Question

4 8 02:55:34) A B с D Outside Diam = 3 in. Outside Diam = 2 in. Outside Diam = 1.5 in. T2 T T 30 in. 18 in. 12 in. Section 2. Problem 4. (25 pts.) A system of bars is loaded torqes as shown. Determine: a) the shear stress in each bar. b) the twist of pont D with respect to point a. Assume that all three bars have a modulus of rigidity of 11,500 ksi. External Torques: T1 = 550 (lb-in.) T2 = 750 (b-in.) T3 = 1000 (b-in.) Shear Stress: The shear stress in bar AB is The shear stress in bar BC is The shear stress in bar CD is (psi) (psi) (psi) Angle of twist: The angle of twist in bar AB is (radians) The angle of twist in bar BC is (radians) The angle of twist in bar CD is (radians) The angle of twist of point with respect to point a is (radians).

Answer 1

3 C, lisin G= 14500 Ks = 11500x103 lblin? Ti= 550 lbrin Bin T2 Ti T2 = 750 ebrin T3 = 1000 db-in 30im T3, 12 in 2 12 in ) IKsiz 1000 db/in? lesia 1 1. 100inz TB - T3 Te - T2 Тост, TA = TB+Te+TD = TXT2+13 = 2300 lb-in Section AB og Tmax = 165 ad3 J = ady 32 LAB - daba TAB= TA= 2300 lb-in 3oin 3 in Ermax - T- &max 16T ad3 I 10= 32TL ad 4 h Tmay 16 X 2300 TXB) 3 433.844 lib/cin) 2 Ø AB= Angle of twist , 32 TAB. LAB ñaha 0.00075 rad. 32 X 2300 x 30 Ax(3)4 x 11500 x103 Seesion BE:- lo-in .. Toe- TA - TB = 2300 - 1000 1300 LBC = 18 in Tmax= 16T 16 X 1300 -- 827.61 lb lín) ? dBC - zin +d3 Tx(2)3 Obe= 32 TB e LBC = 321300 X 18 5d4G AX (2) 4 x 11500 x103 rad. section Coor -0.001295 TED = TA-TB - Tcz 2300 -1000-750 = 550 lbnin LED- Tmax a 16*Ted - 16 4550 deD = 1.5 in od 3 = 1244.945 lb hinta *X[1:53 0 eb 32 TCD LED = 32 x 550x 12 a dua ax(1.5) 4 x 11500X103 0.00115 rad- Angre of twist at end D wort Ado= Ø ABtoBet ped 0.003195 radia 12 in

Shear stress o The Sneas stress in ban A Å is - 433.844 -psi 827.61 psi 22 B C V - ! 1244.945 Psi Angle of twistor Angle of twist in bar AB is 0.00075 radian 0.001295 Be w 97 CD M 0.00115 и of poin't D with respekt to point Ais -0.003195rad. M