# Mechanical Design Applications

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15 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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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

Strain

Stiffness

Stability

Important in compressive members

Stress and strain relationships can be studied
with Mohr’s circle

Often the controlling factor for
functionality

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

Define strain energy density**

V

volume

Put in terms of
s
,
e

1
0
x
Fdx
U
dV
dU

dV
E
U
dU
dV
dV
dU
E
E
x
x
x
x
x
x
2
2
2
1
2
1
2
1
s

s
e
s

e
s
Example

beam in bending

dx
EI
M
U
dx
EI
dA
y
M
EI
y
M
dV
EI
y
M
U
dA
y
I

2
2
)
(
2
2
2
2
2
2
2
2
2
2
2
2
2
)
(
2
2
2
2
2
2
2
2
2
x
f
EI
M
dV
dV
EI
y
M
U
dV
E
U
I
My
x

s
s
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

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

i
i
F
U

Castigliano Example

Beam AB supports a uniformly
deflection at A.

No load acting specifically at point A!

Substitute expressions for M,

M/

Q
A
,
and Q
A

(=0)

We directed Q
A

downward and found

δ
A

to be positive

Defection is in same direction as Q
A

(downward)

Q

EI
wL
A
8
4

EI
wL
dx
x
wx
EI
x
Q
M
wx
x
Q
x
M
Q
U
L
A
A
A
A
A
8
1
)
(
dx
Q
M
EI
M
4
0
2
2
1
2
2
1
A
L
0

A

Stability

Up until now, 2 primary concerns

Strength of a structure

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

Material

failure

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

Where C is as follows:

2
2
L
EI
c
P
crit

C = ¼ ;Le=2L

Fixed
-
free

C = 2; Le=0.7071L

Fixed
-
pinned

C = 1: Le=L

Rounded
-
rounded

Pinned
-
pinned

C = 4; Le=L/2

Fixed
-
fixed

2
2
e
crit
L
EI
P

Buckling

Geometry is crucial to correct analysis

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

Find Euler
-
Johnson tangent point with

For L
e
/
r

< tangent point (“intermediate”), use Johnson’s Equation:

For L
e
/r

> tangent point (“long”), use Euler’s equation:

For L
e
/r

< 10 (“short”), S
cr

=

S
y

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
2

r

e
cr
L
E
S
y
e
S
E
L
2
2

r

2
2
2
4

r

e
y
y
cr
L
E
S
S
S
Special Buckling Cases

Buckling in very long Pipe

2
2
L
EI
c
P
crit

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

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.

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

We can define the load type by the time required to apply the
AL

= time required to apply the load)

Static

“Gray area”

Dynamic

k
m

t
2

t
3

AL
t
t
t
3
2
1

AL
t
t
2
1

AL
t
Stress and deflection due to impact

W

freely falling mass

k

structure with stiffness (usually large)

Assumptions

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

F
e

is the equivalent static force
necessary to create an amount of
deflection equal to

Energy Balance of falling weight, W

s
e
e
e
static
e
W
F
W
s
F
k
F
s
k
k
W
F
h
W

2
1

s
e
s
s
s
s
h
W
F
h
h
W
h
W

2
1
1
2
1
1
2
1
2
1
)
(
2
2
Impact, cont.

Sometimes we know velocity at impact rather than
the height of the fall

An energy balance gives:

s
e
s
s
g
v
W
F
g
v
gh
v

2
2
2
1
1
1
1
2
Pinger Pulse Setup

Pinger

Pressure Pulse in Small Diameter Tubing

1500 Foot Pulse Test