Rigging Math

(Made Simple)

A Primer by

Delbert L. Hall, Ph.D.

ETCP Certified Rigger

ETCP Recognized Trainer

Lesson 8: Simple load on a beam

       In Lesson 6, you learned how to compute the load hanging on a single bridle leg at the end of a truss.  In reality, that seldom occurs.  In most cases the load is somewhere other than directly below the suspension point on the truss.  In this lesson, and the two that follow it, you will learn how to compute the vertical force on each end of a truss, when the loads are spread out in different locations along the truss.  This lesson starts with a single load placed somewhere along the truss. Here is a diagram of the problem:

You might have noticed that I use D1 and D2 in the illustration above instead of H1 and H2 as I have used in previous lessons. Why? Maybe it is not a great reason, but it is because this type of problem does not contain a V1 or V2. This is also true for the problems in Lessons 9 and 10. So, I have substituted “D” for “H” in these lessons. I have also switched the numbering of D1 and D2 also; so do not get confused by this. It does not really make any difference what you label these distances, as long as you understand the concept of which parts you multiply and divide by to solve the problem.

Our desire is to find the vertical force on the two supporting Legs (L1 and L2).  The equations for solving this problem are similar to the equations for finding the center of gravity.  They are:

L1 = (Load x D1) / Span Note: You can also think of it as L1 = Load x (D2/Span) if that is easier for you to remember

L2= (Load x D2) / Span or Note: You can also think of it as L2 = Load x (D1/Span) if that is easier for you to remember

L2 = Load – L1

The trick that I use to remember the method of solving this problem is to remember that to find the tension on one leg, you multiply the load by the distance of the load from the OPPOSITE leg, and then divide by the Span. We used the trick of “multiplying opposites” and then dividing in Lesson 4 on Tension on Bridle Legs, and it can help you with this problem too.

So, let’s work a problem.

Example: If the Span is 20 feet, D1 is 5 feet and D2 is 15 feet (note: D1 + D2 must equal Span), and the Load 1 is 200 lbs, what is the tension on L1 and L2?

L1 = (Load x D1) / Span      or

L1 = (200 x 5) / 20           or

L1 = 1000 / 20                   or

L1 = 50 lbs.      

L2= (Load x D2) / Span       or

L2= (200 x 15) / 20            or

L2= 3000/ 20                     or

L2 = 150 lbs.

L1 + L2 must always equal the Load, which is one way to check your work.  But, since that is true, once you find L1 you can compute L2 by using the simple equation L2 = Load – L1.   Practice these simple problems before going to the next lesson, which puts multiple loads on the truss.

Worksheet

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