Quiz6 - lellnm 3.4-4 A solid circular bet ABC cunslsts of...

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Unformatted text preview: lellnm 3.4-4 A solid circular bet ABC cunslsts of [Wu segments. as shown in the figure. One segment has diameter dl = 50 mm and length L. = 1.25 m: the other segment has diameter :12 = 40 mm and. length l'.2 = LO m. What is the allowable torque 3';th if the shear stress is not to exceed 30 MPa and the angle of twist between the ends of the bar is not to exceed 1.5"? (Assume G = 30 GM.) Salutinn 3.4-4 Bar consisting of two segments d;;50mm d2;40mm '-'-'—"—x_ T A I..=1.25m B Lz=1-Dm C 7,le ; 30 MPa ALLOWABLE. TORQUE BASED LI'PON mun-z OF thsT gem“: t.s==u.02518rad ¢_ EE_ 111+ TL, ::(fl 2) G _ 80 Spa 61,, G!“ Gin G [p] I» d. _ 32T(fl + AUDWABLE. TORQUE BASED Upon SHEAR mm “C 0,1,, d; Segment BC has the smaller diameter and hence the mtdl G larger stress. a.“ = —L1 = 343 N ' m l _ tar _ fm’i‘rmo. 32(d? + d?) =_. _=3I N. Tm: 7rd] annu- ‘6 I” “1 ANGLE 0F TWle GOVFRNS Tflm=34EN-m 4—- Prulrlem 14-12 A prismatic bar A3 of length L and solid circular crass section [diameter d) is loaded by a distributed torque of constant intensity r per unit distance {see figure]. {a} Determine the maximum shear stress rum in the bar. (b) Determine the angle Uftwisl tit between the ends of the bar. Snitlllnn 3.4-12 Bar with distributed torque (a) MAXIMUM SHEAR 5111155? IOT ifiri. Tm, '— rL = m" = — T“ 170" mi“ (b) ANGIJT. OF TWIST ‘ 1rd" ft): = L: .l' = — J P 32 T W = J% = 32 txdx _ I Gt", rrGa“ I = Intensityr of distributed torque ( r. I _ ' _ 32: - 15:13 d=dtameter ‘h— d¢_néniUXdX=1-rfid‘ .— G = shear modulus of elasticity Problem 3.34 A circular bar A3 with ends fixed against rotation has l‘ 25 in' 25 m' l a hole extending for half of its length (see figure}.'I‘h: outer diameter of the bar is d2 = 3.0 in. and the diameter of the hole is d" = 14 in. The total length of the bar is L L 50 in. At what. distance x from Lhe left-hand end of the but should a torque To be applied su that the reactive torques at the support: will be equal? Solution 3.8-! Bar will-I a huh: L ‘- 50 in. Substitute Eq. [1] into Eq. [2) and simplify: U2"25in. d'_I“D[L+L x L L d2 = uutcr dimmer s G “P! 4"?“ In 2"”: 2’91 = 3.0 in. Comm-11mm den = 0 d‘ -= diameter of hole I y L =2.4in. " E13!» _fi TD = Torque applicd at distance I SOLVE ma - 1: Find 1 so [hat TA = T8 1 I x L 4(3 -- E) EQUILIBRIUM 4 I” y _ _ _ 1' I d' — d“ d ‘ TA+TRZIO ‘I‘TA='[E_—u (1) {TE—- 3 '1 l—l—(J) 2 I'm d2 In‘2 L[ (or. )‘ x -— ' '2 + — ‘— 4 d2 SUBSTITUTE NUMERICAL VALUES: 50'. I '.‘ x: m [2+(24in)]=30_l2in. 1— 4 3.0m. 9b” = Aug}: aflwisl at B in —- Pniar moment of mania at left—hand and I” = Polar moment of inertia at right-hand and _ TAM) Tatm) Top: —Lf2] " _ '6!» + THAI _ GI»: _ Tum-'2] GI” {21 Prflhlflm 3.104} A torque I is applied to a thin-walled tube having a cross scdiun in Lhc shape or II regular hexagon with wnmm wall thickness r and side length is {see figure). Obtain formulas for the shear sites: 7 and Lhe rate of twist 9. Solution 3.10-8 Regular hexagon SHEAR STRS T I‘VE 'r = — = . *— m,‘ 95-: ANGLE OF was‘r m1 UNIT LliNlel (RATE or TWIST) J_ 4.4:: _M3.:_9b3t i-d: L, 2 T 7 f _ a 7 i 7 27 7 2'." _ I; _ LensIh a side, 6; 5.195%) _ 96% I = "flinch-ms {radians per unit lengflI) L“ = 6!: FROM APPENDIX D, CASE 25: ,8 = 60" a: = 6 b: A, — "rang: Tm: 3D” 3V3]? ...
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This test prep was uploaded on 04/06/2008 for the course EM 319 taught by Professor Kennethm.liechti during the Fall '08 term at University of Texas at Austin.

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Quiz6 - lellnm 3.4-4 A solid circular bet ABC cunslsts of...

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