this is in the series of providing useful notes for the subject namely "design of machine elements"

Copyright 2011 by Pearson Education, Inc.

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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





Different methods to

solve for beams can be

graphical, mathematical

or computer based.

Singularity functions to

represent the loads is

one of them

(computing)





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




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




Integrals of singularity functions

5




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




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




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




9 Calculating reaction forces R1 and R2 substituting

C1, C2, V and M as 0 @ x=l




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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This chapter is going to be a review of subjects you learned in

mechanics of materials




23

Along axis direction Opposite axis direction

X Y

Z




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




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




28

Mohr circle is a graphical method find the principal stresses

The Mohrs Circle




29

The Mohrs Circle




30

The Mohrs Circle




31

The Mohrs Circle




32




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




34

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




35

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




36

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




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




38 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

Pure bending in Beams

C is taken as +ve initially and

proper sign

applied based on

loading

compression ve and tension +ve




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( / )




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




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





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





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.




45




46




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




48

Torsion

For non-circular sections, max and are

Where Q and K are functions of the Cross Section




49




50




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




52


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




53


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




54


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




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.




56

Stress Concentration




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this is in the series of providing useful notes for the subject namely "design of machine elements"

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