228_Mechanics Homework Mechanics of Materials Solution

228_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 332 3.82 A steel shalt must transmit 150 kW at a speed of 360 rpm. Knowing that G = 77 GPa, design a solid shaft so that the maximum stress will not exceed 50 MPa and the angle of twist in a 2.5m length will not exceed 3°. SOLUTION lsouo‘ W 10: $53 = 6H2 15 (03 H “25%;?" - 3.972%(0‘5‘ N-m Desl‘jn ‘tor 531255 vigthi'i‘ '3‘" 50 MPQ. = 50"}0‘ PG. N=Tg_g_ . C3=2_T=(2\(3.9789¥l03) L 6' "th?s ” 171: traumas] C : 37.00 Mo‘s m : 5o.c,c,mo“ tn" Desfjn “For anal): o‘P +w'l'51L l9ini'} g? ‘-‘ 3° '-' 52.36:!10‘: wad TL _ 2T1. _.‘ CHE} _ (Zl(3.978‘7x(o'5)(2.5} 90: 5:1" “ rec“ 1TGCP Ff77xto*)(52.3exto") c = 35.4ouo‘5m '-‘ 1.5707 “0-; M? Use Dav/jam Value C: 37.00xl53hn '-' 37.0 mm 3 0i = 2C. = 74.0 mm «C PROBLEM 333 3.83 A steel shaft of 1.5-m length and 48-min diameter is to be used to transmit 36 kW between a motor and a machine tool. Knowing that G = 77 GPa, determine the lowest speed of rotation of the shaft at which the maximum stress will not exceed 60 SOLUTION MPa and the angle of twist will not exceed 2.5". P = samba w) (.2 =5”! -— 0.02.qu L: ”Wu G- =77w0" P Tiamvc bases-p 0n WWI-mum shoes: f“: 60 M136; = GOKIO‘ Pa 7::- %¢— ' T = J57; : §c3fz l£(0.02'+)3(6.0xl0:) = 1.230288%“): N'M Tenet): based, an +m‘si dangle (P = 2.5-0 = 43.G33 V10‘3 NJ GP -.-. 11-. .-. T: J50 = “W, 77 0-02? " 77*qu 43.433163) 5: L 2L (”(1.5) = mar/'30 no" U-M Sit-wiper inning jouems T= 1.16730Xl03 M-m = =__E_ 1. miflflnm - M TD Zfl‘FT *2 21T 'Z‘It'UJe'JSOxlo’) ‘- 4‘6” Z i_i;li_ig ...
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