this is in the series of providing useful notes for the subject namely "design of machine elements"
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This is a lecture notes on the design of Machine Elements. Highly useful for Mechanical Engineering Students who are interested to pursue in the application of solid Mechanics.TRANSCRIPT
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
Singularity Functions in beam loadings 1
Since the load is
concentrated or discrete
entities, discontinuous
over the beam length, it
is difficult to represent
discrete functions with
equation valid for entire
beam length
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
Singularity Functions in beam loadings 2
Different methods to
solve for beams can be
graphical, mathematical
or computer based.
Singularity functions to
represent the loads is
one of them
(computing)
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
Singularity Functions in beam loadings 3
Singularity functions are
binomials where x is
the point of interest in
beam, and a is the
place where in x the
load begins to act.
Depending on the type
of load, provides value
of x along the position
as shown in the table
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
Load distribution Function Formula Conditions
Quadratically distributed
load
Unit parabolic function = 0, x a
= (x a)2, x > a
Linearly distributed load Unit ramp function = 0, x a
= (x a), x > a
Uniformly distributed load Unit step function
= 0, x < a
= undefined, x = a
= 1, x > a
Concentrated load Unit impulse function
= 0, x < a
= , x = a
= 0, x > a
Concentrated moment Unit doublet function
= 0, x < a
= indeterminate, x = a
= 0, x > a
4
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
Integrals of singularity functions
5
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6
Integrating the loading function q to get shear V, and
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 and moments included in loading function,
as diagrams should close at beam end
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
7 When x = l, we get 0 for V and M thereby calculating
the reaction forces R1 and R2
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
Formula Conditions
= 0, x a
= (x a)2, x > a
= 0, x a
= (x a), x > a
= 0, x < a
= undefined, x = a
= 1, x > a
Mmax = 88.2
Vmax = -42
R1 = 18
and R2 = 42
Resulting in
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
8
Integrating the loading function q to get shear V, and
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
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
9 Calculating reaction forces R1 and R2 substituting
C1, C2, V and M as 0 @ x=l
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
10 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
where the Vmax is at x=l
Substituting V=0 to get Mmax at X=7.4
Mmax = 147.2
Vmax = -120
R1 = 56.7
and R2 = 123.3
Resulting in
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
11
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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
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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
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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
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This chapter is going to be a review of subjects you learned in
mechanics of materials
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
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Copyright 2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
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Copyright 2011 by Pearson Education, Inc.
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
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Copyright 2011 by Pearson Education, Inc.
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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
23
Along axis direction Opposite axis direction
X Y
Z
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
24 Axis of the stresses is arbitrarily chosen for convenience.
Normal and shear stresses at one point will vary with
direction along the coordinate system.
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 45to the planes of principal normal stresses
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
26 The principal shear
stresses from principal
normal stresses can be
calculated as
Since usually 1>2>3, max is 13 and the
direction of principal
shear stresses are
45to the principal normal planes and are
mutually orthogonal
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28
Mohr circle is a graphical method find the principal stresses
The Mohrs Circle
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The Mohrs Circle
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The Mohrs Circle
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The Mohrs Circle
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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
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Modes of Failure Shear common loading in pins, bolted or riveted construction
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
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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
Modes of Failure
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
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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
Modes of Failure
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
37 Pure bending in Beams
Pure Bending it is rare for beam to be loaded in pure bending. It is useful though to understand the situation
Mostly shear loading and bending moments act together
Applying point loads P equidistant at the simply supported beam, absence of shear loading makes this
pure bending
Assumptions used for analysis
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
38 Pure bending in Beams
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
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
Pure bending in Beams
C is taken as +ve initially and
proper sign
applied based on
loading
compression ve and tension +ve
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
40 Pure bending in Curved Beams
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 = -(-)/LN( / )
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
41 Pure bending in Curved Beams
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
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42 Shear due to transverse loading
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 < 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
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
43 Shear due to transverse loading
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
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Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton
44 Shear due to transverse loading
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.
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47 Torsion
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 shears due to torsional loading
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
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Torsion
For non-circular sections, max and are
Where Q and K are functions of the Cross Section
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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
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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
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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 90hence 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
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Combined bending and torsional stresses
Maximum shear stress and principal stresses that result from this combination
Shear due to transverse loading on B
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
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Spring Rates
55
For a uniform bar in axial tension, =
=
=
For a uniform cross section round bar in torsion,
=
=
=
For a cantilever beam with concentrated point load at end,
=
=
3
3
= 3
3
Note that k for a beam is unique to its manner of support and its loading
distribution, since y depends on the beams deflection equation and point of load application.
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Stress Concentration
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