mechanical advantage of a wedge please help

[CBogue]

If you can, please help
i have a spear looking device with a 20deg angle on one side and a 60 deg angle on the other i need to pull this past a set of keys to acheve a 3500daN total load i have attached a drawing to make up for my lack of good wording ..... lol i am looking for any help possable..

 

[CBogue]

so what im asking for is a formula breakdown and an answer to the load i need... thanks sorry for the lack if good question asking.... lol

 

[TheTick]

Just resolve the components of the normal forces at contact points..

 

[CBogue]

what would the formula be for that?
right now i have loaded the nub looking things with 7300daN each side and pulled the spear out at 2100daN. i cant seam to find any formula so its not trial and error. i haved bought over 150 disk springs and nothing has worked properly. now i need to get custom ones made so i need the actual formula and/or answere. thanks.

 

[desertfox]

Hi CBogue

Have a look at this site it might help:-

http://emweb.unl.edu/negahban/em223/note16/note16.htm

I cannot see where 7300daN or 3500daN 2100daN are on your diagram, on your diagram you have 4500daN and 22daN which I assume you are pulling with 4500daN and the 22daN is the vertical loading your trying to overcome please confirm,also where are the disk springs situated?

desertfox

 

[CBogue]

desertfox (Mechanical) 17 Jun 10 0:18
Hi CBogueHave a look at this site it might help:-

http://emweb.unl.edu/negahban/em223/note16/note16.htmI

cannot see where 7300daN or 3500daN 2100daN are on your diagram, on your diagram you have 4500daN and 22daN which I assume you are pulling with 4500daN and the 22daN is the vertical loading your trying to overcome please confirm,also where are the disk springs situated?desertfox

7300, 2100 daN were actual tests performed in the piece i will change the diagram to state 3500daN and where it says 22daN it should state ??daN lol the springs are placed right at the ??daN(22daN)pushing down untill it bittoms out right where it sits in the diagram if any body could help me break down a simple forumula that would be great... thanks

 

[desertfox]

Hi CBogue

I'll post later but we need the load exerted by the disc springs.

desertfox

 

[CBogue]

ok i might have it, i looked through old books and here is what i got please correct me if i am at all wrong.... thanks again

 

[CBogue]

apparently i dont know how make an attachment lol nor edit lol

 

[desertfox]

Hi CBogue

Looking at your posts again, I assume your trying to load the blocks on the 60 degree angled face so that you need a minimum horizontal force of 3000daN before they move up the slope and allow the spear to pass.
Now your calculations you posted, the coefficient of friction seems very high, normally for steel on steel its about 0.25.
Now to obtain the friction force at the interface then:-

3000daN/cos 30 degrees = 3464.1 daN

therefore the friction force on the slope is:-

3464.1 * .25 = 866.03 daN

you appear to have divided the coefficient of friction into the normal force instead of multiplying unless I mis-understand your post.
Checking a couple of things and I will post later.

desertfox

 

[desertfox]

Have a look at this example to:-

http://books.google.co.uk/books?id=...=onepage&q=resolving forces on wedges&f=false

desertfox

 

[hydtools]

In the calculations for the required nib or block load use only half the force to pull the spear. Each nib retaining the spear carries half the extracting force. The two nibs share the extracting force equally.

Neglecting friction for a moment, the required nib load would be 1500 * tan30 = 866daN. Including the effect of friction which will increase the resistance to extract the spear will decrease the required nib load, even to the point of requiring a force to pull the nib. Especially with the high assumed coefficient of friction. Like desertfox says, the coefficient should be .25 to .30.

It doesn't make sense that the required nib load is greater than the force to extract the spear because of the 60 deg angle. With that 60 deg angle the nib load should be less than the extraction force. Only with an angle less than 45 deg would I expect the nib load to exceed the extraction force. As the angle decreases the extraction force will overcome greater nib or block loading because of the wedge advantage. As the angle increases the advantage goes to the nib or block load to resist the extraction force. To the limit where a 90 deg face will require no nib or block load to resist the extraction force.



Ted

 

[CBogue]

sorry for the long wait for responce, i was away from a computer for some time.. i am trying to figure out what spring to put on top of the nub i have done some tests and came up short.. what i have ran was a 3210lbs spring (6 disk's in series) also an asortment of 581,277 and 360lbs each and i came up short every time. what i need is (3000daN) 6474.6lbs of pull out at the spear. with the 3210lbs spring i acheved 4700LBS pull out. let me know what you think.. thanks again

 

[desertfox]

Hi CBogue

Am I right in assuming that the 3210lbs is the combined load for two nibs?

desertfox

 

[hydtools]

My conversion has 3000daN = 6744lbs.
Your figures are indicating geometry different from what you show. Is the 60deg correct as shown or is the angle 60deg included? Does the angle on the nub match the angle on the spear? Your data indicates the angle to be more towards 36deg, not 60deg as shown.

From your data you need more spring force. 1.43 times more to get 4700 up to 6744lbs.

Ted

 

[hydtools]

CBoque,
The attachment is my free body diagram and general force solution for the nub/key.

Ted

 

[desertfox]

Hi CBogue

Not sure whether you are still looking at this post or not, however I have uploaded a graphical solution to your situation for your reference.
I am in agreement I believe with hydtools both in terms of his diagram and his earlier comments about the nib forces.
If you require further help then you need to respond to some of the comments/questions in the earlier posts.

Regards

desertfox