2005 Pearson Education South Asia Pte Ltd TUTORIAL-4: FAILURE THEORIES 1 THEORIES OF FAILURE Failure in a ductile material is specified by the initiation

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-4: FAILURE THEORIES

1

THEORIES OF FAILURE

Failure in a ductile material is specified by the initiation of yielding.

Failure in a brittle material is specified by fracture.

THEORIES OF FAILURE FOR DUCTILE MATERIALS

Maximum Shear Stress Theory (Tresca Yield Criterion)

The most common cause of yielding of a ductile material such as steel is slipping, which occurs along the contact planes of randomly ordered crystals that make up the material.

This slipping is due to shear stress : 2

minmaxabsmax



2

THEORIES OF FAILURE

The maximum shear stress theory for plane stress can be expressed for any two in-plane principal stresses as s1 and s2 by the following criteria:

Y2

Y1

s1 and s2 have same sign

Y21 s1 and s2 have opposite sign



3

THEORIES OF FAILURE

Maximum Distortion Energy Theory (von Mises Yield Criterion)

s1

s3

s2

von Mises Stress for 3-Dimension

2132

322

21o

s2

s1

von Mises Stress for 2-Dimension

2221

21o

Yield stress, sy , occurs when so = sy



4

THEORIES OF FAILURE

Allowable Stress, sallow & Safety Factor, SF

In designing a component or structure, it is introduced what so-called Allowable Stress, which is defined as

oallow

SFY

O where



5

THEORIES OF FAILURE

THEORIES OF FAILURE FOR BRITTLE MATERIALS

Failure of a brittle material

in tension

45o

Failure of a brittle material

in torsion

If the material is subjected to plane stress, we require that

ultimate1

ultimate2



6

Stress in Shafts Due to Axial Load and Torsion

An axial force of 900 N and a torque of 2.50 N.m are applied to the shaft as shown in the figure. If the shaft has a diameter of 40 mm and the safety factor is 5, determine the minimum yield stress of the material used.

Internal Loadings

(a)(a)(b)

PROBLEM-1

The internal loadings consist of the torque and the axial load is shown in Fig.(b)



7

(a)(b)

Maximum stress Components

Due to torsional load

4(0.02)

.50)(0.02)(

2

2

J

cT

198.9 kPa

Due to axial load

kPa716.22(0.02)

9002

AF

PROBLEM-1



8

The state of stress at point P is defined by these two stress components

Principal Stresses:

2xy

2yy

2,1 22

We get s1 = 767.8 kPa

s2 = – 51.6 kPa

The orientation of the principal plane:

2tan

y

xy1p

2 = – 29o

PROBLEM-1



9

von Mises equivalent stress for 2-D

2221

21o

22o 51.6)(51.6)(767.8)((767.8) = 794.8 kPa

PROBLEM-1

Yield Stress of the shaft material can be found from:

SFY

o

oy SF )( = (5)(794.8) kPa

= 3974 kPa

= 3.974 MPa



10

Solid shaft has a radius of 0.5 cm and made of steel having yield stress of Y = 360 MPa. Determine if the loadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-distortion-energy theory.

PROBLEM-2



11

State of stress in shaft caused by axial force and torque. Since maximum shear stress caused by torque occurs in material at outer surface, we have

MPa165.5kN/cm16.55cm0.52π

cm0.5cmkN3.25

MPa191kN/cm19.10cm0.5π

kN15

24

22

J

Tc

A

Pσ

xy

x

PROBLEM-2



12

Stress components acting on an element of material at pt A. Rather than use Mohr’s circle, principal stresses are obtained using stress-transformation eqns 9-5:

MPa6.286

MPa6.95

1.1915.95

5.16520191

20191

22

2

1

22

22

2,1

σ

σ

σ xyyxyx

PROBLEM-2



13

Maximum-shear-stress theory

Since principal stresses have opposite signs, absolute maximum shear stress occur in the plane, apply Eqn 10-27,

Thus, shear failure occurs by maximum-shear-stress theory.

Fail!,

σσσ Y21

360382.2

?360286.695.6Is

PROBLEM-2



14

Stress in Shaft due to Bending Load and Torsion

T

xy

z

A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a bending moment of 2 kNm and a torque of 2.5 kNm.

Determine:

1. The critical point of the section

2. The stress state of the critical point.

3. The principal stresses and its orientation.

4. Select the material if SF = 6

PROBLEM-3



15

T

xy

z

Analysis to identify the critical point

Maximum shear stresses occur at the peripheral of the section.

Due to the torque T

Due to the bending moment M

Maximum tensile stress occurs at the bottom point (A) of the section.

Conclusion: the bottom point (A) is the critical point

A

PROBLEM-3



16

T

xy

z

Stress components

Due to the torque T

Due to the bending moment M

A

198.9 kPa)(0.02

.50)(0.02)(4

2

2

J

cT

318.3 kPa4

4 (0.02)

2)(2.00)(0.0

zI

cM

PROBLEM-3



17

318.3 kPa

198.9 kPa

Stress state at critical point A

sx = 318.3 kPa txy = 198.9 kPa

Principal stresses

22

22 xyxx

1,2

We get s1 = 413.9 kPa

s2 = – 95.6 kPa

The orientation of the principal plane:

2tan

x

xy1

p2 = 51.33o

25.65o

s1

s2

PROBLEM-3



18

25.65o

s1

s2von Mises equivalent stress for 2-D

2221

21o

22o 95.6)(95.6)(413.9)((413.9)

= 469 kPa

PROBLEM-3

Yield Stress of the shaft material can be found from:

SFY

o

oy SF )( = (6)(469) kPa

= 2814 kPa

= 2.814MPa



19

PROBLEM-4

Steel pipe has inner diameter of 60 mm and outer diameter of 80 mm. If it is subjected to a torsional moment of 8 kN·m and a bending moment of

3.5 kN·m, determine if these loadings cause failure as defined by the maximum-distortion-energy theory. Yield stress for the steel found from a tension test is Y = 250 MPa.



20

Investigate a pt on pipe that is subjected to a state of maximum critical stress.

Torsional and bending moments are uniform throughout the pipe’s length.

At arbitrary section a-a, loadings produce the stress distributions shown.

PROBLEM-4



21

PROBLEM-4



22

PROBLEM-4



23

PROBLEM-4



24

Stress in Shafts Due to Axial Load, Bending and Torsion

A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a compressive force of 2500 N, a bending moment of 800 Nm and a torque of 1500 Nm.

Determine: 1. The stress state of point A.

2. The principal stresses and its orientation

3. Determine the required yield stress if SF = 4.

PROBLEM-5



25

Analysis of the stress components at point A

Due to comprsv load:

Due to torsional load:JcT

A

AF

A'

Due to bending load:z

'A' IcM

(compressive stress)

Stress state at point A

Shear stress: t = tA

Normal stress: s = sA’ + sA”

ts

PROBLEM-5

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2005 Pearson Education South Asia Pte Ltd TUTORIAL-4: FAILURE THEORIES 1 THEORIES OF FAILURE Failure in a ductile material is specified by the initiation