220F09_lec07

220F09_lec07 - 3.6. STATICALLY INDETERMINATE SHAFTS You saw...

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Unformatted text preview: 3.6. STATICALLY INDETERMINATE SHAFTS You saw in Sec. 3.4 that, in order to determine the stresses in a shaft, it was necessary to first calculate the internal torques in the various parts of the shaft. These torques were obtained from statics by drawing the free-body diagram of the portion of shaft located on one side of a given section and writing that the sum of the torques exerted on that portion was zero. There are situations, however, where the internal torques cannot be determined from statics alone. In fact, in such cases the external torques themselves, i.e., the torques exerted on the shaft by the supports and connections, cannot be determined from the free-body diagram of the entire shaft. The equilibrium equations must be complemented by relations involving the deformations of the shaft and obtained by con« sidering the geometry of the problem. Because statics is not sufficient to determine the external and internal torques, the shafts are said to be statically indeterminate. The following example, as well as Sample Prob. 3.5, will show how to analyze statically indeterminate shafts. A circular shaft AB consists of a lO-in.-long, g-insdiameter steel cylinder, in which a 5-in.—long, fi-in.-diameter cavity has been drilled from end B. The shaft is attached to fixed sup- ports at both ends, and a 90 lb - ft torque is applied at its mid- section (Fig. 3.28). Determine the torque exerted on the shaft by each of the supports. Fig. 3.28 Fig. 3.29 Drawing the free—body diagram of the shaft and denoting by TA and T8 the torques exerted by the supports (Fig. 3.29:2), we obtain the equilibrium equation 3+ T3=901b-ft Since this equation is not sufficient to determine the two un- known torques TA and T3, the shaft is statically indeterminate However, TA and TB can be determined if we observe that the total angle of twist of shaft AB must be zero, since both of its ends are restrained. Denoting by gel and qbz, respectively, the angles of twist of portions AC and CB, we write ¢:¢1+¢2=0 From the free—body diagram of a small portion of shaft in- cluding end A (Fig. 3.2939), we note that the internal torque T. in AC is equal to TA; from the free-body diagram of a small portion of shaft including end B (Fig. 3.296), we note that the internal torque T2 in CB is equal to TB. Recalling Eq. (3.16) and observing that portions AC and CB of the shaft are twisted in opposite senses, we write TALI TBLz <5 a] 9% LG J20 Solving for T3, we have L112 T = —T B L2]! A Substituting the numerical data L1=L2=5in. J, 7 %7r(% m4 = 57.6 X 10—3 in4 12 — %7T[(17gin.)4 — (1%in.)4] : 42.6 x 10—31114 we obtain TB = 0740 TA Substituting this expression into the original equilibrium equa- tion, we write 1.740 r, = 90 lb - a TA :51.7lb'ft TB=38.3lb-ft SAMPLE PROBLEM 3.5 A steel shaft and an aluminum tube are connected to a fixed support and to a rigid disk as shown in the cross section. Knowing that the initial stresses are zero, determine the maximum torque T0 that can be applied to the disk if the allowable stresses are 120 MPa in the steel shaft and 70 MPa in the aluminum tube. Use G = 77 GPa for steel and G = 27 GPa for aluminum. SOLUTION Statics. Free Body of Disk. Denoting by T] the torque exerted by the tube on the disk and by T2 the torque exerted by the shaft, we find To : T] + T2 Deformations. Since both the tube and the shaft are connected to the rigid disk, we have TIL. 7"sz ¢1 = (1521 T6] = J262 r. (0.5 m) _ M05 m) (2.003 X 10—61n4)(27 GPa) _ (0.614 x 10‘fi m4)(77 GPa) T2 : 0.8747"I (2) Shearing Stresses. We assume that the requirement ram", 5 70 MPa is 38 r n . . . I 1‘ critical. For the aluminum tube, we have 30 mm I T — Tuhnrljl _ X 10411114) _ N (bl _ I — C] _ 0.038 m i m Aluminum G1: 27 cm Using Eq. (2), we compute the Corresponding Value T2 and the" find the max— 11 : EH38 mm)‘4 — (30 minfi] 7 003 10 G 4 imum shearing stress in the steel shaft. : Li X 7 1n r2 = 0.8741". = 0.874(3690) = 3225 N - m _ T2432 _ (3225 N ' m)(0.025 m) Tm” _ 12 " 0.614 X 10—51514 = 131.3 MPa We note that the allowable steel stress of 120 MPa is exceeded; our assump— tion was wrong. Thus the maximum torque TU will be obtained by making rm... = 120 MPa. We first determine the torque T2. TsteelJZ (120 MPa)(0_6l4 x 10%: m4) T2 3 I _ — 2950 N - m 32 0.025 m From Eq. (2), we have 2950 N - m = 0.8747", T1 = 3375 N - m CI = 77 GPa J1 = £95 "1W4 _ Using Eq. (1), we obtain the maximum permissible torque = 0.61.4 X 10’bm“ do T0=TI+T2=3375N-m+2950N'm TO I 6.325 kN ' m 4 ...
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This note was uploaded on 02/15/2010 for the course ENES 220 taught by Professor Staff during the Spring '04 term at Maryland.

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220F09_lec07 - 3.6. STATICALLY INDETERMINATE SHAFTS You saw...

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