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]

As an engineer I learn to break down by sections
Just my way of sanity.

 

[Gyco]

As an engineer I learn to break down by sections
Just my way of sanity.

Thank you very much for your support but I wanted to ask you if my setting is correct or not.

 

[rb1957]

why "1/2 F" ?

your problem is shown in post #12. If AD is at 32 deg, but the load ad D is at 45 deg, then F is going to be to the right ...
picture it ... the "crane" AD is at 32 deg, the load at 45 deg will collapse the "crane" unless some force acts on the crane to stop it ...
if the crane were at 60 deg, then F would be to the left.

You've guessed F to left, and your sketch is misleading (having the "crane" at 45 deg) ... easier to draw that way, and you probably didn't have the dim'ns then either.

 

[3DDave]

What is the angle ABC and what is the angle between the load at the pulley and AD?

With those two angles one can use the Cross product formula to directly calculate the force in BC.

 

[goutam_freelance]

The loads on hinges B and C are missing

It is easy, and I thought you could do it. Refer below:

 

[goutam_freelance]

I thought that by exploding the structure, the force F is divided into F/2 in the two parts.

You can not arbitrarily divide the force into 2 parts. The laws of equilibrium and vector addition are to be maintained.

Also, you need to reverse the force when transmitting from one to another connected member.

 

[3DDave]

Notice that if theta2 become zero there is no force required to hold the lever in position. If theta 1 becomes zero then an infinite force is required to hold the lever in position.

 

[Gyco]

Notice that if theta2 become zero there is no force required to hold the lever in position. If theta 1 becomes zero then an infinite force is required to hold the lever in position.
View attachment 7365

Do the loads on hinges B and C not exert constraint reactions?

 

[3DDave]

F exerts a load on B. The load on C doesn't much matter except being equal and opposite to that on B.

The load on A will be in line with the lever and be the sum of the dot products of F and FsubD with the lever.

So FsubA = F *cos(theta1) + FsubD* cos(theta2)

From there you can split any of the forces according to any coordinate system orientation you like.

 

[goutam_freelance]

Do the loads on hinges B and C not exert constraint reactions?

Constraint reaction will be equal in magnitude and opposite in direction to the vector sum of the connected member reaction forces. This is from the law of equilibrium of forces.

 

[Gyco]

F exerts a load on B. The load on C doesn't much matter except being equal and opposite to that on B.

The load on A will be in line with the lever and be the sum of the dot products of F and FsubD with the lever.

So FsubA = F *cos(theta1) + FsubD* cos(theta2)

From there you can split any of the forces according to any coordinate system orientation you like.

In conclusion, what are the reactions in A,B,C and D and the Force F worth?

 

[mfgenggear]

In conclusion, what are the reactions in A,B,C and D and the Force F worth?

Hi Gyco
Go back to the video I posted and watch other videos. Generally professors is what I like to refresh with. Did I mention I am a retired gear guy
I generally stick with AGMA specs. That as it may be it is important under stand the concept.
Then what Dave and Goutam posted will make since. Fy and Fx = 0. What are the loads with the given angles and dimensions = equilibrium.
Net 0. Thus static loads.

 

[rb1957]

I think Dave has solved it (I was wrong about the direction of F, it is to the left). Force F is found simply by moments about A ... F*400 = (sqrt(2)*Wg)*d ... which you see easier with Dave's better drawing. d is 3000*sin(13) (there are other ways to get there by constructing the triangle) and 400 is 750*sin(32).

"In conclusion, what are the reactions in A,B,C and D and the Force F worth?" ... since we've solved the problem, you should be able to determine these.

 

[Gyco]

Hi Gyco
Go back to the video I posted and watch other videos. Generally professors is what I like to refresh with. Did I mention I am a retired gear guy
I generally stick with AGMA specs. That as it may be it is important under stand the concept.
Then what Dave and Goutam posted will make since. Fy and Fx = 0. What are the loads with the given angles and dimensions = equilibrium.
Net 0. Thus static loads.

thank you so much for your patience, i will try to do it again.

 

[mfgenggear]

thank you so much for your patience, i will try to do it again.

Also my first comment now I belive where in correct follow the other post. I am OG so
I can make mistakes. The little details are important. I will work on the diagram and see
If we can come up with the same answer.
I am terrible at math. I keep trying till I get right.
That's why I use to write my own programs
As not to get incorrect results.

 

[Gyco]

I think Dave has solved it (I was wrong about the direction of F, it is to the left). Force F is found simply by moments about A ... F*400 = (sqrt(2)*Wg)*d ... which you see easier with Dave's better drawing. d is 3000*sin(13) (there are other ways to get there by constructing the triangle) and 400 is 750*sin(32).

"In conclusion, what are the reactions in A,B,C and D and the Force F worth?" ... since we've solved the problem, you should be able to determine these.

Thank you very much for the solution.

 

[mfgenggear]

Notice that if theta2 become zero there is no force required to hold the lever in position. If theta 1 becomes zero then an infinite force is required to hold the lever in position.
View attachment 7365

Good job Dave, learn something every day

 

[Gyco]

Notice that if theta2 become zero there is no force required to hold the lever in position. If theta 1 becomes zero then an infinite force is required to hold the lever in position.
View attachment 7365

the only thing that doesn't make sense to me is the binding reaction in B since there is a hinge and since it is a regulation tie rod, shouldn't there be an unknown reaction in B?

 

[Gyco]

my doubt is that since the tie rod CB is a regulation one, only a horizontal force acts, therefore in B since it is a hinge there is the reaction Hb. Reasoning like this, however, I have 3 unknowns in 2 equations with the statics.

 

[rb1957]

sorry, but what ? are you interpreting B as a fixed point ? I think everyone else thinks CB is pinned onto AD; how could it be otherwise (unless by design error) ?

I guess, as a school exercise, if it isn't stated then you make that assumption as part of your solution.