Excel to solve rigging problem

[Ralph2]

I would like to make a simple spread sheet to solve frequent rigging problems. But.. school was a long time ago.........

As a typical example
Load weight is 20 Tonnes (44,000 pounds)
Two slings attached to load form angles of 34 and 57 degrees

I want to solve for tension (pull) in each leg of the slings
I want to solve for the component (vector) forces (horizontal and vertical)

Can anyone help me out..

Thanks for your time
Ralph

 

[electricpete]

I assume your angle represents and angle from vertical.
Call you angles alpha and beta.
Call the associated angles T1 and T2.
Call the weight W (44,000 lbf)

Draw free body diagram.

From the vertical forces, we get equation 1:
T1*cos(alpha) + T2*cos(beta) = W

From the horizontal forces, we get equation 2:
T1*sin(alpha) + T2*sin(beta) = 0

Two equations is 2 unknowns. You can solve them by hand. Or else use excel to solve them using one of the techniques on attached spreadsheet (discussed a little bit on the thread on simutaneous equations)


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[electricpete]

Correction: equation 2 should be:
T1*sin(alpha) - T2*sin(beta) = 0

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[Ralph2]

Thanks Electripete..
Sorry but your advice is "really" confusing to me..
Will try to include a PDF sketch of my rigging arrangement
Thanks

 

[Ralph2]

will try again to upload and include my sketch

 

[electricpete]

Your angles are angles from the horizontal. Mine were angles from the vertical. So put the following into the above equations
alpha = 90 - 34
beta = 90 - 57

Also, I assume there are two individual slings are individually attached to the "hook" (or whatever the point of attachment at the top), rather than a single sling allowed to slide through the hook.

Under those assumptions, result should be as I have given above.

Can you try drawing a free body diagrams showin all the forces acting on the lifted piece? As condition of equilibrium, the total horizontal and vertical forces sum to 0.

=========================
Let me start again using different symbols in terms of angles from horizontal (perhaps that makes it easier?).
Theta1 = 34 deg = angle sling1 from horizontal
T1 = tension ion sling1
Theta2 = 57 deg = angle sling2 from horizontal
T2 = tension ion sling2

Vertical forces:
Upward force from sling 1: T1*sin(theta1) from sling1
Upward force from sling2: T2*sin(theta2) from sling2
Downward force: Weight
Total Vertical force = 0 = Upward - Downward
T1*sin(theta1) + T2*sin(theta2) - W = 0 [EQUATION 1]

Horizontal forces:
Righttward force from sling1: T1*cos(theta1) from sling1
Leftward force from sling2: T2*cos(theta2) from sling2
Total Horizontal force = 0 = Rightward - Leftward
T1*cos(theta1) - T2*cos(theta2) = 0 [EQUATION 2]

We have two equations (EQ1 and EQ2) in 2 unknowns (T1 and T2).

Is the remaining question concerned with how to derive these two equations or how to solve them?

Note - Please double-check my results for yourself to make sure I have not made an error.

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[Ralph2]

Thanks again Electricpete..

To solve them....

I was kind of hoping for something as simple as
Cell A1 input total weight
Cell A2 input angle of sling 1 (from horizontal)
Cell A3 input angle of sling 2 (from horizontal)

Cell A5 calculation showing tension in sling 1
Cell A6 calculation showing tension in sling 2

Now, having determined the tension (pull /weight)in each leg I can use the same process (sum of vectors ?)to determine the horizontal and vertical forces.

Cell A8 value of Cell A5
Cell A9 calculation of vertical angle cell 90-cell A2
Cell A10 value of Cell A2 (horizontal angle)

Cell A12 sling 1 vertical force
Cell A13 sling 2 horizontal force

Repeat idea for sling 2

Cell A20 calculation.. Ratio of horizontal forces should represent the center of gravity. If horizontal distance between lift points is known one can should be able to calculate the point at which the crane should be..

Will have to play with this some to prove it out though.. might have to consider the vertical input as well.. But this becomes a refinement after the initial formulas have been proven to work.

Am I making some assumptions that will not work.. i.e. need more known values. I was under the impression that one could split a force into component forces if... one knew the angle of that force. It's just.. that I forget how to do it

 

[electricpete]

Your approach might work - I haven't looked at it closely.

There was an assumption built into my solution that this thing is in static equilibrium with the piece level and lifted. After all, most lifts are arranged to keep the piece level. From the drawing, it looks like the piece would tilt CCW. In that case, everything shifts relative to vertical (and therefore relative to the weight vector). Equation 1 needs to be modified by multiplying W by cos(Phi) where Phi is tilt angle from vertical.

An additional equation is required to solve for that new variable Phi. That new equation would come from setting the moments about the center of gravity equal to 0.

At least that's the way I look at it. That's all I've got time for now. Maybe someone else will jump in

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[Ralph2]

Thanks electicpete...

The load is in equilibrium and the sling angles are those measured while hanging. It has a much heavier mass on one side and the lifting eyes are not on a horizontal plane. My sketch shows only half of the rigging and connects to an I beam. There is an identical sling set 4 feet in from the first. That beam is then raised by 2 hydraulic posts. The total load is estimated to be 40 Tonnes.

However the configuration is just a sample and I would like to make a spreadsheet that would easily resolve future lifts/ issues.

To me the problem is fundamentally the math behind a common tug-of war type of problem..
20 kids from grade 8 have challenged kids from grade 5 and 6. Each grade has its own rope tied at a common spot. Grade 5 can only pull at 56 degrees, grade 6 can only pull at 33 degrees.(from the pull of the grade 8 line)
How many kids does grade 5 need and how many kids does grade 6 need to exactly equal the pull of the grade 8's. (Each kid has the same pulling power)

 

[IRstuff]

There are limits to solutions to rigging general problems. In many cases, you wind up with statically indeterminate conditions.

TTFN

FAQ731-376

 

[yakpol]

Ralph,
After some math conversions formulas for tension in the slings are:

T57=N/cos(57)/(tn(57)+tn(34))
T34=T57*cos(57)/cos(34)

may want to verify static equilibrium afterwards.

cheers

 

[Ralph2]

Thanks yakpol

I appreciate your help and will try to make your formulas work.. I will let you know how I make out.. in a day or so as I am swamped chores at the moment

 

[electricpete]

Attached is a solution of the two equations presented in my post 7 Sep 08 16:06 (assumes everything remains horizontal):
T1 = W*cos(theta2)/sin(theta1+theta2)
T2 = W*cos(theta1)/sin(theta1+theta2)

I hope it is the same as Yakpol's. Anyone wants to compare them, please do and let me know.


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[Ralph2]

I have had.. finally, some time to continue this problem and... it does not work for me..

electicpete solution gives values that are not realistic and yakpol formula need the value of one to solve the other.. as I understand it.
Included is my mini spread sheet rigging.xls

Any further help appreciated
Ralph

 

[electricpete]

Let's see. Test case for my solution. Slings are vertical, theta1=theta2=90. Denominator goes to 0. You're right... doesn't sound right. I'll take another look at what I did.

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[Ralph2]

I played with my rigging problem some more and took a different tact.. I would appreciate anyone looking at this to tell me my logic is right / wrong?? seems to work though...

To recap.. I know the weight of a load and when lifted I can measure the angles of each of the two slings to the horizontal.
The question was... what is the tension in each leg and what is the horizontal pull of each leg.

My reasoning went along these lines.. the vertical pull at the end of each sling is 1/2 the total weight. Knowing this and the angle I can now use c=a/2/sinA to determine the tension and b=a2*cotA to determine the horizontal pull. As my Excel does not seem to have a "cot" function I ended up using b=a/2*1/tanA

electricpete... your solution may have worked had I not erred in using your formula... seems that one needs to convert degrees to radians when using the trig functions..

Included is my spreadsheet rigging1.xls if anyone cares to check its validity

Thanks all

 

[MikeHalloran]

Uh, doesn't Excel like to see angles in RADIANS?



Mike Halloran
Pembroke Pines, FL, USA

 

[Ralph2]

yes... that was one of my errors in applying the formula.. sorry