CIVE 310 Lecture 2, Fall 2011

STRESS _ Fitnuetion Any. angnecring Tigect-, or component . + => Determination ef material and adeguacy i ot physical climensions.. \ Example =. Walls of pressure vessels. Floor of. buildings — Shafts of. It machines . Wings of airplanes, ett... . Feasibitly of Engineering Tegjecls . . 11 Sahely Stability. Corainteining a. stable. shuche [2] Lconomy -———» Strength Cproviding adequate. strength) _ 131. Serviceability —_____» Stiffness Cee ce. dbiffness). CClearances . Déflections . Spans ) .. os _. > * AR = Novmal to Aveo. AA. AB KAR. = ‘Tangential b Area AA. { tntemal forces OA. Cnkinilesmal). AP Cforce vector on AN. [Yq] Stess = Forces , Stresses i Avea. VAN | Normal. Stressex AR : Narmal . Shearing x= lim & = + = 0, Stresses “Stressex? Shearing Sbresses = bm 4% 4A—ro 4A — CActing in a plane -b to Xana in y-dir.) 3 © Cg = -_ os oe a. (Acting ino plane bh ts x and. in Zdiecks | Acting Ke tk rect 02 of. inaplane B Stress in j-ditection Ze axis. . _ i Conversion of Units a USA units S&.L units FORCE Lb_or kips Stress _ psi or ksi 4 as . Clbfin®). — Chips/in®) 1 kg = 2,2 Lb A fee 12 in. ~ 3048" 1 Lb-force ~ 44 N~ Linx 254 em.. _Lpsi x F000 Ta =» 1 ksi 2 FMP. _ kK. =10°PR 5 MR=10°8R ; GR=10°FR. Kip = 10° Lb i Two- Dimensional Sinss -/lane otvwesses _ i Consider an infinitesmal. cube , dk, dy, de. + Jb can-be verified lat Gi = Gy . LG By Laking ZG yig =? Y Sor aguilibrine. =>. ny = Cyx _ ; ee LOtresses in Hane XY are ar shown. i then, let Lny=Typ= T. peal Froblem.s. =>. Biaxial. Stresses : In the absence. { of all shear stresses. and 0% COR)... ora Of or Oe z0 | OL or Oy ae. Gs y= Ce = G2=0. “Thin sheets. — shell clements.§ fabrics... tr EX? yi ‘> Waxial. Shri. 2 dn..Lhe.. absente pe all shear stresses . _ Gpewe 5... Of ser. 5 Oar. Cy = Cxy = Gz = Gz 20. 1 Xs water body. ~ loose. sand .. . > Luaxial Stresses, h the. absente, of all shear stresses and. Lwo axial Stesses , y Axial Ioaded Bars cutting plane 4% X-axis e} member « Prismatic har ot Cross. Sechonal ara =A | iN \ Hl NHS *y Lif o. + Hg ae ! gle dz - Cc p ' d: unitorml if dP Oo * dysdz . a Atributed normal 4 . stress . = On dA O% im Ox A pt 4 ._ Unporyn. Stress o = fore PEN] or [Lb] Area A m> in? Stress Gncentration , 0 Chay pos. wader. concentrated. | | had (Theory). near far 0 Zouhbrum_» P= {odA. ; inn ch 7 [pon Crane > Ory,» (Stress Concentra Ln 20 P 5 a : mete “| ooo P <p Tmax Oman = Cay . _—_ “ay . = Oe é _— Oe Onv. av. Le hear’ Stresses Llontact Areo- Resisting. Area). Tite Pate. “FN jf FP i Ped Together. we cst eae bri ace small d a iglecked - 4 _ Shear fe. =P . are | Shear Ayes = Resishing Aree. =) 4 Shear Stress = Pe _ 7, _ Aabed Double. Shear? mp T=, 2 Aa., ad ‘| | Rawca Areaz 2A so Margin of Safeby ' Design of Aurially Loaded, Members _iitow 2 Sable CFS). Fis. - Ulhmate bead Chimit) _ Ta Allowable, Load Ru. Ou 91 vt eR... ns» Fo -= LU Eimate Stress Chimit) _ i Allowable. Sr: - Why tmplementng 3a * factor in dean ?. Me LS _ eae) oefects . - Approximations in design. . ~ Defects in Wel , connections, dimensions . ~ 2 ffect of load repetition CFatique). - Uncertainty in Loads. ( Live loals, wind K £.@. loads). - SVeman. error tt ' Example. CTRuss F Fete). _ Find. Area 8) menber BC Gen Oj, = 110,000 kPa = C60ksi). | CRecall. 1 ksi ¥MP%). oy = 250,000 ke. ~ (36 ksi). 2 FS 4) = 2,5..5 PS= a. a = 25g Ory > Oey x 1101000 => - = 164,000 kfa.. Fan | Ov= FE = Az E > Ag = a ... Lee Ci, = Czy . oh Age = 2:6 kN = 3.4/5 %10° m* = © 34.15 snd... Crequired, ara) (A) 1 Leng round har 4 diameter: D... _ =» A= 7D, D. | “Brey = 6.6mm... QO Lats | 7 ] r > prided ~ = Lire an bk ; a> O < Oy. ---.¢ a oy order fs sibishy shregth requirement) 8 In this context, rt 1s worth menbonng Lhat cfor design of ComrlessiIn A members, tt ts not enough t check. for strength, bet mem bens tin comp need t Sabisty Stabi 1G against buckling . 4 Shrenglh. repuuirement-: O- r. < OEy. u o Compression /Tembers << ae 4 - wera ees Bag Os Oe. Area. Hl Tey. depends on = Lenglt t, member, Ares 3), cross-section. , Shape L —— Y emss-sectim , and connectivily of member | Homber AB LL Force. = Fag = 2.5. KN7..... CCompreasion. ),..... Assume No buckling 2 Apg= FR = SN 1,524 lO. = 15.24 mm. 02 14,000 KFA. a“ - Round bar aiameler,D= 444 mm.. eq. AS. _ Llse. D= 200 >. Dreg aby. RR “Ing “Jubulaw section, Do =10mm..and. t= fmm oe Din, =Ginm . ! t=fmm. Aorovided-= FB _ _ wpe = 78,540 _ 63, 62mm 4 ae = 15mm Ok. tL . r Dut 210 mm 5 Ace that the sechnd solution. will provide. larger’ inertia -againsk. buck hag, Coverall. buckling), hawever,.one. should. check, pr another failure crilerion, that 7s local Luckhag of Lhin Lubes . ie f Overall Buckling t Me M MARA UTC | Axial Bree and. Axial Mress Qregnms . ab .@ ZxX=o coco enn i ~ZY=o > Vazo L * Jn axially Joaded. members: Nov’ PTL o) -P Veo . . Mzo Or. + at. any Section... Be ©) Pe. Axial —Disqrem - + ~ 2B. SIGN CONVENTION- FOR, INTERNAL. FORCES . Lo xemples n 200° . Ab rod. of variable. eros - ~~ t section huilt ct at one end ts subjected & tree axial ii porces as Shown in the accompanying quires Hind Pre max. normal 1 Stress and F.8. knowing Hot pt, NPD » Typ = 290 MP. ye oO. . Reactions 2 X=0 => Xas 200 KN 180 KN kN ue Cre am 150,005, . Diagvowy = 150. MR CT). y= -110K" =-91,667 KN O,0012 m* Mp,” = 91,669 te), a Og = 200M" — 80,000 KN 2.0025 ™* m = 80 MR CT. : => Omnax,= 150 MR. Notice. OF, = BO MR. ...CSEQMENT). ~~. AC (SEGMENT BD) _ . * Frovided Factor a? Safely CFS) = Otimit . . Cactval - Seamenk QD =» FAS= 230 = 1933 —» Minims FS. O — Segment CD => FS = 226 = 5164. 67 29 nenl™ AS = 290, = 625 . - Seg men AC => AS 220. 3