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# ENGR MECHANICS STATICS CVEN 221

Texas A&M

GPA 3.77

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This 45 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 Lee Lowery in Fall. Since its upload, it has received 52 views. For similar materials see /class/226114/cven-221-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15

CL A55 13 CVEN 221 CLASS NUMBER DATE A 22 PAGE L ggxiig N Egg l W 3 a Z M g mks 33060 39 Raquotv 3M7 3 WWW E 0 GM 30 M M 3R5 2 LfQQ inf 1amp0 ETFE 2 Q gaw9m Zaga mm mm 3w Eggw Rm 0 CVEN 221 CLASS NUMBER DATE 3 2 ag PAGE Z WWWquot 33 S SKET E l y a g g 5 frf 22 mg JSWE 4 375 A R 21 z iquot azltl5 371Wkoom REV375I3W asc g g WM 23 v 1 1 608 CVEN 221 CLASS NUMBER 2 DATE PAGE L F 2 30074 9430ij 0 RF aooNced O TE4 WW39KDE M 3 L magF f Coal CVEN 221 CLASS NUMBER ltquot 4b 5 DATE PAGE 3 F3 Zoo M 3 dsm 517 3 4c 42300 sj39 c CVEN 221 CLASS NUMBER DATE PAGE 2 CVEN 221 CLASS NUMBER DATE PAGE i CVEN 221 CLASS NUMBER DATE PAGE 2 063 EZOOampN 0 3r E 50 7L 5M E a I 39 9 mam g g 5 59046 TQ FZtL 2 97 351m 4 quotR F2 579 8M3 ea Fz ri OZB N 7L E96 1 f quot 2 Fjifj7 125 mlquot Q4LWZ E wax quotEi CLASS lit EXAMPLE PROBLEM 612 A rope and pulley system is used to support a body W as shown in Fig 642a Each pulley is free to rotate One rope is continuous over pulleys A and B the other is continuous over pulley C Determine the tension T in the rope over pulleys A and B required to hold body W in equilibrium if the mass of body W is 175 kg 8584 N c SOLUTION The tension in the cable supporting pulley C is obtained from the equilibrium equation EPy 0 and the freebody diagram for pulley C shown in Fig 642b 131y ZTC w 0 TC 21w mg 175981 8584 N From the free body diagram for pulley B shown in Fig 6 42c T2Fy 2T 8584 0 T 4292 N E 429 N ZQeQ 3 fe gigs M w arses f2 39 i a 175 Qt iggxge M The tension is constant in a contin uous cable over a system of fric tionless pulleys see Example Problem 6 11 281 752 Z 8 a CVEN 221 CLASS NUMBER DATE PAGE Egg 3Te2agrmi Tquot 200 6ng 8 5 s53 10in 52Mamghz39lfm iiiimwg Qt r a W a at fquot sew r AMPLE PROBLEM 613 r 2 armada Qua w laii Jr W it 2 5 wt A beam is loaded and supported as shown in Fig 643a Determine the com ponents of the reactions at supports A and B 800 lb awaawmwmw 1200 lb is quot 3 2 mg is EM Fig 543 union I 1 3 n freebody diagram of the beam is shown in Fig 64317 The action of the pin support at A is represented by force components Ax and Ay The action of the roller support at B is represented by the force By which acts perpendicular to the horizontal surface at B The 200 lb ft distributed load can be represented on the free body diagram by a fictitious resultant R with a line of action at a distance xc from the left support Thus g a R A 2006 12001b xc1926 22ft 5 3 it The resultant R can be used only for calculating external effects reactions For V determinations involving internal surfaces such as those presented later in 1 0 Chapter 8 the actual distributed load must be used The beam is subjected to a coplanar system of parallel forces in the ydirection therefore Ax 0 The two remaining equilibrium equations are available to solve for Ay and By Determination of By LEMA O By25 750 80013 39 120022 0 The resultant R of a distributed load is given by the area under the load diagram and its line of action passes through the centroid of the area of the diagram By 1442 1b By 2 1442 le Ans Determination of Ay L2MB O 1 52358733 8202 308 2 0 Ans Summm39 g moments about point B y y 39 yields a value for force Ay that is Alternatively or as a check independent of the previous deter T2Fy 0 Ay 800 1200 1442 0 un t for force 339 Ay 5581b Ay 5581bT quot282 2amp2 f M s m g r K 1 7 55 E 5 guy 204K 35 cm W EMA s O QXE 3 5035 as GKXffi 3 61 I R33 3 ZQ4K 50K 1x V r t 2MB 20 QJW 353 agt4gt w 19 2 Ear 3 355 2L9 s Q r E Wk 3 4 We M Z g AW EXAMPLE PROBLEM 614 A beam is loaded and supported as shown in Fig 644a Determine the com ponents of the reactions at supports A and B 40 m 45 m b Fig 644 2Cng CVEN 221 CLASS NUMBER DATE PAGE gag a NR 3 w 3 3 w 39 A 399 3 3 if m a fvg Ki 2 5 ZMg w 3 z1 475302 g 3905ngqu a 4 I By 104L2K 39 ZS 0 A y 2442 i848 4 Ax 2 Z 5 545 EXAMPLE PROBLEM 615 A pin connected threebar frame is loaded and supported as shown in Fig 6 45a Determine the reactions at supports A and B Fig 645 284 284 op CVEN 221 CLASS NUMBER DATE PAGE 385153AV a k W 39 ZMWQ A mm wK AE MNXWWBYX OM Ex 3 784 4 m w AFC E4 f 1 353 i wequot EXAMPLE PROBLEM 616 A pinconnected twobar frame is loaded and supported as shown in Fig 64651 Determine the reactions at supports A and B E 4th SOLUTION A freebody diagram of the frame is shown in Fig 6 46b The action of the pin support at A is represented by the single force A at a known angle since bar AC is a twoforce member 01 tan 1 200300 3369quot The action of the pin support at B is represented by force components Bx and By Since the frame is subjected to a general coplanar force system three equilibrium equations are available to solve for the unknown magnitudes of forces A Bx and By Determination of A MMWWW C 317 ber the line of acti n of force A must pass through pins A and C Since bar AC is a oforce mem a Q11mmnn MAMWuquot ALAquot nzQ So D 285 CVEN 221 CLASS NUMBER DATE g 5 PAGE gig l OR E 353 Egg rm MW fa quotx lt W 2855 CVEN 221 CLASS NUMBER DATE PAGE 9 Lg CV 2 42 2M A x ZZHLEM l CDWE Z8 Q CVEN 221 CLASS NUMBER DATE PAGE zmggg 1 mgaf g5 A A y ZZ CKM Z TV 0 221 ZO CV Qv nilZ iwo Eg z 2 33 m g w v 36 3 GOQaDthZ OC gmg f EH 2g m 2 E 1 2 CH C aZ53I EWg Q me axm w o 21 83 A A quot3382 gag CVEN 221 CLASS NUMBER DATE 3 i 98 PAGE Pep w CL A55 13 CVEN 221 CLASS NUMBER DATE A 22 PAGE L ggxiig N Egg l W 3 a Z M g mks 33060 39 Raquotv 3M7 3 WWW E 0 GM 30 M M 3R5 2 LfQQ inf 1amp0 ETFE 2 Q gaw9m Zaga mm mm 3w Eggw Rm 0 CVEN 221 CLASS NUMBER DATE 3 2 ag PAGE Z WWWquot 33 S SKET E l y a g g 5 frf 22 mg JSWE 4 375 A R 21 z iquot azltl5 371Wkoom REV375I3W asc g g WM 23 v 7465 Z 64 CVEN 221 CLASS NUMBER DATE i i Q PAGE as g 3 5 gawligem 5 I 4 4 g I g a 1 g ag bgi X 2q zz zakg 3 443 86 FLEXIBLE CABLES Pg 4 W v w 4 lt2 3 6 M LTOTRL 11 Figure 821 Flexible cable supporting a system of concentrated loads 444 391APTER 3 INTERNAL FORCES IN Ay UCTURAL MEMBERS Ax TI quot7 w 53 2 7am P c p2 Fig 821 d as shown in Fig 821c From this series of freebody diagrams and the equilibrium equation 21 0 it is observed that iiAMPLE PROBLEM 311 A cable supports concentrated loads of 500 lb and 200 lb as shown in Fig 822a If the maximum tension in the cable is 1000 lb determine The support reactions Ax Ay Dx and Dy The tensions T1 T2 and T3 in the three segments of the cable The vertical distances 33 and yc from the level of support A The length L of the cable a b c d Dx I K I T3 T3 D 5 T1 T2 C c at i1 2 g b 5001b 2001b Z f Fig822 SOLUTION a A freebody diagram for the cable is shown in Fig 82217 From the equi librium equation 2MB 0 Ans Since the horizontal component of the tensile force at any point in the cable is constant the maximum The maximum tension of 1000 lb will occur in interval AB of the cable tension in the cable occurs in the Thus from the freebody diagram at point A of the cable Fig 822c segment with the largest angle of Ax V A2 V10002 38242 9240 9241b Ans inclination y Ay Eli50002 2001o 3824 E 382119 I 446 446 a CVEN 221 CLASS NUMBER DATE PAGE w gzzg Z ls V QM xwma kamk SbAmtW ngww 21 9C 23 g gjggwgag H7 5 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 it HE E New CONCEPTUAL EXAMPLE 101 RADIUS OF GYRATION The radius of gyration of an area is used in engineering for determining the buckling strength of columns Two column cross sections with equal areas are shown in Pig CE101 Which cross section offers the greatest resistance to buckling largest radius ofgyration with respect to the x and yaxes shown on the figures UriT Ie 4 inr i 4 in gtl Fig cm 01 SOLUTION The radius of gyration of an area with respect to a given axis is the distancei from the axis at which the area can be conceived to be concentrated and have V the same second moment of area with respect to the axis as does theractual o distributed area For both cross sections A 381 24 in For section a a S vwf s 549 S a CVEN 221 CLASS NUMBER DATE PAGE W is a 2 8 CZquot 7quot3 zmai f 148m KgX LHCEKQ 251E 2 we Hf 2 552 cg 2 I t Eg I Z i he Aa Z Mg a g E ME AQ km T63 ng dzw g m g 234quot 3 1 V Z s g Z 32 m M a m3 afgfmg j Lz fzz4 8X W1 5 8 5 get 7 i Z T 5 m4 mm MWAMV WIWW 560 b CVEN 221 CLASS NUMBER DATE PAGE IE UZEQ fiz 2 4a agga 52 42 wg 5 2 OJOQSMMD 3 gs 55 I 1 Q 2 a 5 x quot M if f V A 7 f a A5234 Q CVEN 221 CLASS NUMBER DATE V PAGE mgz g e w m 101 M Z quot Qquot X 4 cap 9 a x izb MW mwwwmmwwm Z xg mi W 2a ag EZyi g 2 54 03ng aw 75 93 a a mm35 Amb alaszzm 1 a 44584 I CVEN 234 CLASS NUMBER 7 DATE PAGE f g 3151 F33 WEme a W E 121quot 3 55 EC 3 AWE WEE VEE Egg gaigE fw R 83 ZMCW c aMKQ AAM Ag M i e f5 g Q 334 V 2 i 3 Z 2 x W ALv LgtZKR IBC V5 139 8AF 2CALEEB 2 seam Ea E lQ 4 W quotZNij 3amp3 sw m quotw 5w gammy A t 5 4 Mgafg m 85 Agmgm g w E ME 2 w z chp MO L Qwgfi CVEN 221 CLASS NUMBER DATE z Z g O8 PAGE 1 E39EXSO lt3 Mg 6 3 s W 41945 5 r quot I T E a SEQEMAA MES 067 i 5 A 392 W 7 1 L gs DIWSM Egg 1 Hth v R w WADEEAEGS 1 A af 43 A 39 223 3 lt5 H amp gag g O7Eo 798 I 355351 3 g M g 3 Egg 1 2 74 g F 5 q 95 I 147 25 A Ma w 3 AM 1 12am N Eva w 2 22 w JETmME u A quot 53 Him Eggm quot39O a G 98 5 3 g A 235241255 1 141061 ZO Z 3 EN CVEN 221 CLASS NUMBER DATE i E ag PAGE E Tm 39quotN6 5 g M 5 REQTkM QQ l g a m A 164 5 IRMA quot 58 CVEN 221 CLASS NUMBER DATE PAGE y km V r x K Fi V g mam m MEMggf 300 Sam5 635 azf gN a 643 W 3 75 CVEN 221 CLASS NUMBER DATE PAGE W 4V is A 9 EA E A a I 5 f i g 39 N 3 N 23 a m CltI 2mm 2 3 753 m EEQQEM 5 Mam E neeM 9 m a 5 200 m9 2 832 w QMQ MAPLE CVEN 221 CLASS NUMBER DATE PAGE EVER Ex 31 F5 3 Fa W W ww WWW a I v A OW PLOT TEE 5 3m I CVEN 221 CLASS NUMBER DATE 5 Ms Ufm yf im fm m x wa Mmymw lacs 2 E a 66gt a g E E 2 Z i x E a S 2 E E a g i i g g E a i 2 g i i s i 3 i i g Em m PAGEm s i 5 WMJAMMWEM CVEN 221 CLASS NUMBER DATE PAGE ggg gt4 g a 2 m 2 38 7w gm a V 2 Eggi w zf lt4Llt gt 2 er 10 839 857 g fw MAC 3 g w g z Rife Si CVEN 221 CLASS NUMBER DATE PAGE VE Rzmg ng 5 EA IQXZZ

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