PDF vdoc.pub_statics-and-strength-ox + G Google M Gmail File Electronics, Cars, Fa... C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf Pandora One - Liste... My TurboTax® - G... 12.4 HOOKE'S LAW FOR SHEAR Experiments show that as for

normal stress and strain, shearing stress is proportional to shearing strain as long as stress does not exceed the proportional limit. Hooke's law shear may be expressed as T = Gy (12.1) where G is the shear modulus or the modulus of rigidity. The shearing modulus, G, is a constant for a given material. It is Type here to search 100 Example 12.1 Solution: From Eq. (12.2), G= In the test on a steel bar, E= 29.57 x 106 psi and p = 0.303. Find the shear modulus, G. 29.57(106) 2(1 + 0.303) PROBLEMS 12.1 Determine the shear modulus for a magnesium alloy that has a modulus E= 6.5 X 106 psi and Poisson's ra- tio v = 0.340. 12.2 Determine the shear modulus for a steel that has a modulus E = 206 x 106 kN/m² (kPa) and Poisson's ratio v = 0.25. 12.3 Determine Poisson's ratio for a copper alloy that has a modulus E= 105 X 106 kN/m² (kPa) and a shear mod- ulus of G=37.2 X 106 kN/m² (kPa). 12.4 Determine Poisson's ratio for a cast iron that has a modulus of E= 12 X 106 psi and a shear modulus G = 4.8 X 10 psi. 12.5 TORSION OF A CIRCULAR SHAFT EasyBib: Free Biblio... GE News elastic modulus E. It can be shown that the three elastic constants- modulus of elasticity, E, modulus of rigidity, G, and Poisson's ratio, v-are not independent of each other for an isotropic material. An isotropic material has the same properties in all directions. The relationship between elastic constants is given by the equation We consider here the torsion of a circular shaft. Let the bot- tom end of the shaft be fixed and a torque Tbe applied at the top end as shown in Fig. 12.3(a). G= = 11.35 x 106 psi E 2(1 + v) CAD Geometry of Deformation Cutting through the shaft, we draw a free-body diagram of the part of the shaft between plane cross sections J-J and K-K in Fig. 12.3(b). Both cross sections are normal to the axis of the member. We assume that these plane cross sec- tions remain plane after the torque is applied. The plane OABD, which was parallel to the axis of the shaft, moves to a new position OEBD as the top cross section rotates through an angle, (phi), after application of the torque T. The line OA remains straight as it rotates through the angle to the new position OE. The magnitude of the shearing strain at a distance r from the center of the shaft is given by the angle FHG = y expressed in radians. The angle y is small and can be expressed closely in radian measure as Y FG FG == FH L (a) The angle FOG = can be expressed in radians as = FG/FO=FG/r or FG = rp. Combining this with Eq. (a), we obtain. y = (12.2) W Answer ro L (12.3) Homebanking () milConnect TAP T 10:54 AM 3/1/2024 + PDF vdoc.pub_statics-and-strength-ox + G Google M Gmail File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf My TurboTax® - G... Electronics, Cars, Fa... Type here to search D Pandora One - Liste... Solution: From Eq. (12.12), we have for shaft AB dAB = A From Eq. (12.12), we have for shaft dac = For shaft AB, 75 hp=75(0.746)=55.95 kW. From Eq. (12.14), 9550P 9550(55.95) N 28.9 TAB = For shaft BC, because torque is proportional to power, PAB= PROBLEMS H 4BC = TL GJ 180 16 x 18.49 x 10³ N-m # x 41 x 106 N/m² dAB = 0.1319 m = 131.9mm BC 16 x 6.16 x 10³ N-m = 0.0007652 m² #x 41 x 105 N/m² dac 0.09147 m = 91.5 mm For shaft AB, J = md4/32= (131.94/32 = 29.7 x 10 mm, and from Eq. (12.8), we have 18.49 x 10(3 x 10³) 83 x 10³(29.7 x 10) (0.225 x 10-3) = 1.29⁰ TL GJ 180 ↓ 6 m FIGURE 12.13 Where appropriate, use the following values of the shear modulus: Gsteel =12x10° psi (83X10 MPa) and Galuminum= 4x10 psi (28 x 10³ MPa). EasyBib: Free Biblio... GE News TBC= 75 (18.49) 6.16 kN-m 12.5 A solid circular shaft has a diameter of 50 mm. If the torque applied is 2.25 kN-m, what is the maximum shearing stress? C 18.49 x 10³ N-m 18.49 kN-m For shaft BC, J = md/32= m(91.5)4/32 = 6.88 x 10 mm4, and from Eq. (12.8), we have 6.16 x 10(6 x 10³) 83 x 10³(6.88 x 10) (0.647 x 10-3) = 3.66° = 0.002297 m² = 0.225 x 10-³ rad = 0.647 x 10-³ rad - - T- Answer CAD Answer 12.6 A hollow aluminum shaft has an outside diameter of 2.50 in. and an inside diameter of 2.25 in. What is the maximum shearing stress if a torque of 21,500 lb-in. is applied? W Answer 12.7 Find the torque required to produce a maximum shearing stress of 16,500 psi in a solid circular steel shaft with a diameter of 0.875 in. Answer Homebanking () milConnect TAP T 10:55 AM 3/1/2024 + PDF vdoc.pub_statics-and-strength-ox G Google M Gmail File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf My TurboTax® - G... water brake consisting of a simple water pump that is partially filled with liquid. Two of the more popular electrical loading devices are the eddy current brake and the dic generator. Both use magnetic fields to vary the resistance encountered by a metal rotor as it spins within the brake. For those applications where use of a dynamometer is impractical, semors such as the miniature wireless transducer shown in (SB.19(b)) are often used. This device slides onto the rotating shaft and monitors torques produced during actual operation of a machine. The sensor receives torque measurements from shaft mounted strain gauges, converts the signals to digital form, and transmits the data to a computer for recording or display. (Courtesy of Land & Sea. Inc., North Salem, New Hampshire, www.land-and-sea.com) Electronics, Cars, Fa... Type here to search + 10 Pandora One - Liste... 12.8 A circular tube has an outside diameter of 150 mm and an inside diameter of 100 mm. If the allowable shearing stress is 55 MPa, what is the allowable torque? 129 Torques are applied to the solid shaft shown. What is the maximum shearing stress in the shaft, and where does it occur? 12.10 The solid shaft shown is made of steel; what is the an- gle of twist in degrees between A and D 278 CHAPTER TWELVE 12.11 A circular aluminum alloy tube has an outside diame- ter of 4 in, and an inside diameter of 3.4 in. What is the angle of twist in degrees if the torque applied is 30,000 lb-in, and the shaft is 4.5 ft long? 12.12 The electric motor exerts a torque of 18,000 lb-in. on the pulley system shown. If the shaft is solid and the allowable shear stress is 10,000 psi, determine the diameter of shafts AB and BC. 12.13 If the solid steel shaft shown has a diameter of 2.5 in.. determine the angle through which (a) pulley B rotates with respect to motor A and (b) pulley C ro- tates with respect to motor A. 6,000 lb-in. 12,000 lb-in. 18,000 lb-in. De C. 15 in. PROB. 12.12 and PROB. 12.13 12.14 The standard 1/2-hp electric motor found on most washing machines and dishwashers operates at 1750 rpm. If such a motor is connected to a water pump by a 1/4-in.-diamter steel shaft. a. Find the torque carried by the shaft. b. Compute the maximum shear stress developed in the shaft. Find the factor of safety of the shaft for an allow- able shear stress of 8500 psi. 12.15 A length of extra-strong 2-in. pipe (see Table A.8 of the Appendix) is used connect the output shaft of an internal combustion engine to the spindle of a large circular sawblade used to cut logs into boards. If the pipe has an allowable shear stress of 15,600 psi and will operate at 800 rpm, what is the maximum horse- power engine that can be used? 12.16 A power source at Cactuates valves at A and B through a long round steel control rod. A torque of 8.5 kN-m is required to actuate each valve. If the allowable shear stress in the control rod is 80 MPa, (a) what diameters are required for each section of the rod? (b) what is the angle of twist in the rod between the power source at C 165 Nm EasyBib: Free Biblio... GE News 550 Nm 440 N-m G+GG 30 mm 40 mm 0.9 m- 0.6 m PROB. 12.9 and PROB. 12.10 De 200 mm + 12.17 The motor at A through a shaft AB and a set of gears drives the shaft CD. The shear stress in shaft AB is 55 MPa. If the shear stress in CD is also 55 MPa, (a) what is the diameter of shaft CD? (b) what is the initial angle of twist between the motor at A and the pulley at D (Hint: From the shaft at B to the shaft at C through the set of gears, the torques transmitted are propor- tional to the pitch diameters (TB/Tc-DB/Dc), and the angles of rotation are inversely proportional to the pitch diameters (pa/4.- Dc/Ds). The torques and angles of rotation are in opposite directions.) A d = 50 mm 20 mm 55 Nm 1.2 m- 1.25 m1.50 m PROB. 12.17 d₂ D=125 mm ↓ CAD 12.18 A solid steel shaft delivers 5 hp at 30 revolutions per second. If the allowable shearing stress is limited to 95 MPa, find the required diameter of the shaft. 12.19 The allowable stress in a steel shaft with an outside di- ameter of 1.5 in. and an inside diameter of 1.1 in. is 12,000 psi. What hp can the shaft deliver at 1750 rpm? 12.20 A steel shaft with a diameter of 40 mm and length of 1.1 m transmits 60 hp from an electric motor to a compressor. If the allowable shear stress is 50 MPa and the allowable angle of twist is 1.5, what is the mini- mum allowable speed of rotation? 12.21 A motor through a set of gears delivers power to a hol- low steel shaft with an outside diameter of 40 mm and an inside diameter of 30 mm. The shaft is rotating at 900 rpm. Twenty-five percent of the power supplied by the motor is transmitted to A, and 75 percent of the power supplied is transmitted to C. If the allowable shear stress in the shaft is 105 MPa, what is the maxi- mum power that the motor can deliver to the shaft? W Homebanking () milConnect TAP T 10:56 AM 3/1/2024 + PDF vdoc.pub_statics-and-strength-ox G Google M Gmail File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf My TurboTax® - G... profile. The rolling process may also be used to produce knurled and splined forms, as well as special grooves, worms, pinions, and annular or helical nails (SB.20(d)). (Courtesy of Reed Rico, Halden, Massachusetts, www.pccspd.com) Electronics, Cars, Fa... Type here to search + 10 Pandora One - Liste... 282 CHAPTER TWELVE Example 12.10 A flange coupling contains ten 1/2-in.-diameter steel bolts whose allowable shear stress is 14,000 psi. Four of these bolts are located on a 6-in. bolt circle, and six are on a 9-in. bolt circle. Based on shear in the bolts, what is the maximum torque that this coupling can carry? Solution: 12.23 a. For a 1/2-in.-diameter bolt, A-0.196 in.2, and the maximum force that can be exerted by any of the bolts is FTXA-14,000 lb/in.² x 0.196 in.² - 2740 lb From Fig. 12.16, we see that the outside row of bolts must carry the highest force, so Faz-2740 lb, and F is obtained from Eq. (12.16) as Then, from Eq. (12.17), PROBLEMS 12.22 A solid shaft whose diameter is 3.5 in. carries a torque of 950 ft-lb. This shaft is connected to a coupling by the rectangular key shown. Find the required length of the key if its allowable shear stress is 11,200 psi. b. Fai - X -3 in. x Fa 12 7/8 in. 1 G 5/16 in PROB. 12.22 A hollow shaft whose outside diameter is d, -3.00 in. has a uniform wall thickness of t-1/8 in. What is the maximum torque that this shaft can carry without exceeding a torsional shear stress of 20,000 psi? b. The shaft from part (a) is attached to a coupling containing three bolts on a bolt circle whose di- ameter is Dac5 in. If the bolt material has an allowable shear stress of 18,500 psi, what is the minimum diameter (to the nearest 1/16th in.) for these bolts? T-4x (3 in. x 1830 lb) + 6 x (4.5 in. x 2740 lb) - 95,900 in.-lb. 12.24 a. Same as Prob. 12.23(a), except that d, -75 mm, ts-3 mm, and the allowable shear stress for the shaft is 138 MPa. Same as Prob. 12.23(b), except that Dac125 mm, and the allowable shear stress for the bolts is 128 MPa. (Find dg to the nearest I mm.) EasyBib: Free Biblio... GE News 2740 lb 4.5 in. 1830 lb 12.25 Five bolts are made of low-carbon steel having a yield point in shear of 19,200 psi. These bolts are 3/8-in. in diameter and are to be used in a flange coupling that transmits 285 hp at 1250 rpm. If the bolts are to have a factor of safety in shear of 1.5, find the distance, r, from the center of this coupling at which the bolts should be placed. Is this a minimum or maximum distance? C. 12.26 A stationary power plant consists of a diesel engine connected to an electric generator. The diesel produces 875 hp at 550 rpm. a If the diesel's solid output shaft is to be made of steel whose allowable shear stress is 10,500 psi, find, to the nearest 1/16 in., the minimum shaft diameter required. b. If a 4-1/4-in.-diameter solid output shaft is used. find the maximum shear stress in the shaft. The 4-1/4-in. shaft is locked to a bolted flange coupling by a steel key that is 3-1/8 in. long and has an allowable shear stress of 19,500 psi. Com- pute the minimum width of this key. d. The coupling contains four 13/16-in.-diameter bolts on a 7-in. bolt circle. Find the average shear stress in these bolts. Answer 12.27 The coupling shown in Fig. 12.14 is used to join two shafts that transmit 26 hp at 150 rpm. The shaft, cou- pling, and key have dimensions and specifications as follows: d. - 45 mm DH-80 mm Dac- 130 mm nno. bolts-5 ty-70mm - 15 mm b-t- 10 mm Bolt diameter 15 mm Find the stresses in the (a) shaft, (b) keys, (c) bolts, and (d) flanges. (e) If the yield stresses for all parts are 360 MPa and 7,- 250 MPa, what are the fac- tors of safety for the elements in parts (a) through (d)? a, CAD W Homebanking () milConnect TAP T 10:56 AM 3/1/2024 + PDF vdoc.pub_statics-and-strength-ox G Google M Gmail File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf My TurboTax® - G... Electronics, Cars, Fa... Type here to search + 10 Pandora One - Liste... PROB. 12.22 A hollow shaft whose outside diameter is d,- 3.00 in. has a uniform wall thickness of t5-1/8 in. What is the maximum torque that this shaft can carry without exceeding a torsional shear stress of 20,000 psi? b. The shaft from part (a) is attached to a coupling containing three bolts on a bolt circle whose di ameter is Dac5 in. If the bolt material has an allowable shear stress of 18,500 psi, what is the minimum diameter (to the nearest 1/16th in.) for these bolts? 12.23 a. 12.24 a. Same as Prob. 12.23(a), except that d, -75 mm, t-3 mm, and the allowable shear stress for the shaft is 138 MPa. b. Same as Prob. 12.23(b), except that Dec 125 mm, and the allowable shear stress for the bolts is 128 MPa. (Find da to the nearest I mm.) 12.28 Two steel shafts are joined by couplings that are forged integrally with the shaft as shown. The diame- ters of the shaft and bolt circle are d, - 75 mm and Dac 115 mm, the flange has a thickness - 10 mm, and there are six bolts. If the allowable shear stress in the shafts, flanges, and bolts is 93 MPa, (a) determine the bolt diameter required to develop the full shear capacity of the shaft. (b) Determine the average shear and bearing stresses in the flanges. (c) If the shaft is rotating at 100 rpm, determine the horsepower the shaft can transmit. 12.29 Two solid steel shafts are joined by couplings that are forged integrally with the shaft as shown. The diame- ter of the shaft and bolt circle are d,-5 in. and Dac 10 in, the flange has a thickness - 5/8in., and there are eight 1-in.-diameter bolts. If the allowable shear stress in the shaft is 8000 psi, (a) determine the full torque capacity of the shafts based on shear stress. (b) What horsepower can the shafts transmit at 100 rpm? (c) What is the shear stress in the bolts? (d) Determine the bearing stress and maximum shear stress in the flanges. 12.30 A flange coupling assembly contains eight 3/4-in. steel bolts, four located on a 7-in. bolt circle and four on a 10-in. bolt circle. If the allowable shearing stress in these bolts is 14,500 psi, find the maximum horse- power that the coupling can transmit at a speed of 175 rpm. 12.31 The design for a new bolted flange coupling calls for six steel bolts, three located on a 6-in. bolt circle and EasyBib: Free Biblio... GE News find the maximum shear stress in the shaft. The 4-1/4-in. shaft is locked to a bolted flange coupling by a steel key that is 3-1/8 in. long and has an allowable shear stress of 19,500 psi. Com- pute the minimum width of this key. d. The coupling contains four 13/16-in.-diameter bolts on a 7-in. bolt circle. Find the average shear stress in these bolts. 12.27 The coupling shown in Fig. 12.14 is used to join two shafts that transmit 26 hp at 150 rpm. The shaft, cou- pling, and key have dimensions and specifications as follows: d. - 45 mm D-80 mm ty-70mm - 15 mm bt 10 mm Dac- 130 mm n-no. bolts-5 Bolt diameter 15 mm Find the stresses in the (a) shaft, (b) keys, (c) bolts, and (d) flanges. (e) If the yield stresses for all parts are - 360 MPa and 7,- 250 MPa, what are the fac- tors of safety for the elements in parts (a) through (d)? Shear Stresses and Strains: Torsion Bolts, diameter-da Dac PROB. 12.28 and PROB. 12.29 CAD TO three on a 4-in. bolt circle. If the coupling is to trans- mit 105 hp at 1200 rpm, and the allowable shear stress in the bolts is 9750 psi, find, to the nearest 1/16 inch, the required bolt diameter. 283 W Homebanking () milConnect TAP T 10:56 AM 3/1/2024 +