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LECTURE+10+COE-3001-A

# D z 2 z r2 ip 2 dd r1 0

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Unformatted text preview: i.e., dx L TL = ²་  In this case we get total angle of twist: (3- 15) GIp GIp ²་  NOTE THE FORM!: T=kφ (k is an eﬀecbve torsional spring constant, i.e., kT = L ²་  φ expressed In radians 1/30/13 M. Mello/Georgia Tech Aerospace 14 CIRCULAR (HOLLOW) TUBES Ip = Ip = Z 2⇡ 0 Z ⇡4 (r 22 r2 ⇢ 2 ⇢d ⇢ r1 4 r1 ) = ⇡4 (d 32 2 d4 ) 1 (3- 16) CIRCULAR TUBE IN TORSION (far more eﬃcient than solid bars in resisbng torsional load) Alternabve form to (3- 16): ⇡ rt 2 ⇡ dt 2 2 2 Ip = 2 (4r + t ) = 4 (d + t ) (3- 17) ²་  In (3- 17), r is the average radius of the tube, i.e., r=(r1+r2)/2 ²་  In (3- 17), d is the average tube diameter, i.e., d=(d1+d2)/2 ²་  In (3- 17), t is the wall thickness ²་  NOTE: 3- 16 and 3- 17 are exact equabons ²་  FINALLY, note that in the case t<<<r Ip ⇡ 2⇡ r 3 t = ⇡ d3 t/4 (approximate) 1/30/13 M. Mello/Georgia Tech Aerospace 15 CIRCULAR (HOLLOW) TUBES (cont.) All previously derived formulas for torsion, shear stress, and shear strain are applicable to circular hollow tubes provided the appropriate polar moment of inerba is used in accordance with (3- 16, 3- 17, 3- 18). ²་  Note the shear stress distribubon in the tube as depicted in the ﬁgure. The average shear stress is very nearly equal to the maximum shear stress ²་  In the case of a solid bar, the shear stress is maximum at the outer surface and zero at the center. Most material in a solid shaP is stressed signiﬁcantly below the maximum shear stress. ²་  Hollow tubes are thus far more eﬃcient than solid bars in resisbng torsional loads. ²་  Circular tubes advisable and oPen used when weight reducbon and savings of material are cribcal ²་  OPen used for drive shaPs, propeller shaPs, and generator shaPs ²་  Dimensions must be appropriatley selected in order to prevent wrinkling or buckling of the tube 1/30/13 M. Mello/Georgia Tech Aerospace 16 NONUNIFORM TORSION (case I) I. Bar consisbng of prismabc segments with constant torque through out each segment STEP 1: Determine magnitude and direcbon of internal Torque in each segment X TCD = T1 T2 + T3 TCD + T1 + T2 T3 = 0 TCD = 0 X TBC + T1 + T2 = 0 TBC = T1 T2 TBC = 0 X TAB = 0 TAB + T1 = 0 TAB = T1 Note 1: When determining shear stress in each segment, only need magnitudes...
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