# MECH 401 Machine Design

```MECH 401
Mechanical Design Applications
Dr. M. K. O’Malley – Master Notes
Spring 2007
Dr. D. M. McStravick
Rice University
Design Considerations






Stress – Yield Failure or Code Compliance
Deflection
Often the controlling factor for
Strain
functionality
Stiffness
Stability – Important in compressive members
Stress and strain relationships can be studied
with Mohr’s circle
Deflection [Everything’s a Spring]


When loads are applied, we have deflection
Depends on




 Tension
 Compression
 Bending
 Torsion
Cross-section of member
Comparable to pushing on a spring
We can calculate the amount of beam deflection by
various methods
Superposition



Determine effects of individual loads separately and
add the results [see examples 4-2,3,4]
Tables are useful – see A-9
May be applied if



Each effect is linearly related to the load that produces it
A load does not create a condition that affects the result of
Deformations resulting from any specific load are not large
enough to appreciably alter the geometric relations of the
parts of the structural system
Deflection --- Energy Method



There are situations where the tables are insufficient
We can use energy-methods in these circumstances
Define strain energy
s x  Ee x
x1

U   Fdx
0

Define strain energy density**



dU
dV
Put in terms of s, e
V – volume
1
1 sx
  s xe x 
2
2 E
dU

dV
dV  dU
 1 s x2 
dV
U   

2 E 
2
Example – beam in bending
My
s
I
U 
s x2
dV
2E
M 2 y2
U 
dV
2
2 EI
M2
 f ( x)
2
2 EI
I   y 2 dA
2
2
2
2
M y
M y
dV

2
2

2 EI
2 EI
M2
U 
dx
2 EI
U 
M2
 y dAdx
2
2 EI 2
Castigliano’s Theorem
[He was a Grad Student at the Time!!]

Deflection at any point along a beam subjected to n loads may
be expressed as the partial derivative of the strain energy of
the structure WRT the load at that point
U
i 
Fi




We can derive the strain energy equations as we did for
bending
Then we take the partial derivative to determine the deflection
equation
AND if we don’t have a force at the desired point:

Q, work out equations, then set Q = 0
Castigliano Example

Beam AB supports a uniformly
deflection at A.

No load acting specifically at point A!



Q
L
U
M  M
Substitute expressions for M, M/  Q
  
 A,  A 
and QA (=0)
Q A 0 EI  Q A
M ( x)  Q A x  12 wx 2
We directed QA downward and found δA
to be positive
M
Defection is in same direction as QA
 x
(downward)
Q A
A

dx


wL4
A 
8EI
1
A 
EI

L
0

wL4
 wx  x dx 
8EI
1
2
2
Stability

Up until now, 2 primary concerns

Strength of a structure

Material
failure


It’s ability to support a specified load without
experiencing excessive stress
Ability of a structure to support a specified
deformations
Now, look at STABILITY of the structure

It’s ability to support a load without
undergoing a sudden change in configuration
Buckling


Buckling is a mode of failure that does not depend
on stress or strength, but rather on structural
stiffness
Examples:
More buckling examples…
Buckling

The most common problem involving
buckling is the design of columns



Compression members
The analysis of an element in buckling
involves establishing a differential equation(s)
for beam deformation and finding the solution
to the ODE, then determining which solutions
are stable
Euler solved this problem for columns
Euler Column Formula

Pcrit

c 2 EI

L2
Pcrit 
 2 EI
L2e
Where C is as follows:
C = &frac14; ;Le=2L C = 2; Le=0.7071L
Fixed-free
Fixed-pinned
C = 1: Le=L
Rounded-rounded
Pinned-pinned
C = 4; Le=L/2
Fixed-fixed
Buckling

Geometry is crucial to correct analysis
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
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Euler – “long” columns
Johnson – “intermediate” length columns
Determine difference by slenderness ratio
The point is that a designer must be alert to
the possibility of buckling
A structure must not only be strong enough,
but must also be sufficiently rigid
Buckling Stress vs. Slenderness Ratio
Johnson Equation for Buckling
Solving buckling problems
2 2 E

r
Sy
Le

Find Euler-Johnson tangent point with

For Le/r &lt; tangent point (“intermediate”), use Johnson’s Equation:

For Le/r &gt; tangent point (“long”), use Euler’s equation: S 
cr


2
 Le 
Scr  S y  2  
4 E  r 
2
For Le/r &lt; 10 (“short”), Scr = Sy
Sy
 E
 Le 
 
r
2
If length is unknown, predict whether it is “long” or “intermediate”, use the
appropriate equation, then check using the Euler-Johnson tangent point once
you have a numerical solution for the critical strength
2
Special Buckling Cases

Buckling in very long Pipe
Pcrit
c 2 EI

L2
Note Pcrit is inversely related to length squared
A tiny load will cause buckling
L = 10 feet vs. L = 1000 feet:
Pcrit1000/Pcrit10 = 0.0001
•Buckling under hydrostatic Pressure
Pipe in Horizontal Pipe Buckling Diagram

Far End vs. Input Load with Buckling


Buckling Length: Fiberglass vs. Steel

Impact

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
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Impact – Chapter 4
Fatigue – Chapter 6
Examples?
3 categories

Rapidly moving loads of constant magnitude




Driving over a bridge
Explosion, combustion
Direct impact

Pile driver, jack hammer, auto crash
Increasing
Severity
Impact, cont.
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It is difficult to define the time rates of load application
Leads to use of empirically determined stress impact factors
If t is time constant of the system, where
m
k
We can define the load type by the time required to apply the
t  2


Static
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“Gray area”

Dynamic
t AL  3t
1
t  t AL  3t
2
1
t AL  t
2
Stress and deflection due to impact
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W – freely falling mass
k – structure with stiffness (usually large)
Assumptions

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Mass of structure is negligible
Deflections within the mass are negligible
Damping is negligible
Equations are only a GUIDE
h is height of freely falling mass before its release
 is the amount of deflection of the spring/structure
Impact Assumptions
Impact Energy
Balance
Energy balance


Fe is the equivalent static force
necessary to create an amount of
deflection equal to 
Energy Balance of falling weight, W
1
W h     Fe
2
W  k static  ks
Fe  k
 
Fe   W
 s 
Fe 

W s
12
W (h   ) 
W
2 s
12
h  
2 s

   s 1  1 

2h 

 s 

2h 


Fe  W 1  1 
 s 

Impact, cont.


Sometimes we know velocity at impact rather than
the height of the fall
An energy balance gives:
v 2  2 gh
2 

v

   s 1  1 

g s 


v 2 

Fe  W 1  1 

g s 

Pinger Pulse Setup
Pinger
Pressure Pulse in Small Diameter Tubing
1500 Foot Pulse Test
```