σ φ 0

10

Chapter2. Yield Conditions

• Mohr-Column Criteria

τ σ + tanϕ– c=0

→ generalization of Tresca yield criteria →

1 2 2 3 3 1

1 1 1

, ,

2 2 2

k = MAX ⎛ ⎜ ⎝ σ σ − σ − σ σ − σ ⎞ ⎟ ⎠

tan

c c

τ = − σ φ = − μσ

Where c = cohesion

μ

= coefficient of friction

σ

= tensile strength

tan φ

= angle of friction

In concrete failure criteria is controlled by sliding failure σ

τ

τ

max

σ

1

σ

2

τ

σ

φ

σ ε r

c

0

11 Deep beam

Shallow beam

• Failure mode of isotropic material

(

1

,

2

,

3

) 0 f σ σ σ =

(

1

,

2

,

3

) 0 f J J J =

where

σ

1 = the first invariant of the stress tensor

σ

_ij

J

1= the second invariant of the deviatric stress tensor

s

_ij Stress invariant

( σ

_ij

− σδ

_ij

) n

_j

= 0

To find the principal stress 2

r

ε

Sliding failure

Separation failure

Cut-off tensile strength Modified coulomb criteria

Ⅰ

Ⅱ

Separation failure

Sliding failure

Diagonal tension failure mode

균열면에 직각 (separation failure)

3

I

1

σ − σ

1 x

I = σ

(

I

2

= σ

3 x yz zx

I σ τ τ

=

If the co

1 1

2 1

3 1

I I I

σ σ σ σ σ

=

=

=

Deviatio

ij

s

ij

σ =

Where

1

m

3 σ =

ij ij

s = σ s

ij

− s δ

3

s − J s

1

1

0

J =

2

1 J = 2 s

3

1 J = 3 s

2

I

2

σ + σ −

y z

σ σ

+ + =

x y y z

σ σ + σ σ

x xy zx

z y yx

x zy z

τ τ σ τ

τ σ

oordinate are

2 3

2 2 3

2 3

σ σ σ σ σ σ σ

+ +

+ +

on form

m ij

σ δ +

( σ

^x

+ σ

^y

+ σ

m ij

σ δ

−

ij

0 δ =

3

2 3

s − J s − J

(

1

ij ij

6

s s = ⎢⎣ ⎡ σ

x

ij jk ki yx

zx

s

s s s s

s

=

3

0

I =

σ

ii

=

)

z x

σ σ τ

+ −

e chosen to

3 1

σ σ +

) 1

z

3

ij

σ = σ =

= 0

) (

²

x y

σ − σ +

x xy xz

yx y yz

x zy z

s s

s s

s s

2 2

xy yz

τ − τ − τ

coincide wit

1

1 3 I

=

( σ

^y

− σ

^z

)

²

+

12

2

τ

zx

th the princi

( σ σ

z x

)

²

+ −

ipal stress ax

2 2

xy y

τ τ

⎤ + +

⎥⎦

xis

2 2

yz

+ τ

zx

13

1 3

cos sin

R = c ϕ − σ σ + 2 ϕ

or ^{1 3}

R σ σ − 2

=

1 3

cos

1 3

sin

2 c 2

σ σ − = ϕ − σ σ + ϕ

( )

³

( )

1

1 sin 1 sin cos 0

2 σ 2 c

σ + ϕ − − ϕ − ϕ =

1 3

1 sin 2

1 sin 1 sin cos

ϕ σ σ c ϕ

ϕ ϕ

+ − =

− −

Let

2

cos

2

1 sin

1 sin tan 4 2 1 sin

k ϕ π ϕ ϕ

ϕ ϕ

⎛ ⎞ ⎛ ⎞ +

= ⎜ ⎝ − ⎟ ⎠ = ⎜ ⎝ + ⎟ ⎠ = −

( ) * = k σ σ

1

−

3

= 2 c k

: sliding failure At

σ

¹

= f

^A :we expect separation failure

• discontinuity of stress fields

1 3

2

k σ σ − = c k

: sliding

1

f

t

σ =

: separation How to find k & c (or f_t)

1) by compression test

1 3

σ

2 σ σ +

ϕ

ϕ

R

c

a b

σ

3

ϕ

σA

σ

B

σC

τ

A

σ

σ σ^B

σC

Sc

Qc

2)

• by slid

k σ

1

−

σ

₁

= 0

(2)→(1)

f

_c

by tension t

ding failure

3

2c k

σ

− = 0

,

− = σ

₃

f

= 2c k

(*)

test

1

f

t

σ =

- (1)

f

c - (2)

τ

3 fc

σ =

σ 4

π

₋

σ

3

0

σ =

14

3 0

σ =

2

−

ϕ

0

τ

90−ϕ

τ

90° −ϕ

ϕ

τ

ft

fA

σ

σ

15

1

f

t

σ =

,

σ

₃

= 0

-(3) (3)→(1) and (*)

1 t c

k σ = kf = f

∴ ^c

t

k f

= f

• by separation failure

A t

f = f

For sliding failure

1

c A

f f

k <

For separation failure

1

c A

f f

k >

3) by pure shear test

ft

45 0 2

° +ϕ

16

1 3

f

v

σ = − = σ

- (5) (5)→(1)

( )

1 3

1

1 _c

k σ σ − = k + σ = f

( 1 )

1

c

v c v

k f f f f

+ = → = k

+

• For sliding failure if

v A

f < f 1

c A

f f

k <

+

• For separation failure if

1

c A

f f

< k +

In plane stress (member stress)

σ

τ

fv

fA

fv

Separation

sliding

3 0

σ =

III

0

σ =

17 1)

σ

_ΙΙΙ

< σ

_ΙΙ

< σ

_Ι

σ σ

1

=

_Ι,

σ

₃

= 0

• For sliding failure

1 c

k σ = f

- ①

• For separation failure

f

A

σ

_Ι

=

- ②

2)

σ

_ΙΙ

< < 0 σ

_Ι

σ σ

1

=

_Ι,

σ

₃

= σ

_ΙΙ

• For sliding failure

k σ σ

_Ι

−

_ΙΙ

= f

c - ③

• For separation failure

f

A

σ

_Ι

<

- ④ 2)

σ

_ΙΙ

< < 0 σ

_Ι

σ

I

σ

II

σ

III

τ

σ

III

σ

_II

σ

_I

18

1

0

σ =

,

σ

₃

= σ

_ΙΙ

• For sliding failure

f

c

σ

_ΙΙΙ

− =

- ⑤

• For separation failure Non exist

①,③,⑤ = sliding failure only

σ

I

σ

II

σ

III

I

f

c

σ sym

I

f

c

① σ

③

⑤

( , 0) 1

k

19 If

f

_A

= 0

(즉, concrete의 tensile strength =0)

2.1.3 the rupture criteria for concrete

If concrete has to be identified with a modified coulum material, we can conclude the parameter

k

has a value of about 4

1 sin 1 sin 4

k ϕ

ϕ

= + =

−

→

tan ϕ = 0.75

∴

ϕ = ° 37

c

2

f = c k

→

4 f

c

c =

I

f

c

σ

③

II

fc

σ ② or ④

( 1, 0) −

, ( 1)

A A

c c

f f

f k f

⎛ ⎞

⎜ − ⎟

⎝ ⎠

I

f

A

σ =

σ I

σ

II

20

2.1.4 The plastic strength of concrete

f

ce → effective strength of concrete

v

: effectiveness factor

I

fc

σ ( , 0)1

k

3

fc

σ

1 k=

φ=0

φ

τ

σ

τ

σ

brittle

Effective strength of concrete : Effective factor

softening

σ

ε

c c

'

f = ν f

fce

ν

σ

ε

σ

21

The decreasing v-value for increasing strength seems to hold generally - Geometric effects

Stress concentration

Absolute value of dimension - Loading condition

a/d ratio

1) 1978 CEB Model Code

v

= 0.6 2) M. P Nielsen

'

0.7

'

( )

29000

c

ce c

f ⎛ f ⎞ f psi

= ⎜ − ⎟

⎝ ⎠

3) Ramirez

30

'

( )

ce c

f = f psi

4) Schlaich & Weischede

v

=0.67

2.1.3 Failure criteria for concrete

Concrete two phase : cement paste, aggregates

aqq paste

K > K

aqq paste

σ > σ

Coment paste : isotropic plastic material with

ϕ = 0

σ

ε A 0

A 1

0 ' f

f c A 0 = A 1

w

2w

22

ϕ = 0

→

k = 1

Hydraulic test

σ σ

₁

+

₃

= P

1 3 c

k σ σ − = f

1

f

c

P

σ = +

(ideal)

0 P =

f

c

σ = 0 P ≠

f

cc

P σ = +

Eq(2.1.7)

1 3

2c

σ σ − = 2c P σ = +

Failure condition for concrete - Two phase material - For granular material. c=0

σ

τ

f

c

A B

C

C

f

c

σ

τ

f

c

2C

f

cc

σσ

23

c

P

c

2 c

σ = +

for cement paste

a

kP

a

σ =

2 (1 ) 2

c a c a

c

P c kP

P k c kP

σ σ = + σ = + +

= − + +

When

P

_c

= 0

σ

→ max

2c kP

σ = +

(2c is uniaxial compressive strength of the composite material)

f

c

kP

σ = +

(friction angle from aggregates, cohesion from cement paste)

For sufficiently high confining pressure the yield lines in cement paste tend to develop roughly under 45 degree.

① uniaxial compressive strength is not affected by the aggregates

② increase in strength above for cement paste is due to the displacement caused by the aggregate particles of the yield lines in the cement paste.

f

c: real uniaxial strength

f

cc: apparent uniaxial compressive strength

Average stresses

c a

P = P + P

σ

σ

24

(

1 3

)

1

m

2

σ = σ σ +

(

1 3

)

1

m

2

τ = σ σ −

Since

σ σ

₁

=

_m

+ τ

_m,

σ

₃

= σ

_m

− τ

_m

Sub into (2.1.10) and (2.1.8)

1 3 c

k σ σ − = f

,

σ

₁

= f

_A

( k − 1 ) σ

m

+ ( k + 1 ) τ

m

= f

c,

σ

_m

+ τ

_m

= f

_A

Modified coulomb material in

σ

_m-

τ

_m coordinate system

Test result

① thin – walled tube

② Biaxial test

Tensile stress

f

_t

= 0.1 f

_c

σ

3

σ

₁

A B

C

O

25

Law sliding resistance along a OA, BC separation along AB - Shear failure of orthotropic panels

- Sliding resistance of a crack waloven - Effect of softening , transverse pressure - Geometry of structural element and loading

2.1.4 Structural concrete strength

ccf c

ef

tef t t

f vf

c vc

f v f

=

=

=

Softening phenomena is unable to take into account the softening in a rational manner Compressive strength of cracked concrete

Micro crack Macro crack

min 0

e t

A ⋅ = f a u Σ

- ①

Where

A

_e effective area for tension

Concrete tensile Steel tensile Bond

Macro

26 a=

= 2a

_min - ②

since ₀

4

_s

b

A

Σ = d

- ③

③ & ② → ①

0

2

2

e t t b

s e

A f f d

a u u A

A

= =

Σ

Let ^s

e

r A

= A 4

_t

u = λ f

(

λ = 2

for deformed,

λ = 1

for plastic)

4 d

b

β = r

So

4

t b t

a f d

f r β

λ λ

= =

(2.1.54)

Assume _{0 y} ^b _y ^b

t t

d d

l c

f f

σ σ

∝ =

⁰

4

1 4

b s

t s

b t

c d

l f c r

a d f

r

σ λ σ

χ

λ

= = =

1.41

' s

c

r f

χ = σ

(2.1.57)

5) Collins & Mitchell (compression field theory)

'

0.8 170

1 c ce

f f

= ε

+

Reading assignment p81~91 (Mutteni’s book) Refer to Dr. Yun’s paper

By M.P. Nielson

f t

ε

σ

2

σ

2

27

2

c

v f

(in MPa)