rigging math
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TRANSCRIPT
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