Static Failure Theories

LECTURE DATE:07/12/2009

PREPARED BY:EBRU SEMA KOŞAROĞLU & FARAJ KHALIKOV

STATIC FAILURE THEORIES

DUCTILE BRITTLE

TRESCA VON-MISES MOHR- MAXIMUM CRITERIAN THEORY COLOUMB NORMAL STRESS THEORY THEORY(MNST) MSST:MAXIMUM MDET:MAXIMUM SHEAR STRESS DISTORTION ENERGY THEORY THEORY

UNIAXIAL TEST

BRITTLE

푆푌

DUCTILE 푆푌

MAXIMUM NORMAL STRESS THEORY

For maximum normal stress theory, the failure occurs when one of the

principal stresses (휎1, 휎2푎푛푑 휎3) equals to the yield strength.

휎1 > 휎2 > 휎3

Failure occurs when either 휎1=푆푡 or 휎3 = −푆푐 ,where 푆푡 is strength in tension and 푆푐 is strenght in compression. MOHR-COULOMB THEORY

The Coulomb-Mohr theory or internal friction theory assumes that the critical shearing stress is related to internal friction.

MAXIMUM DISTORTION ENERGY THEORY(VON-MISES THEORY)

The maximum distortion energy theory ,also known as the Von Mises theory, was proposed by M.T.Huber in 1904 and further developed by R.von Mises(1913).In this theory failure by yielding occurs when at any point in the body ,the distortion energy per unit volume in a state of combined stress becomes equal to that associated with yielding in a simple tension test.

STRAIN ENERGY

Generally strain energy U is obtained by this equation.

1 U = σ ε 2 ij ij 1 휀 = (휎 − 휈휎 − 휈휎 ) 1 퐸 1 2 3 1 휀 = (−휈휎 + 휎 − 휈휎 ) 2 퐸 1 2 3 1 휀 = (−휈휎 − 휈휎 + 휎 ) 3 퐸 1 2 3 Then, substituting these three equations in to general strain energy equation:

1 1 1 1 1 1 U= 휎 (휎 − 휈휎 − 휈휎 )+ 휎 −휈휎 + 휎 − 휈휎 + 휎 (−휈휎 − 휈휎 + 2 1 퐸 1 2 3 2 2 퐸 1 2 3 2 3 퐸 1 2 휎3)

HYDROSTATIC STRESS

The hydrostatic stress(휎푕 )causes a change in the volume.

휎1 + 휎2 + 휎3 휎 = 푕 3

Strain energy associated with the hydrostatic stress:

1 3 (1 − 2휈) 푈 = [휎 2 + 휎 2 + 휎 2 − 2휈 휎 휎 + 휎 휎 + 휎 휎 = 휎 2 푕 2퐸 푕 푕 푕 푕 푕 푕 푕 푕 푕 2 퐸 푕

Then distortional energy 푈푑 = 푈 − 푈푕

1+휈 From previous equations: 푈 = 휎 2 + 휎 2 + 휎 2 − 휎 휎 − 휎 휎 − 휎 휎 푑 3퐸 1 2 3 1 2 2 3 3 1

Then yielding will occur at this condition:

1 + 휈 푈 = 푆 푑 3퐸 푌

2 2 2 휎푒푓푓 =푆푌 = 휎1 + 휎2 + 휎3 − 휎1휎2 − 휎2휎3 − 휎3휎1

2 2 2 (휎1 − 휎2) + (휎2−휎3) + (휎3 − 휎1) 휎 푒푓푓 = 2

For plane stress condition 휎3 = 0

2 2 휎푒푓푓 = 휎1 + 휎2 + 휎1휎2

If the state stress is in this area then the material will not yield.

For pure shear condition

휎3 = −휎1

2 2 2 2 휎푒푓푓 = 휎1 + 휎3 − 휎1휎3 = 3휎1 = 3휏푚푎푥 =푆푌

휏푚푎푥 = 0.577푆푌

MAXIMUM SHEAR STRESS THEORY

The maximum shearing stress theory is an outgrowth of the experimental observation that a ductile material yields as a result of slip or shear along crystalline planes.

Yielding begins whenever the maximum shear stress in a part equals to the maximum shear stress in a tension test specimen that beings to yield.

SY τ = S = yield strength in shear = max YS 2

σ −σ τ = 1 3 max 2

Then , 푆푌 = 휎1 − 휎3

Elasticity Materials

푆 -brittle 휎 (휎 ) = 푌 where N is safety factor 1 푥 푁

(MNST)

-ductile

(MSST) 휏푚푎푥 = 푆푌푆(yield strength in shear )

휎 −휎 푆 1 3 = 푌 2 2

푆 휎 휎 = 푌 1− 3 푁

푆 (MDET) 휎 = 푌 푒푓푓 푁

Strain Energy:

휎푖

휀푖

1 휎 휀 = 푈 2 푖푗 푖푗

푆 −휎 푁 Design Margin=M= 푌 푒푓푓 푆푌

EXAMPLE:A circular shaft of tensile strength 푆푌=350 MPa is subjected to a combined state of loading defined by bending moment M=8 kN.m and torque T=24kN.m.Calculate the required shaft diameter d in order to achieve a factor of safety N=2.Use a) the maximum shearing stress theory(MSST-Tresca) b)the maximum distortion energy theory(MDET –Von Mises)

휏푥푦

휎 푥

휏푥푦

휎푥

푑 푀푐 푀푑 푇푐 푇푑 c= 휎 = = and 휏 = = 2 푥 퐼 2퐼 푥푦 퐽 2퐽

32푀 16푇 Then,휎 = and 휏 = 푥 휋푑3 푥푦 휋푑3

We need to find principle stresses:

6 a) For maximum shearing stress theory 16 휎 = (푀 ∓ 푀2 + 푇2) 1,2 휋푑3 푆 32 휎 − 휎 = 푌 = 휎 2 + 4휏 2= 푀2 + 푇2 1 2 푁 푥 푥푦 휋푑3

Then, substituting the numerical values of M,T, 푆푌and N: d=113.8mm b) For maximum distortion energy theory

푆 16 휎 = (휎 − 휎 2 + 휎 2 + 휎 2= 휎 2 + 3휏 2= 푌 = 4푀2 + 3푇2 푒푓푓 1 2) 1 2 푥 푥푦 푁 휋푑3

Then, substituting the numerical values of M,T, 푆푌and N: d=109mm

FRACTURE MODES

OPENING

SLIDING

TEARING

Fracture is defined as the separation of a part into two or more pieces.The mechanisms of brittle fracture are the concern of fracture mechanics ,which is based on a stress analysis in the vicinity of a crack or defect of unknown small radius in a part.

MULTIPLE FRACTURES

Stress Intensity Factor: In the fracture mechanics approach a stress intensity factor,퐾퐼, is evaluated. This can be thought of as a measure of the effective local stress at the crack root.

퐾퐼 = 훽휎 휋푎

8 where

휎=normalstress,

푎 훽 = 푔푒표푚푒푡푟푦 푓푎푐푡표푟 푤푕푖푐푕 푑푒푝푒푛푑푠 표푛 , 푎 = 푐푟푎푐푘 푙푒푛푔푡푕 (or half 푤 crack length) ,w=member width(or half width of member)

Fracture Toughness: In a toughness test of a given material, the stress – intensity factor at which a crack will propagate is measured. This is the critical stress intensity factor , referred to as the fracture toughness and denoted by the symbol 퐾퐼퐶.

퐾 N= 퐼퐶 (N=factor of safety) 퐾퐼

EXAMPLE: For a shape with width 12m , crack length 65mm , thickness 30mm and applied loading 50MPa.Find the factor of safety.( 퐾퐼퐶=28.3MPa 푚, 푆푌=240MPa)

50MPa

65mm

30mm 12m 240 푁 = = 4.8 1 50

From the ratio of a/w:

푎 65 × 10−3 = = 0.0054 푤 12

퐾 From table β≅ 1. 훽 = 퐼 휎 휋푎

65 퐾 = 휎 휋푎=50 휋 10−3=15.97MPa 푚 퐼 2

퐾퐼퐶 28.3 For safety factor: 푁2= = = 1.77 퐾퐼 15.97

푁2is better than 푁1 in order to obtain a safe structure; therefore, to have a safe structure, loading should not be used to calculate safety factor.

Reference List:

1. Lecture Notes AEE 361, Demirkan ÇÖKER), 2009, “Static Failure Theories” 2. Static Failure theories, Ansel C. UGURAL & Saul K. Fenster, 2007, “Advanced Strength and Applied Elasticity”, fourth edition. 3. Lecture 5 & 6, The University of Jenessee at Martin School of Engineering, 2009, “Machine Design Notes”. 4. Static Failure theories, J. Keith Nisbett, 2009, “Shigley’s Mechanical Engineering Design”, eighth edition.