Machine Design I

1. [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

2. [68]
 Ductile: and have an identifiable yield strength that is often the same in compression as in tension.
Significant plastic deformation between yield and fracture { 𝜀𝑓 ≥ 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 { 𝜀𝑓 < 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
𝑜𝑟 𝜎1 − 𝜎3 ≤ 𝑆 𝑦 (5 − 1)
Note that this implies that the yield strength in shear is given by:
𝑆 𝑠𝑦 = 0.5 𝑆 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
𝑜𝑟 𝜎1 − 𝜎3 =
𝑆 𝑦
2𝑛
(5 − 3)
For plane stress (where one of the principal stresses is zero), Assuming that 𝜎𝐴 ≥ 𝜎𝐵, there are three
cases to consider in using Eq. (5–1):
Case 1: 𝜎𝐴 ≥ 𝜎𝐵 ≥ 0 . For this case, 𝜎1 =
𝜎𝐴 𝑎𝑛𝑑 𝜎3 = 0. Equation (5–1) reduces to a
yield condition of:
𝜎𝐴 ≥ 𝑆 𝑦 (5 − 4)
Case 2: 𝜎𝐴 ≥ 0 ≥ 𝜎𝐵 . For this case, 𝜎1 =
𝜎𝐴 𝑎𝑛𝑑 𝜎3 = 𝜎𝐵. Equation (5–1) becomes:

3. [69]
𝜎𝐴 − 𝜎𝐵 ≥ 𝑆 𝑦 (5− 5)
Case 3: 0 ≥ 𝜎𝐴 ≥ 𝜎𝐵 . For this case, 𝜎1 = 0 𝑎𝑛𝑑 𝜎3 = 𝜎𝐵. Equation (5–1) reduces to a yield
condition of:
𝜎 𝐵 ≤ −𝑆 𝑦 (5 − 6)
Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the 𝜎𝐴 , 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
𝜎1, 𝜎2, 𝑎𝑛𝑑 𝜎3.
 The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses 𝜎𝑎𝑣 , acting in
each of the same principal directions as in Fig. 5–8a. The formula for 𝜎𝑎𝑣 , is simply:
𝜎𝑎𝑣𝑒 =
𝜎1 + 𝜎2 + 𝜎3
3
( 𝑎)
 If we regard 𝜎𝑎𝑣𝑒 as a component of 𝜎1, 𝜎2, 𝑎𝑛𝑑 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 =
1
2
[ 𝜀1 𝜎1 + 𝜀2 𝜎2 + 𝜀3 𝜎3] ( 𝑏)
 Using Eq. of Hooke's law with substituting Eq.(b) for the principal strains in gives:
𝑢 =
1
2𝐸
[ 𝜎1
2
+ 𝜎2
2
+ 𝜎2
2
− 2𝜐( 𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1)] ( 𝑐)
 The strain energy for producing only volume change 𝑢 𝑣 can be obtained by substituting Eq. (a)

4. [70]
in Eq. (c). The result is:
𝑢 𝑣 =
3𝜎𝑎𝑣
2
2𝐸
(1 − 2𝜐) ( 𝑑)
 If we now substitute the square of Eq. (a) in Eq. (d) and simplify the expression, we get:
𝑢 𝑣 =
1 − 2𝜐
6𝐸
[ 𝜎1
2
+ 𝜎2
2
+ 𝜎2
2
+ 𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1] (5− 7)
 Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (c). This gives:
𝑢 𝑑 = 𝑢 − 𝑢 𝑣 =
1 + 𝜐
3𝐸
[
( 𝜎1 − 𝜎2)2
+ ( 𝜎2 − 𝜎3)2
+ ( 𝜎3 − 𝜎1)2
2
] (5 − 8)
 Note that the distortion energy is zero if 𝜎1 = 𝜎2 = 𝜎3 = 0 .
 For the simple tensile test, at yield, 𝜎1 = 𝑆 𝑦 𝑎𝑛𝑑 𝜎2 = 𝜎3 = 0 and from Eq. (5–8) the distortion
energy is
𝑢 𝑑 =
1 + 𝜐
3𝐸
𝑆 𝑦
2 (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−𝜎2)2
+( 𝜎2 −𝜎3 )2
+( 𝜎3−𝜎1 )2
2
]
1
2⁄
≥ 𝑆 𝑦 (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 − 𝜎1)2
2
]
1
2⁄
(5 − 12)

 For plane stress, let σA and σB be the two nonzero principal stresses. Then from Eq. (5–12), we
get:

5. [71]
σ′
= [ 𝜎𝐴
2
− 𝜎𝐴 𝜎𝐵 + 𝜎𝐵
2]
1
2⁄ (5 − 13)
 Equation (5–13) is a rotated ellipse in the 𝜎𝐴 , 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
[(𝜎𝑥 − 𝜎𝑦)
2
+ (𝜎𝑦 − 𝜎𝑧)
2
+ ( 𝜎𝑧 − 𝜎𝑥)2
+ 6(𝜏 𝑥𝑦
2
+ 𝜏 𝑦𝑧
2
+ 𝜏 𝑧𝑥
2
)]
1
2⁄
(5 − 14)
, and for plane stress,
σ′
=
1
√2
[𝜎𝑥
2
− 𝜎𝑥 𝜎𝑦 + 𝜎𝑦
2
+ 3𝜏 𝑥𝑦
2
]
1
2⁄
(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 𝜎𝑎𝑣𝑒 . 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
+ ( 𝜎2 − 𝜎3)2
+ ( 𝜎3 − 𝜎1)2]
1
2⁄ (5− 16)
 From Eq. (5–16) the octahedral shear stress under the tensile test is:
𝜏 𝑜𝑐𝑡 =
√2
3
𝑆 𝑦 (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:

6. [72]
[
( 𝜎1 − 𝜎2)2
+ ( 𝜎2 − 𝜎3)2
+ ( 𝜎3 − 𝜎1)2
2
]
1
2⁄
≥ 𝑆 𝑦 (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

7. [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 𝜎1 ≥ 𝜎2 ≥ 𝜎3 . The largest circle
connects σ 𝜎1 𝑎𝑛𝑑 𝜎3 , as shown in
Fig. 5–13. The centers of the circles in Fig. 5–13 are 𝐶1, 𝐶2, 𝑎𝑛𝑑 𝐶3 . Triangles 𝑂𝐵𝑖 𝐶𝑖 are similar,
therefore:
𝐵2 𝐶2 − 𝐵1 𝐶1
𝑂𝐶2 − 𝑂𝐶1
=
𝐵3 𝐶3 − 𝐵1 𝐶1
𝑂𝐶3 − 𝑂𝐶1
𝜎1−𝜎3
2
−
𝑆𝑡
2
𝑆𝑡
2
−
𝜎1+𝜎3
2
=
𝑆𝑐
2
−
𝑆𝑡
2
𝑆𝑐
2
+
𝑆𝑡
2
 Cross-multiplying and simplifying reduces this equation to:
𝜎1
𝑆𝑡
−
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 𝜎𝐴 ≥ 𝜎𝐵, there are
three cases to consider in using Eq. (5–1):
1. Case 1: 𝜎𝐴 ≥ 𝜎𝐵 ≥ 0 . For this case, 𝜎1 = 𝜎𝐴 𝑎𝑛𝑑 𝜎3 = 0. Equation (5–21) reduces to a yield
condition of:
𝜎𝐴 ≥ 𝑆𝑡 (5 − 22)
2. Case 2: 𝜎𝐴 ≥ 0 ≥ 𝜎𝐵 . For this case, 𝜎1 = 𝜎𝐴 𝑎𝑛𝑑 𝜎3 = 𝜎𝐵. Equation (5–21) becomes:
𝜎𝐴
𝑆𝑡
−
𝜎 𝐵
𝑆𝑐
≥ 1 (5− 23)
3. Case 3: 0 ≥ 𝜎𝐴 ≥ 𝜎𝐵 . For this case, 𝜎1 = 0 𝑎𝑛𝑑 𝜎3 = 𝜎𝐵. Equation (5–1) reduces to a yield

8. [74]
condition of:
𝜎𝐵 ≤ −𝑆𝑐 (5 − 24)
 Equations (5–22)
to (5–24) are
represented in Fig.
5–14 by the three
lines indicated in
the 𝜎𝐴 , 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
𝑆𝑡
−
𝜎3
𝑆𝑐
=
1
𝑛
(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, 𝜎1 = −𝜎3 = 𝜏 . The torsional yield strength occurs when
𝜏 𝑚𝑎𝑥 = 𝑆𝑠𝑦. Substituting 𝜎1 = −𝜎3 = 𝜏𝑠𝑦 into Eq. (5–21) and simplifying gives:
𝑆𝑠𝑦 =
𝑆 𝑦𝑡 𝑆 𝑦𝑐
𝑆 𝑦𝑡 + 𝑆 𝑦𝑐
(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 𝜎1 ≥ 𝜎2 ≥ 𝜎3.
This theory then predicts that failure occurs whenever:
𝜎1 ≥ 𝑆 𝑢𝑡 𝑜𝑟 𝜎3 ≤ − 𝑆 𝑢𝑐 (5− 27)
 For plane stress, with the principal stresses given by Eq. (3–13), with σA ≥ σB, Eq. (5–28) can be
written as:
𝜎𝐴 ≥ 𝑆 𝑢𝑡 𝑜𝑟 𝜎𝐵 ≤ − 𝑆 𝑢𝑐 (5− 28)

9. [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:

10. [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 𝑇𝐵 = (300 − 50)(4) = 1000 𝑖𝑛 · 𝑙𝑏

11. [77]
Torque on shaft from pulley at C 𝑇𝐶 = (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
𝑅 𝐴𝑦 = 222 𝑙𝑏, 𝑅 𝐴𝑧 = 106 𝑙𝑏, 𝑅 𝐷𝑦 = 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 𝒙𝒚
𝟐 + 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 = (20𝑥103
)/𝑑3
𝜏 𝑚𝑎𝑥 = (5.1x103)/𝑑3
 Now construct Mohr’s circle

12. [78]
C at (10 x 103)/d3
R = (11.2 x 103)/d3
1 = (21.2 x 103)/d3
3 = (-1.2 x 103)/d3
 Use Coulomb-Mohr theory for brittle failure:
"32.1d
8.2
1
d97
2.1
d26
2.21
1
SS
33
uc
3
ut
1





 If making a design recommendation, you would recommend the next largest standard
dimension (16th’s): d = 1.375 in
5–11 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 reading
5–13 Stochastic Analysis: Self reading
5–14 Important Design Equations: Self reading

13. [79]
Problem
5–14
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:

14. [80]
Problem
5–15
The figure shows a crank
loaded by a force F = 190
lbf which causes twisting
and bending of the
Problem 5–15 3 4⁄ in -
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:

15. [81]
Problem
5–27
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.
Use the distortion-energy theory.
SOLUTION:
Problem
5–38
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

16. [82]
the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 60 kpsi.
SOLUTION: