H P T E R Distriuted Forces: Moments of Inerti 47
hpter 9 Distriuted Forces: Moments of Inerti 47 9. Introduction 9. Second Moment, or Moment of Inerti, of n re 9.3 Determintion of the Moment of Inerti of n re Integrtion 9.4 Polr Moment of Inerti 9.5 Rdius of Grtion of n re 9.6 Prllel-is Theorem 9.7 Moments of Inerti of omposite res 9.8 Product of Inerti 9.9 Principl es nd Principl Moments of Inerti 9.0 Mohr s ircle for Moments nd Products of Inerti 9. Moment of Inerti of Mss 9. Prllel-is Theorem 9.3 Moments of Inerti of Thin Pltes 9.4 Determintion of the Moment of Inerti of Three-Dimensionl od Integrtion 9.5 Moments of Inerti of omposite odies 9.6 Moment of Inerti of od with Respect to n ritrr is through. Mss Products of Inerti 9.7 Ellipsoid of Inerti. Principl es of Inerti 9.8 Determintion of the Principl es nd Principl Moments of Inerti of od of ritrr Shpe 9. INTRDUTIN In hp. 5, we nled vrious sstems of forces distriuted over n re or volume. The three min tpes of forces considered were () weights of homogeneous pltes of uniform thickness (Secs. 5.3 through 5.6), () distriuted lods on ems (Sec. 5.8) nd hdrosttic forces (Sec. 5.9), nd (3) weights of homogeneous three-dimensionl odies (Secs. 5.0 nd 5.). In the cse of homogeneous pltes, the mgnitude DW of the weight of n element of plte ws proportionl to the re D of the element. For distriuted lods on ems, the mgnitude DW of ech elementl weight ws represented n element of re D 5 DW under the lod curve; in the cse of hdrosttic forces on sumerged rectngulr surfces, similr procedure ws followed. In the cse of homogeneous three-dimensionl odies, the mgnitude DW of the weight of n element of the od ws proportionl to the volume DV of the element. Thus, in ll cses considered in hp. 5, the distriuted forces were proportionl to the elementl res or volumes ssocited with them. The resultnt of these forces, therefore, could e otined summing the corresponding res or volumes, nd the moment of the resultnt out n given is could e determined computing the first moments of the res or volumes out tht is. In the first prt of this chpter, we consider distriuted forces DF whose mgnitudes depend not onl upon the elements of re D on which these forces ct ut lso upon the distnce from D to some given is. More precisel, the mgnitude of the force per unit re DF/D is ssumed to vr linerl with the distnce to the is. s indicted in the net section, forces of this tpe re found in the stud of the ending of ems nd in prolems involving sumerged non-rectngulr surfces. ssuming tht the elementl forces involved re distriuted over n re nd vr linerl with the distnce to the is, it will e shown tht while the mgnitude of their resultnt R depends upon the first moment Q 5 e d of the re, the loction of the point where R is pplied depends upon the second moment, or moment of inerti, I 5 e d of the sme re with respect to the is. You will lern to compute the moments of inerti of vrious res with respect to given nd es. lso introduced in the first prt of this chpter is the polr moment of inerti J 5 e r d of n re, where r is the distnce from the element of re d to the point. To fcilitte our computtions, reltion will e estlished etween the moment of inerti I of n re with respect to given is nd the moment of inerti I 9 of the sme re with respect to the prllel centroidl 9 is (prllel-is theorem). You will lso stud the trnsformtion of the moments of inerti of given re when the coordinte es re rotted (Secs. 9.9 nd 9.0). In the second prt of the chpter, ou will lern how to determine the moments of inerti of vrious msses with respect to given is. s ou will see in Sec. 9., the moment of inerti of given mss out n is 9 is defined s I 5 e r dm, where r is the distnce from the is 9 to the element of mss dm. Moments of inerti of msses re encountered in dnmics in prolems involving the rottion of rigid od out n is. To fcilitte the computtion
of mss moments of inerti, the prllel-is theorem will e introduced (Sec. 9.). Finll, ou will lern to nle the trnsformtion of moments of inerti of msses when the coordinte es re rotted (Secs. 9.6 through 9.8). MMENTS F INERTI F RES 9. Second Moment, or Moment of Inerti, of n re 473 9. SEND MMENT, R MMENT F INERTI, F N RE In the first prt of this chpter, we consider distriuted forces DF whose mgnitudes DF re proportionl to the elements of re D on which the forces ct nd t the sme time vr linerl with the distnce from D to given is. onsider, for emple, em of uniform cross section which is sujected to two equl nd opposite couples pplied t ech end of the em. Such em is sid to e in pure ending, nd it is shown in mechnics of mterils tht the internl forces in n section of the em re distriuted forces whose mgnitudes DF 5 k D vr linerl with the distnce etween the element of re D nd n is pssing through the centroid of the section. This is, represented the is in Fig. 9., is known s the neutrl is of the section. The forces on one side of the neutrl is re forces of compression, while those on the other side re forces of tension; on the neutrl is itself the forces re ero. The mgnitude of the resultnt R of the elementl forces DF which ct over the entire section is Fig. 9. Δ ΔF = kδ R 5 # k d 5 k # d The lst integrl otined is recognied s the first moment Q of the section out the is; it is equl to nd is thus equl to ero, since the centroid of the section is locted on the is. The sstem of the forces DF thus reduces to couple. The mgnitude M of this couple (ending moment) must e equl to the sum of the moments DM 5 DF 5 k D of the elementl forces. Integrting over the entire section, we otin M 5 # k d 5 k # d The lst integrl is known s the second moment, or moment of inerti, of the em section with respect to the is nd is denoted I. It is otined multipling ech element of re d the squre of its distnce from the is nd integrting over the em section. Since ech product d is positive, regrdless of the sign of, or ero (if is ero), the integrl I will lws e positive. nother emple of second moment, or moment of inerti, of n re is provided the following prolem from hdrosttics: The term second moment is more proper thn the term moment of inerti, since, logicll, the ltter should e used onl to denote integrls of mss (see Sec. 9.). In engineering prctice, however, moment of inerti is used in connection with res s well s msses.
474 Distriuted Forces: Moments of Inerti verticl circulr gte used to close the outlet of lrge reservoir is sumerged under wter s shown in Fig. 9.. Wht is the resultnt of the forces eerted the wter on the gte, nd wht is the moment of the resultnt out the line of intersection of the plne of the gte nd the wter surfce ( is)? If the gte were rectngulr, the resultnt of the forces of pressure could e determined from the pressure curve, s ws done in Sec. 5.9. Since the gte is circulr, however, more generl method must e used. Denoting the depth of n element of re D nd g the specific weight of wter, the pressure t the element is p 5 g, nd the mgnitude of the elementl force eerted on D is DF 5 p D 5 g D. The mgnitude of the resultnt of the elementl forces is thus Δ ΔF = g Δ R 5 # g d 5 g # d Fig. 9. nd cn e otined computing the first moment Q 5 e d of the re of the gte with respect to the is. The moment M of the resultnt must e equl to the sum of the moments DM 5 DF 5 g D of the elementl forces. Integrting over the re of the gte, we hve M 5 # g d 5 g # d Here gin, the integrl otined represents the second moment, or moment of inerti, I of the re with respect to the is. 9.3 DETERMINTIN F THE MMENT F INERTI F N RE Y INTEGRTIN We defined in the preceding section the second moment, or moment of inerti, of n re with respect to the is. Defining in similr w the moment of inerti I of the re with respect to the is, we write (Fig. 9.3) I 5 # d I 5 # d (9.) d = d d d = ( ) d d = d d d d di = d di = d di = d d di = d () () (c) Fig. 9.3
These integrls, known s the rectngulr moments of inerti of the re, cn e more esil evluted if we choose d to e thin strip prllel to one of the coordinte es. To compute I, the strip is chosen prllel to the is, so tht ll of the points of the strip re t the sme distnce from the is (Fig. 9.3); the moment of inerti di of the strip is then otined multipling the re d of the strip. To compute I, the strip is chosen prllel to the is so tht ll of the points of the strip re t the sme distnce from the is (Fig. 9.3c); the moment of inerti di of the strip is d. Moment of Inerti of Rectngulr re. s n emple, let us determine the moment of inerti of rectngle with respect to its se (Fig. 9.4). Dividing the rectngle into strips prllel to the is, we otin d 5 d di 5 d h I 5 # d 5 3 h 3 (9.) 0 9.4 Polr Moment of Inerti d = d h d Fig. 9.4 475 omputing l nd l Using the Sme Elementl Strips. The formul just derived cn e used to determine the moment of inerti di with respect to the is of rectngulr strip which is prllel to the is, such s the strip shown in Fig. 9.3c. Setting 5 d nd h 5 in formul (9.), we write di 5 3 3 d n the other hnd, we hve di 5 d 5 d The sme element cn thus e used to compute the moments of inerti I nd I of given re (Fig. 9.5). Fig. 9.5 d di = 3 d 3 di = d 9.4 PLR MMENT F INERTI n integrl of gret importnce in prolems concerning the torsion of clindricl shfts nd in prolems deling with the rottion of sls is J 5 # r d (9.3) where r is the distnce from to the element of re d (Fig. 9.6). This integrl is the polr moment of inerti of the re with respect to the pole. The polr moment of inerti of given re cn e computed from the rectngulr moments of inerti I nd I of the re if these quntities re lred known. Indeed, noting tht r 5, we write d r J 5 # r d 5 # ( ) d 5 # d # d Fig. 9.6
476 Distriuted Forces: Moments of Inerti tht is, J 5 I I (9.4) 9.5 RDIUS F GYRTIN F N RE () onsider n re which hs moment of inerti I with respect to the is (Fig. 9.7). Let us imgine tht we concentrte this re into thin strip prllel to the is (Fig. 9.7). If the re, thus concentrted, is to hve the sme moment of inerti with respect to the is, the strip should e plced t distnce k from the is, where k is defined the reltion Solving for k, we write I 5 k () k k 5 I (9.5) The distnce k is referred to s the rdius of grtion of the re with respect to the is. In similr w, we cn define the rdii of grtion k nd k (Fig. 9.7c nd d); we write k I 5 k k 5 I (9.6) J 5 k k 5 J (9.7) (c) k If we rewrite Eq. (9.4) in terms of the rdii of grtion, we find tht k 5 k k (9.8) EXMPLE For the rectngle shown in Fig. 9.8, let us compute the rdius of grtion k with respect to its se. Using formuls (9.5) nd (9.), we write k 5 I 5 3h 3 h 5 h 3 k 5 h 3 Fig. 9.7 (d) The rdius of grtion k of the rectngle is shown in Fig. 9.8. It should not e confused with the ordinte 5 h/ of the centroid of the re. While k depends upon the second moment, or moment of inerti, of the re, the ordinte is relted to the first moment of the re. h k Fig. 9.8
SMPLE PRLEM 9. Determine the moment of inerti of tringle with respect to its se. l SLUTIN tringle of se nd height h is drwn; the is is chosen to coincide with the se. differentil strip prllel to the is is chosen to e d. Since ll portions of the strip re t the sme distnce from the is, we write h di 5 d d 5 l d h Using similr tringles, we hve d l 5 h h h l 5 d 5 h h h Integrting di from 5 0 to 5 h, we otin d I 5 # d 5 # h 5 h c h 3 3 4 h 4 d 0 h d 5 h h # h (h 3 ) d 0 0 I 5 h3 SMPLE PRLEM 9. () Determine the centroidl polr moment of inerti of circulr re direct integrtion. () Using the result of prt, determine the moment of inerti of circulr re with respect to dimeter. SLUTIN. Polr Moment of Inerti. n nnulr differentil element of re is chosen to e d. Since ll portions of the differentil re re t the sme distnce from the origin, we write r u du dj 5 u d d 5 pu du r u (pu du) 5 p # u 3 du J 5 # dj 5 # r 0 0 J 5 p r4. Moment of Inerti with Respect to Dimeter. ecuse of the smmetr of the circulr re, we hve I 5 I. We then write J 5 I I 5 I p r4 5 I I dimeter 5 I 5 p 4 r4 477
SMPLE PRLEM 9.3 = k () Determine the moment of inerti of the shded re shown with respect to ech of the coordinte es. (Properties of this re were considered in Smple Pro. 5.4.) () Using the results of prt, determine the rdius of grtion of the shded re with respect to ech of the coordinte es. SLUTIN Referring to Smple Pro. 5.4, we otin the following epressions for the eqution of the curve nd the totl re: 5 5 3 Moment of Inerti I. verticl differentil element of re is chosen to e d. Since ll portions of this element re not t the sme distnce from the is, we must tret the element s thin rectngle. The moment of inerti of the element with respect to the is is then di 5 3 3 d 5 3 3 d 5 3 3 6 6 d d I 5 # di 5 # 0 3 3 6 6 d 5 c 3 7 3 6 7 d 0 I 5 3 Moment of Inerti I. The sme verticl differentil element of re is used. Since ll portions of the element re t the sme distnce from the is, we write di 5 d 5 ( d) 5 d 5 4 d I 5 # di 5 # 0 4 d 5 c 5 5 d 0 I 5 3 5 Rdii of Grtion k nd k. We hve, definition, nd k 5 I 5 3 / /3 5 7 k 5 I 5 3 /5 /3 5 3 5 k 5 7 k 5 3 5 478
SLVING PRLEMS N YUR WN The purpose of this lesson ws to introduce the rectngulr nd polr moments of inerti of res nd the corresponding rdii of grtion. lthough the prolems ou re out to solve m pper to e more pproprite for clculus clss thn for one in mechnics, we hope tht our introductor comments hve convinced ou of the relevnce of the moments of inerti to our stud of vriet of engineering topics.. lculting the rectngulr moments of inerti I nd I. We defined these quntities s I 5 # d I 5 # d (9.) where d is differentil element of re d d. The moments of inerti re the second moments of the re; it is for tht reson tht I, for emple, depends on the perpendiculr distnce to the re d. s ou stud Sec. 9.3, ou should recognie the importnce of crefull defining the shpe nd the orienttion of d. Further, ou should note the following points.. The moments of inerti of most res cn e otined mens of single integrtion. The epressions given in Figs. 9.3 nd c nd Fig. 9.5 cn e used to clculte I nd I. Regrdless of whether ou use single or doule integrtion, e sure to show on our sketch the element d tht ou hve chosen.. The moment of inerti of n re is lws positive, regrdless of the loction of the re with respect to the coordinte es. This is ecuse it is otined integrting the product of d nd the squre of distnce. (Note how this differs from the results for the first moment of the re.) nl when n re is removed (s in the cse for hole) will its moment of inerti e entered in our computtions with minus sign. c. s prtil check of our work, oserve tht the moments of inerti re equl to n re times the squre of length. Thus, ever term in n epression for moment of inerti must e length to the fourth power.. omputing the polr moment of inerti J. We defined J s J 5 # r d (9.3) where r 5. If the given re hs circulr smmetr (s in Smple Pro. 9.), it is possile to epress d s function of r nd to compute J with single integrtion. When the re lcks circulr smmetr, it is usull esier first to clculte I nd I nd then to determine J from J 5 I I (9.4) Lstl, if the eqution of the curve tht ounds the given re is epressed in polr coordintes, then d 5 r dr du nd doule integrtion is required to compute the integrl for J [see Pro. 9.7]. 3. Determining the rdii of grtion k nd k nd the polr rdius of grtion k. These quntities were defined in Sec. 9.5, nd ou should relie tht the cn e determined onl fter the re nd the pproprite moments of inerti hve een computed. It is importnt to rememer tht k is mesured in the direction, while k is mesured in the direction; ou should crefull stud Sec. 9.5 until ou understnd this point. 479
PRLEMS 9. through 9.4 Determine direct integrtion the moment of inerti of the shded re with respect to the is. h 9.5 through 9.8 Determine direct integrtion the moment of inerti of the shded re with respect to the is. Fig. P9. nd P9.5 = k = k = 4h ( ) h Fig. P9. nd P9.6 Fig. P9.3 nd P9.7 Fig. P9.4 nd P9.8 = c( k / ) = c( k / ) 9.9 through 9. Determine direct integrtion the moment of inerti of the shded re with respect to the is. 9. through 9.4 Determine direct integrtion the moment of inerti of the shded re with respect to the is. Fig. P9.9 nd P9. = ke / = k( ) 3 480 Fig. P9.0 nd P9.3 Fig. P9. nd P9.4
9.5 nd 9.6 Determine the moment of inerti nd the rdius of grtion of the shded re shown with respect to the is. Prolems 48 = k / = m = k 3 = k Fig. P9.5 nd P9.7 Fig. P9.6 nd P9.8 9.7 nd 9.8 Determine the moment of inerti nd the rdius of grtion of the shded re shown with respect to the is. 9.9 Determine the moment of inerti nd the rdius of grtion of the shded re shown with respect to the is. h = m + h = c sin k Fig. P9.9 nd P9.0 9.0 Determine the moment of inerti nd the rdius of grtion of the shded re shown with respect to the is. 9. nd 9. Determine the polr moment of inerti nd the polr rdius of grtion of the shded re shown with respect to point P. P P Fig. P9. Fig. P9.
48 Distriuted Forces: Moments of Inerti 9.3 () Determine direct integrtion the polr moment of inerti of the nnulr re shown with respect to point. () Using the result of prt, determine the moment of inerti of the given re with respect to the is. R R 9.4 () Show tht the polr rdius of grtion k of the nnulr re shown is pproimtel equl to the men rdius R m 5 (R R )/ for smll vlues of the thickness t 5 R R. () Determine the percentge error introduced using R m in plce of k for the following vlues of t/r m :,, nd 0. Fig. P9.3 nd P9.4 9.5 nd 9.6 Determine the polr moment of inerti nd the polr rdius of grtion of the shded re shown with respect to point P. = k = c + k r P r P Fig. P9.6 Fig. P9.5 9.7 Determine the polr moment of inerti nd the polr rdius of grtion of the shded re shown with respect to point. 9.8 Determine the polr moment of inerti nd the polr rdius of grtion of the isosceles tringle shown with respect to point. R = + kq Fig. P9.7 q h Fig. P9.8 *9.9 Using the polr moment of inerti of the isosceles tringle of Pro. 9.8, show tht the centroidl polr moment of inerti of circulr re of rdius r is pr 4 /. (Hint: s circulr re is divided into n incresing numer of equl circulr sectors, wht is the pproimte shpe of ech circulr sector?) *9.30 Prove tht the centroidl polr moment of inerti of given re cnnot e smller thn /p. (Hint: ompre the moment of inerti of the given re with the moment of inerti of circle tht hs the sme re nd the sme centroid.)
9.6 PRLLEL-XIS THEREM onsider the moment of inerti I of n re with respect to n is 9 (Fig. 9.9). Denoting the distnce from n element of re d to 9, we write I 5 # d Let us now drw through the centroid of the re n is 9 prllel to 9; this is is clled centroidl is. Denoting 9 9.6 Prllel-is Theorem 483 d ' d ' Fig. 9.9 ' the distnce from the element d to 9, we write 5 9 d, where d is the distnce etween the es 9 nd 9. Sustituting for in the ove integrl, we write I 5 # d 5 # ( d) d 5 # d d # d d # d The first integrl represents the moment of inerti I of the re with respect to the centroidl is 9. The second integrl represents the first moment of the re with respect to 9; since the centroid of the re is locted on tht is, the second integrl must e ero. Finll, we oserve tht the lst integrl is equl to the totl re. Therefore, we hve I 5 I d (9.9) This formul epresses tht the moment of inerti I of n re with respect to n given is 9 is equl to the moment of inerti I of the re with respect to centroidl is 9 prllel to 9 plus the product of the re nd the squre of the distnce d etween the two es. This theorem is known s the prllel-is theorem. Sustituting k for I nd k for I, the theorem cn lso e epressed s k 5 k d (9.0) similr theorem cn e used to relte the polr moment of inerti J of n re out point to the polr moment of inerti J of the sme re out its centroid. Denoting d the distnce etween nd, we write J 5 J d or k 5 k d (9.)
484 Distriuted Forces: Moments of Inerti EXMPLE s n ppliction of the prllel-is theorem, let us determine the moment of inerti I T of circulr re with respect to line tngent to the circle (Fig. 9.0). We found in Smple Pro. 9. tht the moment of inerti of circulr re out centroidl is is I 5 4pr 4. We cn write, therefore, r d = r I T 5 I d 5 4pr 4 (pr )r 5 5 4pr 4 Fig. 9.0 T EXMPLE The prllel-is theorem cn lso e used to determine the centroidl moment of inerti of n re when the moment of inerti of the re with respect to prllel is is known. onsider, for instnce, tringulr re (Fig. 9.). We found in Smple Pro. 9. tht the moment of inerti of tringle with respect to its se 9 is equl to h 3. Using the prllel-is theorem, we write D d' = h 3 d = h 3 Fig. 9. ' h D' ' I 5 I d I 5 I d 5 h 3 h( 3h) 5 36h 3 It should e oserved tht the product d ws sutrcted from the given moment of inerti in order to otin the centroidl moment of inerti of the tringle. Note tht this product is dded when trnsferring from centroidl is to prllel is, ut it should e sutrcted when trnsferring to centroidl is. In other words, the moment of inerti of n re is lws smller with respect to centroidl is thn with respect to n prllel is. Returning to Fig. 9., we oserve tht the moment of inerti of the tringle with respect to the line DD9 (which is drwn through verte) cn e otined writing I DD 5 I d 5 36h 3 h( 3h) 5 4h 3 Note tht I DD9 could not hve een otined directl from I 9. The prllelis theorem cn e pplied onl if one of the two prllel es psses through the centroid of the re. Photo 9. Figure 9.3 tultes dt for smll smple of the rolled-steel shpes tht re redil ville. Shown ove re two emples of wide-flnge shpes tht re commonl used in the construction of uildings. 9.7 MMENTS F INERTI F MPSITE RES onsider composite re mde of severl component res,, 3,... Since the integrl representing the moment of inerti of cn e sudivided into integrls evluted over,, 3,..., the moment of inerti of with respect to given is is otined dding the moments of inerti of the res,, 3,..., with respect to the sme is. The moment of inerti of n re consisting of severl of the common shpes shown in Fig. 9. cn thus e otined using the formuls given in tht figure. efore dding the moments of inerti of the component res, however, the prllel-is theorem m hve to e used to trnsfer ech moment of inerti to the desired is. This is shown in Smple Pros. 9.4 nd 9.5. The properties of the cross sections of vrious structurl shpes re given in Fig. 9.3. s noted in Sec. 9., the moment of inerti of em section out its neutrl is is closel relted to the computtion of the ending moment in tht section of the em. The
Rectngle h ' ' I ' = h 3 I ' = 3 h I = h 3 3 I = 3 3 h J = h( + h ) 9.7 Moments of Inerti of omposite res 485 Tringle h h 3 ' I ' = h 36 3 I = h 3 ircle r I = I = r 4 4 J = r 4 Semicircle r I = I = r 8 4 J = r 4 4 Qurter circle r I = I = r 6 4 J = r 8 4 Ellipse I = 4 3 I = 4 3 J = ( 4 + ) Fig. 9. Moments of inerti of common geometric shpes. determintion of moments of inerti is thus prerequisite to the nlsis nd design of structurl memers. It should e noted tht the rdius of grtion of composite re is not equl to the sum of the rdii of grtion of the component res. In order to determine the rdius of grtion of composite re, it is first necessr to compute the moment of inerti of the re.
Designtion re in Depth in. Width in. is X-X is Y-Y I, in 4 k, in., in. I, in 4 k, in., in. W Shpes (Wide-Flnge Shpes) X Y X W8 76 W6 57 W4 38 W8 3.3 6.8. 9. 8. 6.4 4. 8.00.0 7. 6.77 8.00 330 758 385 0 7.73 6.7 5.87 3.47 5 43. 6.7 37..6.60.55.0 Y S Shpes (mericn Stndrd Shpes) X Y X S8 54.7 S 3.8 S0 5.4 S6.5 6.0 9.3 7.45 3.66 8.0.0 0.0 6.00 6.00 5.00 4.66 3.33 80 7 3.0 7.07 4.83 4.07.45 0.7 9.33 6.73.80.4.00 0.950 0.70 Y Shpes (mericn Stndrd hnnels) X Y X 0.7 0 5.3 8.5 6 8. 6.08 4.48 3.37.39.0 0.0 8.00 6.00.94.60.6.9 9 67.3 3.5 3. 4.6 3.87 3..34 3.86.7.3 0.687 0.797 0.7 0.63 0.536 0.698 0.634 0.57 0.5 Y ngles Y L6 6 L4 4 L3 3 4 L6 4 L5 3 L3 4.0 3.75.44 4.75 3.75.9 35.4 5.5.3 7.3 9.43.09.79. 0.96.9.58 0.953.86.8 0.836.98.74 0.980 35.4 5.5.3 6..55 0.390.79. 0.96.4 0.84 0.569.86.8 0.836 0.98 0.746 0.487 X X Y Fig. 9.3 Properties of rolled-steel shpes (U.S. customr units).* *ourtes of the mericn Institute of Steel onstruction, hicgo, Illinois Nominl depth in inches nd weight in pounds per foot Depth, width, nd thickness in inches 486
is X-X is Y-Y Designtion re mm Depth mm Width mm I k I 0 6 mm 4 mm mm 0 6 mm 4 k mm mm W Shpes (Wide-Flnge Shpes) X Y X W460 3 W40 85 W360 57.8 W00 46. 4400 0800 730 5880 46 47 358 03 79 8 7 03 554 36 60 45.8 96 7 49 88. 63.3 7.9. 5.4 66.3 40.6 39.4 5.3 Y S Shpes (mericn Stndrd Shpes) X Y X S460 8.4 S30 47.3 S50 37.8 S50 8.6 0300 600 480 360 457 305 54 5 5 7 8 84.6 333 90.3 5. 9.6 80 3 03 6. 8.6 3.88.80 0.749 9.0 5.4 4. 7.8 Y Shpes (mericn Stndrd hnnels) X Y X 30 30.8 50.8 00 7. 50. 390 890 70 540 305 54 03 5 74.7 66.0 57.4 48.8 53.7 8.0 3.5 5.45 7 98.3 79.0 59.4.6 0.945 0.545 0.86 0. 8. 5.8 3.6 7.7 6. 4.5 3.0 Y ngles Y L5 5 5.4 L0 0.7 L76 76 6.4 L5 0.7 L7 76.7 L76 5 6.4 700 40 99 3060 40 768 4.7.30 0.5 7.0 3.93 0.454 45.5 30.7 3.5 48.5 40. 4. 47. 30.0. 50.3 44. 4.9 4.7.30 0.5.59.06 0.6 45.5 30.7 3.5 9.0 0.9 4.5 47. 30.0. 4.9 8.9.4 X X Y Fig. 9.3 Properties of rolled-steel shpes (SI units). Nominl depth in millimeters nd mss in kilogrms per meter Depth, width, nd thickness in millimeters 487
9 in. 3 4 in. 4. in. SMPLE PRLEM 9.4 The strength of W4 3 38 rolled-steel em is incresed ttching 9 3 3 4-in. plte to its upper flnge s shown. Determine the moment of inerti nd the rdius of grtion of the composite section with respect to n is which is prllel to the plte nd psses through the centroid of the section. 6.77 in. SLUTIN 7.45 in. d Y ' The origin of the coordintes is plced t the centroid of the wide-flnge shpe, nd the distnce Y to the centroid of the composite section is computed using the methods of hp. 5. The re of the wide-flnge shpe is found referring to Fig. 9.3. The re nd the coordinte of the centroid of the plte re 5 (9 in.)(0.75 in.) 5 6.75 in 5 (4. in.) (0.75 in.) 5 7.45 in. Section re, in, in., in 3 Plte 6.75 7.45 50. Wide-flnge shpe. 0 0 o 5 7.95 o 5 50. Yo 5 o Y(7.95) 5 50. Y 5.79 in. Moment of Inerti. The prllel-is theorem is used to determine the moments of inerti of the wide-flnge shpe nd the plte with respect to the 9 is. This is is centroidl is for the composite section ut not for either of the elements considered seprtel. The vlue of I for the wide-flnge shpe is otined from Fig. 9.3. For the wide-flnge shpe, I 9 5 I Y 5 385 (.)(.79) 5 47.3 in 4 For the plte, I 9 5 I d 5 ( )(9)( 3 4) 3 (6.75)(7.45.79) 5 45. in 4 For the composite re, I 9 5 47.3 45. 5 67.5 in 4 I 9 5 68 in 4 Rdius of Grtion. We hve k 5 I 5 67.5 in4 7.95 in k 5 5.87 in. 488
0 mm 40 mm r = 90 mm SMPLE PRLEM 9.5 Determine the moment of inerti of the shded re with respect to the is. SLUTIN The given re cn e otined sutrcting hlf circle from rectngle. The moments of inerti of the rectngle nd the hlf circle will e computed seprtel. 40 mm ' ' 0 mm = Moment of Inerti of Rectngle. Referring to Fig. 9., we otin I 5 3 h 3 5 3 (40 mm)(0 mm) 3 5 38. 3 0 6 mm 4 0 mm = 38. mm = 8.8 mm Moment of Inerti of Hlf ircle. Referring to Fig. 5.8, we determine the loction of the centroid of the hlf circle with respect to dimeter 9. ' 5 4r (4)(90 mm) ' 5 5 38. mm 3p 3p The distnce from the centroid to the is is 5 0 mm 5 0 mm 38. mm 5 8.8 mm Referring now to Fig. 9., we compute the moment of inerti of the hlf circle with respect to dimeter 9; we lso compute the re of the hlf circle. I 5 8 pr 4 5 8 p(90 mm) 4 5 5.76 3 0 6 mm 4 5 pr 5 p(90 mm) 5.7 3 0 3 mm Using the prllel-is theorem, we otin the vlue of I : I 5 I 5.76 3 0 6 mm 4 5 I (.7 3 0 3 mm )(38. mm) I 5 7.0 3 0 6 mm 4 gin using the prllel-is theorem, we otin the vlue of I : I 5 I 5 7.0 3 0 6 mm 4 (.7 3 0 3 mm )(8.8 mm) 5 9.3 3 0 6 mm 4 Moment of Inerti of Given re. Sutrcting the moment of inerti of the hlf circle from tht of the rectngle, we otin I 5 38. 3 0 6 mm 4 9.3 3 0 6 mm 4 I 5 45.9 3 0 6 mm 4 489
SLVING PRLEMS N YUR WN In this lesson we introduced the prllel-is theorem nd illustrted how it cn e used to simplif the computtion of moments nd polr moments of inerti of composite res. The res tht ou will consider in the following prolems will consist of common shpes nd rolled-steel shpes. You will lso use the prllelis theorem to locte the point of ppliction (the center of pressure) of the resultnt of the hdrosttic forces cting on sumerged plne re.. ppling the prllel-is theorem. In Sec. 9.6 we derived the prllel-is theorem I 5 I d (9.9) which sttes tht the moment of inerti I of n re with respect to given is is equl to the sum of the moment of inerti I of tht re with respect to prllel centroidl is nd the product d, where d is the distnce etween the two es. It is importnt tht ou rememer the following points s ou use the prllelis theorem.. The centroidl moment of inerti I of n re cn e otined sutrcting the product d from the moment of inerti I of the re with respect to prllel is. It follows tht the moment of inerti I is smller thn the moment of inerti I of the sme re with respect to n prllel is.. The prllel-is theorem cn e pplied onl if one of the two es involved is centroidl is. Therefore, s we noted in Emple, to compute the moment of inerti of n re with respect to noncentroidl is when the moment of inerti of the re is known with respect to nother noncentroidl is, it is necessr to first compute the moment of inerti of the re with respect to centroidl is prllel to the two given es.. omputing the moments nd polr moments of inerti of composite res. Smple Pros. 9.4 nd 9.5 illustrte the steps ou should follow to solve prolems of this tpe. s with ll composite-re prolems, ou should show on our sketch the common shpes or rolled-steel shpes tht constitute the vrious elements of the given re, s well s the distnces etween the centroidl es of the elements nd the es out which the moments of inerti re to e computed. In ddition, it is importnt tht the following points e noted.. The moment of inerti of n re is lws positive, regrdless of the loction of the is with respect to which it is computed. s pointed out in the comments for the preceding lesson, it is onl when n re is removed (s in the cse of hole) tht its moment of inerti should e entered in our computtions with minus sign. 490
. The moments of inerti of semiellipse nd qurter ellipse cn e determined dividing the moment of inerti of n ellipse nd 4, respectivel. It should e noted, however, tht the moments of inerti otined in this mnner re with respect to the es of smmetr of the ellipse. To otin the centroidl moments of inerti of these shpes, the prllel-is theorem should e used. Note tht this remrk lso pplies to semicircle nd to qurter circle nd tht the epressions given for these shpes in Fig. 9. re not centroidl moments of inerti. c. To clculte the polr moment of inerti of composite re, ou cn use either the epressions given in Fig. 9. for J or the reltionship J 5 I I (9.4) depending on the shpe of the given re. d. efore computing the centroidl moments of inerti of given re, ou m find it necessr to first locte the centroid of the re using the methods of hpter 5. 3. Locting the point of ppliction of the resultnt of sstem of hdrosttic forces. In Sec. 9. we found tht R 5 g # d 5 g M 5 g # d 5 gi where is the distnce from the is to the centroid of the sumerged plne re. Since R is equivlent to the sstem of elementl hdrosttic forces, it follows tht om : P R 5 M where P is the depth of the point of ppliction of R. Then P (g ) 5 gi or P 5 I In closing, we encourge ou to crefull stud the nottion used in Fig. 9.3 for the rolled-steel shpes, s ou will likel encounter it gin in susequent engineering courses. 49
PRLEMS mm mm 9.3 nd 9.3 Determine the moment of inerti nd the rdius of grtion of the shded re with respect to the is. in. in. in. in. 6 mm in. 8 mm 4 mm in. in. 4 mm in. 6 mm in. 4 mm 4 mm in. in. Fig. P9.3 nd P9.33 Fig. P9.3 nd P9.34 9.33 nd 9.34 Determine the moment of inerti nd the rdius of grtion of the shded re with respect to the is. 9.35 nd 9.36 Determine the moments of inerti of the shded re shown with respect to the nd es when 5 0 mm..5.5 3 Fig. P9.35 Fig. P9.36 d ' 9.37 For the 4000-mm shded re shown, determine the distnce d nd the moment of inerti with respect to the centroidl is prllel to 9 knowing tht the moments of inerti with respect to 9 nd 9 re 3 0 6 mm 4 nd 3.9 3 0 6 mm 4, respectivel, nd tht d 5 5 mm. Fig. P9.37 nd P9.38 d ' 9.38 Determine for the shded region the re nd the moment of inerti with respect to the centroidl is prllel to 9, knowing tht d 5 5 mm nd d 5 5 mm nd tht the moments of inerti with respect to 9 nd 9 re 7.84 3 0 6 mm 4 nd 5.0 3 0 6 mm 4, respectivel. 49
9.39 The shded re is equl to 50 in. Determine its centroidl moments of inerti I nd I, knowing tht I 5 I nd tht the polr moment of inerti of the re out point is J 5 50 in 4. Prolems 493 d 6 in. D Fig. P9.39 nd P9.40 9.40 The polr moments of inerti of the shded re with respect to points,, nd D re, respectivel, J 5 880 in 4, J 5 670 in 4, nd J D 5 4560 in 4. Determine the shded re, its centroidl moment of inerti J, nd the distnce d from to D. 9.4 through 9.44 Determine the moments of inerti I nd I of the re shown with respect to centroidl es respectivel prllel nd perpendiculr to side. mm 8 mm 8 mm mm mm 7 mm 4 mm Fig. P9.4 mm mm.3 in.. in..0 in. 8 mm 5.0 in. 0.5 in. 3.8 in. mm 6 mm mm 6 mm.8 in. 0.9 in..0 in.. in. 3.6 in. 0.5 in. Fig. P9.4 Fig. P9.43 Fig. P9.44 9.45 nd 9.46 Determine the polr moment of inerti of the re shown with respect to () point, () the centroid of the re. 84 mm 54 mm 00 mm 50 mm 7 mm 4 mm 00 mm Fig. P9.45 50 mm Fig. P9.46 Semiellipses
494 Distriuted Forces: Moments of Inerti 9.47 nd 9.48 Determine the polr moment of inerti of the re shown with respect to () point, () the centroid of the re. Semicircle 3 in. 4.5 in. Fig. P9.47 6 in. 6 in. Fig. P9.48 4.5 in. 80 mm 80 mm 0 mm S30 47.3 9.49 Two 0-mm steel pltes re welded to rolled S section s shown. Determine the moments of inerti nd the rdii of grtion of the comined section with respect to the centroidl nd es. 9.50 Two chnnels re welded to rolled W section s shown. Determine the moments of inerti nd the rdii of grtion of the comined section with respect to the centroidl nd es. W8 3 Fig. P9.49 8.5 Fig. P9.50 9.5 To form reinforced o section, two rolled W sections nd two pltes re welded together. Determine the moments of inerti nd the rdii of grtion of the comined section with respect to the centroidl es shown. W00 46. 9.5 Two chnnels re welded to d 3 -in. steel plte s shown. Determine the width d for which the rtio I /I of the centroidl moments of inerti of the section is 6. 6 mm 03 mm 6 in. Fig. P9.5 0 5.3 6 in. d Fig. P9.5
9.53 Two L76 3 76 3 6.4-mm ngles re welded to 50 3.8 chnnel. Determine the moments of inerti of the comined section with respect to centroidl es respectivel prllel nd perpendiculr to the we of the chnnel. L76 76 6.4 in. Prolems 495 0 in. 50.8 Fig. P9.53 9.54 Two L4 3 4 3 -in. ngles re welded to steel plte s shown. Determine the moments of inerti of the comined section with respect to centroidl es respectivel prllel nd perpendiculr to the plte. 9.55 The strength of the rolled W section shown is incresed welding chnnel to its upper flnge. Determine the moments of inerti of the comined section with respect to its centroidl nd es. 9.56 Two L5 3 3 3 -in. ngles re welded to -in. steel plte. Determine the distnce nd the centroidl moments of inerti I nd I of the comined section, knowing tht I 5 4I. Fig. P9.54 50.8 L4 4 W460 3 L5 3 Fig. P9.55 0.5 in. ' 5 in. 5 in. r Fig. P9.56 9.57 nd 9.58 The pnel shown forms the end of trough tht is filled with wter to the line 9. Referring to Sec. 9., determine the depth of the point of ppliction of the resultnt of the hdrosttic forces cting on the pnel (the center of pressure). 9.59 nd *9.60 The pnel shown forms the end of trough tht is filled with wter to the line 9. Referring to Sec. 9., determine the depth of the point of ppliction of the resultnt of the hdrosttic forces cting on the pnel (the center of pressure). ' Fig. P9.57 Qurter ellipse Fig. P9.58 ' h h h ' Fig. P9.59 Prol Fig. P9.60
496 Distriuted Forces: Moments of Inerti 9.6 The cover for 0.5-m-dimeter ccess hole in wter storge tnk is ttched to the tnk with four equll spced olts s shown. Determine the dditionl force on ech olt due to the wter pressure when the center of the cover is locted.4 m elow the wter surfce. 0.3 m D 0.5 m Fig. P9.6 9.6 verticl trpeoidl gte tht is used s n utomtic vlve is held shut two springs ttched to hinges locted long edge. Knowing tht ech spring eerts couple of mgnitude 470 N? m, determine the depth d of wter for which the gte will open.. m 0.5 m h d D E 0.84 m 0.8 m Fig. P9.63 Fig. P9.6 39 mm 39 mm *9.63 Determine the coordinte of the centroid of the volume shown. (Hint: The height of the volume is proportionl to the coordinte; consider n nlog etween this height nd the wter pressure on sumerged surfce.) 64 mm 64 mm Fig. P9.64 *9.64 Determine the coordinte of the centroid of the volume shown; this volume ws otined intersecting n elliptic clinder with n olique plne. (See hint of Pro. 9.63.)
*9.65 Show tht the sstem of hdrosttic forces cting on sumerged plne re cn e reduced to force P t the centroid of the re nd two couples. The force P is perpendiculr to the re nd is of mgnitude P 5 g sin u, where g is the specific weight of the liquid, nd the couples re M 9 5 (gi sin u)i nd M 9 5 (gi sin u)j, where I 5 e 99 d (see Sec. 9.8). Note tht the couples re independent of the depth t which the re is sumerged. 9.8 Product of Inerti 497 ' P M ' q ' Fig. P9.65 M ' *9.66 Show tht the resultnt of the hdrosttic forces cting on sumerged plne re is force P perpendiculr to the re nd of mgnitude P 5 g sin u 5 p, where g is the specific weight of the liquid nd p is the pressure t the centroid of the re. Show tht P is pplied t point P, clled the center of pressure, whose coordintes re P 5 I / nd P 5 I /, where I 5 e d (see Sec. 9.8). Show lso tht the difference of ordintes P is equl to k / nd thus depends upon the depth t which the re is sumerged. P ' P ' P Fig. P9.66 q P *9.8 PRDUT F INERTI The integrl I 5 # d (9.) which is otined multipling ech element d of n re its coordintes nd nd integrting over the re (Fig. 9.4), is known s the product of inerti of the re with respect to the nd es. Unlike the moments of inerti I nd I, the product of inerti I cn e positive, negtive, or ero. When one or oth of the nd es re es of smmetr for the re, the product of inerti I is ero. onsider, for emple, the chnnel section shown in Fig. 9.5. Since this section is smmetricl with respect to the is, we cn ssocite with ech element d of coordintes nd n element d9 of coordintes nd. lerl, the contriutions to I of n pir of elements chosen in this w cncel out, nd the integrl (9.) reduces to ero. prllel-is theorem similr to the one estlished in Sec. 9.6 for moments of inerti cn e derived for products of inerti. onsider n re nd sstem of rectngulr coordintes nd Fig. 9.4 d' Fig. 9.5 d d
498 Distriuted Forces: Moments of Inerti (Fig. 9.6). Through the centroid of the re, of coordintes nd ', we drw two centroidl es nd which re prllel, respectivel, to the nd es. Denoting nd the coordintes of n element of re d with respect to the originl es, nd nd the coordintes of the sme element with respect to the centroidl d es, we write 5 nd 5. Sustituting into (9.), we otin the following epression for the product of inerti I : ' ' I 5 # d 5 # ( )( ) d Fig. 9.6 ' 5 # d # d # d # d The first integrl represents the product of inerti I of the re with respect to the centroidl es nd. The net two integrls represent first moments of the re with respect to the centroidl es; the reduce to ero, since the centroid is locted on these es. Finll, we oserve tht the lst integrl is equl to the totl re. Therefore, we hve I 5 I (9.3) *9.9 PRINIPL XES ND PRINIPL MMENTS F INERTI onsider the re nd the coordinte es nd (Fig. 9.7). ssuming tht the moments nd product of inerti I 5 # d I 5 # d I 5 # d (9.4) of the re re known, we propose to determine the moments nd product of inerti I, I, nd I of with respect to new es nd which re otined rotting the originl es out the origin through n ngle u. We first note the following reltions etween the coordintes, nd, of n element of re d: 5 cos u sin u 5 cos u sin u ' ' sin q d ' ' cos q q Fig. 9.7
Sustituting for in the epression for I, we write I 5 # ( ) d 5 # ( cos u sin u) d 5 cos u # d sin u cos u # d sin u # d Using the reltions (9.4), we write I 5 I cos u I sin u cos u I sin u (9.5) Similrl, we otin for I nd I the epressions I 5 I sin u I sin u cos u I cos u (9.6) I 5 (I I ) sin u cos u I (cos u sin u) (9.7) Reclling the trigonometric reltions sin u 5 sin u cos u cos u 5 cos u sin u nd cos cos u u 5 sin cos u u 5 we cn write (9.5), (9.6), nd (9.7) s follows: I 5 I I I 5 I I I 5 I I I I I I cos u I sin u (9.8) cos u I sin u (9.9) sin u I cos u (9.0) dding (9.8) nd (9.9) we oserve tht I I 5 I I (9.) This result could hve een nticipted, since oth memers of (9.) re equl to the polr moment of inerti J. Equtions (9.8) nd (9.0) re the prmetric equtions of circle. This mens tht if we choose set of rectngulr es nd plot point M of sciss I nd ordinte I for n given vlue of the prmeter u, ll of the points thus otined will lie on circle. To estlish this propert, we eliminte u from Eqs. (9.8) nd (9.0); this is done trnsposing (I I )/ in Eq. (9.8), squring oth memers of Eqs. (9.8) nd (9.0), nd dding. We write Setting I I I I 5 I I I (9.) I ve 5 I I nd R 5 I I I (9.3) we write the identit (9.) in the form (I 9 I ve ) I 99 5 R (9.4) 9.9 Principl es nd Principl Moments of Inerti 499
500 Distriuted Forces: Moments of Inerti which is the eqution of circle of rdius R centered t the point whose nd coordintes re I ve nd 0, respectivel (Fig. 9.8). We I '' oserve tht Eqs. (9.9) nd (9.0) re the prmetric equtions of the I ' sme circle. Furthermore, ecuse of the smmetr of the circle out the horiontl is, the sme result would hve een otined if insted M of plotting M, we hd plotted point N of coordintes I 9 nd I 99 R (Fig. 9.8). This propert will e used in Sec. 9.0. I '' The two points nd where the ove circle intersects the I ' horiontl is (Fig. 9.8) re of specil interest: Point corresponds to the mimum vlue of the moment of inerti I 9, while I min point corresponds to its minimum vlue. In ddition, oth points Ive correspond to ero vlue of the product of inerti I 99. Thus, the vlues u m of the prmeter u which correspond to the points nd cn e otined setting I 99 5 0 in Eq. (9.0). We otin I m () I '' I ve I '' N I ' Fig. 9.8 R () I ' I tn u m 5 (9.5) I I This eqution defines two vlues u m which re 80 prt nd thus two vlues u m which re 90 prt. ne of these vlues corresponds to point in Fig. 9.8 nd to n is through in Fig. 9.7 with respect to which the moment of inerti of the given re is mimum; the other vlue corresponds to point nd to n is through with respect to which the moment of inerti of the re is minimum. The two es thus defined, which re perpendiculr to ech other, re clled the principl es of the re out, nd the corresponding vlues I m nd I min of the moment of inerti re clled the principl moments of inerti of the re out. Since the two vlues u m defined Eq. (9.5) were otined setting I 99 5 0 in Eq. (9.0), it is cler tht the product of inerti of the given re with respect to its principl es is ero. We oserve from Fig. 9.8 tht I m 5 I ve R I min 5 I ve R (9.6) Using the vlues for I ve nd R from formuls (9.3), we write I m,min 5 I I 6 I I I (9.7) Unless it is possile to tell inspection which of the two principl es corresponds to I m nd which corresponds to I min, it is necessr to sustitute one of the vlues of u m into Eq. (9.8) in order to determine which of the two corresponds to the mimum vlue of the moment of inerti of the re out. Referring to Sec. 9.8, we note tht if n re possesses n is of smmetr through point, this is must e principl is of the re out. n the other hnd, principl is does not need to e n is of smmetr; whether or not n re possesses n es of smmetr, it will hve two principl es of inerti out n point. The properties we hve estlished hold for n point locted inside or outside the given re. If the point is chosen to coincide with the centroid of the re, n is through is centroidl is; the two principl es of the re out its centroid re referred to s the principl centroidl es of the re. This reltion cn lso e otined differentiting I 9 in Eq. (9.8) nd setting di 9 /du 5 0.
SMPLE PRLEM 9.6 h Determine the product of inerti of the right tringle shown () with respect to the nd es nd () with respect to centroidl es prllel to the nd es. SLUTIN '. Product of Inerti I. verticl rectngulr strip is chosen s the differentil element of re. Using the prllel-is theorem, we write di 5 di el el d h Since the element is smmetricl with respect to the 9 nd 9 es, we note tht di 99 5 0. From the geometr of the tringle, we otin el d el ' 5 h el 5 d 5 d 5 h d el 5 5 h Integrting di from 5 0 to 5, we otin I 5 # di 5 # el el d 5 # 5 h # 0 0 ( )h d 3 d 5 h c 4 3 3 4 8 d 0 I 5 4 h. Product of Inerti I 99 99. The coordintes of the centroid of the tringle reltive to the nd es re 5 3 5 3h h Using the epression for I otined in prt, we ppl the prllel-is theorem nd write I 5 I 4 h 5 I ( 3)( 3h)( h) I 5 4 h 8 h I 57 h 50
SMPLE PRLEM 9.7 3 in. For the section shown, the moments of inerti with respect to the nd es hve een computed nd re known to e in. I 5 0.38 in 4 I 5 6.97 in 4 4 in. in. in. Determine () the orienttion of the principl es of the section out, () the vlues of the principl moments of inerti of the section out. 3 in. SLUTIN I.5 in..75 in. II.75 in..5 in. III We first compute the product of inerti with respect to the nd es. The re is divided into three rectngles s shown. We note tht the product of inerti I with respect to centroidl es prllel to the nd es is ero for ech rectngle. Using the prllel-is theorem I 5 I, we find tht I reduces to for ech rectngle. Rectngle re, in, in., in., in 4 I.5.5.75 3.8 II.5 0 0 0 III.5.5.75 3.8 o 5 6.56 I 5 o 56.56 in 4. Principl es. Since the mgnitudes of I, I, nd I re known, Eq. (9.5) is used to determine the vlues of u m : tn u m 5 I I I 5 (6.56) 0.38 6.97 53.85 u m 5 75.4 nd 55.4 u m 5 37.7 nd u m 5 7.7 q m = 7.7. Principl Moments of Inerti. Using Eq. (9.7), we write I m,min 5 I I 6 I I I q m = 37.7 5 0.38 6.97 6 0.38 6.97 6.56 I m 5 5.45 in 4 I min 5.897 in 4 Noting tht the elements of the re of the section re more closel distriuted out the is thn out the is, we conclude tht I 5 I m 5 5.45 in 4 nd I 5 I min 5.897 in 4. This conclusion cn e verified sustituting u 5 37.7 into Eqs. (9.8) nd (9.9). 50
SLVING PRLEMS N YUR WN In the prolems for this lesson, ou will continue our work with moments of inerti nd will utilie vrious techniques for computing products of inerti. lthough the prolems re generll strightforwrd, severl items re worth noting.. lculting the product of inerti I integrtion. We defined this quntit s I 5 # d (9.) nd stted tht its vlue cn e positive, negtive, or ero. The product of inerti cn e computed directl from the ove eqution using doule integrtion, or it cn e determined using single integrtion s shown in Smple Pro. 9.6. When ppling the ltter technique nd using the prllel-is theorem, it is importnt to rememer tht el nd el in the eqution di 5 di el el d re the coordintes of the centroid of the element of re d. Thus, if d is not in the first qudrnt, one or oth of these coordintes will e negtive.. lculting the products of inerti of composite res. The cn esil e computed from the products of inerti of their component prts using the prllel-is theorem I 5 I (9.3) The proper technique to use for prolems of this tpe is illustrted in Smple Pros. 9.6 nd 9.7. In ddition to the usul rules for composite-re prolems, it is essentil tht ou rememer the following points.. If either of the centroidl es of component re is n is of smmetr for tht re, the product of inerti I for tht re is ero. Thus, I is ero for component res such s circles, semicircles, rectngles, nd isosceles tringles which possess n is of smmetr prllel to one of the coordinte es.. P creful ttention to the signs of the coordintes nd of ech component re when ou use the prllel-is theorem [Smple Pro. 9.7]. 3. Determining the moments of inerti nd the product of inerti for rotted coordinte es. In Sec. 9.9 we derived Eqs. (9.8), (9.9), nd (9.0), from which the moments of inerti nd the product of inerti cn e computed for coordinte es which hve een rotted out the origin. To ppl these equtions, ou must know set of vlues I, I, nd I for given orienttion of the es, nd ou must rememer tht u is positive for counterclockwise rottions of the es nd negtive for clockwise rottions of the es. 4. omputing the principl moments of inerti. We showed in Sec. 9.9 tht there is prticulr orienttion of the coordinte es for which the moments of inerti ttin their mimum nd minimum vlues, I m nd I min, nd for which the product of inerti is ero. Eqution (9.7) cn e used to compute these vlues, known s the principl moments of inerti of the re out. The corresponding es re referred to s the principl es of the re out, nd their orienttion is defined Eq. (9.5). To determine which of the principl es corresponds to I m nd which corresponds to I min, ou cn either follow the procedure outlined in the tet fter Eq. (9.7) or oserve out which of the two principl es the re is more closel distriuted; tht is corresponds to I min [Smple Pro. 9.7]. 503
PRLEMS 9.67 through 9.70 Determine direct integrtion the product of inerti of the given re with respect to the nd es. + = 4 = k Fig. P9.67 Fig. P9.68 h h h Fig. P9.69 Fig. P9.70 60 mm 0 mm 0 mm 0 mm 60 mm 0 mm 00 mm 0 mm 9.7 through 9.74 Using the prllel-is theorem, determine the product of inerti of the re shown with respect to the centroidl nd es. 60 mm 0 mm 0 mm 0 mm 60 mm 80 mm Fig. P9.7 Fig. P9.7 0.5 in. 0.980 in. 6 in. in. 0.487 in. 6 in. L3 4 3 in. 0.5 in. 504 Fig. P9.73 Fig. P9.74
9.75 through 9.78 Using the prllel-is theorem, determine the product of inerti of the re shown with respect to the centroidl nd es. Prolems 505 9 in. 3 in. 00 mm 8 mm 40 mm 9 in. in. in. 40 mm 8 mm 5 in. 8 mm Fig. P9.75 3 in. 9 in. Fig. P9.76.3 in..0 in. 8.9 mm.7 mm 0.4 in. 5.3 in. 0.5 in. 3.6 in. 0.5 in..5 in. L7 76.7.7 mm 76 mm 44. mm 7 mm Fig. P9.77 Fig. P9.78 9.79 Determine for the qurter ellipse of Pro. 9.67 the moments of inerti nd the product of inerti with respect to new es otined rotting the nd es out () through 458 counterclockwise, () through 308 clockwise. 9.80 Determine the moments of inerti nd the product of inerti of the re of Pro. 9.7 with respect to new centroidl es otined rotting the nd es 308 counterclockwise. 9.8 Determine the moments of inerti nd the product of inerti of the re of Pro. 9.73 with respect to new centroidl es otined rotting the nd es 608 counterclockwise. 9.8 Determine the moments of inerti nd the product of inerti of the re of Pro. 9.75 with respect to new centroidl es otined rotting the nd es 458 clockwise. 9.83 Determine the moments of inerti nd the product of inerti of the L3 3 3 4-in. ngle cross section of Pro. 9.74 with respect to new centroidl es otined rotting the nd es 308 clockwise.
506 Distriuted Forces: Moments of Inerti 9.84 Determine the moments of inerti nd the product of inerti of the L7 3 76 3.7-mm ngle cross section of Pro. 9.78 with respect to new centroidl es otined rotting the nd es 458 counterclockwise. 9.85 For the qurter ellipse of Pro. 9.67, determine the orienttion of the principl es t the origin nd the corresponding vlues of the moments of inerti. 9.86 through 9.88 For the re indicted, determine the orienttion of the principl es t the origin nd the corresponding vlues of the moments of inerti. 9.86 re of Pro. 9.7 9.87 re of Pro. 9.73 9.88 re of Pro. 9.75 9.89 nd 9.90 For the ngle cross section indicted, determine the orienttion of the principl es t the origin nd the corresponding vlues of the moments of inerti. 9.89 The L3 3 3 4-in. ngle cross section of Pro. 9.74 9.90 The L7 3 76 3.7-mm ngle cross section of Pro. 9.78 *9.0 MHR S IRLE FR MMENTS ND PRDUTS F INERTI The circle used in the preceding section to illustrte the reltions eisting etween the moments nd products of inerti of given re with respect to es pssing through fied point ws first introduced the Germn engineer tto Mohr (835 98) nd is known s Mohr s circle. It will e shown tht if the moments nd product of inerti of n re re known with respect to two rectngulr nd es which pss through point, Mohr s circle cn e used to grphicll determine () the principl es nd principl moments of inerti of the re out nd () the moments nd product of inerti of the re with respect to n other pir of rectngulr es 9 nd 9 through. onsider given re nd two rectngulr coordinte es nd (Fig. 9.9). ssuming tht the moments of inerti I nd I nd the product of inerti I re known, we will represent them on digrm plotting point X of coordintes I nd I nd point Y of coordintes I nd I (Fig. 9.9). If I is positive, s ssumed in Fig. 9.9, point X is locted ove the horiontl is nd point Y is locted elow, s shown in Fig. 9.9. If I is negtive, X is locted elow the horiontl is nd Y is locted ove. Joining X nd Y with stright line, we denote the point of intersection of line XY with the horiontl is nd drw the circle of center nd dimeter XY. Noting tht the sciss of nd the rdius of the circle re respectivel equl to the quntities I ve nd R defined the formul (9.3), we conclude tht the circle otined is Mohr s circle for the given re out point. Thus, the scisss of the points nd where the circle intersects the horiontl is represent respectivel the principl moments of inerti I m nd I min of the re. We lso note tht, since tn (X) 5 I /(I I ), the ngle X is equl in mgnitude to one of the ngles u m which stisf Eq. (9.5);
' I I 9.0 Mohr s ircle for Moments nd Products of Inerti 507 q m q ' I min I I '' I ' X' X q q m I I '' I, I () Y Y' I I ' I m Fig. 9.9 () thus, the ngle u m, which defines in Fig. 9.9 the principl is corresponding to point in Fig. 9.9, is equl to hlf of the ngle X of Mohr s circle. We further oserve tht if I. I nd I. 0, s in the cse considered here, the rottion which rings X into is clockwise. lso, under these conditions, the ngle u m otined from Eq. (9.5), which defines the principl is in Fig. 9.9, is negtive; thus, the rottion which rings into is lso clockwise. We conclude tht the senses of rottion in oth prts of Fig. 9.9 re the sme. If clockwise rottion through u m is required to ring X into on Mohr s circle, clockwise rottion through u m will ring into the corresponding principl is in Fig. 9.9. Since Mohr s circle is uniquel defined, the sme circle cn e otined considering the moments nd product of inerti of the re with respect to the rectngulr es 9 nd 9 (Fig. 9.9). The point X9 of coordintes I 9 nd I 99 nd the point Y9 of coordintes I 9 nd I 99 re thus locted on Mohr s circle, nd the ngle X9 in Fig. 9.9 must e equl to twice the ngle 9 in Fig. 9.9. Since, s noted ove, the ngle X is twice the ngle, it follows tht the ngle XX9 in Fig. 9.9 is twice the ngle 9 in Fig. 9.9. The dimeter X9Y9, which defines the moments nd product of inerti I 9, I 9, nd I 99 of the given re with respect to rectngulr es 9 nd 9 forming n ngle u with the nd es cn e otined rotting through n ngle u the dimeter XY which corresponds to the moments nd product of inerti I, I, nd I. We note tht the rottion which rings the dimeter XY into the dimeter X9Y9 in Fig. 9.9 hs the sme sense s the rottion which rings the nd es into the 9 nd 9 es in Fig. 9.9. It should e noted tht the use of Mohr s circle is not limited to grphicl solutions, i.e., to solutions sed on the creful drwing nd mesuring of the vrious prmeters involved. merel sketching Mohr s circle nd using trigonometr, one cn esil derive the vrious reltions required for numericl solution of given prolem (see Smple Pro. 9.8).
SMPLE PRLEM 9.8 L5 0.7 q = 60 For the section shown, the moments nd product of inerti with respect to the nd es re known to e I 5 7.0 3 0 6 mm 4 I 5.59 3 0 6 mm 4 I 5.54 3 0 6 mm 4 Using Mohr s circle, determine () the principl es of the section out, () the vlues of the principl moments of inerti of the section out, (c) the moments nd product of inerti of the section with respect to the 9 nd 9 es which form n ngle of 60 with the nd es. I (0 6 mm 4 ) Y(.59, +.54) I 3.430 0 6 mm 4 E D q m I, I (0 6 mm 4 ) 4.895 0 6 mm 4 Y q m = 3.9 q = 0 f G F I, I Y' X(7.0,.54) X' X q m = 47.8 SLUTIN Drwing Mohr s ircle. We first plot point X of coordintes I 5 7.0, I 5.54, nd point Y of coordintes I 5.59, I 5.54. Joining X nd Y with stright line, we define the center of Mohr s circle. The sciss of, which represents I ve, nd the rdius R of the circle cn e mesured directl or clculted s follows: I ve 5 5 (I I ) 5 (7.0 3 0 6.59 3 0 6 ) 5 4.895 3 0 6 mm 4 D 5 (I I ) 5 (7.0 3 0 6.59 3 0 6 ) 5.305 3 0 6 mm 4 R 5 (D) (DX) 5 (.305 3 0 6 ) (.54 3 0 6 ) 5 3.430 3 0 6 mm 4. Principl es. The principl es of the section correspond to points nd on Mohr s circle, nd the ngle through which we should rotte X to ring it into defines u m. We hve tn u m 5 DX D 5.54.305 5.0 u m 5 47.8 l u m 5 3.9 l Thus, the principl is corresponding to the mimum vlue of the moment of inerti is otined rotting the is through 3.9 counterclockwise; the principl is corresponding to the minimum vlue of the moment of inerti cn e otined rotting the is through the sme ngle.. Principl Moments of Inerti. The principl moments of inerti re represented the scisss of nd. We hve I m 5 5 5 I ve R 5 (4.895 3.430)0 6 mm 4 I m 5 8.33 3 0 6 mm 4 I min 5 5 5 I ve R 5 (4.895 3.430)0 6 mm 4 I min 5.47 3 0 6 mm 4 c. Moments nd Product of Inerti with Respect to the 9 nd 9 es. n Mohr s circle, the points X9 nd Y9, which correspond to the 9 nd 9 es, re otined rotting X nd Y through n ngle u 5 (60 ) 5 0 counterclockwise. The coordintes of X9 nd Y9 ield the desired moments nd product of inerti. Noting tht the ngle tht X9 forms with the horiontl is is f 5 0 47.8 5 7., we write I 9 5 F 5 F 5 4.895 3 0 6 mm 4 (3.430 3 0 6 mm 4 ) cos 7. I 5 5.94 3 0 6 mm 4 I 9 5 G 5 G 5 4.895 3 0 6 mm 4 (3.430 3 0 6 mm 4 ) cos 7. I 5 3.85 3 0 6 mm 4 I 99 5 FX9 5 (3.430 3 0 6 mm 4 ) sin 7. I 5 3.7 3 0 6 mm 4 508
SLVING PRLEMS N YUR WN In the prolems for this lesson, ou will use Mohr s circle to determine the moments nd products of inerti of given re for different orienttions of the coordinte es. lthough in some cses using Mohr s circle m not e s direct s sustituting into the pproprite equtions [Eqs. (9.8) through (9.0)], this method of solution hs the dvntge of providing visul representtion of the reltionships mong the vrious vriles. Further, Mohr s circle shows ll of the vlues of the moments nd products of inerti which re possile for given prolem. Using Mohr s circle. The underling theor ws presented in Sec. 9.9, nd we discussed the ppliction of this method in Sec. 9.0 nd in Smple Pro. 9.8. In the sme prolem, we presented the steps ou should follow to determine the principl es, the principl moments of inerti, nd the moments nd product of inerti with respect to specified orienttion of the coordintes es. When ou use Mohr s circle to solve prolems, it is importnt tht ou rememer the following points.. Mohr s circle is completel defined the quntities R nd I ve, which represent, respectivel, the rdius of the circle nd the distnce from the origin to the center of the circle. These quntities cn e otined from Eqs. (9.3) if the moments nd product of inerti re known for given orienttion of the es. However, Mohr s circle cn e defined other comintions of known vlues [Pros. 9.03, 9.06, nd 9.07]. For these cses, it m e necessr to first mke one or more ssumptions, such s choosing n ritrr loction for the center when I ve is unknown, ssigning reltive mgnitudes to the moments of inerti (for emple, I. I ), or selecting the sign of the product of inerti.. Point X of coordintes (I, I ) nd point Y of coordintes (I, I ) re oth locted on Mohr s circle nd re dimetricll opposite. c. Since moments of inerti must e positive, the entire Mohr s circle must lie to the right of the I is; it follows tht I ve. R for ll cses. d. s the coordinte es re rotted through n ngle U, the ssocited rottion of the dimeter of Mohr s circle is equl to u nd is in the sme sense (clockwise or counterclockwise). We strongl suggest tht the known points on the circumference of the circle e leled with the pproprite cpitl letter, s ws done in Fig. 9.9 nd for the Mohr circles of Smple Pro. 9.8. This will enle ou to determine, for ech vlue of u, the sign of the corresponding product of inerti nd to determine which moment of inerti is ssocited with ech of the coordinte es [Smple Pro. 9.8, prts nd c]. lthough we hve introduced Mohr s circle within the specific contet of the stud of moments nd products of inerti, the Mohr circle technique is lso pplicle to the solution of nlogous ut phsicll different prolems in mechnics of mterils. This multiple use of specific technique is not unique, nd s ou pursue our engineering studies, ou will encounter severl methods of solution which cn e pplied to vriet of prolems. 509
PRLEMS 9.9 Using Mohr s circle, determine for the qurter ellipse of Pro. 9.67 the moments of inerti nd the product of inerti with respect to new es otined rotting the nd es out () through 458 counterclockwise, () through 308 clockwise. 9.9 Using Mohr s circle, determine the moments of inerti nd the product of inerti of the re of Pro. 9.7 with respect to new centroidl es otined rotting the nd es 30 counterclockwise. 9.93 Using Mohr s circle, determine the moments of inerti nd the product of inerti of the re of Pro. 9.73 with respect to new centroidl es otined rotting the nd es 608 counterclockwise. 9.94 Using Mohr s circle, determine the moments of inerti nd the product of inerti of the re of Pro. 9.75 with respect to new centroidl es otined rotting the nd es 458 clockwise. 9.95 Using Mohr s circle, determine the moments of inerti nd the product of inerti of the L3 3 3 4-in. ngle cross section of Pro. 9.74 with respect to new centroidl es otined rotting the nd es 308 clockwise. 9.96 Using Mohr s circle, determine the moments of inerti nd the product of inerti of the L7 3 76 3.7-mm ngle cross section of Pro. 9.78 with respect to new centroidl es otined rotting the nd es 458 counterclockwise. 9.97 For the qurter ellipse of Pro. 9.67, use Mohr s circle to determine the orienttion of the principl es t the origin nd the corresponding vlues of the moments of inerti. 9.98 through 9.0 Using Mohr s circle, determine for the re indicted the orienttion of the principl centroidl es nd the corresponding vlues of the moments of inerti. 9.98 re of Pro. 9.7 9.99 re of Pro. 9.76 9.00 re of Pro. 9.73 9.0 re of Pro. 9.74 9.0 re of Pro. 9.77 (The moments of inerti I nd I of the re of Pro. 9.0 were determined in Pro. 9.44.) 50 9.03 The moments nd product of inerti of n L4 3 3 3 4-in. ngle cross section with respect to two rectngulr es nd through re, respectivel, I 5.33 in 4, I 5.75 in 4, nd I, 0, with the minimum vlue of the moment of inerti of the re with respect to n is through eing I min 5 0.69 in 4. Using Mohr s circle, determine () the product of inerti I of the re, () the orienttion of the principl es, (c) the vlue of I m.
9.04 nd 9.05 Using Mohr s circle, determine for the cross section of the rolled-steel ngle shown the orienttion of the principl centroidl es nd the corresponding vlues of the moments of inerti. (Properties of the cross sections re given in Fig. 9.3.) Prolems 5 6.4 mm 4.9 mm 8.9 mm.7 mm 5 mm L76 5 6.4.4 mm L7 76.7 Fig. P9.04 76 mm 6.4 mm 7 mm 44. mm *9.06 For given re the moments of inerti with respect to two rectngulr centroidl nd es re I 5 00 in 4 nd I 5 300 in 4, respectivel. Knowing tht fter rotting the nd es out the centroid 308 counterclockwise, the moment of inerti reltive to the rotted is is 450 in 4, use Mohr s circle to determine () the orienttion of the principl es, () the principl centroidl moments of inerti. 9.07 It is known tht for given re I 5 48 3 0 6 mm 4 nd I 5 0 3 0 6 mm 4, where the nd es re rectngulr centroidl es. If the is corresponding to the mimum product of inerti is otined rotting the is 67.58 counterclockwise out, use Mohr s circle to determine () the moment of inerti I of the re, () the principl centroidl moments of inerti. 9.08 Using Mohr s circle, show tht for n regulr polgon (such s pentgon) () the moment of inerti with respect to ever is through the centroid is the sme, () the product of inerti with respect to ever pir of rectngulr es through the centroid is ero. 9.09 Using Mohr s circle, prove tht the epression I 9 I 9 I 99 is independent of the orienttion of the 9 nd 9 es, where I 9, I 9, nd I 99 represent the moments nd product of inerti, respectivel, of given re with respect to pir of rectngulr es 9 nd 9 through given point. lso show tht the given epression is equl to the squre of the length of the tngent drwn from the origin of the coordinte sstem to Mohr s circle. 9.0 Using the invrince propert estlished in the preceding prolem, epress the product of inerti I of n re with respect to pir of rectngulr es through in terms of the moments of inerti I nd I of nd the principl moments of inerti I min nd I m of out. Use the formul otined to clculte the product of inerti I of the L3 3 3 4-in. ngle cross section shown in Fig. 9.3, knowing tht its mimum moment of inerti is.57 in 4. Fig. P9.05 76 mm.7 mm
5 Distriuted Forces: Moments of Inerti MMENTS F INERTI F MSSES 9. MMENT F INERTI F MSS onsider smll mss Dm mounted on rod of negligile mss which cn rotte freel out n is 9 (Fig. 9.0). If couple is pplied to the sstem, the rod nd mss, ssumed to e initill t rest, will strt rotting out 9. The detils of this motion will e studied lter in dnmics. t present, we wish onl to indicte tht the time required for the sstem to rech given speed of rottion is proportionl to the mss Dm nd to the squre of the distnce r. The product r Dm provides, therefore, mesure of the inerti of the sstem, i.e., mesure of the resistnce the sstem offers when we tr to set it in motion. For this reson, the product r Dm is clled the moment of inerti of the mss Dm with respect to the is 9. ' ' ' r Δm r Δm Δm r r 3 k m Δm 3 () () (c) Fig. 9.0 onsider now od of mss m which is to e rotted out n is 9 (Fig. 9.0). Dividing the od into elements of mss Dm, Dm, etc., we find tht the od s resistnce to eing rotted is mesured the sum r Dm r Dm.... This sum defines, therefore, the moment of inerti of the od with respect to the is 9. Incresing the numer of elements, we find tht the moment of inerti is equl, in the limit, to the integrl I 5 # r dm (9.8)
The rdius of grtion k of the od with respect to the is 9 is defined the reltion 9. Moment of Inerti of Mss 53 I 5 k m or k 5 I m (9.9) The rdius of grtion k represents, therefore, the distnce t which the entire mss of the od should e concentrted if its moment of inerti with respect to 9 is to remin unchnged (Fig. 9.0c). Whether it is kept in its originl shpe (Fig. 9.0) or whether it is concentrted s shown in Fig. 9.0c, the mss m will rect in the sme w to rottion, or grtion, out 9. If SI units re used, the rdius of grtion k is epressed in meters nd the mss m in kilogrms, nd thus the unit used for the moment of inerti of mss is kg? m. If U.S. customr units re used, the rdius of grtion is epressed in feet nd the mss in slugs (i.e., in l s /ft), nd thus the derived unit used for the moment of inerti of mss is l? ft? s. The moment of inerti of od with respect to coordinte is cn esil e epressed in terms of the coordintes,, of the element of mss dm (Fig. 9.). Noting, for emple, tht the squre of the distnce r from the element dm to the is is, we epress the moment of inerti of the od with respect to the is s I 5 # r dm 5 # ( ) dm r dm Similr epressions cn e otined for the moments of inerti with respect to the nd es. We write Fig. 9. I 5 # ( ) dm I 5 # ( ) dm (9.30) I 5 # ( ) dm It should e kept in mind when converting the moment of inerti of mss from U.S. customr units to SI units tht the se unit pound used in the derived unit l? ft? s is unit of force (not of mss) nd should therefore e converted into newtons. We hve or, since N 5 kg? m/s, l? ft? s 5 (4.45 N)(0.3048 m)( s) 5.356 N? m? s Photo 9. s ou will discuss in our dnmics course, the rottionl ehvior of the cmshft shown is dependent upon the mss moment of inerti of the cmshft with respect to its is of rottion. l? ft? s 5.356 kg? m
54 Distriuted Forces: Moments of Inerti 9. PRLLEL-XIS THEREM ' dm onsider od of mss m. Let e sstem of rectngulr coordintes whose origin is t the ritrr point, nd G999 sstem of prllel centroidl es, i.e., sstem whose origin is t the center of grvit G of the od nd whose es 9, 9, 9 re prllel to the,, nd es, respectivel (Fig. 9.). Denoting,, the coordintes of G with respect to, we write the following reltions etween the coordintes,, of the element dm with respect to nd its coordintes 9, 9, 9 with respect to the centroidl es G999: ' Fig. 9. G ' 5 9 5 9 5 9 (9.3) Referring to Eqs. (9.30), we cn epress the moment of inerti of the od with respect to the is s follows: I 5 # ( ) dm 5 # [( ) ( ) ] dm 5 # ( ) dm # dm # dm ( ) # dm The first integrl in this epression represents the moment of inerti I of the od with respect to the centroidl is 9; the second nd third integrls represent the first moment of the od with respect to the 99 nd 99 plnes, respectivel, nd, since oth plnes contin G, the two integrls re ero; the lst integrl is equl to the totl mss m of the od. We write, therefore, I 5 I m( ) (9.3) nd, similrl, I 5 I m( ) I 5 I m( ) (9.39) ' d ' We esil verif from Fig. 9. tht the sum represents the squre of the distnce etween the nd 9 es. Similrl, nd represent the squres of the distnce etween the nd 9 es nd the nd 9 es, respectivel. Denoting d the distnce etween n ritrr is 9 nd prllel centroidl is 9 (Fig. 9.3), we cn, therefore, write the following generl reltion etween the moment of inerti I of the od with respect to 9 nd its moment of inerti I with respect to 9: G I 5 I md (9.33) Fig. 9.3 Epressing the moments of inerti in terms of the corresponding rdii of grtion, we cn lso write k 5 k d (9.34) where k nd k represent the rdii of grtion of the od out 9 nd 9, respectivel. Note tht the term centroidl is used here to define n is pssing through the center of grvit G of the od, whether or not G coincides with the centroid of the volume of the od.
9.3 MMENTS F INERTI F THIN PLTES onsider thin plte of uniform thickness t, which is mde of homogeneous mteril of densit r (densit 5 mss per unit volume). The mss moment of inerti of the plte with respect to n is 9 contined in the plne of the plte (Fig. 9.4) is Since dm 5 rt d, we write I, mss 5 # r dm 9.3 Moments of Inerti of Thin Pltes 55 I, mss 5 rt # r d ut r represents the distnce of the element of re d to the is ' ' t d r t d r ' t d r ' ' () () (c) Fig. 9.4 9; the integrl is therefore equl to the moment of inerti of the re of the plte with respect to 9. We hve I 9,mss 5 rti 9,re (9.35) Similrl, for n is 9 which is contined in the plne of the plte nd is perpendiculr to 9 (Fig. 9.4), we hve I 9,mss 5 rti 9,re (9.36) onsidering now the is 9 which is perpendiculr to the plte nd psses through the point of intersection of 9 nd 9 (Fig. 9.4c), we write I 9,mss 5 rtj,re (9.37) where J is the polr moment of inerti of the re of the plte with respect to point. Reclling the reltion J 5 I 9 I 9 which eists etween polr nd rectngulr moments of inerti of n re, we write the following reltion etween the mss moments of inerti of thin plte: I 9 5 I 9 I 9 (9.38)
56 Distriuted Forces: Moments of Inerti Rectngulr Plte. In the cse of rectngulr plte of sides nd (Fig. 9.5), we otin the following mss moments of inerti ' with respect to es through the center of grvit of the plte: t Fig. 9.5 d ' t ' Fig. 9.6 ' dm = r r d di = r dm di = di ' + dm = ( r + 4 ) dm di = di ' + dm = ( r + 4 ) dm Fig. 9.7 Determintion of the moment of inerti of od of revolution. r ' r ' ' ' I,mss 5 rti,re 5 rt( 3 ) I,mss 5 rti,re 5 rt( 3 ) serving tht the product rt is equl to the mss m of the plte, we write the mss moments of inerti of thin rectngulr plte s follows: I 5 m I 5 m (9.39) I 5 I I 5 m( ) (9.40) irculr Plte. In the cse of circulr plte, or disk, of rdius r (Fig. 9.6), we write I,mss 5 rti,re 5 rt( 4 pr 4 ) serving tht the product rpr t is equl to the mss m of the plte nd tht I 9 5 I 9, we write the mss moments of inerti of circulr plte s follows: I 5 I 5 4 mr (9.4) I 5 I I 5 mr (9.4) 9.4 DETERMINTIN F THE MMENT F INERTI F THREE-DIMENSINL DY Y INTEGRTIN The moment of inerti of three-dimensionl od is otined evluting the integrl I 5 e r dm. If the od is mde of homogeneous mteril of densit r, the element of mss dm is equl to r dv nd we cn write I 5 re r dv. This integrl depends onl upon the shpe of the od. Thus, in order to compute the moment of inerti of three-dimensionl od, it will generll e necessr to perform triple, or t lest doule, integrtion. However, if the od possesses two plnes of smmetr, it is usull possile to determine the od s moment of inerti with single integrtion choosing s the element of mss dm thin sl which is perpendiculr to the plnes of smmetr. In the cse of odies of revolution, for emple, the element of mss would e thin disk (Fig. 9.7). Using formul (9.4), the moment of inerti of the disk with respect to the is of revolution cn e epressed s indicted in Fig. 9.7. Its moment of inerti with respect to ech of the other two coordinte es is otined using formul (9.4) nd the prllel-is theorem. Integrtion of the epression otined ields the desired moment of inerti of the od. 9.5 MMENTS F INERTI F MPSITE DIES The moments of inerti of few common shpes re shown in Fig. 9.8. For od consisting of severl of these simple shpes, the moment of inerti of the od with respect to given is cn e otined first computing the moments of inerti of its component prts out the desired is nd then dding them together. s ws the cse for res, the rdius of grtion of composite od cnnot e otined dding the rdii of grtion of its componet prts.
9.5 Moments of Inerti of omposite odies 57 Slender rod L G I = I = ml Thin rectngulr plte c G I = m( + c ) I = mc I = m Rectngulr prism c I = m( + c ) I = m(c + ) I = m( + ) Thin disk r I = mr I = I = mr 4 irculr clinder L I = m I = I = m(3 + L ) irculr cone h r 3 I = m 0 3 I = I = m( + h ) 5 4 Sphere I = I = I = m 5 Fig. 9.8 Mss moments of inerti of common geometric shpes.
SMPLE PRLEM 9.9 L Determine the moment of inerti of slender rod of length L nd mss m with respect to n is which is perpendiculr to the rod nd psses through one end of the rod. SLUTIN hoosing the differentil element of mss shown, we write L d dm 5 m L d L I 5 # dm 5 # m L d 5 c m L 0 3 L 3 d 0 I 5 3 ml SMPLE PRLEM 9.0 For the homogeneous rectngulr prism shown, determine the moment of inerti with respect to the is. c SLUTIN d We choose s the differentil element of mss the thin sl shown; thus dm 5 rc d Referring to Sec. 9.3, we find tht the moment of inerti of the element with respect to the 9 is is ' di 5 dm ppling the prllel-is theorem, we otin the mss moment of inerti of the sl with respect to the is. di 5 di dm 5 dm dm 5 ( )rc d Integrting from 5 0 to 5, we otin I 5 # di 5 # 0 ( )rc d 5 rc( 3 ) Since the totl mss of the prism is m 5 rc, we cn write I 5 m( 3 ) I 5 m(4 ) We note tht if the prism is thin, is smll compred to, nd the epression for I reduces to 3m, which is the result otined in Smple Pro. 9.9 when L 5. 58
h SMPLE PRLEM 9. Determine the moment of inerti of right circulr cone with respect to () its longitudinl is, () n is through the pe of the cone nd perpendiculr to its longitudinl is, (c) n is through the centroid of the cone nd perpendiculr to its longitudinl is. ' r d SLUTIN We choose the differentil element of mss shown. r 5 h dm 5 rpr d 5 rp h d h. Moment of Inerti I. Using the epression derived in Sec. 9.3 for thin disk, we compute the mss moment of inerti of the differentil element with respect to the is. di 5 r dm 5 h rp h d 5 4 rp h 4 4 d Integrting from 5 0 to 5 h, we otin I 5 # di 5 # h 0 4 rp h 4 4 d 5 4 h 5 rp h 4 5 5 0 rp 4 h Since the totl mss of the cone is m 5 3rp h, we cn write I 5 0 rp 4 h 5 3 0 ( 3 rp h) 5 3 0 m I 5 3 0 m. Moment of Inerti I. The sme differentil element is used. ppling the prllel-is theorem nd using the epression derived in Sec. 9.3 for thin disk, we write di 5 di dm 5 4 r dm dm 5 ( 4 r ) dm Sustituting the epressions for r nd dm into the eqution, we otin di 5 4 h rp h d 5 rp h 4h 4 d h I 5 # di 5 # rp 0 h 4h 4 d 5 rp h 4h h 5 5 Introducing the totl mss of the cone m, we rewrite I s follows: " I 5 3 5( 4 h ) 3rp h I 5 3 5 m( 4 h ) = 3 h 4 c. Moment of Inerti I 0. We ppl the prllel-is theorem nd write I 5 I m h Solving for I nd reclling tht 5 3 4h, we hve I 5 I m 5 3 5 m( 4 h ) m( 3 4 h) I 5 3 0 m( 4 h ) 59
3 in. in. in. SMPLE PRLEM 9. steel forging consists of 6 3 3 -in. rectngulr prism nd two clinders of dimeter in. nd length 3 in. s shown. Determine the moments of inerti of the forging with respect to the coordinte es, knowing tht the specific weight of steel is 490 l/ft 3. in. in. SLUTIN in. in. in..5 in. 3 in. in. in. in. 6 in. omputtion of Msses Prism V 5 ( in.)( in.)(6 in.) 5 4 in 3 W 5 (4 in3 )(490 l/ft 3 ) 5 6.8 l 78 in 3 /ft 3 6.8 l m 5 3. ft/s 5 0. l? s /ft Ech linder V 5 p( in.) (3 in.) 5 9.4 in 3 W 5 (9.4 in3 )(490 l/ft 3 ) 5.67 l 78 in 3 /ft 3.67 l m 5 3. ft/s 5 0.089 l? s /ft Moments of Inerti. The moments of inerti of ech component re computed from Fig. 9.8, using the prllel-is theorem when necessr. Note tht ll lengths should e epressed in feet. Prism I 5 I 5 (0. l? s /ft)[( 6 ft) ( ft) ] 5 4.88 3 0 3 l? ft? s I 5 (0. l? s /ft)[( ft) ( ft) ] 5 0.977 3 0 3 l? ft? s Ech linder I 5 m m 5 (0.089 l? s /ft)( ft) (0.089 l? s /ft)( ft) 5.59 3 0 3 l? ft? s I 5 m(3 L ) 5 m 5 (0.089 l? s /ft)[3( ft) ( 3 ft) ] (0.089 l? s /ft)(.5 ft) 5 4.7 3 0 3 l? ft? s I 5 m(3 L ) m( ) 5 (0.089 l? s /ft)[3( ft) ( 3 ft) ] (0.089 l? s /ft)[(.5 ft) ( ft) ] 5 6.48 3 0 3 l? ft? s Entire od. dding the vlues otined, I 5 4.88 3 0 3 (.59 3 0 3 ) I 5 0.06 3 0 3 l? ft? s I 5 0.977 3 0 3 (4.7 3 0 3 ) I 5 9.3 3 0 3 l? ft? s I 5 4.88 3 0 3 (6.48 3 0 3 ) I 5 7.84 3 0 3 l? ft? s 50
SMPLE PRLEM 9.3 thin steel plte which is 4 mm thick is cut nd ent to form the mchine prt shown. Knowing tht the densit of steel is 7850 kg/m 3, determine the moments of inerti of the mchine prt with respect to the coordinte es. + _ d = 0. m r = 0.08 m = 0. m = 0.05 m SLUTIN We oserve tht the mchine prt consists of semicirculr plte nd rectngulr plte from which circulr plte hs een removed. omputtion of Msses. Semicirculr Plte V 5 pr t 5 p(0.08 m) (0.004 m) 5 40. 3 0 6 m 3 m 5 rv 5 (7.85 3 0 3 kg/m 3 )(40. 3 0 6 m 3 ) 5 0.356 kg Rectngulr Plte V 5 (0.00 m)(0.60 m)(0.004 m) 5 8 3 0 6 m 3 m 5 rv 5 (7.85 3 0 3 kg/m 3 )(8 3 0 6 m 3 ) 5.005 kg irculr Plte V 3 5 p t 5 p(0.050 m) (0.004 m) 5 3.4 3 0 6 m 3 m 3 5 rv 3 5 (7.85 3 0 3 kg/m 3 )(3.4 3 0 6 m 3 ) 5 0.466 kg c = 0.6 m Moments of Inerti. Using the method presented in Sec. 9.3, we compute the moments of inerti of ech component. Semicirculr Plte. From Fig. 9.8, we oserve tht for circulr plte of mss m nd rdius r I 5 mr I 5 I 5 4 mr ecuse of smmetr, we note tht for semicirculr plte I 5 ( mr ) I 5 I 5 ( 4 mr ) Since the mss of the semicirculr plte is m 5 m, we hve I 5 m r 5 (0.356 kg)(0.08 m) 5.00 3 0 3 kg? m I 5 I 5 4( mr )5 4 m r 5 4(0.356 kg)(0.08 m) 5 0.505 3 0 3 kg? m Rectngulr Plte I 5 m c 5 (.005 kg)(0.6 m) 5.44 3 0 3 kg? m I 5 3 m 5 3(.005 kg)(0. m) 5 3.400 3 0 3 kg? m I 5 I I 5 (.44 3.400)(0 3 ) 5 5.544 3 0 3 kg? m irculr Plte I 5 4 m 3 5 4(0.466 kg)(0.05 m) 5 0.54 3 0 3 kg? m I 5 m 3 m 3 d 5 (0.466 kg)(0.05 m) (0.466 kg)(0. m) 5.774 3 0 3 kg? m I 5 4 m 3 m 3 d 5 4(0.466 kg)(0.05 m) (0.466 kg)(0. m) 5.60 3 0 3 kg? m Entire Mchine Prt I 5 (.00.44 0.54)(0 3 ) kg? m I 5 3.00 3 0 3 kg? m I 5 (0.505 5.544.774)(0 3 ) kg? m I 5 3.8 3 0 3 kg? m I 5 (0.505 3.400.60)(0 3 ) kg? m I 5.9 3 0 3 kg? m 5
SLVING PRLEMS N YUR WN In this lesson we introduced the mss moment of inerti nd the rdius of grtion of three-dimensionl od with respect to given is [Eqs. (9.8) nd (9.9)]. We lso derived prllel-is theorem for use with mss moments of inerti nd discussed the computtion of the mss moments of inerti of thin pltes nd three-dimensionl odies.. omputing mss moments of inerti. The mss moment of inerti I of od with respect to given is cn e clculted directl from the definition given in Eq. (9.8) for simple shpes [Smple Pro. 9.9]. In most cses, however, it is necessr to divide the od into thin sls, compute the moment of inerti of tpicl sl with respect to the given is using the prllel-is theorem if necessr nd integrte the epression otined.. ppling the prllel-is theorem. In Sec. 9. we derived the prllel-is theorem for mss moments of inerti I 5 I md (9.33) which sttes tht the moment of inerti I of od of mss m with respect to given is is equl to the sum of the moment of inerti I of tht od with respect to prllel centroidl is nd the product md, where d is the distnce etween the two es. When the moment of inerti of three-dimensionl od is clculted with respect to one of the coordinte es, d cn e replced the sum of the squres of distnces mesured long the other two coordinte es [Eqs. (9.3) nd (9.3 )]. 3. voiding unit-relted errors. To void errors, it is essentil tht ou e consistent in our use of units. Thus, ll lengths should e epressed in meters or feet, s pproprite, nd for prolems using U.S. customr units, msses should e given in l? s /ft. In ddition, we strongl recommend tht ou include units s ou perform our clcultions [Smple Pros. 9. nd 9.3]. 4. lculting the mss moment of inerti of thin pltes. We showed in Sec. 9.3 tht the mss moment of inerti of thin plte with respect to given is cn e otined multipling the corresponding moment of inerti of the re of the plte the densit r nd the thickness t of the plte [Eqs. (9.35) through (9.37)]. Note tht since the is in Fig. 9.4c is perpendiculr to the plte, I,mss is ssocited with the polr moment of inerti J,re. Insted of clculting directl the moment of inerti of thin plte with respect to specified is, ou m sometimes find it convenient to first compute its moment of inerti with respect to n is prllel to the specified is nd then ppl the prllel-is theorem. Further, to determine the moment of inerti of thin plte with respect to n is perpendiculr to the plte, ou m wish to first determine its moments of inerti with respect to two perpendiculr in-plne es nd then use Eq. (9.38). Finll, rememer tht the mss of plte of re, thickness t, nd densit r is m 5 rt. 5
5. Determining the moment of inerti of od direct single integrtion. We discussed in Sec. 9.4 nd illustrted in Smple Pros. 9.0 nd 9. how single integrtion cn e used to compute the moment of inerti of od tht cn e divided into series of thin, prllel sls. For such cses, ou will often need to epress the mss of the od in terms of the od s densit nd dimensions. ssuming tht the od hs een divided, s in the smple prolems, into thin sls perpendiculr to the is, ou will need to epress the dimensions of ech sl s functions of the vrile.. In the specil cse of od of revolution, the elementl sl is thin disk, nd the equtions given in Fig. 9.7 should e used to determine the moments of inerti of the od [Smple Pro. 9.].. In the generl cse, when the od is not of revolution, the differentil element is not disk, ut thin sl of different shpe, nd the equtions of Fig. 9.7 cnnot e used. See, for emple, Smple Pro. 9.0, where the element ws thin, rectngulr sl. For more comple configurtions, ou m wnt to use one or more of the following equtions, which re sed on Eqs. (9.3) nd (9.3 ) of Sec. 9.. di 5 di ( el di 5 di ( el di 5 di ( el el) dm el) dm el) dm where the primes denote the centroidl es of ech elementl sl, nd where el, el, nd el represent the coordintes of its centroid. The centroidl moments of inerti of the sl re determined in the mnner descried erlier for thin plte: Referring to Fig. 9. on pge 485, clculte the corresponding moments of inerti of the re of the sl nd multipl the result the densit r nd the thickness t of the sl. lso, ssuming tht the od hs een divided into thin sls perpendiculr to the is, rememer tht ou cn otin di 9 dding di 9 nd di 9 insted of computing it directl. Finll, using the geometr of the od, epress the result otined in terms of the single vrile nd integrte in. 6. omputing the moment of inerti of composite od. s stted in Sec. 9.5, the moment of inerti of composite od with respect to specified is is equl to the sum of the moments of its components with respect to tht is. Smple Pros. 9. nd 9.3 illustrte the pproprite method of solution. You must lso rememer tht the moment of inerti of component will e negtive onl if the component is removed (s in the cse of hole). lthough the composite-od prolems in this lesson re reltivel strightforwrd, ou will hve to work crefull to void computtionl errors. In ddition, if some of the moments of inerti tht ou need re not given in Fig. 9.8, ou will hve to derive our own formuls, using the techniques of this lesson. 53
PRLEMS 9. Determine the mss moment of inerti of ring of mss m, cut from thin uniform plte, with respect to () the is 9, () the centroidl is 9 tht is perpendiculr to the plne of the ring. r r 9. thin semicirculr plte hs rdius nd mss m. Determine the mss moment of inerti of the plte with respect to () the centroidl is 9, () the centroidl is 9 tht is perpendiculr to the plte. Fig. P9. ' ' ' r ' ' r Fig. P9. Fig. P9.3 ' 9.3 The qurter ring shown hs mss m nd ws cut from thin, uniform plte. Knowing tht r 5 3 4 r, determine the mss moment of inerti of the qurter ring with respect to () the is 9, () the centroidl is 9 tht is perpendiculr to the plne of the qurter ring. 9.4 The prolic spndrel shown ws cut from thin, uniform plte. Denoting the mss of the spndrel m, determine its mss moment of inerti with respect to () the is 9, () the is DD9 tht is perpendiculr to the spndrel. (Hint: See Smple Pro. 9.3.) ' = k ' D D' 54 Fig. P9.4
9.5 thin plte of mss m ws cut in the shpe of prllelogrm s shown. Determine the mss moment of inerti of the plte with respect to () the is, () the is 9, which is perpendiculr to the plte. Prolems 55 9.6 thin plte of mss m ws cut in the shpe of prllelogrm s shown. Determine the mss moment of inerti of the plte with respect to () the is, () the is 9, which is perpendiculr to the plte. 9.7 thin plte of mss m hs the trpeoidl shpe shown. Determine the mss moment of inerti of the plte with respect to () the is, () the is. ' ' 9.8 thin plte of mss m hs the trpeoidl shpe shown. Determine the mss moment of inerti of the plte with respect to () the centroidl is 9 tht is perpendiculr to the plte, () the is 9 tht is prllel to the is nd is locted t distnce.5 from the plte. Fig. P9.5 nd P9.6 ' ' 9.9 The re shown is revolved out the is to form homogeneous solid of revolution of mss m. Using direct integrtion, epress the mss moment of inerti of the solid with respect to the is in terms of m nd h..5 h Fig. P9.7 nd P9.8 h Fig. P9.9 9.0 Determine direct integrtion the mss moment of inerti with respect to the is of the right circulr clinder shown, ssuming tht it hs uniform densit nd mss m. 9. The re shown is revolved out the is to form homogeneous solid of revolution of mss m. Determine direct integrtion the mss moment of inerti of the solid with respect to () the is, () the is. Epress our nswers in terms of m nd the dimensions of the solid. L L = k Fig. P9.0 h Fig. P9.
56 Distriuted Forces: Moments of Inerti 9. Determine direct integrtion the mss moment of inerti with respect to the is of the tetrhedron shown, ssuming tht it hs uniform densit nd mss m. 9.3 Determine direct integrtion the mss moment of inerti with respect to the is of the tetrhedron shown, ssuming tht it hs uniform densit nd mss m. h *9.4 Determine direct integrtion the mss moment of inerti with respect to the is of the semiellipsoid shown, ssuming tht it hs uniform densit nd mss m. Fig. P9. nd P9.3 + + = c c Fig. P9.4 *9.5 thin steel wire is ent into the shpe shown. Denoting the mss per unit length of the wire m9, determine direct integrtion the mss moment of inerti of the wire with respect to ech of the coordinte es. = ( /3 /3 ) 3/ q h Fig. P9.5 Fig. P9.6 9.6 thin tringulr plte of mss m is welded long its se to lock s shown. Knowing tht the plte forms n ngle u with the is, determine direct integrtion the mss moment of inerti of the plte with respect to () the is, () the is, (c) the is.
9.7 Shown is the cross section of molded flt-elt pulle. Determine its mss moment of inerti nd its rdius of grtion with respect to the is 9. (The densit of rss is 8650 kg/m 3 nd the densit of the fier-reinforced polcronte used is 50 kg/m 3.) Prolems 57 mm Polcronte mm mm ' 8 mm 7 mm 5 mm rss in. 6 9.5 mm 3 in. 8 7.5 mm in. Fig. P9.7 8 ' Neoprene luminum in. 4 in. 9.8 Shown is the cross section of n idler roller. Determine its mss moment of inerti nd its rdius of grtion with respect to the is 9. (The specific weight of rone is 0.30 l/in 3 ; of luminum, 0.00 l/in 3 ; nd of neoprene, 0.045 l/in 3.) Fig. P9.8 3 6 in. rone 9.9 Knowing tht the thin clindricl shell shown is of mss m, thickness t, nd height h, determine the mss moment of inerti of the shell with respect to the is. (Hint: onsider the shell s formed removing clinder of rdius nd height h from clinder of rdius t nd height h; then neglect terms contining t nd t 3 nd keep those terms contining t.) 9.30 The mchine prt shown is formed mchining conicl surfce into circulr clinder. For 5 h, determine the mss moment of inerti nd the rdius of grtion of the mchine prt with respect to the is. h Fig. P9.9 h Fig. P9.30
58 Distriuted Forces: Moments of Inerti 9.3 fter period of use, one of the ldes of shredder hs een worn to the shpe shown nd is of mss 0.8 kg. Knowing tht the ' mss moments of inerti of the lde with respect to the 9 nd 9 es re 0.30 g? m nd 0.680 g? m, respectivel, determine 80 mm G' () the loction of the centroidl is GG9, () the rdius of grtion ' with respect to is GG9. G 9.3 Determine the mss moment of inerti of the 0.9-l mchine component shown with respect to the is 9. Fig. P9.3 0.4 in. ' 9.33 squre hole is centered in nd etends through the luminum mchine component shown. Determine () the vlue of for which the mss moment of inerti of the component with respect to the is 9, which isects the top surfce of the hole, is mimum, () the corresponding vlues of the mss moment of inerti nd the rdius of grtion with respect to the is 9. (The specific weight of luminum is 0.00 l/in 3.) 4. in..4 in. 5 in. ' 4. in.. in..6 in. Fig. P9.3 Fig. P9.33 9.34 The cups nd the rms of n nemometer re fricted from mteril of densit r. Knowing tht the mss moment of inerti of thin, hemisphericl shell of mss m nd thickness t with respect to its centroidl is GG9 is 5m /, determine () the mss moment of inerti of the nemometer with respect to the is 9, () the rtio of to l for which the centroidl moment of inerti of the cups is equl to percent of the moment of inerti of the cups with respect to the is 9. ' G' l d G Fig. P9.34
9.35 nd 9.36 -mm thick piece of sheet steel is cut nd ent into the mchine component shown. Knowing tht the densit of steel is 7850 kg/m 3, determine the mss moment of inerti of the component with respect to ech of the coordinte es. Prolems 59 0 mm 50 mm 50 mm 95 mm 0 mm 50 mm 50 mm Fig. P9.35 Fig. P9.36 350 mm 50 mm 9.37 The cover for n electronic device is formed from sheet luminum tht is 0.05 in. thick. Determine the mss moment of inerti of the cover with respect to ech of the coordinte es. (The specific weight of luminum is 0.00 l/in 3.) 6. in. 3 in..4 in..5 in. 3.5 in. Fig. P9.37 in..5 in. in. Fig. P9.38 9.38 frming nchor is formed of 0.05-in.-thick glvnied steel. Determine the mss moment of inerti of the nchor with respect to ech of the coordinte es. (The specific weight of glvnied steel is 470 l/ft 3.) 0 mm 9.39 susseml for model irplne is fricted from three pieces of.5-mm plwood. Neglecting the mss of the dhesive used to ssemle the three pieces, determine the mss moment of inerti of the susseml with respect to ech of the coordinte es. (The densit of the plwood is 780 kg/m 3.) 300 mm Fig. P9.39
530 Distriuted Forces: Moments of Inerti *9.40 frmer constructs trough welding rectngulr piece of -mm-thick sheet steel to hlf of steel drum. Knowing tht the densit of steel is 7850 kg/m 3 nd tht the thickness of the wlls 0 mm of the drum is.8 mm, determine the mss moment of inerti of the trough with respect to ech of the coordinte es. Neglect 840 mm the mss of the welds. 9.4 The mchine element shown is fricted from steel. Determine the mss moment of inerti of the sseml with respect to () the is, () the is, (c) the is. (The densit of steel is 7850 kg/m 3.) Fig. P9.40 85 mm 40 mm 40 mm 0 mm 0 mm 80 mm 60 mm Fig. P9.4 40 mm 9.4 Determine the mss moment of inerti of the steel mchine element shown with respect to the is. (The specific weight of steel is 490 l/ft 3.) 9 in. 0.9 in. 0.9 in. 0.9 in. 0.6 in..35 in. 3.7 in. 3. in. 0.6 in..4 in.. in. Fig. P9.4 nd P9.43 9.43 Determine the mss moment of inerti of the steel mchine element shown with respect to the is. (The specific weight of steel is 490 l/ft 3.)
9.44 Determine the mss moment of inerti nd the rdius of grtion of the steel mchine element shown with respect to the is. (The densit of steel is 7850 kg/m 3.) Prolems 53 40 mm 60 mm 50 mm 45 mm 38 mm 5 mm 5 mm 45 mm 45 mm Fig. P9.44 9.45 Determine the mss moment of inerti of the steel fiture shown with respect to () the is, () the is, (c) the is. (The densit of steel is 7850 kg/m 3.) 6 mm 40 mm 70 mm 50 mm 38 mm 80 mm 50 mm 4 mm Fig. P9.45 9.46 luminum wire with weight per unit length of 0.033 l/ft is used to form the circle nd the stright memers of the figure shown. Determine the mss moment of inerti of the sseml with respect to ech of the coordinte es. 8 in. 8 in. 8 in. 6 in. 8 in. Fig. P9.46
53 Distriuted Forces: Moments of Inerti. m 9.47 The figure shown is formed of 8-in.-dimeter steel wire. Knowing tht the specific weight of the steel is 490 l/ft 3, determine the mss moment of inerti of the wire with respect to ech of the coordinte es. 8 in.. m 8 in. 8 in. Fig. P9.47 Fig. P9.48. m 9.48 homogeneous wire with mss per unit length of 0.056 kg/m is used to form the figure shown. Determine the mss moment of inerti of the wire with respect to ech of the coordinte es. Fig. 9.9 q L r p dm *9.6 MMENT F INERTI F DY WITH RESPET T N RITRRY XIS THRUGH. MSS PRDUTS F INERTI In this section ou will see how the moment of inerti of od cn e determined with respect to n ritrr is L through the origin (Fig. 9.9) if its moments of inerti with respect to the three coordinte es, s well s certin other quntities to e defined elow, hve lred een determined. The moment of inerti I L of the od with respect to L is equl to e p dm, where p denotes the perpendiculr distnce from the element of mss dm to the is L. If we denote l the unit vector long L nd r the position vector of the element dm, we oserve tht the perpendiculr distnce p is equl to r sin u, which is the mgnitude of the vector product L 3 r. We therefore write I L 5 # p dm 5 # ZL 3 rz dm (9.43) Epressing L 3 r in terms of the rectngulr components of the vector product, we hve I L 5 # [(l l ) (l l ) (l l ) ] dm where the components l, l, l of the unit vector L represent the direction cosines of the is L nd the components,, of r represent the coordintes of the element of mss dm. Epnding the squres nd rerrnging the terms, we write I L 5 l # ( ) dm l # ( ) dm l # ( ) dm l l # dm l l # dm l l # dm (9.44)
Referring to Eqs. (9.30), we note tht the first three integrls in (9.44) represent, respectivel, the moments of inerti I, I, nd I of the od with respect to the coordinte es. The lst three integrls in (9.44), which involve products of coordintes, re clled the products of inerti of the od with respect to the nd es, the nd es, nd the nd es, respectivel. We write 9.7 Ellipsoid of Inerti. Principl es of Inerti 533 I 5 # dm I 5 # dm I 5 # dm (9.45) Rewriting Eq. (9.44) in terms of the integrls defined in Eqs. (9.30) nd (9.45), we hve I L 5 I l I l I l I l l I l l I l l (9.46) We note tht the definition of the products of inerti of mss given in Eqs. (9.45) is n etension of the definition of the product of inerti of n re (Sec. 9.8). Mss products of inerti reduce to ero under the sme conditions of smmetr s do products of inerti of res, nd the prllel-is theorem for mss products of inerti is epressed reltions similr to the formul derived for the product of inerti of n re. Sustituting the epressions for,, nd given in Eqs. (9.3) into Eqs. (9.45), we find tht I 5 I m I 5 I m (9.47) I 5 I m where,, re the coordintes of the center of grvit G of the od nd I, I, I denote the products of inerti of the od with respect to the centroidl es 9, 9, 9 (Fig. 9.). *9.7 ELLIPSID F INERTI. PRINIPL XES F INERTI Let us ssume tht the moment of inerti of the od considered in the preceding section hs een determined with respect to lrge numer of es L through the fied point nd tht point Q hs een plotted on ech is L t distnce Q 5 / I L from. The locus of the points Q thus otined forms surfce (Fig. 9.30). The eqution of tht surfce cn e otined sustituting /(Q) for I L in (9.46) nd then multipling oth sides of the eqution (Q). serving tht (Q)l 5 (Q)l 5 (Q)l 5 where,, denote the rectngulr coordintes of Q, we write I I I I I I 5 (9.48) L Q(,, ) / I L The eqution otined is the eqution of qudric surfce. Since the moment of inerti I L is different from ero for ever is L, no point Q cn e t n infinite distnce from. Thus, the qudric surfce otined is n ellipsoid. This ellipsoid, which defines the Fig. 9.30
534 Distriuted Forces: Moments of Inerti moment of inerti of the od with respect to n is through, is known s the ellipsoid of inerti of the od t. We oserve tht if the es in Fig. 9.30 re rotted, the coefficients of the eqution defining the ellipsoid chnge, since the re ' ' equl to the moments nd products of inerti of the od with respect to the rotted coordinte es. However, the ellipsoid itself remins unffected, since its shpe depends onl upon the distriution of mss in the given od. Suppose tht we choose s coordinte es the principl es,, of the ellipsoid of inerti (Fig. 9.3). The eqution of the ellipsoid with respect to these coordinte es is known to e of the form I 9 9 I 9 9 I 9 9 5 (9.49) which does not contin n products of the coordintes. ompring ' Eqs. (9.48) nd (9.49), we oserve tht the products of inerti of the od with respect to the,, es must e ero. The,, Fig. 9.3 es re known s the principl es of inerti of the od t, nd the coefficients I, I, I re referred to s the principl moments of inerti of the od t. Note tht, given od of ritrr shpe nd point, it is lws possile to find es which re the principl es of inerti of the od t, tht is, es with respect to which the products of inerti of the od re ero. Indeed, whtever the shpe of the od, the moments nd products of inerti of the od with respect to,, nd es through will define n ellipsoid, nd this ellipsoid will hve principl es which, definition, ' ' re the principl es of inerti of the od t. If the principl es of inerti,, re used s coordinte ' es, the epression otined in Eq. (9.46) for the moment of inerti of od with respect to n ritrr is through reduces to I L 5 I l I l I l (9.50) ' Fig. 9.3 Fig. 9.33 ' D The determintion of the principl es of inerti of od of ritrr shpe is somewht involved nd will e discussed in the net section. There re mn cses, however, where these es cn e spotted immeditel. onsider, for instnce, the homogeneous cone of ellipticl se shown in Fig. 9.3; this cone possesses two mutull perpendiculr plnes of smmetr nd. From the definition (9.45), we oserve tht if the nd plnes re chosen to coincide with the two plnes of smmetr, ll of the products of inerti re ero. The,, nd es thus selected re therefore the principl es of inerti of the cone t. In the cse of the homogeneous regulr tetrhedron shown in Fig. 9.33, the line joining the corner to the center D of the opposite fce is principl is of inerti t, nd n line through perpendiculr to D is lso principl is of inerti t. This propert is pprent if we oserve tht rotting the tetrhedron through 0 out D leves its shpe nd mss distriution unchnged. It follows tht the ellipsoid of inerti t lso remins unchnged under this rottion. The ellipsoid, therefore, is od of revolution whose is of revolution is D, nd the line D, s well s n perpendiculr line through, must e principl is of the ellipsoid.
*9.8 DETERMINTIN F THE PRINIPL XES ND PRINIPL MMENTS F INERTI F DY F RITRRY SHPE The method of nlsis descried in this section should e used when the od under considertion hs no ovious propert of smmetr. onsider the ellipsoid of inerti of the od t given point (Fig. 9.34); let r e the rdius vector of point P on the surfce of the ellipsoid nd let n e the unit vector long the norml to tht surfce t P. We oserve tht the onl points where r nd n re colliner re the points P, P, nd P 3, where the principl es intersect the visile portion of the surfce of the ellipsoid, nd the corresponding points on the other side of the ellipsoid. 9.8 Determintion of the Principl es nd Principl Moments of Inerti of od of ritrr Shpe 535 ' P n ' P r P P 3 Fig. 9.34 ' We now recll from clculus tht the direction of the norml to surfce of eqution f (,, ) 5 0 t point P(,, ) is defined the grdient =f of the function f t tht point. To otin the points where the principl es intersect the surfce of the ellipsoid of inerti, we must therefore write tht r nd =f re colliner, =f 5 (K)r (9.5) where K is constnt, r 5 i j k, nd f 5 0f 0 i 0f 0 j 0f 0 k Reclling Eq. (9.48), we note tht the function f(,, ) corresponding to the ellipsoid of inerti is f(,, ) 5 I I I I I I Sustituting for r nd =f into Eq. (9.5) nd equting the coefficients of the unit vectors, we write I I I 5 K I I I 5 K (9.5) I I I 5 K
536 Distriuted Forces: Moments of Inerti Dividing ech term the distnce r from to P, we otin similr equtions involving the direction cosines l, l, nd l : I l I l I l 5 Kl I l I l I l 5 Kl (9.53) I l I l I l 5 Kl Trnsposing the right-hnd memers leds to the following homogeneous liner equtions: (I K)l I l I l 5 0 I l (I K)l I l 5 0 (9.54) I l I l (I K)l 5 0 For this sstem of equtions to hve solution different from l 5 l 5 l 5 0, its discriminnt must e ero: I K I I I I K I 5 0 (9.55) I I I K Epnding this determinnt nd chnging signs, we write K 3 (I I I )K (I I I I I I I I I )K (I I I I I I I I I I I I ) 5 0 (9.56) This is cuic eqution in K, which ields three rel, positive roots K, K, nd K 3. To otin the direction cosines of the principl is corresponding to the root K we sustitute K for K in Eqs. (9.54). Since these equtions re now linerl dependent, onl two of them m e used to determine l, l, nd l. n dditionl eqution m e otined, however, reclling from Sec.. tht the direction cosines must stisf the reltion l l l 5 (9.57) Repeting this procedure with K nd K 3, we otin the direction cosines of the other two principl es. We will now show tht the roots K, K, nd K 3 of Eq. (9.56) re the principl moments of inerti of the given od. Let us sustitute for K in Eqs. (9.53) the root K, nd for l, l, nd l the corresponding vlues (l ), (l ), nd (l ) of the direction cosines; the three equtions will e stisfied. We now multipl (l ), (l ), nd (l ), respectivel, ech term in the first, second, nd third eqution nd dd the equtions otined in this w. We write I (l ) I (l ) I (l ) I (l ) (l ) I (l ) (l ) I (l ) (l ) 5 K [(l ) (l ) (l ) ] Reclling Eq. (9.46), we oserve tht the left-hnd memer of this eqution represents the moment of inerti of the od with respect to the principl is corresponding to K ; it is thus the principl moment of inerti corresponding to tht root. n the other hnd, reclling Eq. (9.57), we note tht the right-hnd memer reduces to K. Thus K itself is the principl moment of inerti. We cn show in the sme fshion tht K nd K 3 re the other two principl moments of inerti of the od.
SMPLE PRLEM 9.4 E D F H c onsider rectngulr prism of mss m nd sides,, c. Determine () the moments nd products of inerti of the prism with respect to the coordinte es shown, () its moment of inerti with respect to the digonl. SLUTIN. Moments nd Products of Inerti with Respect to the oordinte es. Moments of Inerti. Introducing the centroidl es,,, with respect to which the moments of inerti re given in Fig. 9.8, we ppl the prllel-is theorem: I 5 I m( ) 5 m( c ) m( 4 4c ) I 5 3m( c ) E E ' D ' ' D q q q F H c H c ' Similrl, I 5 3 m(c ) I 5 3 m( ) Products of Inerti. ecuse of smmetr, the products of inerti with respect to the centroidl es,, re ero, nd these es re principl es of inerti. Using the prllel-is theorem, we hve I 5 I 99 m 5 0 m( )( ) I 5 4 m Similrl, I 5 4mc I 5 4 mc. Moment of Inerti with Respect to. We recll Eq. (9.46): I 5 I l I l I l I l l I l l I l l where the direction cosines of re l 5 cos u 5 H 5 ( c ) / c l 5 l ( c ) / 5 ( c ) / Sustituting the vlues otined for the moments nd products of inerti nd for the direction cosines into the eqution for I, we hve I 5 c [ 3m( c ) 3m(c ) 3m( )c m m c mc ] I 5 m c c 6 c lterntive Solution. The moment of inerti I cn e otined directl from the principl moments of inerti I, I, I, since the line psses through the centroid. Since the,, es re principl es of inerti, we use Eq. (9.50) to write q q q I 5 I l I l I l 5 c c m ( c ) m (c ) m ( )c d I 5 m c c 6 c 537
SMPLE PRLEM 9.5 If 5 3c nd 5 c for the rectngulr prism of Smple Pro. 9.4, determine () the principl moments of inerti t the origin, () the principl es of inerti t. SLUTIN. Principl Moments of Inerti t the rigin. Sustituting 5 3c nd 5 c into the solution to Smple Pro. 9.4, we hve I 5 5 3mc I 5 0 3 mc I 5 3 3 mc I 5 3 mc I 5 mc I 5 3 4mc Sustituting the vlues of the moments nd products of inerti into Eq. (9.56) nd collecting terms ields K 3 ( 8 3 mc )K ( 3479 44 m c 4 )K 589 54 m 3 c 6 5 0 We then solve for the roots of this eqution; from the discussion in Sec. 9.8, it follows tht these roots re the principl moments of inerti of the od t the origin. K 5 0.568867mc K 5 4.0885mc K 3 5 4.5556mc K 5 0.569mc K 5 4.mc K 3 5 4.56mc. Principl es of Inerti t. To determine the direction of principl is of inerti, we first sustitute the corresponding vlue of K into two of the equtions (9.54); the resulting equtions together with Eq. (9.57) constitute sstem of three equtions from which the direction cosines of the corresponding principl is cn e determined. Thus, we hve for the first principl moment of inerti K : Solving ields ( 5 3 0.568867)mc (l ) 3 mc (l ) 3 4mc (l ) 5 0 3 mc (l ) ( 0 3 0.568867) mc (l ) mc (l ) 5 0 (l ) (l ) (l ) 5 (l ) 5 0.836600 (l ) 5 0.49600 (l ) 5 0.3557 The ngles tht the first principl is of inerti forms with the coordinte es re then (u ) 5 33. (u ) 5 60.3 (u ) 5 76.6 Using the sme set of equtions successivel with K nd K 3, we find tht the ngles ssocited with the second nd third principl moments of inerti t the origin re, respectivel, nd (u ) 5 57.8 (u ) 5 46.6 (u ) 5 98.0 (u ) 3 5 8.8 (u ) 3 5 76. (u ) 3 5 64.3 538
SLVING PRLEMS N YUR WN In this lesson we defined the mss products of inerti I, I, nd I of od nd showed ou how to determine the moments of inerti of tht od with respect to n ritrr is pssing through the origin. You lso lerned how to determine t the origin the principl es of inerti of od nd the corresponding principl moments of inerti.. Determining the mss products of inerti of composite od. The mss products of inerti of composite od with respect to the coordinte es cn e epressed s the sums of the products of inerti of its component prts with respect to those es. For ech component prt, we cn use the prllel-is theorem nd write Eqs. (9.47) I 5 I m I 5 I m I 5 I m where the primes denote the centroidl es of ech component prt nd where,, nd represent the coordintes of its center of grvit. Keep in mind tht the mss products of inerti cn e positive, negtive, or ero, nd e sure to tke into ccount the signs of,, nd.. From the properties of smmetr of component prt, ou cn deduce tht two or ll three of its centroidl mss products of inerti re ero. For instnce, ou cn verif tht for thin plte prllel to the plne; wire ling in plne prllel to the plne; od with plne of smmetr prllel to the plne; nd od with n is of smmetr prllel to the is, the products of inerti I nd I re ero. For rectngulr, circulr, or semicirculr pltes with es of smmetr prllel to the coordinte es; stright wires prllel to coordinte is; circulr nd semicirculr wires with es of smmetr prllel to the coordinte es; nd rectngulr prisms with es of smmetr prllel to the coordinte es, the products of inerti I,I, nd I re ll ero.. Mss products of inerti which re different from ero cn e computed from Eqs. (9.45). lthough, in generl, triple integrtion is required to determine mss product of inerti, single integrtion cn e used if the given od cn e divided into series of thin, prllel sls. The computtions re then similr to those discussed in the previous lesson for moments of inerti. (continued) 539
. omputing the moment of inerti of od with respect to n ritrr is L. n epression for the moment of inerti I L ws derived in Sec. 9.6 nd is given in Eq. (9.46). efore computing I L, ou must first determine the mss moments nd products of inerti of the od with respect to the given coordinte es s well s the direction cosines of the unit vector L long L. 3. lculting the principl moments of inerti of od nd determining its principl es of inerti. You sw in Sec. 9.7 tht it is lws possile to find n orienttion of the coordinte es for which the mss products of inerti re ero. These es re referred to s the principl es of inerti nd the corresponding moments of inerti re known s the principl moments of inerti of the od. In mn cses, the principl es of inerti of od cn e determined from its properties of smmetr. The procedure required to determine the principl moments nd principl es of inerti of od with no ovious propert of smmetr ws discussed in Sec. 9.8 nd ws illustrted in Smple Pro. 9.5. It consists of the following steps.. Epnd the determinnt in Eq. (9.55) nd solve the resulting cuic eqution. The solution cn e otined tril nd error or, preferl, with n dvnced scientific clcultor or with the pproprite computer softwre. The roots K, K, nd K 3 of this eqution re the principl moments of inerti of the od.. To determine the direction of the principl is corresponding to K, sustitute this vlue for K in two of the equtions (9.54) nd solve these equtions together with Eq. (9.57) for the direction cosines of the principl is corresponding to K. c. Repet this procedure with K nd K 3 to determine the directions of the other two principl es. s check of our computtions, ou m wish to verif tht the sclr product of n two of the unit vectors long the three es ou hve otined is ero nd, thus, tht these es re perpendiculr to ech other. 540
PRLEMS 9.49 Determine the mss products of inerti I, I, nd I of the steel fiture shown. (The densit of steel is 7850 kg/m 3.) 6 mm 40 mm 70 mm 50 mm 38 mm 80 mm 50 mm 4 mm Fig. P9.49 9.50 Determine the mss products of inerti I, I, nd I of the steel mchine element shown. (The densit of steel is 7850 kg/m 3.) 6 mm 60 mm 4 mm 0 mm 8 mm r = mm Fig. P9.50 00 mm 0 mm 5 mm 5 mm 0 mm 9.5 nd 9.5 Determine the mss products of inerti I, I, nd I of the cst luminum mchine component shown. (The specific weight of luminum is 0.00 l/in. 3 ) r = 0.55 in. 0.6 in. 5.4 in..4 in. 0.8 in.. in. 4.5 in..4 in..8 in. r = 0.8 in. 3.6 in.. in. 0.3 in. 0.7 in.. in. Fig. P9.5 Fig. P9.5 54
54 Distriuted Forces: Moments of Inerti 9.53 through 9.56 section of sheet steel mm thick is cut nd ent into the mchine component shown. Knowing tht the densit of steel is 7850 kg/m 3, determine the mss products of inerti I, I, nd I of the component. 80 mm 400 mm 00 mm 400 mm 5 mm 300 mm 5 mm Fig. P9.53 Fig. P9.54 95 mm r = 35 mm 5 mm 350 mm 50 mm Fig. P9.55 Fig. P9.56 9.57 nd 9.58 rss wire with weight per unit length w is used to form the figure shown. Determine the mss products of inerti I, I, nd I of the wire figure. 3 3 Fig. P9.57 Fig. P9.58
9.59 The figure shown is formed of.5-mm-dimeter luminum wire. Knowing tht the densit of luminum is 800 kg/m 3, determine the mss products of inerti I, I, nd I of the wire figure. Prolems 543 80 mm 50 mm 300 mm Fig. P9.59 9.60 Thin luminum wire of uniform dimeter is used to form the figure shown. Denoting m9 the mss per unit length of the wire, determine the mss products of inerti I, I, nd I of the wire figure. 9.6 omplete the derivtion of Eqs. (9.47), which epress the prllelis theorem for mss products of inerti. 9.6 For the homogeneous tetrhedron of mss m shown, () determine direct integrtion the mss product of inerti I, () deduce I nd I from the result otined in prt. R Fig. P9.60 R c Fig. P9.6 9.63 The homogeneous circulr clinder shown hs mss m. Determine the mss moment of inerti of the clinder with respect to the line joining the origin nd point tht is locted on the perimeter of the top surfce of the clinder. h Fig. P9.63
544 Distriuted Forces: Moments of Inerti 9.64 The homogeneous circulr cone shown hs mss m. Determine the mss moment of inerti of the cone with respect to the line joining the origin nd point. 3 3 3 Fig. P9.64 9.65 Shown is the mchine element of Pro. 9.4. Determine its mss moment of inerti with respect to the line joining the origin nd point. 40 mm 40 mm 0 mm 80 mm 0 mm 60 mm Fig. P9.65 40 mm 9.66 Determine the mss moment of inerti of the steel fiture of Pros. 9.45 nd 9.49 with respect to the is through the origin tht forms equl ngles with the,, nd es. 9.67 The thin ent plte shown is of uniform densit nd weight W. Determine its mss moment of inerti with respect to the line joining the origin nd point. Fig. P9.67
9.68 piece of sheet steel of thickness t nd specific weight g is cut nd ent into the mchine component shown. Determine the mss moment of inerti of the component with respect to the line joining the origin nd point. Prolems 545 9.69 Determine the mss moment of inerti of the mchine component of Pros. 9.36 nd 9.55 with respect to the is through the origin chrcteried the unit vector L 5 (4i 8j k)/9. 9.70 through 9.7 For the wire figure of the prolem indicted, determine the mss moment of inerti of the figure with respect to the is through the origin chrcteried the unit vector L 5 (3i 6j k)/7. 9.70 Pro. 9.48 9.7 Pro. 9.47 9.7 Pro. 9.46 9.73 For the rectngulr prism shown, determine the vlues of the rtios / nd c/ so tht the ellipsoid of inerti of the prism is sphere when computed () t point, () t point. Fig. P9.68 c c Fig. P9.73 9.74 For the right circulr cone of Smple Pro. 9., determine the vlue of the rtio /h for which the ellipsoid of inerti of the cone is sphere when computed () t the pe of the cone, () t the center of the se of the cone. L L 4 L 4 9.75 For the homogeneous circulr clinder shown, of rdius nd length L, determine the vlue of the rtio /L for which the ellipsoid of inerti of the clinder is sphere when computed () t the centroid of the clinder, () t point. Fig. P9.75 9.76 Given n ritrr od nd three rectngulr es,, nd, prove tht the mss moment of inerti of the od with respect to n one of the three es cnnot e lrger thn the sum of the mss moments of inerti of the od with respect to the other two es. Tht is, prove tht the inequlit I # I I nd the two similr inequlities re stisfied. Further, prove tht I $ I if the od is homogeneous solid of revolution, where is the is of revolution nd is trnsverse is.
546 Distriuted Forces: Moments of Inerti 9.77 onsider cue of mss m nd side. () Show tht the ellipsoid of inerti t the center of the cue is sphere, nd use this propert to determine the moment of inerti of the cue with respect to one of its digonls. () Show tht the ellipsoid of inerti t one of the corners of the cue is n ellipsoid of revolution, nd determine the principl moments of inerti of the cue t tht point. 9.78 Given homogeneous od of mss m nd of ritrr shpe nd three rectngulr es,, nd with origin t, prove tht the sum I I I of the mss moments of inerti of the od cnnot e smller thn the similr sum computed for sphere of the sme mss nd the sme mteril centered t. Further, using the result of Pro. 9.76, prove tht if the od is solid of revolution, where is the is of revolution, its mss moment of inerti I out trnsverse is cnnot e smller thn 3m /0, where is the rdius of the sphere of the sme mss nd the sme mteril. *9.79 The homogeneous circulr clinder shown hs mss m, nd the dimeter of its top surfce forms 45 ngles with the nd es. () Determine the principl mss moments of inerti of the clinder t the origin. () ompute the ngles tht the principl es of inerti t form with the coordinte es. (c) Sketch the clinder, nd show the orienttion of the principl es of inerti reltive to the,, nd es. Fig. P9.79 9.80 through 9.84 For the component descried in the prolem indicted, determine () the principl mss moments of inerti t the origin, () the principl es of inerti t the origin. Sketch the od nd show the orienttion of the principl es of inerti reltive to the,, nd es. *9.80 Pro. 9.65 *9.8 Pros. 9.45 nd 9.49 *9.8 Pro. 9.67 *9.83 Pro. 9.68 *9.84 Pros. 9.48 nd 9.70
REVIEW ND SUMMRY In the first hlf of this chpter, we discussed the determintion of the resultnt R of forces DF distriuted over plne re when the mgnitudes of these forces re proportionl to oth the res D of the elements on which the ct nd the distnces from these elements to given is; we thus hd DF 5 k D. We found tht the mgnitude of the resultnt R is proportionl to the first moment Q 5 e d of the re, while the moment of R out the is is proportionl to the second moment, or moment of inerti, I 5 e d of with respect to the sme is [Sec. 9.]. The rectngulr moments of inerti I nd I of n re [Sec. 9.3] were otined evluting the integrls Rectngulr moments of inerti I 5 # d I 5 # d (9.) These computtions cn e reduced to single integrtions choosing d to e thin strip prllel to one of the coordinte es. We lso recll tht it is possile to compute I nd I from the sme elementl strip (Fig. 9.35) using the formul for the moment of inerti of rectngulr re [Smple Pro. 9.3]. Fig. 9.35 d di = 3 d 3 di = d r d Fig. 9.36 The polr moment of inerti of n re with respect to the pole [Sec. 9.4] ws defined s Polr moment of inerti J 5 # r d (9.3) where r is the distnce from to the element of re d (Fig. 9.36). serving tht r 5, we estlished the reltion J 5 I I (9.4) 547
548 Distriuted Forces: Moments of Inerti The rdius of grtion of n re with respect to the is [Sec. 9.5] ws defined s the distnce k, where I 5 K. With Rdius of grtion similr definitions for the rdii of grtion of with respect to the is nd with respect to, we hd k 5 I k 5 I k 5 J (9.5 9.7) d Fig. 9.37 Prllel-is theorem ' ' omposite res Product of inerti The prllel-is theorem ws presented in Sec. 9.6. It sttes tht the moment of inerti I of n re with respect to n given is 9 (Fig. 9.37) is equl to the moment of inerti I of the re with respect to the centroidl is 9 tht is prllel to 9 plus the product of the re nd the squre of the distnce d etween the two es: I 5 I d (9.9) This formul cn lso e used to determine the moment of inerti I of n re with respect to centroidl is 9 when its moment of inerti I with respect to prllel is 9 is known. In this cse, however, the product d should e sutrcted from the known moment of inerti I. similr reltion holds etween the polr moment of inerti J of n re out point nd the polr moment of inerti J of the sme re out its centroid. Letting d e the distnce etween nd, we hve J 5 J d (9.) The prllel-is theorem cn e used ver effectivel to compute the moment of inerti of composite re with respect to given is [Sec. 9.7]. onsidering ech component re seprtel, we first compute the moment of inerti of ech re with respect to its centroidl is, using the dt provided in Figs. 9. nd 9.3 whenever possile. The prllel-is theorem is then pplied to determine the moment of inerti of ech component re with respect to the desired is, nd the vrious vlues otined re dded [Smple Pros. 9.4 nd 9.5]. Sections 9.8 through 9.0 were devoted to the trnsformtion of the moments of inerti of n re under rottion of the coordinte es. First, we defined the product of inerti of n re s I 5 # d (9.) nd showed tht I 5 0 if the re is smmetricl with respect to either or oth of the coordinte es. We lso derived the prllelis theorem for products of inerti. We hd I 5 I (9.3) where I is the product of inerti of the re with respect to the centroidl es 9 nd 9 which re prllel to the nd is nd nd re the coordintes of the centroid of the re [Sec. 9.8].
' Review nd Summr 549 q ' Fig. 9.38 In Sec. 9.9 we determined the moments nd product of inerti I 9, I 9, nd I 99 of n re with respect to 9 nd 9 es otined rotting the originl nd coordinte es through n ngle u counterclockwise (Fig. 9.38). We epressed I 9, I 9, nd I 99 in terms of the moments nd product of inerti I, I, nd I computed with respect to the originl nd es. We hd Rottion of es I 5 I I I 5 I I I 5 I I I I I I cos u I sin u (9.8) cos u I sin u (9.9) sin u I cos u (9.0) The principl es of the re out were defined s the two es perpendiculr to ech other, with respect to which the moments of inerti of the re re mimum nd minimum. The corresponding vlues of u, denoted u m, were otined from the formul Principl es I tn u m 5 I I (9.5) The corresponding mimum nd minimum vlues of I re clled the principl moments of inerti of the re out ; we hd Principl moments of inerti I m,min 5 I I ; I I I (9.7) We lso noted tht the corresponding vlue of the product of inerti is ero. The trnsformtion of the moments nd product of inerti of n re under rottion of es cn e represented grphicll drwing Mohr s circle [Sec. 9.0]. Given the moments nd product of inerti I, I, nd I of the re with respect to the nd coordinte es, we Mohr s circle
550 Distriuted Forces: Moments of Inerti ' I I q m q ' I min I '' I I ' X' X q q m I I '' I, I Y Y' I I ' I m Fig. 9.39 plot points X (I, I ) nd Y (I, I ) nd drw the line joining these two points (Fig. 9.39). This line is dimeter of Mohr s circle nd thus defines this circle. s the coordinte es re rotted through u, the dimeter rottes through twice tht ngle, nd the coordintes of X9 nd Y9 ield the new vlues I 9, I 9, nd I 99 of the moments nd product of inerti of the re. lso, the ngle u m nd the coordintes of points nd define the principl es nd nd the principl moments of inerti of the re [Smple Pro. 9.8]. Moments of inerti of msses The second hlf of the chpter ws devoted to the determintion of moments of inerti of msses, which re encountered in dnmics in prolems involving the rottion of rigid od out n is. The mss moment of inerti of od with respect to n is 9 (Fig. 9.40) ws defined s I 5 # r dm (9.8) r Δm ' where r is the distnce from 9 to the element of mss [Sec. 9.]. The rdius of grtion of the od ws defined s I k 5 (9.9) m The moments of inerti of od with respect to the coordintes es were epressed s Δm r r 3 Δm 3 Fig. 9.40 I 5 # ( ) dm I 5 # ( ) dm (9.30) I 5 # ( ) dm
We sw tht the prllel-is theorem lso pplies to mss moments of inerti [Sec. 9.]. Thus, the moment of inerti I of od with respect to n ritrr is 9 (Fig. 9.4) cn e epressed s I 5 I md (9.33) where I is the moment of inerti of the od with respect to the centroidl is 9 which is prllel to the is 9, m is the mss of the od, nd d is the distnce etween the two es. Prllel-is theorem Review nd Summr 55 ' d ' ' G t ' ' t ' r ' ' Fig. 9.4 Fig. 9.4 Fig. 9.43 The moments of inerti of thin pltes cn e redil otined from the moments of inerti of their res [Sec. 9.3]. We found tht for rectngulr plte the moments of inerti with respect to the es shown (Fig. 9.4) re I 9 5 m I 9 5 m (9.39) I 9 5 I 9 I 9 5 m( ) (9.40) while for circulr plte (Fig. 9.43) the re I 9 5 I 9 5 4 mr (9.4) I 9 5 I 9 I 9 5 mr (9.4) When od possesses two plnes of smmetr, it is usull possile to use single integrtion to determine its moment of inerti with respect to given is selecting the element of mss dm to e thin plte [Smple Pros. 9.0 nd 9.]. n the other hnd, when od consists of severl common geometric shpes, its moment of inerti with respect to given is cn e otined using the formuls given in Fig. 9.8 together with the prllel-is theorem [Smple Pros. 9. nd 9.3]. In the lst portion of the chpter, we lerned to determine the moment of inerti of od with respect to n ritrr is L which is drwn through the origin [Sec. 9.6]. Denoting l, l, Moments of inerti of thin pltes omposite odies Moment of inerti with respect to n ritrr is
55 Distriuted Forces: Moments of Inerti l the components of the unit vector L long L (Fig. 9.44) nd introducing the products of inerti I 5 # dm I 5 # dm I 5 # dm (9.45) we found tht the moment of inerti of the od with respect to L could e epressed s I L 5 I l I l I l I l l I l l I l l (9.46) ' ' L q r p dm Fig. 9.44 Fig. 9.45 ' Ellipsoid of inerti Principl es of inerti Principl moments of inerti plotting point Q long ech is L t distnce Q 5 I L from [Sec. 9.7], we otined the surfce of n ellipsoid, known s the ellipsoid of inerti of the od t point. The principl es 9, 9, 9 of this ellipsoid (Fig. 9.45) re the principl es of inerti of the od; tht is, the products of inerti I 99, I 99, I 99 of the od with respect to these es re ll ero. There re mn situtions when the principl es of inerti of od cn e deduced from properties of smmetr of the od. hoosing these es to e the coordinte es, we cn then epress I L s I L 5 I 9 l 9 I 9 l 9 I 9 l 9 (9.50) where I 9, I 9, I 9 re the principl moments of inerti of the od t. When the principl es of inerti cnnot e otined oservtion [Sec. 9.7], it is necessr to solve the cuic eqution K 3 (I I I )K (I I I I I I I I I )K (I I I I I I I I I I I I ) 5 0 (9.56) We found [Sec. 9.8] tht the roots K, K, nd K 3 of this eqution re the principl moments of inerti of the given od. The direction cosines (l ), (l ), nd (l ) of the principl is corresponding to the principl moment of inerti K re then determined sustituting K into Eqs. (9.54) nd solving two of these equtions nd Eq. (9.57) simultneousl. The sme procedure is then repeted using K nd K 3 to determine the direction cosines of the other two principl es [Smple Pro. 9.5].
REVIEW PRLEMS 9.85 Determine direct integrtion the moments of inerti of the shded re with respect to the nd es. 9.86 Determine the moments of inerti nd the rdii of grtion of the shded re shown with respect to the nd es. = m = k = / Fig. P9.85 = c Fig. P9.86 9.87 Determine the moment of inerti nd the rdius of grtion of the shded re shown with respect to the is. 9.88 Determine the moments of inerti of the shded re shown with respect to the nd es. 9.89 Determine the polr moment of inerti of the re shown with respect to () point, () the centroid of the re. Fig. P9.87 = k 9.90 To form n unsmmetricl girder, two L76 3 76 3 6.4-mm ngles nd two L5 3 0 3.7-mm ngles re welded to 6-mm steel plte s shown. Determine the moments of inerti of the comined section with respect to its centroidl nd es. L76 76 6.4 Fig. P9.88 54 mm 54 mm Semiellipse 36 mm 540 mm 8 mm Fig. P9.89 6 mm Fig. P9.90 L5 0.7 553
554 Distriuted Forces: Moments of Inerti 9.9 Using the prllel-is theorem, determine the product of inerti of the L5 3 3 3 -in. ngle cross section shown with respect to in. 5 in. 0.746 in. L5 3.74 in. the centroidl nd es. 9.9 For the L5 3 3 3 -in. ngle cross section shown, use Mohr s circle to determine () the moments of inerti nd the product of inerti with respect to new centroidl es otined rotting the nd es 30 clockwise, () the orienttion of the principl es through the centroid nd the corresponding vlues of the moments of inerti. 9.93 piece of thin, uniform sheet metl is cut to form the mchine component shown. Denoting the mss of the component m, determine its mss moment of inerti with respect to () the is, () the is. 3 in. Fig. P9.9 nd P9.9 in. ' ' Fig. P9.93 nd P9.94 0.48 m 9.94 piece of thin, uniform sheet metl is cut to form the mchine component shown. Denoting the mss of the component m, determine its mss moment of inerti with respect to () the is 9, () the is 9, where the 9 nd 9 es re prllel to the is nd lie in plne prllel to nd t distnce ove the plne. 0.76 m Fig. P9.95 9.95 -mm thick piece of sheet steel is cut nd ent into the mchine component shown. Knowing tht the densit of steel is 7850 kg/m 3, determine the mss moment of inerti of the component with respect to ech of the coordinte es. 9.96 Determine the mss moments of inerti nd the rdii of grtion of the steel mchine element shown with respect to the nd es. (The densit of steel is 7850 kg/m 3.) 70 70 44 0 0 44 Fig. P9.96 Dimensions in mm 40 0 0
MPUTER PRLEMS 9. Write computer progrm tht, for n re with known moments nd product of inerti I, I, nd I, cn e used to clculte the moments nd product of inerti I 9, I 9, nd I 99 of the re with respect to es 9 nd 9 otined rotting the originl es counterclockwise through n ngle u. Use this progrm to compute I 9, I 9, nd I 99 for the section of Smple Pro. 9.7 for vlues of u from 0 to 90 using 5 increments. 9. Write computer progrm tht, for n re with known moments nd product of inerti I, I, nd I, cn e used to clculte the orienttion of the principl es of the re nd the corresponding vlues of the principl moments of inerti. Use this progrm to solve () Pro. 9.89, () Smple Pro. 9.7. w n 3 d n 9.3 Mn cross sections cn e pproimted series of rectngles s shown. Write computer progrm tht cn e used to clculte the moments of inerti nd the rdii of grtion of cross sections of this tpe with respect to horiontl nd verticl centroidl es. ppl this progrm to the cross sections shown in () Figs. P9.3 nd P9.33, () Figs. P9.3 nd P9.34, (c) Fig. P9.43, (d) Fig. P9.44. w d 9.4 Mn cross sections cn e pproimted series of rectngles s shown. Write computer progrm tht cn e used to clculte the products of inerti of cross sections of this tpe with respect to horiontl nd verticl centroidl es. Use this progrm to solve () Pro. P9.7, () Pro. P9.75, (c) Pro. 9.77. Fig. P9.3 nd P9.4 w d 9.5 The re shown is revolved out the is to form homogeneous solid of mss m. pproimte the re using series of 400 rectngles of the form cc99, ech of width Dl, nd then write computer progrm tht cn e used to determine the mss moment of inerti of the solid with respect to the is. Use this progrm to solve prt of () Smple Pro. 9., () Pro. 9., ssuming tht in these prolems m 5 kg, 5 00 mm, nd h 5 400 mm. = k n c c' d 9.6 homogeneous wire with weight per unit length of 0.04 l/ft is used to form the figure shown. pproimte the figure using 0 stright line segments, nd then write computer progrm tht cn e used to determine the mss moment of inerti I of the wire with respect to the is. Use this progrm to determine I when () 5 in., L 5 in., h 5 4 in., () 5 in., L 5 7 in., h 5 0 in., (c) 5 5 in., L 5 5 in., h 5 6 in. l Fig. P9.5 l = h( ) Δl ' *9.7 Write computer progrm tht, for od with known mss moments nd products of inerti I, I, I, I, I, nd I, cn e used to clculte the principl mss moments of inerti K, K, nd K 3 of the od t the origin. Use this progrm to solve prt of () Pro. 9.80, () Pro. 9.8, (c) Pro. 9.84. h L 0 *9.8 Etend the computer progrm of Pro. 9.7 to include the computtion of the ngles tht the principl es of inerti t the origin form with the coordinte es. Use this progrm to solve () Pro. 9.80, () Pro. 9.8, (c) Pro. 9.84. Fig. P9.6 L 555
The method of virtul work is prticulrl effective when simple reltion cn e found mong the displcements of the points of ppliction of the vrious forces involved. This is the cse for the scissor lift pltform eing used workers to gin ccess to highw ridge under construction. 556