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B 3 025 703MACHINERY'S DATA SHEETSREVISED AND RE-ARRANGED IN LIBRARY FORM

No. 1 7

Mechanicsand

Strength of MaterialsPRICE 25 CENTS ,

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CONTENTSRules and Formulas in Mechanics 4Table of Forces on Inclined Planes 12Centrifugal Force per Pound of Revolving Mass >. 14Power of Toggle-joints with Equal Arms 19Tables of Height, Velocity and Time of Falling Bodies 20Strength of Materials 23Formulas for the Strength of Flat Plates. 24Factors for Converting Strength of Wire to Strength in Pounds per

Square Inch 32Ratios of Outside Radius to Inside Radius of Thick Cylinders 35Properties of Sections for Punch and Shear Frames 37

The Industrial Press, 49-55 Lafayette Street, New YorkPublishers of MACHINERY

COPYRIGHT. 191O, THE INDUSTRIAL PRESS, NEW YORK


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MACHINERY'S DATA SHEET SERIESCOMPILED FROM MACHINERY'S MONTHLY DATASHEETS AND ARRANGED WITH

EXPLANATORY NOTES

glP^l No. 17 '; 'Mechanics

and

Strength of MaterialsCONTENTS

Rules and Formulas in Mechanics 4Table of Forces on Inclined Planes 12Centrifugal Force per Pound of Revolving Mass 14Power of Toggle-joints with Equal Arms 19Tables of Height, Velocity and Time of Falling feodies 20Strength of Materials 23Formulas for the Strength of Flat Plates 24Factors for Converting Strength of Wire to "Strength in Pounds per

Square Inch 32Ratios of Outside Radius to Inside Radius of Thick Cylinders 35Properties of Sections for Punch and Shear Thames. . . .*'. ..',. 37

\Copyright. 1910, The Industrial Press, Publishers of MACHINERY,49-55 Lafayette Street. New York City


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MECHANICSRules and Formulas in

MechanicsOn pages 4 to 11, inclusive, is given a

collection of rules and formulas in me-chanics, covering the subject in acomprehensive manner. The subjectstreated are work, momentum, energy,centrifugal force, center of gravity, mo-ment of inertia, radius of gyration, fall-ing bodies, accelerated motion in. astraight line, pendulum, inclined plane,friction, moments, safety valve, pronybrake, strap brake, tackle-block, epi-cyclic gears, countershaft, crank andconnecting-rod, toggle-joint, differentialpulley, suspended cable, friction clutch,parallelogram of forces, triangle offorces, polygon of forces, and stresses ina crane. [MACHINERY'S Reference SeriesNo. 5, First Principles of TheoreticalMechanics.]

Forces on Inclined PlanesOn pages 12 and 13 a table is given to

find the force required for moving a bodyalong an inclined plane. The frictionon the plane is not taken into account.The first column in the table gives theper cent of grade, or the rise in feet per100 feet. The second column gives theangle in degrees and minutes corre-sponding to this rise. The third andfourth columns give the sine and co-sine of the angle, and the fifth columnthe pull in pounds required for movingone ton along the inclined surface. Thelast column gives the perpendicular pres-sure on the plane per ton.For example, assume that it is re-

quired to move a body weighing twotons along an inclined plane having arise of 10 feet in 100 feet. What powerwould need to be exerted, the frictionnot being taken into consideration? Tofind the answer from the table, locate

10 in the left-hand column; followingthe line from 10 horizontally, we findthat the power necessary for one ton is199.2 pounds, and hence for two tons,398.4 pounds. Of course, as the frictionmust be taken into consideration, thepower required to pull the load will haveto be increased according to the char-acter of the sliding surfaces.

Calculating1 Centrifugal ForceCentrifugal force, as it is somewhatvaguely called, is not really a force re-

siding in the revolving body and tendingto urge it away from the center; it israther the extraneous force which hasto be exerted to restrain the body in itscircular path, and prevent it from fol-lowing the tangential direction which itconstantly tries to take in accordancewith Newton's law, viz., that a movingbody will continuously follow a straightline, unless acted upon by some outsideforce. This restraining force is directlyproportional to the radius of the circledescribed by the center of gravity orcenter of mass of the revolving body, itsweight (or more strictly, its mass), andthe square of the number of revolutionsper minute. It may be calculated by theformula: F= 0.000341 WRn*in whichF= centrifugal force, in pounds,W= weight of the revolving body, in

pounds,R= radius of circular path describedby the center of mass of the re-volving body, in feet,

n= number of revolutions per minute.The accompanying diagrams shown on

pages 14 to 17, inclusive, were pre-pared to give directly, without calcula-tions, the centrifugal force per one

(Continued on page 18.)

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MACHINERY'S DATA SHEETSRULES AND FORMULAS IN MECHANICS V

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No. 17 MECHANICSRULES AND FORMULAS IN MECHANICS VI

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No. 17 MECHANICS 11RULES AND FORMULAS IN MECHANICSVIM

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12 MACHINERY'S DATA SHEETS No. 17

TABLE OF FORCES ON INCLINED PLANES.


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No. 17 MECHANICS 13

TABLE OF FORCES ON INCLINED PLANES (Continued).


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14 MACHINERY'S DATA SHEETS No. 17CENTRIFUGAL FORCE PER POUND OF REVOLVING MASS I

Contributed by W. H. Acker, MACHINERY'S Data Sheet No. 60. Explanatory note : Page 3.


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No. 17 MECHANICS 15CENTRIFUGAL FORCE PER POUND OF REVOLVING MASS II



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16 MACHINERY'S DATA SHEETS No. 17CENTRIFUGAL FORCE PER POUND OF REVOLVING MASS III

50 100 150CENTRIFUGAL FORCE PER LB. OF REVOLVING MASS

Contributed by W H. Acker. MACHINERY'S Data Sheet No. 60. Explanatory note : Page 3.


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No. 17 MECHANICS 17CENTRIFUGAL FORCE PER POUND OF REVOLVING MASS IV

10 20 30 40 50 60 70 80CENTRIFUGAL FORCE PER LB. OF REVOLVING MASS



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18 MACHINERY'S DATA SHEETS No. 17pound for various radii and revolutionsper minute. The way in which they areused will be readily understood. Onthe graduations at the left, the hori-zontal line corresponding to the givenradius, in feet or inches, is located. Thisline is followed toward the right untilit intersects the diagonal line which cor-responds to the required number ofrevolutions per minute. From this pointof intersection, a vertical line is droppedto the base margin, where the centrifugalforce is directly read in pounds. Thisresult multiplied by the weight of thebody will give the required force for thecase in hand.

It will be observed that the diagonalsfor the rate of revolution are terminatedby a curve at the points where the in-tersections give a velocity of 5,000 feetper minute. This velocity is about thelimit for cast-iron flywheel rims andsimilar parts, whose strength can beshown to be a function of the velocityand the physical properties of the ma-terial only. For speeds greater thanthis, each individual case should be cal-culated from data giving exact informa-tion on every point required.As a simple case, suppose it is re-quired to find the centrifugal force inthe rim of a flywheel used on a gas en-gine. The rim of the flywheel weighs500 pounds, the mean radius is 3 feet,and the speed is 250 R. P. M. Referringto page 17, we find the centrifugal forceper pound at 3 feet radius and 250R. P. M. to be 64. Hence the sum of thecentrifugal forces in the rim is 64 X 500= 32,000 pounds. The force resisted bythe two sections of the flywheel rim,however, is not equal to the sum of allthe radial centrifugal forces, but to theirresultant, which equals 32,000 -r- TT =10,200 pounds, approximately. This lat-ter force is the force resisted by the twosections of the rim at the extremities ofa diameter. A suitable factor of safetyhaving been applied, we readily obtainthe sectional area of rim necessary,kmowing the strength of the material.

Suppose it is required to find thecentrifugal force exerted by the knife ofa wood planer, the knife weighing 28pounds, its center of mass being at aradius of 5 inches, and running at aspeed of 1,800 R. P. M. From page 14we find the centrifugal force per poundto be 460, at 5 inches radius and 1,800R. P. M., and the total centrifugal forceis 28 X 460= 12,880 pounds. Hence thebolts holding the knives in place mustbe of sufficient size to overcome thisforce, allowing a suitable factor ofsafety.The values as obtained from the dia-

grams will be sufficiently accurate forall computations of this kind, as in allcases a large factor of safety is employedin proportioning parts to withstand thestrain. [MACHINERY, September, 1905,Systematic Method of Computing Cen-trifugal Stresses; MACHINERY'S Refer-ence Series No. 40, Flywheels.]

Power of Toggle -joints withEqual ArmsThe toggle-joint, while one of the sim-

plest mechanisms to construct, some-times causes considerable confusionwhen attempts are made to calculate theforces that are obtained through its ap-plication. The table on page 19 is in-tended to obviate some of the difficulties.When the power applied to a toggle-jointwith equal arms, and the angle a asshown in the illustration on page 19, areknown, the resistance in the directionof arrow R may be found from the table.First measure the angle a and locate itin the table in the column headed"Angle." Then find the correspondingcoefficient in the next column. This co-efficient is the ratio of resistance to thepower applied, and if the given poweris multiplied by this coefficient, we ob-tain the resistance, or the power ex-erted in the direction of the arrow R.The friction is neglected in these cal-culations, but in making practical ap-plications of the principle, allowanceshould be made for friction by increas-

( Continued on page 34.)


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No. 17 MECHANICS 19POWERS OF TOGGLE JOINTS WITH EQUAL ARMS.


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20 MACHINERY'S DATA SHEETSHEIGHT, VELOCITY AND TIME OF FALLING BODIES I

No. 17


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No. 17 MECHANICS 21HEIGHT, VELOCITY AND TIME OF FALLING BODIES II


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22 MACHINERY'S DATA SHEETSHEIGHT, VELOCITY AND TIME OF FALLING BODIES III

No. 17


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No. 17 STRENGTH OF MATERIALS 23


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No. 17 STRENGTH OF MATERIALSFORMULAS FOR STRENGTH OF FLAT PLATES II


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No. 17 STRENGTH OF MATERIALS

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No. 17 STRENGTH OF MATERIALS


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32 MACHINERY'S DATA SHEETS No. 17CONVERTING STRENGTH OF WIRE TO STRENGTH PER SQUARE INCH I

Diam.


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No. 17 STRENGTH OF MATERIALS 33CONVERTING STRENGTH OF WIRE TO STRENGTH PER SQUARE INCH II

^Diam.


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34 MACHINERY'S DATA SHEETS No. 17ing the power applied to the toggle-joint.] MACHINERY'S Reference Series No. 22,Calculations of Elements of Machine De-sign, Chapter VII, Toggle-joints.]

Falling- BodiesUnder the influence of gravity alone,all bodies fall to the earth with the

same velocity. ' A body falling freely,from rest to the earth, acquires duringthe first second a velocity of about 32feet per second, at the end of the secondsecond a velocity of about 32 + 32= 64feet, and at the end of the third second,64 + 32= 96 feet per second, etc. Theacceleration due to gravity varies slight-ly according to the location of the place

where it is measured, and is 32.16 feetper second in the vicinity of New York.A definite relation thus exists betweenthe distance a body would fall in a giventime, and the velocity it attains duringthe fall. On pages 20, 21, and 22 aregiven formulas for the relation betweenthe head in feet, the velocity in feet persecond, and the time in seconds, and atable is calculated for heads varyingfrom 0.01 foot to 1,000 feet, giving thevelocities obtained by the falling body atthe end of its fall, and the time re-quired for traversing the given distance.[MACHINERY'S Reference Series No. 5,First Principles of Theoretical Me-chanics, Chapter VIII, Palling Bodies.]

STRENGTH OP MATERIALSTables and Formulas for the

Strength of MaterialsOn page 23 is given a table of the

strength of ordinary materials used inthe mechanical trades. In the lowerpart of the table the elementary formu-las for the strength of materials, to-gether with a formula and values forcalculating the strength of columns sub-jected to bending, are given. [MA-CHINERY, June, 1905, Notes on theStrength of Beams, Plates and Columns;MACHINERY, January, 1908, The ElasticLimit and the Testing of Materials;MACHINERY'S Reference Series No. 22,Calculations of Elements of Machine De-sign, Chapter I, The Factor of Safety.]

Formulas for the Strength ofFlat PlatesThe machine designer is often called

upon to carry out designs consisting inpart of flat surfaces, such as plates sup-ported or fixed at the edges, with orwithout intermediate supports or ribs.Exact formulas for finding the bending

moments of flat plates, and their resist-ance to the stresses created by pressuresnormal to their surfaces, have not beendetermined. The formulas given by dif-ferent authorities are founded on as-sumptions, and they should be consid-ered as approximations only; theyshould be used with caution, as the re-sults obtained are not likely to be veryaccurate. In devising such formulas, allthe assumptions should be made to erron the safe side.

Square PlatesA square cast iron plate fixed or

rigidly heid at the edges and loaded witha uniformly distributed load, or a loadconcentrated at its center, would belikely to fail as shown in Fig. 1. Itwould first fracture along the diagonallines from A to B, and then fail at ornear the fixed edges along lines B B.The plate, of course, might also shearoff along the edges B B, depending uponthe method of loading the span L be-

( Continued on page 36.)


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No. 17 STRENGTH OF MATERIALS 35

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No. 17 STRENGTH OF MATERIALS 37PROPERTIES OF SECTIONS FOR PUNCH AND SHEAR FRAMES


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MACHINERY'S DATA SHEETS No. 17ent hand-books, and as these hand-booksmay not always be easily procurable, thecollection will undoubtedly prove usefulto designers who, at some time or other,may be called upon to lay out a designinvolving square or rectangular flatplates.

Formulas for Strength of FlatCircular PlatesWhen a flat circular plate becomes de-

formed under the action of a load ap-plied normally to its surface, the upperfibers of the material are subjected tocompressive stresses both radially andcircumferentially, while the lower fibersare subjected to tensile stresses both ra-dially and circumferentially. This iscaused by the fact that the plate, whenbending, tends to assume a sphericalshape. The deformation of the platetends to decrease the length or circum-ference of the upper fibers, while it tendsto increase the length of the lower fibers.On pages 29, 30 and 31, several sets offormulas for circular plates are given,selected from different authorities. Theformulas are presented as given by thevarious writers, and then the differentquantities "safe load," "unit stress,"''thickness of plate," etc., as deducedfrom the given formulas, are shown.

Converting 1 the Strength of Wireinto Strength in Poundsper Square InchOn pages 32 and 33 a table is given of

factors for converting the strength ofwire into strength in pounds per squareinch. One column is given headed"Diam.," and one "Factor." The strengthof the wire of a given diameter should bemultiplied by the factor given, in orderto obtain the strength per square inch.For example, if we know that a wire0.035 inch in diameter can sustain aload of 150 pounds before breaking, thenthe strength per square inch of this wireis 150 x 1040 = 156,000 pounds persquare inch. The factor 1040 is foundopposite the diameter 0.035 on page 32.The table is very convenient for calcu-

lations of this kind, as it saves the find-ing of the area of the wire.

*Ratio of Outside Radius to InsideRadius of Thick CylindersOn page 35 a table is given of ratio of

outside to inside radii of thick cylinders.As an example, assume that a cylinderis required of 10-inch inside diameter towithstand a pressure of 2500 pounds persquare inch. As the material is castiron, the allowable stress in this casewould be 6000 pounds per square inch.To solve the problem, locate the allow-able stress per square inch in the left-hand column of the table, and the work-ing pressure at the top of the columns.Then we find that the ratio between theoutside and inside radii in this caseshould be 1.558, and hence the outsidediameter of the cylinder should be 10 X1.558, or about 155/8 inches. The for-mula given in the upper right-hand cor-ner of the table is the one on which thetable is based. [MACHINEEY, July, 1909,Thick Cylinders; MACHINERY'S Refer-ence Series No. 17, Strength of Cylin-ders.]

Properties of Sections for Punchand Shear FramesThe convenience of the tables of prop-

erties of rolled steel sections as pub-lished in the structural steel makers'hand-books impresses anyone who hasoccasion to compute the strength of suchmaterial. Machine frames cannot bestandardized sufficiently to permit ofsimilar tables which would be useful tothe designer of machinery, but it is witha view to the simplifying of the work ofselecting proper sections for punch andshear frames that the table on page 37has been compiled. The values werecomputed by slide rule, and it is believedthat they are accurate enough for allpractical purposes.To illustrate the use of the table, con-sider the punch frame shown diagram-matically in Fig. 3. The "reach," or dis-tance from the center line of punch tothe back of gap, is 24 inches. Assume


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No. 17 STRENGTH OF MATMfALS" ' 39that the maximum pressure, P, tendingto force the jaws apart is that due topunching a 1-inch circular hole in softsteel plate 1 inch thick, or say about157,000 pounds. Consider the section atTX. The action of P is such as to pro-duce a tensile stress on the section tothe left of the neutral axis N, with acompressive stress to the right of N,both due to flexure; and, besides, there

Fig. 3. Typical Punch Frameis a tensile stress distributed uniformlyover the section.

It is usually sufficient to determine themaximum tensile stress to the left of N-Maximum tensile stress = flexure ten-sile stress -{- uniformly distributed ten-sile stress.Flexure tensile stress =

Moment of P about NTensile section modulus of section TXUniformly distributed tensile stress =

PArea of section TX

Assume 3000 pounds per square inchas the allowable maximum tensile stressfor cast iron.Assume that D in Fig. 4 is about 30ID

inches. Then = 10% inches. If the20stress due to flexure only is considered,the required section modulus for tensionwould be

157,000 X (24 + 10%) = 1800, about.3000As no allowance has been made for

the additional tensile stress uniformlydistributed over the section, a sectionmust be selected whose tension modulusis somewhat greater than 1800, say about2500.The table on page 37 gives a great

variety of shapes and proportions, theonly dimensions common to all being thedepth of section, D, and the distance ofthe neutral axis from the extreme ten-sion fiber. This location of the neutralaxis represents average practice and in-sures an economical distribution ofmetal. Generous fillets and rounded cor-ners should, of course, be used on thefinal section.Suppose that a deep narrow section is

desired, similar to that in Fig. 4, inwhich the dimensions are as follows:D => 10 inches,. B = 7 inches, 6 = 4

Fig. 4. Section of Punch Framet

inches, #= 1.98 inch, = h= % inch,2A (area) =26.3 square inches, and &t(section modulus for tension) =78.6.This section, taken from the table, isnot, of course, large enough, but a simi-lar one which is large enough may befound easily as follows:If two sections A and B are similar Inall respects, then


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40 MACHINERY'S DATA SHEETS No. 17Area of AArea of B

(Any dimension of A ) 2, and

( Corresponding dimension of B ) 3Section modulus of ASection modulus of B

(Any dimension of A) 3(Corresponding dimensions of B) 3

Required section modulusModulus of section from the table

(Required!)) 8

(D of section from the table)2500 (Required!)) 8

78.6Hence, = This

(10) 3last equation solved for D gives D= 31.7inches, nearly. As the large section willbe exactly similar to the small one, thearea of the large section is found fromthe equation:Area of required section ( 31.7 ) 2

26.3"

(10) 2and area= 264 square inches, about.7

The neutral axis will be X 31.7, or2011.1 inches from the extreme tensionfiber.This trial section may now be tested

for the maximum stress on it.Flexure tensile stress=157,000 X (24 + 11.D = 2200 pounds2500

per square inch, about.Uniformly distributed tensile stress=157,000

-= 594 pounds per square inch,264

about.The total tensile stress is, therefore,

2794 pounds per square inch.If this result had not been near enoughto the 3000 pounds per square inch as-

sumed, another section could have beenselected and worked out in the same wayto get a closer result. The depth D of

the required section being 31.7 inches ascompared with 10 inches of the similarsection given in the table, each of thedimensions of the required section willbe 3.17 times the corresponding onegiven in the table. Hence D= 31.7, say32 inches, B= 22.2, say 22 inches, b =12.7, say 13 inches, H= 6.3, say 6%inches, t/2= h= 1.98, say 2 inches. Thewebs thicken gradually from the neutralaxis to the tension flange so as to avoidtoo sudden a change in the section. Forselecting the section at TY, Fig. 3, theprocedure would be the same except thatthere would be no uniformly distributedstress to be added as in the case of sec-tion TX.The method here given for determin-

ing the stress on the section TX is theone which has been commonly used. Itis correct for straight beams and untilrecently has been considered approxi-mately correct for curved beams. Inves-tigations by Bach and others, however,indicate that the maximum stresses incurved beams of ductile material arevery much greater than those found bythe method used here; but a quick andaccurate way of applying the new theoryto general sections has not yet been de-vised. Cast iron punch frames whosesections have been determined by the oldanalysis have apparently stood up undertheir loads for years, although the maxi-mum stresses in them as determined bythe new analysis are so great as tothreaten serious injury. Until the newtheory has been more thoroughly verifiedby experiments, however, the old methodwill doubtless be used by most designers.[MACHINERY, November, 1903, Design ofPunch and Riveter Frames; February,1907, Strength of Punch and ShearFrames; August, 1908, The Designing ofMachine Frames; December, 1909, TheDesign of Curved Machine Members un-der Eccentric Load; June, 1910, Limi-tations of the Common Theory of Flex-ure; MACHINERY'S Reference Series No.24, Examples of Calculating Designs,Chapter II, The Designing of MachineFrames.]


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THIS BOOK IS DUE ON THE LAST DATESTAMPED BELOWAN INITIAL FINE OF 25 CENTSWILL BE ASSESSED FOR FAILURE TO RETURNTHIS BOOK ON THE DATE DUE. THE PENALTYWILL INCREASE TO SO CENTS ON THE FOURTHDAY AND TO $1.OO ON THE SEVENTH DAYOVERDUE.

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TJJfl&

UNIVERSITY OF CALIFORNIA LIBRARY


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Wo. 16. Machine Tool Drives. Speedsand Feeds of Machine Tools; Geared orSingle Pulley Drives; Drives for HighSpeed Cutting Tools.No. 17. Strength, of Cylinders For-mulas, Charts, and Diagrams.No. 18. Shop Arithmetic for the Ma-chinist. Tapers Gears; CuttingSpeeds; Feeds; Indexing-; Gearing for Cut-ting Spirals; Angles.No. 19. Use of Formulas in Mechanics.With numerous applications.No. 20. Spiral Gearing1 . Rules, Formu-

las, and Diagrams, etc.No. 21. Measuring1 Tools. History andpment of Standard Measurements;

Si>H;il Calipers; Compasses; MicrometerTools; Protractors, etc.No. 22. Calculation of Elements ofMachine Design. Factor of Safety;Strength of Bolts; Riveted Joints; Keysand K2yways; Toggle-joints.No. 23. Theory of C?ane Design. JibCranes; Calculation of Shaft, Gears, andBearings; Force Required to Move Crane

Trolleys; Pillar Cranes.No. 24. Examples of Calculating* De-signs. Charts in Designing; Punch andRiveter Frames; Shear Frames; Billetand Bar Passes; etc.No. 25. Deep Hole Drilling-. Methodsof Drilling; Construction of Drills.No. 26. Modern Punch, and Die Con-struction. Construction and Use of Sub-press I nes: Moch rn Blanking Di

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Rezistenta materialelor - mechanicsstrengt