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# STRENGTH OF MATERIALS ENSC 2143

OK State

GPA 3.78

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This 0 page Class Notes was uploaded by Ms. Abe Bernier on Sunday November 1, 2015. The Class Notes belongs to ENSC 2143 at Oklahoma State University taught by Bruce Russell in Fall. Since its upload, it has received 28 views. For similar materials see /class/232951/ensc-2143-oklahoma-state-university in Engineering & Applied Science at Oklahoma State University.

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Date Created: 11/01/15

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yield stress is 60 ksi which will 00cm at the lower load concrete crushing or steel yielding and at what total load 1 Draw the P vs A diagram for the concentrically loaded column if the column is initially 20 ft in length Begin the P vs A curve at 00 and continue the curve until the steel yields AND the concrete in the column crushes 9 v I v V s M xi 2 as 3355232 m 51 m Li AV lessens m Wan ttest my nZBwampwp E S 2 L fsg a 4 Mia9 were T Y Ml EEC 5 5 9 wee F1 2 Aegis g Ac C ASEg a 66 a a m2 Eagerlasefl 7 7 a quot33 53w E was agilaeal 3 0235 m 50 SHEETS 5 SQUARES 341238 100 SHEETS 5 SQUARES 30237 200 SHEETS 5 SQUARES 3 0137 quot 200 SHEETS FILLER COMET C vurl r gzwiz A r A m ZQBJQM E a 57 36 c t g Sq jaw waxzsxla frmi quotMom 3 5 C39h r f Ha 9318 LN 1amp5 K figJ gi agfwmw g 5 A58 m awe C O wESEemWEW q 39w quotH 9 ng 2 D 39 I 23 foa Sgtgm gw F2 C i j s 5932 m 31 W m p m C f up 133 392 fkfwr rm Q 55a my MRBM Cowca i fii wufgwmag 9 7938 44 1 O U0 3023 200 SHEETS m 5 301M353 343255 w 50 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9 L A 1 1 I LJ r I J I v 39 quot I r r l v fa 39 57quotquot I J quot quot IL j I I 7 y 0n 1 l 1 v y I l n I i x vL y y w 1 n W a W y x lt x A r T z x J 7 K M Q 4 3M z c a N V f I 4 an w N M a 0 l x A w a m m A y 1 f f 9 l 7 m A U y n V 1 W x 1 V n A r W A II fik 1 NJ lt 3 v 1 y 1 w t w r r g w z 3 amp a w 1 u 3 M A 5 b J M w W M 1 I ll x r m M y W w m m A V w w q H i 7 g a V y 1 1 linlrli n i W 5 x e v a l r v gt I x r w m h w a L X A I M M 4 M m w V A V 7 M 3 u an r A 1 4 gig rzieixi r 9323 15 v E5 U a R E COMET 30 yaw g r k j 01 Solutions 46060 5610 243 PM Page 37 7 2010 Pearson lquottlllL lllUl Inc Upper Saddle Rivet NJ All rights t39csct39vcd liix material is pmtcctctl untlut nll copyright luws its they Clltl cnll39 unisti NU purttmt at this material may be t39t39pmtlttccd in any t39tn m or by any means without purtnisxiun in writing from the publisher 01 53 The twenth shem39 Slims in each 01 the 67mm tlinmutcr bolts and along catch 0139th tour shaded shear planes is not allowcd to uxcuctt 80 Msz and 500 kt a rcspcctivcly Determine the maximum ztxinl force P that can ht applied to Its joint I 100 min 7 Internal Luadings 39l39ltu shunt l39ot39cc dcvelnpcd on L ilCh shout plane 0139 the bolt and the mcmhut39 can be dctcnnincd by writing the force uqttution 0139 equilibrium along the member s axis with reference to the fi eerbocly tlittgrums shown in Figx a and 7 respectively if m l Lith P tl ft I 4 ljt t n3 I V r P4 vcratgu Shear Struss The Ell cit39x ol cztch sltcur plane of thc hulls ilIlLl the memth 17 t a w are 1L T 11006 2821 quotm and ll HUM Mllm respectively Wu obtain t Ib 14 TI r i so It r l tt tl A4 i 282mm n I 9047 N 905 kN commls Ans 39 V P391 l r 39 sun Ht l m l 1h 001 I 30 Mt N 2t kN P a Cl m x I m r a m r x m m I A L4 w J x x K a s in W A M E 5 W x K A a a F w K W H 1 w a a x W a c g Q 4 2 m 4 1 2 3M w c y r m I 1 1 i Q I z m T w 4 r r e w X y e W 1 a x v u I W f a L X W 1 H y 1 4 A i w m a n L 1 x m r I y x H 1 A V x f r I V V I r V i A U g n I V y L 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13 f TD m atmm w V 3 Mquot x v E324 m M4 F r 37 2 6quot F n32 TW i am 1 1 V 1quot Dam Wm Ra mcwl fmA EggVi 53 OUE MW gm Aim AwMW gig Wamm whq MVTMQK Q Zg W 3m 39 Kaix R my H D 2 g a 3 332 a 5 Egsg iw We w 39 H x2353 paw M m Wazoqwe 99 53E 2gt t ALVA a W a 39 3 wmmwix 3 3 Magma 3 T a V a 37 quot is 539 m9 Q 9a4 i 3 w c w39z W x A L m 71 F3 w fam Lag T xf 9 X532 5 29 W2 mwg if 011g 96353 w 2 V m I 9 27 diam y W at 31 rm 7 twp W was far m w mwgfm mwmlmm cat g Engma a mem 56th immkg 75 f 3 A71quot 7 I 5 HHQCS rm Emm g E k L Pa 2 a T 3 J 1 Mr 563 gT T Crow CL w E N 3 Tlvg 6 g 0 53 g m g a 1 19 at f a 0 V W E 40va I c 3 31 M 2 2 L quotquotquot 1 1 4L WGW M I J a 16 w tailkm 29 I zz53 P Gsquot ENQQ g fgm a ma Brand A 2 H if A F W r r nwr a M g m F Exam amp4 FOUAM M ma CW m5 Mommeg fb mggg a Q W3qu vT39 fwmg MN 5 ER mu C 5 To b Wm f a ti a 3 3 k Wk N CD M 98 mm 3 M Q HIV 0 NOTE THAT yamg W i 2amp9 g xx 3 gt 151mw 2 MM 07 WOW m g W a lt3 0 Mg 52 6 A C m 2 P533 0 w w f 9 Q m V T w B V 5 in 1 P in Inf Yquot mo mm AC39TE lt33 M Fw x w 3 53 3 13 mm mm Eggww M M Q g m w mi y k g aid ti zggx Ewgrzgmc a mg f fxy m A1 77 576 L v HS 4 2 mp5 5639 Vem y Cmg gm xc F acgwwiga a Emurzrrza Z39gam5 cw TEE Ng m 650 iv 5 ii a 39 m 7 w 741 4321 JQJEQ Aym u wwz 77 2 Z WZ angs i j Wr ng 7 WW MW w m U M quotquot QMQ wquot W 3 Z 39E A M I Mohr s Circle for Plane Stress The equations for the transformation of stress can be represented geometrically as a circle This Mohr s Circle provides a mathematical representation of the stresses acting on any inclined plane through a two dimensional element The Mohr39s Circle is more convenient and easier to use than the transformation equations it can also be used to provide a graphical method to solve for the state of stress on any plane if consistent sign conventions and geometry are used 1 Sign Conventions Positive a Tension Stresses bl Clockwise Shear 2 Coordinate System Mohr s circle is plotted with normal stresses 0 on the xaxis and shear stresses r on the yaxis Positive stresses are taken as up and to the right 3 Sketch the Element Sketch the element with the known stresses in their proper orientation Note that the geometric solution depends upon the right face being vertical In the figure 0 2 ksi 7x 4 ksi and L1 4 ksi 2 4 its E LA M P Lb l W 4 Locate the Center of the Circle Czia39 1 T O The center of the circle 4 is equal to arm Distances are measured from the origin 61 147 11 Cu 0 07w t quot quotz y Z l 5 Plot the Controlling Point A Azia r 4 Point A is the state of stress on the righthand face The Radius of Mohr s Circle is equal to the distance between Point A and center of the circle AL z 4 8 any a kst Draw the Circle Determine Maximum Stresses 07m quotl kSl 0mm a 139 R i me S Sl rmumln 1R 410 6 7 Solving the General Case See the figure below Stresses on any plane can be easily solved by a Draw a vertical line through A intersecting the oppo ite side of the circle Label this Point B b From Point B draw a second line parallel to the plane where the state of stress is desired This line will pass through the circle at Point D In this example Point D gives the state of stress for the element rotated 45 from the original orientation c Point D gives the state of stress on the plane represented by the line BD Remember that clockwise shears and tension are positive stresses Solving for Principal Stresses and Principal Planes figure below Draw a line from Point B and passing through the intersection of the circle with the aaxis Points P and P39 Note that there are two principal planes located 90 apart Maximum shear stresses and their stress planes are found by constructing lines BS and BS 0quot OZVQ R MLZQV d r 1 Principal d 265quot 6 Stresses 0 39 10 f T4 Original 4 A Stresses 2 1 2 this 39 r 4 i4 1 305 591 Ammo m V TRQL E M mm Hamsxx39urk 3 3 Suiution 33 mhimu 1 a 3 5 sz EjNSC 3H E E Mg W WW S 32 3 93 5mg Hamismy39 2 Sniutiim 39 2 5 amm m mm 131 a w Em mam Nmif Ah f smx i t Wm WWW y WW 1 g2 e I x s E 4a 11 V I m E g 2 z x I 3quot t AW E K v w s 1 Wmf r W i V A W A m A s g V 3 Eiamcxxrmk ii Samiix m SP 01 Fmblcm 2 22 m 1 ii my EE meKi fv x was 5 83223 WEE 5 j 4 lt II A 3 s 3quot 51 y i Q quotV 4 5 m M GK a m z 5 w a g k 3 39 V 9 f mm mm Hamewmk g3 Solution Fre em 3 u e m mam Wham AVE 5 4 33 4amp1 Wm f 01ii i i ma 4 M magw 1 13 iim is 12m mm 9 Mm m alu m 4 be S 4 1 m1 339 lt 1 5 l i g i 3 z w 3 n no a Q 5 9 S 313 3 ENSC 3MB Hemewurk 92 Smiutiun f mk sm 4 3 gzLamx 11 If 2 g 33932 339 mzmm f i 5km ximw 32233 x 23 mltmn 5 325 x u E 5 A g i a 3 w W 3 M M KW is a g o if 39 quotf A z W i e gt w g g s 39 39 Wm w 3 1 6 2 s mm F gt3 4 S NM mm 39 WWW x e G 2 m aw e Hamewerk 3 3 Soiuiiun S 2 3 mi 3535 wzama 3 Nme 1 WW my 1 wk X g 39 V A i y A W g x f Hamewmk 3142 Salutim z Si 2amp3 i s zwcmgc mum mmm 2m a 2 3 12 gram mi g zwm w mgm iu ax g 3quot 1 W W w x R m m 2 may i lt zneu gtrk 7213 Smh on ENEMY 2143 8 3m 3 Friaikam 7395 Fm hi3 pmialmm mad Tammi hfii ligh pg 36 tannin f8 mild Sm i a Rm mx39 i igg 3w 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