. _ r153 -. u :' mfifix‘qvwm

THE DESIGN OF MACHiva
FOUNDATIONS

Thesis for the Degree of M. S.
MICHIGAN STATE COLLEGE
David Owen Van Sfrien
1951

Thhlstoeertlfgthatthe

thesis entitled

THE DESIGN OF MACHINERY FOUNDATIONS

presented by

DAVID own VA! STRIKE

he: been accepted towards fulfillment
of the requirements for

_&§.-_degree mwGINmG

ghéna,

Major professor

Due April 25 , 1221

 

 

 

 

 

 

THE DESIGN OF MACHINiRY FOUNDATIONS

by

David Owen Van Strien

A THESIS

Submitted to the School of Graduate Studies of Michigan
State College of Agriculture and Applied Science
in partial fulfillment of the requirements

for the degree of
MASTER OF SCIENCE

Department of Civil Engineering
1951

THES‘S.

‘
'\

ACKNOWLEDGEMENT

The writer wishes to acknowledge his
indebtedness to rrofessor C. M. Cade
for his assistance in the preparation

of this report.

”355373

 

 

TABLE OF CONTENTS

I — Introduction

II - The design of foundations to distribute the
loads to the supporting medium. (considering
the loads to be static at the time when the
maximum dynamic loads occur)

A.

B.

Ce

Load resultant predominately down

1. Safe soil pressures

"U

m
a;

m

H

6
6

2. Resultant through the center of gravity ll

3. Resultant not thru the center of graVity 14

a. Foundation rectangular in cross-
section

b. Foundation not rectangular in cross-
section.

Horizontal component of resultant load
significant

Upward components of forces are significant

III - Determination of such forces acting on a
machine foundation as gear thrust and belt
pull

IV - Determination of forces dugbthe torque of a
motor which may act on a machine foundation.

V - Theory behind internal, unbalanced, mechanical
forces, originating in the machine itself, wnich
may act on a machine foundation

A.
B.
C.

In purely rotative machines
In reciprocating machines

Discussion of the theory of vibrations as
applied to machines

VI - Making allowances for internal mechanical forces
in the design of foundations for machinery

A.

In the design of an elastic framework to
to support a machine

l4

17
24

29

32

39

48

48

1. Design of beam supports to prevent
synchronism

2. Design of column supports to prevent
synchronism

B. In the design of foundations to prevent or
minimize the transmission of vibrations

1. Design of a massive foundation to act as
an inertia damper

2. Design of resilient mountings to support
a machine

VII - General Foundation Design Details
A. Materials (Concrete)
B. Reinforcing-steel
C. Anchor bolts
D. Other details

VIII - Various design considerations depending on the
type of machine ‘

IX - Detailed design 0f electrical machinery founda-
tions making use of some of the principles here-
tofore discussed

A. Foundation for a large electric motor where

foundation is acted upon by static and torque
loads

B. Concrete foundation for an electric motor in
which belt pull and static loads must be con-
sidered

C. Design considerations for the above motor
when it is to be mounted on an elastic
framework

X - Appendix
A. Bibliography

B. List of Manufacturers of vibration analyzing
instruments.

C. List of manufacturers of insulating materials

51
53
55

59
69
69
7O
7O
71

79

91

91

99

E‘I‘RODUCT ION

 

All too often the design of machine foundations is
left to chance instead of to science.

It is the Job of the structural engineer and architect
to design a building with.walls, floors, and roof capable of
supporting the loads that will be imposed upon it and also
will house the machinery which is required. The mechanical
engineer in turn very precisely designs these machines to
as efficiently as possible carry out their intended Job.

The plant lay-out man arranges the machines in the completed
building, taking great care that the flow of material is

as efficient as possible as it passes from machine to machine.
No one however, gives much scientific thought to the design
of the foundation which must support the machine. This is
often true because the person in authority does not realize
the extremely important part that an adequate foundation may
play in.the Operation of the entire plant.

It is astounding to note the lack of written information
on the subject of machine foundation design. Unfortunately,
it has too often been the practice to place a block of con-
crete in the ground, set a machine on it, and, because, as
luck would have it, the concrete was massive enough, the soil
was good, and nothing of a delicate character was around to
shake, the foundation was a success. When the installation
of another foundation, for a similar machine, was required,

“the original drawings were used again. This time, perhaps,

the soil was not so good and the result was excessive vi-
bration. A third foundation was required, so remembering
the unhappy experience of number two, the foundation size
was increased to be "safe". It seems evident that the
evolution of foundation design so often has followed this
process and that seldom has an evaluation of the items
envolved been made.

It might be observed that the manufacturers of the
machinery should be the ones to give advise as to the
foundation requirements for their machine. In many cases
they may give valuable aid, however, more often they will
evade the responsibility and give such advise as one maker
of engines gave: "make the foundation large enough to en-
compass the anchor bolts and deep enough to carry down to
supporting soil, and the mass of concrete will be suffi-
cient". You can see that this leaves much to be desired,
especially when it is considered that the engine is one
having unbalanced inertia forces. As often as not the
machine for which you are designing a foundation is an old
-one for which no design data exists.

It is to familiarize myself with the items involved
in the complete and successful design of a machine
foundation that I have undertaken this investigation.

Each different type of machine offers an individual
problem to the foundation designer. Some machines require
only provisions for the distribution of their static weight

upon the supporting medium. Other types of machinery

require provisions to handle forces which arise from the
motion of the moving parts or from gear thrust or belt pull.
In some types of machinery a provision for the damping or
isolation of vibration is of paramount importance. Thus, a
treatise of this kind cannot possibly cover adequately the
complete field of machinery foundation design. As a result
only a few of what I consider the most important features of
design have been included with some accompanying examples,
only two of which represent complete designs.

It is admitted that the dynamic forces in some machines
are so complicated that they cannot be determined mathemati-
cally. In these cases the foundations must be designed by
empirical methods based on past successes. These things are
pointed out in the following pages.

As a preliminary to the problem at hand it might be well
to list the various purposes for which a machine foundation
is required:

1. A foundation is required to maintain the
machinery fast in position and level with all
parts in true alignment.

2. A foundation is required to transmit the dead
weight of the machine to the ground or other
supporting medium in such a manner that the
safe bearing pressure of the ground or other
supporting medium is not exceeded, thereby,
preventing settlement or deformation of the

supports.

3. A foundation may be required to transmit the
live load of the machine to the ground or
supporting medium.

4. A foundation may be required to overcome dyna-
mically the effects of free inertia forces and
couples by balancing such kinetic reactions with-
in the foundation itself so as to cause them to
wholly or partially cancel.

5. A foundation may be required to prevent sychron-
ism with adjoining machines or structures, by
prOper location or distribution of loads, by
correct prOportioning of structural members, or
by isolation.

6. A foundation is required to absorb as far as
possible the residual vibrations set up in the
machinery so as to transmit as little as
possible to the Surrounding ground.

According to the type of machine and the nature of its
supporting medium, one or the other of the above requirements
will assume primary importance. Thus in the case of an
electric motor generating set, leveling and alignment are of
more importance than the absorption of vibration. While in
considering a suitable foundation for a diesel engine, the
problem would certainly involve a study of the dynamic forces
involved and a consideration of the vibrations if any.
Certainly a heavier foundation would result than for the

electric motor.

II. THE DISTRIBUTION OF LOADS TO THE SUPPORTING MEDIUM

Various requirements of a machine foundation have been
stated in the previous article. A discussion of these re-
quirements follows in detail, beginning with the problems
incurring in the distribution of the loads to the support-
ing soil.

It seems in order, before we discuss just how the loads
are transmitted to the ground, that some time be spent on

the capability of the soil to support the imposed loads.
A-l Safe soilgpressures

It is obvious that different soils and soil conditions
will have different safe bearing pressures, and that any em-
pirical rules regarding these have to be followed with cau-
tion. It is with this word of warning that Table I is of-
fered. This table gives the safe bearing powers of various
soils as advised by two different authorities. These values
have been established by experimentation and observation.
Note that Mr. CroftS' figures are similar to those given by
the New York City Building code. For machine foundations
having vibration it appears that he has cut his figures in
half. In regard to the table, Mr. Croft points out that by
"non vibrating"machinery he refers to practically no vibra-

tion, such as motor generators, sychronous condensers,

smoothly running motors and nicely running steam turbines.**

Most machines with excessive vibrations are set on
foundations which rest directly on the soil. It is common
knowledge that a portion of the soil will vibrate with the
foundation. The soil must, of course, support the machine
and foundation with a certain minimum deflection, but an-
other extremely important factor to consider is that the
vibrations occuring must not be transmitted thru the soil to
the detriment of other machines or structures.

As one can see, it is imperative that the designer know
something about the soil upon which the machine is to be
supported. Part of the answer is to run static soil tests on
the prOposed soil. Thus, bearing pressure values might be
obtained which multiplied by a safety factor, will at least
give us an allowable which if not exceeded will provide for
the first aforementioned requirement.

Something more, however, must be known about the soil
before we can be sure that it will not transmit vibrations
from or to the machine. A trial and error method is some—
times rather an expensive way to arrive at this information.

**Terazagi, an eminent soils engineer has
this to say about soil loading when they
‘are not static: "For vibrating machines
and machines with noticable dynamic load-
ings a factor of safety must be included
when using soil pressure tables, to take
care of the uncertainity of the reactions
of various soils to dynamic loads." He

does not, however, suggest what safety
factors to use.

It must be understood that in dealing with these things
we are treading on a more or less uneXplored ground (on vib-
ration in soils). Studies have been made, most of them in
Europe, and are now proceeding but much information must
still be left to chance.

It is well known that water-logged or wet soil is prone
to transmit vibrations readily. There are many instances
reported in engineering literature where excessive ground
water has increased the transmission of vibrations thru wet
sand and to a lesser extent thru well saturated clay. I
have noticed this fact to be true when using a surveying in-
strument along a railroad track. When a train passes, the
vibrations are quickly damped out by a dry well drained
ballast or sub-grade, but if the ground is swampy and badly
drained,the instrument will vibrate excessively, and settle
from its original position.

The fundamental principle which should be deduced from
such instances is that to help prevent the transmission of
vibrations thru the soils from or to the foundation the
ground water should be lowered below that of the foundation
walls and footings, by providing adequate drainage.

Of course, the above provision will not keep all soils
from vibrating excessively, under all frequences of vibra-
tions. In many instances, however, it will make a sufficient
difference so that additional provisions need not be

considered.

If it doesn't the next step must be either to change,
the Operating frequency of the machine or to isolate the

foundation from the soil in some way.

TABLE I

#/ft? by Terrel Croft "Machinery Foundations and Erection"

 

For Vibrat- For Non-

 

Kind of Soil ing Vibrating

Machines Machines
The hardest in thick layers

in natural bed 40,000 80,000
Rock - Equal to best ashlar masonary 25,000 50,000
Equal to best brick masonary 15,000 30,000
Equal to poor brick masonary 5,000 10,000
In thick beds, always dry 4,000 8,000
Clay - In thick beds moderately dry 2,000 4,000
Soft 1,000 2,000
Hardpan 8,000 16,000
Gravel 6,000 12,000
Dry, compact and well cemented 4,000 8,000
Sand - Clean and dry 2,000 4,000
Wet Sand 2,000 4,000
Sand and Clay mixed or in layers 2,000 4,000
Quicksand, alluvial soils 500 1,000

 

New York City Building Code - 1936

 

Kind of Soil #/rt. 2
Hard Rock 80,000
Medium Rock 30,000
Soft Rock 16,000
Hard dry clay 8,000
Firm clay 4,000
Wet clay 2,000
Hardpan 20,000
Gravel 12,000
Coarse Sand 8,000
Fine and Dry Sand 6,000
Wet Sand 4,000
Sand Clay Mix 4,000

 

‘-1o-

A-2. Foundations in which the axis of loads

passes through the center of gravity of
the area of the base.

The first type of machine foundation which we will
take up will be one of the simplest. The axis of loads
passes through the center of the area of the base and the
loads which must be considered are static, and the problem
is not complicated by belt pull or gear thrust as would be
the case with an electric motor which usually drives some-
thing. A good example of this type of foundation would be
a syncronous converter. Illustrated in figure 1.

Requirements of this type of foundation:

 

 

1. The size must be sufficient to

 

 

 

include the anchor bolts.

2. Area of the base must be great

 

 

 

 

 

 

 

 

 

 

enough to transmit the dead load

 

weight of the machine plus the weight

 

of foundation to the ground without
exceeding its safe bearing pressure.

3. The foundation must support the

 

‘9

 

K E

 

lei * ‘1 machine in its desired position. If
[ in an exposed position it must extend

 

 

below the frost line. Very often the
foundation can be made hollow to pro-
Figure l
vide space inside for appertenances
to the machine.
In a foundation of this type the design is usually one

of expediency, as the soil is usually able to support the

- 11 -

load involved. The problem sometimes becomes one of de-
creasing the bearing area.

Figure 2 shows a diagram of the loading of such a
'foundation, a study of which indicates that the cross sect-
ional area must be symetrioal with c. g. of the machine and

the unit pressure "p" must not exceed the soil pressure

A.

specified for static loads on the type of soil on which the

foundation is to be constructed.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d y In this design we make use of the
T simple formula:
= P g WEI + Wf 1
6 f K W ( )
1 Where "f" is the allowable soil
plan
le pressure, Wm is the weight of the
{"4 2 machine, Wf is the weight of the
s 5L foundatibn, b and d the plan di-
mensions of the foundation.
We can safely assume that the
Figure 2.

pressures are the same at any point
under the foundation.
Numerical example:~ (Illustrating above formula)

Design a concrete foundation for a syncronouscmnverter,
whose weight is 4000#, in an eXposed location, where the
frost depth is 4'. The base of the machine must be 28" above
the ground. The soil is a soft clay. Figure 3 shows the

location of the anchor bolts.

- 12 -

...___ 410'.___u

(7 2 qf——Anchor bolt holes

|Machine Base *
plate

' . i
my} 4”

 

 

 

 

 

 

 

 

Figure 3
Solution:

This machine would require a base of at least 54" x 42"
Usual practice being to allow at least a couple of inches
around the edge of the base plate. The concrete foundation
would then be 60" x 48" x 72".

Then applying formula (1) with:

Wf : 15O#£w' x 5' x 4' x 6' 2 18,000#

- 4000 8000 _ '

Wm: 4ooo#

 

 

 

K/
f1

 

 

 

 

 

Wf = lCOOO#

 

Figure 4
This pressure is probably well within the allowable
even for a soft clay. (For a vibrating machine it would be
close to the allowable.)
3332; the details of a foundation of this type, such as re-

inforcing - anchor bolts, etc. will be taken up later,
(Part 7) as they apply to nearly every type of foundation.

- 13 -

 

 

A-3(a) Foundations in which the loads do not
pass through.the center of foundation.

A good example of this type of foundation would be one
designed to carry the loads from an electric motor driving
4a machine by means of a belt.

In figure 5 T 2 load due to belt tension (strictly
speaking,the resultant of pulls from the tight and slack
sides), Wm = the weight of the motor, wf = the weight of the
foundation.

The resultant "R" of all the forces acting on the soil
is shown below. It is obvious that R does not act through
the center of the foundation and that the pressures on the
soil would not be constant but would vary in direct prOpor-
tion to their distance from the axis.

With this type of loading it is imperative that R does
not fall outside the base line and it is.usually suggested
by foundation experts that the resultant fall within the
middle One-third of the base. Of course the three require-

ments mentioned in the previous article hold for this one too.

 

fl
7‘
l

T

 

 

 

 

 

 

!
..__1

_J—_ib' AflDDLE}g’

 

 

Figure 5

-14-

After finding the resultant of the loads acting on the
soil and making the weight of the foundation sufficient to
keep R within the middle one-third of the base, the next step
is to calculate the resulting pressures on the ground.

If we resolve the resultant force, R into its vertical
and horizontal components at the point where it intersects
the base line, (figure 6) we see that the downward pressure
on the soil is Rv I a force due to a couple vae. The hori-
zontal component of R causes a horizontal load Rh which tends
to slide the foundation on the soil.

The downward unit pressure on the soil therefore could be
considered as a uniformly distributed pressure Rv/bc i the
varying pressure Mc/I.

p = RV/bct.Rvex/I (2)
flhggg: RV: vertical component of the resultant of loads, R
b,c : the dimensions of the foundation base.

e = the distance from the center of the base to the
point where the vector R intersects the base line

x = the distance from the center of the base to the
point where we are concerned with the pressure.

I : moment of inertia of the base about an axis

through the center of the base, 1/12 cb3 for
a rectangular foundation.

The maximum pressure will be at "n" and of a magnitude:

Pb): Rv/bc t 6Rve/cb2 (3)

It is obvious that the greatest unit pressure occuring,
should be below that which the supporting soil can safely
carry. There is, however, another important consideration.
The foundation is bound to settle a small amount, depending

on the pressure and the type of soil; practically speaking

- 15 -

PLAN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m ./7
09.
' U
r
m
\ n
i b \. 7
an,
9/;
2|
.9 :
B, I 1
' ,9
l 3:
1
m ( Pveb
—T fec
L
762‘U/
pressures

 

 

 

 

_T

SO/L 1025530958

Figure 6

-15-

the settlement will be prOportional to the pressures. It
will be seen, therefore, that if the settlement of every
portion of the foundation is to be about equal the pressures
must also be about equal. Very little trouble will be ex-
perienced if the maximum pressure is kept below twice the
minimum pressure. -

The problem Just illustrated assumes that the force T
acted centrally to the foundation block. Actually the belt
pull, T, acts ecoentrically since it is usual for the pulley
of the motor to be overhung. The foundation has therefore
to resist an additional overturning moment equal to the pro-
duct of the distance from the pully center to the center of
the armature, and the force T.

This problem is also simplified to the extent of our
assuming a rectangular foundation base, which in a great many
cases is not possible.

The next article deals with a foundation in which the
base is not a rectangle.

A-3 (b) Foundations in which the base of the

foundation is not rectangular

For the foundation problem in which the base is not
rectangular, we can use the same theory developed for a
short strut subjected to an eccentric load. The problem is
to determine the pressure at various points due to the sc-
centric load Rv. From the sketch (figure 7) it is seen that
xex and Y-Y are the centroidal axis. If it is assumed that

a plane before bending is a plane after bending, Hookes law

- 17 _

holds and pressures will vary over the cross section as the

ordinates of the plane (efgh).

9

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

*efj A (x)? 3:
. I d0 ‘ :
[— Q3” 3 4 f’A
- 1 - T 3 7,19
Z a
heal/on 0/19,, E 5
-———-4J W J
9 C 4
C 1 b 4
a
T e 1mm" , b ,
snaawasavx“X' (L
a b

Figure 7.

These pressures are represented in figure 7 as vectors
drawn to the plane abcd. The stress at A is : p = a 4by + cx (1)
depending on the values of the constants a,b, and c, which
in turn are dependent upon the loading and dimensions of the
cross section. The three equations of equilibrium will be
used to find these constants:

l. Sumation Mx :0
2. Sumation Hy,y = O
3. Sumation F, = 0
By condition 3, Rv :V/fpdA, where dA is the differential area

at any point A. substituting the value for p noted above (1).

RV : a/dA + b/di o c/ydA

since X-X and Y-Y are centroidal axis /di = O and o/ydA = 0

therefore:
RV Z a/dA CI‘ 8. '3 P/A

Using the second equilibrium condition 2: My : 0 and calling

the moment of the force Rv about the Y axis My.

My z/PX dA
My = a/di + b/xedA + c/‘xydA

The first term on the right hand side of the equation = O
and therefore: _

My = ny + ony (5)
where Iy .J/i2 dA - the moment of inertia about the Y axis.
and Ixy =//xydA = the product of inertia.

By the third condition of equilibrium MX = 0 and

Mx FJ/pydA

Mx = a ydA + b/xydA + S/y2dA

MX = bIXy + c1x (7)
Equations 5 and 7 can be solved simultaneously for b and c:

(Multiply 5 by Ix and 7 by Ixy then subtract 7 from 5)

Mny = nyIX 4 cIXyIx

MxIxy = bey2 L chny

Mny - MxIxy = b(Iny-Ixy2)

 

b = IVMIX " B’lexy
Iny - Ixfz

In a similar way:

x Y- KY

-19...

Placing the values found above for a,b, and c in equation (1):

 

.. f _ J'
= P M IX MXIXY IiXI Ii Ix (4)
S — 4- LI I I 2— X + . _ y
A x y ' xy x y Ky

Where: My 3 Pey , Mx = Pex , and ey and ex are the eccentrici-
ties measured from the x and y axis respectively.

4 The stress 8 is now completely defined at any point A on
the cross-section x (see figure 7) in terms of the dimensions
and the load.

Some convention for signs must be used when applying the
above equation-(4):. A tensile force F is positive, and the
values of x and y in the upper right hand quadrant are positive;
therefore, M? is positive if ey is negative.

In a particular problem, the problem of calculating the
stress s at some point A is as follows:

(1) Determine the centroidal axis of the cross-section.

(2) Determine the moments of inertia about the centroi-

. dal axis, Ix and Iy.

(3) Determine the product of inertia of the cross-

section'Ixy.

(4) Calculate my : Fey and Mg : Pex

(5) Determine the stress 3 at the selected point A,

using formula (4)

In special cases of loading and cross sections, the

formula just developed is somewhat simplified:

(1) Where base area has two axis of symmetry (rect-

angles, circles, I sections, etc.)

- 20 _

For these, the product of inertia Ixy'becomes

zero and formula (4) become":

3 g +M x.Y + Myx (5)
A Ix 1y

9

(2) Where the cross-section has two axis of symmetry
and the load is eccentric on the x-x axis:

In this case, equation (5) may be used, but

{x becomes zero because ex is zero and Mx =

Pex 3 P.0 : 0. Thus, equation (5) becomes:

-P Mx
3 K‘WJIL; . (6)

In most practical problems the designer is usually in-
terested only in the maximum and minimum soil pressures. Then
x becomes c, the distance from the centroidal axis to the
edge of the foundation.

An example follows where it is desired to find the maxi-
mum and minimum soil pressures under a certain foundation

with the resultant load on the x - x axis.
Numerical Example:

With the loadings as shown on (figure 8) find the maxi-
mum and minimum unit pressures which will be imposed on the

soil by the foundation.

_ 21 -

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n , , . .
r———3-——aT+-22——w+———&8———~-——-3———+~
EICD C)
Y. — iw‘ - - e4"- . Y
\n
- 52 e 58 l
”7 “G we
# ‘
30le 244000
lé?fif }[
T
x
Figure 8
Solution:

1. Find the weight of the foundation.
150# x 3(15+12+1o) = 16,6so#
2. Find the centroidal axis.

Use method of moments taking moments about "mn".

A = 37 sq.‘

37? 3 12 x 9.5 + 10 X 5.5 ‘ 15 X 1.5
377 = 114 + 55 + 22.5
§ 3 222:: : 5.2

37

3. Find Ix (The moment of inertia of a rectangle

about its base is : 1/3 bh3)

% x 5 x 5.23 = 234
-% x 4 x 5.83 : gég
+ 494
'__22_
+ 462 ft?
% x 3 x 2.23 = 10.65
% x 2 x 2.83 2 14,65
- 25030

4. Find ex (the distance from the center of area
to the point where the resultant load
Wt acts.)
I. ll

Take moments about mn .

. 70,650 x yl : 30,000 x 1.5 + 16,650 x 5.2 + 24,000 x 9.5

y : 222l§§9 = 5.08
70,650
ex = 5.2-5.08 = .12'

5. Mi : ex . Wt .12 x 70,650 = 8479’#

6. Determine the stress at the maximum and min-

imum points by using formula (4).

 

 

. y
min. A Ix
Sbex.= zgggig— * 84726; 5-2 = 1912 + 94: 2006#/ft.2

 

- 702550 _ 8478 X .8 - _ 2
Smin.’ 37 459 5 " 1912 ' 105 * AQQZ#/ft.

2-23-

B. When the horizontal component of the resultant

~load, acting on the foundation is significant;

In the previous articles, it was suggested that the
resultant of all the forces acting on the foundation should
fall within the middle third of the foundation base, and
that the vertical pressures on the soil should not be too
great or of too great variation. A consideration now arises
as to the horizontal component of the resultant load, if
there be one. Where a horizontal component does exist, the
only thing which keeps the foundation from slipping on the
soil or its supporting medium is their frictional resist-
ance, unless, of course, it is supported also horizontally
as by a floor slab.

The weight of the foundation must be sufficient to
prevent slipping. If the horizontal component of the load
acts through the c.g. the required weight is found quite
simply. The assumption is made that the load is distri-
buted uniformily over the base area.1 Therefore, the hori-
zontal force F is resisted by the product of the coefficient
of friction and the sum of the weights of the machine and

the foundation, which can be assumed to act at the c.g. or:
P 2"! (Wf + Wm) (7)

From which the required weight of foundation plus machine

becomes:

Wf 4 Wm=P//U (8)

-24-

It must be noted, however, that in many cases the hori-
zontal component does not act through the c.g. If this be
the case, an expression must be derived for the required
weight to resist this turning tendency and added to the above,
in order to find the necessary weight of the foundation.

Derivation of expression for Wf required to resist

 

turnigg.
Figure 9 shows a cross section of a foundation with a

horizontal load P acting at a distance 6 from the c. g. The
weight of the machine and foundation will be assumed to be

a, uniformily supported over the

 

entire base, therefore, the

e L] frictional resistance of each

 

€952— lP infinitesimal area dA will be

 

the same. If the coefficient
o/A/

of friction between the soil

 

 

 

and the concrete is f1 ,
Figure 9 then the resisting force is,
Where A is the area of base: /£(Wf + wm).%i
The resisting moment of the force acting on the increment

area dA is:
p[/1(wf+wm)9%] (A=areaofbase)

If the resisting moments of all these forces are summed up,
we have an expression for the resistance to pure turning)

which the friction between the foundation and soil offers:

- 25 -

Resisting M Z/ap (wf + wm) g4.
‘%‘(Wf "’ Wm) 210 dA.

Equating this to the moment causing the turning and sub-

stituting K for fjodA:
P6 '4— (Wf 4' Wm) K.
From this expression the required weight to prevent pure

turning becomes:

w w = 2.6.5... A 9
f... m fi'K ()

To this must be added the weight required to prevent slip-

ping in the direction of P (formula 8) and the total weight

required to resist Fe is:

-PA P (10)
Wf+Wm-/u-—eK-— vi

The value for "K" in the above formula must be found
from calculus. First the area is divided into 8 sections
as shown in figure 10. Then K is found for the "a" sections
and called Ka’ and for the “b" sections and called Kb .
The sum of Ka and Kb is the K in formula (10).

 

 

 

 

 

 

Figure 10

- 25 -

 

 

 

 

0’9
HgJ/G 1579 //b
¢Cbée
(a {.seefigfla) “(fa/”70%) =50 b/nge

0

=§lba/2‘36'§73+2'24 ’09/€i§,,7fl§—)]¢
=§b3 21 (330% *yl/Oy/,£§%%g) 50f J/n ¢=g ana/ c05¢ :‘Cé'

sage-Lsgewe/fieg) = 356/2!va we as

Kb (Seef/gzl/b)= ‘/ of; e/e =34-la3/ Zécoggj‘f/Oy {fig/34))
00{ amp =26»- and came-=54

a = saws-e e + semi-iii} =

[A ‘3’03/2/ 92% fill/09 (354’)

-27-

4 bac p3log c-ai §3log c$b J
K Ka + Kb n — [ + (c-a + abc +

= 3 2 2 2 c-b
- 2 b31°$ c+a a3 103 c+b ]
K - 3[2abc + -2- (3:5) + 5 (3:5) (11)

Where the foundation base is square or side I'equals side'w;

a equals b and K now becomes:

=2 3 a3105 a 2-1 )1
K 3[2s r5+22 (a( -1

K = §e3 (2.828 4 log 5.82) = e3(2828 - 1.761)

bun)

K 3.06a3 but a = g so K = fiagé W3 = .38?!3

8
Putting this into the expression for the required weight to
resist the turning moment we have:
Wf = 2; (l.+ Tjgfi) - Wm (12)

This of course is only true when the foundation cross-section
is square. For other relationships between the side dimens-
ions the general formula (11) must be used for K.
The coefficient of friction between the concrete and the soil:

There is a great deal to be said in regard to a coef-
ficient of friction between a concrete foundation and soils.
Many things enter into its determination for which there is
not time here to discuss. Certainly the type of soil and its
water content are of utmost importance. Table II gives
values for the coefficient ([1) which have been found by ex-
perimentation and represent probably the most reliable

general values obtainable.

- 28 -

TABLE II

Values of (/1) for Earth Foundations*

 

 

(,a ) with
Material safety safety
f“ factor factor
on gravel 0.5 2.5 0.20
Concrete
or on sand 0.4 2.5 0.16
Masonry

on clay 0.3 2.5 0.12

 

* From "Low Dams" Water Resources Committee - National
Resources Board.

C. When there exists an upward component of

force on the foundations.

It is possible with a gear or belt driven machine or
other type of similar mechanism, that an upward thrust may
exist which would cause the foundation weight to be so un-
evenly distributed, as to be reflected in settlement on one
side. The possibility may also occasionally arise where
the weight of the foundation block must be made sufficiently

large so as to keep the machine from being lifted.

- 29 _

0o

 

_‘
I7

 

 

 

 

 

A

m

 

 

 

 

e .
4.1 w;
._a;__.i

a; ,

L——z eJ.

Figure 12

 

 

 

 

 

 

 

 

An eXpression for the required weight of foundation to
keep the machine from lifting may be obtained by applying
the principle of equilibrium; namely, that the summation of
moments of all the forces acting on a body about any point
are zero.

Figure 12 shows a machine and foundation with an upward
component of force PV. The moments of the downward forces
Wf and Wm, about a, (the point about which the foundation
would pivot) must at least equal the moment of the force Pv
about the same point same point if equilibrium is going to
result.

fodf+VImxdm=vadp

and transposing, the required foundation weight:

f df

 

- 3o -

In regard to the first possibility mentioned above,
it has been cited in previous articles, that to insure
against the possibility of unequal settlement it is usually
assumed that the resultant of forces acting on the ground
should pass through the middle 1/3 of the foundation. If
this is to be provided the resultant of all the forces act-
ing on the soil "R" must be at leastgirom the point “a"
(figure 12).

As the moment of the resultant R about a, must be

equal to the sum of the moments of all of the other forces

about a,
Hg = Wm x dm 7 Wf x df - Pv x dp

the resultant
R = Wm + Wf - Pv
(Wm + Wf - Pv)% = wm.dm +1Wf x df - Pv.dp

and transposing the required foundation weight.

W - v Apr m m J

- 31 _

III - THE DETERMINATION OF FORCES SUCH AS GEAR THRUSP AND

BELT TENSION WHICH MAY ACT ON A MACHINE FOUNDATION

The easily derived formulas given below can be used in

the determination of gear thrust or belt pull on a

foundation:
F = 10 OO . P (gear drives) (14)
5 N.D
f
F = O 000 .P (belt drives) (15)
b N.Df

Wherein: F - tangential force on the gear teeth in pounds.

P = the power which is being transmitted in horse-
power

N = the speed of the gear or pulley in rpm.
Df = the diameter of the gear or pulley in feet.

63“
II

the pull of the machine due to the belt, in
pounds.

For derivation of these formulas see any machine design

hand-bOOK.

- 32 -

IV - LOADINGS DUE TO TORQUE OF A MOTOR WHICH MAY CHANGE
THE LOADING OF A FOUNDATION.

In the design of a foundation to support a rotating
motor, we have in addition to the static weight Of the machine,
the forces transmitted to the machine supports arising from
the torque of the motor itself.

An investigation of just what this torque is and how it
may be calculated from known data, follows.

When a force "P" drives a body through a distance "s"

against an equal and Opposite resistance "Q" work is per-

formed.
wkI'Ps = Qs

(In the case of an electric motor the force which drives the
armature is "P" and the equal and Opposite force Q will cause
a torque, which must be resisted by the machine supports.)
The work noted above will be in foot pounds if P is in
pounds and s is in feet. If this work is done in time t, the

average velocity will equal the displacement s over t.

vav. ' S/t
The rate Of doing work would be P9 or QV. In our case
the velocity is that of a pt on a circular path or:
. where: r is radius at
Vav. 8 Z 77’r-N which P acts and N is

revolutions per minute.

work done per minute , 2,, = Parr/“N

,0

One horse-power : 33,000'#/minute Hp = 3—3‘7‘3‘50 =
P
HP :3 ____1___ = P.2 rN a PrN and P 3‘ HP 2 2 (16)
33,000 33,000 5252.1 Nr

-33..

The above relationship will allow us to solve for the
force P as the horse-power, the revolutions per minute and
the value "r" are usually known or easily obtainable from
the manufacturer or by tests.

A-problem follows which will illustrate the use of
the above to find the additional loadings on the foundation

due to torque. (see problem 9A)

- 34 _

V - THE THEORY BEHIND INTERNAL UNBALANCED MECHANICAL FORCE§,
ORGINATINATING IN THE MACHINE ITSELFL WHICH MAY ACT ON
A MACHINE FOUNDATION.

In the preceding articles the loads acting on the found-
ation were assumed to be-gradually applied, or static like
the load on a building foundation. In many machine foundations
however, resistance to loads caused by moving bodies must be
provided. A study of these loads and the design of found-
ations to provide for them follows:

A study of the internal or unbalanced, mechanical forces
which may act on a machine shows that they may be due to

either rotating or reciprocating masses.

A - In purely rotative machines the internal forces,
(usually small) which do act, are due to a lack of perfect
balance in the rotating parts. If the center of gravity of
[the rotating mass is at the exact center of rotation there

can be no internal forces.

, F

Figure 13

Figure 13 shows the rotative parts of a machine in three
positions. It is rotating about the longitudinal axis of the

shaft and the center of gravity is atG. The machine is con-

-35...

tinually acted upon by a centrifugal force F caused by the
unbalanced mass. This force, varying in direction, must be
resisted by the bearings, and ultimately by the foundation.

If weight is added to the rotative part on the side
Opposite the point G, the centrifugal force would be decreased.
It is thus possible to balance, to a certain extent the rota-
tive parts of many machines. The extent of balance in various
parts however, is usually a question mark in the mind of the
foundation designer.

If there exists such an unbalance,the load which is
normally delivered to the foundation from the machine will be
increased by the amount of this centrifugal force. This
force in itself might not be too critical if it were not for
the fact that it occurs periodically thus giving rise to vi-
brations which may or may not be harmful. (vibrations and

their treatment will be discussed later.)

B - In reciprocating machinery there are usually forces
of the type occuring in rotating machine, as well asforces
due to the acceleration and deceleration of reciprocating
parts.

In order to study these forces, consider a single
cylinder horizontal engine of the corliss type. (see figure
14).

- 35 _

' Crank lo/n

    

p/S {on

   
 

00/790 by 1190’

  

.[Mq0. / confine ewgmoe

Figure 14

The moving parts of the engine possess mass and there-
fore inertia and when they are accelerated or retarded give
rise to inertia forces in accordance with Newton's funda-
mental law:

Force = mass x Acceleration

The summation of these fOrces at a given instant is
therefore the force transmitted to the foundation at that
particular instant.

These forces are in general approximately harmonic
and therefore periodic. The resultant must of necessity
also be periodic.

In order to express the relationships involved in a
mathematical manner let:

.W : wt of eccentric rotating parts.

W2 wt of reciprocating parts

H
II

length of connecting rod between centers of pins.

-37-

r : crank radius

we

up: angular velocity of crank;

In regard to the rotating parts:

rmxf is the normal acceleration for uniform Speed.

Then Fl (inertia force) : .glrQDF
and this force is directed radially outward from the center
of rotation. I

In regard to the reciprocating parts:

The maximum acceleration occurs at the ends of the
strokes where the direction of motion is reversed. For a
connecting rod of finite length, acceleration depends on a

ratiOr of connecting rod length to the crank radius.

In this case it is easily shown that the acceleration
at the out end of stroke is rcn? (lei) and at the in end
of stroke is rco2 (l+%)

The inertia force due to this acceleration or retard-

' W
ation is F2 =._§ r002(1£%)
8

Therefore, the maximum kinetic reaction exerted by the

engine would be: W 2 ,
- 'l “’2 2 +1
F1 + F2 .. S r to + __ rd.) (1 ‘q (17)

8
The magnitude of the internal forces then depends (1)
In purely rotative machines on the weight, speed and dis-

tance of the center of gravity of the rotating mass from

_ 38 _

the axis of rotation. (2) In reciprocating machines, on
speed of the machine, the weight of the rotating masses,

the weight of the connecting rod, the ratio of crank length
to connecting-rod length, the radius of the center of

gravity of the rotative weights, and on the degree of effect-
iveness of the counter balancing.

Now to incorporate provisions in the design of a found-
ation to take care of these internal forces is a rather
complicated and partially unsolvable problem. The advice
given by most authorities is to make allowance for them by
a sufficient factor of safety.

The designer can usually do a little better than this.
He can in many cases calculate these forces approximately
and make allowances for this approximation; with a sufficient

factor of safety.
C - THEORY BEHIND VIBRATIONS IN MACHINE FOUNDATIONS

A Vibration may be defined as a periodic motion which
changes direction twice during a complete cycle, and repeats
itself after a certain interval of time (its period). The
simplest kind of periodic motion is simple harmonic motion.

Simple harmonic motion may be defined as the motion of
a point in a straight line such that the acceleration of the
point is prOportional to the distance, x, of the point from
some fixed origin, 0, in the line, and is directed toward 0.

(It is a special case of rectilinear motion).

-39—

Examples of simple harmonic motion are:

l. The motion of a weight attached to the lower end
of a helical spring.

2. The motion of a cross—head of a steam engine close-
ly approximates a harmonic motion if the ratio of.
length of connecting-rod to that of the cranx is
large.

3. The motion of an oscillating pendulum if the arc
is small.

In fact many vibrational motions so common to engineer-
ing problems may be assumed without serious error to be
simple harmonic motions.

The harmonic motion of one vibrating mechanism may differ
considerably from the harmonic motion of another vibrating
mechanism. The differences are described by such terms as:

Amplitude: A: The magnitude or the vibratory motion
from the equilibrium point to the extreme position.

Frequency; The number of cycles of motion per unit time
(measured in cycles per second (f)) (or radians per second 90)).

Period P: The time it takes to complete a cycle
(measured in seconds).

For our purpose vibrations may be conSidered to con-
sist or two types: free damped vibrations and forced vibra-
tions.

Free damped vibrations are the same as those described
above except that they are affected by forces such as air

resistance, internal friction of the vibrating material,

-40..

’ )
)r
r.)

.iIIIx ll!‘il : ‘l
Ill"

friction between sliding surfaces, etc. These forces are
always present and act as sources of damping forces.

Forced vibrations are the result of a periodic dis-
turbing force acting on a vibrating body and exist wherever
there are moving parts in machines.

The frequency of vibration of the body subjected to a
continuously acting period force will be that of the dis-
turbing force. Therefore, f vibrations are to be minimized,
-the forced vibrations must be brought under control. This
and the problem of making sure that the condition of re-
sonance, (when the natural frequency of the foundation,
supporting soil, or supporting structure is nearly the same
as the frequency of the periodic force) does not exist, are
the two problems facing the foundation designer in regards
to vibrations.

From The Theory of Vibrations (Forced vibrations with

viscous damping.)

A = A- ..._ I (183.)
)(H 32-2)? (2c. 60¢)"

 

 

Where:
A = amplitude
13 Z deflection due to static load of the same magni-
tude as the unbalanced periodic force of engine. (W)
2 frequency of unbalanced periodic force of engine.

undamped natural frequency of elastic system.

o‘o8

coefficient of damping; C‘=l, is point of critical
damping where damping is so great as to prevent

vibration.

- 41 _

The right hand term above is called the magnification
factor. Figure 15 shows curves with various values of c.

It should be noted that they all have a sharp peak where
cu

’

.p—— /.

’0

'The values for the natural frequency of various sup-
porting structures, at least those of a relatively simple
nature, can be obtained rather easily as we will demonstrate
later.

The above is also approximately true for a machine on
elastic subsoil, and it can be seen from the curve that the
amplitude of the vibration is greatly increased if its
natural frequency is close to the forced frequency of the
unbalanced forces of the machine, and of necessity this
condition must be avoided.

The value of c will vary for different soils or sup-
porting media. If we assume 0 = 0.25 as an average soils,
we can see that if we avoid 622? values between 0.5 and 1.25
the amplitude will not be greatly increased. 2

From "Mechanical Vibrations" by DenHartog it is seen

that the undamped natural frequency or an elastic system:

{/0=/g)and fn :55 94 (l9)
fn = 3.14423 cycles per second (19a)
fn a led/ég~ cycles per minute. (19b)

-42-

55

W

A/A Ma mfioa‘l’lon Factor

P

1

9

 

 

 

 

 

 

 

 

 

a
‘ u
C=ral5i%\\
/

 

0d):

 

 

 

crazdFDVk

 

 

 

 

 

 

 

 

 

 

 

 

 

fiF-VJE 1-5 . E
Rm %

 

«oer-“mam 0F DAMDINq waves—

Figure 15

-43..

Note: This formula (19b) may be used for computing the
horizontal translational natural frequency as well, if for
the deflection an apparent deflection is substituted,

The torsional natural frequency is not unlike the

above.

- Kt
Where:

Kt : torsional spring rate

I moment of inertia of mounted equipment

(in slug feet sqd.)

Kt is figured by multiplying the translational spring
rate (load reqd. per foot of linear deflection) xthe
distance from the elastic center (in feet) squared for each
individual mounting and summing the products, giving the
torsional spring rate in (pounds feet/radians) for the
entire installation. The elastic center is a point about
which a couple would cause rotary movement of the mounted
equipment and is sometimes known as the center of elastic
resistance.

Having arrived at this natural frequency the trans-
missability formula given applies fully. (recognized that
figures on moment of inertia are not always available,
but such data can commonly be arrived at by observing tor-

sional resonance in a trial installation.)

Thus, it is seen that if a value forA is known we
can predict the undamped natural frequency from Ibrmula

(19b), and the magnification factor from figure 15. £5 ,

-44..

the static deflection for soils should be determined by
static load tests on the soil at the bottom of the found-
ation in the field.

.The undamped natural frequency for some typical soils
are given in Table III, which was taken from an article

by Larkin. @

_ 45 _

TABLE III

THE UNDAMPSD NATURAL Fasauawcx 0? some TYPICAL 501L%§

 

Undamped Natural

 

Material Loading #/ft2 Frequency - CPM.
Clay (wet) 2000 540
Clay (wet) 4000 380
Clay (dry hard) 2000 660
Clay (dry hard) 8000 330
Gravel 2000 1330
Gravel 8000 660
Sandstone 2000 4200
Sandstone 8000 2100

 

Causes of Vibration: The causes of vibration Spring
from "free inertia forces" due to the unbalance of the
mOVing components of a machine.. A not-perfectly balanced
motor shaft is a cause of simple vertical translation vib-
ration of the first order ( f-= l x shaft rpm ).

The eccentricity in bearings can cause vibration. A
motor driven by common 60 cycle A.C. current will deliver
a constantly changing motor torque, with peaks and valleys
causing a torsional oscillation or vibration of '73200 c.p.m.

See the discuSSion of unbalanced internal mecnanical

forces (page 35).

947--

VI - MAKING.ALLOWANCES FOR INTERNAL MECHANICAL FORCES
IN THE DESIGN OF FOUNDATIONS FOR MACHINERY

A - Designing of an Elastic Framework for the
Support of Machinery.

It is sometimes necessary in the design of machinery
foundations to support a machine on an elastic framework.
The dimensions of the component parts of this framework can-
not be determined alone from the strength required to support
the static weight of the machine.

The framework must be designed by taking into account
these static loads, the kinetic forces due to the moving
parts of the machine, and Just as important, the possibility
of vibrations.

’When any machine is supported on an elastic framework,
its motion causes vibrations. These will become excessive
when the speed of the machine becomes nearly equal to the
natural frequency of the parts of the supporting structure.

It is, therefore, necessary in the design of the frame-
work, to be certain that the pr0portions are such that the
natural frequencies are not near the operating speed of the
machine.

A—l. Design of Beam Supports., To explain the design
of beams, suppose a machine of weight W is located at the
center of a simple beam of Span L. (figure 16) If for a

first approximation,

- 4g -

A(be

5;"
[EL .
____4.

 

La

 

Figure 10

the weight or the beam is neglected, the deflection at

the center of the span is:

A ; wt3

48 31

the natural period P of free vibration of the beam:

(I9)
- TT‘A
P - 2.¢4§:~

2
where A is in inches and g = 386 "/sec
Inserting into this the expression for deflection, P

/ 3
: 2 WL
P 7T £8 EIg

If N represents the revolutions per minute of the

becomes:

machine, then as the frequency in cycles per second - the
revolutions per second: H
N360f‘ orN=-6—O-
- P
The speed at which the amplitude of vibration of the

beam will be a maximum or the critical speed is:

-49-

N .. 194821

critical ' 77 WL: 8 (21)

where E, I, and L are in inches, W is in pounds and N is
in revolutions per minute.

The critical value of N must be calculated by the above
formula and if the Operating speeds are anywhere close to
the resulting N critical, the design must be changed.

An approximate method for including the weight of the
beam in a formula for N critical follows:

The deflection now consists of two parts. The part due

to the weight of the machine W.

A1: WL?’

EI

and the part due to the uniformly distributed weight of the

4
_ wL
A 2 ‘ BBZFEI'"

the total deflection at the center of the Span is therefore:

3
A .__ A,+A2:WEI (Wi-SWL)

which is the same as for a single concentrated load of amount

beam wL,

(W +‘3WL). Assuming the natural frequency of vibration of
the beam is the same for a single concentrated load as when
the load is partly uniformly distributed the formula for

critical speed becomes:

48131 I
- O 8
Ncritical ‘ gfp/f;;, g wL) L: (213)

 

 

- 50 -

(The correct solution, given by S. Timoshenko shows that
5/8 should be replaced by 17/35)

In the design of a beam to avoid synchronism a work-
ing condition must be assumed, (say N critical = 10 N), and
the size of the beam determined accordingly. The procedure
could be to first, neglect the weight of the beam and solve
for 1. Second, using this I, calculate the dimensions and
weight of the beam. Then as a check, the critical speed
could be redetermined using (W + 17/35 wL) for the concen-

trated load at the center.

A-2. Design of Column Supports. To eXplain the pro-

 

cedure in the design of columns to prevent the synchronism
discussed above, consider a weight W supported on a column

of length L. (figure 17)

0’g‘x 2
IQ quw .—-1

 

 

T
l

 

7’7” W

Figure 17

For a first assumption neglect its weight and consider

the column as deflecting as a cantilever beam under the

“551‘

action of the components of kinetic forces of the machine
acting horizontally. If Fx represents the horizontal com-
ponents acting on one column, the deflection laterally of

its upper end will be:

A: F.L3
“‘3 13':

Because Fx is pr0portional to (3 we know the motion
is simple harmonic and the fundamental equation may be

used.

(K is the disturbing force at unit distance from upright
position). In this case K is the value of H for A = 1.

K=EEI

The period of harmonic motion is: (m being the mass of the

m
P = 2TE/gfl

Therefore:

body)

and the frequency is

=___1 _I_{_1 3155
f 15 277 m ' 277 w

then the critical Speed for the column becomes

Ncritical ’ %§ / %%%S = 325 '£13 (22)

WL

where E, I, and L are expressed in inch unit s and-W in

pounds.

_ 52 _

If the weight of the column is included in the calcu-
lations, an approximate formula is found as follows: The

deflection of a cantilever beam under uniform load is:

: WL
8E1

and the total deflection under both a concentrated load W

and a uniform load wl.will be:

3 4 3
£3_: WL + WL : L 1
3E 8E1 ‘3EI (W + 3/8 wL)
which is the same as a concentrated load of (W +-3/8 wL).
Assuming that the frequency of the column is the same for

a single concentrated load as when part of the load is un-

iformly distributed, we have for N critical approximately:

a

N : 599/ AEEIE» 22
critical 77 I?(W + 3/8 wL) ( a)

 

 

(By Timoshenkos' rigorous analysis the 3/8 wL becomes
33/140 wL)

In the design of a column the procedure for determin-
ing that it will not synchronize will be similar to that

described above for the beam.

B - Designing_Machine Foundations to Prevent or
Minimize the Transmission of Vibrations.
The prevention of vibration in the parts of a machine,
its structural members, or in the structure of the building
is important; in eliminating excessive wear, in reducing

repeated stresses that are likely to cause the failure of a

-53.—

member by fatigue, and in reducing objectionable noise.

One of the problems which therefore faces the machine
foundation designer is to find a way of reducing the vib-
rations which may be present. There are several possibil-
ities to consider in attacking this problem:

1. By Balancing. An attempt might be made to balance

 

the machine and thus remove the exciting force. This, how-
ever, is the job of the machine designer and usually has
been done to the best of his ability, although it must be
understood as mentioned on a previous page that it is not
always practicable or possible to completely balance a
machine. In any event this is a condition over which the
foundation designer has no direct control.

2. By Tuning . In certain instances excessive vib-

 

rations in a machine could be prevented by so changing the
design of the machine, that it will not Operate near its
critical or resonant Speed. An example of how this workfiin
one instance is given in Mr. DenHartogs' book - "Mechanical
Vibrations". Excessive vibrations were eliminated in this
case by replacing the existing 17 bucket runner of a hyd-
raulic turbine by a 16 bucket runner, thus changing the
interval of time between the impulses of two adjacent guide
vanes. As in number 1, the foundation designer seldom may
resort to expediencies of this sort.

3. By Damping. It is sometimes possible to control
vibrations by damping or by the introduction of frictional

forces in order to reduce their amplitudes. Damping has

-54..

little effect except in the neighborhood of the resonant
frequency. An example of damping is the use of shock
absorbers as friction dampers on an automobile, to limit
the resonant vibrations of the body of the car induced by
road irregularities. In a machine anchored rigidly to a
massive fixed foundation, the mass of the foundation, acts
'as an inertia damper to limit the amplitude of the Vibrations.
4. By Isolationp A method may be applied by whiCh the
vibration is so isolated that the periodic force reaction
on the soil or supporting medium is reduced. The usual
method of isolation is to use some form of elastic suspension

of the vibrating body.

We Shall limit this discussion to methods 3 and 4 above.

B-l. Massive Foundations as Inertia Dampers.

The use of heavy foundations is the Simplest and most
primitive means of providing resistance to the kinetic re-
actions ariSlng from the mOVing parts of a macnine. Obvious-
ly this type of deSign cannot be used in many cases, but where
they can be used they serve effeCtively to decrease the
amplitudes of the foundations forced vibrations. They are
never efficient, however, from the standeint of power losses
and strain on the machine and foundation, often cauSlng 7
failure of the metal and disinte gratlon of concrete founda-
tions, by fatigue failure. I

It must be noted that in this type of design the theory
Will not be to attain an amplitude of zero, because such an

amplitude would require an infinitely large mass, and unless

—5b-

a small amount or vibration is allowed strains and result-
ing damage are likely to occur.

It is unscientific as well as uneconomical to mount a
precisely designed machined machine on a foundation whose
design has been mostly guess work. Although this seems to
be the prevalent method of attack, a more scientific method
of approach to the problem is now given, illustrated by the
same Single cylinder engine discussed in article 5—B. hqzt.

The periodic inertia force representing the sumation of
all the forces caused by rotating and reciprocating parts,
is resisted jointly by the inertia of the masses to which it
is transmitted including the mass of the machine, the mass
of the foundation, and the mass of that part of the soil or
sub-foundation, which may be assumed to act as a unit with it.

Thus it is seen that the practice of anchoring a machine
to a massive foundation is based on the principle of using
the relatively small accelerations set up by a large mass,
namely foundation and underpining, to balance the large ac-
celerations of relatively small masses, namely the moving
parts of the machine.’ For this reason the soil on which a
foundation rests usually adds greatly to its effectiveness,
for the mass of soil is also accelerated Just as far as the
disturbance transmitted to it by the foundation extends.

Let K denOte the ratio of the soil accelerated to the
mass of the foundation prOper, and let(a) denOte the average
acceleration for the entire mass set in motion. Then the
condition for equilibrium againSt horizontal translation in

the present case is: (from for. 17) 1r59

- 55 _

Wm " “if * K”f .a - 3L1 w2 + 1'2. r002 (1 + l) (23)
8 8 g q

Since the motion of the foundation is periodic with the
same periOd as the engine speed, it is a suffiCiently close
approximation to assume that it is harmonic. If the ampli-
tude of this harmonic motion is 2b then a = be}; Substi-
tuting this value, canceling out the common term “5; and
solving for the required weight of the foundation, Wf we
have:

2 2

(Dz CU - CL) 1
w + W + KW ~b-—— ' W r - + W r .__ -‘ ,_i
( m f f) 8 l g 2 8 (1 + q

wf (b + Kb) = er 4» War (1 + é) - me

w
w a r w w 1 + l - m
1' b(l+K')'[1+ 2) J 11K

(24)

The horizontal reaction applied to the bedplate of the
engine is accompanied by a vertical overturning couple act-
ing on the foundation in the plane of the motion. In this
case however the effect of this couple is unimportant in com-
parison with the lateral motion due to the horizontal re-
action.

Example: In an engine of the Corliss type, having the
appr0ximate dimensions given below, the lateral motion must
be limited to 0.005“ in either direction from rest position.
Find the required foundation weight.

Solution: Substitute in formula 24 above, (assume K =10)

—b‘(..

Weight of rotating parts = 150#
Weight of reciprocating parts = 400#
Total weight of engine a 12 tons '
Speed = 120 rpm

length of connecting rod = 5'

length of stroke='20"

10" , 1/60 24000
Wf = .OOSW(1 + 10) [15W 4' 4UU# (l 'i‘ /10) "' l + 10

181.82 (150 + 466.67) - 2181.82

: 112,171 4 2182 = 109,989# say 55 tons.

By the method Just described we can determine the mass
of block required for a certain allowable amplitude. The
uncertainty involved is in the amount of subsoil that should
be included in the mass subJect to vibration. This is vari-
able due to the type of soil, the unit loading, whether the
foundation sets on or in the subsoil, and other factors.
Certainly the choice of'K would require some experience.

The theory does Show us that the amplitude of vibration
varies directly as the unbalanced inertia force, and in-
vercely as the mass subject to vibration. Therefore, it is
clear that any means of increasing the mass subJect to vib-
ration will reduce the amplitude of vibration, such as tamp-

ing, the use of piles, etc.

-58..

B-2. Isolation by Suspension of the Machine.

A more modern method of dealing with the kinetic forces
involved in moving machinery is to mount the machine on some
type of resilient mounting, or to mount the machine plus
some massive foundation on a resilient mounting.

It is now necessary to get a little background before
moving into the actual design procedure.

In studying this type of mounting, we shall confine
ourselves to dealing with forces which originate within the
machine itself. There are two types of problems with which
we are concerned.

First the reduction of transmitted vibrations in the
form of a wave motion, and second, the consideration of im-
pact shock, which may occur at such infrequent intervals that
it can hardly be classed as a wave motion.

A resiliently mounted piece of equipment has six degrees
of freedom, that is, motion of a greater or less amplitude
is possible in as many directions. Three of these directions
are of translational form in three separate planes; three
are of a rotational nature about three separate axis.

These two types of vibration can be exemplified by a
motor mounted on resilient mounts. The translational type
motion could take place in a plane thru the shaft of the
motor. The most obvious evidence of the rotational type of
vibration is about the shaft axis of the motor. The word
freedom used in this sense should be understood to be purely

relative and accordingly, motion might be brought about with

-59...

much less effort in one direction than in another. More-
over, the fact that a degree of freedom exists in a given
direction does not in any sense imply that forces or
moments actually prevail to cause movement of this charact-
er.

Naturally, the support will sustain a steady deflec-
tion due to the weight of the machine supported; then as
vibratory impulses are set up, oscillating motion occurs
on both sides of this staticly deflected position. The
effect of a vibratory force exerted from within is to ac-
celerate the machine in the instantaneous direction of the
force and it is resisted but little by the comparatively
soft resilient mountings. But, before the distance thru
which the machine moves becomes very great, the vibrating
force has changed direction; accordingly the only part of
the disturbing force transmitted to the frame is that
small amount involved in deflecting the resilient mount-
ings a distance corresponding to the motion of the machine.

Vibratory forces are transmitted directly from the
body generating the forces to the supporting foundation,
if the two are rigidly attached. However, if the same
vibratory force or forces could be made to appear in a
body completely free in space, the force would be resisted
solely by the inertia of the body and the body would vib-
rate only slightly. In practice this free condition is
approximated by placing the body on resilient mountings.

The mountings must be sufficiently flexible so that

_ 60 -

the vibratory forces are resisted by the inertia of the
body and thus reduce the oscillatory motion of the body.
Although a large portion of the vibration may be

absorbed by the mountings, small forces will act on the
supporting structure. This portion is a function of the
ratio between the disturbing frequency and the natural
frequency of the isolated system. The effectiveness of
the resilient mounting can be obtained from this relation-

‘

snip.

 

T = g l
Transmissibility factor (disturbing frequencx)2 _

natural frequency

- 1 (24)

Solution of the above will give the portion of the
vibrating force transmitted as compared to that if a
solid support was used. This percentage subtracted from
100 will give the efficiency of the mounting.

For effective isolation of vibration the ratio between
the disturbing frequency and the natural frequency must be
greater than ‘VFE- . Isolation becomes more efficient as
the ratio becomes greater. As the ratio of F/fn becomes
smaller thanM/E-, a magnification of forces will occur.

This condition, spoken of as resonance, would be in-
finitely worse than if no insolators were used, so it is
important to keep out of this dangerous range.

The natural frequency should be less than 1/2 pre-

ferably 1/3 of the forced frequency.

- 51 _

At ratio of 2:1 it is possible to obtain 66% eff.

At ratio of 4:1 it is possible to obtain 93% eff.

We are now prepared to pick an isolator and complete
the formula. The disturbing frequency (discussed) of the
equipment is usually the operating frequency of greatest
amplitude. This can be found from the number of revolu-
tions per minute or the number of impacts per minute.

The natural frequency of the isolator depends on its

deflection under static load, and this can be found by

= . /1 '
fcmp 168 2:;-

For example: a certain isolator defects 1/16"

formula (19b)

 

o = l _ 108 _ Ocm
. . f 188 75555 — .2s0 - 75 p

if the Operating frequency is 1500 cpm, the frequency
ratio would be 1500/750 cn12.

The transmissibility (formula 24) would be 0.33. There-
'fore, the vibration absorption would be 67%.

- 52 _

TABLE IV

RELATIONS BETWEEN DISTURBING FREQUENCY AND DEFLECTION

__

 

 

Disturbing Reqd. Deflection of Isolation Medium
frequency For 65% eff. For 93% eff.
cpm
4240 1/100 in. l/32 in.
1220 1/10 in. 3/8 in.
752 _ f 1/4 in. l in.
432 3/4 in. 3 in.

 

The next step is to select the prOper mounting. The
merits of each type should be Studied, and the best one for
your problem chosen.

Some of those in greatest use today are: Cork, Balsa
wood, Felt, Rubber, Timber, and Steel Springs.

Various considerations limit the choice. From a study
of Table IV, it is evident that low frequency disturbances
and high vibration require large deflections of the elastic
medium. Deflections of a" are difficult to obtain with
organic materials unless one goes to complicated construct-
ions. Therefore, we see that rubber, cork, felt and like
materials have a definitely limited region in which they can

work*

* (Rubber shear loaded mountings are good to get large
deflections.)

- 53 -

Steel springs particularly of the coil spring type also
.readily provide the large deflections necessary for low fre-
quency disturbances. They are easily adaptable to all pur-
poses because their elastic prOperties can be accurately pre-
determined and controlled thru a wide range of dimensions
and combinations. It should be kept in mind that it is
necessary to design a spring suspension in such a manner
that all six degrees of freedom are provided for.

Certain of tne materials mentioned above are not too
well suited for machinery foundation isolation. Rubber is
adversely affected by oil; it is expensive; it may take a
permanent set; is not permanently elastic and is likely to
become hardened due to exposure; it has therefore not found
much use on foundation isolation for permanently placed
machinery.

33;; may in time take a permanent set; it is not water-
proof; it is absorbent; it may be attacked by insects.

Untreated granulated cork may in time pack down and un-
less confined tends to flow.

Timber is not waterproof, when used alone in small
quantities; it does not have sufficient deformation to make
it useful under most services; it may be affected adversely
by oil. However, in the past, treated timber has been widely
used for cushions under forging hammer foundations.

It is apparent from the above that choosing an isolat-
ing material calls for a certain amount of experience and

wisdom.

A simple example in which a choice of materials has to
be made, follows: '

Example: A motor generator set Operates at 1750 rpm.

We have a choice between three mountings, that undergo static
deflections in all mounting points of 1/32 in., l/l6 in. and
3/32 in. respectively. Which is the best choice?

By the use of formulas (19b) and (24) it is apparent
that the first is critical, the second provides 65% isola-
tion and the third provides 85% isolation. Obviously the
third will be the best.

It is here necessary to point out a very important fact,
oftentimes forgotten in the design of isolations for a found-
ation, namely that the isolating layer may not be inserted
at any arbitrary place, for, as already mentioned the ampli-
tude of vibration depends on the ratio of the mass of the
moving parts to that of the foundation and fixed parts.

Since inserting a resilient layer diminishes the effective
mass of the foundation, these layers should, therefore, be
placed at such a depth that a machine will still be attached
to a sufficiently heavy foundation mass. Moreover, insert-
ing a resilient layer has the effect of raising the center

of gravity of the machine and therefore affects its stability,
which must also be taken into account in determining the

position of such a layer.
Impact Shock Isolation

In dealing with the reduction of transmitted vibrations

most discussions are concerned mostly with the general

_65..

principle of vibration isolation which deals with vibration
in the form of wave motion, and do not consider separately
the effect of an impact shock. A heavy impact blow may occur
only at distant or infrequent intervals, and in itself may

be damaging to the mechanical equipment and foundations,
although its frequency is so low or irregular, that it cannot
be treated as a wave motion.

In shock protection, the fundamental principle is to
increase the period over which the impact forces are applied.
Rubber in compression is recommended where shock protection
is paramount, because of the continuously smooth deflection
curve which presents no opportunity for shock to be re-created.

The Fabreek Products Company of Boston, manufacturers
of a pad made of layers of rubber and impregnated cotton duck,
that have been vulcanized together, offers the following
Table V, and I quote their example of the design for pads

under a typical shock mounting.

__ 55 -

Stiffness Coefficients

TABLE‘V

Fabreek Products Company

 

#per sqLAinch, per inch

of deflection

 

Loading in us; ‘50 100 150 200 400

14 ply 1/4" thick 45,000 50,000 54,000 58,000 71,000
1? ply 9/32" thick 36,000 40,000 45,000 46,000 57,000
21 ply 11/32" thick 28,000 52,000 o4,000 h7,000 44,000
31 ply 1/2" thick 18,000 20,000 22,000 25,000 29,000
39 ply 5/8" thick 14,000 16,000 18,000 20,000 2o,000
Double 39 1-1/4" 7,000 8,000 9,000 10,000 11,000

 

From the Fabreeka Company Bulletin:
"The force of an impact is measured by the work done;
that is if a 2000# weight drops 2‘ the work done is
2000 x 2 = 4000'#. To find the force this blow ex-
erts it is necessary to know the distance in which it
is brought to rest. Assuming l/l6" (or .0052') then'

4000/.0052 : 770,000# is the dynamic load.

If the 2000# wt. dropping 2" is a board hammer and the
(l/lSJCLuset by tie hammer,is Lie reflection
deflectionAof the work being done, then the dynamic

load on the base of the anvil is 7 70,000#. If area

is 2000 sq." the unit loading would be 400 psi. If
two Fabreeka pads each 5/8" thick are placed under this
base the deflection of the pads is .060 or .050 feet
which added to

.0052 ' .0102'. The dynamic load is

now = 392,000#, decreasing it 49%

it...

000
.0102

- 57 -

The above assumes a rigid foundation. 0n soft
marshy soil the necessity of pads would be re-
duced, because the ground would be absorbing some

of impact.

Also pads increase efficiency due to damping effect."

I
O\
U)

I

VII - GENERAL FOUNDATION DESIGN DETAILS

After the dimensions of a foundation base have been
determined by some means or other depending upon the loadings
and allowable pressures on the soil, or supporting medium,
there still remain some questions which must be cleared up
before the foundation design may be termed complete. Such
details as foundation material, reinforcing steel, anchor
bolts and plates, grouting, etc. must be considered. In a
rather brief treatiSe of this kind none of these t0pics can
be covered in more than a fragmentary fashion, but they are

important and must be mentioned if but briefly.

A - Materials

 

Machinery foundations, at least those where some mass
is required, are made almost universally of concrete. It
is important to specify a concrete with a low volume change
and high unit strength. It is also important to get a
good strong uniform concrete; as the constant vibration of
the attached machine will subject the concrete to a rather
severe life.

In regard to the shape of the foundation, it will often
be found that a simple forming Job will be more economical
than a complicated forming Jab which may require a lot less
concrete. To provide additional bearing area than is avail-
able when the block is made about the size of the base
plate, it will be found most economical to "step" the found-

ation, as in any wall or column foundation.

-60..

B - Reinforcing Bars

With the exception of a cantilever footing slab or on
foundations which are hollow, there are no calculated
stresses which must be carried by reinforcing steel, but it
is my belief that its use in nominal quantities is necessary
to prevent cracking. Cracks which would cause no concern
in ordinary concrete construction are often serious in
machinery foundations, especially where there exist vibra-
tions, which may cause progressive cracking. Iarkinfizdvises
using deformed intermediate grade steel reinforcing bars a
or 5/8 inches in diameter spaced on 12" centers, both hori-
zontally and vertically near all faces of the foundation

block. Other authors try to prove the steel is not necessary,

but I think that the general practice is to put in steel bars.
C - Anchor Bolts

Anchor bolts are usually made up locally and therefore
do not receive the constant improvement and development
which a manufactured product does. The calculation of
stresses and the design of bolts is a relatively simple pro-
cedure as they are usually only required to take a tension
load. A minimum diameter bolt would be 5/8" as a smaller
one could be twisted in tightening the nut.' The size of
bolt required for a certain machine will usually be speci-
fied by the manufacturer. A few important things about

anchor bolt installation will now be mentioned.

- 70'-

For heavy machines the anchor bolts are usually made
removable so that it is not necessary to lower the machine
over the bolts. Of course, this necessitates some kind of
pockets. For most light machines the bolts are not made
removable, but are installed in such a way as to allow for
a certain amount of springing to align with the holes in
the machine bases, because a carpenter cannot set bolts to
the tolerances usually held on drilled holes or castings.

A good design has some provision such as a pipe at the
t0p with the bolt inside to provide for a small amount of
springing. There should be some means of holding the pipe
as well as the bolt in place while the concrete is being
placed.

The only other recommendation is that the bolts should
be relatively long so as to offer a certain amount of re-

siliency against breaking.

D - Grouting

,For obvious reasons the foundation is not built quite
as high as the elevation to which the base of the machine
must set. The difference, (3/4" to 1%") is made up by
leveling the machine and pouring or forcing a 1:1 mixture
of portland cement grout between the foundation and the
machine base. The grout is required to take the horizontal
cloads into the foundation, as well as to level up the
machine. Other materials are sometimes used for grouting,

but portland cement is by far the most widely used.

 

E - Design of Concrete Mixtures

Obncrete is a mixture of cement, fine aggregate,
coarse aggregate and water. It is a generally accepted fact
that the strength depends entirely upon the ratio of water
to cement, provided (1) that the cement and aggregates used
are suitable, (2) that the amounts of aggregates are correct
to give a workable mix, (3) that prOper conditions are main-
tained during the hardening period.

Pr0portioning by weight is admittedly the most accurate
because of bulking of the aggregate, however, proportioning
by volume is by far the most popular method due to the fact
that expensive weighing equipment is not required, and it
can be done with reasonable accuracy if allowance is made for
bulking. In view of this fact, in the following discussion
when pr0portions are mentioned they will be given by volumkg
so that a 1:2:4 concrete implies: one cubic foot of cement
to 2 of fine, and 4 of coarse aggregate.

The first and most important ingredient of concrete is
portland cement. All of the manufacturers of cement in this
country turn out a good product so all that need be said is
that a cement be used which is not more than a year old and
has been kept dry. In certain instances it may be wise to
build the foundation using High-early strength portland
cement which would allow its early use. At 72 hours it
would have the strength ordinarily obtainable at 28 days.
The water must be clean and free from oil, and alkali,

organic matter or other deleterious substances.

- 72 -

The fine aggregate, usually sand, should be well grad-
ed and free from organic matter, clay and silt. (See ASTM
specification.)

The coarse aggregate, washed gravel or crushed stone,
should be clean, hard, and durable; free from alkali organic
or other deleterious matter. Ocarse aggregate includes all
particles larger than 1/4 inch. It is suggested that for
foundations the maximum size be one inch. As with fine ag-
gregate it is important that they be well graded. (See ASTM
Specification.)

Concrete Mixtures. In general the design of a concrete
mix«consists in choosing a water-cement ratio, and then
selecting the pr0portions of fine and coarse aggregate so
that the combination will be workable. The choice of a
water-cement ratio in general requires the consideration of
strength, exposure, and class of structure.

The selection of a pr0per water-cement ratio for any
desired strength under average Job conditions may safely be
made from the specification for strength of the American
Concrete Institute, which follows the straight line empiri-
cal formula S0 = 1000(7 - % R). Where Sc is the compressive
strength of the concrete in 28 days (in pounds per square
inch), and R is the water-cement ratio (U. S. gal. water
per sack of cement (94#)). Thus, an ultimate compressive
strength of 3000 psi would require a water-cement ratio at 6.
‘ To select the necessary water-cement ratio, however, is
more a matter of Judgement, as the actual compressive

strength required in a machine foundation is probably a

_73..

rather small figure. It is noted, however, that as the
compressive strength of the concrete increases so does its
density and its ability to withstand vibrations without dis-
integration. As a result if one chooses a strength require-
ment somewhere between 2500 and 3000 psi, (the lower for
machines relatively free from vibration or impact forces

and the higher for machines with excessive vibrations or
impact), the result should be satisfactory.

A workable concrete mix is one which can be worked
into all the corners of the forms without excessive pud-
Idling, without segregation of the ingredients, and without
water collecting on the t0p surface. Workability is in-
fluenced by the amount of each ingredient and by the nature
of the aggregates.

Some measure of the workability of a concrete mixture
is obtained by making a slump test. (See ASTM specification).
For most concrete machine foundations a slump of from 3 to 5
inches is satisfactory.

Having selected a water-cement ratio and knowing the
approximate slump desired, the next step is to determine
the prOper amounts and pr0portions of the aggregates. This
can best be done by making several trial mixes, recording
the proportions and choosing the most workable. A balance
must be maintained between the coarse aggregate which.will
up to a certain point reduce the cement factor and beyond
that will produce an under-sanded harsh concretefi difficult
to placeJ and the fine aggregate which produces a smoother

mix but which used in excess, makes the concrete~Ux>expensive.

For these small trial mixes l/lO of a bag of cement
(9.4#) may be conveniently used and the aggregates sur-
face dried so that corrections for moisture will not be
required.

All aggregates under Job conditions contain more or
less moisture which must be calculated in with the mixing
water. The amount of moisture may be found by drying a
sample of the aggregate to a constant weight. If w 2 weight
of the damp sample, w' = weight after drying and p = the

percentage of total moisture:

: 00 _E__:_;E'
P 1 w

The moisture which becomes a part of the mixing water
is this percentage p (above) minus the percentage which
will be absorbed by the aggregates, about 1% for average
sand pebbles and crushed limestone, 0.5% for traprock and
granite.

Moisture in sand and even in coarse aggregate has the
effect of bulking the aggregate, therefore calculations
must be made to determine the "bulking factor" or relation
between the damp volume and the dry volume of‘a given
quantity of dry aggregate. Essential computations are given

in the example below:

-75..

 

Item ‘ Sand Coarse

 

ggregate
A. Weight of damp samples, oz. 35.0 32.5
B. Weight of ovendry sample, oz. 33.0 31.7
'0. Weight of water in damp sample,oz.
(A - B) 2.0 0.8
D. Per cent of total moisture in
terms of dry aggregate (C/B)X100) 6.0 2.5
E. Per cent of absorption (assumed) 1.0 1.0
F. Per cent of surface moisture
(D -— E) 5.0 1.5
G. Weight per cu. ft. damp, loose, lb. 97.2 94.4
H. Weight of surface-dry aggregate in
1 cu. ft. of damp loose material,
lb. (0- x(100/100+F‘) 92.6 93.0
1. Weight of water in 1 cu. ft. of
damp loose material, lb.(G—H) 4.6 1.4
J, Weight per cu. ft. of surface-dry
compact aggregate (by~test) 112.0 99.0
K. Bulking factor (J/H) 1.21 1.06

 

The correction for bulking is made by-adding proportion-
ately larger amounts of the bulked aggregate to secure any
desired actual volume of dry, compact aggregate. Thus, if
a dry, compact mix of 1:2&:3% is to be used with the above
aggregates, the pr0portion of sand, based on loose volume,
will be 2% x 1.21 : 2.7, and of coarse aggregate
3% x 1.06 = 3.7, giving a field mix of l:2.7:3.7.

- 75 _

If the water-cement ratio for the above mix is to be
7 gal. per bag, the amount of water to be added at the
mixer is determined as follows: Since 1 gal. of water weighs
8-1/3 1b., the amount of surface moisture in 1 cu. ft. of
damp, loose sand is 4.6/8.33 = 0.55 gal., and in the coarse
aggregate 1.4/8.33 = 0.17 gal. In each one-bag batch the
amount of free or surface water in the sand is 0.55 X 2.7 =
1.48 gal., and in the coarse aggregate 0.17 X 3.7 = 0.63
gal. The amount of water to be added to each 1-bag batch
is, therefore, 7 - 1.48 - 0.63 = 4.89 gal., or approximately
5 gal.

When the proportions have been decided upon the deter-
mination of yield and quantities required is the next task.

The yield is the sum of the absolute volumes of the
cement, sand, and coarse aggregate, and the volume of water.
For the mix computed in the above example, the yield for a
one-bag mix would be computed as follows:
There is 0.487 cu. ft. of sclids in 1 cu. ft. of cement.
The specific gravity of the common aggregates usually aver-
ages 2.65. One cu. foot of dry sand contains ll2.0/2.65
x 62.4 = 0.68 cu. ft. of solids, and one cu. ft. of dry
coarse aggregate contains 99.0/2.65 x 62.5 3 0.60 cu. ft.
of solids. For a 1-bag batch, which contains 7 gal.
(or 7x0.134 = 0.938 cu. ft.) of water, including the surface
water in the aggregates, the yield will be 0.487+2%x0.68+

392x0.60+o.938 =- 5.06 cu.‘

- 77 -

If 200 cu. feet of concrete were required for the job

the quantities required would be determined as follows:

No. of 1-bag batches required.= (200x27/5.06) = 1068

Cement

Sand = (1068 x 2.7) / 27
(1068 x 3.7) / 27
(1068 x 5)

Coarse aggregate

Water - at mixer

To provide prOper curing conditions the

1068 bags

108 cu. yds.

148 cu. yds.

5340 gal.

concrete should

be protected against premature drying out for at least one

week.

- 78 -

VIII - VARIOUS DESIGN CONSIDERATIONS DEPENDING ON THE
TYPE OF MACHINE. '

In a study of this kind, one can well generalize up to

a certain point,

but then he finds that each machine is an

individual problem. With this in mind the following rather

incomplete remarks on the specific requirements of individ-

ual machines have been included, together with the references

for a more complete discussion.

A - Machine Tools (3K)

1. The principle requirement is rigidity, in order

to maintain accuracy of Operation. (requires

massive foundation)

(3.)

B - Hammers

Planers : Planer- tables are very sensi—
tive and may easily be warped enough to
make close machining impossible. It is,
therefore, important that the foundation
be rigid and that the planer“ bed bear
uniformly upon the foundation. No anchor
bolts should be used. 0n large foundat-
ions, leveling blocks are usually used.
longitudinal steel is a necessity both in
tOp and bottom because of the length of

the foundation.

@NZDQE

1. Besides the weight of the foundation and the

hammer, the forces caused by the drOpping of

the hammer (all of these vertically) must be

- 79 -

distributed to the soil.

Excessive noise_and shock likely to endanger
the building, and produce nervous fatigue,
must be guarded against. Usually by insula-
tion. See article B-2

Most hammers being used today are of what is
known as a high ratio construction. That is
the weight of the anvil‘to the weight of the
hammer is high. (15:1 or greater). These re-
quire only a light simple foundation, with
usually some isolation material to minimize

the transmission of jar and noise.

C - Electrical Machinery 67C)

1.

The elimination of vibration is an important
aspect in the design of any high speed.machine.
(see article B-2)

It should be noted that in many cases the el-
ectrical machine is only a small part of com-
plex machine. The nature of the other part

may greatly alter the foundation required.

Some provision must often be made in the found-
ation for ventilation, particularly in large
capacity machines.

Table VI gives some typical figures on success-
ful foundations for rotary electrical machinery
foundations (taken from "Factory Installation

Work" - A. J. Coker.)

Q

-80...

TABLE'VI

Machine Description Speed Ratio of Total

r.p.m. Foundation Pressure
Wt. to on ground
Machine wt. Tons/sg.

 

 

 

75 KW belt driven generator 850 2.1 0.34
100 KW belt driven generator 625 3.6. 0.51
20 hp belt-drive motor 775 5.1 0.32
45 hp belt—drive motor 950 4.7 0.37
500 KW Synchronous convertor 500 0.51 0.29
900 KW Synchronous convertor 250 0.67 0.35
D - Textile looms
l. The reduction of shock and noise is of para-
mount importance. Often a whole bank of looms
is insulated together.
E - Steam and Other Reciprocating Engines

 

1.

Besides the static loads, it is necessary to
design the foundation to take care of periodic
inertia forces occasioned by the rotating and
reciprocating parts of the engine. (see article
5 B). The eccentric rotating parts would in-
clude the crank pin, crank cheeks, and the
crank end of the connecting rod. The recipro-
cating parts include the piston head, wrist pin,

piston rod, crosshead, and the reciprocating end

- 81 -

of the connecting rod.

The general requirements for the foundations

for a reciprocating engine are somewhat more

exacting than those for rotative machines.

Most engine foundations are designed using one

of the three following empirical methods:

(a) The weight of the foundation is based on

' the weight of the machine. (Its weight

should be 3 to 5 times the weight of the
machine.)

(b) The foundation depth is based on cylinder
diameter (the other dimensions determined
by the size of the bed plate). This method

is used only for simple Corliss engines.

A formula given by Wm. E. Ninds where:

Lnf

depth of foundation in feet

K = constant depending on steam
pressures (see Table VII)
di 2 diameter of cylinder in inches.

- 82 _

TABLE VII

 

 

Steam Pressure K
100 0.895
110 0.938
120 0.98
130 1.02
140 1.06
150 1.10
160 1.13
170 1.17
180 1.20
190 1.24
200 1.27

 

(o)

The foundation weight is based on both
the weight and speed of the supporting
machine.

A formula is given by E. W. Roberts:

Wf= KNExN

where:
Wf : weight of foundation in pounds

WE 3 weight of the engine in pounds
N 2 Speed of engine in rev. per min.

K = a constant as given in Table VIII

- g3 -

TABLE VIII

 

 

Type of Machine K
4-cylinder vertical gas engine 0.130
3-cylinder vertical gas engine 0.150
2—cylinder vertical gas engine 0.175
Single-crank double-acting tandem 0.320
Double-crank double—acting tandem 0.190
Single cylinder, horizontal semi diesel 0.300
2 - cylinder horizontal semi diesel 0.240
3 - cylinder horizontal semi diesel 0.230
4 - cylinder horizontal semi diesel 0.225
2 cycle horizontal semi diesel . 0.230
4 cylinder vertical diesel engine 0.177

 

2. Steam engine foundations can be designed with
safety using one of the above rules. It is
usually suggested that they be made mono-
lithically.

3. Compressgg foundations can be designed accord-
ing to rule (a) above. ‘

4. Reciprocating Pump foundation design depends
entirely on the type of pump. Generally speak-
ing much less foundation is required for pumps

than for steam engines occupying the same space.

-84-

F...

Direct acting duplex pumps require the small-
est foundation as the plunger motion is almost
balanced within the machine: Single-cylinder
pumps should be supplied with deeper founda-
tions their depth may be computed from formula
given in (b) above. Crank - and - fly wheel
pumps require foundations fully as heavy as a
steam engine of comparable size, because of
the lack of balance in their reciprocating
parts.

Motor Generator Sets

 

A small motor generator set, with a bed plate,
often requires no more foundation than the ordin-
ary floor.

Electric Motors Driving Fans for Ventilation.

 

Because a ventilating fan must of necessity be
located directly in a ventilating duct where noise
and vibration may be easily carried throughout the
building, the set must be carefully isolated to
prevent the transmission of said vibration.

Steam Turbine Units W®®

Steam turbine units are purely rotative, self con-
tained machines, and are not normally subject to
unbalanced forces. In the design of a foundation
for a steam turbine, however, one should antici-
pate certain abnormal Operating conditions such as
short circuits, bad synchronizing, broken blades,

water or mud slugs, etc.

-RR-

It is usually wise to make the foundation struct-
urally independent of its building to prevent vib-
ration transmission.

An excellent answer to the problem of design of .
steam turbine foundations is an idea originated by
N. W. Akimoff of the General Machinery Foundations
Company, Philadelphia.

The result of this type of mounting is to free the
platform from its support allowing the machine and
platform to vibrate as a unit with small amplitude
about an axis thru a fixed point of support, with—
out straining either the machine or its support as
would be the case with a rigid anchorage. An equal-
ly important result is that the possibility of
forced vibrations being set up in adjacent struct-
ures by means of synchronism is likewise prevented.
Explanation: The usual practice in designing,
machinery foundations consists in anchoring the
machine rigidly to the foundation which is as massive
as can conveniently be obtained. The foundation
design suggested by Akimhoff is based on the prin-
ciple that it is desirable to have a certain flex-
ibility in the foundation rather than extreme rigid-
ity. In this design the machine itself and the cap
slab which supports it are permitted a limited amp-
litude of vibration within definitely assigned
limits. These limits may be made as small as desir-

ed and without in any way decreasing the strength of

-86-

the structure under static loading. The machine
frame and the foundation may be lightened by in-
creasing the allowable working stresses to the
values ordinarily used for dead loads. The arrange:
ment consists in mounting the turbo—unit on a
structural steel platform which resembles a deck-
plate girder. The turbo generator is bolted down
rigidly to the platform, but the platform itself is
supported at only three points, on the super-
structure. Of these three points of s pport one is
rigid and two are resilient. The rigid support is
placed at the turbine end. It consists of a rocker
plate which is so designed as to allow the plat-
form freedom to pivot slightly about this point.
However, the rigid support-effectively anchors the
platform against any tendency to slide transversely
or longitudinally.

The two resilient supports consist of vertical and
transverse Springs of large capacity, designed so
that there is no possibility of their motion being
violent.

The design includes vertical and transverse stops so
in case of a short circuit or other accident the
platform will come to a solid bearing.

With this three point design the level of the plat-
form may be maintained by shimming up one or at

most two of the supports.

-R7-

Mr. Rathbone in an article "Turbine Foundations"
in the A.S.M.E. transactions has this to say,
among other things, about turbine foundations:
"Much may be gained as an aide to future
(designs by vibration studies on existing
structures, now made possible thru the
use of improved seismic instruments."
NOTE: A list of manufacturers of seismic in-
struments appears on page 107.
Diesel Engines @@(3
The design of a foundation and a suitable isolation
mounting for a diesel engine, or for that matter
any multicylinder engine, presents a rather formi-
dable problem to one unexperienced in its design.
There may be unbalanced lateral forces due to un-
balanced revolving masses, there may be torques act-
ing to rock the engine in a transverse direction and
couples acting to rock the engine longitudinally.
All of these may act together giving to the system
six different modes of vibration each with its own
frequency. For a description of the design proced-
ure for diesel engines see "Design of Diesel Engine
Foundations" by Kenneth H. Iarkin in A.S.M.E. trans-
actions, March, 1942.
In this same article Mr. Larkin gives some average
values for foundation yardage required for diesel

engines. His values, given in Table IX below were

_ 88 -

obtained from recommendations of several engine_
manufacturers. The values are all within 20% of

those recommended and most are within 10%.

TABLE IX

Average Values for Reqd. Foundation Yardage for Diesel Engines

 

No. of Cylinders cu. yds./hp No. of Engines Tabulated

 

3 0.141 7
4 0.120 10
5 0.108 20
6 0.100 20
7 0.096 14
8 0.091 14

 

These yardages are the result of years of experience
with damping engine vibration by the use of a mass of con-
crete. They are based on a hard firm subsoil.

It is interesting to note that the above table gives no
relationship between foundation size, speed or cylinder size,
which items are bound to influence the yardage required.

I also take the liberty of quoting Mr. Larkin in regard
to empirical dimensions.

"The depth of foundation should be not less
than 4 to 5 times the stroke of the engine,
with a minimum of 5 feet. The width of the

base should be at least equal to the vertical

- 89 -

height from the bottom of the foundation to

the center of the shaft, and should be in-
creased for engines having unbalanced hori-
zontal forces". --- "The length of the base

is usually so large that no consideration is
required in this direction, except that the
length should be so adjusted that the center
of gravity of the total weight on the subsoil
coincides with the c.g. of the area of contact
with the subsoil. All of the rules are strictly
empirical and when necessary should give way to

more important considerations".

- 90 _

IX - DETAIL DESIGN OF FOUNDATIONS FOR CERTAIN R.CHINES

There follows two complete foundation designs in which
some of the principles heretofore discussed will be put to
use. It has already been emphasized that each.type of
machine has its individual problems, although they have much
in common. There is space here only to single out one or

two of the many problems for a more complete analysis.

A - Design

Design a reinforced concrete foundation for a large con-
stant speed, non-reversing snycronous motor (shaft drive.).
The motor has a rated horse—power of 1200 at 200 rpm. It can
Operate at a 33-1/3 overload. Figure 18 shows dimensions of
the base plate and also the points of application of the
loads.

Weights of the component parts are listed below:

Bedplate 27,000#

Rotor 35,000#

Frame 30,000#

Details 2,500#

Pedestals and Shaft 6,400#
6,400#

 

Total Weight 107,3oo#

First step - To choose dimensions of concrete foundation.
It is usually desirable to allow for the possibility of
errors and thus the dimensions on the design should be slight-

ly larger all around than the bedplate. It is also a good

.01.

idea to bevel the exposed corners so that there will be less
chance of their being damaged and thus marring the appearance
of the foundation. Corners with 900 angles are much easier
to form than obtuse or acute angles so sides if possible
should be vertical. The plan dimensions of the foundation
must completely encompass the bed plate and two inches will
be left all around. The vertical dimension must be great A
enough to set the foundation on solid soil. In this case a
minimum depth will be assumed satisfactory. The soil in

this case is a thick bed of clay, it will be drained and kept
moderately dry but allowance should be made for the possi-
bility of a small amount of moisture in it.

4000 #/sq. ft. is a good allowable on this soil

(see Table I).

The dimensions choosen are shown in figure 19. These
will be checked to see if the resulting soil pressures are
satisfactory. The requirements are that the maximum pressure
must not be too high, that the variation in soil pressures
must not be too great, and that the resultant of loads on the
soil must pass through the middle 1/3 of the foundation.
Second step: To calculate the loads at various_points on

the foundation.

The static loads due to the weights of the component
parts of the machine are given on page 91. The distribution
of these along with the load caused by the torque of the

motor must now be determined.

- 92 -

Bafiwfiflig)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. o e i
\
{
._/.’8~__..J “/26; z {0
. 8.9 ‘6
Q \
J; l
{D 09
V
m - BotOr ) )
tb 7 w
' J it
it :0
.1 L ..
3? <3 ,;
L 9
p— 619" _ ~ 6:41——-’*
M54” 7

 

 

 

 

 

 

 

 

I

 

 

 

»

 

 

   

 

 

I.
l

\

Bedbé;;:>

 

Sketch showing 1200 HP Syncronous Motor

Figure 18

-93-

 

1 N

1%04

 

- 2-’0"- r250;

denier

 

 

I I

8‘2

 

 

 

834"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e————7#fl
PLAN
Ts
$ ‘0
.N i
l _
v /t/'- 8" r
SEC T/ON

Chosen' dimensions for foundation

Figure 19

-94..

.

(A) To find the torque load to be a plied at points

Band C (see figure 20)

 

 

 

 

Figure 20

The motor is rated at 1200 hp. but there is a possi-
bility of a 33-1/3 % overload, so in calculating the torque

load the motor will be considered as developing 1600 hp.

Applying formula 16, page 33; P,the load caused by the
torque is:

P : H413. X5212.1 -_- 1600 X 5252.1 = ,
N x P 200 x76.33 66407

One half of this amount or 3320# is applied at both B and 0

one up and one down, along with the loads

caused by the weights of the motore parts.

 

 

load at A:

Wf weight of concrete foundation l20,750#
weight of bedplate 27,000#
weight of details 2LSOO#
Total at A 150,250#

-95....

Load at B:

 

 

% weight Of frame 15,000#
% torque load 6,640f
Total at B 21,640#
load at C:

a weight of frame 15,000
a torque load -6,640f
Total at C . 8,360#

From figure 21 it is seen that the weight of the rotor must

be distributed properly to D and E.

 

 

 

 

 

 

 

 

 

dance”
V
A
*+—-467 r- 967 1
##33 1
Figure 21
Rd = 9.67/14.35 . 35,000 = 23,500
Re = 4.67/14.33 . 35,000 = 11,400
load at D:
PrOportional weight of rotor 23,600#
Weight of pedestal 6,400i
Total at D 30,000#

-96-

load at E:

 

PrOportional weight of rotor 11,400#
Weight of pedestal 5 409i
17,800#

Total at E

Third step - To calculate the stress at the maximum
and minimum points.

(A) The centroidal axis are the geometrical axis of

the foundat ion.

(B) Find: Ix and 1y.
Ix = 1/12 tn; = 1/12 . 14.67 . 16333 = 5323.6
Iy = 1/12 bh3 : 1/12 . 16.33 . 14.673:4296.3

 

 

 

«133

1::
~§
Q
8.
, lm
a7, ‘— 0 fl
#0

 

 

 

 

 

 

 

 

 

K >——-;X
.5? I’I__T’ C1
7 C
21, 5'40“ ‘4 #37 83.63j' I
-w /“0' ' . 1%) “’h.
. . “9
50,000,. 0 ' ‘0
"an___1_ 1
0 m f n
. /46i
Figure 22

' 97 -

 

(C) T

he product of inertia - 0.

(D) Find the resultant of all the loads wt and the

point

T

228,050 .

Take
228,050 x

228,050 x
x

8x

(E) c

MX

MX

through which it acts.
he resultant is the sum of all the loads:
150,250
8,360
21,640
30,000
17L000

 

Wt = 228,050#

moments about m-n:
y ‘ 21,640 . 5.67 + 8.360 . 567 + 150.250 . 8.17
4 30,000 . l - 17.800 . 15.33 3 1.700.517

Y = 1,700,517
y = 7.457
9y 2 0.71

moments about O-p:
. 150,250 . 7.35 + 21,640 .'1 + 8,360 . 13.67
+ 30,000 . 7.33 + 17.800 . 7.33
: 1.587.628
6.962
0.37

alculate My and Mx

= P 8y : ‘2889050 . (-.37) 106,578

-288,050 . (-.71) = 204,516

=Pex

- 98 -

(F)

Calculate max. and min. soil pressures:

pmax

m

= P/A + Mxy/I + Myx/I
A = 16.33 . 14.67 = 239.56
—288,O5O/239.56 + 204,516 . (-8.l7)/5.324 +

1063578 o (‘7.33)/4:296
-952 - 314 - 182 = -1448 psf.

in : -952 + 204,516 . 8.17/5324 + 106,578 .

7.33/4296 = -952 t 314 + 182 = -456 psf.

(G) Conclusions:

(H)

1.

The calculated soil pressures are well below the
allowable.

The maximum soil pressure does not exceed twice
the average. (a good indication of a satisfactory
variation).

The resultant is obvious within the middle 1/3 of

the foundation.

For details of foundations in general see article 7.

B - Desggn

 

Design a reinforced concrete foundation block to be

set directly on hard clay to support the electric motor

shown in figure 23. The motor and bedplate weigh 1120#.

The motor develops 30 hp at 1500 rpm. The dimensions are

shown in figure 23.

-99-

 

 

 

 

 

Wm.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P——— 31‘I'——H
Figure 23

First step - Find the loading due to the belt pull; (assume
that the belt takes off at an angle as shown in figure 23)

applying formula 15, page 32.

- L:______3_ ..
Fb _ “‘0000 31:0 0 _ 720’?
1500 x-l—e-

The components of the belt pull in x and y

directions are:

Fbx 5%.in = M : 70556

- 100 -

Second step. Assume horizontal dimensions as in figure 24.

 

 

 

‘r——2/

 

r+—————3¥s

 

 

 

 

a: 319“

 

 

Figure 24

Third step.‘ Find the weight of foundation necessary to
prevent sliding on the soil:
Apply formula 10, page 26 which expresses the required

weight to prevent slipping.

 

P A P _
wf- fl Ka-Z -wm
P : 705#
e = 21/12 1.75'
A = 3.75 . 3.5 = 13.1 sq'
/% : (from Table II, page 29) 0.12
Wm = 1220#
K = % (2abc + % log 3;: r %3 log gég)

(formula 11, page 28)
where: a is 1/2 of the short dimension of the foundation
3 1.75 and b is l/2 the long dimension of the foundation
8 1.875.

- 101 -

 

 

 

2 2
c =‘/F;E’+ b2 - 1.75 4 1.875 : 2.54

 

3
— 2 v " 1- 1’87 2°54 ‘- 10.15 1'

 

1.25 2.54 + 1.875
2 l°5(2.5T- 1.875))

 

_ 2 . 4.29i » - 4.41

K - 3(16.670 + 3.29 log( .79) + 2.05 log (—:33§))
K = §416.678 + 329 log 5.43 . 2.68 log 6.64)

K : %(l6.678 + 3.29 x 1.69 + 2.68 x 1.89)

K : §(16.s7s + 5.556 + 5.073) = §(27.317)

K : 18.21

Inserting these values in the formula above:

705X1.75 (05 - 1220

w
r O.12xl8.2l * 0.12

565 + 5875 - 1220

wf 5230# (Necessary to prevent slipping)

Third step - Find the weight of foundation necessary to
prevent unequal settlement.

Referring to article 2—C, page 29, we shall now make
use of formula 13 to solve for the foundation weight which
will insure that the resultant of loads will be within the

middle 1/3 of the foundation base.

- 102 -

Pv (3dp - 1) - Wm(3dm - 1)

wt ” 3df - 1

 

Where PV (as previously computed) = 141#

 

 

dp = 3.5'
1 = 3.5'
Wm = 1220#
am = 1075'
(if = 1.75'

W = 141 x - - 1220 x l - ‘7

f 3 X 1.75 - 3.5
= 141 x 7 - 1220 x 1.75 z: 281 - 2140
1.75 1.75
Wf = “660/?

Which means that the weight of the motor is sufficient to

prevent unequal settlement.

Fourth step — Design the foundation block:

 

The weight of the foundation, having chosen the cross-
sectional dimensions, must be 5230#. Or in cu. ft. of con-
crete required: ,

5230/150 = 34.8 cu. ft.
The cross-sectional area = 13.1 sq. ft. so the required
foundation height is: '

34.8/13.1 - 2.66'

Detailed design: see article 7-A,B,C, and D.

-103...

10.

ll.

l2.

13.
14.

APPENDIX A
BIBLIOGRAPHY

Factory Installation Work; A. J. Coker, 1941
Chemical Publishing Co., Inc. New York, N. Y.

Foundations, Abutments and Footings; H001 and Kinne
McGraw-Hill Book Company, Inc. New York and London

Mechanical Vibrations; DenHartog, 1947, D. VanNostrand
Company, New York.

Vibration Problems in Engineering; Timoshenko, 1928,
D. VanNostrand Company , New York.

Benefits of Vibration Control; 5. Rosenzweig, Diesel
Power, June, 1945.

Analitical Mechanics for Engineers; Seeley and Ensign,
John Wiley and Sons, Inc. New York.

Machinery Foundations and Erections; Croft, 1923,
McGraw Hill Book Company, Ins., New York and London.

Control of Vibration Can Increase Production EffiCiency;
John Parma Jr., Steel, Feb. 17, 1947 - p. 94-96,
Feb. 24 - p. 76, March 3, - p. 116.

Turbine Vibration and Balancing; Rathbone, ASME
Transactions, 1929, 267-284,

Mechanical Vibration of Penstocks of Hydraulic Turbine
Installations; DenHartog, ASME, 1929

Some Special Fundamentals for Solid Resilient Mountings;
P. C. Rocke, Electric Manufacturer, Jan. 1943.

Machine Tools Must Have Good Foundations; J. Bayuk,
Factory Management and Maintenance.

Misalignment;_E. L. Parry, Electrical Review, Aug. 1943.
Effectiveness of Shear Stressed Rubber Compounds in

Isolating Machine Vibrations; Madden, ASHE Transactions,
Aug. 1943.

104

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.
27.
28.
29.
30.

31.
32.

Rubber Mountings Minimize Impact from Jolt Mouldings
Machines; Iron Age, Feb. 20, 1947.

Simplified Method for Design of Vibrating Isolating
Suspensions; ASME Transactions, Nov. 1947.

Vibration Effects Reduced by Flexible Mountings;
S. Rosenzweig, Industry and Power, June 1946, p. 69

Anti-Vibration Foundations for Heavy Machines; Iron Age,
Sept. 1946.

Turbine-Generator Foundation Design; Engineering News-
Record, Sept. 1946.

Design of Diesel Engine Foundations; Kenneth H. Larkin,
ASME Transactions, March, 1942.

Influence of Damping in Elastic Mounting of Vibrating
Machines; E. H. Hull, ASME Transactions, 1931,
Paper APM, 53-12.

Diesel Engineering; J. W. Anderson, McGraw-Hill Book
Company, Inc. New York, N. Y. 1938, p. 264.

Noise and Vibration Engineering; S. E. Slocum, Van-
Nostrand Co., Inc., New York, N. Y. 1931.

Isolating Vibrations in Mechanical Equipment; E. N.
Kemler and C. R. Freberg, Heating and Ventilating,
June, 1945.

rlain Concrete; Bauer
McGraw-Hill Book Co., New York, N. Y.

Steam Turbine Principles and Practice; Croft.

Terry Instruction Book

General Electric Co. Instruction Book, #82,215

Westinghouse Company; Book #5180

Turbine Foundations; T. C. Rathbone, ASME Transactions,
1941.

Anchor Bolts; Power, June, 1942.

Materials and Construction for Resisting Shock;
American Machinest, Sept. 9, 1915.

105

33.

39.

40.

41.

42.

Mount Motors on Strong Rigid Supports; Factory Manage-
ment and Maintenance, July, 1950.

Prime Movers Committee Report; 1921, "The Three Point
Support Semi-Rigid Steam Turbine Foundation".
(National Electric Light Association). .

Design and Construction of Heat Engines; Wm. E. Ninde.

Gas Engine Handbook; E. W. Roberts.

Turbine Foundations; T. C. Rathbone, ASME Transactions,
1941.

Stopping Vibrations due to Unbalanced Engines; Bigelow
Power, Sept. 11, 1928.

Several Methods to Prevent and Correct Noises Made by
Fans; Heat, Piping and Air Conditioning, Aug. 1929.

Vibrations of Frames of Electrical Machines; Den Hartog
ASME Transactions, Sept. Dec. 1928.

Vibrations Effects on Foundations and Masonry Structures;
Engineering News-Record, May 29, 1930.

Foundigentals of Soil Mechanics; Donald Taylor, Wiley,
19 .

106

APPENDIX B

Vibration problems related to operating machinery and
mechanisms can often be best solved by the use of vibra-

tion recording analyzing systems or measuring instruments.

Manufacturers of these instruments include:

Aeroquip Corporation -- Jackson, Michigan

Brush Development Company -- Cleveland, Ohio
Consolidated Engineering Corporation -- Pasadina, Calif.
General Electric Company -- Cambridge, Mass.

MB Manufacturing Company -- New Haven, Conn.

Western Electric Company -- New York, N. Y.

Westinghouse Electric Corporation -- Pittsburg, Pa.

107

APPENDIX C

Manufacturers of isolating materials, most of which
welcome requests for information and are prompt to offer
their engineering services if required have been listed

below with their addresses:

American Felt Company -- Glenville, Conn.

L. N. Barry Co.'1nc., (Rubber Products), Cambridge, Mass.

Electron Corp., (confined air column), Freeport, New York.

Fabreeka Products Company (rubber and cotton duck pads),
Boston, Mass.

Felter Company, Inc., (felt), Boston, Mass.

Firestone Industrial Product Co., (rubber), Akron, Ohio.

International Balsa Corp., (Balsa) Jersey City, N. J.

B. F. Goodrich Company, (isolators, rubber), Akron, Ohio.

Korfund Company, Inc., (isolation), Long Island, N. Y.'

Lord Manufacturing Company, Erie, Pa.

MB Manufacturing Company, New Haven, Conn.

U. S. Rubber Co., Meck Goods Division, New York, N. Y.

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