5 eg- ce- engineering properties of rocks prt.pps

8/15/2019 5 EG- CE- Engineering Properties of Rocks PRT.pps

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Dr. Abdelrahman Abueladas

Lecture 5

Faculty of Engineering

Civil Engineering

Engineering Geology

CE

Engineering Properties of rocks & Rocks deformation


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Rock properties are an essential part of the exploration , design ,

construction and during service life of the project.Rock classification which provide rock names and geologic

characteristics for most engineering applications.

Intact Rocks:

Is a rock containing no discontinuities, such as joints and beddings, it is called rock material!

Rock mass :

It is a mass of rock interrupted b" discontinuities , with each

constituent discrete block having intact rock .

Intact Rock

#a" be described b" standard geologic terms such as $1% rock

name , $&% mineralog" ,$'% (exture $)% degree and kind of

cementation, and $5% weathering.

Engineering Properties of Rocks

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5

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(he t"pical intrusive igneous rock will have larger cr"stals than

extrusive igneous rocks, while sharing similar cr"stalline interlocking

textures and same composition of silicate mineral .

The metamorphic rock name provides information about

mineralog" and degree of foliation if present .

sedimentary rock name ma" impl" a certain ph"sical features

while leaving others undefined .*andstones and shale are defined b" their predominant discrete

grains.+ limestone is defined b" its composition rather then grain sie. It

ma" be composed of cemented individual grains of calcite derived

from wave action ,or ma" be a dense cr"stalline limestone from

precipitation of a lim" mud on a sea or lake bottom.Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5

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-iffering amount of cla" minerals, silt or sand sied grains of

uart and fossils ma" be present in a given specimen, but this will

not alter the rock name, but all affecting the engineering properties .-escriptive adjectives such as /finel" cr"stalline ,

$orgillaceous% or sand" $arenaceous% are applied.

0eologic data are informative, but it dos not provide the engineerwith the uantitative data that are needed. Rock is usuall"

anisotropic due to precipitation environment or due to tectonic a

activities afterwards.


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Rock strength

(he strength and elasticit" of intact rock are used in the design

of dams and pressure tunnels , prediction of amount of

deformations and their rate in openings made in highl" stressed

highl" elastic rocks ,and the operation and performance of tunnel

boring machines.

Properties and inde!es that defines intact rock properties

are listed "elo#:

Rock t"pe *trength2olor 3ardness *onic 4elocit"

0rain sie -urabilit" oung #odulus

(exture 6 7abric 8orosit" 8oisson s ratio

9eathering -ensit" 8ermeabilit"


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$trength: is defined as the applied stress that cause rock failure or rupture%

the applied stress may "e compressive% shear or tensile

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5:

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'nia!ial Compressive $trength ('n confined)*


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(he factors that affect the compressive strength of intact rock

are(+* Confining pressure and (,* Rate of stress application*uggested rate of stress application is =.5 to 1.= #pa>sec.

(he specimen used is at lest 5)mm diameter with a height todiameter ratio?&.5'.=.

(he following table present @niaxial compressive strength of nine

common rock t"pes from the three major classes of rock .

A(he smaller crystal size in basalt compared with granite is a primar" reason for the higher mean strength and max strength of

basalt .

A(he presence of gas voids in some basalt cause significant

reduction strength .In granite, cr"stal sie is primar" strength factor.

BoteC #pa $#ega 8ascal%? million 8ascal $1,===,=== pa%

1 8a ? 1.)5=';;D1=E) psi $ 8ounds per suare inch%


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2lass ofRocks

Igneous #etamorphic *edimentar"

*trength 0ranite Gasalt 0neissH

*chist uartiteH

#arbleH

Limestone *andstone *hale

+v.*trength

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AThe degree of foliation affect the average strength and range for

metamorphic rock as for gneiss and schist data.AJther factors such as variation of mineralogy , crystal interlocking , and

orientation of foliations with applied strength, affect significantl" the

strength of metamorphic rock .$gneiss, schist, Ketc%.

A(he absence of foliation and strong bonding between uart particles

cause ver" high intact rock strength of uart.

$edimentary rock C Limestone ma" range in strength from strong for highl"

siliceous rocks ,to low for ver" cla"e" $argillaceous% or shal" limestone.

(he strength of sandstone is affected b" degree and type of cementation

and the proportion of clay and silt present in the sandstone

*trength of shale a function of clay mineral type of cementing agent

,orientation of bedding planes relative to the applied stress .


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Intact rock can be classified according to the compressive strength as

follows.

*trength categories @nconfined compressive*trength #pa

4. high $strong% &5=

3igh to 4. high $strong% 1==&5=

$#edium or #oderate%tohigh

5=1==

9eak to #oderate &55=

4. low $weak% 1&5


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Tensile $trength(ensile strength of intact rock is the least common determined rock

strength propert". (ensile or extension strength is needed in the

following casesC*lope stabilit".

Roof spans of underground excavation.

(unnels and mines.

(ensile strength of rock is controlled b" the same factors that governs

compressive and shear strength, i.e. composition, texture, grain size, kindand amounting of cementing material, and moisture content . (ensile

strength is much lower than other two strengths.


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2ore specimens with lengthtodiameter ratios $L>-% of between

& to &.5 are placed in a compression loading machine with the

load platens situated diametricall" across the specimen. (he

maximum load $8% to fracture the specimen is recorded and usedto calculate the split tensile strength.


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.echanical Properties of Rocks

Rock propertiesC mass densit", porosit", and

permeabilit" *tress

#ohrMs circle

*train

Nlasticit" of rocks

Rock propertiesC strength

Nngineering classification of intact rocks

Rock mass properties


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A Rock properties:A $pecific gravityC the ratio between the mass and that of eual

volume of water $i.e. the ratio of mass densit" and water

densit"%.

A 'nit #eight gamma?$specific gravit"%x$unit weight of water%

unit weight of water? :&.) pcf $lbs>ft'%

for most rocks, gamma ? 1&= to &== pcf.

A Porosity n measures the relative amount of void space

$containing liuids and or gases%.

porosit"?$void space%>$total volume%

A Permea"ility measures the rate at which fluids will flow

through a saturated materials. 9e will discuss themeasurements of permeabilit" later in the lecture of

0roundwater.


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A $tressA *tress is force per unit area acting on a plane at

an" point within a material. (here are three t"pesof stressesC

A compressive stressC eual forces that act towards a

point from opposite directionsA tensile stressC eual forces that pull awa" from

each other.

A shear stressC eual forces that act in opposite

directions but offset from each other to act as a

couple.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 51<


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A Principal stresses

A Jn an" plane within a solid, there are stresses

acting normal to the plane $either compressional

or tensional, called normal stresses% and shear

stresses acting parallel to the plane. +t an" point

within a solid, it is possible to find three mutuall"

perpendicular principal stresses which are

maximum, intermediate, and minimum. Jn the

planes perpendicular to the principal stresses

$called principal planes%, there are not shearstresses.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 51F

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A .ohr/s circleA *uppose we wish to measure stresses $both normal and

shear% acting on an" given plane besides the principalstresses. In general, this is a three dimensional problem

and can be done using mathematical tensors and vectors.

A In a special case where we can assume that theintermediate and minimum stresses are eual $for example

below the ground surface%, we can work in t#o

dimensions. #ohrMs circle provides a simple, graphical

method to find the normal and shear stresses on inclined planes from principal planes using the maximum and

minimum principal stresses.

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A $trainA (he application of stress to a material causes it to

deform. (he amount of deformation is called

strain.

A axial strainC deformation along the direction ofloading ∆L>L.

A lateral strainC the lateral extension perpendicular to

the direction of loading, ∆G>G.

A 8oissonMs ratio ? $lateral strain%>$axial strain%.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&1

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A Elasticity of rocks

A *ome of the deformation of a rock under stress will be recovered when the load is removed. (he

recoverable deformation is called elastic and the

nonrecoverable part is called plastic deformation.

8lastic behavior involves continuous deformationafter some critical value of stress has been reached.

A 2ommonl", the elastic deformation of rock is

directl" proportional to the applied load. (he ratio of

the stress and the strain is called modulus of

elasticit".

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&&

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A R k ti t th


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A Rock properties: strengthA Rock strength indicates the level of stress needed to cause failure.

A compressive strength is the compressive stress reuired to break a rock sample.

(he unit is pounds per suare inch $psi% or newtons per suare meter $pascals%.

A unconfined (unia!ial* compression test:

A the rock sample is unconfined at its side while the load is applied verticall" until

failure occurs. In this case, the compressive strength is called unconfined

compressive strength $uniaxial compressive strength%.

Aconfined compression test:

A 7or design of underground structure $such as tunnels, mining, waste repositor"%,

we need to take into account of the confining pressure at depth. (his is done at

laborator" b" socalled triaxial compression test. (he failure curve constructed

using #ohrMs circle after a series of tests gives the shear strength $cohesion% and

internal friction $angle of shearing resistance% of the rock $or soil% sample. (hiswill be further discussed on #ohr2oulomb failure criterion in the next lecture on

*oil #echanics.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&'

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A Engineering classification of intact rocksA Intact rock is internall" continuous, intact, and free from weakness

planes such as jointing, bedding, and shearing.

A (he standard engineering classification of intact rocks is based on the

unia!ial compressive strength $+ through N% and the modulus of

elasticity, developed b" -eere and #iller $1F::%.

A (he unia!ial compressive strength is divided into five categoriesC +

through N for ver" high to ver" low level of strength, ranging fromabove '&,=== to below ),=== psi.

A Rock classification also involves the modulus of elasticit". #ore

specificall", the modulus ratio is used, which is the ratio of the

modulus of elasticit" to the unconfined compressive strength. (hreemodulus ratio categories are 3 $high% for O5==, # $medium% for &==

5==, and L $low% for &==.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&)

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Classification of intact rocks


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Classification of intact rocksIgneous rocks:

*trong when consisting interlocking network of cr"stals $which explains the small range for

granite%.

7or cr"stalline rocks, the smaller grain sie gives higher strength $the average and maximum

strength of basalt is higher than granite%.

Nxtrusive rocks have variable strength, because of possible vesicular, p"roclastic textures.

$edimentary rocks:

0imestone% dolomiteC cr"stalline texture, thus generall" strong, but variable $fossils%.

$andstone: wide range depending on the degree of cementation.

$haleC variable because of bedding.

.etamorphic rocks:

*trength increases in some cases because of compaction and recr"stalliation.

$chists have wide variation because of foliation.

1uart2ite: strong because of interlocking silica cr"stals and absence of foliation.

.ar"le: similar to limestone or dolomite and smaller strength range.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&5

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A Rock mass properties

A (he strength and deformation properties of intact rocks cannot be directl"

applied to the overall rock mass in the field situation. (he strength and

behavior of a rock mass are largel" controlled b" the nature of its

discontinuities or #eakness planes. -iscontinuities act to lower the strength

of the rock mass. (he rock mass tends to fail along existing weakness planes

rather than develop new fracture within intact solid rocks.

A Nxamples of rock mass discontinuities includeC

A sedimentar"C bedding planes, sedimentar" structure $mud cracks, ripple marks, cross

beds, etc.%

A structuralC faults, joints, fissures

A metamorphicC foliation

A igneousC cooling joints, flow contacts, intrusive contacts, dikes, sills

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A(he deformation is the change of shape and sie of

a material under loading.

A(he elastic deformation is the part or the kind of

deformation that can be recoverable, i.e., after theload is removed, the material changes back to its

original shape and sie,A(he part or kind of deformation that cannot be

recovered is the plastic or ductile deformation.

Rock -eformation

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$TRI3

Change in shape or si2e of an o"4ect inresponse to an applied stress5 Deformation

Three Types of $train

6Elastic

6-uctile (Plastic*

67rittle (Rupture*

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5&<

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Th T f $t i


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Three Types of $train+8Elastic -eformation

+ temporar" change in shape or sie that is recovered

when the applied stress is removed.

,8-uctile (Plastic* -eformation

A+ permanent change in shape or sie that is not

recovered when the stress is removed. Ai.e. it flows orbends

98 Rupture is a kind of 7rittle -eformation

(he loss of cohesion of a bod" under the influence ofdeforming stress.

A i.e. it breaks!

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$h . d l ( " k th l tt G* CE


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$hear .odulus (some "ooks use the letter G*

(he shear modulus describes how difficult it is to deform a cube of the

material under an applied shearing force.

7or example, imagine "ou have a cube of material firml" cemented toa table top. Bow, push on one of the top edges of the material parallel to

the table top.If the material has a small shear modulus, "ou will be able to deform

the cube in the direction "ou are pushing it so that the cube will take on

the shape of a parallelogram.If the material has a large shear modulus, it will take a large force

applied in this direction to deform the cube.0ases and fluids can not support shear forces. (hat is, the" have shear

modulus of ero.7rom the relation given above, notice that this implies that fluids and

gases do not allow the propagation of the shear motion carried b" the

seismic *waves.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5''

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7rom the euations given above notice that this implies that CE


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7rom the euations given above, notice that this implies that

fluids and gases do not allow the propagation of

* waves.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5')

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7ulk .odulus ; CE


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7ulk .odulus ; Imagine "ou have a small cube of the material making up the medium

and that "ou subject this cube to pressure b" sueeing it on all sides.If the material is not ver" stiff, "ou can image that it would be possible

to sueee the material in this cube into a smaller cube.(he bulk modulus describes the ratio of the pressure applied to the

cube to the amount of volume change that the cube undergoes.If k is ver" large, then the material is ver" stiff, meaning that it doesnMt

compress ver" much even under large pressures.If P is small, then a small pressure can compress the material

b" large amounts.7or example, gases have ver"

small Gulk #odulus .*olids and liuids have large Gulk #odulus

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$ i i l it t i l< h i tiCE


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$eismic velocity vs material


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Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5)1

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=.=='≈Ultimate strain

ε

+*(#

σ

Initial modulus

Secant modulus

$tress8$train Relationship

tangent modulus

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5)&

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$ecant .odulusC *ecant modulus is the slope of a line

drawn from the origin of the stressstrain diagram and

intersecting the curve at the point of interest.. *ecant

modulus is commonl" denoted b" Ns.

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Tangent .odulusC (angent modulus is defined as

the slope of a line tangent to the stressstrain curve

at a point of interest. (angent modulus can havedifferent values depending on the point at which it

is determined.

Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5))

E!ample: CE


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+xial Load$8% PB

2ompression 3

*train S ?3> ho

*tress T?8>+

#8a

= = = =

;: =.) .==& 15.1

1&' =.< .==) &).5

1:' 1.& .==: '&.)

1F; 1.: .==< 'F.&

&&; &.= .=1 )5

&== &.) .=1& 'F.;

1;: &.; .=1'5 '5

p(he following data were obtained from unconfined compressive test on intact rock

sample.

-o?


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2alculation of 8oissonWs ratio when =>?1.&mm, -?=.1&mm

S1 ? 3>ho?1.&>&==?=.==:

Sh ? ->- ?=.1&>S1 ?.==15>.==:?=.&5

Es [email protected]

.a! unia!ial [email protected]


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);

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)<

-ynamic Elastic .oduli CE


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y

oungWs modulus N, shear modulus 0, and 8oissonWs ratio Y

ma" be obtained b" d"namic methods, b" rapid application of

stress. (his can be achieved b" subjecting the sample to

ultrasonic compression and shear wave pulses, throughtransducers attached to both ends of the sample. (he pulse is

emitted from one end and received from the other end. (he

velocit" of the wave is calculated from the travel time $t% and

length of sample $L%4?L>t

(he shear wave velocit" 4s is about &>' 4 p the compression or

8wave velocit"

%$&

&M

%$

%)'$

&&

&&

&

&&

&&

&

s p

s p

d

sd

s p

s p

sd

V V

V V atio s !oisson

V k " #odulus$hear

V V

V V V k % #odulus&oung

−

−

=

=

−

−

=

µ

ρ

ρ


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5=

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Q ? mass densit"

4p? 8 or compression wave velocit"

4s? * or shear wave velocit"k? constant, depending on units used

Nxample '.&

+ granite rock sample was tested using nonZdestructive methods

using compressive pwave and shear swave.

(he following results were obtained C

-iameter of sample ? := mm, length ? 15= mm

Gulk densit" ? &.:)' g>cm'

8wave travel time through sample ? &.F= x 1=5 sec* Z wave travel time through sample ? 5.)5 x 1=5 sec

Gulk densit"% ? &.:)= g>cm'

-etermine C $a% Nd, 0d, Yd


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7ulk Elastic Properties

(he bulk elastic properties of a material determine how much

it will compress under a given amount of external pressure.

(he ratio of the change in pressure to the fractional volumecompression is called the bulk modulus of the material.

(he reciprocal of the bulk modulus is called the compressibilit" of the substance. (he amount of compression of solids and liuids

is seen to be ver" small.

(he bulk modulus of a solid influences the speed of sound and other mechanical waves in the material


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Tria!ial Compression TestCE


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Tria!ial Compression Test

In a triaxial compression test, the direction of the load is

called the maximum principal direction and the direction of

the confining pressure applied is the minimum principaldirection.


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Mohr’s Circle for Stress StatesCE


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Introduced b" Jtto #ohr in 1


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g $ % p p p,

and is eual to one half the angle between the line (xy and the Taxis as shown in the

schematic below,

(he angle [ p defines the principal directions where the onl" stresses are normal

stresses. (hese stresses are called principal stresses and are found from the originalstresses $expressed in the x, y, z directions% via,

Lx"


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The method for dra#ing .ohr/s Circle is as follo#s:

1 9e draw a coordinate s"stem with the x axis representing the normal stresses and the "

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1. 9e draw a coordinate s"stem with the xaxis representing the normal stresses, and the "

axis representing the shear stresses.

&. @sing the values from a given structural element $-iagram 1%, we graph two initial

points. 8oint \ with coordinates$ %, and 8oint with coordinates $ % as

shown in -iagram &.

'. 9e now connect points + and G. (he line connecting points + and G

intersects the xaxis. (his is the center of #ohrMs 2ircle.

-iagram 1

-iagram &

5: -r. +bdelrahman +bueladas

2ivil Nngineering 2lass

(h f ll i ti b d t l l t th t i # h i l

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(he location of the center of #ohrMs 2ircle is 2 ?

from the origin.

(he radius of the circle is given b"

R =

(he following euations can be used to calculate the stresses in #ohr circle


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50 MPa

50 MPa

25 MPa

25 MPa

80 MPa

A

B

80 MPa

25 MPa

25 MPa

x

y

Draw the Mohr’s circle of the stress element shown below. Determinin

the !rinci!le stresses an" maxim#m shear stresses.

$hat we %now&

'x( )80 MPa

'y( 50 MPa

*xy ( 25 M!a

Coor"inates of !oints&

A+)80,25-

B+50,)25-

in" c / Answer & )5

in" 1 / Answer & 3.

'( 54. M!a

'2( )84. M!a

*max ( 3.5 M!a


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#ohrMs 2ircle can be used to transform stresses from one coordinate set to another.CE


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*uppose that the normal and shear stresses, T x, T " , and ( xy, are obtained at a point o in the

bod", e!pressed #ith respect to the coordinates XY e #ish to find the stresses e!pressed

in the ne# coordinate set X'Y' % rotated an angle D from XY % as

(o do this we proceed as followsC

A-raw #ohrMs circle for the given stress state $T x, T y, and ] xy/ shown below%.

A-raw the line ( xy across the circle from $T x, ] xy% to $T y, ] xy%.ARotate the line L xy "y ,D (t#ice as much as the angle "et#een XY and

X'Y' *.

A(he stresses in the ne# coordinates $T x) , T y) , and ] x)y) % are then read off the

circle.Dr. Abdelrahman Abueladas Civil Engineering Class Engineering Geology Lecture 5

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+lso, we can use the following euation to calculate the stress at an" inclined plane. GutCE


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, g " p

"ou showed now the right angle and the sign of the angle before "ou substitute in these

euations

AE!ample

5 eg- ce- engineering properties of rocks prt.pps

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