points for torques (moments) are better than others in equilibrium situations. I think a good point in this case would be point O (see drawing below) defined as the point of intersection with line BC. That puts it in the plane of the boom and gravity vectors and makes the calculation of lever arms easier.
• Use unit vector notation to write the position vectors of point G and the center of the rectangle and the two forces of gravity. Then write the net torque as $$\vec {\tau}_{\text{net}}=\sum_{i=1}^2 \vec {r}_i\times \vec F_i.$$ Less room for error.
• The above calculation pertains only to tipping about BC. A second similar calculation for tipping about AB will be needed with the reference point on line AB.
• At fixed $\theta$, what is the critical angle $\varphi_{\text{crit}}$, where one should shift from one expression to the other?

 

[haruspex]

what is the range of the angles phi for which the crane is still stable?

Are you required to analyse it in that way? To me it seems easier to consider the horizontal envelope within which a given load must remain.

 

[eitan77]

Are you required to analyse it in that way? To me it seems easier to consider the horizontal envelope within which a given load must remain.

Yes, that is what is required of me.

 

[Lnewqban]

Homework Statement: suggest a design of a truck-based crane. The boom of the crane is 50 m long, and it shall be able to lift a weight of 10 tons.
what is the optimal way to attach the boom to the crane platform to reduce the risk of the crane turning over and what is the range of the angles phi for which the crane is still stable?

The wording of the problem seems to be incomplete.
How can the question about “the range of the angles phi for which the crane is still stable?” could be answered without the data about the weight of the crane’s counter-weight, of the truck, and dimensions?

In real life, a truck-crane with a telescopic 50-meter boom is a machine of big dimensions and able to lift 50 tons, but never sideways and at a shallow angle with the horizon.

Copied from:
https://en.m.wikipedia.org/wiki/Crane_(machine)#Truck-mounted

“Most cranes of this type also have moving counterweights for stabilization beyond that provided by the outriggers. Loads suspended directly aft are the most stable, since most of the weight of the crane acts as a counterweight.”

 

[kuruman]

The wording of the problem seems to be incomplete.
How can the question about “the range of the angles phi for which the crane is still stable?” could be answered without the data about the weight of the crane’s counter-weight, of the truck, and dimensions?

We are not given a lot of parameters, those you mentioned and whole lot more such as the location of the truck's CM, the mass of the boom and so on. Although "design" is mentioned in the statement of the problem, I think it should be taken with a grain of salt. The key hint (I think) is

what is the optimal way to attach the boom to the crane platform to reduce the risk of the crane turning over and what is the range of the angles phi for which the crane is still stable?

OP has already figured out that attaching the boom in the middle of the rectangular footprint as indicated in post #1 provides symmetric front-back instability and corrected that in post #8. What remains now is finding the range of angles $\varphi$ for which the crane is still stable. My interpretation of that is to generate a family of allowed values of $\varphi$ at fixed $\theta$. This is not about design, it is about analyzing an equilibrium situation when the pivot stays fixed while the net torque rotates relative to it.

 

[haruspex]

Yes, that is what is required of me.

Ok, but it still seems to me that the easiest is approach is to figure out that envelope (geometrically simple) then deduce the ranges for $\phi$ in terms of the other variables.

 

[eitan77]

Ok, but it still seems to me that the easiest is approach is to figure out that envelope (geometrically simple) then deduce the ranges for $\phi$ in terms of the other variables.

So does that mean you think the solution I gave in #12 is incorrect?
After going through it again, I'm starting to be convinced that the solution is actually correct.

 

[haruspex]

So does that mean you think the solution I gave in #12 is incorrect?
After going through it again, I'm starting to be convinced that the solution is actually correct.

You have a typo in (3), h instead of l.
All equations down to "$(3)\rightarrow$" are ok but should really be inequalities expressing constraints. Otherwise it is unclear which side of that constitutes stability.

Beyond that, I do not understand your logic. You took the condition that it is about to tip about AB and combined that with the condition that it is about to tip about BC. So the equation that resulted only applies when it is about to tip about both. Is that interesting? And what does it mean when you turn that into an inequality? That will not be a bound on not tipping. It might be a bound on whether it tips one way rather than another.

You should have four separate inequalities, one being the condition that it does not tip about AB, one that it does not tip about BC, then CD and DA likewise. All four must be met for stability.
At that point you can see if you can rewrite them as bounds on $\phi$.

 

[eitan77]

You have a typo in (3), h instead of l.

Thank you!

All equations down to "$(3)\rightarrow$" are ok but should really be inequalities expressing constraints. Otherwise it is unclear which side of that constitutes stability.

I will be more clear, the situation I described in the calculation refers to angles $\phi$ in the quarter of the positive plane ( $0<= \phi < \pi /2$ )
and $0<= \theta < \pi /2$ .

$\theta _m$ - is the maximal $\theta$ (same with $\phi$ )

I am not letting $\theta$ be negative and since the problem is symmetric around the x-axis I calculate $\phi_m$ only on the quarter positive plane and I use it as $-\phi _m < \phi < \phi _m$.

Beyond that, I do not understand your logic. You took the condition that it is about to tip about AB and combined that with the condition that it is about to tip about BC. So the equation that resulted only applies when it is about to tip about both. Is that interesting? And what does it mean when you turn that into an inequality? That will not be a bound on not tipping. It might be a bound on whether it tips one way rather than another.

Maybe I am wrong about that, but following on from what I wrote above, I guess the rotation axis can be somewhere between AB and BC.

You should have four separate inequalities, one being the condition that it does not tip about AB, one that it does not tip about BC, then CD and DA likewise. All four must be met for stability.
At that point you can see if you can rewrite them as bounds on $\phi$.

following on from all that I wrote above, I do not think I need to check it for CD and DA (since the symmetry).

but I think I should calculate $\theta _m$ in the condition where $\phi =0$ as well ,

hence:
$(3)\rightarrow$ $cos(\theta_m1 ) = (l/2) *[(mg+2W) / 50W]$

if $\phi$ is not zero then we can use the expression from # 12 :

$cos(\theta _m2) = (h/l) * [(mg+W)/(50Wsin(\phi _m) )$

now $\theta _m$ is the max {$\theta _m1$,$\theta _m2$}

please let me know if I was clear enough and if that makes sense to you.

 

[haruspex]

the rotation axis can be somewhere between AB and BC.

No, it will be AB or BC. Treat the two cases separately.

 

[eitan77]

No, it will be AB or BC. Treat the two cases separately.

In that case

I am not sure how I should continue from here.

 

[haruspex]

As I wrote before, you should be writing inequalities, not equalities. If you end up with an equation $\phi=…$, how do you know whether it is stable for $\phi$ larger than that or smaller $\phi$? And that's also why it is not obvious how to continue.

I don't understand why you considered $\phi=\pi/2$ and solved for $\theta$. In post #1 you wrote that you want the ranges for $\phi$. Nothing there about ranges for $\theta$. Have you posted the problem exactly as given to you?