# Machine Design Book Summary

```Singularity Functions in beam loadings
1
concentrated or discrete
entities, discontinuous
over the beam length, it
is difficult to represent
discrete functions with
equation valid for entire
beam length
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
2
Different methods to
solve for beams can be
graphical, mathematical
or computer based.
Singularity functions to
one of them
(computing)
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
3
Singularity functions are
binomials &lt;&gt; where x is
the point of interest in
beam, and a is the
place where in x the
Depending on the type
of x along the position
as shown in the table
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
4
Function
Formula
Conditions
= 0, x  a
Unit parabolic function
= (x – a)2, x &gt; a
= 0, x  a
Unit ramp function
= (x – a), x &gt; a
= 0, x &lt; a
Unit step function
= undefined, x = a
= 1, x &gt; a
= 0, x &lt; a
Unit impulse function
= , x = a
= 0, x &gt; a
= 0, x &lt; a
Concentrated moment
Unit doublet function
= indeterminate, x = a
= 0, x &gt; a
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
5
Integrals of singularity functions
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
6
integrating shear (V) to get momentum (M),
When x is either 0 or l, we get 0 for V and M thereby giving 4 boundary conditions
• When x = 0, we get 0 for V and M thereby
calculating the constants C1 and C2 as 0
• C1 and C2 will always be 0, if reaction forces
as diagrams should close at beam end
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
When x = l, we get 0 for V and M thereby calculating 7
the reaction forces R1 and R2
Formula
Conditions
= 0, x  a
= (x – a)2, x &gt; a
= 0, x  a
= (x – a), x &gt; a
= 0, x &lt; a
= undefined, x = a
= 1, x &gt; a
Substituting C1 and C2 as 0
and R1 and R2 in equations
for V and M for different
values of x between 0 and l
We can find the graph
Vmax is at x=l
Substituting V=0 to get Mmax
Resulting in
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Mmax = 88.2
Vmax = -42
R1 = 18
and R2 = 42
Copyright &copy;2011 by Pearson Education, Inc.
8
integrating shear (V) to get momentum (M),
When x is either 0 or l, we get 0 for V and M
thereby giving 4 boundary conditions
When x = 0, we get 0 for V and M
thereby calculating the constants C1 and
C2 as 0, similar to previous situation
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Calculating reaction forces R1 and R2 substituting
C1, C2, V and M as 0 @ x=l
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
9
Copyright &copy;2011 by Pearson Education, Inc.
Substituting C1 and C2 as 0 and R1 and R2 in 10
equations for V and M for different values of
x between 0 and l, We can find the graph
where the Vmax is at x=l
Substituting V=0 to get Mmax at X=7.4
Resulting in
Mmax = 147.2
Vmax = -120
R1 = 56.7
and R2 = 123.3
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
11
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
13
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
15
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
17
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
18
This chapter is going to be a review of subjects you learned in
mechanics of materials
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
23
Z
X
Y
Along axis direction
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Opposite axis direction
Copyright &copy;2011 by Pearson Education, Inc.
• Axis of the stresses is arbitrarily chosen for convenience.
Normal and shear stresses at one point will vary with
direction along the coordinate system.
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• There will be planes where the shear stress is 0, and the
normal stresses acting on these planes are principal
stresses, and planes as principal planes
• There will be planes where the shear stress components
are 0, and the normal stresses acting on these planes are
principal stresses, and planes as principal planes
• Direction of surface normals to the
planes is principal axes .
• And the normal stresses acting in
these directions are principal normal
stresses
• The principal shear stresses, act on
planes at 45&deg;to the planes of
principal normal stresses
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
• The principal shear
stresses from principal
normal stresses can be
calculated as
26
• Since usually 1&gt;2&gt;3,
max is 13 and the
direction of principal
shear stresses are
45&deg;to the principal
normal planes and are
mutually orthogonal
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
28
The Mohr’s Circle
• Mohr circle is a graphical method find the principal stresses
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
The Mohr’s Circle
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
29
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The Mohr’s Circle
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Robert L. Norton
30
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The Mohr’s Circle
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Robert L. Norton
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
33
Modes of Failure
• Tension - When object is in tension there is only one stress
• More common cables, struts, bolts and any axially loaded members
• Applied normal
stress for axial
tension is
• Change in
length
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
34
Modes of Failure
• Direct Shear is when there is no bending, there can be shear with
bending too
• For direct shear, there is no gap between the blade
and the vice, which is very rare. when there is a
small gap, most common, there is a moment
creating bending stresses and shear
• This is preferred
because, the area
of shear is doubled
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
35
Modes of Failure
• Direct Bearing – the pin in hole can fail other than shear. Which is when the
surfaces of the pin and hole or subjected to bearing stresses (compressive in
nature) crushes the pin and hole than shear it.
• Direct Shear is when there is no bending, there can be shear with bending too
• Projected area of
contact in case of tight fit
• In case of
loose/clearance fit, it
approximated to
• Length and dia needs to
be carefully selected to
avoid bearing failure
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
36
Modes of Failure
• Tearout Failure – another
failure can be the tearout of
material surrounding the hole.
Happens when hole is placed
close to the edge.
• Provide sufficient material
around the holes to prevent this
mode of failure
• One pin diameter of material
between the edge and the hole
is good starting point for
calculations
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Pure bending in Beams
37
• Pure Bending – it is rare for beam to be loaded in pure
bending. It is useful though to understand the situation
together
• Applying point loads P equidistant at the simply
pure bending
• Assumptions used for analysis
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Pure bending in Beams
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
38
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Pure bending in Beams
39
• N-N along the neutral axis, no change in length
• A-A shortents (in compression) and B-B lengthens (in tension
• Bending stress is 0 at N-N and is linearly proportional to distance y away from
N-N
• Where M is the bending moment and I is the area moment of Inertia of
the beam cross section at the neutral plane, y the distance from N-N
• The max stress at outer plane is
• Where c is distance from neutral plane (it should be same both sections
if the beam is symmetrical about the neutral axis
• C is taken as +ve
initially and
proper sign
applied based on
compression –ve
and tension +ve
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Pure bending in Curved Beams
40
• Machines have curved beams like in C clamps, hooks etc, with a ROC) the
first 6 assumptions still apply
• If it has a significant curvature the neutral axis will not be coincident with the
centroidal axis and the shift e is found from
• Where A is area, rc is ROC of centroidal axis and r is the ROC of the
differential area dA
• For a rectangular beam this can be 𝒆 = 𝒓𝒄 -(𝒓𝒐 -𝒓𝒊 )/LN(𝒓𝒐 / 𝒓𝒊 )
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Pure bending in Curved Beams
41
• Stress distribution is not linear but hyperbolic. Sign convention is +ve moment
straightens the beam (tension inside and compression outside)
• For pure bending loads stresses at inner and outer surface is
• And if a force is applied on the curved beam then
• The stresses will be
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
42
• Common case is both shear and bending moment on the beam.
• Fig shows a point loaded beam shear and moment diagram
• Cutting our segment P of width dx around A, cut from outer side
at c upto depth y1 it is seen that the M(x1) to left &lt; M(x2) to right
and the difference being dM
• Similarly for stresses (can be seen in fig b) since the stresses
are proportion to moment
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
43
• Similarly for stresses (can be seen in fig b) since the stresses
are proportion to moment. This stress imbalance is countered
by shear stress  component
• Stress acting on left hand side of p at a distance y from neutral
axis is stress times the differential area dA at that point
• The total force acting on the left hand side will then be
• Similarly for right hand side
• Shear force on the top face is
• Where bdx is the area of the top face of the element
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
44
• For equilibrium, the forces acting on p is 0
• Gives an expression for shear stress as a
function of change in momentum wrt x
• Since slope of dM/dx is the magnitude of
the shear function V (dV/dx = d2M/dx2)
• Assigning the integral as Q, then
• Shear stresses vary across y
• And becomes 0 when c=y1
• And maximum at neutral axis
• A common rule of thumb is the shear stress due to transverse
loading in a beam will be small enough to ignore if the length
to depth ratio of the beam is 10 or more. Short beams below
that ratio should be investigated for both transverse shear
and bending.
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
46
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Torsion
47
• Isotropic, within elastic limits, pure torsion normal
to its axis on straight bar
• Fixed on one end and torsion applied at other,
bar twists about its long axis
• Element taken anywhere on the outer surface
• Shear stress varies from 0 at the center
maximum on the outer element, as shown in ‘b’
• Max shear stress at outer radius r is
• Where J is the polar moment of inertia of the CS
and the angular deflection  is
• Where G is shear modulus, l is length and K is J
which depends on the cross section
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
Torsion
48
• For non-circular sections, max and  are
• Where Q and K are functions of the Cross Section
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
51
Combined bending and torsional stresses
• Most bars have combined stresses creating both normal and shear stresses. So
these need to combined at certain locations to find the principal stresses and
maximum shear stress
• Limiting to bending of cantilever and in torsion
• The shear and moment diagrams will be similar
to a cantilever beam loaded at its end
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
52
Combined bending and torsional stresses
• Torque applied at the end is Tmax = Fa = 8000 lb-in and is uniform over its length
• Most heavily loaded CS is at the wall
where M, T and  are maximum
• Looking at the CS diagram, bending
normal stress is maximum at outer and 0
in the neutral axis, and shear it is reverse
• Shear due to torsion is proportional to
radius and is 0 at the center. While
bending is along Y axis, torsion is
uniformly distributed at outer fiber
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
53
Combined bending and torsional stresses
• Choosing 2 points A and B at the outer end to get worst combination of stresses
• A will have max bending and max torsion, and stresses in
element A is shown in fig ‘b’
• The normal stress x acts on the X face in X direction while
torsional shear xz acts on x face in +Z direction
• At B, the torsional shear has same mag of point A but the
direction is 90&deg;hence xy
• At B, the transverse shear stress is also maximum which is xy
and both act in –y direction
• Maximum normal bending and torsional
shear stress on A
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
54
Combined bending and torsional stresses
• Maximum shear stress and principal stresses that result from this combination
• Total shear stress on B is the sum of torsion
and transverse
• While point A has higher principal stress,
Point B has higher shear which is principal.
• Both points need to be checked
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
55
Spring Rates
For a uniform bar in axial tension,
𝑘=
𝐹
𝑦
=
𝐹
𝐹𝐿
𝐴𝐸
=
𝐴𝐸
𝐿
For a uniform cross − section round bar in torsion,
𝑘=
𝑇

=
𝑇
𝑇𝐿
𝐽𝐺
𝐺𝐽
=
𝐿
For a cantilever beam with concentrated point load at end,
𝑘=
𝐹
𝑦
=
𝐹
𝐹𝑙3
3𝐸𝐼
3𝐸𝐼
= 3
𝑙
Note that k for a beam is unique to its manner of support and its loading
distribution, since y depends on the beam’s deflection equation and
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
56
Stress Concentration
Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition
Robert L. Norton
Copyright &copy;2011 by Pearson Education, Inc.