Laterally Loaded Pile – Brom’s Approach
Name : Putra Mahendra Kusuma Aji
Class : Civil Engineering KKI 2020
NPM : 2006517695
PROBLEM 1:
A 4.8m long 0.4m PC square pile (Class C) is to resist a lateral load in a free-head condition; the lateral load is to be
applied at ground surface level. The soil is sand layer with  = 35 and unsat = 17 kN/m3, and the groundwater table
was not observed during the soil investigation. Determine the ultimate lateral capacity HU.
Kp (passive lateral pressure coefficient) = (1+sin (’))/(1-sin(’)) = (1+(sin(35))/(1-(sin(35))= 3,69
L/d = Length/Diameter = 4.8/0.4 = 12
Hu = (75) ∗ (3.69) ∗ (0.4) ∗ (17) = 291.1 kN
HU / Kpd3’ (Ultimate lateral resitance)= 75
f = 0.82 (Hu / ’dKp)0.5 = 0.82 (291,07/(17*0.4*3.69))0.5 = 2.8 m
Mmax = Hu (e + 2/3  f) = 291,07 (e + 2/3* 2.8) = 542.0 kNm
Check whether Mmax less than Multimate : is Mmax < Multimate : 542.0 > 215.8
Yield Moment = Myield / Kp’d4 = Multimate / Kp’d4 = 215.8/(3.69*17*.4^3) = 134.4
Laterally Loaded Pile – Brom’s Approach
HU / Kpd3’ = 50 kN
Hu = (HU / Kpd3’)*Kp*^3 * Yunsat = 55 ∗ (3,69) ∗ (0,4)^3 ∗ (17) = 200,8 𝑘𝑁
Class C
Therefore, using Broom’s approach, we were able to find the ultimate lateral capacity would be Hu which is
200.8 kN.
Laterally Loaded Pile – Brom’s Approach
PROBLEM 2:
A 4.8m long 0.4m PC square pile (Class C) is to resist a lateral load in a free-head condition; the lateral load is to be
applied at ground surface level. The soil is sand layer with  = 35 and sat = 18 kN/m3, and the groundwater table was
observed at the ground surface. Determine the ultimate lateral capacity HU.
DIFFERENCE IN GROUND WATER TABLE
Kp (passive lateral pressure coefficient) = (1+sin (’))/(1-sin(’)) = (1+(sin(35))/(1-(sin(35))= 3,69
L/d = Length/Diameter = 4.8/0.4 = 12
Hu = (75) ∗ (3.69) ∗ (0.4) ∗ (18 − 9,81) = 140,23 𝑘𝑁
HU / Kpd3’ (Ultimate lateral resitance)= 75 kN
f = 0.82 (Hu / ’dKp)0.5 = 0.82 (291,07/(17*0.4*3.69))0.5 = 2.8 m
Mmax = Hu (e + 2/3  f) = 291,07 (e + 2/3* 2.8) = 542.0 kNm
Check whether Mmax less than Multimate : is Mmax < Multimate : 542 > 215,8
Yield Moment = Myield / Kp’d4 = Multimate / Kp’d4 = 215.8/(3.69*(18.9,81)*.4^3) = 278.9
Laterally Loaded Pile – Brom’s Approach
HU / Kpd3’ = 65 kN
Hu = (HU / Kpd3’)*Kp*^3 * Yunsat = 65 ∗ (3,69) ∗ (0,4)^3 ∗ (18-9,81) = 125.7 𝑘𝑁
Class C
Therefore, using Broom’s approach, we were able to find the ultimate lateral capacity would be Hu which is
125.7 kN, compared ot number 1’s 200,8 𝑘𝑁 (control).
The impact of grondwater on the pile lateral capacity causes the decrease of ultimate lateral capacity. This is
because as the water increases, the depth of the undrained soil decreases in height.
Compare and discuss the effect of groundwater on pile lateral capacity HU as suggested by Problem 1 and Problem 2.
Laterally Loaded Pile – Brom’s Approach
PROBLEM 3:
A 4.8m long 0.4m PC square pile (Class C) is to resist a lateral load in a free-head condition; the lateral load is to be
applied 1.0 m above ground surface level. The soil is sand layer with  = 35 and unsat = 17 kN/m3, and the
groundwater table was not observed during the soil investigation. Determine the ultimate lateral capacity HU.
DIFFERENCE IN ECCENTRICITY
Kp (passive lateral pressure coefficient) = (1+sin (’))/(1-sin(’)) = (1+(sin(35))/(1-(sin(35))= 3,69
L/d = Length/Diameter = 4.8/0.4 = 12
Hu = 72,5 ∗ (3,69) ∗ (0,4)^3∗ (18 − 9,81) = 140,23 𝑘𝑁
HU / Kpd3’ (Ultimate lateral resitance)= 75 kN
f = 0.82 (Hu / ’dKp)0.5 = 0.82 (291,07/(17*0.4*3.69))0.5 = 2.8 m
Mmax = Hu (e + 2/3  f) = 291,07 (e + 2/3* 2.8) = 542.0 kNm
Check whether Mmax less than Multimate : is Mmax < Multimate : 542 > 215,8
Yield Moment = Myield / Kp’d4 = Multimate / Kp’d4 = 215.8/(3.69*17*.4^3) = 134.4
Laterally Loaded Pile – Brom’s Approach
HU / Kpd3’ = 40 kN
Hu = (HU / Kpd3’)*Kp*^3 * Yunsat = 40 ∗ (3,69) ∗ (0,4)^3 ∗ (17) = 160.588 kN
Class C
Therefore, using Broom’s approach, we were able to find the ultimate lateral capacity would be Hu which is
160.6 kN, compared to number 1’s controlled variable at 200,8 𝑘𝑁.
The impact of eccentricity causes the ultimate lateral resistance to be lower in value, causing the Hu to be lower
than expected, this is because the excentricity of the line would skew towards the free headed. Therefore,
increasing the eccentricity, according the Broom’s approach, would reduce the amount of ultimate lateral
capacity of the pile.
Compare and discuss the effect of load eccentricity above ground surface on ultimate lateral capacity HU as suggested
by Problem 1 and Problem 3.
Laterally Loaded Pile – Brom’s Approach
PROBLEM 4:
A 15m long, 0.4m PC square pile (Class A2) is to resist a lateral load; the pile top is in a fixed-head condition. The
soil is clay with cU = 50 kPa. Determine the ultimate lateral capacity HU.
DIFFERENT IN CLASS
Data
cU = 50 kPa
l = 15 m
d = 0.4 m
class A2
MU / cud3 = (8.25*8.80665)/(50*(0.4)^3) = 41.37
HU / cud2 (ultimate lateral resistance) = 30
HU = (HU / cud2) * (cud2) = (30)*(50*(0.4)^2) = 240 kN
Laterally Loaded Pile – Brom’s Approach
PROBLEM 5:
A 15m long, 0.4m PC square pile (Class B) is to resist a lateral load; the pile top is in a fixed-head condition. The
soil is clay with cU = 50 kPa. Determine the ultimate lateral capacity HU.
DIFFERENT IN CLASS
Data
cU = 50 kPa
l = 15 m
d = 0.4 m
Class B
MU / cud3 = (13.5*9.80665) / (50*(0.4)^3) = 25.28
HU / cud2 = 18
HU = 18 * 50 * 0.4^2 = 144 kN
Compare and discuss the pile ultimate strength on pile lateral capacity HU as suggested by Problem 4 and Problem 5.
The Hu problem 4 (Class A2) = 240 kN and problem 5 (Class B) = 144 kN, this would mean that the same
pile with the lower grade class would have lower ultimate strength capacity. For structures that needs to give
higher strength, capacity, it would need a higher grade of pile.