I L LU ST RAT E D S OU RC E 8 0 0 K of ME C H A N I CAL C O M I 0 N E N T S

SECTION 30

NOMOGRAMS 30-2

Arc Length Versus Central Angle 30-3

Chordal Height and length of Chord 30-4

Forces in Toggle Joint with Equal Arms 30-5

Radical Defections of Rotating Disks 30-6

Power Capacity of Spur Gears 30-7

Alignment Chart for Face Gears 30- 10

Linear to Angular Conversion of Gear-Tooth Index Error 30-12

Determine Parallel Axis Moment of Inertia 30- 13

Moment of Inertia of a Prism about the Axis aa 30- 1 5

Chart for Transferring Moment of Inertia 30- 16

Rotary Motion 30- 1 7

Accelerated linear Motion 30- 1 8

Theoretical Capacity of Gear Pumps & Motors 30- 19

Nomogram for Piston Pumps 30-20

Weight and Volume 30-2 1

Volumes of Spherical Segments 30-22

Volumes in Horizontal Round Tanks with Flat Ends 30-23

Another Shortcut to Torsion-Bar Design 30-24

Nomogram for Angles in Constructed Shapes

30-2

Nomogram for Angles in Constructed Shares H. F. Bariffi

FROM THE ALTITUDE, h, and the dif- ing the Sb-scale at 22 in. Also, con- ferences (a-d) and (b-6) in necting 25 in. on the outer right hand lengths of corresponding sides of up- (b-c)-scale and 17 in. on the h-scale per and lower bases, the dihedral given value of ( a d ) on the inner gives a line crossing the S.-scale at angle A can be calculated. left hand scale. Connect the inter- 21.1 in. A line connecting 21.1 in,

To calculate Sa connect the given section of this line and the diagonal on the inner right hand S. scale with value of (a-d) on the right hand T-line with the (b--r)-scale. Project 22 in. on the diagonal s b scale inter- outer scale with the value of h on the this line to the left hand scale and sects the Q-line at the indicated point. h-scale and read the value of S,. In read the angle A . A line connecting this oint with 28 like manner, the vaIue of S. can be in. on the inner left [and ( a d ) - calculated from (6--c) and h. Con- EXAMPLE: When h = 17 in., (6-c) scale crosses the diagonal T-line at the nect the value of S. to the value of SI = 25 in. and ( a d ) = 28 in., find indicated point. A line connecting on the diagonal scale and project to I., s b and A. Connecting 28 in. on this point with 2 5 in. on the (b--c)- the Q-ine. Connect the intersection the outer right hand (a-d)-scale and scale passes through the point 112.1 of this line and the Q-line with the 17 in. on the h-scale gives a line cross- deg on the A scale.

45 J

(0-d) (b-c) COS A= - -

._ +- - I

6 .-29

15--21

Nomograms 30-3

Arc Length Versus Central Angle

(Angle of Bend, Length, and Radius)

1 R

- -

b l q - 2 ; 9) m -

- - - - -

- - 3 ;4 -

I - 2 - -

h

Draw a straight line through the two known points. The answer wil l be found

Example: For a 6-in. radius and 45deg. bend, length of arc is 4.7 in. at the intersection of this line with the third scale.

30-4

Chordal Height and Length of chord 4 - -

4 y - - - - - - - - - - - - - - - - - - - t i - - - - - - - - -3 - - - - - - - - -

'4 - - / * / = - - - - - - - - - - - - -

- -

- ' 5 - - - - - - - - c -

- - - - - - 4 - - - - - - - L - - - - - - - - - 25- - - - - - - -

- - - - - - - - - 9

IO Draw a straight line through the two known points. The answer will be found

Example: Leflgth of chord is 3 in., and radius of circle is 4 in. The height h of at the intersection of this line with the third scale.

the chord is 0.29 in.

Nomograms 30-5

Forces in Toggle Joint with Equal Arms

P - 8 F-a -----

---------- - -------------- s _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ - _ _ _ - - - -

Sin in.

200

-

100 800 -s 80

60 50 40

30

-

-

01 0.2 03 0.4 0506 0.8 I 2 3 4 5 6 8 1 0 h in in.

Example: Use mutually perpendicular lines drawn on tracing cloth or celluloid. In the example given for S = 10 in. and h = 1 in., a force F of 10 lb. exerts pressures P of 25 lb. each.

4.3.077.10 - d ( 1 . 5 K J + 7.5K) 5.0 -.

R - where: K= - -

40 /

/

- J/' - / - 0.10 - -

0.05 0.04

0.03

0.02

a3 8 4 F O . 0 1 _.

,s. -.- 1 /I/. c /

/

2 0.005 u" 0.004 z 0.003

/

- .c - j 0.02 - f 0.001 a-

- 0.0005 0.0004

0.0003

0.0002 -

- -

-

- - - r a w - - - r H m - - - - lop00 -

- 5

- i -. i

- a

- - - - 20,000 - - - - - - - - - -30,OWJ - - - - - - --4O.o00 - - - T 50,OOO - - - - - - -

60.000

1.0 - 70,000 - -

0.0001 -

0.9

- -80,000 - - -

Nomograms 30-7

Power Capacity of Spur Gears Charles Tiplitz

MAXIMUM RATED HORSEPOWER that can safely be trans- mitted by a gear depends upon whether i t runs for short periods or continuously. Capacity may be based on tooth strength if the gear is run only periodically; durability or wear governs rated horsepower for con- tinuous running.

Checking strength and surface durability of gears

can be a lengthy procedure. The following charts simplify the work and give values accurate to 5 to 10%. They are based on AGMA standards for strength and durability of spur gears.

Strength Nomograph is used first. Apart from the

Strength of Spur Gears. Based on AGMA 220.01

Diametral Circulor P i t c h p i t c h p i t c h dio.- in.

120 100 80

60 50 40

20

N (Teeth)

200* I50

I O -L

40

30

4

3

1.5

.C

0.6 If

r v m - $ 7 IO

-----

X 0.4 T

, O a 2 i OJ

0.07--

0.04--

0.02 --

0.01 -- 0.007--

-_ O.G04--

F o c e w id t h- in .

2 0.0007-

0.4 02 0.0002--

0.1

-- _ - -- -I-

i 00001

30-8

usual design constants only two of the following three need be known: pitch, number of teeth and pitch diameter. To use the charts connect the two known factors by a straight line, cutting the third scale. From this point on the scale continue drawing straight lines through known factors, cutting the pivot scales. Be- twecn the double pivot scales the line should be drawn

parallel to thc adjacent lines.

Durability nomograph must be entered on scale X a t the same value that was cut on the X scale on the strength chart. Both pinion and gear should be checked if made of different materials and the smaller of the values obtained should be used.

Strength X

0.04

0.0 2

0.007 O.O1 f

0,04 I

f 0.0007 0.0004

0.00 0 2 I

f 0.000 I

1 0.00001

Spur Gears (cont.) P i v o t sco le

D i a m e t r o l pi tch

100 70

4' \ =:E 2 I L

\ '.

P e o h horsepower

0.1

i T o o t h f a r m

foctor I Pressure angle No.

IO

.&-- , *

R a t e d horsepower -$. 1000

200

I O 0

70

40

20

Service f o c t o r

Service and shock (Eric losea gear ing)

I O

7

4

O.!

+

2

I

0 7 I I

0.4

4 0.1 $007

f 0.04 0.0 2

0.0 I

0.007

omograms 30-9

Surface Durability of Spur Gears. Based on AG In 0 - o c

+ 00,

, > c \ E a-

0 0 0 0 0 0 000 0 0 0 ok N -

+

a-

s 0

30- 10

-

-

Alignment Chart for Face Gears

180

I70

160

150 . 140

130

120

1 IO

100 95 90

85

80

75

70 5 65 2 60 f

E

0

L

B. Bloomfield

1:

--

--

THE MAXIMUM PRACTICAL DIAMETER for face gears is that diameter at which the teeth become pointed. The limiting inside diameter is the value at which tooth trimming occurs. This is always larger than the diameter for which the

operating pressure angle is zero. The two alignment charts that follow can be used to find the maximum OD and the minimum ID if the numbers of teeth in the face gear and pinion are known. They eliminate lengthy calculations.

55 .o : D

--50 $ ._ 67 +

l 4 5 0

40

35

--a

-- 25

-- 20 -. 19 ~. 18

50 45 40

35

30

25

14

13

12

C H A R T 1

I--

A

Nomograms

+150

--la

-_

30- 11

5

la--

130--

FOR BOTH CHARTS the pinions are assumed to be s ur gears

gear and pinion are assumed to intersect at right angles. Bo:h should be used onlv for tooth ratios of 1.5 to 1 or

larger. Smaller ratios require pinion modifications not

face gear diameter is found by dividing the factor from the chart bv the diametral pitch of the pinion.

of standard AGMA proportions, and the axes o P the face allowed for in these data. For both charts, the appropriate

1:

CHART II

- - I 3 0

--

-- 120

Example:

120--

c

- 'I 90 Rnd ths maximum &dc diameter and the -3 7 - 7 0

minimum inside diameter of 06xa gear with 70 tcelh'thqt will mate with o 20 bob stondad pinion wtase pessure wle is

30- 1 2

Linear to Angular Conversion of Gear-Tooth Index Error For pitch diameters up to 200 inch, chart quickly converts index error from ten-thousandths of an inch to seconds or arc.

Harold R. Ronan JR.

0.000

kl EXAMPLES 1 Pitch dio of geor=t41 in Index error=0001 in Read error converted t o 3 sec on scole B 2 Pi tch dio=41 in

Index error=0001 in Read error converted t o IO sec on scole A

- -----A-sco/e ftor 0 fo /oo In dol

\

\

65 4 165 50- 150 1

145

I40

" t

Nomograms 30-13

Determine Parallel Axis Moment of Inertia Herbert F. Bariffi

ONE METHOD of determining the moment of inertia of a mass through its center of gravity is to suspend the mass from a knife-edge. Such an arrange- ment, using a piece of keystock for a knife-edge, is shown in Fig. 1. If this compound gravity pendulum, which is in effect what the setup amounts to, is set swinging and the arc of swing is limited to about 6 deg double ampli- tude, the moment of inertia about the support axis is equal to

(1;

where I , = moinetit of inertia ahout suppor t

W = weight of mass. lb t = oscillation period, sec L = distancr herwcen center of gravity

To find the moment of inertia about center of gravity the following equa- tion can be used.

axis

and support a x i s , in.

where 8 = acceleration of gravity, ft per sec2

Substituting Z, value from Eq (1) -into E9 ( 2 )

Simplifying

This Eq (3) will give values of IC( , in in. Ib sec.

The nomogram illustrated can be used to solve for Z(:(, with value: of t . L and W known. To use this nomo-

I,, = /VI , [O 02.533 t - 0 OOZScYXLI ( 3 )

gram start with a value t on the inner left-hand A scale, pass through a value of L on the A curve to the right-hand, or turning axis. Then, follow a straight line from the turning point through a value of W on the A diagonal, and return to the left-hand axis, reading Zcq on the outer scale. By using the right-hand Z and t scales, and working to the left over the B c u r x for L, the

Fig. 1-Connecting heystock.

rod sumended from

same problem can be solved with larger values of t .

EXAMPLE: Given a pulley of 8 Ib weight, L equals 10 in., t equals 1.5 sec. Find Zcg.

SOLUTION: Starting at t equals 14 on inner A scale of left-hand axis, follow the dotted line through curve A at L equals 10 to the turning line; go back through 1V equals 8 on diag- onal A to I , equals 2.5 in. Ib sec 2.

It should be noted that the effects of L and t are screened in the equation. However, I , , is directly proportional to W , and this fact permits the un- limited extension of the nomograms use.

Suppose, in the above example, W had been 880 Ib. It would appear that this value is too large for the W scale, but Eq (3) can be written:

!eo_ = ~ _ _ WL [0.02533f2 - 0 02588 L] k k

So, in this case, assume R equals 110, and use the nomogram as above with W equaling 8, and read I c g / k equals 2.5. Now merely multiply this value by 110 to find 275 in. Ib sec. This is the true value of Icg .

An unusual feature of this norno- gram is the relation between the values of t and L ; these quantities cannot be assumed at random for trial of nomo- gram accuracy. For example, a scant passing through t = 1.81 and L = 30 also passes through L r 2, thus sug- gesting that two different suspension lengths will satisfy the same result. One of these lengths is an extraneous value as the previously outlined experi- ment will indicate. Furthermore, t is always larger than 0.31936 \ / L when inch and second units are used

30- 14

Nomogram to Determine Parallel Axis Moment of Inertia

100

9.5

9 0

8 5

8 0

7 5

NU 7 0 W 0

n - 65 f

0 60

5 5

3

c L

- cc 0 4 5 0 p1

: 4 5 1

4 0 -

35-

3 0 -

2 5 -

20-

1 5 -

I O -

05-

0 - A sca le s

0 -- 2.0

1.0 -- 2.5

0.5 -

- 1.2 - - 1.4 -- 3.0 - 1.6 - -- 3.5 1.8 -

2.0 -- 4.0 2.1 - 2.2 -- 4.5 N

6

2.4 - n 0' - 5.0 -, c

-

Q 2.3 -

.- 2 .5 - 2.6 -_ I 5.5 2:7 - 2.8 --

2.9 -

3.0 -- 6.5

6.0

3.1 - 3.2 -

3.3 -- 7.5

3.4 -

3.5 -

- 7.0

- 8.0

- 0.5

- 9.0

- 9.5

4.0 -- 'O.0

B scales

Nomograms 30- 1 5

Moment of Inertia of a Prism about the Axis aa

aJ f

m C C L

._ -

.-

r-

30- 16

Chart for Transferring Moment of Inertia

I = I , + wx2 30

35 - To use chart, draw two mutually perpendicular lines on a sheet of trans-

40- parent material. For example, the cross- lines show that, when the weight of the

45- maSS is 12 lb., its moment of inertia Io about a given axis is 30 1b.-in. squared

50- -0 and the distance to another parallel axis

-5 is 2.5 in.; then the moment of inertia I 55- about the second axis is 105 1b.-in. squared. 60 - - IO

- 15 65 - 2 70-

-

.- VI -20 z c -25 2

v1 ._ - aJ - F 75- 2 h 80-

," 85-

G 90-

JZ +

3 0

n 2 .- c -45 0

N 2

+ C

E -60 0

- 70

- 75

- a0 - 85 - 90

1 " " ' ~ " ' I ~ ~ ' l ' " I ' " I I " .

0 :: a 3 g g s

- hl m - + a u J s

. T A b-A'b i log g g 9 2 z S" -

z E g a $ o u

30- 1 8

Accelerated linear Motion

- - = - E - - - G 2s - V V 32.16F T2 2s T W -

1,000 900

5 00

400

3 00

200

G - - L- 100

F Lb. W Lb. ft. per sec. 7 persec. - 10

* = turning point V = velocity at time T, in ft. per sec. S = distance passed through, in f t . T = time during which force acts, in sec. F = accelerating force, in lb.

W = weight of moving body, in lb. G = constant acceleration, in ft. per sec.

Nomograms 30- 1 9

Theoretical Capacity of Gear Pumps and Motors A. E. Maine

711rs NOMOGRAM is based on the equation EXAMPLE-At what speed must a gear pump operate to deliver 50 gal/min assuming 2 in. pitch dia gears, 1) in. tooth width, and each g a r has 10 teeth.

Line I connects the pitch dia and number of teeth. From the point representing 50 gal/rnin, Line I1 is drawn through the intersection of Line I and the reference line. Since Range 1 was used to establish the left end of Line 11, the required speed is read at 4350 rpm.

F = 0.0036 D2Sw/P where F = flow g m

D = pitch $a of gear, in. S = speed, r m UJ = width orgear teeth, in. P = gear tooth factor

-- _-

--

--

-- N 2000--4000 -_

.+ 70--35 - u 0 c 0 e

_-

500- - 1000

Number of teeth 14 13 12 II I O 9 8 e7

30-20

Nomogram for Piston Pumps U! ayo,+s

egg:$ ,o 3 2 R 0 2ecot-e m * r) C V Y -

0 I

9 L

.n = m a . * a In t *) N - 0 N ut s+arvo!a UO+S!d

c - 2 E .-

1 9

- 3.5 C c 'n0.32 - + -32 .C _ _

0.30 - .p r --.a09 d .- a08 5 Sfee /0285- - - - - - - * - r"

0.27 -

t .- z -I

0.10 p Brass 0 3 / - - - - - - - - - t - -c 4- 0 = - 3

Casfiron P 3 0 Rol//edz;"c}Q253 - - i25*: 9 1; - - 2.7

- 2.5 2 -- 0.06 2 -- 0.05 2 0.22 - +

3 - - 2.2

- 2 - - O M = 0.20 - - - -

-- 0.030 0.17 - - 1.7 - -- a025 - 1.5 a15 - - 0.020

0.12 - 0.0 I5 1: i 0.010 0.10 -

30-22

Volumes of Spherical Segments C. P. Nachod

T H I S nomogram is designed for calculating graphically the volume

of a spherical segment such as would be used for the rounded ends of tanks.

The chart is based on the equation

i n which V = volume of the w e n t h = height of segment 7 = radius of sphere d = diameter of sphere

The nomogram gives h up to a hemi- sphere for d up to 10. For a greater range of values, the volume of the spherical segment can be found by pro- portion since the volumes of similar segments are proportional to the cubes of their diameters. For example, let the diameter d of the sphere equal 15 in. and h equal 6 in., v' can be found as follows:

h 6 4 d 15 10 _ = _ = _

Chart shows for d equal to 10, and IL

Then for d equal to 15, and h equal equal to 4, that V equals 184 cu. in.

to 6

Scales are sliown for drawing more h lilies b o that the precise one needed for any calculation can be drawn.

260 - d, = diameter o f sphere

240 - - w o

200 - - I , 500

When h and d are each multiplied by 10, t h e -1.300 volume i s multiplied by 1.000

170-

160 --1.200

I5O --I, I 00

140 - -1,000 +2 130- 4-

120--900 L C .-

0 ''0--800

- 0

70--500 (3 60 -

-400 50 - 40 --300

30 --200 20 -

0 I 2 3 4 5 Scale for h Lines

5

4.5

4

cn 3.5

3

@

3 Q

I L

9,

cn

-

2.5

2

1.5

1

D

Nomograms 30-23

Volumes in Horizontal Round Tanks with Flat Ends

Notes: Shift decimal point on volume scale two points for a one-point shift on diameter scale; one point for a one-point shift on length scale.

Ezample: Tank is 6 ft. in diameter and 15 ft. long. H = 0.9 ft. HID = 0.15. Join 0.15 on From point of intersection with turning line, draw line to

If D had been 0.6 ft., H 0.09 ft., H / D scale with 6 on diameter scale. 15 ft. on the length scale. and length the same, the answer would be 3.00 gal.

The volume scale shows 300 gal.

30-24

- -~

- - -

- - -

Circular section -

- -

\ z 3

- 2 , Rectangular section ,

41 252 Base points ;; - 3

a Z, CT II 8 , - v) 0 0

I

Another Shortcut to Torsion-Bar Design This nomograph supplies values for a mass moment of inertia of torsion bars-or any round or rectangular bar.

D. A. Derse

Answers read directly off the nomograph are tor torsion bars, fixed at one end, made of materlal with a density of 0.280 Ib/in For other materials, multiply moment of inertia I by dcnsityi0.280. If the bar isn't fixed at either end, double the value of I .

Example

Find moment of inertia of a circular torsion bar with diameter d = 0.50 in., length L = 26 in. Draw two lines intersecting at right angles on a piece of transparent paper or plastic. Line up the cross on the three knowns on the nomograph and read off I = 5.8 x 10' Ib-in.-seca.

SYMBOLS

a = Half width of rectangular bar, in. b = Half height of rectangular bar, in. d = Diameter of round bar, in. I = Effective mass moment of inertia, lb-in.-seca L = Length of bar between restraint and plane

M = Mass of bar, Ib-secZ/in. where torque is applied, in.

Table of ContentsSection 30. NomogramsNomogram for Angles in Constructed Shapes Arc Length Versus Central AngleChordal Height and Length of ChordForces in Toggle Joint with Equal ArmsRadical Defections of Rotating DisksPower Capacity of Spur GearsAlignment Chart for Face GearsLinear to Angular Conversion of Gear-Tooth Index ErrorDetermine Parallel Axis Moment of InertiaMoment of Inertia of a Prism about the Axis aaChart for Transferring Moment of InertiaRotary MotionAccelerated Linear MotionTheoretical Capacity of Gear Pumps & MotorsNomogram for Piston PumpsWeight and VolumeVolumes of Spherical SegmentsVolumes in Horizontal Round Tanks with Flat EndsAnother Shortcut to Torsion-Bar Design