chapter 5 - islamic university of...

Chapter 5 Failures Resulting from Static Loading

Chapter Outline

Dr. Mohammad Suliman Abuhaiba, PE

April 19, 2014 2

Stress Concentration

Failure Theories for Ductile Materials

Maximum-Shear-Stress Theory

Distortion-Energy Theory

Coulomb-Mohr Theory

Failure Theories for Brittle Materials

Maximum-Normal-Stress Theory

Modifications of the Mohr Theory

Selection of Failure Criteria

Stress Concentration

Localized increase

of stress near

discontinuities

Kt = Theoretical (Geometric) Stress

Concentration

Factor


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(3.48)

Appendix A–15 & A–16

Peterson’s Stress-Concentration Factors

Theoretical Stress

Concentration Factor


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Figure A–15–1: Bar in

tension or simple

compression with a

transverse hole. σ0 =

F/A, A = (w − d)t and t

is the thickness.

Theoretical Stress

Concentration Factor


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Figure A–15–9:

Round shaft with

shoulder fillet in

bending. σ0 =

Mc/I, c = d/2 & I

= πd4/64.

Stress Concentration for Static

and Ductile Conditions

With static loads and ductile materials:

Highest stressed fibers yield (cold work)

Load is shared with next fibers

Cold working is localized

Overall part does not see damage unless

ultimate strength is exceeded

SC effect is commonly ignored for static

loads on ductile materials


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Stress Concentration for Static

and Ductile Conditions

SC must be included for dynamic

loading

SC must be included for brittle

materials, since localized yielding may

reach brittle failure rather than cold-

working and sharing the load.


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Uniaxial stress element

(tension test)

Multi-axial stress

element One strength, multiple stresses

How to compare stress state

to single strength?

Need for Static Failure Theories


Strength Sn

Stress

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Need for Static Failure Theories

Failure theories propose

appropriate means of comparing

multi-axial stress states to single

strength


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Maximum Normal Stress

Theory (MNS)

Yielding begins when max principal stress

in a stress element exceeds the yield

strength.

Use Mohr’s circle to find principal stresses

Compare largest principal stress to yield

strength


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Theory is

unsafe in 4th

quadrant

Not safe to

use for ductile

materials


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Max Normal Stress Theory (MNS)

Max Shear Stress Theory (MSS)

Yielding begins when tmax in a stress

element exceeds tmax in a tension test specimen of same material when that

specimen begins to yield.

For a tension test specimen, tmax = 1 /2.

At yielding, 1 = Sy and tmax = Sy /2

Yielding begins when max shear stress in a

stress element exceeds Sy/2


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Mohr’s circle: find max shear stress.

Compare max shear stress to Sy/2.

Order principal stresses: 1 ≥ 2 ≥ 3

Incorporate a design factor n


max

/ 2ySn

t

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Consider a plane stress state

A & B = two non-zero principal stresses

Order them with zero principal stress such

that 1 ≥ 2 ≥ 3

Assuming A ≥ B , three cases to consider:

Case 1: A ≥ B ≥ 0

Case 2: A ≥ 0 ≥ B

Case 3: 0 ≥ A ≥ B


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Case 1: A ≥ B ≥ 0

1 = A and 3 = 0

Eq. (5–1) reduces to A ≥ Sy

Case 2: A ≥ 0 ≥ B

1 = A and 3 = B

Eq. (5–1) reduces to A − B ≥ Sy

Case 3: 0 ≥ A ≥ B

1 = 0 and 3 = B

Eq. (5–1) reduces to B ≤ −Sy


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1. A ≥ B ≥ 0, A ≥ Sy

2. A ≥ 0 ≥ B, A − B ≥ Sy

3. 0 ≥ A ≥ B, B ≤ −Sy


Fig. 5–7

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Conservative in

all quadrants

Used for design

situations


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Distortion Energy (DE) Failure Theory

Other Names:

Octahedral Shear Stress

Shear Energy

Von Mises

Von Mises – Hencky


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Ductile materials stressed hydrostatically

exhibited yield strengths greatly in excess

of expected values.

Strain energy is divided into:

1. hydrostatic volume changing energy

2. angular distortion energy

Yielding is primarily affected by distortion

energy


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Yielding occurs when distortion strain energy

per unit volume reaches distortion strain

energy per unit volume for yield in simple

tension or compression of same material.

Dr. Mohammad Suliman Abuhaiba, PE Fig. 5–8

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Hydrostatic stress is average of principal

stresses

Strain energy per unit volume,


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Substituting av for 1, 2, and 3, volume

Strain energy is obtained:


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Tension test specimen at yield has 1 = Sy

and 2 = 3 =0

Applying to Eq. (5–8), distortion energy for

tension test specimen is

DE theory predicts failure when DE (Eq. 5.8)

exceeds DE of tension test specimen (Eq.

5.9)


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Von Mises Stress (VMS)

Von Mises stress

For plane stress, simplifies to

In terms of xyz components, in 3D


April 19, 2014 24

Distortion Energy Theory With VMS

Von Mises Stress can be thought of as a

single, effective stress for the entire general

state of stress in a stress element.

Distortion Energy failure theory simply

compares von Mises stress to yield strength.

Introducing a design factor,


ySn

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DE Theory Compared to

Experimental Data

Plot VMS on

principal stress

axes to

compare to

experimental

data

DE curve is

typical of data


Fig. 5–15

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DE Theory Compared to

Experimental Data

Typically equates to a 50% reliability from

a design perspective

Commonly used for analysis situations

MSS theory useful for design situations

where higher reliability is desired


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Shear Strength Predictions

For pure shear

loading, Mohr’s

circle shows that A

= −B = t

Intersection of pure

shear load line with

failure curve

indicates shear

strength has been

reached Dr. Mohammad Suliman Abuhaiba, PE

Fig. 5–9

April 19, 2014 28

For MSS theory, intersecting pure shear load

line with failure line [Eq. (5–5)] results in


Fig. 5–9

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For DE theory, intersection pure shear load line with

failure curve [Eq. (5–11)] gives


Fig. 5–9

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Selection of

Failure Criteria


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Example 5-1

A hot-rolled steel has a yield strength of Syt =

Syc = 100 kpsi and a true strain at fracture of

εf = 0.55. Estimate the factor of safety for the

following principal stress states:

a. σx = 70 kpsi, σy = 70 kpsi, τxy = 0 kpsi

b. σx = 60 kpsi, σy = 40 kpsi, τxy = −15 kpsi

c. σx = 0 kpsi, σy = 40 kpsi, τxy = 45 kpsi

d. σx = −40 kpsi, σy = −60 kpsi, τxy = 15 kpsi

e. σ1 = 30 kpsi, σ2 = 30 kpsi, σ3 = 30 kpsi


Some materials have compressive strengths

different from tensile strengths

Mohr theory is based on three simple tests:

1. Tension

2. Compression

3. shear

Mohr Theory


Fig. 5−12

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Coulomb-Mohr Theory

Coulomb-Mohr theory

simplifies to linear failure

envelope using only

tension and

compression tests


Fig. 5−13

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Dr. Mohammad Suliman Abuhaiba, PE Fig. 5−13

April 19, 2014 35

Coulomb-Mohr Theory

Incorporating factor of safety

For ductile material, use tensile and

compressive yield strengths

For brittle material, use tensile and

compressive ultimate strengths


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Coulomb-Mohr Theory

Consider three cases

Case 1: A ≥ B ≥ 0 , 1 = A and 3 = 0

Eq. (5−22) reduces to

Case 2: A ≥ 0 ≥ B , 1 = A and 3 = B

Eq. (5-22) reduces to

Case 3: 0 ≥ A ≥ B , 1 = 0 and 3 = B

Eq. (5−22) reduces to


April 19, 2014 37

Coulomb-Mohr Theory


Fig. 5−14

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Coulomb-Mohr Theory

Intersect pure shear load line with the

failure line to determine the shear

strength


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Coulomb-Mohr Theory

Example 5-2

A 25-mm-diameter shaft is statically torqued

to 230 N・m. It is made of cast 195-T6 aluminum, with a yield strength in tension of

160 MPa and a yield strength in

compression of 170 MPa. It is machined to

final diameter. Estimate the factor of safety

of the shaft.


Example 5-3 A certain force F applied at D near the end

of the 15-in lever shown in Fig. 5–16, which is

quite similar to a socket wrench, results in

certain stresses in the cantilevered bar

OABC. This bar (OABC) is of AISI 1035 steel,

forged and heat-treated so that it has a

minimum (ASTM) yield strength of 81 kpsi. We

presume that this component would be of

no value after yielding. Thus the force F

required to initiate yielding can be regarded

as the strength of the component part. Find

this force. Dr. Mohammad Suliman Abuhaiba, PE

Example 5-3


Fig. 5−16

Example 5-4

The cantilevered tube shown in Fig. 5–17 is

to be made of 2014 aluminum alloy

treated to obtain a specified minimum

yield strength of 276 MPa. We wish to

select a stock-size tube from Table A–8

using a design factor nd = 4. The bending

load is F = 1.75 kN, the axial tension is P =

9.0 kN, and the torsion is T = 72 N・m. What is the realized factor of safety?


Example 5-4


Fig. 5−17

Failure Theories for Brittle Materials

Experimental data indicates some

differences in failure for brittle materials.

Failure criteria is generally ultimate

fracture rather than yielding

Compressive strengths are usually larger

than tensile strengths


April 19, 2014 45


Failure occurs when max principal stress in

a stress element exceeds the strength.

Predicts failure when

For plane stress,

Incorporating design factor,


April 19, 2014 46

Unsafe in part of

fourth quadrant

Not

recommended for

use

historical

comparison


Fig. 5−18

April 19, 2014 47


Failure equations dependent on quadrant

Quadrant condition Failure criteria Equation No.

Brittle Coulomb-Mohr


April 19, 2014 48

Failure equations dependent on quadrant

Brittle Coulomb-Mohr


Fig. 5−14

April 19, 2014 49

Brittle Failure Experimental Data

Coulomb-Mohr is

conservative in

4th quadrant

Modified Mohr

criteria adjusts to

better fit the

data in the 4th

quadrant


Fig. 5−19

April 19, 2014 50

Quadrant condition Failure criteria Equation No.

Modified-Mohr


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Example 5-5

Consider the wrench in Ex. 5–3, Fig. 5–16, as

made of cast iron, machined to dimension.

The force F required to fracture this part can

be regarded as the strength of the

component part. If the material is ASTM

grade 30 cast iron, find the force F with

a. Coulomb-Mohr failure model.

b. Modified Mohr failure model.


Example 5-5


Fig. 5−16

Selection of

Failure Criteria


April 19, 2014 54

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