Reading 1

68

I\I\echonicol I\I\echonicol Engineering Design

One of the the main objec objectiv tives es of this this book book is to to descri describe be how specifi specificc machin machinee compon component entss functi function on and how how to design design or speci specify fy them them so that that they they functio function n safely safely withou withoutt failin failing g struct structura urally lly.. Althou Although gh earlie earlierr discus discussio sion n

has descri described bed struct structura urall

streng strength th

in tenus tenus of 

load load or stress stress versus versus streng strength, th, failur failuree of funct function ion for struct structura urall reason reasonss may arise arise from from other factors factors such as excessive excessive deformatio deformations ns or deflecti deflections. ons. Here Here it is is ass assum umed ed that that the reade readerr has has compl complet eted ed basi basicc cour course sess in static staticss of rigid rigid bodies bodies and mechan mechanics ics of mater material ialss and is quite quite famili familiar ar with with the analy analysis sis of loads, loads, and the stre stresse ssess and defor deformat mation ionss associ associate ated d with with the basic load load states states of simple simple prism prismatic atic elem elemen ents ts.. In this this chapt chapter er and and Chap Chap.. 4 we will will revi review ew and and exte extend nd thes thesee topi topics cs brie briefl fly. y. Comp Comple lete te deri deriva vati tion onss will will not be prese present nted ed here here,, and and the the read reader er is urge urged d to retu return rn to basic textbook textbookss and notes on these these subjects. subjects. This This chapter chapter begins begins with with a review review of equili equilibri brium um and free-b free-body ody diagra diagrams ms associ associate ated d with with load load-c -car arry ryin ing g comp compon onen ents ts.. One One must must unde unders rsta tand nd the the natu nature re of forc forces es befo before re attemp attemptin ting g to perfor perform m an exte extensi nsive ve stress stress or defle deflectio ction n analys analysis is of a mecha mechanic nical al pone ponent nt.. An extr extrem emel ely y

usef useful ul too/ too/ in hand handlin ling g

singularity ty functions. functions. employs Macaulay or singulari

disc discon onti tinu nuou ouss

Sing Singul ular arit ity y

load loadin ing g

func functi tion onss

comcom-

of stru struct ctur ures es

are are descr describ ibed ed

in

Sec. 3-3 as appl applied ied to the shear shear forces forces and bend bending ing momen moments ts in beam beams. s. In Chap. Chap. 4, the the use use of sing singul ular arit ity y func functi tion onss will will be expa expand nded ed to show show thei theirr real real powe powerr in hand handli ling ng deflec deflectio tions ns of comp complex lex geome geometry try and static staticall ally y indete indetermi rminat natee proble problems ms.. Mach Machin inee comp compon onen ents ts tran transm smit it forc forces es and and motio motion n from from one one poin pointt to anoth another er.. The The transm transmiss ission ion of forc forcee can be envis envision ioned ed as a flow flow or force force distri distribut bution ion that that can be furfurther ther visua visualiz lized ed by isolat isolating ing intern internal al surfac surfaces es within within the comp compone onent. nt. over over a surfac surfacee leads leads to the the concept concept of stress stress,, stress stress compon component ents, s, tions tions (Mohr's (Mohr's circle circle)) for all all possibl possiblee surfac surfaces es at a point. point.

Force Force distri distribut buted ed

and stress stress transf transform ormss-

The The remain remainde derr of the the chapt chapter er is dev devot oted ed to the the stre stress sses es asso associ ciat ated ed with with the the basi basicc loadin loading g of pris prismat matic ic elemen elements, ts, such such as unifor uniform m loadin loading, g, bendin bending, g, and torsion torsion,, and topics topics with with major major design design ramifica ramificatio tions ns such such as stress stress concen concentra tratio tions, ns,

thinthin- and thickthick-wal walled led

pressurize pressurized d cylinders, cylinders, rotating rings, rings, press and shrink shrink fits, thermal thermal stresses, stresses, curved curved beams, beams, and COntactstresses.

3-] 3-]

Equilibrium Equilibrium and Free-Body Free-Body Diagrams Diagrams  Eq uilib ui lib riu m The word s.yslem. will be used used to deno denote te any any isolated  part part or porti portion on of a mach machin inee

or

S(~c S(~c~u ~u.r .ree-lll lllcl clud udll lllg lg.. all all of it if ~es ~esir ired ed-t -tha hatt we wish wish (0 stud study. y. A syste system, m, unde underr this this d.ef.t d.ef.toUl oUlOn, On,may m ay consis consistt of a partic particle, le, severa severall Partic Particles les,, a pan of a rigid rigid body, body, an entire entire ngld ngld body, body, or even severa severall rigid rigid bodies bodies.. [~we [~we assume assume that that the system system to be studie studied d is moti motionl onless ess or, at at most, most, has constan constantt veIOC veIOC~ty ~ty,,lh~n.t l h~n.th~ h~ system system has zero zero accele accelerat ration ion.. Under Under this this conditio condition n the system system is said said static equilibriu equilibrium m is also ~obe 11 1 eqlllhbrllf~ll: T.he phrase static also used used to imply imply that that the system system IS til rest. For eqUihbnum the fa d . thOI ,fces an moments acung on the system balance such

(3-1) (3-2) force and h which States States that the Slim of all force t e Slim of all mome1/l vectors vectors acti system system in equilib equilibrium rium is zero. zero. acting ng upon upon a

Load and Sness Analysis Analysis

69

 Free-Body Diagrams We can greatly greatly simplify simplify the analysisof analysisof a very complex complex structure structure or machine machine by successively successively free-body diagrams diagrams.. isolating each element and studying and analyzing it by the use of free-body When all the members have been treated in this manner, the knowledge can be assembled to yield information information concerning concerning the behavior of the total system. system. Thus, free-body free-body diagramming is essentially a means of breaking a complicated complicated problem into manageable manageable segments, segments, analyzing analyzing these simple problems, problems, and then, usually, putting the the information information together together again. Usin Using g freefree-bo body dy diagr diagram amss for for force force analy analysi siss serv serves es the the foll follow owin ing g impo import rtant ant purposes: • The diagra diagram m establ establishe ishess the direct directions ions of reference reference axes, axes, provides provides a place to record record the dimensi dimensions ons of the subsyste subsystem m and the magnitude magnitudess and directi directions ons of the known known forces, forces, and helps in assuming assuming the directions directions of unknown unknown forces. • The diagram simplifies simplifies your thinking thinking because it provides provides a place place to store one thought thought while proceeding proceeding to the next. next. The diagram diagram provide providess a means of commun communica icating ting your your thought thoughtss clearl clearly y and unambiguously biguously to other people. • Careful Careful and complete complete construction construction of the diagram diagram clarifies clarifies fuzzy thinking by bringing out various various points points that that are not always always apparent apparent in the statem statement ent or in the geomet geometry ry of the total problem. Thus, the diagram aids in understanding all facets of the problem. • The diagram diagram helps helps in the planni planning ng of a logical logical attack attack on the problem problem and in setting setting up the mathematical mathematical relations. relations. • The The diag diagra ram m helps helps in reco recordi rding ng prog progre ress ss in the soluti solution on and and in illust illustra rati ting ng the the methods methods used. The diagram allows others to follow your reasoning, reasoning, showing all forces.

EXAMPL EXAMPLE E 3-1

Solution

Figure Figure 3-la 3-la shows shows a simplifi simplified ed renditio rendition n of a gear gear reducer reducer where where the input input and output output A B an a n d C D co, co , W  o, shafts are rotating at constant speeds and respectively. The input and output torques (torsional moments) are T , =240 lbf . in and T o > respectively. The shafts are supported supported in the housing housing by bearing bearingss at A, B, C, and D. The pitch radii of gears G[ and G2 are "t =0.75 in and r z =1.5 in, in, respective respectively. ly. Draw the free-body diagrams diagrams of  each member member and determi determine ne the net reaction reaction forces forces and moment momentss at all points. points. First, we will list all simplifying simplifying assumptions. assumptions. 1

Gears G1 and G2 are simple spur gears with a standard standard pressure angle ¢=20°

2

(see Sec. 13-5). The bearin bearings gs are self self-ali -aligni gning ng and the shaf shafts ts can be conside considered red to be simply simply

3 4 5

supported. The weig weight ht of each memb member er is negligi negligible ble.. Fric Fricti tion on is is negl negligi igibl ble. e. The The moun mounti ting ng bolt boltss at at E, F, H, and I  are the same size.

The separate free-body diagrams of the members are shown in Figs. 3-Ib-d. Note that action and reactio reaction, n, is used extensi Newton Newton's 's third third law, law, called called the law of action extensivel vely y where where each member member mates. mates. The force transmit transmitted ted between between the spur gears gears is not tangentia tangentiall but at the pressure angle ¢. Thus, N  =F  Ian ¢.

70

Iv\echanical

Engineering

Design

. > 

fa)

G eM

, , , , , ,

/R

t;

m1uce:r (1))

R A • 

Gear box

~"24{)lbf'in

(rllllpUl Yiafl

I

F ',

(d) Output .haft

Figure 3-1 (0) Gaor reducer,

lb-dl free-body

d ia g ro m s .

D i og r am s o r e n o t d ro w n t o s c al e .

Summing moments about the x axis of shaft AB in Fig. 3-1d  gives

LM,

=F(0.75) -  240 =0

F ~  320 fbi The normal force is N  = = 320 tan 20° =116.51bf. Using the equilibrium equations for Figs. 3-1c and d, the reader should verify that: = 192 Ibf  R c z : :: : =192 lbf  R It < . : : : : :  69.91bf, R By  =128 lbf  R Bz = 46.6 Ibf, R

AI   R ,

cy Dy  = = 128 lbf  R D, ::::: 46.61bf, and To :::::4 80 lbf . in. The direction of the output 69."9lbf  R torque To is opposite W o because it is the resistive load on the system Opposing the motion W  _ o

Note in Fig. 3-lb the net force from the bearing reactions is zero whereas the net

momentaboul1hexaxis is 2.25 (192) + 2.25 (128) =:: 720 Ibf. in. This value is the same 240 + 480 = 720 lbf . in, as shown in Fig. 3-la. The reaction forces Rf., Rf", R  , and Rr, from the mounting bolts cannot be determined from the equilibrium equations as there are too many unknowns. Only three equations are y available, F  :::= F 7 . =  M ", :::= O . In case you were wondering about assumption 5, here is where we will use it (see Sec. 8--12).The gear box tends to rotate about the .r axis because of a pure torsional moment of 720 lbf . in. The bolt forces must provide as T;

+ T  :::= H  Q 

L

L

L

Load end Stress Analysis

71

an equal but opposite torsional moment. The center of rotation relative to the bolts lies at the centroid of the bolt cross-sectional

areas. Thus if the bolt areas are equal: the center

of rotation is at the center of the four bolts, a distance of  J(4/2)2

=

=

=

+ (5/2)2

=3.202 in

R), and each boll force RH  R, from each bolt; the bolt forces are equal (RE  =RF  is perpendicular to the line from the bolr to the center of rotation. This gives a net torque from the four bolts of  4R(3.202)

3-2

=720. Thus, RE 

=

RF 

=

Shear Force and Bending Moments Figure 3-2a shows a beam supported

by reactions

RH 

=

R/  =56.22 lbf 

in Beams

Ri and R2 and loaded by the con-

cenrrated forces F v , F 2  , and F3 . If the beam is cut at some section located at x =Xl and the left-hand portion is removed as a free body, an internal shear force V  and bending moment M  must act on the cut surface to ensure equilibrium (see Fig. 3-2b). The shear

force is obtained by summing the forces on the isolated section. The bending moment is the sum of the moments of the forces to the left of the section taken about an axis through the isolated section. The sign conventions used for bending moment and shear force in this book are shown in Fig. 3-3. Shear force and bending moment are related by the equation dM  v ~ dx

Sometimes q(x)

the bending is caused by a distributed

is called the load intensity with units afforce

Figure3-2 Free-bodydiogrom of simplysupported beam with V ond

M 

shown in positive directions.

(b)

(0)

Figure 3-3 Sign conventions for berlding and shear

Positive bending

Negative bending

Positive shear

Negative shear

I Figure3-4

q(x)

Distributed lood on beam.

~ ,

(3-31 load q(x), as shown in Fig. 3-4;

per unit length and is positive in the

72

fVlechonicol Engineering De~ign

positive y direction. It can be shown that differentiating

Eq. (3-3) results in

dV d 2 M  dx = dx2  =q

(3-4)

Normally the applied distributed load is directed downward Fig. 3-6). In this case, W =-o -

and labeled w (e.g., see

Equations (3-3) and (3--4) reveal additional relations if they are integrated. if we integrate between, say, XA and XB, we obtain

Thus,

(3-5) which states that the change in shear force from A to B is equal to the area of the loading diagram between XA and xe. L n a similar manner,

(3-61 which states that the change in moment from A to B is equal to the area of the shear.  force diagram between XA and  Xe.

Table 3-1 Singularity ffv\ocQuloyt} Concentrated moment

Functions

x¥:-o 

{X-O)-2=O 

(unit doublet)

{x- 0)-2  =±oo  /(X-O}-2dx=

Concentraled force

(x_o)-l 

xi=o 

(X-O)-l=O 

(unit impulse)

t

x= a

X= a 

(X_o}-l =+00

dx=  {x-o}D 

 /(X_O)-l

Unit slep (x_o)D= 

L :--+-----'--, ,

 j

0 x 1

(7 3),

so always order your principal stresses. Do this in any computer code you generate and you'll always generate

3-8

Tmax·

Elastic Strain Normal

strain E is defined and discussed

in Sec. 2-1 for the tensile specimen

and is

given by Bq. (2-2) as e ::::: 8/  I, where 8 is the total elongation of the bar within the length I. Hooke's law for the tensile specimen is given by Eq. (2-3) as a = E, where the constant

[3-17}

E  is called Young's modulus or the modulus of elasticity.

'For development of this equation and further elaboration of  three-dimensional stress transformations

see:

Richard G. Budynas, Adml1ced Strength and Applied siress Anaiysu. 2nd ed., McGraw-Hill, New York, 1999. pp. 46--78. 2Nole the difference belween this notation and that for a shear stress, say, not accepted practice. but it is used here to emphasize the distinction.

"[xy· 

The use of the shilling mark is

When a material is placed in tension, there exists n?t only.an axial st~ain, bU~also negative strain (contraction) perpendicular to the axial st~am. Assuillm.g a h?ear, isotropic material, this lateral strain is proportional to the axial strain. If the axial direction is x, then the lateral strains are {O y =(O z =-VE....The constant of proportionality v is called Poisson's ratio, which is .about 0.3 for most structural metals.

homogeneous,

See Table A-5 for values of  v for common matenals. If the axial stress is in the x direction, then from Eq. (3-17)

",

 €... =

(3-18)

E 

For a stress element undergoing a ..., 0y, and a z simultaneously, are given by

If 

"&

 ,,~

yDf4

Put.,

cU6lf-/N 

j   I

rtll'lY ~.y

THE'7V fHA/US

ol?IVS6  J J 7 /lC1f  $U'~'()$(TION. Fo lS"A~/tG. '.AD' IV' I,S (;I.. ""  TO (J   filS

the normal strains

E x= ~[ax-v(ay+az)]

1

(O y =E  [a y ~ v( a x

+ az)

]

.

t

(3 19) -

 / 

1

(o z = E  [ a z - v ( a . . . + a y )] '"

 

Shear strain y is the change in a right angle of a stress element when subjected-to pure shear stress, and Hooke's law for shear is given by

r=Gy

13-20)

where the constant G is the shear modulus of elasticity or modulus of rigidity.

It can be shown for a linear, Isotropic, homogeneous stants are related to each other by

material, the three elastic con./  

 E~2G(l+")

3-9

/ 

(3-21)

Uniformly Distributed Stresses The assumption of a uniform distribution of stress is frequently made in design. The result is then often called pure tension, pure compression, or pure shear, depending upon how the externalload is applied to the body under study. The word simple is sometimes used instead of  pure to indicate that there are no other complicating effects. The tension rod is typical. Here a tension load F  is applied through pins at the ends of  the bar. The assumption of uniform stress means that if we cut the bar at a section remote from the ends and remove one piece, we can replace its effect by applying a uniformty distributed force of magnitude aA to the cut end. So the stress a is said to be uniformly distributed. It is calculated from the equation

F  a= -   A

(3-22)

This assumption of uniform stress distribution requires that: • The bar be straight and of a homogeneous

material

The line of action of the force contains the centroid of the section The section be taken remote from the ends and from any discontinuity change in cross section

or abrupt

F

s, b

_ c

load and Stress Analysis

85

For simple compression, Eq. (3-22) is applicable with F  normally being considered a negative quantity. Also, a slender bar in compression may fail by buckling, and this possibility must be eliminated from consideration before Eq. (3-22) is used.' Use of the equation F 

r= -

 A

(3-231

for a body, say, a bolt, in shear assumes a uniform stress distribution too. It is very difficult in practice to obtain a uniform distribution of shear stress. The equation is included because occasions do arise in which this assumption is utilized.

Normal Stresses for Beams in Bending The equations for the normal bending stresses in straight beams are based on the fol tJ  lowing assumptions: StiI lIKf"cH 1.6N¥1Ijrl,tt:¥l1'FS$ prl 

/"7

1

The beam is subjected to pure bending. This means that the shear force is zero, and that no torsion or axial loads are present. The material is isotropic and homogeneous. The material obeys Hooke's law. The beam is initially straight with a cross section that is constant throughout the

5 6

beam length. The beam has an axis of symmetry in the plane of bending. The proportions of the beam are such that it would fail by bending rather than by

7

crushing, wrinkling, or sidewise buckling. Plane cross sections of the beam remain plane during bending.

In Fig. 3-13 we visualize a portion of a straight beam acted upon by a positive bending moment M  shown by the curved arrow showing the physical action of the moment together with a straight arrow indicating the moment vector. The x axis is coincident with the neutral axis of the section, and the .rz plane, which contains the neutral axes of all cross sections, is called the neutral plane. Elements of the beam coincident with this plane have zero stress. The location of the neutral axis with respect to the cross section is coincident with the centroidai axis of the cross section.

g ur.3-13 n ig h t b ,

dl ng

'

earn In positive

lSee Sec. 4--1 1.

86

M .ecnonicol Engineering Design

Figure 3-14

Compression

=r------,~

Bending sresses according to

Eq 13-241.

Neutral axis. Cenlmidal axis

C 

-.1_<

Tension

The bending stress varies linearly with the distance from the neutr~l axis, y, and is gi~n~ "  My ax =--/-

.(3-24(

where I  is the second moment of area about the

J 

1= The stress distribution given by Eq. (3-24)

z

axis. That is

ldA

, (3-25)

is shown in Fig. 3-14.

The maxi.mu~ magm-

tude of the bending stress will occur where y has the greatest magnitude. D e s ig n at in g

a

max as the maximum magnitude of the bending stress, and e as the maximum magnitude of  y Me 

-r-

umax =

(3-260)

Equation (3-24) can still be used to ascertain as to whether um ax is tensile or compressive. Equation (3-200) is often written as

amax

M 

= -

(3-26b)

Z

where Z :: lie is called the section modulus.

EXAMPLE 3-5 A beam having a T section with the dimensions shown in Fig. 3-15

is subjected

to a

bending moment of  1600 N . m that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses. Solution The area of the composite section is A

= 1956 mm-. Now divide the T section into two

rectangles. numbered I and 2, and Sum the moments of these areas about the top edge. W e then have 1956e,

and hence

CI

= 12(75)(6)

= = 32.99 mrn Therefore

C z 

==

+

12(88)(56)

100 _ 32.99

==

67.01

mrn.

Next we calculate the second moment of area of each rectangle about its own centroidal axis. Using Table A-18, we find for the top rectangle

I  / 

II 

I

= T2bh3 = = 12(75) 123 = = 1.080

X

104 m m "

• Load and Stress Analysis

87

,

Figure 3-15 Dimensions in millimeters.

I

~ I ~ ' " ~ 'I

,

I " T

-

,

I Tc

.

I

,

.2

00

c

U For the bottom rectangle, we have

We now employ the paralfel-axis theorem  to obtain the second moment of area of the composite

figure about its own centroidal axis. This theorem states

t,=le g

+  Ad 2 

where le g is the second moment of area about its own centroidal axis and Iz is the second moment of area about any parallel axis a distance gle, the distance

d  removed, For the top rectan-

is

d l  =32.99 - 6 =26.99 mm and for the bottom rectangle,

d2 Using the parallel-axis

I  = [1.080

=

=

67.01 ~44

23.01 mm

theorem for both rectangles, X

104

we now find that

+ 12(75)26.99 2] + [6.815

x lOs

+ 12(88)23.012)

= 1.907 x 1& mm " Finally, the maximum

tensile stress. which occurs at the top surface, is found to be 3

1600(32.99)101.907(10 6)

Answer

Similarly, the maximum compressive

Answer

o=

__ Me 2 = I 

27.68(106)

Pa =27.68 MPa

stress at the lower surface is found to be

1600(67.01)10-

3

6

1.907(10

~

)

=

-56.22(10

6

) Pa

=

-56.22 MPa

I\I\e