HW#2 - NRE -_ 3651 608 aw 8L 6011pth Problem 1 Determine...

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Unformatted text preview: NRE -_ 3651 608 aw 8L 6011pth Problem 1 Determine the resultant internal'ioadings acting on the cross seetion at C of the machine shaft shown in Fig. 1—5a. The shaft is supported by bearings at A and B, which exert only vertical forces on the shaft. ; 225 N (800 N/m)(0. 150 m) = 120 N 50 mm 50 mm (b) (a) Fig. 1—5 Soiution We will solve this'probiern using segment AC of the shaft. Support Reactions. A free~body diagram of the entire shaft is shown in Fig. 1—51). Since segment AC is to be considered, only the reaction at A has to be determined. Why? 5+ 2 MB : 0; “Ay(0.400 m) + 120 N(0.125 m) m 225 N(0.100 m) : 0 A = “18.75 N y The negative sign for A), indicates that A}, acts in the opposite sense to that shown on the free-body diagram. - Free-Body Diagram. Passing an imaginary section perpendicular to the axis of the shaft through C yields the free-body diagram of segment AC shown in Fig. 1—5c. Equations of Equilibrium. Lszxzo; NC=O Ans. +T2Fy=0; —18.75N—4ON—VC:0 () ° = ch—sssN Am _L+E MC = 0; MC + 40 N(0.025 in) + 18.75 N(0.2_50 m) = 0 MC : —5.69 N - in An}; Pmkfi em 9. The gears attached to the fixed—end steel shaft are subjected to the torques shown in Fig. 5—20a. If the shear modulus of elasticity is 80 GPa and the shaft has a diameter of lfi'mm, determine the displacement of the tooth P on gear A. The shaft toms freely within the bearing at B. Ta; = 130 N-m fig: 150 N-m 150 N-m Solution Internal Torque. By inspection, the torques in segments AC, CD, and DE are different yet constant throughout each segment. Free-body diagrams of appropriate segments of the shaft along with the calculated internal torques t are shown in Fig. 5—2015. Using the right-hand rule and the established Sign convention that positive torque is directed away frorn the sectioned end of I' the shaft, we have _ TAG = +150N-m TCD = —130N-m TDE = ~170N'm 150N111 These results are also shown 0n the torque diagram, Fig: 540C. Angle of Twist The polar moment of inertia for the shaft is “i _ 3' =13 .003) = Harmoqm“ Applying 5-16 to each segment and adding the results algebraically, we have Tm“) __ TL i (+150 N '- m)(0.4 m) 4“ ’ 75 + (~130 N - m) (0.3 m) 5 -s;;.q=t(to-‘q}m4[30(109) N/m2)] + (—170N'm)(0.5 m) _r '0 “3‘ C; - tramp—q) m4[80(109) N/m2)] ' m (C) Since the answer is negative, by the right-hand rule the thumb is directed toward the end E of the shaft, and therefore gear A will rotate as shown in Fig. 5—20d. The displacement of tooth P on gear A is SP 2 qur = {.1161 rad)(100 mm) : lb-lmm Ans. Remember that this analysis is valid only if the shear stress does not exceed the proportional limit of the material. Probkm 3 I'Ihe member shown in Fig. 8—5a has a rectangular cross sectiOn. Determine 5 the state of stress that the loading produces at point C. [6.45 kN ' I Solution Intemal Loadings. The support reactions on the member have been : determined and are shown in Fig. 8—515. If the left segment AC of the mem— ‘ her is considered, Fig. 8—5c, the resultant internal loadings at the section consist of a normal force, a shear force, and a bending moment. Solving, 7‘ N = 16.45 kN V = 21.93 kN M = 32.89 kN-m To SJQQOFK’ feaddoflfi m Cb) Cong-reef 3:} EEC = 0:. Ax ~ (-35‘X125‘ \ch «x- 0) +72 if :0: Ag-(thzswfi + (gm C13 +5 Eng-o = 075 margin ’— moss») — Aécsm C3) Normal Force 64.5 MPa CFC: 1.32 MPa 66:63.15 MPa Shear Force Bending Moment (e) ' (0 fig. 8~5 (cont) Stress Components. Normal force. The uniform normal-stress distribution acting over the cross section is produced by the normal force, Fig. 8—5d. At point C, P 16.45 kN “C i Z i (0.050 m)(0.250 m) _ 1'32 MP3- Shear force. Here the area A’ = 0, since point C is located at the top of the member. Thus Q 2 FA' 2 0 and for C, Fig. 8—52, the shear stress TC=0 Bending moment. Point C is located at y = C = 125 mm from the neutral axis, so the normal stress at C, Fig. 8—5f, is Me _ (32.89 kN‘ m)(0.125 m) 0' : — = 63.15 MPa C [50.050 m)(0.250)3] Superposition} The shear stress is zero. Adding the normal stresses determined above gives a compressive stress at C having a value of crc : 1.32 MPa + 63.15 MI’a = 64.5 MPa Ans. ' This result, acting on an element at C; is shown in Fig. $55K "55km ' We COL Shown RACKS cu éiamfcr oi \5m- (on wkod arc-«M: magfivbée and orien+c~¥ion crew ‘MO'XXMW nowm _ glass ark \ooiNr A? 303 WWGr ‘1‘: 4w: MafiMgm shear Shes: Mm? 8001b ([4 in.) = 11200 lb-in. 3001b 500 [b )6 800 1b (10 in.) :3000 lb-in. “r 500 lb (141m): 7000'lb-in 800 lb 500 lb (13) elation . Fig- 3-6 Mama! Loading; The rod is sectioned through point A. Using the free— pdy diagram of segment AB, Fig. 8—6b, the resultant internal loadings can ‘der the equal but opposite resultant? acting on AC, Fig. 8260- 800 lb 112001b-in. _‘ 21.13 ksi Normal force Shear force Bending moment Bending moment Torsional moment Combined loading (500 1b) (800 lb) (8000 lb-in.) (7000 ib-in.) (11200 lb-in.) ' (c) (a) (e) m (g) (h) S tress Components. Normal force. The normal—stress distribution is shown in Fig. 86d. For point A, we have 0.604 ksi + 16.90 ksi P 5001b - _ . 0.283 ksi+2Ll3 ksi “A : X : 7“0‘75 my a 283 981 _ 0283 km Shearforce. The shear-stress distribution is shown in Fig. 8456. For point or 17.5 ksi A, Q is determined from the shaded semicircular area. Using the table $2M ksi on the inside front cover, we have _ 4(0.75 in.) 1 , 2 _ 3 Q = y'A' = %- —7r(0.75 1n.) 2 0.28131n so that Fig- 34‘ “on” _ VQ " 8001b(0.2813 in3) T ’ M = 604 si = 0,604 ksi A It [$740.75 in.)4]2(0.75 in.) p .Bending moments. For the 8000-”: - in. component, point A lies on the neutral axis, Fig. 8~6f, so the normal stress is 014:0 For the 700071b - in. moment, c = 0.75 in., so the normal stress at point A, Fig. 8—6g, is MC 7000 lb ~ in.(0.75 in.) _ . 2 w = ——1——~—‘—-4—— = 21126p51: 21.13 ks: I [31r(0.751n.) ] Torsional Moment. At point A, p A = c = 0.75 in, Fig. 8—6h. Thus the shear stress is Te 11 200 lb - in.(0.75 in.) . _ TA = 7 : ———1-—w—IT— : 16901 p31 = 16.901991 . [517(035 1n.) ] , Superposition. When the above results are superimposed, it is seen that an element of material at A is subjected to both normal and shear stress components, Fig. 8—611 “A [_._.._. 2- L \/___> mukbi s1 ,11 :: 3L2 GMax 3 3L1 ‘23}; .6“ =— 99-3" A “(1%: a : (it) + (as) \m = 2L0 L ...
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This note was uploaded on 04/11/2008 for the course AAE 352 taught by Professor Chen during the Spring '08 term at Purdue.

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HW#2 - NRE -_ 3651 608 aw 8L 6011pth Problem 1 Determine...

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