# Vectors and determining plate thickness

1. Sep 8, 2014

### harvez_007

Hi,

I have two 1.5 lifting lugs supported in the roof of a concrete tunnel 3.5m apart at a height of 7m.

The lugs require load testing and as per AS4991 require load testing to double the W.L.L therefore 3 tonne.

My work has a 5 tonne RUD lug readily available therefore was thinking of using that, however the plate to mount the lug on is required.

I have designed the plate for the 5 tonne - 5 x 1.5 (strength) x 1.5 (dynamic) x 9.81 = 111kN

The angle the 5 tonne lug will be at is 75.9 degrees.

Therefore the load in the direction of pull will be 111kN / cos75.9 = 455kN ??? I am not sure if this is correct.

M* = N* x l (I have assumed a plate 400mm x 400mm - therefore lever arm = 0.2m)
= 455kN x 0.2
= 91kN.m

øMs = øfy x [b(t^2)]/4

Let M* = øMs

t > √[(4xM*x10^6)/(øxfyxb)]

∴t > 82mm

Due to the large thickness I am getting I am not sure if the process I am using is correct. I do understand I am only load testing to 3 tonne however designing for 5 tonne - is this OK? I am happy to increase the size of the plate however could someone please ensure I am using the correct method for calculating this.

Any help is appreciated in advance.

Regards,
Harvez

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2. Sep 9, 2014

### SteamKing

Staff Emeritus
Designing lifting lugs is a little more complicated than scribbling a few formulas down on a piece of paper and plugging numbers into them.

The geometry of the lug must match the design of the shackles being used, in order to put the shackle pins in bearing stress only and not bending stress. You also have no overall sketch showing the load to be lifted, its weight and center of gravity, the location of the lugs, a calculation showing the expected forces in the shackles, etc.

My advice: Start over and proceed in a more systematic fashion.