Soil Dynamics Students

8/3/2019 Soil Dynamics Students

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Dhaka University of Engineering Technology

Department of Civil Engineering

Class Lecture on

Soil DynamicsCE - 6112

Course Teacher

Dr. Md. Mokhlesur Rahman

Professor Department of Civil Engineering




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Course Teacher:Dr. Md. Mokhlesur RahmanProfessor Department of Civil EngineeringDUET

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 earthretaining structures. The study of Soil dynamics include the machine foundations, impact loadings, dynamicsoil properties, slope stability, bearing capacity, settlement, vibratory compaction, pile driving analysis andfield testing, ground anchor systems, seismic design parameters, liquefaction, sheet pile walls and laboratorytesting.

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 8.6E (10 20ergs) 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 theearth’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 siderapidly moves relative to the other side of the fault.




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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 isusually expressed in hertz (i.e. cycles per second).

The period and the frequency are interrelated as,

Free vibration: Free vibration occur under the influence of forces inherent in the system itself, without anyexternal force. However, to start free vibrations, some external force or natural disturbance is required. Oncestarted, 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 theabsence of any externally applied force, the frequency with which it vibrates is its natural frequency.




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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,

Where, Degree of freedom (n): The number of independent co-ordinates required to describe the motion of a system iscalled a degree of freedom. A system may in general have several degrees of freedom; such a system is called amulti degree freedom system.

k

(a) (b)

1

k 2

k 3 z 2

z 3

z 1

m1

m2

m3

k 1 k 2

xn = 2

n =3

n = 1

(d)(c)

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 issystematic 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 principalmodes is mode equal to unity, the motion is then called the normal mode of vibration.

∞




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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 toavoid 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 theotherwise increased effects at resonance. Various damping mechanism are available, either inherent in thevibrating 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 usuallyleads 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:

Mass

StaticCondition

Mass

(a) (b)

z = 0

Mass

(c)

z

Mass

(d)

zz

z = max

Mass

(e)

z

z = max z

DoubleAmplitude

Mass

kz + kz stat

mg=W

(e)

k k k k k +z

-z

z stat

Figure - 2: Spring Mass System

a) Un-stretched Spring b) Equilibrium Positionc) Mass in Oscillating Positiond) Mass in maximum Downward Positione) Mass in upward positionf) Free body diagram of mass corresponding to (c)




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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

The static deflection z stat of the spring is,

Where,

( is defined as force per unit deflection)

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

0

, 0 ……………………... (1)

Where, g acceleration due to gravity z Displacement

t time

m mass




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In order to solve the equation (1), we get

……………………... (2)

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

Now,

= = ……………………...(3)

From equation (1) and (3) we can write,

z 0

, z 0

, ,

(-)

(+)

D i s p l a c e m e n t , z

Time

One Cycle

Peak to Peak Amplitude

z

z

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 torepeat itself is called the period of the vibration. When the time required to complete one cycle of motionis 2, the time period T of this motion can be written as22

Frequency,




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Example: - 1A mass supported by a spring has a static deflection of 0.5 mm, Determine its natural frequency of

vibration.

Solution: Given,

g 9810 mm/sec 2

z stat 0.5 mm

. 22.32 (Ans)

Example: - 2 Find the fundamental frequency of vibration of a ve rtical 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, L20 m , Flexural rigidity, EI 125 10 2 N‐m 2 .

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

Now, frequency,

1.1 (Ans)

W = 1000 Kg

EI = 125×10 2 N-m 2

L = 20 m




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Example: - 3Determine the spring constant for the system of springs shown in figure

k k

c

x x

k

k

(a) (b)

1 2

1

2

1 2

a b

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 and of respectively.

The deflection of points a and b are of and respectively.

Therefore, deflection of point c is.

=

=

Hence, the resulting equivalent spring constant at c is,

If x 1 = x 2 = x and k 1 k 2 k ,then 2

On the application of a unit load in figure b the total deflection is, Hence, equivalent spring constant,

If k 1 k 2 k , then




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Example: - 4Write the equation of motion for the systems shown in figure (7) and determine the natural frequencies.

k k

o

x x

k 3

1 2

m

A B

Figure - 7: Equivalent Spring Constant Solution: On application of unit load at c for spring k 1 and k 2 the deflection

=

Hence, equivalent spring constant for k 1 and k 2,

Total deflection is

=

Hence, the equivalent spring constant,

We have, z




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Where,

A & B = Arbitrary constant

Example: - 5

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

Also,

r ? f

rcos? lcos f

l+r

s

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

Find the equation of velocity and acceleration.

Solution:

11

If we expand, the right hand side, by the binomial theorem we get,

1 ……

or, 1 ……

Substituting in the expression for S, we get,

1 1

1 ……




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The series in brackets contains and even powers of .

Now, 1 2 1 2

1 2 22

1 2 24

2 24

Now,

1 1 22 24……If is very small, and higher powers can be neglected.

141 2

Putting that is a function of and 2 where n = 1, 2, 3 …………….

14 12

Velocity,

sin2 2

Acceleration,

2




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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 torelative velocity, when it is low. At higher relative velocities, the damping force may be proportional to thesquare 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 is proportional to velocity, it is termed viscous damping,

Thus………….(1)

Where c is a constant of proportionality, referred to as the viscous damping co-efficient.

k c

rigid mass, m

x

x

Figure - 9: Spring mass Dashpot system

Figure (9) shows a single degree of freedom oscillator to which is added a dashpot that induces the dampingforces. From the force body diagram the equation of motion is

, 0 ………….(2)

If we define a critical viscous damping co-efficient,

is,

2√ ...………..(3)

And a damping ratio,...………..(4)

2√ 2 2

,2 where,




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So equation (2) becomes, 2 0 ...………..(5)

The general solution of this equation is,

...………..(6)

Where A & B are constants to be determined from the initial conditions at

0and

0.

,are the roots of the auxiliary equation,

, √ 1 The three cases,

D = 1.0 for Critically Damped ConditionD>=1.0, Over Damped ConditionD<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.




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If, D = 1,

The if roots are equal.

By the initial condition i.e. 0 and 00, we get the value of A & B. gives aperiodic 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 √ 1

√ 1

Where,

& are complex and conjugate roots.

For the initial conditions 0 and 00, the solution of equation ( 7) gives,

1 1

or, 1

Where, 1 and is the phage angle.

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

(-)

(+)

Time

Point of Tangency

Maximum

Figure - 11: Damped Free Oscillation




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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 dampingcoefficient (Cc).

Mathematically , ,

Figure - 12: Energy absorption of Materials

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

Deformation

F o r c e




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

(-)

(+)

Time x

A x

0

1.0

-1.0

-0.5

0.5

0.5 1.0 1.5 2.0 2.5 3.0 3.5

D = 0.2

D = 1 D = 2

D = 5

D = 8

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 sameslope; this point does not correspond to the maximum excursion for the mean position, which occurs when0, the difference is small for small values of the damping ratio, D.

Natural frequency or Un-damped natural frequency ( ) and Damped frequency ( ) and operatingfrequency ( ) 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 naturalfrequency ( ).

k ck

System mass, m System mass, m

Figure - 15: Spring mass with Dashpot system




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Mathematically,

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 ( ) of the exciting force. The operating frequency ( isindependent of the natural frequency ( ) of the system. But when, , the response of the system isinfinite. The condition is known as resonance. Damped frequency may be defined as

√ 1 .

Where, D = Damping factor Always damped frequency is less than the un-damped natural frequency

.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 is close to the natural un-damped frequency , 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 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 periodicmotion.

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 forcedvibration.

Where,

D = Damping factor √




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(-)

(+)

Time x A x

0

1.0

-1.0

-0.5

0.5

0.5 1.0 1.5 2.0 2.5 3.0 3.5

D=0.2

D=1 D=2

D=5

D= 8

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 0 and 0 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 1is one, i.e. the case where the operating

frequency of the system is equal to natural frequency . At this case, the system reaches at resonancecondition where higher or uncontrolled velocity of the system is created. For machine foundation design, thecase is unexpected.

Critical damping may be defined as the damping factor 1is one, where actual damping

and critical damping are equal. At this case, the system returns to its equilibrium position in the shortestpossible time. For machine foundation design, this case is expected.

So the above two situations are opposite to each other.

2√ Logarithmic Decrement and the experimental Determination of Damping:

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

…………… (1)

In which and are two successive peak amplitudes, from a displacement verses time trace willenable the damping constant to be estimated.

∞




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(-)

(+)

Time X

T

32.5000

50.7454

Figure - 17: Amplitude of Viscously Damped vibration

Here is an arbitrary cycle number.

If occurs at time , then from oscillatory damping motion equation, we have

1 …………….. (2)

Where,

√

At a time interval of one periodic later,

……………. (3)

1 .…………….. (4)

The values of the sine functions are equal when the time is increased by the period, , so that the ratio of thedisplacements becomes,

…………….(5)

So,

…..………… (6)




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For small values of , the equation (6) can be expressed in series form,

2 1 21 2

Hence,

2……………..(7)

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

Consider first the ratio of after cycles,

Now, ,,………………

From which is found,

……………..(8)

For the case, when

2 0.693

And,

20.693

Therefore, 0.11. If 0.1 1.1 0.2 0.55

Depends on materials, for soil If 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 afree-vibration test, the system is displaced from its equilibrium position and a record of the amplitude of displacement is made. Then we have,

2 1 2 1 2

……………………. (1)




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……………………. (2)

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

Figure - 18: Determination of damping ratio from forced vibration

The amplitude of the motion is

..………………….. (3)

Putting, frequency ratio, 1, We have,

………………………(4)

Putting, the frequency ratio , when the amplitude of motion 0.707,

We get, .

,.

,√ √

, 1 2 4 8

, 21 21 80

, , 21 2 41 241 8

21 2√ 4 16164 32

1 22√ 1




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Now, 4√ 1 4 If is very small.Now,

2.1 Since, 2

So,

4 2

,

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

Forced Harmonic Vibration with Viscous Damping

The harmonic force acts on the system which has a spring cons tant and a viscous dampingconstant as shown in figure (19). The circular frequency of the force application is , and the amplitude isconstant

.

k c

rigid mass, m

x

x

Figure - 19: Forced Harmonic Vibration with Viscous Damping

The equation of motion is found from the free body and is written as ……………………….(1)

The solution of the equation, Re = Real

0

is, ……………………….(2)

Where, A is an arbitrary constant.




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Equation (2) substituting into equation (1), we have

, . .

Where, and √ The static deflection is and phase angle is,

The phase angel is the angle, by which the response lags behind the disturbin g fore,

.

.

The amplitude of the motion is

.

,

,

1.0




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

M a g n i f i c a t i o n f a c t o r ,

Frequency ratio,

Frequency ratio,




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Significance of the above figures:

• The damping ratio has a large influence on the amplitude and phase angle in the frequencyregion near resonance when 1.

• The maximum magnification factor occurs for

1when damping is present.

• When the damping is small,

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 springforce.

• When 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 1,the phase angle approaches 180°.

• The exciting force then almost equals to the inertia force.• The amplitude approaches the static displacement when 1.

• The amplitude becomes small when

1.

• The phase angle is very sensitive to the ratio in the region of near resonance for small damping.

• When √1 2D,then reaches a maximum.

• The amplitude at resonance is found to be A .

Forced Vibration with Viscous Damping (Derivation for maximum magnification)

A spring-mass-dashpot system under the action of a force of excitation such that,

…………………… (1)

Where,Frequency of force of excitation.

The equation of motion for free-body, …………………….(2)

The solution of the equation is,

…………………….(3)Where,




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Equation (3) substituting into equation (2), we have

.And,

, .

Where, √

,

For maximum value of magnification factor,

0

, . . . . . 0

, 0 4 18 0

, 12 ,√ 1 2

,√1 2 , √1 2

Where, 2 2 Now,

1 11 2 4 And 2 4 41 24 8




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So,

.

√

So,

12 12

, 1 22

, √ 1 2

, for maximum vibration, which is always avoided.

xFk Static de lectio

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

M a g n i f i c a t i o n f a c t o r ,

Frequency ratio,




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

Effect of frequency ratio, for a particular case

When

0,

If, 0, 1 If, 1, ∞ If ∞ , 0

At 1,resonance occurs and amplitudes tend to infinity.

The introduction of damping reduces the amplitudes to finite values.

The phase angle is zero if

1; the displacement, is in the phase with the exciting force, and °

if 1.

Frequency ratio,




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Effect of Damping factor:

As the damping increases, the peak of the magnification factor shifts slightly to the left. This is dueto the fact that the maximum amplitudes occur in damped vibrations when the forcing frequency equalsthe system’s damped natural frequency ,which slightly smaller than the un-damped naturalfrequency

.

For 1,the phase angle is90°for all values of damping, except when 0 For 1,the phase angle is less then 90° For 1,the phase angle is greater than 90°

The maximum amplitude of motion at 1and 0,is . Force transmitted to foundation:

The force transmitted to the foundation by the spring and damper system is given by:

………………..(1)

And after substitution from equation

. .

And,

k c

rigid mass, m

x

Figure- 22: Force transmission by spring damper




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We have,

. . ………………….(2)

And as

ω . .√ . 2

,ω . .√ .

, 2 ………………… (3)

.. .. . .

, . . . . . …………………(4)

The amplitude of the force transmitted is given by

.

.

, . .

The ratio is referred to as the transmissibility of the force.

,




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Figure-23: Plot of force transmission and frequency ratio

Now, the ratio 1, when

1 2 .1 2 . , 1 1

, 1 1 0,2

, 0,√ 2

,√ 2 1

If 1, 1

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

If D is negligible and √ 2, then 1.

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

F o r c e

t r a n s m

i s s i b i l i t y , T

Frequency ratio




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If D is negligible and √ 2, then 1 1

Replacing, and

2, we have

12 1

Where

If is small, the is large.

√ 1 20 1 20

1√ 2 0.707

0 sin 0is a static case.

Which means that the maximum response is the static resp onse.

1 0.6

Where,

2




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Example: - 6A machine of mass 100 kg is supported directly on springs which have a total stiffness of 2000 KN/m. Anunbalanced rotating mass results in a disturbing force of 200N at a speed of 3000 rpm. Assuming a dampingfactor of D = 0.10 determine the amplitude of vibration, the transmissibility, and the force transmitted to thefoundation.

Solution: The static deflection of the system is 100 9.812000 100.49 100.49

The natural frequency of the single degree of freedom system is

22.5 The ration of forcing to the natural frequency, then becomes

.2.22

The amplitude of vibration,

. . . . . 0.12

The transmissibility,

. . . .. . .0.2

The transmitted force is then

0.277 200 5 .

Example: - 7A machine of mass 100 kg is supported directly on springs which have a total stiffness of 100 KN/m. Anunbalanced rotating mass results in a disturbing force of 200N at a speed of 3000 rpm. Assuming a dampingfactor of D = 0.10 determine the amplitude of vibration, the transmissibility, and the force transmitted to thefoundation.

Solution: The static deflection of the system is

100 9.81100 109.8 109.81

The natural frequency of the single degree of freedom system is

5.03 The ration of forcing to the natural frequency, then becomes

. 9.94




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The amplitude of vibration, . . . . . 0.10

The transmissibility,

. . . .. . .0.022

The transmitted force is then 0.0227 200 4. .

Example : - 8

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

Solution: Let, 1

12 , 9512

, 95 2,95 2 ……………………………(1)

Again,

12 , 7512

, 75 2,75 2 ……………………………(2)




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Equation we get,

1.60

,1.60

, w 1.67 kg .

Again, 95 2

, .. 60.65 / .

Exercise - 1: A body 65 kg weighing is suspended from a spring which deflects 1.57 cm under the load. It is subjected to adamping 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 withrespect to the disturbing force.

Solution:

.. 40615

Natural frequency of the un-damped,

f 398 Hz An

Again, ω ω 1 D or, 2π f 2πf 1 D or, f f 1 D3.98 1 0.253.85 HzAns

The ratio of successive amplitudes

. .5.06 5

Amplitude . . . . .27.2

Phase Angle

... 0.87545°




Page 36 of 102

Tensional Resonant Column Test

The shearing strain on a circular cross section in a tensional resonant column test varies from zero at thecentre 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.

J

x

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,

θ ,

θ ,

14

θ




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For Clay shearing strain amplitude up to 1%Sand shearing strain amplitude up to 5% (for 40psi or 276KN/m 2 confining pressure.)

Cyclic tri-axial compression tests

Cyclic test permit evolution of modulus, either

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 cyclicsoil testing, the cyclic tri-axial compression test was developed first.

Figure - 25: Cyclic shear test

In this test cylindrical tri-axial samples are initially consolidate under a cell pressure, resulting in stressshown by condition 1, figure (26)

Condition 1

Condition 2

Figure - 26: Stress Condition




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In principle, the sample is then subject to an increase in axial stress and a simultaneous reduction

in the cell pressure by an equal amount (condition -2, Figure-26). The normal stress on the 45° plane throughthe sample is not changed but a shearing stress of is developed on that plane. The axial stress and cell

pressure are then simultaneously reversed by 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 bymaintaining the cell pressure at a constant value and cycling the axial stress by σ as shown in figure (27).

Figure - 27: Constant Cell Pressure




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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 andtested un-drained. If samples are partially saturated or tested with drainage, it is necessary to utilize axial andlateral stress control to simulate earthquake loading. In the many versions of the cyclic tri-axial test, theconfiguration of the specimen is standard but the loading and control equipment are variable. Most currentlyused 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, , and damping ratio are oftenmeasured in the cyclic tri-axial test (Figure - 29) by performing strain controlled tests. These tests areperformed in essentially the same manner as the stress controlled test, however, a servo system is used toapply cycles of controlled deformation.

V e r t i c a l L o a d

V e r t i c a l D e f o r m a t i o n

P o r e P r e s s u r e




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

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

For strained strain controlled tests, shear modulus is computed from, in which is Poisson’s

ratio.

The cyclic tri-axial test has limitations among which are:

1. Shearing strain measurements below 10percent 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.

Vertical Stress,

Vertical Strain

∆

∆

12




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Dynamic Soil Properties

The major dynamics properties are:

• Shear strength (S u) evaluated in terms of strain rates and stress - strain characteristics• Dynamic moduli, Young modulus E(E v,E h), Shear modulus G(G max ) and Constrained modulus• Poisson’s ratio ν(vh, νhh, νhv)• Damping (D)

E v and E h are equal for isotropic behavior of soil. Normally, E v and E h are not equal because of fabric,characteristics of grains, particles orientation in a grain mass of soil and geometric history of loading on soilmass. Always, E v > E h .

The majority of the numerous analytical methods presently available for assessing the response of soildeposits or soil structure due to earthquake, explosion or machine loading require and accurate assessment of maximum shear modulus (G max ) in the field, where G max is defined as the shear modulus at shearing strainamplitude 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.

The is defined as the ratio of horizontal strain to vertical strain at shearing strain amplitudes less or equalthan 0.001 percent.

ν

0.001%

Figure - 30: Strain Stress curve




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Dynamic soil testThe 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 propagationUp hole or down hole wave propagationSurface wave propagationBlock Vibration test C u, D, E, G, C ψ Cyclic Plate load test C u,E, G

Laboratory Test

Cyclic tri-axial test E,G,Wave Propagation test/wave velocity method E,G, ν

Resonant column test E,G,DCyclic Simple shear testCyclic Tensional Simple shear testX-ray diffraction analysis/Ultra sonic pulse testAir pollution methodWater submergence methodWet compaction method




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Single amplitude axial strain test:

(-)

(+)

Time

(-)

(+)

TimeDouble Amplitude

0.001% Strain

Figure - 31: Strain and stress have same direction

The stiffness at which 0.001% strain is called young modulus, ∆∆

∆ 0

0




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30%

0.421

Small Stiffness

Granular Soil

Stiffness is increased withincreased vertical stress.1

0.50

Unique Relationship

Figure - 32: Shear modulus and Shear strain relationship

E o r G




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Bore hole wave propagation:

In the method, the velocity of wave propagation from one surface boring to a second subsurface boring ismeasured. At least two bore holes are required, one for the impulse and one or more for sensors. As shown infigure 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 knowndistance 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




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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.

/

Now,Shear modulus = Young modulus = 21

Where, ν = Poisson’s ratio of soilρ = mass density of soilv = 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-holemethod, 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 isused in the bore hole.

S2

S1

R

R

Up bore-hole Down bore-hole

S = SourceR = Receiver

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




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Up bore-hole method Down bore-hole method

Shear modulus

Young modulus 21 21 Where,

ν = Poisson’s ratio of soilρ = mass density of soilv = velocity of shear wave

Expander Pump

Rubber Enpander Back Plate

3 componentgeophone

Recorder

WoodenHammer

Triger Geophone Weight

Figure - 35: Equipment and instruments of down hole survey




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Surface wave propagation method:

The Rayleigh wave (R-wave) travels in a zone close to the surface. An electronic or other harmonicvibrator can be used to generate a steady state R-wave and the ground surface can be deformed as shown infigure-36.

x

S S

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 thehorizontal plate of the oscilloscope is fixed 30 cm away from the oscillator along a ray drawn so that thesensing axis of the geophone is vertical. A similar geophone connected to the vertical plates of theoscilloscope, is moved along this ray, away from the oscillator. The sensing axis of the geophone is keptvertical until the Lissajous figure on the oscilloscope screen becomes a circle. However, if the phase angle isdeferent 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 soilsextend to infinite depths and the Lissajous figure is a circle, the wave length, λ of propagating waves is given

by, 4 Velocity of shear waves, In which is the frequency of vibration at which the wave length has been measured. and of soil mediumare calculated as follows

21

Where,ρ = mass density of soil

= velocity of shear waves ν = Poisson’s ratio of soil




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Value of Poisson’s ratio of soil0.05 for clay0.03 - 0.35 for sand0.15 - 0.25 for rock

If compression waves are used, the

In which= velocity of shear waves

S = DistanceT = corresponding time of travel of wave

Then Elastic modulus is determined by:

Block Resonance Test

Block resonance test for determining modulus and damping values. A standard block 1.5×0.75×0.70 m high iscast either at the surface or in a pit 4.5×2.75 m at a suitable depth (figure - 37) and is excited in both horizontaland vertical modes.

For H Test

Motor Oscillator Assembly

For V Test

AccelerationPicks Up

Depth to be

selectedConcrete (M50)

1 m min

1 m min

4.50 m

2.75 m

Figure - 37: Block Resonant test setup




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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 itgenerates 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 aconstant frequency. The oscillator frequency is increased in steps of small values, say, from 1 cycle up to themaximum frequency of the oscillator, and the signals are recorded. The same procedure is repeated for thevarious dynamic force values. All force level and frequency; the dynamic force should not exceed 20% of totalmass of the block and motor-oscillator assembly.

In case of forced-vertical-vibration tests, the amplitude of vibration, at a given frequency is given by

In which represents the vertical acceleration of vibration in

/.

The coefficient of elastic uniform compression of soil is given by,

In which,

√ .

In forced vertical vibration tests, the value of damping co-efficient, D of soil is given by the followingequation,

Figure - 38: Determination Damping ratio from forced vibration test




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Logarithmic decrement,

√

In which,

,Two frequencies on the amplitude frequency plot at which the amplitude is equal to

√

= Maximum amplitude

= 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 areusually made of steel and are 25 mm thick and 150 mm to 762 mm in diameter. To conduct a test, a hole isexcavated 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 untilthe settlement becomes negligible. The final settlement is recorded by dial gauges. Then the load is removedand the plate is allowed to rebound. At the end of the rebounding period, the settlement of the plate isrecorded. 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,




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1.13√ Kg/cm 2

1 √1.13

2 1

In which P = Corresponding load intensity (Kg/cm 2)

S c = Elastic rebound (cm)

A = Contact Area

Magnitude of C z can be obtained from the plot of q vs. S e from figure

Factors affecting stress-deformation and strength characteri stics 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, 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 besubjected to deformations of the type shown in figure - 40.




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x x

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 theshearing chamber to deform the soil uniformly.

Shearing Chamber

Plan View

Soil Sample

End Plate Rotation Soil Deformation

Elevation

Figure - 41: Schematic diagram illustrating rotation of hinged end plates and soil deformation in oscillatory simple shear

0




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Test data from simple shear tests have been analyzed to determine shear parameters, soil moduli anddamping.

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.

Typical SoilElement

x x

Base Motion

Rock

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 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 axialcompression or equal to major principal stress during lateral compression.




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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 until a limiting strain, • Modulus beyond strain, • Strain,

When the direction of strain is reversed, behavior is again determined by the modulus until a strain changeof 2has developed and the modulus again controls the behavior. This pattern then continuousthroughout the cycle.




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Pendulum Loading Apparatus:

Point

N e w A p p a r a t u s R a d i o u s 1 8 '

P i l o t A p p a r a t u s R a d i o u s 7 '

Load Gauge

Test Specimen

Adjustable reaction

Upper Cylinder

Deformation gauge

Spring, k = 250lb/inch, 6 inchlong, 3 inch dia

Hydraulic cylinders3 in bore, 3 in stroke

Lower Cylinder

Figure - 43: Pendulum Loading Apparatus:

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

The apparatus utilizes the energy of a pendulum which, when released from a selected height, strikes aspring connected to the piston rod of a hydraulic (lower) cylinder. The lower cylinder, in turn, is connectedhydraulically to an upper cylinder, which is mounted within a loading frame. The time of loading for apendulum loading apparatus is proportional to the square root of the weight of the pendulum and is inverselyproportional to the square root of the spring constant. In addition, the maximum force is proportional to thefirst 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 0.015was 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 ametal ring. The strain introduced in the gauges was then in direct proportion to the load. These load gaugescan be calibrated under static load and can be used in a dynamic test. Similarly a deformation gauge wasconstructed on a cantilever metal strip with electric resistance strain gauges, mounted on one end while theother end rested on an unmovable support. The strain introduced in the cantilever was a measure of thedeformation of the soil sample.




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30 50%LL = 37 - 59% PL = 20 - 27%

S t r e s s ( K g / c m

² )

Stress1

2

4

3

00.04 0.08 0.12 0.16 0.20 0.24

Strain

Shear failure at 0.02 sec

Time, sec

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 thantheir 10 - min static strength.

• Modulus of deformation, defined as the slope of a line drawn from the origin through the point on thestress - deformation curve and corresponding to a stress of one-half the strength, was about two timesin 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 completeliquefaction of sands.

This aspect of the problem is of great practical importance.




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Differences between dynamic and static test:

Dynamic Static1 It is monotonic or cyclic loading test with relatively

large strain rate.It is monotonic or cyclic loading test withrelatively 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-frequencyor 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.




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Definitions of several types of stiffness

Secant Young’s modulus,

Tangent Young’s modulus

Secant Young’s modulus at

Equivalent Young’s modulus

q max

E sec

0

Emax

Axial Strain

D e v i a t o r S t r e s s , q q 0

1

1

1

2qSA

E eq

E tan

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

2




Page 60 of 102

: Young’s modulus and shear modulus obtained from in situ elastic shear wave velocity , whichare 21

Where,

q max

0

E

q1

max2

Figure - 46 : Plot of Stress vs Axial strain




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Monotonic loading for over consolidated soils and cyclic presheareared soils, Drained

Cyclic Loading

-510

-410

-310

-210

-110

-610

G

Gmax

E

E maxeq

Limit of elasticrespose in

monotoric loading.

Decay Curve

Monotonic Loading

Peak

Undrained cyclic loadingof saturated loose sandand M.C soft clay

Residul

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

E eq

1

E sec

0

Figure - 48: Variation of shear stress verses shear strain

,




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

Geq

G f

Gmax

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

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




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x

J

J0

x

Rigid mass Specimen; non-rigid distributed mass

Driving force

Weightless Spring

Specimen; non-rigid distributed mass

Fixed Support

Deshpot

Weightless Spring

Rigid mass

(a) (b)

(c) (d)

Driving force

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 sine wave but by adding a

mass with mass polar moment , at the top of the specimen as in figure (50-b)the variation of θ along thesample becomes nearly linear. The end effects to obtain uniform strain distribution through the length of thespecimen. The apparatus configuration in figure (50-c) can be described as the “spring base” model. For acondition 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 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 ispartially restrained by a spring and dashpot which in turn reacts against an inertial mass. For 1tests,the inertial mass is balanced by a counter weight, but by changing the counter weight, axial load can beapplied to the specimen.

14 ∞ 0.50

, ,14




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Machine foundation

Machine foundations are subjected to the dynamic forces caused by the machine. These dynamicforces are transmitted to the foundation supporting the machine. These loads may result from various causessuch 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 responseis a pulsating curve. It vanishes before the next pulse. The speed is usually between 60 to 150 blowsper minute.

3. High speed machines, such as turbines and rotary compressors. The speed of such machines is veryhigh; 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. Thefoundation 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 thanthat 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 restson 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.

Pedestal Box

Base Slab Wall

Top Slab

Columns

Frame

(a) Block Type (b) Box Type

(c) Wall Type (d) Framed Type

Figure - 51: Types of Machine Foundation.




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Suitability of various types of Machine Foundation

Machines which produce periodical and impulsive forces at low speeds are generallyprovided with a block type foundation. Framed type foundations are generally used for themachines working at high speeds and for those of the rotating type. Some machines which inducevery 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 besatisfied:

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, thelimiting 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 4 for frequencies less than 30 hertz

and 125 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 atright angles in the shaft should be least equal to the distance between the centre of the shaftand 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 topof 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 toensure 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 (

q) 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 betweenthe 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 filledwith 1 : 2 cement mortar.




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12. A number of similar machines can be created on individual pedestals on a common raft. Theanalysis for such machines can be made assuming that each foundation acts independentlywith an area of foundation equal to that obtained by dividing up the raft into sectionscorresponding to separate machines.

Reinforcement and Construction Details:

1. The reinforcement in the concrete block should not be less than 25 kg/m 3. For machinesrequiring special design consideration of foundations, such as machine pumping explosivegases, the minimum reinforcement is 40 kg/m 3.

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 beplaced 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/mm 2.8. 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 sometimesrecommend the mass of foundation required for the machines. However, the mass recommended aregenerally empirical rust based largely on experience.

Vibration isolation and control:

Vibrations may cause harmful effects on the adjoining structures and machines. Besides, thesevibrations cause annoyance to the persons working in the area around the machine. However, if thefrequency ratio is kept outside the critical range of 0.4 and 1.5, and the amplitude is within thepermissible 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). Thefollowing measures are generally adopted.

1. The machine foundation should be located away from the adjoining structures. This is knownas geometric isolation. The amplitude of surface waves (R-waves) reduces with an increase indistance. A considerable reduction in the amplitude is achieved by locating the foundation at agreat 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 frequencymachines to make it a multiple degree freedom system and to change the natural frequency. Inreciprocating machines, the vibrations are considerably reduced by counterbalancing theexciting forces by attaching counterweights to the sides of the crank.




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3. Vibrations are considerably reduced by placing absorbers, such as rubber mountings, felts andcorks between the machine and the base.

4. If an auxiliary mass with a spring is attached to the machine foundation, the system becomes atwo-degree-freedom system. The method is especially effective when the system is inresonance.

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 operatingfrequency.

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.

Mathematical problems and Solutions of Machine Foundation

Example: - 9A mass is supported by a spring. The static deflection of the spring due to the mass is 0.04 cm. Find

the natural frequency of vibration.

Solution: Given,

g 981 cm/sec 2

z stat 0.04 mm

. 24.94 (Ans)

Example: - 10The weight of a machine foundation is 30 kN and the spring constant, k is 6×10 3 kN/m. Determine thenormal frequency of vibration and the period of oscillation.

Solution:

. 7.05

1

17.050.142




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Example: - 11A machine foundation is subjected to a maximum vertical force 2500 . Given: weight of

machine + foundation = 18000 kg; spring constant 70 10/; operating frequency of the machine,40 . Assuming that the foundation can be idealized as a mass spring system, determinea) The natural frequency of the system, and

b) The maximum vertical displacement of the foundation.

Solution:

70 101800098161.8 .

Maximum amplitude

1 250070 1014061.80.06

Example: - 12A machine foundation weighs 60 kN and has a spring constant k = 11 000 kN/m. Assuming the dampingcoefficient, c, of the system as equal to 200 kN-sce/m, determine

a) Whether the system is over damped, under damped or critically damped. b) The logarithmic decrement.c) The ratio of two successive amplitude.d) Damped natural frequency.

Solution:

Critical damping coefficient is

2√ 2 11000. 518.76

Damping factor,

.0.386 1, So, Under damped condition

Logarithmic decrement, 2 12 3.14159 01 0.3862.63

The ratio of peak amplitude,

.

13.87

Damped natural frequency

√1

.6.75

6.75√1 3.866.23




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Example: - 13The weight of machine foundation including the machine is 15 kN. The spring constant of the supporting soilhas been found out to be 11 10. The foundation is subjected to a vertical forced vibration of the type40, where 150 /.If the damping factor D of the system is 0.2 determine

a) The un damped natural frequency of the foundation.

b) The amplitude of motionc) The damped natural frequency.d) Effect of amplitude of motion if damping is ignored.

Solution:

.14.8

2 2 14.8 93

. .0.00021

√ 1 14.8√1 0.214.50

. .0.00022

The effect of ignoring the damping is to increase the amplitude of motion by about 8% which is not verysignificant in this case.

Example: - 14

A square concrete foundation block is to be designed to support a centrifugal pump of weight equal to10.4 tonnes and operating frequency equal to 12 hertz. The supporting soil consists of dense silty sand γ 1.8 t/m. The specification required the vertical mode of vibration that the resonant and operatingfrequencies be separated by a factor of at least 2, and the maximum vibration velocity is 2 mm/sec. themaximum vertical out of balance force at the operating frequency is estimated to be 15 kN. The results of

cyclic plate load test on the silty sand with a 0.5 m×0.5m plate give elastic uniform compression

10 /.

It has been proposed to use a concrete foundation block 3m×3m×1m. determine whether this proposalwould satisfy the specification requirements.

Solution:

The weight of footing + pump = 3 3 1 2400




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54 10. 9 10/ . 24.83 10

158 /or

25.2 Ratio of natural and operating frequencies = .2.1

The specification is satisfied.

For the amplitude of motion

Operating angular frequency,

12 2

. 24 100.024

The maximum vibration velocity at 0,

12 2 0.024 1

Example: - 15A rectangular concrete foundation of size 16 ft × 7 ft × 3 ft is to be constructed for a vertical single-cycliccompressor. The following data are available.Operating frequency of compressor = 450 rpm, Constant amplitude force of excitation, 10. Thecoefficient of elastic uniform compression, 350 /.Required

a) The natural frequency of the system b) The vertical amplitude (maximum)

Solution:

Total weight = Wt of foundation block + compressor = 16×7×3×150 + 10900 = 61300 lb force

.1904

194 / .




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31 1860

Operating angular frequency,

2 47.1 / .

.16.1 100.0002

Example: - 16A compressor is supported on a concrete block foundation of size 2.5 2.5 and 1.5 thick which is locatedon the surface of a sandy clay soil. The compressor has a total weight of 120 and an operating frequency of 10 hertz. The total weight of the oscillating mass in the compressor is 2.5 kN at an eccentricity of 15 cm. Thedominant vibration is vertical. Field tests give a value of 134 10/for the spring constant,Determine

a) The damped natural frequency

b)

The maximum vertical amplitude at the operating frequency.

Assuming the damping factor 0.54, and the unit weight of concrete = 24 /.

Solution: Total weight of foundation block + compressor = 2.5×2.5×1.5×24 + 120 = 345 kN-force

. 35.2kg-sec/m

Natural angular frequency

. 61.7 /.

Damped natural frequency, is

√1 61.7√1 0.5451.93 /

The vertical amplitude is

.. 0.255 /.

10 2 62.8 /.

..1.018

1.036




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.. .. . . . .0.00093

Example: - 17Determine the natural frequency of a machine foundation having a base area

2 2and a mass of 15

Mg, including the mass of the machine. Taking 4 1043.Solution:

103.28 /.

.16.43 Example: - 18The natural frequency of a machine foundation is 4 hertz. Determine its magnification at the operatingfrequency of 8 hertz. Take damping factor (D) as 0.30.

Solution: 2

M . 0.31

Example: - 19The exciting force of a machine is 100 kN. Determine the transmitted force if the natural frequency of themachine foundation is 3.0 Hz. Take D = 0.40 and the operating frequency as 5 Hz.

Solution: | | 12 M . 0.45

| | 100 0.45 12 0.475




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Example: - 20Determine the coefficient of uniform compression if a vibration tests on a block 1 1 gave aresonance frequency of 30 Hz in the vertical direction. The mass of the oscillator used was 60 kg.

Solution:

Mass of foundation block

1 1 12400 2400

Total mass = 2400 + 60 = 2460 kg.

8.74 10/

Example: - 21A 2.50 Mg vertical compressor foundation system is operated at 40 Hz. The soil at the site is medium stiff clay4 10/. Determine the natural frequency and the magnification factor, assuming

0.2 . The base area is 2.5 . Take D = 0.

Solution:

Total mass = 2.5 + 0.2×2.5 = 3.0 Mg. = 3×10 3 kg.

.29.06

M . . 1.12




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Example: - 22In a test block of the size 1 .5 1.0 0.75 , resonance occurs at a frequency of 20 cycles per second in the vertical vibration. Determine the coefficient of elastic uniform compression if themass of oscillator is 70 kg and the force produced by it at 15 cycles per second is 1000N. Also compute themaximum amplitude at 15 cycles per second.

Solution:

2 2 20 40

Mass of oscillator = 70 kg.

Mass of block = 1.51.0 0.75 2400

Total Mass = 70 + 2700 = 2770 kg

Contact Area =

1.5 1 1.5

40 .

29.16 1029.16 10/

Maximum Amplitude

5.23 100.052

Example: - 23A foundation block of weight 30 kN rests on a soil for which the stiffness may be assumed as 25000 kN/m,The machine is vibrated vertically by and exciting force 3.030 .find the natural frequency, naturalperiod, natural circular frequency and the amplitude of vertical displacement. The damping factor is 0.5.

Solution:

The exciting force is

3.030.

. 90.42 /.

. 14.39




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.0.069

.0.33

M . . .

1.05

Static deflection, . 0.012

Amplitude A 1.05 0.012 0




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LiquefactionQuick Sand condition:

A soil under critical hydraulic gradient will be unstable and is sail to be in a ‘quick’ condition. By thisdefinition any granular soil may be a “quick sand” but soil with high permeability i.e. gravels and coarse sandsrequire 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

0 , Where,

Thus, when the pore pressure equals the total pressure on a plane, a quick condition exists and the porepressure can only equal the total pressure when

∆ 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 willtemporarily liquefy. This phenomenon is termed as liquefaction. In the situation just described, four criteriawere 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, thestructure is somewhat unstable. The grain size is such that the pore pressure can “float” the grains. The resultis 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 flowlaterally. This phenomenon has been observed to occur in several fairly recent earthquakes. It also sometimesoccurs 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 upwardpropagation of shear waves in given by as stress ratio,

Where,

20⁄ 10⁄

1 0.015 1.




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.

Acceleration due to gravity,

0.15 6.0

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

Maximum resisting shear stress ratio,

0.06760.225. for 0.04 mm ≤ ≤ 0.60 mm and

0.06760.05for 0.6 mm ≤ ≤ 1.50 mm and

Where,= mean particle (50% finer) diameter of soil in mm.

The normalized SPT N‐value obtained through the co‐rellation factor for overburden

pressure,

Defined as .. Where,

The effective overburden pressure = kgf/cm 2

After so determining the induced stress and shear strength of a soil element, the liquefaction potential of asand deposit is evaluated in terms of factor of safety, defined as,

,

If the factor of safety is less than on

1, the liquefaction is said to take place, otherwise liquefaction does

not occur.

The required N-value ( i.e. un-corrected N-values) obtained by

14.7930.225.

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




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Depth(m)

σ0 = σv

(KPa) σ0’ = σv’ CN γd

Induce stress ratioτmax / σv’

N req F 1 = 1.0 F 1 = 1.25 F 1 = 1.5

01

2345

SPT (N-Value)

Depth (m)

Liquefaction Potentiallyof Subsoil at alocation

F >1.01 1

max

0(Induced)

max, 1

0(resistance)

N req

N1

max0

max, 10,

Figure – 52: Plot of and with SPT




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




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Acceleration spectra & Velocity spectra:

For any given ground motion, values of the spectral velocity, and the spectral acceleration, for asingle-degree-of-freedom structure having a period are related approximately by the equation,

0

0

0.5

1.0

2.0

1.5

0.5 1.0 1.5 2.0 2.5 3.00

1

2

4

3

0.5 1.0 1.5 2.0 2.5 3.0

Damping Factor

0.01

0.2

0.05

M a x i m u m

A c c e l e r a t i o n , g

Natural Periods

M a x i m u m

V e l o c i t y ,

f t / s e c

Natural Periods

Damping Factor

0.01

0.2

0.05

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

The maximum base shear, for a multistory structure subjected t oa given base motion can be estimatedfrom the equation

.

Where,W = The weight of the structure

= Spectral acceleration corresponding to the natural period of the structure (SDOF)

= 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 earthquake2. Distance of the site from the source of energy release3. Geologic characteristics of the rocks along the wave transmission path from source to site4. Source mechanism of the earthquake5. Wave interference effects related to the direction and speed of fault rupturing6. Local soil conditions at the site




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Influence of soil conditions on ground motion characteristics:

0

0.2

0.3

0.5

0.4

0.6

0.7

2 5 10 20 50 3001

Mean for Rock sites, M = 6.6s

Deep Alluvium or Mean for ImperialValley, M = 6.8s

Closest Horizontal Distance

P e a k H o r i z o n t a l A c c e l e r a t i o n

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,

Closest Horizontal Distancefrom Zone of Energy Release

P e a k G r o u n d

A c c e l e r a t i o n

H o r i z o n t a l

M = 5.0 M = 6.6

M = 5.6

M = 7.6

M = 8.5

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

Magnitude.




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Approximate relationship between maximum acceleration on rock and other soils:

Maximum Acceleration on Rock, g

M a x i m u m

A c c e l e r a t i o n , g Soft to medium

stiff clay and sand

Rock Stiff Soil

DeepCohesionlessSoil

0.2

0.3

0.5

0.4

0.6

0.7

0.1

0.1 0.2 0.3 0.3 0.4 0.50 0.6 0.7

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 accelerationdeveloped on different types of soil do not differ appreciably, particularly at acceleration levels less than about0.3 to 0.4 g. Even at higher acceleration levels on rock of the order of 0.7 g, accelerations on deposits of anydepth 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 foundationconditions except soft to medium clays are plotted together, it may not be possible to differentiate betweenacceleration levels for rock and different site conditions.

1.4 1.5

Maximum Ground Velocity:

Geologic Condition ⁄

Rock 55 cm/sec/g

Stiff soils (<200) 110 cm/sec/g

Deep stiff soils (>200) 135 cm/sec/g




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0

1

10

50

30

80

100

3 10 30 100 500 10001

Distance from zone of Energy Release, Km

P e a k V e l o c i t y , c m

/ s e c

Soil Sites

Rock Sites

M = 6.5 Mean Value

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

Response Spectra:

0

1

2

4

3

0.5 1.0 1.5 2.0 2.5 3.0

S p e c t r a l A c c e l e r a t i o n

M a x i m u m

G r o u n d

A c c e l e r a t i o n

Periods, (s)

5% Damping

Normalized AccelerationResponse Spectrum

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




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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.




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Selection of Ground Motions for Design

Differentiate between two earthquake motions that are related but can be significantly different from eachother:

1) The maximum free-field earthquake ground motions which a structure should be able to withstand withan acceptable margin of safety.

2) The level of shaking in evaluating the safety of a structure. This motion, which is often called the “DesignEarthquake 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 and2. 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 beendetermined.

2. Peak accelerations for different soil conditions may be obtained by correcting the mean peak acceleration for rock sites.

3. Values of 1standard deviation acceleration can be determined from relationship11.4 1.5

4. For sites within about 50 Km/s, values of maximum ground velocity.

For rock sites:

55 / /

For stiff sites: 110 / /

5. Values of maximum spectral acceleration,

For mean spectral shape: 2.7

For 1spectral shape: 3.4




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6. Design values recommended by the seismic review panel:Maximum acceleration = 0.75gMaximum velocity = 0.35 cm/secMaximum 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 and3. 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 overburdenpressure, the effective stress becomes zero, the sand loses its strength completely, and it develops aliquefied state.

Where,

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.

(a) Idealized field loading condition.




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S h e a r S t r e s s

( + )

( - )

Time, s

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 confiningpressure and the sand will readily begin to undergo large deformations will shear strains, they mayexceed ± 20 percent or more. If the sand will undergo vertically unlimited deformations withoutmobilizing 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 becomemore compact with a resulting transfer of stress to the pore water and a reduction in stress on the soilgrains. As a result, the soil grain structure rebounds to the extent required keeping the volumeconstant, and this interplay of volume reduction and soil structure rebound determines the magnitudeof the increase in pore water pressure in the soil. The basic phenomenon is illustrated schematically infigure – 64. The mechanism can be quantified so that the pore pressure increases due to any givensequence 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 reboundcharacteristics of the sand due to stress reduction.

Pressure

V o i d

R a t i o

AC

B

e 0

Compression Curve

e = Equivalent volumevhange of grain structuredue to cyclic strainapplications duringdrainage loading

Induced PorePressure

f 0

u

= effective pressureinitial and final stages

f 0,

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

(b) Shear stress variation determined by response analysis.




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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 thecyclic stress is reapplied on the next stress cycle, or if the sand is subjected to monotonic loading, the soil willtend to dilate, the pore pressure will drop if the sand is un-drained, and the soil will ultimately develop enoughresistance to withstand the applied stress.

However, it will have to undergo some degree of deformation to develop the resistance, and as thecyclic 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 soilwill be able to with stand any number of cycles of given stress without further increase in maximumdeformation. This type of behavior is termed “Cyclic mobility” and it is considerably less serious thanliquefaction, it’s significance depending on the magnitude of the limiting strain. However, that once the cyclicstress applications stops if they have to a zero stress condition, there will be a residual pore water pressure inthe soil equal to the overburden pressure and they will inevitably lead to an upward flow of water in the soilwhich 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 conditions6. The geometry of the particular situation7. 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 properties2. Environmental factors and3. 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 modulus2. Damping characteristic3. Unit weight4. Grain characteristics5. Relative density6. Soil structure




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Environmental factors:

1. Method of soil formation2. Seismic history3. Geologic history (aging, cementation)4. Lateral earth pressure co-efficient

,,

5. Depth of water table6. Effective confining pressure

Earthquake characteristics:

1. Intensity of ground shaking2. Duration of ground shaking

Some of these factors cannot be determined directly, but then effects can be included in the evaluationprocedure by performing cyclic loading tests on undisturbed samples or by measuring the liquefactioncharacteristics 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 differentdepths in the deposit and conversion of the irregular stress histories to equivalent number or uniformstress 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 aground response analysis involving the unit weight of the soils, the dynamic moduli and the soildamping 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 samplesconducted at different confining pressures, or by correction of these properties with some measurablein-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 appropriatelycorrected to be representative of field conditions. By this means the soil type, the in-plane the seismicand geologic histories of the deposit of the initial effective stress conditions are approximately takeninto account.




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

Figure – 65: Zone of liquefaction

3. A composition of the shear stresses included by the earthquake with those required to causesliquefaction, to determine whether any zone exists within the deposit where liquefaction can beexpected 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 dueprimarily 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 stresson the soil element would be

.

Where,The maximum ground surface acceleration, and

= The unit weight of the soil.

Because the soil column behave as a deformable body, the actual shear stress at depth, h, addetermined by a ground response analysis will be less than and might be expressed by




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Maximum shear stress,

Where, is a stress reduction <1.

max r

max d

m

h

(a) (c)(b)

D e p t h

Figure – 66: Procedure for determining maximum sh ear stress

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

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

.




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

Earthquake magnitude Number of significant stress cycles,

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 = 50%

Initial void ratio = 0.68

Initial confining pressure, 5.0 /

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 nosignificant deformation.

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

The number of stress cycles required to calculate the sample to liquefy depends on the magnitude of theapplied 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 causing liquefaction in

the number of stress cycles representative of the design earthquake.




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

(-)

(+)

S h e a r S t r e s s

S h e a r S t r a i n

P o r e W a t e r P r e s s u r e

R e s p o n s e

(a) Pore Water Pressure Response

(-)

(+)

(b) Applied Cyclic Shear Stress

(c) Shear Strain Response

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

Number of Cycles tocause failure (log scale)

P e a k C y c l i c

S h e a r

S t r e s s , , K

g / c m ² Medium Imontercy Sand

Initial effective confiningPressure, = 2.0 Kg/cm²

Strength

0.1

h0.2

0.3

0.4

1 10 100 1000

20 cyclic

v

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




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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 and3. 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 consolidatedunder an ambient confining pressure, , and then subjected to cyclic deviator stress application .Test dataare similar in form to those obtained in cyclic stress ratio, which causes liquefaction or cyclic mobility in

eh desired number of cycles. The cyclic stress ratio 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 thelaboratory by the expression (Seed, 1979).

2

= Cell pressure

Where values of are approximately.

0.57 0.400.9 1 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 deposit2. Soil structure or grain arrangement3. Cementation at grain contacts, which increases with the age of the deposit.




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Example: - 24

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. 0.16. Given115 ,0.93 ,

h = 20' h = 16'w

Solution:

Let,

62.5 Total overburden pressure = 20 115 2300 Effective pressure

4 115 16115 62.51300

Cyclic stress ratio developed by earthquake,

0.65

0.65 0.15230013000.93 0

Case study - 2

Number of Cycles of liquefaction

C y c l i c

S h e a r S t r e s s R a t i o ,

0.32

0.1

0.2

0.3

0.6

2 20 100

A0.4

0.5

0.7

0.8

10 50

15

Tube Sample

0.5

B

Block Sample

2

105

103




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For a critical depth of about 20 ft:

20'

10'

20 120 2400

10 120 10

The cyclic stress ratio developed by the earthquake (M = 7.5)

0.65

0.65 0.25240018000.93 0

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

0.57

0.57 0.40

0.57 0.32 0.18 0.20It 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 isfound to be,

0.57 0.5 0.28 0

Not to be liquefied the sand.

And the factor of safety against liquefaction is

. .. 1.40




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Questions

Soil Dynamics1. What is soil dynamics?2.

What are the nature/sour3.

At what Richter Scales, 4.

What are situation when5. What are the problems in6. Define periodic motion a7. What is resonance?8. What is dumping?9. What are the principal o10. How will you minimize 11. Sketch the different posia spring m12. Define by figure for forcinitial equations.13.

Define spring constant.14. A mass supported by a svibration.15. Write down the starting 16. Define the damping con17. Draw the sinusoidal curv18. Write down the name fo• z, ωn, C, , Cc, D, , K, Ax, Fd, Ø, ωnd, sinØ, f n 19. Write down the expressi20.

Define the following tera.

Damping factorb. Critical frequencyc. Critical damping21. With initial and final exp22. Write down the relations23. What are ranging values24. Write down Richter Scalearthquake.25. Write down the relationsa.

and b.

D and f for bandwidth c. r and Dd. D and x26. Define harmonic force w27. Define magnitude factor28. Draw a graph for magnit29. Write down the significa30. Draw the graph phase an31. Write down the 4 signifi




Page 100 of 102

32. What are the effect of moccurred? Sketch proper33. What are effects of D onwith variation of34. Define transmission of f35.

Define transmissibility fdifferent damping.36.

What are values of frequ37. Practice the Examples.38. Define natural frequencyDynamic Soil Properties

1. Write down the name of2. Define isotropic soil and3. Define shear modulus an4.

What are field tests for s5.

What are laboratory test 6. Write down name and sy7. Draw the figure for cros8. What are four sources usproperties?9. What the calculations ar10. Draw a figure for up hol11. Draw a schematic diagra12. How will you calculate f13.

What are the calculation14.

Draw the R‐wave and de15.

What do you mean by lifigure for dynamics soil 16. How will you calculate s17. Draw the figure for bloc18. How will you calculate t19. Draw the graph for cycli20. How will you calculate t21. What are the factors affe22. Draw and define the bili23.

Draw the figures for pen24.

What are short comings 25. What are the differences26. With graph define severa27. Draw a graph for linear 28. Define resonant column 29. Draw the schematically 30. Define torsional resonan31. Draw the figures for tors




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32. Define cyclic tri‐axial co33. Draw the mohr‐circle at 34. Show that vertical load, cycles of load.35. What are dynamics para36.

Draw the loop for comprparameters from figure.37. Why is the average sheaMachine Foundation

1. Mention the characterist2. Mention the name of thr3. What are the measures g4. What are the specificatiofoundation?5.

What are the design criteLiquefaction

1. Define the parameters for liquefaction; mention the equation of stress ratio.2. With the limitation of fine particle size express the maximum resisting shear stress ratio equations.3. Define the normalized SPT N - value.4. Define factor of safety againstliquefaction.5. State the phenomenon `q6. What is liquefaction pot7. Draw the graph structural damage intensity verses depth of soil for various numbers of stories.8. What are the two tropics of most general interest concentrating liquefaction?9. What are main characteristics of the ground motions? Draw their graphs for liquefaction.10. Draw the graphs for maximum acceleration and velocity verses natural periods for various damping

factors.11. When and why liquefact12. What are the simple guid13. What are the equations for maximum base shear for multistory structure?14. What are factors affecting earthquake ground motions for liquefaction?15. Draw the relations between maximum acceleration verses maximum acceleration on rock for various

types of soil.

16. Draw the graph

verses period for various types of soil (5% damping).

17. How will you select of ground motion for design earthquake.18. Write down the important five factors affecting design earthquake motion.19. What are two aspects for specification of design spectra?20. What are the simple guidelines of design spectra of earthquake during liquefaction?21. What are specifications of design acceleration for liquefactio?22. Write down the three causes of soil liquefaction.23. By the graph show the schematic illustration of mechanism of pore pressure generation during cyclic

loading during earthquake for liquefaction.




24. Define cycle of mobility25. When is the cyclic stress applications stop during liquefaction?26. What are the factors on which liquefaction depended?27. What are the soil proper 28. What are environmental factors considered for evaluating liquefaction?29. By the graph show the zone of liquefaction.

30. Why is average shear stress 65% of maximum shear stress?31. Show by the figure the procedure for determining maximum shear stress induced by earthquake.32. What are the numbers of significant stress cycles for corresponding earthquake magnitude?33. Mention geotechnical properties of loose Monterey sand. What is the number cyclic (normally) stress

required such a soil failed during liquefaction?34. Show the graphical variation of pore pressure; shear strain and shear stress with number of cyclic

applied stress when liquefaction occurred.35. What are the graphical relation between peak cyclic shear stress and number of cyclic for Monterey

sand when effective confining pressure = 2.0 kg/cm 2?36. What are the difficulties in simple shear testing for liquefaction?37. What are the remedies during simple shear testing for liquefaction?38. What are the factors influenced on sand deposit for the action of cyclic loading characteristics?

Math

1. Compute the cyclic streswater table, 7 ft below tacceleration 0.18g, sarespectively.2.

At unknown weight w atIf 2 kg weight is added tspring constant k.3. A body 65 kg is suspenddamping effect adjusted frequency of un‐dampedsuccessive amplitudes i25 kg and a frequency eq

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