hw1sol - 1.1 Two solid cylindrical rods AB and BC are...

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Unformatted text preview: 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as Problem 1 -1 shown. Detennine the magnitude of the force P for which the tensile stress in red AB is twice the magnitude of the compressive stress in red BC. 31:). ‘I L - AM: {-93% 3.141;.“ - 301m. n C2130} - P 6-3:. AM. 60*? -.- 3.9333—o.mq7? 7 06 8E; -Equwl"3flg. EM +9 2 63‘ 0.323er: amazes -o.I41471°‘J 1”: “5"" “‘1" “a _ - 1.11 Therig'id harm; is supported bytheiruss systemshown. Knowin thaxthe Problem 1 .1"! member CG is a solid cirmfl'ar rod‘chJS-iin. diameter, determine the nonfisl stress in CG. use“: Par+;o|n EFGCB cue m‘i‘FM-rkoJY, +4 21:7 =0: % Ena” 3-3200 = a' F“ = $000 it. Using beam 5135-“ .1 FM: Ecol-h {)‘MF:0= —(~=n§55+m(§ga= 0 Fan; = Fae- : m «FL. *Cmss Edits-via} and. o-F Melamine-e CG: Am = EH"- : 37540.15)" 2 GENE—Ff :13 “Wind s-l'rE'hs in C6. f: G. . .. 6’ :- 42: r __°°° = “5539 . w A} DAV-“‘7‘? 'PS '- Problem 1.38 .fl. {P m wm. 1'; +{Pm 46*](o.é) hit-".3, casso°Jfo.5W-U-'ans~Fn 300N054) = 9 : LSGLJQS P — CL'H‘FG'Q Fan = ,3 Fab 2 h'EIEES F’ = [Lar‘ESSKEGr-Ws'lr fizqmgxm‘w _ romm‘ = fl _ qufloa F8 3.45 1J6 Three steel bélts are to be usual to attach the steel plate shown to 3. Wooden beam. Knowing that the plate Will support a l 10 RN load, that the ultimate shearing stress for the steel used is 360 MPa, and that a factor ofsafety «3.35 is desirad, ' determine the required diameter of the bolts. For curl: bolt, ’P= 1%”- : 35.“? k1) Ream”! P _— (-p.s.w=(s.35)ca_e.ga-r3-=mash: .. 1"; _ Pu HEEL 1:" I: ' Ear” ‘_ ml" HEP _ (93:11.55:qu __ -3 d= 11"; - WEBeowrafi‘} "' 2.9.2::(9 "" J=20-8 MM “'3 Pmblem 1_55 1.55 thMmE-m-damfluplnhmcdnd NEW phmmdatfimflfl. meiugthflflnukhflflmflngflrmiflflflhflhuafl macfim‘udthmflumfinmmflmifififlmmmofhmfinka imningflnudfl,imminefiaaflmbkhadPifmnwnlqufanfdyuflflia dull" I . '1'an x But-J an dad“: shear In Pin A... #- =3Eal1=3$fmmar = 5.9.2“. 1'94]... P-= fiefi = 3.12 an? H Band on Jayme 51155.“ in Pius 1.4 B and E}. A: :Efil‘ = "EEGJJRE‘ — “3+ 1.: no" m" Flu 'l' it-§"E= (RIKIDDIID: ”3J9" ‘3“) = 15"}!!01' N 7' E5.- 3-u P = HF“. = 3.9“?!1'0’ H Basra! on cnmpruzssf'un in jalnki BD- : FM- on: ml: A: (maaaXama‘; =.- lea-no' an" an' 25““ =LIMA_I_“ ‘ lawn“) Zil'fflmsfl P 7" fiffin = Hrmno‘ N mm”: we a: 1? .-... ”1116+. P= 3.?2vio‘N P?- 3.11 RN "1 Pmblem 155 . 1.65 Tmmudmnumbmufiflx llflmunmmmWWl mimm jninndbyihnfinuflcfluudmufspliueshwn. Knuwlmfllfitthcmimmnaflmmhk Mmindrglmdnplimkfimm dalmn'linnflwlu'fiaIuiallnade unbeaafilyuppliod A. = {anaemic-n. Hum: TJHo" M" B = aohlza': "ro‘ J}: T no" Jc’s‘oo wag-9‘} {vi-n 70' can ?D‘ = “J's-yam" pa P=1Laahu ...
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