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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
freebody 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 freebody 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 lOin.long, ginsdiameter
steel cylinder, in which a 5in.—long, ﬁin.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=901bft 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 freebody 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=38.3lbft 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=3375Nm+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.
 Spring '04
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