Failure CriteriaAE1108-II: Aerospace Mechanics of Materials

Aerospace Structures& Materials

Faculty of Aerospace Engineering

Dr. Calvin RansDr. Sofia Teixeira De Freitas

Consider a Uniaxial Test Specimen

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Consider a Uniaxial Test Specimen

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Linear Elastic Part of Test

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Initial Plasticity and Strain Hardening

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Necking

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Final Failure

Images taken from the following youtube video:https://www.youtube.com/watch?v=D8U4G5kcpcM

Fracture SurfaceWhy failure at an angle?Why at an angle of ±45°?

Recall the Mohr Circle for a Uniaxial Test

σx = P/Ax

y

Ductile Failure

• What happens on ±45° plane?• Maximum shear stress

• Yielding occurs when

• Tresca Yield Criterion

max 2yield

Oσ

τ

C

P/(2A)

σy = 0

Px

y

τmax

σx = P/A

θ = -45°

1max 2

1 yield

(ccw)

(cw)

What if σy ≠ 0?

Oσ

σx

τmax = σx/2

σy = 0

CO

σσx

τmax = σx/4

σy = σx/2

C

σx σx

σy = σx/2

Lower max shear

????

(ccw)

(cw)

τ

(ccw)

(cw)

τ

Role of Out-of-Plane Principal Stress

σy

σx

x

y

z

τxy

Plane Stress State Out-of-Plane Direction (z-axis)

0z

0zx zy

Direction where shear stress = 0(definition of principal stress direction)

z 0 Principal Stress

How Can We Visualize?

O σC

σ1

σ2

xy-plane

xz-plane

yz-plane

σ3 = 0

σy

σxx

y

z

σy

z

y

σy

σx

y

x

σx

x

z

τmax1

τmax3

τmax2

3-D Mohr Circle(ccw)

(cw)

τ

3 Possible Values for Max Shear

O σC

σ1

σ2

xy-plane

xz-plane

yz-plane

σ3 = 0

σy

σxx

y

z

τmax1

τmax3

τmax2

1 2max1

2 3max 2

1 3max3

2

2

2

τmax is the largest of these three

(ccw)

(cw)

τ

What if σy ≠ 0?

Oσ

σx

τmax = σx/2

σy = 0

C O σσx

τmax = σx/2

σy = σx/2

σx σx

σy = σx/2

τ= σx/4

(ccw)

(cw)

τ

(ccw)

(cw)

τ

Tresca Yield Criterion

σ1

σ2+σyield

+σyield

-σyield

-σyield

O σ

τ

σ1

σ2

max 2yield

Oσ

τ

σ1

σ2

O

σ

τ

σ1

σ2

Oσ

τ

σ1

σ2

Other Failure Criteria

• Von Mises yield criterion (ductile materials)• Brittle failure criterion (brittle materials)• Tsai-Wu-Hill failure criterion (composite materials)• Many, many more!!!!

In this course we will limit ourselves to Tresca yield criterion

18Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Tresca Yield Criterion ExampleFor the given plane stress state at a critical point in a 2024-T3 Al structure, determine if yielding has occurred according to the Tresca Yield Criterion

σyield = 345 MPa

x

τ

200 MPa

= 75 MPa

-100 MPa

Solution: Step 1: Find principal stresses (Mohr’s circle)

σ

-100 200

-75

75

2 275 200 50 168R MPa

1

2

3

218118

0 (plane stress)

ave

ave

R MPaR MPa

MPa

R

(cw)

τ (ccw)

A

A

B

B

σave = 50MPa

19Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Tresca Failure Criterion ExampleFor the given plane stress state at a critical point in a 2024-T3 Al structure, determine if yielding has occurred according to the Tresca Criterion

σyield = 345 MPa

x

τ

200 MPa

= 75 MPa

-100 MPa

Solution: Step 2: Determine Max Shear Stress

1 2 3218 118 0MPa MPa MPa

1 3max3

218 0 1092 2

MPa

1 2max1

218 118 1682 2

MPa

2 3max 2

118 0 592 2

MPa

Oσ

τ

σ1σ2

20Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Tresca Failure Criterion Example

Step 3: Tresca Criterion

168 345 / 2 172.5

169 172.5MPa MPa satisfied, failure does not occur

max 2yield

1

2

3

max

218118

0 (plane stress)168

MPaMPa

MPaMPa

σ1

σ2

345

345

-345

-345

(345, 345)

(-345, -345) (218, -118)

σyield = 345 MPa

But it is pretty darn close!