Rigging Math

(Made Simple)

A Primer by

Delbert L. Hall, Ph.D.

ETCP Certified Rigger

ETCP Recognized Trainer

Lesson 10: Complex Load on a Beam

       In Lesson 9, you learned how to compute the tensions on L1 and L2 when there were multiple loads on the beam (truss).   However, in all of those problems the loads were between L1 and L2.   In this lesson you will learn how to compute the loads when one or more of the loads are cantilevered outside of the legs. Here is a diagram of the problem:

The equations for solving this problem are similar to the ones you learned in Lesson 9, except in how they deal with the load that is cantilevered outside of the leg that is opposite the leg that we are computing the tension on.   The equations for solving this problem are:

L1 = ((Load1 x D1) + (Load2 x D2) – (Load 3 x D3)) / Span

L2 = (Load1 + Load 2 + Load3) – L1

Before we begin, let’s note that, as in the previous lesson, the Span has no relevance to D1, D2, or D3.  But, as in the previous lesson, all of the distances (D1, D2, an D3) will be measured from L2.  I also want to explain the “– (Load 3 x D3)” part of this equation. 

Just like in Lesson 7, you need to think of this rig as a teeter-totter or a lever where the fulcrum is at the point where the Leg attaches to the truss.   When we are computing L1, the fulcrum is where L2 attaches to the truss.  If Loads 1 and 2 did not exist, the L1 end of the truss would pivot up (at the point where L2 attached to the truss) because Load 3 would not have anything counterbalancing it on the other side (and the L1 cable, chain, or round sling would not provide any resistance).  This would tell you that there is a positive force on L2, but a negative force on L1.  (Remember: downward forces are “positive” and upward forces are “negative.”) Because this is true, we need to include this negative force when we calculate the load on L1.  When all of the loads were between the two Legs, you only had positive forces, so all of the forces were “added.”  Since we now have a negative force, we need to “subtract” it.  That is why we have “– (Load 3 x D3) “ in this equation.  “– (Load 3 x D3)” is the negative force on L1.  Got it?

Example: If the Span is 40 feet; Load 1 is 200 lbs. and D1 is 30 feet; Load 2 is 150 lbs. and D2 is 10 feet; and Load 3 is 100 lbs. and D3 is 10 feet; what is the tension on L1 and L2?

L1 = ((Load1 x D1) + (Load2 x D2) – (Load 3 x D3)) / Span         or

L1 = ((200 x 30) + (150 x 10) – (100 x 10)) / 40                         or

L1 = (6000 + 1500 – 1000) / 40                                                 or

L1 = 6500 / 40                                                                         or

L1 = 162.5 lbs.   

L2 = (Load1 + Load 2 + Load3) – L1        or

L2= (200 + 150 + 100)  - 162.5              or

L2= 450 - 162.5                                      or

L2 = 287.5 lbs.

The most common mistake in solving these types of problems is adding instead of subtracting for the Load that is cantilevered.    With a little practice, you can master these equations.

Worksheet

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