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Words 15647

Pages 63

Soil Dynamics:

Soil Dynamics is the branch of soil mechanics which deals with the engineering properties and behavior of soil under dynamic stress, including the analysis of the stability of earth supported and earth retaining structures. The study of Soil dynamics include the machine foundations, impact loadings, dynamic soil properties, slope stability, bearing capacity, settlement, vibratory compaction, pile driving analysis and field testing, ground anchor systems, seismic design parameters, liquefaction, sheet pile walls and laboratory testing.

Nature/sources of types of dynamic loading: * Dynamic loads on foundation and soil structure may act due to * Earthquake * Bomb blast * Operation of reciprocating and rotary machines and hammers * Construction operation such as pile driving * Quarrying * Fast moving traffic including landing aircraft * Wind * Loading due to wave action of water * Etc * The nature of each of these loads is quite different from the nature of the loads in the other cases. * Earthquakes constitute the single most important source of dynamic loads on structures and foundation. * Every earth quake is associated with a certain amount of energy released at its source and can be assigned a magnitude (m) which is just a number. * Table gives an idea of the energy associated with a particular magnitude M (Richter) | 5.0 | 6.0 | 6.33 | 6.5 | 7.0 | 7.5 | 8.0 | 8.6 | E (1020ergs) | 0.08 | 2.5 | 8.00 | 14.1 | 80 | 446 | 2500 | 20000 |

Earthquake:

The vibration of earth that accompanies an earthquake is one of the most terrifying natural phenomena known. From geological point of view, earthquakes provide the evidences of the instability of the earth’s crust and a logical starting point for any examination of the dynamics of the earth.

Most earthquakes take place along faults in the upper 25 miles of the earth's surface when one side rapidly moves relative to the other side of the fault.

Due to ground motion during an earthquake * Footing may settle * Building may tilt * Soils may liquefy * lose ability to support structures * light structures may float

Problems of dynamic loading of soils and soil structures: * Earthquake, ground vibration, wave propagation through soil * Dynamic stress, deformation and strength characteristics * Earth pressure problems and retaining walls * Dynamic bearing capacity and design of shallow footing * Embankments under earthquake loading * Piles foundation under dynamic loads * Liquefaction of soils * Machine foundation

Basic concepts:

For a proper understanding and appreciation of the different aspects of design of foundation and soil structures subjected to dynamic loads, it is necessary to be familiar with the simple theoretical concepts of harmonic vibration.

Basic Definitions:

Vibration or Oscillation: It is the time dependent, repeated motion of translational or rotational type.

Periodic motion: It is the motion which repeats itself periodically in equal time intervals.

Period (T): The time elapsed in which the motion repeats itself is called the period of motion or simply period.

Cycle: The motion completed in the period is called the cycle of motion or simply cycle.

Frequency (f): The number of cycles of motion in a unit of time is known as the frequency of vibration. It is usually expressed in hertz (i.e. cycles per second).

The period T and the frequency f are interrelated as, T=1f

Free vibration: Free vibration occur under the influence of forces inherent in the system itself, without any external force. However, to start free vibrations, some external force or natural disturbance is required. Once started, the vibrations continue without an external force.

Forced vibration: Forced vibrations occur under the influence of a continuous external force.

Natural frequency: If an elastic system vibrates under the action of forces inherent in the system and in the absence of any externally applied force, the frequency with which it vibrates is its natural frequency.

Resonance: When the frequency of the exciting force is equal to one of the natural frequencies of the system, the amplitudes of motion become excessively large. This condition is known as resonance.

Damping: The resistance to motion which develops due to friction and other causes is known as damping. Viscous damping is a type of damping in which the damping force is proportional to the velocity.

It is expressed as, F=cdzdt

Where,

c= damping co-efficie dzdt= Velocity

Degree of freedom (n): The number of independent co-ordinates required to describe the motion of a system is called a degree of freedom. A system may in general have several degrees of freedom; such a system is called a multi degree freedom system. n=∞ Figure - 1: System with different degree of freedom (a) One degree of freedom n=1;

(b) Two degree of freedom n=2; (c) Three degree of freedom n=3;

(d) Infinite degree of freedom n=∞.

Principal modes of vibrations: A system with more than one degree of freedom vibrates in complex modes. However, if each point in the system follows a definite pattern of common natural frequency, the mode is systematic and orderly and is known as the principal mode of vibration. A system with n degrees of freedom has a principal modes and n natural frequencies.

Normal mode of vibrations: When the amplitude of some point of the system vibrating in one of the principal modes is mode equal to unity, the motion is then called the normal mode of vibration.

Resonance: When the frequency of the exciting force is equal to one of the natural frequencies of the system, the amplitudes of motion become excessively large. This condition is known as resonance. It is important to avoid or control or minimize this situation. In this condition a large magnitude of force and amplitude of motion can be generated which is destructive to the structure.

Minimization or Control of resonance:

It is important; therefore, to avoid or minimize these situations by either avoiding the equalization of the forcing and natural frequencies by use of appropriate damping mechanism that will reduce the size of the otherwise increased effects at resonance. Various damping mechanism are available, either inherent in the vibrating system or especially designed into the system. Reference will be later to specific mechanisms, but for the moment it is sufficient that damping loads to the dissipation of energy per cycle of motion and usually leads to a reduction or decay in amplitude of the motion.

Methods of avoid resonance: * Isolate resonant component * Change exciting frequency using springs, pads, pneumatic, suspending components * Use Vibration absorption * Increase system damping * Reduce forcing amplitude * Avoid forcing a system at a natural frequency

Un-damped free vibration of a spring mass system:

Figure - 2: Spring Mass System a) Un-stretched Spring b) Equilibrium Position c) Mass in Oscillating Position d) Mass in maximum Downward Position e) Mass in upward position f) Free body diagram of mass corresponding to (c)

Un-damped Free vibration of a spring mass system:

Figure - 3: Free Vibration of a mass-spring system

Figure - 3 shows a foundation resisting on a spring. Let the spring represent the elastic properties of soil. If the area of the foundation is equal to A, the intensity of load transmitted to the sub grade q=WA

The static deflection zstat of the spring is, zstat=Wk=mgk

Where,

k = Spring constant (k is defined as force per unit deflection)

W = Weight of mass zstat = Static Deflection

If the foundation is disturbed from its static equilibrium position, the system will vibrate. According to Newton’s second law of motion,

Wgz+kz=0

or, z+kmz=0 ……………………... (1)

Where,

g = acceleration due to gravity z = Displacement z=d2zdt2=Acceleration t = time m = mass = Wg

In order to solve the equation (1), we get z=Asin(ωnt)+Bcos(ωnt) ……………………... (2)

Where, A and B are arbitrary constant and ωn is the circular natural frequency of system.

Now, z=dzdt=ωnAcos(ωnt) - ωnBsin(ωnt) z=d2zdt2=-ωn2Asin(ωnt)-ωn2Bcos(ωnt) = -ωn2Asin(ωnt)+Bcos(ωnt) = -ωn2z z=Asin(ωnt)+Bcos(ωnt) ……………………...(3)

From equation (1) and (3) we can write, -ωn2z +kmz=0 or, -ωn2+kmz=0 or, ωn2= km or, ωn= km ωnt Figure - 4: Plot of Displacement, amplitude and cycle for the free vibration of mass spring system

From figure - 4 the magnitude of maximum displacement is equal to z. This usually referred to as the single amplitude. The peak to peak displacement amplitude is equal to 2z. The time required for the motion to repeat itself is called the period of the vibration. When the time required to complete one cycle of motion is ωnTn=2π, the time period T of this motion can be written as

Tn=2πωn=2πmk

Frequency, f= 1T =12πkm=12πkgmg= 12πgmgk= 12πkgzstat

Example: 1

A mass supported by a spring has a static deflection of 0.5 mm, Determine its natural frequency of vibration.

Solution:

Given, g = 9810 mm/sec2 zstat = 0.5 mm f=12πgzstat=12π98100.5=22.32 Hz (Ans)

Example: 2

Find the fundamental frequency of vibration of a vertical cantilever as shown in figure (5) that supports a mass m which is large relative to total mass of the cantilever. Data - mass, m = 1000 kg, length, L= 20 m, Flexural rigidity, EI = 125×102 N-m2.

L = 20 m

EI = 125×102 N-m2

W = 1000 Kg

Figure - 5: Vertical Cantilever supporting mass

Solution:

Distribution mass of cantilever can be ignored. The lateral stiffness, k of the cantilever at the level of m is, k=3EIL3

Now, frequency, f= 12πkm=12π3EImL3=12π3×125×1031000×203=1.1 Hz (Ans)

Example: 3

Determine the spring constant for the system of springs shown in figure

Figure - 6: Equivalent Spring Constant

Solution:

Let us consider that a unit load is applied at c.

It is shared at a and b in the ratio of x2x1+x2 and of x1x1+x2 respectively.

The deflection of points a and b are of x2x1+x2×1k1 and x1x1+x2×1k2 respectively.

Therefore, deflection of point c is. x1x1+x2×1k2+ x2x1+x2×1k1-x1x1+x2×1k2x2x1+x2

= 1(x1+x2)2x12k2+x1x2k2+x22k1-x1x2k2

= 1(x1+x2)2x12k2+x22k1

Hence, the resulting equivalent spring constant at c is, keqv=(x1+x2)2x12k2+x22k1

If x1 = x2 = x and k1 = k2 = k, then keqv=2k

On the application of a unit load in figure b the total deflection is, 1k2+1k1=k1+k2k1k2

Hence, equivalent spring constant, keqv=k1k2k1+k2

If k1 = k2 = k, then keqv=k2

Example: 4

Write the equation of motion for the systems shown in figure (7) and determine the natural frequencies.

Figure - 7: Equivalent Spring Constant

Solution:

On application of unit load at c for spring k1 and k2 the deflection

= 1(x+x)2x2k1+x2k2 =x2(2x)21k1+1k2 =14k2+k1k1k2

Hence, equivalent spring constant for k1 and k2, k'eqv=4k1k2k1+k2

Total deflection is 1k'eqv+1k3=k2+k14k1k2+1k3 = k3(k1+k2)+4k1k24k1k2k3

Hence, the equivalent spring constant, keqv=4k1k2k3k3(k1+k2)+4k1k2

We have, z=dzdt=ωnAcos(ωnt) - ωnBsin(ωnt)

Where, ωn=keqvm=4k1k2k3mk3(k1+k2)+4k1k2 A & B = Arbitrary constant

Example: 5

We will consider the motion of the piston of a reciprocating machine on a soil foundation. The soil displacement of the position from the extreme position is

S=l+r-rcosθ-lcosθ

Also, lsinϕ=rsinθ ϕ lcosϕ rcosθ θ

Figure - 8: Motion of a Piston of a reciprocating machine

Find the equation of velocity and acceleration.

Solution: cosϕ=(1-sin2ϕ)12= 1-rlsin2θ12 sinϕ=rlsinθ

If we expand, the right hand side, by the binomial theorem we get, cosϕ=1-12rl2sin2θ - 18rl4sin4θ-116rl6sin6θ ……………….. or, 1- cosϕ=12rl2sin2θ+ 18rl4sin4θ+116rl6sin6θ ………………..

Substituting cosϕ in the expression for S, we get, S=r1-cosθ+ l(1-cosϕ) =r1-cosθ+12rlsin2θ+ 18rl3sin4θ+116rl5sin6θ ………………..

The series in brackets contains sin2θ and even powers of sinθ.

Now, sin2θ=121-cos2θ sin4θ=sin2θ2=141-cos2θ2 =141-2cos2θ+cos22θ =141-2cos2θ+12+12cos4θ =1432-2cos2θ+12cos4θ

Now, S=r1-cosθ+14rl1-cos2θ+ 132rl332-2cos2θ+12cos4θ………………..

If rl is very small, rl2 and higher powers can be neglected.

S=r1-cosθ+lr4l1-cos2θ

Putting θ= ωt that S is a function of cosωt and cos2nωt where n = 1, 2, 3 …………….

S=r1-cosωt+lr4l1-cos2ωt

Velocity,

S=dsdt=rωsinωt+r2lsin2ωt

Acceleration,

S=d2sdt2=rω2cosωt+rlcos2ωt

Damped Free Vibration or Free Vibration with Viscous Damping

Field frictional resistance to motion of a body produces a viscous damping force that is directly proportional to relative velocity, when it is low. At higher relative velocities, the damping force may be proportional to the square of the velocity. Internal frictional resistance of materials, associated with the internal molecular structure, will also lead to decay of vibration.

If the force of damping Fd is proportional to velocity, it is termed viscous damping,

Thus Fd=cx ………….(1)

Where c is a constant of proportionality, referred to as the viscous damping co-efficient. mx Figure - 9: Spring mass Dashpot system

Figure (9) shows a single degree of freedom oscillator to which is added a dashpot that induces the damping forces. From the force body diagram the equation of motion is mx+ cx+ kx=0 or, x+ cmx+ kmx=0 ………….(2)

If we define a critical viscous damping co-efficient, cc is, cc=2km ...………..(3)

And a damping ratio, D= c cc ...………..(4) cD= cc=2km= 2mkm=2ωnm or, cm=2Dωn where, ωn2=km

So equation (2) becomes, x+ 2Dωnx+ωn2x=0 ...………..(5)

The general solution of this equation is, x=Aer1t+ Ber2t ...………..(6)

Where A & B are constants to be determined from the initial conditions at x(0) and x(0). r1,r2 are the roots of the auxiliary equation, r1,2= -ωnD±D2-1

The three cases,

D = 1.0 for Critically Damped Condition

D>=1.0, Over Damped Condition

D<1.0, Under Damped Condition

Figure - 10: Free Vibration of a spring mass Dashpot system (a) Over Damped case, (b) Critically Damped Case, (c) Under-Damped case.

If, D = 1, r1=r2=-ωn

The x=(A+Bt)e-ωnt if roots are equal.

By the initial condition i.e. x0=Ax and x0= 0, we get the value of A & B. x=Axe-ωt gives a periodic motion or non-oscillatory motion i.e. D=1 or D>1 are case of over Damped condition.

For damping, the motion or velocity comes rapidly to zero.

If D=1, the system is under damped, i.e. gives oscillatory motion.

Then

r1=-ωnD±D2-1 r2=-ωnD-D2-1

Where, r1 & r2 are complex and conjugate roots.

For the initial conditions x0=Ax and x0= 0, the solution of equation (7) gives, x=Axe-ωnDt(1-D2)sin∅cosωn(1-D2)t+cos∅sinωn(1-D2) or, x=Axe-ωnDt(1-D2)sinωn(1-D2)t+∅

Where, sin∅= (1-D2) and cos∅=D and ∅ is the phage angle.

The motion is oscillatory with exponentially decaying amplitude, the general nature of which shown in the following figure (11)

xAx

Figure - 11: Damped Free Oscillation

Damping ratio (D) or Damping factor (D) and its significance:

Damping factor may be defined as The ratio of the actual damping coefficient (C) to the critical damping coefficient (Cc).

Mathematically, D=Actual DampingCritical Damping=CCc

Force

or, D=Energy absorbedEnergy applied

Deformation

Figure - 12: Energy absorption of Materials

As usual case, D<1, for soil, D= 0.1~0.5

If D>1, the system is over damped and non-oscillatory and motion is a periodic motion.

If D=1, the system is critically damped and non oscillatory and motion is non periodic.

If D<1, the system is under damped and oscillatory and motion is periodic.

Only under-damped systems are of practical importance in the design of machine foundation.

(a) (b)

Figure - 13: (a) Un-damped Vibration (Amplitude vs Time); (b) Damped Vibration (Amplitude vs Time)

Since for over damped, the motion is a periodic, the system returns to its original position in the minimum time, when critically damped.

Figure - 14: Free vibration with various value of D

The envelope curve and the oscillatory curve have a common point where the curve have the same slope; this point does not correspond to the maximum excursion for the mean position, which occurs when x=0, the difference is small for small values of the damping ratio, D.

Natural frequency or Un-damped natural frequency (ωn) and Damped frequency (ωnd) and operating frequency (ω) and their significance:

An elastic system vibrates under the action of forces inherent itself in the system and in the absence of any externally applied force, the frequency with which vibrates is its natural frequency or un-damped natural frequency (ωn).

Figure - 15: Spring mass with Dashpot system

Mathematically, ωn= km

Where,

k = Stiffness of spring and M = Mass of System

Vibrations that occur under the excitation of externally applied forces are termed forced vibrations. Forced vibrations occur at the operating frequency (ω) of the exciting force. The operating frequency (ω) is independent of the natural frequency (ωn) of the system. But when, ω= ωn, the response of the system is infinite. The condition is known as resonance. Damped frequency (ωnd) may be defined as ωnd=ωn1-D2.

Where, D = Damping factor =Actual DampingCritical Damping=CCc

Always damped frequency (ωnd) is less than the un-damped natural frequency (ωn). However for small values of D, the influence of D is not great. As an ideal un-damped system is non-existent, damping always exists and the response is finite. However, when operating frequency (ω) is close to the natural un-damped frequency(ωn), the response is very high. To avoid this condition, the operating frequency should not be close to the natural frequency. For a safe design, the frequency ratio (ωωn) is normally kept outside the critical range of 0.40 to 1.50.

Non Oscillatory and critically damped motion, Damped force vibration and their significance:

When D>1, the system is over-damped, the motion of the system is called non-oscillatory or a periodic motion.

When D=1, the system is critically damped, the motion of the system is called critically damped motion.

When D<1, the system is under-damped, the motion of the system varies sinusoidal is called damped forced vibration.

Where, D = Damping factor =Actual DampingCritical Damping=CCc = C2ωnm= C2km

D=∞

Figure - 16: Spring mass with dashpot system and free vibration with various value of D

It is not intended to investigate the algebra of these above cases as they are rarely of interest in machine foundation problems.

Figure (16) shows the behavior of these cases for the initial conditions x0= Ax and x0=0.

In the case of critical damping, the mass, m returns to its equilibrium position in the shortest possible time. Critical frequency may be defined as the frequency factor ωωn=1 is one, i.e. the case where the operating frequency (ω) of the system is equal to natural frequency(ωn). At this case, the system reaches at resonance condition where higher or uncontrolled velocity of the system is created. For machine foundation design, the case is unexpected.

Critical damping may be defined as the damping factor D=CCc=1 is one, where actual damping and critical damping are equal. At this case, the system returns to its equilibrium position in the shortest possible time. For machine foundation design, this case is expected.

So the above two situations are opposite to each other. ωn= km Cc=2km

Logarithmic Decrement and the experimental Determination of Damping:

Logarithmic decrement is a measure of the decay of successive maximum amplitude of viscously damped vibration and is expressed as

δ= logexmxm+1 …………… (1)

In which xm and xm+1 are two successive peak amplitudes, from a displacement verses time trace will enable the damping constant to be estimated. xm+1 xm

Figure - 17: Amplitude of Viscously Damped vibration

Here m is an arbitrary cycle number.

If xm occurs at time t, then from oscillatory damping motion equation, we have xm=Axe-ωnDt(1-D2)sinωn(1-D2)t+∅ …………….. (2)

Where, ∅=tan-1 1-D2D

At a time interval of one periodic later, T=t+ 2πωn(1-D2) ……………. (3) xm+1=Axe-ωnDt - 2πD(1-D2)(1-D2)sinωn1-D2(t+T)+∅ .…………….. (4)

The values of the sine functions are equal when the time is increased by the period, T, so that the ratio of the displacements becomes, xmxm+1= e 2πD(1-D2) …………….(5)

So,

δ= logexmxm+1=logee 2πD(1-D2)=2πD(1-D2) …..………… (6)

For small values of D, the equation (6) can be expressed in series form, δ= 2πD(1-D2)=2πD(1-D2)-12= 2πD1+ D22+…≈ 2πD

Hence,

δ=2πD ……………..(7)

In practice, when interpreting a vibration trace or record, it is better to find the number of cycles required for the amplitude to reduce to half its value.

Consider first the ratio of x0xn= 21 after n cycles,

Now,

x0xn= x0x1, x1x2,x2x3………………xn-1xn= eδn=eδn

From which is found, δ= 1nlnx0xn ……………..(8)

For the case, when x0xn= 21 lnx0xn=ln2=0.693

And, nδ ≈2πnD=lnx0xn=0.693

Therefore,

nD=0.11= 0.6932π

If D=0.1 n=1.1

D=0.2 n=0.55

D Depends on materials, for soil If D=0.1∼0.5.

Determination of Viscous Damping (by Bandwidth Method)

Damping can be determined from either a free vibration or a forced vibration test on a system. In a free-vibration test, the system is displaced from its equilibrium position and a record of the amplitude of displacement is made. Then we have, δ= 2πD(1-D2)=2πD(1-D2)-12= 2πD1+ D22+…≈ 2πD D= δ2π=12πlogexmxm+1 ……………………. (1)

D= δ2π=12nπlogex0xn ……………………. (2)

In a forced-vibration test, the system may be excited with a constant force of excitation and varying frequencies. Figure (18) shows a resonance curves.

Figure - 18: Determination of damping ratio from forced vibration

The amplitude of the motion is Axxs= 11-ωωn22+2Dωωn2 ..………………….. (3)

Putting, frequency ratio, ωωn=1, We have, Axxs=12D ………………………(4)

Putting, the frequency ratio r=Axxs=ffn, when the amplitude of motion =0.707×12D=Axxs,

We get, 0.7072D=11-r22+2Dr2 or, 0.7072D=11-r22+2Dr2 or, 122D=11-2r2+ r4+4D2r2 or, 1-2r2+ r4+4D2r2=8D2 or, r4-2r21-2D2+1-8D2=0 or, r21,2 =1221-2D2±41-2D22-41-8D2 =1221-2D2±4-16D2+16D4-4+32D2 =1-2D2±2d1+D2

Now,

r22- r12=4D1+D2≈4D If D is very small.

Now,

r22- r12= f22-f12fn2=f1+f2fn f1-f2fn= 2f2+f1fn.1 Since, f1+f2fn=2

So, 4D=2f2+f1fn or, D= 12f2-f1fn

This method for determining viscous damping is known as the bandwidth method.

Forced Harmonic Vibration with Viscous Damping

The harmonic force F0cosωt acts on the system which has a spring constant k and a viscous damping constant c as shown in figure (19). The circular frequency of the force application is ω, and the amplitude is constant F0

.mx

F0cosωt

F0cosωt

Figure - 19: Forced Harmonic Vibration with Viscous Damping

The equation of motion is found from the free body and is written as mx+ cx+ kx=F0cosωt=ReF0eiωt ……………………….(1)

The solution of the equation, Re = Real mx+ cx+ kx=0 is, x=Aeiωt ……………………….(2)

Where, A is an arbitrary constant.

Equation (2) substituting into equation (1), we have x=ReF0eiωtmiω2+ciω+k or, x=ReF0keiωt1-ωωn2+ i.2D.ωωn

Where, ωn= km and D= c2km

The static deflection is xs= F0k and phase angle is, ∅= tan-12Dωωn1-ωωn2

The phase angel is the angle, by which the response lags behind the disturbing fore, x=Rexsei(ωt-ϕ)1-ωωn22+ 2D.ωωn2 =Rexs(cosωt-ϕ)1-ωωn22+ 2D.ωωn2

The amplitude of the motion is Ax=xs1-ωωn22+ 2D.ωωn2 Axxs=M=Magnificaiton factor Axxs=fωωn, D ϕ=fωωn, D Axxs=1.0 for static case.

Magnification factor, M= Ax xs

Frequency ratio, ω ωn

Frequency ratio, ω ωn

Figure - 20: Magnification factor and phase angle in forced Vibration

Significance of the above figures: * The damping ratio D has a large influence on the amplitude Ax and phase angle ϕ in the frequency region near resonance when ωωn≈1. * The maximum magnification factor (M) occurs for ωωn<1 when damping is present. * When the damping is small, D≪1, the amplitude and the phase angle are almost independent of D. * The damping and inertia forces are then very small, so that the exciting force is almost equal to spring force. * When ωωn is close to unity, the damping force and exciting force are almost equal. The spring force and inertia force are almost balanced. * When the frequency ratio ωωn≫1, the phase angle (ϕ) approaches 180°. * The exciting force then almost equals to the inertia force. * The amplitude approaches the static displacement when ωωn≪1. * The amplitude becomes small when ωωn≫1. * The phase angle (ϕ) is very sensitive to the ratio ωωn in the region of near resonance for small damping. * When ωωn=1-2D2, then Axxs reaches a maximum. * The amplitude at resonance is found to be Ax=xs2D.

Forced Vibration with Viscous Damping (Derivation for maximum magnification)

A spring-mass-dashpot system under the action of a force of excitation F such that, F=F0sinωtsinωt …………………… (1)

Where,

ω= Frequency of force of excitation.

The equation of motion for free-body, mx+ cx+ kx=F0sinωtsinωt …………………….(2)

The solution of the equation is, x=Axsinωt-ϕ …………………….(3)

Where,

∅=Phase Angle

Equation (3) substituting into equation (2), we have Ax=F0k1-ωωn22+ 2D.ωωn2 And, ∅= tan-12Dωωn1-ωωn2 or, AxF0k=11-ωωn22+ 2D.ωωn2

Where, D=Damping Factor= c2km ωn=Natural frequency= km M=Magnification factor=fωωn,D

For maximum value of magnification factor, Mmax ∂m∂ωωn=0 or, 1×12×1-ωωn22+ 2D.ωωn2×21-ωωn-2ωωn.1+2.2D.ωωn 1-ωωn22+ 2D.ωωn22=0 or, 0-41-ωωn2×ωωn+8D2ωωn=0 or, 1-ωωn2=2D2 or, ωωn= 1-2D2 or, fmfn=1-2D2 or, fm=fn1-2D2

Where,

ω=2πfm and ωn=2πfn

Now,

1-ωωn22=1-1-2D22=4D4

And 2Dωωn2=4D2ωωn2=4D21-2D2=4D2-8D4

So,

Mmax=11-ωωn22+ 2D.ωωn2 = 14D4+4D2-8D4 = 14D21-D2 =121-D2=fD

So,

Mmax=121-D2=fD or, ωωn= 1-2D2=fD or, fm=fn1-2D2

fm≈fn, for maximum vibration, which is always avoided.

xs=F0k=Static deflection

Magnification factor, M= Ax xs

Frequency ratio, ω ωn

Figure - 21(a): Plot of Magnification factor vs Frequency ratio in forced Vibration

Frequency ratio, ω ωn

Figure - 21(b): Plot of phase angle vs frequency ratio in forced Vibration

Effect of frequency ratio, r= ωωn for a particular case

When D=0,

If, r=0, M=1

If, r=1, M=∞

If r=∞, M=0

At r=1, resonance occurs and amplitudes tend to infinity.

The introduction of damping reduces the amplitudes to finite values.

The phase angle ϕ is zero if r<1; the displacement, x is in the phase with the exciting force, F0 and ϕ=180° if r>1.

Effect of Damping factor:

As the damping increases, the peak of the magnification factor M shifts slightly to the left. This is due to the fact that the maximum amplitudes occur in damped vibrations when the forcing frequency ω equals the system’s damped natural frequency ωnd, which slightly smaller than the un-damped natural frequency(ωn).

For r=1, the phase angle ∅ is 90° for all values of damping, D except when D=0.

For r<1, the phase angle is less then 90°

For r>1, the phase angle is greater than 90°

The maximum amplitude of motion at r=1 and D>0, is Ax=F0cω .

Force transmitted to foundation:

The force FT transmitted to the foundation by the spring and damper system is given by: FT= cx+ kx ………………..(1)

And after substitution from equation x=ReF0keiωt1-ωωn2+ i.2D.ωωn

And, xs=F0k

F0cosωt

Figure- 22: Force transmission by spring damper

We have, FT=Rexsk+iωt1-ωωn2+ i.2D.ωωn ………………….(2)

And as ckω= c.2.ω2km.km=2cccωωn or, ckω= c.2.ω2km.km or, ckω= 2Dωωn ………………… (3)

FT=Re.F0k.k 1+i.ck.ωeiωt1-ωωn2+ i.2D.ωωn or, FT=Re.F0 1+i.2D.ωωneiωt1-ωωn2+ i.2D.ωωn …………………(4)

The amplitude of the force transmitted is given by FT=F0 1+2D.ωωn21-ωωn22+ 2D.ωωn2 or, FTF0= 1+2D.ωωn21-ωωn22+ 2D.ωωn2

The ratio FTF0 is referred to as the transmissibility of the force. TF= FTF0=fωωn,D

Frequency ratio, ωωn

Force transmissibility, TF=FTF0

Figure-23: Plot of force transmission and frequency ratio

Now, the ratio TF= FTF0=1, when

1-ωωn22+ 2D.ωωn2=1+2D.ωωn2 or, 1-ωωn2= ±1 or, ωωn2=1±1=0,2 or, ωωn=0,2 or, ωωn>2 then TF<1

If D≪1, then TF≪1

Some damping is needed when ω must pass through the resonant condition, where a magnification will occur.

If D is negligible and ωωn>2, then TF≪1.

Some damping is needed when ω must pass through the resonant condition, where a magnification will occur.

If D is negligible and ωωn>2, then

TF=1ωωn2-1

Replacing, ωn2=gxs and ω=2πf, we have

TF=12πf2xsg-1

Where xs=ωk

If k is small, the xs is large. fm=fn1-2D2=0 1-2D2=0

D= 12=0.707

ω=0

F0sinωt=0 is a static case.

Which means that the maximum response is the static response.

f=Axcd cd=1-0.6zb Where, z=Depth of foundation from ground level b=a1+a22=Average of width and langth of foundation

Example-6:

A machine of mass 100 kg is supported directly on springs which have a total stiffness of 2000 KN/m. An unbalanced rotating mass results in a disturbing force of 200N at a speed of 3000 rpm. Assuming a damping factor of D = 0.10 determine the amplitude of vibration, the transmissibility, and the force transmitted to the foundation.

Solution:

The static deflection of the system is xs= wk= mgk= 100×9.812000×103=0.49×10-3 m=0.49 mm

The natural frequency of the single degree of freedom system is fn= 12πkm =12π2000×103100=22.5 Hz

The ration of forcing to the natural frequency, then becomes ωωn=ffn= 300060×122.5=2.22

The amplitude of vibration, Ax=xs1-ωωn22+ 2D.ωωn2=0.491-2.2222+2×0.10×2.222= 0.124 mm Ans.

The transmissibility, TF=FTF0= 1+2D.ωωn21-ωωn22+ 2D.ωωn2= 1+2×0.1×2.2221-2.2222+ 2×0.1×2.222=0.277 Ans.

The transmitted force is then FT= TF×F0=0.277×200=55 N Ans.

Example-7:

A machine of mass 100 kg is supported directly on springs which have a total stiffness of 100 KN/m. An unbalanced rotating mass results in a disturbing force of 200N at a speed of 3000 rpm. Assuming a damping factor of D = 0.10 determine the amplitude of vibration, the transmissibility, and the force transmitted to the foundation.

Solution:

The static deflection of the system is xs= wk= mgk= 100×9.81100×103=9.8×10-3 m=9.81 mm

The natural frequency of the single degree of freedom system is fn= 12πkm =12π100×103100=5.03 Hz

The ration of forcing to the natural frequency, then becomes ωωn=ffn= 300060×15.03=9.94

The amplitude of vibration, Ax=xs1-ωωn22+ 2D.ωωn2=9.811-9.9422+2×0.10×9.942= 0.10 mm Ans.

The transmissibility, TF=FTF0= 1+2D.ωωn21-ωωn22+ 2D.ωωn2= 1+2×0.1×9.9421-9.942+ 2×0.10×9.942=0.0227 Ans.

The transmitted force is then FT= TF×F0=0.0227×200=4.55 N Ans.

Example-8:

An unknown weight W attached to the end of an unknown spring k has a natural frequency of 95 cpm. If 1 kg weight is added to W, the natural frequency is lowered t o75 cpm. Determine the weight W and spring constant K.

Solution:

Let,

m=w kg m'=w+1kg fn=12πkm or, 95=12πkm or, km=95×2π or, km=95×2π2 ……………………………(1)

Again,

fn'=12πkm' or, 75=12πkm' or, km'=75×2π or, km'=75×2π2 ……………………………(2)

Equation (1)(2) we get, m'm= 95×2π275×2π2=1.60 or, w+1w=1.60 or, w=1.67 kg Ans.

Again, km= 95×2π2 or, k= 1.67 95×2π29.81×1000=60.65 KN/m Ans.

Exercise - 1:

A body 65 kg weighing is suspended from a spring which deflects 1.57 cm under the load. It is subjected to a damping effect adjusted to a value 0.25 times that required for critical damping. Find the natural frequency of the un-damped and damped vibrations and the latter case; determine the ratio of successive amplitudes. If the body is subjected to a periodic disturbing force with a maximum value of 25 kg and a frequency equal to 0.75 times the natural un-damped frequency, find the amplitude of forced vibration and the phase difference with respect to the disturbing force.

Solution:

k= F0xs= 65×9.81×1001.57=40615 Nm

Natural frequency of the un-damped,

fn=12πkm = 12π4061565=398 Hz (Ans)

Again,

ωnd= ωn1-D2 or, 2π fn= 2πfn1-D2 or, fnd= fn1-D2=3.98×1-0.252=3.85 Hz Ans

The ratio of successive amplitudes xmxm+1=e2πD1-D2= e2π0.251-0.252=5.06 ≈5 Ans Amplitude Ax=xs1-ωωn22+ 2D.ωωn2=15.71-0.7522+2×0.25×0.752=27.2 mm Ans

Phase Angle ∅= tan-12Dωωn1-ωωn2= tan-120.250.751-0.752= tan-10.875=45° (Ans)

Tensional Resonant Column Test

The shearing strain on a circular cross section in a tensional resonant column test varies from zero at the centre to a maximum at the outer edge.

To study the influence of shearing strain amplitude on shear modulus and damping a hollow cylinder apparatus shown in figure (24) with a configuration.

θ J,t)

≪14 sine wave θ J,t) θ Figure - 24: Tensional Resonant column test apparatus

The average shearing strain on any horizontal cross section is not greatly different form the maximum or minimum and shearing strain is uniform along the height of the specimen.

Figure (24) also increased the torque capacity of the device to produce, large shearing strain amplitudes,

For

Clay shearing strain amplitude up to 1%

Sand shearing strain amplitude up to 5% (for 40psi or 276KN/m2 confining pressure.)

Cyclic tri-axial compression tests

Cyclic test permit evolution of modulus, either E or G as appropriate for the specific test configuration and material damping. The field condition to be reproduced in a cyclic loading test is shown conceptually in figure (25) but because apparatus that could produce these stress conditions did not exist in the early stages of cyclic soil testing, the cyclic tri-axial compression test was developed first. τ τ σv k0σv τ τ σv k0σv

Figure - 25: Cyclic shear test

In this test cylindrical tri-axial samples are initially consolidate under a cell pressure, σa resulting in stress shown by condition 1, figure (26) τmax σa σa σ0±σdc2 σ0±σdc2 σa-σdc2 σa-σdc2 Figure - 26: Stress Condition

In principle, the sample is then subject to an increase in axial stress σdc2 and a simultaneous reduction in the cell pressure by an equal amount (condition -2, Figure-26). The normal stress on the 45° plane through the sample is not changed but a shearing stress of σdc2 is developed on that plane. The axial stress and cell pressure are then simultaneously reversed by σdc2 so that the shearing stress reverses on the 45° plane while the normal stress remains the same. These stress conditions are intended to be similar to those experienced on a horizontal plane in an element of soil in the field. For convenience the test is normally performed by maintaining the cell pressure at a constant value and cycling the axial stress by ±σdc as shown in figure (27).

σ3 σ3 σ3 σ3 ±σdc

Figure - 27: Constant Cell Pressure

Pore Pressure

Vertical Deformation

Vertical Load

Figure - 28: Vertical load, vertical deformation and pore pressure record as a function of the number of cycles of load

The technique results in essentially the same stress conditions as long as the sample is saturated and tested un-drained. If samples are partially saturated or tested with drainage, it is necessary to utilize axial and lateral stress control to simulate earthquake loading. In the many versions of the cyclic tri-axial test, the configuration of the specimen is standard but the loading and control equipment are variable. Most currently used apparatus are stress controlled devices in which a cyclic axial load is applied to an un-drained specimen.

Vertical load, vertical deformation and pore pressure are recorded as a function of the number of cycles of load (figure - 28). Some of the more common load control systems are the pneumatic, hydraulic, electro hydraulic and the pneumatic hydraulic.

In addition to liquefaction characteristics of soils, Young’s modulus, E, and damping ratio D are often measured in the cyclic tri-axial test (Figure - 29) by performing strain controlled tests. These tests are performed in essentially the same manner as the stress controlled test, however, a servo system is used to apply cycles of controlled deformation.

Vertical Stress,

Vertical Strain

-εv

+εv

Compression

Tension

-∆σv

+∆σv

D= Energy AbsorbedEnerdy Applied= 12π

Figure - 29: Stress-Strain plot of strain controlled test

Young’s modulus E is determined from the ratio of the applied axial stress to axial strain.

For strained strain controlled tests, shear modulus G is computed from, G= E21+γ in which γ is Poisson’s ratio.

The cyclic tri-axial test has limitations among which are: 1. Shearing strain measurements below 10-2 percent are difficult to achieve. 2. The extension and compression phases of each cycle produce different results, therefore hysteresis loops are not symmetric in strain controlled tests and samples tend to neck in stress controlled tests. 3. Void ratio redistribution occurs within the specimen during cyclic testing. 4. Stress concentrations occur at the cap and base of the specimen and the major principal stress changes direction by 90° during test.

Dynamic Soil Properties The major dynamics properties are: * Shear strength (Su) evaluated in terms of strain rates and stress - strain characteristics * Dynamic moduli, Young modulus E(Ev,Eh), Shear modulus G(Gmax) and Constrained modulus * Poisson’s ratio ν(vh, νhh, νhv) * Damping (D) Ev and Eh are equal for isotropic behavior of soil. Normally, Ev and Eh are not equal because of fabric, characteristics of grains, particles orientation in a grain mass of soil and geometric history of loading on soil mass. Always, Ev > Eh .

The majority of the numerous analytical methods presently available for assessing the response of soil deposits or soil structure due to earthquake, explosion or machine loading require and accurate assessment of maximum shear modulus (Gmax) in the field, where Gmax is defined as the shear modulus at shearing strain amplitude less or equal than 0.001 percent.

Also E is defined as the elastic Young’s modulus at strain amplitudes less or equal than 0.001 percent.

G=E21+ν Ev= σvΔεh

The ν is defined as the ratio of horizontal strain to vertical strain at shearing strain amplitudes less or equal than 0.001 percent. νvh= ΔεhΔεv σv εv

Figure - 30: Strain Stress curve

Dynamic soil test

The following tests are applied in laboratory or field for measuring the soil parameters: Field test Parameter Obtained

Wave propagation test E,G,ν

Cross bore hole wave propagation

Up hole or down hole wave propagation

Surface wave propagation

Block Vibration test Cu, D, E, G, Cψ

Cyclic Plate load test Cu,E, G Laboratory Test

Cyclic tri-axial test E,G,

Wave Propagation test/wave velocity method E,G,ν

Resonant column test E,G,D

Cyclic Simple shear test

Cyclic Tensional Simple shear test

X-ray diffraction analysis/Ultra sonic pulse test

Air pollution method

Water submergence method

Wet compaction method

Single amplitude axial strain test:

∆σv ≠0 σh σh=0 σh εv

εv σv Figure - 31: Strain and stress have same direction

The stiffness at which 0.001% strain is called young modulus, Ev= ∆σv∆εv

E or G logεalogν log(E) log(σv) EeqEmax or GGmax ε or ν

Figure - 32: Shear modulus and Shear strain relationship

Bore hole wave propagation:

In the method, the velocity of wave propagation from one surface boring to a second subsurface boring is measured. At least two bore holes are required, one for the impulse and one or more for sensors. As shown in figure the impulse rod is struck on top, causing an impulse to travel down the rod to the soil at the bottom of the hole. The shearing between the rod and the soil creates shear waves that travel horizontally through the soil, to the vertical motion sensor in the second hole, the time required for a shear wave to traverse this known distance is measured. There are four sources of major concern is conducting cross-bore hole shear test: * The bore holes * The seismic sources * The seismic receiver * The recording and timing equipment

Figure - 33: Schematic diagram of cross hole seismic survey technique

Major criteria for a seismic source are: * It must be capable of generating predominantly one kind of wave. * It must be capable of repeating desired characteristics at a predetermined energy level. Travel time =t= t2-t1 Distance =d Velocity= dt cm/s Now, Shear modulus = G= ρv2 Young modulus = E=2ρv2(1+ν) Where, ν = Poisson’s ratio of soil ρ = mass density of soil v = velocity of shear wave

Up-hole or down-hole wave propagation method Up-hole and down-hole tests can be performed by using only one bore hole. In the up-hole method, the sensor is placed at the surface and shear waves are generated at various depths within the bore hole. In the down-hole method, the excitation is applied at the surface and one or more sensors are place at different depth (figure - 34) within the hole. Both the up-hole and the down-hole methods give average values of wave velocities for the soil between the excitation and the sensor is one sensor is used or between the sensors, if more than one is used in the bore hole.

Figure - 34: Schematic diagram of up bore-hole and down bore-hole technique for wave propagation method

Up bore-hole method Down bore-hole method vu= v1+v22 vd= v1'+v2'2 D= 12f2-f1fn Shear modulus G=ρvu2 G= ρvd2 Young modulus E=2ρvu2(1+ν) E=2ρvd2(1+ν) Where, ν = Poisson’s ratio of soil ρ = mass density of soil v = velocity of shear wave

Figure - 35: Equipment and instruments of down hole survey

Surface wave propagation method: The Rayleigh wave (R-wave) travels in a zone close to the surface. An electronic or other harmonic vibrator can be used to generate a steady state R-wave and the ground surface can be deformed as shown in figure-36. Q0sinωt

Figure - 36: Deformed shape of half space surface One ray is down away from the centre line of the oscillator. One of the geophones connected to the horizontal plate of the oscilloscope is fixed 30 cm away from the oscillator along a ray drawn so that the sensing axis of the geophone is vertical. A similar geophone connected to the vertical plates of the oscilloscope, is moved along this ray, away from the oscillator. The sensing axis of the geophone is kept vertical until the Lissajous figure on the oscilloscope screen becomes a circle. However, if the phase angle is deferent than 90°, the Lissajous figure is an ellipse, and for zero phase angle it is a straight line. The distance, S between the two geophone is measured. This distance is then measure of the wave length of the generated R-wave. The test is repeated with the oscillator’s other frequencies of operation. In cases where uniform soils extend to infinite depths and the Lissajous figure is a circle, the wave length, λ of propagating waves is given by, λ=4S Velocity of shear waves, vs= λf In which f is the frequency of vibration at which the wave length has been measured. E and G of soil medium are calculated as follows E=2ρvu2(1+ν) G=ρvu2 D= 12f2-f1fn Where, ρ = mass density of soil vs = velocity of shear waves ν = Poisson’s ratio of soil Value of Poisson’s ratio of soil 0.05 for clay 0.03 - 0.35 for sand 0.15 - 0.25 for rock If compression waves are used, the vc= St In which vc = velocity of shear waves S = Distance T = corresponding time of travel of wave Then Elastic modulus E is determined by: E= vcρ1+ν1-2ν1+ν Block Resonance Test Block resonance test for determining modulus and damping values. A standard block 1.5×0.75×0.70 m high is cast either at the surface or in a pit 4.5×2.75 m at a suitable depth (figure - 37) and is excited in both horizontal and vertical modes. Figure - 37: Block Resonant test setup Forced Vertical Vibration Test: For the vertical vibration mode, two acceleration pickups are fixed on top of the block as shown in figure - 37, so that they can sense vertical motion of the block. The mechanical oscillator, which works on the principle of eccentric masses mounted on two shafts rotating in opposite directions, is mounted on the block so that it generates purely vertical sinusoidal vibrations. The line of action of the vibrating force passes through the centre of gravity of the block. After a suitable dynamic force value is chosen, the oscillator is operated at a constant frequency. The oscillator frequency is increased in steps of small values, say, from 1 cycle up to the maximum frequency of the oscillator, and the signals are recorded. The same procedure is repeated for the various dynamic force values. All force level and frequency; the dynamic force should not exceed 20% of total mass of the block and motor-oscillator assembly. In case of forced-vertical-vibration tests, the amplitude of vibration, Az at a given frequency f (hertz) is given by Az= az4π2f2 In which az represents the vertical acceleration of vibration in mm/sec2. The coefficient of elastic uniform compression cu of soil is given by, cu=4π2fnz2MA cu defined as the slope of stress verses elastic settlement curve. In which, fnz=natural frequency i nvertical vibrations M=mass of the block, oscillator and motor or the system A=Contact area of the block with the soil. E= cu 1-ν2A1.13 G= E21+ν In forced vertical vibration tests, the value of damping co-efficient, D of soil is given by the following equation, D= 2πD2fnz Figure - 38: Determination Damping ratio from forced vibration test Logarithmic decrement, δ= 2πD1-D2 In which, f1,f2= Two frequencies on the amplitude frequency plot at which the amplitude is equal to xm2 xm = Maximum amplitude fnz = frequency at which amplitude is maximum, i.e. resonant frequency.

Cyclic Plate Load Test

The cyclic field plate load test is similar to the plate bearing test conducted in the field for evaluation of the allowable bearing capacity of soil for foundation design purposes. The plates used for tests in the field are usually made of steel and are 25 mm thick and 150 mm to 762 mm in diameter. To conduct a test, a hole is excavated to the desired depth. The plate is placed at the center of the hole, and load is applied to the plate in steps-about one-fourth to one-fifth of estimated ultimate load-by a jack. Each step load is kept constant until the settlement becomes negligible. The final settlement is recorded by dial gauges. Then the load is removed and the plate is allowed to rebound. At the end of the rebounding period, the settlement of the plate is recorded. Following that, the load on the plate is increases to reach a magnitude of the next proposed stage of loading. The process of settlement recording is then repeated. Figure - 39: Nature of load settlement diagram for cyclic plate load test

Figure shows the nature of the plot of q versus settlement (S) obtained from a cyclic plate load test.

Note that soil pressure, q= Loads on the plate, QArea of the plate, A

Cz= PSe=1.13E(1-μ2)A Kg/cm2

E= Cz(1-γ2)√A1.13

G= E2(1+μ)

In which

P = Corresponding load intensity (Kg/cm2) Sc = Elastic rebound (cm) A = Contact Area

Magnitude of Cz can be obtained from the plot of q vs. Se from figure Factors affecting stress-deformation and strength characteristics of cohesive soils under pulsating loads or Factors affecting the dynamic properties of cohesive soils - * Type of soil and its properties (for example - water content, γd and state of disturbance) * Initial static (sustained) stress level * Magnitude of dynamic stress * Number of repetitions of dynamic stress * Frequency of loading * Shape of wave form of loading * One directional or two directional loading Oscillatory simple shear test and its shortcomings * The simple shear device consists essentially of a simple box, an arrangement for applying a cyclic load to the soil and an electronic recording system. * The box of Roscoe, which contains a square sample with a side length of 6 cm and a thickness of about 2 cm, is provided with two fixed side walls and two hinged end walls so that the sample may be subjected to deformations of the type shown in figure - 40. σ σ σ τ=0 -τ

+τ

Figure - 40: Idealized Stress condition for element of soil below ground surface during an earthquake. A schematic diagram in figure - 40 illustrates how the end walls rotate simultaneously at eh ends of the shearing chamber to deform the soil uniformly. Figure - 41: Schematic diagram illustrating rotation of hinged end plates and soil deformation in oscillatory simple shear Test data from simple shear tests have been analyzed to determine shear parameters, soil moduli and damping. Shortcomings * Stress in a tri-axial compression test does not adequately simulate the field loading condition. * During the earthquake, the normal stresses on this plane remain constant while cyclic shear stresses are induced during the period of shaking.

Figure - 41: Field Condition of soil Field condition differs: * In the field, there is a cyclic reorientation of the principal stress directions. The major principal stress is initially vertical and rotates through some angle θ, to the right and left of its initial position. In a tri-axial compression test, the major principal stress can act only in either the vertical or horizontal direction. * In the field, the soil element is initially consolidated to k0 condition. * In the field, deformations are presumed to occur under plane strain condition, while in a tri-axial compression test, the intermediate principal stress is either equal to minor principal stress during axial compression or equal to major principal stress during lateral compression.

Bilinear Model:

Figure - 42: Stress Strain curve of Soil and Bilinear model

The shear-stress-strain relationship may as shown in figure 42.

The soil exhibit nonlinear stress-strain characteristics from the very beginning of the loading cycle. For purposes of analysis, this behavior may be represented by a bilinear model in figure - 42.

The bilinear model is defined by three parameters * Modulus G1until a limiting strain, γγ * Modulus G2 beyond strain, γγ * Strain, γγ

When the direction of strain is reversed, behavior is again determined by the modulus G1 until a strain change of 2γγ has developed and the modulus G2 again controls the behavior. This pattern then continuous throughout the cycle.

Pendulum Loading Apparatus:

Figure - 43: Pendulum Loading Apparatus:

Three pieces of equipment were designed for this purpose. One of the pieces of equipment is the pendulum loading apparatus shown in figure - 43.

The apparatus utilizes the energy of a pendulum which, when released from a selected height, strikes a spring connected to the piston rod of a hydraulic (lower) cylinder. The lower cylinder, in turn, is connected hydraulically to an upper cylinder, which is mounted within a loading frame. The time of loading for a pendulum loading apparatus is proportional to the square root of the weight of the pendulum and is inversely proportional to the square root of the spring constant. In addition, the maximum force is proportional to the first power of the distance the pendulum is pulled back, to the square root of the spring constant, and to the square root of the weight of eh pendulum. This apparatus, with a time of loading of between 0.05 and 0.015 was found to be best suited for performing fast transient test.

The load gauge used with this equipment consisted of electric resistance strain gauge, mounted on a metal ring. The strain introduced in the gauges was then in direct proportion to the load. These load gauges can be calibrated under static load and can be used in a dynamic test. Similarly a deformation gauge was constructed on a cantilever metal strip with electric resistance strain gauges, mounted on one end while the other end rested on an unmovable support. The strain introduced in the cantilever was a measure of the deformation of the soil sample.

ωn=30-50% LL = 37 - 59% PL = 20 - 27%

Figure - 44: Time vs stress and strain in an unconfined transient test on Cambridge clay.

From the typical test data presented above, it may be concluded that: * The strength of clays loaded to failure in about 0.02 s is approximately 1.5 to 2.0 times greater than their 10 - min static strength. * Modulus of deformation, defined as the slope of a line drawn from the origin through the point on the stress - deformation curve and corresponding to a stress of one-half the strength, was about two times in the transient test.

These investigations suffer from the following short comings:

* The dynamic load was not superimposed on a static load. * At best, the transient loading is adopted in the investigations represents only one cycle of earthquake loading. Sometimes there may be as many as 100 peaks in an actual earthquake. * Finally, the sands were tested while dry and dense. The effect of dynamic loading on saturated loose sands may induce large pore pressures resulting in loss of strength and consequent partial or complete liquefaction of sands.

This aspect of the problem is of great practical importance.

Differences between dynamic and static test:

| Dynamic | Static | 1 | It is monotonic or cyclic loading test with relatively large strain rate. | It is monotonic or cyclic loading test with relatively slow strain rate. | 2 | Inertia effect is considered. | Inertia effect is not considered | 3 | Acceleration is a major controlling factor. | Acceleration is not a major controlling factor | 4 | The stiffens are found. | The stiffens are not found. | 5 | Testing procedure are complex. | Testing procedure is simple. |

The term “Dynamic loading” is more general and it means “monotonic or cyclic” loading at such a * Relatively large strain rate as the effect of inertia cannot be ignored. Therefore “Dynamic (loading) test” refer to conditions where the acceleration is a major. * Controlling factor and the stiffness of specimen is determined by the dynamic properties of specimen or system including a specimen (i.e. the wave velocity within the specimen or the resonant-frequency or natural frequency of the system). * Also, the modifier ‘Static’ should not be equated exclusively with “monotonic virgin loading at a relatively slow strain rate”.

The term “Static (loading) test” should be defined as “monotonic or cyclic loading at a * Relatively slow strain rate where the effect of inertia can be ignored. * Hence, “Static (loading) test” refer to those in which both the stresses (and / or loads) and the strains (and / or deformations) are measured under the conditions without discernible effects of inertia.

Definitions of several types of stiffness

Secant Young’s modulus, Esec= q-q0ε1

Tangent Young’s modulus Etan= dqdε1

Secant Young’s modulus at q-q0= qmax-q02 Eso= Esec at q-q0=qmax-q02

Equivalent Young’s modulus Eeq= qsaSAε1

2ε1SA

ε1

Figure - 45: Plot of Deviator stress vs Axial strain

When the initial portion of the stress-strain curve is linear and the strain is fully recoverable, we can see that

Esec= Etan= Eeq=Emax

Ef and Gs: Young’s modulus and shear modulus obtained from in situ elastic shear wave velocity vs, which are Ef= 21+νGf Gf= ρvs2= γtgvs2

Where,

ρ=mass density of soil or rock γt= Unit weight of soil or rock g=Acceleration of gravity=980 cm/sec2 vc=Velocity of Shear waves q ε1

Figure - 46: Plot of Stress vs Axial strain

logε, logγ

Figure - 47: Plot of Linear elastic, elastic weak plastic, elastic-obvious plastic cyclic loading

εSA σ Figure - 48: Variation of shear stress verses shear strain

In the engineering practice of soil dynamics, the in situ relationship between the equivalent shear modulus, Geq and single amplitude shear strain, γSA under cyclic loading conditions is estimated using the following methodology.

Figure - 49: A method used to estimate the in-situ stiffness as a function of strain for dynamic loading

Resonant Column Tests

The Resonant column test for determining modulus and damping characteristics of soil is based on the theory of compression waves or shear waves propagation in prismatic rods. In a resonant column apparatus the exciting frequency is adjusted until the specimen experiences resonance. The modulus is computed from the resonant frequency and the geometric properties of the specimen and driving apparatus. Damping is determined by turning off the driving power at resonance and recording the decaying vibrations from which a logarithmic decrement is calculated. Alternative methods of damping measurement include determining damping from the shape of the resonance curve or determining a “resonant factor” from driving coil current measurements. Several versions of the resonant column test are possible using different end conditions to constrain the specimen. Some common end conditions are shown in figure (50).

14 sine wave

JJ0= ∞

JJ0= 0.50 θJt θJt θJ,t θJ,t

14 sine wave θ θ

Figure - 50: Some common end conditions of resonant column test.

Each configuration requires slightly different driving equipment and methods of data interpretation. The “fixed free” apparatus is the simplest configuration in terms of equipment and interpretation as figure (50-a). Figure (50-a) the distribution of angular rotation, θ, along the specimen is a 14 sine wave but by adding a mass with mass polar moment J0, at the top of the specimen as in figure (50-b)the variation of θ along the sample becomes nearly linear. The end effects to obtain uniform strain distribution through the length of the specimen. The apparatus configuration in figure (50-c) can be described as the “spring base” model. For a condition where the spring is weak compared to the specimen, the configuration of figure (50-c) could be called “free free”.

A mode will occur at mid height of the specimen and the rotation distribution would be a 12 sine wave. By adding end masses the rotation distribution can also be made nearly linear. To study the influences of anisotropic stress conditions on shear modulus and damping. Figure (50-d) has fixed base and a top cap that is partially restrained by a spring and dashpot which in turn reacts against an inertial mass. For k0=1.0 tests, the inertial mass is balanced by a counter weight, but by changing the counter weight, axial load can be applied to the specimen.

Machine foundation

Machine foundations are subjected to the dynamic forces caused by the machine. These dynamic forces are transmitted to the foundation supporting the machine. These loads may result from various causes such as vibratory motion of machines, movement of vehicles, impact of hammers, earthquakes, winds, waves, nuclear blasts, mine explosions, and pile driving.

There are three types of machine foundation 1. Machine which produce a periodic unbalanced force, such as reciprocating engines and compressors. The speed of such machines is generally less than 600 r.p.m. In these machines, the rotary motion of the crank is converted into the translatory motion. The unbalanced force varies sinusoidally. 2. Machine which produce impact load, such as forge hammers and punch presses. In these machines, the dynamic force attains a peak value in a very short time and then dies out gradually. The response is a pulsating curve. It vanishes before the next pulse. The speed is usually between 60 to 150 blows per minute. 3. High speed machines, such as turbines and rotary compressors. The speed of such machines is very high; sometimes, it is even more than 3000 r.p.m.

Types of Machine foundation 4. Block Type: This type of machine foundation consists of a pedestal resting on a footing. The foundation has a large mass and a small natural frequency. 5. Box Type: The foundation consists of a hollow concrete block. The mass of the foundation is less than that in the block type and the natural frequency is increased. 6. Wall Type: A wall type of foundation consists of a pair of walls having a top slab. The machine rests on the top slab. 7. Framed Type. This type of foundation consists of vertical columns having a horizontal frame at their tops. The machine is supported on the frame.

Figure - 51: Types of Machine Foundation.

Suitability of various types of Machine Foundation Machines which produce periodical and impulsive forces at low speeds are generally provided with a block type foundation. Framed type foundations are generally used for the machines working at high speeds and for those of the rotating type. Some machines which induce very little dynamic forces, such as lathes, need not be provided with a machine foundation. Such machines may be directly bolted to the floor.

Design Criteria For Foundation's of Reciprocating Machine:

The following design anions for the foundation of reciprocating machines should be satisfied: 1. The machine foundation should be isolated at all levels from the adjoining foundations.

2. The natural frequency of the foundation-soil system should be higher than the highest disturbing frequency and the frequency ratio should be less than 0.4, as far as possible. However, if it is not possible to satisfy above criterion, the natural frequency should be lower than the lowest disturbing frequency and the frequency and ratio should not be less than 1.50. 3. The amplitude of vibration should the within the permissible limits. For most sells, the limiting amplitude for low speed machines u usually taken as 200 micron (0.2 mm). According to another criterion, the amplitude in mm is limited to 4f for frequencies less than 30 hertz and 125f2 for higher frequencies, where f is frequency in hertz (cycles/scc). 4. Concrete block foundations should be used. However, when the soil is not suitable to support block foundation, cellular foundation may be used. 5. The size of the block in plan should be larger than the bed plate of the machine. There should be a minimum all-round clearance of 15 mm. The total width of the foundation measured at right angles in the shaft should be least equal to the distance between the centre of the shaft and the bottom of the foundation. 6. The eccentricity of the foundation system along X-X and Y-Y axes should not exceed 5% of the length of the corresponding side of the contact area. 7. The combined centre of gravity of' machine end foundation should be as much below the top of foundation as possible. In no case, it should be above the top of foundation. 8. The depth of foundation should be sufficient in provide the required bearing capacity and to ensure stability against rotation in the vertical plane. 9. The stresses in the soil below the foundations should not exceed 80% of the allowable stress under static leads. The base pressure is limited to half the normal allowable pressure (qna) in extreme cases. 10. Where it is not practicable to design a foundation to give satisfactory dynamic response, the transmitted vibrations may be reduced by providing anti-vibration mountings either between the machine and the foundation or between the foundation and the supporting system. 11. The machine should be anchored to the foundation block using a base plate and anchor bolts. Bolt holes should be backfilled with concrete and the space below the plate should be filled with 1 : 2 cement mortar. 12. A number of similar machines can be created on individual pedestals on a common raft. The analysis for such machines can be made assuming that each foundation acts independently with an area of foundation equal to that obtained by dividing up the raft into sections corresponding to separate machines.

Reinforcement and Construction Details: 1. The reinforcement in the concrete block should not be less than 25 kg/m3. For machines requiring special design consideration of foundations, such as machine pumping explosive gases, the minimum reinforcement is 40 kg/m3. 2. Steel reinforcement around all and openings shall be at least equal to 0.5 to 0.75% of the cross-sectional area of the pit or opening. 3. The reinforcement shall run in all the three directions. 4. The minimum reinforcement shall usually consist of 12 mm bars at 200 to 250 mm spacing excluding both vertically and horizontally near all faces of the foundation block. The ends of all bats should always be hooked. 5. If the height of the foundation block exceeds one mate, shrinkage reinforcement shall be placed a suitable spacing in all the three directions. 6. The cover should be a minimum of 75 mm at the bottom and 50 mm on sides and the top. 7. The concrete shall be at least M15 with a characteristic strength of 15 N/mm2. 8. The foundation block should be preferably cast in a single, continuous operation. In case of very thick blocks (exceeding 5 m), construction joints can be provided.

Mass of Foundation

Heavy foundations eliminate excessive vibrations. Manufactures of machines sometimes recommend the mass of foundation required for the machines. However, the mass recommended are generally empirical rust based largely on experience.

Vibration isolation and control:

Vibrations may cause harmful effects on the adjoining structures and machines. Besides, these vibrations cause annoyance to the persons working in the area around the machine. However, if the frequency ratio is kept outside the critical range of 0.4 and 1.5, and the amplitude is within the permissible limits, the harmful effects are considerably reduced, especially if the system is damped.

Transmission of vibrations am be controlled and tile detrimental effects considerably reduced by isolating either the source (active isolation) or by protecting the receiver (Passive isolation). The following measures are generally adopted. 1. The machine foundation should be located away from the adjoining structures. This is known as geometric isolation. The amplitude of surface waves (R-waves) reduces with an increase in distance. A considerable reduction in the amplitude is achieved by locating the foundation at a great depth, as the R-waves also reduce considerably with an increase in depth. 2. Additional masses known as dampers are attached to the foundation of high frequency machines to make it a multiple degree freedom system and to change the natural frequency. In reciprocating machines, the vibrations are considerably reduced by counterbalancing the exciting forces by attaching counterweights to the sides of the crank. 3. Vibrations are considerably reduced by placing absorbers, such as rubber mountings, felts and corks between the machine and the base. 4. If an auxiliary mass with a spring is attached to the machine foundation, the system becomes a two-degree-freedom system. The method is especially effective when the system is in resonance. 5. If the strength of the soil is increased by chemical or cement stabilization, it increases the natural frequency of the system. The method is useful for machines of low operating frequency. 6. The natural frequency of the system is modified by making structural changes in foundation, such as connecting the adjoining foundations, changing the base area or mass of foundation or use of attached slabs. 7. The propagation of waves can be reduced by providing sheet piles, screens or trenches.

Liquefaction

Quick Sand condition:

A soil under critical hydraulic gradient will be unstable and is sail to be in a ‘quick’ condition. By this definition any granular soil may be a “quick sand” but soil with high permeability i.e. gravels and coarse sands require large quantities of water to maintain a critical hydraulic gradient. Quick sand conditions are therefore usually confined to fine grained sands. At a quick condition σ'= 0=σ-U or, σ=U

Where,

σ=Total stress σ'=Effective Stress U=Pore water Pressure

Thus, when the pore pressure equals the total pressure on a plane, a quick condition exists and the pore pressure can only equal the total pressure when ∆h>0, which is a flow condition.

Liquefaction:

When a fine or medium fine, saturated, loose sand deposit is subjected to a sudden shock the mass will temporarily liquefy. This phenomenon is termed as liquefaction. In the situation just described, four criteria were given: a particular sand, loose state, saturation and sudden shock.

The shock temporary increases the pro pressure. The total stress is not large when the soil is loose - also, the structure is somewhat unstable. The grain size is such that the pore pressure can “float” the grains. The result is a temporary liquefaction of sand mass until pore drainage occurs. During this time lag the very viscous sand-after mixture has little shear strength to support any structures on it and, if not confined, may flow laterally. This phenomenon has been observed to occur in several fairly recent earthquakes. It also sometimes occurs during pile driving i.e. when the pile has great penetration for several of the hammer blows. Liquefaction can be readily observed in the laboratory by building a “quick sand” tank.

Liquefaction analysis based on method by Ishihara (1993).

The cyclic shear stress induced at any point in level ground during an earthquake due to upward propagation of shear waves in given by as stress ratio, τmaxσv'=amaxg×rd×σvσv' Where, τmax=maximum shear stress induced on a given soil element at a depth, z. σv=γz=total overburden pressure acting on the element. σv'= zγsat- z1γw=effective stress, where γsat=20KNm3 and γw=10KNm3 γd=1-0.015z=stress reduction co-efficient to allow for deformability of the soil column, γd<1. amax=the peak normalized acceleration on the ground surface caused by earthquake loading. g= Acceleration due to gravity, amaxg=0.15 for M=6.0 Richter Scale

Co-relation between the cyclic strength and SPT N-value for clean sand.

Maximum resisting shear stress ratio, τmaxσv'=0.0676N12+0.225log100.35D50 for 0.04 mm ≤ D50 ≤ 0.60 mm and τmaxσv'=0.0676N12+0.05 for 0.6 mm ≤ D50 ≤ 1.50 mm and

Where,

D50 = mean particle (50% finer) diameter of soil in mm. N1= The normalized SPT N-value obtained through the co-rellation factor for overburden pressure, CN

Defined as N1=CNN CN= 1.70σv'+0.70

Where,

σv'= The effective overburden pressure = kgf/cm2

After so determining the induced stress and shear strength of a soil element, the liquefaction potential of a sand deposit is evaluated in terms of factor of safety, F1 defined as, F1= Shear StrengthInduced Stress= τmax,1σv'τmaxσv'

If the factor of safety is less than on F1≤1, the liquefaction is said to take place, otherwise liquefaction does not occur.

The required N-value (Nreq i.e. un-corrected N-values) obtained by Nreq= N1CN=14.793τmaxσv'F1-0.225log100.35D50

If N1≤Nreq at a given depth, liquefaction is likely to occur in soil at that depth,

Depth(m) | σ0 = σv(KPa) | σ0’ = σv’ | CN | γd | Induce stress ratioτmax/ σv’ | Nreq | | | | | | | F1 = 1.0 | F1 = 1.25 | F1 = 1.5 | 0 | | | | | | | | | 1 | | | | | | | | | 2 | | | | | | | | | 3 | | | | | | | | | 4 | | | | | | | | | 5 | | | | | | | | |

Figure – 52: Plot of Nreq and N1 with SPT

Relationship between structural damage intensity and soil depth in earthquake of 1967.

Figure – 53: Plot of Structural Damage Intensity verses Depth of Soil.

For 3 to 5 story buildings damage was many times greater where soil depths ranged from 30 to 50m than for soil depths over 100m. For 5 to 9 story buildings, the structural damage intensity was slightly higher for soil depths of 50 to 70 m than for other depths of soil, but buildings over 10 stories high, the structural damage intensity was several hundred percent higher where soil depths exceed 160m than for soil depths below 140 m. It would appear that the depth and characteristics of the underlying soil deposits had a significant effect on the characteristics of ground motion and the resulting building damage, although building characteristics are also very much involved in determining the damage patterns. The magnitude of the influence of local soil conditions on the characteristics of earthquake ground motions than thereby on building damage merits their careful consideration is seismic design. Because this monograph cannot review all aspects of soil behavior during earthquakes, it will concentrate on the two topics of most general interest: * The influence of soil conditions on the characteristics of earthquake ground motions and * Methods of evaluating the liquefaction potential of soil deposits. Main Characteristics of the ground motions: Such as

* Maximum ground acceleration * Maximum ground velocity * Maximum ground displacement * Duration of significant ground shaking

Figure – 54: Acceleration, Velocity and Displacement relationship with time

Acceleration spectra & Velocity spectra:

For any given ground motion, values of the spectral velocity, Sv and the spectral acceleration, Sa for a single-degree-of-freedom structure having a period T are related approximately by the equation, Sv= T2πSa for 0<T<S sec

Figure – 55: Acceleration, Velocity and Damping relationship with natural periods

The maximum base shear, Vmax for a multistory structure subjected t oa given base motion can be estimated from the equation Vmax=W.Sag

Where,

W = The weight of the structure Sa = Spectral acceleration corresponding to the natural period of the structure (SDOF) g = The acceleration of gravity, (SDOF)

Factors affecting Earthquake ground motions:

The characteristics of earthquake ground motion at any site are influenced by a number of factors including: 1. Magnitude of the earthquake 2. Distance of the site from the source of energy release 3. Geologic characteristics of the rocks along the wave transmission path from source to site 4. Source mechanism of the earthquake 5. Wave interference effects related to the direction and speed of fault rupturing 6. Local soil conditions at the site

Influence of soil conditions on ground motion characteristics:

Figure – 56: Plot of Peak Horizontal Acceleration with Closest Horizontal Distance of Sources.

Comparison of attenuation curves for rock sites and Imperial Valley (Deep alluvial) Earthquake.

This compassion in figure (56) indicates that at comparable,

Figure – 57: Plot of Peak Horizontal Acceleration verses Closest Horizontal Distance of Sources with Earthquake Magnitude.

Approximate relationship between maximum acceleration on rock and other soils:

Figure – 58: Plot of relationship between maximum acceleration on rock and other soils

It may be seen that apart from deposits involving soft to medium stiff clay, values of peak acceleration developed on different types of soil do not differ appreciably, particularly at acceleration levels less than about 0.3 to 0.4 g. Even at higher acceleration levels on rock of the order of 0.7 g, accelerations on deposits of any depth which do not involve soft to medium stiff clays are likely to be only about 25% less than these on rock.

For Engineering practice and for most practical purposes it may well be considered that peak acceleration values on rock and stiff soils of any depth are about the same. In fact, if data for all foundation conditions except soft to medium clays are plotted together, it may not be possible to differentiate between acceleration levels for rock and different site conditions. amaxmean+1amaxmean≈1.4 to 1.5 Maximum Ground Velocity:

Geologic Condition Vmaxamax Rock 55 cm/sec/g Stiff soils (<200) 110 cm/sec/g Deep stiff soils (>200) 135 cm/sec/g

Figure – 59: Plot of relationship between Peak Velocity and Energy Release on rock and other soils

Response Spectra:

Figure – 60: Plot of Normalized Acceleration Response Spectra for 5% Damping

Figure – 61: Effect of soil type on average spectral acceleration at 5% damping.

Figure – 62: Normalized spectral shapes for various types of soil and rock.

Selection of Ground Motions for Design

Differentiate between two earthquake motions that are related but can be significantly different from each other: 1) The maximum free-field earthquake ground motions which a structure should be able to withstand with an acceptable margin of safety. 2) The level of shaking in evaluating the safety of a structure. This motion, which is often called the “Design Earthquake Motion”, depends on many factors: (a) Method of analysis into which it will be incorporated (b) Conservatism of the analysis procedure (c) Level of damping, taking into account the acceptable level of damage (d) Depth of embedment of the structure in the ground (e) Effect of soil-structure interaction, if they are not included directly in the analysis procedure (f) Effect of spatial variations in ground motion if they are not included directly in the analyses procedure (g) Material properties (h) Combination of loadings or component of ground motion (i) Strength characteristics of the structure (j) Ductility of the structure

Specifications of Design Spectra:

Some aspects of the analysis, to develop a time history which has: 1. The general characteristics of a reasonable earthquake motion and 2. A response spectrum that just, envelops the specified spectrum shape.

Simple guideline: 1. Once magnitude and distance to the source of energy release of possible earthquakes have been determined. 2. Peak accelerations for different soil conditions may be obtained by correcting the mean peak acceleration for rock sites. 3. Values of mean+1 standard deviation acceleration can be determined from relationship amaxmean+1amaxmean=1.4 to 1.5 4. For sites within about 50 Km/s, values of maximum ground velocity.

For rock sites: Vmaxamax=55cm/sec/g

For stiff sites: Vmaxamax=110cm/sec/g 5. Values of maximum spectral acceleration,

For mean spectral shape: Samax=2.7amax

For mean+1 spectral shape: Samax=3.4amax 6. Design values recommended by the seismic review panel:

Maximum acceleration = 0.75g

Maximum velocity = 0.35 cm/sec

Maximum spectral acceleration = 1.75g

Specification of design acceleragram

Main characteristics of the design acceleragram: 1. A peak acceleragram of 0.33g. 2. Response spectrum close to the mean +1 standard deviation spectrum shape for rock records and 3. Duration of about 16 seconds.

Causes of Soil Liquefaction: 1. If a saturated sand is subjected to ground vibrations, it tends to compact and decrease in volume; if drainage is unable to occur, the tendency to decrease in volume results in an increase in pore water pressure, and if the pore water pressure builds up to the point at which it is equal to the overburden pressure, the effective stress becomes zero, the sand loses its strength completely, and it develops a liquefied state. σ'= σ-U

Where,

σ=Total stress σ'=Effective Stress U=Pore water Pressure 2. The basic cause of liquefaction is saturated cohesion less soils during earthquakes is the buildup of excess hydraulic pressure due to the application of cyclic shear stresses induced by the ground motions.

τ τ σ0' k0σ0' τ τ σ0'

k0σ0'

(a) Idealized field loading condition.

(b) Shear stress variation determined by response analysis.

Figure – 63: Cyclic shear stress on a soil element during ground shaking. 3. If the sand is loose, the pore pressure will increase suddenly to a value equal to the applied confining pressure and the sand will readily begin to undergo large deformations will shear strains, they may exceed ± 20 percent or more. If the sand will undergo vertically unlimited deformations without mobilizing significant resistance to deformation, it can be said to be liquefied. 4. As a consequence of the applied cyclic stress, the structure of the cohesion less soil tends to become more compact with a resulting transfer of stress to the pore water and a reduction in stress on the soil grains. As a result, the soil grain structure rebounds to the extent required keeping the volume constant, and this interplay of volume reduction and soil structure rebound determines the magnitude of the increase in pore water pressure in the soil. The basic phenomenon is illustrated schematically in figure – 64. The mechanism can be quantified so that the pore pressure increases due to any given sequence of stress applications can be computed from knowledge of the stress-strain characteristics, the volume change characteristics of the sand under cyclic strain conditions and the rebound characteristics of the sand due to stress reduction.

Figure – 64: Schematic illustration of mechanism of pore pressure generation during cyclic loading

Cyclic mobility:

For dense sand, it may develops a residual pore water pressure, on completion of a full stress cycle, which is equal to eh confining pressure (a peak cyclic stress to pore pressure ratio of 100%), but when the cyclic stress is reapplied on the next stress cycle, or if the sand is subjected to monotonic loading, the soil will tend to dilate, the pore pressure will drop if the sand is un-drained, and the soil will ultimately develop enough resistance to withstand the applied stress. However, it will have to undergo some degree of deformation to develop the resistance, and as the cyclic loading continues the amount of deformation required to produce a stable condition may increase. Ultimately, however, for any cyclic loading condition, there appears to be a cyclic strain level at which the soil will be able to with stand any number of cycles of given stress without further increase in maximum deformation. This type of behavior is termed “Cyclic mobility” and it is considerably less serious than liquefaction, it’s significance depending on the magnitude of the limiting strain. However, that once the cyclic stress applications stops if they have to a zero stress condition, there will be a residual pore water pressure in the soil equal to the overburden pressure and they will inevitably lead to an upward flow of water in the soil which could have deleterious consequences for overlying layers.

Factors on which liquefaction will depends: 1. The extent to which the necessary hydraulic gradient can be developed and maintained. 2. Will be determined by the compaction characteristics of the sand. 3. The nature of ground deformations. 4. The permeability of the sand. 5. The boundary drainage conditions 6. The geometry of the particular situation 7. The duration of the induced vibrations

General method of evaluating liquefaction potential

The liquefaction potential of any given soil deposit is determined by a combination 1. Soil properties 2. Environmental factors and 3. Characteristics of the earthquake to which it may subjected

Specific factors which any liquefaction evaluation should desirably take into account include the following:

Soil Properties: 1. Dynamic shear modulus 2. Damping characteristic 3. Unit weight 4. Grain characteristics 5. Relative density 6. Soil structure

Environmental factors: 1. Method of soil formation 2. Seismic history 3. Geologic history (aging, cementation) 4. Lateral earth pressure co-efficient k0,ka, kp 5. Depth of water table 6. Effective confining pressure

Earthquake characteristics: 1. Intensity of ground shaking 2. Duration of ground shaking

Some of these factors cannot be determined directly, but then effects can be included in the evaluation procedure by performing cyclic loading tests on undisturbed samples or by measuring the liquefaction characteristics of the soil by mean of some in-situ test procedure.

With the recognition of this fact, the basis evaluation procedure involves: 1. A determination of the cyclic shear stress induced by the earthquake ground motions at different depths in the deposit and conversion of the irregular stress histories to equivalent number or uniform stress cycles. By means the intensity of ground shaking, the duration of shaking and the variation of induced shear stresses with depth are taken into account. The determination may be made either by a ground response analysis involving the unit weight of the soils, the dynamic moduli and the soil damping characteristics. A plot of the induced equivalent uniform shear stress level as a function of depth as shown in figure - 65. 2. A determination by means of laboratory cyclic loading test on representative undisturbed samples conducted at different confining pressures, or by correction of these properties with some measurable in-situ characteristics of the cyclic shear stresses. Other cyclic load simple shear tests or cyclic load tri-axial compression tests may be used for this purpose, provided the test results are appropriately corrected to be representative of field conditions. By this means the soil type, the in-plane the seismic and geologic histories of the deposit of the initial effective stress conditions are approximately taken into account.

The stresses required to cause liquefaction can then be plotted as a function of depth as shown in figure – 65.

Figure – 65: Zone of liquefaction 3. A composition of the shear stresses included by the earthquake with those required to causes liquefaction, to determine whether any zone exists within the deposit where liquefaction can be expected to occur that is, where induced stresses exceed those required to cause liquefaction.

Specified procedure for evaluating stress induced by Earthquake:

The shear stresses developed at any point in a soil deposit during an earthquake appear to be due primarily to the vertical propagation of shear waves in the deposit. This leads to a simplified procedure for evaluating the induced shear stress (Seed and Indris, 1971)

If the soil column above a soil element at depth, h behaved as a rigid body, the maximum shear stress on the soil element would be τmaxr= γhg.amax

Where,

amax= The maximum ground surface acceleration, and γ = The unit weight of the soil.

Because the soil column behave as a deformable body, the actual shear stress at depth, h, τmaxd ad determined by a ground response analysis will be less than τmaxr and might be expressed by τmaxd= rdτmaxr

Maximum shear stress, rd= τmaxdτmaxr

Where, rd is a stress reduction <1. τmaxr= γhg.amax

amax

Figure – 66: Procedure for determining maximum shear stress

Vibrations of τmaxr, τmaxd and rd in figure (66) the value of rd will decrease from the value of 1 at the ground surface to much lower values at large depth.

Figure – 67: Range of shear stress reduction factor rd for the deformation nature of soil

Computation as of the value of rd for a wide variety of earthquake motions and soil conditions having sand in the upper soft have shown that rd generally falls within the rouge of values shown in figure (67).

It may be seen that in the upper 30 or 40 ft, the scatter of the results is not great and for any of the deposits, the error involved in using the average values shown by the dashed line would generally be less than about 50%.

Thus to a depth of about 40 ft, a reasonably accurate assessment of the maximum shear stress developed during an earthquake can be made from eth relationship τmax= γhg.amax.rd

Where, rd are taken from the dashed line in figure - (67), the critical depth for development of liquefaction, if it is going to occur, will normally be in the depth covered by this relationship.

The actual time history of shear stress at any point in a soil deposit during an earthquake will have an irregular from such as that shown in figure (68). τavg ≅0.65 τmax

Figure – 68: Time history of shear stress during earthquake

By appropriate weighting of the individual stress cycles, based on laboratory test date, this determination can readily be made. τavg ≅0.65 τmax τavg ≅0.65γhgamax.rd

The appropriate number of significant stress cycles Nc will developed of ground shaking and thus on the magnitude of the earthquake. Representative number of stress cycles is as follows:

Earthquake magnitude Number of significant stress cycles, Nc 5 ¼ 2-3 6 5 6 ¾ 10 7 ½ 15 8 ½ 26

Determination of cyclic stress levels causing liquefaction from laboratory test data.

Cyclic simple shear test

Typical results of a cyclic simple shear test on a simple of loose sand are shown in figure (69).

Loose Monterey Sand

Initial relative density = Dr=50%

Initial void ratio = ei=0.68

Initial confining pressure, σv=5.0 Kg/cm2

Soil failure occurs at 20 cyclic (Normally)

In the early stages of cyclic stress applications, the pore water pressures build up in the sample but there is no significant deformation.

After a number of applications, the pore pressure suddenly jumps to value equal to the vertical confining pressure, reducing the effective stress to zero and the same time the sample begins to undergo large cyclic deformations. This denotes the onset of liquefaction.

The number of stress cycles required to calculate the sample to liquefy depends on the magnitude of the applied shear stress and the initial vertical effective pressure under which the sample is consolidated.

Typical results of a series of tests on identical samples of sand at a relative density of 50% are shown in figure (70).

From plots of this type it is readily possible to read off the cyclic shear stress ratio τhσv' causing liquefaction in the number of stress cycles representative of the design earthquake.

When tests are performed on dense samples of sand, the onset of liquefaction or cyclic mobility is not so abrupt and a critical condition is normally considered to develop when the pore pressure ratio builds up to a value of 100% and the cyclic shear strain is ±5%.

Figure - 69: Record a typical cyclic loading test on loose sand-simple shear conditions.

Figure -70: Typical form of the relationship between cyclic shear stress and the number of cyclic cause failure simple shear conditions.

Difficulties in sample shear testing’s: 1. Selection of representative samples. 2. Avoidance of stress concentrations in the samples and the maintenance of uniform stresses and strains during the conduct of the test and 3. Obtaining undisturbed samples of sand for the test program and maintaining a high level of un-disturbance while the samples are setup in the test equipment.

Remedy:

1. Careful and skilled test techniques. 2. Technique of freezing the soil prior to sampling for undisturbed samples. 3. Significant degree of judgment to allow for the possible effects of sample disturbance. Cyclic Tri-axial Compression Test:

In the performance of test tests to represent level ground conditions in the field. Samples are first consolidated under an ambient confining pressure, σ0, and then subjected to cyclic deviator stress application σdc. Test data are similar in form to those obtained in cyclic stress ratio, σdc2σ0 which causes liquefaction or cyclic mobility in eh desired number of cycles. The cyclic stress ratio τhσv' causing liquefaction under multidimensional shearing conditions in the field is related to the cyclic stress ratio causing liquefaction of a tri-axial test sample in the laboratory by the expression (Seed, 1979). τhσv'field=Crσdc2σ0field σdc=Deviatoric σ0 = Cell pressure

Where values of Cr are approximately.

Cr= 0.57 for k0=0.400.9 to 1 for k0=1.0

Effects of sample disturbance on cyclic loading test data:

The cyclic loading characteristics of a natural sand deposit are strongly influenced by 1. Relative density of the deposit 2. Soil structure or grain arrangement 3. Cementation at grain contacts, which increases with the age of the deposit.

Example:

Find the cyclic stress ratio developed by earthquake at depth of soil 20 ft below from existing ground level if water table 4 ft below the ground and peak ground acceleration≈ 0.16g i.e. amax ≅0.16g. Given γsat=115 pcf, γd=0.93 pcf,

Solution:

Let, γw=62.5 pcf

Total overburden pressure = σ0 ≅20×115= hγsat=2300 pcf

Effective pressure σv' ≅h×γsat+hwγsat- γw =4×115+16×115-62.5=1300 psf

Cyclic stress ratio developed by earthquake, τhσv'd≅0.65×amaxg×σ0σ0' ×rd

=0.65×0.15×23001300×0.93=0.17

Case study - 2 γd=103 pcf γd=105 pcf σdc2σ3 For a critical depth of about 20 ft:

σ0≅h×γsat=20×120=2400 psf σ0' ≅h×γsat+ h×γ'= h×γsat+ h×γsat- γw=10×120+10×60 =1800 psf

The cyclic stress ratio developed by the earthquake (M = 7.5) τavσ0'd≅0.65×amaxg×σ0σ0' ×γd

=0.65×0.25×24001800×0.93=0.20

For, M = 7.5 causes approximately is equivalent stress cycles.

From the data on curve (A) in figure------the cyclic stress ratio, causing liquefaction in 15 cycles is τavσ0'f≅0.57σdc2σ3f=Crσdc2σ3f-triaxial Cr=0.57 for k0=0.40 =0.57 ×0.32=0.18<0.20

It appears that the sand could liquefy in the design earthquake.

For the test results by curve (B), the cyclic stress ratio required to cause liquefaction of the sand in 15 cycles is found to be, τavσ0'f=Crσdc2σ3f=0.57 ×0.5=0.28>0.20

Not to be liquefied the sand.

And the factor of safety against liquefaction is

F.S= τavσ0'fτavσ0'd=0.280.20=1.40

Free Essay

...Soil investigation is often neglected or rejected by most clients on the basis of cost, despite the fact that the cost of carrying out a soil investigation is very little compared to the cost of the project Soil investigation is done for various purposes. In engineering, soil investigation is very necessary. It is essential to investigate the soil of the selected plot on which a structure will be constructed. Based on soil investigation a soil report is prepared for the purpose of designing the building foundation. When an engineer designs building foundation he/she must carefully read the report and design the foundation based on the data provided in the report. Soil investigation is required for the following purposes - To know the allowable bearing capacity of foundation for proposed building. To know the depth and type of foundation for the proposed building. To know the allowable passive resistance for the foundation of proposed building. To know the type, grading and nature of soil. To know the ground water level. Typical steps of soil investigation Soil investigation involves following steps – Details planning for the sequence of operations. Collecting the samples of soil from the plot. Determining the soil characteristics by conducting field tests. Study the condition of ground water level. Collecting ground water sample for chemical analysis. Soil exploration. Testing all collected samples in the laboratory. Preparation of......

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...Soil erosion poses a significant threat to the environment and agriculture. The repeated loss of fertile topsoil negatively affects the long term sustainability of natural systems. Agricultural productivity faces a significant decline as a result of soil erosion Kusimi et al., (2015). Agricultural land across the world has either been lost or is rapidly experiencing degradation as a result of soil erosion. According to Arekhi et al., (2012), close to 40% of the world’s agricultural land is degraded, this includes 65% for Africa, 74% and 45% for North and South America respectively. Notable is that soil loss by erosion is an ongoing process, it was earlier reported by (Dudal 1981) that, across the globe , approximately 6,000,000 ha of fertile...

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...Maeve Upton 14310368 C. ‘In order to understand the geography of soil all one needs is a good map of the solid geology.’ Abstract: It would be naïve to assume that a good map of solid geology is the only resource needed to understand the geography of soils. The geography of soil does not depend solely on the solid geology of the biosphere and lithosphere. When one studies the geography of soil it is important to look at the properties of soils including the parent material which is usually the dominating underlying bedrock. However, one must take into account the factors that affect soil development and the processes in soils that can produce variations. For examples, climate, topography, time, biological agents such as animals and human interference. Pedology provides us with a soil classification system that can be used to determine types of soil but throughout history it has been...

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...Erosion Modelling Soil erosion is a significant environmental process that degrades the soil in which we rely on for food, fuel, clean water, carbon storage, and as a substrate for buildings and infrastructure (Quinton 2011). It is the disruption of the soil mantle – the pedosphere, or the underlying rock base – the lithosphere by the action of matter of external geomorphic factors, such as water, snow, ice, air, weathered debris, organisms and man (Zachar 1982). Both abiotic and biotic forms of erosion forms patterns that are typical for a particular area such as climate, relief, nature of the surface, activity of the organism, and activity of man (Zachar 1982). It is the degradation or aggradations of the Earth’s surface by the movement of soil material by wind, rain, overland flow and gravity (ASSIGNMENT). Problems with Erosion The movement of sediment and associated pollutants over the landscape and into water bodies is of increasing concern with respect to pollution control and environmental protection. With the expected change in climate over the coming decades, there is a need to predict how environmental problems associated with sediment are likely to be affected so that appropriate management systems can be put in place (Morgan & Nearing). Erosion can impact the productivity of agricultural, post-mining and native systems and is a sign of land degradation (ASSIGNMENT). Soil erosion acts a mechanism for transferring pollutants to surface waters and reduces...

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...ecosystem services … . So severe that the residual forest can no longer qualify as forest in any practical sense of the world.” Alan Graigner (1980, AS quoted in Saxena and Nautiyal, 1997) asserts that selective logging does not “lead to forest clearance and does not constitute deforestation”, whereas Norman Myers (1980, 1993) thinks that logging is crucial because, although it may only affect a small proportion of trees per hectare, it damages wide areas and is the precursor of penetration by the forest farmers. For the purpose of this study, the FAO’s latest definitions (1993) will be used. The FAO defines forests as “ecosystems with a minimum of 10% crown cover of trees and/or bamboo, generally associated with wild flora, fauna, and natural soil conditions, and not subject to agricultural practices” and deforestation as a “change of land use with a depletion of tree crown cover to less than 10% crown cover”. “A livelihood comprises the assets (natural, physical, human, financial and social capital), activities, and the access to...

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...Soil Conservation Working Group Report This report provided content for the Wisconsin Initiative on Climate Change Impacts first report, Wisconsin’s Changing Climate: Impacts and Adaptation, released in February 2011. THE WISCONSIN INITIATIVE ON CLIMATE CHANGE IMPACTS 1st Adaptive Assessment Report Contribution of the Soil Conservation Working Group July 2010 Contour stripcropping in central Wisconsin Photo by Ron Nichols, USDA Natural Resources Conservation Service Participants of Working Group William L. Bland, Professor, Department of Soil Science, University of Wisconsin-Madison (Working Group Chair and lead author) Kelly R. Maynard, M.S. Agroecology, University of Wisconsin-Madison (Project Assistant) Jeremy Balousek, P.E., Urban Conservation Engineer, Dane County Land and Water Resources Department Denny Caneff, Executive Director, River Alliance of Wisconsin, Inc. Laura W. Good, Associate Scientist, Department of Soil Science, University of Wisconson-Madison Kevin Kirsch, Water Resource Engineer, Wisconsin Department of Natural Resources Patrick Murphy, State Resource Conservationist, Natural Resources Conservation Service John M. Norman, Emeritus Professor of Soil science, Department of Soil Science, University of Wisconsin-Madison James VandenBrook, Water Quality Section Chief, Wisconsin Department of Agriculture, Trade, and Consumer Protection Sara Walling, Water Quality Specialist, Wisconsin Department of Agriculture, Trade,......

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...and sampling and recommended methods for static and dynamic sounding by European Group Subcommittee (1968). The probe consists of a cased screwed onto the lower end of the rod. The rods are of 16mm diameter HY steel each of length 120 cm. The rods are connected to each other by 25 mm outer diameter couplings. These couplings provide the lateral support to the rods so as to prevent buckling during driving. Driving is performed with a small hammer of 5 kg in weight and falling vertically through a fixed height of 30 cm along a guide rod. The total number of blows required for the pointer to penetrate a distance of 30 cm is recorded and used as a measure of the consistency of cohesive soil and the packing of granular soil. The relationship between Mackintosh Probe and Safe Pressure is as follows: - P = (2860 + 550 (R - 40)1/2) x 0.04788 kN/m2 for blows > 40 P = Refer Chart for blows < 40 Where, P = safe pressure (kN/m2) R = Mackintosh Probe Penetration resistance in blows/0.3m For more information or site appointment, please call us or drop us a message here. The site investigation is the one thing that must be done before starting the construction of the building. This is because the soil condition at the site need to be identifies to determine the suitable foundation use for the building. As we know, soil play a main role to support the load that come from...

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...available to assign value to biodiversity. This will be done in terms of its ecosystem services, functional value, economic value and social value. A lot of the time ecosystem services and economic value overlap. An example of this shall be discussed later on. The reason for this is because ecosystem services from nature are free, for example bees as pollinators. However when human interaction is needed for these services, that is when a monetary value is added. Ecosystem services can be separated into four categories; 1. Provisioning: Regulating and cultural ecosystem services 2. Regulating: Control of water availability 3. Cultural: Aesthetic and spiritual value of biodiversity 4. Supporting: Contributions to the maintenance of ‘healthy soil’ (Thomas Bolger, Biodiversity in our lives) These so called services are essentially priceless but a global estimation of economic values of $33 trillion has been made. It has also been estimated that earthworms in Ireland contribute up to €723 million per annum to livestock production. The earthworm may seem insignificant but it a keystone species and holds great value for the environment. (Thomas bulger, Biodiversity in our lives) Ecosystem services represent the benefits human populations gain directly or indirectly from ecosystems functions as mentioned in the four above. The question is, will impoverished ecosystems function less efficiency than more species rich ecosystems? If so, their ability to deliver ecosystem services......

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...Waves Waves are generated by wind blowing over the sea. The characteristics of waves are determined by the strength of the wind, its duration and fetch (distance a wave travels). The stronger the wind the greater the friction on the surface of the sea and therefore the bigger the wave. Constructive waves Constructive waves are flat and low in height and have a long wave length. Their strong swash carries material up the beach, forming a berm. They have a low frequency of between 6 and 8 waves per minute. The wave energy dissipates over a wide area which results in a weak backwash. Destructive waves Destructive waves have a large wave height and short wave length. They have tall breakers that have a high downward force and a strong backwash. Their frequency is high with between 13 and 15 waves per minute. Their strong downward energy helps erode beach material and cliffs. The strong backwash results in narrow beach profiles. Tides Tides are the rise and fall of sea levels. Tides are caused by the gravitational pull of the moon and to a lesser extent the sun. When the earth, the moon and sun are aligned the gravitational pull is at it’s greatest. This creates a Spring tide. A Spring tide results in a high, high tide and low, low tide. This creates a high tidal range (difference between the highest and lowest tide) and results in stronger tidal currents than normal. Spring tides usually occur twice a month when there is a full moon. When the sun and moon are at a right angle to...

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...Soil Fertility, Evaluation and Nutrient Management Major contributors to increased agricultural production * Fertiliser use * Plant breeding * Other cultural practises * (planting date, crop density, rotations etc) * Weed and pest control * Irrigation Future Challenges * World population continues to grow * Decrease in productive land due to industrial, residential, transport use and soil degradation * 2 degree increase in temperature over 100 years * Extreme weather conditions * Advances in agricultural production needed to feed population Factors for crop yield * Plant growth and yield are affected by over 50 factors * We cannot control many of the climate factors * Soil and crop factors can be managed for maximum and economical production * Table: major factors affecting yield potential The Law of the minimum * Crop yield is determined by the most limiting factor * Crop yield can only be increased by eliminating the most limiting factor * Yield cannot be increased by increasing the supply of other nutrients and factors * Challenge is to identify the limiting factors and eliminate them Sources of mineral nutrients for plant uptake * Plants take up mineral nutrients from soil solution * Amounts of nutrients in solution are small * Soil solution nutrients are replenished by there in soil/solid form Table: Nutrient dynamics in the soil that affect nutrient supply to plants The......

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...University of Phoenix Material Soil and Glaciers Worksheet From Visualizing Earth Science, by Merali, Z., and Skinner, B. J, 2009, Hoboken, NJ: Wiley. Copyright 2009 by Wiley. Adapted with permission. Part 1 Size grades of soil are named sand, silt, and clay, which includes colloids. Size grades are defined using the metric system. Use Figure 4.8 from the textbook to fill in the following chart. Specify the type and size and description of the particle. In some cases, particle size will be less than some value or greater than another value. For instance, gravel is greater than 2.0 mm. |Name |Size |Description | |Gravel |>2.0 mm |Gravel is very small, irregular pieces of rock and stone. Gravel is more rough and rocky than sand, and | | | |smaller than stones. The word gravel comes from the French word gravele, "gravel or sand," which in turn | | | |comes from grave, "seashore or sand ("Gravel ", 2015). | |Sand |0.05 |sedimentary material, finer than a granule and coarser than silt, with grains between 0.06 and 2.0 | | | |millimeters in diameter ("Sand", 2003-2015). | |Silt |0.0002 |Silt is......

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...ENVIRONMENTAL CHALLENGES IN NORTHERN NIGERIA: THE WAY FORWARD A position paper submitted to Northern Delegates at the National Conference Abuja By Yusuf Abdullahi Rigasa (PhD) yusuf.rigasa@gmail.com An Associate Chief Lecturer at the Department of Environmental Science Kaduna Polytechnic, currently on secondment to National Oil Spill Detection and Response Agency, NOSDRA, Federal Ministry of Environment Abuja. 2014 Introduction Northern Nigeria was a British protectorate which lasted from 1900 until 1914 and covered the northern part of what is now Nigeria. The protectorate spanned 255,000 miles (410,000 km) and included the states of the Sokoto Caliphate and the Kano emirate and parts of the former Bornu Empire, conquered in 1902. The protectorate was ended in 1914, when it was unified with Southern Nigerian Protectorate and Lagos Colony, to become Northern Province of the colony and protectorate of Nigeria or the Northern region. The Northern Region was one of Nigeria's federating units. It was created before independence in 1960, with its capital at Kaduna. In 1962, it acquired the territory of the British Northern Cameroons, who voted to become part of Nigeria. In 1967 the region was split into states - Benue-Plateau State, Kano State, Kwara State, North-Central State, North-Eastern State and North-Western State. Currently, the region comprises of 19 states and Federal Capital Territory Abuja. The climatic conditions in the northern part of Nigeria......

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...1. A. Soil Formation Weathering break up the surface of parent rocks into small particles. Air and water enter the spaces between the particles and chemical changes take pace which resut in the production of chemical substances. Bacteria and plant life soon appear. When plant and animal organisms die, they decay and produce a substance called humus. This is very important to so fertility. Bacteria play a vital role into the decomposition of plants and animals remains. The end product of these mechanical, chemical, and biological processes is called soil which is one of the world’s most important natural resources. B. Composition of Soil All soil contain mineral matter, organic matter, air, water and living organisms, especially bacteria. If any one of these is seriously reduced in amount, or removed frm a soil, ten the soil deteriorates. Soil is an important component of the physical environment of many ecosystem. The main components of a typical soild can be divided into two: Organic components and Inorganic components. Organic Components - include include living organism such as baceteria and fungi which serve as the decomposers;a very important parts of the recycling and decay processes. It also include many invertebrates animal such as insects and worms. Finally, there are many plants or parts and worms. Finally, there are many plants or parts of palnts such as roots and seeds. Inorganic compoents – Inorganic components include mineral, water and air.......

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