Quiz SM202 Torsion Lucas Montogue X Problems Problem 1 (Hibbeler, 2014, w/ permission) The 20-mm diameter steel shaft (G = 75 GPa) is subjected to the torques shown. Determine the absolute value of the angle of twist at end B. A) ๐๐ = 1.85o B) ๐๐ = 3.66o C) ๐๐ = 5.73o D) ๐๐ = 7.54o Problem 2 (Beer et al., 2012, w/ permission) The aluminum rod AB (G = 27 GPa) is bonded to the brass rod (G = 39 GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at end A. A) ๐๐ = 2.09o B) ๐๐ = 4.13o C) ๐๐ = 6.02o D) ๐๐ = 8.10o © 2019 Montogue Quiz 1 Problem 3A (Gere & Goodno, 2009, w/ permission) A hollow aluminum tube used in a roof structure has an outside diameter ๐๐2 = 104 mm and an inside diameter ๐๐1 = 82 mm (see figure). The tube is 2.75 m long, and the aluminum has shear modulus G = 28 GPa. If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 48 MPa? A) ๐๐ = 3.75o B) ๐๐ = 5.20o C) ๐๐ = 6.75o D) ๐๐ = 8.20o Problem 3B What diameter d is required for a solid shaft (see preceding figure) to resist the same torque with the same maximum stress? A) ๐๐ = 66.5 mm B) ๐๐ = 77.3 mm C) ๐๐ = 88.4 mm D) ๐๐ = 99.2 mm Problem 3C What is the ratio ๐๐ of the weight of the hollow tube to the weight of the solid shaft? A) ๐๐ = 0.524 B) ๐๐ = 0.636 C) ๐๐ = 0.745 D) ๐๐ = 0.811 Problem 4A (Beer et al., 2012, w/ permission) A torque of magnitude 4 kNโm is applied at end A of the 2.5-m long composite shaft shown. Knowing that the shear modulus is 77 GPa for steel and 27 GPa for aluminum, determine the maximum shear stress at the steel core. A) ๐๐st = 12.4 MPa B) ๐๐st = 32.4 MPa C) ๐๐st = 53.6 MPa D) ๐๐st = 73.6 MPa © 2019 Montogue Quiz 2 Problem 4B Determine the angle of twist at A. A) ๐๐ = 3.09o B) ๐๐ = 5.07o C) ๐๐ = 7.05o D) ๐๐ = 9.03o Problem 5A (Philpot, 2013, w/ permission) The torsional assembly of the next figure consists of a cold-rolled stainless steel tube connected to a solid cold-rolled brass segment at flange C. The assembly is securely fastened to rigid supports at A and D. Stainless steel tube (1) and (2) has an outside diameter of 3.50 in., a wall thickness of 0.120 in., and a shear modulus of G = 12,500 ksi. The solid brass segment (3) has a diameter of 2.00 in. and a shear modulus of G = 5600 ksi. A concentrated torque of TB = 6 kip-ft is applied to the stainless steel pipe at B. Determine the maximum shear stress magnitude in the stainless steel tube. A) ๐๐max = 7.40 ksi B) ๐๐max = 9.81 ksi C) ๐๐max = 14.4 ksi D) ๐๐max = 27.2 ksi Problem 5B Determine the rotation angle of flange C. A) ๐๐ = 2.41o B) ๐๐ = 4.22o C) ๐๐ = 6.33o D) ๐๐ = 8.14o Problem 6A (Beer et al., 2012) A solid shaft and a hollow shaft are made of the same material and are of the same weight and length. Assume that the angle of twist is also the same for each shaft. Denoting by n the ratio ๐๐1 ⁄๐๐2 , where ๐๐1 and ๐๐2 are the inner and outer radii of the hollow shaft, respectively, show that the ratio of torques Tsolid/Thollow is such that Ts 1 − n2 = 1 + n2 Th Problem 6B Suppose now that the angle of twist is the same in each shaft. Show that the ratio of torques now becomes Ts 1 − n 2 = Th 1 + n 2 © 2019 Montogue Quiz 3 Problem 7 (Beer et al., 2012, w/ permission) An annular plate of thickness t and modulus G is used to connect shaft AB of radius ๐๐1 to the tube CD of radius ๐๐2 . Knowing that a torque T is applied to end A of shaft AB and that end D of tube CD is fixed, show that the angle through which end B of the shaft rotates with respect to end C of the tube is = φBC T ๏ฃซ1 1๏ฃถ ๏ฃฌ − ๏ฃท 4π Gt ๏ฃญ r12 r22 ๏ฃธ Problem 8A (Gere & Goodno, 2009, w/ permission) A tapered bar of solid circular section is twisted by torques T applied at the ends. The diameter of the bar varies linearly from ๐๐๐ด๐ด at the left-hand end to ๐๐๐ต๐ต at the right-hand end, with ๐๐๐ต๐ต assumed to be greater than ๐๐๐ด๐ด . Derive a formula for the angle of twist of the bar. Problem 8B For what ratio ๐๐๐ต๐ต /๐๐๐ด๐ด will the angle of twist of the tapered bar be one-half the angle of twist of a prismatic bar of diameter ๐๐๐ด๐ด ? The prismatic bar is made of the same material, has the same length, and is subjected to the same torque as the tapered bar. A) ๐๐๐ต๐ต ⁄๐๐๐ด๐ด = 1.13 B) ๐๐๐ต๐ต ⁄๐๐๐ด๐ด = 1.45 C) ๐๐๐ต๐ต ⁄๐๐๐ด๐ด = 1.72 D) ๐๐๐ต๐ต ⁄๐๐๐ด๐ด = 2.04 Problem 9A (Gere & Goodno, 2012, w/ permission) A prismatic bar AB of solid cross-section (diameter d) is loaded by a distributed torque (see figure). The torque per unit distance, is denoted by t(x) and varies linearly from a maximum value at A to zero at end B. The length of the bar is L, and the shear modulus of elasticity of the material is G. Determine the maximum shear stress in the bar. © 2019 Montogue Quiz 4 A) ๐๐max = 2๐ก๐ก๐ด๐ด ๐ฟ๐ฟ ⁄(๐๐๐๐3 ) B) ๐๐max = 4๐ก๐ก๐ด๐ด ๐ฟ๐ฟ⁄(๐๐๐๐3 ) C) ๐๐max = 8๐ก๐ก๐ด๐ด ๐ฟ๐ฟ⁄(๐๐๐๐3 ) D) ๐๐max = 16๐ก๐ก๐ด๐ด ๐ฟ๐ฟ⁄(๐๐๐๐3 ) Problem 9B Determine the angle of twist ๐๐ between the ends of the bar. A) ๐๐ = 16๐ก๐ก๐ด๐ด ๐ฟ๐ฟ2 ⁄(3๐๐๐๐๐๐4 ) B) ๐๐ = 16๐ก๐ก๐ด๐ด ๐ฟ๐ฟ2 ⁄(๐๐๐๐๐๐4 ) C) ๐๐ = 32๐ก๐ก๐ด๐ด ๐ฟ๐ฟ2 ⁄(3๐๐๐๐๐๐4 ) D) ๐๐ = 32๐ก๐ก๐ด๐ด ๐ฟ๐ฟ2 ⁄(๐๐๐๐๐๐4 ) Problem 10 (Hibbeler, 2014, w/ permission) The shafts have elliptical and circular cross-sections and are to be made from the same amount of a similar material. Determine the percent increase in the maximum shear stress compared to the circular shaft when both shafts are subjected to the same torque and have the same length. A) % increase in maximum shear stress = 23.3% B) % increase in maximum shear stress = 41.4% C) % increase in maximum shear stress = 62.1% D) % increase in maximum shear stress = 80.2% Problem 11 (Hibbeler, 2014, w/ permission) It is intended to manufacture a circular bar to resist torque. However, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor of k, as shown. Determine the factor by which the maximum shear stress is increased. A) ๐๐ = 1⁄2๐๐ B) ๐๐ = 1⁄๐๐ C) ๐๐ = 1⁄2๐๐ 2 D) ๐๐ = 1⁄๐๐ 2 © 2019 Montogue Quiz 5 X SOLUTIONS P.1 c Solution The torque diagram for the shaft is provided below. The total angle of twist is the sum of the angle of twist associated with each shaft segment; that is, φ =∑ i Ti Li Gi J i Performing the summation and using the proper signs, we obtain φ= = ∴φ TBC LBC TCD LCD TDA LDA + + GJ GJ GJ 1 × ( −80 × 0.8 − 60 × 0.6 − 90 × 0.2 ) ๏ฃซπ 9 4๏ฃถ ( 75 ×10 ) × ๏ฃฌ๏ฃญ 2 × 0.01 ๏ฃท๏ฃธ ∴φ =−0.1 rad 5.73o ∴φ = The angle of twist at free end B is about 5.7 degrees. J The correct answer is C. P.2 c Solution The structure in question is illustrated below. The polar moments of inertia of segments AB, BC, and CD are, respectively, π J AB = × 0.0184 = 1.65 × 10−7 m 4 2 π J BC =× 0.034 = 1.27 × 10−6 m 4 2 π J CD = × ( 0.034 − 0.024 ) = 1.02 × 10−6 m 4 2 Let TBC be the torque developed in shaft BC. Taking moments about the xaxis, we have ΣM =0 → 800 + 1600 − TBC =0 ∴ T= 2400 N ⋅ m BC © 2019 Montogue Quiz 6 Since no torque is applied at C, torque TCD at the hollow segment of the brass rod is such that T= T= 2400 N ⋅ m CD BC The angle of twist at A is given by the sum of the angles of rotation in segments AB, BC, and CD, as follows, φ= = ∴φ TAB LAB TBC LBC TCD LCD + + J AB GAB J BC GBC J CDGCD 800 × 0.4 2400 × 0.375 + −7 9 (1.65 ×10 ) × ( 27 ×10 ) (1.27 ×10−6 ) × ( 39 ×109 ) + 2400 × 0.25 0.105 rad = (1.02 ×10−6 ) × ( 39 ×109 ) φ = 6.02o The end A of the structure will twist by about 6 degrees. J The correct answer is C. P.3 c Solution Part A: The angle of twist for a cylindrical member under pure torsion is φ= TL GJ Manipulating the expression above gives 2L ๏ฃซ Td 2 ๏ฃถ 2 L =τ max × ๏ฃท× J๏ณ๏ฃธ Gd 2 Gd 2 ๏ฃญ๏ฑ2๏ฒ φ =๏ฃฌ τ max where ๐๐max is the shear stress at the outermost region of the cross-section, located a distance ๐๐2 /2 from the center. In the present case, ๐๐max = 48 MPa; substituting this and other pertaining variables yields φ= τ max × 2L 2 × 2750 48 × 0.0907 rad =5.20o = = Gd 2 28, 000 ×104 Under a shear stress of 48 MPa, the bar will rotate by 5.2 degrees. J The correct answer is B. Part B: If we were to replace the hollow shaft with a solid shaft to resist the same torque and with the same maximum stress, such a stress would be determined as Tr τ max = = J d 2 16T = π πd3 32d 4 T× In the hollow section, the same stress is calculated as τ max = 16Td 22 π ( d 24 − d14 ) Equating the two previous equations and solving for d, it follows that d 24 − d14 ) ( 16 T d 2 16 T 3 →d = = d2 π d 3 π ( d 24 − d14 ) 1 ๏ฃซ d 24 − d14 ๏ฃถ 3 ∴d = ๏ฃฌ ๏ฃท ๏ฃญ d2 ๏ฃธ © 2019 Montogue Quiz 7 Substituting ๐๐1 = 82 mm and ๐๐2 = 104 mm, we find that 1 ๏ฃซ 1044 − 824 ๏ฃถ 3 = d ๏ฃฌ= 88.4 mm ๏ฃท ๏ฃญ 104 ๏ฃธ The diameter of the solid shaft is just above 8.8 centimeters. J The correct answer is C. Part C: Recall that the weight of each shaft is proportional to its crosssectional area; consequently, a comparison of weights is obtained by dividing the cross-sectional area of one section by that of the other. For the hollow section, π AH =× (1042 − 822 ) = 3213.9 mm 2 4 while for the solid section, π AS =× 88.42 = 6137.5 mm 2 4 so that the ratio of weights, ๐๐, is determined as AH 3213.9 = = 0.524 AS 6137.5 = ρ That is to say, the hollow structure necessitates only about half as much material as the solid structure to withstand the same torque with the same maximum stress. J The correct answer is A. P.4 c Solution Part A: The polar moments of inertia of the steel core, ๐ฝ๐ฝ1 , and the aluminum jacket, ๐ฝ๐ฝ2 , are respectively π 8.35 × 10−7 m 4 J1 = × 0.027 4 = 2 π 1.80 × 10−6 m 4 J2 = × ( 0.0364 − 0.027 4 ) = 2 Let ๐๐1 be the torque exerted on the steel core and ๐๐2 be the torque imparted on the aluminum jacket. We can propose the relation T1 + T2 = T (I) where T = 4 kNโm is the torque applied at end A. Since this is the only equation we can obtain from statics, a compatibility equation is needed. Because the core and the jacket are concentric and perfectly fitted, the angle of twist is the same for both members, i.e., φ1 = φ2 Using the torque-twist relationship, we have T1 L T L T T2 = 2 → 1 = (II) J1Gst J 2GAl J1Gst J 2GAl where L cancels out because the core and sleeve have the same length. Substituting the pertaining variables, we obtain T1 T2 T1 T2 = → = −7 J1Gst J 2GAl (8.35 ×10 ) × 77 (1.80 ×10−6 ) × 27 1.32T2 ∴ T1 = Backsubstituting into equation (I) gives © 2019 Montogue Quiz 8 T1 + T2 =T → (1.32T2 ) + T2 = 4000 ∴ T= 2 4000 = 1724.1 N ⋅ m 2.32 and, consequently, T= 4000 − T2= 4000 − 1724.1= 2275.9 N ⋅ m 1 Finally, the maximum shear stress in the steel core, ๐๐st , is determined as = τ st T1r1 2275.9 × 0.027 = = 73.6 MPa J1 (8.35 ×10−7 ) If need be, we can also determine the stress in the aluminum sleeve using ๐๐2 above. J The correct answer is D. Part B: The angle of twist in question can be determined via either of the following equations, = φ1 T1L1 T2 L2 = ; φ2 Gst J1 GAl J 2 since both expressions should yield the same result, in accordance with the compatibility equation. Let us make use of the equation to the left. Substituting the pertaining variables brings to φ1 = T1L1 = Gst J1 2275.9 × 2.5 = 0.0885 rad ( 77 ×109 ) × (8.35 ×10−7 ) 5.07 o ∴φ = J The correct answer is B. P.5 c Solution Part A: Consider first the polar moments of inertia of the stainless steel tube, ๐ฝ๐ฝ1 , and the solid brass tube, ๐ฝ๐ฝ3 , π J1 = × ( 3.54 − 3.264 ) = 3.64 in.4 32 π J 3 = × 2.04 =1.57 in.4 32 Consider a free-body diagram cut around section B, as shown. From equilibrium of moments, we can write ΣM = 0 → −T1 + T2 + TB = 0 ∴ T1 = T2 + TB (I) Similarly, an equilibrium of moments around section C yields ΣM = 0 → −T2 + T3 = 0 ∴ T3 = T2 (II) © 2019 Montogue Quiz 9 Since these are all the equations we can glean from statics, one equation of compatibility is in order. Noting that the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero; that is, φ1 + φ2 + φ3 = 0 Using the torsion formula, we have T1L1 T2 L2 T3 L3 0 + + = J1G1 J 2G2 J 3G3 Using equations (I) and (II), we can replace ๐๐1 and ๐๐3 , (T2 + TB ) L1 + T2 L2 J1G1 ∴ J 2G2 + T2 L3 = 0 J 3G3 T2 L1 TB L1 T2 L2 T2 L3 + + + = 0 J1G1 J1G1 J 2G2 J 3G3 ๏ฃซ L L ๏ฃถ L T L ∴ T2 ๏ฃฌ 1 + 2 + 3 ๏ฃท = − B 1 J1G1 ๏ฃญ J1G1 J 2G2 J 3G3 ๏ฃธ TB L1 J1G1 ∴ T2 = − ๏ฃซ L1 L ๏ฃถ L + 2 + 3 ๏ฃท ๏ฃฌ ๏ฃญ J1G1 J 2G2 J 3G3 ๏ฃธ giving However, ๐ฝ๐ฝ1 = ๐ฝ๐ฝ2 and ๐บ๐บ1 = ๐บ๐บ2 , which simplifies the equation a bit further, TB L1 ( 6 × 12 ) × 42 J1G1 3.64 × 12,500 T2 = − = − = −15.4 kip-in 1 24 L3 1 × ( 42 + 30 ) + ( L1 + L2 ) + 3.64 × 12,500 1.57 × 5600 J1G1 J 3G3 We can then return to equations (I) and (II) and determine the remaining torques, T1 = T2 + TB → T1 = −15.4 + ( 6 × 12 ) = 56.6 kip-in. and T3 = T2 = −15.4 kip-in. The maximum shear stress magnitudes in the three segments are = τ1 T1r1 56.6 × 3.5 2 = = 27.2 ksi 3.64 J1 = τ2 T2 r2 15.4 × 3.5 2 = = 7.40 ksi 3.64 J2 = τ3 T3r3 15.4 × 2.0 2 = = 9.81 ksi J3 1.57 We were asked to determine the maximum shear stress magnitude in the stainless steel tube; accordingly, such a stress is ๐๐1 = ๐๐max = 27.2 ksi. J The correct answer is D. © 2019 Montogue Quiz 10 Part B: The angle of rotation in member (1) can be defined by the difference in the rotation at ends A and B. Mathematically, φ1 =φB − φ A → φB =φ1 + φ A (III) Similarly, the angle of rotation of segment (2) is given by φ2 = φC − φB → φC = φ2 + φB Substituting from equation (III), we obtain φC = φ2 + φB → φC = φ2 + (φ1 + φ A ) Since joint A is restrained from rotating, the equation simplifies to φC = φ2 + φ1 + φ A → φC =φ1 + φ2 (IV) Now, the angle of twist associated with member (1) is 56.6 × 42 T1L1 = = 0.0522 rad J1G1 3.64 × 12,500 φ1 = while that of segment (2) is 15.4 × 30 3.64 × 12,500 TL J 2G2 2 2 = − = −0.0102 rad φ2 = Substituting these results into the equation for ๐๐๐ถ๐ถ , the desired angle of twist is calculated as φC = 0.0522 − 0.0102 = 0.0420 rad ∴φ = 2.41o J The correct answer is A. P.6 c Solution Part A: Since the solid shaft and the hollow shaft are of same weight and length, their cross-sectional areas must also be the same. This means that As = Ah → π × c 2 = π × ( c22 − c12 ) where ๐๐ is the radius of the solid shaft, ๐๐1 is the inner radius of the hollow shaft, and ๐๐2 is the outer radius of the hollow shaft. We can adjust this expression as ๏ฃฎ ๏ฃซ c ๏ฃถ2 ๏ฃน π × c × ๏ฃฏ1 − ๏ฃฌ 1 ๏ฃท ๏ฃบ π × c =π × ( c − c ) → π × c = ๏ฃฏ๏ฃฐ ๏ฃญ c2 ๏ฃธ ๏ฃบ๏ฃป ๏ฃฎ ๏ฃซ c ๏ฃถ2 ๏ฃน 2 2 ∴ c = c2 × ๏ฃฏ1 − ๏ฃฌ 1 ๏ฃท ๏ฃบ ๏ฃฏ๏ฃฐ ๏ฃญ c2 ๏ฃธ ๏ฃบ๏ฃป 2 2 2 2 2 1 2 2 However, ๐๐1 /๐๐2 = ๐๐, so that c 2 = c22 × (1 − n 2 ) (I) Now, the maximum shear stress in the solid shaft is τs = Ts c (II) Js where Js is the polar moment of inertia, which is given by Js = π 2 c4 © 2019 Montogue Quiz 11 From equation (I), we get π π J s = c 4 → J s = × c24 (1 − n 2 ) 2 2 2 Proceeding similarly with the hollow shaft, we have τh = Th c2 (III) Jh where π J h = × ( c24 − c14 ) 2 ๏ฃฎ ๏ฃซ c ๏ฃถ4 ๏ฃน ∴ J h = × c ๏ฃฏ1 − ๏ฃฌ 1 ๏ฃท ๏ฃบ 2 ๏ฃฏ๏ฃฐ ๏ฃญ c2 ๏ฃธ ๏ฃบ๏ฃป π ∴ Jh = 4 2 π 2 × c24 (1 − n 4 ) The shear stress for both shafts should be the same, so that, equating (II) and (III), we obtain Ts c Th c2 T Jc = → s= s 2 Js Jh Th J h c Substituting the expressions obtained for the polar moments, we get Ts J s c2 T = → s= Th J h c Th π 2 c24 (1 − n 2 ) c2 π 2 2 c24 (1 − n 4 ) c From equation (I), ๐๐ = ๐๐2 √1 − ๐๐2; that is, Ts = Th π π c24 (1 − n 2 ) c2 c24 (1 − n 2 ) c2 T 2 2 = → s π 4 π T 4 4 h c2 (1 − n ) c c2 (1 − n 4 ) c2 1 − n 2 2 2 2 2 ( ) Noting that (1 − ๐๐ 4 ) = (1 − ๐๐ 2 )(1 + ๐๐2 ) and canceling common terms in the numerator and the denominator, it follows that π c24 (1 − n 2 ) c2 2 1 − n2 ) ( Ts Ts 2 = → = Th π 4 Th (1 − n 2 )(1 + n 2 ) 1 − n 2 4 2 c2 (1 − n ) c2 1 − n 2 2 ) ( ∴ Ts 1 − n2 = 2 Th 1+ n This concludes the proof. Part B: The angle of rotation can be obtained with the torque-twist relationship, φ= Tr J Since the angles of twist of the solid shaft, ๐๐๐ ๐ , and of the hollow shaft, ๐๐โ , must be the same, we can write © 2019 Montogue Quiz 12 φ= φh → s ∴ Ts L T L = h Js G Jh G Ts J s = Th J h Substituting the polar moments from previous results gives π c24 (1 − n 2 ) 2 1 − n2 ) ( Ts J s Ts 2 = = → = Th J h Th π 4 1 − n4 ) 4 ( c2 (1 − n ) 2 2 Noting that (1 − ๐๐4 ) = (1 − ๐๐ 2 )(1 + ๐๐2 ), we conclude that 1 − n2 ) 1 − n2 ) ( ( Ts Ts = = 4 → = Th (1 − n ) Th (1 − n2 ) (1 + n2 ) 2 ∴ 2 Ts 1 − n 2 = Th 1 + n 2 as we intended to show. P.7 c Solution Consider the following elemental strip along the circumference of the rod. The inner circle is the cross-section of segment AB, while the outer circle is the section of BC. The force per unit length on the bar is given by the product of shear stress and thickness, F= τ × t Consider now a free-body diagram for the bar. Summing moments about the center of the bar, we obtain τ t × 2π r × r − T = 0 → τ = T (I) 2π tr 2 The equation above provides the maximum shear stress in the structure. The next figure shows the geometry of an elemental strip taken from the structure. © 2019 Montogue Quiz 13 Here, d๐๐ is the angle of twist of the strip, d๐ฟ๐ฟ is the circumferential displacement, and ๐พ๐พ is the shear strain. From the stress-strain relationship for shear, we can write τ = Gγ → γ = τ G Substituting ๐๐ from equation (I) gives γ= τ G 1 T T × →γ = (II) 2 G 2π tr 2π tr 2G = Since ACC’ is a right triangle, we see that tan dφ = dδ r For small angles, however, we can use the approximation tan ๐๐๐๐ ≈ ๐๐๐๐. Accordingly, dφ ≈ dδ → dδ = rdφ r Because BCC’ is also a right triangle, we have tan γ = dδ dr Using the approximation tan ๐พ๐พ ≈ ๐พ๐พ yields γ= dδ dr Substituting d๐ฟ๐ฟ = ๐๐๐๐๐๐, the equation becomes γ= dδ dφ dr = r → dφ = γ dr dr r At this point, we can replace the shear strain with equation (II), dr T dr = dφ γ= r 2π tG r 3 To obtain the angle of twist ๐๐๐ต๐ต๐ต๐ต of end B with respect to end C, all we have to do is integrate the expression above from r1 to r2; that is, r = r2 r2 dr T T ๏ฃฎ 1 ๏ฃน → φBC= − φBC= 3 ∫ 2π tG r1 r 2π tG ๏ฃฏ๏ฃฐ 2r 2 ๏ฃบ๏ฃป r =r1 φBC ∴= T ๏ฃซ1 1๏ฃถ ๏ฃฌ − ๏ฃท 4π tG ๏ฃญ r12 r22 ๏ฃธ as expected. © 2019 Montogue Quiz 14 P.8 c Solution Part A: We begin by setting up an expression that models the variation of the diameter d with distance x from the end A, dB − d A x L d ( x= ) dA + Using this expression, we can establish the polar moment of inertia as πd4 π d − dA ๏ฃถ ๏ฃซ J ( x) = = × ๏ฃฌ dA + B x๏ฃท 32 32 ๏ฃญ L ๏ฃธ 4 Substituting this relation in the torsion formula, the angle of twist is determined as = φ Tdx L ∫= GJ ( x ) 0 32T πG ∫ dx L 0 dB − d A ๏ฃซ ๏ฃฌ dA + L ๏ฃญ ๏ฃถ x๏ฃท ๏ฃธ 4 The integration above can be carried out with the standard result dx ∫ ( a + bx ) 4 = − 1 3b ( a + bx ) 3 or, in the present case, ∫ dx L 0 dB − d A ๏ฃถ ๏ฃซ x๏ฃท ๏ฃฌ dA + L ๏ฃญ ๏ฃธ 4 = ๏ฃซ 1 1 ๏ฃถ L L L − = − ๏ฃฌ ๏ฃท 3d A3 ( d B − d A ) 3d B3 ( d B − d A ) 3 ( d B − d A ) ๏ฃญ d A3 d B3 ๏ฃธ Returning to the expression for ๐๐ with this result, we conclude that φ = ๏ฃซ 1 32TL 1 ๏ฃถ ๏ฃฌ 3− 3๏ฃท 3π G ( d B − d A ) ๏ฃญ d A d B ๏ฃธ This is the desired equation for the angle of twist of the tapered bar. As pointed out by Gere & Goodno, a convenient form in which to write this result is φ= TL ๏ฃซ α 2 + α + 1 ๏ฃถ ๏ฃฌ ๏ฃท GJ A ๏ฃญ 3α 3 ๏ฃธ where dB π d A4 = α = ; JA dA 32 That is, ๐ผ๐ผ is the ratio of end diameters dB/dA and JA is the polar moment of inertia of the cross-section at A. In the special case ๐ผ๐ผ = 1, the diameters are equal and the equation reduces to = φ TL ๏ฃซ 12 + 1 + 1 ๏ฃถ TL = ๏ฃฌ ๏ฃท GJ A ๏ฃญ 3 × 13 ๏ฃธ GJ A ๏ฑ๏ด๏ฒ๏ด๏ณ =1 That is to say, the torque-twist relationship becomes the torsion formula as applied to a solid shaft of constant cross-section, in accordance with our expectations. For values of ๐ผ๐ผ greater than 1, the angle of rotation decreases because the larger diameter at end B produces an increase in the torsional stiffness (relatively to a prismatic bar). Part B: The angle of twist of the hypothetical tapered bar, ๐๐1 , is given by the relation derived in the previous part, © 2019 Montogue Quiz 15 TL ๏ฃซ α 2 + α + 1 ๏ฃถ φ1 = ๏ฃฌ ๏ฃท GJ A ๏ฃญ 3α 3 ๏ฃธ where ๐ผ๐ผ = dB/dA. For a prismatic bar, the angle of twist, ๐๐2 , is determined as φ2 = TL GJ A According to the problem statement, we shall have ๐๐1 = (1⁄2)๐๐2. Substituting from the two previous equations, we have TL ๏ฃซ α 2 + α + 1 ๏ฃถ 1 TL 1 = φ1 φ2 → ๏ฃฌ ๏ฃท= GJ A ๏ฃญ 3α 3 ๏ฃธ 2 GJ A 2 ∴ 2α 2 + 2α + 2 = 3α 3 ∴ 3α 3 − 2α 2 − 2α − 2 = 0 This is a third-degree polynomial, and as such can be solved by numerical methods. Using a CAS such as Mathematica, we verify that the only real solution is ๐ผ๐ผ = ๐๐๐ต๐ต ⁄๐๐๐ด๐ด ≈ 1.45. That is to say, the larger cross-section of the tapered bar should have 1.45 times the diameter of the smaller cross-section if the bar is to have half the angle of twist of its prismatic counterpart. J The correct answer is B. P.9 c Solution Part A: According to the torsion formula, shear stress is proportional to the applied torque. In the present case, the shear stress reaches a maximum of ๐ก๐ก๐ด๐ด ๐ฟ๐ฟ ⁄2 when x = L, that is, at section A. The maximum shear stress is, accordingly, = τ max 16Tmax 16 × ( t A L 2 ) = πd3 πd3 8t A L ∴ τ max = πd3 J The correct answer is C. Part B: Let T(x) be the torque at a distance x from end B. From the geometry of the previous figure, we see that, in this loading configuration, the torque is T ( x) = t ( x ) x tA x x tA x2 = × = 2 L 2 2L Now, the infinitesimal angle of twist d๐๐ is given by dφ = T ( x ) dx GJ which, substituting T(x) and the polar moment J = ๐๐๐๐ 4⁄32, becomes © 2019 Montogue Quiz 16 L tA d x 2 dx → φ φ = = ∫0 ∫ 0 2 LGJ φ tA x3 × ๏ฃซπd4 ๏ฃถ 3 2 LG × ๏ฃฌ ๏ฃท ๏ฃญ 32 ๏ฃธ x=L x =0 16t A L3 ∴φ = 3π G L d 4 16t L2 ∴φ = A 4 3π Gd Clearly, the angle of twist is proportional to the intensity of the distributed torque and the length of the beam. J The correct answer is A. P.10 c Solution Since the elliptical and circular shafts are made of the same amount of material, their cross-sectional areas must be the same. Mathematically, π × b × 2b =π c 2 → c =2b (Recall that the area of an elliptical section is A = ๐๐๐๐๐๐, where a is the semimajor axis and b is the semi-minor axis). The circular shaft has a maximum shear stress (๐๐max )๐ถ๐ถ such that Tr T × 2b = = J π × 2b 4 2 (τ max )= C ( T 2π b3 ) while the elliptical member has a maximum shear stress (๐๐max )๐ธ๐ธ given by = (τ max )E 2T 2T T = = 2 2 π ab π × 2b × b π b3 Let ๐๐ be the percentage increase of stress in the elliptical section relative to the circular section, which, in mathematical terms, equals ρ= (τ max )E − (τ max )C (τ max )C Substituting the results obtained above, it follows that T T − 3 (τ max )E − (τ max )C π b 2π b3 = = = ρ T (τ max )C 2π b3 T ๏ฃซ 1 ๏ฃถ 1− ๏ฃท 3 ๏ฃฌ πb ๏ฃญ 2๏ฃธ = 0.414 1 T 3 2 πb 41.4% ∴ρ= That is, the elliptical section, albeit being made of the same amount of material, will have a maximum shear stress over 40% greater than an equivalent circular section. This problem illustrates the greater efficiency of the circular shaft in carrying torsional loads relatively to other types of geometry. J The correct answer is B. P.11 c Solution For the circular shaft, the maximum shear stress is given by the wellknown result © 2019 Montogue Quiz 17 d Tr 2 = 16T = (τ max )= C J π ๏ฃซ d ๏ฃถ4 π d 3 ×๏ฃฌ ๏ฃท 2 ๏ฃญ2๏ฃธ T× Knowing that a = d/2 is the semi-major axis and a = kd/2 is the semi-minor axis of the elliptical cross-section, the corresponding shear stress is determined as 2T → (τ max ) E = π ab 2 (τ max )E = 2T ๏ฃซ d ๏ฃถ ๏ฃซ kd ๏ฃถ π ×๏ฃฌ ๏ฃท×๏ฃฌ ๏ฃท ๏ฃญ2๏ฃธ ๏ฃญ 2 ๏ฃธ 2 16T π k 2d 3 = The factor f of increase in maximum shear stress follows as 16T = f (τ max ) E π k 2 d 3 = 16T (τ max )C 1 → f =2 k πd3 Note that, for the particular case in which k = 1, the ellipse restores a circular shape and f = 1, as one would expect. J The correct answer is D. X ANSWER SUMMARY Problem 1 Problem 2 3A 3B 3C 4A 4B 5A 5B 6A 6B Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 8A 8B 9A 9B Problem 8 Problem 9 Problem 10 Problem 11 C C B C A D B D A Open-ended pb. Open-ended pb. Open-ended pb. Open-ended pb. B C A B D X REFERENCES • • • • BEER, F., JOHNSTON, E., DEWOLF, J., and MAZUREK, D. (2012). Mechanics of Materials. 6th edition. New York: McGraw-Hill. GERE, J. and GOODNO, B. (2009). Mechanics of Materials. 7th edition. Stamford: Cengage Learning. HIBBELER, R. (2014). Mechanics of Materials. 9th edition. Upper Saddle River: Pearson. PHILPOT, T. (2013). Mechanics of Materials: An Integrated Learning System. 3rd edition. Hoboken: John Wiley and Sons. © 2019 Montogue Quiz 18 Got any questions related to this quiz? We can help! Send a message to contact@montogue.com and we’ll answer your question as soon as possible. © 2019 Montogue Quiz 19