Rigging Angles Trouble !! 1

[jdkuhndog]

I have a Lifting Beam that has two vertical chains coming off of it and one horizontal chain coming off it as well.

I have a load which i will be lifting and then tilting up a little bit with the horizontal chain. Let me layout with words where all the connections are.


Lug 1 - this will be the origin. This will be where the vertical chains will be attached (there are actually 2 lugs, but we can just assume there is one lug with one chain for now).

Lug 2 - this is where the horizontal chain is attached. It is located 51" to the left of Lug 1 and 50" above Lug 1.

CG - Center of Gravity. this is located 1" to the left of Lug 1 and 50" below Lug 1.

T1 = Tension in chain coming from Lug 1 (its divided in two chains, but let's ignore that for now). Let's assume it is vertical even though there will be a slight angle to it.

T2 = Tension in chain coming from Lug 2. This force is directed up to the left at an angle from the horziontal of 15 degrees.

W = 100# at the CG.

This is what I did:

Sum Forces in Y direction
EQ1: T1+T2*sin(15) = 100#

Sum Moments about Lug 1
EQ2: T2*sin(15)*51"+T2*cos(15)*50" = (100#)*(1")

Solve Equations:

T2 = 1.6#, T1=99.6#

So, I was trying to determine the load in T2 and it comes out to 1.6#...but then I ask myself, what vertical force at Lug 2 would be required to counter the moment about Lug 1 from the Weight at the CG...so, I do this calculation:

T2y = vertical component at Lug 2

T2y*51" = (100#)*(1")

Solve for T2y:

T2y = 1.96#

So, if I need a vertical component at Lug 2 of 1.96# and my chain is at a 15 degree angle, then the tension in the chaing would be:

T2 = 1.96#/sin(15) = 7.6#

Why is this so much higher when I look at it that way?? What am I doing wrong ?!?!?

obviously, the weights and spacing and angles were used as an example...actual weight is around 25Ton - so this is causing big headaches for me !

Thanks for any help !

 

[unclesyd]

Take a look at this website. They reference several regulations, USA, that might cover you situation.
Usually when the calculations aren’t straightforward there is a problem that is not covered by jurisdictional authorities.

 

[jdkuhndog]

what website?

 

[unclesyd]

Sorry about that.

http://www.tandemloc.com/

 

[jdkuhndog]

unclesyd,

nice website...but not much help in my question.

anyone?

thanks,

jdkuhndog

 

[dvd]

I think you need to look at your equation

"Sum Moments about Lug 1
EQ2: T2*sin(15)*51"+T2*cos(15)*50" = (100#)*(1")"

taking counterclockwise moments positive, it should read:

Sum Moments about Lug 1
EQ2: T2*sin(15)*51"-T2*cos(15)*50" = (100#)*(1")

which gives a negative tension in your chain, and that means you need to replace the chain with a compression member.

 

[jdkuhndog]

dvd,

Lug 2 is to the left and above Lug 1 and T2 is directed up and to the right...the vertical component and the horizontal component of T2 both produce a cw moment about Lug 1. The CG is to the left of Lug 1 and W is directed down which produces a ccw moment about Lug 1...therefore the addition of the cw moments produced by the two T2 components should be equal to the ccw moment created by W.

jdkuhndog

 

[GregLocock]

Taking a step back, the function of T2 is to pull the load away from directly under lug 1

The moment of the weight about lug 1 is 1"*W

The moment arm of T2 about Lug 1 is roughly 12" (accurately 51*sin(15))

So I'd expect T2 to be about 1/12*W, or in other words your second approach was on the right lines.

I think your error is a sign error in EQ2

T2*sin(15)*51"-T2*cos(15)*50" = (100#)*(1")





Cheers

Greg Locock

 

[desertfox]

Hi jdkuhndog

Firstly in your last post that T2 is up to the right and in your original post you say its up to the left at 15 degrees.
However if it acts to the right then I agree with your original figure of 1.626 Tons.
I think your error in your second method is that you assume
that the moment from the CofG is balanced by a vertical moment at T2 which cannot be the case as there will be a
horizontal force at T2 as the chain is at 15 degrees to the horizontal.
If you take your 7.6 Tons chain tension and resolve themhorizontally and vertically I cannot get the sum of moment to equal zero.

ie:- 7.6 tons * cos 15 = 7.341 tons
7.6 tons * sin 15 = 1.967 tons


now assuming equilibrium = 7.341* 50" + 1.967 * 51"= 100*1"

= 467.367 = 100

I think you had it right the first time and I assume also that T2 acting to the left was a typo error in your question.


regards desertfox

 

[jdkuhndog]

oops ! There was a typo in the original post. T2 is directed back toward the lift beam which would be up and to the right.

dvd - sorry i didn't notice the typo previously, not i understand why you wrote the equation that way.

dersertfox - thanks, that deeper insight into the 2nd equation exposes it as erroneous !

I appreciate it !

jdkuhndog

 

[desertfox]

Hi jdkuhndog

Your welcome, glad I could be of help.

regards desertfox