Ram weight – Drop height

500 kN´s påle

The capacity of the pile is both a geotechnical and a

structural issue

Geotechnical CapacityThe geotechnical capacity of a driven pile

is the ability of the surrounding soil and/or the rock to withstand loads without harmful movements.

Geotechnical capacityThe rule of thumb in Sweden for driven piles

to set less than 10mm/10bl

Concrete piles: 1 kN per cm2

Steel piles: 13 kN per cm2

Geotechnical capacityOften much higher capacities is used, for

concrete piles up to 1,6kN/cm2 and steel piles 22kN/cm2. This require that the pile is driven to solid rock in order to verify capacity.

To verify capacity 5% to 25% of the piles are normally tested.

Ram weightOne condition in order to install the pile to

the desired capacity is to have sufficient ram weight. A rule of thumb in Sweden is that for micro piles it is recommended that the piston in hydraulic hammers has a weight at least 2 times the pile weight per meter.

Ram weight – Drop heightIn order to mobilize available capacity the

pile final set for the test blow has to be a few millimeters.

This require both drop height and ram weight.

Rule of thumb:

Drop height 7-8 % of the length of the pile

Ram weight 1-1,5 % of desired capacity

Ram weight- Drop heightThe relationship between impact velocity and

drop height

Kinetic Energy = Potential Energy

mv2/2 = mgH

H = v2/2g or

v = √2gH

To drop one kg on the toe from one meter is equally painful as dropping 10kg from the same height. Only the the 10kg hurts for a longer time ;-)

Drop heightExample 1: Calculation of Z (EA/c) steel pipe

pile 140mm x 8mm

A (Area): 33 cm2 (0,0033 m2)

E (Elasticitetsmodul): 210000 MPa

c (wavespeed): 5120 m/s

Z=210000*106*0,0033/5120=135000 Ns/m=

135 kNs/m

Ram weight-Drop heightIf a pile is struck by a impact velocity of 3

m/s, the force in the pile will be:

F = v*Z = 3*135 = 405 kN

To achieve this partical velocity (force) the drop height has to be:

H = v2/2g = 32/2*10 = 0,45 m

Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher

Ram weight –Drop heightIf the pile is struck by an impact velocity of 6

m/s the force in the pile will be:

F = 6*135 = 810 kN

To achieve this partical velocity (force) the drop height has to be:

H = v2/2g = 62/2*10 = 1,8 m

Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher

Ram weight-Drop height

Piles driven to solid bed rock the downward traveling compression wave is superimposed which results in higher stresses than that induced by the hammer (specially on short piles)

Ram weight – Drop heightThe stresses in micro piles are commonly

very close to the yeild stress for the test blow. Important that the pile is cut correct prior to test and that the hammer is lined up correctly.

Ram weight – Drop heightExample 2: Calculation of Z (EA/c) for a

concrete pile (side=235mm)

A (Area): 552 cm2 (0,0552 m2)

E (Elastic modulus): 40000 MPa

c (Wave speed): 3900 m/s

Z=40000*106*0,0552/3900=566000 Ns/m=

566 kNs/m

Ram weight – Drop heightIf the pile is struck by an impact velocity of 3

m/s the force in the pile will be:

F=3*566=1698 kN

To achieve this partical velocity (force) the drop height has to be:

H=v2/2g = 32/2*10 = 0,45 m

Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher

Ram weight – Drop heightIf the pile is struck with a drop height of 1,2m

the partical velocity will be:

V= √2gH = √2*10*1,2 = 4,9 m/s

The force in the pile will than be (energy losses are neglected):

F=4,9*566=2773 kN, which is equal to a stress of 2,773/0,0552 = 50 MPa.

The pile will only withstand a few of these blows!

Ram weight – Drop heightWarning!

Modern machinery are often equiped with accelerating hammers. These hammers can quite easily over stress the pile.