Franglais:
shep532:
As shown in the OP picture, that load will easily slide in any direction. This two straps are only providing a downward force of about 700daN (700 kg) to try and stop 4000kg sliding.
That ignores the weight of the load itself doesnt it? Thatll be 4,000kg for a 4,000kg mass.
engineeringtoolbox.com/frict … d_778.html
This shows a co-efficient of friction of 0.25 to 0.5 for wood against wood. Assuming the lower figure youd need a down force (normal reaction) of 16,000kg to stop movement due to friction alone during a 1g stop. 4,000kg from weight leaves 12,000kg down force needed from the straps. What angle do they pass over the load? 45degrees?? Agreed, straps should be attached to the chassis, not just the rave. Chains are better of course, as others have said, but probably not allowed due to load damage risks. My suggestion of using straps to pull load backwards goes some way to addressing this Id say.
Feel free to question and correct my workings and assumptions. Wont be the first time Ive got it all wrong.
For dry wood on wood a coefficient of friction of 0.3 would be about right. Yes it can be higher or lower but 0.3 is a suitable starting point.
The BS EN 12195-1:2010 standard tells us an acceleration force of 0.8 in a Forward direction.
0.8 of payload (single coil) is 3200kg
The friction of 0.3 means the friction will stop 0.3 of payload or 1200kg from sliding
That leaves 2000kg for the straps to stop sliding by applying a downward force
If the strap has an STF of 350daN, using a ‘K’ factor of 1.8 as suggested in the German VDI2700 Standard that’d mean a total pretension of 630daN
with a strap angle of 45 degrees only 0.71 of the total pretension would be applied to the load = 447.3daN
Multiply the coefficient of friction by our total effective pretension - 447.3 x 0.3 = 134.19 daN
Divide our 2000kg by 134.19 = 14.90.
Multiply by a safety factor of 1.25 = 18.63 or 19 straps.
The above is for overstrapping only.
Of course we could BLOCK the load in a forward direction using timber and some 100mm nails. According to the European Code of Practice a single 100mm galvanised nail will stop 0.32t sliding forwards at 0.3 coefficient of friction so that’d need 12 nails through a 50mm timber and with 50mm into the vehicle deck.
Now our timber and nails have stopped it sliding forward we’d still need 8 overstraps to stop the load sliding sideways.
If using spring lashings it’d need two at LC2500daN to the front and one to the rear. These would only stop forwards and backwards so we’d still need 8 overstraps to stop sideways or five 100mm nails and a length of timber and just 2 overstraps to prevent tipping.
No matter which way you look at it - no matter what recognised calculations or standards are used, the picture in the OP shows a load nowhere near secured to withstand potential forces that may act on the load under braking or swerving/cornering.
By putting them through the centre then around each end to each side pulling in opposite directions in a sort of X looking from above. Which would obviously also have the effect of helping to stop side ways movement too and at least a lot better than what’s shown there.
