HW6 - (1.)Determine the maximum and minimum combined normal...

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Unformatted text preview: (1.)Determine the maximum and minimum combined normal stresses across the reduced section for the bar loaded as shown. Take the thickness to be P 0.5 in.. h. = [.0 in.. h: = 2.0 in.. and P = 37501bf. (’2) If lhe allowable combined normal stress in the steel clamp is [2 ksi. determine the largest clamp force F that can be a lied b A the I . r. = 0.375 tn. and :2 = 0.1875 in. PP y c amp Take (3) The L-shaped member has a square cross section 30 mm on a side and is acted upon by the forces shown. Normal strains of 300 u. and —400 n are measnred at points A and B. respectively. Determine P and e. Take E = 7'0 GPa. Do h and I. enter into the solution of the problem? Explain. (4') A solid circular-cross—seclion bar is acted upon by the two loads shown. Determine the max, shear stress and the max prune. stress at {at point A. the top of the bar. and (b) point B. on the side of the bar. Take the radius of thebartobeZOmm.L = 0.3 m.a.ndP= 600N. ME 423 Hw SOLUTWN r p Y m? “h 6—— 20 h H :h, A, in K U it {@UmeR’tum ZR: =‘=> [7x1- P zmz=D => 7142— - Pfos.) W? Y EL dxx— —' Ii + A _ P(o.€)Y + P — 7:; (05%|? (0‘)” r (2 F), + '2 P I 2 P (4y +1) P: 375'!) {L I uh..— A’l' y: +0.5: 0;»: 8/”: Bojowlby,‘; AT y: I'o‘y 0:0»:H4fr’fS—JOUO7DJ'; i Y 301m Mfigwm *‘-;Cy)u< -/3'§rl DETERm we c arm—Ea I1): I. -0. 75A (aw-wa- dzv s / y - (/.a)(o.7s-?— (f-o-0.?7s-)(o.7r—o.r97r) 0.?2?$ + 0.0366? 0 7s- -— 0.35-1.51 0.765c'2 _ 0.3-?qu ' Y -—.. .— —- DETERV‘MNE I: (Afiou‘r GENTRo'D) TI: (0.75%, 3'75)g -+- (O.7$)f0.?7§lf{0-Q%F"aué a - i ‘4'” [72.6675717'0 375954'6, (375' f‘flfsw Lil—{Ziawsy IE“: _!—. _Lg ‘-' 0. 0032¢55v + 0-000“? + 0.0039147 +0 .0/45‘6/ = 0.02774 M4 EdulLkBmum: FX=F M3= — F ['25 +6.0—owsr) 7 -12.5>3r]7 '2 CONT”, F— 2.?35’Fy 650:: "T ‘1'— I? P :— (075)@. 37:) #6— 037396: . [375) 2, 335' F“ y ______,____ + 0.02779 '3 F (2.50??? +102.o(5- y) PM): 0;), ace/Kr AT y: -o.éés-g_fy=o.33s- 53 ET 70xfo‘? fa Ewmufimum ,‘ Fx 7- f: m?‘ Pa 001M155an S‘r'fiagyrr P Fby 6y»: "T "' Pf?“ ——-’ a 4 ii. ) P-P 9,03) #(0103‘, ( a“ 3 ) f ’P 0000‘; 4.7gmo'5" A”? A1 y: - {s-flm ram" ' f—E—(-—J— $00!: 84 3 5, E 00004 @flfwd, r 0/)“ " L ( e 0.0”)“ $5 7 CB -- -— E 043504 .J- ‘vs'k£a r (MM _ F 2 €A4 fa: .- E #010909 .. 4.: _ P z [(30ka 70*”)! 0.000, 7 PT- ?ls'oN j arggmyfi; ELQerafilumi _ Tl’lr‘* I'}=O 3”” 7.. F), '2 '2? T2: O M”: -- PL My :— 0 I442 = -7PL _ ‘2PL7 _ 2(6ooK039Y Cfxx ’ I? '— WCouz)‘, "‘4- : 2.?é45’x10'y Due Tb m2 #PLV— _ (gogl/o.3)r* f: 3’ " '- “($0124 2 :-7_(;2o;¢(03y- Due To mwaug (my) 2P6); __ o M" A 6”? — Igba - [ MAS: AT '5' 3 171341-7124 (0.04) :— j,27325*xr0‘ Pa? (4) GMT, AT Pow-r" A”. 6” = (2.8c48xro%_oz)= 57 3MP # 4? (SW: ~(7J62xma/fomz): [4. 3214412: 6 _ 57-? \/—5-—7-—*'*—“ —‘ ‘3 a “2 Z i ( 2 ) +1495 2 .60. waO o/ - - ,Y _ (7 {@20r103)(o oz) + {27525xroc * —/3./MP4 PM?“ ...
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HW6 - (1.)Determine the maximum and minimum combined normal...

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