Rigging math

Rigging Math

Calculating our WLL from catalog WLL

Basically

WLL=UL(d)

Basically

WLL=UL(d)

Working Load Limit

Ultimate Load

Design Factor

But More Like

WLL=UL(d)(e)

Working Load Limit

Ultimate Load

Design Factor

Efficiency

UL(Force Ratio)

And Actually

WLL=UL(d)(e)

My Working Load Limit

Ultimate Load

My Design Factor

Efficiency

UL(Force Ratio)

Catalog WLL/(d)

Vendor Design Factor

Or

WLL=(Catalog WLL/Vendor Design Factor)(Force Ratio)(Design Factor)(Efficiency)

…But I can’t remember ever using precisely that formula.

Terms

• WLL– Working Load Limit– Safe Working Load– Maximum Allowed Force

– A derived strength value based on the Ultimate Load and the Design Factor

Terms

• UL– Ultimate Load– Failure Load– Minimum Failure Load– Breaking Strength– Minimum Breaking Strength

– The applied force that causes failure

Terms

• Design Factor– Safety Factor… “saferty factor”

• A mathematical value applied to provide appropriate capacity for unknown influences

• Usually expressed as a ratio, i.e. “10:1”

Terms

• Efficiency

• The nature of the application may require the load capacity to be reduced

• Some things that change efficiency– D/d Ratio– Type of termination

Terms

• Force Ratio

• The nature of the application may require the applied load to be increased

• Something causing a force ratio– Accelerations• 16 FPM Chain hoists add 25% on stop and start

Work In Steps

• Remember– Is this really the load?– Do I get full capacity?– Am I comparing apples & apples• Apples = Breaking Strength• Oranges = Vendor Working Load Limit• Peaches = My Working Load Limit

– Do Not Exceed Vendor Working Load Limit

Work In Steps

• To derive your WLL for a given component– Multiply UL by your Design Factor• Include efficiency calculation if necessary

• To find UL– Look it up in the vendor information– Multiply vendor WLL by their Design Factor

Example

• If the catalog WLL of a given ¼” screw pin anchor shackle is ½ Ton, what is the maximum applied force I can apply using a 10:1 Design Factor?

Example

• ¼” Shackle has a WLL of 1/2t or 1000#• If you follow the asterisk you see this:

• “Maximum Ultimate Strength is 6 times the Working Load”

Example

• Vendor WLL=1000#• Vendor Design Factor = 6:1• Ultimate load = 6000#

• My Design Factor = 10:1• My WLL= 6000#/10• My WLL= 600#

Strip their design factor to get to

breaking strength and then apply our more conservative design factor to get

to our WLL.

Example

• What is the maximum load I can apply to a sling of ¼” 6x19 IWRC Cable terminated at each end with wire rope clips used in a “basket” over a 1 ½ Schedule 40 pipe.

Example

• Find the cable breaking strength

• cable breaking strength is 2.94t or 5880#• Since breaking strength is given there is no

additional calculation

Example

• Find the termination efficiency

• termination efficiency is “80% for 1/8”-7/8” sizes”

Example• Find the efficiency of the cable around the

pipe• Pipe diameter

is 1.9”• Cable diameter

is 0.25”• D/d ratio is

7.6:1

• Efficiency is roughly 82%

Example

• Find the capacity change due to sling configuration

• Basket is SWL x 2

Example

• Cable Breaking Strength = 5880#• Termination Efficiency = 80%• Sling breaking strength

• 5880#(0.8)= 4704#

• Strength lost due to pipe diameter = 82%• 4707#(0.82)= 3857#

• Strength gained from configuration = 200%• 3857#(2)=7714#

• Apply our Design Factor of 10:1• 7714#(0.1)= 771#

Rigging Math

Calculating our WLL from catalog WLL