Pulley friction on overhead crane 1

[Marinel]

I need to adjust the safety trip at 110% of the nominal load ( 130 tons ).
This means that I need to load the overhead crane with 143 Tons. With this load on the floor I start to rise the load with the crane, but I measure less than those 143 Tons ( say 135 Tons). This means that we can never achive the value os 143 Tons unless we stop the rising movement.
I want to know what to do.
In the atached file you can see the overall arrangement of the pulleys in the overhead crane and also values achived on the real test ( separate sheet). The load has 130 tons.
In order to check the maths involved I developed several equations to show the different forces involved and the efficiency of the system ( note that f is the friction force in each pulley).

 

[chicopee]

On the average allow 2% of the rope load over that sheave for ball and roller bearings with rope making a 180 degree turn.
Actual losses can be as high as 4-1/2% to as low as 0.9% depending on the type of bushing.
General equation for force at the drum to raise/lower the load is P= (weight of load + weight of block,hooks etc.)/(R) where
R= Sigma(1-u)^+or-(m+n);+to raise load' - to lower load; m= no of 180 degrees turns between drum and block;n=number of parts; u= coefficient of friction and assume 0.02. Sigma is the greek symbol for summation. Equation is from the courtesy of Shapiro's Crane and Derricks.

 

[Marinel]

Thanks.