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by: Jaylan Rath


Jaylan Rath
Texas A&M
GPA 3.77


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Class Notes
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This 37 page Class Notes was uploaded by Jaylan Rath on Wednesday October 21, 2015. The Class Notes belongs to CVEN 221 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/226126/cven-221-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15
CVEN 221 CLASS NUMBER DATE PAGE 7 Cy 2 ME g C 3 GQQEQE Mf Z QElt lt x 7 g yX m CV 2M1 73C 2 213 quot quot 399 EV gt 39 7 Q E 5 2 1sz MAEMEZ m3 84183 5C3ng CVEN 221 CLASS NUMBER DATE PAGE 5m VA Lug M 5m E a ww 0 DZ LHMM 8de g A x gap g 3 356ng L IL km g 2 i Z LMZI 3085quot x m A Eb QCBeDBSrMAH m 2amp5 I 2 A a 332 aa w 3amp3 AMA CVEN 221 CLASS NUMBER DATE PAGE j g mM Www me r w w ZR 33 WZAAM OA VAL 62244 AWEwad ZMQVT 2 O 2 iZQC7LZ MAL 2 A fgwmfa Li Z f W j g f v E r 3 s A z s V x i k W 6 j 39 r f r a Ix 1 7 ifr f A h 1 f g f 2 I g E J z I i f 5 V39 1 7 i 39 39 ZVVO 1 w 24 HGJ a an Emmi 3 a 32 WLMLZ7 C X gt 1037301 N4 2 03 LZ 43 E W o3lt m xgampgwlt2 A wgi EXAMPLE PROBLEM 414 The magnitude of force F in Fig 428 is 500 N Determine the scalar compo nent of the moment at point 0 about line OC quotSOL UTION The moment of force F about line OC can be determined by using a twostep 6 0 process The zcoordinate of point A that will be needed to write a vector equa q 1 tion for force F is 25 800 tan 20 2912 mm The force F can be written in quotquot 2 W 39 I WMWWW r m Cartesian vector form as Wye Mm N E E q 850i 800i 4588k 33886i 31893j 18291k N F 500 l V 8502 8002 45882 The position vector two can be written in Cartesian vector form as rigo 04850i 0750k m The moment of force F about pomt 0 15 then given by the expresswn The moment of force F about Point i i k C is given by the vector cross O850 0 0750 product Mo r x F where r is a 338 86 31s93 18291 position vector from point 0 to 39 any point on the line of action of Mor13ox F 23920i 9867j 27109k N m force F A unit vector eoc along line DC with a sense from O to C is eoc 1000i g wear The scalar component of moment Mo along line CC is then determined by us mg the dOt prOduCt quotgag g 5 The dot product Mo eoc gives the MOC M0 eoc 23920i Mam 100i projection or scalar component of 239O N In E 239 N m Ans moment Mo along line OC Alternatively the scalar moment component Moe can be determined in a sin gle step by using the expression 3m eny enz 0 MOC rx ry rz 0850 O 0750 x Py P 33886 31893 18291 23920 N In E 239 N m Ans 142 148 CHAPTER 4 RIGID BODIES 44 EQUIVALENT FORCEMOMENT SYSTEMS A couple is a system of forces Whose resultant force is zero but Whose resultant moment about a point is not zero A couple can be repre sented by two equal noncollinear parallel forces of opposite sense as shown in Fig 430a Since the two forces are equal parallel and of op posite sense the sum of the forces in any direction is zero Therefore a couple tends only to rotate the body on which it acts The moment of a couple is defined as the sum of the moments of the pair of forces that comprise the couple For the two forces F1 and F2 shown in Fig 430a the moments of the couple about points A and B in the plane of 00 the couple are Lg MA leld MB lFildE However Therefore a e MAZMBZFd which indicates that the magnitude of the moment of a couple about a point in the plane of the couple is equal to the magnitude of one of the forces multiplied by the perpendicular distance between the forces Other characteristics of a couple can be determined by consider ing two parallel forces in space such as those shown in Fig 4 30a The sum of the moments of the two forces about any point 0 is MO 1391XF1 I 2XF2 or since F2 equals F1 M0 1 1 XF rzx FD r1 r2xF1 rABXF1 where rAB is the position vector from any point B on F2 to any point jg A on F1 Therefore from the definition of the vector cross product v g Mo mm X F1 IABHFllSiII or en Pld en 4 9 Where d is the perpendicular distance between die forces of the coupl and en is a unit vector perpendicular to the plane of the couple see i Fig 4 3019 with its sense in the direction specified for moments by m 39A righthand rule Equation 419 indicates that the moment of a coupler i I t w does not depend on the location of the moment center 0 Thus the moment of a couple about any point is the same which indicates that W39W 39 a couple is a free vector EXAMPLE PROBLEM 416 Two parallel forces of opposite sense F1 75i 1001 200k N and F2 75i lOOj 200k N act at points A and O of a body as shown in Fig 4 33 Determine the moment of the couple and the perpendicular distance between the two forces Fig 433 SOLUTION The position vector rAo can be written in Cartesian vector form as rAo 0075i 0090139 0150k m The moment of the couple is obtained from the expression i k Mo W0 xF1 0075 0090 0150 75 100 200 3300i 3751quot 1425k N m The magnitude of moment Mo is MO IMO V MEX May ME 33002 3752 14252 35igN m The magnitude of force F1 is The moment of force F1 about point 0 is given by the vector cross product Mo r x P where r is a position vector from point 0 to any point on the line of action of force F1 i a t My 5W aaav bmfya F1 W VFix Fig Pi 752 1002 2002 23585 N The distance d is obtained by using the definition of a moment Thus 3614 d M F 23585 015323 m 2 1532 mm 152 45 RESOLUTION OFA FORCE INTO A FORCE AND A E COUPL G qu H Wm M Doc 158 CHAPTER 4 RIGID BODIES EQUIVALENT FORCEMOMENT SYSTEMS w w w 5 procedure for39xtgansformmg a force mto a force F and a cou i J quot gl C is illustrated in the gfollowing example 1 g mew a a r 80000 IQ Fl VW X i 39WMMWM WW EXAMPLE PROBLEM 419 A BOOlb force FA is applied to a bracket at point A as shown in Fig 437a Re place the force FA by a force F0 and a couple C at point 0 Ema a C 10690 inlb c E w i g k f 900M30lt guts M CVEN 221 CLASS NUMBER 397 DATE PAGE WA RQELAQEFW J 1 E W 3 fQ W Sm wi W Mfr3 W W g g 2 w Ega w Q0 g3 TEgt 2324 m ZN g 3 W Z QGQCX X 3 EggWV vv wawgt hgtmw wWWWMWMWyWW AWWWW fif CVEN 221 L 39 C ASS NUMBER 7 DATE 2 12 PAGE 3W gimxg Reskgcg Ru 302ng quot mamng 439 g MMWWU 5 lt vryzwwsrw39tnmigzw w 3 35 45 am A a 4353 MB 1ng lt5Qa109 w 2 000 Zixema DATE PAGE Fir3 W Wquot M 2w QQG NW 83 50c j a BC 139 7 quotM I A i fg i g W ma a WW x g w Mi w I A quot L g 3 6 5 QQQQWZ Q 00 H j010 I 7 lg mm Qq 6w 5 v4 3 a x g w 3 6quot t 3 I J 539 g r 17 f r g lt f1 g uv4va39m m I x CVEN 221 CLASS NUMBER 2 3 g E WW 2 K V mswzmmwmavawmmwwisa m DATE PAGE F Chapter 9 ECTION 9H INTRODUCTION 92 CHARACTERISTICS OF COULOMB FRICTION 9 ANALYSIS OF SYSTEMS INVOLVING DRY FRICTION 94 ROLLING RESISTANCE S U M M A R Y 94 INTRODUCTION 471 AL 0L CVEN 221 CLASS NUMBER DATE PAGE a 2 q 8w 6323 3 2 7N ng i kiwm mam gm v HAZE 05 1 353mm QES VGRMG Momng 18 e0 05m1 agmy LAW 9 5m Wig ngk f Q C 35 2 a mfm wf m 47W CVEN 221 CLASS NUMBER DATE PAGE is 2 O 2 M5 9 M 394 2 63 352 i I M a FMBZZ ng ma 1 m ng mwm Zit rm 5 x gggm Q ng x CD 671 947l444rggr w ME RM 42 17 6 CVEN 221 CLASS NUMBER DATE PAGE 7uj CVEN 221 CLASS NUMBER DATE PAGE T Q lt A J wga wmv Egag E f WW T 4 12 18 i n x Ex PE fgaAwa mA VEN 221 CLASS NUMBER DATE i 2 Q PAGE m I Ewwf f m w m2 A 3 g H H 515 M1 s 5 f g NEE g I bi m 57 5 30 0t 57 z 3 W Z 517jww C iagivgg m w 3 I 2 www 12 31 7 Q 39 6 E g b E La 3 3 3 Egg quot11k E E g ILI 674 Q 5 WJWW fquot 3993 F Vg 3W1 3 i y g i 0 W E NM CVEN 221 CLASS NUMBER DATE PAGE 1 f39 g a 3 4 Pmgpxg gpugKg e D g Q if K EM iii ED 3 5 D Z M3 6 TEMQVQ J f u 2 M 3 gmwgw a gmzwgw 2 CAka 35 ff 3 2ng m Q g ZPV m if 7 LWMw E553 CVEN 221 CLASS NUMBER DATE PAGE Z MA 6 g g 8 m ng CK CVEN 221 CLASS NUMBER DATE PAGE i I 15 V 339 I 3M 26 21 asfgi EEg lt 2x1 g 5 KN 43033242 Rg 1231 3 1 2 2761 a i 21 a 4ng 4am xmg WEE m 6 i 3 k 392 WWampW WWmmemwmmmltwmm a 139 u my I f 4 ZM 2 O Q3va 33 m3glt3gtfz4OOCZ i 1sz0 E aw i3 25 m 3735 M 3 3 M 3 739 3 1 a 73 H 2 OGi CVEN 221 CLASS NUMBER DATE PAGE g a i WW WWWW W M w I 333 r g K quot E a at g E a 3ng Efr gym a a WMMMWMWM WKMWI rm Chapter 5 DISTRIBUTED FORCES CENTROIDS AND CENTER OF GRAVITY 55 THEOREMS OF PAPPUS AND GULDINUS 56 DISTRIBUTED LOADS ON BEAMS 57 FORCES ON SUBMERGED SURFACES S U M M A R Y 51 INTRODUCTION 52 CENTER OF GRAVITY AND CENTER OF MASS 53 CENTROIDS OF VOLUMES AREAS AND LINES 5 4 CENTROIDS OF COMPOSITE BODIES 51 INTRODUCTION wind load Figure 51 M loads Examples of distributed 189 dtstrlbuted 21m 39XAMPLE PROBLEM 51 Locate the center of gravity for the six particles shown in Fig 55 if WA 50 lb W3 25 lb WC 30 lb WD 35 lb WE 20 1b and Wp 40 lb Equation 1 Sums moments about the y direction direction lt direction Fig 55 a SOLUTION The center of gravity for the system of particles is located by using the princi ple of moments for a parallel system of gravitational forces W and their re sultant W The total weight W for the six particles is WWAW3WCWDWEWp50253035204O2001b Summing moments for the individual particles yields Myz EWixi WAxA WBJCB chc WDxD WEJCE prl 500 250 300 353 203 403 285 in 39 lb sz ZWiIi WAyA Ways chc WDxD WEIE WEIF 500 253 303 350 206 406 525 in 39 lb Mxy Ewizi WAZA 39139 W323 WCZC WDxD WEZE W122 500 250 304 354 204 400 340 in lb The principle of moments states that the sum of the moments of the weights Wl equals the product of the total weight W and the corresponding coordinate of the center of gravity G Thus EWixi 285 WxG 2Win x5 W 2 00 1425 in Ans EWi 525 ya 2W 1 ya W 200 2 625 m Ans Ewizi 340 WzG 2Wz 2G W 200 1700 in Ans axis with the weights pointing down 2 Equation 2 Sums moments about the x axis with the weights pointing down 2 Equation 3 Sums moments about the y axis with the weights pointing in the xaxis See figure below The resultant weight W of a system of particles is the sum of the weights of the individual particles furthermore the first moment of the resultant is the algebraic sum of the first moments of the individ ual particles about any axis or plane c194 Intermediate Problems 55 Locate the center of gravity for the five particles shown in Fig 55 if WA 251b W3 35 1b WC 15 1b WD28 1b and WE161b m 3 t 10 256wa3 33 M we Lag c 7 I gwi r 2 ngggg f Me a We Wwmx I 2 23 6 195 EXAMPLE PROBLEM 52 Locate the centroid of the rectangular area shown in Fig 57a 3 Ciquot A as c Fig 57 SOLUTION Symmetry considerations require that the centroid of a rectangular area be lo cated at the center of the rectangle Thus for the rectangular area shown in Fig 5 7a xc 172 and yc 112 These results are established by integration in the following fashion For the differential element of area shown in Fig 5 7b dA b dy The element dA is located a distance y from the xaxis there fore the moment of the area about the x axis is 7112 A h 2h The principle of moments states that Ayc j y dA Mx gnaw 4 Therefore M bhzz h AbhE Ans yc In a similar manner using an element of area dA h dx the moment of the area about the y axis is 11172 b xzb My Ldi Lxhdx h i O 2 The principle of moments states that icigccfxcm1v1y Therefore M1 111222 b c L Z Ans A bk 2 The element of area 11 b dy used to calculate Mx was not used to cal culate My since all parts of the horizontal strip are located at different distances x from the y axis As a result of this example it is now known that xc b 2 for the element of area dA b Ely shown in Fig 57b This result will be used frequently in later examples to simplify the integrations w If an area has an axis of symmetry the centroid is located on that axis and if the area has two axes of symmetry the centroid is located quotat the point of intersection of the two 199 quot CVEN 221 CLASS NUMBER DATE 3 PAGE J 90 Zbo 4699 E 3 309 OC7 7L0 7 2952 2 qui Z W Bcogt1 w 00034 R m gmmwg a r ZOO fNZii 74 gooiooa4 267 443 53 g 2 4C iN39i E Zao i Z M MA 2 2m x V 2st 2 MLVGL A mm 2 gQX X 3 Zim CVEN 221 CLASS NUMBER DATE PAGE i Now find what would have to be used to replace the previous force at the origin Note now that moments have to be added Previously no moments were required since the single force caused the same moments at its new location as the original set of forces However replacing it at the origin did not cause the proper original moments and they had to be added as shown CVEN 221 CLASS NUMBER DATE PAGE i I j 4 quot quot 52 W5quot 39 gmg am a CM M x 21gt A M A l wigw i a G 5 W 1 2 T4 A M u I 39 v a 5 i W if i j m M D M 4 at if 53 V W 3 f g 5 gt g Q quot 9 i a quot a 35 mf g39 fgww r H A 5 V 1 I E 3 Z M Ag 5a 5 W V wwp M z j A A 3 H m aw W A 332mg m E CVEN 221 CLASS NUMBER DATE PAGE 2 Wm a 3quot mltogtag ggggt Wv D 65quot quot39 M 5 XOI39N quot3 88GM mquot 3 2 ME 4 M j K m M a 3 ngff mu i O W W g fquot Racgg Vat Mfma QWN Y E Z Mgim 941 25XO 7 HQ 035i2 me l g m w w jg r O7 m Hq 010l397I J 25 H 907MW7 Q gg gggigw g V AF g g i mmsr m 219235974 wmvwm wawmww mm zlt Lam fmf Z 1 A 444 A 2 1 r22ad m22le 5 wi za gy 5 5 4111 OlN CVEN 221 CLASS NUMBER DATE PAGE 2 RE F E3 55 39 6 393 WK Wgtggt gag i 5 j 89Q 828 mga jwa q 4 0 mg kg g m w QEKFEgN39E W


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