Calculating the maximum load a lorry can lift

[sharris]

Please help with this query to find how much load a dumper truck can lift without it becoming unstable and toppling.

Homework Statement

A lorry has a volume of 42 cubic metres and can hold 78,000kg sand. The overall length of the vehicle is 9.5m with a pivot 1.2m from an end.

The bed of the lorry is raised to deposit the sand.

What is the maximum load that can be lifted at the opposite end of the beam to deposit the sand?

Homework Equations

The Attempt at a Solution

The line of action/ force acting through the centre of mass = 78000kg x 10 = 780KN (approx) which acts 4.75m along the beam.

There is a clockwise moment as the centre of mass acts to the left of the pivot.

The moment between the pivot and line of action is = 780KN x 3.55m = 2769 KNm

I'm not sure how to proceed to calculate the maximum loadto prevent the lorry tipping on it's side.

 

[Delphi51]

Welcome to PF.
The question "how much load a dumper truck can lift without it becoming unstable and toppling" doesn't make sense. I can't think of any way it could topple. The center of mass of the load will always be between the front and back wheels unless the wheels are very badly placed (and no information about their placement is given).
Also no reason why it would tip to the side in the unlikely event that the load lifts sideways.

What is the maximum load that can be lifted at the opposite end of the beam to deposit the sand?

does make some sense. You could calculate the lift force that must be provided to lift that maximum load. It will be less than the weight of the load due to the leverage.

 

[sharris]

Thank you for your response Delphi51.

This is a real application that I have not perhaps explained fully. The ram lifting a tipper that is loaded fails causing the ram to fall back to the chassis, along with the tipper bed. The loaded tipper then bounces off the chassis onto the floor.

Assuming that the vehicle is well maintained and that the ground on which the vehicle is operating is even, I would like to calculate how much can be loaded into the tipper so that it can be safely operated, without the above event happening. (Unfortunately, it has not been a singular event).

Any guidance in how to calculate this would be much appreciated.

 

[Delphi51]

Okay, kind of what I was thinking. Carry on with your torque calc.
Knowing the maximum force F the ram can handle, you can calculate the weight of the load that can just be lifted.
Torque of ram = torque of load
F*a = W*b
where a and b are the distances of the force and the weight from the pivot.

 

[sharris]

Thank you again Delphi51.

So, if the ram limit is 45 tonnes, tipper length is 9.6m

The value for F would be 45000kg x 10 = 450KN
W is the load to be found
a = distance from pivot to load = 4.8 +3.6m = 8.4m
b = distance from pivot to centre of mass = 4.8m - 1.2m =3.6m

Substituting

W = 450 x 8.4/ 3.6 = 1050 KN = 105 000kg

Eek, this is more than the weight of the load. Now I am confused. What have I done wrong?

 

[Delphi51]

F*a = W*b
F*8.4 = 780000*3.6
F = 780000*3.6/8.4
F = 334 286 N
Remember that is Newtons, not Kilograms.
I don't see how you got the 8.4 from the given information. It seems to me the hydraulic lift would be near the end of the load, which is 9.5 - 1.2 = 8.3. I guess 8.4 makes sense.

 

[sharris]

When I looked at the calculation again after your input and substituted values I realized that the actual length of the lorry was 9.6m. The distance from the pivot to the ram was calculated as 9.6m -1.2m to get 8.4m.

Thank you Delphi51 for your time, and patience. Much appreciated.