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Unformatted text preview: 31205.5 STATICALLY INDETERMINATE TORQUELOADED MEMBERS 223 *5.7 THINWALLED TUBES HAVING CLOSED
CROSS SECTIONS Thinwalled tubes of noncircular shape are often used to construct light
weight frameworks such as those used in aircraft. In some applications,
they may be subjected to a torsional loading. In this section we will
analyze the effects of applying a torque to a thinwalled tube having a
closed cross section, that is, a tube that does not have any breaks or slits
along its length. Such a tube, having a constant yet arbitrary
crosssectional shape, is shown in Fig. 5—30a. For the analysis we will
assume that the walls have a variable thickness I. Since the walls are thin,
we will be able to obtain an approximate solution for the shear stress by
assuming that this stress is uniformly distributed across the thickness of
the tube. In other words, we will be able to determine the average shear
stress in the tube at any given point. Before we do this, however, we will first discuss some preliminary concepts regarding the action of shear
stress over the cross section. Shear Flow. Shown in Figs. 5—30a and 5—30b is a small element of the
tube having a finite length s and differential width dx. At one end the
element has a thickness tA, and at the other end the thickness is t3. Due
to the applied torque T, shear Stress is developed on the front face of
the element. Specifically, at end A the Shear stress is 134, and at end B it
is 1'5. These stresses can be related by noting that equivalent shear
Stresses TA and 7,; must also act on the longitudinal sides of the element,
shown shaded in Fig. 5—301). Since these Sides have constant thicknesses
tA and t3, the forces acting on them are dFA = rA(tA dx) and dFB =
1303 dx). Force equilibrium requires these forces to be of equal magni
tude but Opposite direction, so that TAIA = ‘7ng This important result states that the product of the average longitudi
nal shear stress times the thickness of the tube is the same at each point
an the tube’s crosssectional area. This product is called shear ﬂow,* q,
and in general terms we can express it as Since q is constant over the cross section, the largest average shear stress
will occur where the tube’s thickness is the smallest. *The terminology “flow” is used since q is analogous to water ﬂowing through an open
channel of rectangular cross section having a constant depth and variable width w. Although the water’s velocity v at each point along the channel will be different (like ravg),
the flow q = vy will be constant. 224 CH. 5 TORSION If a differential element having a thickness t, length ds, and width
six is isolated from the tube, Fig. 5—30c, it is seen that the colored area
over which the average shear stress acts is dA =t ds. Hence, dF =
ravgt ds = q ds, or q = dF/ds. In other words, the shear ﬂow, which is con
stant over the crosssectional area, measures the force per unit length along
the tube’s crosssectional area. It is important to note that the shear—stress components shown in
Fig. 5—306 are the only ones acting on the tube. Components acting in
the other direction, as shown in Fig. 5—30d, cannot exist. This is because
the top and bottom faces of the element are at the inner and outer walls
of the tube, and these boundaries must be free of stress. Instead, as noted
above, the applied torque causes the shear ﬂow and the average stress
always to be directed tangent to the wall of the tube, such that it contributes S‘mss‘fﬁ:$°““dary to the resultant torque T. Average Shear Stress. The average shear stress, Tavg, acting on the
shaded area dA = 1‘ as of the differential element shown in Fig. 5505,
can be related to the torque T by considering the torque produced by
smwﬁee boundary the shear stress about a selected point 0 within the tube’s boundary,
(bottom) Fig. 5A3Oe. As shown, the shear stress develops a force 05F = 1an (M =
(d) 111ng as) on the element. This force acts tangent to the centerline of the tube’s wall, and since the moment arm is h, the torque is at = h(dF) = tea3,3: ds)
For the entire cross section, we require T: fl; h 7“,: ds Here the “line integral” indicates that integration is performed around
(6) the entire boundary of the area. Since the shear ﬂow q = ravgt is constant,
these terms together can be factored out of the integral, so that T= Tavgi§ h ds A graphical simplification can be made for evaluating the integral
by noting that the mean area, shown by the colored triangle in Fig. 5—30e,
is dAm = (1/2)h ds. Thus, T x 2%: f dAm = zfavgmm Fig. 5—30 SEC. 5.7 THINrWALLED TUBES HAVING CLOSED CROSS SECTIONS 225 Solving for raw, we have (5—18) Here 1an = the average shear stress acting over the thickness of the tube
T 2 the resultant internal torque at the cross section, which is found
using the method of sections and the equations of equilibrium
t = the thickness of the tube where 1'an is to be determined
Am = the mean area enclosed within the boundary of the centerline of
the tube’s thickness. Am, is shown shaded in Fig. 5430]”. Since 9' = Tavgf, we can determine the shear flow throughout the
cross section using the equation (5—19) Angle of Twist. The angle of twist of a thinwalled tube of length L
can be determined using energy methods, and the development of the
necessary equation is given as a problem later in the text.* If the
material behaves in a linearelastic manner and G is the shear modulus,
then this angle qb, given in radians, can be expressed as TL d
¢= 4,4316 lTS (5—20) Here the integration must be performed around the entire boundary of
the tube’s cross—sectional area. _ ~IMP_ORTANT POINTS " ° Shear flow q is the product of the tube’s thickness and the average
Ishearstt'es's. This value is constant at all points along the tube’s cross
. section. As a result, the largest average shear stress on the cross section
_ occUrs where the tube’s thickness is smallest. ' 17f Both shear. flow and the average shear stress act tangent to the wall of
. thertube at all points and in a direction so as to contribute to the resultant
__ torque“. . ' *See Prob. 14w23. SEC. 5.7 THIN~WALLED TUBES HAVING CLOSED CROSS SECTIONS 227 EXAMPLE 5—15 The tube is made of CSOIOO bronze and has a rectangular cross
section as shown in Fig. 5—32a. If it is subjected to the two torques,
determine the average shear stress in the tube at points A and B. Also,
what is the angle of twist of end C? The tube is fixed at E. SOLUTION Average Shear Stress. If the tube is sectioned through points A and
B, the resulting freebody diagram is shown in Fig. 5—323). The
internal torque is 35 N  m. AS shown in Fig. 5—32d, the area Am is Am = (0.035 m)(0.057 m) = 0.00200 m2
Applying Eq. 5m18 for point A, IA = 5 mm, so that
T 35 N  m
TA = 2mm I 2(0005 m)(0.00200 m2) = 175 MP3 A“ And for point B, :3 = 3 mm, and therefore 7 = T =..__35N—'mnn
3 2mm 2(0003 m)(0.00200 m2) These results are shown on elements of material located at points
A and B, Fig. 5—326. Note carefully how the 35—N  m torque in Fig.
5—3213 creates these stresses on the colorshaded faces of each element. = 2.92 MPa Ans. Angle of Twist. From the free—body diagrams in Fig. 5—32?) and 5—32c,
the internal torques in regions DE and CD are 35 N  m and 60 N  m,
respectively. Following the Sign convention outlined in Sec. 5.4, these
torques are both positive. Thus, Eq. 5—20 becomes TL d5
4’ = 2 4,4310 i7
60 N  m(0.5 m) 57 mm
4(000200 m2)2(38(109) mez) (
3iN  m(1.5 m)
4(000200 m2)2(38(109) N/mz) [
= 629(10—3) rad 5mm + 2 (57 mm
5 mm 228 CH. 5 TORSION EXAMPLE 5—16 A square aluminum tube has the dimensions shown in Fig. 5—33a.
Determine the average shear stress in the tube at point A if it is
subjected to a torque of 85 lb  ft. Also compute the angle of twist due
to this loading. Take Ga; = 380(103) ksi. SOLUTION Average Shear Stress. By inspection, the internal resultant torque at
the cross section where point A is located is T= 85 lb  ft. From
Fig. 5—33b, the area Am, shown shaded, is Am = (2.5 in.)(2.5 in.) = 6.25 in2 Applying Eq. 5—18,
_ T _ 85 lb ' ft(12 in./ft)
7M ‘ 2mm ‘ 2(0.5 in.)(6.25 inz)
Since t is constant except at the corners, the average shear stress is the
same at all points on the cross section. It is shown acting on an element
located at point A in Fig. 5—33c. Note that Taps. acts upward on the colorshaded face, since it contributes to the internal resultant torque
T at the section. = 163 psi Angle of Twist. The angle of. twist caused by T is determined from
Eq. 5—20; i.e., 95: TL ii: 85lbft(12in./ft)(5ft)(12in./ft)jg d3
4A3..G r 4(6.25 in2)2[3.80(106) lb/inz] (0.5 in.) = 0.206(10"3) in"1 j£ ds Here the integral represents the length around the centerline boundary
of the tube, Fig. 5—333). Thus, a, = 0.206(10—3) in‘1[4(2.5 in.)] : 2.06(10'3) rad Ans SEC. 5.7 THINWALLED TUBES HAVING CLOSED CROSS SECTIONS 229 EXAMPLE 5—17 A thin tube is made from three 5mm—thick A—36 steel plates such that
it has a cross section that is triangular as shown in Fig. 5—3461.
Determine the maximum torque T to which it can be subjected, if the
allowable shear stress is Tangw = 90 MPa and the tube is restricted to
twist no more than qb = 2(10‘3) rad. SOLUTION The area Am is shown shaded in Fig. 5—34b. It is Am = 1—(200 mm)(200 mm sin 60°) 2
= 17.32003) mmzatr6 mzt'mmz) m 1732(10’3) m2
The greatest average shear stress occurs at points where the tube’s thickness is smallest, which is along the sides and not at the corners.
Applying Eq. 5—18,with t = 0.005 m, yields T
W = 2:2,“; 90(106) Nlmz 2 201005 m)(17.32(10—3) n12) (b)
T = 15.6 kN  m Fig 544
Also, from Eq. 5—20, we have
d) = TL it Q
4Ai,G :
T 3 m) ds
”902 rad : 4(17.32(10—3) £1)2[75(109) N/mz] jg (0.005 m) 300.0 I T% 415 The integral represents the sum of the dimensions along the three sides
of the centerline boundary. Thus, 300.0 = T[3(0.20 [11)]
T = 500 N  In Ans. By comparison, the application of torque is restricted due to the angle
of twist. ...
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