ME311 Machine Design

W Dornfeld26Sep2019 Fairfield University

School of Engineering

Lecture 4: Stress Concentrations; Static Failure

Stress Concentration

We saw that in a curved beam, the stress was distorted from the uniform distribution of a straight beam, and higher at the inside radius.

A similar distortion happens near holes, notches, and changes in section:

The effect varies with the loading type: tension, bending, or torsion.

Stress Concentration“Stress Concentration Factors” relate the maximum stress to

the average stress in the minimum cross section of the part.

375.00.2

75.0==

bd

psiK

psi

nomc

nom

216096025.2

9603125.0

300

25.0)75.00.2(

300

max =×=×=

==

−

=

σσ

σ

Hamrock Fig. 6.2aHamrock’s plots tell

you the nominal stress.

Which has the bigger Kc?

150 100 150 100

r r

M M

=

=

=

=

cKh

r

rh

H

FigUse _______:.

=

=

=

=

cKh

r

rh

H

FigUse _______:.

Predicting Failure

σ

τ

100

50

σ

τ

60

80-40

We are given some material that is tensile tested and yields at Sy=100ksi.

It has a Mohr’s circle that looks like this:

We want to use this material in an application where it sees this loading:

It has a Mohr’s circle that looks like this:

80

40

y

x

τ=0

The Max Principal stress is less than 100ksi, but will it yield?

Failure Theory

Failure Theory addresses how to translate a real, multi-axial state of stress into something that can be compared with a simple uniaxial (tensile) test result.

For ductile materials, there are two prevailing theories:

1. Max Shear Stress Theory (MSST) or Tresca Theory

2. Distortion Energy Theory (DET) or von Mises criterion.

Max Shear Stress

MSST says "Yield occurs whenever the maximum shear stress in an element exceeds the max shear at yield in a tensile test"

In other words, the Mohr's circle lies within the shear bounds (upper and lower) of the tensile test.

σ

τ

100

50

Max Shear Stress

How did we do?

σ

100

50

τ

60

80-40

Doesn’t look good.

MSST predicts yielding when σ1 - σ3 ≥ Sy. (~Eqn. 6.8)

σ1 is Max principal and σ3 is Min principal.

In our case, σ1 - σ3 = [80 – (-40)] = 120 > Sy of 100.

Another way to say is that τmax ≤ Sy / 2

Max Shear Stress

Here is a representation of MSST in terms of limits to Max and Min principal stresses:

σσσσ 2

σσσσ 1

Syt

Syt

σ1σ2

σ1σ2

σ1σ2

σ1σ2

σ2σ1σ1&σ2

σ2σ1

Inside the “box” is safe.

This is the pure shear line, where

σ1 = -σ2.

Distortion Energy

DET observes that materials can take enormous hydrostatic pressures (e.g., rocks deep in the earth that see pressures well above their compressive strength) and not fracture – suggesting that it must be distortion that causes failure.

DET essentially computes the total strain energy and subtracts the volume change energy to get the distortion energy.

In terms of principal stresses:

where σe is the von Mises equivalent stress.

2

31

2

32

2

21 )()()(2

1σσσσσσσ −+−+−=e

Note that for hydrostatic loading, σ1=σ2=σ3, and σe=0!

Distortion Energy

In the more common case of biaxial principal stresses (σ3=0), this reduces to:

In our example, this is:

This exceeds Sy = 100ksi, so DET also predicts yielding.

ksi

e

8.105200,11

160032006400

)40()40)(80(8022

==

++=

−+−−=σ

2

221

2

1σσσσσ +−=e

Distortion Energy - Simplified

If you have only direct biaxial stress (σz=0), you can

calculate DET directly from σx, σy, and τxy without getting

the principal stresses:

And, if you have only bending and torsion (σy=0), this

reduces to:

(Hamrock buries this in Ex. 7.4 on page 170.)

2223 yxyxyxe τσσσσσ +−+=

223 yxxe τσσ +=

The Failure Theories, overlaid

MSST is a little more conservative than DET.

Brittle Theories

Max Normal Stress Theory

Brittle materials can take more compression than tension, so have different failure bounds.

Internal Friction Theory & Modified Mohr Theory

The Brittle theories use Ultimate Strengths because they fracture rather than yield, and both Compressive and Tensile strengths because they are different.

Remember the Crank Arm ?

1200 in

lb T

orq

ue

1800 in.lbMoment

300 lb Shear

1200 in

lb

To

rqu

e

300 lb Shear

A

BShearPure

psi

psiA

V

psiJ

Tr

psiI

Mc

BENDINGxzTOTAL

BENDING

xz

x

→

=−=

====

====

====

582,13

905325.1

1200

4)75.0(3

)300)(4(

3

4

487,1403106.0

450

32)75.0(

)375.0)(1200(

460,4301553.0

675

64)75.0(

)375.0)(1800(

2

4

4

τττ

π

τ

π

τ

π

σA

B

y

xz

At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.

At “B": Torsion T = 1200 in.lb.Shear V = 300 lb

C

Note that at location “C”, the two shears would ADD to 15,392 psi.

Stresses in a Crank Arm

The stresses at A occur at the same point, so can use Mohr.

psiJ

Tr

psiI

Mc

xz

x

487,14

460,43

==

==

τ

σ

A

y

xz

For Xstress = 43.5, Ystress = 0.0, XYShear = -14.5, Angle = -16.85, Stress1 =

47.85, Stress2 = 0.00, Stress3 =-4.39, ShearMax = 26.12

X,-T

Y,T

-30

-20

-10

0

10

20

30

-10 0 10 20 30 40 50 60

Face Stresses

Sh

ea

r S

tre

ss

The Mohr circle gives:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi

What are the DET stresses at “A” on the Crank Arm?

psiJ

Tr

psiI

Mc

xz

x

487,14

460,43

==

==

τ

σA

At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.

The Mohr circle for “A” gave:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi

ksi

yxyxyxe

18.504.25186.6298.1888

)487.14(30046.43322222

==+=

+−+=+−+= τσσσσσ

ksi

e

18.502518192102285

)39.4()39.4)(8.47(8.47222

331

2

1

==++=

−+−−=+−= σσσσσ

Using the Principal stresses:

Using the Direct stresses:

What are the MSST stresses at “A” on the Crank Arm?

psiJ

Tr

psiI

Mc

xz

x

487,14

460,43

==

==

τ

σA

At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.

The Mohr circle for “A” gave:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi

Using the Principal stresses (our only option with MSST):

ksiMSST 24.52)39.4(85.4731 =−−=−= σσ

This number is slightly higher than the von Mises stress of 50.2 – consistent with the MSST being more conservative.

What are the MSST & DET stresses at “C” on the Crank Arm?

ShearPure

psi

psiA

V

BENDINGxzTOTAL

BENDING

→

=+=

==

392,15

9053

4

τττ

τ

C

At “C": Torsion T = 1200 in.lb.Shear V = 300 lb

“C” sees pure shear, with σx & σy both zero.

σ

τ

15.4

15.4

So σ1 = 15.4 & σ3 = -15.4

ksi

MSST

8.30)4.15(4.1531

=−−=

−= σσ

ksi

e

7.265.711)4.15(3

)4.15()4.15)(4.15(4.15

2

22

2

331

2

1

===

−+−−=

+−= σσσσσ

ksi

yxyxyxe

7.265.711)4.15(3

)4.15(3000

3

2

222

222

===

+−+=

+−+= τσσσσσ

DET using the Principal stresses:

DET using the Direct stresses:

Factors of SafetyThe Factor of Safety, ns, for a component is the ratio of the stress allowed by the component’s material, divided by the maximum stress predicted (or measured). We usually use the yield strength, Sy, as the allowable stress for ductile materials. See Hamrock’s Appendix A for Sy.

maxσ

Syns =

Using the Failure Theories, we will generally compute ns either by:

e

s

Syn

σ

=

For DET (von Mises)max

31

5.0

)(

τ

σσ

Syn

or

Syn

s

s

=

−

=

or

For MSST

Crank Arm Safety Factors

This assumes a material with Sy = 100ksi.Can you find a material with Sy = 100ksi?

C

A

99.12.50

100

91.124.52

100

==

==

ADET

AMSST

n

n

75.37.26

100

25.38.30

100

==

==

CDET

CMSST

n

n

Static Safety Factors

A bar of AISI 1040 Steel sees a bending and twisting load, resulting in a bending stress of 15 ksi and a shear stress of 6.5 ksi at its most highly stressed location.

Find the MSS and the DET factors of safety against yielding.

Calculate both FoS’s using both Direct stresses and Principal stresses.

Selecting Factors of SafetyHamrock gives guidelines for selecting Safety Factors in Section 1.4.1.

),(),,( EDnCBAnn sysxs ×=

Quality of materials,

processing, maintenance, & inspection

Environment (Load) control

Quality of analysis &

testDanger to humans

$$ impact

1.1 ≤ nsx ≤ 3.95; 1.0 ≤ nsy ≤ 1.6; so 1.1 ≤ ns ≤ 6.32