Machine building exercise 1

[Gyco]

The figure (attached) shows the arm of a lift in the particular position in which while the adjustment tie rod CB is horizontal, it also happens that the section of the lifting cable DE is horizontal. With the measurement dimensions and with the mass M = 500 kg to be lifted, the loads on the hinges A, B, C and on the pin D of the pulley are calculated; also choose the profile and dimensions of the arm AD.

exercise 2 construction solution: Make a construction drawing of the connection between the arm and the pulley (D) and between the arm and the hydraulic piston (B)
My hypothetical solution is the following:
equilibrium condition of the point of material of mass M, or T = Mg directed upwards, applied to the center of gravity. Then considering the pulley for the 3rd principle of dynamics there is tension of the wire acting in the downward direction equal to T.
To derive the reactions of the hinges I thought of calculating the moment with respect to the intersection of the points A and C (bottom point) in order to eliminate 3 reactions out of 4 in the hinges A and C. What does not make sense to me is that I cannot calculate the distance between C and D for reactions in D. I hope that someone can give me a hand in the resolution.

 

[mfgenggear]

Well there is missing dimensions .
This project can be broken down with trigonometry. It needs two lengths or angle
And a length.
Can not draw this in cad with out those dimensions.
You can draw it. But with the missing dimension added cad would give fast results
Then it would be a breeze.

 

[3DDave]

In this configuration you don't need the horizontal distance between and D and can calculate the vertical distance between C and D.

Torque is the cross product of the force vector and the radial vector. From the weight being applied by gravity on the mass M and the balanced force from the drum E acting on the upper end of the arm, you can determine the torque that is being applied to the arm. Force F provides an opposing torque.

 

[Gyco]

This is an exam assignment on machine construction and the professor told me that F can be obtained immediately, by making some assumptions. But I don't understand how?

 

[rb1957]

"400" and "750" allow you to calculate the angle of AD. Which is not 45deg, which is a problem. In the initial position the cable supporting the weight is vertical and horizontal (given); since the tension is the same on both sides of the wheel D, then the slope should be 45 deg, particularly as A is clearly pinned. Then DE is not horizontal.
but AD is not a two force member, it is a three force member (look it up, to see what this means).
Since the load at D is at 45 deg, and we have this load F applied, we can figure our the reactions at A.
Doesn't matter ... this is only a drafting exercise.
The location of C is somewhere left of B, you also don't need a radius for D and E, assume something, equal would be a convenient guess.
But you are going to develop loads later, so review the strike out above ...

 

[3DDave]

Instead of 400 they could have used 375 and made everyone a bit happier.

 

[Gyco]

this is how i thought to set it up

 

[3DDave]

What's the vector sum of the forces on the pulley? Since the torque has to be zero, the vector sum acts through the pivot.

 

[Gyco]

Is the identification of the forces at play correct? Could you guide me to the resolution.

 

[mfgenggear]

this is how i thought to set it up
View attachment 7298View attachment 7299

What are the loads on C and D

 

[Gyco]

What are the loads on C and D

These are the constraint reactions of the hinges

 

[goutam_freelance]

Did this a long long time ago..

 

[Gyco]

Did this a long long time ago..
View attachment 7327

The loads on hinges B and C are missing

 

[mfgenggear]

These are the constraint reactions of the hinges

Loads on the crane

 

[Gyco]

Loads on the crane

so what would the free scheme with the forces be? In the text it says to calculate the loads on the hinges A,B,C and on the pin D

 

[mfgenggear]

Here are examples

 

[Gyco]

Here are examples

Unfortunately there is nothing similar online and I find it very difficult to solve this exercise.

 

[mfgenggear]

Examples 8 thru 11 depicts Mas at center mass of the crane x y and z, which is part off solution but not all

 

[Gyco]

Examples 8 thru 11 depicts Mas at center mass of the crane x y and z, which is part off solution but not all

In the above solution the CB section with the constraint reactions is missing and then I thought that by exploding the structure the force F is divided into F/2 in the two parts.

 

[Gyco]

I thought of solving it this way but I have 4 unknowns in 3 equations that cannot be solved.The problem is to find F which is unknown.