SHEAR STRENGTH OF SOIL (4).pptx

7/23/2019 SHEAR STRENGTH OF SOIL (4).pptx

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EAG 345 –GEOTECHNICAL

ANALYSIS

By: Dr Mohd Ashraf MohamadIsmai

!iii" Mohr#Co$om% of S&r'ss



Mohr Circles & Failure Envelope

Y

σc

σc

σc

GL

As loading progresses,Mohr circle becomeslarger…

.. and fnally ailure occurswhen Mohr circle touchesthe envelope

∆σ



Shear ailure mechanism

At ailure, shear stress along the ailure surace (τreaches the shear strength (τ .

σ τ

τ

σ τ

τ



σΖ

σ!

τzx

Soil element

())

*+A

5) *+A

4) *+A

Mohr Circle of stress

Suppose a cubical sample o soil is sub"ected tothe stresses shown in the above fgure

#e would li$e to $now what the stresses at anypoint within the sample due to the applied

stresses.



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


% ($&a

'

($&a

)

*)

+))

+. he frst step is to plot a normal stress a!is(abscissa and shear stress a!is (ordinate



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

-. Assign a sign convention. #e will use anti*cloc$wise shear and compression as ve

% ($&a

'

($&a

( ), x zx

σ τ

( ), z zx

σ τ A !()),4)"

B !5),#

4)"



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

/. &lot the stresses. &oint A (+)),0)represents the stresses on the hori1ontalplane while point 2 (),*0) represents thestresses on the vertical plane. 3oin A2

A (+)),0)

2 (),*0)

% ($&a

'

($&a

( ), z zx

σ τ

( ), x zx

σ τ



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

0. #ith point ) as origin and 4A or 42 as radiusdraw a circle.

A (+)),0)

2 (),*0)

% ($&a

'

($&a4



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

. he ma"or principal stress is the value o thenormal stress at point +. he minor principalstress is the value o the normal stress atpoint /.

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

% ($&a

'

($&a



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

Alternatively5 the principal stresses are related to the stresscomponent '1 '! %1! by5

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

22

1

2

2

3

2 2

2 2

z x z x zx

z x z x zx

σ σ σ σ σ τ

σ σ σ σ σ τ

+ − = + + ÷

+ − = − + ÷

% ($&a

'($&a



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

Alternatively5

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

maxτ

1 3max

2

σ σ τ

−=

% ($&a

'

($&a



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

. #e will now determine the pole o Mohr6scircle. 7raw a line through A to represent thehori1ontal plane and a line through 2 torepresent the vertical plane. he intersectiono these planes is the pole, denoted by &.

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

% ($&a

'

($&a



σΖ

σ!

τzx

())

*+A

5) *+A

4) *+A


)

*)

+)

)

8. #e will now determine the pole o Mohr6s circle.7raw a line through A to represent thehori1ontal plane and a line through 2 torepresent the vertical plane. he intersection o

these planes is the pole, denoted by &.

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

&

% ($&a

'($&a



&


D C

BA

-

E

.

z σ

zxτ

zxτ

xσ

% ($&a

'($&a

A5 ( ), z xy

σ τ −

25 ( ), x xy

σ τ A ( ), z xy

σ τ −

2 ( ), x xy

σ τ

&oint A on the Mohr6s circle represents the stresses on the plane A2. Sothe line A& is drawn parallel to A2. &oint & become the &ole (& in thiscase

(&ole

9 we need to fnd stresses on a plane :;, we draw a line rom thepole parallel to :;, he point o intersection o this line with the

Mohr6s circle is < he coordinates o < give the stresses on the plane :;

/o' m'&hod of 01di12 s&r'ss's ao12 a

+a1'

.

<( ),n n

σ τ

1σ 3

σ



σ6

2

'

3

'

1 σ σ +'

3σ '

1σ

PD = Pole w.r.t. plane

θ

(σ’, τ )

Orientation of Failure Plane

φ6

’1

’1

’3

’3

’

τ

’1

’1

’3

’3

’

τ

;ailure

envelope

(90 = θ)

hereore,

90 = θ + φ6 > θ

θ = 45 + φ62



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+))

?. @et6s fnd the direction o the ma"or principalstress plane. A line is drawn rom the pole, & tothe point representing the principal stress, i.e…point +. the angle +&A is the inclination o thema"or principal plane to the hori1ontal and it ispositive as shown (anti*cloc$wise direction to

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

&

% ($&a

'

($&a

Horio1&a+a1'

Maor+ri1i+a

+a1'



σΖ

σ!

τzx

())*/a

5) */a

4) */a

Mohr Circle of Stress

)

*)

+))

A (+)),0)

2 (),*0)

+/

3σ 1σ

&

3σ 1σ

% ($&a

'

($&a

ψ

Horio1&a+a1'

Maor +ri1i+a

+a1'



σΖ

σ!

τzx

())*+A

5) *+A

4) *+A


)

*)

+)

)

A (+)),0)

2 (),*0)

+/

3 27.8σ = 1 122.2σ =

&

Alternatively5 the angle between the ma"or principal stress planeand the hori1ontal plane ( is5

1

tan zx

x

τ ψ

σ σ =

−

% ($&a

'($&a

2ψ

ψ



σΖ

σ!

τzx

())

*/a

5) */a

4) */a


)

*)

+)

)

A (+)),0)

2 (),*0)

+/

3σ 1σ

&

B. o determine the stresses on any planeoriented at C rom the hori1ontal plane(cloc$wise rom the hori1ontal plane isnegative, draw a line rom the pole, & at an

angle C to the hori1ontal plane, A& here we

*C

θ τ

θ σ ( ),θ θ σ τ

% ($&a

'($&a

C




)

*)

+)

)

A (+)),0)

2 (),*0)

+/

3σ 1σ

&*C


1 3 1 3

1 3

co!22 2

!"n 2

2

θ

θ

σ σ σ σ σ θ

σ σ τ θ

+ −= +

−=

( ),θ θ σ τ

% ($&a

'($&a

σΖ

σ!

τzx

())

* /a

5) */a

4) */a

θ τ

θ σ

C




)

*)

+)

)

A (+)),0)

2 (),*0)

+/

3σ 1σ

&*C


co! 2 !"n 22 2

!"n 2 co! 2

2

z x z x

zx

z x

zx

θ

θ

σ σ σ σ σ θ τ θ

σ σ τ θ τ θ

+ −= + +

−= −

( ),θ θ σ τ

% ($&a

'($&a

σΖ

σ!

τzx

())

* /a

5) */a

4) */a

θ τ

θ σ

C



Example 1

A sample o soil ().+ m ! ).+ m is sub"ected to the orces shown in

;ig. +. 7eterminea '+ ,'/ and

b he ma!imum shear stressc he stresses on a plane oriented at /)o countercloc$wise rom the

ma"or principal stress plane

σΖ

σ!

τzx

5 *N

3 *N

( *N

;ig. +



Example 1





σΖ

σ!

τzx

5 *N

3 *N

( *N

;ig. +



Example 1

/ro'd$r':

here are two approaches to solve this problem. You can either useMohr6s circle or the appropriate eDuation. 2oth approaches will beused here.

(+ ;ind the area5 Area5

2#.1 #.1 #.#1 m A = × =

(- Ealculate the stress5

$orce 55## %Pa

area #.#1

33## %Pa

#.#11

1## %Pa#.#1

z

x

zx

σ

σ

τ

= = =

= =

= =

σΖ

σ!

τzx

))$&a

/))$&a

+))$&a

;ig. +



Example 1

(/ 7raw Mohr6s circle and e!tract '+ ,'/ and %ma!

A()),+))

2 (/)),*+))

%

'

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))



Example 1


A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

1

3

max

54# &Pa

2# &Pa

14# &Pa

σ

σ

τ

==

=

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))



Example 1

(0 7raw the pole on Mohr6s circle. he pole o Mohr6s circle is shown

by point & in the ;ig. below

A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))

/



Example 1





σΖ

σ!

τzx

5 *N

3 *N

( *N

;ig. +



Example 1

( 7etermine . 7raw line rom & to '+ and measure the angle

between the hori1ontal plane and this line.

A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))

#22.5ψ =

Horio1&a/a1'

Maor +ri1i+as&r'ss +a1'



Example 1

( Alternatively the angle between A4E can be used to determine

the

A ()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))

C

##45

22.52 2

θ ψ = = =

CO



Example 1

(8 7etermine the stresses on a plane inclined at /)o rom the ma"orprincipal stress plane. 7raw a line M6 and F6 through & with an

inclination o /)o rom the ma"or principal stress plane, angle E&F. he coordinate at point F is (0?), +-)

A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+))

-)

)

/))

*+))

*-))

*/))

+)

)

-)

)

/)

)

0)

)

)

)

8)

)

/)o

C

C

/N6

Coordi1a&' a& +oi1& N is!47),(8)"

M6 /a1' i1i1'd a&3)) from &h'

maor +ri1i+as&r'ss +a1'



Example 1 (Alternative solution)


A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

1

3

max

54# &Pa

2# &Pa

14# &Pa

σ

σ

τ

==

=

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))




Alternatively, we can use analytical solution rom the describedeDuation e!plained beore5

A()),+))

2 (/)),*

+))

3σ 1σ

maxτ

%

'

1

3

max

54# &Pa

2# &Pa

14# &Pa

σ

σ

τ

=

=

=

+))

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))




Analytical

solution5

2

2

1

22

3

2 2

2 2

z x z x zx

z x z x zx

σ σ σ σ σ τ

σ σ σ σ σ τ

+ − = + + ÷

+ − = − + ÷

1 3

max

2

σ σ τ

−=




Ehec$

:Duilibrium5

8895o

5)) */a

()) */a

3)) */a

()) */a54(94*/a

!" Y

!";

(

83

@ength o -*/ > ).+ m

@ength o /*+ > ).+ ! (tan --.o >).)0+0 m

@ength o +*- > ).+ H (cos --.o >



Example 1 (Alternative solution)Ehec$:Duilibrium5

8895o

5)) */a

()) */a

3)) */a

()) */a54(94*/a

!" Y

!";

(

83

#

#

# ( 3## #.#414 1## #.1 541.4 #.1#82 co!)22.5 * #

# ( 1## #.#414 5## #.1 541.4 #.1#82 !"n)22.5 * #

x

y

F

F

= − × − × + × × =

= − × − × + × × =∑∑





A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+)

)

-))

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))

#

22.5ψ =

Horio1&a/a1'

Maor +ri1i+a

s&r'ss +a1'

1

1##tan #.414 22.5

541.4 3##

o zx

x

τ ψ

σ σ

= = = =

− −




A()),+))

2 (/)),*+))

3σ 1σ

maxτ

%

'

+))

-)

)

/))

*+))

*-))

*/))

+))

-))

/))

0))

))

8))

/)o

C

C

/N6

Coordi1a&' a& +oi1& N is!47),(8)"

M6 /a1' i1i1'd a&3)) from &h'

maor +ri1i+as&r'ss +a1'


( )





1 3 1 3

1 3

co!22 2

!"n22

θ

θ

σ σ σ σ σ θ

σ σ τ θ

+ −= +

−=

( )

( )

541.4 258. 541.4 258.co! 2 3# 47#.7 &Pa

2 2541.4 258.

!"n 2 3# 122.5 &Pa2

θ

θ

σ

τ

+ −= + =

−= =

H !



Homeor!

()35*N<m8

=8(*N<m8

4(4*N<m8

4(4*N<m8

;or the stressed soil element shown in ;ig. abovedetermine5

a Ma"or principal stressb Minor principal stressc Formal and shear stresses on the plane A:

2y using graphical and analytical solution

Ieer to5 &rinciples o Geotechnical :ngineering (2ra"a, M. 7as

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