{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test1 - \r C(2 A 1/2” x 8” AISI 1018 Cold Drawn steel...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: \r C, (2) A 1/2” x 8” AISI 1018 Cold Drawn steel bar (Yield strength is 54 kpsi) is cantilevered to support a static load of 10,000 lbs in the x direction and 1,000 lbs in the y direction as shown in the figure below. The bar is connected to the support using two V2” — 13 UNC SAE grade 5 bolts and one 3A” — 10 UNC SAE grade 5 bolt, respectively at points A, B, and C. Find the factor of safety against bolt shear and strength of member only. (Assume that the threaded part of the bolt is small) (3 5 points) A A '2 A( H (fi' N 10,000Ibs °3 30—} _> (:5 Q "3 i1,000|bs @209 «1“ N = x v < 10" , l $/Z////////////////7////Z///////d 1/2" f FWMMWJ%:WWM,&FMwKW%F?M%fi ;_ fl\Y,+A’L)(L + ’93 X3 ' “WT p11. +1613 ) y Pr Z +A X +0333“ {/9633 (6) (" ’9433(4)*(""Z§.{§):39/2 _: l I l a, ‘ZM (“l +03 “963+ /‘?63+44//a? {Am 1 7’4 booojm) #013,000)“, 3 9,2): 90003158337 Fri/many {/1 aw)” a :I4,88¢;/5_W ill moon I ‘ ‘ ) @3844“? 1W" Z”: .3. :: 19393 stc Iojooo Nu X A4; 3394/ . ’ /000 p , )9? pm :EL 2 H ~— '— J M <9 Z12 A6 $7244 'WWM 1”.” <9 T” I 1 ~ 1 -13”: 7Q : T n3 T 209,394») (2.5:?) )’ Z 1 (2. HEY/943) + (N33 WWW/«9133‘ ('99)?) N0 W /9 CAFGZ C51“ 5% flf'c' 0a w ('J 8’5“”)301 /r9/yp\, Mom «(£th fig . Qéam ‘fflof 056% 9x Z303» r46; @WQ/f "' ' [9 / \ 7139M SAM/r aw WM mm L» “’ffl482<3»* a 596/3 dW IL: NMWQM, 0 law“: 07,386.54, pg 9. m bWA) 27% z 1!, 939.4“ Ié) 22.2.63 : 25130323233 51%” ”b94344“ : 35” V <577)(dp$000>: /» 7:) fl:/~7 [4% [(7 22/ €33,1f WELLS—.345 J. Mm 210.3% Z00?” C+ %m M f: 2/.33 - 2.2579] I: 1877/2; m9 :9 m :04 88¢,¢,7>(2.§882 + /87/2 0/ V} -; 2,058.72 + 3,200 /0,0<90 .r[8_.:-~S—\7fl (1) In the figure below, two steel machine parts are clamped together with a single 1/2” — l3UNC grade 5 bolt with rolled threads. The two parts are subjected to a separating force that fluctuates between 1,000 lbs and 7,000 lbs. Find the factor of safety assuming a reused connection and the threaded part of the bolt is very small compared to the unthreaded part. (35 points) Wfiw M- m - R“ (M) 3/ KL: 9i: (r/953M30a‘10‘) =. Z19 45“ HO" )L/(jn { 1 KM _ .577 W'Ed K7717: BOA’Oth 1' ________________. __,_.__.‘.______._.-——-—— 2&1 /.II5+'D-d.) (DAB. 3 1% I’HFLZLLI-E '.7.fl~5’) .(Msé +1) “1) j) KT]: 1» 7n r)(-7r—-f _ Km": ZgaL, 1}0‘ IL/wx -: KM;- :3 Km 3 “le Km; : E5;— [2 war?“ 'b/AL ) Crfi2 .1533 Km'iflbfifi‘}, 1 \ K‘Ikik; 5““? 5” => Se:(JZSHS’)(-YOH)(H/o);'.§/-Q‘ z,“- I {:13 [A t {d —} / -135? Pa: 3000”) Rm: ((2700,; J Ref/wed wmwcé’dm => H ;.75 Re SP : (.7S)(./L,/q)(?5) = ?'04’6:KP1|~ Fmrm $0M: 3—H) K/:3 34¢ a W Maw/m met *-=> ”V:— £111....” 0.2.3123 __ Qoflecrr/UJ { _.._J )= Kym/“I .IZfB 3 (300 0) (L534) + 9000 $5!)an km 9 _ f2.7x10‘nggo\l; 5K 3’25” ”a < Fj- __.,_.___.._‘___,__,__..‘. w-“ MAX _ ___..... V K“ KIM 2.?9: )3 I'2.7UO‘ 2-,) Sepamxfifgn dog; rwf (Wax/r «— ”2; ~.—_—._.E:__.._. -. 3943-443"; Km A 5 { :2 2 3 ——-—~—- 37w.“ 14» + “M ==> J/m /. :6"? s a - ' __.g.,LL._M.: M = 9,791,403 n,> fl 1+ ke H— 296% . Km I?.7 sppmm dawn of will»? r m; ~_ 16:32:33 112 t ”23% :2) F‘JJZM'E [JAE/Lg MW aka hr Squaw/4529}; S/nmmfie/ygr {Ufa—4.36)! ..,...____...____.-—-—— (3) A C — clamp shown below uses a 1/2” — UNC square thread and a collar of 5/8” mean diameter. The friction coefficients are estimated to be .15 for the screw and .12 for the collar. Find the force required at the end of a 5” handle to develop a 600 lb clamping force. (30 points) Fro/m fabt’e. (-2) “=13” {=;_ - c0769 a _ ’3 M} - (1- ; :'L’6’§’ (' 7": de le {9 2. ( MN+ coma.m + Wind; ”11m; CUM/m... (,u. 2 06». 3’0 =0 CESAR-1d _ 2:) 7": (600){-4(/S) IT(-L,cn‘)(-/Y}+-076i9 ”i ‘\_/ 2 [71-905) - ('07‘ ”(451] \ - +- (600) -I2 "“3 - \ T”... 1i.33 + 22.: Z :9 T: 50.83154)“ = Fat = FM rm_______ 50.173 _. ""r /o_ [(7 lb ...
View Full Document

{[ snackBarMessage ]}