# Torsion of Closed Thin Wall (CTW) Sections

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1 9 orsion of losed hin Wall (W) Secions 9 1

2 Lecure 9: ORSION OF LOSED HIN WALL (W) SEIONS ALE OF ONENS Page 9.1 Inroducion losed W Secions Examples losed Recangular ube Slied Recangular ube A Hybrid ross Secion Reciaion Problem Example

3 9.2 LOSED W SEIONS 9.1. Inroducion his Lecure coninues he reamen of hin Wall (W) secions in orsion. We proceed now o he case of losed hin Wall (W) secions. hese are much more effecive in resising orsion because a cell shear flow circui can be esablished o ake up he inernal orque losed W Secions A closed hin wall (W) secion is one in which uninerruped circuis for a shear flow q can be esablished. his shear flow, which is an inernal force per uni lengh angenial o he wall, resiss he applied orque. Some examples of closed W secions are depiced in Figure 9.1. ube box aircraf fuselage space shule fuselage car frame (unibody consrucion) wing orque box Figure 9.1. losed hin wall secion examples. Such secions are much more effecive in resising orque han simularly configured open W secions, someimes by orders of magniude. 1 his effeciveness is paricularly imporan for aircraf wings because i deermines he fluer and divergence speeds of he fligh envelope. If he wall hickness is much smaller han he oher dimensions he shear flow q can be assumed o be uniform across he wall hickness. he wall shear sress magniude is hen simply τ = q/. his approximaion is sufficienly accurae for engineering purposes if he hickness does no exceed, say, 20% of he smalles cross secion dimension. 2 losed W secions are classified ino single cell o mulicellular, depending on wheher here is only one shear flow circui or several. he necessary noaion for he single cell case is inroduced in Figure 9.2. Poins along he wall midline are defined by s, called he arclengh or perimeer coordinae. he wall hickness may vary as he conour is raversed, i.e. = (s), bu he maerial is assumed o remain he same. he sress analysis of he single cell case follows from elemenary saics and is worked ou on pages of he eer-johnson-dewolf exbook. 3 Here we only summarize he major seps. Le he disance from he midline o he cener of applicaion of he orque 4 be denoed by h, as shown in Figure 6(a). hen = d = q(hds) = q hds= q 2 da E = q(2a E ) = 2 qa E, (9.1) s 1 Exercise 4.3 of HW#4 quanifies his saemen for a wing orque box. his is also obvious in he relaaive srengh of he wo cross secions esed in Experimenal Lab 1. 2 Discrepancies from he exac soluion can be readily included in he safey facor agains orsion failure. 3 F. eer, E. R. Johnson and J.. DeWolf, Mechanics of Maerials, McGraw-Hill, 4h ed., his is called he shear cener in a more advanced reamen of orsion. 9 3

4 Lecure 9: ORSION OF LOSED HIN WALL (W) SEIONS (a) y = (s) (c) ds z q ds O s Shear flow q = 2A E Enclosed area A E h Figure 9.2. orsion of single-cell closed W secion. in which A E is he enclosed area defined by he wall midline as illusraed in Figure 6(c). I is imporan no o confuse his wih he cross secion area, also called he maerial area or he wall area, which is ha shaded in grey in Figure 9.2(a). Solving (9.1) for q yields he shear flow and sress formulas q = 2A E, τ = q = 2 A E (9.2) Since q is consan along he conour, plainly he maximum shear sress occurs when he hickness is minimum: τ max = q = (9.3) min 2 min A E I can be shown 5 ha he wis-angle rae is given by φ = dφ dx = q 2GA E ds = 4GA 2 E ds = GJ, J = 4A2 E ds (9.4) If he hickness is uniform he conour inegral in (9.4) reduces o p/, where p = ds is he wall midline perimeer lengh, and he wis-rae expression simplifies o φ = dφ dx = qp 2A E = GJ, in which J = 4A2 E. (9.5) p 5 Obaining his relaion requires energy mehods, which is graduae level maerial. 9 4

5 9.3 EXAMPLES z (a) y A q 2 q q 1 1 D (c) A q _ 1 q 2 q 2 q _ 2 q q 3 2 q 1 D ell flows q 3 Wall flows Figure 9.3. he orsion of mulicellular closed W secions is no covered in his course. he analysis of a mulicellular closed W secion under orsion, skeched in Figure 9.3, is far more elaborae because he problem is saically indeerminae, and is no considered in his course Examples (a) (c) d/2 Same d and as in (a) d d/2 d hin cu d d Figure 9.4. Shaf W cross secions for examples in 9.3. he folowing hree problems were given in Miderm Exam 2, Fall 2005 (he hird one was a onus Quesion) losed Recangular ube he cross secion of his shaf is shown in Figure 9.4(a). Dimensions are d = 2 in and = 1/8 in, d being a midline dimension. Same hickness all around. he shear modulus is G = psi. he shaf is subjeced o a orque of = 1600 in-lb, which is uniform along a lengh of L = 60 in. Find (a) he maximum shear sress τ max in psi (sign unimporan), and he wis angle φ in boh radians and degrees over he lengh L. 9 5

6 Lecure 9: ORSION OF LOSED HIN WALL (W) SEIONS Soluion. Using W heory: (a) Enclosed area: A E = 1 2 d2 = 2in 2. Shear flow: q = /(2A E ) = 1600/(2 2) = 400 lbs/in. Shear sress: τ = q/ = 400/(1/8) = 3200 psi. his is also τ max because he hickness is uniform all around. he simplified formula (9.5) for J may be used because is consan all around. Perimeer: p = d d + d d = 3d = 6 in. orsional consan: J = 4A2 E /p = 1/3 in4. wis angle over lengh L: φ = L/(G J) = rad = Slied Recangular ube he recangular ube secion of he previous example is cu along he lengh as skeched in Figure 9.4. he widh of he cu is negligible. Required: (a) he maximum shear sress τ max in psi (sign unimporan) he wis angle φ in boh radians and degrees over he lengh L; (c) compare he shear sress and wis angle o hose found in he previous example. Soluion. Using OW heory: (a) Recify ino narrow recangle wih hickness = 1/8 in and long dimension b = p = d d + d d = 3d = 6 in. orsional consan: J α J β 1 3 (3d) 3 = 1/256 in 4. Max shear sress: τ max = /J α = psi. wis angle over lengh L: φ = L/(G J) = rad 400. (c) Maximum shear sress raio: 51200/3200=16. wis angle raio: 400 / A Hybrid ross Secion Figure 9.4(c) shows a hybrid W (HW) secion, in which wo fins of he dimensions shown are aached o he recangular ube of Figure 9.4(c). he maerial of he fins and ube is he same. Required: (a) he maximum shear sress τ max in psi (sign unimporan) he wis angle φ in boh radians and degrees over he lengh L; (c) compare he shear sress and wis angle o hose found in he W example. Soluion. Use decomposiion and apply W and OW heory as appropriae. Sar by decomposing he orque ino he porions aken by he ube and he fins: = ube + fin + fin = ube + 2 fin. (9.6) Nex compue he orsional componen for hese componens, using he approriae heory. For he ube, J ube = 4A 2 E /p = 1/6 in4 from Example in For a fin, J fin = 1 3 d3 = (1/8)3 = 1/768 in 4. he oal J is J ube + 2J fin = (65/384) in 4. Since G is he same for ube and fins, he orque porions are given by he J-raios ube = J ube J = = in-lb, fin = J fin J = = in-lb 130 (9.7) (a) For he maximum shear sress, apply he W and OW formulas for he ube and fin, 9 6

7 9.3 EXAMPLES (a) b 1 A / b 1 b 1 decomposiion A 1 W b 2 b 3 2 OW OW D E D E respecively: b 1 = 200 mm, b 2 = b 3 = 400 mm, 1 = 2 = 3 = 6mm G = 80 GPa (seel), = mm-n, L = 2514 mm Figure 9.5. A-shaped hybrid W secion under orsion for Example 9.2: (a) cross secion, decomposiion ino hree porions. τ max,ube = ube 2A E = (1/8) τ max, fin = fin J fin = (1/8) 1/768 τ max = max(3151, 1182) = 3151 psi = 3151 psi, = 1182 psi, in ube. (9.8) (c) For he wis angle we use he oal J compued above: φ = L GJ = (65/384) = rad = (9.9) omparing he above resuls o hose for he ube of Figure 9.4(a), i is obvious ha he addiion of he fins has no reduced he maximum shear and he wis angle by much (from 3200 psi o 3151 psi, and from 8.80 o 8.66, respecively). his corroboraes he well known fac ha OW componens are relaively ineffecive in resising orsion Reciaion Problem Example his is given as Problem#3ofReciaion #4. he A-shaped cross secion of a HW prismaic shaf is shown in Figure 9.5(a). I is subjeced o uniform orque over a disance L. he riangle A forms he W porion, whereas legs D and E collecively form he OW porion. Daa is given in meric unis: b 1 = 200 mm, b 2 = b 3 = 400 mm, 1 = 2 = 3 = 6mm, G = 80 GPa (seel), = mm-n, L = 2514 mm, 9 7

8 Lecure 9: ORSION OF LOSED HIN WALL (W) SEIONS in which he shear modulus G is he same for he whole secion. Find: (1) maximum overall shear sress τ max in MPa, and (2) wis angle φ over lengh L in boh radians and degrees. Soluion. Decompose ino hree porions as shown in Figure 9.5. Porion 1 is he W riangular box A. Porions 2 and 3 are he legs D and E, respecively. he oal orque is accordingly decomposed as = in which i denoes he orque aken by he i h porion. o find hese proceed as follows. ompue he J β of porion 1 using W heory: A E = 1 2 base heigh = 1 2 b 1 (b 1 sin 60 ) = 1 4 b2 1 3 = mm 2, J β1 = 4 A2 E ds/ = 4b1 4(3/16) b 1 / 1 + b 1 / 1 + b 1 /( 1 /2) = 3 16 b3 1 1 = mm 4. Here A E denoes he enclosed area of he W porion. Nex compue he J β of porions 2 and 3 using OW heory. Since he legs are idenical: J β2 = J β3 = 1 3 b = 1 3 b = mm4. Here β 1 is used for simpliciy since he b/ raios of he OW recangles exceed 10. he oal 3 J β for he whole secion is J β = J β1 + J β2 + J β3 = mm 4 ecause G is he same for all porions, he raios J βi /J β, i = 1, 2, 3, deermine how much orque is aken up by each porion: 1 = = mm-n, 6 2 = 3 = = mm-n For sress calculaions we may ake J α2 = J β2 and J α3 = J β3 since α β 1 when he b/ of an 3 OW recangle exceeds 10. he maximum shear sresses over each porion are 1 τ max1 = 2A E ( 1 /2) = MPa, τ max2 = τ max3 = 2 2 = 3 3 J α2 J α3 whence he overall maximum shear is τ max = max (τ max1,τ max2,τ max3 ) = MPa = MPa his occurs over segmen of he W porion because ha segmen has he minimum hickness ( 1 /2 = 3 mm) when raversing ha circui. he wis rae of he whole secion is dφ dx = = rad/mm GJ β Since G, and J β are consan over he shaf lengh L, he end-o-end wis angle is φ = dφ dx L = L GJ β = rad =

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