CEP Presentation

Theories of failure

Over view ●

Stress- strain curve

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Hardening types

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Combined stresses

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Principal stresses, Hydrostatic, Deviatoric and octahedral stresses.

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Failure theories

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Yield function

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Yield criteria

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Flow rule

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Various material models

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Finite element example

Uniaxial loading ●

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Simple tension

For some materials yield point is so poorly defined that it is taked as 0.2 percent of the permanent strain. A few materials such as annealed mild steel, exhibit a sharp drop in yield after the upper yield point B is reached, this is because of the luder bands. True stress can be obtained from the nominal stresses by considering no volumetric change. A o L o= AL

Uniaxial loading ●

True stress

P PL σ= = =σn (1+ εn) A A o Lo L

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True strain

ε=∫ dl / l= ln ( Lo

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L ) Lo

Hardening Types ●

Kinematic Hardening: Elastic range remains constant

σ C =2 σ o −σ B

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Isotropic Hardening: yield in tension is same as yield in compression

σ C =−σ B

Hardening types ●

Mixed hardening: Yield in tension and compression are independent

σB

σo

σ C =σo σ C =2 σ o −σ B

σo ●

σ Actual

Actual Hardening is above the mixed hardening yield in compression

Stress strain idealized curves ●

Perfectly linear elastic

Elastic Perfectly plastic M

σ

σ σ=E ε

E

σ <σ o then ε= σ E

E

ε

ε ●

Rigid perfectly plastic

σ

Bilinear hardening material M

ε=0 for σ< σ o

ε

σo σ≥σo then ε= +λ E

ε>0 for σ >σ o

M

σ H E

σ <σ o then ε= σ E

σ≥σo then ε=

ε

σ o (σ−σ o) + E H

Emperical equations for stress strain curves ●

Ludwick equation

σ=σo +mε ●

n

Ramberg Osgood equation

ε= σ +k ( σ ) E E

n

Tangent modulus e

d ε=d ε + d ε

p

σ

dσ dσ p dε = ;d ε = p E E

Ep

e

d ε=

dσ dσ + p E E

dσ dσ dσ d ε= t = + p E E E

1 1 1 = + p t E E E

E dε

e

dε

dε

p

ε

Multi directional loads P

P

P

T Q

Q R P ●

Unit stress:

T P

P

Δ Fi ΔF P= lim ( )= lim ( ) Δ A Δ A Δ A→0 Δ A →0 j The unit stress is not normal to the plane. The value of unit stress is referred to a particular plane.

Stress at a point ●

Representation of stress at a point

σ11 σ 12 σ 13 σ xx τ xy τ xz ( σ 21 σ 22 σ 23)=( τ yx σ yy τ yz )=σ ij σ 31 σ 32 σ 33 τ zx τ zy σ zz

Stresses on an arbitrary plane ●

l, m and n be the direction cosines of the normal acting on plane ABC, then

AoB=m Δ A ; AoC=n Δ A ; BoC=l Δ A

B

σs σn

σi

x−direction , Δ A S X =σ x . l . Δ A+ m. Δ A τ xy +n . Δ A τ zx S X =l . σ x +m . τ xy +n. τ zx

n

σ2i σ3i A

σ 1i

S y =l . τ xy +m . σ yy +n . τ yz

o C

S z=l . τ xz +m . τ yz +n . σ zz

S j=σij . ni Cauchy's stress

Stresses on an arbitrary plane S n =l . S x +m . S y +n . S z

S n =l2 σ x +m2 σ y +n2 σ z +2(lm τ xy +mn τ zy +nl τ zx ) 2

2

2

S s=S −S n

Principal stresses ●

Consider a plane on which the resultant stress is perpendicular to the plane B Sy Sx

S

S X =l . S

S y =m . S S z=n . S

Sz

C A

Substituting the above values in the Cauchy's stress equation we get l . S=l . σ xx +m . τ xy + n . τ zx ⇒ l.(σ xx −S )+m . τ xy +n . τ zx=0 m . S=l . τ xy +m. σ yy +n . τ yz ⇒l . τ xy +m.(σ yy −S )+n . τ yz =0 n . S=l . τ xz +m . τ yz +n . σ zz ⇒ l . τ xz +m. τ yz +n .(σ zz−S)=0 In indical notation l i (σ ij −δ ij S)=0

Prinicpal stresses and Invariants ●

For a solution to be non- trivial, determinant of the three equations is zero

|

|

σ xx −S τ xy τ xz |σij −δij . S|= τ yx σ yy −S τ yz =0 τ zx τ zy σ zz −S 3

2

S −I 1. S + I 2.S −I 3 =0

Where I 1=σ xx +σ yy + σ zz 2

2

2

I 2=τ xy +τ yz +τ zx −(σ x .σ y +σ y . σ z +σ z .σ x ) 2

2

2

I 3=σ x . σ y . σ z +2 τ xy . τ yz . τ zx −(σ x . τ yz +σ y τ zx +σ z . τ xy )

at a point on any plane the values of these invariants doesn't change.

Principal stresses and Invariants S 3 −I 1. S 2 +I 2.S −I 3 =0 ●

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The cubic equation has three real roots and consequently three principal stresses σ 1, σ 2 and σ 3 . From the pricipal stress values we can get the eigen vectors l,m and n, if in addition l 2 +m 2 +n 2=1 If σ 1=σ 2=σ3 → Hydrostatic stress (any three perpendicular directions are principal)

σ 1≠σ 2≠σ3 → all principal directions are unique and orthogonal. If σ 1=σ 2≠σ3 → one principal direction will be unique, but the other If

two directions can be any two directions orthogonal to first. ●

If

σ 1, σ 2 and σ 3 are the co-ordinate axes then

I 1=σ 1 +σ 2 +σ 3 ; I 2=−(σ 1. σ2 + σ2. σ 3 +σ3. σ 1); I 3=σ1. σ 2. σ 3

Octahedral shear stresses l=m=n=

1 √(3)

1 S n =l σ x +m σ y + n σ z + 2(lm τ xy + mn τ zy + nl τ zx )= (σ1 + σ2 + σ3 ) 3 2

2

2

1 2 2 2 1 1 2 2 2 2 2 S s= (σ1 +σ2 +σ3 )− (σ1 + σ2 +σ3 ) = ((σ 1−σ2 ) +(σ2 −σ 3) +(σ 3 −σ 1) ) 3 9 9 2 1 2 2 2 S s= ((σ 1−σm ) +(σ 2−σm ) +(σ 3−σm ) ) 3 1 ( ) 1 2 2 2 τ oct = ((σ 1−σ m ) +(σ 2−σ m ) +(σ3 −σ m ) ) 2 √(3)

τ oct

1 ( ) (2) 2 √ = (I +3 I ) 2

3

1

2

Hydrostatic and deviatoric stress tensor ●

Stress tensor can be divided in to hydrostatic part and deviatoric part.

σm 0 0 pij =σm δij ( 0 σ m 0 ) 0 0 σm

where σ m=

(σ x + σ y + σ z ) (σ 1 +σ 2 +σ 3) = 3 3

The value of hydrostatic tensor is same for any streess state at a point

σ xx −σ m τ xy τ xz S ij =σij −Pij =( τ yx σ yy−σ m τ yz ) τ zx τ zy σ zz −σ m principal deviatoric stresses can be found out similarly like the previous one

Pricipal deviatoric stresses S 3 −J 1. S2 + J 2.S −J 3=0

Where

J 1=0

1 1 2 2 2 2 2 2 2 J 2= [(σ x −σ y ) +(σ y −σ z ) +(σ z −σ x ) ]+(τ xy +τ yz +τ xz )= (I 1 +3 I 2) 6 3 1 3 J 3 =(σ1 −σ m)(σ 2−σ m )(σ 3−σ m)= (3 I 1 +9 I 1 I 2+ 27 I 3) 27

in terms of pricipal deviatoric stresses J 2=−( S1 S 2 + S 2 S3 + S 3 S 1)

J 3 =S 1 S 2 S 3

Haigh-Westergaard Stress Space ●

How to geomertrically represent a stress state? a) Considering six indepenedent stresses as six components of positional coordinates. b) Use the pricipal stresses.