Statically Indeterminate Truck - Finding reaction forces at supports

[Ashferico]

Homework Statement

A truck is supported at its two axles and its stabiliser legs. It has a weight of 59kN and carries a load on a lifting arm of 45kN. What are the reaction forces at each wheel and stabiliser leg for static equilibrium? The problem is statically indeterminate because there are more unknowns than equilibrium equations.

Relevant Equations

Vertical Forces in Y axis = 0
Horizontal Forces in X axis = 0
Moments about a point = 0
Compatibility of Deformations/Displacements/Deflections
Force-Displacement: Displacement = (Force x Initial Length)/(Area x Young's Modulus)

If you do the sum of vertical forces and sum of moments, you're always left with an unknown.

Vertical Forces: Ray + Rby + Rcy = 59kN + 45kN (1)
Sum of Moments about A: (59 x 1.85) - (Rby x 3.7) - (Rcy x 4.8) + (45 x 6.5) = 0 (2)

If you make a formula for Rby using Rcy and substitute back into equation (1), you will still have Ray as an unknown hence statically indeterminate.

From my understanding statically indeterminate problems require that you somehow relate the displacements/deflection of supports to each other. Hence forming another equation and eliminating an unknown.

I know for beam problems you find a point on the beam that you KNOW the deflection is equal to zero. Then you calculate the deflection of the beam at that point IF ONLY one load or reaction force is acting. Then you do this for each load/reaction force and sum them all equal to zero. You eliminate unknowns and then substitute back into the equilibrium equations.

However, I don't understand how to apply this to my truck scenario. I've been trying to do it for 3 days. I would love if the smart guys from Physics Forums could point me in the right direction.

Thank you :)

 

[jbriggs444]

Let us review your understanding of truck suspension.

Is the front axle sprung or unsprung?
Is the rear axle sprung or unsprung?
Are the stabilizer legs sprung or unsprung?

If the driver extends the stabilizer legs until they are just touching the ground and then begins to lift the payload, do you see a way to solve the problem?

 

[Ashferico]

Let us review your understanding of truck suspension.

Is the front axle sprung or unsprung?
Is the rear axle sprung or unsprung?
Are the stabilizer legs sprung or unsprung?

If the driver extends the stabilizer legs until they are just touching the ground and then begins to lift the payload, do you see a way to solve the problem?

Hi J Briggs,

Thank you for the response. I'm not super clued up about suspensions. I would say that both axles and stabilizer legs are sprung with the same stiffness.

With the last question, does that mean the stabilizer reaction force is zero? If so, I guess that makes it possible. I'm not quite sure.

 

[jbriggs444]

Thank you for the response. I'm not super clued up about suspensions. I would say that both axles and stabilizer legs are sprung with the same stiffness.

In fact, the stabilizer legs are rigidly mounted to the frame while the tires, wheels and axles are connected to the frame through springs.

The wheels have "give". The stabilizer legs do not.

You will probably need a simplifying assumption about the spring constants for the front and rear axles. One way of approaching that would be to assume that the travel for both front and rear springs is identical under a typical load. That would let you determine the ratio of the two spring constants.

Edit: The diagram can be used to determine the front/rear load ratio before the payload is lifted. It is a very simple one.

 

[haruspex]

both axles and stabilizer legs are sprung with the same stiffness.

That seems unlikely. There are reasons for springs on the axles, but not on the stabilizers.
I would take the legs as rigid but the two axles with the same spring constant. The one constant can represent both the springs and the elasticity of the tyres.
Next, you need variables for the small amounts by which the vehicle will rise at the axles when the load is applied. There is a geometric relationship between them. Then relate them also to the new forces on them.

 

[Ashferico]

Edit: The diagram can be used to determine the front/rear load ratio before the payload is lifted. It is a very simple one.

Thank you! What more information on my diagram do I need to calculate the reaction forces when payload is lifted?

That seems unlikely. There are reasons for springs on the axles, but not on the stabilizers.
I would take the legs as rigid but the two axles with the same spring constant. The one constant can represent both the springs and the elasticity of the tyres.
Next, you need variables for the small amounts by which the vehicle will rise at the axles when the load is applied. There is a geometric relationship between them. Then relate them also to the new forces on them.

Thank you! So I think I have applied what you mentioned. Is this what you had in mind?

 

[haruspex]

Thank you! What more information on my diagram do I need to calculate the reaction forces when payload is lifted?
Thank you! So I think I have applied what you mentioned. Is this what you had in mind?
View attachment 259850

Your first equation there is wrong. Think through the similar triangles again.

I always strongly advise working purely algebraically, only plugging in numbers at the end. Create variables for the distances.
Had you done that, I think you would have spotted the mistake.

 

[Ashferico]

Your first equation there is wrong. Think through the similar triangles again.

I always strongly advise working purely algebraically, only plugging in numbers at the end. Create variables for the distances.
Had you done that, I think you would have spotted the mistake.

Yes, you're right. I've corrected it now.

Thank you for your help!

 

[Ashferico]

Hello again,

I have to calculate the required force of the linear actuator that drives the stabilizer leg. The 79.3kN I calculated before, if that was spread over 2 stabilizers, it would approximately 40kN per leg. If it was angled at 60 degrees like on the diagram, does that mean linear actuator force is: 40/sin(60) = 46kN

Have I got this wrong?

Kind regards,

Ashley