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3
Soil exploration
In this chapter some of the most effective or popular methods for soil exploration, or
soil investigations in the field will be described.
3.1
Cone Penetration Test
A simple, but very effective method of soil investigation consists of pushing a steel
rod into the soil, and then measuring the force during the penetration, as a function of depth. This force consists of the reaction of the soil at the point (the cone resistance),
and the friction along the circumference of the rods. The method was developed in
the 1930’s in the Netherlands. It was mainly intended as an exploration tool, to give
an indication of the soil structure, and as a modelling tool for the design of a pile
foundation. This sounding test , cone penetration test , or simply CPT, has been
developed from a simple tool, that was pushed into the ground by hand or a manual
pressure device, into a sophisticated electronic measuring device, with an advanced
hydraulic loading system. The load is often provided by the weight of a heavy truck.
Originally the CPT was a purely mechanical test, as shown schematically in Figure3-1. The instrument consists of three movable parts, with a common central axis. The
upper part is connected, by a screw thread, to a hollow rod, that reaches to the soil
surface, using extension rods of 1 meter length. The procedure was that pressure was
alternately exerted upon the central axis or the outer rods. When pushing on the
internal axis at first only the cone is pushed into the ground, over a distance of 35
mm. The other two parts do not move with respect to the soil (by the friction of the
soil), so that the force represents the cone resistance only. When pushing the
instrument beyond a distance of 35 mm the second part, the friction sleeve , moves
with the cone, so that in this stage the force consists of the cone resistance plus the
friction along the friction sleeve. The upper part of the instrument is still stationary in
this stage. If it is assumed that the cone resistance is still the same as before, the
sleeve friction can be determined by subtraction. If in the next step the force is
exerted on the outer rods, the cone remains stationary and the system is compressed
to its original state, but at a greater depth (10 cm). The diameter of the lowest part of
the sleeve, which is attached to the cone and moves with it, was sometimes reduced,
to ensure that in the first stage only point resistance is measured.
Modern versions of the CPT use an electrical cone, see Figure 3-2. Both the cone
resistance and the friction are measured continuously, using a system of strain gauges
in the interior of the cone. The instrument again consists of three parts, that are
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20 Soil mechanics
separated by thin rings of rubber. The very sensitive strain gauges can measure the
forces on the lower two parts of the instrument independently.
Figure 3-1. Mechanical CPT. Figure 3-2. Electrical cone.
The results of a cone penetration test give a good insight into the layered structure of the soil. Clay layers have a much smaller cone resistance than sand. A typical cone
resistance qc for a sand layer is 5 MPa or 10 MPa, or even higher, whereas the cone
resistance of soft clay layers is below 1 MPa. If the local friction is also measured the
difference is even more pronounced. The ratio of friction to cone resistance for clays
is much higher than for sand. In sands the friction usually is only about 1 % of the
cone resistance, whereas in clays this ratio usually is 3 % to 5 %. Higher values (8 %
– 10 %) may suggest a layer of peat. In peat the friction usually is substantial, but it
has a very small cone resistance.
Recent developments are to install additional measuring devices in the cone, such apore pressure meter. This type of cone is denoted as a piezocone. A small chamber
inside the cone is connected to the pores in the soil by a number of tiny holes in the
cone. This enables to measure the local pore water pressure. This pressure is
determined by the actual pore water pressure in the soil, but also by the penetration
of the cone in the soil, at least in materials of low permeability. In a very dense clay
the material may have a tendency to expand, which will lead to and under pressure in
the water, with respect to the hydrostatic pressure. This enables to distinguish very
thin layers of clay. In measuring the cone resistance or the friction such thin layers
are not observed, because of the averaging procedure in measuring forces.
An example of the results of as cone penetration test is shown in Figure 3-3. At a
depth of 7 meter a sand layer of about 2 meter thickness can be observed At a depth
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3. Soil exploration 21
of 18 meter the top of a thick sand layer is found. The low values above the first sand
layer, and between the two sand layers indicate soft soil, probably clay. A simple
building (a house) can be founded on the top sand layer, provided that the presence
of this layer is general. A single CPT is insufficient to conclude the existence of this
layer everywhere, having it observed in 3 CPT’s at practically the same depth (and atabout the same thickness) usually is sufficient evidence of its general existence. A
heavy foundation, for a large building, usually requires a foundation reaching into
the deep sand.
Soil type Friction ratio Cone resistance qc
sand, medium – coarse
sand, fine – medium
sand, fine
sand, silty
sand, clayey
sandy clay or loam
silt
clay, silty
clay
clay, peaty
peat
0.4%
0.6%
0.8%
1.1%
1.4%
1.8%
2.2%
2.5%
3.3%
5.0%
8.1%
5 - 30 MPa
5 - 10 MPa
0.5 - 2 MPa
0.1 - 1 MPa
Table 3-3. Friction ratio and cone resistance
Figure 3-3. Result of CPT.
The cone penetration test is in the Netherlands also used as a model test for
estimating the bearing capacity of pile foundations. In the west of the country
generally about 10 m to 20 m of soft soil layers lay on top of a stiff sand layer, which
is excellent for using pile foundations. The bearing capacity of a pile depends mostly
on the capacity of the sand layer. The capacity of the tip of the pile with area A can
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22 Soil mechanics
be estimated by:
F tip = qc A (3.1)
3.2
Vane test
The shear strength of soils can be measured reasonably accurately in situ using the
vane test . In this test a small instrument in the shape of a vane is pushed into the
ground, through a system of rods, just as in the cone penetration test. The vane is
connected, by a central steel axis, to a screw at the top of the rods. This screw can be
rotated, so that the soil in a cylindrical element of soil is sheared along its surface,
against the soil outside the cylinder. Measuring the moment necessary for the
rotation enables to determine the average shear stress along the boundary, which is
about equal to the (undrained) shear strength of the soil. The vane test is very popular
in Scandinavian countries, where the soil very often consists of thick layers of clay of
reasonable strength.
Figure 3-4. Vane test.
3.3
Standard Penetration Test
In many parts of the world, especially in Anglo-Saxon countries, the properties of the
soil are often determined by using a Standard Penetration Test , or SPT. In this test a
sampling tube is driven into a borehole in the ground using a standardized
hammering weight. The actual test consists of measuring the number of blows
needed to achieve a penetration of 300 mm (1 foot) into the ground. This is denoted
as N, the blow count , the number of blows per foot. An advantage of the SPT is that
no heavy equipment is needed, as for instance in the CPT, which has to be pushed
into the ground statically, and thus requires a large counter weight. Another
advantage of the SPT is that immediately provides a soil sample. The sample is not
of the best quality, but at least there is a sample. The reproducibility of the SPT
usually is not so very good, and the difference between sand and clay is not sopronounced as it is in the CPT. It is also not possible to immediately derive the shear
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3. Soil exploration 23
strength from the blow count.
For many projects the initial soil data often may be restricted to a series of SPT-
results. Then it is useful to know that a characteristic blow count for sand is N = 20 ,
and that for soft clay the value may be N = 5 , or even lower, down to N = 1 . A first
indication can be obtained from Table 3-4, derived from Terzaghi & Peck. Manyresearchers have tried to obtain a correlation with the CPT, but their results are not
very consistent.
Sand Clay
N Density N Consistency
< 4 Very Loose < 2 Very soft
4-10 Loose 2-4 Soft
10-30 Normal 4-8 Normal
30-50 Dense 8-15 Stiff
> 50 Very dense 15-30 Very stiff
>30 Hard
Table 3-4: Interpretation of SPT according to Terzaghi &Peck.
3.4
Soil sampling
For many engineering projects it is very useful to take a sample of the soil, and to
investigate its properties in the laboratory. The investigation may be a visual
inspection (which indicates the type of materials: sand, clay or peat), a chemicalanalysis, or a mechanical test, such as a compression test or a triaxial test.
Figure 3-5. SPT.
A simple method to take a sample is to drive a tube into the ground, and then
recovering the tube with the soil in it. The tube may be about 1 meter long, see
Figure 3-5, and may have a valve at its bottom, to prevent loosing the sample. The
tube may be brought into the soil by driving it into the ground using a falling weight,
or a hammer. An advantage of this method is that it does not require heavy
equipment. It is possible to take a sample in a terrain that is inaccessible to heavyvehicles. The sample is somewhat disturbed, of course, during the sampling process,
but even so a good impression of the composition of the soil can be obtained The
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24 Soil mechanics
sample is not very well suited for a refined test, however, as the initial state of stress
is disturbed, and perhaps also the density. To take a deep sample the sampling tube
may be of smaller diameter than the borehole, which is supported and deepened by a
special boring tube.
An alternative method is to push the sampler into the ground, by using hydraulicequipment, mounted on a heavy truck. In this case the sampling process is somewhat
more careful, and the disturbance of the sample is less. Due to friction of the sample
with the wall of the sampling tube, however, the samples are not undisturbed.
Figure 3-6. Begemann sampler.
Various institutes have developed systems in which the sample to be taken is almost
undisturbed. A completely undisturbed sample is impossible, but some procedures
come very close. Some methods are, for instance, to take a very large block of soil,
and use the inner part only, or freezing a block of sand, and then cutting a sample
from the frozen soil. Good quality samples can also be obtained using the Begemann
sampler, developed at GeoDelft, see Figure 3-6. This sampler consists of two steel
tubes, that are being pushed into the soil together. The sample is cut by the outer
tube, which immediately widens behind the cutting edge, and the sample is
surrounded by a nylon stocking, that initially is rolled up on the inner tube. The end
of the stocking is attached to a plate at the top of the future sample, so that, when the
tubes are pushed down, the stocking gradually displace downward the stocking is
gradually stripped off the inner tube. The final result is a very long soil sample (for
instance 20 meter long), enclosed by a nylon stocking. Around the stocking the
sample is supported by a heavy fluid (of unit weight ! " 15 kN/m3), that simulates
the original lateral support of the soil. This fluid also reduces the friction along the
circumference of the sample. The samples produced by this sampler are of high
quality. Very thin layers of all sorts of materials can be identified, including loosesand. The quality of the samples is good enough to be used for accurate laboratory
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3. Soil exploration 25
testing, in compression tests or triaxial tests. The results of a boring may be
presented in the form of a color photograph of one half of the sample, cut along its
length. That the thin layers are not disturbed near the boundary confirms that there is
very little friction.
It may be interesting to note that samples can also be taken from the bottom of thesea. One possible method is by using a diving bell, in which the air pressure is kept at
the same level as the water pressure. From this diving bell a sample can be taken by
the operators, or they can make a cone penetration test. Another method is to use a
heavy frame, that is submerged in the water from a ship. Using a remote control
system a cone can be made to penetrate the soil, or a sample can be taken. This
method can even be used in water depths of 1000 meter, or more. An example of a
continuous Begemann boring is shown in Figure 3-7.
Figure 3-7. Begemann sample.
Investigating the sea bottom is of special interest in offshore engineering, of course.For the production of oil and gas from the sea bottom large platforms are
constructed, which usually need a pile foundation to withstand the extreme wave
load conditions during a storm. The piles usually are steel tubular piles, of large
diameter (one meter or more), and very large length (50 meter or more). These piles
derive their bearing capacity mostly from the friction along the shaft, and not from
the point resistance (as most piles in Western Netherlands). It is of great importance
to predict the maximum shearing resistance along the pile shaft. This can be
measured very well by a cone penetration test, from the bottom of the sea. Even
though this is a costly operation, it gives very valuable information about the soilstructure, and it gives numerical values for the cone resistance and the friction, as a
function of depth
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Direct shear test
A direct shear test is a laboratory test used by geotechnical engineers to find theshear strength parameters of soil. The U.S. and U.K. standards defining how the
test should be performed are ASTM D 3080 and BS 1377-7:1990 respectively.
The test is performed on three or four specimens from a relatively undisturbed soilsample. A specimen is placed in a shear box which has two stacked rings to hold
the sample; the contact between the two rings is at approximately the mid-height ofthe sample. A confining stress is applied vertically to the specimen, and the upper
ring is pulled laterally until the sample fails, or through a specified strain. The load
applied and the strain induced is recorded at frequent intervals to determine astress-strain curve for the confining stress.
Direct Shear tests can be performed under several conditions. The sample isnormally saturated before the test is run, but can be run at the in-situ moisturecontent. The rate of strain can be varied to create a test of undrained or drained
conditions, depending whether the strain is applied slowly enough for water in thesample to prevent pore-water pressure buildup.
Several specimens are tested at varying confining stresses to determine the shear
strength parameters, the soil cohesion (c) and the angle of internal friction(commonly friction angle) (φ). The results of the tests on each specimen are
plotted on a graph with the peak (or residual) stress on the x-axis and the confiningstress on the y-axis. The y-intercept of the curve which fits the test results is thecohesion, and the slope of the line or curve is the friction angle.
http://en.wikipedia.org/wiki/Geotechnical_engineeringhttp://en.wikipedia.org/wiki/Geotechnical_engineeringhttp://en.wikipedia.org/wiki/Shear_strength_%28soil%29http://en.wikipedia.org/wiki/Soilhttp://en.wikipedia.org/wiki/Soilhttp://en.wikipedia.org/wiki/ASTMhttp://en.wikipedia.org/wiki/ASTMhttp://en.wikipedia.org/wiki/British_Standardshttp://en.wikipedia.org/wiki/British_Standardshttp://en.wikipedia.org/wiki/Strain_%28materials_science%29http://en.wikipedia.org/wiki/Stress-strain_curvehttp://en.wikipedia.org/wiki/Friction#Angle_of_frictionhttp://en.wikipedia.org/wiki/Friction#Angle_of_frictionhttp://en.wikipedia.org/wiki/Stress-strain_curvehttp://en.wikipedia.org/wiki/Strain_%28materials_science%29http://en.wikipedia.org/wiki/British_Standardshttp://en.wikipedia.org/wiki/ASTMhttp://en.wikipedia.org/wiki/Soilhttp://en.wikipedia.org/wiki/Shear_strength_%28soil%29http://en.wikipedia.org/wiki/Geotechnical_engineering
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Exploratory DrillingExploratory Drilling
Dheeraj Kumar
Soil ExplorationSoil Exploration
Failing Truck Mounted Rig All-Terrain Rig
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Soil ExplorationSoil Exploration
Water Boring from Bargefor Bridge Crossing
Wireline Rig
Soil ExplorationSoil Exploration
Track Mounted Rig
Track Rig
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Drilling & Sampling of Soil & Rock Drilling & Sampling of Soil & Rock
Objectives: Objectives:
Recognize various drilling techniquesRecognize various drilling techniques
Be familiar with undisturbed vs. disturbedBe familiar with undisturbed vs. disturbed
sampling methodssampling methods
List rock exploration methodsList rock exploration methods
Familiarity with core bits & barrelsFamiliarity with core bits & barrels
Observations to be made during drilling, includingObservations to be made during drilling, including
Rock Quality Designation (RQD).Rock Quality Designation (RQD).
Appreciate role of geologic mapping in obtaining Appreciate role of geologic mapping in obtaininginformation on rock masses.information on rock masses.
Soil & Rock DrillingSoil & Rock Drilling
Soil ExplorationSoil Exploration
Advancing a Boring/Borehole Advancing a Boring/Borehole
Sampling (Drive, Undisturbed)Sampling (Drive, Undisturbed)
Transport of Samples to LaboratoryTransport of Samples to Laboratory
Rock ExplorationRock Exploration Rock Coring and NonRock Coring and Non--coring Techniquescoring Techniques
Logging of Rock Core (CR and RQD)Logging of Rock Core (CR and RQD)
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Selection of Drilling methodsSelection of Drilling methods
Depth of drilling:Depth of drilling: all drilling methods have certainall drilling methods have certain
limitationslimitations;; Sample recovery:Sample recovery: type of samples desired, i.e. soil,type of samples desired, i.e. soil,
groundwater, disturbed or undisturbed, frequency ofgroundwater, disturbed or undisturbed, frequency of
sampling, yield estimation;sampling, yield estimation;
TargetTarget lithologylithology:: well installation completed inwell installation completed in
unconsolidated or consolidated formationunconsolidated or consolidated formation..
Health and SafetyHealth and Safety
Level of contaminationLevel of contamination
High yield of formation may produce high pressures;High yield of formation may produce high pressures;
Underground fire hazards in gaseous areasUnderground fire hazards in gaseous areas
Selection of Drilling methodsSelection of Drilling methods
Access and Noise Access and Noise
Terrain roughness;Terrain roughness;
Space and height limitations;Space and height limitations;
Municipal noise ordinanceMunicipal noise ordinance
Disposal of drilling fluids and cuttingsDisposal of drilling fluids and cuttings
Contaminated cuttings and groundwater may haveContaminated cuttings and groundwater may have
to be handled as hazardous wastes and beto be handled as hazardous wastes and betransported to landfill or special waste disposaltransported to landfill or special waste disposal
facilities.facilities.
Lithology and Aquifer characteristicsLithology and Aquifer characteristics
Soil type (sand, clay, boulders)Soil type (sand, clay, boulders)
Depth to water tableDepth to water table
Depth to bedrock Depth to bedrock
CostCost
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Drilling MethodsDrilling Methods
Methods which do not use circulationMethods which do not use circulation
(drilling) fluids(drilling) fluids
Displacement boringDisplacement boring
Driven wellsDriven wells
SolidSolid--stem augerstem auger
HollowHollow--stem augerstem auger
Sonic drillingSonic drilling
Drilling MethodsDrilling Methods
Methods which use circulation (drilling) fluids toMethods which use circulation (drilling) fluids to
carry drill cuttings to the surfacecarry drill cuttings to the surface
1. Rotary Drilling1. Rotary Drilling
Rotary (direct) DrillingRotary (direct) Drilling
Reverse Circulation Rotary Drilling (RC)Reverse Circulation Rotary Drilling (RC)
DualDual--wall Reverse Circulation Drillingwall Reverse Circulation Drilling
2. Percussion Drilling2. Percussion Drilling
CableCable--tool percussiontool percussion
Air percussion down Air percussion down--thethe--hole hammerhole hammer
Air percussion casing hammer Air percussion casing hammer
ODEX percussion downODEX percussion down--thethe--hole hammerhole hammer
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Drilling MethodsDrilling Methods
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© Kaplan AEC Education, 2008
Chapter 9 Strength of Materials
KEY TERMS AND DEFINITIONS
combined loads
Complex loading consisting of axial loads, shear force, bending moments, and torsionalmoments acting simultaneously on a system.
equilibrium equations for plane (2-D) systems
ΣF = 0 in two independent directionsΣM = 0 about any arbitrarily selected point
flexural stress (σ)
The bending moment, M, divided by the section modulus, S, of the section.
homogeneous material
Material with the same composition throughout.
internal member forces
P = axial force perpendicular to section
V = shear force tangential to sectionM = bending moment about rectangular axis
T = torsional moment (torque) about polar axis
isotropic material
Material with the same mechanical properties in all directions.
linearly elastic materialMaterial that obeys Hooke’s law (linear) and for which the residual deformation is zero upon
removal of a force (elastic).
modulus of elasticity (E)
The constant of proportionality between stress and strain. E equals stress divided by strain, which
can be calculated as the slope of the initial linear portion of a stress-strain diagram.
Mohr’s circle
A semi-graphical method of transforming states of stress or strain at a point in an element subjectto combined loads.
normal stress (σ)
The axial force, P, divided by the area, A, of the section.
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2
Poisson’s ratio
The ratio of lateral strain to longitudinal strain resulting from a member subjected to an axial
force.
positive faceIn a body under stress, a plane area under load with a normal outward stress in tension.
section modulus
The ratio of the moment of inertia of a beam cross-section to the distance from the neutral axis to
the farthest structural fiber.
section properties of an area
Section properties normally are calculated with respect to the centroid of an area, which is the
point about which the first moment of an area is zero.A = area of cross-section
I = rectangular moment of inertia, computed as the second moment of an areaabout an axis
S = section modulus—the moment of inertia, I, divided by the distance from theneutral (centroidal) axis to the farthest structural fiber.
J = polar moment of inertial, computed as the second moment of an area about a point
r = radius of gyration, computed as the square root of the moment of inertia divided by the cross-sectional area.
shear flow
In a hollow, thin-walled shaft under torsion, the product of wall thickness and shear stress.
shear stress (τ)
The tangential force, V, divided by the area, A, of the section.
strain
The ratio of the change in a dimension under a deforming force to the original dimension.
stress
Force per unit of area.
superposition principle
A complex loading system can be divided into a series of simple loads with each being analyzed
separately. These can be combined to obtain the solution of the complex loading. Superposition
applies only to linear systems—those whose behavior is governed by linear algebraic ordifferential equations.
torsion
The twisting deformation of a long member under a load.
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3
yield
The point on a plot of stress versus strain where the relationship is no longer linear and strain
increases rapidly.
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Guidelines for proper use of
Plate elements
In structural analysis using finite element method, the analysis model is created by dividing the entire
structure into finite elements. This procedure is known as finite element discretization. We then establish
the relation between applied load and displacement for the structure defined by the assemblage of
elements. This means that the stiffness of the overall structure thus depends on the accuracy of the
stiffness of individual finite elements.
In the case of beam element widely used in analysis practice, the element formulation is based on
theoretically exact equations, thereby resulting in an exact solution. However, in the case of other
elements (2-dimensional and 3-dimensional elements) some errors are included due to adopted
assumptions in theoretical formulation.
Many types of plate elements used to model two-dimensional structures such as slabs and walls have been
developed. However, none of the plate elements provide the exact solution for all cases. In the majority of
cases, adequate assumptions should be adopted to approach the exact solution.
The statement of “no plate elements provide the exact solution” means that no such a plate element
currently exists, which gives acceptable solutions for all types of loads, boundary conditions and material
properties even for two-dimensional plane domains. It also means that if the configuration of the plate
element deviates from the shape assumed in the formulation of the element stiffness, the margin of error
increases. In the case of two or three-dimensional structures, it is necessary to analyze the structural
model so as to minimize the discretization error.
The following outlines some guidelines (practical considerations) to avoid situations, which cause errors
in application of plate elements and to minimize possible errors:
Performance of a Plate Element
The relation between element load and nodal displacement is defined by the element stiffness matrix. The
element stiffness matrix of each element used in structure discretization is assembled in global stiffness
matrix defining the equilibrium equation of the entire structure. The equilibrium equation retains the
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Guidelines for the proper use of plate elements
following form:
[ ]{ } { }FuK =
where,
[ ]K = global stiffness matrix{ }u = vectors of unknown nodal displacements { }F = vectors of applied nodal loads
Once we obtain the vectors of nodal displacements from the above equation, strains and stresses at any
point in the structure are approximated through interpolation.
From the above consideration, we can summarize the performance criteria for plate elements in the
following two points:
• How well does the element stiffness perform?
• How well does it interpolate the strains and stresses at any point?
The above two criteria are related to the element formulation, which are directly processed by the analysis
program. However, the element type and shape selected by the user can seriously affect the accuracy of
analysis results. Therefore, the guidelines will explain the selection of appropriate elements from the
several plate elements provided by the program. We will then examine the major shape factors affecting
the efficiency of the elements. Lastly, main considerations for using the plate elements are described.
Selection of Plate Elements
The plate element is an element whose thickness is much smaller than its length and effectively resists the
applied loads by the combination of in-plane (membrane) stiffness and out-of-plane (bending) stiffness.
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Guidelines for the proper use of plate elements
Figure 1. Classification of Plate Elements
Plate elements are classified in thin and thick plates depending on the thickness-to-length ratio, as shown
in Fig.1. If the plate thickness is relatively small (thickness-to-length ratio < 1/10) then the plate is
considered thin and the effect of shear deformations can be ignored. In this case the bending stiffness
becomes dominant. If the ratio of thickness-to-length is larger than one-tenth (1/10) the plate is
considered thick and both bending and shear deformations must be accounted for.
In a thin plate, the margin of error resulted by neglecting the shear deformation effect is as little as 2%. If
the thickness is very small, it is more efficient to use plane stress (membrane) element instead of the plate
element. On the contrary, if the ratio is very large, it is desirable to use three-dimensional solid elements.
Also, the user can separately input in-plane and out-of-plane thicknesses, where the former is the
membrane thickness and the later is the bending thickness. As such, the stiffness for each behavior can
be adjusted by the respective thicknesses.
To evaluate the efficiency of the plate elements provided by MIDAS, a typical model (1/4 circular plate)
is considered as shown in Fig. 2(a). Figs. 2(b) to (d) illustrate the displacement convergence of thick and
thin plates for different thickness-to-length ratios and element meshes. The positive and negative signs of
the errors represent that the calculated value is larger or smaller than the theoretical value respectively.
a) Adopted Element Meshes and Location of Measured Displacement for Test Model
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Guidelines for the proper use of plate elements
b) Displacement Convergence for 50/1t/L =
c) Displacement Convergence for 10/1t/L =
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Guidelines for the proper use of plate elements
d) Displacement Convergence for 5/1t/L =
Figure 2. Displacement Convergences for Thick Plate and Thin Plate
As shown in Figs. 2(b) and (c), if the thickness-to-length ratio is equal to or less than one-tenth (1/10), the
thin plate element produces more accurate results than the thick plate. However, the thick element
produces better results if the ratio is larger than 1/10. In the case of the thick plate (t/L=1/5), we can
observe that the errors associated with the thin plate element are much larger than those produced by the
thick plate element, regardless of the mesh density
Shape Factors Affecting the Performance of Plate Elements
The major element shape factor, which has a great impact on the analysis results, is the Jacobian
Determinant, which affects the element stiffness performance. Other shape factors such as aspect ratio,
skew angle, taper, warping, etc. are the factors, which affect to a large extent the interpolation of various
results at any point within the element domain based on the computed nodal displacements.
If strains and stresses at a particular point cannot be properly calculated based on the computed nodal
displacements, it is referred to as ‘interpolation failure’. This is closely related to locking phenomenon
whereby the structure’s behavior suddenly becomes stiff, as a result of excessive stiffness calculated for a
particular deformation state. Such a locking phenomenon will likely occur when the shape factors such as
aspect ratio, skew angle, taper and warping are very poor.
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Guidelines for the proper use of plate elements
(1) Jacobian Determinant J
For an isoparametric element, the stiffness is calculated by numerical integration using natural
coordinates. Transformation into natural coordinates is illustrated in Fig. 3. After transformation, the
arbitrarily shaped element is mapped into a square master element with the length of 2. Here, Jacobian
determinant J denotes the ratio of the original element’s area to the master element’s area (see Fig. 3).
Figure 3. Mapping of Isoparametric Element to Square Element
It should be noted that the shape of the four-node quadrilateral element must be of convex form. In
other words, the element sides should not intersect each other (see Fig. 4). If the configuration of the
quadrilateral element is not convex, the value of J is 0 or negative, and the element stiffness has a
0 or negative value. As a reference, the ratio of maximum and minimum values of Jacobian
determinant should satisfy the condition 2/ ≤minmax
JJ . The farther the interior angles in a
four-node quadrilateral element are from 90°, the larger the value.
Figure 4. Relation between Element Shape and Jacobian Determinant
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Guidelines for the proper use of plate elements
(2) Aspect Ratio Λ
Aspect ratio as shown in Fig. 5 is the length ratio of the shortest side to the longest side. The bestelement shape is a square whose aspect ratio is 1, and the aspect ratio becomes much smaller than 1
for a poorer shape. Therefore, when stress evaluation is important the aspect ratio should not be over
one-third, and when the deformation (displacement) evaluation is important it should not be over
one-fifth. It should be noted that the results of nonlinear analysis are more sensitive to the change of
aspect ratio compared to the results of linear analysis.
Figure 5. Aspect Ratio
(3) Skew Angle α
Skew angle denotes the degree of angular deviation from a rectangle. The best element shape is a
square whose skew angle is 0, and the skew angle is farther away from 0 for a poorer shape. To
obtain accurate analysis results, it is recommended that the skew angle be maintained no less than
45°, and all the interior angles of the quadrilateral element should remain in between 45° and 135°.
Figure 6. Skew Angle
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Guidelines for the proper use of plate elements
(4) Taper τ
Taper represents the measure of geometric deviation of the element form from a rectangle.Accordingly, the best value of taper is equal to 1. The value of taper farther from 1 indicates a poor
element shape
Figure 7. Taper
(5) Warping ω
Warping measures the degree of deviation (plane distortion) of four nodes from a plane. Whileaspect ratio, skew angle and taper evaluate in-plane offset, warping evaluates out-of-plane offset.
The magnitude of warping should not exceed one-hundredth. Especially, it is necessary to be
cautious with quadrilateral elements created in the vicinity of the intersections (connections) of
curved surfaces.
Figure 8. Warping
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Guidelines for the proper use of plate elements
Items to be considered when Using Plate Elements
• For any element shape, the use of quadrilateral elements is better than triangular elements.
However, when the quadrilateral shape deviates from the permitted limit range, the use of
triangular elements is preferable.
• Since bending stresses exist in plate elements, the top and the bottom surfaces are distinguished.
Therefore, the normal direction of the adjacent plate elements must be coincident in order to
align the top and the bottom surfaces of the contiguous elements.
• Based on the nodal degrees of freedom (DOF), elements are classified as elasticity elements with
nodal DOFs, which include only translational displacements and structural elements with nodal
DOFs, which include translational and rotational displacements. Hinges are formed at the
nodes where these two types of elements meet. The plate element can be used in combination
with three-dimensional element (solid element). In such a case, moment should be transmitted
at the connections of the elements by including additional plate elements shared by both types of
elements or using rigid links or rigid beam elements.
• The plate element does not have the rotational stiffness about normal direction (drilling).
Therefore, it is necessary to restrain the drilling degree of freedom (RZ) or to put a fictitious
rotational spring with small stiffness to avoid singularity error. If a beam is connected
perpendicular to the plate element without the drilling DOF, an additional beam element should
be introduced within the plate element to transmit torque. MIDAS automatically assigns
fictitious springs with appropriate stiffness to the plate elements.
• If triangular elements are used in a symmetric structure, the mesh layout should be also kept
symmetrical in order to obtain symmetrical analysis results. Note that MIDAS contains a
function, which produces the total results based on a ½ or ¼ model analysis.
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•College of Engineering •Department of Mechanical Engineering
Tenth Edition
CHAPTER
2c
Engineering Mechanics Statics
FORCE VECTORS
by
Dr. Ibrahim A. Assakkaf
SPRING 2007
ENES 110 – Statics
Department of Mechanical Engineering
University of Maryland, Baltimore County
Slide No. 1Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
OBJECTIVESLecture’s Objectives:
Students will be able to :
a) Represent a position vector in Cartesian
coordinate form, from given geometry.
b) Represent a force vector directed along
a line.
In-Class Activities:
• Reading quiz
• Applications /
Relevance
• Write position vectors
• Write a force vector
• Concept quiz
• Group Problem (Ex. 3)
• Attention quiz
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Slide No. 2Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
READING QUIZ
1. A position vector, rPQ, is obtained by
A) Coordinates of Q minus coordinates of P
B) Coordinates of P minus coordinates of Q
C) Coordinates of Q minus coordinates of the origin
D) Coordinates of the origin minus coordinates of P
2. A force of magnitude F , directed along a unit
vector u, is given by F = ______ .
A) F (u)
B) u / F
C) F / u
D) F + u
E) F – u
Slide No. 3Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
APPLICATIONS
Wing strut
How can we
represent the force
along the wing
strut in a 3-D
Cartesian vector
form?
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Slide No. 4Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Position Vectors
A position vector is defined asa fixed vector that locates a
point in space relative to
another point.
Consider two points, A & B, in
3-D space. Let their coordinates
be ( x A, y A, z A) and ( x B, y B, z B ),
respectively.
The position vector directed from A to B, r AB , is defined as
r AB = {( x B – x A ) i + ( y B – y A ) j + ( z B – z A ) k }m
Please note that B is the ending point and A is the starting point.
So ALWAYS subtract the “tail” coordinates from the “tip”
coordinates!
Slide No. 5Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
Position Vectors
Example 1
An elastic band is
attached to points A
and B as shown in the
figure. Determine its
length and its direction
measured from A
toward B.
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Slide No. 6Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Position Vectors
Example 1 (cont’d)
– Establish the
coordinates of points A
and B:
– Find r = r B – r
A:
( ) ( )3,2,2 3,0,1 −− B A
( )3,0,1 − A
( )3,2,2− B
[ ] [ ] ( )[ ]
{ }m623 330212
k ji
k jir
++−=
−−+−+−−=
Slide No. 7Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
Position Vectors
Example 1 (cont’d)
– The magnitude of r represents the length of
the rubber band:
– Formulating a unit vector in the direction of r,we have
( ) ( ) ( ) m7623 222 =++−== r r
{ }m623 k jir ++−=
k jir
u7
6
7
2
7
3++
−==
r
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Slide No. 8Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Position Vectors
Example 1 (cont’d)The components of this
unit vector yield the
coordinate direction
angles:
The angles are
measured as shown
o1
o1
o1
0.317
6cos
4.737
2cos
1157
3cos
=⎟ ⎠
⎞⎜⎝
⎛ =
=⎟ ⎠
⎞⎜⎝
⎛ =
=⎟ ⎠
⎞⎜⎝
⎛ −=
−
−
−
γ
β
α
Slide No. 9Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
Force Vector Directed along a Line
If a force is directed along a line,
then we can represent the force
vector in Cartesian Coordinates by
using a unit vector and the force
magnitude. So we need to:
a) Find the position vector, r AB , along two points on
that line. b) Find the unit vector describing the line’s direction,
u AB = (r AB/r AB).
c) Multiply the unit vector by the magnitude of the
force, F = F u AB .
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Slide No. 10Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Force Vector Directed along a Line
Example 2
Given: 400 lb force along the
cable DA.
Find: The force F DA in the
Cartesian vector form.
Plan:
1. Find the position vector r DA and the unit vector u DA .
2. Obtain the force vector as F DA = 400 lb u DA .
Slide No. 11Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
Force Vector Directed along a Line
Example 2 (cont’d)
– The figure shows that
when relating D to A,
we will have to go -2 ft
in the x-direction, -6 ft
in the y-direction, and
+14 ft in the z-
direction. Hence,
{ }ft1462 k jir +−−= DA
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Slide No. 12Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Force Vector Directed along a Line
Example 2 (cont’d)The figure shows that when relating D
to A, we will have to go -2 ft in the x-
direction, -6 ft in the y-direction, and
+14 ft in the z-direction. Hence,
r DA = {-2 i – 6 j + 14 k} ft.
We can also find r DA by subtracting
the coordinates of D from the
coordinates of A.
r DA
= (22 + 62 + 142)0.5 = 15.36 ft
u DA = rDA/r DA and F DA = 400 u DA lb
F DA = 400{(-2 i – 6 j + 14 k)/15.36} lb
= {-52.1 i – 156 j + 365 k} lb
( )0,6,2 D
( )14,0,0 A
Slide No. 13Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k k a f_ S P 0 7
CONCEPT QUIZ
1. P and Q are two points in a 3-D space. How are the position
vectors rPQ and rQP related?
A) rPQ = rQP B) rPQ = - rQP
C) rPQ = 1/rQP D) rPQ = 2 rQP
2. If F and r are force vector and position vectors, respectively,
in SI units, what are the units of the expression (r * (F / F )) ?
A) Newton B) Dimensionless
C) Meter D) Newton - Meter
E) The expression is algebraically illegal.
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Slide No. 14Chapter 2c. FORCE VECTORS
E N E S 1 1 0 © A s s a k
k a f_ S P 0 7
Example 3
Given: Two forces are acting on
a pipe as shown in the
figure.
Find: The magnitude and the
coordinate direction
angles of the resultant
force.
Plan:
1) Find the forces along CA and CB in the Cartesian vector form.
2) Add the two forces to get the resultant force, F R.
3) Determine the magnitude and the coordinate angles of F R.
To be discussed and solved in class
Slide No. 15Chapter 2c. FORCE VECTORS
s s a k k a f_ S P 0 7
Example 3 (cont’d)To be discussed and solved in class