[67]

CHAPTER 5 Failures Resulting from Static Loading:5–1 Static Strength 5–7 Maximum-Normal-Stress Theory for Brittle Materials 5–2 Stress Concentration 5–8 Modifications of the Mohr Theory for Brittle Materials 5–3 Failure Theories 5–91 Selection of Failure Criteria

5–4 Maximum 5–10 Introduction to Fracture Mechanics 5–5 Distortion 5–11 Stochastic Analysis

5–6 Coulomb 5–12 Important Design Equations

Strength is a property or characteristic of a mechanical element. Failure – any change in a machine part which makes it unable to perform its intended

function. We will normally use a yield failure criteria for ductile materials. The ductile failure

theories presented are based on yield. In this chapter our attention is focused on the predictability of permanent distortion or

separation. In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufficiently to accomplish his or her purposes.

5–1 Static Strength: A static load is a stationary force or couple applied to a member. To be stationary, the force or

couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner.

You can now appreciate the following four design categories: Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design.The part is made in large enough quantities that a moderate series of tests is feasible.The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing.The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it.

5–2 Stress Concentration:see Sec. 3–13

5–3 Failure Theories:Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine

element fails.

Structural metal behavior is typically classified as being ductile

or brittle, although under special situations, a material normally

considered ductile can fail in a brittle manner. Static failure can

be classified (as shown in the following sketch) into:

Ductile

Brittle

Ductile: and have an identifiable yield strength that is often the

[68]

same in compression as in tension. Significant plastic deformation between yield and

fracture { ε f ≥0.05}. The generally accepted theories for ductile materials (yield criteria)

are:

a. Maximum shear stress (MSS).

b. Distortion energy (DE).

c. Ductile Coulomb-Mohr (DCM).

Brittle:

Do not exhibit an identifiable yield strength, and are typically classified by ultimate

tensile and compressive strengths. Yield ~= fracture { ε f<0.05}. The generally

accepted theories for brittle materials (fracture criteria) are:

a. Maximum normal stress (MNS).

b. Brittle Coulomb-Mohr (BCM).

c. Modified Mohr (MM).

5–4 Maximum-Shear-Stress Theory for Ductile Materials:It predicts that yielding begins whenever the maximum shear stress (MSS) in any element equals or

exceeds the maximum shear stress in a tension test specimen of the same material when that specimen

begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Thus, for a general

state of stress, the maximum-shear-stress theory predicts yielding when:

τ max=σ1−σ3

2≤S y

2∨σ1−σ3≤ Sy (5−1 )

Note that this implies that the yield strength in shear is given by:

Ssy=0.5S y (5−2 )

, which, as we will see later is about 15 percent low (conservative). For design purposes, Eq. (5–1) can

be modified to incorporate a factor of safety, n. Thus,

τ max=S y

2n∨σ1−σ3=

S y

2n(5−3 )

For plane stress (where one of the principal stresses is zero), Assuming that σ A≥σB, there are three cases

to consider in using Eq. (5–1):

Case 1: σ A≥σB≥0 . For this case, σ 1=σ A∧σ 3=0.

Equation (5–1) reduces to a yield condition of:

σ A≥S y (5−4 )

Case 2: σ A≥0≥σ B . For this case, σ 1=σ A∧σ 3=σ B

.

Equation (5–1) becomes:

[69]

σ A−σB≥S y (5−5 )

Case 3: 0≥σ A≥σ B . For this case, σ 1=0∧σ3=σB. Equation (5–1) reduces to a yield condition of:

σ B≤−Sy (5−6 )

Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the σ A , σB plane.

5–5 Distortion-Energy (DE) Theory for Ductile Materials: It predicts that yielding occurs when the distortion strain energy per unit volume reaches or

exceeds the distortion strain energy per unit volume for yield in simple tension or compression

of the same material.

The distortion-energy (DE) theory originated from the observation that ductile materials stressed

hydrostatically exhibited yield strengths greatly in excess of the values given by the simple

tension test. Therefore it was postulated that yielding was not a simple tensile or compressive

phenomenon at all, but, rather, that it was related somehow to the angular distortion of the

stressed element.

To develop the theory, note, in Fig. 5–8a, the unit volume subjected to any three-dimensional

stress state designated by the stresses σ 1, σ2 ,∧σ3.

The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses σ av,

acting in each of the same principal directions as in Fig. 5–8a. The formula for σ av, is simply:

σ ave=σ1+σ2+σ3

3(a )

If we regard σ ave as a component of σ 1, σ2 ,∧σ3, then this component can be subtracted from

them, resulting in the stress state shown in Fig. 5–8c. This element is subjected to pure angular

distortion, that is, no volume change.

The strain energy per unit volume for the element shown in Fig. 5–8a is:

u=12

[ε1σ1+ε2σ2+ε3σ 3 ] (b )

Using Eq. of Hooke's law with substituting Eq.(b) for the principal strains in gives:

u= 12 E [σ1

2+σ 22+σ2

2−2υ (σ1σ2+σ2σ3+σ 3σ 1) ] (c )

The strain energy for producing only volume change uv can be obtained by substituting Eq. (a)

in Eq. (c). The result is:

uv=3σ av

2

2 E(1−2υ) (d )

If we now substitute the square of Eq. (a) in Eq. (d) and simplify the expression, we get:

[70]

uv=1−2υ

6 E[σ1

2+σ22+σ 2

2+σ1σ2+σ2σ3+σ3σ1 ] (5−7 )

Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (c). This gives:

ud=u−uv=1+υ3E [ (σ1−σ2 )2+(σ 2−σ3 )2+(σ3−σ1 )2

2 ] (5−8 )

Note that the distortion energy is zero if σ 1=σ2=σ3=0 .

For the simple tensile test, at yield, σ 1=S y∧σ2=σ 3=0 and from Eq. (5–8) the distortion energy

is

ud=1+υ3 E

S y2 (5−9 )

So for the general state of

stress given by Eq. (5–8),

yield is predicted if Eq. (5–

8) equals or exceeds Eq. (5–

9). This gives:

[ (σ1−σ 2 )2+ (σ2−σ3 )2+(σ 3−σ1 )2

2 ]12

≥ Sy (5−10 )

Thus, the left of Eq. (5–10) can be thought of as a single, equivalent, or effective stress. This

effective stress is usually called the Von Mises stress, σ′, named after Dr. R. Von Mises, who

contributed to the theory. Thus Eq. (5–10), for yield, can be written as:

σ '≥S y (5−11)

, where the von Mises stress is:

σ '=[ (σ1−σ2 )2+(σ2−σ3 )2+(σ3−σ 1)2

2 ]12

(5−12 )

For plane stress, let σA and σB be the two nonzero principal stresses. Then from Eq. (5–12), we

get:

σ '=[σ A2 −σ AσB+σ B

2 ]12 (5−13 )

[71]

Equation (5–13) is a rotated ellipse in the σ A , σB plane, as shown in Fig. 5–9 with '=S y . The

dotted lines in the figure represent the MSS theory, which can be seen to be more restrictive,

hence, more conservative.

Using xyz components of three-dimensional stress, the von Mises stress can be written as:

σ '= 1√2

[ (σ x−σ y)2+(σ y−σ z )

2+(σ z−σ x)2+6 (τ xy2 +τ yz

2 +τ zx2 ) ]

12 (5−14 )

, and for plane stress,

σ '= 1√2

[σx2−σx σ y+σ y

2+3 τ xy2 ]

12 (5−15 )

The distortion-energy theory is also called:

1. The von Mises or von Mises–Hencky theory

2. The shear-energy theory

3. The octahedral-shear-stress theory

Octahedral-Shear-Stress Theory: Failure is assumed to occur whenever the

octahedral shear stress for any stress state equals

or exceeds the octahedral shear stress for the

simple tension-test specimen at failure.

Consider an isolated element in which the

normal stresses on each surface are equal to the

hydrostatic stress σ ave. There are eight surfaces

symmetric to the principal directions that contain this stress. This forms an octahedron as shown in

Fig. 5–10. The shear stresses on these surfaces are equal and are called the octahedral shear stresses

(Fig. 5–10 has only one of the octahedral surfaces labeled). Through coordinate transformations the

octahedral shear stress is given by:

τ oct=[ (σ1−σ2 )2+(σ 2−σ3 )2+(σ3−σ1 )2 ]12 (5−16 )

From Eq. (5–16) the octahedral shear stress under the tensile test is:

τ oct=√23Sy (5−17 )

When, for the general stress case, Eq. (5–16) is equal or greater than Eq. (5–17), yield is predicted.

This reduces to:

[72]

[ (σ1−σ 2 )2+ (σ2−σ3 )2+(σ 3−σ1 )2

2 ]12

≥ Sy (5−18 )

Equation (5-11) can be expressed as a design equation by:

σ '=S y

n(5−19 )

Thus, the shear yield strength predicted by the distortion-energy theory is Ssy=0.577S y.

EXAMPLE 1:A hand cranking lever, as shown in next Figure is used to

start a truck engine by applying a force F = 400 N. The

material of the cranking lever is 30C8 for which yield

strength = 320 MPa; Ultimate tensile strength = 500 MPa;

Young’s modulus = 205 GPa; Modulus of rigidity = 84

GPa and poisson’s ratio = 0.3. Assuming factor of safety

to be 4 based on yield strength, design the diameter ‘d’ of the lever at section X-X near the guide bush

using : 1. Maximum distortion energy theory; and 2. Maximum shear stress theory. [Ans. 28.2 mm;

28.34 mm]

5–6 Coulomb-Mohr Theory for Ductile Materials: Not all materials have compressive strengths equal to their corresponding tensile values. For

[73]

example, the yield strength of

magnesium alloys in compression may

be as little as 50 percent of their yield

strength in tension.

A variation of Mohr’s theory, called

the Coulomb-Mohr theory or the

internal-friction theory, assumes that

the boundary BCD in Fig. 5–12 is

straight. With this assumption only the

tensile and compressive strengths are

necessary. Consider the conventional

ordering of the principal stresses such

that σ 1≥σ2≥σ3. The largest circle

connects σ σ 1∧σ3 , as shown in Fig.

5–13. The centers of the circles in Fig. 5–13 are C1,C2,∧C3 . Triangles OBiC i are similar, therefore:

B2C2−B1C1

OC 2−OC1

=B3C3−B1C1

OC3−OC1

σ1−σ3

2−S t

2S t

2−σ1+σ3

2

=

Sc2

−S t

2Sc

2+S t

2

Cross-multiplying and simplifying reduces this equation to:

σ1

S t

−σ3

Sc

=1 (5−21 )

, where either yield strength or ultimate strength can be used.

For plane stress (where one of the principal stresses is zero), Assuming that σ A≥σB, there are

three cases to consider in using Eq. (5–1):

1. Case 1: σ A≥σB≥0 . For this case, σ 1=σ A∧σ 3=0. Equation (5–21) reduces to a yield condition

of:

σ A≥St (5−22 )

2. Case 2: σ A≥0≥σ B . For this case, σ 1=σ A∧σ 3=σ B. Equation (5–21) becomes:

σ A

S t

−σ B

Sc≥1 (5−23 )

3. Case 3: 0≥σ A≥σ B . For this case, σ 1=0∧σ3=σB. Equation (5–1) reduces to a yield condition of:

[74]

σ B≤−Sc (5−24 )

Equations (5–22)

to (5–24) are

represented in Fig.

5–14 by the three

lines indicated in

the σ A , σB plane.

For design

equations,

incorporating the

factor of safety n,

divide all strengths by n. For example, Eq. (5–22) as a design equation can be written as:

σ1

S t

−σ3

Sc

=1n

(5−25 )

Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce

it from Eq. (5–21). For pure shear τ , σ 1=−σ3=τ . The torsional yield strength occurs when τ max=Ssy

. Substituting σ 1=−σ3=τ sy into Eq. (5–21) and simplifying gives:

Ssy=S yt S yc

Syt+Syc

(5−26 )

5–8 Maximum-Normal-Stress Theory for Brittle Materials: The maximum-normal-stress (MNS) theory states that failure occurs whenever one of the three

principal stresses equals or exceeds the strength.

Again we arrange the principal stresses for a general stress state in the ordered form σ 1≥σ2≥σ3. This

theory then predicts that failure occurs whenever:

σ 1≥Sut∨σ3≤−Suc (5−27 )

For plane stress, with the principal stresses given by Eq. (3–13), with σ A≥σB, Eq. (5–28) can be

written as:

σ A≥Sut∨σ B≤−Suc (5−28 )

[75]

, which is plotted in Fig. 5–18a. As before, the failure criteria equations can be converted to design

equations as:

The load lines

are shown in

Fig. 5–18b.

5–9 Modifications of the Mohr Theory for Brittle Materials:

The equations provided for the theories will be restricted to plane stress and be of the design type

incorporating the factor of safety. On the basis of observed data for the fourth quadrant, the modified Mohr

theory expands the fourth quadrant as shown in Fig. 5–19.

Brittle-Coulomb-Mohr:

[76]

Modified Mohr:

EXAMPLE:

Given:

Shaft of ASTM G25 cast iron subject

to loading shown

From Table A-24

Sut = 26 kpsi

Suc = 97 kpsi

Find: For a factor of safety of n = 2.8, what

should the diameter of the shaft (d) be?

Solution:

First, we need to find the forces acting on the

shaft Torque on shaft from pulley at B T B=(300−50)(4)=1000in·lb

[77]

Torque on shaft from pulley at C

T C=(360−27)(3)=1000 in·lb

Shaft is in static equilibrium; note that shaft is free to

move along the x-axis (bearings). Draw a FBD reaction

forces at points of attachment to show constrained motion.

Use statics to solve for reactions forces

RAy=222lb ,R Az=106 lb ,RDy=127 lb

OK, now we know all the forces. The problem gives us a

factor of safety, but unlike our last example, we aren’t told

specific places (elements) at which to look for failure! We are

going to have to calculate stresses. What do we need? Axial

forces, bending moments, and torques. We need to find our

moments… HOW? Shear-Moment diagrams will give us the

forces and moments along the shaft. Failure will likely occur

where the max values are seen

Moment in the x-z plane : Failure is going to occur at either B or C,

since these are locations where maximum moments are seen.

M=√M xy2 +M xz

2

We found the following:

MB x-y = 1780 in·lb

MB x-z = 848 in·lb

MC x-y = 762 in·lb

MC x-z = 1690 in·lb

Calculating the magnitudes with

MB = 1971.7 in·lb

MC = 1853.8 in·lb

Since the overall max moment is at B, we will expect failure there,

and use MB in our stress calculations. If we had been told the

location of interest, we would essentially start here.

σ max=(20 x103)/d3

τ max=¿5.1x103¿ /d3

Now construct Mohr’s circleC at (10 x 103)/d3

[78]

R = (11.2 x 103)/d3 1 = (21.2 x 103)/d33 = (-1.2 x 103)/d3

Use Coulomb-Mohr theory for brittle failure:σ 1

Sut−σ3

Suc=1η

21 .2

26d3+1.2

97d3=1

2 .8d=1.32 {} } } {¿

¿

If making a design recommendation, you would recommend the next largest standard dimension (16th’s): d = 1.375 in5–11 Selection of Failure Criteria:

Figure 5–21 provides a summary flow-chart for the selection of an effective procedure for analyzing

or predicting failures from static loading for brittle or ductile behavior.

5–12 Introduction to Fracture Mechanics: Self reading5–13 Stochastic Analysis: Self reading5–14 Important Design Equations: Self reading

[79]

Problem5–14

This problem illustrates that the

factor of safety for a machine

element depends on the particular

point selected for analysis. Here you

are to compute factors of safety,

based upon the distortion-energy

theory, for stress elements at A and

B of the member shown in the

figure. This bar is made of AISI

1006 cold-drawn steel and is loaded

by the forces F = 0.55 kN, P = 8.0

kN, and T = 30 N · m.

SOLUTION:

[80]

Problem5–15

The figure shows a crank loaded by a force F = 190 lbf which causes twisting and bending of the

Problem 5–15 3/ 4in -diameter shaft fixed to a support at the origin of the reference system. In

actuality, the support may be an inertia which we wish to rotate, but for the

purposes of a strength analysis we can consider this to be a statics problem.

The material of the shaft AB is hot-rolled AISI 1018 steel (Table A–20).

Using the maximum-shear-stress theory, find the factor of safety based on

the stress at point A.

SOLUTION:

Problem5–27

The figure is a schematic drawing of a

countershaft that supports two V-belt

pulleys. For each pulley, the belt

tensions are parallel. For pulley A

consider the loose belt tension is 15

percent of the tension on the tight side.

A cold-drawn UNS G10180 steel shaft

of uniform diameter is to be selected for

this application. For a static analysis

with a factor of safety of 3.0, determine

the minimum preferred size diameter.

[81]

Use the distortion-energy theory.

SOLUTION:

Problem5–38

Two steel tubes are shrink-fitted together where the nominal diameters are 1.50, 1.75, and 2.00 in.

Careful measurement before fitting revealed that the diametral interference between the tubes to be

0.00246 in. After the fit, the assembly is subjected to a torque of 8000 lbf · in and a bending-

moment of 6000 lbf · in. Assuming no slipping between the cylinders, analyze the outer cylinder at

the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 60 kpsi.

[82]

SOLUTION: