PROF. DR.-ING. E. VEES UND PARTNER BAUGRUNDINSTITUT GMBH
PLATE LOAD TEST IN ROAD AND E ARTHWORKS CONSTRUCTION
ACCORDING TO DIN 18134 TESTING PROCEDURES, TESTING EQUIPMENT , THEORETICAL B ACKGROUN ACKGROUND D
Waldenbucher Straße 19 Postfach 20 03 39
[email protected]
70771 Leinfelden-Echterdingen, Germany 70752 Leinfelden-Echterdingen, Germany
Telefon 0049 Telefax 0049
(711) (711)
79 73 50- 0 79 73 50-20
Prof. Dr.-Ing. E. Vees und Partner
page 2
THE PLATE LOAD TEST IN ROAD AND E ARTHWORKS CONSTRUCTION
In Central Europe, road construction and pavement design are mainly based on the deformation modulus E VV determined by the PLATE LOAD TEST. The deformation modulus E VV can be understood as a modulus of elasticity. The more compressible a soil, the lower is the deformation modulus. THE PLATE LOAD TEST is described in DIN 18134 and with certain modifications in ASTM D 1195 and ASTM D 1196. The following description and evaluation of the test follows the German Standard DIN 18134.
EQUIPMENT AND TEST PROCEDURE The load is applied to a circular rigid steel bearing plate by a hydraulic jack in several steps. The settlement under each load step is recorded. The following sketch shows the principle of the test.
F F
= load
s D
= settlement = diameter of the plate
s
D
Fig. 1: Principle of plate plate load test The diameter D of the plate is generally 0.30 m. For very coarse grained material also plates with diameter D = 0.60 m and D = 0.762 m are used. The load is applied in 6 load increments of equal size. Under each load step the settlement must come to a noticeable end (< 0.02 mm/minute). After the maximum load is reached the unloading procedure procedure can begin. After that, the plate is reloaded in 5 steps. A loaded truck, an excavator or a roller usually serve as counterweight for the hydraulic jack. This is shown in the next figures.
Prof. Dr.-Ing. E. Vees und Partner
Fig. 2: Plate load test equipment. equipment. Excavator serves as counterweight. counterweight.
Fig. 3: Bearing plate (0.30 m diameter) diameter) with hydraulic jack ass assembly embly and beam with dial gage to determine plate settlement.
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Prof. Dr.-Ing. E. Vees und Partner
page 4
The DEFORMATION MODULI E VV are calculated from the first loading curve ( E VV11) and from the reloading curve ( E VV22) according to the following equation: equation: E v = 0.75 ⋅ D D ⋅ ∆σ / / ∆ s
E v ∆σ ∆ s
D
= deformation modulus = load increment = settlement increment = diameter of the plate, generally 0.30 m
For this calculation ∆σ and and ∆ s are usually taken from the load span between 0.3 σ max max and 0.7 σ max max.
The basis of the given equation is Boussinesq’s theory of the relationship between the modulus of elasticity and the settlement of a circular rigid plate with the diameter D. The derivation of the equation is shown in the appendix.
As an example the result of a plate load load test is given in the following table table:: Plate Diameter: 300 mm F
Load Pressure
[kN]
σ0
Settlement of the Plate
[kN/m2]
[1/100 mm]
5.65
80
115
11.31
160
209
17.67
250
287
23.33
330
325
29.69
420
380
35.34
500
421
17.67
250
395
8.84
125
360
0
0
259
5.65
80
311
11.31
160
353
17.67
250
378
23.33 29.69
330 420
398 413
FIRST LOADING
UNLOADING in steps to half of the preceding load
RELOADING to the last but one step
Prof. Dr.-Ing. E. Vees und Partner
page 5
The result of the test is plotted in a pressure-settlement diagram: diagram:
Pressure 0
100
σ0
in kN/m²
200
300
400
0 0.3 ⋅ σ0, max
0.7 ⋅ σ0, max
0.5
1.0 m m n i t n e m e l t t e S
1.5 1.95 2.0 f i r s t l o a d i n g
2.5
3.0
r e el l o o a d a i n n g
3.45 3.5
4.0 4.05
unl o oa d i i ng
4.5
EVALUATION OF THE GIVEN EXAMPLE: FIRST LOADING ( E E V V1 1) ∆σ
= 350 – 150 =
200 kN/m2
∆ s
= 3.50 3.50 – 1.95 =
1.55 mm = 0.00155 m
E V V1 1
= 0.75 0.30 200 200 / 0.00155 = 29032 kN/m2
=
2
∆σ
= 350 –150
200 kN/m2
∆ s
= 4.05 – 3.45 =
29.0 MN/m
RELOADING ( E E V V2 2)
E V V2 2
=
0.60 mm = 0.00060 m
=
0.75 0.30 200 / 0.00060
=
75000 kN/m2 =
2
75.0 MN/m
500
Prof. Dr.-Ing. E. Vees und Partner
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DEFORMATION RATIO: E V 2
=
E V 1
75.0 29.0
= 2.59
Generally in road construction the following values of E VV22 are required: subgrade:
E V V2 2 ≥
45 MN/m²
surface of sub-base layer:
E V V2 2 ≥ 120
MN/m²
For fine grained (cohesive) soils the deformation modulus E V2 which can be accomplished by compacting soils, depends on the index of consistency I C. Approximately the following relation has been found: Ev2
Ic
consistency
> 15
> 0.8
stiff
> 20
> 0.9
stiff
> 30
> 1.0
very stiff
> 45
> 1.2
very stiff to hard
2
[MN/m ]
Hard consistency of cohesive soils is encountered rarely. A deformation modulus of E V V2 2 ≥ 45
MN/m² as usually required for the subgrade under pavements can nearly always be obtained by soil stabilization.
REQUIREMENTS FOR THE DEFORMATION RATIO FOR COMPACTED SOILS E vv22 /E vv11
≤ 2.0 ≤ 2.2
fine grained soils to 2.6
coarse grained soils
≤ 3.0
mixed grained soils
≤ 4.0
rockfill material
Higher ratios than the given values are an indication that the soil had not been compacted properly.
Leinfelden-Echterdingen LeinfeldenEchterdingen,, November, 30 th, 2004
gez. Prof. Dr.-Ing. E. Vees
Prof. Dr.-Ing. E. Vees und Partner
A1
APPENDIX EVALUATION OF THE PLATE LOAD TEST ACCORDING TO DIN 18134 THEORETICAL B ACKGROUN ACKGROUND D
Under a circular flexible load the stress is distributed in the ground underneath the plate. According to BOUSSINESQ’S theoretical approach this stress distribution can be described by pressure bulbs as shown in the following figure:
D = 2 R
pressure σ 0
flexible circular load
0 0.9 σ 0 0.7 σ 0 1
0.5 σ 0 0.3 σ 0
2
3 0.1 σ 0 4 /R
Fig. A1: Contours of constant vertical stress beneath a uniformity loaded circular area Stress in the ground causes settlement. The settlement of a rigid plate approximately corresponds to the settlement of the so called CHARACTERISTIC POINT C of of a flexible circular load.
C
0.845 R
R
R
Fig. A2: Definition of the characteristic point of flexible load on a circular area.
Prof. Dr.-Ing. E. Vees und Partner
A2
The settlement s of this characteristic point C can can be calculated from the distribution curve of the vertical stress σ and the modulus of deformation E V V:
σ 0
0
σ
1
=
s
E V
z 1
⋅ ∫ σ dz = 0
1
E V
⋅ A
(1)
= stress area (see left figure) 1
According to Schultze Schultze / Horn1 the solution of equation (1) is:
s
s
= (1 − µ ) ⋅
F
2
⋅
2 E V R
(2)
= settlement
µ
= Poisson’s ratio F = resultant force R = radius of circular load = radius of bearing plate
With
σ 0
we obtain
s
or
E V
1
=
F
⋅
π R
2
= (1 − µ ) ⋅
2
= (1 − µ
2
⋅ ⋅
σ 0 π R
2 E V
(3)
) π ⋅ R ⋅ σ
0
2
s
(4)
Schultze, E. und Horn, A.: Setzungsberechnung in: Grundbautaschenbuch, herausgegeben von U. Smoltczyk, 5. Auflage, Teil 1, 1996, S. 225-254.
Prof. Dr.-Ing. E. Vees und Partner
A3
For µ = 0.25 equation (4) becomes: E V
= 1.5 ⋅ R ⋅
σ 0
s
Hence, for a given load increment ∆σ and a measured settlement increment ∆ s the deforma can be defined as: tion modulus E VV can
or
D = 2R
= 1.5 ⋅ R ⋅
E V
= 0.75 ⋅ D ⋅
= Diameter of the the bearing bearing plate.
Leinfelden-Echterdingen LeinfeldenEchterdingen,, November, 30 th, 2004 gez. Prof. Dr.-Ing. E. Vees
∆σ ∆ s
E V
∆σ ∆ s
