Middleton W.M. (ed.) Reference Data for Engineers: Radio, Electronics, Computer and Communications

49-1

4

TABLE

8.

NEC ALLOWABLE CURRENT-CARRYING CAPACITIES,

IN

AMPERES,

OF

CONDUCTORS

Aluminum

or

Copper-Clad Aluminum

Copper-Conductor Insulation

Conductor Insulation

RH, RHW, TA, TBS,

RH, RHW,

RUH

,

SA, FEP,

RUH

(14

-

2)

FEPB

,

(12

-

2),

TA, TBS,

RUW THW, RHH, RUW THW, SA,RHH,

Size

(14

-

2)

THWN, THHN,

(12

-

2)

THWN, THHN

,

AWG T,TW,UF XHHW XHHW* TFET T,TW,UF XHHW XHHW*

- -

-

40

14 15 15

12 20 20

10

30 30 40f 75

30$ 15 15 2%

25 25 308

25$ 55

8 40

45

50

95 30 40 40

6 55

65 70 120 40

50 55

4 70

85 90 145 55

65 70

3 80

100 105 170 65

75 80

2 95

115 120 195 75

90 95

1 110

130 140 220 85

100 110

0

125

150 155 250

100

120

125

00

145 175 185 280

115 135 145

000

165 200 210 315 130 155 165

0000

195 230 235 370 155 180 185

Correction Factors

for Higher Room

Temperatures

"C

OF

40 104 0.82

0.88 0.91

-

0.82

0.88 0.91

45 113

0.71

0.82 0.87

-

0.71 0.82

0.87

50

122 0.58

0.75 0.82

-

0.58

0.75 0.82

55 131 0.41

0.67 0.76

-

0.41

0.67 0.76

60 140

-

0.58

0.71 .95

-

0.58 0.71

*

Dry locations only.

t

Nickel

or

nickel-coated copper only.

$

For types FEP,

FEPB,

RHH, THHN, and XHHW, sizes

14, 12,

10 shall be the same

as

designated

for

RH, RHW, etc.

0

For types RHH, THHN, and XHHW, sizes

12

and

10

shall be the same

as

designated

for

RH, RHW, etc.

Notes:

1.

Not more than three conductors in raceway

or

cable.

2.

Based on room temperature

of

30

OC

(86

OF).

See correction factors for higher temperatures.

3.

Derating factors-more than three conductors in raceway

or

cable:

Number

of

conductors

4-6 7-24 25-42 >42

%

of

current capacity

80 70 60 50

(H), Wiring should be shielded from the direct heat

radiation of high-temperature parts such as elec-

tron tubes and power resistors.

(I)

Wire identification may be required for conve-

nience in manufacture, installation, or servicing.

As

a matter of expediency, most electronic equip-

ment employs the smallest conveniently handled wire

size (usually 20 to 24

AWG)

for most wiring, with the

larger conductors being installed only for circuits carry-

ing currents greater than that permitted for the "gener-

al-use" wire size. With the trend toward compact

solid-state integrated-circuit equipment, smaller wire

sizes are being used. However, the reduction in wiring

bundle size may be small unless the conductor insula-

tion thickness (determined by voltage considerations)

can be reduced.

Table 11 gives recommended current ratings for

copper and some aluminum based on a

45

"C

(40 "C for

wires smaller than 22

AWG)

conductor temperature rise

due to load current. Table

11

may be used for the

temperature conditions given in Table 12.

A

60

"C

ambient temperature around the wiring

(20

"C

internal temperature rise from

40

"C

[IO4

OF]

TABLE

9.

NEC MOTOR STARTING-

CURRENT DATA*

Kilovolt-Amperes per

Horsepower

with

Code Letter Locked Rotor

~~~

A

0

-

3.14

B

3.15

-

3.54

C

3.55

-

3.99

D

4.00

-

4.49

E

4.50

-

4.99

F

5.00

-

5.59

G

5.60

-

6.29

H

6.30

-

7.09

J

7.10

-

1.99

K

8.00

-

8.99

L

9.00

-

9.99

M

10.00

-

11.19

N

11.20

-

12.49

P

12.50

-

13.99

R

14.00

-

15.99

S

16.00

-

17.99

T

18.00

-

19.99

U

20.00

-

22.39

V

22.40

up

*

Locked-rotor currents of motors are

useful

in

determin-

ing branch-circuit overcurrent protection requirements and

voltage drop

at

start. These values are indicated

by

code

letters

on

motor nameplate.

Note

NEMA standard.

ambient around the equipment) is typical of some

electronic equipment. If higher ambient temperatures,

high power, or compact designs with electron tubes or

magnetic-core components (except for very low power)

are a factor, the temperature in the wiring space should

be specifically determined.

“Wiring confined” ratings are based

on

15

or more

wires in a bundle, with the sum of all the actual load

currents of the bundled wires

not

exceeding

20%

of the

permitted “wiring confined” sum total carrying capaci-

ty of the bundled wires. These ratings approximate

60%

of the free-air ratings (with some variations due to

rounding). They should be used for wire in harnesses,

cable, conduit, and general chassis conditions. Bundles

of

fewer than

15

wires may have the allowable sum of

the load currents increased as the bundle approaches the

single-wire condition.

RESISTANCE CHANGE WITH

TEMPERATURE

The resistance of most conductor materials changes

with temperature. Table

11

shows the copper-wire

resistance at

100

“C. Correction factors must be applied

to determine the resistance at other temperatures, or for

other materials. Thus, from Table 11 determine the

copper-wire resistance at 100 “C (multiply by conver-

sion factor

m

of

Table 13 for other materials). Use the

equation

R,

=

R,m[l

+

K(t

-

loo)]

where,

R,

=

resistance at desired temperature

t,

R,

=

resistance at 100 “C for copper (Table ll),

TABLE 10. NEC MOTOR FULL-LOAD RUNNING CURRENTS

IN

AMPERES (USUAL CONDITIONS

AND

SPEEDS)

Single-phase AC 3-Phase ACT DC

Horsepower

115

V

230

V*

115

V

230

V*

460

V

120

v

240

V

I

1

-

- - -

6

4.4 2.2

4

3.1 1.6

3

4.1 2.0

-

5.8 2.9

7.2 3.6

1

9.8 4.9 4 2

1

5.4 2.7

1

13.8 6.9 5.6 2.8 1.4 7.6 3.8

2

1 16

8

7.2

3.6

1.8 9.5

4.7

1;

20

10

10.4

5.2

2.6 13.2

6.6

2 24 12 13.6

6.8 3.4

17 8.5

3 34 17

-

9.6

4.8

25 12.2

5 56

28

-

15.2

7.6

40 20

1;

80

40

-

22

11 58

29

10

100

50

-

28

14 76

38

- - -

2

4

*

For

208

V, multiply by

1.1;

for

200

V,

multiply

by

1.15.

t

Induction tvoe. sauirrel

cage

and

wound

rotor.

TABLE 11. RECOMMENDED CURRENT RATINGS (CONTINUOUS

DUTY)

FOR

ELECTRONIC EQUIPMENT

AND

CHASSIS WIRING*

Maximum Current in Amperes

Wire Size Copper Conductor Copper Wire Aluminum Wire

(100

"C) Nominal

Circular Resistance Wiring in Wiring Wiring in Wiring

AWG Mils

(Ohms/1000

ft) Free Air Confined Free Air Confined

32 63.2 188.0 0.53 0.32

30 100.5 116.0 0.86 0.52

28

26

24

22

20

18

16

14

12

10

8

6

4

2

1

0

00

000

0000

159.8

254.1

404.0

642.4

1022

1624

2583

4107

6530

10 380

16 510

26 250

41 740

66 370

83 690

105 500

133

100

167 800

211 600

72.0

45.2

28.4

22.0

13.7

6.50

5.15

3.20

2.02

1.31

0.734

0.459

0.290

0.185

0.151

0.117

0.092

0.074

0.059

1.4

2.2

3.5

7.0

11.0

16

22

32

41

55

73

101

135

181

21

1

245

283

328

380

0.83

1.3

2.1

5.0

7.5

10

13

17

23

33

46

60

80

100

125

150

175

200

225

60

83

108

152

174

202

235

266

303

36

50

66

82

105

123

145

162

190

*

See Table

12.

m

=

material factor for

100

"C resistance value

K

=

correction factor (Table 13),

(Table 13),

t

=

desired temperature ("C).

WIRE

IDENTIFICATION

In a complex wiring assembly, or if both ends of a

wire cannot be seen from one station. a means of

TABLE 12. TEMPERATURE CONDITIONS

FOR

TABLE 11

Maximum Maximum

Allowable Ambient

Conductor Temperature Typical Conductor

Temperature Around Wire and Insulation

"C "C

105 60

Bare

or tinned copper or

aluminum; polyvinyl-

chloride insulation

200 155

Silver-coated copper; FEP

or PTFE with FEP jacket

insulation*

PTFE insulation*

260 215

Nickel-coated copper;

*

FEP

=

Fluorinated ethylene propylene.

PTFE

=

Polytetrafluoroethylene.

identifying each lead simplifies manufacture, installa-

tion, and servicing. Common identification methods

are:

(A)

Tag each end of a lead with an assigned designa-

tion

(an

alternative method is to print the designation at

frequent intervals along the wire insulation).

(B)

Color code the wires. The wire insulation may

be a solid color, color stripes may be spiraled around

the

wire,

or the name

of

the color (or its numerical code

equivalent) may be stamped at frequent intervals along

the wire.

TABLE

13.

TEMPERATURE

AND

MATERIALS

CORRECTION

FACTORS

Material Correction

Conductor Material Factor

rn

Factor

K

Soft copper

Hard copper

Copper-clad steel:

30%

conductivity

40%

conductivity

Aluminum

Nickel

Nickel-clad copper:

10%

30%

Silver

1

.oo

1.03

3.47

2.56

1.64

5.28

1.07

1.35

0.94

0.0039

0.0038

0.0044

0.0041

0.0039

0.0047

0.0038

0.0036

0.0038

MISCELLANEOUS DATA

49-1

7

Color Coding

The commonly used colors and their numerical codes

are:

0

Black

5

Green

1

Brown

6

Blue

2

Red

7

Violet (purple)

3

Orange

8

Gray (slate)

4

Yellow

9

White

While spiral stripes can be applied

on

top of any

basic insulation color, under less favorable viewing

conditions it is difficult to distinguish some colors from

the basic insulation color. Identification may be slow

and subject

to

error. The preferred combination consists

of one or two (sometimes three) colored stripes

on

a

white basic insulation. To minimize identification er-

rors, the first stripe is made wider than the second (or

third), and some rules require that the second stripe be

of higher numerical code than the first stripe. If the

required variety of wire color codes is not great, the

preceding guides should be followed.

Table 14 gives a standard color code used to distin-

guish by function the various leads in electronic cir-

cuits.

In

manufacturing practice, it is preferred that, at any

harness breakout point, all wires of the same color code

be connected to the same terminal at that location.

When this rule and the wire color coding of Table

14

are

both applicable, additional tracers may be used to

supplement the primary coding of Table

14.

DIAMETER

OF

CIRCLE

ENCLOSING A GIVEN

NUMBER

OF

SMALLER

CIRCLES*

Four of many possible compact arrangements of

circles within a circle are shown in Fig.

3.

To determine

the diameter of the smallest enclosing circle for a

particular number of enclosed circles all

of

the same

size, three factors that influence the size of the enclos-

ing circle should be considered, as follows.

Arrangement

of

Center or

Core Circles

The four most common arrangements of center or

core circles

are

shown in cross section in Fig.

3.

It

may

seem that Fig.

3A

would require the smallest enclosing

circle for a given number of enclosed circles, but this is

not always the case since the most compact arrangement

will depend in part

on

the number of circles to be

enclosed.

TABLE 14.

COLORS

FOR

WIRE

IDENTIFICATION

BY

FUNCTION

Identifi-

Function Color cation

No.

Grounds, grounded Black

0

Power supply B

+

Red

2

Screen grids Orange 3

Cathodes and transistor Yellow

4

Control grids and tran- Green

5

elements

Heaters or filaments Brown

1

emitters*

t

sistor basest

sistor collectors*t

Anodes (plates) and tran- Blue

6

Power supply, negative

(-)

Violet (purple)

7

Ac power lines Gray (slate)

8

*

Applies to diodes, semiconductor elements, photoelec-

tric cells, mercury-arc rectifiers, and other elements with

operation similar to vacuum tubes and transistors.

t

Applies

to

all types

of

gas tubes with operation similar to

vacuum tubes.

Diameter

of

Enclosing Circle

When Outer Layer

of

Circles

Is

Complete

Successive, complete “layers” of circles may be

placed around each

of

the central cores of 1,

2,

3,

or

4

circles. The number of circles contained in arrange-

ments of complete “layers” around a central core of

circles, as well as the diameter of the enclosing circle,

may be obtained from Table

15.

Thus, for example,

Fig.

3A

has a total of

18

circles arranged in two

complete “layers” around a central core consisting of

one circle; this agrees with the data shown in the left

half of Table

15

for

n

=

2.

To determine the diameter of the enclosing circle, the

data in the right half of Table

15

are

used. Thus, for

n

=

2

and an

“A”

pattern, diameter

D

is

5

times the

diameter,

d,

of the enclosed circles.

Diameter

of

Enclosing Circle

When Outer Layer

of

Circles

Is

Not Complete

In

most cases, it

is

possible to reduce the size of the

enclosing circle

from

that required

if

the outer layer

A

B

C

D

-

*

Dutka,

J.

“How Many Wires Can Be Packed Into a

Circular Conduit.

”

Machinery’s Handbook.

New York In-

dustrial Press, Inc.,

1956.

Fig.

3.

Arrangements

of

circles within a circle.

(Reprinted

with permission from

Machinery’s Handbook,

17th Edition,

Industrial Press,

Inc.,

New York.)

49-1

8

TABLE

15.

NUMBER

OF

CIRCLES CONTAINED

IN

COMPLETE

LAYERS

OF

CIRCLES

AND

DIAMETER

OF

ENCLOSING

CIRCLE

Number of Circles

in

Center Pattern

1

2 3 4

1

2 3 4

Arrangement

of

Circles

in

Center Pattern (see Fig.

3)

No.

Complete

Over

Core,

n

Layers

“A”

“B”

“C”

“D”

“A”

“B”

“C” “D”

Number of Circles,

N,

Enclosed

Diameter,

D,

of

Enclosing Circle*

0

1

2 3 4 d

2d

2.155d 2.414d

1

7 10 12 14 3d 4d 4.055d 4.386d

2 19 24 21 30 5d

6d

6.033d 6.379d

3 37

44

48 52 7d

8d 8.024d

8.375d

4 61 70 75

80

9d

1

Od

10.018d 10.373d

5

91 102 108 114

1

Id

12d

12.015d 12.372d

n

t

*

Diameter

D

is given in terms

of

d,

the

diameter

of

the enclosed circles.

t

For

n

complete layers over core, the number of enclosed circles

N

for

“A”

center pattern is

3n2

+

3n

+

1;

for “B,”

3n2

+

5n

+

2;

for “C,”

3n2

+

6n

+

3;

for “D,”

3nZ

+

7n

+

4;

while the diameter

D

of the enclosing circle for

“A”

center pattern is

(2n

+

1)d;

for “B,”

(2n

+

2)d;

for “C,”

[l

+

2(nz

+

n

+

$”‘Id;

and for “D,”

[l

+

(4n2

+

5.644n

+

2)”2]d.

Reprinted with permission from

Machinery’s Handbook,

17th

edition. New York: Industrial Press, Inc.

were complete. Thus, for example, Fig.

3B

shows that

the central core consisting of two circles is surrounded

by one complete layer of eight circles and one partial

outer layer of four circles

so

that the total number

of

circles enclosed is

14.

If the outer layer were complete,

then (from Table

15)

the total number of enclosed

circles would be

24

and the diameter of the enclosing

circle would be

64

however, since the outer layer is

composed of only four circles out of a possible

14

for a

complete second layer, a smaller diameter of enclosing

circle may be used. Table

16

shows that for a total of

14

enclosed circles arranged in a

“B”

pattern with the

outer layer of circles incomplete, the diameter for the

enclosing circle is

4.606d.

Table

16

can be used to determine the smallest

enclosing circle for a given number

of

circles to be

enclosed by direct comparison of the

“A,”

“B,”

and

“C”

columns. For data outside the range of Table

16,

use the equations in Dr.

Dutka’s

article.*

Approximate Equation When

Number

of

Enclosed Circles

is

Large

When a large number of circles

is

to be enclosed, the

arrangement of the center circles has little effect on the

diameter

of

the enclosing circle. For numbers

of

circles

greater than

10

000,

the diameter of the enclosing circle

may be calculated within

2

percent from the equation

D

=

d[l

+

(N/0.907)”2]

where,

D

=

diameter of enclosing circle,

d

=

diameter of enclosed circles,

N

=

number of enclosed circles.

TORQUE AND HORSEPOWER

Torque varies directly with power and inversely with

rotating speed of the shaft, or

T

=

KPIN

where,

T

=

torque in pound-inches,

P

=

horsepower,

N

=

revolutionshninute,

K

=

63

000

(constant).

TRANSMISSION-LINE

SAG

CALCULATIONS*

For transmission-line work, with towers on the same

or slightly different levels, the cables are assumed to

*

Dutka,

J.

“How

Many Wires Can Be Packed Into

a

Circular Conduit.”

Machinery’s Handbook.

New York: In-

dustrial Press, Inc.,

1956.

*

Reprinted

by

permission from

Transmission Towers.

Pittsburgh, Pa.:

American

Bridge Co.,

1923;

p.

70.

TABLE

16.

FACTORS

FOR

DETERMINING DIAMETER,

D,

OF SMALLEST

ENCLOSING

CIRCLE FOR VARIOUS

NUMBERS,

N,

OF

ENCLOSED

CIRCLES*

Center Circle Pattern

~~ ~~

Center Circle Pattern

“A”

“B”

“C?’

“A”

“B”

“C” “A”

“B”

“C”

No.

N

Diameter Factor

K

N

Diameter Factor

K

N

Diameter Factor

K

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

3

4.464

5

6.292

7

2

2.732

3.646

4

4.606

5.359

5.583

6

6.196

6.568

7.083

-

2.155

3.309

4.055

5.163

5.619

6.033

6.773

7.110

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

7

I

7.928

8.211

8.21 1

8.211

8.21

1

8.211

9

9.718

7.083

7.245

I.

245

7.557

7.551

8

8.550

8.810

8.937

9.185

9.544

7.110

7.429

1.429

8.024

8.572

9.083

9.326

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

9.718

10.165

11

11.392

9.544

9.660

9.888

10

10.539

10.644

10.849

11.149

11.440

9.326

10.018

10.238

10.452

10.866

11.214

*

The diameter,

D,

of the enclosing circle is equal to the diameter factor,

K,

multiplied

by

d,

the diameter of the enclosed circles,

or

D

=

K

X

d.

For example, if the number of circles to be enclosed,

N,

is

12,

and the center circle arrangement is “C,” then for

d

=

1;

inches,

D

=

4.005

X

1;

=

6.083

inches. Reprinted with permission fromMuchineryS

Handbook,

17th

Edition. New York

Industrial Press, Inc.

take the form of a parabola instead of their actual form

of a catenary. The error is negligible, and the computa-

tions are much simplified. In calculating sags, the

changes in cables due

to

variations in load and tempera-

ture must be considered.

where,

L

=

length

of

span in feet,

L,

=

length

of

cable in feet,

S

=

sag

of

cable at center

of

span in feet,

Supports

at

Same

Level

Refer to Fig.

4.

The equations used in calculating

sags are

H

=

WL2/8S

S

=

WL2/8H

=

[(L,

-

L)3L/8]”*

L,

=

L

+

8S2/3L

Fig.

4.

Supports at same elevation.

49-20

REFERENCE

DATA

FOR ENGINEERS

H

=

tension in cable at center of span in pounds

=

horizontal component

of

the tension at any

point,

H

L

W

=

weight of cable in pounds per lineal foot.

If cables are subject to wind and ice loads,

W

=

the

algebraic sum of the loads. That is, for ice on cables,

W

=

weight of cables plus weight of ice; for wind on

bare or ice-covered cables,

W

=

the square root of the

sum

of

the squares of the vertical and horizontal loads.

For any intermediate point at a distance

x

from the

center

of

the span, the sag is

Fig. 5. Supports at different elevations.

s,

=

S(1

-

421L2)

Supports at Different Levels

Refer to Fig.

5.

S

=

So

=

WL2 cosaI8T

=

WL2/8T cosa

S,

=

WLf2I8H

S2

=

WL?/8H

L1/2

=

L12

-

(hH

cosalWL)

L2/2

=

L12

+

(hH cosalWL)

L,

=

L

+

i[(S,21L,)

+

(S?/LZ)l

where,

W

=

weight of cable in pounds per lineal foot

between supports or in direction

of

Lo,

T

=

tension in cable direction parallel with line

between supports.

The change,

E,

in length of cable, L,, for varying

temperature is found by multiplying the number

of

degrees,

n,

by the length

of

the cable in feet times the

coefficient of linear expansion per foot per degree

Fahrenheit,

c.

This

is

l=L,Xnxc

A

short approximate method for determining sags

under varying temperatures and loadings that is close

enough for all ordinary line work is as follows:

(A)

Determine sag

of

cable with maximum stress

under maximum load of lowest temperature occurring at

the time

of

maximum load, and find length

of

cable

with this sag.

(B)

Find length of cable at the temperature for which

the sag is required.

(C)

Assume a certain reduced tension in the cable at

the temperature and under the loading combination for

which the sag is required; then find the decrease in

length of the cable due to the decrease

of

the stress from

its maximum.

(D)

Combine the algebraic sum of

(B)

and

(C)

with

(A)

to get the length of the cable under the desired

conditions; from this length, the sag and tension can be

determined.

(E)

If

this tension agrees with that assumed in

(C),

the sag in

(D)

is correct. If it does not agree, another

assumption

of

tension in

(C)

must be made and the

process repeated until

(C)

and

(D)

agree.

STRUCTURAL STANDARDS

FOR STEEL RADIO TOWERS*

Material

(A)

Structural steel shall conform to American Soci-

ety for Testing Materials

Standard SpeciJcations for

Steel for Bridges

and

Buildings,

Serial Designation

A-7,

as amended to date.

(B)

Steel pipe shall conform to American Society

for Testing Materials standard specifications either for

electric-resistance welded steel pipe, Grade A

or

Grade

B,

Serial Designation A-135, or for welded and seam-

less steel pipe, Grade A or Grade

B,

Serial Designation

A-53, each

as

amended to date.

Loading

(A)

20-pound design: Structures up to

600

feet in

height (unless they

are

to

be

located within city limits)

shall be designed for a horizontal wind pressure of

20

-

*

Abstracted from

American Standard Minimum Design

Loads in

Buildings

and

Other Structures,

A58.1-1955.

New

York: American National Standards Institute, Inc. Also from

Electronic Industries Association Standard TR-116, October

1949. Sections on manufacture and workmanship, finish, and

plans and marking

of

the standard are not reproduced here.

The section “Wind Velocities and Pressures’’ is not part of

the

standard.

poundslfoot’ on flat surfaces and

13.3

poundslfoot’ on

cylindrical surfaces.

(B)

30-pound design: Structures more than

600

feet

in height and those of any height to be located within

city limits shall be designed for a horizontal wind

pressure of 30 poundslfoot’ on flat surfaces and

20

poundslfoot’ on cylindrical surfaces.

(C)

Other designs: Certain structures may be de-

signed to resist loads greater than those described in

(A)

and

(B).

Figure

1

of American Standard A58.1-1955

shows sections of the United States where greater wind

pressures may occur. In all such cases, the pressure on

cylindrical surfaces shall be computed as being of that

specified for flat surfaces.

(D)

For open-face (latticed) structures

of

square

cross section, the wind pressure normal to one face

shall be applied to

2.20

times the normal projected area

of all members in one face, or

2.40

times the normal

projected area

of

one face for wind applied to one

comer. For open-face (latticed) structures of triangular

cross section, the wind pressure normal to one face

shall be applied to

2.00

times the normal projected area

of all members in one face, or

1.50

times the normal

projected area for wind parallel to one face. For

closed-face (solid) structures, the wind pressure shall be

applied to

1.00

times the normal projected area for

square or rectangular shape,

0.80

for hexagonal or

octagonal shape, and

0.60

for round or elliptical shape.

(E)

Provisions shall be made for all supplementary

loadings caused by the attachment of guys, antennas,

transmission and power lines, ladders, etc. The pressure

shall be as described for the respective designs and shall

be applied to the projected area of the construction.

(F)

The total load specified above shall be applied to

the structure in the directions that will cause the

maximum stress in the various members.

(G)

The dead weight of the structure, and all

materials attached thereto, shall be included.

Unit Stresses

(A)

All parts of the structure shall be

so

designed

that the unit stresses resulting from the specified loads

shall not exceed the following values:

Axial tension on net section

=

20

000

Axial compression on gross section:

poundslinch’

For members with values of LIR not greater than

120.

=

17

000

-

0.485L’IR‘ poundslinch’

For members with values of LIR greater than

120,

poundslinch2

-

18

000

-

1

+

L’118 OOOR

where L

=

unbraced length of the member, and

R

=

corresponding radius of gyration, both in

inches.

Maximum LIR for main leg members

=

140

Maximum LIR for other compression members

Maximum LIR for members with no calculated

Bending on extreme fibres

=

20

000

poundslinch’

Single shear on bolts

=

13

500

poundslinch’

Double shear on bolts

=

27

000

poundslinch’

Bearing on bolts (single shear)

=

30

000

Bearing on bolts (double shear)

=

30

000

Tension on bolts and other threaded parts, on

with calculated stress

=

200

stress

=

250

poundslinch’

nominal area at root of thread

=

16

000

poundslinch’

Members subject to both axial and bending stresses

shall be

so

designed that the calculated unit axial stress

divided by the allowable unit axial stress, plus the

calculated unit bending stress, divided by the allowable

unit bending stress, shall not exceed unity.

(B)

Minimum thickness of material for structural

members:

Pinted structural angles and plates

=

&

inch

Hot-dip galvanized structural angles and plates

=

Other structural members to mill minimum for

(C)

Where materials of higher quality than specified

under “Material” above are used, the above unit

stresses may be modified. The modified unit stresses

must provide the same factor of safety based on the

yield point of the materials.

inch

standard shapes.

Foundations

(A)

Standard foundations shall be designed for a soil

pressure not to exceed

4000

poundslfoot’ under the

specified loading.

In

uplift, the foundations shall be

designed to resist 100 percent more than the specified

loading, assuming that the base of the pier will engage

the frustum of an inverted pyramid of earth whose sides

form an angle of

30

degrees with the vertical.

Earth

shall be considered to weigh

100

pounds/foot3 and

concrete 140 poundslfoot3.

(B)

Foundation plans shall ordinarily show standard

foundations as defined in

(A).

Where the actual soil

conditions are not normal, requiring some modification

in the standard design, and complete soil information is

provided to the manufacturer by the purchaser, the

foundation plan shall show the required design.

(C)

Under conditions requiring special engineering

such as pile construction, roof installations, etc., the

manufacturer shall provide the necessary information

SO

that proper foundations can be designed by the purchas-

er’s engineer or architect.

(D)

In the design of guy anchors subject to submer-

sion, the upward pressure

of

the water should be taken

into account.

Wind Velocities and Pressures

Data regarding wind velocities and pressures are

contained in Table 17.

VIBRATION AND

SHOCK

ISOLATION

Symbols

b

=

damping factor

d

=

static deflection in inches

E

=

relative transmissibility

=

(force transmitted by isolators)/(force

transmitted by rigid mountings)

F

=

force in pounds

Fo

=

peak force in pounds

f

=

frequency in hertz

fo

=

resonant frequency of system in hertz

G

=

acceleration of gravity

g

=

peak acceleration in dimensionless

=

386

inches per second’

gravitational units

Xo/G

(-I)”~, vector operator

stiffness constant; force required to compress

or extend isolators unit distance in pounds

per inch

coefficient of viscous damping in pounds per

inch per second

time in seconds

weight in pounds

displacement from equilibrium position in

inches

peak displacement in inches

velocity in inches per second

dxldt

peak velocity in inches per second

acceleration in inches per second’

d2xldt2

peak acceleration in inches per second’

phase angle in radians

angular velocity in radians per second

2Tf

Equations

The following relations apply to simple harmonic

motion in systems with one degree of freedom. Al-

though actual vibration is usually more complex, the

equations provide useful approximations for practical

purposes.

TABLE

17.

WIND

VELOCITIES AND

PRESSURES

~~

Pressure

P

(pounds/foot2)

Indicated Velocity Projected Areast

(miledhour)

Actual Cylindrical

Velocity V,* 3-cup

4-cup

Surfaces Flat Surfaces

(miledhour)

Anemometer Anemometer

(P

=

0.0025V2)

(P

=

O.O042V,2)

10

9 10 0.25

0.42

20

20 23

1

.o

1.7

40 42

50

4.0 6.7

50

54

64

6.3

10.5

60 65 77

9.0

15.1

70 76 91 12.3 20.6

80 88 105 16.0 26.8

90 99 119

20.3

34.0

100 110 132 25.0 42.0

110 121 146 30.3 50.8

120 133 160 36.0 60.5

130 144 173 42.3 71.0

140 155 187 49.0 82.3

30 31 36 2.3 3.8

150 167

20

1 56.3 94.5

*

Although wind velocities are measured with cup anemometers, all data published by the

US

Weather Bureau since January

1932 include instrumental corrections and

are

actual velocities. Prior to 1932, indicated velocities were published.

In

calculating pressures on structures, the “fastest-single-mile velocities” published by the Weather Bureau should he multiplied

by a gust factor of 1.3 to obtain the maximum instantaneous actual velocities. See “Wind-Velocity and Temperature Extremes in

North America” in this chapter for fastest-single-mile records at various places in the United States and Canada.

t

The American Bridge Company equations given here are based on a ratio of 25/42 for pressures

on

cylindrical and flat surfaces,

respectively, while the Electronic Industries Association specifies a ratio of 213. The actual ratio varies in a complex manner with

Reynolds number, shape, and size of the exposed object.

MISCELLANEOUS

F

=

W(Z/G)

F,

=

wg

DATA

49-23

x

=

Xo

sin(ot

+

4)

(Eq.

3)

(E¶.

4)

x,

=

9.77 g/f2

Xo

=

oxo

=

6.28fX0

=

61.4g/f

X,

=

w2X,

=

39.5f2X,

=

386s

(Eq.

5)

(Eq.

6)

When damping is neglected

fo

=

3.13/di/z

(Eq.

12)

E

=

9.77/(df2

-

9.77)

(Eq.

13)

Acceleration

r

-

j(k/o)

r

+

j[(oW/G)

-

k/w]

The intensity of vibratory forces is often defined in

terms of

g

values. From Eq.

2,

it is apparent, for

example, that a peak acceleration of

log

on a body will

result in a reactionary force by the body equal to

10

If

an object is mounted on vibration isolators, the

accelerations

of

the vehicle are transmitted to the object

(or vice versa) in an amplitude and phase that depend on

the elastic flexing of the isolators

in

the direction

in

which the accelerations (dynamic forces) are applied.

E=

1

fo

=

3.13

(k/W)'/*

(Eq.

8)

b

=

9.77r/(kW)"2

(Eq.

9)

times its weight.

For critical damping,

b

=

1.

Neglecting dissipation

(b

=

O),

or atf6

=

(2)"2

for

any degree of damping

PEAK DISPLACEMENT AMPLITUDE,

Xo.

IN

INCHES

FREQUENCY,

I,

IN

HERTZ

Fig.

6.

Relation

of

frequency and

peak

values of velocity,

displacement,

and acceleration.

Middleton W.M. (ed.) Reference Data for Engineers: Radio, Electronics, Computer and Communications

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