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CT2 MOD1-4 - Combined notes
Construction Technology 2 (Substructure) (300721)
Western Sydney University
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year 2 semester 2Preview text
Module 1.
the Strength of Foundations
We tend to take the foundation for granted. Our everyday experience is that the earth is solid and stable. Nothing could be further from the truth.
Figure 1 Earthquake damage (from secretsofthefed) Somewhere in the world, an earthquake has just happened or people are coping with the aftermath of a seismic event.............! It is only the catastrophic events that reach the news bulletins, but even these events remind us that the ground can move and sometimes violently.
Australia is located in the middle of a tectonic plate, the Indo-Australian plate, so we don’t experience earthquakes on a daily basis. Even the earthquakes that have struck Australia, like the Newcastle earthquake, are mild by world standards.
In Australia, we still experience foundation problems, but they are the result of much slower and milder effects, such as water ingress, poor soils and less frequently, over-loading. These effects can still cause loss of life a nd livelihood, as we saw with the Thredbo landslide.
Hence, major structures must be designed by geotechnical engineers; minor structures, like houses, must be certified by consulting engineers.
The sub-structure can comprise roughly 20% of the total building cost, so it is important that the substructure is designed with care. Defects in the substructure can be very expensive to
rectify. Once a building is completed, the superstructure obscures the substructure. Even during the rectification process, the superstructure loads must be supported........
It is important that construction managers understand basic geotechnical theory so that they don’t build in defects that can only be fixed at great effort and great cost. In coming lectures, we will look in detail at the structural action of each footing system and how this affects the resulting construction system.
In this lecture, we start with the basics of geomechanics; we will investigate what happens to the soil under load.
When a building fails, it seems to settle into the ground. In other words, that the soil is compressed. But in reality...............
Foundation soils are relatively incompressible. Footings and retaining walls mainly fail when the soil particles are shifted sideways and upwards.
In fact, a building can be designed to “float” on top of a very poor foundation soil. The foundation soil under the World Trade Centre (“Ground Zero”) in New York, was so poor that the entire complex of seven buildings on the site (including the two main towers), was floating in a gigantic caisson.
Figure 2 Ground Zero (2010) (from daniellemenendez) The soil is said to shear when it moves sideways. One layer of the soil moves to the side in relation to the main bulk of the soil. This behaviour occurs in all ground-coupled construction systems, including footing beams, slabs and retaining walls.
1 Prediction of Shear Strength
The shearing process does not involve just one mechanism, but rather two. The shear strength of most real soils depends on both components, but initially we will be examining the mechanisms, as if they were separate.
Soils resist loads by cohesion and friction between the soil particles.
The shearing effect in soils has to be resisted by something in the soil. If there was no shearing resistance in the soil, then the foundation soil would continue to “flow” sideways until the footing failed. This can actually happen in an earthquake or after flooding. The soil is liquefied. The bearing resistance is reduced to zero.
Cohesion is important in clayey soils. Cohesion results from:
Chemical bonding amongst the soil particles and between the soil particles and water in the pores. The type of chemical bonding depends on the moisture content, electrolyte concentration and the compaction of the soil.
Clay particles are flat platelets. The faces of the clay particles are negatively charged and the edges are positively charged. The most stable form of bonding occurs when the positive edges contact the negative faces (“flocculated”).
Figure 4 Flocculent clay structure
The arrangement of electrostatic charges is shown in a close-up of the flocculated structure, in Figure 5.
Figure 5 Electrostatic charges on clay platelets
Capillary forces (“Apparent cohesion”) within the soil pores. Menisci develop within the soil voids that are capable of binding the soil together. Capillary forces are optimised at a particular moisture content.
The meniscus is an example of surface tension. A water insect walking on water demonstrates this. The water insect indents the pond surface slightly... going through the surface. Like a trampoline, the water surface tries to spring back, pushing the water insect upwards.
Figure 6 Apparent cohesion in soil
The meniscus works like an arch. If the soil is subjected to a force or stress, then the meniscus will spread the force around the void.. than allowing the soil particles to drift apart or compress.
If the moisture content is too low, the meniscus cannot develop.... If it is too high, the pore is saturated with water. The meniscus arch cannot develop.
1 Theories of Shear Strength
We can use the example of a trench to illustrate the concepts of friction and cohesion. The two types of shear resistance have been developed into two theories.
Two theories were advanced to explain the shearing behaviour of foundations. They were developed to explain why trenches fail during excavation. The first theory accounted for the dramatic cave-ins that we often hear about in the news.
The Coulomb Theory
Coulomb Theory states that catastrophic failure occurs when the supported weight exceeds the shear capacity of soil.
This theory dates back to 1776. It’s as old as Australia!
The Coulomb theory emphasises the cohesive nature of soils. It is most applicable for clayey soils. In the case of a trench, you keep digging and digging..... the trench will just suddenly collapse.
Figure 9 Coulomb trench
The weight on the soil has reached a limit, that is related to the bonding within the soil. It’s like a breaking strain within the soil. If further load is added to the soil (ie. over-burden), the soil bonds are broken and the wall of the trench collapses.
Load can superimposed on the trench. Load on the top of the trench (building materials and equipment), vibration and increased groundwater moisture can tip the balance and cause collapse.
The best idea is to always prop the trench as you excavate. One advantage of clayey soils is that the propping does not need to be continuous. One pass of excavation can be undertaken, the trench propped and then excavation continued below, as shown in Figure 10. For a firm
soil (i. not too moist and not too sandy), you can continue to excavate downwards indefinitely. It doesn’t matter how deep you excavate, as long as you prop periodically, as you continue downwards. You merely have to make sure that the props are strong enough to take the lateral pressure that the soil exerts.
Figure 10 Excavation shoring (from tk-steelcom.com)
The Mohr Theory
Mohr Theory states that progressive failure occurs so that the soil adopts an angle of repose.
The Mohr theory was formulated much later than the Coulomb Theory in 1900. It is applicable to sandy soils.
The sandy particles are not bonded in the first place so that the trench will fail at any depth. As the trench is excavated, sand falls into the hole.
Builders mistakenly think that this is the reason that you shore up the trench, to keep the trench clear of debris. This neglects the other form of catastrophic failure, modelled by the Coulomb Theory, which is more dangerous.
Figure 11 Mohr trench
1 Revision of Engineering Units
Geotechnics uses just two basic units:
Mass [kg]
Length [m]
All other units are derived from these two basic quantities:
Weight
Weight is different from Mass. An object has the same Mass, wherever it is in the Universe. 10 kilograms is still 10 kilograms on the Moon. Weight, on the other hand, depends on gravity. A object will weigh 5 times more on Earth than it does on the Moon. Weight is a force. It is the quantity that engineers are interested in. A force measures the impact that a load has on a sub-structure.
There are many uncertainties in foundation engineering, so it doesn’t make much difference if we use 9 m/s 2 (accurate) or 10 m/s 2 (approximate) for gravity. So, a 10 kilogram mass has the Weight: 10 kg x 10 m/s 2 = 100 Newtons [N]
In foundation engineering, a 10 kilogram mass is insignificant. We tend to work in tonnes (1,000 kilograms), rather than kilograms: 1000 kg x 10 m/s 2 = 10000 Newtons [N] 1 tonne x 10 m/s 2 = 10 kiloNewtons [kN]
Pressure or Stress
Weight is often distributed across a surface in foundations (e. the pressure on the base of the footing from the superstructure loads).
Pressure is a distributed form of force. A 1 tonne weight on an area, 1 metre square, is: 10 kiloNewtons / 1 m x 1 m = 10 kiloNewtons per metre squared or 10 kiloPascals [kN/m 2 or kPa]
Loads are also transmitted through a structure or through the foundation in a distributed form............. the cross-sectional area of the material or members of the structure. A distributed force is not called pressure when it is internal. It is known as Stress.
Density or Unit Weight
Unlike the superstructure, sub-structural loads are not concentrated. They usually found in the form of bulk volumes; e. over-burden.
To calculate sub-structural loads, we have to start with the Density of the material. This is a measure of the mass in a unit volume (1m x 1m x 1m). For instance, one cubic metre of soil could weigh 2 tonnes or 2,000 kilograms. Hence, the density of this soil would be: 2000 kg / 1m x 1m x 1m = 2000 kg per metre cubed [kg/m 3 ]
Engineers are more interested in the force in a volume; just as they are more interested in Weight rather than Mass. Instead of density, they prefer to work with Unit Weight. The mass in a unit volume (density) is converted to a weight per unit volume.
2000 kg/m 3 x 10 m/s 2 = 20,000 Newtons per cubic metre [N/m 3 ] or 20 kiloNewtons per cubic metre [kN/m 3 ]
Internal Friction in Soils
The term “σf tan φ" is a measure of the frictional resistance of the soil. The frictional force is related to the normal force or stress, σf, on the failure plane by the coefficient
“tan φ” of the normal force,
The term, normal force (or stress), refers to the pressure or force at right angles to the slip surface. It is the pressure that is pushing the shearing strips of soil together. The term “normal” is used instead of the vertical force (or pressure) because the slip surface is not always horizontal.
The frictional force depends on: i. the roughness of the failure surface measured by the coefficient of friction “tan φ”.
Granular soils have a high coefficient of friction; hence sandy soils have substantial frictional resistance to failure. The frictional resistance is also related to the angle of repose that a material takes up when it flows onto a heap.
Figure 14 (from allanblock) A sandy soil has a higher angle of repose than a silty or clayey soil. Clayey soils have smooth particles and sandy soils have rough particles.
ii. normal stress “σf”. The frictional strength of the soil is also affected if the soil particles are forced together. The frictional resistance increases with the load placed on the soil.
Densely packed foundations are much more stable to support a building superstructure; t hat is why builders put so much effort into compaction of the foundation material.
2 Measurement of Shear Strength
In Experiment One, we will be simulating the shear strength of a soil with polymer strips. Our simulated tests are able to demonstrate the shearing behaviour much more simply than the relevant geotechnical tests.
2 Shear Box
In a geotechnical laboratory, the strength of a soil, undisturbed or remoulded, can be measured using a shear box apparatus, shown in Figure 9. Different weights can be placed on a loading cradle to simulate different overburden pressures on the shearing soil.
The soil is progressively sheared by a piston. The Shear Box determines the shear stress at failure for a soil sample at different normal stresses. A line of best fit is drawn to the data points.
Figure 15 Direct shear apparatus (from Craig [1992], p. 105)
“c” is derived from the Shear Box experiment. It is equal to the value of the shear strength when it intercepts the y-axis of the Mohr-Coulomb curve, as shown in Figure 14. In other words, when there is no external load to create a frictional effect.
Figure 16 Mohr-Coulomb diagram
Module 2
Bearing Capacity of Shallow Footings
In the first lecture, we examined how applied loads cause shearing of the soil. The ground resists these forces through friction and cohesion between the soil particles. In other words, we were investigating the process at a very small scale.
In this lecture, we will investigating how the ground and a footing act together at a large scale; How the shearing effects on a small scale play out when we are considering an entire footing.
Next week, we will be looking at the actual design and construction of shallow footing systems.
1 Mechanisms of Failure
Footings fail by local or general bearing failure.
The maximum bearing stresses occur where the soil is being subjected to the most work ie. where it is being sheared.
Local Bearing Failure: Local Bearing Failure is confined to the region immediately beneath the footing. Loads are not transferred outward into the foundation material. The peripheral soil is not highly loaded. Local Bearing Failure can be caused by weak foundation soils or inadequate footings.
Figure 1 Local bearing failure
- The soil is too compressible. Soil structure can be disturbed by seismic activity or vibration. With compressible foundation soil, the soil is compacted under the centre of the footing. The surrounding soil is not disturbed. In an earthquake, the usual structure of the soil can be disturbed by vibration. The soil behaves as if it is highly compressible.
Figure 2 Anchorage earthquake, 1964 (from luirig.altervista)
- The footing is too flexible. Under overload conditions, the footing transfers load to the soil directly below. If the outstands of the footing is too flexible, they fold away from the peripheral soil. Little load is transferred to the periphery. In a sense, the footing acts as if it has a smaller effective width. This form of failure occurs when the footing hasn’t been properly designed for the actual loads.
General Bearing Failure: where the soil bulges to one side of the structure. This form of failure occurs with rigid footings or incompressible soils.
Figure 3 General bearing failure
This form of failure is demonstrated by the Transcona gain silo collapse in 1913 (shown in Figure 4). The failure mechanism is idealized in Figure 3. The shearing is assumed to occur
1 Zones of Bearing Failure
Figure 5 Modes of bearing failure (from Das, 1993:p. 468)
The soil under a footing is not subjected to the same conditions throughout its depth. Terzaghi identified three distinct zones under a footing.
Zone I (Elastic) soil is constrained to settle under the footing. The zone coincides with the edge of the footing at an angle of φ. In this zone, the soil is constrained on all sides. On the top, the footing bears down on it and below, the soil in Zones II locks the soil in place. This is the zone most affected during Local Bearing Failure.
Zone II (Radial Shear) The soil is sheared plastically sideways and upward. Maximum shearing occurs at the slip surface. The maximum shearing action happens towards the edge of the footing, in Zones II. In these regions, the soil is actively sheared. The maximum shearing stresses occur on the lower edge of the zones. When the footing reaches its bearing capacity, the soil will shear along this. A slip plane is created, which is roughly circular.
The bearing capacity of the foundation is determined by the cohesive and frictional characteristics along the slip plane.
The location of the slip plane can vary. If there is a weak strata in the foundation, it will be preferentially sheared and the slip plane will not be circular.
Zone III (Passive) The soil is moved upward, without deformation. The soil is not sheared during the footing failure.
Superstructure loads are distributed sideways and down from the footing. Hence, the shearing stresses rapidly diminish as we move from Zone II to Zone III. In Zone III, the soil is not being moved in an arc, just passively lifted.
The soil in Zone III can still contribute to the bearing capacity of a footing. It represents dead weight that the soil in Zone II must overcome in order to fail.
Expansion of the Footing
If you expand the size of the footing, you enlarge the size of the zones.
The footing expansion will affect the footing in two ways:
Zone II: The size of the slip surface is larger. More frictional and cohesive forces can be built up on the surface, before it fails.
Zone III: More soil must be moved passively carried upwards and sideways, as the footing fails.
Figure 6 Expansion of footing
Zone I is slightly increased when the footing is expanded. Since the zone is merely compressed, its enlargement does not affect the strength of the footing.
CT2 MOD1-4 - Combined notes
Course: Construction Technology 2 (Substructure) (300721)
University: Western Sydney University
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