Lecture8 - LECTURE Third Edition SHAFTS POWER STRESS...

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Unformatted text preview: LECTURE Third Edition SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 8 Chapter 3.7 – 3.8 3.13 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 1 ENES 220 ©Assakkaf Power Transmission Work of a Force – A force does work only when the particle to which the force is applied moves. Work = U = Pd (1) P d 1 Slide No. 2 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission ENES 220 ©Assakkaf Work in Two and Three Dimension U = P· d = (P cos φ) d = Px dx + Py dy P P cos φ φ (2) d Slide No. 3 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission ENES 220 ©Assakkaf Work of a Couple – The work of a couple is defined as the magnitude of the couple C times the angular movement of the body. U1→2 = C∆θ rr dU = C ⋅ dθ (3) 2 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 4 ENES 220 ©Assakkaf Power Transmission by Torsional Shaft The power is defined as the time rate of doing work, that is dU dθ dθ =C⋅ =T dt t dt = Tω (4) ω = angular velocity of the shaft in radians per minute LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 5 ENES 220 ©Assakkaf Power Transmission by Torsional Shaft – But ω = 2π f, where f = frequency. The unit of frequency is 1/s and is called hertz (Hz). – If this is the case, then the power is given by P = 2πfT or T= (5) P 2πf 3 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 6 ENES 220 ©Assakkaf Power Transmission by Torsional Shaft – Units of Power SI US Customary watt (1 N·m/s) hp (33,000 ft·lb/min) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 7 ENES 220 ©Assakkaf Power Transmission by Torsional Shaft – Some useful relations 1 −1 1 s= Hz 60 60 1 hp = 550 ft ⋅ lb/s = 6600 in ⋅ lb/s 1 rpm = rpm = revolution per minute 4 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 8 ENES 220 ©Assakkaf Example 5 What size of shaft should be used for a rotor of 5-hp motor operating at 3600 rpm if the shearing stress is not to exceed 8500 psi in the shaft? 1 hp = 6600 in 5 hp = P P = 5(6600) = 33,000 in ⋅ lb/s LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 9 ENES 220 ©Assakkaf Example 5 (cont’d) 1 Hz f = (3600 rpm ) = 60 Hz = 60/s 60 rpm 33,000 P = = 87.54 lb ⋅ in T= 2πf 2π (60 ) Let J denotes the polar moment of area, and c the maximum radius, therefore, τ= Tc JT ⇒= J cτ (6) 5 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 10 ENES 220 ©Assakkaf Example 5 (cont’d) – Evaluating the term J/c in Eq. 6, yields 14 πc 1 J2 = = π 2c 3 2 c c – Therefore, 1 2 3 J T 87.54 = 0.001030 in 3 πc= = = 2 c τ 8500 ⇒ c = 0.1872 in ⇒ Shaft size (dia) = 2c = 0.375 in LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 11 ENES 220 ©Assakkaf The derivation of the torsion formula, τ max = Tc J (7) assumed a circular shaft with uniform cross-section loaded through rigid end plates. 6 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 12 ENES 220 ©Assakkaf Stress Concentrations in Circular Shafts The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and cross-section discontinuities can cause stress concentrations. Experimental or numerically determined concentration factors are applied as τ max = K Tc J (8) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 13 ENES 220 ©Assakkaf 7 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 14 ENES 220 ©Assakkaf Example 6 The stepped shaft shown is to rotate at 900 rpm as it transmits power from a turbine to a generator. The grade of steel specified in the design has an allowable shearing stress of 8 ksi. (a) For preliminary design shown, determine the maximum power that can be transmitted. (b) If in the final design the radius of the fillet is increased so that r = 15/16 in., what will be the percent change , relative to the preliminary design in the power? LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 15 ENES 220 ©Assakkaf Example 6 (cont’d) 8 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 16 ENES 220 ©Assakkaf Example 6 (cont’d) (a) Preliminary Design: Using Fig. 3.32, and knowing that the following are given: D = 7.50 in., d = 3.75 in., r = 9/16 in. = 0.5625 in. Therefore, D 7.50 = =2 d 3.75 and r 0.5625 = = 0.15 3.75 d • Hence, from Fig. 3.32 K = 1.33 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 17 ENES 220 ©Assakkaf Example 6 (cont’d) K = 1.33 9 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 18 ENES 220 ©Assakkaf Example 6 (cont’d) Using Eq. 40, τ max = K Tc J or T = J τ max cK 4 1 1 3.75 4 J = πc 4 = π = 19.414 in 2 22 J τ max 19.414(8) ∴T = = = 62.3 kip - in (3.75 / 2)1.33 cK From Eq. 37, the power is given by 900 6 P = 2πfT = 2π (62.3) = 5,869.9 in - kip/s = 5.87 × 10 in - lb/s 60 5.87 × 106 Pa = = 889.3 hp 6600 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 19 ENES 220 ©Assakkaf Example 6 (cont’d) – Final Design: Using Fig. 3.32, and knowing that the following are given: D = 7.50 in., d = 3.75 in., r = 15/16 in. = 0.9375 in. Therefore, D 7.50 = =2 d 3.75 and r 0.9375 = = 0.25 3.75 d • Hence, from Fig. 3.32 K = 1.20 10 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 20 ENES 220 ©Assakkaf Example 6 (cont’d) K = 1.20 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular Shafts Slide No. 21 ENES 220 ©Assakkaf Example 6 (cont’d) Using Eq. 40, τ max = K Tc J or T = J τ max cK 4 1 1 3.75 4 J = πc 4 = π = 19.414 in 2 22 19.414(8) J τ max = = 69 kip - in ∴T = (3.75 / 2)1.20 cK From Eq. 37, the power is given by 900 6 P = 2πfT = 2π (69.0 ) = 6,503.1 in - kip/s = 6.50 ×10 in - lb/s 60 6.50 × 106 985 − 889.3 = 985 hp Change in P = 100 = 11% Pb = 6600 889.3 11 Slide No. 22 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members Bar of Rectangular Cross Section The maximum shearing stress and the angle of twist for a uniform bar of rectangular cross section, and subjected to pure torsion T are given by τ max = T τmax a b T L φ= T k1ab 2 (9) TL k 2 ab 3G (10) The coefficients k1 and k2 can be obtained from Table 1. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 23 ENES 220 ©Assakkaf Torsion of Noncircular Members Table 1. Coefficients for Rectangular Bars in Torsion a/b k2 1.0 Beer and Johnston, 2002 k1 0.208 0.1406 1.2 0.219 0.1661 1.5 0.231 0.1958 2.0 0.246 0.229 2.5 0.258 0.249 3.0 0.267 0.263 4.0 0.282 0.281 5.0 0.291 0.291 10.0 0.312 0.312 ∞ 0.333 0.333 12 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 24 ENES 220 ©Assakkaf Torsion of Noncircular Members Thin-Walled Hollow Shafts – It was indicated earlier that the determination of the stresses in noncircular members generally requires the use of advanced mathematical methods. – In the case of thin-walled hollow noncircular shaft (Fig. 16), however, a good approximation of the distribution of stresses can be obtained. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 25 ENES 220 ©Assakkaf Torsion of Noncircular Members Thin-Walled Hollow Shafts Figure 16 Center line or Mean Perimeter t t τ τ Ab 13 Slide No. 26 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members Thin-Walled Hollow Shafts – The shearing stress τ at any given point of the wall may be expressed in terms of the torque T as t τ τ= Ab T 2tAb (11) Ab = area bounded by center line LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 27 ENES 220 ©Assakkaf Torsion of Noncircular Members Thin-Walled Hollow Shafts – The shearing stress τ of Eq. 11 represents the average value of the shearing stress across the wall. – However, for elastic deformations the distribution of the stress across the wall may be assumed uniform, and Eq. 11 will give the actual value of the shearing stress at a given point of the wall. 14 Slide No. 28 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members Thin-Walled Hollow Shafts The angle of twist of a thin-walled shaft of length L and modulus of rigidity G is given by t τ L φ= TL ds 4 Ab2G ∫ t (12) Where the integral is computed along the center line of the wall section. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 29 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 Using τall = 40 MPa, determine the largest torque which may be applied to each of the brass bars and to the brass tube shown. Note that the two solid bars have the same cross-sectional area, and that the square bar and square tube have the same outside dimensions 15 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 30 ENES 220 ©Assakkaf Torsion of Noncircular Members Figure 17 T1 T2 m m m m 64 25 mm (1) t = 6 mm 40 mm 40 40 mm 40 m m T3 (3) (2) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 31 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 (cont’d) 1. Bar with Square Cross Section: For a solid bar of rectangular cross section, the maximum shearing stress is given by Eq. 9: T τ max = k1ab 2 where the coefficient k1 is obtained from Table 1, therefore a = b = 0.040 m a = 1.00 b k1 = 0.208 16 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 32 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 (cont’d) For τmax = τall = 40 MPa, we have τ max = T1 k1ab 2 40 = T1 ⇒ T1 = 532 N ⋅ m 2 0.208(0.04 )(0.04 ) 2. Bar with Rectangular Cross Section: a 0.064 = = 2.56 b 0.025 By interpolation, Table 1 gives : k1 = 0.259 a = 0.064 m b = 0.025 m LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 33 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 (cont’d) τ max = T2 k1ab 2 40 = T2 ⇒ T2 = 414 N ⋅ m 2 0.259(0.064)(0,025) 3. Square Tube: For a tube of thickness t, the shearing stress is given by Eq. 11 as τ= T 2tAb 17 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 34 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 (cont’d) where Ab is the area bounded by the center line of the cross section, therefore, Ab = (0.034 )(0.034) = 1.156 × 10 −3 m 2 40 mm 34 mm t = 6 mm 34 mm 40 mm LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 35 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 7 (cont’d) τ = τall = 40 MPa and t = 0.006 m. Substituting these value into Eq. 11 gives τ= T 2tAb 40 = T3 2(0.006 ) 1.156 × 10 −3 ( ) ∴ T3 = 555 N ⋅ m 18 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 36 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 8 Structural aluminum tubing of 2.5 × 4-in. rectangular cross section was fabricated by extrusion. Determine the shearing stress in each of the four walls of a portion of such tubing when it is subjected to a torque of 24 kip·in., assuming (a) a uniform 0.160in. wall thickness (Figure 18a), (b) that, as a result of defective fabrication, walls AB LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 37 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 8 (cont’d) and AC are 0.120-in thick, and walls BD and CD are 0.200-in thick (Fig. 18b) A 2.5 in. 4 in. A B 0.160 in. 2.5 in. 0.160 in. C D (a) 4 in. B 0.120 in. 0.200 in. C Figure 18 D (b) 19 Slide No. 38 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members Example 8 (cont’d) (a) Tubing of Uniform Wall Thickness: Figure 19 A The area bounded by the center line (Fig. 19) is given by 3.84 in. B 0.160 in. 2.34 in. Ab = (3.84)(2.34 ) = 8.986 in 2 0.160 in. C D LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) Slide No. 39 ENES 220 ©Assakkaf Torsion of Noncircular Members Example 8 (cont’d) Since the thickness of each of the four walls is t = 0.160 in., we find from Eq. 11 that the shearing stress in each wall is τ= T 24 = = 8.35 ksi 2tAb 2(0.160 )(8.986) (b) Tubing with Variable Wall Thickness: Observing that the area Ab bounded by the center line is the same as in Part a, and substituting t = 0.120 in. and t = 0.200 in. into 20 Slide No. 40 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THIN-WALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members Example 8 (cont’d) • Eq. 11, the following values for the shearing stresses are obtained: τ AB = τ AC = 24 = 11.13 ksi 2(0.120 )(8.986 ) τ BD = τ CD = 24 = 6.68 ksi 2(0.200)(8.986 ) and • Note that the stress in a given wall depends only upon its thickness t. 21 ...
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