plate load test procedure _din 18134
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PROF. DR.-ING. E. VEES UND PARTNER B A U G R U N D I N S T I T U T G M B H
Waldenbucher Straße 19 70771 Leinfelden-Echterdingen, Germany Telefon 0049 (711) 79 73 50- 0 Postfach 20 03 39 70752 Leinfelden-Echterdingen, Germany Telefax 0049 (711) 79 73 50-20 info@geotechnik-vees.de
PLATE LOAD TEST
IN ROAD AND EARTHWORKS CONSTRUCTION ACCORDING TO DIN 18134
TESTING PROCEDURES, TESTING EQUIPMENT,
THEORETICAL BACKGROUND
Prof. Dr.-Ing. E. Vees und Partner page 2
THE PLATE LOAD TEST IN ROAD AND EARTHWORKS CONSTRUCTION In Central Europe, road construction and pavement design are mainly based on the deforma-tion modulus EV determined by the PLATE LOAD TEST. The deformation modulus EV can be understood as a modulus of elasticity. The more compressible a soil, the lower is the defor-mation modulus. THE PLATE LOAD TEST is described in DIN 18134 and with certain modifica-tions 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.
Fig. 1: Principle of 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 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.
∆s
D
FF = load ∆s = settlement D = diameter of the plate
Prof. Dr.-Ing. E. Vees und Partner page 3
Fig. 2: Plate load test equipment. Excavator serves as counterweight.
Fig. 3: Bearing plate (0.30 m diameter) with hydraulic jack assembly and beam with dial gage to determine plate settlement.
Prof. Dr.-Ing. E. Vees und Partner page 4
The DEFORMATION MODULI EV are calculated from the first loading curve (EV1) and from the reloading curve (EV2) according to the following equation:
Ev = 0.75 ⋅ D ⋅ ∆σ / ∆s Ev = deformation modulus ∆σ = load increment ∆s = settlement increment D = diameter of the plate, generally 0.30 m For this calculation ∆σ and ∆s are usually taken from the load span between 0.3 σmax and 0.7 σmax.
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 test is given in the following table:
Plate Diameter: 300 mm
F Load Pressure σ0 Settlement of the Plate
[kN] [kN/m2] [1/100 mm]
FIRST LOADING
5.65 80 115
11.31 160 209
17.67 250 287
23.33 330 325
29.69 420 380
35.34 500 421
UNLOADING in steps to half of the preceding load
17.67 250 395
8.84 125 360
0 0 259
RELOADING to the last but one step
5.65 80 311
11.31 160 353
17.67 250 378
23.33 330 398
29.69 420 413
Prof. Dr.-Ing. E. Vees und Partner page 5 The result of the test is plotted in a pressure-settlement diagram:
Pressure σ0 in kN/m² Se
ttlem
ent i
n m
m
0 100 200 300 400 500
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
0
first loading
reloading
unloading
0.3 ⋅ σ0, max 0.7 ⋅ σ0, max
1.95
3.45
4.05
EVALUATION OF THE GIVEN EXAMPLE: FIRST LOADING (EV1) ∆σ = 350 – 150 = 200 kN/m2 ∆s = 3.50 – 1.95 = 1.55 mm = 0.00155 m EV1 = 0.75 ⋅ 0.30 ⋅ 200 / 0.00155 = 29032 kN/m2 = 29.0 MN/m2 RELOADING (EV2) ∆σ = 350 –150 = 200 kN/m2 ∆s = 4.05 – 3.45 = 0.60 mm = 0.00060 m EV2 = 0.75 ⋅ 0.30 ⋅ 200 / 0.00060 = 75000 kN/m2 = 75.0 MN/m2
Prof. Dr.-Ing. E. Vees und Partner page 6 DEFORMATION RATIO:
59.20.290.75
1
2 ==V
V
EE
Generally in road construction the following values of EV2 are required: subgrade: EV2 ≥ 45 MN/m² surface of sub-base layer: EV2 ≥ 120 MN/m² For fine grained (cohesive) soils the deformation modulus EV2 which can be accomplished by compacting soils, depends on the index of consistency IC. Approximately the following relation has been found:
Ev2 [MN/m2]
Ic consistency
> 15 > 0.8 stiff
> 20 > 0.9 stiff
> 30 > 1.0 very stiff
> 45 > 1.2 very stiff to hard
Hard consistency of cohesive soils is encountered rarely. A deformation modulus of EV2 ≥ 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 Ev2/Ev1 ≤ 2.0 fine grained soils ≤ 2.2 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, November, 30th, 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 BACKGROUND
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:
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 corre-sponds to the settlement of the so called CHARACTERISTIC POINT C of a flexible circular load.
Fig. A2: Definition of the characteristic point of flexible load on a circular area.
C
R R
0.845 R
flexible circular load
D = 2 R
pressure σ0
0.7 σ0
0.5 σ0
0.3 σ0
0.1 σ0
0.9 σ0
0
1
2
3
4
z/R
Prof. Dr.-Ing. E. Vees und Partner A2 The settlement s of this characteristic point C can be calculated from the distribution curve of the vertical stress σ and the modulus of deformation EV:
According to Schultze / Horn1 the solution of equation (1) is:
( )RE
FsV ⋅
⋅−=2
1 2µ (2)
s = settlement µ = Poisson’s ratio F = resultant force R = radius of circular load = radius of bearing plate
With 20 RF⋅
=π
σ
we obtain ( )VE
Rs
21 02 ⋅⋅
⋅−=πσ
µ (3)
or ( )s
REV02
21
σπµ ⋅⋅−= (4)
1Schultze, E. und Horn, A.: Setzungsberechnung in: Grundbautaschenbuch, herausgegeben von
U. Smoltczyk, 5. Auflage, Teil 1, 1996, S. 225-254.
σ0 σ 0
z1
z
A
AE
dzE
sV
z
V
⋅=⋅= ∫11 1
0
σ (1)
A = stress area (see left figure)
Prof. Dr.-Ing. E. Vees und Partner A3 For µ = 0.25 equation (4) becomes:
sREV
05.1σ⋅⋅=
Hence, for a given load increment ∆σ and a measured settlement increment ∆s the deforma-tion modulus EV can be defined as:
sREV ∆
∆⋅⋅=
σ5.1
or s
DEV ∆∆
⋅⋅=σ75.0
D = 2R = Diameter of the bearing plate. Leinfelden-Echterdingen, November, 30th, 2004 gez. Prof. Dr.-Ing. E. Vees
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