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SECTION 1– DESIGN FOR SIMPLE STRESSES TENSION, COMPRESSION, SHEAR DESIGN PROBLEMS 1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of 8000 lb. Let h = 1.5b . If the load is repeated but not reversed, determine the dimensions of the section with the design based on (a) ultimate strength, (b) yield strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005 in., what should be the dimensions of the cross section? Problems 1 – 3. Solution: For AISI C1045 steel, as rolled (Table AT 7) su = 96 ksi s y = 59 ksi E = 30× 106 psi F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd = (a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 96,000 8000 = 6 1.5b 2 5 b = 0.577 in say in . 8 2 SECTION 1– DESIGN FOR SIMPLE STRESSES h = 1.5b = 15 in 16 (b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 59,000 8000 = 3 1.5b 2 9 in . b = 0.521 in say 16 27 h = 1.5b = in 32 (c) Elongation = δ = FL AE where, δ = 0.005 in F = 8000 lb E = 30×106 psi L = 15 in A = 1.5b 2 then, FL δ= AE (8000)(15) 0.005 = (1.5b 2 )(30 ×106 ) 3 b = 0.730 in say in . 4 1 h = 1.5b = 1 in 8 2. The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35 018. Solution: For malleable iron, ASTM A47-52, grade 35 018(Table AT 6) su = 55 ksi s y = 36.5 ksi E = 25×10 6 psi 3 SECTION 1– DESIGN FOR SIMPLE STRESSES F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd = (a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 55,000 8000 = 6 1.5b 2 7 b = 0.763 in say in . 8 5 h = 1.5b = 1 in 16 (b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 36,500 8000 = 3 1.5b 2 11 b = 0.622 in say in . 16 1 h = 1.5b = 1 in 32 (c) Elongation = δ = FL AE where, δ = 0.005 in F = 8000 lb E = 25×10 6 psi L = 15 in A = 1.5b 2 then, 4 SECTION 1– DESIGN FOR SIMPLE STRESSES δ= FL AE 0.005 = (8000)(15) (1.5b )(25 ×10 ) 2 b = 0.8 in say h = 1.5b = 1 3. 6 7 in . 8 5 in 16 The same as 1 except that the material is gray iron, ASTM 30. Solution: For ASTM 30 (Table AT 6) su = 30 ksi , no s y E = 14.5 ×106 psi Note: since there is no s y for brittle materials. Solve only for (a) and (c) F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd = (a) Based on ultimate strength N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1) s F sd = u = N A 30,000 8000 = 7. 5 1.5b 2 3 b = 1.1547 in say 1 in . 16 25 h = 1.5b = 1 in 32 FL (c) Elongation = δ = AE where, δ = 0.005 in F = 8000 lb E = 14.5 ×106 psi 5 SECTION 1– DESIGN FOR SIMPLE STRESSES L = 15 in A = 1.5b 2 then, FL δ= AE 0.005 = (8000)(15) (1.5b )(14.5 ×10 ) 2 6 b = 1.050 in say 1 h = 1.5b = 1 4. 1 in . 16 19 in 32 A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on (a) ultimate strength, (b) yield strength. Solution: From Fig. AF 2 for AISI 3140, OQT 1000 F su = 152.5 ksi s y = 132.5 ksi F = force = π (20)2 (125) = 39,270 lb = 39.27 kips 4 From Table 1.1, page 20 Nu = 8 Ny = 4 (a) Based on ultimate strength N F A= u su π 2 (8)(39.27 ) d = 4 152.5 5 d = 1.62 in say 1 in 8 (b) Based on yield strength NyF A= sy π 4 d2 = (4)(39.27 ) 132.5 6 SECTION 1– DESIGN FOR SIMPLE STRESSES 1 d = 1.23 in say 1 in 4 5. A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if Do = 2 Di . Solution: su = 65 ksi Nu = 8 F = 1500 kips A= π (D 4 2 o ) − Di2 = π (4D 4 2 i ) − Di2 = 3πDi2 4 3πDi2 N u F (8)(1500 ) A= = = 4 su 65 7 Di = 8.85 in say 8 in 8 3 7 Do = 2 Di = 2 8 = 17 in 4 8 6. A short compression member with Do = 2 Di is to support a dead load of 25 tons. The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside diameters on the basis of (a) yield strength, (b) ultimate strength. Solution: From Table AT 7 for 4130, WQT 1100 F su = 127 ksi s y = 114 ksi From Table 1.1 page 20, for dead load N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 Area, A = π (D 4 2 o ) − Di2 = π (4D 4 2 i ) − Di2 = 3πDi2 4 F = 25 tons = 50 kips (a) Based on yield strength 3πDi2 N y F (2 )(50) A= = = 4 sy 114 7 SECTION 1– DESIGN FOR SIMPLE STRESSES 5 in 8 5 1 Do = 2 Di = 2 = 1 in 4 8 (b) Based on ultimate strength 3πDi2 N u F (4 )(50) A= = = 4 su 127 7 Di = 0.82 in say in 8 3 7 Do = 2 Di = 2 = 1 in 4 8 Di = 0.61 in say 7. A round, steel tension member, 55 in. long, is subjected to a maximum load of 7000 lb. (a) What should be its diameter if the total elongation is not to exceed 0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if the load is gradually applied and repeated (not reversed). Solution: (a) δ = FL FL or A = AE δE where, F = 7000 lb L = 55 in δ = 0.030 in E = 30× 10 6 psi A= π 4 d2 = (7000)(55) (0.030)(30 ×106 ) 3 in 4 (b) For gradually applied and repeated (not reversed) load Ny = 3 d = 0.74 in say sy = N yF A = (3)(7000) = 47,534 psi π (0.75)2 4 s y ≈ 48 ksi say C1015 normalized condition ( s y = 48 ksi ) 8. A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin is in double shear under the action of the centrifugal force, determine the diameter 8 SECTION 1– DESIGN FOR SIMPLE STRESSES needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in. from the axis of rotation. Solution: From Table AT 3, for B138-A, ½ hard sus = 48 ksi W F = ω 2r g where W = 0.332 lb g = 32.2 fps 2 2π n 2π (10,000 ) ω= = = 1047 rad sec 60 60 r = 12 in W 0.332 F = ω 2r = (1047)2 (1) = 11,300 lb = 11.3 kips g 32.2 From Table 1.1, page 20 N = 3 ~ 4 , say 4 N F A= u su π (4)(11.3) 2 d 2 = for double shear 48 4 25 d = 0.774 in say in 32 CHECK PROBLEMS 9. The link shown is made of AISIC1020 annealed steel, with b = 3 in and 4 1 h = 1 in . (a) What force will cause breakage? (b) For a design factor of 4 based 2 on the ultimate strength, what is the maximum allowable load? (c) If N = 2.5 based on the yield strength, what is the allowable load? Problem 9. 9 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: For AISI C1020 annealed steel, from Table AT 7 su = 57 ksi s y = 42 ksi (a) F = su A 3 1 A = bh = 1 = 1.125 in 2 4 2 F = (57 )(1.125) = 64 kips s A (b) F = u Nu Nu = 4 3 1 A = bh = 1 = 1.125 in 2 4 2 (57 )(1.125) = 16 kips F= 4 (c) F = sy A Ny N y = 2. 5 3 1 A = bh = 1 = 1.125 in 2 4 2 (42 )(1.125) = 18.9 kips F= 2 10. A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq. in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until the tensile stress is 80 % of the yield strength, as determined by measuring the total elongation. What should be the total elongation? Solution: sL δ= E from Table AT 7 for cold-finished B1113 s y = 72 ksi then, s = 0.80 s y = 0.8(72 ) = 57.6 ksi E = 30 ×106 psi = 30,000 ksi sL (57.6)(5) δ= = = 0.0096 in E 30,000 10 SECTION 1– DESIGN FOR SIMPLE STRESSES 11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center of rotation. The axis of the bolts is normal to the plane in which the centrifugal force acts and the bolt is in double shear. At what speed will the bolt shear in two if it is made of AISI B1113, cold finish? Solution: From Table AT 7, sus = 62 ksi = 62,000 psi 2 1 3 A = 2 (π ) = 0.2209 in 2 4 8 W F = ω 2 r = sus A g 4 ω 2 (14) = (62,000)(0.2209) 32.2 ω = 88.74 rad sec 2π n ω= = 88.74 60 n = 847 rpm 12. How many ¾-in. holes could be punched in one stroke in annealed steel plate of AISI C1040, 3/16-in. thick, by a force of 60 tons? Solution: For AISI C1040, from Figure AF 1 su = 80 ksi sus = 0.75su = 0.75(80) ksi = 60 ksi 3 3 A = π d t = π = 0.4418 in2 4 16 F = 60 tons = 120 kips n = number of holes n= 13. F 120 = = 5 holes Asus (0.4415)(60 ) What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400 lb. and the allowable bearing pressure is 200 psi of the projected area? Solution: pDL = W where p = 200 psi D = 4 in 11 SECTION 1– DESIGN FOR SIMPLE STRESSES W = 6400 lb (200)(4)L = 6400 L = 8 in BENDING STRESSES DESIGN PROBLEMS 14. A lever keyed to a shaft is L = 15 in long and has a rectangular cross section of h = 3t . A 2000-lb load is gradually applied and reversed at the end as shown; the material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a) What should be the dimensions of a section at a = 13 in ? (b) at b = 4 in ? (c) What should be the size where the load is applied? Problem 14. Solution: For AISI C1020, as rolled, Table AT 7 su = 65 ksi s y = 49 ksi Design factors for gradually applied and reversed load Nu = 8 Ny = 4 th 3 , moment of inertial 12 but h = 3t h4 I= 36 I= Moment Diagram (Load Upward) 12 SECTION 1– DESIGN FOR SIMPLE STRESSES Based on ultimate strength s s= u Nu (a) s = Mc Fac = I I h 2 F = 2000 lbs = 2 kips (2)(13) h 65 2 s= = 4 8 h 36 h = 3.86 in h 3.86 t= = = 1.29 in 3 3 say 1 h = 4.5 in = 4 in 2 1 t = 1.5 in = 1 in 2 c= (b) s = Mc Fbc = I I h 2 F = 2000 lbs = 2 kips (2)(4) h 65 2 s= = 4 8 h 36 h = 2.61 in h 2.61 t= = = 0.87 in 3 3 say h = 3 in t = 1 in c= (c) 13 SECTION 1– DESIGN FOR SIMPLE STRESSES 3 − h 4. 5 − 3 = 4 13 − 4 h = 2.33 in 1 − t 1.5 − 1 = 4 13 − 4 t = 0.78 in say 5 h = 2.625 in or h = 2 in 8 15. A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel, SAE 080. The cross section is rectangular (let h ≈ 3b ). (a) Determine the dimensions for N = 3 based on the yield strength. (b) Compute the maximum deflection for these dimensions. (c) What size may be used if the maximum deflection is not to exceed 0.03 in.? Solution: For cast steel, SAE 080 (Table AT 6) s y = 40 ksi E = 30× 106 psi 14 SECTION 1– DESIGN FOR SIMPLE STRESSES From Table AT 2 FL (4)(54 ) Max. M = = = 54 kips − in 4 4 bh 3 I= 12 but h = 3b h4 I= 36 (a) s = c= sy Ny = Mc I h 2 (54) h 40 2 = 3 h4 36 h = 4.18 in h 4.18 b= = = 1.39 in 3 3 h 4. 5 1 1 = 1.5 in = 1 in say h = 4 in , b = = 2 3 3 2 FL3 (b) δ = = 48 EI (c) δ = (4000)(54)3 = 0.0384 in 3 6 (1.5 )(4.5 ) 48(30 ×10 ) 12 FL3 h4 48E 36 3 ( 4000)(54 ) (36 ) 0.03 = 48(30 ×106 )(h 4 ) h = 4.79 in h 4.79 b= = = 1.60 in 3 3 1 h 5.25 3 say h = 5.25 in = 5 in , b = = = 1.75 in = 1 in 4 3 3 4 15 SECTION 1– DESIGN FOR SIMPLE STRESSES 16. The same as 15, except that the beam is to have a circular cross section. Solution: s Mc (a) s = y = Ny I I= πd4 64 d c= 2 d M 32 M 2 s= 4 = πd πd3 64 40 32(54 ) = 3 πd3 d = 3.46 in 1 say d = 3 in 2 (b) δ = I= FL3 48 EI πd4 64 3 64 FL3 64(4000 )(54) δ= = = 0.0594 in 48 E (π d 4 ) 48(30 × 106 )(π )(3.5)4 (c) δ = 64 FL3 48 E (π d 4 ) 64(4000)(54 ) 0.03 = 48(30 ×106 )(π )d 4 d = 4.15 in 1 say d = 4 in 4 3 17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of C1020 structural steel. (a) Basing your calculations on the ultimate strength, determine the dimensions of the rectangular cross section for h = 2b . (b) Determine the dimensions based on yield strength. (c) Determine the dimensions using the principle of “limit design.” 16 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: From Table AT 7 and Table 1.1 su = 65 ksi s y = 48 ksi N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 FL (6)(48) = = 72 in − kips 4 4 Mc s= I h c= 2 bh 3 I= 12 h but b = 2 4 h I= 24 h M 12M 2 s = 4 = 3 h h 24 M= (a) Based on ultimate strength s 12 M s= u = 3 Nu h 65 12(72) = 4 h3 h = 3.76 in 17 SECTION 1– DESIGN FOR SIMPLE STRESSES b= h 3.76 = = 1.88 in 2 2 3 h 3.75 7 say h = 3.75 in = 3 in , b = = = 1.875 in = 1 in 4 2 2 8 (b) Based on yield strength s y 12 M s= = 3 Ny h 48 12(72 ) = 2 h3 h = 3.30 in h 3.30 b= = = 1.65 in 2 2 h 3. 5 1 3 say h = 3.5 in = 3 in , b = = = 1.75 in = 1 in 2 2 2 4 (c) Limit design (Eq. 1.6) bh 2 4 h 2 h 2 72 = (48) 4 h = 2.29 in h 2.29 b= = = 1.145 in 2 2 1 h 2. 5 1 say h = 2.5 in = 2 in , b = = = 1.25 in = 1 in 2 2 2 4 M = sy 18. The bar shown is subjected to two vertical loads, F1 and F2 , of 3000 lb. each, that are L = 10 in apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4 based on the ultimate strength; h = 3b . Determine the dimensions h and b if the bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A4752, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and moment diagrams approximately to scale. 18 SECTION 1– DESIGN FOR SIMPLE STRESSES Problems18, 19. Solution: F1 = F2 = R1 = R2 = 3000 lb Moment Diagram M = R1a = (3000)(3) = 9000 lbs − in = 9 kips − in N = factor of safety = 4 based on su bh 3 12 h c= 2 h 3 h h4 3 I= = 12 36 I= (a) For gray cast iron, SAE 111 su = 30 ksi , Table AT 6 h M s Mc 18M 2 s= u = = 4 = 3 N I h h 36 30 18(9 ) s= = 3 4 h h = 2.78 in h 2.78 b= = = 0.93 in 3 3 say h = 3.5 in , b = 1 in (b) For malleable cast iron, ASTM A47-52, grade 35 018 19 SECTION 1– DESIGN FOR SIMPLE STRESSES su = 55 ksi , Table AT 6 h M s Mc 18M 2 s= u = = 4 = 3 N I h h 36 55 18(9 ) s= = 3 4 h h = 2.28 in h 2.28 b= = = 0.76 in 3 3 1 3 say h = 2 in , b = in 4 4 (c) For AISI C1040, as rolled su = 90 ksi , Fig. AF 1 h M s Mc 18M 2 s= u = = 4 = 3 N I h h 36 90 18(9 ) s= = 3 4 h h = 1.93 in h 1.93 b= = = 0.64 in 3 3 7 5 say h = 1 in , b = in 8 8 19. The same as 18, except that F1 acts up ( F2 acts down). Solution: [∑ M A =0 ] R1 = R2 = 1875 lb 20 SECTION 1– DESIGN FOR SIMPLE STRESSES Shear Diagram Moment Diagram M = maximum moment = 5625 lb-in = 5.625 kips-in (a) For gray cast iron su 18M = 3 N h 30 18(5.625) = 4 h3 h = 2.38 in h 2.38 b= = = 0.79 in 3 3 1 3 say h = 2 in , b = in 4 4 (b) For malleable cast iron s= su 18M = 3 N h 55 18(5.625) = 4 h3 h = 1.95 in h 1.95 b= = = 0.65 in 3 3 7 5 say h = 1 in , b = in 8 8 s= 21 SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For AISI C1040, as rolled su 18M = 3 N h 90 18(5.625) = 4 h3 h = 1.65 in h 1.65 b= = = 0.55 in 3 3 1 1 say h = 1 in , b = in 2 2 s= 20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb. at θ = 0 . Let d = 3 in , L = 10 in and h = 3b . Determine the dimensions of the section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron, ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic reasons, the pins at A, B, and C are to be the same size. What should be their diameter if the material is AISI C1035, as rolled, and the mounting is such that each is in double shear? Use the basic dimensions from (c) as needed. (e) What sectional dimensions would be used for the C1035 steel if the principle of “limit design” governs in (c)? Problems 20, 21. Solution: 22 SECTION 1– DESIGN FOR SIMPLE STRESSES [∑ M [∑ M A B =0 =0 ] 3RB = 13(2500 ) ] RB = 10,833 lb 3RA = 10(2500) RA = 8333 lb Shear Diagram Moment Diagram M = (2500 )(10) = 25,000 lb − in = 25 kips − in h = 3b bh 3 I= 12 h4 I= 36 h c= 2 h M Mc 18M 2 s= = 4 = 3 I h h 36 (a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(25) = 6 h3 23 SECTION 1– DESIGN FOR SIMPLE STRESSES h = 5.13 in h b = = 1.71 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h 52 18(25) = 4 h3 h = 3.26 in h b = = 1.09 in 3 3 1 say h = 3 in , b = 1 in 4 4 (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength s 18M s= u = 3 N h 85 18(25) = 4 h3 h = 2.77 in h b = = 0.92 in 3 say h = 3 in , b = 1 in (d) For AISI C1035, as rolled s su = 64 ksi N = 4 , RB = 10.833 kips s R s s = su = B N A π π A = 2 D 2 = D 2 4 2 64 10.833 = ss = π 2 4 D 2 D = 0.657 in 24 SECTION 1– DESIGN FOR SIMPLE STRESSES 11 in 16 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi say D = b= h 3 h 2 h 3 M = 25 = (55) 4 h = 1.76 in h b = = 0.59 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8 The same as 20, except that θ = 30o . Pin B takes all the horizontal thrust. 21. Solution: FV = F cos θ [∑ M [∑ M A B =0 =0 ] 3RB = 13FV ] 3RB = 13(2500 ) cos 30 RB = 9382 lb 3RA = 10 FV 3RA = 10(2500) cos 30 RA = 7217 lb Shear Diagram 25 SECTION 1– DESIGN FOR SIMPLE STRESSES Moment Diagram M = (2165)(10 ) = 21,650 lb − in = 21.65 kips − in 18M s= 3 h (a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(21.65) = 6 h3 h = 4.89 in h b = = 1.63 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h ( 52 18 21.65) = 4 h3 h = 3.11 in h b = = 1.04 in 3 say h = 3 in , b = 1 in (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength 26 SECTION 1– DESIGN FOR SIMPLE STRESSES su 18M = 3 N h 85 18(21.65) = 4 h3 h = 2.64 in h b = = 0.88 in 3 5 7 say h = 2 in , b = in 8 8 s= (d) For AISI C1035, as rolled s su = 64 ksi N = 4 , RBV = 9382 lb RBH = FH = F sin θ = 2500 sin 30 = 1250 lb 2 2 RB2 = RBV + RBH = (9382) + (1250 ) 2 2 RB = 9465 lb s R s s = su = B N A π π A = 2 D 2 = D 2 4 2 64 9.465 ss = = 4 π D2 2 D = 0.614 in 5 say D = in 8 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi b= h 3 h 2 h 3 M = 21.65 = (55) 4 h = 1.68 in h b = = 0.56 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8 27 SECTION 1– DESIGN FOR SIMPLE STRESSES 22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually applied, repeated loads (in phase), one of 2000 lb. at e = 10 in from the free end, and one of 1000 lb at the free end. (a) Determine the dimensions of the cross section if b = c ≈ 3a . (b) The same as (a) except that the top of the tee is below. Problem 22. Solution: For cast iron, ASTM 50 su = 50 ksi , suc = 164 ksi For gradually applied, repeated load N = 7 ~ 8 , say 8 M = F1d + F2 (d + e ) where: F1 = 2000 lb F2 = 1000 lb d = 30 − 10 = 20 in d + e = 30 in M = (2000 )(20 ) + (1000)(30 ) = 70,000 lb − in = 70 kips − in Mc I Solving for I , moment of inertia s= (3a )(a ) a + (3a )(a ) 5a = [(3a )(a ) + (3a )(a )]y 2 y= 2 3a 2 28 SECTION 1– DESIGN FOR SIMPLE STRESSES I= (3a )(a )3 + (3a )(a )(a 2 ) + (a )(3a )3 + (3a )(a )(a 2 ) = 17a 4 12 12 2 (a) 3a 2 5a cc = 2 Based on tension s Mct st = u = N I (70) 3a 50 2 = 8 17a 4 2 a = 1.255 in Based on compression s Mcc sc = uc = N I (70) 5a 164 2 = 8 17 a 4 2 a = 1.001 in Therefore a = 1.255 in 1 Or say a = 1 in 4 And b = c = 3a = 3(1.25) = 3.75 in ct = 29 SECTION 1– DESIGN FOR SIMPLE STRESSES 3 Or b = c = 3 in 4 (b) If the top of the tee is below 5a 2 3a cc = 2 17a 4 I= 2 M = 70 kips − in ct = Based on tension s Mct st = u = N I (70) 5a 50 2 = 8 17 a 4 2 a = 1.488 in Based on compression s Mcc sc = uc = N I (70) 3a 164 2 = 8 17a 4 2 a = 0.845 in Therefore a = 1.488 in 1 Or say a = 1 in 2 1 And b = c = 3a = 4 in 2 CHECK PROBLEMS 30 SECTION 1– DESIGN FOR SIMPLE STRESSES 23. An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3 in. and is subjected to two loads; F1 and F2 = 2F1 ; F1 is 5 in. from one end and F2 is 5 in. from the other ends. The beam is 25 in. long; flange width is b = 2.509 in ; I x = 2.9 in 4 . Determine (a) the approximate values of the load to cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum deflection caused by the safe loads. Problems 23 – 25. Solution: [∑ M [∑ F V A =0 =0 ] ] 5 F1 + 20(2 F1 ) = 25RB RB = 1.8 F1 F1 + 2 F1 = RA + RB RA = 3F1 − 1.8F1 = 1.2 F1 Shear Diagram Moment Diagram 31 SECTION 1– DESIGN FOR SIMPLE STRESSES M = 9F1 = maximum moment For AISI C1020, as rolled s y = 48 ksi Mc I d 3 where c = = = 1.5 in 2 2 (9 F1 )(1.5) s y = 48 = 2.9 F1 = 10.31 kips F2 = 2 F1 = 20.62 kips (a) s y = sy Mc N I 48 (9 F1 )(1.5) s= = 3 2.9 F1 = 3.44 kips F2 = 2 F1 = 6.88 kips (b) s = (c) = L 25 = = 9.96 < 15 (page 34) b 2.509 sc = 20 ksi ( page 34, i1.24) Mc I (9 F1 )(1.5) 20 = 2.9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc = (d) For maximum deflection, by method of superposition, Table AT 2 3 y max Fb′ a( L + b′ ) 2 = , a > b′ 3EIL 3 or 3 y max Fa b′(L + a ) 2 = , b′ > a 3EIL 3 32 SECTION 1– DESIGN FOR SIMPLE STRESSES y max caused by F1 3 F a b′(L + a1 ) 2 y max1 = 1 1 1 , b1′ > a1 3EIL 3 where E = 30,000 ksi a1 = 5 in b1′ = 20 in L = 25 in I = 2.9 in 4 3 y max1 F1 (5) 20(25 + 5) 2 = = 0.0022 F1 3(30,000 )(2.9 )(25) 3 y max caused by F2 3 F b′ a (L + b2′ ) 2 y max 2 = 2 2 2 , a2 > b2′ 3EIL 3 where b2′ = 5 in a2 = 20 in 3 y max 2 2 F1 (5) 20(25 + 5) 2 = = 0.0043F1 3(30,000 )(2.9 )(25) 3 Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.31) = 0.067 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.44) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30 ) = 0.028 in 24. The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated. Solution: For aluminum alloy, 2024-T4, heat treated s y = 47 ksi (a) s y = Mc I 33 SECTION 1– DESIGN FOR SIMPLE STRESSES s y = 47 = (9 F1 )(1.5) 2. 9 F1 = 10.10 kips F2 = 2 F1 = 20.20 kips sy Mc N I 47 (9 F1 )(1.5) s= = 3 2.9 F1 = 3.36 kips F2 = 2 F1 = 6.72 kips (b) s = (c) = L 25 = = 9.96 < 15 (page 34) b 2.509 sc = 20 ksi ( page 34, i1.24) Mc I ( 9 F1 )(1.5) 20 = 2.9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc = (d) Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.10 ) = 0.066 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.36) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30 ) = 0.028 in 25. A light I-beam is 80 in. long, simply supported, and carries a static load at the midpoint. The cross section has a depth of d = 4 in , a flange width of b = 2.66 in , and I x = 6.0 in 4 (see figure). (a) What load will the beam support if it is made of C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or not the safe load should be less. 34 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: (a) For C1020, as rolled, su = 65 ksi Consider flange buckling L 80 = = 30 b 2.66 L since 15 < < 40 b 22.5 22.5 sc = = = 15 ksi 2 2 ( 30 ) L 1 + 1800 1 + 1800 b Mc s= I d 4 c = = = 2 in 2 2 From Table AT 2 FL F (80) M= = = 20 F 4 4 Mc s = sc = I ( 20 F )(2) 15 = 6 F = 2.25 kips , safe load (b) Considering stress owing to the weight of the beam wL2 (Table AT 2) 8 where w = 7.7 lb ft add’l M = 35 SECTION 1– DESIGN FOR SIMPLE STRESSES wL2 7.7 (80 ) = = 513 lb − in = 0.513 kips − in 8 12 8 M = 20 F + 0.513 = total moment Mc s = sc = I (20 F + 0.513)(2 ) 15 = 6 F = 2.224 kips Therefore, the safe load should be less. 2 add’l M = 26. What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley? The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension and forces of sawing.) Solution: t = 0.0265 = 0.01325 in 2 r = 13.75 + 0.01325 = 13.76325 in Using Eq. (1.4) page 11 (Text) Ec s= r where E = 30× 106 psi c= (30 ×10 )(0.01325) = 28,881 psi s= 6 13.76325 27. A cantilever beam of rectangular cross section is tapered so that the depth varies uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and the length 30 in. What safe load, acting repeated with minor shock, may be applied to the free end? The material is AISI C1020, as rolled. Solution: For AISI C1020, as rolled su = 65 ksi (Table AT 7) Designing based on ultimate strength, N = 6 , for repeated, minor shock load 36 SECTION 1– DESIGN FOR SIMPLE STRESSES su 65 = = 10.8 ksi N 6 Loading Diagram s= 4 −1 h − 1 = 30 x h = 0.10 x + 1 wh 3 I= 12 h c= 2 M = Fx (Fx ) h Mc 3Fx 2 = 6 Fx = 3Fx = = 2 2 3 I 2h h wh (0.10 x + 1)2 12 Differentiating with respect to x then equate to zero to solve for x giving maximum stress. (0.10 x + 1)2 (1) − 2( x )(0.10 x + 1)(0.10 ) ds = 3F =0 dx (0.10 x + 1)4 0.10 x + 1 − 2(0.10 x ) = 0 x = 10 in h = 0.10(10 ) + 1 = 2 in s 3Fx s= u = 2 N h 3F (10 ) 10.8 = (2)2 F = 1.44 kips s= TORSIONAL STRESSES DESIGN PROBLEMS 37 SECTION 1– DESIGN FOR SIMPLE STRESSES 28. A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What should be the diameter of the pump shaft if it is made of AISI C1045 as rolled? Consider the load as gradually repeated. Solution: For C1045 as rolled, s y = 59 ksi sus = 72 ksi Designing based on ultimate strength s s = us , N = 6 (Table 1.1) N 72 s= = 12 ksi 6 33,000hp 33,000(15) Torque, T = = = 45 ft − lb = 540 in − lb = 0.540 in − kips 2π n 2π (1750) For diameter, 16T s= π d3 16(0.540) 12 = πd3 d = 0.612 in 5 say d = in 8 29. A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid shaft? (b) If the shaft is hollow, Do = 2 Di , what size is required? (c) What is the weight per foot of length of each of these shafts? Which is the lighter? By what percentage? (d) Which shaft is the more rigid? Compute the torsional deflection of each for a length of 10 ft. Solution: 33,000hp 33,000(2500 ) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 1137, annealed s y = 50 ksi (Table AT 8) s ys = 0.6s y = 30 ksi Designing based on yield strength N = 3 for medium shock, one direction 38 SECTION 1– DESIGN FOR SIMPLE STRESSES Design stress s 30 = 10 ksi s = ys = N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276) 10 = π D3 D = 5.20 in 1 say D = 5 in 4 s= (b) J = [ π (Do4 − Di4 ) π (2 Di )4 − Di4 = 32 Do 2 Di c= = = Di 2 2 TDi 32T s= = 4 15π Di 15π Di3 32 32(276 ) 10 = 15π Di3 Di = 2.66 in 32 ] = 15π D 4 i 32 Do = 2 Di = 5.32 in say 5 Di = 2 in 8 1 Do = 5 in 4 (c) Density, ρ = 0.284 lb in 3 (Table AT 7) 39 SECTION 1– DESIGN FOR SIMPLE STRESSES For solid shaft w = weight per foot of length π 2 w = 12 ρ D 2 = 3πρD 2 = 3π (0.284)(5.25) = 73.8 lb ft 4 For hollow shaft π 2 2 w = 12 ρ Do2 − Di2 = 3πρ Do2 − Di2 = 3π (0.284)(5.25) − (2.625) = 55.3 lb ft 4 Therefore hollow shaft is lighter 73.8 − 55.3 Percentage lightness = (100% ) = 33.5% 55.3 ( ) ( [ ) ] (d) Torsional Deflection TL JG where L = 10 ft = 120 in θ= G = 11.5 ×103 ksi For solid shaft, J = θ= π D4 32 (276)(120 ) 180 o = 0.039 rad = (0.039) = 2.2 π π 4 3 (5.25) (11.5 × 10 ) 32 For hollow shaft, J = θ= π (Do4 − Di4 ) 32 (276 )(120) 180 o = 0.041 rad = (0.041) = 2.4 π 4 4 3 π [(5.25) − (2.625) ](11.5 × 10 ) 32 Therefore, solid shaft is more rigid, 2.2o < 2.4o 30. The same as 29, except that the material is AISI 4340, OQT 1200 F. Solution: 33,000hp 33,000(2500 ) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 4340, OQT 1200 F s y = 130 ksi s ys = 0.6s y = 0.6(130 ) = 78 ksi Designing based on yield strength 40 SECTION 1– DESIGN FOR SIMPLE STRESSES N = 3 for mild shock Design stress s 78 = 26 ksi s = ys = N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276) 26 = π D3 D = 3.78 in 3 say D = 3 in 4 s= (b) J = [ π (Do4 − Di4 ) π (2 Di )4 − Di4 = 32 Do 2 Di c= = = Di 2 2 TDi 32T s= = 4 15π Di 15π Di3 32 32(276) 26 = 15π Di3 Di = 1.93 in 32 ] = 15π D 4 i 32 Do = 2 Di = 3.86 in say Di = 2 in Do = 4 in (c) Density, ρ = 0.284 lb in 3 (Table AT 7) 41 SECTION 1– DESIGN FOR SIMPLE STRESSES For solid shaft w = weight per foot of length π 2 w = 12 ρ D 2 = 3πρD 2 = 3π (0.284)(3.75) = 37.6 lb ft 4 For hollow shaft π 2 2 w = 12 ρ Do2 − Di2 = 3πρ Do2 − Di2 = 3π (0.284)(4) − (2) = 32.1 lb ft 4 Therefore hollow shaft is lighter 37.6 − 32.1 Percentage lightness = (100% ) = 17.1% 32.1 ( ) ( [ ) ] (d) Torsional Deflection TL JG where L = 10 ft = 120 in θ= G = 11.5 ×103 ksi For solid shaft, J = π D4 32 (276)(120) 180 o θ= = 0.148 rad = (0.148) = 8.48 π 4 3 π (3.75) (11.5 × 10 ) 32 For hollow shaft, J = θ= π (Do4 − Di4 ) 32 (276)(120 ) 180 o = 0.122 rad = (0.122 ) = 6.99 π π 4 4 3 [(4) − (2 ) ](11.5 × 10 ) 32 Therefore, hollow shaft is more rigid, 6.99o < 8.48o . 31. A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should be its diameter if the deflection is not to exceed 1o in 20 D ? (b) If deflection is primary what kind of steel would be satisfactory? Solution: 33,000hp 33,000(40 ) (a) T = = = 420 ft − lb = 5.04 in − kips 2π n 2π (500 ) G = 11.5 ×103 ksi L = 20 D 42 SECTION 1– DESIGN FOR SIMPLE STRESSES θ = 1o = θ= π π 180 TL JG rad (5.04)(20 D ) = π D4 11.5 ×103 32 D = 1.72 in 3 say D = 1 in 4 180 ( (b) s = ) 16T 16(5.04 ) = = 4.8 ksi π D 3 π (1.75)3 Based on yield strength N =3 s ys = Ns = (3)(4.8) = 14.4 ksi s ys 14.4 = 24 ksi 0.6 0.6 Use C1117 normalized steel s y = 35 ksi sy = 32. = A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 inlb. For medium shock, what should be its size? Solution: For AISI 1118 cold-finish s y = 75 ksi s ys = 0.6s y = 45 ksi N = 3 for medium shock s T s = ys = N Z′ where, h = b 2b 2 h 2b 3 Z′ = = (Table AT 1) 9 9 T = 1200 in − lb = 1.2 in − kips 45 1.2(9) s= = 3 2b 3 b = h = 0.71 in 3 say b = h = in 4 43 SECTION 1– DESIGN FOR SIMPLE STRESSES CHECK PROBLEMS 33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a ½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending neglected). (b) What will be the corresponding design factor if the shaft is made of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that is characteristics of this machine, do you thick the design is safe enough? Solution: For AISI C1020, as rolled sus = 49 ksi F = sus (π Dt ) 15 where D = in 16 1 t = in 2 15 1 F = 49(π ) = 72.2 kips 16 2 T = Fr 3 where r = in 4 3 T = (72.2 ) = 54.2 in − kips 4 16T π d3 where d = 3.5 in 16(54.2) s= = 6.44 ksi 3 π (3.5) (a) s = (b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = 34. sus 64 = = 9.94 , the design is safe ( N ≈ 10 ) s 6.44 The same as 33, except that the shaft diameter is 2 ¾ in. Solution: 44 SECTION 1– DESIGN FOR SIMPLE STRESSES d = 2.75 in 16T π d3 16(54.2 ) s= = 13.3 ksi 3 π (2.75) (a) s = (b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = sus 64 = = 4.8 , the design is not safe ( N < 10 ) s 13.3 A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the torsional stress in the shaft (stress from bending and propeller thrust are not considered). (b) Compute the factor of safety. Does it look risky? 35. Solution: For Monel shaft, s us = 98 ksi (Table AT 3) N = 3 ~ 4 , for dead load, based on ultimate strength Tc J π Do4 − Di4 π (13.5)4 − (6.5)4 J= = = 3086 in 4 32 32 Do 13.5 c= = = 6.75 in 2 2 33,000hp 33,000(10,000) T= = = 262,606 ft − lb = 3152 in − kips 2π n 2π (200) (3152)(6.75) = 6.9 ksi s= 3086 (b) Factor of safety, (a) s = ( N= ) [ ] sus 98 = = 14.2 , not risky s 6. 9 45 SECTION 1– DESIGN FOR SIMPLE STRESSES STRESS ANALYSIS DESIGN PROBLEMS 36. A hook is attached to a plate as shown and supports a static load of 12,000 lb. The material is to be AISI C1020, as rolled. (a) Set up strength equations for dimensions d , D , h , and t . Assume that the bending in the plate is negligible. (b) Determine the minimum permissible value of these dimensions. In estimating the strength of the nut, let D1 = 1.2d . (c) Choose standard fractional dimensions which you think would be satisfactory. Problems 36 – 38. Solution: s = axial stress s s = shear stress (a) s= F 4F = 2 1 πd2 πd 4 Equation (1) d = 4F πs 46 SECTION 1– DESIGN FOR SIMPLE STRESSES s= F 1 π D 2 − D12 4 ( ) Equation (2) D = ss = 4F 4F 4F = = 2 2 2 2 2 π D − D1 π D − 1.44d 2 π D − (1.2d ) ( ) [ ] ( 4F + 1.44d 2 πs F F = π D1h 1.2π dh Equation (3) h = ss = = F 1.2π ds s F π Dt Equation (4) t = F π Dss (b) Designing based on ultimate strength, Table AT 7, AISI C1020, as rolled su = 65 ksi sus = 49 ksi N = 3 ~ 4 say 4, design factor for static load s 65 = 16 ksi s= u = N 4 s 49 s s = us = = 12 ksi N 4 F = 12,000 lb = 12 kips From Equation (1) 4F 4(12 ) d= = = 0.98 in πs π (16 ) From Equation (2) 4F 4(12 ) 2 D= + 1.44d 2 = + 1.44(0.98) = 1.53 in πs π (16) From Equation (3) F 12 h= = = 0.27 in 1.2π ds s 1.2π (0.98)(12 ) From Equation (4) F 12 t= = = 0.21 in π Dss π (1.53)(12 ) 47 ) SECTION 1– DESIGN FOR SIMPLE STRESSES (c) Standard fractional dimensions d = 1 in 1 D = 1 in 2 1 h = in 4 1 t = in 4 37. The same as 36, except that a shock load of 4000 lb. is repeatedly applied. Solution: (a) Same as 36. (b) N = 10 ~ 15 for shock load, based on ultimate strength say N = 15 , others the same. s 65 s= u = = 4 ksi N 15 s 49 s s = us = = 3 ksi N 15 F = 4000 lb = 4 kips From Equation (1) 4F 4(4 ) d= = = 1.13 in πs π (4) From Equation (2) 4F 4(4) 2 D= + 1.44d 2 = + 1.44(1.13) = 1.76 in πs π (4 ) From Equation (3) F 4 h= = = 0.31 in 1.2π ds s 1.2π (1.13)(3) From Equation (4) F 4 t= = = 0.24 in π Dss π (1.76)(3) 48 SECTION 1– DESIGN FOR SIMPLE STRESSES (c) Standard fractional dimensions 1 d = 1 in 8 3 D = 1 in 4 3 h = in 8 1 t = in 4 38. The connection between the plate and hook, as shown, is to support a load F . Determine the value of dimensions D , h , and t in terms of d if the connection is to be as strong as the rod of diameter d . Assume that D1 = 1.2d , sus = 0.75su , and that bending in the plate is negligible. Solution: s= F 1 πd2 4 1 F = π d 2s 4 1 s (1) F = π d 2 u 4 N 49 SECTION 1– DESIGN FOR SIMPLE STRESSES s= F F = 1 1 π D 2 − D12 π D 2 − 1.44d 2 4 4 1 F = π (D 2 − 1.44d 2 )s 4 1 s (2) F = π D 2 − 1.44d 2 u 4 N F F ss = = π D1h 1.2π dh F = 1.2π dhss ( ) ( ( ) ) s 0.75su F = 1.2π dh us = 1.2π dh N N 5s (3) F = 0.9π dh u N F ss = π Dt F = π Dtss s 0.75su F = π Dt us = π Dt N N s (4) F = 0.75π Dt u N Equate (2) and (1) 1 s 1 s F = π D 2 − 1.44d 2 u = π d 2 u 4 N 4 N D 2 = 2.44d 2 D = 1.562d Equate (3) and (1) s 1 s F = 0.9π dh u = π d 2 u N 4 N d h= = 0.278d 4(0.9) Equate (4) and (1) s 1 s F = 0.75π Dt u = π d 2 u N 4 N s 1 s F = 0.75π (1.562d )(t ) u = π d 2 u N 4 N d t= = 0.214d 4(0.75)(1.562 ) ( ) 50 SECTION 1– DESIGN FOR SIMPLE STRESSES 39. (a) For the connection shown, set up strength equations representing the various methods by which it might fail. Neglect bending effects. (b) Design this connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as rolled. The load is repeated and reversed with mild shock. Make the connection equally strong on the basis of yield strengths in tension, shear, and compression. Problems 39, 40 Solution: (a) s s = F 1 5 π D 2 4 4F 5π s s Equation (1) D = s= F t (b − 2 D ) Equation (2) b = s= F + 2D ts F 5 Dt Equation (3) t = F 5Ds (b) For AISI C1020, as rolled s y = 48 ksi (Table AT 7) s ys = 0.6s y = 28 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 48 s= = 12 ksi 4 28 ss = = 7 ksi 4 From Equation (1) 51 SECTION 1– DESIGN FOR SIMPLE STRESSES D= 4F 5π s s where F = 2500 lb = 2.5 kips D= 4F 4(2.5) 5 in = = 0.30 in say 5π s s 5π (7 ) 16 From Equation (3) F 2.5 5 t= = = 0.13 in say in 5Ds 32 5 5 (12 ) 16 From Equation (2) F 2. 5 5 b = + 2D = + 2 = 1.96 in say 2 in ts 5 16 (12) 32 40. The same as 39, except that the material is 2024-T4, aluminum alloy. Solution: (a) Same as 39. (b) ) For 2024-T4, aluminum alloy s y = 47 ksi (Table AT 3) s ys = 0.55s y = 25 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 47 = 12 ksi s= 4 25 ss = = 6 ksi 4 From Equation (1) 4F D= 5π s s where F = 2500 lb = 2.5 kips D= 4F 4(2.5) 3 = = 0.33 in say in 5π s s 5π (6) 8 From Equation (3) 52 SECTION 1– DESIGN FOR SIMPLE STRESSES t= F 2.5 1 = = 0.11 in say in 8 5Ds 3 5 (12) 8 From Equation (2) F 2. 5 1 3 b = + 2D = + 2 = 2.42 in say 2 in ts 2 1 8 (12) 8 41. (a) For the connection shown, set up strength equations representing the various methods by which it might fail. (b) Design this connection for a load of 8000 lb. Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the plates. Let the load be repeatedly applied with minor shock in one direction and make the connection equally strong on the basis of ultimate strengths in tension, shear, and compression. Problem 41. Solution: (a) F sP = t (b − D ) 3 F sP = 4 t (b − 2 D ) or s sR = F 1 4 πD 2 (2 ) 4 53 Equation (1) Equation (2) SECTION 1– DESIGN FOR SIMPLE STRESSES sR = F 4 Dt Equation (3) (b) For AISI C1015, as rolled suR = 61 ksi , susR = 0.75suR = 45 ksi For AISI C1020, as rolled suP = 65 ksi N = 6 , based on ultimate strength s 65 s P = uP = = 10.8 ksi N 6 s 61 s R = uR = = 10.1 ksi N 6 s 45 = 7.5 ksi s sR = usR = N 6 F = 8000 lb = 8 kips Solving for D F s sR = 2π D 2 7 F 8 in = = 0.412 in say 16 2π s sR 2π (7.5) Solving for t F sR = 4 Dt F 8 1 t= = = 0.453 in say in 4 Ds R 2 7 4 (10.1) 16 Solving for b F Using s P = t (b − D ) F 8 7 b= +D= + = 1.92 in say 2 in ts P 16 1 (10.8) 2 3 F 4 Using s P = t (b − 2 D ) D= 54 SECTION 1– DESIGN FOR SIMPLE STRESSES 3F 3(8) 7 + 2D = + 2 = 1.99 in say 2 in 4ts P 16 1 4 (10.8) 2 Therefore b = 2 in 7 D = in 16 1 t = in 2 b= 42. Give the strength equations for the connection shown, including that for the shear of the plate by the cotter. Problems 42 – 44. Solution: Axial Stresses s= F 1 π D12 4 s= 55 = 4F π D12 F (L − D2 )e Equation (1) Equation (2) SECTION 1– DESIGN FOR SIMPLE STRESSES s= s= s= F D2 e Equation (3) F 1 π a 2 − D22 4 ( F 1 π D22 − D2e 4 ) = = 4F Equation (4) π a 2 − D22 ( 4F Equation (5) π D − 4 D2e 2 2 Shear Stresses ss = F 2eb ss = F 2(L − D2 + e )t 56 ) Equation (6) Equation (7) SECTION 1– DESIGN FOR SIMPLE STRESSES 43. ss = F π at Equation (8) ss = F π D1m Equation (9) ss = F 2 D2 h Equation (10) A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate by means of a cotter that is made of as-rolled C1020, in the manner shown. (a) Determine all dimensions of this joint if it is to withstand a reversed shock load F = 10 kips , basing the design on yield strengths. (b) If all fits are free-running fits, decide upon tolerances and allowances. Solution: (See figure of Prob. 42) 7 t = in = 0.875 in , s sy = 0.6s y 8 For steel rod, AISI C1035, as rolled s y1 = 55 ksi s sy1 = 33 ksi For plate and cotter, AISI C1020, as rolled s y2 = 48 ksi s sy2 = 28 ksi N = 5 ~ 7 based on yield strength say N = 7 From Equation (1) (Prob. 42) sy 4F s= 1 = N π D12 55 4(10) = 7 π D12 D1 = 1.27 in 1 say D1 = 1 in 4 57 SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (9) ssy F ss = 1 = N π D1m 33 10 = 7 1 π 1 m 4 m = 0.54 in 9 say m = in 16 From Equation (3) sy F s= 1 = N D2e 55 10 s= = 7 D2e D2 e = 1.273 From Equation (5) sy 4F s= 1 = 2 N π D2 − 4 D2e 55 4(10 ) = 2 7 π D2 − 4(1.273) D2 = 1.80 in 3 say D2 = 1 in 4 and D2 e = 1.273 3 1 e = 1.273 4 e = 0.73 in 3 say e = in 4 By further adjustment 5 Say D2 = 2 in , e = in 8 From Equation (8) s sy F ss = 2 = N π at 28 10 = 7 π a (0.875) a = 0.91 in say a = 1 in 58 SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (4) sy 4F s= 2 = N π a 2 − D22 48 4(10 ) = 7 π a 2 − 22 a = 2.42 in 1 say a = 2 in 2 1 use a = 2 in 2 From Equation (7) ssy F ss = 2 = N 2( L − D2 + e )t 28 10 = 5 7 2 L − 2 + (0.875) 8 L = 2.80 in say L = 3 in From Equation (6) s sy F ss = 2 = N 2eb 28 10 = 7 5 2 b 8 b = 2 in From Equation (10) s sy F ss = 2 = N 2 D2 h 28 10 = 7 2(2 )h 5 h = 0.625 in = in 8 Summary of Dimensions L = 3 in 5 h = in 8 b = 2 in 7 t = in 8 ( ( ) ) 59 SECTION 1– DESIGN FOR SIMPLE STRESSES 9 in 16 1 a = 2 in 2 1 D1 = 1 in 4 D2 = 2 in 5 e = in 8 m= (b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3 ± 0.010 in h = 0.625 ± 0.010 in t = 0.875 ± 0.010 in m = 0.5625 ± 0.010 in a = 2.500 ± 0.010 in D1 = 1.25 ± 0.010 in For Free Running Fits (RC 7) Table 3.1 Female Male + 0.0030 − 0.0040 b = 2.0 in b = 2.0 in − 0.0000 − 0.0058 allowance = 0.0040 in + 0.0030 − 0.0040 D2 = 2.0 in D2 = 2.0 in − 0.0000 − 0.0058 allowance = 0.0040 in + 0.0016 − 0.0020 e = 0.625 in e = 0.625 in − 0.0000 − 0.0030 allowance = 0.0020 in 44. A 1-in. ( D1 ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a) Determine all the dimensions for this connection so that all parts have the same ultimate strength as the rod. The load F reverses direction. (b) Decide upon tolerances and allowances for loose-running fits. Solution: (Refer to Prob. 42) (a) For AISI C1035, as rolled su1 = 85 ksi sus1 = 64 ksi For AISI C1020, as rolled 60 SECTION 1– DESIGN FOR SIMPLE STRESSES su2 = 65 ksi sus2 = 48 ksi Ultimate strength Use Equation (1) 1 1 2 Fu = su1 π D12 = (85) π (1) = 66.8 kips 4 4 Equation (9) Fu = sus1 π D1m 66.8 = (64 )(π )(1)m m = 0.33 in 3 say m = in 8 From Equation (3) Fu = su1 D2e 66.8 = (85)D2e D2 e = 0.7859 From Equation (5) 1 Fu = su1 π D22 − D2e 4 1 66.8 = (85) π D22 − 0.7859 4 D2 = 1.42 in 3 say D2 = 1 in 8 3 D2 e = 1 e = 0.7859 8 e = 0.57 in 9 say e = in 16 From Equation (4) 1 Fu = su2 π a 2 − D22 4 2 1 3 66.8 = (65) π a 2 −1 4 8 a = 1.79 in 3 say a = 1 in 4 From Equation (8) ( ) 61 SECTION 1– DESIGN FOR SIMPLE STRESSES Fu = sus2 π at 66.8 = (48)(π )(a )(1) a = 0.44 in 1 say a = in 2 3 use a = 1 in 4 From Equation (2) Fu = su2 ( L − D2 )e 3 9 66.8 = (65) L − 1 8 16 L = 3.20 in 1 say L = 3 in 4 From Equation (7) Fu = 2 sus2 (L − D2 − e )t 3 9 66.8 = 2(48) L − 1 − (1) 8 16 L = 1.51 in 1 say L = 1 in 2 1 use L = 3 in 4 From Equation (6) Fu = 2 sus1 eb 9 66.8 = 2(64 ) b 16 b = 0.93 in say b = 1 in From Equation (10) Fu = 2 sus1 D2 h 3 66.8 = 2(64 ) 1 h 8 h = 0.38 in 3 say h = in 8 Dimensions 1 L = 3 in 4 62 SECTION 1– DESIGN FOR SIMPLE STRESSES 3 in 8 b = 1 in t = 1 in 3 m = in 8 3 a = 1 in 4 D1 = 1 in 3 D2 = 1 in 8 9 e = in 16 h= (b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3.25 ± 0.010 in h = 0.375 ± 0.010 in t = 1.000 ± 0.010 in m = 0.375 ± 0.010 in a = 1.75 ± 0.010 in D1 = 1.000 ± 0.010 in For Loose Running Fits (RC 8) Table 3.1 Female Male + 0.0035 − 0.0045 b = 1.0 in b = 1.0 in − 0.0000 − 0.0065 allowance = 0.0045 in + 0.0040 − 0.0050 D2 = 1.375 in D2 = 1.375 in − 0.0000 − 0.0075 allowance = 0.0050 in + 0.0028 − 0.0035 e = 0.5625 in e = 0.5625 in − 0.0000 − 0.0051 allowance = 0.0035 in 45. Give all the simple strength equations for the connection shown. (b) Determine the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the connection will be equally strong in tension, shear, and compression. Base the calculations on ultimate strengths and assume sus = 0.75su . 63 SECTION 1– DESIGN FOR SIMPLE STRESSES Problems 45 – 47. Solution: (a) Neglecting bending 1 Equation (1): F = s π D 2 4 1 Equation (2): F = ss 2 π c 2 4 Equation (3): F = s (2bc ) Equation (4): F = s (ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s (d − c )a su s and s s = us N N Therefore s s = 0.75s Equate (2) and (1) 1 1 F = ss 2 π c 2 = s π D 2 4 4 (b) s = 1 1 0.75s c 2 = s D 2 2 4 c = 0.8165 D Equate (3) and (1) 1 F = s (2bc ) = s π D 2 4 1 2b(0.8165D ) = π D 2 4 b = 0.4810 D 64 SECTION 1– DESIGN FOR SIMPLE STRESSES Equate (4) and (1) 1 F = sac = s π D 2 4 1 a(0.8165 D ) = π D 2 4 a = 0.9619 D Equate (5) and (1) 1 F = s[2(d − c )b] = s π D 2 4 1 2(d − 0.8165 D )(0.4810 ) = π D 2 4 d = 1.6329 D Equate (6) and (1) 1 F = s s (4mb ) = s π D 2 4 1 0.75(4m )(0.4810 D ) = π D 2 4 m = 0.5443D Equate (7) and (1) 1 F = ss (2nb ) = s π D 2 4 1 0.75(2n )(0.4810 D ) = π D 2 4 n = 1.0886 D Equate (8) and (1) 1 F = s (d − c )a = s π D 2 4 (1.6329 − D − 0.8165D )a = 1 π D 2 4 a = 0.9620 D Summary a = 0.9620 D b = 0.4810 D c = 0.8165 D d = 1.6329 D m = 0.5443D n = 1.0886 D 65 SECTION 1– DESIGN FOR SIMPLE STRESSES 46. The same as 45, except that the calculations are to be based on yield strengths. Let s sy = 0.6s y . Solution: (Refer to Prob. 45) (a) Neglecting bending 1 Equation (1): F = s π D 2 4 1 Equation (2): F = ss 2 π c 2 4 Equation (3): F = s (2bc ) Equation (4): F = s (ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s (d − c )a (b) s = sy and s s = s sy N N Therefore s s = 0.6 s Equate (2) and (1) 1 1 F = ss 2 π c 2 = s π D 2 4 4 1 1 0.6 s c 2 = s D 2 2 4 c = 0.9129 D Equate (3) and (1) 1 F = s (2bc ) = s π D 2 4 1 2b(0.9129 D ) = π D 2 4 b = 0.4302 D Equate (4) and (1) 1 F = sac = s π D 2 4 1 a(0.9129 D ) = π D 2 4 a = 0.8603D 66 SECTION 1– DESIGN FOR SIMPLE STRESSES Equate (5) and (1) 1 F = s[2(d − c )b] = s π D 2 4 1 2(d − 0.9129 D )(0.4302) = π D 2 4 d = 1.8257 D Equate (6) and (1) 1 F = s s (4mb ) = s π D 2 4 1 0.6(4m )(0.4302 D ) = π D 2 4 m = 0.7607 D Equate (7) and (1) 1 F = ss (2nb ) = s π D 2 4 1 0.6(2n )(0.4302 D ) = π D 2 4 n = 1.5214 D Equate (8) and (1) 1 F = s (d − c )a = s π D 2 4 (1.8257 − D − 0.9129 D )a = 1 π D 2 4 a = 0.8604 D Summary a = 0.8604 D b = 0.4302 D c = 0.9129 D d = 1.8257 D m = 0.7607 D n = 1.5214 D 47. Design a connection similar to the one shown for a gradually applied and reversed load of 12 kips. Base design stresses on yield strengths and let the material be AISI C1040 steel, annealed. Examine the computed dimensions for proportion, making changes that you deem advisable. Solution: (See figure in Prob. 45 and refer to Prob. 46) 67 SECTION 1– DESIGN FOR SIMPLE STRESSES N = 4 based on yield strength for gradually applied and reversed load. For AISI C1040, annealed s y = 47 ksi (Fig. AF 7) s sy = 0.6s y = 28 ksi sy 47 = 11.75 ksi N 4 1 F = s π D 2 4 1 12 = 11.75 π D 2 4 D = 1.14 in 1 say D = 1 in 8 1 a = 0.8604 D = 0.86041 = 0.97 in 8 but a > D 1 say a = 1 in 4 1 b = 0.43021 = 0.48 in 8 1 say b = in 2 1 c = 0.91291 = 1.030 in 8 say c = 1 in 1 d = 1.82571 = 2.05 in 8 say d = 2 in 1 m = 0.76071 = 0.86 in 8 7 say m = in 8 1 n = 1.52141 = 1.71 in 8 3 say n = 1 in 4 s= = Dimension: 1 a = 1 in 4 68 SECTION 1– DESIGN FOR SIMPLE STRESSES 1 in 2 c = 1 in d = 2 in 7 m = in 8 3 n = 1 in 4 1 D = 1 in 8 b= 48. Give all the strength equations for the union of rods shown. Problems 48 – 68. Solution: 1 F = s π d 2 4 Equation (1) F = s s (π ad ) Equation (2) F = ss (2tc ) Equation (3) 69 SECTION 1– DESIGN FOR SIMPLE STRESSES F = ss [2(D − e )b] Equation (4) Equation (5) F = set F = s (D − e )t Equation (6) 1 F = s π k 2 − e2 4 ( ) Equation (7) 1 F = s π m 2 − e 2 − (m − e )t 4 ( ) F = s s (2ef ) Equation (9) 70 Equation (8) SECTION 1– DESIGN FOR SIMPLE STRESSES 1 F = s π e 2 − et 4 Equation (10) 49-68. Design a union-of-rods joint similar to that shown for a reversing load and material given in the accompanying table. The taper of cotter is to be ½ in. in 12 in. (see 172). (a) Using design stresses based on yield strengths determine all dimensions to satisfy the necessary strength equations. (b) Modify dimensions as necessary for good proportions, being careful not to weaken the joint. (c) Decide upon tolerances and allowances for loose fits. (d) Sketch to scale each part of the joint showing all dimensions needed for manufacture, with tolerances and allowances. Prob. No. Load, lb. AISI No., As Rolled 49 50 51 52 3000 3500 4000 4500 1020 1030 1117 1020 52 54 55 56 5000 5500 6000 6500 1015 1035 1040 1020 57 58 59 60 7000 7500 8000 8500 1015 1118 1022 1035 61 62 63 64 9000 9500 10,000 10,500 1040 1117 1035 1022 65 66 67 68 11,000 11,500 12,000 12,500 1137 1035 1045 1030 71 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: (For Prob. 49 only) (a) For AISI 1020, as rolled s y = 48 ksi s ys = 0.6s y = 0.6(48) = 28.8 ksi For reversing load, N = 4 based on yield strength s 48 = 12 ksi s= y = N 4 s 28.8 = 7.2 ksi s s = ys = N 4 F = 3000 lb = 3 kips Equation (1) 1 F = s π d 2 4 1 3 = 12 π d 2 4 d = 0.5642 in 9 say d = in 16 Equation (2) F = s s (π ad ) 9 3 = 7.2(π a ) 16 a = 0.236 in 1 say a = in 4 Equation (5) F = set 3 = 12et et = 0.25 Equation (10) 1 F = s π e 2 − et 4 1 3 = 12 π e 2 − 0.25 4 e = 0.798 in 13 say e = in 16 et = 0.25 72 SECTION 1– DESIGN FOR SIMPLE STRESSES 13 t = 0.25 16 t = 0..308 in 5 say t = in 16 Equation (6) F = s (D − e )t 13 5 3 = 12 D − 16 16 D = 1.6125 in 5 say D = 1 in 8 Equation (4) F = ss [2(D − e )b] 5 13 3 = 7.2 21 − b 8 16 b = 0.256 in 1 say b = in 4 Equation (7) 1 F = s π k 2 − e2 4 2 1 13 3 = 12 π k 2 − 4 16 k = 0.989 in say k = 1 in Equation (9) F = s s (2ef ) ( ) 13 3 = 7.2(2) f 16 f = 0.256 in 1 say f = in 4 Equation (8) 1 F = s π m 2 − e 2 − (m − e )t 4 ( ) 2 1 13 5 13 3 = 12 π m 2 − − m − 16 16 16 4 73 SECTION 1– DESIGN FOR SIMPLE STRESSES 0.25 = 0.7854m 2 − 0.5185 − 0.3125m + 0.2539 0.7854m 2 − 0.3125m − 0.5146 = 0 m 2 − 0.3979m − 0.6552 = 0 m = 1.032 in say m = 1 in Equation (3) F = ss (2tc ) 5 3 = 7.2(2) c 16 c = 0.667 in 11 say c = in 16 DIMENSIONS: 9 d = in 16 1 a = in 4 1 b = in 4 11 c = in 16 1 f = in 4 13 e = in 16 5 t = in 16 k = 1 in 5 D = 1 in 8 m = 1 in (b) Modified dimensions 9 in 16 1 a = in 4 3 b = in 4 11 c = in 16 d= 74 SECTION 1– DESIGN FOR SIMPLE STRESSES 1 in 2 13 e = in 16 5 t = in 16 k = 1 in 5 D = 1 in 8 1 m = 1 in 4 f = (c) Tolerances and allowances No fit, ± 0.010 in d = 0.5625 ± 0.010 in a = 0.250 ± 0.010 in f = 0.500 ± 0.010 in D = 1.625 ± 0.010 in k = 1.000 ± 0.010 in m = 1.250 ± 0.010 in Fits, Table 3.1, loose-running fits, say RC 8 Female + 0.0035 in − 0.0000 allowance = 0.0045 in + 0.0028 c = 0.6875 in − 0.0000 allowance = 0.0035 in + 0.0035 e = 0.8125 in − 0.0000 allowance = 0.0045 + 0.0022 t = 0.3125 in − 0.0000 allowance = 0.0030 in b = 0.750 Male b = 0.750 − 0.0045 in − 0.0065 c = 0.6875 − 0.0035 in − 0.0051 e = 0.8125 − 0.0045 in − 0.0065 t = 0.3125 75 − 0.0030 in − 0.0040 SECTION 1– DESIGN FOR SIMPLE STRESSES (d) ROD COTTER 76 SECTION 1– DESIGN FOR SIMPLE STRESSES SOCKET CHECK PROBLEMS 69. 1 1 The connection shown has the following dimensions: d = 1 in , D = 2 in , 4 2 1 5 1 D1 = 1 in , h = in , t = in ; it supports a load of 15 kips. Compute the tensile, 2 8 2 compressive, and shear stresses induced in the connection. What is the corresponding design factor based on the yield strength if the rod and nut are made of AISI C1045, as rolled, and the plate is structural steel (1020)? 77 SECTION 1– DESIGN FOR SIMPLE STRESSES Problem 69. Solution: Tensile Stresses F 15 = = 12.22 ksi (1) s1 = 2 1 2 1 1 πd π 1 4 4 4 F 15 (2) s 2 = = = 8.4 ksi 2 1 2 1 1 π D1 π 1 4 4 2 Compressive Stress F 15 (3) s3 = = = 4.78 ksi 2 2 1 2 2 1 1 1 π (D − D1 ) π 2 − 1 4 4 2 2 Shear Stresses F 15 (4) s s4 = = = 3.82 ksi π Dt 1 1 π 2 2 2 F 15 (5) s s5 = = = 5.09 ksi π D1h 1 5 π 1 2 8 For AISI C1045, as rolled (rod and nut) s y1 = 59 ksi s ys1 = 0.6 s y = 0.6(59 ) = 35.4 ksi For structural steel plate (1020) s y2 = 48 ksi s ys1 = 0.6 s y = 0.6(48) = 28.8 ksi Solving for design factor 78 SECTION 1– DESIGN FOR SIMPLE STRESSES (1) N1 = (2) N 2 = (3) N 3 = (4) N 4 = (5) N 5 = s y1 s1 s y1 s2 s y2 s3 s ys2 s s4 s ys1 s s5 = 59 = 4.83 12.22 = 59 = 6.95 8.49 = 48 = 10.04 4.78 = 28.8 = 7.54 3.82 = 35.4 = 6.96 5.09 The corresponding design factor is N = 4.83 70. 3 7 3 in , t = in , b = 3 in , and let the load, which is applied 4 16 4 centrally so that it tends to pull the plates apart, be 15 kips. (a) Compute the stresses in the various parts of the connection. (b) If the material is AISI C1020, as rolled, what is the design factor of the connection based on yield strengths? In the figure, let D = Problem 70. Solution: (a) Tensile stresses F 15 s1 = = = 11.43 ksi t (b − D ) 7 3 3 3 − 16 4 4 3 3 (15) F 4 4 s2 = = = 11.43 ksi t (b − 2 D ) 7 3 3 3 − 2 16 4 4 Compressive bearing stress F 15 s3 = = = 11.43 ksi 4 Dt 3 7 4 4 16 79 SECTION 1– DESIGN FOR SIMPLE STRESSES Shearing stress F 15 ss 4 = = = 4.24 ksi 2 1 2 3 4 π D (2) π (2 ) 4 4 (b) For AISI C1020, as rolled s y = 48 ksi s ys = 0.6s y = 28.8 ksi N= sy s or N = Using N = s ys ss sy s s 48 = 4.2 N= y = s 11.43 s Using N = ys ss s ys 28.8 N= = = 6.8 s s 4.24 Therefore the design factor is N = 4.2 71. For the connection shown, let a = 15 9 3 1 in , b = in , c = in , d = 1 in , 16 16 4 2 3 15 in , m = n = in . The material is AISI C1040, annealed (see Fig. AF 1). 4 16 (a) For a load of 7500 lb., compute the various tensile, compressive, and shear stresses. Determine the factor of safety based on (b) ultimate strength, (c) yield strengths. D= Problem 71. Solution: (a) Tensile stresses 80 SECTION 1– DESIGN FOR SIMPLE STRESSES s1 = F 1 π D2 4 = 7.5 1 3 π 4 4 2 = 16.98 ksi F 7. 5 = = 8.89 ksi 2b(d − c ) 9 1 3 2 1 − 16 2 4 F 7. 5 s3 = = = 10.67 ksi a(d − c ) 15 1 3 1 − 16 2 4 Compressive Stresses (Bearing) F 7.5 s4 = = = 8.89 ksi 2bc 9 3 2 16 4 F 7. 5 s5 = = = 10.67 ksi ac 15 3 16 4 Shearing Stresses F 7.5 s s6 = = = 3.56 ksi 4mb 15 9 4 16 16 F 7.5 s s7 = = = 7.11 ksi 2nb 15 9 2 16 16 For AISI C1040, annealed,Fig. AF 1 s y = 47 ksi s2 = su = 79 ksi s ys = 0.6s y = 28 ksi sus = 0.6su = 47.4 ksi (b) Based on ultimate strength s 79 N= u = = 4.65 s1 16.98 (c) Based on yield strength sy 47 N= = = 2.77 s1 16.98 72. The upper head of a 60,000-lb. tensile-testing machine is supported by two steel rods, one of which A is shown. These rods A are attached to the head B by split rings C. The test specimen is attached to the upper head B so that the tensile force 81 SECTION 1– DESIGN FOR SIMPLE STRESSES in the specimen pulls down on the head and exerts a compressive force on the rods A. When the machine is exerting the full load, compute (a) the compressive stress in the rods, (b) the bearing stress between the rods and the rings, (c) the shearing stress in the rings, Problem 72. Solution: F = 60,000 lbs (a) sc = (b) sb = (c) sc = (60,000 2) 2 1 1 π 3 − (3)2 4 2 (60,000 2 ) 2 1 2 1 π (4) − 3 4 2 = 11,753 psi = 11.75 ksi = 10,186 psi = 10.19 ksi (60,000 2) = 3,183 psi = 3.18 ksi π (3)(1) DEFORMATIONS 73. A load of 22,000 lb. is gradually applied to a 2-in. round rod, 10 ft. long. The total elongation is observed to be 0.03 in. If the stretching is entirely elastic, (a) what is the modulus of elasticity, and (b) what material would you judge it to be, wrought iron or stainless steel (from information available in the tables)? (c) How much energy is absorbed by the rod? (d) Suppose that the material is aluminum alloy 3003-H14; compute its elongation for the same load. Is this within elastic action? Solution: F = 22,000 lbs D = 2 in L = 10 ft = 120 in δ = 0.03 in 82 SECTION 1– DESIGN FOR SIMPLE STRESSES (a) δ = E= FL EA FL 4 FL 4(22,000 )(120) = = = 28 × 10 6 psi 2 2 δA δπ D (0.03)(π )(2) (b) Use both stainless steel, Table AT 4, E = 28× 10 6 psi and wrought iron , Table AT 7, E = 28× 10 6 psi . 1 1 (c) Energy absorbed = Fδ = (22,000 )(0.03) = 330 lb − in 2 2 (d) For Aluminum alloy, 3003-H14 E = 10× 106 psi s y = 21 ksi FL 4 FL 4(22,000 )(120) = = = 0.084 in 2 2 EA Eπ D 10 × 10 6 (π )(2 ) 4F 4(22,000 ) s= = = 7003 psi = 7.0 ksi < s y , within the elastic limit. 2 πD (π )(2)2 δ= 74. ( ) The same as 73, except that F = 88 kips and total δ = 0.112 in . Is the computation for part (d) valid? Explain. Solution: (a) F = 88 kips δ = 0.112 in FL 4 FL 4(88,000 )(120) E= = = = 30 ×10 6 psi 2 2 δA δπ D (0.112)(π )(2) (b) Use wrought steel, Table AT 4, E = 30× 106 psi 1 1 (c) Energy absorbed = Fδ = (88,000 )(0.112) = 4928 lb − in 2 2 (d) For Aluminum alloy, 3003-H14 E = 10× 106 psi s y = 21 ksi δ= FL 4 FL 4(88,000)(120 ) = = = 0.336 in 2 2 EA Eπ D 10 × 10 6 (π )(2 ) ( ) 83 SECTION 1– DESIGN FOR SIMPLE STRESSES 4F 4(88,000 ) = = 28,011 psi = 28.0 ksi > s y , not within the elastic limit, therefore 2 πD (π )(2)2 not valid. s= 75. (a) A square bar of SAE 1020, as rolled, is to carry a tensile load of 40 kips. The bar is to be 4 ft. long. A design factor of 5 based on the ultimate stress is desired. Moreover, the total deformation should not exceed 0.024 in. What should be the dimensions of the section? (b) Using SAE 1045, as rolled, but with the other data the same, find the dimensions. (c) Using SAE 4640, OQT 1000 F, but with other data the same as in (a), find the dimensions. Is there a change in dimensions as compared with part (b)? Explain the difference or the lack of difference in the answers. Solution: L = 4 ft = 48 in (a) For SAE 1020, as rolled su = 65 ksi , E = 30,000 ksi s F s= u = N A 65 40 = 5 x2 x = 1.754 in FL δ= EA (40)(48) 0.024 = (30,000)x 2 x = 1.633 in 3 Therefore say x = 1 in 4 (b) For SAE 1045, as rolled su = 96 ksi , E = 30,000 ksi s F s= u = N A 96 40 = 5 x2 x = 1.443 in FL δ= EA (40)(48) 0.024 = (30,000)x 2 x = 1.633 in 84 SECTION 1– DESIGN FOR SIMPLE STRESSES 5 Therefore say x = 1 in 8 (c) For SAE 4640, as rolled su = 152 ksi , E = 30,000 ksi s F s= u = N A 152 40 = 2 5 x x = 1.15 in FL δ= EA (40)(48) 0.024 = (30,000)x 2 x = 1.633 in 5 Therefore say x = 1 in 8 There is lack of difference in the answers due to same dimensions required to satisfy the required elongation. 76. The steel rails on a railroad track are laid when the temperature is 40 F. The rails are welded together and held in place by the ties so that no expansion is possible due to temperature changes. What will be the stress in the rails when heated by the sun to 120 F (i1.29)? Solution: s δ α L∆t = = E L L For steel α = 0.000007 in in − F E = 30× 106 psi s = α ∆tE = (0.000007 )(120 − 40 ) 30 × 10 6 s = 16,800 psi ( 77. ) Two steel rivets are inserted in a riveted connection. One rivet connects plates that have a total thickness of 2 in., while the other connects plates with a total thickness of 3 in. If it is assumed that, after heading, the rivets cool from 600 F and that the coefficient of expansion as given in the Text applies, compute the stresses in each rivet after it has cooled to a temperature of 70 F, (no external load). See i1.29. Also assume that the plates are not deformed under load. Is such a stress likely? Why is the actual stress smaller? 85 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: s = α ∆tE For steel α = 0.000007 in in − F E = 30× 106 psi s = (0.000007 )( 600 − 70 )(30,000 ) = 111.30 ksi The stress is unlikely because it is near the ultimate strength of steel. Actual stress must be smaller to allow for safety. 78. Three flat plates are assembled as shown; the center one B of chromium steel, AISI 5140 OQT 1000 F, and the outer two A and C of aluminum alloy 3003-H14, are fastened together so that they will stretch equal amounts. The steel plate is 2 x ½ in., the aluminum plates are each 2 x 1/8 in., L = 30 in ., and the load is 24,000 lb. Determine (a) the stress in each plate, (b) the total elongation, (c) the energy absorbed by the steel plate if the load is gradually applied, (d) the energy absorbed by the aluminum plate. (e) What will be the stress in each plate if in addition to the load of 24,000 lb. the temperature of the assembly is increased by 100 F? Problem 78, 79. Solution: For chromium steel, AISI 5140 OQT 1000F (Table AT 7) α1 = 0.000007 in in − F E1 = 30 × 10 6 psi = 30,000 ksi For aluminum alloy, 3003-H14 (Table AT 3) α 2 = 0.0000129 in in − F E2 = 10 × 106 psi = 10,000 ksi (a) PA = PC PA + PB + PC = F (1) 2 PA + PB = F = 24 kips 1 A2 = (2) = 0.25 in 2 8 1 A1 = (2 ) = 1 in 2 2 86 SECTION 1– DESIGN FOR SIMPLE STRESSES δ A = δB PA L PL = B A2 E2 A1 E1 PA PB = (0.25)(10,000) (1)(30,000) (2) PB = 12 PA (1) 2 PA + 12 PA = 24 kips PA = 1.714 kips PB = 12(1.714) = 20.568 kips Stresses: Aluminum plate P 1.714 s A = sC = A = = 6.856 ksi A2 0.25 Chromium steel plate P 20.568 sB = B = = 20.568 ksi A1 1 (1.714)(30) = 0.021 in PA L = A2 E2 (0.25)(10,000 ) 1 1 (c) Energy absorbed by steel plate = PBδ = (20.568)(0.021) = 0.216 kips − in 2 2 1 1 (d) Energy absorbed by aluminum plate = PAδ = (1.714 )(0.021) = 0.018 kips − in 2 2 (e) 2 PA + PB = F = 24 kips δ TA + δ A = δ TB + δ B (b) δ = δT δT δT δT A = α 2 L∆ t B = α1 L∆t A = (0.0000129 )(30)(100 ) = 0.0387 in B = (0.000007 )(30)(100 ) = 0.021 in PA L PA (30) = = 0.012 PA A2 E2 (0.25)(10,000) PL PB (30) δB = B = = 0.001PB A1E1 (1)(30,000 ) Then 0.0387 + 0.012 PA = 0.021 + 0.001PB 0.0177 + 0.012 PA = 0.001(24 − 2 PA ) 0.0177 + 0.012 PA = 0.024 − 0.002 PA δA = 87 SECTION 1– DESIGN FOR SIMPLE STRESSES PA = 0.45 kips PB = 24 − 2(0.45) = 23.1 kips Stresses: Aluminum plate P 0.45 sA = A = = 1.8 ksi A2 0.25 Chromium steel plate P 23.1 sB = B = = 23.1 ksi A1 1 79. The same as 78, except that the outer plates are aluminum bronze, B150-1, annealed. Solution: For aluminum bronze, B150-1, annealed (Table AT 3) E2 = 15,000 ksi α 2 = 0.0000092 in in − F (a) (1) 2 PA + PB = F = 24 kips δ A = δB PA L PL = B A2 E2 A1 E1 PA PB = (0.25)(15,000) (1)(30,000) (2) PB = 8 PA 2 PA + 8PA = 24 kips PA = 2.4 kips PB = 8(2.4) = 19.2 kips Stresses: Aluminum plate P 2.4 s A = sC = A = = 9.6 ksi A2 0.25 Chromium steel plate P 19.2 sB = B = = 19.2 ksi A1 1 88 SECTION 1– DESIGN FOR SIMPLE STRESSES PA L (2.4)(30) = 0.019 in = A2 E2 (0.25)(15,000 ) 1 1 (c) Energy absorbed by steel plate = PBδ = (19.2 )(0.019 ) = 0.182 kips − in 2 2 1 1 (d) Energy absorbed by aluminum plate = PAδ = (2.4)(0.019 ) = 0.023 kips − in 2 2 (e) 2 PA + PB = F = 24 kips δ TA + δ A = δ TB + δ B (b) δ = δ T = α 2 L∆ t δ T = α1 L∆t δ T = (0.0000092 )(30)(100 ) = 0.0276 in A B A δ T = (0.000007 )(30)(100 ) = 0.021 in B PA L PA (30 ) = = 0.008 PA A2 E2 (0.25)(15,000) PL PB (30) δB = B = = 0.001PB A1E1 (1)(30,000 ) Then 0.0276 + 0.008PA = 0.021 + 0.001PB 0.0066 + 0.008PA = 0.001(24 − 2 PA ) 0.0066 + 0.008PA = 0.024 − 0.002 PA PA = 1.74 kips PB = 24 − 2(1.74 ) = 20.52 kips Stresses: Aluminum plate P 1.74 sA = A = = 6.96 ksi A2 0.25 Chromium steel plate P 20.52 sB = B = = 20.52 ksi A1 1 δA = 80. A machine part shown is made of AISI C1040, annealed steel; L1 = 15 in ., 3 1 L2 = 6 in ., D1 = in ., and D2 = in . Determine (a) the elongation due to a force 4 2 F = 6000 lb ., (b) the energy absorbed by each section of the part if the load is gradually applied. 89 SECTION 1– DESIGN FOR SIMPLE STRESSES Problems 80, 81 Solution: For AISI C1040, annealed steel E = 30× 106 psi (a) δ = δ1 + δ 2 FL (6000)(15) = 0.0068 in δ1 = 1 = A1 E π 3 2 6 30 × 10 4 4 FL (6000)(6) = 0.0061 in δ2 = 2 = A2 E π 1 2 6 30 × 10 4 2 δ = δ1 + δ 2 = 0.0068 + 0.0061 = 0.0129 in ( ) ( ) (b) Energy absorbed 1 1 U1 = Fδ1 = (6000 )(0.0068) = 20.4 lb = in 2 2 1 1 U 2 = Fδ 2 = (6000 )(0.0061) = 18.3 lb = in 2 2 81. A rod as shown is made of AISI 2340 steel, OQT 1000 F, and has the following 7 3 dimensions: L1 = 20 in ., L2 = 12 in ., D1 = in ., and D2 = in . The unit strain at 8 4 point A is measured with a strain gage and found to be 0.0025 in./in. Determine (a) the total elongation, and (b) the force on the rod. Solution: For steel E = 30000 ksi δ F (a) 2 = ε = L2 A2 E 90 SECTION 1– DESIGN FOR SIMPLE STRESSES F = ε A2 E 2 A D δ1 = = ε 2 L1 = ε 2 L1 A1E A1 D1 D 2 3 δ T = δ1 + δ 2 = ε 2 L1 + L2 = 0.0025 D1 7 ε A2 EL1 2 4 (20) + 12 = 0.067 in 8 2 π 3 (b) F = ε A2 E = 0.0025 (30,000) = 33.13 kips 4 4 82. A rigid bar H is supported as shown in a horizontal position by the two rods (aluminum 2024 T4, and steel AISI 1045, as rolled), whose ends were both in contact with H before loading was applied. The ground and block B are also to be considered rigid. What must be the cross-sectional area of the steel rod if, for the assembly, N = 2 based on the yield strengths? Problem 82. Solution: For aluminum 2024-T4 (Table AT 3) s y1 = 47 ksi , E1 = 10,600 ksi For steel AISI 1045, as rolled (Table AT 7) s y2 = 59 ksi , E2 = 30,000 ksi [∑ M G =0 ] R1 (24 ) + R2 (12) = 24(20 ) 2 R1 + R2 = 40 Equation (1) 91 SECTION 1– DESIGN FOR SIMPLE STRESSES δ1 = δ2 24 12 δ1 = 2δ 2 RL δ1 = 1 1 E1 A1 RL δ2 = 2 2 E2 A2 L1 = 8 ft = 96 in L2 = 12 ft = 144 in A1 = 0.5 in 2 δ1 = 2δ 2 R1 L1 2 R2 L2 = E1 A1 E2 A2 R1 (96) 2R2 (144) = (10,600)(0.5) (30,000)A2 0.53R2 R1 = A2 But s 2 = R2 s y2 = A2 N R2 59 = = 29.5 A2 2 R1 = 0.53(29.5) = 15.64 kips R1 s y1 = A1 N R1 47 = 0. 5 2 R1 = 11.75 kips use R1 = 11.75 kips R2 = 40 − 2(11.75) = 16.5 kips R 16.5 A2 = 2 = = 0.56 in 2 29.5 29.5 s1 = 92 SECTION 1– DESIGN FOR SIMPLE STRESSES The bar shown supports a static load F = 2.5 kips with θ = 0 ; d = 3 in ., 3 L = 10 in ., h = 2 in . b = 1 in . It is made of AISI 1035, as rolled. (a) How far 4 does point C move upon gradual application of the load if the movement of A and B is negligible? (b) How much energy is absorbed? 83. Problem 83. Solution: [∑ M [∑ M A B =0 =0 ] dRB = (d + L )F ] 3RB = (3 + 10 )(2.5) RB = 10.83 kips dRA = LF 3RA = 10(2.5) RA = 8.33 kips 93 SECTION 1– DESIGN FOR SIMPLE STRESSES M = RA x − RB x − 3 d2y M = EI 2 = 8.33 x − 10.83 x − 3 dy dy 2 EI = 4.165 x 2 − 5.415 x − 3 + C1 dy 3 EIy = 1.388 x 3 − 1.805 x − 3 + C1 x + C2 When x = 0 , y = 0 3 EI (0 ) = 1.388(0) − 1.805 0 + C1 (0 ) + C2 3 C2 = 0 When x = 3 , y = 0 3 EI (0 ) = 1.388(3) − 1.805 0 + C1 (3) + 0 3 C1 = −12.492 3 EIy = 1.388 x 3 − 1.805 x − 3 − 12.492 x When x = d + L = 13 in 3 EIy = 1.388(13) − 1.805 10 − 12.492(13) = 1082 For AISI 1035, as rolled , E = 30,000 ksi 3 bh 3 (1)(2.75) = = 1.7331 in 4 12 12 EIy = 1082 (30,000)(1.7331)y = 1082 y = 0.021 in , upward. 3 I= 94 SECTION 1– DESIGN FOR SIMPLE STRESSES PRESSURE VESSELS 84. A storage tank for air, 36 in. in diameter, is to withstand an internal pressure of 200 psi with a design factor of 4 based on su . The steel has the strength equivalent of C1020 annealed and the welded joints should have a relative strength (efficiency) of 90 %. Determine a suitable plate thickness. Compute the stress on a diametral section and compare it with the longitudinal stress. Solution: For C1020 annealed su = 57 ksi su 57 = = 14.25 ksi N 4 Solving for plate thickness pD s= 2 tη p = 200 psi = 0.2 ksi D = 36 in (5.2)(36) s = 14.25 = 2 t (0.9) t = 0.281 in 5 say t = in 16 Stress on diametral section (0.2 )(36) = 6.40 ksi pD s= = 4 tη 5 4 (0.9) 16 Stress on longitudinal section (0.2 )(36) = 12.80 ksi pD s= = 2 tη 5 2 (0.9) 16 Stress on diametral section < stress on longitudinal section s= 85. A spherical air tank stores air at 3000 psig. The tank is to have an inside diameter of 7 in. (a) What should be the wall thickness and weight of the tank if it is made of 301, ¼-hard, stainless steel, with a design factor of 1.5 based on the yield strength and a joint efficiency of 90 %. (b) Compute the wall thickness and weight if annealed titanium (B265, gr. 5) is used? (c) What is the additional saving in weight if the titanium is hardened? Can you think of circumstances for which the higher cost of titanium would be justified? 95 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: (a) For 301, ¼ hard, stainless steel s y = 75 ksi (Table AT 4) sy 75 = 50 ksi N 1.5 p = 3000 psi = 3 ksi pD s= 4 tη (3)(7) 50 = 4t (0.90) t = 0.117 in s= = γ = 0.286 lb in 3 2 W = 4π r 2t γ = π D 2t γ = π (7 ) (0.117 )(0.286 ) = 5.2 lb (b) For annealed titanium B265, gr. 5 s y = 130 ksi (Table AT 3) sy 130 = 86.67 ksi N 1.5 p = 3000 psi = 3 ksi pD s= 4 tη (3)(7) 86.67 = 4t (0.90) t = 0.061 in s= = γ = 0.160 lb in 3 2 W = 4π r 2t γ = π D 2t γ = π (7 ) (0.061)(0.160) = 1.5 lb (c) For hardened titanium s y = 158 ksi (Table AT 3) sy 158 = 105 ksi N 1.5 p = 3000 psi = 3 ksi pD s= 4 tη (3)(7 ) 105 = 4t (0.90 ) s= = 96 SECTION 1– DESIGN FOR SIMPLE STRESSES t = 0.056 in γ = 0.160 lb in 3 2 W = 4π r 2t γ = π D 2t γ = π (7 ) (0.056)(0.160 ) = 1.38 lb 1.50 − 1.38 (100%) = 8% 1.50 Circumstances: less in weight and small thickness. Savings in weight = 86. Decide upon a material and estimate a safe wall thickness of a cylindrical vessel to contain helium at –300 F and 2750 psi. The welded joint should have a relative strength ≥ 87 %, and the initial computations are to be for a 12-in.-diameter, 30ft.-long tank. (Note: Mechanical properties of metals at this low temperature are not available in the Text. Refer to INCO Nickel Topics, vol. 16, no. 7, 1963, or elsewhere.) Solution: From Kent’s Handbook, Table 8 Material – Hot Rolled Nickel At – 300 F, su = 100 ksi , N = 4 (Table 1.1) s 100 s= u = = 25 ksi N 4 pD s= 2 tη p = 2750 psi = 2.75 ksi D = 12 in η = 87% (2.75)(12) s = 25 = 2 t (0.87 ) t = 0.759 in 3 say t = in 4 CONTACT STRESSES 87. (a) A 0.75-in. diameter roller is in contact with a plate-cam surface whose width is 0.5-in. The maximum load is 2.5 kips where the radius of curvature of the cam surface is 3.333 in. Compute the Hertz compressive stress. (b) The same as (a) except that the follower has a plane flat face. (c) The same as (a) except that the roller runs in a grooved face and contacts the concave surface. (d) What is the maximum shear stress for part (a) and how far below the surface does it exist? Solution: 97 SECTION 1– DESIGN FOR SIMPLE STRESSES (a) 2r1 = 0.75 in , r1 = 0.375 in r2 = 3.333 in F = 2.5 kips b = 0.5 in 1 1 1 2 + 0 . 35 F r1 r2 sc max = 1 1 b + E1 E2 E = 30,000 ksi 1 sc max 1 2 1 + 0.35(2.5) 0.375 3.333 = = 279 ksi 2 0.5 30,000 1 (b) sc max 1 2 1 ( ) 0 . 35 2 . 5 + 3 . 333 3 . 333 = 126 ksi = 2 0.5 30,000 (c) sc max 1 2 1 ( ) 0 . 35 2 . 5 − 0.375 3.333 = = 249 ksi 2 0.5 30,000 1 (d) Maximum shear stress s s max = 0.3sc max = 0.3(279 ) = 84 ksi Location: 1 1 2 4 sc max 1 − µ 2 + 4(279 ) 1 − 0.32 E1 E2 30,000 w= = = 0.023 in 1 1 1 1 + + 0.375 3.333 r1 r2 ( 88. ) ( ) Two 20o involute teeth are in contact along a “line” where the radii of curvature of the profiles are respectively 1.03 and 3.42 in. The face width of the gears is 3 in. If the maximum permissible contact stress for carburized teeth is 200 ksi, what normal load may these teeth support? 98 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: r1 = 1.03 in r2 = 3.42 in b = 3 in sc max = 200 ksi 1 1 1 2 + 0 . 35 F r1 r2 sc max = 1 1 b + E1 E2 E = 30,000 ksi 1 1 2 1 + 0.35 F 1.03 3.42 sc max = 200 = 2 3 30,000 F = 18 kips TOLERANCES AND ALLOWANCES 89. The pin for a yoke connection has a diameter of D of ¾ in., a total length of 2 ½ in., with a head that is 1 ¼ in. in diameter and 3/8 in. thick. The tolerance on D (both pin and hole) is 0.003 in., with an allowance of 0.001 in., basic-hole system. Sketch the pin showing all dimensions with appropriate tolerances. Solution: D = 0.75 in For pin + 0.000 D = 0.749 in − 0.003 For hole + 0.003 D = 0.750 in − 0.000 Sketch 99 SECTION 1– DESIGN FOR SIMPLE STRESSES 90. A shaft with a nominal diameter of 8 in. is to fit in a hole. Specify the allowance, tolerances, and the limit diameters of the shaft and hole on a sketch for: (a) a close sliding fit, (b) a precision-running fit, (c) medium-running fit, (d) a loose-running fit. Solution: D = 8 in (a) For close-sliding fit, RC 1 Hole, in +0.0008 -0.0000 Shaft, in - 0.0006 -0.0012 Allowance = 0.0006 in With tolerances, + 0.0008 Hole D = 8.0000 in − 0.0000 + 0.0000 Shaft D = 7.9994 in − 0.0006 Limit dimension, Hole D = 8.0000 to 8.0008 in Shaft D = 7.9994 to 7.9988 in Sketch 100 SECTION 1– DESIGN FOR SIMPLE STRESSES (b) For a precision-running fit, RC 3 Hole, in Shaft, in +0.0012 -0.0020 -0.0000 -0.0032 Allowance = 0.0020 in With tolerances, + 0.0012 Hole D = 8.0000 in − 0.0000 + 0.0000 Shaft D = 7.9980 in − 0.0012 Limit dimension, Hole D = 8.0000 to 8.0012 in Shaft D = 7.9980 to 7.9968 in Sketch (c) For medium-running fit, RC5, RC 6. Say RC 5 Hole, in +0.0018 -0.0000 Shaft, in -0.0040 -0.0058 Allowance = 0.0040 in With tolerances, + 0.0018 Hole D = 8.0000 in − 0.0000 + 0.0000 Shaft D = 7.9960 in − 0.0018 Limit dimension, Hole D = 8.0000 to 8.0018 in Shaft D = 7.9960 to 7.9942 in Sketch 101 SECTION 1– DESIGN FOR SIMPLE STRESSES (d) For loose-running fit, RC 8, RC 9. Say RC 8 Hole, in +0.0070 -0.0000 Shaft, in -0.0100 -0.0145 Allowance = 0.010 in With tolerances, + 0.0070 Hole D = 8.0000 in − 0.0000 + 0.0000 Shaft D = 7.9900 in − 0.0045 Limit dimension, Hole D = 8.0000 to 8.0070 in Shaft D = 7.9900 to 7.9855 in Sketch 91. The same as 90, except that the nominal diameter is 4 in. Solution: D = 4 in (a) For close-sliding fit, RC 1 102 SECTION 1– DESIGN FOR SIMPLE STRESSES Hole, in +0.0006 -0.0000 Shaft, in -0.0005 -0.0009 Allowance = 0.0005 in With tolerances, + 0.0006 Hole D = 4.0000 in − 0.0000 + 0.0000 Shaft D = 3.9995 in − 0.0004 Limit dimension, Hole D = 4.0000 to 4.0006 in Shaft D = 3.9995 to 3.9991 in Sketch (b) For a precision-running fit, RC 3 Hole, in Shaft, in +0.0009 -0.0014 -0.0000 -0.0023 Allowance = 0.0014 in With tolerances, + 0.0009 Hole D = 4.0000 in − 0.0000 + 0.0000 Shaft D = 3.9986 in − 0.0009 Limit dimension, Hole D = 4.0000 to 4.0009 in Shaft D = 3.9986 to 3.9977 in Sketch 103 SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For medium-running fit, RC5, RC 6. Say RC 6 Hole, in +0.0022 -0.0000 Shaft, in -0.0030 -0.0052 Allowance = 0.0030 in With tolerances, + 0.0022 Hole D = 4.0000 in − 0.0000 + 0.0000 Shaft D = 3.9970 in − 0.0022 Limit dimension, Hole D = 4.0000 to 4.0022 in Shaft D = 3.9970 to 7.9948 in Sketch (d) For loose-running fit, RC 8, RC 9. Say RC 9 Hole, in +0.0090 -0.0000 Shaft, in -0.0100 -0.0150 Allowance = 0.0100 in 104 SECTION 1– DESIGN FOR SIMPLE STRESSES With tolerances, + 0.0090 in − 0.0000 + 0.0000 Shaft D = 3.9900 in − 0.0050 Limit dimension, Hole D = 4.0000 to 4.0090 in Shaft D = 3.9900 to 3.9850 in Sketch Hole D = 4.0000 92. A cast-iron gear is to be shrunk onto a 3-in, steel shaft. (a) Determine the tolerance and the maximum, minimum, and average interferences of metal for class FN 1 fit. (b) Sketch and dimension the shaft and hole with proper tolerances. (c) Compute the stresses by the method given in the Text (i3.8) for the maximum and minimum interferences of metal. Solution: D = 3 in (a) For class FN 1 fit, Table 3.2 Tolerances Hole, in Shaft, in +0.0007 +0.0019 -0.0000 +0.0014 Max. interference = 0.0019 in Min. interference = 0.0014 – 0.0007 = 0.0007 in Ave. interference = 0.5(0.0019 + 0.0007) = 0.0013 in (b) 105 SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For maximum interference Ei D E = 23,000 ksi (23,000 )(0.0019 ) = 14.6 ksi s= 3 s= For minimum interference s= 93. (23,000 )(0.0007 ) = 5.4 ksi 3 The same as 92, except that the gear hub is C1035 steel and class of fit is FN 3. Solution: D = 3 in (a) For class FN 3fit, Table 3.2 Tolerances Hole, in Shaft, in +0.0012 +0.0037 -0.0000 +0.0030 Max. interference = 0.0037 in Min. interference = 0.0030 – 0.0012 = 0.0018 in Ave. interference = 0.5(0.0037 + 0.0018) = 0.0028 in (b) 106 SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For C1035 steel, E = 30,000 ksi For maximum interference Ei D ( 30,000)(0.0037 ) s= = 37 ksi 3 s= For minimum interference s= 94. (30,000)(0.0018) = 18 ksi 3 For a No. 7 ball bearing, the New Departure Handbook states that the maximum bore should be 1.3780 in. and the minimum, 1.3775 in.; for average conditions, the shaft should have a maximum diameter of 1.3784 in. and a minimum of 1.3779 in. (a) Determine the corresponding tolerances and allowances. (b) What class of fit is this? (c) New Departure states: “. . . bearing bores are held uniformly close, . . . averaging within 1.3778 in. to 1.3776 in.” What will be the maximum and minimum interference of metal with these diameters (if maximum and minimum sizes are deliberately chosen for assembly)? Solution: (a) Tolerances: For No. 7 ball bearing Bore, 1.3780 – 1.3775 in = 0.0005 in + 0.0005 Therefore, D = 1.3775 in − 0.0000 107 SECTION 1– DESIGN FOR SIMPLE STRESSES Shaft, 1.3784 – 1.3775 = 0.0009 in 1.3779 – 1.3775 = 0.0004 in + 0.0000 D = 1.37845 in − 0.0005 Tolerances Hole, in +0.0005 -0.0000 Shaft, in +0.0009 +0.0004 Allowance = 0 – 0.0009 in = - 0.0009 in (b) Since allowance is < 0. It is a force and shrink fil class. (c) 1.3778 – 1.3775 = 0.0003 in 1.3776 – 1.3775 = 0.0001 in New tolerances Hole, in +0.0005 -0.0000 Shaft, in +0.0003 +0.0001 Maximum interference = 0.0003 in Minimum interference = 0.0000 in (since 0.0001 – 0.0005 = - 0.0004 < 0) 95. For a roller bearing having a bore of 65 mm. an SKF catalog states that the largest diameter should be 2.5591 in. and the smallest, 2.5585 in. If this bearing is to be used in a gear transmission, it is recommended for the shaft (where the bearing fits) to have a maximum diameter of 2.5600 in. and a minimum of 2.5595 in. (a) Determine the tolerances and allowances (or interferences of metal) for this installation. (b) What class of fit would this be? Solution: (a) 65 mm = 2.5591 in 2.5591 – 2.5585 = 0.0006 in 2.5600 – 2.5585 = 0.0015 in 2.5595 – 2.5585 = 0.0010 in 108 SECTION 1– DESIGN FOR SIMPLE STRESSES Tolerances Hole, in +0.0006 -0.0000 Shaft, in +0.0015 +0.0010 Maximum interference = 0.0015 in Minimum interference = 0.0010 – 0.0006 = 0.0004 in (b) Class of fit, Force and shrink fit TOLERANCES, STATISTICAL CONSIDERATION 96. (a) A machine tool is capable if machining parts so that the standard deviation of one critical dimension is 0.0006 in. What minimum tolerance may be specified for this dimension if it is expected that practically all of the production be acceptable? Assume that it is possible to “center the process.” (b) The same as (a), except that it has been decided to tolerate approximately 4.56 % scrap. Solution: (a) T = NS = 6σ = 6(0.0006 ) = 0.0036 in (b) A= 0.0456 = 0.0228 in 2 109 SECTION 1– DESIGN FOR SIMPLE STRESSES From Table 3.3 z = 2. 0 σ z = 2σ T = 2 z = 4σ T = 4(0.0006 ) = 0.0024 in 97. A pin and the hole into which it fits have a nominal diameter of 1 ½ in. The pin tolerance has been set to 0.002 in., the bore tolerance at 0.003 in., and the allowance at 0.001 in., basic hole system. The parts are to be a natural spread of 0.0015 in. for the pin and 0.002 in. for the hole. Assuming that the processes are centered, determine the expected minimum clearance and the maximum clearance. What is the most frequent clearance? Solution: 1.5015 – 1.4980 = 0.0035 in NS 0.0015 σ 1 ( pin ) = 1 = = 0.00025 in 6 6 NS 0.0020 σ 2 (hole ) = 2 = = 0.00033 in 6 6 σ D2 = σ 12 + σ 22 = (0.00025)2 + (0.00033)2 σ D = 0.00041 in Natural Spread of Difference = 6σ D = 6(0.00041) = 0.00246 in 110 SECTION 1– DESIGN FOR SIMPLE STRESSES Expected minimum clearance = 0.00227 in Expected maximum clearance = 0.00473 in Most frequent clearance = 0.0035 in 98. A rod and the hole into which it fits has a nominal diameter of 2 in. The tolerances are 0.003 in. for both rod and hole, and the allowance as 0..001 in., basic hole system. The natural spread of the process of manufacturing the hole is 0.002 in., and for the rod, 0.0015 in. What are the probable maximum and minimum clearances, provided that the tolerances are met, but assuming that the processes might simultaneously operate at their extreme permissible position? Solution: NS1 0.0015 = = 0.00025 in 6 6 NS 0.0020 σ 2 ( pin ) = 2 = = 0.00033 in 6 6 σ D2 = σ 12 + σ 22 = (0.00025)2 + (0.00033)2 σ D = 0.00041 in σ 1 (rod ) = 111 SECTION 1– DESIGN FOR SIMPLE STRESSES NS = 6σ D = 6(0.00041) = 0.00246 in 2.0020 – 1.99675 = 0.00525 in Probable maximum clearance = 0.00648 in Probable minimum clearance = 0.00402 in 99. It is desired that the clearance in a 4-in. bearing neither exceed 0.004 in. nor be less than 0.002 in. Assume that the natural spread of the processes by which the journal and the bearing surfaces are finished is the same. (a) What should be the natural spread of these processes? (b) Assuming this natural spread to be equal to the tolerance, determine the corresponding allowance. (c) If the foregoing conditions are not practical decide upon practical tolerances and allowances for the computed natural spread. Solution: (a) NS = 6σ D NS 0.002 σD = = = 0.00033 in 6 6 (b) σ D = 2σ σ 0.00033 σ= D = = 0.000233 in 2 2 112 SECTION 1– DESIGN FOR SIMPLE STRESSES NS = 6σ = 6(0.000233) = 0.0014 in Tolerance = 0.0014 in Corresponing allowance = 0.0016 in (c) From Fig. 3.4, T > NS Tolerance = T = 1.3NS = 1.3(0.0014 ) = 0.0018 in Allowance = 0.003 – 0.0018 = 0.0012 in 100. A 4-in, journal-bearing assembly is made for class RC 6 fit. Assume that the natural spread of the manufacturing process will be about 75 % of the tolerance. Compute the probable maximum and minimum clearances (which occur when the processes are not centered) and compare with the allowance. Make a sketch of the journal and hole properly dimensioned. 113 SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: From Table RC 6, D = 4 in Hole Shaft +0.0022 -0.0030 -0.0000 -0.0052 or + 0.0022 in − 0.0000 + 0.0000 D (shaft ) = 3.997 in − 0.0022 D(hole ) = 4.000 NS = 6σ 0.00165 σ= = 0.00028 in 6 σ D = 2σ = 2 (0.00028) = 0.0004 in NS D = 6σ D = 6(0.0004 ) = 0.0024 in 114 SECTION 1– DESIGN FOR SIMPLE STRESSES Maximum clearance = 0.00912 in Minimum clearance = 0.00336 in Sketch 101. The same as 100, except that class RC 3 fit is used. Solution: From Table RC 3, Table 3.1, D = 4 in Hole Shaft +0.0009 -0.0014 -0.0000 -0.0023 or + 0.0009 in − 0.0000 + 0.0000 D (shaft ) = 3.9986 in − 0.0009 D(hole ) = 4.000 115 SECTION 1– DESIGN FOR SIMPLE STRESSES NS = 6σ 0.000675 σ= = 0.0001125 in 6 σ D = 2σ = 2 (0.0001125) = 0.00016 in NS D = 6σ D = 6(0.00016 ) = 0.00096 in Maximum clearance = 0.001595 in Minimum clearance = 0.002555 in Sketch 102. It is desired that the running clearance for a 3-in. bearing be between approximately 0.003 in. and 0.007 in. The natural spread of the processes of finishing the journal and bearing are expected to be virtually the same ( σ 1 = σ 2 ). Decide upon a suitable tolerance and allowance with a sketch properly dimensioned (to a ten thousandth). (Suggestion: compute first a theoretical natural spread for bearing and journal from the given spread of the clearances. Let the tolerances be approximately equal to this computed NS, and assume that manufacturing processes are available that produce an actual NS of about 70 % of this computed NS.) Check for processes being off center but within ± 3σ limits so that virtually no scrap is manufactured. Solution: 116 SECTION 1– DESIGN FOR SIMPLE STRESSES σ D = 0.00067 in σ 12 + σ 22 = σ D2 2 2σ 12 = (0.00067 ) σ 2 = 0.00047 in NS = 0.70 NS1 NS = 0.70(6)(0.00047 ) = 0.00197 in T = NS1 = (6)(0.00047 ) = 0.00282 in 117 SECTION 1– DESIGN FOR SIMPLE STRESSES For processes off-center running clearance = 0.00385 in to 0.00785 in running clearance = 0.00215 in to 0.00615 in since allowance = 0.00218 in ≈ 0.00215 in, it is checked. 103. If the tolerances shown are maintained during manufacture, say with the processes centered, what would be the approximate overall tolerances and limit dimensions? 118 SECTION 1– DESIGN FOR SIMPLE STRESSES Problem 103. Solution: T1 = 4.004 − 4.000 = 0.004 in T2 = 5.008 − 5.000 = 0.008 in T3 = 6.707 − 6.700 = 0.007 in T 2 = T12 + T22 + T32 = (0.004) + (0.008) + (0.007 ) T = 0.0114 in 2 2 2 Limit dimensions 4.000 to 4.0114 in 5.000 to 5.0114 in 6.700 to 6.7114 in 104. If a cylindrical part needs to have the following tolerances, what process would you recommend for finishing the surface in each instance? (a) 0.05 in., (b) 0.01 in., (c) 0.005 in., (d) 0.001 in., (e) 0.0001 in., (f) 0.00005 in.? Solution: Use fi. 3.9, page 95., Text. (a) 0.05 in Surface finishes = 500 or greater Processes: 1. Flame cutting-machine 2. Rough turning 3. Contour sawing 4. Rough grinding 5. Shaping and planning 6. Drilling 7. Milling – high speed steel 8. Boring (b) 0.01 in Surface finishes = 63 to 250 119 SECTION 1– DESIGN FOR SIMPLE STRESSES Processes: 1. Contour sawing 2. Rough grinding 3. Shaping and planning 4. Drilling 5. Milling – high speed steel 6. Finish turning 7. Broaching 8. Boring 9. Reaming 10. Commercial grinding 11. Barrel finishing (c) 0.005 in Surface finishes = 32 to 125 Processes: 1. Shaping and planning 2. Drilling 3. Milling – high speed steel 4. Finish turning 5. Broaching 6. Boring 7. Reaming 8. Commercial grinding 9. Milling – carbides 10. Gear shaping 11. Barrel finishing 12. Honing (d) 0.001 in Surface finishes = 8 to 32 Processes: 1. Finish turning 2. Broaching 3. Boring 4. Reaming 5. Commercial griniding 6. Milling – carbides 7. Gear shaping 8. Barrel finishing 9. Roller burnishing 10. Diamond turning 120 SECTION 1– DESIGN FOR SIMPLE STRESSES 11. Diamond and precision boring 12. Precision finish grinding 13. Honing 14. Production lapping 15. Superfinishing (e) 0.0001 in Surface finishes = 1 to 8 µin, rms Processes: 1. Barrel finishing 2. Roller burnishing 3. Diamond turning 4. Diamong and precision boring 5. Precision finish grinding 6. Honing 7. Production lapping 8. Superfinishing (f) 0.00005 in Surface finishes = 0 to 2 µin. Processes: 1. Honing 2. Production lapping 3. Superfinishing 105. If it cost $100 to finish a certain surface to 500 microinches rms, what would be the approximate cost to finish it to the following roughness: (a) 125, (b) 32, (c) 8, (d) 2 µin. rms? Solution: From Fig. 3.9 Relative cost of 500 µin rms = 1.75 (a) Relative cost of 125 µin rms = 3 3 Cost = $100 = $171 1.75 121 SECTION 1– DESIGN FOR SIMPLE STRESSES (b) Relative cost of 32 µin rms = 5 5 Cost = $100 = $286 1.75 (c) Relative cost of 8 µin rms = 7.75 7.75 Cost = $100 = $443 1.75 (d) Relative cost of 2 µin rms = 11.5 11.5 Cost = $100 = $657 1.75 DATA LACKING – DESIGNER’S DECISIONS* * Properties of rolled structural sections are found in various handbooks. 106-125. Design a bell crank, similar to the one shown, to carry a mild shock load. The mechanical advantage ( L1 L2 = F2 F1 ), the force F1 , the length L1 , and the material are given in the accompanying table. (a) Make all significant dimensions, including tolerances and allowances. One approach could be to compute dimensions of the yoke connections first; t should be a little less than a . An assumption for the shaft may be that, on occasion, the torque for F1 is transmitted through the shaft (ignoring bending for local convenience). (b) Check all dimensions for good proportion; modify as desirable. (c) Sketch to scale each part, showing all dimensions with tolerances necessary to manufacture. Prob. No. Load F1 L1 AISI No. As Mech, Rolled Advantage 106 107 108 109 700 650 600 550 12 14 15 18 C1020 C1020 C1022 C1035 1.5 2 2.5 3 110 111 112 113 500 800 750 750 20 12 14 14 C1040 C1020 C1020 C1020 4 1.5 2 2.5 114 115 116 117 650 600 900 850 18 20 12 14 C1035 C1040 C1020 C1020 3 4 1.5 2 122 SECTION 1– DESIGN FOR SIMPLE STRESSES 118 119 120 121 800 750 700 1000 15 18 20 12 C1022 C1035 C1040 C1020 2.5 3 4 1.5 122 123 124 125 950 900 850 800 14 15 18 20 C1020 C1022 C1035 C1040 2 2.5 3 4 Problems 106 to 125. Solution: F1 = 700 lb = 0.7 kip L1 = 12 in MA = 1.5 For AISI C1020 as rolled (Table AT 7) su = 65 ksi sus = 49 ksi Designing based on ultimate strength N = 6 (Table 1.1) mild shock, one direction su 65 = = 10 ksi N 6 s 49 s s = us = = 8 ksi N 6 s= Consider yoke connection A 123 SECTION 1– DESIGN FOR SIMPLE STRESSES ss = F1 1 2 π d12 4 0.7 8= 1 π d12 2 d1 = 0.24 in 1 say d1 = in 4 = F1 1 π d12 2 F1 a1d1 0. 7 10 = 1 a1 4 a1 = 0.28 in 5 say a1 = in 16 since t1 < a1 1 say t1 = in 4 F1 s= D1t1 0. 7 10 = 1 D1 4 D1 = 0.28 in 5 say D1 = in 16 Consider yoke connection B. L1 = 1.5 L2 12 L2 = = 8 in 1.5 F2 = 1.5 F1 F2 = 1.5(0.7 ) = 1.05 kip F2 F2 ss = = 1 1 π d 22 2 π d 22 2 4 s= 124 SECTION 1– DESIGN FOR SIMPLE STRESSES 1.05 1 π d 22 2 d 2 = 0.29 in 5 say d 2 = in 16 8= F2 a2 d 2 1.05 10 = 5 a2 16 a2 = 0.34 in 3 say a2 = in 8 since t 2 < a2 5 say t 2 = in 16 F1 s= D1t1 1.05 10 = 5 D2 16 D2 = 0.34 in 3 say D2 = in 8 For shaft diameter Assume torque, T1 = F1L1 = (0.70 )(12) = 8.4 in − kips 16T1 ss = π d s3 16(8.4) 8= π d s3 3 d s = 1.75 in = 1 in 4 Tolerances and allowances, consider RC 4 (Table 3.1) Hole Shaft + 0.0006 + 0.0000 d1 = 0.2500 in d1 = 0.2495 in − 0.0000 − 0.0006 allowance = 0.0005 in s= 125 SECTION 1– DESIGN FOR SIMPLE STRESSES + 0.0006 + 0.0000 in d 2 = 0.3120 in − 0.0000 − 0.0006 allowance = 0.0005 in + 0.0010 + 0.0000 d s = 1.7490 in d s = 1.7500 in − 0.0010 − 0.0000 allowance = 0.0010 in d 2 = 0.3125 Female Male + 0.0006 + 0.0000 a1 = 0.3125 in a1 = 0.3120 in − 0.0000 − 0.0006 allowance = 0.0005 in + 0.0006 + 0.0000 a2 = 0.3750 in a2 = 0.3745 in − 0.0000 − 0.0006 allowance = 0.0005 in (b) For good proporion use the following dimension D1 = D2 = 1 in 3 d1 = d 2 = in 4 3 t1 = t 2 = in 4 a1 = a2 = 1 in d s = 2 in Tolerances and allowances, consider RC 4 (Table 3.1) Hole Shaft + 0.0008 + 0.0000 d1 = 0.7500 in d1 = 0.7492 in − 0.0000 − 0.0008 allowance = 0.0008 in + 0.0008 + 0.0000 d 2 = 0.7500 in d 2 = 0.7492 in − 0.0000 − 0.0008 allowance = 0.0008 in + 0.0012 + 0.0000 d s = 2.0000 in d s = 1.9988 in − 0.0000 − 0.0012 allowance = 0.0012 in Female Male + 0.0008 + 0.0000 a1 = 1.0000 in a1 = 0.9992 in − 0.0008 − 0.0000 allowance = 0.0008 in + 0.0008 + 0.0000 a2 = 1.0000 in a2 = 0.9992 in − 0.0000 − 0.0008 126 SECTION 1– DESIGN FOR SIMPLE STRESSES allowance = 0.0008 in (c) Sketch 126. A simple beam 12 ft. long is to support a concentrated load of 10 kips at the midpoint with a design factor of at least 2.5 based on yield strength. (a) What is the size and weight of the lightest steel (C1020, as rolled) I-beam that can be used? (b) Compute its maximum deflection. (c) What size beam should be used if the deflection is not to exceed ¼ in.? Solution: N = 2. 5 (a) For AISI C1020, as rolled, Table AT 7 s y = 48 ksi s= Mc s y = I N 127 SECTION 1– DESIGN FOR SIMPLE STRESSES 48 = 19 ksi 2.5 FL (10 )(144) M= = = 360 lb − in 4 4 M s= Z Z = section modulus M 360 Z= = = 18.95 in 3 s 19 From Table B-3, Strength of Material by F. Singer, 2nd Edition Select 10I35 Section Index Unsupported length = 12 ft Weight per foot = 35 lb Section Modulus = Z = 29.2 in 3 s= I = 145.8 in 4 , moment of inertia Size (Depth) = 10.0 in Weight of beam = (35)(12) = 420 lb FL3 48 EI E = 30,000 ksi (b) δ = δ= (10 )(144 )3 = 0.142 in 48(30,000 )(145.8) FL3 48 EI (10)(144 )3 0.25 = 48(30,000)I I = 82.9 in 4 From Table B-3, Strength of Material by F. Singer, 2nd Edition Select 10I35 Section Index Unsupported length = 12 ft Weight per foot = 35 lb I = 145.8 in 4 , moment of inertia Size (Depth) = 10.0 in (c) δ = 127. A 10-in., 35-lb. I-beam is used as a simple beam, supported on 18-ft. centers, and carrying a total uniformly distributed load of 6000 lb. Determine the maximum stress and the maximum deflection. Solution: 128 SECTION 1– DESIGN FOR SIMPLE STRESSES 6000 + 630 = 30.7 lb in (18)(12) Table B.3, From Strength of Materials, F.L. Singer, 2nd Edition For 10-in., 35-lb. I-beam I = 145.8 in 4 Z = 29.2 in 3 w= M max Z 2 wL2 (30.7 )(216 ) M max = = = 179,042 lb − in = 179 kips − in 8 8 179 s max = = 6.13 ksi 29.2 5FL3 δ max = 384 EI E = 30,000 ksi F = wL = (30.7 )(216) = 6631 lbs = 6.631 kips s max = 5(6.631)(216 ) = 0.20 in 384(30,000)(145.8) 3 δ max = 128. The same as 127, except the beam is a cantilever. Solution: wL2 (30.7 )(216) = = 716,170 lb − in = 716.17 kips − in 8 2 M 716.17 = max = = 24.53 ksi Z 29.2 2 M max = s max 129 SECTION 1– DESIGN FOR SIMPLE STRESSES w= 6000 + 630 = 30.7 lb in (18)(12) (30.7 )(216) = 1.91 in wL4 = 8EI 8(30,000 )(145.8) 4 δ max = 129. Two equal angles, placed back to back as shown, act as a simple beam and are to support a load of F = 2,000 lb .; L = 40 in .; a = 15 in . What size angles should be used if the maximum stress is not to exceed 20 ksi? The stress due to the weight of the angles is negligible. Problems 129, 130. Solution: Table AT 2 M s= Z Fab M= L a = 15 in L = 40 in b = L − a = 40 − 15 = 25 in (2 )(15)(25) = 18.75 kips − in M= 40 M 18.75 Z= = = 0.9375 in 3 s 20 1 For each angles, Z = (0.9375) = 0.46875 in 3 2 From Strength of Materials, F.L. Singer, 2nd Edition Table B-5 Say size 3” x 3”, thickness = ¼ in I Z = = 0.58 in3 c 130. The same as 129, except that a rolled T-section is to be used. Solution: From Table AT 1, No. 6 130 SECTION 1– DESIGN FOR SIMPLE STRESSES aH 2 + bt 2 2(aH + bt ) c2 = H − c1 c1 = Bt 3 ah 3 + (Bt )d 2 + + (ah )e 2 12 12 say a = t h=B B = 4t H = h+t = B+t b = B−a 2 t (B + t ) + (B − t )t 2 c1 = 2[t (B + t ) + (B − t )t ] Ix = t (B + t ) + (B − t )t 2 t (5t ) + (3t )t 2 = = 1.75t 2(2 Bt ) 4 4t 2 c2 = H − c1 H = B + t = 5t c2 = 5t − 1.75t = 3.25t 2 2 c1 = Ix = ( ) Bt 3 ah 3 + (Bt )d 2 + + (ah )e 2 12 12 d = c1 − t = 1.75t − 0.5t = 1.25t 2 a=t h = B = 4t h e = c2 − = 3.25t − 0.5(4t ) = 1.25t 2 (4t )t 3 + (4t )t (1.25t )2 + t (4t )3 + t (4t )(1.25t )2 = 18.17t 4 Ix = 12 12 Mc2 s= I M = 18.75 kips − in 131 SECTION 1– DESIGN FOR SIMPLE STRESSES s = 20 = (18.75)(3.25t ) 18.17t 4 t = 0.55 in 9 say t = in 16 1 9 B = 4t = 4 = 2 in 4 16 1 9 H = B + t = 2 + = 2.8125 in 4 16 say H = 3 in ″ ″ 1 9 Size: 2 × 3′′ × T section 4 16 - end - 132 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS VARYING STRESSES – NO CONCENTRATION DESIGN PROBLEMS 141. The maximum pressure of air in a 20-in. cylinder (double-acting air compressor) is 125 psig. What should be the diameter of the piston rod if it is made of AISI 3140, OQT at 1000 F, and if there are no stress raisers and no column action? Let N = 1.75 ; indefinite life desired. How does your answer compare with that obtained for 4? Solution: For AISI 3140, OQT 1000 F su = 153 ksi s y = 134 ksi sn = 0.5su = 0.5(153) = 76.5 ksi For axial loading, with size factor sn = 0.5su = (0.8)(0.85)(76.5) = 52 ksi Soderberg line 1 sm sa = + N s y sn For double-acting π 2 Fmax = F = pA = (125) (20 ) = 39,270 lb = 39.27 kips 4 Fmin = − F = −39.27 kips sm = 0 4 F 4(39.27 ) 50 sa = = = 2 π d2 πd2 d 50 2 1 1 d = = 0+ N 1.75 52 d = 1.2972 in 5 say d = 1 in 16 comparative to Problem 4. 142. A link as shown is to be made of AISI 2330, WQT 1000 F. The load F = 5 kips is repeated and reversed. For the time being, ignore stress concentrations. (a) If its surface is machined, what should be its diameter for N = 1.40 . (b) The same as (a), except that the surface is mirror polished. What would be the percentage saving in weight? (c) The same as (a), except that the surface is as forged. Page 1 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Prob. 142 – 144 Solution: For AISI 2330, WQT 1000 F su = 105 ksi s y = 85 ksi sn = 0.5su = 0.5(105) = 52.5 ksi sm = 0 4F 4(5) 20 sa = = = 2 2 πd πd π d2 Soderberg line 1 sm sa = + N s y sn 1 s = 0+ a N sn s sa = n N Size factor = 0.85 Factor for axial loading = 0.80 (a) Machined surface Surface factor = 0.85 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(0.85)(52.5) ksi = 30.345 ksi 20 30.345 sa = = 2 πD 1 .4 D = 0.542 in 9 say D = in 16 (b) Mirror polished surface Surface factor = 1.00 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(1.00)(52.5) ksi = 35.7 ksi 20 35.7 sa = = 2 πD 1 .4 Page 2 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS D = 0.5 in 2 2 9 1 − 16 2 Savings in weight = 2 (100% ) = 21% 9 16 (c) As forged surface Surface factor = 0.40 (Fig. AF 5) sn = 0.5su = (0.80)(0.85)(0.40 )(52.5) ksi = 14.28 ksi 20 14.28 sa = = π D2 1 .4 D = 0.79 in 3 say D = in 4 143. The same as 142, except that, because of a corrosive environment, the link is made from cold-drawn silicon bronze B and the number of reversals of the load is expected to be less than 3 x 107. Solution: For cold-drawn silicon bronze, Type B. sn = 30 ksi at 3 x 108 s y = 69 ksi su = 93.75 ksi 0.085 3 × 108 sn at 3 x 10 = (30 ) = 36.5 ksi 7 3 × 10 sn = (0.80)(0.85)(36.5) = 24.82 ksi 20 24.82 sa = = 2 πD 1 .4 D = 0.60 in 5 say D = in 8 7 144. The same as 142, except that the link is made of aluminum alloy 2024-T4 with a minimum life of 107 cycles. Solution: For AA 2024-T4 s y = 47 ksi su = 68 ksi sn = 20 ksi at 5 x108 Page 3 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 0.085 5 × 108 sn at 10 (20 ) = 27.9 ksi 7 10 sn = (0.80)(0.85)(27.9) = 19 ksi 20 19 sa = = 2 π D 1 .4 D = 0.685 in 11 say D = in 16 7 145. A shaft supported as a simple beam, 18 in. long, is made of carburized AISI 3120 steel (Table AT 10). With the shaft rotating, a steady load of 2000 lb. is appliled midway between the bearings. The surfaces are ground. Indefinite life is desired with N = 1.6 based on endurance strength. What should be its diameter if there are no surface discontinuities? Solution: For AISI 3120 steel, carburized sn = 90 ksi s y = 100 ksi su = 141 ksi Size Factor = 0.85 Surface factor (ground) = 0.88 sn = (0.85)(0.88)(90) = 67.32 ksi sm = 0 32 M sa = π D3 FL (2000 )(18) M= = = 9000 in − lb = 9.0 in − kips 4 4 Soderberg line 1 sm sa = + N s y sn Page 4 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 s = 0+ a N sn s sa = n N 32(9 ) 67.32 = πD3 1 .6 D = 1.2964 in 1 say D = 1 in 4 146. (a) A lever as shown with a rectangular section is to be designed for indefinite life and a reversed load of F = 900 lb . Find the dimensions of a section without discontinuity where b = 2.8t and L = 14 in . for a design factor of N = 2 . The material is AISI C1020, as rolled, with an as-forged surface. (b) compute the dimensions at a section where e = 4 in . Problems 146, 147 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi s n = 0.5su = 32.5 ksi Surface factor (as forged) = 0.55 (a) sm = 0 Mc sa = I 3 tb3 t (2.8t ) I= = = 1.8293t 4 12 12 b 2.8t c= = = 1.4t 2 2 M = FL = (900)(14) = 12,600 in − lb = 12.6 in − kips (12.6)(1.4t ) = 9.643 sa = 1.8293t 4 t3 sn = (0.85)(0.55)(32.5) = 15.20 ksi Page 5 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Soderberg line 1 sm sa = + N s y sn 1 s = 0+ a N sn s sa = n N 9.643 15.20 = t3 2 t = 1.08 in b = 2.8t = 2.8(1.08) = 3.0 in 1 say t = 1 in , b = 3.0 in 16 (b) M = Fe = (900)(4 ) = 3,600 in − lb = 3.6 in − kips (3.6)(1.4t ) = 2.755 sa = 18293t 4 t3 2.755 15.20 = t3 2 t = 0.713 in b = 2.8t = 2.8(0.713) = 1.996 in 23 say t = in , b = 2 in 32 147. The same as 146, except that the reversal of the load are not expected to exceed 105 (Table AT 10). Solution: sn = 32.5 ksi 0.085 106 sn at 10 = (32.5) 5 = 39.5 ksi 10 sn = (0.85)(0.55)(39.5) = 18.5 ksi 5 sn N 9.643 18.5 = t3 2 (a) sa = Page 6 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS t = 1.014 in b = 2.8t = 2.8(1.014) = 2.839 in 13 say t = 1 in , b = 2 in 16 sn N 2.755 18.5 = t3 2 (b) sa = t = 0.6678 in b = 2.8t = 2.8(0.6678) = 1.870 in 11 7 say t = in , b = 1 in 16 8 148. A shaft is to be subjected to a maximum reversed torque of 15,000 in-lb. It is machined from AISI 3140 steel, OQT 1000 F (Fig. AF 2). What should be its diameter for N = 1.75 ? Solution: For AISI 3140 steel, OQT 1000 F su = 152 ksi s y = 134 ksi sn = 0.5su = 76 ksi For machined surface, Surface factor = 0.78 Size factor = 0.85 sns = (0.6)(0.85)(0.78)(134) = 53.3 ksi s ys = 0.6 s y = 0.6(134 ) = 80.4 ksi 1 sms sas = + N s ys sns sms = 0 16T sas = πD3 T = 15 in − kips 16(15) 240 sas = = πD3 πD3 1 s = 0 + as N sns s sas = ns N Page 7 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 240 53.3 = πD 3 1.75 D = 1.3587 in 3 say D = 1 in 8 149. The same as 148, except that the shaft is hollow with the outside diameter twice the inside diameter. Solution: Do = 2 Di 16TDo 16(15)(2 Di ) 32 sas = = = 4 4 4 4 π (Do − Di ) π (2 Di ) − Di πDi3 s sas = ns N 32 53.3 = πDi3 1.75 [ ] Di = 0.694 in 11 3 say Di = in , Do = 1 in 16 8 150. The link shown is machined from AISI 1035 steel, as rolled, and subjected to a repeated tensile load that varies from zero to 10 kips; h = 2b . (a) Determine these dimensions for N = 1.40 (Soderberg) at a section without stress concentration. (b) How much would these dimensions be decreased if the surfaces of the link were mirror polished? Problems 150, 151, 158. Solution: For AISI 1035, steel as rolled su = 85 ksi s y = 55 ksi sn = 0.5su = 42.5 ksi Page 8 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 (10 + 0 ) = 5 kips 2 1 Fa = (10 − 0 ) = 5 kips 2 F 5 10 sm = m = = 2 2 bh 1.5b 3b F 5 10 sa = a = = 2 2 bh 1.5b 3b (a) Soderberg line 1 sm sa = + N s y sn For machined surface, Factor = 0.88 Fm = Size factor = 0.85 sn = (0.80)(0.85)(0.88)(42.5) = 25.4 ksi 1 10 10 = 2 + 2 1.40 3b (55) 3b (25.4 ) b = 0.5182 in 9 say b = in 16 27 h = 1.5b = in 32 (b) Mirror polished, Factor = 1.00 Size factor = 0.85 sn = (0.80)(0.85)(1.00)(42.5) = 28.9 ksi 1 10 10 = 2 + 2 1.40 3b (55) 3b (28.9 ) b = 0.4963 in 1 say b = in 2 3 h = 1.5b = in 4 151. The same as 150, except that the link operates in brine solution. (Note: The corroding effect of the solution takes precedence over surface finish.) Page 9 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: Table AT 10, in brine, AISI 1035, sn = 24.6 ksi s y = 58 ksi sn = (0.80)(0.85)(24.6) = 16.73 ksi 1 10 10 = 2 + 2 1.40 3b (55) 3b (16.73) b = 0.60 in 5 say b = in 8 15 h = 1.5b = in 16 152. The simple beam shown, 30-in. long ( = a + L + d ), is made of AISI C1022 steel, as rolled, left a forged. At a = 10 in , F1 = 3000 lb. is a dead load. At d = 10 in , F2 = 2400 lb. is repeated, reversed load. For N = 1.5 , indefinite life, and h = 3b , determine b and h . (Ignore stress concentration). Problem 152, 153 Solution: For AISI C1022, as rolled su = 72 ksi s y = 52 ksi sn = 0.5su = 36 ksi For as forged surface Figure AF 5, factor = 0.52 Size factor = 0.85 sn = (0.85)(0.52)(36) = 16 ksi Loading: Page 10 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS ∑M A =0 10(3000) + 20(2400 ) = 30 R2 R2 = 2600 lb ∑F V =0 R1 + R2 = F1 + F2 R1 + 2600 = 3000 + 2400 R1 = 2800 lb Shear Diagram M C1 = (2800 )(10 ) = 28,000 in − lb = 28 in − kips M D1 = (2600 )(10 ) = 26,000 in − lb = 26 in − kips Then Loading ∑M A =0 10(3000) + 30 R2 = 20(2400 ) R2 = 600 lb ∑F V =0 R1 + F2 = F1 + R2 R1 + 2400 = 3000 + 600 R1 = 1200 lb Page 11 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Shear Diagram M C2 = (1200 )(10 ) = 12,000 in − lb = 12 in − kips M D2 = (600 )(10 ) = 6,000 in − lb = 6 in − kips Then using M max = M C1 = 28 in − kips M min = M C2 = 12 in − kips 1 (M max + M min ) = 1 (28 + 12) = 20 in − kips 2 2 1 1 M a = (M max − M min ) = (28 − 12 ) = 8 in − kips 2 2 M c M c sm = m , sa = a I I 3 3 bh b(3b ) I= = = 2.25b 4 12 12 h c = = 1.5b 2 Mm Ma sm = , sa = 3 1.5b 1.5b3 1 sm sa = + N s y sn Mm = 20 8 1 1.5b3 1.5b3 = + 1.5 52 16 b = 0.96 in say b = 1 in h = 3b = 3 in 153. The same as 152, except that the cycles of F2 will not exceed 100,000 and all surfaces are machined. Solution: Page 12 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 0.085 106 ( ) sn at 10 cycles = 36 5 = 43.8 ksi 10 su = 72 ksi Machined surface, factor = 0.90 sn = (0.85)(0.90)(43.8) = 33.5 ksi 5 20 8 1 1.5b3 1.5b3 = + 1.5 52 33.5 b = 0.8543 in 7 say b = in 8 5 h = 3b = 2 in 8 154. A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable torque whose maximum value is 6283 in-lb. For N = 1.5 on the Soderberg criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum. Solution: For AISI 1020, cold-finished su = 78 ksi s y = 66 ksi sn = 0.5su = 39 ksi size factor = 0.85 sns = (0.6)(0.85)(39) = 20 ksi s ys = 0.6 s y = 0.6(66 ) = 40 ksi 1 sms sas = + N s ys sns (a) Reversed torque sms = 0 16T sas = πD3 T = 6283 in − lb 16(6283) 32,000 32 sas = = psi = 3 ksi 3 3 πD D D 1 sas = 0+ N sns Page 13 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 32 3 1 D = 0+ 1.5 20 D = 1.34 in 3 say D = 1 in 8 (b) Tmin = 0 , Tmax = 6283 in − lb 1 Tm = (6283) = 3141 in − lb 2 1 Ta = (6283) = 3141 in − lb 2 16(3141) 16,000 16 = psi = 3 ksi 3 3 πD D D 16(3141) 16,000 16 = sas = psi = 3 ksi 3 3 πD D D 16 16 1 D3 D 3 = + 1.5 40 20 D = 1.22 in 1 say D = 1 in 4 sms = (c) Tmin = 3141 in − lb , Tmax = 6283 in − lb 1 Tm = (6283 + 3141) = 4712 in − lb 2 1 Ta = (6283 − 3141) = 1571 in − lb 2 16(4712 ) 24,000 24 = psi = 3 ksi 3 3 πD D D 16(1571) 8,000 8 sas = psi = 3 ksi = 3 3 πD D D 24 8 1 D3 D 3 + = 1.5 40 20 D = 1.145 in 5 say D = 1 in 32 sms = Page 14 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS CHECK PROBLEMS 155. A simple beam 2 ft. long is made of AISI C1045 steel, as rolled. The dimensions of the beam, which is set on edge, are 1 in. x 3 in. At the midpoint is a repeated, reversed load of 4000 lb. What is the factor of safety? Solution: For AISI C1045, as rolled su = 96 ksi s y = 59 ksi sn = 0.5su = 0.5(96) = 48 ksi size factor = 0.85 sn = (0.85)(48) = 40.8 ksi 1 sm sa = + N s y sn sm = 0 6M sa = 2 bh h = 3 in b = 1 in FL (4000 )(24 ) M= = = 24,000 in − lb = 24 in − kips 4 4 6(24 ) sa = = 16 ksi (1)(3)2 1 16 = 0+ N 40.8 N = 2.55 156. The same as 155, except that the material is normalized and tempered cast steel, SAE 080. Solution: Table AT 6 sn′ = 35 ksi s y = 40 ksi sn = (0.85)(35) = 29.75 ksi 1 16 = 0+ N 29.75 N = 1.86 157. A 1 ½-in. shaft is made of AISI 1045 steel, as rolled. For N = 2 , what repeated and reversed torque can the shaft sustain indefinitely? Page 15 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: For AISI 1045, as rolled su = 96 ksi s y = 59 ksi sn′ = 0.5su = 0.5(96) = 48 ksi sns = (0.6)(0.85)(48) = 24.48 ksi s ys = 0.6 s y = (0.6 )(59 ) = 35.4 ksi 1 sms sas = + N s ys sns sms = 0 1 s = 0 + as 2 24.48 sas = 12.24 ksi 16T sas = = 12.24 πD 3 T = 8 in − kips VARIABLE STRESSES WITH STRESS CONCENTRATIONS DESIGN PROBLEMS 158. The load on the link shown (150) is a maximum of 10 kips, repeated and reversed. The link is forged from AISI C020, as rolled, and it has a ¼ in-hole drilled on the center line of the wide side. Let h = 2b and N = 1.5 . Determine b and h at the hole (no column action) (a) for indefinite life, (b) for 50,000 repetitions (no reversal) of the maximum load, (c) for indefinite life but with a ground and polished surface. In this case, compute the maximum stress. Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi For as forged surface Surface factor = 0.55 Size factor = 0.85 sn = (0.80)(0.85)(0.55)(32.5) = 12.2 ksi 1 sm K f sa = + N sy sn Page 16 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Fig. AF 8, b h > 1 Assume K t = 3.5 d 1 Figure AF 7, r = = in = 0.125 in 2 8 a = 0.01 in 1 1 q= = = 0.926 a 0.01 1+ 1+ r 0.125 ( ) K f = q K t − 1 + 1 = 0.926(3.5 − 1) + 1 = 3.3 sm = 0 F 10 = b(h − d ) b(2b − 0.25) K s 1 (a) = 0+ f a N sn 1 (3.3)(10) = 0+ 1.5 b(2b − 0.25)(12.2) 2 2b − 0.25b = 4.06 b 2 − 0.125b − 2.03 = 0 b = 1.489 in 1 say b = 1 in , h = 2b = 3 in 2 sa = (b) For 50,000 repetitions or 50,000 cycles 106 sn = (12.2 ) 4 5 ×10 (log K ) 3 0.085 = 15.74 ksi (log 3.3 ) 3 f n (5 ×104 ) = = 2.0 log K 10log 3.3 10 f 1 K fl sa = N sn 1 (2.0)(10) = 1.5 b(2b − 0.25)(15.74) 2b 2 − 0.25b = 1.906 b 2 − 0.125b − 0.953 = 0 b = 1.04 in 1 1 say b = 1 in , h = 2b = 2 in 16 8 K fl = (c) For indefinite life, ground and polished surface Surface factor = 0.90 Page 17 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi 1 K f sa = N sn 1 (3.3)(10) = 1.5 b(2b − 0.25)(20) b 2 − 0.125b − 1.2375 = 0 b = 1.18 in 3 3 say b = 1 in , h = 2b = 2 in 16 8 Kf F Maximum stress = b(h − d ) b h > 1 , d h = 0.25 2.375 = 0.105 Figure AF 8 K t = 3.5 K f = q (K t − 1) + 1 = 0.926(3.5 − 1) + 1 = 3.315 smax = 159. (3.315)(10) = 13.14 ksi 1.1875(2.375 − 0.25) A connecting link as shown, except that there is a 1/8-in. radial hole drilled through it at the center section. It is machined from AISI 2330, WQT 1000 F, and it is subjected to a repeated, reversed axial load whose maximum value is 5 kips. For N = 1.5 , determine the diameter of the link at the hole (a) for indefinite life; (b) for a life of 105 repetitions (no column action). (c) In the link found in (a) what is the maximum tensile stress? Problem 159 Solution: For AISI 2330, WQT 1000 F su = 135 ksi s y = 126 ksi sn = 0.5su = 0.5(135) = 67.5 ksi For machined surface, Fig. AF 7, surface factor = 0.80 Size factor = 0.85 sn = (0.80)(0.85)(0.80)(67.5) = 36.72 ksi Page 18 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 sm K f sa = + N sy sn Fig. AF 8, b h > 1 Assume K t = 2.5 d 1 Figure AF 7, r = = in = 0.0625 in 2 16 a = 0.0025 in 1 1 q= = = 0.96 a 0.0025 1+ 1+ r 0.0625 K f = q(K t − 1) + 1 = 0.96(2.5 − 1) + 1 = 2.44 (a) Indefinite life, K f = 2.44 sn = 36.72 ksi sm = 0 F 4F 4(5) 20 sa = = = = 2 2 2 πD πD − 4 Dd 1 πD − 0.5 D πD 2 − 4 D − Dd 4 8 K s 1 = 0+ f a N sn 1 (2.44)(20) = 1.5 36.72(πD 2 − 0.5D ) πD 2 − 0.5D = 2.00 D = 0.88 in 7 say D = in 8 (b) For a life of 105 repetitions or cycles 106 sn = (36.72 ) 5 10 (log K ) 3 0.085 = 44.66 ksi (log 2.4 ) 3 f n (105 ) K fl = log K f = 10log 2.44 10 1 K fl sa = N sn 1 (1.81)(20) = 1.5 44.66(πD 2 − 0.5 D ) πD 2 − 0.5D = 1.216 D = 0.71 in Page 19 of 62 = 1.81 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS say D = 3 in 4 4K f F πD 2 − 0.5D 7 d 0.125 D = in , = = 0.14 8 D 0.875 Figure AF 8 K t = 2.6 (c) smax = K f = q(K t − 1) + 1 = 0.96(2.6 − 1) + 1 = 2.54 smax = 160. 4(2.54)(5) 2 7 7 π − 0.5 8 8 = 25.82 ksi A machine part of uniform thickness t = b 2.5 is shaped as shown and machined all over from AISI C1020, as rolled. The design is for indefinite life for a load repeated from 1750 lb to 3500 lb. Let d = b . (a) For a design factor of 1.8 (Soderberg), what should be the dimensions of the part? (b) What is the maximum tensile stress in the part designed? Problems 160, 161 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn′ = 0.5su = 0.5(65) = 32.5 ksi For machined surface Surface factor = 0.90 Size factor = 0.85 sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi 1 sm K f sa = + N sy sn (a) For flat plate with fillets Figure AF 9 b d r= = 3 3 Page 20 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS r 1 = = 0.333 d 3 h 2b = =2 d b K t = 1.65 a = 0.01 in 1 q= ≈ 1.0 a 1+ r K f ≈ K t = 1.65 Fm bt Fa sa = bt b t= 2 .5 1 Fm = (3500 + 1750 ) = 2625 lb 2 1 Fa = (3500 − 1750 ) = 875 lb 2 2625 6562.5 sm = = b2 b b 2 .5 875 2187.5 sa = = b b2 b 2 .5 1 6562.5 (1.65)(2187.5) = + 1.8 48,000b 2 20,000b 2 b = 0.7556 in or b = 0.75 in b 0.75 t= = = 0.3 in 2 .5 2 .5 sm = For flat plate with central hole Fig. AF 8, b h > 1 , d h = b 2b = 1 2 Assume K f ≈ K t = 2.9 Fm Fm F = = m (h − d )t (2b − b )t bt Fa Fa F sa = = = a (h − d )t (2b − b )t bt sm = Page 21 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 2625 6562.5 = b b2 b 2 .5 875 2187.5 sa = = b b2 b 2 .5 1 6562.5 (2.9)(2187.5) = + 1.8 48,000b 2 20,000b 2 b = 0.904 in 15 or b = 0.9375 in = in 16 b 3 t= = in 2 .5 8 15 d = b = in 16 15 3 15 use b = in , t = in , d = in 16 8 16 sm = (b) smax = sm + K f sa d 15 = in 2 32 1 q= = 0.98 0.01 1+ 15 32 K t = 2.9 r= K f = q(K t − 1) + 1 = 0.98(2.9 − 1) + 1 = 2.86 Fm 6562.5 6562.5 = = = 7467 psi 2 bt b2 15 16 F 2187.5 2187.5 sa = a = = = 2489 psi 2 bt b2 15 16 smax = sm + K f sa = 7467 + (2.86 )(2489 ) = 14,586 psi sm = 162. The beam shown has a circular cross section and supports a load F that varies from 1000 lb to 3000 lb; it is machined from AISI C1020 steel, as rolled. Determine the diameter D if r = 0.2 D and N = 2 ; indefinite life. Page 22 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Problems 162 – 164. Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn′ = 0.5su = 0.5(65) = 32.5 ksi For machined surface Surface factor = 0.90 Size factor = 0.85 sn = (0.85)(0.90 )(32.5) = 24.86 ksi ∑M A =0 12 F = 24 B F = 2B F B= 2 F A= B= 2 At discontinuity 6F M= = 3F 2 M max = 3(3000) in − lb = 9000 in − lb = 9 in − kips M min = 3(1000) in − lb = 3000 in − lb = 3 in − kips 1 M m = (9 + 3) = 6 in − kips 2 1 M a = (9 − 3) = 3 in − kips 2 32 M s= πD3 Figure AF 12 D d =1.5d d =1.5 r d = 0.2d d = 0.2 K t = 1.42 assume K f ≈ K t = 1.42 Page 23 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 sm K f sa = + N sy sn 1 (32 )(6 ) (1.42 )(32 )(3) = + 2 48πD 3 24.86πD 3 D = 1.821 in 13 say D = 1 in 16 At maximum moment 12 F M= = 6F 2 M max = 6(3000) in − lb = 18000 in − lb = 18 in − kips M min = 6(1000) in − lb = 6000 in − lb = 6 in − kips 1 M m = (18 + 6 ) = 12 in − kips 2 1 M a = (18 − 6 ) = 6 in − kips 2 32 M s= πD3 K f = 1.00 1 sm K f sa = + N sy sn 1 (32 )(12 ) (1.0 )(32 )(6 ) = + 2 48πD 3 24.86πD 3 D = 1.4368 in 13 Therefore use D = 1 in 16 164. The shaft shown is machined from C1040, OQT 1000 F (Fig. AF 1). It is subjected to a torque that varies from zero to 10,000 in-lb. ( F = 0 ). Let r = 0.2 D and N = 2 . Compute D . What is the maximum torsional stress in the shaft? Solution: For C1040, OQT 1000 F su = 104 ksi s y = 72 ksi Page 24 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS sn′ = 0.5su = 0.5(104) = 52 ksi For machined surface Surface factor = 0.85 Size factor = 0.85 sns = (0.60)(0.85)(0.85)(52) = 22.5 ksi 1 Ta = Tm = (10,000 ) = 5000 in − lb = 5 in − kips 2 s ys = 0.6 s y = 0.6(72 ) = 43.2 ksi sms = sas = 16T πD3 K fs sas 1 sms = + N s ys sns Figure AF 12 D d =1.5d d =1.5 r d = 0.2d d = 0.2 K ts = 1.2 assume K fs ≈ K ts = 1.2 (16)(5) + (1.2)(16)(5) 1 = 2 43.2πD 3 22.5πD 3 D = 1.5734 in 9 say D = 1 in 16 smax = sm + K f sa smax = 165. (16)(5) 9 π 1 16 3 + (1.2)(16)(5) = 14.686 ksi 9 π 1 16 3 An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the proportions shown, with a fillet radius r ≈ 0.25 D ; F varies from 400 lb to 1200 lb.; the supports are to the left of BB not shown. Let N = 2 (Soderberg line). (a) At the fillet, compute D and the maximum tensile stress. (b) Compute D at section BB. (c) Specify suitable dimensions keeping the given proportions, would a smaller diameter be permissible if the fillet were shot-peened? Problems 165 – 167 Page 25 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Solution: For AISI 1144, OQT 1000 F su = 118 ksi s y = 83 ksi sn′ = 0.5su = 59 ksi For machined surface Surface factor = 0.83 Size factor = 0.85 sn = (0.85)(0.83)(59) = 41.62 ksi (a) At the fillet D d =1.5d d =1.5 r d = 0.25d d = 0.25 K t = 1.35 assume K f ≈ K t = 1.35 M = 6F M max = 6(1200) in − lb = 7200 in − lb = 7.2 in − kips M min = 6(400) in − lb = 2400 in − lb = 2.4 in − kips 1 M m = (7.2 + 2.4 ) = 4.8 in − kips 2 1 M a = (7.2 − 2.4 ) = 2.4 in − kips 2 32 M s= πD3 1 sm K f sa = + N sy sn 1 (32 )(4.8) (1.35)(32 )(2.4 ) = + 2 83πD 3 41.62πD 3 D = 1.4034 in 7 say D = 1 in 16 (b) At section BB, M = 30 F M max = 30(1200) in − lb = 36000 in − lb = 36 in − kips M min = 30(400) in − lb = 12000 in − lb = 12 in − kips 1 M m = (7.2 + 2.4 ) = 4.8 in − kips 2 1 M a = (7.2 − 2.4 ) = 2.4 in − kips 2 Page 26 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 32 M πD3 K f = 1 .0 s= 1 sm K f sa = + N sy sn (32)(36) + (1.0)(32)(12) 1 = 2 83π (1.5 D )3 41.62π (1.5 D )3 D = 1.6335 in 11 say D = 1 in 16 (c) Specified dimension: D = 2 in , 1.5 D = 3 in A smaller diameter is permissible if the fillet were shot-peened because of increased fatigue strength. 166. A pure torque varying from 5 in-kips to 15 in-kips is applied at section C. ( F = 0 ) of the machined shaft shown. The fillet radius r = D 8 and the torque passes through the profile keyway at C. The material is AISI 1050, OQT 1100 F, and N = 1.6 . (a) What should be the diameter? (b) If the fillet radius were increased to D 4 would it be reasonable to use a smaller D ? Solution: Tmax = 15 in − kips Tmin = 5 in − kips 1 Tm = (15 + 5) = 10 in − kips 2 1 Ta = (15 − 5) = 5 in − kips 2 For AISI 1050, OQT 1100 F su = 101 ksi s y = 58.5 ksi sn = 0.5su = 0.5(101) = 50.5 ksi Page 27 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS For machined surface Surface factor = 0.85 Size factor = 0.85 sns = (0.60)(0.85)(0.85)(50.5) = 21.9 ksi (a) At the fillet r d =r D = =1 8 D d =1.5 K ts = 1.3 assume K fs ≈ K ts = 1.3 At the key profile K fs = 1.6 use K fs = 1.6 s ys = 0.6 s y = 0.6(58.5) = 35.1 ksi 1 sms K fs sas = + N s ys sns (16)(10) + (1.6)(16)(5) 1 = 1.6 35.1πD 3 21.9πD 3 D = 1.7433 in 3 say D = 1 in 4 (b) r = D 4 r D = 0.25 D d =1.5 Figure AF 12 K ts = 1.18 K fs ≈ K ts = 1.18 < 1.6 Therefore, smaller D is not reasonable. 170. The beam shown is made of AISI C1020 steel, as rolled; e = 8 in . The load F is repeated from zero to a maximum of 1400 lb. Assume that the stress concentration at the point of application of F is not decisive. Determine the depth h and width t if h ≈ 4t ; N = 1.5 ± 0.1 for Soderberg line. Iteration is necessary because K f depends on the dimensions. Start by assuming a logical K f for a logical h (Fig. AF 11), with a final check of K f . Considerable estimation inevitable. Page 28 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Problem 170 Solution: 1 F 2 At the hole A= B= F M = eB = (8) = 4 F 2 M max = 4 F M min = 0 1 M m = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips 2 1 M a = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips 2 Mc s= I (h − 2d )3 t I= 12 1 d = in = 0.5 in 2 1 11 c = 1 + = 1.75 in 2 22 For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi Size factor = 0.85 sn = (0.85)(32.5) = 27.62 ksi Fig. AF 7, c d = 1.75 0.5 = 3.5 > 0.5 Page 29 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Assume K t = 3.5 1 1 r = = 0.25 in 2 2 a = 0.010 in 1 1 q= = = 0.962 a 0.010 1+ 1+ r 0.25 K f = q (K t − 1) + 1 = 0.962(3.5 − 1) + 1 = 3.4 1 sm K f sa = + N sy sn 1 12(2.8)(1.75) (3.4 )(12 )(2.8)(1.75) = + 3 1.5 48(h − 2d )3 t 27.62(h − 2d ) t (h − 2d )3 t = 12.70 [h − 2(0.50)]3 t = 12.70 (4t − 1)3 t = 12.70 t = 0.8627 in 7 say t = in 8 h = 4t = 3.5 in 1 1 1 h > 1 + 1 + in 2 2 2 h > 3.5 in Figure AF 11, h d > 10 h = 10d = 10(0.50) = 5 in 1 d = 2 = 0 .5 b 5 −1 1 2 2 Therefore K t = 3.5 , K f = 3.4 1 Use h = 5 in , t = 1 in 4 171. Design a crank similar to that shown with a design factor of 1.6 ± 0.16 based on the modified Goodman line. The crank is to be forged with certain surfaces milled as shown and two ¼-in. holes. It is estimated that the material must be of the order of AISI 8630, WQT 1100 F. The length L = 17 in. , a = 5 in. , and the load varies form + 15 kips to –9 kips. (a) Compute the dimensions at section AB with h = 3b . Check the safety of the edges (forged surfaces). (Iteration involves; one could first make calculations for forged surfaces and then check safety at holes.) (b) Without redesigning but otherwise considering relevant factors , Page 30 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS quantitatively discuss actions that might be taken to reduce the size; holes must remain as located. Problems 171-174. Solution: (a) AISI 8630, WQT 1100 F su = 96 ksi sn = 0.5su = 0.5(96) = 48 ksi Size factor = 0.85 As-forged surface (Fig. AF I) Surface factor = 0.4 sn = (0.85)(0.42)(48) = 17 ksi Milled surface (Machined) Surface factor = 0.85 sn = (0.85)(0.85)(48) = 34.68 ksi At AB, machined 1 sm K f sa = + N su sn Figure AF 11 1 b = in = 0.5 in 2 1 d = in = 0.25 in 4 d 0.25 = in = 0.5 b 0 .5 Assume K f = 3.50 q = 0.998 K f = q (K t − 1) + 1 = 0.998(3.5 − 1) + 1 = 3.495 Mc I 3 ( h − 2d ) b I= 12 s= Page 31 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 1 1 h 1 1 1 1 − + = h − 1 + = (4h − 4 + 1) = (4h − 3) 2 2 2 4 2 4 8 8 h = 3b 1 M (4h − 3) 8 s= 3 1 h − 2 4 b 12 3 M (12b − 3) 2 s= (3b − 0.5)3 b 4.5M (4b − 1) s= (3b − 0.5)3 b M = F (L − a ) M max = (15)(17 − 5) = 180 in − kips c= M min = (− 9)(17 − 5) = 108 in − kips 1 M m = (180 − 108) = 36 in − kips 2 1 M a = (180 + 108) = 144 in − kips 2 1 sm K f sa = + N su sn 1 4.5(36 )(4b − 1) (3.495)(4.5)(144 )(4b − 1) = + 3 1.6 96(3b − 0.5)3 b 34.68(3b − 0.5) b (4b − 1) = 1 (3b − 0.5)3 b 107.2 (3b − 0.5)3 b = 107.2 (4b − 1) b = 2.6 in 5 say b = 2 in 8 7 h = 3b = 7 in 8 Checking at the edges (as forged) M max = (15)(17 ) = 255 in − kips M min = (− 9)(17 ) = −153 in − kips 1 M m = (255 − 153) = 51 in − kips 2 Page 32 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 M a = (255 + 153) = 204 in − kips 2 6M 6M 2 M s= 2 = 3 = 3 bh 9b 3b K f ≈ 1 .0 1 sm K f sa = + N su sn 1 2(51) (1.0)(2)(204) = 3 + 1.6 3b (96) 3b3 (17 ) b = 2.373 in 3 say b = 2 in 8 5 3 since b = 2 in > 2 in , ∴ safe. 8 8 (c) Action: reduce number of repetitions of load. CHECK PROBLEMS 173. For the crank shown, L = 15 in , a = 3 in , d = 4.5 in , b = 1.5 in . It is as forged from AISI 8630, WQT 1100 F, except for machined areas indicated. The load F varies from +5 kips to –3 kips. The crank has been designed without detailed attention to factors that affect its endurance strength. In section AB only, compute the factor of safety by the Soderberg criterion. Suppose it were desired to improve the margin of safety, with significant changes of dimensions prohibited, what various steps could be taken? What are your particular recommendations? Solution: For as forged surface sn = 17 ksi For machined surface sn = 34.68 ksi s n = 72 ksi In section AB, machined Page 33 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS M = F (L − a ) M max = (+ 5)(15 − 3) = 60 in − kips M min = (− 3)(15 − 3) = −36 in − kips 1 M m = (60 − 36 ) = 12 in − kips 2 1 M a = (60 + 36 ) = 48 in − kips 2 d = h = 4.5 in , b = 1.5 in h =3 b 4.5M (4b − 1) s= (3b − 0.5)3 b 4.5(12 )[4(1.5) − 1] sm = = 2.8125 ksi [3(1.5) − 0.5]3 (1.5) 4.5(48)[4(1.5) − 1] sa = = 11.25 ksi [3(1.5) − 0.5]3 (1.5) 1 sm K f sa = + N sy sn K f = 3.495 from Problem 171. 1 2.8125 (3.495)(11.25) = + N 72 34.68 N = 0.85 < 1 , unsafe To increase the margin of safety 1. reduce the number of repetitions of loads 2. shot-peening 3. good surface roughness Recommendation: No. 1, reducing the number of repetitions of loads. 175. The link shown is made of AISI C1020, as rolled, machined all over. It is loaded 3 9 5 in tension by pins in the D = in holes in the ends; a = in , t = in , 8 16 16 1 h = 1 in . Considering sections at A, B, and C, determine the maximum safe 8 axial load for N = 2 and indefinite life (a) if it is repeated and reversed; (b) if it is repeated varying from zero to maximum; (c) if it is repeatedly varies or F = −W to F = 3W . (d) Using the results from (a) and (b), determine the ratio of the endurance strength for a repeated load to that for a reversed load (Soderberg line). Page 34 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Problems 175 - 178 Solution: For AISI C1020, as rolled su = 65 ksi s y = 48 ksi sn = 0.5su = 0.5(65) = 32.5 ksi Size factor = 0.85 For machined all over Surface factor = 0.90 sn = (0.85)(0.90)(0.80)(32.5) = 20 ksi 1 sm K f sa = + N sy sn at A, Figure AF 8 9 b = in 16 1 h = 1 in 8 3 d = D = in 8 5 t = in 16 3 d = 8 = 0.33 h 11 8 9 b 16 = = 0.5 1 h 1 8 K tA = 3.6 d 3 r = = in 2 16 a = 0.01 in Page 35 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 0.95 a 0.01 1+ 1+ 3 r 16 ( ) k fA = q ktA − 1 + 1 = 0.95(3.6 − 1) + 1 = 3.47 q= 1 = F F 64 F = = (h − d )t 1 1 − 3 5 15 8 8 16 1 64 Fm 3.47(64)Fa = + 2 15(48) 15(20) 8 Fm + 1.48 Fa at A 1= 45 s= At B Figure AF 9 9 d = a = in 16 1 h = 1 in 8 3 r = in 16 5 t = in 16 3 r 16 = = 0.33 9 d 16 1 1 h = 8 =2 9 d 16 K tB = 1.63 a = 0.01 in 1 1 q= = = 0.95 a 0.01 1+ 1+ 3 r 16 k fB = q (ktB − 1) + 1 = 0.95(1.63 − 1) + 1 = 1.6 s= F F 256 F = = 9 5 dt 45 16 16 Page 36 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 256 Fm 1.6(256)Fa = + 2 45(48) 45(20) 32 1= Fm + 0.455 Fa at B 135 at C, Figure AF 8, b >1 h 1 in 8 9 h = a = in 16 1 d = 8 = 0.22 h 9 16 K tC = 3.5 d 1 r = = in 2 16 a = 0.01 in 1 1 q= = = 0.862 a 0.01 1+ 1+ 1 r 16 k fC = q (ktC − 1) + 1 = 0.862(3.5 − 1) + 1 = 3.2 D= F F 256 F = = (h − d )t 9 − 1 5 35 16 8 16 1 256 Fm 3.2(256)Fa = + 2 35(48) 35(20) 32 1= Fm + 1.17 Fa at C 105 s= Equations 8 Fm + 1.48 Fa 45 32 At B, 1 = Fm + 0.455 Fa 135 32 At C, 1 = Fm + 1.17 Fa 105 At A, 1 = Page 37 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS (a) Repeated and reversed load Fm = 0 Fa = F use at A 8 1= Fm + 1.48 Fa 45 8 1 = (0 ) + 1.48 Fa 45 F = 0.676 kip (b) Fm = Fa = F 8 at A, 1 = F + 1.48 F 45 F = 0.603 kip 32 F + 0.455 F at B, 1 = 135 F = 1.480 kips 32 at C, 1 = F + 1.17 F 105 F = 0.678 kip use F = 0.603 kip (c) Fmin = −W , Fmax = 3W 1 Fm = (3W − W ) = W 2 1 Fa = (3W + W ) = 2W 2 8 at A, 1 = W + 1.48(2W ) 45 W = 0.319 kip 32 at B, 1 = W + 0.455(2W ) 135 W = 0.884 kip 32 at C, 1 = W + 1.17(2W ) 105 W = 0.378 kip use W = 0.319 kip Fmax = 0.957 kip Page 38 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS (d) Ratio = 179. F (b ) 0.603 = = 0.892 F (a ) 0.676 A steel rod shown, AISI 2320, hot rolled, has been machined to the following 3 1 dimensions: D = 1 in. , c = in. , e = in. A semicircular groove at the 4 8 1 1 midsection has r = in. ; for radial hole, a = in. An axial load of 5 kips is 8 4 repeated and reversed ( M = 0 ). Compute the factor of safety (Soderberg) and make a judgement on its suitability (consider statistical variations of endurance strength – i4.4). What steps may be taken to improve the design factor? Problems 179-183 Solution: AISI 2320 hot-rolled (Table AT 10) su = 96 ksi s y = 51 ksi sn = 48 ksi Size factor = 0.85 Surface factor = 0.85 (machined) sn = (0.80)(0.85)(0.85)(48) = 27.74 ksi 1 sm K f sa = + N sy sn sm = 0 , reversed sa = s 1 K f sa = N sn s sa = n NK f at the fillet, Figure AF 12 1 r = e = in 8 Page 39 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS d =c= 3 in 4 D = 1 in 1 r 8 = = 0.17 d 3 4 D 1 = = 1.3 d 3 4 K t = 1.55 a = 0.010 in 1 1 q= = = 0.926 a 0.010 1+ 1+ 1 r 8 K f = q (K t − 1) + 1 = 0.926(1.55 − 1) + 1 = 1.51 sa = s = N= 4(5) 3 π 4 2 = 11.32 ksi sn 27.74 = = 1.62 sa K f (11.32)(1.51) At the groove, Figure AF 14 1 3 d = b = D − 2r = 1 in − 2 in = in 8 4 D = 1 in 1 r = in 8 1 r 8 = = 0.17 d 3 4 D 1 = = 1.3 d 3 4 K t = 1.75 a = 0.010 in Page 40 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 0.926 a 0.010 1+ 1+ 1 r 8 ( ) K f = q K t − 1 + 1 = 0.926(1.75 − 1) + 1 = 1.7 q= 1 = 4F 4(5) = = 11.32 ksi 2 2 πd 3 π 4 s 27.74 N= n = = 1.44 sa K f (11.32)(1.7 ) sa = s = At the hole, Figure AF8 D = h = 1 in 1 d = a = in 4 1 D 4 = = 0.25 h 1 K t = 2.44 a = 0.010 in 1 1 q= = = 0.926 a 0.010 1+ 1+ 1 r 8 K f = q(K t − 1) + 1 = 0.926(2.44 − 1) + 1 = 2.33 sa = s = F πD 4 N= 2 − Dd = 5 π (1) 2 1 − (1) 4 4 = 9.34 ksi 27.74 sn = = 1.27 sa K f (9.34)(2.33) Factor of safety is 1.27 From i4.4 s = 0.76sn sn N= = 1.32 min > 1.27 0.76 sn Therefore, dimensions are not suitable. Steps to be taken: 1. Reduce number of cycle to failure Page 41 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 2. Good surface condition 3. Presetting 186. A stock stud that supports a roller follower on a needle bearing for a cam is 5 7 3 made as shown, where a = in , b = in , c = in . The nature of the junction 8 16 4 of the diameters at B is not defined. Assume that the inside corner is sharp. The material of the stud is AISI 2317, OQT 1000 F. Estimate the safe, repeated load F for N = 2 . The radial capacity of the needle bearing is given as 1170 lb. at 2000 rpm for a 2500-hr life. See Fig. 20.9, p. 532, Text. Problem 186 Solution: AISI 2317, OQT 1000 F su = 106 ksi s y = 71 ksi sn = 0.5su = 53 ksi Size factor = 0.85 sn = (0.85)(53) = 45 ksi Figure AF 12 5 d = a = in 8 3 D = c = in 4 r d ≈ 0 , sharp corner 3 D 4 = = 1.2 d 5 8 Assume K t = 2.7 K f ≈ K t = 2 .7 s= 32 M π a3 7 M = Fb = F = 0.4375 F 16 Page 42 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 5 in = 0.625 in 8 32(0.4375)F s= = 18.25 F π (0.625)3 sm = sa = s = 18.25F 1 sm K f sa = + N sy sn a= 1 18.25 F (2.7 )(18.25 F ) = + 2 71 45 F = 0.370 kip = 370 lb < less than radial capacity of the needle bearing. Ok. 187. The link shown is made of AISI C1035 steel, as rolled, with the following 3 7 1 1 dimensions a = in. , b = in. , c = 1 in. , d = in. , L = 12 in. , r = in. The 8 8 2 16 axial load F varies from 3000 lb to 5000 lb and is applied by pins in the holes. (a) What are the factors of safety at points A, B, and C if the link is machined all over? What are the maximum stresses at these points? Problems 187, 188 Solution: AISI C1035, as rolled su = 85 ksi s y = 55 ksi sn = 0.5su = 42.5 ksi size factor = 0.85 sn = (0.6)(0.85)(42.5) = 21.68 ksi 1 sm K f sa = + N sy sn 1 (5 + 3) = 4 kips 2 1 Fa = (5 − 3) = 1 kip 2 Fm = Page 43 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS (a) at A, Figure AF 9 1 r = in 16 3 d = a = in 8 7 h = b = in 8 1 r 16 = = 0.17 3 d 8 7 h 8 = = 2.33 d 3 8 K t = 1.9 a = 0.010 in 1 1 q= = = 0.862 a 0.010 1+ 1+ 1 r 16 K f = q(K t − 1) + 1 = 0.862(1.9 − 1) + 1 = 1.78 s= F ac 4 = 10.67 ksi 3 (1) 8 1 sa = = 2.67 ksi 3 (1) 8 1 10.67 (1.78)(2.67 ) = + N 55 21.68 N = 2.42 sm = At B, same as A, K f = 1.78 F (b − a )c 4 sm = = 8 ksi 7 3 − (1) 8 8 s= Page 44 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 = 2 ksi 7 3 − (1) 8 8 1 8 (1.78)(2 ) = + N 55 21.68 N = 3.23 sa = At C, Figure AF 8 1 d = in 2 h = c = 1 in b h >1 1 d 2 = = 0.5 h 1 K t = 2.2 a = 0.010 in d 1 r = = in = 0.25 in 2 4 1 1 q= = = 0.962 a 0.010 1+ 1+ r 0.25 K f = q(K t − 1) + 1 = 0.962(2.2 − 1) + 1 = 2.15 F (b − a )(c − d ) 4 sm = = 16 ksi 7 3 1 − 1 − 8 8 2 1 sm = = 4 ksi 7 3 1 − 1 − 8 8 2 1 16 (2.15)(4 ) = + N 55 21.68 N = 1.45 s= (b) Maximum stresses at A s A = sm + K f sa = 10.67 + 1.78(2.67 ) = 15.42 ksi at B sB = sm + K f sa = 8 + 1.78(2 ) = 11.56 ksi Page 45 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS at C sC = sm + K f sa = 16 + 2.15(4 ) = 24.6 ksi IMPACT PROBLEMS 189. A wrought-iron bar is 1in. in diameter and 5 ft. long. (a) What will be the stress and elongation if the bar supports a static load of 5000 lb? Compute the stress and elongation if a 5000 lb. weight falls freely 0.05 in. and strikes a stop at the end of the bar. (b) The same as (a), except that the bar is aluminum alloy 3003H14. Solution: D = 1 in. , L = 5 ft For wrought iron, E = 28× 106 psi (a) elongation F = 5000 lb (5000)(5)(12) = 0.01364 in FL = δ= π AE (1)2 28 ×106 4 Stress and elongation h = 0.05 in W = 5000 lb L = 5 ft = 60 in ( ) 1 W W 2hEA 2 s = + 1 + A A LW 1 2 2 6 π 2 ( 0 . 05 ) 28 × 10 ( 1 ) 5000 5000 4 = 24,741 psi 1 + s= + π 2 π 2 (60)(5000) (1) (1) 4 4 sL (24,741)(60 ) δ= = = 0.053 in E 28 × 106 ( (b) Aluminum alloy 3003-H14 E = 10× 106 psi F = 5000 lb FL (5000)(5)(12) = 0.038 in = δ= AE π (1)2 10 ×106 4 Stress and elongation h = 0.05 in ( Page 46 of 62 ) ) SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS W = 5000 lb L = 5 ft = 60 in 1 W W 2hEA 2 s = + 1 + A A LW 1 2 2 6 π 2 0 . 05 10 × 10 1 ( ) ( ) 5000 5000 4 = 18,475 psi 1 + s= + π 2 π 2 (60)(5000) (1) (1) 4 4 sL (18,475)(60 ) δ= = = 0.111 in E 10 × 106 ( 190. ) What should be the diameter of a rod 5 ft. long, made of an aluminum alloy 2024-T4, if it is to resist the impact of a weight of W = 500 lb dropped through a distance of 2 in.? The maximum computed stress is to be 20 ksi. Solution: For aluminum alloy, 2024-T4 E = 10.6 × 106 psi W = 500 lb h = 2 in L = 5 ft = 60 in s = 20 ksi = 20,000 psi 1 W W 2hEA 2 s = + 1 + A A LW ( ) 1 5000 5000 2(2 ) 10.6 × 106 A 2 20,000 = + 1+ (60)(500) A A 1 40 A = 1 + (1 + 1413 A)2 A= πD 2 4 = 0.9332 D = 1.09 in , say D = 1 191. 1 in 16 A rock drill has the heads of the cylinder bolted on by 7/8-in. bolts somewhat as shown. The grip of the bolt is 4 in. (a) If the shank of the bolt is turned down to the minor diameter of the coarse-thread screw, 0.7387 in., what energy may each bolt absorb if the stress is not to exceed 25 ksi? (b) Short bolts used as described above sometimes fail under repeated shock loads. It was found in one instance that if long bolts, running from head to head, were used, service failures were eliminated. How much more energy will the bolt 21 in. long absorb for a stress of Page 47 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 25 ksi. That the bolt 4 in. long? As before, let the bolt be turned down to the minor diameter. The effect of the threads on the strength is to be neglected. Problem 191 Solution: 2 s2 ( AL ) = s AL U= 2E 2E 2 πD (a) A = 4 L = 4 in D = 0.7387 in E = 30×106 psi s = 25 ksi = 25,000 psi (25,000)2 π (0.7387 )2 (4) 4 2 30 × 106 U= ( ) = 17.86 in − lb (b) L = 21 in (25,000)2 π (0.7387 )2 (21) 4 = 93.75 in − lb 2 30 × 106 ∆U = 93.75 − 17.86 = 75.89 in − lb U= 192. ( ) As seen in the figure, an 8.05-lb body A moving down with a constant acceleration of 12 fps2, having started from rest at point C. If A is attached to a steel wire, W & M gage 8 (0.162 in. diameter) and if for some reason the sheave D is instantly stopped, what stress is induced in the wire? Problems 192, 193 Solution: s 2 AL U= 2E 1 2 1 U = mv = m(2ah ) = mah = maL 2 2 Page 48 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS s 2 AL = maL 2E 2maE 2WaE s2 = = A gA W = 8.05 lb a = 12 fps 2 g = 32 fps 2 b = 12 fps 2 E = 30×106 psi A= πD 2 4 8WaE 8(8.05)(12)(30 × 106 ) s2 = = πD 2 g π (0.162)2 (32) s = 93,741 psi 193. The hoist A shown, weighing 5000 lb. and moving at a constant v = 4 fps is attached to a 2 in. wire rope that has a metal area of 1.6 sq. in. and a modulus E = 12× 106 psi . When h = 100 ft , the sheave D is instantly stopped by a brake (since this is impossible, it represents the worst conceivable condition). Assuming that the stretching is elastic, compute the maximum stress in the rope. Solution: s 2 AL 2E 1 W 2 v U = mv 2 = 2 2g U= s 2 AL W 2 = v 2E 2g Wv 2 E gAL W = 5000 lb s2 = Page 49 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS v = 4 fps E = 12× 106 psi A = 1.6 in 2 L = h = 100 ft g = 32 fps 2 s2 = (5000)(4)2 (12 ×106 ) (32)(1.6)(100) s = 13,693 psi 194. A coarse-thread steel bolt, ¾ in. in diameter, with 2 in. of threaded and 3 in. of unthreaded shank, receives an impact caused by a falling 500-lb weight. The area at the root of the thread is 0.334 sq. in. and the effects of threads are to be neglected. (a) What amount of energy in in-lb. could be absorbed if the maximum calculated stress is 10 ksi? (b) From what distance h could the weight be dropped for this maximum stress? (c) How much energy could be absorbed at the same maximum stress if the unthreaded shank were turned down to the root diameter. Solution: s 2 AL U= 2E (a) U = U1 + U 2 s12 A1L1 2E 2 s AL U2 = 2 2 2 2E A1 = 0.334 in 2 U1 = A2 = π (0.75) = 0.442 in 2 4 s1 = 10,000 psi s A (10,000 )(0.334 ) s2 = 1 1 = = 7556 psi A2 0.442 L1 = 2 in L2 = 3 in E = 30 × 10 6 psi U1 = (10,000)2 (0.334)(2) = 1.113 in − lb U2 = (7556)2 (0.442)(3) = 1.262 in − lb 2(30 × 106 ) 2(30 ×106 ) Page 50 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS U = U1 + U 2 = 1.113 + 1.262 = 2.375 in − lb 1 W 2hEA 2 (b) s = 1 + 1 + A LW 1 2 W 2hE s = 1 + 1 + A1 L L W 1 + 2 A1 A2 1 W 2hEA1 A2 2 s = 1 + 1 + A1 W ( A2 L1 + A1L2 ) W = 500 lb A1 = 0.334 in 2 A2 = 0.442 in 2 L1 = 2 in L2 = 3 in E = 30×106 psi s = 10,000 psi ( ) 1 2h 30 × 106 (0.334 )(0.442 ) 2 500 10,000 = 1 + 1 + 0.334 500[(0.442 )(2 ) + (0.334 )(3)] h = 0.0033 in s 2 AL 2E A = 0.334 in 2 L = 5 in E = 30×106 psi s = 10,000 psi (c) U = 2 ( 10,000) (0.334)(5) U= = 2.783 in − lb 2(30 ×106 ) 196. A part of a machine that weighs 1000 lb. raised and lowered by 1 ½-in. steel rod that has Acme threads on one end (see i8.18 Text, for minor diameter). The length of the rod is 10 ft. and the upper 4 ft are threaded. As the part being Page 51 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS lowered it sticks, then falls freely a distance of 1/8 in. (a) Compute the maximum stress in the rod. (b) What would be the maximum stress in the rod if the lower end had been turned down to the root diameter? Solution: 1 W 2hEA 2 s= 1 + 1 + A LW 1 2 W 2hE s = 1 + 1 + A1 L L W 1 + 2 A1 A2 1 W 2hEA1 A2 2 s = 1 + 1 + A1 W ( A2 L1 + A1L2 ) 1 see i8.18 , D2 = 1 in , D1 = 1.25 in 2 2 π (1.25) A1 = = 1.227 in 2 4 π (1.5)2 A2 = = 1.767 in 2 4 L1 = 4 in L2 = 6 in 1 h = in = 0.125 in 8 W = 1000 lb E = 30×106 psi ( ) 1 2(0.125) 30 × 106 (1.227 )(1.767 ) 2 1000 s= 1 + 1 + = 28,186 psi 1.227 1000[(1.767 )(4 ) + (1.227 )(6 )] 1 W 2hEA 2 (b) s = 1 + 1 + A LW 2 A = A1 = 1.227 in L = L1 + L2 = 10 in Page 52 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS ( ) 1 2(0.125) 30 × 106 (1.227 ) 2 1000 s= 1 + 1 + = 25,552 psi 1.227 10(1000 ) 197. A weight W of 50 lb is moving on a smooth horizontal surface with a velocity of 2 fps when it strikes head-on the end of a ¾-in. round steel rod, 6 ft. long. Compute the maximum stress in the rod. What design factor based on yield strength is indicated for AISI 1010, cold drawn? Solution: 1 2 2 Wv E s= g AL1 + We o W W We = b 3 Wb = ρAL ρ = 0.284 lb in3 π 3 2 2 = 0.442 in 4 4 L = 6 ft = 72 in Wb = (0.284)(0.442)(72) = 9.038 lb 9.038 = 3.013 lb We = 3 W = 50 lb v = 2 fps A= g o = 32 fps 2 E = 30×106 psi L = 6 ft 1 2 2 6 (50)(2) (30 ×10 ) = 8166 psi s= (32)(0.442)(6)1 + 3.013 50 For AISI 1010, cold drawn s y = 55 ksi = 55,000 psi N= s y 55,000 = = 6.74 s 8166 Page 53 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 199. A rigid weight of 100 lb is dropped a distance of 25 in. upon the center of a 12 in., 50-lb. I-beam ( I x = 301.6 in 4 ) that is simply supported on supports 10 ft apart. Compute the maximum stress in the I-beam both with and without allowing for the beam’s weight. Solution: Without beams weight y s = sst yst FL3 48EI F 48EI k= = 3 y L y= 1 W 2hk 2 y =δ = 1 + 1 + k W 6 E = 30×10 psi L = 10 ft = 120 in I = 301.6 in 4 48(30 × 106 )(301.6) = 251,333 lb in (120)3 W = 100 lb h = 25 in k= 1 100 2(25)(251,333) 2 y= 1 + 1 + = 0.1415 in 100 251,333 WL3 (100)(120) yst = = = 0.0004 in 48EI 48(30 × 106 )(301.6) Mc sst = I WL (100 )(120 ) M= = = 3000 in − lb 4 4 h 12 c= = = 6 in 2 2 (3000)(6) = 59.68 psi sst = 301.6 0.1415 s = (59.68) = 21,112 psi 0.0004 3 Page 54 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS with mass of beam 1 2h 2 y = yst + yst 1 + yst h - correction factor = 1 W 1+ e W 17Wb 35 Wb = (50 lb ft )(10 ft ) = 500 lb 17(500 ) We = = 243 lb 35 1 h - correction factor = = 0.292 243 1+ 100 1 2(25)(0.292 ) 2 y = 0.00041 + 1 + = 0.0764 in 0 . 0004 y 0.0764 s = sst = (59.68) = 11,400 psi yst 0.0004 We = 201. A 3000 lb. automobile (here considered rigid) strikes the midpoint of a guard rail that is an 8-in. 23-lb. I-beam, 40 ft. long; I = 64.2 in4 . Made of AISI C1020, as rolled, the I-beam is simply supported on rigid posts at its ends. (a) What level velocity of the automobile results in stressing the I-beam to the tensile yield strength? Compare results observed by including and neglecting the beam’s mass. Solution: For AISI C1020, as rolled s y = 48 ksi = 48,000 psi Fδ Wv 2 = 2 2 go F 48 EI k= = 3 δ L Mc FLc s= = I 4I 4 Is F= Lc Fδ F 2 L3 16 I 2 s 2 L3 s 2 IL = = 2 2 = 2 96 EI L c (96 EI ) 6 Ec 2 Page 55 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS neglecting mass of beam Fδ s 2 IL Wv 2 = = 2 6 Ec 2 2 go 3Wv 2 Ec 2 2 g o IL W = 3000 lb g o = 32 fps 2 h 8 c = = = 4 in 2 2 E = 30×106 psi I = 64.2 in4 L = 40 ft s = s y = 48 ksi = 48,000 psi s2 = ( ) 3Wv 2 Ec 2 3(3000 )v 2 30 × 106 (4 ) s = (48,000 ) = = 2 g o IL 32(64.2 )(40 ) v = 6.62 fps Including mass of beam 2 2 3Wv Ec 1 s2 = 2 g o IL 1 + We W 17Wb We = 35 Wb = (23 lb ft )(40 ft ) = 920 lb 17(920 ) We = = 447 lb 35 2 2 2 2 2 2 2 6 3 Wv Ec 3 ( 3000 ) v ( 30 × 10 ) ( 4 ) 1 s 2 = (48,000) = = 447 2 go IL 32(64.2)(40) 1 + 3000 v = 7.10 fps 2 DATA LACKING – DESIGNER’S DECISIONS 202. A simple beam is struck midway between supports by a 32.2-lb. weight that has fallen 20 in. The length of the beam is 12 ft. If the stress is not to exceed 20 ksi, what size I-beam should be used? Solution: Page 56 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 2h 2 y = yst + yst 1 + yst y s = sst yst h = 20 in s = 20,000 psi WL3 yst = 48EI 1 y 96 EIh 2 = 1 + 1 + yst WL3 with correction factor 1 2 96 EIh 1 y = 1 + 1 + 3 We yst WL 1+ W Mc WLd sst = = I 8I 17 wL We = 35 1 2 WLd 96 EIh 1 s= 1 + 1 + 8I WL3 1 + 17 wL 35W W = 32.2 lb h = 20 in L = 12 ft = 144 in E = 30×106 psi s= (32.2)(144)d 8I 1 2 1 96 30 × 106 (I )(20) 1 + 1 + (32.2)(144)3 1 + 17(w)(12) 35(32.2) ( ) 1 579.6d 1 2 s= 1 + 1 + 599 I I 1 + 0.181w From The Engineer’s Manual By Ralph G. Hudson, S.B. Page 57 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS Use 3”, 5.7 lb, I = 2.5 in4 1 2 579.6(3) 1 s= 1 + 1 + 599(2.5) = 19,600 psi < 20,000 psi 2.5 1 + 0.181(5.7 ) Therefore use 3-in depth, 5.7-lb I-beam ( I = 2.5 in4 ) 204. A 10-in., 25.4-lb.., I-bean, AISI 1020, as rolled, is 10 ft. long and is simply supported at the ends shown. There is a static load of F1 = 10 kips , 4 ft from the left end, and a repeated reversed load of F2 = 10 kips , 3 ft from the right end. It is desired to make two attachments to the beam through holes as shown. No significant load is supported by these attachments, but the holes cause stress concentration. Will it be safe to make these attachments as planned? Determine the factor of safety at the point of maximum moment and at points of stress concentration. Problem 204 Solution: Mass of beam negligible For AISI C1020, as rolled s y = 48 ksi su = 65 ksi (∑ M A = 0) 4 F1 + (10 − 3)F2 = 10 B 1 B = (4 F1 + 7 F2 ) 10 (∑ M B = 0) 3F2 + (10 − 4)F1 = 10 A 1 A = (6 F1 + 3F2 ) 10 F1 = 10 kips F2 = −10 to 10 kips 1 Bmin = [4(10 ) + 7(− 10 )] = −3 kips 10 Page 58 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 1 [4(10) + 7(10)] = 11 kips 10 1 Amin = [6(10 ) + 3(− 10 )] = 3 kips 10 1 Amax = [6(10 ) + 7(30 )] = 9 kips 10 Figure AF 11, 1 e = 1 in , 2 1 d = in 4 1 1 c = 1 + 2 = 1.625 in 2 4 h = 10 in h 10 1 b = − e = − 1 = 3.5 in 2 2 2 d 0.25 = = 0.07 b 3 .5 e 1.50 = = 6 > 0 .5 d 0.25 Use K t = 3.0 1 q= = 0.926 0.010 1+ 1 8 K f = q (K t − 1) + 1 = 0.926(3 − 1) + 1 = 2.85 Bmax = sn = 0.5su = 0.5(65) = 32.5 ksi size factor = 0.85 sn = 0.85(32.5) = 27.6 ksi left hole, M = (2)A M max = 2(9) = 18 ft − kips M min = 2(3) = 6 ft − kips Mc s= I 1 M m = (18 + 6 ) = 12 ft − kips = 144 in − kips 2 1 M a = (18 − 6 ) = 6 ft − kips = 72 in − kips 2 c = 1.625 in I = 122.1 in 4 (Tables) Page 59 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS sm = sa (144)(1.625) = 1.92 ksi 122.1 (72 )(1.625) = 0.96 ksi = 122.1 1 sm K f sa = + N sy sn 1 1.92 (2.85)(0.96 ) = + N 48 27.6 N = 7 .2 right hole , M = (1.5)B M max = 1.5(11) = 16.5 ft − kips M min = 1.5(− 3) = −4.5 ft − kips Mc s= I 1 M m = (16.5 − 4.5) = 6 ft − kips = 72 in − kips 2 1 M a = (16.5 + 4.5) = 10.5 ft − kips = 126 in − kips 2 c = 1.625 in I = 122.1 in 4 (Tables) (72)(1.625) = 0.96 ksi sm = 122.1 (126)(1.625) = 1.68 ksi sa = 122.1 1 sm K f sa = + N sy sn 1 0.96 (2.85)(1.68) = + N 48 27.6 N = 5.67 at maximum moment, or at , F2 M max = 3(11) = 33 ft − kips M min = 3(− 3) = −9 ft − kips Mc s= I 1 M m = (33 − 9 ) = 12 ft − kips = 144 in − kips 2 1 M a = (33 + 9 ) = 21 ft − kips = 252 in − kips 2 Page 60 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS 10 = 5 in 2 I = 122.1 in 4 (Tables) (144)(5) = 5.90 ksi sm = 122.1 ( 252 )(5) sa = = 10.32 ksi 122.1 K f = 1 .0 c= 1 sm K f sa = + N sy sn 1 5.90 (1.0 )(10.32 ) = + N 48 27.6 N =2 Since the design factor at the holes is much larger than at the point of maximum moment, it is safe to make these attachment as planned. 205. The runway of a crane consists of L = 20 ft. lengths of 15-in., 42.9-lb. I-beams, as shown, each section being supported at its ends; AISI C1020, as rolled. The wheels of the crane are 9 ft apart, and the maximum load expected is F = 10,000 lb on each wheel. Neglecting the weight of the beam, find the design factor (a) based on variable stresses for 105 cycles, (b) based on the ultimate strength. (Hint. Since the maximum moment will occur under the wheel, assume the wheels at some distance x from the point of support, and determine the dM reaction, R1 as a function of x ; = 0 gives position for a maximum bending dx moment.) Problem 205. Solution: ∑ M R2 = 0 ( ) (L − x )F + (L − x − a )F = LR1 (2 L − 2 x − a )F R = 1 L x M = R1 x = (2 L − 2 x − a )F L Page 61 of 62 SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS dM F = [(2 L − 2 x − a ) + x(− 2 )] = 0 dx L 2L − 2x − a − 2x = 0 1 a x = L− 2 2 2 a a L− F L− 2 2 2 L − L − a − a F = M max = L 2 2L L = 20 ft = 240 in a = 9 ft = 108 in F = 10,000 lb = 10 kips 2 108 240 − (10 ) 2 M max = = 720.75 in − kips 2(240 ) For 15-in., 42.9 lb, I-beam I = 441.8 in4 15 c = = 7.5 in 2 Mc (720.75)(7.5) smax = = = 12.24 ksi I 441.8 For AISI C1029, as rolled su = 65 ksi sn = 0.5su = 32.5 ksi size factor = 0.85 sn = 0.85(32.5) = 27.6 ksi (a) at 105 cycles 0.085 106 sn = 27.6 5 = 34 ksi 10 s 34 N= n = =7 s 12.24 (b) N = su 65 = = 5.31 s 12.24 - end - Page 62 of 62 SECTION 3 – SCREW FASTENINGS SIMPLE TENSION INCLUDING TIGHTENING STRESSES DESIGN PROBLEMS 221. A 5000-lb. gear box is provided with a steel (as rolled B1113) eyebolt for use in moving it. What size bolt should be used: (a) if UNC threads are used? (b) If UNF threads are used? (c) If the 8-thread series is used? Explain the basis of your choice of design factor. Solution: B1113, as rolled s y = 45 ksi (Table AT-7) Fe = 5000 lb i5.6, sd = sy 6 1 ( As ) 2 3 D < in 4 3 4 sd = 0.4 sy For D ≥ in sd = 0.4(45,000) = 18,000 psi As = Fe 5000 = = 0.2778 sq.in. sd 18 ,000 Table AT 14 and Table 5.1 (a) UNC Threads 3 4 Use D = in , As = 0.334 sq.in. (b) UNF Threads 3 4 Use D = in , As = 0.373sq.in. (c) 8-Thread Series Use D = 1 in , As = 0.606sq.in. Page 1 of 42 SECTION 3 – SCREW FASTENINGS 222. A motor weighing 2 tons is lifted by a wrought-iron eye bolt which is screwed into the frame. Decide upon a design factor and determine the size of the eyebolt if (a) UNC threads are used, (b) UNF threads are used. Note: Fine threads are not recommended for brittle materials. Solution: Table AT-7 Wrought iron, s y = 25 ksi sd = 0.4 sy = 0.4(25) = 10 ksi = 10,000 psi As = Fe 2(2000) = = 0.4 sq.in. sd 10,000 Table AT 17 (a) UNC Threads 7 8 Use D = in , As = 0.462 sq.in. (b) UNF Threads 7 8 Use D = in , As = 0.509 sq.in. 224. A wall bracket, Fig. 8-13, Text, is loaded so that the two top bolts that fasten it to the wall are each subjected to a tensile load of 710 lb. The bolts are to be cold forged from AISI C1020 steel with UNC threads, Neglecting the effect of shearing stresses, determine the diameter of these bolts if they are well tightened. Figure 8-13 Page 2 of 42 SECTION 3 – SCREW FASTENINGS Solution: cold forged, AISI C1020 s y = 66 ksi (Table AT-7) Fe = 710 lb s 3 3 Fe = y ( As ) 2 D < in 4 6 3 66,000 ( As ) 2 710 = 6 3 As = 0.161 sq.in. , D < in 4 Table AT 14 , UNC Threads Use D = 225. 9 in , As = 0.1820 sq.in. 16 A connection similar to Fig. 5.9, Text, is subjected to an external load Fe of 1250 lb. The bolt is made from cold-finished AISI B1113 steel with UNC threads. (a) Determine the diameter of the bolt if it is well tightened. (b) Compute the initial tension and corresponding approximate tightening torque if si = 0.85s y (i5.8). Figure 5.9 Page 3 of 42 SECTION 3 – SCREW FASTENINGS Solution: Cold-finished AISI B1113 Table A-7, s y = 72 ksi Fe = 1250 lb s 3 (a) Fe = y ( As ) 2 6 3 72,000 1250 = ( As ) 2 6 3 in 4 Table AT 14 , UNC Threads As = 0.2214 sq.in. , D < Use D = 5 in , As = 0.2260 sq.in. 8 (b) si = 0.85s y = 0.85(72,000 ) = 61,200 psi Initial Tension Fi = si As = (61,200 )(0.2260 ) = 13,831 lb Tightening torque T = CDFi 5 T = 0.2 DFi = 0.2 (13,831) = 1729 in − lb 8 226. The cylinder head of a 10 x 18 in. Freon compressor is attached by 10 stud bolts made of SAE Grade 5. The cylinder pressure is 200 psi. (a) What size bolts should be used? (b) What approximate tightening torque should be needed to induce a tightening stress si of 0.9 times the proof stress? Solution: Page 4 of 42 SECTION 3 – SCREW FASTENINGS Table 5.2 SAE Grade 5 Assume s y = 88 ksi π 2 200 (10 ) 4 (a) Fe = = 1571 lb 10 s 3 3 Fe = y ( As ) 2 , D < in 6 4 3 88,000 1571 = ( As ) 2 6 3 As = 0.2255 sq.in. , D < in 4 Table AT 14 , UNC Threads Use D = 5 in , As = 0.2260 sq.in. 8 (b) T = CDFi C = 0. 2 s i = 0. 9 s p s p = 85 ksi , (Table 5.2) si = 0.9(85,000 ) = 76,500 psi Fi = si As = (76,500 )(0.2260 ) = 17,289 lb Tightening torque 5 T = 0.2 DFi = 0.2 (17,289) = 2161 in − lb 8 227. The American Steel Flange Standard specifies that 8 bolts are to be used on flanges for 4-in. pipe where the steam or water pressure is 1500 psi. It is also specified that, in calculating the bolt load, the outside diameter of the gasket, which is 6 3/16 in., should be used. Determine (a) the diameter of the UNC bolts if they are well-tightened and made of ASTM 354 BD (Table 5-2), (b) the approximate torque to tighten the nuts if the initial stress is 90 % of the proof stress. The Standard specifies that 1 1/4 –in. bolts with 8 th./in. be used (these bolts are also subjected to bending). How does your answer compare? Solution: Table 5.2, ASTM 354 BD s p = 120 ksi s y = 125 ksi Page 5 of 42 SECTION 3 – SCREW FASTENINGS 2 π 3 1500 6 4 16 = 5638 lb Fe = 8 sy (a) Fe = 3 ( As ) 2 , D< 6 3 125,000 5638 = ( As ) 2 6 As = 0.4184 sq.in. , Table AT 14 , UNC Threads Use D = D> 3 in 4 7 in , As = 0.4620 sq.in. 8 3 in 4 use sd = 50,000 psi § 5.6, ASTM A354 BD F 5638 As = e = = 0.1128 sq.in. sd 50 ,000 Table AT 14 , UNC Threads 3 4 Use D = in , As = 0.334 sq.in. (b) T = 0.2 DFi s i = 0. 9 s p si = 0.9(120,000) = 108,000 psi Fi = si As = (108,000)(0.3340) = 36,072 lb Tightening torque 3 T = 0.2DFi = 0.2 (36,072) = 5411 in − lb 4 1 D < 1 in as specified by the standard. 4 CHECK PROBLEMS 228. A cap screw, ¾ in.-10-UNC-2, with a hexagonal head that is 9/16 in. thick, carries a tensile load of 3000 lb. If the material is AISI 1015, cold drawn, find the factor of safety based on ultimate strengths of (a) the threaded shank, (b) the head against being sheared off, and (c) the bearing surface under the head. (d) Is there any need to consider the strength of standard cap-screw heads in design? Page 6 of 42 SECTION 3 – SCREW FASTENINGS Solution: For ¾ in. UNC, Table AT 14, As = 0.334 sq.in. Head: 1 A = 1 in. 8 For AISI 1015, cold drawn su = 77 ksi , sus = 58 ksi F 3000 = = 8982 psi As 0.334 s 77,000 N= u = = 8.57 sd 8982 (a) s = (b) s s = F πDt 9 in 16 3000 ss = = 2264 psi 3 9 π 4 16 s 58,000 N = us = = 25.6 ss 2264 t= (c) θ = 360 o = 30 o 12 2 1 1 1 A A 1 8 Area = 6(2) tan θ = 6(2) tan 30 = 1.096 sq.in. 2 2 2 2 2 Page 7 of 42 SECTION 3 – SCREW FASTENINGS sb = N= F = Area − Ab 3000 1.096 − π 3 2 = 4586 psi 4 4 su 77,000 = = 16.8 sb 4586 (d) No need to consider the strength of standard cap-screw heads since its factor of safety is very much higher than for the threaded shank. 229. A bolt, 1 1/8 in.-7-UNC-2, is subjected to a tensile load of 10,000 lb. The head has a thickness of ¾ in. and the nut a thickness of 1 in. If the material is SAE grade 2 (Table 5.2), find the design factor as based on ultimate stresses (a) of the threaded shank, (b) of the head against being sheared off, and (c) of the bearing surface under the head. The bolt head is finished. (d) Is there any need to consider the strength of standard bolt heads in design? Solution: 1 For SAE grade 2 (Table 5.2), D = 1 in 8 su = 55 ksi , sus = 0.75su For 1 1/8 in.-7-UNC-2 (Table AT 14) As = 0.763 sq.in. A =1 11 in 16 F = 10,000 lb. F 10,000 = = 13,106 psi As 0.763 s 55,000 N= u = = 4.2 sd 13,106 (a) s = (b) s s = t= F πDt 3 in 4 Page 8 of 42 SECTION 3 – SCREW FASTENINGS 10,000 = 3773 psi 1 3 π 1 8 4 0.75(55,000 ) s N = us = = 11 ss 3773 ss = (c) θ = 360 o = 30 o 12 2 11 1 1 A A 1 16 Area = 6(2) tan θ = 6(2) tan 30 = 2.4661 sq.in. 2 2 2 2 2 F 10,000 sb = = = 6793 psi 2 Area − Ab π 1 2.4661 − 1 4 8 s 55,000 N= u = = 8.1 sb 6793 (d) No need to consider the strength of standard bolt head in design since its factor of safety is higher than for the threaded shank. 230. An axial force is applied to a regular nut which of course tends to shear the threads on the screw. (a) What is the ratio of the force necessary to shear the threads (all threads initially in intimate contact) to the force necessary to pull the bolt in two? Use coarse threads, a 1 ½ -in. bolt, and assume that sus = 0.75su . The head thickness is 1 in. and the nut thickness is 1 5/16 in. (b) Is failure of the thread by shear likely in this bolt? Solution: 1 ½ - in. UNC As = 1.405 sq.in. (a) Fs = shear force = susπDt 1 D = 1 in. 2 5 t = 1 in. 16 Page 9 of 42 SECTION 3 – SCREW FASTENINGS sus = 0.75su 1 5 Fs = (0.75su )(π )1 1 = 4.6388su 2 16 F = su As = 1.405su 4.6388su = 3.3 Ratio = 1.405su (b) Ratio > 1, failure by shear is not likely to occur. 231. For bolted structural joints, specifications suggest that ½-in. bolts (highstrength material) be tightened to an initial tension of Fi = 12,500 lb . What should be the approximate tightening torque? How does your answer compare with T = 90 ft − lb ., which is the value in the specification? Solution: 1 T = 0.2 DFi = 0.2 (12,500) = 1250 in − lb 2 T = 90 ft − lb = 1080 in − lb < 1250 in − lb o.k. 232. One method of estimating the initial tensile stress in a tightened bolt is to turn the nut until it is snug, but with no significant stress in the bolt. Then the nut is turned through a predetermined angle that induces a certain unit strain corresponding to the desired stress. A ¾ - in. bolt of the type shown in Fig. 5.4, Text, is turned down until, for practical purposes, the diameter of the entire shank is the minor diameter. The material is AISI 4140, OQT 1200 oF. The grip is 5 in. and the effective strain length is estimated to be 5.3 in. If the initial tensile stress at the root diameter is to be about 75 % of the yield strength, through what angle should the nut be turned after it is just snug? The threads are UNC and the parts being bolted are assumed to be rigid. Solution: For ¾ in., UNC Dr = 0.6273 in As = 0.334 sq.in. Th in. = 10 AISI 4140, OQT 1200 oF Page 10 of 42 SECTION 3 – SCREW FASTENINGS s y = 115 ksi s = 0.75(115) = 86.25 ksi sL E L = 5.3 in δ= pitch, p = 1 in = 0.10 in 10 δ (360o ) p sL θ= ( 360 o ) pE (86,250)(5.3) (360o ) = 55o θ= (0.10)(30 ×106 ) θ= 233. When both ends of a bolt are accessible for micrometer measurements, the total elongation δ caused by tightening can be determined by measuring lengths before and after tightening. In order to reduce this total elongation to unit elongation, thence to stress, the effective strain length for the bolt must be known. For a 1 ¼-in steel bolt, threaded for its full length, 8-thread series, the effective strain length has been found by experiment to be Le = 0.97G + 1.1 in. , where G is the grip (by W.A. McDonald, North Carolina State College). Let the bolt material be AISI 8742, OQT 1000 oF. (a) It is desired that the initial tensile stress be about 0.7 s y . What total elongation should be obtained for a grip length of 4.8 in.? (b) Investigate the approximate tightening torque for the specified condition. How could this torque be obtained? Solution: 1 ¼ in., 8-thread series Table 5.1 Dr = 1.0966 in As = 1.000 sq.in. Th in. = 8 AISI 8742, OQT 1000 oF s y = 147 ksi (a) si = 0.70 s y si = 0.70(147 ) = 102.9 ksi = 102,900 psi Page 11 of 42 SECTION 3 – SCREW FASTENINGS si Le E Le = 0.97G + 1.1 in. G = 4.8 in Le = 0.97(4.8) + 1.1 in. = 5.756 in δ= δ= si Le (102,900 )(5.756 ) = = 0.01975 in E 30 × 106 (b) θ = δ p = 64TL πDr4G G = 11.5 ×10 6 psi 1 p = in = 0.125 in 8 0.01975 64T (5.756 ) = θ= 4 0.125 π (1.0966) (11.5 ×106 ) T = 22,408 in − lb ELASTIC CONSIDERATIONS 235. The member C shown is part of a swivel connection that is to be clamped by a 1-in. bolt D to the member B, which has large dimensions in the plane perpendicular to the paper. Both B and C are aluminum alloy 2024-T4, HT aged. The bolt is made of AISI C1113, cold-drawn steel; consider the unthreaded shank to be 2 in. long; it is well tightened with a torque of 250 ftlb.; UNC threads, unlubricated. (a) Estimate the initial tension by equation (5.2), assume elastic action, and compute the bolt elongation and the total deformation of B and C. Let the effective strain length be 2 in. (b) After tightening an external axial force Fe of 5000 lb. is applied to member C. Determine the total normal stresses in the bolt and in B and C. (c) Determine the load required to “open” the connection. Draw a diagram similar to Fig. 5.6, Text, locating points A, B, D and M. Page 12 of 42 SECTION 3 – SCREW FASTENINGS Prob. 235, 236 Solution: For aluminum alloy, 2024-T4 HT aged, E = 10.6 ×106 psi s y = 47 ksi For AISI C1113, cold-drawn steel, E = 30×106 psi s y = 72 ksi (a) T = 0.2 DFi D = 1 in. T = 250 ft − lb = 3000 in − lb Fi = 15,000 lb Deformations: L = 2 in. Table AT 14, 1-in. UNC Bolt, As = 0.66 sq.in. π 2 Ab = (1) = 0.785 sq.in. 4 Bolt: FL (15,000)(2) = 0.00127 in δi = i = Ab Ei (0.785)(30 × 10 6 ) Member B and C FL δc = i Ac Ec π π Ac = De2 − D 2 4 4 h De = (Nut or head width across flats) + 2 Table AT 14 1 A = 1 in 2 1 2 1 De = 1 + = 2 in. 2 2 2 π 2 π 2 Ac = De − D 4 4 π 2 Ac = (2.5) − (1)2 = 4.1234 sq.in. 4 FL (15,000)(2) = 0.000686 in. δc = i = Ac Ec (4.1234 )(10.6 × 10 6 ) (b) Fe = 5000 lb [ Page 13 of 42 ] SECTION 3 – SCREW FASTENINGS kb δb 0.000686 = Fe = 5000 ∆Fb = Fe 0.000686 + 0.00127 kb + k c δb + δc ∆Fb = 1754 lb Bolt: Ft = Fi + ∆Fb = 15,000 + 1754 = 16,754 lb F 16,754 sb = t = = 29,132 psi As 0.606 Member B and C kc Fc = Fi − Fe kb + k c δc Fc = Fi − Fe δ δ + b c 0.00127 Fc = 15,000 − 5000 = 11,754 lb 0.00127 + 0.000686 F 11,754 sc = c = = 2851 psi Ac 4.1234 (c) Fo = opening load δ + δc Fo = Fi i δi Fig. 5.6 237. 0.00127 + 0.000686 = 15,000 = 23,102 lb 0 . 00127 A 1-in. steel bolt is used to clamp two aluminum (2014-T6, HT aged) plates together as shown by Fig. 5.9, Text. The aluminum plates have a total thickness of 2 in. and an equivalent diameter of 2 in. The bolt is heated to a temperature of 200 oF, the inserted in the aluminum plates, which are at 80 oF, and tightened so as to have a tensile tightening stress of 30 ksi in the unthreaded shank while steel at 200 oF. What is the tensile stress in the bolt after assembly has cooled to 80 oF? The deformations are elastic. Figure 5.9 Page 14 of 42 SECTION 3 – SCREW FASTENINGS Solution: For aluminum 2014-T6 E = 10.6 × 106 psi s b = 30,000 psi π 2 Fi = s b Ab = (30,000 ) (1) = 23,562 lb 4 Steel bolt. Eb = 30 × 10 6 psi s L (30,000 )(2 ) δi = b = = 0.002 in. Eb 30 × 10 6 δc = Ac = Ac = Fi L . Ac E c π 4 π De2 − [(2) 4 2 π 4 D2 ] − (1) = 2.3562 sq.in. 2 Ec = 10.6 × 10 6 psi FL (23,562)(2) = 0.001887 in δc = i = Ac Ec (2.3562) 10.6 × 10 6 L′ = L − δ c = 2 − 0.001887 = 1.998113 in. ∆L = αL′∆t α = 0.000007 in. (in. − F ) for steel ∆L = (0.000007 )(1.998113)(80 − 200 ) = −0.001678 in. δ i′ = δ i + ∆L = 0.002 − 0.001678 = 0.000322 in. ( Page 15 of 42 ) SECTION 3 – SCREW FASTENINGS sb′ L Eb δ i′ = sb′ (2 ) 30 × 106 sb′ = 4830 psi 0.000322 = 238. A 1 1/8-in. steel bolt A passes through a yellow brass (B36-8) tube B as shown. The length of the tube is 30 in. (virtually the unthreaded bolt length), the threads on the bolt are UNC, and the tube’s cross-sectional area is 2 sq. in. After the nut is snug it is tightened ¼ turn. (a) What normal stresses will be produced in the bolt and in the tube? Assume that washers, nut, and head are rigid. (b) What are the stresses if an axial load of 5 kips is now applied to the bolts end? Compute the bolt load that just results in a zero stress in the tube. Prob. 238 Solution: For Yellow brass, B36-8, E = 15× 10 6 psi Steel bolt E = 30× 106 psi Table AT 14, 1 1/8 in., UNC Dr = 0.9497 in. As = 0.763 sq.in. Th in = 7 L = 30 in. θ= δi p 1 p = in. 7 1 θ = turn 4 1 1 1 δ i = = in. 4 7 28 Page 16 of 42 SECTION 3 – SCREW FASTENINGS δi = Fi L Ab Eb 1 Fi (30) = 28 π 1 2 6 1 30 × 10 4 8 Fi = 35,500 lb ( (a) Bolt: sb = ) Fi 35,500 = = 46,527 psi As 0.763 Fi Ac Ac = 2 sq.in. F 35,500 sc = i = = 17,750 psi Ac 2 Tube: sc = (b) Fe = 5000 lb ( ) Ac Ec (2 ) 15 × 106 kc = = = 1,000,000 lb in L 30 2 π 1 6 1 30 × 10 Ab Eb 4 8 kb = = = 994,000 lb in L 30 ( ) Bolts: kb Fe Ft = Fi + kb + k c 994,000 Ft = 35,500 + (5000 ) = 38,000 lb 994,000 + 1,000,000 F 38,000 st = t = = 49,800 psi As 0.763 Tube: kc Fe Fc = Fi − kb + kc 1,000,000 Fc = 35,500 − (5000 ) = 33,000 lb 994,000 + 1,000,000 F 33,000 sc = c = = 16,500 psi Ac 2 For zero stress in the tube Page 17 of 42 SECTION 3 – SCREW FASTENINGS k +k Fo = b c kc 994,000 + 1,000,000 Fi = (35,500) = 70,787 lb 1,000,000 ENDURANCE STRENGTH DESIGN PROBLEMS 239. As shown diagrammatically, a bearing is supported in a pillow block attached to an overhead beam by two cap screws, each of which, it may be assumed, carried half the total bearing load. This load acts vertically downward, varying from 0 to 1500 lb. The screws are to be made of AISI C1118, as rolled, and they are tightened to give an initial stress of about si = 0.5s y . The pillow block is made of class-20 cast iron. Assume that the effective length of screw is equal to the thickness t , as shown, and that the head and beam are rigid (overly conservative?). The equivalent diameter of the compression area may be taken as twice the bolt diameter. For a design factor of 1.75, determine the size of the screw: (a) from the Soderberg line, (b) from the modified Goodman line. (c) What size do you recommend using? Problem 239 Solution: For AISI C1118, as rolled s y = 46 ksi su = 75 ksi si = 0.5s y si = 0.5(46) = 23 ksi = 23,000 psi Fi = si As = 23 As kip = 23,000 As lb kb Fe ∆Fb = k + k b c Ab Eb kb = Lb Page 18 of 42 SECTION 3 – SCREW FASTENINGS Eb = 30× 10 6 psi (steel) Lb = t Ac Ec Lc For cast-iron class 20 Ec = 9.6 × 106 psi Lc = t kc = Ac = π 4 De2 − π 4 D2 π D2 4 De = 2 D Ab = Ac = π (2 D )2 − π D 2 = 3π D 2 = 3 Ab 4 AE kb = b b Lb ( 4 4 ) Ab 30× 10 6 t 3 A 9.6 × 106 kc = b t kb 30 ×10 6 300 = = 6 6 kb + k c 30 × 10 + 3(9.6 × 10 ) 588 ∆Fb1 = 0 kb = ( ) kb 300 1500 Fe = ∆Fb 2 = = 383 lb 588 2 k b + kc 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 23,000 As + (383 + 0) = 23,000 As + 192 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (383 − 0) = 192 lb 2 2 s n = 0.5su = 0.5(75,000 psi ) = 37,500 psi For axial loading with size factor s n = (0.8)(0.85)(37,500 psi ) = 25,500 psi N = 1.75 sm = Fm 23,000 As 192 192 = + = 23,000 + As As As As Page 19 of 42 SECTION 3 – SCREW FASTENINGS sa = Fa 192 = As As Table AT 12, K f = 1.8 (a) Soderberg line 1 sm K f sa = + N sy sn 192 192 23,000 + (1.8) As As 1 = + 1.75 46,000 25,500 As = 0.2482 sq.in. Table AT 14, UNC 3 Use D = in. , As = 0.334 sq.in. 4 (b) Modifies Goodman line 1 sm K f sa = + N su sn 192 192 23,000 + (1.8) As As 1 + = 1.75 75,000 25,500 As = 0.0609 sq.in. Table AT 14, UNC 3 Use D = in. , As = 0.0775 sq.in. 8 (c) Recommended, D = 240. 3 in. − UNC 4 A connection similar to Fig. 5.9, Text, is subjected to an external load that varied from 0 to 1250 lb. The bolt is cold forged from AISI B1113 steel; UNC threads.The aluminum parts C (3003 H14) have a total thickness of 1 ½ in. and an external diameter of 2 D . It is desired that the connection not open for an external load of 1.5Fe . Determine (a) the initial tensile load on the bolt, (b) the bolt diameter for N = 2 based on the Soderberg line. Fig. 5.9 Page 20 of 42 SECTION 3 – SCREW FASTENINGS Solution: kc lb (a) Fi = QFe k + k b c Q = 1.5 AE kb = b b Lb Ab = π D2 4 Eb = 30 × 10 6 psi 1 Lb = 1 in. 2 Ac Ec kc = Lc Ac = π De2 − 4 De = 2 D Ac = π π 4 D2 (2 D )2 − π D 2 = 3π D 2 = 3 Ab 4 4 4 6 Ec = 10 × 10 psi (3003-H14 aluminum) 1 Lc = 1 in. 2 A 30 × 10 6 kb = b 1 1 2 3 A 10 × 10 6 kc = b 1 1 2 kc 3 10 × 10 6 = = 0.5 kb + k c 30 ×10 6 + 3 10 × 10 6 ( ) ( ) ( Page 21 of 42 ( ) ) SECTION 3 – SCREW FASTENINGS Fe = 1250 lb kc lb Fi = QFe k + k b c Fi = (1.5)(1250 )(0.5) = 937.5 lb (b) For AISI B1113 steel, cold forged su = 83 ksi s y = 72 ksi s n = 0.5su = 0.5(83) = 41.5 ksi = 41,500 psi For axial loading with size factor s n = (0.8)(0.85)(41,500 psi ) = 28,220 psi kb Fe ∆Fb = kb + k c ∆Fb1 = 0 kb 30 × 106 Fe = (1250 ) = 625 lb ∆Fb 2 = 6 6 30 × 10 + 3(10 × 10 ) kb + k c 1 (∆Fb 2 + ∆Fb1 ) = 937.5 + 1 (625 + 0) = 1250 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (625 − 0) = 312.5 lb 2 2 F 1250 sm = m = As As F 312.5 sa = a = As As Fm = Fi + Soderberg line, K f = 1.8 Table AT 12 1 sm K f sa = + N sy sn 1 1250 (1.8)(312.5) = + 2 72,000 As 28,220 As As = 0.07459 sq.in. Table AT 14, UNC Page 22 of 42 SECTION 3 – SCREW FASTENINGS 3 in. , As = 0.0775 sq.in. 8 Use D = 243. This problem concerns the Freon compressor of 226: size, 10 x 18 in.; 10 studs, UNC; made of C1118, as rolled; 200 psi gas pressure. The initial tension in the bolts, assumed to be equally loaded, is such that a cylinder pressure of 300 psi is required for the joint to be on the opening. The bolted parts are cast steel and for the first calculations, it will be satisfactorily to assume the equivalent diameter of the compressed parts to be twice the bolt size. (a) For N = 2 on the Soderberg criterion, what bolt size is required? (b) Compute the torque required for the specified initial tension. Solution: π (10 ) Fo = 300 = 2356 lb. 4 10 kc Fi = Fo kb + k c 2 kb = Ab Eb Lb π D2 4 Eb = 30 × 10 6 psi Ab = Lb = L AE kc = c c Lc Ac = π De2 − 4 De = 2 D Ac = π π 4 D2 (2 D )2 − π D 2 = 3π D 2 = 3 Ab 4 4 4 6 Cast Steel, Ec = 30 × 10 psi Lc = L Ab (30× 10 6 ) L 3 Ab 30 × 106 kc = = 3kb L kc 3kb = (2356 ) = 1767 lb Fi = Fo kb + k c kb + 3kb kb = ( Page 23 of 42 ) SECTION 3 – SCREW FASTENINGS kb Fe (a) ∆Fb = kb + k c ∆Fb1 = 0 kb ∆Fb 2 = k b + kc k b π (10 )2 (200 ) Fe = = 393 lb 10 kb + 3kb 4 1 (∆Fb 2 + ∆Fb1 ) = 1767 + 1 (393 + 0 ) = 1964 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (393 − 0) = 196 lb 2 2 F 1964 sm = m = As As F 196 sa = a = As As For C1118, as rolled su = 75 ksi Fm = Fi + s y = 46 ksi s n = 0.5su = 0.5(75) = 37.5 ksi = 37,500 psi For axial loading with size factor s n = (0.8)(0.85)(37,500 psi ) = 25,500 psi K f = 1.8 Table AT 12 1 sm K f sa = + N sy sn 1 1964 (1.8)(196) = + 2 46,000 As 25,500 As As = 0.1131 sq.in. Table AT 14, UNC 1 Use D = in. , As = 0.1419 sq.in. 2 (b) T = 0.2 DFi 1 T = 0.2 (1767 ) = 176.7 in − lb. 2 245. A cast-iron (class 35) Diesel-engine cylinder head is held on 8 stud bolts with UNC threads. These bolts are made of AISI 3140 steel, OQT 1000 oF (Fig. AF2). Assume that the compressed material has an equivalent diameter twice Page 24 of 42 SECTION 3 – SCREW FASTENINGS the bolt size. The maximum cylinder pressure is 750 psi and the bore of the engine is 8 in. Let the initial bolt load be such that a cylinder pressure of 1500 psi brings the joint to the point of opening. For a design factor of 2, determine the bolt diameter (a) using the Soderberg equation, (b) using the Goodman equation. (c) What approximate torque will be required to induce the desired initial stress? (d) Determine the ratio of the initial stress to the yield strength. Considering the lessons of experience (i5.8), what initial stress would you recommend? Using this value, what factor of safety is computed from the Soderberg equation? Solution: π (8) Fo = 1500 = 9425 lb. 4 8 kc Fi = Fo kb + k c 2 kb = Ab Eb Lb π D2 4 Eb = 30 × 10 6 psi Ab = Lb = L AE kc = c c Lc Ac = π De2 − 4 De = 2 D Ac = π π 4 D2 (2 D )2 − π D 2 = 3π D 2 = 3 Ab 4 4 4 6 Ec = 14.5 × 10 psi , for cast-iron (class 35) Lc = L ( ) Ab 30× 10 6 L 3 Ab 14.5 × 106 kc = L kc 3(14.5 × 10 6 ) = (9425) = 5578 lb Fi = Fo 6 6 30 × 10 + 3(14.5 × 10 ) kb + k c kb = ( Page 25 of 42 ) SECTION 3 – SCREW FASTENINGS kb Fe ∆Fb = kb + k c ∆Fb1 = 0 kb π (8)2 (750 ) 30 × 10 6 Fe = ∆Fb 2 = = 1923 lb 6 6 8 30 × 10 + 3(14.5 × 10 ) 4 kb + k c 1 (∆Fb 2 + ∆Fb1 ) = 5578 + 1 (1923 + 0) = 6540 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (1923 − 0) = 962 lb 2 2 Fm = Fi + Fm 6540 = As As F 962 sa = a = As As (a) For AISI 3140 steel, OQT 1000 oF su = 153 ksi s y = 134 ksi sm = s n = 0.5su = 0.5(153) = 76.5 ksi = 76,500 psi For axial loading with size factor s n = (0.8)(0.85)(76,500 psi ) = 52,000 psi Table AT 12, K f = 3.3 (hardened) Soderberg Equation 1 sm K f sa = + N sy sn 1 6540 (3.3)(962) = + 2 134,000 As 52,000 As As = 0.2197 sq.in. Table AT 14, UNC 5 Use D = in. , As = 0.226 sq.in. 8 (b) Goodman Equation 1 sm K f sa = + N su sn Page 26 of 42 SECTION 3 – SCREW FASTENINGS (3.3)(962) 1 6540 = + 2 153,000 As 52,000 As As = 0.2076 sq.in. Table AT 14, UNC 5 Use D = in. , As = 0.226 sq.in. 8 (c) T = 0.2 DFi 5 T = 0.2 (5578) = 697 in − lb. 8 Fi 5578 = = 24,681 psi As 0.226 s 24,681 Ratio = i = = 0.184 s y 134,000 (d) si = i5.8 si = 0.85s y = 0.85(134,000) = 113,900 psi Factor of safety Fi = si As = (113,900 )(0.226 ) = 25,742 lb 1 Fm = 25,742 + (1923) = 26,704 lb 2 1 Fa = (1923) = 962 lb 2 Fm 26,704 sm = = = 118,159 psi As 0.226 F 962 sa = a = = 4257 psi As 0.226 Soderberg Equation 1 sm K f sa = + N sy sn 1 118,159 (3.3)(4257 ) = + N 134,000 52,000 N = 0.87 246. A 30,000-lb. body is to be mounted on a shaker (vibrator). The shaker will exert a harmonic force of F = 30,000 sin 2π t f lb. on the body where f cps is the frequency and t sec. is the time. The frequency can be varied from 5 to 10,000 cps. The harmonic force will exert a tensile load on the bolts that Page 27 of 42 SECTION 3 – SCREW FASTENINGS attach the body to the shaker when F is positive. Determine the minimum number of ½-in.-UNF bolts that must be used for N = 2 based on Soderberg line. The material of the bolts is to be AISI 8630, WQT 1100 oF; the material of the body that is to be vibrated is aluminum alloy, 2014-T6 and the joint is not to open for an external force that is 1.25 times the maximum force exerted by the shaker. It may be assumed that the equivalent diameter of the material in compression is twice the bolt diameter. Solution: Fe min = 0 Fe max = 30,000 lb kc Fi = QFe k + k b c Q = 1.25 AE kb = b b Lb Ab = π D2 4 Eb = 30 × 10 6 psi Lb = L AE kc = c c Lc Ac = π De2 − 4 De = 2 D Ac = π π 4 D2 (2 D )2 − π D 2 = 3π D 2 = 3 Ab 4 4 4 6 Ec = 10.6 × 10 psi , (Aluminum 2014-T6) Lc = L Page 28 of 42 SECTION 3 – SCREW FASTENINGS Ab Eb Ab (30× 106 ) kb = = Lb L Ac Ec 3 Ab (10.6 × 106 ) kc = = Lc L kc Fi = QFe kb + k c ( ) 3 10.6 × 10 6 Fi = (1.25)(30,000 ) = 19, 296 lb 6 6 30 × 10 + 3 10 . 6 × 10 ( ) kb Fe ∆Fb = k + k b c ∆Fb1 = 0 kb 30 × 106 Fe = (30,000 ) = 14,563 lb ∆Fb 2 = 6 6 k + k 30 × 10 + 3 ( 10 . 6 × 10 ) b c 1 (∆Fb 2 + ∆Fb1 ) = 19,296 + 1 (14,563 + 0) = 26,578 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (14,563 − 0 ) = 7282 lb 2 2 Fm = Fi + Fm nAs F sa = a nAs For ½-in.-UNF (Table AT 14) As = 0.1419 sq.in. F 26,578 187,300 sm = m = = nAs 0.1419n n F 7282 51,318 sa = a = = nAs 0.1419n n sm = For AISI 8630, WQT 1100 oF K f = 3.3 su = 137 ksi s y = 125 ksi s n = 0.5su = 0.5(137 ) = 68.5 ksi = 68,500 psi For axial loading with size factor Page 29 of 42 SECTION 3 – SCREW FASTENINGS s n = (0.8)(0.85)(68,500 psi ) = 46,580 psi Soderberg Equation, N = 2 1 sm K f sa = + N sy sn 1 187,300 (3.3)(51,318) = + 2 125,000n 46,580n n = 10.3 Minimum number of bolts = 10 bolts 248. The maximum external load on the cap bolts of an automotive connecting rod end, imposed by inertia forces at top dead center, is taken to be 4000 lb.; the minimum load is zero at bottom dead center. The material is AISI 4140, OQT 1100oF (qualifying for SAE grade 5); assume that s n′ = 0.45su . The grip for through bolts is 1.5 in. For design purposed, let each bolt take half the load, 3 and use an equivalent De = 1 in. for the connected parts. The threads extend 8 a negligible amount into the grip. For the initial computation, use an opening load Fo = 1.75 Fe . Considering the manner in which the bolt is loaded, we decide that a design factor of 1.4 (Soderberg) should be quite adequate. (a) Does a 5/16-24 UNF satisfy this situation? If not, what size do you recommend? (b) Experience suggests that, in situations such as this, an initial stress of the order suggested in i5.8, Text, is good insurance against fatigue failure. Decide upon such an si and recomputed N . How does it change? Would you be concerned about the safety in this case? Consider the variation of si as a consequences of the use of torque wrench and also the stress relaxation with time (due to seating and other factors), and discuss. Compute the required tightening torque for each si . Solution: Fo = 1.75Fe = 1.75(4000) = 7000 lb kc Fi = Fe kb + kc AE kb = b b Lb Ab = π D2 4 Eb = 30 × 10 6 psi Page 30 of 42 SECTION 3 – SCREW FASTENINGS Lb = 1.5 in. AE kc = c c Lc Ac = π 4 De2 − π 3 π 4 2 D2 π π 2 2 1 − D = 1.485 − D = 1.485 − Ab 4 8 4 4 Ec = 10.6 × 106 psi , (Aluminum 2014-T6) Ac = Lc = 1.5 in. As ≈ Ab Ab Eb Ab (30 × 106 ) kb = = Lb 1.5 Ac Ec (1.485 − As )(30 × 106 ) kc = = Lc 1.5 1.485 − As Fi = 7000 = 7000 − 4714 As 1.485 kb Fe ∆Fb = kb + k c ∆Fb1 = 0 kb A Fe = s (4,000 ) = 2694 As ∆Fb 2 = 1.485 kb + k c 1 (∆Fb 2 + ∆Fb1 ) = 7000 − 4714 As + 1 (2694 As + 0) = 7000 − 3367 As 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As 2 2 Fm = Fi + Table 5.2 su = 120 ksi s y = 88 ksi s n′ = 0.45su = 0.45(120 ) = 54 ksi = 54,000 psi F 7000 sm = m = − 3367 As As F s a = a = 1347 As Page 31 of 42 SECTION 3 – SCREW FASTENINGS K f = 3.3 (hardened, Table AT 12) Soderberg Equation, N = 1.4 1 sm K f sa = + N sy sn 1 7000 3367 (3.3)(1347 ) = − + 1.4 88,000 As 88,000 54,000 As = 0.1187 sq.in. 7 Table At14, we D = in , As = 0.1187 sq.in. 16 (a) 5/16-24 UNF will not satisfy the situation. Instead use D= 7 in , 16 As = 0.1187 sq.in. (b) i5.8, Text si = 0.85s y = 0.85(88) = 74.8 ksi = 74,800 psi Fi = si As = (74,800)(0.1187 ) = 8879 lb 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 8879 + (2694 As + 0 ) = 8879 + 1347 As 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As 2 2 F 8879 sm = m = + 1347 As As F s a = a = 1347 As 1 sm K f sa = + N sy sn 8879 + 1347 1 0.1187 + (3.3)(1347 ) = N 88,000 54,000 N = 1.06 , it decreases N > 1 , therefore, safe. Considering variation of si , si tends to exceeds the limiting stress therefore reduces the factor of safety. While stress relaxation tends to reduce the limiting stress approaching the si and causing lower design factor. (c) Fi = 7000 − 4714 As = 7000 − 4714(0.1187 ) = 6440 lb Page 32 of 42 SECTION 3 – SCREW FASTENINGS 7 T = 0.2 DFi = 0.2 (6440) = 564 in − lb 16 at Fi = 8879 lb 7 T = 0.2 DFi = 0.2 (8879 ) = 777 in − lb 16 CHECK PROBLEMS 249. A 1-in. steel bolt A (normalized AISI 1137, cold-rolled threads) passes through a yellow brass tube B (B36-8, ½ hard) as shown. The tube length is 30 in., its cross-sectional area is 2 sq. in. and the UNC bolt threads extend a negligible amount below the nut. The steel washers are ¼ in. thick and are assumed not to bend (clearances are exaggerated). The nut is turned ¼ turn. (a) If an external tensile axial load, varying from 0 to 5 kips, is repeatedly applied to the bolt, what is the factor of safety of the bolt by the Soderberg criterion? (b) What is the external load on the bolt at the instant that the load on the tube becomes zero. Problem 249, 250 Solution: For 1-in. UNC As = 0.606 sq.in. Dr = 0.8466 in. Th in. = 8 1 1 = in. Th in 8 δ 1 = p 4 1 1 1 δ = = in. 4 8 32 p= Page 33 of 42 SECTION 3 – SCREW FASTENINGS 1 Lb = 30 + 2 = 30.5 in. 4 Eb = 30× 10 6 psi 1 Fi (30.5) = δ= π 32 (1)2 30 ×106 4 Fi = 24,141 lb ( ) kb ∆Fb = Fe kb + k c π Ab Eb 4 = Lb AE kc = c c Lc Ac = 2 sq.in. Lc = 30 in. (1)2 (30 ×106 ) kb = 30.5 = 772,525 Ec = 15× 106 psi (Yellow Brass) ( ) 2 15 ×10 6 kc = = 1,000,000 30 (a) ∆Fb1 = 0 772,525 ∆Fb 2 = (5000 ) = 2179 lb 772,525 + 1,000,000 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 24,141 + (2179 + 0 ) = 25,230 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2179 − 0) = 1090 lb 2 2 Fm 25,230 sm = = = 38,227 psi As 0.606 F 1090 sa = a = = 1799 psi As 0.606 For normalized AISI 1137, cold-rolled thread su = 98 ksi s y = 58 ksi s n = 0.5su = 0.5(98) = 49 ksi = 49,000 psi K f = 1.4 (Table AT 12) Page 34 of 42 SECTION 3 – SCREW FASTENINGS For axial loading, s n = 0.8(49,000 ) = 39,200 psi 1 sm K f sa = + N sy sn 1 38,227 (1.4 )(1799) = + N 58,000 39,200 N = 1.38 kc (b) Fi = Fo kb + k c 1,000,000 24,141 = Fo 772,525 + 1,000,000 Fo = 42,790 lb 250. A ¾-in. fine-thread bolt, made of AISI 1117, cold drawn, with rolled threads, passes through a yellow brass tube and two steel washers, as shown. The tube is 4 in. long, 7/8 in. internal diameter, 1 ¼-in. external diameter. The washers are each ¼-in. thick. The unthreaded part of the bolt is 3 in. long. Assume that there is no stretching of the bolt inside the nut in finding its k . The unlubricated bolt is tightened by a torque of 1800 in-lb. The external load, varying from 0 to 4 kips, is axially applied to the washers an indefinite number of times. (a) Compute the factor of safety of the bolt by the Soderberg criterion. Is there any danger of failure of the bolt? (b) What pull must be exerted by the washers to remove all load from the brass tube? Solution: T = 0.2 DFi 3 1800 = 0.2 Fi 4 Fi = 12,000 lb kb ∆Fb = Fe kb + k c 1 1 1 = + kb kb1 kb 2 AE kb1 = b b Lb1 Lb1 = 3 in. Ab = π 3 2 = 0.4418 sq.in. 4 4 Page 35 of 42 SECTION 3 – SCREW FASTENINGS Eb = 30× 10 6 psi (0.4418)(30 ×106 ) = 4,418,000 kb1 = 3 As Eb Lb 2 For ¾-in. UNF (Table AT 14) As = 0.373 sq.in. kb 2 = 1 Lb 2 = 4 + 2 − 3 = 1.5 in. 2 (0.373) 30 ×106 = 7,460,000 kb 2 = 1.5 1 1 1 = + kb kb1 kb 2 1 1 1 = + kb 4,418,000 7,460,000 kb = 2,774,733 AE kc = c c Lc ( ) 2 2 π 1 7 1 − = 0.6259 sq.in. 4 4 8 Ec = 15× 106 psi Ac = Lc = 4 in. kc = (0.6259)(15 ×106 ) = 2,347,125 4 ∆Fb1 = 0 2,774,733 ∆Fb 2 = (4000) = 2167 lb 2,774,733 + 2,347,125 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 12,000 + (2167 + 0) = 13,084 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (2167 − 0) = 1084 lb 2 2 F 13,084 sm = m = = 35,078 psi As 0.373 F 1084 sa = a = = 2906 psi As 0.373 For AISI 111, cold drawn, rolled threads s n = 40 ksi Page 36 of 42 SECTION 3 – SCREW FASTENINGS s y = 68 ksi K f = 1.4 s n = 0.8(40) = 32 ksi = 32,000 psi , axial loading (a) 1 sm K f sa = + N sy sn 1 35,078 (1.4 )(2906 ) = + N 68,000 32,000 N = 1.56 kc (b) Fi = Fo kb + k c 2,347,125 12,000 = Fo 2,774,733 + 2,347,125 Fo = 26,186 lb A coupling bolt (i5.13, Text) is used to connect two parts made of cast-iron, 251. class 35. The diameter of the coarse-thread bolt is ½-in.; its grip is 2 in., which is also nearly the unthreaded length. The bolt tightened to have an initial tension of 4000 lb. The parts support an external load Fe that tends to separate them and it varies from zero to 5000 lb. What is the factor of safety, (Soderberg)? Solution: Fi = 4000 lb kb ∆Fb = Fe kb + k c AE kb = b b Lb π 1 2 = 0.19635 sq.in. (unthreaded length) 4 2 Eb = 30× 10 6 psi Ab = Lb = 2 in. kb = (0.19635)(30 ×106 ) = 2,945,250 2 Table AT 14, UNC Page 37 of 42 SECTION 3 – SCREW FASTENINGS 1 in. 2 As = 0.1419 sq.in. 3 A = in. 4 h De = A + 2 h = 2 in. 3 2 3 De = + = 1 in. 4 2 4 AE kc = c c Lc D= Ac = π 4 De2 − π 4 D2 = 2 2 π 3 1 1 − = 2.209 sq.in. 4 4 2 Ec = 14.5 × 106 psi , (Cast iron, class 35) Lc = 2 in. kc ( 2.209 )(14.5 × 106 ) = = 16,015,250 2 ∆Fb1 = 0 2,945,250 ∆Fb 2 = (5000 ) = 777 lb 2,945,250 + 16,015,250 1 1 Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 4000 + (777 + 0 ) = 4389 lb 2 2 1 1 Fa = (∆Fb 2 − ∆Fb1 ) = (777 − 0) = 389 lb 2 2 Fm 4389 sm = = = 30,930 psi As 0.1419 F 389 sa = a = = 2741 psi As 0.1419 1 For ASTM 354 BC (Table 5.2), D = in. 2 su = 125 ksi s y = 109 ksi s n = 0.5su For axial loading s n = (0.8)(0.5)(125) = 50 ksi = 50,000 psi K f = 1.8 Soderberg Line Page 38 of 42 SECTION 3 – SCREW FASTENINGS 1 sm K f sa = + N sy sn 1 30,930 (1.8)(2741) = + N 109,000 50,000 N = 2. 6 252. The cap on the end of a connecting rod (automotive engine) is held on by two 5/16-in. bolts that are forged integrally with the main connecting rod. These bolts have UNF threads with a 5/8-in. on an unthreaded length of virtually 5/8 in. The nuts are to be tightened with a torque of 20 ft-lb. and the maximum external load on one bolt is expected to be 2330 lb. Let the equivalent diameter of the connected parts be ¾ in. (a) Estimate the maximum force on the bolt. (b) Compute the opening load. Is this satisfactory? (c) If the bolt material is AISI 4140, OQT 1000 oF, what is the factor of safety based on the Soderberg criterion? Solution: T = 20 ft − lb = 240 in − lb T = 0.2 DFi 5 240 = 0.2 Fi 16 Fi = 3840 lb kb ∆Fb = Fe kb + k c AE kb = b b Lb π5 2 = 0.0767 sq.in. (unthreaded length) 4 16 Eb = 30× 10 6 psi 5 Lb = in. 8 (0.0767) 30 ×106 = 3,681,600 kb = 5 8 AE kc = c c Lc Ab = ( π π ) 2 2 π 3 5 − = 0.3651 sq.in. 4 4 16 Ec = 30× 10 6 psi , (Cast iron, class 35) Ac = 4 De2 − 4 Page 39 of 42 D2 = SECTION 3 – SCREW FASTENINGS 5 in. 8 ( 0.3651) 30 × 10 6 kc = = 17,524,800 5 8 3,681,600 ∆Fb = (2330) = 405 lb 3,681,600 + 17,524,800 Lc = ( ) (a) Fmax = Fi + ∆Fb = 3840 + 405 = 4245 lb kc (b) Fi = Fo k + k b c 17,524,800 3840 = Fo 3,681,600 + 17,524,80 Fo = 4647 lb < Fmax ∆Fb 405 = 3840 + = 4042 lb 2 2 ∆F 405 Fa = b = = 202 lb 2 2 For AISI 4140, OQT 1000 oF su = 170 ksi (c) Fm = Fi + s y = 155 ksi Table AT 12, K f = 2.6 s n = 0.5su For axial loading s n = (0.8)(0.5)(170) = 68 ksi = 68,000 psi Soderberg Line 1 sm K f sa = + N sy sn For 5/16-in.-UNF, Table AT 14, As = 0.0580 sq.in. F 4042 sm = m = = 69,690 psi As 0.0580 F 202 sa = a = = 3843 psi As 0.0580 1 sm K f sa = + N sy sn Page 40 of 42 SECTION 3 – SCREW FASTENINGS 1 69,690 (2.6 )(3483) = + N 155,000 68,000 N = 1.72 SET SCREWS 254. A 6-in. pulley is fastened to a 1 ¼ in. shaft by a set screw. If a net tangential force of 75 lb, is applied to the surface of the pulley, what size screw should be used when the load is steady? Solution: 6 Tangential force = (75 lb ) = 365 lb 1.25 Assume tangential force = holding force Table 5.3, use Screw size 8, Holding force = 385 lb. 255. An eccentric is to be connected to a 3-in. shaft by a setscrew. The center of the eccentric is 1 ¼ in. from the center of the shaft when a tensile force of 1000 lb. is applied to the eccentric rod perpendicular to the line of centers. What size set screw should be used for a design factor of 6? Solution: 1.25 = 833 lb Tangential force = (1000 lb ) 32 Holding force = (6)(833) = 5000 lb Table 5.3, use Screw size ¾ in. 256. A lever 16 in. long is to be fastened to a 2-in. shaft. A load of 40 lb. is to be applied normal to the lever at its end. What size of set screw should be used for a design factor of 5? Solution: Torque = (16)(40) = 640 in − lb 2(640 ) Tangential force = = 640 lb 2 Holding force = (5)(640 ) = 3200 lb Table 5.3, use Screw size 9/16 in. 257. A 12-in. gear is mounted on a 2-in. shaft and is held in place by a 7/16 in. setscrew. For a design factor of 3, what would be the tangential load that could be applied to the teeth and what horsepower could be transmitted by the screw. Solution: Page 41 of 42 SECTION 3 – SCREW FASTENINGS Table 5.3, 7/16 in. Holding force = 2500 lb 2500 Tangential force = = 833 lb 3 2 Tangential load on gear = 833 = 139 lb 12 Assume vm = 4500 fpm (139)(4500) = 19 hp Hp transmitted = 33,000 - end - Page 42 of 42 SECTION 4 - SPRINGS HELICAL COMPRESSION SPRINGS DESIGN – LIGHT, MEDIUM SERVICE 271. A solenoid brake (Fig. 18.2, Text) is to be actuated by a helical compression spring. The spring should have a free length of approximately 18 in. and is to exert a maximum force of 2850 lb. when compressed to a length of 15 in. The outside diameter must not exceed 7 in. Using oil-tempered wire, design a spring for this brake, (wire diameter, coil diameter, number of active coils, pitch, pitch angle, “solid stress”). General Electric used a spring made of 1 in. wire, with an outside diameter of 6 in., and 11 ½ free coils for a similar application. Solution: For oil tempered wire, Table AT 17 146 su = 0.19 ksi , [0.032 < Dw < 0.5] Dw “solid stress” = 0.6 s u design stress, (average service) ssd = 0.324 su 0.324(146 ) 47.304 ssd = = ksi Dw0.19 Dw0.19 Dw + Dm ≤ 7 F = 2850 lb = 2.85 kips 8FDm 47.304 = ss = K 3 Dw0.19 πDw say K = 1.3 8(2.85)(7 − Dw ) 47.304 ss = 1.3 = πDw3 Dw0.19 Dw = 1.062 in > 0.5 in 47.304 use ssd = ksi = 54 ksi (0.5)0.19 8(2.85)(7 − Dw ) ss = 1.3 = 54 πDw3 Dw = 1.015 in say Dw = 1.0 in 8(2.85)Dm ss = 1.3 = 54 3 π (1) Dm = 5.72 in Page 1 of 70 SECTION 4 - SPRINGS say Dm = 5.0 in OD = Dm + Dw = 5.0 + 1.0 = 6 in < 7 in. D 5 .0 C= m = =5 Dw 1.0 δ = Free length – Compressed length = 18 in – 15 in = 3 in. δ= 8FC 3 N c GDw G = 10,500 ksi , Dw > 3 in 8 8(2.85)(5) N c (10,500)(1) N c = 11.05 say N c = 11.5 3 δ =3= 8(2.85)(5) (11.5) = 3.12 in (10,500)(1) Free length = 15 + 3.12 = 18.12 in 3 δ= At 2.85 kips 8FDm ss = K 3 πDw C =5 4C − 1 0.615 4(5) − 1 0.615 K= + = + = 1.3105 4C − 4 C 4(5) − 4 5 8(2.85)(5) ss = 1.3105 = 47.55 ksi 3 π (1) Permissible solid stress 0.6(146 ) sso = 0.6 su = ksi = 99.93 ksi (0.5)0.19 F k= δ using or let δ T = Free length – Solid height 47.55 99.93 = 3.12 δT δ T = 6.56 in δ T = Free length – Solid height = (P − Dw )N c 6.56 = (P − 1)(11.5) P = 1.570 in Page 2 of 70 SECTION 4 - SPRINGS 1 use P = 1 in 2 Pitch angle, 1.5 P o o λ = tan −1 = tan −1 = 5.5 < 12 , o.k. πD π ( 5 ) For actual solid stress δ T = (1.5 − 1)(11.5) = 5.75 in. 47.55 sso = 3.12 5.75 sso = 87.63 ksi < 99.93 ksi , ok Summary of answer: Dw = wire diameter = 1 in. Dm = coil diameter = 5 in. N c = no. of active coils = 11 1/2 P = pitch = 1 ½ in. γ = pitch angle = 5.5o sso = solid stress = 87.63 ksi 272. A coil spring is to be used for the front spring of a automobile. The spring is to have a rate of 400 lb./in., an inside diameter of 4 3/64 in., and a free length of 14 1/8 in., with squared-and-ground ends. The material is to be oiltempered chrome vanadium steel. Decide upon the diameter of the wire and the number of free coils for a design load of F = 1500 lb . Be sure “solid stress” is all right. How much is the pitch angle? Solution: Table AT 17 Cr-V steel 168 su = 0.166 ksi , [0.032 < Dw 0.437] Dw average service ssd = 0.324su 0.324(168) 54.432 ssd = = 0.166 ksi Dw0.166 Dw Max “solid stress” = 0.6su Page 3 of 70 SECTION 4 - SPRINGS 3 in = 4.046875 in 64 Dm = Dw + 4.046875 in ID = Dm − Dw = 4 8FDm = ssd ss = K 3 πDw Assume K = 1.3 F = 1500 lb = 1.5 kips ssd = 8(1.5)(Dw + 4.046875) 54.432 = 1.3 0.166 Dw πDw3 Dw = 0.747 in > 0.437 in use 54.432 ssd = ksi = 62.45 ksi (0.437 )0.19 8(1.5)(Dw + 4.046875) ssd = 1.3 = 62.45 πDw3 Dw = 0.724 in 3 use Dw = in 4 3 3 51 Dm = + 4 = 4 in 4 64 64 8FDm ss = K 3 πDw 4C − 1 0.615 K= + 4C − 4 C 51 4 Dm 64 C= = ≈ 6.4 Dw 3 4 4(6.4) − 1 0.615 K= + = 1.235 4(6.4) − 4 6.4 51 8(1.5) 4 64 = 53.64 ksi < 62.45 ksi , (o.k.) ss = 1.235 3 3 π 4 δ= 8FC 3 N c GDw Page 4 of 70 SECTION 4 - SPRINGS 3 G = 10,500 ksi , D w > in 8 F 1500 δ= = = 3.75 in k 400 3 8(1.5)(6.4 ) N c δ = 3.75 = (10,500) 3 4 N c = 9.4 Table AT 16, Total coils = N c + 2 = 9.4 + 2 = 11.4 for square and grounded end. Summary of answer: Dw = wire diameter = ¾ in. No. of free coils = 11.4 To check for solid stress. Permissible solid stress = 0.6(168) = 115.65 ksi (0.437 )0.166 Free length = PN c + 2 Dw 3 Solid height = Dw ( N c + 2 ) = (11.4 ) = 8.55 in 4 1 14 − 8.55 = 78.74 ksi < 115.65 ksi (safe) Solid stress = (53.64 ) 8 3.75 Pitch: 1 PN c + 2 Dw = 14 in 8 1 3 P(9.4 ) + 2 = 14 8 4 11 P = 1.343 in = 1 in 32 Pitch angle, 11 1 −1 P −1 = tan 32 = 5.1o < 12o , o.k. λ = tan πD π 4 51 64 Page 5 of 70 SECTION 4 - SPRINGS 273. A coiled compression spring is to fit inside a cylinder 5/8 in. in diameter. For one position of the piston, the spring is to exert a pressure on the piston equivalent to 5 psi of piston area, and in this position, the overall length of the spring must not exceed (but may be less than) 2 in. A pressure of 46 psi on the piston is to compress the spring ¾ in. from the position described above. Design a spring for medium service. Specify the cheapest suitable material, number of total and active coils for square-and-ground ends, and investigate the pitch angle, and “solid stress”. Solution: 8FDm ss = K 3 D π w OD = Dm + Dw = Dm + 1.5 Dw = 5 D in − w 8 2 5 in 8 2 π 5 F1 = (5) = 1.534 lb 4 8 2 π 5 F2 = (46 + 5) = 15.647 lb 4 8 Using hard-drawn spring wire, Cost Index = 1 ssd = 0.324su (0.85) 140 su = 0.19 ksi Dw , [0.028 < Dw < 0.625] 70 ksi Dw0.19 140 38.556 ssd = 0.324(0.85) 0.19 = Dw Dw0.19 Max “solid stress” = 8FC 38.556 38,556 = ss = K ksi = 0.19 psi 2 0.19 Dw Dw πDw 8(15.647 )C K = 38,556 Dw1.81 π K (39.845)C = 38,556 Dw1.81 Dm + 1.5Dw = 0.625 CDw + 1.5Dw = 0.625 0.625 Dw = C + 1 .5 Page 6 of 70 SECTION 4 - SPRINGS K= 4C − 1 0.615 + 4C − 4 C 1.81 4C − 1 0.615 0.625 + (39.845)C = 38,556 C 4C − 4 C + 1 .5 4C − 1 0.615 1.81 + (C + 1.5) C = 413.3 C 4C − 4 C = 7.035 0.625 0.625 = = 0.0732 in Dw = C + 1.5 7.035 + 1.5 Table AT 15, Dw = 0.0720 in , W & M 15 Dm = 7.035(0.0720) = 0.5065 in For N c δ 2 − δ1 = 8(F2 − F1 )C 3 N c GDw G = 11.5 × 106 psi 3 8(15.647 − 1.534)(7.035) N c δ 2 − δ1 = = 4 11.5 × 106 (0.0720) N c = 15.8 Table AT 16, Total coils = N c + 2 = 15.8 + 2 = 17.8 Solid height = ( N c + 2)Dw = (15.8 + 2)(0.0720) = 1.28 in Free length = PN c + 2 Dw Free length = 2 + δ1 3 ( δ1 = ) 8(F1 )C 3 N c GDw 8(1.534 )(7.035) (15.8) = 0.082 in. 11.5 × 106 (0.0720) Free length = 2 + 0.082 = 2.082 in 3 δ1 = δ2 = ( ) 8(F2 )C 3 N c GDw 8(15.647 )(7.035) (15.8) = 0.832 in. 11.5 × 106 (0.0720) Solid Height ≤ Free Length - δ 2 Solid Height ≤ 2.082 − 0.832 in Solid Height ≤ 1.25 in But Solid Height > 1.25 in. 3 δ2 = ( ) Therefore change material to Oil-tempered spring wire, Cost Index = 1.5 Page 7 of 70 SECTION 4 - SPRINGS Table AT 17 146 su = 0.19 ksi , 0.028 < Dw < 0.5 Dw Max “solid stress” = ssd = 0.324 87.5 ksi D w0.19 146 47.304 = Dw0.19 Dw0.19 8FC 47.304 47,304 = ss = K ksi = psi 2 0.19 0.19 π D D D w w w ( ) 8 15.647 C K = 47,304 Dw1.81 π K (39.845)C = 47,304 Dw1.81 0.625 Dw = C + 1 .5 4C − 1 0.615 K= + 4C − 4 C 1.81 4C − 1 0.615 0.625 + (39.845)C = 47,304 C 4C − 4 C + 1 .5 4C − 1 0.615 1.81 + (C + 1.5) C = 507.1 C 4C − 4 C = 7.684 0.625 0.625 Dw = = = 0.0680 in C + 1.5 7.684 + 1.5 Table AT 15, Dw = 0.0625 in , W & M 16 Dm = 7.684(0.0625) = 0.48025 in 15 say Dm = = 0.46875 in 32 D 0.46875 C= m = = 7 .5 Dw 0.0625 8FC ss = K 2 πDw 4C − 1 0.615 4(7.5) − 1 0.615 K= + = + = 1.1974 4C − 4 C 4(7.5) − 4 7.5 8(15.647 )(7.5) ss = 1.1974 = 91,600 psi = 91.6 ksi 2 π (0.0625) For N c Page 8 of 70 SECTION 4 - SPRINGS δ 2 − δ1 = 8(F2 − F1 )C 3 N c GDw G = 11.5 × 106 psi 3 8(15.647 − 1.534)(7.5) N c = 4 11.5 × 106 (0.0625) N c = 11.32 Table AT 16, squared and ground ends Total coils = N c + 2 = 11.32 + 2 = 13.32 Solid height = ( N c + 2)Dw = (11.32 + 2)(0.0625) = 0.8325 in Free length = PN c + 2 Dw Free length = 2 + δ1 3 δ 2 − δ1 = δ1 = ( ) 8(F1 )C 3 N c GDw 8(1.534 )(7.5) (11.32) = 0.082 in. 11.5 × 106 (0.0625) Free length = 2 + 0.082 = 2.082 in = P(11.32) + 2(0.0625) 3 δ1 = ( ) P = 0.1729 in ≈ 11 in 64 Pitch angle, 0.1729 P o o = tan −1 λ = tan −1 = 6.7 < 12 , o.k. 0 . 46875 πD π ( ) Solid stress 2 − 0.8325 sso = (91.6 ) = 142.6 ksi 0.75 87.5 Permissible solid stress = = 148.8 ksi > 137.5 ksi , safe. (0.0625)0.19 Summary of answer: Suitable material = Oil-Tempered Spring Wire Total Coils = 13.32 Active Coils, N c = 11.32 274. A helical spring is to fit about a 11/16-in. rod with a free length of 2 ¾ in. or less. A maximum load of 8 lb. is to produce a deflection of 1 ¾ in. The spring is expected to be compressed less than 5000 times during its life, but is subjected to relatively high temperatures and corrosive atmosphere. Select a material and determine the necessary wire size, mean coil diameter, and number of coils. Meet all conditions advised by Text. Page 9 of 70 SECTION 4 - SPRINGS Solution: For 5000 cycles < 104 cycles, use light service Use stainless steel, type 302 (Cr-Ni), ASTM A313 – for relative high temperature and corrosive atmosphere, Table AT 17. ssd = 0.32su (i) 170 ksi , [0.01 < Dw < 0.13] Dw0.14 97 su = 0.41 ksi , [0.13 < Dw < 0.375] Dw Maximum “solid” so = 0.47 su su = 8FDm ss = K 3 π D w F = 8 lb 11 D Dm − Dw = + w 16 2 Dm − 1.5Dw = 0.6875 in CDw − 1.5 Dw = 0.6875 0.6875 Dw = C − 1 .5 4C − 1 0.615 K= + 4C − 4 C 170 assume su = 0.14 ksi Dw 0.32(170 ) 54.4 54,400 ssd = = 0.14 ksi = psi 0.14 Dw Dw Dw0.14 4C − 1 0.615 8(8)C + = 54,400 D1w.86 C π 4C − 4 1.86 4C − 1 0.615 64C 0.6875 + = 54,400 C π 4C − 4 C − 1 .5 4C − 1 0.615 1.86 + (C − 1.5) C = 1330 C 4C − 4 C = 12.919 0.6875 Dw = = 0.0602 in 12.919 − 1.5 Use Table AT 15, Dw = 0.0625 in , 16 W & M Dm = 12.919(0.0602) = 0.8074 in 25 say Dm = in = 0.78125 in 32 Page 10 of 70 SECTION 4 - SPRINGS 11 16 0.78125 − 0.0625 > 0.6875 0.71875 > 0.6875 Dm − Dw > 0.71875 − 0.6875 = 0.03125 = C= Dw , o.k. 2 Dm 0.71875 = = 12.5 Dw 0.0625 [0.0625 < 0.13], therefore, su = 170 ksi is o.k. Dw0.14 8FC ss = K 2 π D w 4(12.5) − 1 0.615 K= + = 1.1144 4(12.5) − 4 12.5 8(8)(12.5) ss = 1.1144 = 72,648 psi 2 π (0.0625) 8(F )C 3 N c δ= GDw G = 10.6 × 106 psi 3 8(8)(12.5) N c 4 10.6 × 106 (0.0625) N c = 9.3 To check for solid stress and pitch Minimum solid height = Dw N c = (0.0625)(9.3) = 0.58125 in 3 δ =1 = ( ) (72,648) 2 3 − 0.58125 4 = 90,000 psi = 90 ksi 3 1 4 (0.47 )(170) = 117.8 ksi > 90 ksi , o.k. Permissible solid stress = (0.0625)0.14 Free length = PN c , minimum 3 P (9.3) = 2 4 P = 0.2957 in Pitch angle, 0.2957 P = tan −1 = 7.5o < 12o , o.k. λ = tan −1 πD π (0.71825) Summary of answer Material, Stainless Steel, Cr-Ni. ASTM A313 Solid stress = Page 11 of 70 SECTION 4 - SPRINGS Dw = 0.0625 in , 16 W & M 25 Dm = in 32 N c = 9.3 275. In order to isolate vibrations, helical compression springs are used to support a machine. The static load on each spring is 3500 lb., under which the deflection should be about 0.5 in. The solid deflection should be about 1 in. and the outside coil diameter should not exceed 6 in. Recommend a spring for this application; include scale, wire size, static stress, material, number of coils, solid stress, and pitch of coils. Solution: Use Music wire (The best material) Table AT 17 190 su = 0.154 ksi , [0.004 < Dw < 0.192] Dw Maximum “solid” sso = 0.5su Light service, ssd = 0.405su 0.405(190 ) 76.95 76,950 ssd = = 0.154 ksi = 0.154 psi 0.154 Dw Dw Dw 8FC ss = K 2 πDw F = 3500 lb OD = Dm + Dw = 6 in (C + 1)Dw = 6 6 Dw = C +1 76,950 4C − 1 0.615 8(3500 )C = ss = + C 6 6 0.154 4C − 4 π C + 1 C + 1 4C − 1 0.615 1.846 + = 235.9 C (C + 1) C 4C − 4 C = 5.635 6 Dw = = 0.9043 in > 0.192 in 5.635 + 1 76.950 use ss = = 99,216 psi (0.192)0.154 [ Page 12 of 70 ] SECTION 4 - SPRINGS 4C − 1 0.615 8(3500 )C = 99,216 ss = + C 6 4C − 4 π C + 1 4C − 1 0.615 2 + C (C + 1) = 400.8 C 4C − 4 C = 6.205 6 = 0.8328 in Dw = 6.205 + 1 13 Say Dw = in = 0.8125 in 16 Dm = (6.205)(0.8125) = 5.042 in Say Dm = 5 in D 5 C= m = = 6.154 Dw 0.8125 4(6.154) − 1 0.615 K= + = 1.2455 4(6.154) − 4 6.154 8FC ss = K 2 πDw 8(3500)(6.154) ss = 1.2455 = 103,481 psi > 99,216 psi , not o.k. 2 π (0.8125) Use Dm = 4.5 in D 4 .5 C= m = = 5.5385 Dw 0.8125 4(5.5385) − 1 0.615 K= + = 1.2763 45.5385 − 4 5.5385 8(3500)(5.5385) ss = 1.2763 = 95,435 psi > 99,216 psi , o.k. 2 π (0.8125) To check for solid stress (0.5)(190) = 122.488 ksi = 122,488 psi Permissible solid stress = (0.192 )0.154 1 Solid stress = (95,435) = 190,870 psi > 122,488 psi , not ok 0 .5 Use 0 .5 ssd = 122,488 = 61,244 psi 1 . Page 13 of 70 SECTION 4 - SPRINGS 4C − 1 0.615 8(3500 )C = 61,244 ss = + C 6 4C − 4 π C + 1 4C − 1 0.615 2 + C (C + 1) = 247.4 C 4C − 4 C = 5 .1 6 = 0.9836 in Dw = 5 .1 + 1 Say Dw = 1.0 in Dm = (5.1)(1.0) = 5.1 in Say Dm = 5 in D 5 C = m = =5 Dw 1 8FC ss = K 2 πDw 4(5) − 1 0.615 K= + = 1.3105 4(5) − 4 5 8(3500)(5) ss = 1.3105 = 58,400 psi > 61,244 psi o.k. 2 ( ) 1 . 0 π Use Dw = 1.0 in , Dm = 5 in 1 Solid stress = (58,400 ) = 116,800 psi < 122,488 psi , o.k. 0 .5 8(F )C 3 N c δ= GDw (Table AT 17) G = 12× 106 psi 8(3500 )(5) N c δ = 0.5 = 12 ×106 (1.0) N c = 1.7143 say N c = 1.75 Free length – Solid length = Solid Deflection PN c − Dw N c = 1 in P(1.75) − (1)(1.75) = 1 9 P = 1.5714 in ≈ 1 in 16 Pitch angle, 3 ( Page 14 of 70 ) SECTION 4 - SPRINGS 9 1 P = tan −1 16 = 5.68o < 12o , o.k. λ = tan −1 πD π (5) Summary of answer. F 3500 Scale, k = = = 7000 lb in δ 0 .5 Wire size, Dw = 1.0 in Material = Music Wire Solid sress = 116,800 psi 9 Pitch of stress = P = 1 in 16 CHECK PROBLEMS – LIGHT, MEDIUM SERVICE 276. The front spring of an automobile has a total of 9 ½ coils, 7 3/8 active coils (square-and-ground ends), an inside diameter of 4 3/64 in., and a free length of 14 ¼ in. It is made of SAE 9255 steel wire, OQT 1000oF, with a diameter of 43/64 in. Compute (a) the rate (scale) of the spring; (b) the “solid stress” and compare with a permissible value (is a stop needed to prevent solid compression?). (c) Can 95 % of the solid stress be repeated 105 times without danger of failure? Would you advise shot peening of the spring? Solution: 8FC 3 N c GDw 43 3 Dw = in > in 64 8 6 G = 10.5 × 10 psi D C= m Dw Dm − Dw = ID 43 3 Dm − = 4 in 64 64 23 Dm = 4 in 32 23 4 C = 32 = 7.0233 43 64 (a) δ = Page 15 of 70 SECTION 4 - SPRINGS Nc = 7 3 8 43 (10.5 ×10 ) 64 6 F GD k = rate = = 3 w = δ 8C N c = 345 lb in 3 3 8(7.0233) 7 8 (b) “Solid Stress” 43 1 Solid height = N c = (Dw )(Total Coils ) = 9 = 6.3828 in 64 2 Solid deflection = Free length – Solid height = 14 ¼ - 6.3828 = 7.8672 in. Solid Force = Fso = 7.8672(345) = 2714 lb 8F C Solid Stress = K so2 πDw 4C − 1 0.615 K= + 4C − 4 C 4(7.0233) − 1 0.615 K= + = 1.212 4(7.0233) − 4 7.0233 8(2714)(7.0233) ss = 1.212 = 130,322 psi 2 43 π 64 Permissible value, ss = s ys = 0.6 sy , [Dw > 0.5 in] SAE 9255, OQT 1000 oF s y = 160 ksi su = 180 ksi , s ys = 0.6(160 ) = 96 ksi = 96,000 psi < 130,322 psi Therefore a stop is needed to prevent solid compression. (c) ssd = 0.324su (105 cycles) ssd = 0.324(180) = 58.32 ksi 0.95sso = 0.95(130,322) = 123,800 psi = 123.8 ksi > 58.32 ksi There is a danger of failure, shot peening is advisable s ys = 1.25(96,000 ) = 120,000 psi ≈ 0.95sso 277. An oil-tempered steel helical compression spring has a wire size of No. 3 W & M, a spring index of 4.13, 30 active coils, a pitch of 0.317 in., ground-andsquared ends; medium service. (a) What maximum load is permitted if the recommended stress is not exceeded (static approach)? Compute (b) the Page 16 of 70 SECTION 4 - SPRINGS corresponding deflection, (c) “solid stress,”. (d) pitch angle, (e) scale, (f) the energy absorbed by the spring from a deflection of 0.25 in. to that of the working load. (g) Is there any danger of this spring buckling? (h) What maximum load could be used if the spring were shot peened? Solution: Table AT 17, oil-tempered 146 su = 0.19 ksi , [0.032 < Dw < 0.5] Dw 87.5 Maximum “solid” sso = 0.19 ksi Dw ssd = 0.324su (medium service) Table AT 15, No. 3 W & M Dw = 0.2437 in C = 4.13 Dm = CDw = 4.13(0.2437 ) = 1.0 in 8F C (a) ss = K s 2 πDw 4C − 1 0.615 K= + 4C − 4 C 4(4.13) − 1 0.615 K= + = 1.3885 4(4.13) − 4 4.13 0.324(146 ) ss = ssd = = 61.858 ksi = 61,858 psi (0.2437 )0.19 8(F )(4.13) ss = 61,858 = 1.3885 2 π (0.2437 ) F = 252 lb 8FC 3 N c δ= GDw G = 11.5 × 106 psi N c = 30 8(252)(4.13) (30) = 1.52 in 11.5 × 106 (0.2437 ) 3 δ= ( ) (c) For solid stress . Square-and-ground end) Free length = PN c + 2 Dw = (0.317 )(30) + 2(0.2437 ) = 9.9974 in Page 17 of 70 SECTION 4 - SPRINGS Solid height = Dw N c + 2 Dw = (30 + 2)(0.2437 ) = 7.7984 in Solid deflection = 9.9974 – 7.7984 = 2.199 in. 2.199 Solid stress = (61,858) = 89,491 psi 1.52 Maximum “solid” sso = 87.5 87.5 ksi = ksi = 114.4 ksi > 89.491 ksi , o.k. safe 0.19 Dw (0.2437 )0.19 0.317 P = tan −1 = 5.76o < 12o , o.k. πD π (1) F 252 (e) scale = k = = = 166 lb in δ 1.52 1 (f) U s = k (δ 22 − δ12 ) 2 k = 166 lb in δ1 = 0.25 in δ 2 = 1.52 in 1 2 2 U s = (166 ) (1.52 ) − (0.25) = 186.6 in − lb 2 (d) λ = tan −1 [ ] (g) i 6.18 Free length = 9.9974 in Mean Diameter = Dm = 1.0 in Free length 9.9974 = = 9.9974 > 4 Mean Diameter 1 .0 There is a danger for spring buckling (h) Shot peened, Table AT 17 ssd = (61,858)(1.25) = 77,322 psi 8(F )(4.13) ss = 77,322 = 1.3885 2 π (0.2437 ) F = 314 lb 280. It is desired to isolate a furnace, weighing 47,300 lb., from the surroundings by mounting it on helical springs. Under the weight, the springs should deflect approximately 1 in., and at least 2 in. before becoming solid. It has been decided to use springs having a wire diameter of 1 in., an outside diameter of 5 3/8 in., 4.3 free coils. Determine (a) the number of springs to be used, (b) the stress caused by the weight, (c) the “solid stress”. (d) What steel should be used? Solution: Page 18 of 70 SECTION 4 - SPRINGS Dw = 1 in 3 Dm + Dw = 5 in 8 3 Dm = 4 in 8 3 4 D C = m = 8 = 4.375 Dw 1 8FC 3 N c (a) δ = GDw Assume N c = 4.3 G = 10.5 × 106 psi , Dw > 3 in 8 8F (4.375) (4.3) 10.5 × 106 (1) F = 3645 lb W 47,300 No. of springs = = = 13 F 3645 3 δ = 1.0 = ( ) W 47,300 = = 3638 lb 13 13 8F C ss = K s 2 πDw (b) F = 4C − 1 0.615 + 4C − 4 C 4(4.375) − 1 0.615 K= + = 1.3628 4(4.375) − 4 4.375 K= 8(3638)(4.375) ss = 1.3628 = 55,235 psi π (1.0)2 2 (c) “Solid Stress” = ss = 55,235 = 110,470 psi 1 (d) s ys ≈ 110,470 psi s ys 110,470 = 184,117 psi = 184.117 ksi 0.6 0.6 From Table AT 7, Use AISI 8760, OQT 800 oF, s y = 200 ksi sy = = Page 19 of 70 SECTION 4 - SPRINGS VARYING STRESS APPROACH DESIGN PROBLEMS 282. A spring, subjected to a load varying from 100 lb. to 250 lb., is to be made of oiltempered, cold-wound wire. Determine the diameter of the wire and the mean diameter of the coil for a design factor of 1.25 based on Wahl’s line. The spring index is to be at least 5. Conform to good practice, showing checks for all significant parameters. Let the free length be between 6 and 8. Solution: Fmax = 250 lb Fmin = 100 lb 1 1 Fm = (Fmax + Fmin ) = (250 + 100 ) = 175 lb = 0.175 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (250 − 100 ) = 75 lb = 0.075 kip 2 2 Wahl’s line 1 sms − sas 2sas = + N s ys sno 8 KFa Dm 8 KFa C = sas = πDw3 πDw2 8 KFm Dm 8 KFmC sms = = K cπDw3 K cπDw2 C =5 4C − 1 0.615 K= + 4C − 4 C 4(5) − 1 0.615 K= + = 1.31 4(5) − 4 5 Fig. AF 15, C = 5 K c = 1.19 For oil-tempered wire, 87.5 s ys = 0.19 Dw , [0.032 < Dw < 0.5] 47 Dw0.1 , [0.041 < Dw < 0.15] 30 sno = 0.34 Dw , [0.15 < Dw < 0.625] sno = N = 1.25 Page 20 of 70 SECTION 4 - SPRINGS 8(1.31)(0.075)(5) 1.251 = πDw2 Dw2 8(1.31)(0.175)(5) 2.453 sms = = (1.19)πDw2 Dw2 30 say sno = 0.34 ksi Dw 1 sms − sas 2sas = + N s ys sno sas = 2.453 − 1.251 1.251 2 2 2 Dw 1 + Dw = 1.25 87.5 30 0.19 0.34 Dw Dw 1 1 1 = + 1.81 1.25 72.8 Dw 11.99 Dw1.66 Dw = 0.2857 in > 0.15 in Table AT 15, use No. 1, W & M Dw = 0.2830 in Dm = CDw = 5(0.2830) = 1.415 in 7 say Dm = 1 in 16 Check for Free length 6 in < Free length < 8 in 7 Free length = 4 Dm = 41 = 5.75 in 16 Increase Dm 1 Dm = 1 in 2 1 Free length = 4 Dm = 41 = 6 in , o.k. 2 Summary of answer Dw = 0.2830 in 1 Dm = 1 in 2 283. A carbon-steel spring is to be subjected to a load that varies from 500 to 1200 lb. The outside diameter should be between 3.5 and 4 in., the spring index between 5 to 10; approximate scale of 500 lb./in. Choose a steel and for a design factor of 1.4 by the Wahl line, find the wire diameter. Also determine the number of active Page 21 of 70 SECTION 4 - SPRINGS coils and the free length for squared-and-ground ends. Conform to the general conditions specified in the Text. Solution: For carbon steel, Table AT 17 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Fmax = 1200 lb Fmin = 500 lb 1 1 Fm = (Fmax + Fmin ) = (1200 + 500 ) = 850 lb = 0.85 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (1200 − 500 ) = 350 lb = 0.35 kip 2 2 OD = 3.5 ~ 4.0 in C = 5 ~ 10 Wahl’s line 1 sms − sas 2sas + = N s ys sno Figure AF 15, C = 5 ~ 10 Assume K = 1.2 , K c = 1.125 8 KFa Dm sas = πDw3 8 KFm Dm sms = K cπDw3 OD ≈ 3.75 in Dm = 3.75 − Dw 8(1.2 )(0.35)(3.75 − Dw ) 1.0695(3.75 − Dw ) = sas = πDw3 Dw3 8(1.2 )(0.85)(3.75 − Dw ) 2.3088(3.75 − Dw ) sms = = (1.125)πDw3 Dw3 Dw 91 0.1 Dw 1 3.75 − Dw 3.75 − Dw = + 2.9 1.4 73.4285 Dw 22.9079 Dw2.85 Dw = 0.6171 in > 0.25 in Use 1 = 1.4 3.75 − Dw 2(1.0695) Dw3 + 49 0.15 Dw (2.3088 − 1.0695) 3.75 −3 Dw Page 22 of 70 SECTION 4 - SPRINGS 91 = 104.53 ksi (0.25)0.1 49 sno = = 60.33 ksi (0.25)0.15 s ys = 3.75 − Dw 2(1.0695) Dw3 + 60.33 (2.3088 − 1.0695) 3.75 −3 Dw Dw 1 = 1 .4 104.53 1 3.75 − Dw 3.75 − Dw = + 1.4 84.346 Dw3 28.205 Dw3 1 3.75 − Dw = 1.4 21.137 Dw3 Dw = 0.5935 in use 19 Dw = in 32 3 Dm + Dw ≈ 3 in 4 19 3 = 3 in Dm + 32 4 5 Dm = 3 in 32 5 3 Dm 32 C= = = 5.316 Dw 19 32 . o.k. 19 Wire Diameter Dw = in , Carbon Steel 32 Number of coils: 8FC 3 N c δ= GDw 3 G = 10.5 × 106 psi = 10,500 ksi , Dw > in 8 F GDw =k = 3 δ 8C N c (10.5 ×10 ) 19 32 6 8(5.316 ) N c N c = 10.4 Table AT 16, square-and-ground ends 500 = Page 23 of 70 3 SECTION 4 - SPRINGS Free length = PN c + 2 Dw Solid height = Dw N c + 2 Dw Total Coils = N c + 2 19 Solid height = Dw N c + 2 Dw = (10.4 + 2 ) = 7.3625 in 32 F 1200 δ= = = 2.4 in k 500 Min. Free length = 2.4 + 7.3625 in = 9.7625 in Use Free length = 10 in To check for pitch angle. Free length = PN c + 2 Dw 19 P(10.4 ) + 2 = 10 32 P = 0.8474 in P 0.8474 = 4.885o < 12o , o.k. = tan −1 λ = tan −1 πDm π 3 5 32 Solid stress: δ T = solid deflection = 10 − 7.3625 = 2.6375 in F = kδ T = (500)(2.6375) = 1319 lb 4C − 1 0.615 K= + 4C − 4 C 4(5.316) − 1 0.615 K= + = 1.29 4(5.316) − 4 5.316 5 8(1.29)(1319) 3 8 KFDm 32 = 23,033 psi = 23.033 ksi < s (= 104.53 ksi ) ss = = ys 3 3 πDw 19 π 32 284. A helical compression spring, made of oil-tempered, cold-wound carbon steel, is to be subjected to a working load varying from 100 to 300 lb. for an indefinite time (severe). A mean coil diameter of 2 in. should be satisfactory. (a) Using the static approach, compute a wire diameter. (b) For this wire size, compute the factor of safety as given by the Wahl line. Solution: Page 24 of 70 SECTION 4 - SPRINGS Table AT 16, For carbon steel, 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = 0.1 ksi 91 Dw Max. “solid” s ys = 0.1 ksi Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Dm = 2 in. Fmax = 300 lb Fmin = 100 lb (a) F = 300 lb = 0.3 kip severe service, ssd = 0.263su = (0.263)(182) = 47.866 ksi Dw0.1 8F C ss = K s 2 πDw 4C − 1 0.615 K= + 4C − 4 C Dm C= Dw D 2 Dw = m = C C 4C − 1 0.615 8(0.3)(2 ) 47.866 ss = + = C 2 3 2 0.1 4C − 4 π C C 4C − 1 0.615 2.9 4C − 4 + C C = 233.84 C = 6.075 2 Dw = = 0.3292 in > 0.25 in 6.075 47.866 Therefore use ssd = = 54.984 ksi (0.25)0.1 Page 25 of 70 Dw0.1 SECTION 4 - SPRINGS 4C − 1 0.615 8(0.3)(2 ) ss = + = 54.984 C 2 3 4C − 4 π C 4C − 1 0.615 3 4C − 4 + C C = 287.9 C = 6.136 2 = 0.3259 in Dw = 6.136 21 say Dw = in 64 91 = 104.53 ksi (0.25)0.1 49 sno = = 60.33 ksi (0.25)0.15 1 1 Fm = (Fmax + Fmin ) = (300 + 100 ) = 200 lb = 0.2 kip 2 2 1 1 Fa = (Fmax − Fmin ) = (300 − 100 ) = 100 lb = 0.1 kip 2 2 Dm 2 C= = = 6.095 Dw 21 64 Figure AF 15 K c = 1.15 K = 1.25 K 8Fm Dm sms = K c πDw3 (b) s ys = sms 1.25 8(0.2 )(2) = = 31.34 ksi 1.15 21 3 π 64 sas = 8 KFa Dm πDw3 Page 26 of 70 SECTION 4 - SPRINGS 8(0.1)(2) sas = 1.25 = 18.02 ksi 21 3 π 64 Wahl’s line 1 sms − sas 2sas = + N s ys sno 1 31.34 − 18.02 2(18.02 ) = + N 104.53 60.33 N = 1.38 285. A helical spring of hard-drawn wire with a mean diameter of 1 ½ in. and squareand-ground ends is to be subjected to a maximum load of 325 lb. (a) Compute the wire diameter for average service. (b) How many total coils are required if the scale is 800 lb./in.? (c) For a minimum load of 100 lb., what is the factor of safety according to Wahl line? Would it be safe for an indefinite life? Solution: Table AT 17, Hard-drawn wire, 140 su = 0.19 ksi , [0.028 < Dw < 0.625] Dw 70 Maximum “solid” ss = s ys = 0.19 ksi Dw (0.9)(47 ) ksi , [0.041 < D < 0.15] sno = w Dw0.1 (0.9)(30) ksi , [0.15 < D < 0.625] sno = w Dw0.34 Average service (a) ssd = 0.85(0.324 )su = 0.2754 su = F = 325 lb = 0.325 kip 1 Dm = 1 in 2 8 FDm ss = K 3 πDw Page 27 of 70 0.2754(140 ) 38.556 = ksi Dw0.19 Dw0.19 SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 1 .5 Dw = C K= 4C − 1 0.615 8(0.325)(1.5) 38.556 ss = + = 3 0.19 C 4C − 4 1.5 1.5 π C C 4C − 1 0.615 2.81 + C = 97.05 C 4C − 4 C = 4.586 1 .5 1 .5 Dw = = = 0.3271 in < 0.625 in C 4.586 21 Dw = in 64 Dm 1 .5 = = 4.57 Dw 21 64 4(4.57 ) − 1 0.615 K= + = 1.345 4(4.57 ) − 4 4.57 (b) C = 8FC 3 N c GDw F GD =k = 3 w δ 8C N c k = 800 lb in = 0.8 kip in (11,500) 21 64 0.8 = 3 84.57 N c N c = 6.2 δ= (c) s ys = 70 0.19 21 64 (0.9)(30) = 86.5 ksi = 39.44 ksi , Dw > 0.15 in 0.19 21 64 1 Fm = (325 + 100 ) = 212.5 lb = 0.2125 kip 2 sno = Page 28 of 70 SECTION 4 - SPRINGS 1 (325 − 100) = 112.5 lb = 0.1125 kip 2 K c = 1.212 , Fig. AF 15 K = 1.345 K 8Fm Dm 1.345 8(0.2125)(1.5) sms = = = 25.5 ksi 3 K c πDw3 1.212 21 π 64 8Fm Dm 8(0.1125)(1.5) sas = K = 1.345 = 16.36 ksi 3 3 21 πDw π 64 1 sms − sas 2sas 25.5 − 16.36 2(16.36) + = + = 86.5 39.44 N s ys sno Fa = N = 1.07 < 1.15[N min ] Not safe for indefinite life. 286. A helical spring is to be subjected to a maximum load of 200 lb. (a) Determine the wire size suitable for medium service if the material is carbon steel ASTM A230; C = 6 . Determine the factor of safety of this spring according to the Wahl line (b) If the minimum force is 150 lb., (c) if the minimum force is 100 lb., (d) if the minimum force is 25 lb. Solution: For carbon steel ASTM A230 Table AT 17 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw Medium Service ssd = 0.324su 182 58.968 58,968 (a) ssd = 0.324 0.1 = ksi = psi 0.1 Dw Dw0.1 Dw Page 29 of 70 SECTION 4 - SPRINGS 8 FDm ss = K 3 πDw 4C − 1 0.615 K= + 4C − 4 C 4(6) − 1 0.615 K= + = 1.2525 4(6) − 4 6 F = 200 lb 8(200)(6) 58,968 ss = 1.2525 = 2 Dw0.1 πDw Dw = 0.2371 in Table At 15, use Dw = 0.2437 in , No. 3 W & M Dw = 0.2437 in < 0.25 in , o.k. Factor of safety. 91 91 s ys = 0.1 ksi = ksi = 104.8 ksi Dw (0.2437 )0.1 49 49 sno = 0.15 ksi = ksi = 60.56 ksi Dw (0.2437 ) (a) Fm = 1 (200 + 150) = 175 lb = 0.175 kip 2 1 (200 − 150) = 25 lb = 0.025 kip 2 Figure AF 15, K c = 1.156 Fa = K 8FmC 1.2525 8(0.175)(6 ) = 48.8 ksi = K c πDw2 1.156 π (0.2437 )2 8(0.025)(6 ) 8F C sas = K a 2 = 1.2525 = 8.1 ksi 2 πDw π (0.2437 ) 1 sms − sas 2sas = + N s ys sno 1 48.8 − 8.1 2(8.1) = + N 104.8 60.56 N = 1.525 sms = (b) Fm = Fa = 1 (200 + 100) = 150 lb = 0.15 kip 2 1 (200 − 100) = 50 lb = 0.05 kip 2 Page 30 of 70 SECTION 4 - SPRINGS Figure AF 15, K c = 1.156 8FmC 1.2525 8(0.15)(6 ) = = 41.8 ksi 2 2 πDw 1.156 π (0.2437 ) 8(0.05)(6) 8F C sas = K a 2 = 1.2525 = 16.11 ksi 2 πDw π (0.2437 ) 1 sms − sas 2sas + = N s ys sno 1 41.8 − 16.11 2(16.11) + = N 104.8 60.56 N = 1.287 sms = K Kc (c) Fm = 1 (200 + 25) = 112.5 lb = 0.1125 kip 2 1 (200 − 25) = 87.5 lb = 0.0875 kip 2 Figure AF 15, K c = 1.156 Fa = K 8FmC 1.2525 8(0.1125)(6) = 31.36 ksi = K c πDw2 1.156 π (0.2437 )2 8(0.0875)(6) 8F C sas = K a 2 = 1.2525 = 28.2 ksi 2 D π ( ) 0 . 2437 π w 1 sms − sas 2sas = + N s ys sno 1 31.36 − 28.20 2(28.20 ) = + N 104.8 60.56 N = 1.04 sms = CHECK PROBLEMS A Diesel valve spring is made of 3/8-in. chrome-vanadium steel wire, shot-peened; inside diameter is 3 in., 7 active coils, free length is 7 3/8 in., solid length is 4 1/8 in., length with valve closed, 6 ¼ in., length when open, 5 1/8 in. (a) Compute the spring constant and the factor of safety as defined by the Wahl criterion (see § 6.13, Text). (b) Is there any danger of damage to the spring if it is compressed solid? (c) What is the natural frequency? If this spring is used on a 4-stroke Diesel engine at 450 rpm, is there any danger of surge? Compute the change of stored energy between working lengths. Solution: For chrome-vanadium steel wire, shot-peened, Table AT 17 Page 31 of 70 SECTION 4 - SPRINGS (1.25)(168) ksi , [0.032 < D w < 0.437 ] Dw0.166 (1.25)(100 ) ksi , [0.032 < D < 0.437] s ys = w Dw0.166 (1.25)(49) ksi , [0.028 < D < 0.5] sno = w Dw0.15 3 Dw = in = 0.375 in 8 (1.25)(100) ksi = 147.1 ksi s ys = (0.375)0.166 (1.25)(49) ksi = 70.96 ksi sno = (0.375)0.15 su = 8FC 3 N c GDw F GD =k = 3 w δ 8C N c (a) δ = G = 11.5 × 106 psi Nc = 7 Dw = 0.375 in Dm − Dw = ID = 3 in Dm = 3.375 in D 3.375 C= m = =9 Dw 0.375 k = spring constant GDw ( 11.5 × 106 )(0.375) k= 3 = = 105.64 lb in 3 8C N c 8(9 ) (7 ) 3 1 δ1 = 7 − 4 = 3.25 in 8 8 F1 = kδ1 = (105.64)(3.25) = 343.33 lb 3 1 δ 2 = 7 − 6 = 1.125 in 8 4 F2 = kδ 2 = (105.64)(1.125) = 118.85 lb 1 Fm = (343.33 + 118.85) = 231.09 lb = 0.231 kip 2 1 Fa = (343.33 − 118.85) = 112.24 lb = 0.11224 kip 2 K 8FmC sms = K c πDw2 Page 32 of 70 SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 4(9) − 1 0.615 K= + = 1.162 4(9 ) − 4 9 Figure AF 15, K c = 1.10 K= 8FmC 1.162 8(0.231)(9) = = 39.8 ksi 2 2 πDw 1.10 π (0.375) 8(0.11224 )(9) 8F C sas = K a 2 = 1.162 = 21.3 ksi 2 πDw π (0.375) 1 sms − sas 2sas = + N s ys sno 1 39.8 − 21.3 2(21.3) = + N 147.1 70.96 N = 1.377 sms = K Kc (b) max. “solid” ss = s ys = 147.1 ksi Min. Solid Height = Dw N c = (0.375)(7 ) = 2.625 in 3 Solid deflection = 7 − 2.625 = 4.75 in. 8 F = kδ = (105.64)(4.75) = 501.8 lb = 0.5018 kip 8(0.5018)(9) 8 FC Solid stress = ss = K 2 = 1.162 = 95 ksi < 147.1 ksi 2 πDw π (0.375) There is no danger of damage (c) Natural frequency For steel 14,050 Dw φ= cps N c Dm2 14,050 φ= cps N c C 2 Dw 14,050 φ= cps = 66 cps (7 )(9)2 (0.375) 2π For 450 rpm, φ = 450 = 47 cps 60 66 = 1.4 < 12 , there is danger of surging. 47 Page 33 of 70 SECTION 4 - SPRINGS (d) U s = [ ] 1 1 2 2 k δ12 − δ 22 = (105.64 ) (3.25) − (1.125) = 491 in − lb 2 2 289. ( ) A helical spring is hot wound from 5/8-in. carbon-steel wire with an outside diameter of 3 ¼ in. A force of 3060 lb. is required to compress the spring 1 ¾ in to the solid heigh. In service the spring is compressed so that its deformation varies form ½ in. to1 1/8 in. (a) What is the factor of safety by the Wahl criterion? (b) Is the “solid stress” safe? Compute (c) the pitch angle, (d) the change of stored energy between the working lengths, (e) the factor of safety if the spring is peened? Solution: For hot-wound carbon steel wire 5 Dw = in 8 Table AT 17 91 s ys = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 s ys = ksi = 104.5 ksi , Dw > 0.25 in. (0.25)0.1 49 sno = 0.15 ksi , [0.093 < Dw < 0.25] Dw 49 sno = ksi = 60.33 ksi , Dw > 0.25 in. (0.25)0.15 Permissible solid stress = ss = ss = 117 ksi , [Dw > 0.375 in.] § 6.3 Dw0.31 117 ksi = 35.4 ksi (0.625)0.31 3060 = 1748.6 lb in δ 1.75 1 F1 = kδ1 = (1748.6 ) = 874.3 lb 2 1 F2 = kδ 2 = (1748.6 )1 = 1967.2 lb 8 1 Fm = (1967.2 + 874.3) = 1420.7 lb = 1.4207 kip 2 1 Fa = (1967.2 − 874.3) = 546.4 lb = 0.5464 kip 2 (a) k = F = Page 34 of 70 SECTION 4 - SPRINGS 5 in = 0.625 in 8 1 Dm + Dw = 3 in 4 Dm = 2.625 in D 2.625 C= m = = 4 .2 Dw 0.625 4C − 1 0.615 K= + 4C − 4 C 4(4.2) − 1 0.615 K= + = 1.3808 4(4.2) − 4 4.2 K c = 1.234 Dw = K 8FmC 1.3808 8(1.4207 )(4.2) = 43.5 ksi = K c πDw2 1.234 π (0.625)2 8(0.5464)(4.2) 8F C sas = K a 2 = 1.3808 = 20.7 ksi 2 πDw π (0.625) 1 sms − sas 2sas = + N s ys sno 1 43.5 − 20.7 2(20.7 ) + = N 104.5 60.33 N = 1.106 sms = (b) Permissible solid stress = 135.4 ksi F = 3.060 kip 8(3.060)(4.2) 8 FC = 115.7 ksi < 135.4 ksi , safe Solid stress, ss = K 2 = 1.3808 2 πDw π (0.625) 3 (c) Solid deflection = 1 in 4 (P − Dw )N c = 1.75 in 8FC 3 N c δ= GDw G = 10.5 × 106 psi , hot-wound F GD k= = 3w δ 8C N c Page 35 of 70 SECTION 4 - SPRINGS (10.5 ×10 )(0.625) 6 1748.6 = 8(4.2 ) N c 3 N c = 6.332 (P − 0.625)(6.332) = 1.75 P = 0.9014 in Pitch angle P P tan λ = = πDm πCDw λ = tan −1 0.9014 P = tan −1 = 6.24o πCDw π (4.2)(0.625) [ ] 1 1 2 2 k δ12 − δ 22 = (1748.6 )(1.125) − (0.5) = 888 in − lb 2 2 (e) When peened s ys = 12.5(104.5) = 130.6 ksi ( (d) U s = ) sno = 1.25(60.33) = 75.4 ksi 1 43.5 − 20.7 2(20.7 ) = + N 130.6 75.4 N = 1.38 ENERGY STORAGE 293. A 10-lb. body falls 10 in. and then strikes a helical spring. Design a harddrawn carbon steel spring that will absorb this shock occasionally without permanent damage. Determine appropriate values of wire diameter, coil diameter, pitch, free length, closed length, and the maximum stress under the specified conditions, and scale. Let C = 7 . Solution: For hard-drawn carbon steel, Table AT 17 182 su = 0.1 ksi , [0.093 < Dw < 0.25] Dw 91 Max. “solid” ss = 0.1 ksi , [0.093 < Dw < 0.25] Dw 36.855 ssd = (0.50 )(0.405)su = ksi Dw0.1 Us = ss2V 4 K 2G Page 36 of 70 SECTION 4 - SPRINGS 4C − 1 0.615 + 4C − 4 C 4(7 ) − 1 0.615 K= + = 1.213 4(7 ) − 4 7 K= πD 2 V ≈ w (πDm )N c 4 π 2 Dw2 Dm N c V= 4 Dm = CDw V= π 2CDw3 N c 4 8FC 3 N c δ= GDw U s = W (h + δ ) 8 FC ss = K 2 πDw s πD 2 F= s w 8KC s πD 2 8C 3 N c δ = s w 8KC GDw s sπDwC 2 N c KG 2 2 3 ssπDwC 2 N c s s π CD w N c U s = W h + = KG 16 K 2G Wh Nc = 2 2 2 ss π CDw s sπDwC 2W − 16 K 2G KG 36.855 when ss = ksi Dw0.1 Wh Nc = 2 2 2.8 (36.855) π CDw − 36.855πD w0.9C 2W 16 K 2G KG ( 0.010)(10) Nc = (36.855)2 π 2 (7)Dw2.8 − 36.855πD w0.9 (7 )2 (0.010) 2 (1.213)(11,500) 16(1.213) (11,500) 0.10 Nc = 2.8 0.3466 Dw − 0.004067 D w0.9 δ= Page 37 of 70 SECTION 4 - SPRINGS combination of Dw and N c Gage No. W & M 12 11 10 9 8 7 6 5 4 3 Dw 0.1055 0.1205 0.1350 0.1483 0.1620 0.1770 0.1920 0.2070 0.2253 0.2437 Nc 991.2 312.1 166.1 108.0 75.2 53.7 40.2 31 23.4 18.1 Dw N c 105 37.6 22.4 16.0 12.2 9.5 7.7 6.4 5.3 4.4 Use Dw = 0.2437 in < 0.25 in , N c = 18.1 45 Dm = 7 Dw = 7(0.2437 ) = 1.7059 in = 1 in 64 0.9 2 0.9 2 s sπDwC 2 N c 36.855πD w C N c 36.855π (0.2437 ) (7 ) (18.1) = = = 2.066 in (1.213)(11,500) KG KG 36.855 ss = = 42.44 ksi (0.2437 )0.1 91 91 sso = 0.1 = = 104.8 ksi Dw (0.2437 )0.1 Solid deflection 104.8 = (2.066 ) = 5.1 in 42.44 (P − Dw )N c = 5.1 δ= (P − 0.2437)(18.1) = 5.1 P = 0.5255 in 17 say P = = 0.53125 in 32 Minimum Solid Height = Dw N c = (0.2437 )(18.1) = 4.41 in Assume squared and ground end Solid height = Dw N c + 2 Dw = (0.2437 )(18.1) + 2(0.2437 ) = 5.0 in Solid deflection = (0.53125 − 0.2437 )(18.1) = 5.2 in Free length = 5.0 in + 5.2 in = 10.2in Summary of answer: Dw = 0.2437 in , No. 3 W & M Page 38 of 70 SECTION 4 - SPRINGS Dm = 1 45 in 64 17 in 32 Free length = 10.2 in Closed length = 5 in Maximum stress = 42.44 ksi P= 294. A helical spring, of hard-drawn steel wire, is to absorb 75 in-lb of energy without being stressed beyond the recommended value of average service. Let C = 6 . Decide upon satisfactory dimensions; Dw , Dm , N c , free length, pitch angle, solid stress, volume of metal, possibility of spring buckling. Solution: For hard-drawn steel wire, shock load, average service 140 su = 0.19 ksi , [0.028 < Dw < 0.625] Dw 70 Max. “solid” ss = 0.19 ksi , [0.028 < Dw < 0.625] Dw 140 19.278 ssd = (0.50)(0.85)(0.324)su = (0.50)(0.85)(0.324) 0.19 = 0.19 ksi Dw Dw s 2V s 2π 2 Dw3 CN c Us = s 2 = s 4K G 16 K 2G C=6 4C − 1 0.615 K= + 4C − 4 C 4(6) − 1 0.615 K= + = 1.2525 4(6) − 4 6 U s = 75 in − lb = 0.075 in − kip 19.278 π 2 Dw3 (6)N c U s = 0.075 = 0.19 2 Dw 16(1.2525) (11,500) 0.9837 = Dw2.62 N c Table AT-15 W&M Page 39 of 70 Dw Nc Dw N c SECTION 4 - SPRINGS 9 8 7 6 5 4 3 2 1 0 2-0 3-0 4-0 5-0 0.1483 0.1620 0.1770 0.1920 0.207 0.2253 0.2437 0.2625 0.2830 0.3065 0.3310 0.3625 0.3938 0.4305 146 116 92 74 61 49 40 32.7 26.9 21.8 17.8 14.0 11.3 8.95 Use Dw = 0.4305 in , 5-0 W & M Nc ≈ 9 9 Dm = 6(0.4305) = 2.583 in ≈ 2 in 16 19.278 ss = = 22.63 ksi (0.4305)0.19 70 Max. Solid Stress = sso = = 82.16 ksi (0.4305)0.19 s sπDwC 2 N c (22.63)(π )(0.4305)(6) (9) = = 0.6885 in KG (1.2525)(11,500) 82.16 Solid deflection = (0.6885) = 2.5 in 22.63 (P − Dw )N c = 2.5 (P − 0.4305)(9) = 2.5 P = 0.7083 in 45 say P = = 0.703125 in 64 Solid deflection = (0.703125 − 0.4305)(9) = 2.453625 in 2.453625 Solid stress = 22.63 = 80.65 ksi 0.6885 2 δ= 7 Minimum Solid Height = Dw N c = (0.4305)(9 ) = 3.8745 in ≈ 3 in 8 45 21 Minimum Free Length = PN c = (9 ) = 6.328125 in ≈ 6 in 64 64 Pitch Angle Page 40 of 70 21.65 18.79 16.28 14.21 12.63 11.04 9.75 8.58 7.61 6.68 5.89 5.075 4.45 3.85 SECTION 4 - SPRINGS 45 P = tan −1 64 = 5o < 12o λ = tan −1 πDm π 2 9 16 Volume π (0.4305)2 9 πDw2 3 (πDm )N c = V ≈ π 2 (9 ) = 10.55 in 4 4 16 Summary of answer: Dw = 0.4305 in , No. 5-0 W & M 9 Dm = 2 in 16 Nc = 9 21 Free length = 6 in 64 Pitch Angle = λ = 5o Solid Stress = 80.65 ksi Volume of metal = 10.55 in3 Possibility of spring buckling 21 6 64 = 2.47 < 4 , no possibility 9 2 16 CONCENTRIC HELICAL SPRINGS 297. Two concentric helical springs are to be subjected to a load that varies from a maximum of 235 lb. to a maximum of 50 lb. They are to fit inside a 1 5/8 in. cylinder. The maximum deflection is to be ¾ in., and the deflection when compressed solid is to be approximately 1 in. Using the “static approach” for severe service (maximum load), determine the wire diameter, mean coil diameter, number of coils, solid length, and free length of both springs. (Start with oil-tempered wire and assume a diametral clearance between the outer D spring and the cylinder of w , assume a similar clearance between springs. 2 Search for a suitable spring index and wire size.) Solution: For oil-tempered wire Table AT 17 Page 41 of 70 SECTION 4 - SPRINGS 146 ksi , [0.032 < Dw < 0.5] Dw0.19 87.5 Max. “solid” ss = 0.19 ksi , [0.032 < Dw < 0.5] Dw Severe service 0.263(146 ) 38.398 ssd = 0.263su = = ksi Dw0.19 Dw0.19 F = 235 lb = 0.235 kip δo = δi su = 8FoCo3 N co 8Fi Ci3 N i = GDwo GDwi Assume, Co = Ci 3GDwo Fo = 32C 3 N co 3GDwi Fi = 32C 3 N ci 8F C sso = K o2 πDwo 8F C ssi = K i 2 πDwi D D C = mo = mi Dwo Dwi 4C − 1 0.615 K= + 4C − 4 C D Dmo − Dwo − wi = Dmi + Dwi 2 Dmo − Dmi = Dwo + 1.5Dwi D 1.625 − wo = Dmo + Dwo 2 Dmo + 1.5Dwo = 1.625 CDwo + 1.5Dwo = 1.625 1.625 Dwo = C + 1 .5 1.625C Dmo = C + 1 .5 CDwo − CDwi = Dwo + 1.5Dwi (C − 1)Dwo = (C + 1.5)Dwi Page 42 of 70 SECTION 4 - SPRINGS 1.625(C − 1) (C + 1.5)2 1.625C (C − 1) Dmi = (C + 1.5)2 Dwi = 8F C 38.398 sso = K o2 = 0.19 ksi Dwo πDwo 1.81 15.08Dwo Fo = KC 8F C 38.398 ssi = K i 2 = 0.19 ksi Dwi πDwi 1.81 15.08Dwi KC Fo + Fi = F = 0.235 kip Fi = 1.81 1.81 15.08Dwo 15.08Dwi + = 0.235 KC KC 1.81 1.81 15.08 Dwo + 15.08 Dwi = 0.235 KC 1.81 1.81 1.625 1.625(C − 1) 15.08 = 0.235 KC + 15.08 C + 1 .5 C + 1.5 1.81 ( 1 C − 1) 4C − 1 0.615 154.52 + = 0.235 + C 1.81 3.62 C (C + 1.5) 4C − 4 (C + 1.5) C = 5.328 1.625(5.328 − 1) Dwi = = 0.1509 in (5.328 + 1.5)2 1.625 Dwo = = 0.2380 in 5.328 + 1.5 Table AT 15, use Dwi = 0.1620 in , No. 8 W & M and Dwo = 0.2625 in , No. 2 W & M 13 Dmo = CDwo = (5.328)(0.2625) = 1.3986 in ≈ 1 in 32 7 Dmi = CDwi = (5.328)(0.1620 ) = 0.8631 in ≈ in 8 7 Dmi 8 Ci = = = 5.401 Dwi 0.1620 Co = Dmo Dwo 13 1 32 = = 5.357 0.2625 Page 43 of 70 SECTION 4 - SPRINGS 1.81 15.08Dwo K o Co 4(5.357 ) − 1 0.615 Ko = + = 1.287 4(5.357 ) − 1 5.357 Fo = 15.08(0.2625) = 0.194 kip (1.287 )(5.357 ) 1.81 Fo = 1.81 15.08Dwi K i Ci 4(5.401) − 1 0.615 Ki = + = 1.2843 4(5.401) − 1 5.401 15.08(0.1620) Fi = = 0.081 kip (1.2843)(5.401) Fo + Fi = 0.194 + 0.071 = 0.275 kip > 0.235 kip , ok 3GDwo Fo = 32C 3 N co 3(11,500)(0.2625) 0.194 = 3 32(5.357 ) N co N co = 9.5 3GDwi Fi = 32C 3 N ci 3(11,500)(0.1620) 0.071 = 3 32(5.401) N ci N ci = 15.6 87.5 Max. solid stress, sss = 0.19 ksi , Dw 87.5 ssso = = 112.82 ksi (0.2625)0.19 87.5 sssi = = 123.65 ksi (0.1620)0.19 Stress 38.398 ssi = = 54.26 ksi (0.1620)0.19 38.398 sso = = 49.51 ksi (0.2625)0.19 Fi = Solid stress 1 sso = 49.51 = 66.01 ksi < 112.82 ksi 0.75 Page 44 of 70 SECTION 4 - SPRINGS 1 ssi = 54.26 = 72.35 ksi < 123.65 ksi 0.75 Solid length Dwo N co = (0.2625)(9.5) = 2.5 in Dwi N ci = (0.1620 )(15.6) = 2.53 in assume solid length = 3 in Dwi ( N ci + xi ) = (0.1620)(15.6 + xi ) = 3 in xi = 2.92 Total coils = 15.6 + 2.92 = 18.52 Dwo ( N co + xo ) = (0.2625)(9.5 + xo ) = 3 in xo = 1.93 Total coils = 9.5 + 1.93 = 11.43 Free Length = 3 in + 1 in = 4 in Summary of answer: Outside wire. Dwo = 0.2625 in , No. 2 W & M 13 Dmo = 1 in 32 N to = 11.43 Solid length = 3 in Free length = 4 in Inside wire. Dwi = 0.1620 in , No. 8 W & M 7 Dmi = in 8 N ti = 18.52 Solid length = 3 in Free length = 4 in 298. Two concentric, helical compression springs are used on a freight car. The larger spring has an outside diameter of 7 in., a free length of 7 1/8 in., and is made of a 1 ¾ in. steel bar. The smaller has an outside diameter of 4 1/8 in., a free length of 6 13/16 in. , and is made of 7/8 in. steel bar. The solid height of each spring is 5 ¼ in. and the forces required to compress them solid are 15,530 lb. and 7,000 lb., respectively. The working load on the two springs is 11,350 lb. Determine (a) the number of free coils in each spring, (b) the stress in each spring when compressed solid, (c) the stresses induced by the working Page 45 of 70 SECTION 4 - SPRINGS load. Notice that the outer spring deflects 5/16 in. before the inner one takes a load. (d) What energy is absorbed while changing deflection from that at the working load to that when the springs are compressed “solid”? Solution: ODo = 7 in 3 Dwo = 1 in 8 1 FLo = 7 in 8 1 ODi = 4 in 8 7 Dwi = in 8 13 FLi = 6 in 16 1 (a) Solid height = Dw N T = 5 in 4 5.25 N To = = 3.82 1.375 5.25 =6 N Ti = 0.875 (b) Fo = 15,530 lb Fi = 7000 lb 8 FC ss = K 2 πDw 3 Dmo = 7 − 1 = 5.625 in 8 Dmo 5.625 Co = = = 4.091 Dwo 1.375 4(4.091) − 1 0.615 Ko = + = 1.393 4(4.091) − 4 4.091 1 7 Dmi = 4 − = 3.25 in 8 8 D 3.25 Ci = mi = = 3.714 Dwi 0.875 4(3.714) − 1 0.615 Ki = + = 1.442 4(3.714) − 4 3.714 Page 46 of 70 SECTION 4 - SPRINGS Solid stress 8(15,530)(4.091) sso = 1.393 = 119,203 psi 2 ( ) π 1 . 375 8(7000 )(3.714 ) ssi = 1.442 = 124,689 psi 2 π (0.875) (b) Stresses induced by working load Fi + Fo = 11,350 lb 15,530 ko = = 8283 lb in 1 1 7 − 5 4 8 7000 ko = = 4480 lb in 1 13 6 − 5 4 16 5 δ o − δ i = = 0.3125 in 16 Fi = kiδ i = 4480δ i Fo = koδ o = 8283(0.3125 + δ i ) Fi + Fo = koδ o = 4480δ i + 8283(0.3125 + δ i ) = 11,350 lb δ i = 0.6865 in δ o = 0.3125 + 0.6865 = 0.9990 in Fi = (4480)(0.6865) = 3076 lb Fo = (8283)(0.9990) = 8275 lb Stresses 8(8275)(4.091) sso = 1.393 = 63,516 psi 2 π (1.375) 8(3076)(3.714) ssi = 1.442 = 54,792 psi 2 π (0.875) (d) Energy 1 U so = ko (δ o22 − δ o21 ) 2 1 1 δ o 2 = 7 − 5 = 1.875 in 8 4 δ o1 = 0.9990 in 1 2 2 U so = (8283)(1.875) − (0.999 ) = 10,427 in − lb 2 [ Page 47 of 70 ] SECTION 4 - SPRINGS 1 ki δ i22 − δ i21 2 13 1 δ i 2 = 6 − 5 = 1.5625 in 16 4 δ o1 = 0.6865 in 1 2 2 U si = (4480 )(1.5625) − (0.6865) = 4,413 in − lb 2 U si = ( ) [ ] TORSION-BAR SPRINGS 299. A torsion-bar similar to that shown is to be used for the front spring of an automobile. Its rate should be 400 lb./in. of deflection of the end of the arm which is e = 10 in. long. It is made of AISI 9261,OQT 900 oF, and the maximum repeated load is 1500 lb. perpendicular to the centerline of the arm. The support is such that bending of the bar is negligible. (a) Determine its diameter and length so that no permanent set occurs due to a 30 % overload (limited by a stop). Use s ys = 0.6 s y , but check with equation (c) § 6.3, Text, if appropriate. (b) Determine the factor of safety according to the Soderberg criterion if the load varies from 1200 lb. to 1500 lb.; minimum r d = 0.1 , D d = 3 . (c) The same as (b) except that the bar is shot-peened all over. What other steps may be taken to improve the fatigue strength? Problem 299, 300 Solution: e = 10 in For AISI 9261, OQT 900 oF s y = 192 ksi su = 215 ksi s ys = 0.6 s y = 115.2 ksi = 115,200 psi Page 48 of 70 SECTION 4 - SPRINGS 16T πd3 T = Fe = (1.3)(1500)(10) = 19,500 in − lb 16(19,500 ) 115,200 = π d3 d = 0.95 in use d = 1 in 117 117 § 6.3 ss = 0.3 = 0.3 = 117 ksi ≈ s ys Dw (1) (a) ss = (b) Soderberg Criterion 1 sms K f sas = + N s ys sns sns = (0.6)(0.5)(215) = 64.5 ksi Figure AF 12, r d = 0.1 , D d = 3 K t = 1.45 K f ≈ K t = 1.45 1 (1500 + 1200) = 1350 lb 2 Tm = (1350)(10) = 13,500 in − lb = 13.5 in − kips 16(13.5) sms = = 68.8 ksi π (1)3 Fm = 1 (1500 − 1200) = 150 lb 2 Ta = (150)(10) = 1500 in − lb = 1.5 in − kips 16(1.5) sas = = 7.64 ksi π (1)3 1 68.8 (1.45)(7.64 ) = + N 115.2 64.5 N = 1.30 Fa = (c) Shot-peened s ys = 1.25(115.2 ) = 144 ksi sns = 1.25(64.5) = 80.6 ksi 1 68.8 (1.45)(7.64 ) = + N 144 80.6 N = 1.625 Page 49 of 70 SECTION 4 - SPRINGS 300. A solid steel torsion bar is loaded through a 10 in. arm as shown. The load F perpendicular to the center-line of the arm varies from 500 to 1000 lb., 7 200,000 cycles. The bar is d = in. in diameter and 30 in. long; let D d = 3 ; 8 r d = 0.1 ; (a) Determine the maximum stress in the bar, the angular deflection, and the scale (lb./in.) where F is applied. The support is such that bending of the bar is negligible. (b) Select a material and heat treatment for this bar for a minimum N = 1.2 , Soderberg criterion. Problem 299, 300 Solution: Fig. AF 12, K f = 1.45 1 (1000 + 500) = 750 lb 2 1 Fa = (1000 − 500 ) = 250 lb 2 Tm = (750)(10) = 7500 in − lb = 7.5 in − kips Fm = Ta = (250)(10) = 2500 in − lb = 2.5 in − kips K fl = n (log K f ) 3 Kf = (200,000)(log1.45) 3 1.33 1.45 16T πd3 16(7.5) sms = = 57 ksi 3 7 π 8 16(2.5) sas = = 19 ksi 3 7 π 8 ss = (a) smax = sms + K fl sas = 57 + (1.33)(19 ) = 82.27 ksi Page 50 of 70 SECTION 4 - SPRINGS TL 64TL = JG π d 4G 64(500 )(10 )(30 ) θ min = = 0.4533 rad 4 7 6 π 11.5 ×10 8 64(1000 )(10 )(30 ) θ max = = 0.9066 rad 4 7 6 π 11.5 ×10 8 F 1000 scale = = = 110.3 lb in θ e (0.9066)(10) θ= (c) ( ) ( ) 1 sms K f sas = + N s ys sns 10 6 sns = (0.6 )(0.5)su 200,000 s ys = 0.6 s y 0.085 = 0.344 su 1 57 (1.33)(19) = + 1.2 0.6s y 0.344su Use AISI 8760, OQT 800 oF s y = 200 ksi su = 220 ksi N = 1.24 HELICAL SPRINGS – NON CIRCULAR SECTION 301. A spring is to be designed of square oil-tempered steel wire and subjected to a repeated maximum load of 325 lb.; mean coil diameter, 1 ½ in.; deflection, 13/32 in. Determine (a) the wire size for average service, (b) the required number of active coils, (c) the solid height, free length, and pitch (the ends are squared and ground, the “solid stress” must be satisfactory, and the pitch angle not excessive). (d) What amount of energy is stored when the load is 325 lb.? Express in in-lb. and Btu. Solution: For oil-tempered wire, 146 su = 0.19 , [0.032 < Dw < 0.5] Dw Max. “solid” ss = 0.6su Page 51 of 70 SECTION 4 - SPRINGS (a) average service, b = Dw , t = b K FD (3b + 1.8t ) 2.4 FDm ss= q m 2 2 = Kq 2b t b3 s sd = 0.324su , average service 2.4 FDm 0.324(146 ) Kq = 2b 3 b 0.19 F = 0.325 kip 1 Dm = 1 in 2 K q = 1.25 (assumed) 2.4(0.325)(1.5) 0.324(146 ) 1.25 = b3 b 0.19 b = 0.2902 in Table AT 15, use b = 0.313 in , # 1 wire size D 1 .5 C= m = = 4 .8 b 0.313 Figure AF 15, K q = 1.275 (b) δ = 2.45 FDm3 N c 2.45FDm3 N c = Gt 3 (b − 0.56t ) 0.44Gb 4 13 2.45(0.325)(1.5) N c = 32 0.44(11,500 )(0.313)4 N c = 7.34 3 (c) Solid height = b( N c + 2) = 0.313(7.34 + 2) = 2.92 in Free length = PN c + 2b F = 0.325 lb 2.4(0.325)(1.5) 2.4 FDm ss = Kq = 1.275 = 48.65 ksi 3 3 b (0.313) 0.6(146 ) solid stress = = = 109.2 ksi (0.313)0.19 109.2 13 solid deflection = = = 0.91 in 48.65 32 (P − b )N c = 0.91 (P − 0.313)(7.34) = 0.91 P = 0.437 in Page 52 of 70 SECTION 4 - SPRINGS use P = 7 in 16 27 7 Free length = PN c + 2b = (7.34 ) + 2(0.313) = 3.837 in ≈ 3 in 32 16 7 P 16 tan λ = = πDm π (1.5) λ = 5.3o < 10o (d) U s = Us = 1 2 1 1 13 kδ = Fδ = (0.325) = 0.066 in − kip = 66 in − lb 2 2 2 32 66 = 0.085 Btu 778 302. A coil spring, of hard-drawn carbon steel, is to deflect 1 in. under a load of 100 lb. The outside coil diameter is to be 1 in. Compute the number of active coils, (a) if the wire is round, 5/32 in. in diameter, (b) if the wire is square, 5/32 in. on the side, (c) if the wire is rectangular 1/8 x 3/16 in., long dimension parallel to the axis, (d) If the wire is rectangular 3/16 x 1/8 in., short dimension parallel to the axis. (e) What is the maximum stress in each of the above springs under the 100-lb load? (f) What is the ratio of the approximate volumes, square- or rectangular-wire to round wire spring? Solution: Dm + Dw = 1 in 5 (a) Dw = in 32 5 27 Dm = 1 − = in 32 32 27 Dm 32 C= = = 5.4 Dw 5 32 8FC 3 N c δ= GDw 8(100 )(5.4 ) N c 5 11.5 × 10 6 32 N c = 14.3 3 1= ( Page 53 of 70 ) SECTION 4 - SPRINGS 5 in 32 5 27 Dm = 1 − = in 32 32 2.45 FD m3 N c δ= 0.44Gb 4 (b) Square, b = 3 27 2.45(100) N c 32 1= 4 6 5 0.44 11.5 × 10 32 N c = 20.5 ( ) 3 1 in , t = in 16 8 1 7 Dm = 1 − t = 1 − = in 8 8 3 2.45FD m N c δ= 3 Gt (b − 0.56t ) (c) b = 3 7 2.45(100 ) N c 8 1= 3 1 3 1 11.5 × 10 6 − 0.56 8 16 8 ( ) N c = 16.1 1 3 in , t = in 8 16 3 13 Dm = 1 − t = 1 − = in 16 16 3 2.45FD m N c δ= 3 Gt (b − 0.56t ) (d) b = 3 13 2.45(100 ) N c 16 1= 3 3 1 3 11.5 × 10 6 − 0.56 16 8 16 ( ) N c = 11.5 (e) Maximum Stress Page 54 of 70 SECTION 4 - SPRINGS 8 FC π Dw3 4(5.4) − 1 0.615 K= + = 1.284 4(5.4) − 4 5.4 For (a) ss = K 8(100)(5.4) ss = 1.284 = 72,320 psi 5 3 π 32 FD m (3b + 1.8t ) 2.4 FD m = Kq 2 2 2b t b3 27 Dm 32 C= = = 5.4 b 5 32 K q = 1.25 For (b) ss = K q 27 2.4(100) 32 = 66,355 psi ss = 1.25 3 5 32 FD m (3b + 1.8t ) For (c) ss = K q 2b 2t 2 7 D 8 C = m = =7 t 1 8 K q = 1 .1 7 (100) 8 3 3 + 1.8 1 = 68,992 psi ss = 1.1 2 3 1 2 16 8 2 16 8 FD m (3b + 1.8t ) For (d) ss = K q 2b 2t 2 13 Dm 16 C= = = 4.33 t 3 16 Page 55 of 70 SECTION 4 - SPRINGS K q = 1 .2 13 (100) 16 3 1 + 1.8 3 = 63,232 psi ss = 1.2 2 1 3 2 8 16 2 8 16 (e) Ratio of the approximate volumes For (a) Round wire π Va = Dw2 (π Dm )N c 4 π 5 2 27 Va = π (14.3) = 0.727 in 3 4 32 32 For (b) Square wire Vb = b 2 (π Dm )N c 2 5 27 Vb = π (20.5) = 1.327 in 3 32 32 For (c) rectangular wire Vc = bt (π Dm )N c 3 1 7 Vc = π (16.1) = 1.037 in3 16 8 8 For (d) rectangular wire Vd = bt (π Dm )N c 1 3 13 Vd = π (11.5) = 0.688 in 3 8 16 16 Ratio of volume Square to round wire V 1.327 = b = = 1.825 Va 0.727 Rectangular to round wire (long dimension parallel to the axis) V 1.037 = c = = 1.426 Va 0.727 Rectangular to round wire (short dimension parallel to the axis) V 0.688 = d = = 0.946 Va 0.727 Page 56 of 70 SECTION 4 - SPRINGS TENSION SPRINGS 305. Design two tension springs for a spring balance with a capacity of 200 lb. Each spring supports a maximum load of 100 lb. The outside diameter must not exceed 1 ¼ in. and the total length including end loops must not exceed 9 ½ in. Select a material and determine the dimension, including wire diameter, number of coils, and free length. Solution: Table AT 17, assume oil tempered wire 146 su = 0.19 ksi Dw 87.5 s ys = 0.19 ksi Dw 0.8(87.5) 70 ssd = = 0.19 ksi , [0.032 < Dw < 0.5] 0.19 Dw Dw F = Fi + kδ 8 K cFDm ss = π Dw3 2r D C= m = m Dw Dw 8FaC 3 N c G Dw GD k = 3w 8C N 3 G D 8F C N c kδ = 3 w a = Fa 8C N GDw Fa = 100 lb = 0.10 kip Figure AF 15, assume K c = 1.2 8 K cFi Dm 8 K cFa Dm ss = + π Dw3 π Dw3 8 K cFaC s s = K c si + π Dw2 OD = Dm + Dw = 1.25 in 1.25 Dw = C +1 § 6.21, assume si = 18 ksi ssd = ss δ= Page 57 of 70 SECTION 4 - SPRINGS 70 8(1.2 )(0.1)C = (1.2 )(18) + 0.19 Dw π Dw2 70(C + 1) (1.25)0.19 0.19 8(1.2 )(0.1)C (C + 1) π (1.25)2 2 = 21.6 + 67.1(C + 1) 0.19 = 21.6 + 0.1956C (C + 1) 2 67.1(C + 1) − 0.1956C (C + 1) = 21.6 C = 6 .7 1.25 1.25 = = 0.1623 in Dw = C + 1 6 .7 + 1 Table AT 15, use Dw = 0.1620 in , 8 W & M 0.19 2 Dm = CDw = (6.7 )(0.1620 ) = 1.085 in say Dm = 1.0 in D 1 .0 C= m = = 6.17 Dw 0.1620 si = 17.7 ksi To check, Fig. AF 15, K c = 1.15 8(1.15)(0.10 )(6.17 ) ss = 1.15(17.7 ) + = 89.20 ksi π (0.1620)2 70 ssd = = 98.92 ksi > 89.20 ksi , o.k. (0.1620)0.19 Total length = Dw N c + 2(Dm + Dw ) 9.5 = (0.162)N c + 2(1.0 + 0.162) N c = 44.3 coils Free length = Dw N c = (0.1620)(44.3) = 7.18 in Summary of answer: Material, oil-tempered wire Dw = 0.1620 in , 8 W & M N c = 44.3 coils Free length = 7.18 in. 306. Two helical tension springs are to be used in scales for weighing milk. The capacity of the scales is 30 lb., each spring carries 15 lb. with a deflection of 3 9/16 in. The springs are made of No. 14, W & M steel wire, outside diameter, 29/32 in. (a) how many coils should each spring have? (b) What is the maximum stress in the wire? What material should be used? Page 58 of 70 SECTION 4 - SPRINGS Solution: kδ = 15 lb 9 δ = 3 in 16 (a) Table AT 15, No. 14 W &M Dw = 0.0800 in 29 − 0.0800 = 0.82625 in 32 D 0.82625 C= m = = 10.328 Dw 0.0800 Dm = OD − Dw = δ= 8(kδ )C 3 N c GDw 9 8(15)(10.328) N c = 16 11.5 ×106 (0.080) N c = 24.8 3 3 ( ) (b) F = Fi + kδ Fi = π si Dw3 8Dm § 6.21, C = 10.328 si = 11,272 psi Fi = π (11,272)(0.08)3 = 2.743 lb 8(0.82625) F = 2.743 + 15 = 17.743 lb Figure AF 15, K c = 1.09 8 K c FDm 8(1.09 )(17.743)(0.82625) ss = = = 79,476 psi πDw3 π (0.080)3 s 79,476 s ys ≈ s = = 99,345 psi = 99.345 ksi 0 .8 0 .8 Table AT 17, use Hard drawn wire 70 70 s ys = 0.19 = = 113 ksi > 99.345 ksi Dw (0.080)0.19 307. A tension spring for a gas-control lever is made of Dw = 0.078 in steel wire; inside diameter, 0.609 in.; number of coils, 55; free length including end loops, 5 9/16 in. When the spring is extended to a length of 6 5/16 in., it must exert a force 5 ½ lb.; it must extend to (a) the initial tension, (b) the stress in the spring caused by the initial tension (compare with the recommended Page 59 of 70 SECTION 4 - SPRINGS maximum values), (c) the stress caused by the 5 ½-lb load, (d) the maximum stress. What material should be used? (e) What energy is absorbed from the point where the load is the initial tension until the spring’s length is 6 5/16 in.? (Data courtesy Worthington Corporation.) Solution: Dw = 0.078 in Dm − Dw = 0.609 in Dm = 0.609 + 0.078 = 0.687 in D 0.687 C= m = = 8 .8 Dw 0.078 N c = 55 8FC 3 N c GDw 1 F = 5 lb 2 5 9 δ = 6 − 5 = 0.75 in 16 16 3 8(kδ )(8.8) (5) δ = 0.75 = 11.5 × 106 (0.078) kδ = 2.244 lb δ= ( ) (a) Fi = F − kδ = 5.5 − 2.244 = 3.256 lb 8 F C 8(3.256 )(8.8) (b) si = i 2 = = 12,000 psi π Dw π (0.078)2 § 6.21, C = 8.8 si = 13,300 psi > 12,000 psi , ok (c) F = 5.5 lb 8 K c FC ss = π Dw2 C = 8 .8 Figure AF 15 K c = 1.1 8(1.1)(5.5)(8.8) ss = = 22,284 psi π (0.078)2 (d) maximum stress kδ 2.244 k= = = 2.992 lb in δ 0.75 Page 60 of 70 SECTION 4 - SPRINGS F = kδ ′ 5 9 δ ′ = 9 − 5 = 3.75 in 16 16 F = Fi + kδ ′ = 3.256 + (2.992)(3.75) = 14.476 lb 8 K c FC 8(1.1)(14.476 )(8.8) ss = = = 58,651 psi π Dw2 π (0.078)2 Table AT 16 s 58,651 s ys ≈ s = = 73,300 psi = 73.3 ksi 0 .8 0 .8 Table AT 17, use Hard drawn wire 70 70 s ys = 0.19 = = 113.658 ksi > 73.3 ksi Dw (0.078)0.19 (e) U s = 1 2 1 2 kδ = (2.992 )(0.75) = 0.8415 in − lb 2 2 TORSION SPRINGS 308. A carbon-steel (ASTM A230) torsion spring is to resist a force of 55 lb. at a radius of 2 in.; the mean diameter is to be 2 ½ in. Compute (a) the diameter of the wire for average service, (b) the number of coils for a deflection of 180o under the given torque, (c) the energy the spring has absorbed when the force is 55 lb. Solution: T = M = Fa F = 55 lb a = 2 in T = M = (55)(2) = 110 in − lb Dm = 2.5 in 182 Table AT 17, su = 0.1 ksi , [0.093 < Dw < 0.25] Dw Average service 182 117.936 117,936 sd = (1.6 )(0.405)su = 0.648 0.1 = ksi = psi 0.1 Dw Dw0.1 Dw KMc I For round wire, assume K c = K ci = 1.08 , Table AT 18 D c= w 2 (a) ss = Page 61 of 70 SECTION 4 - SPRINGS I π Dw3 = c 32 (1.08)(110)(32) = 117,936 ss = π Dw3 Dw0.1 Dw = 0.2060 in < 0.25 in Table AT 15, use Dw = 0.2070 in , No. 5 W & M r D 2 = 9.66 > 9 , ok To check: = m = c Dw 0.2070 Table AT 18, K = 1.08 (1.08)(110)(32) = 136,430 psi ss = π (0.2070 )3 117,936 ssd = = 138,054 psi > 136,430 psi (0.2070)0.1 Therefore, use No. 5 W & M, Dw = 0.2070 in MπDm N c EI 6 E = 30× 10 psi (b) θ = I= π Dw4 64 θ = 180o = π 64 MDm N c θ= EDw4 64(110 )(2 )N c π= (30 ×106 )(0.2070)4 N c = 12.29 1 1 (c) U s = Tθ = (110 )(π ) = 172.8 in − lb 2 2 312. A pivoted roller follower is held in contact with the cam by a torsion spring. The moment exerted by the spring varies from 20 lb-in to 50 lb-in. as the follower oscillates through 30o. The spring is made of AISI 6152 steel, OQT 1000 oF. What should be the value of Dw , Dm , and N c if the factor of safety is 1.75 based on the Soderberg line? Would this be a conservative or risky approach? Solution: AISI 6152, OQT 1000 oF su = 184 ksi Page 62 of 70 SECTION 4 - SPRINGS s y = 173 ksi sn = 0.5su = 92 ksi 1 M m = (50 + 20 ) = 35 lb − in 2 1 M a = (50 − 20 ) = 15 lb − in 2 assume K = 1.08 32 KM m 32(1.08)(35) 385 sm = = = 3 psi πDw3 πDw3 Dw 32 KM a 32(1.08)(15) 165 sa = = = 3 psi πDw3 πDw3 Dw 1 s m sa = + N s y sn 1 385 165 = + 3 1.75 173,000 Dw 92,000 Dw3 Dw = 0.1916 in Table AT 15, use Dw = 0.1920 in , No. 6 W & M To solve for K 32(35)K sm = = 50,369 K psi π (0.1920)3 32(15)K sa = = 21,587 K psi π (0.1920)3 1 50,369 K 20,587 K = + 1.75 173,000 92,000 K = 1.0868 Table AT 18 K ci = K = 1.0868 r Dm = = 9.32 > 9 , ok c Dw Dm = 9.32(0.1920) = 1.7894 in 7 use Dm = 1 in = 1.875 in 8 ∆MπDm N c 64∆MDm N c ∆θ = = EI EDw4 30π 64(50 − 20 )(1.875)N c = 180 (30 ×106 )(0.1920)4 N c = 5.93 Summary of answer: Dw = 0.1920 in , No. 6 W & M Page 63 of 70 SECTION 4 - SPRINGS 7 Dm = 1 in 8 N c = 5.93 , N > 1.4 , therefore conservative. FLAT AND LEAF SPRINGS 315. A cantilever flat spring of uniform strength, Fig. 6.20, Text, is to absorb an energy impact of 500 ft-lb. Let the thickness of the steel, AISI 1095, OQT 900 o F, be ½ in. and let the maximum stress be half of the yield strength. (a) Find the width b of the spring at the widest point in terms of the length L . Determine values of b for lengths of 36 in., 48 in., 60 in., and 72 in. (b) Determine the deflection of the spring for each set of values found in (a). Solution. Fig. 6/20 6 FL bh 2 6 FL3 δ= Ebh3 AISI 1095, OQT 900 oF, s y = 104 ksi , Table AT 9 sB = s B = 0.5s y = 0.5(104 ) = 52 ksi = 52,000 psi 1 Fδ 2 s bh 2 F= B 6L s bh 2 L3 s B L2 = δ = 6 B 3 Eh 6 L Ebh Us = 1 s bh 2 s B L2 1 s B2 bhL = U s = B 2 6 L Eh 12 E U s = 500 ft − lb = 6000 in − lb Page 64 of 70 SECTION 4 - SPRINGS (52,000)2 b 1 L 1 2 12 30 × 106 bL = 1598 in 2 6000 = b= 1598 in 2 L L = 36 in , b = 1598 in 2 = 44.4 in 36 in L = 48 in , b = 1598 in 2 = 33.3 in 48 in L = 60 in , b = 1598 in 2 = 26.6 in 60 in L = 72 in , b = 1598 in 2 = 22.2 in 72 in (b) δ = s B L2 Eh 2 ( 52,000 )(36 ) L = 36 in , δ = = 4.4928 in 1 30 × 10 2 2 ( 52,000 )(48) L = 48 in , δ = = 7.9872 in 6 1 30 ×10 2 2 ( 52,000 )(60 ) L = 60 in , δ = = 12.48 in 6 1 30 ×10 2 2 ( 52,000)(72 ) L = 72 in , δ = = 17.9712 in 6 1 30 × 10 2 ( 317. 6 ) ( ) ( ) ( ) One of the carbon contacts on a circuit breaker is mounted on the free end of a phosphor-bronze beam ( µ = 0.35 ). This beam has the shape of the beam 9 1 1 shown in Fig. 6.24, Text, with b = 1 in. , b′ = in. , L = 4 in. , and h = in. 16 2 16 When the contacts are closed, the beam deflects ¾ in. Compute (a) the force on the contacts, (b) the maximum stress. Solution: Page 65 of 70 SECTION 4 - SPRINGS Figure 6.24 6 FL 3WL s= 2 = bh bh 2 K FL3 1 − µ 2 K WL3 1 − µ 2 δ= 1 = 1 3EI 6 EI 9 b′ 16 = = 0.5625 b 1 Figure 6.25, K1 = 1.14 ( ) ( ) (a) Force on contacts = F K FL3 1 − µ 2 δ= 1 3EI E = 16× 106 psi (phosphor bronze) ( ) bh3 12 3 4 K FL 1 − µ 2 δ= 1 Ebh 3 3 2 3 4(1.14 )F (4.5) 1 − (0.35) = 3 4 (16 ×106 )(1) 161 F = 8 lb as a beam, I = ( ) [ (b) s = 318. ] 6 FL 6(8)(4.5) = = 55,296 psi 2 bh 2 1 (1) 16 A cantilever leaf spring 26 in. long is to support a load of 175 lb. The construction is similar to that shown in Fig. 6.22 (a), Text. The leaves are to be 2 in. wide, 3/16 in. thick; SAE 9255 steel, OQT 1000 oF; 107 cycles (§ 6.26). (a) How many leaves should be used if the surfaces are left as rolled? (b) The same as (a) except that the leaves are machined and the surfaces are not decarburized. (c) The same as (b), except that the surface is peened all over. (d) Which of these springs absorbs the most energy? Compute for each: (e) What are the load and deflection of the spring in (b) when the maximum stress is the standard-test yields strength? Solution: Figure 6.22 (a) 6 FL sA = 2 bh Page 66 of 70 SECTION 4 - SPRINGS 6 FL3 Ebh3 b = N1b′ F = 175 lb b′ = 2 in 3 h = in 16 L = 26 in § 6.26, SAE 9255, OQT 1000 oF su = 180 ksi s y = 160 ksi δA = 3 in = 0.1875 in 16 sd = 83.75 ksi t= (a) As rolled, Figure AF 5 Surface factor = 0.275 sd = 0.275(83.75) = 23 ksi = 23,000 psi 6 FL sA = 2 bh 6(175)(26 ) 23,000 = 2 3 N 1 (2 ) 16 N1 = 16.88 say N1 = 17 (b) Machined, Figure AF 5 Surface factor = 0.75 sd = 0.75(83.75) = 62.8 ksi = 62,800 psi 6 FL sA = 2 bh 6(175)(26 ) 62,800 = 2 3 N 1 (2 ) 16 N1 = 6.2 say N1 = 7 (c) Peened surface, (b) Page 67 of 70 SECTION 4 - SPRINGS sd = 1.25(62.8) = 78.5 ksi = 78,500 psi 6 FL sA = 2 bh 6(175)(26 ) 78,500 = 2 3 N 1 (2 ) 16 N1 = 4.95 say N1 = 5 1 Fδ 2 F = 175 lb (d) U s = 6 FL3 EN1b′h 3 For (a) N1 = 17 δ= 6(175)(26 ) 3 δ= = 2.745 in 3 3 30 × 10 (17 )(2 ) 16 1 U s = (175)(2.745) = 240 in − lb 2 For (b) N1 = 7 ( 6 ) 6(175)(26 ) 3 δ= = 6.666 in 3 3 30 ×10 6 (7 )(2 ) 16 1 U s = (175)(6.666 ) = 583 in − lb 2 For (c) N1 = 5 ( ) 6(175)(26 ) 3 δ= = 9.332 in 3 3 30 × 10 (5)(2 ) 16 1 U s = (175)(9.332 ) = 817 in − lb 2 ( 6 ) answer – spring (c) (e) sd = s y = 160 ksi , N1 = 7 (b) Page 68 of 70 SECTION 4 - SPRINGS 6 FL 6 FL = 2 bh N1b′h 2 6 F (26 ) 160,000 = 2 (7 )(2) 3 16 load F = 505 lb sd = 6(505)(26 ) 3 δ= 3 30 × 10 (7 )(2 ) 16 ( 319. 6 ) 3 = 19.24 in The rear spring of an automobile has 9 leaves, each with an average thickness of 0.242 in. and a width of 2 in.; material is SAE 9261, OQT 1000 oF. The length of the spring is 56 in. and the total weight on the spring is 1300 lb. Assume the spring to have the form shown in Fig. 6.22 (b), Text. Determine (a) the rate of the spring, (b) the maximum stress caused by the dead weight. (c) What approximate repeated maximum force (0 to Fmax ) would cause impending fatigue in 105 cycles, the number of applications of the maximum load expected during the ordinary life of a car? (If the leaves are cold rolled to induce a residual compressive stress on the surfaces, the endurance limit as su 2 should be conservative.) Solution: Figure 6.22 (b) 3FL sA = 2bh 2 3FL3 δA = 8Ebh 3 F = 1300 lb h = 0.242 in N1 = 9 b′ = 2 in L = 56 in (a) Rate , k = k= F δA = ( 8Ebh3 3L3 ) 8 EN1b′h 3 8 30 × 10 6 (9 )(2 )(0.242 ) = = 116.21lb in 3 3L3 3(56 ) (b) s A = 3 3FL 3(1300 )(56 ) = = 103,590 psi 2 2 2 N1b′h 2(9 )(2 )(0.242 ) Page 69 of 70 SECTION 4 - SPRINGS (c) SAE 9261. OQT 1000 oF su = 192 ksi 192 sn = = 96 ksi 2 3FL sA = 2 N1b′h 2 3F (56 ) 96,000 = 2 2(9 )(2 )(0.242 ) F = 1200 lb 321. The front spring of an automobile is made similar to Fig. 6.23, Text. The average thickness for each of the 6 leaves, 0.213 in.; material is SAE 9255, OQT 1000 oF. The load caused by the weight of the car is 775 lb. (a) What stress is caused by a force of twice the dead weight? (b) What load would stress the spring to the yield strength? Solution: Figure 6.23 6 FL 3WL 3WL s= 2 = = 2 bh bh N1b′h 2 W = 775 lb , N1 = 6 , b′ = 2 in , h = 0.213 in 36 in = 18 in L= 2 (a) W = 2(775) = 1550 lb 3(1550 )(18) s= = 153,740 psi (6)(2)(0.213)2 (b) SAE 9255, OQT 1000 oF s y = 160 ksi 6 FL 3WL 3WL = = 2 2 bh bh N1b′h 2 3W (18) 160,000 = (6)(2)(0.213)2 W = 1613 lb s= - end - Page 70 of 70 SECTION 5 – COLUMNS DESIGN PROBLEMS 334. A round steel rod made of structural steel, AISI C1020, as rolled, is to be used as a column, centrally loaded with 10 kips; N = 3 . Determine the diameter for (a) L = 25 in. , (b) L = 50 in. (c) The same as (a) and (b) except that the material is AISI 8640, OQT 1000 F. Is there any advantage in using this material rather than structural steel? Solution: For AISI C1020,as rolled s y = 48 ksi F = 10 kips N =3 (a) Le = L = 25 in. Consider first J.B. Johnson 2 Le sy k Fc = NF = s y A1 − 2 4π E 2 πD A= 4 D k= 4 E = 30 × 103 ksi 2 (48) 25 D 2 (3)(10) = (48) πD 1 − 2 4 3 4 4π 30 ×10 4 30 = 12πD 2 1 − 2 2 π D 48 30 = 12πD 2 − ( π D = 1.096 in 1 say D = 1 in = 1.0625 in 16 Page 1 of 18 ) SECTION 5 – COLUMNS Le 25 = = 94 < 120 ∴ o.k. k 1.0625 4 (b) Le = L = 50 in. Consider Euler’s Equation π 2 EA Fc = NF = 2 Le k πD 2 π 2 30 ×103 4 (3)(10 ) = 2 50 D 4 3 4 30 = 0.1875π D D = 1.507 in 1 say D = 1 in = 1.5 in 2 Le 50 = = 133 > 120 ∴ o.k. k 1 .5 4 ( ) (c) For AISI 8640, OQT 1000 F s y = 150 ksi 1 Le 2π 2 E 2 = k s y ( ) 1 Le 2π 2 30 × 103 2 = = 62.83 k 150 For (a) Le = L = 25 in. Consider first J.B. Johnson 2 Le sy k Fc = NF = s y A1 − 2 4π E Page 2 of 18 SECTION 5 – COLUMNS 2 (150) 25 D 2 (3)(10) = (150) πD 1 − 2 4 3 4 4π 30 ×10 12.5 30 = 37.5πD 2 1 − 2 2 π D 468.75 30 = 37.5πD 2 − ( ) π D = 1.23 in say D = 1.25 in Le 25 = = 80 > 62.83 ∴ use Euler’s equation k 1.25 4 π 2 EA Fc = NF = 2 Le k πD 2 π 2 30 ×103 4 (3)(10) = 2 25 D 4 3 4 30 = 0.75π D D = 1.0657 in 1 say D = 1 in = 1.0625 in 16 Le 25 = = 94 > 62.83 ∴ ok k 1.0625 4 For (b) Le = L = 50 in. Consider Euler’s Equation π 2 EA Fc = NF = 2 Le k ( Page 3 of 18 ) SECTION 5 – COLUMNS πD 2 4 π 2 (30 ×103 ) (3)(10) = 2 50 D 4 3 4 30 = 0.1875π D D = 1.507 in 1 say D = 1 in = 1.5 in 2 Le 50 = = 133 > 62.83 ∴ o.k. k 1 .5 4 There is no advantage. 335. A hollow circular column, made of AISI C1020, structural steel, as rolled, is to support a load of 10,000 lb. Let L = 40 in , Di = 0.75Do , and N = 3 . Determine Do by (a) using either Euler’s or the parabolic equation; (b) using the straightline equation. (c) What factor of safety is given by the secant formula for the dimensions found in (a)? Solution: For AISI C1020, as rolled s y = 48 ksi Le = L = 40 in F = 10,000 lb = 10 kips N =3 Di = 0.75Do I A π (Do4 − Di4 ) 4 I= = π Do4 − (0.75Do ) = 0.033556 Do4 64 π (Do2 − Di2 ) π Do2 − (0.75Do )2 A= = = 0.343612 Do2 4 4 k= [ [ k= 0.033556 Do4 = 0.3125Do 0.343612 Do2 Page 4 of 18 ] ] SECTION 5 – COLUMNS (a) Consider parabolic equation 2 Le sy k Fc = NF = s y A1 − 2 4π E 25 (48) 0.3125Do (3)(10) = (48)(0.343612)Do2 1 − 2 4 3 4π 30 × 10 30 = 16.493376 Do2 − 10.9519 Do = 1.576 in 9 say Do = 1 in = 1.5625 in 16 Le 40 = = 82 < 120 ∴ o.k. k 0.3125(1.5625) ( (b) Straight-line equation F L = 16,000 − 70 A k 10,000 40 = 16,000 − 70 2 0.343612 Do 0.3125 Do 10,000 = 5498 Do2 − 3078 Do Do = 1.6574 in 5 say Do = 1 in = 1.625 in 8 Le 40 = = 78.8 < 120 ∴ o.k. k 0.3125(1.625) (c) Secant formula NF ec Le NF sy = 1 + sec A k 2 2k EA Do = 1.5625 in k = 0.3125Do = 0.4883 in A = 0.343612 Do2 = 0.8389 in 2 Page 5 of 18 ) 2 SECTION 5 – COLUMNS ec = 0.25 , (i7.8) k2 48 = N (10) 40 1 + 0.25 sec 0.8389 2(0.4883) [ ( 48 = 11.92 N 1 + 0.25 sec 0.81645 N N = 2.289 336. )] A column is to be built up of ½-in., AISI C1020, rolled-steel plates, into a square box-section. It is 6 ft long and centrally loaded to 80,000 lb. (a) Determine the size of section for N = 2.74 . (b) Compute N from the secant formula for the size found and compare with 2.74. Solution: For AISI C1020, rolled-steel plate s y = 48 ksi 4 4 b 4 (b − 1) b 4 − (b − 1) − = 12 12 12 2 2 A = b − (b − 1) I= 4 k= 10 N (30 ×103 )(0.8389) I b 4 − (b − 1) = 2 A 12 b 2 − (b − 1) [ ] Le = L = 6 ft = 72 in F = 80,000 lb = 80 kips (a) N = 2.74 Consider J.B. Johnson 2 Le sy k NF = s y A1 − 2 4π E 2 72 (48) (2.74)(80) = (48)A 1 − 2 k 3 4π (30 ×10 ) 10.085 A 219.2 = 48 A − k2 Page 6 of 18 SECTION 5 – COLUMNS try b = 3.23 in 4 4 ( 3.23) − (3.23 − 1) k= 2 2 = 1.1331 in 12[(3.23) − (3.23 − 1) ] 2 2 2 A = b 2 − (b − 1) = (3.23) − (3.23 − 1) = 5.46 in 2 10.085(5.46 ) 219.2 = 48(5.46 ) − = 219.2 ok (1.1331)2 Therefore use b = 3.23 in Le 72 = = 63.54 < 120 ∴ o.k. k 1.1331 1 b = 3.23 in or b = 3 in 4 (b) s y = NF ec Le 1 + sec A k 2 2k NF EA ec = 0.25 , (i7.8) k2 48 = 72 N (80) 1 + 0.25 sec 5.46 2(1.1331) [ ( 80 N (30 ×103 )(5.46) 48 = 14.652 N 1 + 0.25 sec 0.70214 N N = 2.2 < 2.74 337. )] A column is to be made of ½-in structural steel plates (AISI 1020, as rolled), welded into an I-section as shown in Table AT 1 with G = H . The column, 15 ft long, is to support a load of 125 kips. (a) Determine the cross-sectional dimensions from the straight-line equation. (Using either Johnson’s or Euler’s equation, compute the equivalent stress and the factor of safety. (c) Compute N from the secant formula. Solution: For AISI C1020, as rolled s y = 48 ksi Le = L = 15 ft = 180 in F = 125 kips Page 7 of 18 SECTION 5 – COLUMNS Table AT 1. G=H A = GH − gh = H 2 − (H − 0.5)(H − 1) = H 2 − H 2 − 1.5H + 0.5 = 1.5H − 0.5 = 0.5(3H − 1) ( k= ) 3 1 GH 3 − gh3 1 H 4 − (H − 0.5)(H − 1) = = 12 GH − gh 12 0.5(3H − 1) (a) Straight-line equation L F = 16,000 A1 − 0.0044 k (180) 125,000 = 16,000 A1 − 0.0044 k 0.792 7.8125 = A1 − k use H = 7.37 in (7.37 )4 − (7.37 − 0.5)(7.37 − 1)3 6(3(7.37 ) − 1) A = 0.5[3(7.37 ) − 1] = 10.555 in k= = 3.04527 in 0.792 7.8125 ≈ 10.5551 − = 7.81 3.04527 Therefore use H = 7.37 in 3 Or H = 7 in = 7.375 in 8 L 180 = 59 < 120 (b) Consider J.B. Johnson, e = k 3.04527 s se = y N F 125 se = = = 13.8 ksi 2 2 Le 180 sy (48) k 3.04527 A 1− 10.555 1 − 4π 2 E 4π 2 30 ×103 ( Page 8 of 18 ) 3 H 4 − (H − 0.5)(H − 1) 6(3H − 1) SECTION 5 – COLUMNS N= sy 48 = = 3.48 se 13.8 (c) From secant formula NF ec Le NF sy = 1 + sec A k 2 2k EA ec = 0.25 , (i7.8) k2 48 = 180 N (125) 1 + 0.25 sec 10.555 2(3.04527 ) [ ( 48 = 11.843 N 1 + 0.25 sec 0.5872 N N = 2 .8 338. 125 N 3 (30 ×10 )(10.555) )] The link shown is to be designed for N = 2.5 to support an axial compressive load that varies from 0 to 15 kips; L = 20 in ; Material AISI 1030, as rolled. (a) Determine the diameter considering buckling only. (b) Determine the diameter considering varying stresses and using the Soderberg line (perhaps too conservative). Estimate an appropriate strength-reduction factor (see Fig. AF 6). (c) Keeping in mind that the stress is always compressive, do you think that the answer from (a) will do? Discuss. Problem 338. Solution: For AISI C1030, as rolled s y = 51 ksi su = 80 ksi 1 1 Le 2π 2 E 2 2π 2 30 ×103 2 = = = 108 k s y 51 L = 20 in N = 2 .5 (a) F = 15 kips Consider J.B. Johnson ( Page 9 of 18 ) SECTION 5 – COLUMNS 2 Le s y k NF = s y A1 − 2 4π E D k= 4 πD 2 A= 4 Le = L = 20 in 2 20 (51) D 2 (2.5)(15) = (51) πD 1 − 2 4 3 4 4π (30 ×10 ) 2.72 37.5 = 12.75π D 2 1 − 2 2 π D 34.68 37.5 = 12.75π D 2 − π D = 1.101 in 3 say D = 1 in = 1.1875 in 16 Le 20 = = 68 < 108 ∴ o.k. k 1.1875 4 (b) Variable stresses sn = 0.5su = 0.5(80) = 40 ksi Size factor = 0.85 sn = 0.85(40) = 34 ksi K f = 2.8 (Figure AF 6) 1 sm K f sa = + N sy sn F = 0 to 15 kips Fm = Fa = 7.5 kips sem = sea (2.8)sem 1 s = em + 2.5 51 34 sem = 3.923 ksi Page 10 of 18 SECTION 5 – COLUMNS 2 Le s em k Fm = sem A1 − 2 4π E 2 20 (51) D 2 πD 1 − 2 4 3 7.5 = (3.923) 4 4π (30 ×10 ) 2.72 7.5 = 0.98π D 2 1 − 2 2 π D 2.67 7.5 = 0.98π D 2 − π D = 1.65 in 5 say D = 1 in = 1.625 in 8 Le 20 = = 49 < 108 ∴ o.k. k 1.625 4 (c) The answer in (a) will not do because it is lower than (b) 339. The connecting link for a machine (see figure) is subjected to a load that varies fro + 450 (tension) to –250 lb. The cross section is to have the proportions G = 0.4 H , t = 0.1H , fillet radius r ≈ 0.05 H ; L = 10 in ; material, AISI C1020, as rolled. (a) Considering buckling only, determine the dimensions for a design factor of 2.5. (b) For the dimension found compute the factor of safety from the Soderberg criterion. Problems 339, 340 Solution: For AISI C1030, as rolled s y = 48 ksi su = 65 ksi Page 11 of 18 SECTION 5 – COLUMNS Table AT 1 G = 0 .4 H t = 0.1H r ≈ 0.05 H A = GH − gh g = G − t = 0.4 H − 0.1H = 0.3H h = H − 2(0.1H ) = 0.8 H A = (0.4 H )(H ) − (0.3H )(0.8H ) = 0.16 H 2 k= 3 3 1 GH 3 − gh3 1 (0.4 H )(H ) − (0.3H )(0.8 H ) = = 0.35824 H 12 GH − gh 12 0.16 H 2 ( ) (a) Consider J.B. Johnson 2 Le s y k NF = s y A1 − 2 4π E F = 350 lb = 0.35 kip Le = 10 in 2 10 48 2 ( 0 . 35824 H ) 2 (2.5)(0.35) = (48) 0.16 H 1 − 2 3 4π 30 × 10 0.875 = 7.68 H 2 − 0.2425 H = 0.3815 in Le 10 = = 73 < 120 ∴ ok k (0.35824)(0.3815) 15 say H = in = 0.46875 in 32 3 G = 0.4 H = 0.1875 in = in 16 3 t = 0.1H = 0.046875 in = in 64 ( Page 12 of 18 ) ( ) SECTION 5 – COLUMNS (b) with H = 0.46875 in 2 A = 0.16(0.46875) = 0.0352 in 2 k = 0.35824(0.46875) = 0.1679 in − 350 Fmin 0.0352 A smin = s e = = = −11,600 psi = −11.6 ksi 2 2 Le 10 s y 48 1 − k 1 − 0.1679 4π 2 E 4π 2 30 × 103 F + 450 smax = max = = +12,800 psi = +12.8 psi A 0.0352 1 sm = (12.8 − 11.6 ) = 0.6 ksi 2 1 sa = (12.8 + 11.6 ) = 12.1 ksi 2 su = 0.5sn = 0.5(65) = 32.5 ksi Size factor = 0.85 su = 0.85(32.5) ksi = 27.62 ksi ( ) Figure AF 9, r = 0.05H = 0.05(0.46875) = 0.023 h = 1.5H = 15.(0.46875) = 0.7031 in d = H = 0.4688 in r 0.05 H = = 0.05 d H h 1 .5 H = = 1 .5 d H K t = 2.65 1 1 q= = = 0.70 0.01 0.01 1+ 1+ r 0.023 K f = 0.70(2.65 − 1) + 1 = 2.2 1 sm K f sq = + N sy sn 1 0.6 (2.2 )(12.1) = + N 48 27.62 N = 1.024 CHECK PROBLEMS 341. The link shown is subjected to an axial compressive load of 15 kips. Made of AISI C1030, as rolled, it has sectional length of 20 in. Assume a loose fit with the Page 13 of 18 SECTION 5 – COLUMNS pins. What is (a) the critical load for this column, (b) the design factor, (c) the equivalent stress under a load of 15 kips? What material does the secant formula indicate as satisfactory for the foregoing load, when (e) ec k 2 = 0.25 , (f) L e= e . 400 Problem 341, 342 Solution: For AISI C1030, as rolled s y = 51 ksi b = 0.75 in h = 1.75 in A = bh = (0.75)(1.75) = 1.3125 in 2 For loose fit bh3 I= 12 I bh 3 h 1.75 k= = = = = 0.5052 in A 12bh 12 12 Le 20 = = 39.6 < 108 for AISI C1030, as rolled k 0.5052 use J.B. Johnson equation 2 2 Le 20 sy 51 k 0.5052 (a) Fc = s y A 1 − = (51)(1.3125) 1 − = 62.42 kips 4π 2 (30,000) 4π 2 E (b) Fc = NF F 62.42 N= c = = 4.16 F 15 s 51 (c) se = y = = 12.26 ksi N 4.16 F 15 (d) Actual s = = = 11.43 ksi A 1.3125 Page 14 of 18 SECTION 5 – COLUMNS Secant Formula NF ec Le NF sy = 1 + sec A k 2 2k EA ec (e) 2 = 0.25 k 62.42 20 62.42 sy = 1 + 0.25 sec = 64.4 ksi 3 1.3125 2(0.5052) 30 × 10 (1.3125) use AISI C1020, cold drawn, s y = 66 ksi ( ) Le 20 = = 0.05 in 400 400 h 1.75 c= = = 0.875 in 2 2 ec (0.05)(0.875) = = 0.1714 k2 (0.5052)2 (f) e = sy = 62.42 20 1 + 0.1714 sec 1.3125 2(0.5052 ) 62.42 = 59.12 ksi 3 30 × 10 (1.3125) ( ) use AISI C1045, cold drawn, s y = 59 ksi 343. A schedule-40, 4-in. pipe is used as a column. Some of its properties are: Do = 4.5 in , Di = 4.026 in , I = 3.174 sq.in. , L = 15 ft ; material equivalent to AISI C1015, as rolled. The total load to be carried is 200 kips. (a) What minimum number of these columns should be used if a design factor of 2.5 is desired and the load evenly distributed among them? For the approximately fixed ends, use Le = 0.65L as recommended by AISC. (b) What is the equivalent stress in the column? Solution: For AISI C1015, as rolled s y = 45.5 ksi 1 1 Le 2π 2 E 2 2π 2 30 × 103 2 = = = 114 k s y 45.5 L = 15 ft = 180 in Le = 0.65L = 0.65(180) = 117 in Le 117 = = 77.5 < 114 k 1.509 ( Page 15 of 18 ) SECTION 5 – COLUMNS Use J.B. Johnson equation 2 Le sy 2 s y A k (45.5)(3.174) 45.5(77.5) (a) F = 1− = 1 − 2 = 44.4 kips N 4π 2 E 2.5 4π (30,000) No. of columns 200 = = 4.5 say 5 columns 44.4 F A (b) se = 2 Le s y 1 − k 4π 2 E 200 F= = 40 kips 5 40 3.174 se = = 16.4 ksi 2 77.5 45.5 k 1 − 4π 2 (30,000) 344. A generally loaded column is a 10-in. x 49 lb., wide-flange I-beam whose properties are (see figure); k x = 4.35 in , k y = 2.54 in , area A = 14.4 sq.in. , I x = 272.9 in 4 , I y = 93.0 in 4 ; length L = 30 ft , material AISI 1022, as rolled. Let the ends be a “little” fixed with Le = 0.8L and determine the critical load (a) according to the Johndon or the Euler equation; (b) according to the secant formula if ec k 2 is assumed to be 0.25. Page 16 of 18 SECTION 5 – COLUMNS Solution: For AISI C1022, as rolled s y = 52 ksi 1 1 Le 2π 2 E 2 2π 2 (30 ×103 ) 2 = = = 107 k s y 52 (a) k = 2.54 in I = 93.0 in 4 Le = 0.8(30)(12) = 288 in Le 288 = = 113.4 > 107 k 2.54 Use Euler’s Equation π 2 EA π 2 (30,000)(14.4) Fc = = = 332 kips 2 (113.4)2 Le k (b) Secant formula NF ec Le NF sy = 1 + sec A k 2 2k EA 113.4 F 52 = c 1 + 0.25 sec 14.4 2 F 52 = c 1 + 0.25 sec 0.0863 14.4 Fc = 273 kips { 348. [ Fc 3 (30 ×10 )(14.4) Fc ]} A 4 x 3 x ½-in. angle is used as a flat-ended column, 5 ft. long, with the resultant load passing through the centroid G (see figure); k x = 1.25 in , k y = 0.86 in , ku = 1.37 in , kv = 0.64 in , A = 3.25 sq.in. Find the safe load if N = 2.8 and the material is (a) structural steel, (b) magnesium alloy AZ 91C (i7.12.\, Text), (c) magnesium alloy AZ 80A, (d) magnesium alloy AZ 80A as before, but use the Johnson formula and compare. Page 17 of 18 SECTION 5 – COLUMNS Solution: L (5)(12 ) Le = = = 30 in 2 2 k = k min = 0.64 in Le 30 = = 46.875 k 0.64 (a) Structural steel, s y = 48 ksi Le = 46.875 < 120 k use J.B. Johnson 2 Le sy 2 s y A k (48)(3.25) 48(46.875) F= 1− = 1 − 2 = 50.75 kips N 4π 2 E 2.8 4π (30,000 ) (b) magnesium alloy AZ 91C NF C = 2 A Le C k 1+ 6 64.4 × 10 C = 57,000 (2.8)(F ) = 57,000 psi 2 3.25 57,000(46.875) 1+ 64.4 ×106 F = 22,467 lb = 22.467 kips (c) magnesium alloy AZ 80A C = 82,900 (2.8)(F ) = 82,900 psi 2 3.25 82,900(46.875) 1+ 64.4 × 106 F = 25,134 lb = 25.134 kips (d) By J.B. Johnson For magnesium alloy AZ 80A, s y = 36 ksi 2 Le s y s y A (36)(3.25) 1 − 36(46.875)2 = 39 kips > 25.134 kips k F= 1− 2 = 2 N 4π E 2.8 4π (30,000) - end - Page 18 of 18 SECTION 6 – COMBINED STRESSES ECCENTRIC LOADING (NORMAL STRESSES) DESIGN PROBLEM 361. It is necessary to shape a certain link as shown in order to prevent interference with another part of the machine. It is to support a steady tensile load of 2500 lb. with a design factor of 2 based on the yield strength. The bottom edge of the midsection is displaced upward a distance a = 2 ½ in. above the line of action of the load. For AISI C1022, as rolled, and h ≈ 3b, what should be h and b? Solution: F = 2500 lb Ny = 2 a = 2.5 in F Fec σ= ± A I For AISI C1022, as rolled, sy = 52 ksi (Table AT7). σ= sy Ny = 52 = 26 ksi = 26,000 psi 2 A = bh = 3b2 bh 3 b(3b )2 I= = = 2.25b 4 12 12 h c = = 1.5b 2 h e = a + = a + 1.5b = 2.5 + 1.5b 2 F Fec σ= ± A I (2.5 + 1.5b)(1.5b) 1 26,000 = 2500 2 + 2.25b 4 3b 2.5 + 1.5b 1 26,000 = 2500 2 + 1.5b 3 3b By trial and error method: 5 b = 0.625 in = in 8 7 h = 3b = 3(0.625) in = 1.875 in = 1 in 8 Page 1 of 133 SECTION 6 – COMBINED STRESSES 362. A tensile load on a link as described in 361 varies from 0 to 3000 lb.; it is machined from AISI 1045, as rolled, and the lower edge of the link is a = 0.5 in. above the center line of the pins; h ≈ 3b. Determine the dimensions of the link for N = 2 based on the Soderberg line. Solution: Soderberg Line: 1 sm sa = + N s y sn For AISI 1045, as rolled (Table AT 7). sy = 59 ksi su = 96 ksi sn′ = 0.5su = 48 ksi Size factor = 0.85 Load factor (axial) = 0.80 sn = 0.85(0.80)(48) = 32.64 ksi 1 (3000 lb + 0 ) = 1500 lb 2 1 Fa = (3000 lb − 0 ) = 1500 lb 2 h e = a + = a + 1.5b = 0.5 + 1.5b 2 1 ec sm = Fm + A I (0.5 + 1.5b )(1.5b) 1 sm = 1500 2 + 2.24b 3 3b 0.5 + 1.5b 1 sm = 1500 2 + 1.5b 3 3b 1 ec sa = Fa + A I (0.5 + 1.5b)(1.5b ) 1 sa = 1500 2 + 3 b 2.24b 3 0.5 + 1.5b 1 sa = 1500 2 + 1.5b 3 3b 1 sm sa = + N s y sn Fm = Page 2 of 133 SECTION 6 – COMBINED STRESSES 1 1500 1500 1 0.5 + 1.5b = + + 2 59,000 32,640 3b 2 1.5b3 1 0.5 + 1.5b + = 7.00 2 3b 1.5b 3 By trial and error method: b = 0.53 in h = 3b = 3(0.53) in = 1.59 in Use b x h = 9/16 in x 1 5/8 in 363. The same as 362, except that the load continuously reverses, 3 kips to -3 kips. Solution: 1 Fm = (3 − 3) = 0 lb 2 1 Fa = (3 + 3) = 3 kips 2 1 ec sm = Fm + A I (0.5 + 1.5b )(1.5b) 1 sm = 1500 2 + 2.24b 3 3b 0.5 + 1.5b 1 sm = (0) 2 + =0 1.5b 3 3b 1 ec sa = Fa + A I (0.5 + 1.5b)(1.5b) 1 s a = 3 2 + 2.24b 3 3b 0.5 + 1.5b 1 sa = 3 2 + 1.5b 3 3b 1 sm sa = + N s y sn 1 3 1 0.5 + 1.5b =0+ 2 + 2 32.64 3b 1.5b 3 1 0.5 + 1.5b + = 5.44 2 3b 1.5b 3 By trial and error method: b = 0.5905 in Page 3 of 133 SECTION 6 – COMBINED STRESSES h = 3b = 3(0.5905) in = 1.7715 in Use b x h = 5/8 in x 1 13/16 in 364. A circular column (See Fig. 8.3, Text), the material of which is SAE 1020, as rolled, is to have a length of 9 ft. and support an eccentric load of 16 kips at a distance of 3 in. from the center line. Let N = 3. (a) What should be the outside diameter Do if the column is hollow and Di = 0.75Do? (b) What should be the diameter if the column is solid? Solution: a. Try J.B. Johnson 2 Le sy F k = se 1 − 2 A 4π E For SAE 1020, as rolled, s y = 48 ksi E = 30×10 6 psi Transition point 1 ( ) 1 Le 2π 2 E 2 2π 2 30 ×10 6 2 = = = 111 k sy 48000 Le = 2L L = 9 ft = 108 in Le = 2L = 2(108) = 216 in I A π Do4 − Di4 π Do4 − (0.75Do )4 0.6836πDo4 I= = = 64 64 64 k= ) [ ( A= π (D 4 2 o ) − Di2 = Page 4 of 133 π 4 ] [D 2 o ] − (0.75Do )2 = 0.109375πDo2 SECTION 6 – COMBINED STRESSES 0.6836πDo4 64 k= = 0.3125Do 0.109375πDo2 16 46.5642 F 2 0.109375πDo Do2 A se = = = 2 2 Le 216 1 − 19.363 s 48 y Do2 0.3125Do 1 − k 4π 2 E 1 − 4π 2 30 ×103 ( σ = se + Fec I Do = 0.5Do 2 e = 3 in F = 16 kips c= sy 48 = 16 ksi N 3 46.5642 Do2 16(3)(0.5Do ) σ = 16 = + 19.363 0.6836Do4 1 − 64 Do2 σ= = σ = 16 = 46.5642 Do2 19.363 1 − Do2 + 715.22 Do3 By trial and error method Do = 3.23 in k = 0.3125(3.23) = 1.0094 in Le 2L 216 = = = 214 > 111 k k 1.0094 Therefore use Euler’s equation π 2E se = 2 L N e k Page 5 of 133 ) SECTION 6 – COMBINED STRESSES se = π 2 (30,000) 216 3 0.3125Do Fec σ = se + I 16 = 0.20658Do2 + 2 = 0.20658Do2 ksi 16(3)(0.5Do ) 715.22 = 0.20658Do2 + 4 0.6836πDo Do3 64 Do = 3.802 in Di = 0.75Do = 0.75(3.802) = 2.8515 in To check: k = 0.3125Do = 0.3125(3.802) in = 1.188125 in Le 216 in = = 182 > 111 k 1.188125 in Use Do = 3 13/16 in, Di = 2 13/16 in b. For solid, also using Euler’s equation. I A πD 4 I= 64 πD 2 A= 4 k= πD 4 k= se = se = 64 = 1 D = 0.25D πD 2 4 4 π 2E 2 L N e k π 2 (30,000) 216 3 0.25D Fec σ = se + I 2 = 0.1322D 2 ksi 16(3)(0.5D ) 489 = 0.1322D 2 + 3 4 πD D 64 By trial and error method. 16 = 0.1322D 2 + Page 6 of 133 SECTION 6 – COMBINED STRESSES D = 3.221 in k = 0.25D = 0.25(3.221) in = 0.80525 in Le 216 in = = 268 > 111 k 0.80525 in Use D = 3 ¼ in. 365. The same as 364, except that the length is 15 ft. Solution: Euler’s Equation: Le = 2L = 2(15)(12 ) = 360 in σ = 16 ksi e = 3 in 0.6836πDo4 64 A = 0.109375πDo2 k = 0.3125Do a. I = se = se = π 2E L N e k 2 π 2 (30,000) 360 3 0.3125Do Fec σ = se + I 16 = 0.07437Do2 + 2 = 0.07437Do2 ksi 16(3)(0.5Do ) 715.22 = 0.07437Do2 + 4 0.6836πDo Do3 64 Do = 3.624 in Di = 0.75Do = 0.75(3.624 ) = 2.718 in To check: k = 0.3125Do = 0.3125(3.624) in = 1.1325 in Le 216 in = = 191 > 111 k 1.1325 in Use Do = 3 5/8 in, Di = 2 5/8 in Page 7 of 133 SECTION 6 – COMBINED STRESSES b. I = A= πD 4 64 πD 2 4 k = 0.25D se = se = π 2E 2 L N e k 2 π (30,000) 360 3 0.25D Fec σ = se + I 2 16 = 0.0476D 2 + = 0.0476D 2 ksi 16(3)(0.5D ) 489 = 0.0476D 2 + 3 4 πD D 64 By trial and error method. D = 3.158 in use 3 3/16 iin k = 0.25D = 0.25(3.158) in = 0.7895 in Le 216 in = = 274 > 111 k 0.7895 in Use D = 3 3/16 in. 366. A link similar to one shown is to be designed for: steady load F = 8 kips, L = 20 in. θ = 30o; aluminum alloy 2024-T4; N = 2.6 on the yield strength. It seems desirable for the dimension b not to exceed 1 3/8 in. Determine b and h and check their proportions for reasonableness. The support is made so that the pin at B carries the entire horizontal component of F. Page 8 of 133 SECTION 6 – COMBINED STRESSES Solution: Aluminum alloy (2024-T4), sy = 47 ksi RBH = F sin 30o FL cos 30o d F (L + d )cos 30o RBV = d M = RAd = FL cos 30o Mc s2 = I h c= 2 bh 3 I= 12 6M 6FL cos 30o s2 = 2 = bh bh 2 F sin 30o s1 = bh σ t = s1 + s2 RA = F sin 30o 6FL cos 30o + bh bh 2 s 47 σt = y = = 18 ksi N 2. 6 F = 8 kips 3 b = 1 in = 1.375 in 8 L = 20 in σt = σt = F sin 30o 6FL cos 30o + bh bh 2 Page 9 of 133 SECTION 6 – COMBINED STRESSES 6(8)(20)cos 30o 1.375h 1.375h 2 24.75h 2 − 4h − 831.4 = 0 18 = (8)sin 30o + 7 h = 5.877 in ≈ 5 in 8 7 5 in h = 8 = 4.27 b 1 3 in 8 7 3 Therefore, use h = 5 in , b = 1 in 8 8 367. A column 15 ft. long is to support a load F2 = 50,000 lb. Acting at a distance of e = 8 in. from the axis of the column as shown (with F1 = 0). Select a suitable I-beam for a design factor of 3 based on yield strength. The upper end of the column is free. See handbook for the properties of rolled sections. Solution: Use C1020, structural steel, sy = 48 ksi Secant Formula NF ec L NF 1 + 2 sec e sy = A k 2 EI F = F2 = 50,000 lbs = 50 kips e = 8 in N =3 E = 30,000 ksi depth c= 2 Le = 2L = 2(15)(12 ) = 360 in Page 10 of 133 SECTION 6 – COMBINED STRESSES (8) depth ( )( ) 3(50) 360 3 50 2 sec 48 ≥ 1 + A k2 2 (30,000)(I ) rd From Strength of Materials, 3 Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select Wide-Flange Sections by trial and error. Then selecting W360 x 51, properties are A = 6450 mm2 = 10 in2 Depth = 355 mm = 14 in k = 148 mm = 5.83 in I = 1.41 x 108 mm4 = 338.8 in4 Substitute, (8) 14 (3)(50) 3(50) 2 sec 360 48 ≥ 1 + 2 10 (5.83) 2 (30,000)(338.8) 48 ≥ 47.08 Therefore suitable wide flange I-beam is W14 x 34 lb. (English units) A = 10 in2 Depth = 14 in k = 5.83 in I = 338.8 in4 368. The same as 367, except that F1 = 50,000 lb. Solution: Use C1020, structural steel, sy = 48 ksi N =3 E = 30,000 ksi Le = 2L = 2(15)(12 ) = 360 in Transition Point 1 1 Le 2π 2E 2 2π 2 (30,000) 2 = = = 111 k sy 48 F = F1 = 50,000 lbs = 50 kips Page 11 of 133 SECTION 6 – COMBINED STRESSES Check J.B. Johnson Formula 2 Le sy k Fc = NF1 = sy A 1 − 2 4π E NF1 sy = 2 Le sy k A 1 − 2 4π E 3(50 ) 48 ≥ 2 360 48 k A 1 − 2 4π (30,000) 150 48 ≥ 5.2549 A1 − k2 From Strength of Materials, 3rd Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select Wide-Flange Sections by trial and error. Then selecting W310 x 21, properties are A = 2690 mm2 = 4.17 in2 k = 117 mm = 4.61 in Substitute, 150 48 ≥ 5.2549 4.17 1 − 2 (4.61) 48 ≥ 150 5.2549 4.17 1 − 2 (4.61) 48 ≥ 47.8 Check for validity of JB Johnson Formula Le 360 = = 78.1 < 111 k 4.61 Therefore, JB Johnson formula is valid and suitable wide flange I-beam is W12 x 14 lb. (English units) A = 4.17 in2 k = 4.61 in Page 12 of 133 SECTION 6 – COMBINED STRESSES CHECK PROBLEMS 369. A cam press, similar to that of Fig. 19-1, Text, exerts a force of 10 kips at a distance of 7 in. from the inside edge of the plates that make up the frame. If these plates are 1 in. thick and the horizontal section has a depth of 6 in., what will be the maximum stress in this section? Solution: σ= F Fec + A I F = 10 kips 6 e = 7 in + in = 10 in 2 6 c = in = 3 in 2 A = 2(1)(6 ) = 12 in 2 I=2 (1)(6 )3 12 = 36 in 4 F Fec + A I 10 (10)(10)(3) σ= + = 9.2 ksi 12 36 σ= 370. A manufacturer decides to market a line of aluminum alloy (6061-T6) C-clamps, (see Fig. 8.4, Text). One frame has a T-section with the following dimensions (letters as in Table AT 1): H = 1 1/16 , B = 17/32, a = 1/8, and t = 1/8. The center line of the screw is 2 3/8 in. from the inside face of the frame. (a) For N = 3 on the yield strength, what is the Page 13 of 133 SECTION 6 – COMBINED STRESSES capacity of the clamp (gripping force)? (b) Above what approximate load will a permanent deformation of the clamp occur? Solution: AA 6061-T6, sy = 40 ksi (Table AT 3) See Fig. 8.4, (C-clamp) T-section F Fe′c + A I 1 e′c σ = F + A I From Table AT 1 (T-section) A = Bt + a(H − t ) σ= I= Bt 2 ah 2 + (Bt )d 2 + + (ah )e 2 12 12 H = 1.0625 in B = 0.53125 in a = 0.125 in t = 0.125 in h = H − t = 1.0625 − 0.125 = 0.9375 in For c1 : h t c1 (Bt + ha ) = t + (ha ) + (Bt ) 2 2 0.9375 c1 [(0.53125)(0.125) + (0.9375)(0.125)] = 0.125 + (0.9375)(0.125) 2 0.125 + (0.53125)(0.125) 2 c1 = 0.4016 in c 2 = H − c1 = 1.0625 − 0.4016 = 0.6609 in Page 14 of 133 SECTION 6 – COMBINED STRESSES t 0.125 d = c1 − = 0.4016 − = 0.3391 in 2 2 h 0.9375 e = t + − c1 = 0.125 + 0.4016 = 0.19215 in 2 2 A = Bt + a(H − t ) A = (0.53125)(0.125) + (0.125)(1.0625 − 0.125) = 0.1836 in 2 Bt 2 ah 2 + (Bt )d 2 + + (ah )e 2 12 12 (0.53125)(0.125)2 (0.125)(0.9375)2 I= + (0.53125)(0.125)(0.3391)2 + 12 12 2 4 + (0.125)(0.9375)(0.19215) = 0.02063 in I= c = c1 = 0.4016 in e′ = 2.375 + 0.4016 = 2.7766 in sy 40 = 13 ksi N 3 (2.7766)(0.4016) 1 σ = 13 = F + 0.02063 0.1836 F = 0.218 kips = 218 lbs a. σ = = b. σ = sy = 40 ksi (2.7766)(0.4016) 1 + 0.02063 0.1836 F = 0.672 kips = 672 lbs σ = 40 = F 371. A C-frame (Fig. 8.5 Text) of a hand-screw press is made of annealed cast steel, ASTM A27-58 and has a section similar to that shown. The force F acts normal to the plane of the section at a distance of 12 in. from the inside face. The various dimensions of the sections are: a = 3 in., b = 6 in., h = 5 in., d = e= f = 1 in. Determine the force F for N = 6 based on the ultimate strength. Page 15 of 133 SECTION 6 – COMBINED STRESSES Solution: a = 3 in b = 6 in h = 5 in d = e = f = 1 in A = fa + dh + (b − f − d )e A = (1)(3) + (1)(5) + (6 − 1 − 1)(1) = 12 in 2 For c1 : f d b c1 A = hd + (b − f − d ) + af b − 2 2 2 1 1 6 c1 (12) = (5)(1) + (6 − 1 − 1) + (3)(1) 6 − 2 2 2 c1 = 2.583 in c 2 = b − c1 = 6 − 2.583 = 3.417 in Page 16 of 133 SECTION 6 – COMBINED STRESSES j= 6 −1 + 1 b − f −d b− f +d − 2.583 = 0.417 in − (c1 − d ) = − c1 = 2 2 2 f 1 = 3.417 − = 2.917 in 2 2 d 1 m = c1 − = 2.583 − = 2.083 in 2 2 k = c2 − af 3 e(b − f − d )3 hd 3 + afk 2 + + e(b − f − d ) j 2 + + hdm 2 12 12 12 (3)(1)3 (1)(6 − 1 − 1)3 (5)(1)3 I= + (3)(1)(2.917 )2 + + (1)(6 − 1 − 1)(0.417 )2 + + (5)(1)(2.083)2 12 12 12 I = 53.92 in 4 I= For ASTM A27-58 Annealed Cast Steel su = 60 ksi σ= su 60 = = 10 ksi N 6 Page 17 of 133 SECTION 6 – COMBINED STRESSES 1 e′c + A I c = c1 = 2.583 in e′ = 12 + 2.583 = 14.583 in σ = F 1 (14.583)(2.583) 10 = F + 53.92 12 F = 12.789 kips = 12,789 lbs In the link shown (366), let b = ½ in., h = 2 in., d = 2 in., L = 18 in., and θ = 60o. The clearance at the pins A and B are such that B resists the entire horizontal component of F; material is AISI C1020, as rolled. What may be the value of F for N = 3 based on the yield strength? 372. Solution: Refer to Prob. 366. F sinθ 6FL cos θ + σt = bh bh 2 sinθ 6L cos θ + σ t = F bh 2 bh For AISI C1020, as rolled, sy = 48 ksi . σ= sy N = 48 = 16 ksi 3 b = 0.5 in h = 2 in L = 18 in θ = 60o sin 60 6(18)cos 60 16 = F + (0.5)(2)2 (0.5)(2) F = 0.574 kips = 574 lbs 373. The link shown is subjected to a steady load F1 = 2.1 kips; b = 0.5 in., h = a = d = 2 in., L = 18 in.; material AISI 1040, cold drawn (10% work). The dimensions are such that all of the horizontal reaction from F2 occurs at A; and F2 varies from 0 to a maximum, acting Page 18 of 133 SECTION 6 – COMBINED STRESSES towards the right. For N = 1.5 based on the Soderberg line, what is the maximum value of F2? Assume that the stress concentration at the holes can be neglected. Solution: b = 0.5 in h = a = d = 2 in L = 18 in F1 = 2.1 kips ∑M A =0 (a + L + d )E = aF1 + hF2 E= aF1 + hF2 a+L+d ∑F V =0 Ay = F1 − E ∑F H =0 Ax = F2 For F2 = 0 : (2 )(2.1) + 0 = 0.191 kip E= 2 + 18 + 2 Ay = F1 − E = 2.1 − 0.191 = 1.909 kips Ax = F2 = 0 M = aAy = (L + d )E M= (L + d )(aF1 + hF2 ) a+L+d F2 = 0 Page 19 of 133 SECTION 6 – COMBINED STRESSES M= (18 + 2)[(2)(2.1) + 0] = 3.82 in − kips 2 + 18 + 2 Let F2 = max F2 Mmax + 3.82 2 ( 18 + 2)[2(2.1) + 2F2 ] Mmax = = 1.82(2.1 + F2 ) 2 + 18 + 2 1.82(2.1 + F2 ) + 3.82 Mm = = 3.82 + 0.91F2 2 Mm = Mmax − 3.82 2 1.82(2.1 + F2 ) − 3.82 Ma = = 0.91F2 2 Ma = Fm = Fa = 0.5F2 For SAE AISI 1040 Cold Drawn (10% Work) sy = 85 ksi sn′ = 54 ksi sn (bending ) = sn′ × size factor = (54)(0.85) = 45.9 ksi se = sm + se = sy sn sa M c F = m + m N I A sy sy Ma c Fa + + A sn I 85 (1.82 )(2.1 + F2 )(1) 0.5F2 85 (0.91F2 )(1) 0.5F2 = + + + 1 1.5 (0.5)(2) 45.9 1 (0.5)(2) 3 3 F2 = 3.785 kips 376. A free-end column as shown, L = 12 ft. long, is made of 10-in. pipe, schedule 40, (Do = 10.75 in., Di = 10.02 in., k = 3.67 in., Am = 11.908 in2., I = 160.7 in4., Z = 29.9 in3.). The load completely reverses and e = 15 in.; N = 3; material is similar to AISI C1015, as rolled. (a) Using the equivalent-stress approach, compute the safe (static) load as a column only. (b) Judging the varying loading by the Soderberg criterion, compute the safe maximum load. (c) Determine the safe load from the secant formula. (d) Specify what you consider to be a reasonable safe loading. Page 20 of 133 SECTION 6 – COMBINED STRESSES Solution: For AISI C1015, as rolled. sy = 45.5 ksi su = 61 ksi sn = 0.5su = 0.5(61) = 30.5 ksi Do = 10.75 in Di = 10.02 in L = 12 ft k = 3.67 in Am = 11.908 in 2 I = 160.07 in 4 Z = 29.9 in3 N =3 a. As a column only (static) Le = 2L = 2(12)(12) = 288 in Le 288 in = = 78.5 < 120 k 3.67 in E = 30,000 ksi Use J.B. Johnson Formula: F s1 = 2 Le sy k A 1 − 2 4π E Page 21 of 133 SECTION 6 – COMBINED STRESSES F s1 = (45.5)(78.5)2 11.9081 − 2 4π (30,000) F s1 = 9.09 Fe s2 = Z e = 15 in F (15) F s2 = = 29.9 2 σ = s1 + s2 45.5 F F = + 3 9.09 2 F = 24.863 kips = 24,863 lbs b. Varying load: Fmax = F , Fmin = −F F −F =0 2 F − (− F ) Fa = =F 2 s se = sm + y sa sn Axial load factor = 0.80 sn = sn′ × size factor × load factor = (30.5)(0.85)(0.80) = 20.74 ksi Fm = F e F s F e F se = m + m + y a + a A sn Z A Z sy 45.5 F (15) F se = = (0) + + = 2.25F N 20.74 29.9 1.908 45.5 = 2.25F 3 F = 6.74 kips = 6740 lbs c. Secant Formula sy = Z= NF ec L 1 + 2 sec e A k 2 I c Page 22 of 133 NF EI SECTION 6 – COMBINED STRESSES I 160.07 = = 5.354 in Z 29.9 ec (15)(5.354) = = 5.963 k2 (3.67 )2 c= Le NF 288 3F = = 0.11382 F 2 EI 2 (30,000)(160.07 ) NF 3F = = 0.252F A 11.908 NF ec L NF 1 + 2 sec e sy = A k 2 EI [ ( sy = 45.5 = 0.252F 1 + 5.963 sec 0.11382 F )] F = 22.5 kips = 22,500 lbs d. 6740 lbs. 377. A bracket is attached as shown (367) onto a 14-in. x 193-lb., wide flange I-beam (A = 56.73 sq. in., depth = 15.5 in., flange width = 15.710 in., Imax = 2402.4 in4., Imin = 930.1 in4., kmin = 4.05 in.). The member is an eccentrically loaded column, 40 ft. long, with no central load (F1 = 0) and no restraint at the top. For e = 12 in. and N = 4, what may be the value of F2? Solution: Using secant formula: NF ec L NF 1 + 2 sec e sy = A k 2 EI I = Imax = 2402.4 in 4 E = 30,000 ksi e = 12 in k = kmax = 6.50 in Le = 2L = 2(40)(12) = 960 in Page 23 of 133 SECTION 6 – COMBINED STRESSES Le = 148 k N=4 A = 56.73 in 2 depth 15.5 c= = = 7.75 in 2 2 For C1020, as rolled, structural steel, sy = 48 ksi sy = 48 = 4F (12)(7.75) 460 sec 1 + 2 56.73 2 (6.50) [ ( 48 = 0.0705F 1 + 2.2 sec 0.11308 F F = F2 = 104.9 kips = 104,900 lbs 378. 4F (30,000)(2402.4) )] A 14-in. x 193-lb., wide flange I-beam is used as a column with one end free (A = 56.73 sq. in., depth = 15.5 in., Imax = 2402.4 in.4, Imin = 930.1 in.4, kmin = 4.05 in., length L = 40 ft.). If a load F2 is supported as shown on a bracket at an eccentricity e = 4 in. (with F1 = 0), what may be its value for a design factor of 4? Flange width = 15.71 in. Solution: Using secant formula: L NF NF ec 1 + 2 sec e sy = A k 2 EI F = F2 A = 56.73 in 2 e = 4 in flange width 15.71 c= = = 7.855 in 2 2 k = kmin = 4.05 in E = 30,000 ksi Page 24 of 133 SECTION 6 – COMBINED STRESSES I = Imin = 930.1 in 4 Le = 2L = 2(40 )(12) = 960 in N=4 s y = 48 ksi s y = 48 = 4F2 (4)(7.855) 960 1 + sec 2 56.73 2 (4.05) ( 48 = 0.0705F2 1 + 0.554 sec 0.18174 F2 4F2 (30,000)(930.1) ) F2 = 68.88 kips = 68,880 lbs 379. The same as 378, except that F1 = 0.5 F2. Solution: s L F ec σ = y = 2 1 + 2 sec e N A k 2 NF2 EI F1 + A F1 = 0.5F2 F 0.5F2 48 = 2 1 + 0.554 sec 0.18174 F2 + 4 56.73 56.73 12 = 0.01763F2 1 + 0.554 sec 0.18174 F2 + 0.008814F2 ( ) ( ) 12 = 0.026444F2 + 0.009767F2 sec 0.18174 F2 By Trial and error: F2 = 68.56 kips F1 = 0.5F2 = 34.28 kips 380. The cast-steel link (SAE 080) shown (solid lines) is subjected to a steady axial tensile load and was originally made with a rectangular cross section, h = 2 in., b = ½ in., but was found to be too weak. Someone decided to strengthen it by using a T-section (dotted addition), with h and b as given above. (a) Will this change increase the strength? Explain. (b) What tensile load could each link carry with N = 3 based on yield? Page 25 of 133 SECTION 6 – COMBINED STRESSES Solution: For SAE 080, s y = 40 ksi (a) This change will not increase the strength because of increased bending action that tends to add additional stress. s y 40 (b) σ = = = 13.3 ksi N 3 Rectangular cross section: F F σ= = A bh F 13.3 = (0.5)(2 ) F = 13.3 kips T-section: h = 2 in , b = 0.5 in A = b(h − b ) + bh = 0.5(2 − 0.5) + 0.5(2) = 1.75 in 2 b 1 c1 A = b(h − b ) (h − b ) + b + bh 2 2 0. 5 1 c1 (1.75) = 0.5(2 − 0.5) (2 − 0.5) + 0.5 + 0.5(2) 2 2 c1 = 0.6786 in c 2 = h − c1 = 2 − 0.6786 = 1.3214 in 1 (h − b) = 1.3214 − 1 (2 − 0.5) = 0.5714 in 2 2 b 0.5 f = c1 − = 0.6786 − = 0.4286 in 2 2 d = c2 − b(h − b )3 hb 3 + b(h − b )d 2 + + hbf 2 12 12 0.5(2 − 0.5)3 2(0.5)3 I= + 0.5(2 − 0.5)(0.5714)2 + + 2(0.5)(0.4286)2 12 12 I= I = 0.59 in 4 Page 26 of 133 SECTION 6 – COMBINED STRESSES F Fec + A I c = c1 = 0.6786 in e = c = 0.6786 in σ= (0.6786)(0.6786) 1 13.3 = F + 0.59 1.75 F = 6.941 kips COPLANAR SHEAR STRESSES 381. The figure shows a plate riveted to a vertical surface by 5 rivets. The material of the plate and rivets is SAE 1020, as rolled. The load F = 5000 lb., b = 3 in., θ = 0, and c = 5 in.; let a = 3D. Determine the diameter D of the rivets and the thickness of plate for a design factor of 3 based in yield strengths. Solution: θ =0 For SAE 1020, as rolled. s y = 48 ksi s sy = 0.6s y N=3 Page 27 of 133 SECTION 6 – COMBINED STRESSES 1 2 F 2 R = F12 + 5 F2 F1 = a 2a F1 = 2F2 [∑ M C =0 ] 2F1 (2a ) + 2F2 (a) = F (c ) + F sinθ (2a − b) θ =0 a = 3D b = 3 in c = 5 in F = 5 kips F1 = 2F2 4(2F2 )(3D ) + 2F2 (3D ) = (5)(5) + 5 sin 0 o (6D − 3) 30F2 D = 25 5 F2 = 6D 5 5 F1 = 2F2 = 2 = 6D 3D A= τ= τ= π 4 D2 R A s sy N = 0.6s y N Page 28 of 133 SECTION 6 – COMBINED STRESSES 0.6 sy N = R A 1 2 F 2 2 F1 + 5 0.6 sy = N A 1 0.6(48) = 3 5 2 5 2 2 + 3D 5 π 4 D2 1 2.778 2 2 + 1 D 9. 6 = 0.7854D 2 By trial and error method. 5 D = 0.625 in = in 8 For thickness of plate, t . A = Dt R σ= A 1 1 2 2 5 2 5 2 2 5 5 2 R = + = + = 2.85 kips 3D 5 3(0.625) 5 s R σ= y = N Dt 48 2.85 = 3 0.625t 1 t = 0.285 in = in 4 382. The same as 383, except that θ = 30o. Page 29 of 133 SECTION 6 – COMBINED STRESSES Solution: 1 2 2 F F R = F12 + − 2F1 cos120o 5 5 1 2 F 2 2 F R = F1 + + 2F1 cos 60o 5 5 MC = 0 [∑ ] 2F1 (2a ) + 2F2 (a) = F (c ) + F sinθ (2a − b) ( ) 4(2F2 )(3D ) + 2F2 (3D ) = 5 cos 30 o (5) + 5 sin 30 o (6D − 3) 30F2 D = 21.65 + 7.5(2D − 1) = 15D + 14.15 15D + 14.15 F2 = 30D 0.472 F2 = 0.5 + D 0.472 0.944 F1 = 2F2 = 2 0.5 + =1+ D D R τ= A s 0.6s y τ = sy = N N 0.6 sy N = R A 1 2 F 2 2 F o F1 + + 2F1 cos 60 0.6 sy 5 5 = N A Page 30 of 133 SECTION 6 – COMBINED STRESSES 1 0.6(48) = 3 0.944 2 5 2 0.944 5 2 o + + 21 + cos 60 1 + D 5 D 5 π 2 D 4 1 0.944 2 2 0.944 o 1 + + 1 + 2 1 + cos 60 D D 9. 6 = 2 0.7854 D By trial and error method. D = 0.641 in 5 Say D = 0.625 in = in (same as 381). 8 For t . 1 0.944 2 2 0.944 o R = 1 + + 1 + 21 + cos 60 D D 1 0.944 2 2 0.944 o R = 1 + + 1 + 21 + cos 60 = 3.1325 in 0.625 0.625 R σ= A s R σ= y = N Dt 48 3.1325 = 3 0.625t 5 t = 0.31325 in = in 16 383. Design a riveted connection, similar to that shown, to support a steady vertical load of F = 1500 lb. when L = 18 in. and θ = 0o. Let the maximum spacing of the rivets, horizontally and vertically, be 6D, where D is the diameter of the rivet; SAE 1020, as rolled, is used for all parts; N = 2.5 based on yield. The assembly will be such that there is virtually no twisting of the channel. The dimensions to determine at this time are: rivet diameter and minimum thickness of the plate. Page 31 of 133 SECTION 6 – COMBINED STRESSES Solution: θ = 0 o , a = b = 6D 1 (6D ) = 3D 2 F R = F1 + 4 R τ= A For SAE 1020, as rolled, s y = 48 ksi . c= N = 2.5 s sy = 0.6s y τ= s sy N = 0.6s y N = 0.6(48) = 11.52 ksi 2.5 4F1c = F (L + c ) 4F1 (3D ) = 1.5(18 + 3D ) 0.375(6 + D ) F1 = D 0.375(6 + D ) 1.5 2.25 R= + = + 0.75 D 4 D 1 A = πD 2 4 Page 32 of 133 SECTION 6 – COMBINED STRESSES τ= R A 2.25 + 0.75 11.52 = D 1 2 πD 4 11 D = 0.6875 in = in 16 sy R N A′ 2.25 + 0.75 48 = D 2.5 Dt 2.25 + 0.75 0 . 6875 19.2 = 0.6875t 5 t = in 16 384. The same as 383, except that θ = 45o. σ= = Solution: τ = 11.52 ksi , σ = 19.2 ksi Page 33 of 133 SECTION 6 – COMBINED STRESSES 2 F F R + + 2F1 cos 45 o 4 4 4F1 (3D ) = F cos θ (L + 3D ) = 1.5 cos 45 o (18 + 3D ) 0.2652(6 + D ) F1 = d 2 = F12 2 2 0.2652(6 + D ) 1.5 0.2652(6 + D ) 1.5 o R = + + 2 cos 45 D D 4 4 2 2 0.2652(6 + D ) 6+D R2 = + 0.140625 + 0.140625 D D 2 0.84375 6+D R = 0.0703 + 0.28125 + D D R τ= A 2 1 2 2 0.84375 6+D + 0.28125 0.0703 + D D 11.52 = 1 2 πD 4 D = 0.594 in 19 say D = in = 0.59375 in 32 1 2 2 0.84375 6 + 0.59375 R = 0.0703 + 0.28125 = 3.221 kips + 0.59375 0.59375 R σ= Dt R 19.2 = Dt 3.221 19.2 = 0.59375t t = 0.2815 in say t = 385. 1 in . 4 The plate shown (381) is made of SAE 1020 steel, as rolled, and held in place by five ¾ in. rivets that are made of SAE 1022 steel, as rolled. The thickness of the plate is ½ in., a = 2 ½ in., c = 6 in., b = 4 in., and θ = 0. Find the value of F for a design factor of 5 based on the ultimate strength. Page 34 of 133 SECTION 6 – COMBINED STRESSES Solution: Plate, SAE 1020, as rolled (Table AT 7) su = 65 ksi su 65 = = 13 ksi N 5 R = σπDt 3 1 R = (13)(π ) = 15.3 kips 4 2 σ= Rivets, SAE 1022, as rolled (Table AT 7) s su = 54 ksi s 54 τ = su = = 10.8 ksi N 5 3 2 π πD 2 4 R =τ = 10.8 = 4.77 ksi 4 4 use R = 4.77 ksi From 381. 1 2 F 2 R = F12 + 5 F1 = 2F2 2F1 (2a ) + 2F2 (a ) = Fc 2(2F2 )(2 )(2.5) + 2F2 (2.5) = F (6) 25F2 = 6F F2 = 0.24F F1 = 2(0.24F ) = 0.48F 2 R = (0.48F ) 2 2 F + = (4.77 )2 5 F = 9.173 kips = 9,173 lbs Page 35 of 133 SECTION 6 – COMBINED STRESSES 386. The same as 385, except that θ = 90o. Solution: R = 4.77 kips F R = F1 + 5 F1 = 2F2 2F1 (2a) + 2F2 (a ) = F (2a − b) 2(2F2 )(2 )(2.5) + 2F2 (2.5) = F [2(2.5) − 4] 25F2 = F F2 = 0.04F F1 = 0.08F F R = 4.77 = 0.08F + 5 F = 17 ,000 lbs Page 36 of 133 SECTION 6 – COMBINED STRESSES 387. The plate shown is made of AISI 1020 steel, as rolled, and is fastened to an I-beam (AISI 1020, as rolled) by three rivets that are made of a steel equivalent to AISI C1015, cold drawn. The thickness of the plate and of the flanges of the I-beam is ½ in., the diameter of the rivets is ¾ in., a = 8.5 in., b = 11.5 in. and c = 4.5 in., d = 4 in. For F2 = 0, calculate the value of F1 for N = 2.5 based on yield strength. Solution: 3 D = in 4 1 t = in 2 a = 8.5 in b = 11.5 in c = 4.5 in d = 4 in Plate, AISI 1020 Steel, as rolled, s y = 48 ksi Rivet, AISI C1015, cols drawn, s sy = 0.6(63) = 37.8 ksi sy 48 = 19.2 ksi N 2.5 s 37.8 τ = sy = = 15.12 ksi N 2.5 σ= = 3 1 R = σ (πDt ) = 19.2(π ) = 22.6 kips 4 2 πD 2 R = τ 4 2 π 3 = 15.12 = 6.68 kips 4 4 πD 2 Use R = τ 4 2 π 3 = 15.12 = 6.68 kips 4 4 F2 = 0 , Page 37 of 133 SECTION 6 – COMBINED STRESSES d 2 2 4 2 ρ = 2.5 in 2 c 3 2 ρ2 = + 4. 5 3 2 ρ 2 = + F′ F = 2c ρ 3 F 2c F′ = ρ 3 c 2c F1 a + = 2Fρ + F ′ 3 3 2(4.5) F 4. 5 3 F1 8.5 + = 2F (2.5) + 3 2. 5 F1 = 0.86F F F= 1 0.86 c3 cos θ = 2 ρ 4. 5 3 cos θ = = 0.60 2. 5 2 F F R 2 = F 2 + 1 + 2F 1 cos θ 3 3 2 2 F F F F R = 1 + 1 + 2 1 1 (0.60) 0.86 3 0.86 3 2 Page 38 of 133 SECTION 6 – COMBINED STRESSES R = 1.389F1 R = 6.68 kips = 1.389F1 F1 = 4.8 kips . 388. The same as 387, except that F1 = 0, and the value of F2 is calculated. Solution: R = 6.68 kips R=F+ F2 3 Fρ (2c 3) c 2c F2 b − = F + 2F ′ρ 3 3 F′ = 4.5 2(4.5) 2F (2.5)2 F2 11.5 − + =F 3 3 2(4.5) 3 F2 = 0.7167F Page 39 of 133 SECTION 6 – COMBINED STRESSES 0.7167F = 1.24F 3 6.68 = 1.24F F = 5.387 kips R =F + NORMAL STRESSES WITH SHEAR DESIGN PROBLEMS The bracket shown is held in place by three bolts as shown. Let a = 5 ¼ in., θ = 30o, F = 1500 lb.; bolt material is equivalent to C1022, as rolled. (a) Compute the size of the bolts by equation (5.1), Text. (b) Assuming that the connecting parts are virtually rigid and that the initial stress in the bolts is about 0.7sy, compute the factor of safety by (i) the maximum shear stress theory, (ii) the octahedral shear theory. (c) Compute the maximum normal stress. 389. Solution: 3 (a) Eq. 5-1, Fe = [∑ M corner =0 s y As 2 6 3 , D < in 4 ] 2FA (9 ) + FB (3) = F sinθ (3) + F cos θ (a ) ( ) ( ) 18FA + 3FB = (1500) sin 30 o (3) + (1500) cos 30 o (5.25) 18FA + 3FB = 9070 lbs FA FB = 9 3 FA = 3FB Page 40 of 133 SECTION 6 – COMBINED STRESSES 18FA + FA = 9070 lbs FA = 477.4 lbs Fe = FA For C1022, as rolled, s y = 52,000 psi 3 Fe = s y As 2 6 3 52,000 As 2 Fe = 477.4 = 6 2 As = 0.1448 in Select Say D = 1 in , UNC, As = 0.1419 in 2 2 Fe + si A 477.4 st = + 0.7(52,000) = 39,764 psi 0.1419 (b) st = ss = ( ) F cos θ 3 (1500) cos 30 o 3 = = 3052 psi As 0.1419 (i) Maximum shear theory 1 1 s = N s y 2 ss + s 2 y 2 2 2 = 39,764 + 3052 52,000 52,000 1 2 2 2 N = 1.293 (ii) Octahedral shear theory 1 1 s = N s y 2 ss + s 3 y 2 1 2 2 2 2 = 39,764 + 3052 52,000 52,000 3 N = 1.296 (c) Maximum normal stress = 39,764 psi. 390. For the mounted bracket shown, determine the rivet diameter (all same size) for N = 3, the design being for the external loading (initial stress ignored); F = 2.3 kips, θ = 0, c = 17 in., a = 1 ½ in., b = 14 ½ in.; rivet material is AISI 1015, as rolled. Compute for (a) the maximum shear theory, (b) the maximum normal stress theory, (c) the octahedral shear theory. Page 41 of 133 SECTION 6 – COMBINED STRESSES Solution: F2 F = 1 a a+b F2 F1 = 1.5 1.5 + 14.5 F2 = 0.09375F1 2F1 (a + b) + F2 (a ) = Fc 2F1 (1.5 + 14.5) + 0.09375F1 (1.5) = (2.3)(17 ) F1 = 1.2165 kips F1 1.2165 = ksi A A F 2.3 0.7667 ss = = = ksi 3 A 3A A For AISI 4015, as rolled. s y = 45.5 ksi s= (a) Maximum shear theory 1 1 s = N sy ss 2 + s ys s ys = 0.5sy = 0.5(45.5) = 22.75 ksi 2 Page 42 of 133 2 SECTION 6 – COMBINED STRESSES 1 2 2 1 1.2165 0.7667 2 = + 3 45.5A 22.75A A = 0.1291 in 2 πD 2 = 0.1291 in 2 4 D = 0.4054 in 3 say D = in 8 A= 1 1 2 2 1.2165 1.2165 2 0.7667 2 1.5869 s s (b) σ = + + s s2 = + + = 2 2 2A A 2 A A 1 σ = N sy σ= sy N 1.5869 45.5 = A 3 A = 0.1046 in 2 A= πD 2 = 0.1046 in 2 4 D = 0.365 in 3 say D = in 8 (c) s ys = sy 3 = 45.5 3 = 26.27 ksi 1 2 1 1.2165 0.7667 2 = + 3 45.5A 26.27 A A = 0.11874 in 2 A= πD 2 = 0.11874 in 2 4 D = 0.3888 in 3 say D = in 8 Page 43 of 133 SECTION 6 – COMBINED STRESSES 392. The same as 390, except that the two top rivets are 2 in. long and the bottom rivet is 1 ¼ in. long. Solution: δ2 a δ2 1.5 = = δ1 a+b δ1 1.5 + 14.5 δ 2 = 0.09375δ1 F2 (1.25) = 0.09375(2)F1 F2 = 0.15F1 2F1 (a + b) + F2 a = Fc 2F1 (16) + 0.15F1 (1.5) = (2.3)(17 ) F1 = 1.2133 kips F1 1.2133 = ksi A A F 2.3 0.7667 ss = = = ksi 3 A 3A A For AISI 4015, as rolled. s y = 45.5 ksi s= Page 44 of 133 SECTION 6 – COMBINED STRESSES (b) Maximum shear theory 1 1 s = N sy ss 2 + s ys s ys = 0.5sy = 0.5(45.5) = 22.75 ksi 2 2 1 2 2 1 1.2133 0.7667 2 = + 3 45.5 A 22.75A A = 0.1289 in 2 πD 2 = 0.1289 in 2 4 D = 0.4051 in 3 say D = in 8 A= 1 1 2 2 1.2133 1.2133 2 0.7667 2 1.5843 s s + (b) σ = + + s s2 = + = 2 2 2A A 2 A A 1 σ = N sy σ= sy N 1.5843 45.5 = A 3 A = 0.1045 in 2 πD 2 = 0.1045 in 2 4 D = 0.3648 in 3 say D = in 8 A= (c) s ys = sy 3 = 45.5 3 = 26.27 ksi 1 2 1 1.2133 0.7667 2 = + 3 45.5 A 26.27 A A = 0.1186 in 2 Page 45 of 133 SECTION 6 – COMBINED STRESSES πD 2 = 0.1186 in 2 4 D = 0.3886 in 3 say D = in 8 A= 393. The same as 390, except that the load is applied vertically at B instead of at A; let AB = 8 in. The two top rivets are 12 in. apart. Solution: b ρ = 6 + 3 2 2 2 14.5 3 ρ = 7.705 in 2 ρ 2 = 62 + F1 ρ = F2 2b 3 2b F 2(14.5) F1 F2 = 1 = = 1.2546F1 3 ρ 3 7.705 Page 46 of 133 SECTION 6 – COMBINED STRESSES 2b F (8) = 2F1 ρ + F2 3 2(14.5) 3 (2.3)(8) = 2F1 (7.705) + 1.2546F1 F1 = 0.6682 kips cos θ = 6 ρ = 6 = 0.7787 7.705 2 F F R 2 = F12 + + 2F1 cos θ 3 3 2 2.3 2.3 R = (0.6682) + + 2(0.6682) (0.7787 ) 3 3 R = 1.3536 kips R 1.3536 ss = = ksi A A 2 2 1 1 s (a) = N sy 2 ss + s ys 2 2 From Problem 390. 1.2165 s= ksi A s y = 45.5 ksi s ys = 22.75 ksi N=3 1 2 2 1 1.2165 1.3536 2 = + 3 45.5A 22.75A A = 0.1957 in 2 πD 2 = 0.1957 in 2 4 D = 0.5 in 1 say D = in 2 A= 1 1 2 2 1.2165 1.2165 2 1.3536 2 2.10 s s (b) σ = + + s s2 = + + = 2 2 2A A 2 A A 1 σ = N sy Page 47 of 133 SECTION 6 – COMBINED STRESSES sy σ= N 2.10 45.5 = A 3 A = 0.1385 in 2 A= πD 2 = 0.1385 in 2 4 D = 0.42 in 7 say D = in 16 (c) s ys = 26.27 ksi 1 2 1 1.2165 1.3536 2 = + 3 45.5A 26.27 A A = 0.1742 in 2 πD 2 = 0.1742 in 2 4 D = 0.471 in 1 say D = in 2 A= 394. The bracket shown is made of SAE 1020, as rolled, and the rivets are SAE 1015, cold drawn. The force F = 20 kips, L = 7 in., and θ = 60o. Let the design factor (on yield) be 2. (a) Determine the thickness t of the arm. (b) Compute the rivet diameter by both maximum shear and octahedral shear theories and specify a standard size. (c) Decide upon a proper spacing of rivets and sketch the bracket approximately to scale. Is some adjustment of dimensions desirable? Give suggestions, if any. (No additional calculations unless your instructor asks for a complete design.) Solution: Bracket: SAE 1020, as rolled, s y = 48 ksi Rivets: SAE 1015, cold drawn, s y = 63 ksi Page 48 of 133 SECTION 6 – COMBINED STRESSES N=2 (a) Bracket. (F cosθ )(L ) 4 F sinθ 2 s= + A I A = 4t t (4)3 I= = 5.333t 12 ( ) 48 20 sin 60 o 20 cos 60 o (7 )(2) = + N 2 4t 5.333t t = 1.275 in 1 say t = 1 in 4 s= sy = (b) F2 F1 = 2 6 F1 = 3F2 ( ) ( ) 3(3F )(6) + 2F (2 ) = (20)(cos 60 )(7 ) + (20)(sin 60 )(3) 3F1 (6) + 2F2 (2) = F cos 60o (L ) + F sin 60 o (3) o 2 2 F2 = 2.10 kips F1 = 3F2 = 3(2.10) = 6.31 kips F cos θ 20 cos 60 2 = = 5A 5A A F1 F sin 60 6.31 20 sin 60 9.774 s= + = + = A 5A A 5A A ss = 1 1 s = N sy 2 ss + s ys 2 2 s y = 63 ksi Max. shear: s ys = 0.5sy = 0.5(63) = 31.5 ksi 1 2 2 1 9.774 2 2 = + 2 63A 31.5A A = 0.3353 in 2 Page 49 of 133 o SECTION 6 – COMBINED STRESSES A= πD 2 = 0.3353 in 2 4 D = 0.653 in 3 say D = in 4 Octahedral shear, s ys = sy 3 63 = 3 1 2 2 1 9.774 2 2 = + 2 63A 36.37 A A = 0.3292 in 2 A= πD 2 = 0.3292 in 2 4 D = 0.6474 in 3 say D = in 4 (c) Spacing F cos θ 5(S − D )t t = 1.5 in , s y = 48 ksi σ= D = 0.75 in F = 20 kips sy F cos θ N 5(S − D )t 48 20 cos 60 = 2 5(S − 0.75)(1.5) S = 0.806 in 7 use S = in adjust to 2 in 8 σ= = Page 50 of 133 = 36.37 ksi SECTION 6 – COMBINED STRESSES Adjust spacing to 2 in from 7/8 in as shown. CHECK PROBLEMS 396. (a) If the rivets supporting the brackets of 390 are 5/8 in. in diameter, θ = 0, c = 14 in. a = 2 in., and b = 18 in., what are the maximum tensile and shear stresses in the rivets induced by a load of F = 10 kips. (b) For rivets of naval brass, ¼ hard, compute the factor of safety by maximum shear and octahedral shear theories (initial tension ignored). Solution: Page 51 of 133 SECTION 6 – COMBINED STRESSES F2 F = 1 a a+b F2 F = 1 2 2 + 18 F2 = 0.1F1 2F1 (a + b) + F2 (a ) = Fc 2F1 (2 + 18) + 0.1F1 (2) = (10)(14) F1 = 3.8425 kips F1 F 3.8425 = 1 = = 12.525 ksi π π A (0.625)2 D2 4 4 F F 10 ss = = = = 10.865 ksi 3A π 2 π 2 3 D 3 (0.625) 4 4 s= 1 1 s 2 2 2 12.525 2 2 2 (a) τ = + s s = + (10.865) = 12.541 ksi 2 2 1 1 2 2 12.525 12.525 2 2 s s 2 σ = + + s s2 = + + (10.865) = 18.804 ksi 2 2 2 2 (b) Naval Brass, ¼ hard, s y = 48 ksi . 1 1 s = N sy ss 2 + s ys Max. shear theory; s ys = 0.5sy = 0.5(48) = 24 ksi 2 Page 52 of 133 2 SECTION 6 – COMBINED STRESSES 1 2 2 1 12.525 10.865 2 = + N 48 24 N = 1.914 Octahedral shear theory; s ys = sy 3 = 48 3 = 27.71 ksi 1 2 2 1 12.525 10.865 2 = + N 48 27.71 N = 2.123 397. The same as 396, except that the two top rivets are ¾ in. in diameter and the bottom one is ½ in. in diameter. Solution: δ2 = δ1 a+b a δ2 = δ1 a+b F2 a F1 = A2 a + b A1 a Page 53 of 133 SECTION 6 – COMBINED STRESSES 2 a D2 F2 = F1 a + b D1 2 2 0. 5 F2 = F1 = 0.0444F1 2 + 18 0.75 2F1 (a + b) + F2 (a ) = Fc 2F1 (2 + 18) + 0.0444F1 (2) = (10)(14) F1 = 3.492 kips F1 3.492 = = 7.9043 ksi π A1 (0.75)2 4 F 10 ss = = = 9.26 ksi 2 A1 + A2 π π 2 2 2 (0.75) + (0.50) 4 4 s= 1 1 s 2 2 2 7.9043 2 2 2 (a) τ = + s s = + (9.26) = 10.068 ksi 2 2 1 1 2 2 7.9043 7.9043 2 2 s s 2 σ = + + s s2 = + + (9.26) = 14.02 ksi 2 2 2 2 1 1 s (b) = N sy ss 2 + s ys Max. shear theory; s ys = 0.5sy = 0.5(48) = 24 ksi 2 2 1 2 2 1 7.9043 9.26 2 = + N 48 24 N = 2.384 Octahedral shear theory; s ys = 1 2 2 1 7.9043 9.26 2 = + N 48 27.71 N = 2.684 Page 54 of 133 sy 3 = 48 3 = 27.71 ksi SECTION 6 – COMBINED STRESSES 398. What static load F may be supported by the ¾-in. rivets shown, made of cold-finished C1015, with N = 3; θ = 0, a = 1 ½, b = 9, c = 14, f = 9, g = 12 in.? Count on no help from friction and ignore the initial tension. Check by both maximum shear and octahedral shear theories. Solution: For cold-finished, C1015, s y = 63 ksi . F1 F = 2 a+b a F1 F = 2 1.5 + 9 1.5 F1 = 7F2 θ = 0o 3F1 (a + b) + F2 a = Fc 3(7F2 )(1.5 + 9 ) + F2 (1.5) = F (14) F2 = 0.0631F F1 = 7F2 = 0.4417F s= F1 0.4417F = =F A π (0.75)2 4 Page 55 of 133 SECTION 6 – COMBINED STRESSES ss = F F = = 0.5659F 4A π 2 ( ) 4 0.75 4 1 1 s = N sy 2 2 ss 2 + s ys Max. shear theory; s ys = 0.5sy = 0.5(63) = 31.5 ksi 1 2 2 1 F 0.5659F 2 = + 3 63 31.5 F = 13.9 kips Octahedral shear theory; s ys = sy 3 = 63 3 = 36.37 ksi 1 2 2 1 F 0.5659F 2 = + 3 63 36.37 F = 15 kips 399. The 2-in., UNC cap screw shown has been subjected to a tightening torque of 20 in-kips. The force F = 12 kips, θ = 60o, and Q = 0; L = 24 in., a = 20 in., b = 15 in.; screw material is AISI C1137 as rolled. (a) What is the approximate initial tightening load? (b) What is the increase in this load caused by the external force F if the bar is 8 in. wide and 2 in. thick and the unthreaded shank of the screw is 2 in. long? (See §5.9, Text.) (c) What are the maximum tensile and shear stresses in the bolt? (d) Compute the factor of safety from maximum normal stress, maximum shear, and octahedral shear theories. Solution: (a) T = CDFi C = 0.2 D = 2 in T = 20 in − kips 20 = 0.2(2)Fi Page 56 of 133 SECTION 6 – COMBINED STRESSES Fi = 50 kips (b) θ = 60 o Fe (L − a ) = F cos 60 o (a ) ( ) ( ) Fe (24 − 20) = 12 cos 60 o (20) Fe = 30 kips kb ∆Fb = Fe kb + kc AE kb = s Lb AE kc = c Lc For 2”-UNC As = 2.50 in 2 Width across flat = 3 in. 2.5E kb = = 1.25E 2 π kc = 4 (3)2 E 2 = 3.5343E 1.25E ∆Fb = (30) = 7.84 kips 1.25E + 3.5343E F + ∆Fb 50 + 7.84 (c) s = i = = 23.14 ksi As 2.50 ss = F sinθ 12 sin 60o = = 4.16 ksi As 2.5 1 1 s 2 2 23.14 2 2 2 τ = + ss2 = + ( 4 . 16 ) = 12.3 ksi 2 2 1 1 2 2 23.14 23.14 2 2 s s 2 2 σ = + + s s = + + (4.16) = 23.87 ksi 2 2 2 2 (d) For C1137, as rolled, s y = 55 ksi σ= sy N 23.87 = 55 N Page 57 of 133 SECTION 6 – COMBINED STRESSES N = 2.304 Maximum Shear: 0.5sy τ= N 0.5(55) 12.3 = N N = 2.236 Octahedral shear, s ys = sy 3 = 55 3 = 31.75 ksi 1 1 s = N sy 2 2 ss 2 + s ys 1 2 2 1 23.14 4.16 2 = + N 55 31.75 N = 2.269 400. The plate shown is attached by three ½-in., UNC cap screws that are made of ASTM A325, heat-treated bolt material; L = 26 in., a = 6 in., b = 4 in., θ = 0. The shear on the screws is across the threads and they have been tightened to an initial tension of 0.6sp (sp = proof stress, §5.8, Text). Which screw is subjected to (a) the largest force, (b) the largest stress? What safe static load can be supported by the screws for N = 1.5 based on the Hencky-Mises criterion? Solution: For ASTM A325, Heat-Treated, ½-in. UNC s y = 88 ksi s p = 85 ksi si = 0.6s p = 0.6(85) = 51 ksi Page 58 of 133 SECTION 6 – COMBINED STRESSES 2 b a 2 3 ρ = 2.83 in 2 4 2 2 6 3 2 ρ 2 = + = + (a) Largest force, at A (b) Largest stress, at A F2 F1 ρ (2a 3) ρF1 2.83F1 F2 = = = 0.7075F1 (2a 3) 2(6 ) 3 = 2a 2a 2F2 ρ + F1 = F L + 3 3 2(6) 2(6 ) 2(0.7075)(F1 )(2.83) + F1 = F 26 + 3 3 F1 = 3.75F F F R = F1 + = 3.75F + = 4.1F 3 3 For ½ in UNC, As = 0.1419 in 2 R 4.1F ss = = = 29F A 0.1419 s = si = 51 ksi By Hencky-Mises Criterion. 1 1− µ (s x + s y )+ (1 + µ ) s x − sy σ= 2 2 Page 59 of 133 2 2 + s s2 SECTION 6 – COMBINED STRESSES s x = s = 51 ksi sy = 0 s s = 29F µ = 0.3 1 1− µ (s x + sy ) + (1 + µ ) s x − sy σ= = N 2 2 sy 2 2 2 + ss 1 2 2 88 1 − 0.3 = (51) + (1 + 0.3) 51 + (29F )2 1.5 2 2 F = 0.63 kips = 630 lbs 401. The same as 400, except that the cap screw A is ¾ in. in diameter. Solution: A1 = As1 = 0.3340 in 2 (3/4 in UNC) A2 = As2 = 0.1419 in 2 (1/2 in UNC) F2 F1 = A2 ρ A1 (2a 3) F2 F1 = (0.1419 )(2.83) (0.3340)[2(6) 3] F2 = 0.3006F1 2a 2a 2F2 ρ + F1 = F L + 3 3 2(6) 2(6) 2(0.3006F1 )(2.83) + F1 = F 26 + 3 3 F1 = 5.262F F2 = 0.3006F1 = 0.3006(5.262F ) = 1.582F For ¾ in UNC. F F R = F1 + = 5.262F + = 5.6F 3 3 Page 60 of 133 SECTION 6 – COMBINED STRESSES ss = 5.6F = 16.8F ksi 0.3340 For ½ in. UNC, 1 2 2 F F R = F22 + − 2F2 cos θ 3 3 a3 63 cos θ = = = 0.7067 ρ 2.83 1 2 2 F F R = (1.582F )2 + − 2(1.582F ) (0.7067 ) = 1.367F 3 3 1.367F ss = = 9.6F ksi 0.1419 (a) Max. force, at ¾ in. (b) Max. stress, at ¾ in. s s = 16.8F ksi s = si = 51 ksi 1 2 2 88 1 − 0.3 σ= = = (51) + (1 + 0.3) 51 + (16.8F )2 N 1.5 2 2 sy F = 1.09 kips = 1090 lbs NORMAL STRESSES WITH TORSION DESIGN PROBLEMS 402. A section of a machined shaft is subjected to a maximum bending moment of 70,000 inlb., a torque of 50,000 in-lb., and an end thrust of 25,000 lb. The unsupported length is 3 ft. and the material is AISI C1030, normalized. Since the computations are to be as though the stresses were steady, use N = 3.3. Compute the diameter from both the maximum-shear and the octahedral-shear theories and specify a standard size. Solution: wL2 2 M1 = 70,000 in − lb L = 3 ft = 36 in M = M1 + Page 61 of 133 SECTION 6 – COMBINED STRESSES w = ρA ρ = 0.284 lb in 3 πD 2 A= 4 0.284πD 2 w= = 0.223D 2 lb in 4 0.223D 2 (36 )2 M = 70,000 + = 70,000 + 144.5D 2 2 F = 25,000 lb F Mc s= + A I D c= 2 πD 2 A= 4 πD 4 I= 64 4F 32M s= 2 + πD πD 3 4(25,000) 32 70,000 + 144.5D 2 s= + πD 2 πD 3 ( ) 31,831 713,014 1472 + + D D2 D3 713,014 31,831 1472 s= + + D D3 D2 s= ss = 16T 16(50,000) 254,6548 = = πD 3 πD 3 D3 For AISI C1030, normalized, s y = 47 ksi = 47 ,000 psi 1 1 s = N sy 2 ss + s ys 2 2 Maximum shear s ys = 0.5sy = 0.5(47 ) = 23.5 ksi = 23,500 psi 1 713,014 31,831 1472 2 2 2 + + 1 D 3 D + 254,648 D2 = 23,500D 3 3.3 47 ,000 By trial and error, Page 62 of 133 SECTION 6 – COMBINED STRESSES D = 4.125 in Octahedral Shear sy 47 s ys = = = 27.14 ksi = 27 ,140 psi 3 3 1 713,014 31,831 1472 2 2 2 + + 1 D 3 D + 254,648 D2 = 27 ,140D 3 3.3 47 ,000 By trial and error, D = 4.125 in 1 use D = 4 in standard 4 403. The same as 402, except that the unsupported length is 15 ft. Do not overlook the moment due to the weight of the shaft, which acts in the same sense as the given bending moment. Solution: L = 15 ft = 180 in M = 70,000 + 0.223D 2 (180)2 = 70,000 + 3612.6D 2 2 F = 25,000 lb F Mc s= + A I 4F 32M s= 2 + πD πD 3 4(25,000) 32 70,000 + 3612.6D 2 s= + πD 2 πD 3 ( ) 31,831 713,014 36,798 + + D D2 D3 713,014 31,831 36,798 s= + + D D3 D2 s= Maximum shear 1 713,014 31,831 36,798 2 2 2 + + 1 D 3 D + 254,648 D2 = 23,500D 3 3.3 47 ,000 By trial and error, D = 5.125 in Page 63 of 133 SECTION 6 – COMBINED STRESSES Octahedral Shear 1 713,014 31,831 36,798 2 2 2 + + 3 2 1 D D + 254,648 D = 27 ,140D 3 3.3 47 ,000 By trial and error, D = 5.0625 in 1 use D = 5 in standard 4 404. A shaft is to be made in two sections, I and II, of diameters D1 and D2, somewhat as shown, machined from AISI 1045, annealed. It is expected that a = 8 in., b = 24 in., L = 20 in., and the load Q = 2 kips, so seldom repeated that the design is for steady load. The factor of safety is to be 2.2 on the basis of the octahedral-shear theory and closely the same in each section. The ends A and B are restrained from twisting, but they are designed to support the balancing reactions from Q without other moments. Decide upon standard size for D1 and D2. Solution: T = QL = (2)(20 ) = 40 in − kips T1 + T2 = T = 40 in − kips θ1 = θ 2 T1L1 T2 L2 = J1G J 2 G T1a Tb = 24 4 πD1 πD2 32 32 8T1 24T2 = 4 D14 D2 Page 64 of 133 SECTION 6 – COMBINED STRESSES 4 D T1 = 3T2 1 D2 T1 + T2 = T = 40 in − kips D T2 3 1 D 2 4 + 1 = 40 40 T2 = D 4 3 1 + 1 D 2 4 4 D D 40(3) 1 120 1 D2 D2 T1 = = 4 D D 4 1 3 + 1 3 1 + 1 D 2 D2 A + B = Q = 2 kips aA = bB 8A = 24B A = 3B 3B + B = 2 B = 0.5 kips A = 1.5 kips 4 D 4 D 16120 1 1920 1 D 2 16T D2 s s1 = 31 = = πD1 D 4 D 4 3 3 1 πD1 3 + 1 πD1 3 1 + 1 D 2 D 2 16T2 16(40) 640 s s2 = = = 3 4 πD 2 D D 4 3 3 1 πD2 3 + 1 πD2 3 1 + 1 D 2 D 2 32M1 32 Aa 32(1.5)(8) 384 s1 = = = = 3 πD13 πD13 πD13 πD1 32M 2 32Bb 32(0.5)(24) 384 s2 = = = = 3 πD 23 πD23 πD 23 πD2 For AISI 1045, annealed, s y = 55 ksi , N = 2.2 Octahedral Shear Page 65 of 133 SECTION 6 – COMBINED STRESSES 1 1 s = N sy s ys = sy 3 2 ss + s ys = 55 3 2 2 = 31.75 ksi 1 1 s1 = N sy 2 s s1 + s ys 2 2 1 2 2 4 D 1920 1 2 D2 1 384 = + 2.2 55πD13 D 4 3 1 31.75πD1 3 + 1 D 2 1 2 2 4 19.25 D1 2 D 1 2.2224 2 = + 2.2 D13 D 4 3 1 D1 3 + 1 D 2 1 1 s2 = N sy 2 s s2 + s ys 2 2 1 2 1 384 640 + = 3 2.2 55πD 2 D 3 1 31.75πD 2 3 D 2 4 + 1 1 2 2 2 1 2.2224 6.42 + = 2.2 D 23 D 4 3 1 D 2 3 + 1 D 2 Page 66 of 133 2 2 SECTION 6 – COMBINED STRESSES By trial and error, D Trial 1 D2 1 1.25 1.37 1.40 D1 D2 2.268 2.372 2.398 2.403 1.820 1.730 1.714 1.712 D1 D2 1.25 1.37 1.40 1.40 Actual 1 3 Use standard D1 = 2 in , D 2 = 1 in 2 4 1 But D1 > D 2 , use D1 = 2 in 2 1 D 2 = D1 + 2r = 1.3D1 = 1.3 2 = 3.25 in 2 1 say D 2 = 3 in 2 405. The shaft shown overhangs a bearing on the right and has the following dimensions: a = 5 in., b = ½ in., and e = 10 in. The material is AISI C1040, annealed. This shaft is subjected to a torque T = 10,000 in-lb., forces F1 = 10,000 lb., and F2 = 20,000 lb. Using a staticdesign approach, determine the diameter D for N = 2.5, with computations from the maximum-shear and octahedral-shear theories. Solution: ρ = 0.284 lb in 3 π w = 0.284 D 2 = 0.223D 2 4 2 we 0.223D 2 (10 )2 M= = = 11.15D 2 2 2 Bending due to F1 and load [ ] 32(F1a + M ) 32 (10,000)(5) + 11.15D 2 509,296 114 = = + πD 3 πD 3 D3 D Bending due to F2 s1 = Page 67 of 133 SECTION 6 – COMBINED STRESSES 1 32(20,000 ) 32F2 b 2 = 101,859 s2 = = 3 3 πD πD D3 Tension, 4(20,000 ) 25,465 s3 = = πD 2 D2 since s1 > s 2 s = s3 + s1 − s2 25,465 509,296 114 101,859 + + − s= D D2 D3 D3 407,437 25,465 114 s= + + D D3 D2 16T 16(10,000) 50,930 = = πD 3 πD 3 D3 N = 2.5 ss = 1 1 s = N sy ss 2 + s ys For AISI C1040, annealed (Fig. AF 1) s y = 48 ksi 2 2 Maximum shear, s ys = 0.5sy = 0.5(48) = 24 ksi 1 407,437 25,465 114 2 2 2 + + 1 D 3 D + 50,930 D2 = 24,000D 3 2.5 48,000 D = 2.95 in Octahedral Shear sy 48 s ys = = = 27.71 ksi 3 3 1 407,437 25,465 114 2 2 2 + + 1 D 3 D + 50,930 D2 = 27,710D 3 2.5 48,000 D = 2.95 in use D = 3.0 in Page 68 of 133 SECTION 6 – COMBINED STRESSES 406. The same as 405, except that F2 = 0. Solution F2 = 0 s 2 = 0 , s3 = 0 509,296 114 s = s1 = + D D3 50,930 ss = D3 N = 2.5 Maximum shear, s ys = 0.5sy = 0.5(48) = 24 ksi 1 509,296 114 2 + 1 D 3 D + 50,930 = 24,000D 3 2.5 48,000 2 2 D = 3 in Octahedral Shear sy 48 s ys = = = 27.71 ksi 3 3 1 509,296 114 2 2 2 + 1 D 3 D + 50,930 = 27 ,710D 3 2.5 48,000 D = 2.997 in use D = 3.0 in CHECK PROBLEMS 407. The shaft shown overhangs a bearing at the right and has the following dimensions: D = 2 in., a = 4 in., b = ¾ in., c = 2 in., d = 6 in., e = 8 in., r = ¼ in. This shaft is subjected to a torque T = 8000 in-lb. and forces F1 = 8000 lb., and F2 = 16,000 lb. Determine the maximum-shear and normal stresses, and the octahedral-shear stress: (a) at points A and B (θ = 45o), (b) at points M and N, (c) at point G. Page 69 of 133 SECTION 6 – COMBINED STRESSES Solution: Bending due to F2 : 32F2 b 32(16,000 )(0.75) s2 = = = 15,279 psi πD 3 π (2)3 Tension 4F 4(16,000) s3 = 22 = = 5093 psi πD π (2)2 Bending due to F1 and weight of beam wx 2 32 F1a + 2 s1 = πD 3 x from the free end. ss = 16T 16(8000) = = 5093 psi πD 3 π (2)3 w = 0.223D 2 = 0.223(2)2 = 0.892 lb in (A) x = e − c = 8 − 2 = 6 in 0.892(6)2 32 (8000)(4) + 2 s1 = = 40,764 psi 3 π (2 ) at A s = s3 + (s1 − s 2 ) sin 45o = 5093 + (40,764 − 15,279) sin 45o = 23,114 psi Max. Shear 1 1 s 2 2 23,114 2 2 2 τ = + s s2 = + (5093) = 12,629 psi 2 2 Max. Normal 1 1 2 2 23,114 23,114 2 2 s s 2 2 τ = + + s s = + + (5093) = 24,186 psi 2 2 2 2 Octahedral Shear Stress Page 70 of 133 SECTION 6 – COMBINED STRESSES τo = 2 2 s + 3s s2 3 ( 1 2 ) = [ 2 (23,114)2 + 3(5093)2 3 1 2 ] = 11,662 psi At B: s = (s1 − s2 )sin 45o − s3 = (40,764 − 15,279) sin 45o − 5093 = 12,928 psi Max. Shear 1 1 s 2 2 12,928 2 2 2 τ = + s s2 = + (5093) = 8229 psi 2 2 Max. Normal 1 1 2 2 12,928 12,928 2 2 s s 2 σ = + + s s2 = + + (5093) = 14,693 psi 2 2 2 2 Octahedral Shear Stress 1 2 2 2 τo = s + 3s s2 2 = (12,928)2 + 3(5093)2 3 3 ( [ ) 1 2 ] = 7 ,378 psi (B) 1 = 7.75 in 4 0.892(7.75)2 32(8000)(4 ) + 2 x =e −c =8− s1 = π (2)3 = 40,778 psi at M s = s3 + s1 − s2 = 5093 + 40,778 − 15,279 = 30,592 psi Max. Shear 1 1 s 2 2 30,592 2 2 2 τ = + s s2 = + (5093) = 16,122 psi 2 2 Max. Normal 1 1 2 2 30,592 30,592 2 2 s s 2 2 σ = + + s s = + + (5093) = 31,418 psi 2 2 2 2 Octahedral Shear Stress 1 2 2 2 τo = s + 3s s2 2 = (30,592)2 + 3(5093)2 3 3 ( ) [ 1 2 ] = 15,009 psi At N: s = s1 − s 2 − s3 = 40,778 − 15,279 − 5093 = 20,406 psi Max. Shear Page 71 of 133 SECTION 6 – COMBINED STRESSES 1 1 s 2 2 20,406 2 2 2 2 τ = + s s = + (5093) = 11,403 psi 2 2 Max. Normal 1 1 2 2 20,406 20,406 2 2 s s 2 τ = + + s s2 = + + (5093) = 21,606 psi 2 2 2 2 Octahedral Shear Stress 1 1 2 2 2 τo = s + 3s s2 2 = (20,406)2 + 3(5093)2 2 = 10,480 psi 3 3 ( [ ) ] (C) At G. x = e − c = 8 − 2 = 6 in 0.892(2 )2 32 (8000)(4) + 2 s1 = = 40,746 psi 3 π (2) s = s3 + (s1 − s 2 ) cos 30 o = 5093 + (40,746 − 15,279) cos 30 o = 17 ,826 psi Max. Shear 1 1 s 2 2 17 ,826 2 2 2 τ = + s s2 = + (5093) = 10,265 psi 2 2 Max. Normal 1 1 2 2 17 ,826 17 ,826 2 2 s s 2 σ = + + s s2 = + + (5093) = 19,178 psi 2 2 2 2 Octahedral Shear Stress 1 2 2 2 τo = s + 3s s2 2 = (17,826)2 + 3(5093)2 3 3 ( 411. ) [ 1 2 ] = 9376 psi A 4-in. shaft carries an axial thrust of 20 kips. The maximum bending moment is 2/3 of the twisting moment; material is AISI 8630, WQT 1100 F, and N =3. Use the steady stress approach and compute the horsepower that may be transmitted at 2000 rpm? Solution: For AISI 8630, WQT 1100oF, s y = 125 ksi 32M 4F + πD 3 πD 2 16T ss = 3 πD 2 M= T 3 F = 20 kips s= Page 72 of 133 SECTION 6 – COMBINED STRESSES D = 4 in 2 32 T 3 4(20 ) s= + = 0.1061T + 1.59155 π (4 )3 π (4)2 16T ss = = 0.0796T π (4)3 By maximum shear. s ys = 0.5s y = 0.5(125) = 62.5 ksi 1 1 s = N sy 2 2 ss 2 + s ys 1 2 2 1 0.1061T + 1.59155 0.0796T 2 = + 3 125 62.5 T = 213 in − kips = 213,000 in − lb Tn (213,000 )(2000) hp = = = 6762 hp 63,000 63,000 412. The same as 411, except that the shaft is hollow with an inside diameter of 2 ½ in. Solution: 32MDo 4F s= + 4 4 2 π Do − Di π D o − D i2 ( ) ( ) 2 32 T (4) 4(20) 3 s= + = 0.1252T + 2.612 4 4 π (4) − (2.5) π (4)2 − (2.5)2 16TDo 16T (4) ss = = = 0.0939T 4 4 π Do − Di π (4)4 − (2.5)4 [ ] [ ( ) [ ] ] 1 1 s = N sy 2 2 ss 2 + s ys 1 2 2 1 0.1252T + 2.612 0.0939T 2 = + 3 125 62.5 T = 177 in − kips = 177,000 in − lb Tn (177,000)(2000) hp = = = 5619 hp 63,000 63,000 413. A hollow, alloyed-steel shaft, AISI 4130, OQT 1100 F, has an OD of 3 ¼ in. and an ID of 2 ½ in. It is transmitting 1500 hp at 1200 rpm, and at the same time is withstanding a maximum bending moment of 40,000 in-lb. and an axial compressive force F = 10 kips. The length of the shaft between bearings is 10 ft. Using a steady stress approach, Page 73 of 133 SECTION 6 – COMBINED STRESSES determine (a) the maximum shearing stress in the shaft, (b) the maximum normal stress, (c) the factor of safety in each case. (d) Also compute N from the octahedral –shear theory. Solution: For alloy-steel shaft, AISI 4130, WQT 1100oF, s y = 114 ksi (Table AT 7) 63,000hp 63,000(1500 ) = = 78,750 in − lb n 1200 1 Do = 3 in 4 1 Di = 2 in 2 π π w = 0.284 Do2 − Di2 = 0.284 (3.25)2 − (2.5)2 = 0.962 lb in 4 4 2 wL M = M1 + (Table AT 8) 8 L = 10 ft = 120 in T= ( [ ) ] 0.962(120)2 = 41,732 in − lb 8 32MDo 4F s= + 4 4 2 π Do − Di π Do − Di2 32(41,732)(3.25) 4(10,000) s= + = 22,000 psi 4 4 π (3.25) − (2.5) π (3.25)2 − (2.5)2 16TDo 16(78,750)(3.25) ss = = = 17 ,978 psi 4 4 π Do − Di π (3.25)4 − (2.5)4 M = 40,000 + ( ) ( [ ) ] [ ) [ ( ] ] 1 2 1 s 2 22,000 2 2 2 (A) τ = + s s2 = + (17 ,978) = 21,076 psi 2 2 1 1 2 2 22,000 22,000 2 2 s s 2 2 (B) σ = + + s s = + + (17 ,978) = 32,076 psi 2 2 2 2 sys (C) N = τ = 0. 5s y τ = 0.5(114) = 2.704 21.076 sy 114 N= = = 3.554 σ 32.076 1 1 s (D) = N sy s ys = sy 3 = 2 ss + s ys 114 3 2 2 = 65.82 ksi Page 74 of 133 SECTION 6 – COMBINED STRESSES 1 2 2 1 22.0 17.978 2 = + N 114 65.82 N = 2.99 VARYING STRESSES COMBINED DESIGN PROBLEMS 414. The force F on the lever in the illustration (in the plane of the lever) varies from a maximum of 424.2 lb. to a minimum of -141.4 lb.; L = 20 in., a = 15 in., D2 = 1.2D1, r = 0.125D1, θ = 45o; the material is cold-drawn SAE 1040, 10% worked, the design factor N = 1.5. Compute the diameter D1 using the Soderberg-line approach with both the maximum-shear and octahedral-shear theories; indefinite life. Solution: For SAE 040, 10% Worked s n′ = 54 ksi s y = 85 ksi s n 54 1 = = sy 85 1.574 s ns s n 1 = = s ys s y 1.574 SF = 0.85 Strength Reduction Factors With r d = r D1 = 0.125 D d = D 2 D1 = 1.2 Fig. AF 12 K t = 1.54 K ts = 1.27 Page 75 of 133 SECTION 6 – COMBINED STRESSES Assume q = 1 K f = 1 + q (K t − 1) = 1 + (1)(K t − 1) = K t K f = 1.54 K fs = 1.27 Forces: Fmax = 424.2 lb Fmin = −141.4 lb 1 (Fmax + Fmax ) = 1 (424.2 − 141.4) = 141.4 lb 2 2 1 1 Fa = (Fmax − Fmax ) = (424.2 + 141.4) = 282.8 lb 2 2 Fm = πD 2 = 0.223D 2 w = 0.284 4 wL2 wL2 M = M1 + = FL + 2 2 0.233D 2 (20 )2 M m = (141.4)(20) + = 2828 + 44.6D 2 2 0.233D 2 (20)2 Ma = (282.8)(20) + = 5656 + 44.6D 2 2 T = (F cos θ )a ( ) = (282.8)(cos 45 )(15) = 3000 in − lb Tm = (141.4) cos 45o (15) = 1500 in − lb Ta o 32M πD 3 32 2828 + 44.6D 2 28,806 454 sm = = + 3 D πD D3 32 5656 + 44.6D 2 57 ,612 454 sa = = + D πD 3 D3 s= ( ) ( ) 16T πD 3 16(1500) 7640 sms = = 3 πD 3 D 16(3000 ) 15,280 sas = = πD 3 D3 ss = Page 76 of 133 SECTION 6 – COMBINED STRESSES s es = K fs sas sns sms + s ys SF 1 7640 1.27 15,280 27,684 + = 1.574 D 3 0.85 D 3 D3 K f sa s se = n s m + sy SF s es = se = 1 28,806 454 1.54 57,612 454 122,681 1111 + + + + = 1.574 D 3 D 0.85 D 3 D D D3 1 2 2 2 1 se ses = + N sn s ns Maximum shear, sns = 0.5sn = 0.5(54,000) = 27,000 psi 1 122,681 1111 2 2 2 + 1 D 3 D + 27 ,684 = 1.5 54,000 27,000D 3 D = 1.5625 in = 1 9 in 16 Octahedral Shear. sns = sn 3 = 54,000 3 = 31,177 psi 1 122,681 1111 2 2 2 + 1 D 3 D + 27,684 = 1.5 54,000 31,177D 3 D = 1.55 in 9 say D = 1 in 16 5 say D = 1 in 8 417. A hollow steel shaft, SAE 1045, as rolled, has an inside diameter of one half of the outside diameter and is transmitting 1600 hp at 600 rpm. The maximum bending moment is 40,000in-lb. Determine the diameter for N = 3 by both the maximum-shear and octahedral shear theories. Specify a standard size. Use the Soderberg line for obtaining the equivalent stresses. Page 77 of 133 SECTION 6 – COMBINED STRESSES Solution: For SAE 1045, as rolled, s y = 59 ksi , su = 96 ksi sn = 0.5su = 48 ksi sn sns 48 1 = = = sy sys 59 1.229 Assume K f = K fs = 1 Do = 2Di For bending: sm = 0 sa = s s 1 se = n sm + K f sa = (0) + (1)s = s sy 1.229 32MDo 32(40,000)(2Di ) 54,325 = = 4 4 Di3 π Do − D i π (2Di )4 − Di4 For torsion: 63,000hp 63,000(1600) T= = = 168,000 in − lb n 600 s s es = ns sms + K fs sas sys se = s = ( [ ) ] sas = 0 s ms = s sns 1 16TDo 1 16(168,000)(2Di ) 92,825 = sms = = 4 4 s ys 1.229 π Do − Di 1.229 π (2Di )4 − Di4 D i3 Maximum shear, sns = 0.5sn = 0.5(48,000) = 24,000 psi ses = ( [ ) 1 2 2 2 1 54,325 92,825 = + 1.5 48,000Di3 24,000D i3 Di = 2.295 in say Di = 2.25 in , Do = 2Di = 4.5 in Octahedral Shear. sns = sn 3 = 48,000 3 = 27,713 psi 1 2 2 2 1 54,325 92,825 = + 1.5 48,000Di3 27,713Di3 Di = 2.20 in say Di = 2.25 in , Do = 2Di = 4.5 in Page 78 of 133 ] SECTION 6 – COMBINED STRESSES 1 1 Standard Size Di = 2 in , Do = 4 in 4 2 418. A section of a shaft without a keyway is subjected to a bending moment that varies sinusoidally from 30 to 15 then to 30 in-kips during two revolutions, and to a torque that varies similarly and in phase from 25 to 15 to 25 in-kips; there is also a constant axial force of 40 kips; the material is AISI 2340, OQT 1000 F; N = 1.5. Determine the diameter by (a) the maximum-shear-stress theory; (b) the octahedral-shear-stress theory. Solution: For AISI 2340, OQT 1000oF. s y = 120 ksi , su = 137 ksi s n = 0.5su = 0.5(137 ) = 68.5 ksi sn sns 68.5 1 = = = sy sys 120 1.752 Assume K f = K fs = 1 1 (30 + 15) = 22.5 in − kips 2 1 Ma = (30 − 15) = 7.5 in − kips 2 1 Tm = (25 + 15) = 20 in − kips 2 1 Ta = (25 − 15) = 5 in − kips 2 Mm = 32M m 4F 32(22.5) 4(40) 229 51 + = + = 3 + 2 3 D D πD πD 2 πD 3 πD 2 32Ma 32(7.5) 76 sa = = = 3 D πD 3 πD 3 sm = 16Tm 16(20) 102 = = 3 D πD 3 πD 3 16Ta 16(5) 25.5 sas = = = 3 πD 3 πD 3 D sms = se = Kf sn 1 229 51 1 76 220 29 sm + sa = + + + = sy SF 1.752 D 3 D 2 0.85 D 3 D 3 D 2 ses = K fs sns 1 102 1 25.5 88 sms + sas = + = s ys SF 1.752 D 3 0.85 D 3 D 3 (a) Maximum shear, sns = 0.5sn = 0.5(68.5) = 34.25 psi Page 79 of 133 SECTION 6 – COMBINED STRESSES 1 220 29 + 1 D 3 D 2 = 1.5 68.5 2 2 2 88 + 3 34.25D D = 1.93 in say D = 2 in , Octahedral Shear. sns = sn 3 = 68.5 3 = 39.55 psi 1 220 29 + 1 D 3 D 2 = 1.5 68.5 2 2 2 88 + 3 39.55D D = 1.909 in say D = 2 in 419. The same as 418, except that the shaft has a profile keyway at the point of maximum moment. Solution: K f = 1.6 K fs = 1.3 Kf sn 1 229 51 1.6 76 274 29 sm + sa = + + + = sy SF 1.752 D 3 D 2 0.85 D 3 D 3 D 2 K fs s 1 102 1.3 25.5 97 ses = ns sms + sas = + = s ys SF 1.752 D 3 0.85 D 3 D 3 se = (a) Maximum shear, sns = 0.5sn = 0.5(68.5) = 34.25 psi 1 274 29 + 1 D 3 D 2 = 1.5 68.5 2 2 2 + 97 34.25D 3 D = 2.04 in say D = 2 in (b) Octahedral Shear. sns = Page 80 of 133 sn 3 = 68.5 3 = 39.55 psi SECTION 6 – COMBINED STRESSES 1 274 29 + 1 D 3 D 2 = 1.5 68.5 2 2 2 97 + 3 39.55D D = 2.02 in say D = 2 in CHECK PROBLEMS 420. A 2-in. shaft made from AISI 1144, elevated temperature drawn, transmits 200 hp at 600 rpm. In addition to the data on the figure, the reactions are B = 4.62 kips and E = 1.68 kips. Compute the factor of safety by the maximum-shear and octahedral-shear theories. Solution: For AISI 1144, Elevated Temperature, drawn, s y = 83 ksi , su = 118 ksi sn = 0.5su = 0.5(118) = 59 ksi sn 59 1 = = sy 83 1.407 MB = (2.1)(10) = 21 in − kips , MC = (1.68)(10) = 16.8 in − kips 63,000hp 63,000(200) T= = = 21,000 in − lbs = 21 in − kips n 600 Table AT 13 K f = 2.0 , K fs = 1.6 se = sn sm + K f sa sy Mm = 0 , s m = 0 Ma = Mm 32(16.8) 32MC s e = K f sa = K f = 2.0 = 42.8 ksi 3 3 πD π (2) s s es = ns sms + K fs sas sys Tm = T , Ta = 0 Page 81 of 133 SECTION 6 – COMBINED STRESSES ses = sns s 16T 1 16(21) sms = ns 3 = = 9.5 ksi s ys sys πD 1.407 π (2)3 Maximum shear, sns = 0.5sn = 0.5(59 ) = 29.5 psi 1 1 se = N sn 2 ses + sns 2 2 1 2 2 1 42.8 9.5 2 = + N 59 29.5 N = 1.26 Octahedral Shear. s ns = 0.577 sn = 0.577(59 ) = 34.05 psi 1 2 2 1 42.8 9.5 2 = + N 59 34.05 N = 1.26 421. In the figure (399), the bar supports a static load Q = 3000 lb. acting down; L = 16 in., a = 12 in., b = 7 in. The force F = 2500 lb. is produced by a rotating unbalanced weight and is therefore repeated and reversed in both the horizontal and the vertical directions. The 1-in. cap screw, with cut UNC threads, is made of AISI C1137, annealed, and it has been subjected to a tightening torque of 4600 in-lb. The thickness of the bar is 2 in. (a) Compute the factor of safety for the load reversing in the vertical direction, and (b) in the horizontal direction (maximum-shear theory), with the conservative assumption that friction offers no resistance. Solution: For AISI C1137, annealed, s y = 50 ksi , su = 85 ksi sn = 0.5su = 0.5(85) = 42.5 ksi K f = 2.8 (Table AT 12) sn sns 42.5 1 = = = sy sny 50 1.1765 T = 0.2DFi 4600 = 0.2(1)(Fi ) Page 82 of 133 SECTION 6 – COMBINED STRESSES Fi = 23,000 lbs = 23 kips For 1-in cap screws, UNC As = 0.606 in 2 Nut: A = 1.5 in kb ∆Fb = Fe kb + kc AE AE kb = s , kc = c Le Le kb As 0.606 = = = 0.2554 k b + k c As + Ac 0.606 + π (1.5)2 4 Q = 3000 lb = 3 kips F = 2500 lb = 2.5 kips (a) Vertical, Moment at Edge = 0, Q > F (a − b)Fe max = (Q + F )(b) (12 − 7 )Fe max = (3.0 + 2.5)(7 ) Fe max = 7.7 kips (a − b)Fe min = (Q − F )(b) (12 − 7 )Fe min = (3.0 − 2.5)(7 ) Fe min = 0.7 kips ∆Fb max = (7.7 )(0.2554) = 1.97 kips ∆Fb min = (0.7 )(0.2554) = 0.18 kips Fb max = Fi + ∆Fbmax = 23 + 1.97 = 24.97 kips Fb min = Fi + ∆Fb min = 23 + 0.18 = 23.18 kips 1 (Fb max + Fb min ) = 1 (24.97 + 23.18) = 24.1 kips 2 2 1 1 Fa = (Fb max − Fb min ) = (24.97 − 23.18) = 0.9 kip 2 2 Fm = Fm 24.1 = = 39.8 ksi As 0.606 F 0.9 sa = a = = 1.5 ksi As 0.606 sm = Page 83 of 133 SECTION 6 – COMBINED STRESSES 1 sm K f sa = + N sy sn say K f = 2.8 , SF = 0.85 , Factor for tension = 0.80 sn = (0.85)(0.80)(42.5) = 28.9 ksi 1 39.8 2.8(1.5) = + N 50 28.9 N = 1.06 (b) Horizontal: (a − b)Fe = Qb (12 − 7 )Fe = (3)(7 ) Fe = 4.2 kips ∆Fb = (4.2 )(0.2554 ) = 1.073 kips Fb = Fi + ∆Fb = 23 + 1.073 = 24.1 kips F 24.1 s= b = = 39.8 ksi A 0.606 sm = s sa = 0 K f sa s 1 se = n sm + = (39.8) + 0 = 33.83 ksi sy SF 1.1765 Shear: Fm = 0 1 Fa = (Fmax − Fmin ) = 2.5 kips 2 s ms = 0 Fa 2.5 = = 4.13 ksi As 0.606 K fs sas s (1.0)(4.13) = 4.86 ksi ses = ns sms + =0+ s ys SF 0.85 sas = 1 1 se = N sn ses 2 + sns sns = 0.5sn = 0.5(42.5) = 21.25 ksi , maximum shear 2 2 1 2 2 1 33.83 4.86 2 = + N 42.5 21.25 N = 1.21 Page 84 of 133 SECTION 6 – COMBINED STRESSES 422. The load Q, as seen (404), acts on the arm C and varies from 0 to 3 kips. The ends A and B of the shaft are restrained from turning through an angle but are supported to take the reactions A and B without other moments. The shaft is machined from AISI 1045, as rolled; D1 = 2, D2 = 2.5, L = 15, a = 10, b = 20 in. For calculation purposes, assume that the shaft size changes at the section of application of Q. Determine the factor of safety in accordance with the maximum-shear and octahedral-shear theories. Investigate both sections I and II. Would you judge the design to be 100% reliable? Solution: T = QL Tmax = (3)(15) = 45 in − kips T1 + T2 = 45 in − kips T1a T2 b = J J T1a T2 b = D14 D 24 T1 (10) T2 (20) = (2 )4 (2.5)4 T1 = 0.8192T2 0.8192T2 + T2 = 45 in − kips T2 = 24.74 in − kips T1 = 0.8192T2 = 0.8192(24.74) = 20.27 in − kips 1 1 Ta1 = Tm1 = T1 = (20.27 ) = 10.14 in − kips 2 2 1 1 Ta 2 = Tm 2 = T2 = (24.74 ) = 12.37 in − kips 2 2 16T πD 3 16T 16(10.14) sms1 = sas1 = 31 = = 6.46 ksi πD1 π (2)3 sms = sas = Page 85 of 133 SECTION 6 – COMBINED STRESSES sms 2 = sas 2 = 16T2 16(12.37 ) = = 4.03 ksi πD 23 π (2.5)3 A + B = Q = 3 kips Aa = Bb A(10) = B(20) A = 2B 2B + B = 3 kips B = 1 kip A = 2 kips M = Aa = Bb = (2 )(10) = 20 in − kips Mmax = M = 20 in − kips 1 M m = Ma = Mmax = 10 in − kips 2 32M sm = sa = πD 3 32(10 ) sm1 = sa1 = = 12.73 ksi π (2)3 32(10) sm 2 = sa 2 = = 6.52 ksi π (2.5)3 Use (1) sms = 6.46 ksi , sm = 12.73 ksi sas = 6.46 ksi , sa = 12.73 ksi r = 0.15D1 r D D 2.5 = 0.15 , = 2 = = 1.25 D1 d D1 2.0 K t = 1.5 , K ts = 1.25 (Figure AF 12) r = 0.15D1 = 0.15(2 ) = 0.30 1 = = 0.968 a 0.01 1+ 1+ r 0. 3 K f = q (K t − 1) + 1 = 0.968(1.5 − 1) + 1 = 1.484 q= 1 K fs = q (K ts − 1) + 1 = 0.968(1.25 − 1) + 1 = 1.242 Profile Keyway K f = 1.6 , K fs = 1.3 Net Page 86 of 133 SECTION 6 – COMBINED STRESSES K f = (1.484)(1.6 )(1 − 0.20) = 1.9 K fs = (1.242)(1.3)(1 − 0.20 ) = 1.3 For AISI 1045, as rolled, s y = 59 ksi , su = 96 ksi sn = 0.5su = 48 ksi sns sn 48 1 = = = s ys sy 59 1.229 SF = 0.85 , RF = 0.85 se = K f sa sn 1 sm + = (12.73) + 1.9(12.73) = 43.84 ksi sy (SF )(RF ) 1.229 (0.85)(0.85) ses = K fs sas sns 1 s ms + = (6.46) + 1.3(6.46) = 16.88 ksi s ys (SF )(RF ) 1.229 (0.85)(0.85) Maximum shear, sns = 0.5sn = 0.5(48) = 24 psi 1 1 se = N sn 2 ses + sns 2 2 1 2 2 1 43.84 16.88 2 = + N 48 24 N = 0.87 Octahedral Shear. sns = 0.577 sn = 0.577(48) = 27.7 psi 1 2 2 1 43.84 16.88 2 = + N 48 27.7 N = 0.91 Not 100% reliable, N < 100. 423. A rotating shaft overhangs a bearing, as seen in the illustration. A ¼-in. hole is drilled at AB. The horizontal force F2 varies in phase with the shaft rotation from 0 to 5 kips, but its line of action does not move. A steady torque T = 8 in-kips is applied at the end of the shaft; D = 2, D2 = 2.5, a = 2, b = 5, e = 0.5, r = ¼ in. The material is AISI C1040, annealed. What steady vertical load F1 can be added as shown if the design factor is to be 2.5 from the octahedral-shear theory? Assume that the cycling of F2 may be such that the worst stress condition occurs at the hole. Page 87 of 133 SECTION 6 – COMBINED STRESSES Solution: AISI C1040, annealed, s y = 48 ksi (Fig. AF 1), su = 80 ksi sn = 0.5su = 40 ksi sn 40 1 = = sy 48 1.2 For hole: d D = 0.25 2 = 0.125 K t = 2.2 , K ts = 1.6 a = 0.01 (annealed) 0.25 r= = 0.125 2 1 1 q= = = 0.926 a 0.01 1+ 1+ r 0.125 K f = q (K t − 1) + 1 = 0.926(2.2 − 1) + 1 = 2.11 K fs = q (K ts − 1) + 1 = 0.926(1.6 − 1) + 1 = 1.56 At hole s = s3 + s1 − s2 Bending F2 : Mc M s2 = = I I c F2 e (5)(0.5) s2 = = = 4.04 ksi 3 2 3 πD dD π (2) (0.25)(2 )2 − − 32 6 32 6 F2 ( 5) s3 = = = 1.89 ksi πD 2 π (2)2 − dD − (0.25)(2) 4 4 F1b (F1 )(5) s1 = = = 8.08F1 3 2 3 πD dD π (2) (0.25)(2)2 − − 32 6 32 6 Page 88 of 133 SECTION 6 – COMBINED STRESSES ss = T πD 3 dD − 16 6 2 = π (2 ) 3 16 − (8) (0.25)(2 )2 = 5.70 ksi 6 smin = s3 + s1 − s2 = 1.89 + 8.08F1 − 4.04 = 8.08F1 − 2.15 smax = s1 = 8.08F1 1 1 sm = (smax + smin ) = (8.08F1 + 8.08F1 − 2.15) = 8.08F1 − 1.08 ksi 2 2 1 1 sa = (smax − smin ) = (8.08F1 − 8.08F1 + 2.15) = 1.08 ksi 2 2 se = 6.74F1 − 1.78 s s es = ns sms + K fs sas sys s ms = s s sas = 0 1 (5.7 ) = 4.75 ksi 1.2 N = 2.5 ses = Octahedral Shear Theory s sns = n = 0.577 sn = 0.577(40) = 23.08 ksi 3 1 1 se = N sn 2 ses + sns 2 2 1 2 2 1 6.74F1 − 1.78 4.75 2 = + 2.5 40 23.08 F1 = 2.3 kips POWER SCREWS 424. Design a square-thread screw for a screw jack, similar to that shown, which is to raise and support a load of 5 tons. The maximum lift is to be 18 in. The material is AISI C1035, as rolled, and N ≈3.3 based on the yield strength. Page 89 of 133 SECTION 6 – COMBINED STRESSES Solution: AISI C1035, as rolled, s y = 55 ksi sy 55 = 16.6 ksi N 3.3 F = (5)(2) = 10 kips F 10 A= = = 0.6034 in 2 s 16.6 πD 2 A = r = 0.6034 in 2 4 Dr = 0.876 in say 1 ¼ in, Dr = 1.000 in L = 18 in Le = 2L = 36 in 1 1 k = Dr = (1.000) = 0.125 in 8 8 Le 36 = = 288 > 40 k 0.125 Transition: s= = 1 1 Le 2π 2E 2 2π 2 (30,000) 2 = = = 104 k sy 55 L Use column formula, Eulers e > 104 k π 2EI F= 2 NLe 10,000 = I= π 2 (30 × 10 6 )I 3.3(36)2 πDr4 = 0.14444 64 Dr = 1.31 in use 1 ¾ in, Dr = 1.400 in Page 90 of 133 SECTION 6 – COMBINED STRESSES 425. (a) For the screw of 424, what length of threads h will be needed for a bearing pressure of 1800 psi? (b) Complete the design of the jack. Let the base be cast iron and the threads integral with the base. Devise a method of turning the screw with a round steel rod as a lever and fix the details of a nonrotating cap on which the load rests. (c) What should be the diameter of the rod used to turn the screw? If a man exerts a pull of 150 lb. at the end, how long must the rod be? Solution: (a) Th/in = 2.5 Dr = 1.40 in 1 = 0.4 in 2.5 Lead λ = tan −1 πD m 1 Dm = (1.75 + 1.40 ) = 1.575 in 2 0.40 o λ = tan −1 = 4.62 ( ) π 1 . 575 Lead Pitch = f = 0.15 tan β = f = 0.15 β = 8.53o F cos(β + λ ) s= (Do − Dr )L 10,000 cos(8.53 + 4.62 ) (1.75 − 1.40)L L = 16.30 in 1800 = h = L tan λ = 16.30 tan 4.62 o = 1.32 in say h = 1.5 in Page 91 of 133 SECTION 6 – COMBINED STRESSES (b) Assume ASTM 20. sus = 32 ksi , su = 20 ksi , N = 5 32 = 6.4 ksi 5 20 s= = 4 ksi 5 ss = F πDh 10 6.4 = πD(1.5) D = 0.33 in ss = Dr = 1.4 in > 0.33 in 3 Do = 1 in 4 Use proportions from figure based on diameter. Method: Manual, normal pull. 7 (c) D = in (Based on proportion) 8 FDm (10 )(1.575) tan(8.53 + 4.62) = 1.84 in − kips T= tan(β + λ ) = 2 2 T = F ′a F ′ = 150 lb = 0.15 kips 1.84 = 0.15a a = 12.3 in 426. A screw jack, with a 1 ¼-in. square thread, supports a load of 6000 lb. The material of the screw is AISI C1022, as rolled, and the coefficient of friction for the threads is about 0.15. The maximum extension of the screw from the base is 15 in. (a) Considering the ends of the screw restrained so that Le = L, find the equivalent stress and the design factor. (b) If the load on the jack is such that it may sway, the screw probably acts as a column with one end free and the other fixe. What is the equivalent stress and the factor of safety in this instance? (c) What force must be exerted at the end of a 20-in. lever to raise the load? (d) Find the number of threads and the length h of the threaded portion in the cast-iron base for a pressure of 500 psi on the threads. (e) What torque is necessary to lower the load? Page 92 of 133 SECTION 6 – COMBINED STRESSES Solution: From Table AT 7, AISI C1022, as rolled, sy = 52 ksi F = 6000 lb = 6 kips For 1 ¼ in square thread, Dr = 1.0 in, Th/in. = 3.5 f = 0.15 (a) With Le = L = 15 in 1 1 k = Dr = (1.0) = 0.125 in 8 8 Le 15 = = 120 k 0.125 Transition for AISI C1020; 1 1 Le 2π 2E 2 2π 2 (30,000) 2 = = = 107 k sy 52 Le Use column formula, Eulers > 107 k π 2EA Fc = NF = (Le k )2 sd = F π 2E = A N (Le k )2 Equivalent stress F 4F sd = = 2 A πDr 4(6) sd = = 7.64 ksi π (1.0)2 Design factor π 2E sd = N (Le k )2 Page 93 of 133 SECTION 6 – COMBINED STRESSES 7.64 = π 2 (30,000) N (120)2 N = 2.69 (b) With Le = 2L = 30 in Le 30 = = 240 k 0.125 Transition for AISI C1020; 1 1 Le 2π 2E 2 2π 2 (30,000) 2 = = = 107 k sy 52 Le Use column formula, Eulers > 107 k π 2EA Fc = NF = (Le k )2 sd = F π 2E = A N (Le k )2 Equivalent stress F 4F sd = = 2 A πDr 4(6) sd = = 7.64 ksi π (1.0)2 Design factor π 2E sd = N (Le k )2 7.64 = π 2 (30,000) N (240)2 N = 0.673 not safe (c) For force exerted at the end of 20-in. lever to raise the load = Fa WDm T= tan(β + λ ) 2 1 Lead = Pc = = 0.2857 in 3. 5 1 Dm = (1.25 + 1.00) = 1.125 in 2 Lead 0.2857 λ = tan−1 = tan−1 = 4.62o πDm π (1.125) f = tan β = 0.15 β = 8.53o W = 6000 lb Page 94 of 133 SECTION 6 – COMBINED STRESSES WDm tan(β + λ ) 2 6000(1.125) Fa (20) = tan(8.53 + 4.62) 2 Fa = 39.43 lb T = Fa a = (d) Let p = pressure = 500 psi, W = 6000 lb, Do = 1.25 in, Di = 1.00 in. Nt = number of threads, h = length of threaded portion. 4W p= 2 π Do − Dr2 Nt 4(6000 ) 500 = π (1.25)2 − (1.00 )2 Nt Nt = 27 Then h = Nt Pc = (27 )(0.2857 ) = 7.7 in ( ) [ ] (e) Torque necessary to lower the load. WDm T= tan(β − λ ) 2 (6000)(1.125) tan(8.53 − 4.62) T= 2 T = 230.7 in − lb. 427. A square-thread screw, 2 in. in diameter, is used to exert a force of 24,000 lb. in a shaftstraightening press. The maximum unsupported length of the screw is 16 in. and the material is AISI C1040, annealed. (a) What is the equivalent compressive stress in the screw? Is this a satisfactory value? (b) What torque is necessary to turn the screw against the load for f = 0.15? (c) What is the efficiency of the screw? (d) What torque is necessary to lower the load? Solution: For 2 in. square thread screw, Do = 2 in, Dr = 1.612 in, Th/in. = 2.25 from Table 8.1 W = 24,000 lb = 24 kips, L = 16 in (a) For unsupported length, Le = L = 16 in. For AISI C1040, annealed, Figure AF-1, sy = 47.5 ksi Transition, 1 1 k = Dr = (1.612 ) = 0.2015 in 8 8 1 1 Le 2π 2E 2 2π 2 (30,000) 2 = = = 112 k sy 47.5 Then Le 16 = = 79.4 < 112 k 0.2015 Use column formula, JB Johnson Formula, Page 95 of 133 Le < 112 k SECTION 6 – COMBINED STRESSES sy (Le k )2 F = se 1 − A 4π 2E se = 4W sy (Le k )2 4π 2E 4(24) se = 47.5(79.4)2 π (1.612)2 1 − 2 4π (30,000) πDr2 1 − se = 15.74 ksi s 47.5 = 3.0 satisfactory N= y = se 15.74 (b) Torque to turn the screw against the load WDm tan(β + λ ) 2 1 Lead = Pc = = 0.4445 in 2.25 1 Dm = (2.00 + 1.612 ) = 1.806 in 2 Lead 0.4445 λ = tan−1 = tan−1 = 4.48o πDm π (1.806) f = tan β = 0.15 T= β = 8.53o W = 24,000 lb WDm tan(β + λ ) 2 24,000(1.806) T= tan(8.53 + 4.48) 2 T = 5008 in − lb T= (c) Torque necessary to lower the load. WDm T= tan(β − λ ) 2 24,000(1.806) T= tan(8.53 − 4.48) 2 T =1535 in − lb. 428. (a) A jack with a 2-in., square-thread screw is supporting a load of 20 kips. A single thread is used and the coefficient of friction may be as low as 0.10 or as high as 0.15. Page 96 of 133 SECTION 6 – COMBINED STRESSES Will this screw always be self-locking? What torque is necessary to raise the load? What torque is necessary to lower the load? (b) The same as (a) except that a double thread is used. (c) The same as (a) except that a triple thread is used. Solution: Table 8.1, 2 in. square thread, Do = 2 in, Dr = 1.612 in, Th/in = 2.25 (a) Self-locking? And Torque necessary to raise the load. 1 Dm = (2.00 + 1.612 ) = 1.806 in 2 1 Lead = Pc = = 0.4445 in 2.25 Lead 0.4445 λ = tan−1 = tan−1 = 4.48o πDm π (1.806) If f = 0.10 f = tan β = 0.10 β = 5.71o If f = 0.15 f = tan β = 0.15 β = 8.53o Since β is always greater than λ, the screw is always self-locking. WDm tan(β + λ ) 2 W = 20 kips WDm tan(β + λ ) T= 2 20(1.806) T= tan(8.53 + 4.48) 2 T = 4.173 in − kips Torque necessary to lower the load. WDm T= tan(β − λ ) 2 20(1.806) T= tan(8.53 − 4.48) 2 T = 1.279 in − kips. T= (b) Self-locking? And Torque necessary to raise the load. 2 Lead = 2Pc = = 0.8889 in 2.25 Lead 0.8889 λ = tan−1 = tan−1 = 8.904o πDm π (1.806) If f = 0.10 f = tan β = 0.10 β = 5.71o If f = 0.15 Page 97 of 133 SECTION 6 – COMBINED STRESSES f = tan β = 0.15 β = 8.53o Since β is always less than λ, the screw is always not self-locking. WDm tan(β + λ ) 2 W = 20 kips WDm tan(β + λ ) T= 2 20(1.806) T= tan(8.53 + 8.904) 2 T = 5.671 in − kips Torque necessary to lower the load = 0 T= (c) Self-locking? And Torque necessary to raise the load. 3 Lead = 3Pc = = 1.3333 in 2.25 Lead 1.3333 λ = tan−1 = tan−1 = 13.224o πDm π (1.806 ) If f = 0.10 f = tan β = 0.10 β = 5.71o If f = 0.15 f = tan β = 0.15 β = 8.53o Since β is always less than λ, the screw is always not self-locking. WDm tan(β + λ ) 2 W = 20 kips WDm T= tan(β + λ ) 2 20(1.806) T= tan(8.53 + 13.224) 2 T = 7.207 in − kips Torque necessary to lower the load = 0 T= 429. The conditions for a self-locking screw are given in §8.23, Text. Assume that the coefficient of friction is equal to the tangent of the lead angle and show that the efficiency of a self-locking screw is always less than 50%. Solution: tan λ λ e= ≈ tan(β + λ ) β + λ For self-locking, β > λ, then β + λ > 2λ Then, Page 98 of 133 SECTION 6 – COMBINED STRESSES e< λ 2λ e < 0.50 e < 50% CURVED BEAMS 430. It is necessary to bend a certain link somewhat as shown in order to prevent interference with another part of the machine. It is estimated that sufficient clearance will be provided if the center line of the link is displaced e = 3 in. from the line of action of F, with a radius of curvature of R ≈ 5.5 in., L = 10 in., material is wrought aluminum alloy 2014 T6; N = 2 on the basis of the maximum shear stress; F = 2500 lb. with the number of repetitions not exceeding 106. (a) If the section is round, what should be its diameter D? (b) If the link is bend to form cold, will the residual stresses be helpful or damaging? Discuss. Solution: (a) Table AT3. Wrought aluminum alloy 2014 T6 sn = 18 ksi @5 ×108 cycles sy = 60 ksi At 106 cycles 106 sn = sn′ nc 0.09 0.09 10 6 18 = sn′ 8 5 ×10 sn′ = 31.49 ksi With size factor. sn = 0.85sn′ = 0.85(31.49) = 26.77 ksi sn 26.77 = = 13.38 ksi N 2 Equation: F K Mc s= + c A I s= Page 99 of 133 SECTION 6 – COMBINED STRESSES A= I= πD 2 4 πD 4 64 D c= 2 M = Fe D K c (Fe ) 4F 2 s= 2 + πD πD 4 64 s= 4F 32K c Fe + πD 2 πD 3 Using Trial and error and Table AT 18: r 2R 2(5.5) 11 = = = c D D D 4(2.5) 32K c (2.5)(3) + 13.38 = πD 2 πD 3 By trial and error D = 1.92 in r 11 = = 6.0908 c 1.92 Table AT 18: Kc = 1.152 4(2.5) 32(1.152)(2.5)(3) s= + = 13.30 ksi ≈ 13.38 ksi π (1.92)2 π (1.92)3 Use D = 2 in. (b) Residual stress is helpful due to a decrease in total stress on tension side. 431. The same as 430, except that the section is rectangular with h ≈ 3b; see figure. Solution: (a) Table AT3. Wrought aluminum alloy 2014 T6 sn = 18 ksi @5 ×108 cycles sy = 60 ksi At 106 cycles Page 100 of 133 SECTION 6 – COMBINED STRESSES 106 sn = sn′ nc 0.09 0.09 10 6 18 = sn′ 8 5 × 10 sn′ = 31.49 ksi With size factor. sn = 0.85sn′ = 0.85(31.49) = 26.77 ksi s 26.77 s= n = = 13.38 ksi N 2 Equation: F K Mc s= + c A I A = bh = b(3b) = 3b 2 bh 3 b(3b)3 = = 2.25b 4 12 12 h c = = 0.5h = 1.5b 2 M = Fe F K (Fe )(1.5b) s= 2 + c 3b 2.25b 4 I= s= F K Fe + c 3 2 3b 1.5b Using Trial and error and Table AT 18: r 2R 2(5.5) 11 11 = = = = c h h h 3b 2.5 K (2.5)(3) 13.38 = 2 + c 3b 1.5b3 By trial and error b = 0.787 in r 11 = = 4.66 c 3(0.787 ) Table AT 18: Kc = 1.1736 2.5 (1.1736 )(2.5)(3) s= + = 13.38 ksi 2 3(0.787 ) 1.5(0.787 )3 Use b = 7/8 in. h = 3b = 2 5/8 in (b) Residual stress is helpful due to a decrease in total stress on tension side. 432. A hook is to be designed similar to that shown to support a maximum load F = 2500 lb. that will be repeated an indefinite number of times; the horizontal section is to be circular of radius c and the inside radius a is 1 ½ in. (a) Determine the diameter of the Page 101 of 133 SECTION 6 – COMBINED STRESSES horizontal section for N = 2 based on the Soderberg line, if the material is AISI 4130, WQT 1100 F. (b) Calculate the value of the static load that produces incipient yielding. Solution: (a) For AISI 4130, WQT 1100 F, Table AT 7 sy = 114 ksi, su = 127 ksi, sn’ = su/2 for reversed bending sn = SFsn′ = 0.85sn′ = 0.85(su 2) Soderberg line: 1 sm K f sa = + N sy sn sm = sa = s repeated load 2 K f = 1.0 s2 1 s = + N 2 sy 0.85(su 2 ) 1 1 1 = + s N 2sy 0.85su 1 1 1 s = + 2 2(114) 0.85(127 ) sd = 36.63 ksi For curved beam F K Mc s= + c A I a = 1.5 in A= π (2c )2 = πc 2 4 F = 2500 lb = 2.5 kips M = F (a + c ) I= π (2c )4 64 Page 102 of 133 = πc 4 4 SECTION 6 – COMBINED STRESSES Table AT 18, r =a+c r a + c 1. 5 + c = = c c c Substitute: 2.5 K (2.5)(1.5 + c )c 36.63 = 2 + c πc πc 4 4 2.5 10K c (1.5 + c ) 36.63 = 2 + πc πc 3 By trial and error: c = 0.633 r 1.5 + 0.633 = = 3.37 , K c = 1.293 c 0.633 2.5 10(1.293)(1.5 + 0.633) 36.63 = s = + 2 π (0.633) π (0.633)3 36.63 = s ≈ 36.60 ksi Use c = 11/16 = 0.6875 in Diameter = 2c = 1.375 in = 1 3/8 in (b) Static load that produces incipient yielding. sd = sy = 114 ksi F K Mc s= + c A I K (F )(1.5 + c )c F 114 = 2 + c πc πc 4 4 F K (F )(1.5 + c ) 114 = 2 + c πc πc 3 r 1.5 + 0.6875 = = 3.18 , K c = 1.312 c 0.6875 1.312(F )(1.5 + 0.6875) F 114 = + 2 π (0.6875) π (0.6875)3 F = 32.71 kips 433. The same as 432, except that the hook is expected to be subjected to 100,000 repetitions of the maximum load. Page 103 of 133 SECTION 6 – COMBINED STRESSES Solution: (a) For AISI 4130, WQT 1100 F, Table AT 7 sy = 114 ksi, su = 127 ksi, sn’ = su/2 for reversed bending At 100,000 repetitions 10 6 sn = 0.85(su 2 ) n c Soderberg line: 1 sm K f sa = + N sy sn sm = sa = 0.085 106 = 0.85(su 2) 100 , 000 s repeated load 2 K f = 1.0 s2 1 s = + N 2 sy 0.5169su 1 1 1 s = + N 2sy 1.0338su 1 1 1 s = + 2 2(114 ) 1.0338(127 ) sd = 41.66 ksi For curved beam F K Mc s= + c A I a = 1.5 in π (2c )2 = πc 2 4 F = 2500 lb = 2.5 kips M = F (a + c ) A= I= π (2c )4 = πc 4 64 4 Table AT 18, r =a+c r a + c 1. 5 + c = = c c c Substitute: 2.5 K (2.5)(1.5 + c )c 41.66 = 2 + c πc πc 4 4 2.5 10K c (1.5 + c ) 41.66 = 2 + πc πc 3 By trial and error: c = 0.601 r 1.5 + 0.601 = = 3.5 , K c = 1.28 c 0.601 Page 104 of 133 0.085 = 0.5169su SECTION 6 – COMBINED STRESSES 2. 5 10(1.28)(1.5 + 0.601) + 2 π (0.601) π (0.601)3 41.66 = s ≈ 41.64 ksi Use c =5/8 Diameter = 2c = 1.25 in = 1 1/4 in 41.66 = s = (b) Static load that produces incipient yielding. sd = sy = 114 ksi F K Mc s= + c A I K (F )(1.5 + c )c F 114 = 2 + c πc πc 4 4 F K (F )(1.5 + c ) 114 = 2 + c πc πc 3 r 1.5 + 0.625 = = 3.4 , K c = 1.29 c 0.625 1.29(F )(1.5 + 0.625) F 114 = + 2 ( ) π 0.625 π (0.625)3 F = 25.97 kips 434. A hook, similar to that shown with a horizontal circular section of diameter 2c, is to be designed for a capacity of 2000 lb. maximum, a load that may be applied an indefinite number of times. A value of a = 2 in. should be satisfactory for the radius of curvature of the inside of the hook. Let N = 1.8 based on the modified Goodman line. At the outset of design, the engineer decided to try AISI C1040, OQT 1100 F. (a) Compute the diameter of the horizontal section, (b) If the 45o circular section is made the same diameter, what is its design factor (modified Goodman)? Could this section be made smaller or should it be larger? Solution: (a) For AISI C1040, OQT 1100 F, Figure AF 1 su = 100 ksi, sn’ = su/2 for reversed bending sn = SF x sn’ = 0.85(0.5)(100) = 42.5 ksi Kf = 1.0 Modified Goodman line: Page 105 of 133 SECTION 6 – COMBINED STRESSES 1 sm K f sa = + N su sn s sm = sa = repeated load 2 K f = 1.0 s2 s2 1 = + 1.8 100 42.5 sd = 33.14 ksi For curved beam F K Mc s= + c A I a = 2.0 in A= π (2c )2 = πc 2 4 F = 2000 lb = 2.0 kips M = F (a + c ) I= π (2c )4 = πc 4 64 4 Table AT 18, r =a+c r a + c 2. 0 + c = = c c c Substitute: 2.0 K (2.0)(2.0 + c )c 33.14 = 2 + c πc πc 4 4 2.0 8K (2.0 + c ) 33.14 = 2 + c 3 πc πc By trial and error: c = 0.639 r 2.0 + 0.639 = 4.13 , K c = 1.224 = c 0.639 2. 0 8(1.224)(2.0 + 0.639) 33.14 = s = + 2 π (0.639) π (0.639)3 33.14 = s ≈ 33.08 ksi Use c = 11/16 in Diameter = 2c = 1.375 in = 1 3/8 in (b) sus = 0.6su = 0.6 x 100 ksi = 60 ksi sns = 0.6sn = 0.6 x 42.5 ksi = 25.5 ksi Equivalent stress (Modified Goodman Line) s s se = m n + K f sa su s s ses = ms ns + K f sas sus Page 106 of 133 SECTION 6 – COMBINED STRESSES 1 2 2 2 1 se ses = + N sn sns s sm = sa = 2 s sms = sas = s 2 F cos 45 K c Mc F cos 45 K c F (a + c )(cos 45)c s= + = + A I πc 2 πc 4 4 F cos 45 4K c F (a + c )(cos 45) + s= πc 2 πc 3 F sin 45 F sin 45 ss = = A πc 2 11 c = in = 0.6875 in (assuming constant diameter) 16 r a + c 2.0 + 0.6875 = = = 3.91 c c 0.6875 Table AT 18, K c = 1.239 (2.0 )cos 45 4(1.239)(2.0)(2.0 + 0.6875)(cos 45) s= + π (0.6875)2 π (0.6875)3 s = 19.40 ksi (2.0 )sin 45 ss = = 0.95 ksi π (0.6875)2 Then s s se = m n + K f sa su 19.40 42.5 ss se = n + K f = + 1 = 13.82 ksi 2 su 2 100 s s ses = ms ns + K f sas sus ses = 0.85 25.5 ss sns + K f = + 1 = 0.68 ksi 2 sus 2 60 2 2 1 se ses = + N sn sns 1 2 2 1 13.82 0.68 2 = + N 42.5 25.5 N = 3.06 Since N > 1.8, this section could be made smaller. Page 107 of 133 SECTION 6 – COMBINED STRESSES 435. A C-frame hand press is made of annealed cast steel (A27-58) and has a modified Isection, as shown. The dimensions of a 45o section CD are: a = 3, b = 6, h = 4, t = 1 in., radius r = 1 in.; also g = 12 in.; and the maximum force is F = 17 kips, repeated a relatively few times in the life of the press. (a) Applying the straight-beam formula to the 45o section, compute the maximum and minimum normal stresses. (b) Do the same, applying the curved-beam formula. (c) By what theory would you judge this section to have been designed? If the radius r were increased several times over, as it could have been done, would the stress have been materially reduced? Give reasons for your conclusions. Solution: (a) Straight-beam formula Consider only normal stresses, relatively static. F cos 45 Mc s= ± A I c M = F g − r + 2 + r cos 45 2 A = ht + at + (b − 2t )t t t b − 2t ht + (b − 2t )(t ) + t + at b − t + 2 2 2 c2 = ht + (b − 2t )t + at Page 108 of 133 SECTION 6 – COMBINED STRESSES ht 2 t bt + (b − 2t ) + at b − 2 2 2 c2 = ht + (b − 2t )t + at c1 = b − c 2 (4)(1)2 1 (6 )(1) + [6 − 2(1)] + (3)(1) 6 − 2 2 2 c2 = = 2.77273 in (4 )(1) + [6 − 2(1)](1) + (3)(1) c1 = b − c 2 = 6 − 2.7723 = 3.22727 in I = I + Ad 2 A1 = ht A2 = (b − 2t )t A3 = at ht 3 t I1 = + ht c 2 − 12 2 I2 = t (b − 2t )3 b + (b − 2t )(t ) − c2 12 2 at 3 t I3 = + at c1 − 12 2 I1 = 2 2 2 2 (4)(1)3 ( )( ) 1 + 4 1 2.77273 − = 21 in 4 12 2 2 (1)[6 − 2(1)]3 [ ( )]( ) 6 I2 = + 6 − 2 1 1 − 2.77273 = 5.54 in 4 12 (3)(1) 3 2 2 1 + (3)(1) 3.22727 − = 22.564 in 4 12 2 I = 21 + 5.54 + 22.564 = 49.104 in 4 Then F cos 45 Mc 2 smax = + A I F cos 45 Mc1 smin = − A I A = (4 )(1) + (3)(1) + [6 − 2(1)](1) = 11 in 2 I3 = 2.77273 M = 17 12 − 1 + + 1 cos 45 = 215.686 in − kips 2 17 cos 45 (215.686)(2.77273) smax = + = 13.27 ksi in tension 11 49.104 17 cos 45 (215.686)(2.77273) smin = − = −13.08 ksi = 13.08 ksi in compression 11 49.104 (b) Curved-beam formula Page 109 of 133 SECTION 6 – COMBINED STRESSES F cos 45 K ci Mc 2 + A I F cos 45 K co Mc1 smin = − A I Using Table AT18 r Z = −1 + [b1 log e (r + c1 ) − (t − b1 )log e (r + c 4 ) + (b − t )log e (r − c3 ) − blog e (r − c 2 )] A smax = r = 1 + 2.77273 = 3.77273 in c 2 = 2.77273 in c1 = 3.22727 in c 4 = 3.22727 − 1 = 2.22727 in c3 = 2.77273 − 1 = 1.77273 in b1 = 3 in t = 1 in b = 4 in c 1 + Z (r + c ) I Kc = Arc 3.77273 3log e (3.77273 + 3.22727 ) − (1 − 3)log e (3.77273 + 2.22727) Z = −1 + 11 + (4 − 1)log e (3.77273 − 1.77273) − 4 log e (3.77273 − 2.77273) Z = 2.944455 c = −c2 c2 2.77273 (49.104) 1 + I 1 + Z (r − c 2 ) 2.944455(3.77273 − 2.77273) K ci = = (11)(3.77273)(2.77273) Arc 2 K ci = 0.8286 c = c1 c1 3.22727 (49.104) 1 + I 1 + Z (r + c1 ) 2.944455(3.77273 + 3.22727 ) K co = = (11)(3.77273)(3.22727 ) Arc1 K co = 0.424 F cos 45 K ci Mc 2 smax = + A I F cos 45 K co Mc1 smin = − A I 17 cos 45 (0.8286)(215.686)(2.77273) smax = + = 11.18 ksi in tension 11 49.104 Page 110 of 133 SECTION 6 – COMBINED STRESSES smin = 17 cos 45 (0.424)(215.686)(2.77273) − = −4.07 ksi = 4.07 ksi in compression 11 49.104 (c) This section must be designed based on straight beam formula. Maximum stress is higher. Increasing the radius r. Table A-18. r = 2 + 2.77273 = 4.77273 in c 2 = 2.77273 in c1 = 3.22727 in c 4 = 3.22727 − 1 = 2.22727 in c3 = 2.77273 − 1 = 1.77273 in b1 = 3 in t = 1 in b = 4 in c 1 + Z (r + c ) I Kc = Arc 4.77273 3log e (4.77273 + 3.22727 ) − (1 − 3)log e (4.77273 + 2.22727 ) Z = −1 + 11 + (4 − 1)log e (4.77273 − 1.77273) − 4 log e (4.77273 − 2.77273) Z = 3.622343 c = −c2 c2 2.77273 1 + I 1 + (49.104) ( ) Z r − c 3 . 622343 ( 4 . 77273 − 2 . 77273 ) 2 K ci = = ( )( )( ) Arc 2 11 4.77273 2.77273 K ci = 0.4664 c = c1 c1 3.22727 (49.104) 1 + I 1 + Z ( r + c ) 3.622343(4.77273 + 3.22727 ) 1 K co = = (11)(4.77273)(3.22727) Arc1 K co = 0.3221 F cos 45 K ci Mc 2 smax = + A I F cos 45 K co Mc1 smin = − A I 17 cos 45 (0.4664)(215.686)(2.77273) smax = + = 6.77 ksi in tension 11 49.104 17 cos 45 (0.3221)(215.686)(2.77273) smin = − = −2.83 ksi = 2.83 ksi in compression 11 49.104 Page 111 of 133 SECTION 6 – COMBINED STRESSES 436. The stress is reduced using by increasing the radius r in Curved Beam Formula. Reason: As the radius r increased the stress factor for curved beam decreases thence the maximum stress is reduced. A heavy C-clamp, similar to the figure, is made of normalized cast steel (A27-58) and has a T-section where t= 7/16 in.; q= 2 ¾ , a =1 ¾ in. What is the safe capacity if N = 2 based on yield? Solution: F K Mc s = + ci i A I Table AT 1 2 3t 7 A = 4t + t (4.5t ) = 10.5t 2 = 10.5 = 2.009766 in 2 2 16 (t ) 4.5t + 3 t 2 2 3 + (4t − t ) t 2 2 = 2.035714t c1 = 3 3 2t 4.5t + t + (4t − t ) t 2 2 7 c1 = 2.035714 = 0.890625 in 16 c 2 = 4.5t + 1.5t − c1 = 6t − 2.035714t = 3.964286t 7 c 2 = 3.964286 = 1.734375 in 16 Table AT 18 r = a + c1 = 1.75 + 0.890625 = 2.640625 in r a + c1 2.640625 = = = 2.965 c c1 0.890625 K ci = 1.4212 M = F (q + ci ) = F (2.75 + 0.890625) = 3.640625F For Normalized cast steel, A27-58, sy = 36 ksi Moment of Inertia Page 112 of 133 SECTION 6 – COMBINED STRESSES 3 (4t ) 3 t 2 2 3 2 + (4t ) 3 t 2.035714t − 3 t + (t )(4.5t ) + (t )(4.5t ) 3.964286t − 4.5 t I= 12 4 12 2 2 4 7 I = 31.861607t = 31.861607 = 1.167293 in 4 16 4 F K ci Mc i + A I sy 36 F (1.4212 )(3.640625)(F )(0.890625) s= = = + N 2 2.009766 1.167293 F = 4.049 kips = 4049 lb s= 437. The same as 436, except that the section is trapezoidal with b = ¾ in. (see figure). Ignore the effect of resounding off the corners. Solution: F K Mc s = + ci i A I From other sources. 2 1 (b + 2b)(3b ) = 4.5b 2 = 4.5 3 = 2.53125 in 2 2 4 3b 2b + 2b 4 43 c1 = = b = = 1 in 3 b + 2b 3 34 4 5 5 3 c 2 = 3b − b = b = = 1.25 in 3 3 3 4 A= 4 (3b)3 [b 2 + 4b(2b) + (2b)2 ] 3 4 I= = 3.25b = 3.25 = 1.02832 in 4 36(b + 2b) 4 Table AT 18 3.25b4 Z = −1 + 2r b − a (r + c 2 ) × log e r + c 2 − (b − a ) a + (a + b)c c r − c1 Page 113 of 133 SECTION 6 – COMBINED STRESSES r = a + c1 = 1.75 + 1 = 2.75 in a = b = 0.75 in b = 2b = 2(0.75) = 1.50 in c = 3b = 3(0.75) = 2.25 in Z = −1 + 2(2.75) 1.50 − 0.75 2.75 + 1.25 (2.75 + 1.25) × log e − (1.50 − 0.75) 0.75 + (0.75 + 1.5)(2.25) 2.25 2.75 − 1 Z = 0.05627 c1 1 (1.02832) 1 + I 1 + Z (r − c1 ) 0.05627(2.75 − 1) K ci = = = 1.6479 (2.53125)(2.75)(1) Arc1 M = F (q + c1 ) = F (2.75 + 1) = 3.75F For Normalized cast steel, A27-58, sy = 36 ksi F K ci Mc i + A I sy 36 F (1.6479)(3.75)(F )(1) s= = = + N 2 2.53125 1.02832 F = 2.810 kips = 2810 lb s= THICK-SHELL CYLINDERS; INTERFERENCE FITS 438. Special welded steel pipe, equivalent in strength to SAE 1022, as rolled, is subjected to an internal pressure of 8000 psi. The internal diameter is to be 4 ½ in. and the factor of safety is to be 3, including an allowance for the weld. (a) Find the thickness of the pipe according to the distortion-energy theory. (b) Using this thickness find the maximum normal and shear stresses and the corresponding safety factors. (c) Compute the thickness from the thin-shell formula and from the Barlow formula. Solution: 4.5 ri = = 2.25 in , N = 3 , pi = 8000 psi 2 SAE 1022, as rolled, sy = 52 ksi (a) Distortion-Energy Theory 1 2 1 − 1 in t = ri 1 − 3pi s sy 52 s = = = 17.333 ksi = 17 ,333 psi N 3 Page 114 of 133 SECTION 6 – COMBINED STRESSES 1 2 1 − 1 = 2.774 in t = 2.25 1 − 3 × 8000 17 ,333 (b) Maximum normal stress p r 2 + r 2 − 2p r 2 σ ti = i o 2 i 2 o o ro − ri ( ) ( ) pi ro2 + ri2 ro2 − ri 2 ri = 2.25 in ro = 2.25 + 2.774 = 5.024 in σ ti = ( ) 8000 5.024 2 + 2.25 2 = 12,014 psi 5.0242 − 2.252 s 52,000 = 4.33 N= y = σ ti 12,014 Maximum shear stress r 2 (p − p ) τ = o 2 i 2o ro − ri σ ti = τ= τ= ro2 pi ro2 − ri 2 (5.024)2 (8000) = 10,007 psi 5.0242 − 2.252 sy 52,000 N= = = 2.60 2τ 2(10,007 ) (c) From thin-shell formula p r (8000)(2.25) t= i i = = 1.0385 in st 17,333 From Barlow formula pr p (r + t ) t= i o = i i st st pr (8000)(2.25) t= i i = = 1.929 in st − pi 17,333 − 8000 439. The internal diameter of the cast-steel cylinder, SAE 0030, of a hydraulic press is 12 in. The internal working pressure is 6000 psi, N = 2.5. Find the thickness of the cylinder walls (a) from the maximum-shear-stress theory, (b) from the octahedral-shear theory. (c) Compute the thickness from the thin-shell and Barlow formulas. What do you recommend? Solution: Page 115 of 133 SECTION 6 – COMBINED STRESSES Table AT 6. SAE 0030 = A27-58, sy = 35 ksi (a) Maximum shear theory r 2 (p − p ) s τ = o 2 i 2o = y ro − ri 2N 12 ri = = 6 in 2 pi = 6,000 psi = 6 ksi po = 0 ksi ro2 (6 − 0 ) 35 = 2 2 2(2.5) ro − (6) ro = 15.8745 in t = ro − ri = 15.8745 − 6 = 9.8745 in (b) Octahedral Sheat Theory 12 1 − 1 t = ri 3pi 1 − s sy 35 = 14 ksi s= = N 2.5 12 1 t = (6 ) − 1 = 5.8195 in 3 (6) 1 − 14 (c) Thin shell formula p r sy st = i i = t N (6)(6) = 35 t 2.5 t = 2.5714 in Barlow formula sy pr st = i o = t N 6(6 + t ) 35 = t 2.5 t = 4.5 in Recommended: Maximum shear theory , t = 9.8745 in thick. Page 116 of 133 SECTION 6 – COMBINED STRESSES 440. The same as 439, except a higher-strength material is selected. Try cast-steel SAE 0105. Solution: Table AT 6. SAE 0105 = A148-58, sy = 85 ksi (a) Maximum shear theory r 2 (p − p ) s τ = o 2 i 2o = y ro − ri 2N 12 ri = = 6 in 2 pi = 6,000 psi = 6 ksi po = 0 ksi ro2 (6 − 0 ) 85 = 2 2 2(2.5) ro − (6) ro = 7.459 in t = ro − ri = 7.459 − 6 = 1.459 in (b) Octahedral Sheat Theory 12 1 − 1 t = ri 3pi 1 − s sy 85 s= = = 34 ksi N 2.5 12 1 − 1 = 1.2005 in t = (6) 3 (6 ) 1 − 34 (c) Thin shell formula p r sy st = i i = t N (6)(6) = 85 t 2.5 t = 1.0588 in Barlow formula sy pr st = i o = t N 6(6 + t ) 85 = t 2.5 t = 1.2857 in Recommended: Maximum shear theory , t = 1.459 in thick. Page 117 of 133 SECTION 6 – COMBINED STRESSES 441. A 2 ½ in. heavy-wall pipe has the following dimensions: OD = 2.875, ID = 1.771, t = 0.552 in.; inside surface area per foot of length = 66.82 in.2, outside surface area per foot of length = 108.43 in.2. The material is chromium-molybdenum alloy, for which the permissible tangential tensile stress is 15 ksi at temperatures between 700 – 800 F. (a) Compute the maximum internal working pressure for this pipe from Lame’s formula, by the maximum-shear and octahedral-shear theories. (b) What is the stress at an external fiber? (c) A higher design stress would be permitted for an external pressure alone. Nevertheless, compute the external pressure corresponding to a maximum tangential stress of 15 ksi. Solution: OD 2.875 ro = = = 1.4375 in 2 2 ID 1.771 ri = = = 0.8855 in 2 2 t = 0.552 in (a) Lame’s Equation p r 2 + r 2 − 2p r 2 σ ti = i o 2 i 2 o o = s ro − ri ( ) [ ] pi (1.4375)2 + (0.8855)2 − 0 (1.4375)2 − (0.8855)2 pi = 6.7477 ksi Maximum shear theory r 2 (p − p ) s τ = o 2 i 2o = ro − ri 2 15 = (1.4375)2 (pi ) 15 = 2 2 (1.4375) − (0.8855) 2 pi = 4.654 ksi Octahedral shear theory 12 1 t = ri − 1 3pi 1 − s 12 1 0.552 = (0.8855) − 1 3pi 1 − 15 pi = 5.374 ksi (b) Stress at external fiber, pi = 4.654 ksi σ to = ( 2pi ri 2 − po ro2 + ri2 ro2 − ri2 Page 118 of 133 ) SECTION 6 – COMBINED STRESSES 2(4.654)(0.8855)2 − 0 σ to = = 5.592 ksi (1.4375)2 − (0.8855)2 (c) External pressure alone. p r 2 + r 2 − 2p r 2 σ ti = i o 2 i 2 o o = s ro − ri ( ) 0 − 2po (1.4375)2 (1.4375)2 − (0.8855)2 po = 4.654 ksi − 15 = 442. A cast-steel hub is to be shrunk on a 1.5-in., SAE 1035, as-rolled, steel shaft. The equivalent diameter of the hub is 2.5 in., its length is 4 in. (a) What must be the interference of metal if the holding power of this fit is equal to the torsional yield strength of the shaft? Use Baugher’s recommendations. (b) What are the corresponding tangential and radial stresses in the hub? Solution: Table AT 7, SAE 1035, as rolled, sy = 55 ksi. sys = 0.6 sy = 33 ksi Es = 30,000 ksi µs = 0.3 For hub, Cast steel, Eh = 30,000 ksi, µh ~ 0.3 (a) Interference of metal For solid shaft, same E and µ. 2 Ei Di 1 − pi = 2Di Do Di = 1.5 in Do = 2.5 in L = 4 in For pi: fp πD 2L T= i i 2 But πDi3 sys T= 16 Then πDi3 sys fpiπDi2L = 16 f = 0.1 as per Baugher’s recommendation Di sys (1.5)(33) pi = = = 15.46875 ksi 8 fL 8(0.1)(4 ) Then Page 119 of 133 SECTION 6 – COMBINED STRESSES pi = 2 Ei Di 1 − 2Di Do (30,000)(i ) 1.5 2 15.46875 = 1 − 2(1.5) 2.5 i = 0.002417 in - answer. (b) Tangential and radial stresses in the hub Tangential stress Ei Di 1 + σ th = 2Di Do σ th = 2 (30,000)(0.002417 ) 1.5 2 = 32.87 ksi 1 + 2(1.5) 2.5 Radial stress σ rh = − pi = −15.46875 ksi 443. The same as 442, except that the hub is ASTM 20, cast iron. Will the resulting tensile stresses be safe for cast iron? Solution: Table AT 6, ASTM 20, cast iron, suc = 83 ksi, su = 20 ksi (hub) Table AT 7, SAE 1035, as rolled, sy = 55 ksi. sys = 0.6 sy = 33 ksi (a) Interference of metal For hub of cast iron and the shaft is steel. D 2 Ei 1 − i Do pi = 2 D Di 3 + µ + (1 − µ ) i Do Di = 1.5 in Do = 2.5 in L = 4 in E = 30,000 ksi µ = 0.27 For pi: fp πD 2L T= i i 2 But πDi3 sys T= 16 Page 120 of 133 SECTION 6 – COMBINED STRESSES Then πDi3 sys = fpiπDi2L 16 f = 0.1 as per Baugher’s recommendation Di sys (1.5)(33) pi = = = 15.46875 ksi 8 fL 8(0.1)(4 ) Then D 2 Ei 1 − i Do pi = 2 Di Di 3 + µ + (1 − µ ) Do 2 (30,000)(i )1 − 1.5 2.5 15.46875 = 2 (1.5)3 + 0.27 + (1 − 0.27 ) 1.5 2.5 i = 0.004269 in - answer. (b) Tangential and radial stresses in the hub Tangential stress D 2 Ei 1 + i Do σ th = 2 D Di 3 + µ + (1 − µ ) i Do 2 (30,000)(0.004269)1 + 1.5 2.5 = 32.87 ksi σ th = 2 (1.5)3 + 0.27 + (1 − 0.27 ) 1.5 2.5 (30,000)(0.002417 ) 1.5 2 σ th = = 32.87 ksi 1 + 2(1.5) 2.5 > 20 ksi. Not safe for cast iron ASTM 20. Radial stress σ rh = − pi = −15.46875 ksi 444. A cast-steel gear is pressed onto a 2-in. shaft made of AISI 3140, OQT 1000 F. The equivalent hub diameter is 4 in., and the hub length is 4 in. (a) What are the maximum tangential and radial stresses in the hub caused by a class FN 2 interference fit? Compute for the apparent maximum value of i (but recall the probability of this event). Page 121 of 133 SECTION 6 – COMBINED STRESSES (b) What axial force F in tons will be required to press the gear on the shaft if f1 is assumed to be 0.2? (c) What torque may the force fit safely transmit? (d) Is the holding capacity of this fit large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft? Solution: Cast steel, E = 30 x 106 psi, µ = 0.27 or approximately 0.3 AISI 3140, OQT 1000 F, E = 30 x 106 psi, µ = 0.3, sy = 133 ksi (Fig. AF 2). Di = 2 in, Do = 4 in, L = 4 in. For Class FN 2 interference fit. Table 3.2, page 85, 2 in diameter. Maximum value of i = 0.0027 – 0.0000 = 0.0027 in (a) For same material and same Poisson’s ratio Tangential stress Ei Di 1 + σ th = 2Di Do 2 (30 ×10 )(0.0027) 1 + 2 = 6 σ th 2(2) = 25,313 psi 4 2 Radial stress 2 Ei Di 1 − σ rh = − pi = − 2Di Do 2 6 30 ×10 (0.0027 ) 2 σ rh = − 1 − = −15,188 psi 2(2) 4 ( ) (b) Axial force F in tons. f p πD L F = 1 i i tons 2000 ( 0.2 )(15,188)(π )(2)(4) = 38.17 tons F= 2000 (c) Torque safely transmit. fp πD 2L T= i i 2 f = 0.1 as recommended by Baugher. (0.1)(15,188)(π )(2)2 (4) T= = 38,172 in − lb 2 (d) With simple torsional stress of 0.6sys. ss = 0.6sys = 0.6(0.6sy ) = 0.6(0.6 )(133) = 47.88 ksi = 47 ,880 psi ssπDi3 (47 ,880)(π )(2)3 = = 72,210 psi 16 16 No. The holding capacity of this fit is not large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft. T= Page 122 of 133 SECTION 6 – COMBINED STRESSES 445. The same as 444, except that a class FN 4 fit is investigated and the computation is made for the average i. Solution: Cast steel, E = 30 x 106 psi, µ = 0.27 or approximately 0.3 AISI 3140, OQT 1000 F, E = 30 x 106 psi, µ = 0.3, sy = 133 ksi (Fig. AF 2). Di = 2 in, Do = 4 in, L = 4 in. For Class FN 4 interference fit. Table 3.2, page 85, 2 in diameter. Maximum value of i = 0.0042 – 0.0000 = 0.0042 in Minimum value of i = 0.0035 – 0.0012 = 0.0023 in Average value of i = 0.5 (0.0042 + 0.0023) = 0.00325 in (a) For same material and same Poisson’s ratio Tangential stress Ei Di 1 + σ th = 2Di Do 2 (30 ×10 )(0.00325) 1+ 2 = 6 σ th 2(2 ) = 30,469 psi 4 2 Radial stress 2 Ei Di 1 − σ rh = − pi = − 2Di Do 2 6 30 ×10 (0.00325) 2 σ rh = − 1 − = −18,281 psi 2(2 ) 4 ( ) (b) Axial force F in tons. f p πD L F = 1 i i tons 2000 ( 0.2)(18,281)(π )(2 )(4) = 45.95 tons F= 2000 (c) Torque safely transmit. fp πD 2L T= i i 2 f = 0.1 as recommended by Baugher. (0.1)(18,281)(π )(2)2 (4) T= = 45,945 in − lb 2 (d) With simple torsional stress of 0.6sys. ss = 0.6sys = 0.6(0.6sy ) = 0.6(0.6 )(133) = 47.88 ksi = 47 ,880 psi ssπDi3 (47 ,880)(π )(2)3 = = 72,210 psi 16 16 No. The holding capacity of this fit is not large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft. T= Page 123 of 133 SECTION 6 – COMBINED STRESSES 446. A No. 217 ball bearing has a bore of 3.3465 in., a width of 1.1024 in., and the inner race is approximately 3/8 in. thick. This bearing is to be mounted on a solid shaft with i = 0.0014. (a) Calculate the maximum radial and tangential stresses in the race. (b) Estimate the force required to press the bearing onto the shaft. Solution: Di = 3.3465 in, Do = 3.3465 + 2(3/8) = 4.0965 in, i = 0.0014 in. (a) Maximum radial stress in the race 2 Ei Di 1 − σ rh = − pi = − 2Di Do 2 6 30 ×10 (0.0014) 3.3465 σ rh = − = −2,087 psi 1 − 2(3.3465) 4.0965 Tangential stress 2 Ei Di 1 + σ th = 2Di Do 2 6 30 ×10 (0.0014) 3.3465 σ th = 1 + = 10,463 psi 2(3.3465) 4.0965 (b) Force required to press the bearing onto the shaft f p πD L F = 1 i i tons , use f1 = 0.175 on the average 2000 (0.175)(2,087 )(π )(2)(4 ) = 4.59 tons F= 2000 ( ( ) ) 447. A steel disk of diameter Do and thickness L = 4 in. is to be pressed onto a 2-in. steel shaft. The parts are manufactured with class FN 5 fit, but assembled parts are selected so as to give approximately the average interference. What will be the maximum radial and tangential stresses in the disk if (a) Do is infinitely large; (b) Do = 10 in.; (c) Do = 4 in.; (d) Do = 2.5 in.? Solution: (a) Maximum radial stress if Do → ∞ . 2 Ei Di 1 − σ rh = − pi = − 2Di Do Ei σ rh = − pi = − 2Di (30 ×10 )(0.005) = −37,500 psi 6 σ rh = − 2(2 ) Maximum tangential stress if Do → ∞ . σ th = Ei Di 1 + 2Di Do Page 124 of 133 2 SECTION 6 – COMBINED STRESSES σ th = Ei 2Di (30 ×10 )(0.005) = 37,500 psi 6 σ th = 2(2 ) (b) Maximum radial stress if Do = 10 in . 2 Ei Di 1 − σ rh = − pi = − 2Di Do 2 30 ×10 6 (0.005) 2 σ rh = − 1 − = −36,000 psi 2(2 ) 10 Maximum tangential stress if Do = 10 in . ( ) 2 Ei Di 1 + σ th = 2Di Do 6 30 ×10 (0.005) 2 σ th = 1 + 10 = 39,000 psi 2(2 ) (c) Maximum radial stress if Do = 2.5 in . ( ) 2 Ei Di 1 − σ rh = − pi = − 2Di Do 2 30 ×106 (0.005) 2 σ rh = − = −13,500 psi 1 − 2(2 ) 2.5 Maximum tangential stress if Do = 2.5 in . ( ) 2 Ei Di 1 + σ th = 2Di Do 6 30 ×10 (0.005) 2 σ th = 1 + 2.5 = 61,500 psi 2(2 ) ( 448. ) A steel cylinder is to have an inside diameter of 3 in. and pi = 30,000 psi. (a) Calculate the tangential stresses at the inner and outer surfaces if the outside diameter is 6 in. (b) It was decided to make the cylinder in two parts, the inner cylinder with D1 = 3 in. and Di = 4.5 in., the outer cylinder with Di = 4.5 in. and Do = 6 in. (see figure). The two cylinders were shrunk together with i = 0.003 in. Calculate the pressure at the interface and the tangential stresses at the inner and outer surfaces of each cylinder. (Suggestion: first derive an equation for the interface pressure). Page 125 of 133 SECTION 6 – COMBINED STRESSES Solution: (a) Tangential stresses at the inner and outer surface. Di = 3 in, ri = 1.5 in, pi = 30,000 psi Do = 6 in, ro = 3 in, po = 0 p r 2 + r 2 − 2p r 2 σ ti = i o 2 i 2 o o ro − ri ( ) σ ti = (30,000)[(3)2 + (1.5)2 ]− 0 = 50,000 ksi (3)2 − (1.5)2 σ to = 2pi ri 2 − po ro2 + ri2 ro2 − ri2 ( ) 2(30,000 )(1.5)2 − 0 = 20,000 ksi (3)2 − (1.5)2 (b) Pressure at the interface, tangential stresses at the inner and outer surface of each cylinder. σ + µ h pi σ ts + µ s pi i = 2( δ h + δ s ) = Di th − E E h s σ to = Eh = E s , µh = µs ( ) σ th = pi ro2 + ri2 ro2 − ri 2 σ ts = 2 p1r12 − pi ri2 + r12 ri 2 − r12 ( ) µp σ µp D σ i = Di th + i − ts − i = i (σ th − σ ts ) E E E E E D p r2 +r2 p r2 +r2 2p r 2 i = i i 2o 2i + i 2i 21 − 2 1 1 2 E ro − ri ri − r1 ri − r1 p1 = 30,000 psi, ro = 3 in, ri = 2.25 in, r1 = 1.5 in ( Page 126 of 133 ) ( ) SECTION 6 – COMBINED STRESSES Pressure at the interface, pi. Ei 2 p1r12 + Di ri2 − r12 pi = 2 2 ro + ri r2 +r2 + i2 12 2 2 ro − ri ri − r1 (30 ×10 )(0.003) + 2(30,000)(1.5) 2 6 4.5 (2.25)2 − (1.5)2 = 20,000 + 48,000 3.571429 + 2.6 (3)2 + (2.25)2 + (2.25)2 + (1.5)2 2 2 2 2 (3) − (2.25) (2.25) − (1.5) pi = 11,018.5 psi Tangential stresses: Inner cylinder: Inner surface: p r 2 + r 2 − 2p r 2 σ ti = 1 i 2 1 2 i i ri − r1 pi = ( ) [ ] 30,000 (2.25)2 + (1.5)2 − 2(11,018.5)(2.25)2 = 38,333.4 psi (2.25)2 − (1.5)2 Outer surface: 2p r 2 − p r 2 + r 2 σ to = 1 1 2 i 2i 1 ri − r1 σ ti = ( σ to = ) [ ] 2(30,000)(2.25)2 − 11,018.5 (2.25)2 + (1.5)2 = 79,351.9 psi (2.25)2 − (1.5)2 Outer cylinder: Inner surface: p r 2 + r 2 − 2p r 2 σ ti = i o 2 i 2 o o ro − ri ( ) [ ] 11,018.5 (3)2 + (2.25)2 − 2(0 )(3)2 = 39,351.8 psi (3)2 − (2.25)2 Outer surface: 2p r 2 − p r 2 + r 2 σ to = i i 2 o 2o i ro − ri σ ti = ( σ to = 449. ) [ ] 2(11,018.5)(2.25)2 − 0 (3)2 + (2.25)2 = 28,333.3 psi (3)2 − (2.25)2 A phosphor-bronze (B139C) bushing has an ID = ¾ in., an OD = 1 ¼ in., and a length of 2 in. It is to be pressed into a cast-steel cylinder that has an outside diameter of 2 ½ in. An ASA class FN 2 fit is to be used with selective assembly to give approximately the interference i = 0.0016 in. Calculate (a) pi, (b) the maximum tangential stress in the steel Page 127 of 133 SECTION 6 – COMBINED STRESSES cylinder, (c) the force required to press bushing into the cylinder, (d) the decrease of the inside diameter of the bushing. Solution: Phosphor Bronze B139C, Es = 16 x 106 psi (Table AT3), µs = 0.36 (other reference). Cast steel, Eh = 30 x 106 psi , µh = 0.27 (Table AT 6) σ + µ h pi σ ts + µ s pi i = 2( δ h + δ s ) = Di th − Eh Es ( ) σ th = pi ro2 + ri2 ro2 − ri 2 σ ts = − pi ri2 + r12 ri2 − r12 ( ) σ µ p σ µp i = Di th + h i − ts − s i Eh Es Es Eh p r2 +r2 µ p p r2 + r2 µ p i = Di i o2 i 2 + h i + i i 2 12 − s i Eh Es E s ri − r1 E h ro − ri (a) pi i Di pi = 2 2 ro + ri ri2 + r12 µ µ + + h− s 2 2 2 2 E h ro − ri E s ri − r1 Eh Es ( ( ) ) ( ( ( ) 2.25 = 1.125 in 2 1.25 ri = = 0.625 in 2 0.75 r1 = = 0.375 in 2 ro = Page 128 of 133 ( ) ) ) SECTION 6 – COMBINED STRESSES L = 2 in Di = 1.25 in pi = (1.125)2 + (0.625)2 30 ×10 6 [(1.125)2 − (0.625)2 ] 0.0016 1.25 ( 0.625)2 + (0.375)2 0.27 0.36 + + − 2 2 6 6 30 ×10 16 ×10 6 16 ×10 (0.625) − (0.375) [ ] −3 1.28 ×10 6.309524 ×10 + 1.328125 ×10 −7 + 0.9 ×10 −8 − 2.25 ×10−8 pi = 7,017 psi (b) Maximum tangential stress in the steel cylinder. p r2 + r2 σ th = i 2o 2i ro − ri pi = −8 ( ) (7,017 )[(1.125)2 + (0.625)2 ] = 13,282 psi (1.125)2 − (0.625)2 σ th = (c) F f1piπDi L tons , use f1 = 0.175 on the average 2000 (0.175)(7,017 )(π )(1.25)(4) = 4.82 tons F= 2000 (d) Decrease of the inside diameter of the bushing. The bushing is phosphor bronze. Subscript is “s” as in shaft. σ + µ s pi δ s = − ts Es F= ( ) σ ts = − pi ri2 + r12 ri2 − r12 σ ts = − 7,017 (0.625)2 + (0.375)2 = −14,911 psi (0.625)2 − (0.375)2 [ ] − 14,911 + 0.36(7,017 ) = 0.000774 in 16 ×10 6 δ s = − DESIGN PROJECTS DESIGN PROJECTS 450. A jib crane similar to the one shown is to be designed for a capacity of F = ___ (say, 1 to 3 tons). The load F can be swung through 360o; L ≈ 10 ft., b ≈ 8.5 ft., c ≈ 2 ft. The moment on the jib is balanced by a couple QQ on the post, the forces Q acting at supporting bearings. The crane will be fastened to the floor by 6 equally spaced bolts on a D1 = 30-in. bolt circle; outside diameter of base D2 = 36 in. (a) Choose a pipe size (handbooks) for the column such that the maximum equivalent stress does not exceed 12 ksi. (b) Choose an I-beam for the jib such that the maximum stress does not exceed 12 ksi. (c) Compute the maximum external load on a base bolt and decide upon the size. Page 129 of 133 SECTION 6 – COMBINED STRESSES (d) Complete other details as required by the instructor, such as: computing Q and choosing bearings (ball or roller?), the internal construction and assembly in this vicinity, detail sketches giving full information. 451. Design an air-operated punch press similar to the one shown. Let the force at the punch be 12 tons, (or other capacity as specified by the instructor), the depth of throat to the inside edge of the frame be 25 in., the diameter and stroke of the piston about 8 in. by 8 in., the mechanical advantage of the lever about 7, and the diameter of the punch 3/16 in. Determine first the horizontal section of the frame, and locate and design the cylinder. Then determine the relative arrangement of the various links and make a force analysis, from which the design of certain parts follows. Determine the actual distance of movement of the punch (not less than about 1 in.). The illustration will assist the student in settling upon the proportions of parts for which strength calculations cannot be made. Page 130 of 133 SECTION 6 – COMBINED STRESSES 452. Design a screw press similar to that shown for a load of ___ (say, 3) tons on the screw. The depth of the throat g is to be ___ (10) in. and the height of the throat h is to be ___ (15) in. (The instructor will assign the data.) The order of procedure may be as follows: (a) Find the diameter of the screw. If Le/k > 40, check as a column. If the top of the screw is squared off for a handwheel or handle, check this section for twisting. The equation for pivot friction, if desired, is in §18.10, Text. (b) Decide upon the diameter of the handwheel or the length of handle (if one is needed to obtain the maximum pressure), assuming that the maximum force to be exerted by a man is about 150 lb. Dimensions of handwheels may be found in handbooks. The handle may be designed by the flexure formula. (c) Design the frame. The shape of the section of the frame will depend on the material used. A T-section is suitable for cast iron (say N = 6 on the ultimate strength), a hollow box or modified I-section is suitable for cast steel. The 45o section CD of the frame should be safe as a curved beam. See Table AT 18. In this connection, it will be well to make the radius r as large as practicable, since the larger r the less the stresses from a given load. Compute the dimensions of the vertical section. It is a good plan to keep t and a the same in all sections. (d) Design the bushing if one is used. The height b depends upon the number of threads in contact, which in turn depends upon the bearing pressure used in design. (Say half-hard yellow brass?) Compute the outside Page 131 of 133 SECTION 6 – COMBINED STRESSES bushing diameter, the diameter and thickness of the collar, and decide upon dimensions to be used. (e) Fix the location and number of bolts to be used to fasten the frame to the base plate, and determine their size. Use a common bolt material. (f) Decide upon all other details of design. Make a separate sketch of each part of the machine showing thereon all dimensions necessary for manufacture. It is suggested that, first, all materials be tentatively decided upon, after which design stresses may be chosen. See that design stresses for the various parts bear a logical relation to one another. It is not necessary to follow this procedure in detail. It is likely that one will have to leave certain details unfinished from time to time, because these details depend on parts of the design not yet completed. Make sure that all parts can be assembled after they are made. Notice that the plate on the lower end of the screw must be connected in such a manner that the screw may turn while the plate does not. 453. Design a jib crane, as suggested by the illustration, to lift a load of W of ___ tons. The maximum radius of swing is to be about ___ ft. (The instructor will assign data). Suggested procedure: (a) From catalogues, select a hoist to suit the purpose, giving reasons for your choice, and noting significant dimensions. Of course, in the end, the hoist trolley has to match the size of I-beam used. (b) Let the angle that the diagonal tension rod makes with the horizontal be about 20o to 25o, and decide upon the dimensions H and L. Note that the point G does not necessarily have to be at the extreme position of the load. As a matter of fact, some advantage may result from having G inside the outermost position of the load. Make the force analyses (including weight of hoist as part of load) for (1) the condition of maximum column action, (2) the condition of maximum bending moment on the beam, and (3) the condition for maximum force on the hinge B (to be used for the design of this hinge). (c) Find the size of I-beam such that the maximum stress for any position of the load falls within the limits of 12 and 15 ksi, usually by assuming a standard beam and checking the stress. According to the arrangement of parts, it may be necessary to design the connection at G between the rod and the beam first. With the details of this connection known and Page 132 of 133 SECTION 6 – COMBINED STRESSES with the details of beam assumed, the location of point G, the point of application of the force T, can be determined. The bending moment of a section a minute distance to the right of G is W(x – dx). A minute distance to the left of G, the bending moment is W(x + dx) – Txe – Tydx; that is, the moment changes suddenly at G by the amount Txe. (d) Determine the size of diagonal support, including details of connections. (e) Design the connections at each end of the diagonal and the hinge at C. Settle upon the details including the method of attaching the hinge to the vertical surface, which may be wideflange beam. (f) Design the hinge at B and the connection to the I-beam; also the details of the method of attaching the hinge to the vertical surface. Where material is not specified, make your choice clear. There should be no doubt as to your design stresses or design factor. Show a neat large sketch, fully dimensioned, of each part separately. It is unlikely that too much detail will be shown. - E n d - Page 133 of 133 SECTION 7 – SHAFT DESIGN 471. A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed (profile keyway) midway between the bearings. The pressure angle of the gear teeth φ = 20 o ; N = 1.5 based on the octahedral shear stress theory with varying stresses. (a) Neglecting the radial component R of the tooth load W , determine the shaft diameter. (b) Considering both the tangential and the radial components, compute the shaft diameters. (c) Is the difference in the results of the parts (a) and (b) enough to change your choice of the shaft size? Problem 471. Solution: For SAE 1035, as rolled s y = 55 ksi su = 85 ksi sn = 0.5su = 0.5(85) = 42.5 ksi A = W cos φ 63,000hp 63,000(30 ) T= = = 6300 in − lb n 300 AD T= 2 A(12 ) 6300 = 2 A = 1050 lb A = W cos φ 1050 = W cos 20 W = 1118 lb Shear stress 16T 16(6300 ) ss = = π d3 π d3 100,800 ss = sms = π d3 sas = 0 Page 1 of 76 SECTION 7 – SHAFT DESIGN bending stress From Table AT 2 FL M= 4 (a) Negligible R : AL (1050 )(16 ) M= = = 4200 in − lb 4 4 32 M 32(4200 ) 134,400 s= = = π d3 π d3 π d3 sm = 0 134,400 sa = s = π d3 K f sa s se = n s m + sy SF For profile keyway K f = 2 .0 K fs = 1.6 SF = 0.85 K s (2.0)(134,400) = 100,661 se = f a = SF (0.85) π d 3 d3 K fs sas s ses = ns sms + s ys SF ( ) sns sn 42.5 1 = = = s ys s y 55 1.294 sns 1 100,800 24,796 = sms = 3 s ys d3 1.294 π d Octahedral-shear theory ses = 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 100,661 24,796 = + 3 1.5 42,500d 0.577 42,500d 3 d = 1.569 in ( Page 2 of 76 1 ) 2 2 SECTION 7 – SHAFT DESIGN use d = 1 11 in 16 (b) Considering both radial and tangential component. WL (1118)(16 ) M= = = 4472 in − lb 4 4 32 M 32(4472 ) 143,104 s= = = π d3 π d3 π d3 sm = 0 143,104 sa = s = π d3 K s (2.0)(143,104) = 107,180 se = f a = SF (0.85)(π d 3 ) d3 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 107,180 24,796 = + 3 1.5 42,500d 0.577 42,500d 3 d = 1.597 in 11 use d = 1 in 16 ( 1 ) 2 2 (c) The difference in the results of the parts (a) and (b) is not enough to change the choice of the shaft size. 472. A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100 and back to 200 hp in each revolution at a speed of 600 rpm. The power is received by a 20-in. spur gear A and delivered by a 10-in. spur gear C. The tangential forces have each been converted into a force ( A and C shown) and a couple (not shown). The radial component R of the tooth load is to be ignored in the initial design. Let 2 and, considering varying stresses with the maximum shear theory, compute the shaft diameter. Problems 472 – 474 Page 3 of 76 SECTION 7 – SHAFT DESIGN Solution: For AISI 1141, cold-finished s y = 90 ksi sn = 50 ksi sn 1 = s y 1.8 SF = 0.85 63,000hp T= n 63,000(200 ) Tmax = = 21,000 in − lb 600 63,000(100 ) Tmin = = 10,500 in − lb 600 1 1 Tm = (Tmax + Tmin ) = (21,000 + 10,500 ) = 15,750 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (21,000 − 10,500 ) = 5,250 in − lb 2 2 16T ss = πd3 16(15,750 ) 252,000 sms = = π d3 π d3 16(5250 ) 24,000 = sas = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2 .0 K fs = 1.6 sns sn 1 = = s ys s y 1.8 1 252,000 (1.6)(84,000) 94,894 + ses = = 3 0.85π d 3 d3 1.8 π d Bending stress, negligible radial load T = 21,000 in − lb at 200 hp For A: 20 A = T 2 A(10 ) = 21,000 Page 4 of 76 SECTION 7 – SHAFT DESIGN A = 2100 lb at 200 hp For C: 10 C = T 2 C (5) = 21,000 C = 4200 lb at 200 hp [∑ M ] = 0 A(10 ) + D(25) = C (15) at 200 hp (2100)(10) + D(25) = (4200)(15) D = 1680 lb [∑ F B ] =0 A+C = B + D at 200 hp 2100 + 4200 = B + 1680 B = 4620 lb At 200 hp: A = 2100 lb , B = 4620 lb , C = 4200 lb , D = 1680 lb Shear Diagram V Maximum moment at B M = (2100)(10 ) = 21,000 in − lb 32 M 32(21,000 ) 672,000 s= = = π d3 π d3 π d3 sm = 0 672,000 sa = s = π d3 K f sa (2.0)(672,000) = 503,304 s se = n s m + = 0+ sy SF 0.85π d 3 d3 Page 5 of 76 SECTION 7 – SHAFT DESIGN 94,894 d3 Maximum Shear Theory ses = 1 2 2 2 1 se ses = + N sn 0.5sn 2 1 503,304 94,894 = + 3 2 50,000d 0.5 50,000d 3 d = 2.78 in 3 use d = 2 in 4 ( 475. 1 ) 2 2 A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W, which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and back to 10 hp during each revolution of S. The design is to account for the varying stresses, with calculations based on the octahedral shear stress theory. Let N = 1.8 and compute the shaft diameter, using only the tangential driving loads for the first design. Problem 475 – 477 Solution. For AISI 1137, cold drawn s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi sn sns 51.5 1 = = = s y s ys 93 1.806 63,000hp T= n 5 in. E n= (2000 rpm) = 667 rpm 15 in. A Page 6 of 76 SECTION 7 – SHAFT DESIGN 63,000(100 ) = 9450 in − lb 667 63,000(10 ) Tmin = = 945 in − lb 667 1 1 Tm = (Tmax + Tmin ) = (9450 + 945) = 5197.5 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (9450 − 945) = 4252.5 in − lb 2 2 16T ss = πd3 16(5197.5) 83,160 sms = = π d3 π d3 16(4252.5) 68,040 = sas = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2 .0 Tmax = K fs = 1.6 SF = 0.85 1 83,160 (1.6)(68,040) 55,425 + ses = = 3 0.85π d 3 d3 1.806 π d Bending stress, using only tangential loads For 100 hp: T = 9450 in − lb 15 A = T 2 A(7.5) = 9450 A = 1260 lb For C: 10 C = T 2 Page 7 of 76 SECTION 7 – SHAFT DESIGN C (5) = 9450 C = 1890 lb [∑ M B ] = 0 6 A + 20 D = 14C 6(1260) + 20 D = 14(1890) D = 945 lb [∑ F ] =0 A+C = B + D 1260 + 1890 = B + 945 B = 2205 lb Shear diagram V Maximum moment at B M = (1260)(6) = 7560 in − lb 32 M 32(7560 ) 241,920 s= = = π d3 π d3 π d3 sm = 0 241,920 sa = s = π d3 K f sa (2.0 )(241,920 ) 181,189 s se = n s m + = = sy SF 0.85π d 3 d3 55,425 d3 Octahedral Shear Theory ses = 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 181,189 55,425 = + 3 2 51,500d 0.577 51,500d 3 ( d = 1.997 in use d = 2 in Page 8 of 76 1 ) 2 2 SECTION 7 – SHAFT DESIGN 478. A shaft made of AISI 1137, cold rolled, for a forage harvester is shown. Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the roller chain to the cutter exerts a force vertically upwards, and the V-belt to the blower at C exerts a force vertically upwards. At maximum operating conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered to the cutter and 10 hp to the blower. The two sections of the shaft are joined by a flexible coupling at D and the various wheels are keyed (sled-runner keyway) to the shafts. Allowing for the varying stresses on the basis of the von Mises-Hencky theory of failure, decide upon the diameters of the shafts. Choose a design factor that would include an allowance for rough loading. Problem 478. Solution: For AISI 1137, cold rolled s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi sn sns 51.5 1 = = = s y s ys 93 1.806 Pulley, 63,000hp 63,000(35) TA = = = 5188 in − lb n 425 For flat-belt 2T 4(5188) FA = F1 + F2 = 2(F2 − F1 ) = 2 A = = 692 lb 30 DA Sprocket, 63,000hp 63,000(25) TB = = = 3706 in − lb n 425 For chain, 2T 2(3706 ) FB = B = = 741 lb DB 10 Sheave, 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425 Page 9 of 76 SECTION 7 – SHAFT DESIGN For V-belt, 2T FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C DC Consider shaft ABD. 35 hp Shaft ABD [∑ M D' =0 ] (6 + 8 + 4)FA = (8 + 4)A'+4FB 18(692) = 12 A'+4(741) A' = 791 lb [∑ F V =0 ] FA + D′ = FB + A′ 692 + D′ = 741 + 791 D′ = 840 lb Shear Diagram Maximum M at A’. M = (6)(692) = 4152 in − lb. 32 M 32(4152 ) 132,864 s= = = π d3 π d3 π d3 sm = 0 sa = s = 132,864 π d3 Page 10 of 76 3(1482 ) = = 445 lb 10 SECTION 7 – SHAFT DESIGN K f sa sn sm + sy SF For sled-runner keyway (Table AT 13) K f = 1 .6 se = K fs = 1.6 SF = 0.85 K f sa (1.60)(132,864) = 79,610 s se = n s m + = 0+ sy SF 0.85π d 3 d3 at A’ T = TA = 5188 in − lb 16T 16(5188) 83,008 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 1 83,000 14,630 = ses = 3 d3 1.806 π d Choose a design factor of 2.0 N = 2 .0 von Mises-Hencky theory of failure (Octahedral shear theory) 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 79,610 14,630 = + 3 2 51,500d 0.577 51,500d 3 d = 1.48 in 1 use d = 1 in 2 ( 1 ) 2 2 Consider shaft D-C 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425 For V-belt, 2T 3(1482 ) FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C = = 445 lb 10 DC Page 11 of 76 SECTION 7 – SHAFT DESIGN [∑ M C' =0 ] 8 D′′ = 3FC 8 D′′ = 3(445) D′′ = 167 lb [∑ F V =0 ] C ′ = D′′ + FC C ′ = 167 + 445 C ′ = 612 lb Shear Diagram M = (167 )(8) = 1336 in − lb 32 M 32(1336 ) 42,752 s= = = π d3 π d3 π d3 sm = 0 , sa = s K f sa (1.60)(42,752) = 25,616 s se = n s m + = 0+ sy SF 0.85π d 3 d3 at C’, TC = 1482 in − lb 16T 16(1482 ) 23,712 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 4180 1 23,712 +0 = 3 ses = 3 d 1.806 π d 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 25,616 4180 = + 3 2 51,500d 0.577 51,500d 3 ( Page 12 of 76 1 ) 2 2 SECTION 7 – SHAFT DESIGN d = 1.011 in use d = 1 in 479. A shaft for a punch press is supported by bearings D and E (with L = 24 in.) and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44in. pulley at B, the belt being at 45o with the vertical. An 8-in. gear at A delivers the power horizontally to the right for punching operation. A 1500-lb flywheel at C has a radius of gyration of 18 in. During punching, the shaft slows and energy for punching comes from the loss of kinetic energy of the flywheel in addition to the 25 hp constantly received via the belt. A reasonable assumption for design purposes would be that the power to A doubles during punching, 25 hp from the belt, 25 hp from the flywheel. The phase relations are such that a particular point in the section where the maximum moment occurs is subjected to alternating tension and compression. Sled-runner keyways are used for A, B, and C; material is colddrawn AISI 1137, use a design factor of N = 2.5 with the octahedral shear theory and account for the varying stresses. Determine the shaft diameters. Problems 479-480 Solution: Flat-Belt Drive (B) 63,000hp 63,000(25) TB = = = 6300 in − lb n 250 2T 4(6300) FB = F1 + F2 = 2(F1 − F2 ) = 2 B = = 573 lb 44 DB Gear A, Doubled hp 63,000hp 63,000(25 + 25) TA = = = 12,600 in − lb n 250 2T 2(12,600 ) FA = A = = 3150 lb DA 8 Loading: Page 13 of 76 SECTION 7 – SHAFT DESIGN Vertical: BV = FB cos 45 = 573 cos 45 = 405 lb [∑ M D =0 ] 6(1500) + 8BV = 24 EV 6(1500) + 8(405) = 24 EV EV = 510 lb [∑ F V =0 ] 1500 + EV = DV + BV 1500 + 510 = DV + 405 DV = 1605 lb Shear Diagram Page 14 of 76 SECTION 7 – SHAFT DESIGN M DV = (6 )(1500 ) = 9000 in − lb M BV = (16 )(510 ) = 8160 in − lb M AV = (5)(510 ) = 2550 in − lb Horizontal: Bh = FB sin 45 = 573 sin 45 = 405 lb [∑ M D =0 ] 8 Bh + 24 Eh = 19 FA 8(405) + 24 Eh = 19(3150) Eh = 2359 lb [∑ F h =0 ] Dh + Bh + Eh = FA Dh + 405 + 2359 = 3150 Dh = 386 lb Shear Diagram Page 15 of 76 SECTION 7 – SHAFT DESIGN M Dh = 0 in − lb M Bh = (8)(386 ) = 3088 in − lb M Ah = (5)(2359 ) = 11,795 in − lb MA = MB = (M ) + (M ) (M ) + (M ) 2 Ah 2 Bh 2 = (11,795)2 + (2550)2 2 = (3088)2 + (8160)2 AV BV = 12,068 in − lb = 8725 in − lb M D = 9000 in − lb Therefore M max = 12,068 in − lb 32 M 32(12,068) 386,176 s= = = π d3 π d3 π d3 Maximum moment subjected to alternating tension and compression sm = 0 386,176 sa = s = π d3 K f sa sn sm + sy SF For AISI 1137, cold-drawn, s y = 93 ksi se = su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi For sled-runner keyway (Table AT 13) K f = 1 .6 K fs = 1.6 SF = 0.85 (1.60)(386,176 ) = 231,386 se = 0 + 0.85π d 3 d3 Page 16 of 76 SECTION 7 – SHAFT DESIGN At A, 50 hp max. and 25 hp min. 50 hp 63,000hp 63,000(25 + 25) TA = = = 12,600 in − lb n 250 2T 2(12,600 ) FA = A = = 3150 lb DA 8 16T 16(12,600 ) 201,600 ss max = = = π d3 π d3 π d3 25 hp 63,000hp 63,000(25) TA = = = 6,300 in − lb n 250 2T 2(6,300 ) FA = A = = 1575 lb DA 8 16T 16(6,300 ) 100,800 ss min = = = π d3 π d3 π d3 sms = 1 ,800 151,200 = (ss max + ss min ) = 1 201,600 + 100 3 2 2 πd π d3 sas = 1 ,800 50,400 = (ss max − ss min ) = 1 201,600 − 100 3 2 2 πd π d3 ses = K fs sas sns sms + s ys SF 1 151,200 (1.6)(50,400) 56,848 + ses = = 3 0.85π d 3 d3 1.806 π d 1 2 2 2 1 se ses = + N sn 0.577 sn 2 1 231,386 56,848 = + 3 2 51,500d 0.577 51,500d 3 d = 2.14 in 3 say d = 2 in 16 ( THRUST LOADS Page 17 of 76 1 ) 2 2 SECTION 7 – SHAFT DESIGN 481. A cold-drawn monel propeller shaft for a launch is to transmit 400 hp at 1500 rpm without being subjected to a significant bending moment; and Le k < 40 . The efficiency of the propeller is 70 % at 30 knots (1.152 mph/knot). Consider that the number of repetitions of the maximum power at the given speed is 2x 105. Let N = 2 based on the maximum shear theory with varying stress. Compute the shaft diameter. Solution: For cold-drawn monel shaft, Table AT 10 s y = 75 ksi sn = 42 ksi at 108 at 2 x 105 0.085 108 sn ≈ 42 = 71.23 ksi 5 2 × 10 sn sns 71.23 1 = = = s y s ys 75 1.053 63,000hp 63,000(400 ) = = 16,800 in − lb n 1500 16T 16(16,800 ) 268,800 ss = = = π D3 π D3 π D3 s ms = ss T= s as = 0 K fs sas sns sms + s ys SF SF = 0.85 assume K f = K fs = 1.0 ses = 81,255 1 268,800 + 0 = ses = 3 D3 1.053 π D Fvm = ηhp 33,000 vm = (30 knots )(1.152 mph knot )(5280 ft mi )(1 hr 60 min ) = 3041 fpm F (3041) = (0.70)(400) 33,000 F = 3040 lb 4F 4(3040 ) 12,160 s= = = 2 πD π D2 π D2 sm= s sa = 0 Page 18 of 76 SECTION 7 – SHAFT DESIGN se = K f sa sn sm + sy SF 3676 1 12,160 +0 = 2 se = 2 D 1.053 π D Maximum Shear Theory 1 2 2 2 1 se ses = + N sn 0.5sn 2 1 3676 81,255 = + 2 2 71,230 D 0.5 71,230 D 3 ( 1 ) 2 2 1 2 2 1 1 2.2815 2 = + 2 19.377 D 2 D 3 By trial and error 11 D = 1.66 in = 1 in 16 482. A shaft receives 300 hp while rotating at 600 rpm, through a pair of bevel gears, and it delivers this power via a flexible coupling at the other end. The shaft is designed with the average forces ( at the midpoint of the bevel-gear face); the tangential driving force is F , G = 580 lb , Q = 926 lb ; which are the rectangular components of the total reaction between the teeth; Dm = 24 in , L = 36 in , a = 10 in . Let the material be AISI C1045, cold drawn; N = 2 . Considering varying stresses and using the octahedral shear theory, determine the shaft diameter. Problems 482, 485, 486. Solution: For AISI C1045, cold drawn s y = 85 ksi su = 100 ksi Page 19 of 76 SECTION 7 – SHAFT DESIGN sn = 0.5su = 0.5(100) = 50 ksi SF = 0.85 sn sns 50 1 = = = s y s ys 85 1.7 63,000hp 63,000(300 ) T= = = 31,500 in − lb 600 600 16T 16(31,500 ) 504,000 ss = = = π D3 π D3 π D3 s ms = ss s as = 0 K fs sas s ses = ns sms + s ys SF 94,370 1 504,000 + 0 = ses = 3 D3 1.7 π D D F m = T 2 24 F = 31,500 2 F = 2625 lb Vertical: D 24 Q m = 926 = 11,112 in − lb 2 2 G = 580 lb [∑ M B ] =0 − Av (36 ) + QDm − G (10 ) = 0 2 QDm = G (10 ) + Av (36 ) 2 11,112 = 580(10) + Av (36) Av = 148 lb Page 20 of 76 SECTION 7 – SHAFT DESIGN [∑ F =0] v Av + Bv = 580 lb 148 + Bv = 580 lb Bv = 432 lb Shear Diagram Moment Diagram M Cv = 11,112 in − lb M Bv = 5328 in − lb Horizontal: [∑ M B =0 ] Ah (36) = (2625)(10) Ah = 729 lb [∑ F h =0 ] Bh = Ah + F Bh = 725 + 2625 Page 21 of 76 SECTION 7 – SHAFT DESIGN Bh = 3354 lb Shear Diagram M Ch = 0 M Bh = (36 )(729 ) = 26,244 in − lb Maximum M M = MB = (M ) + (M ) 2 Bh (26,244)2 + (5328)2 = 26,780 in − lb 32(26,780 ) 4(926 ) 856,960 3,704 = + = + 2 BV = 32 M 4Q + 3 π D π D2 π D3 π D2 π D3 4Q 32 M 3704 856,960 smin = − = − π D 2 π D3 π D 2 π D3 1 sm = (smax + smin ) 2 1 856,960 3704 3704 856,960 3704 = sm = + + − π D2 π D2 π D 3 π D 2 2 π D3 1 sa = (smax − smin ) 2 856,960 sa = π D3 K f sa s se = n s m + sy SF smax = assume K f = 1.0 at B 1 3704 1.0 856,960 964 320,916 + = 2 + se = 2 3 D3 1.7 π D 0.85 π D D Octahedral Shear Theory 1 2 2 2 1 se ses = + N sn 0.577 sn Page 22 of 76 π D2 SECTION 7 – SHAFT DESIGN 1 2 694 320,916 1 2 2 2 2 + 2 3 1 94,370 6.42 3.27 1 D + = = D + 3 + 3 3 2 2 50,000 D D 0.577(50,000)D 72 D By trial and error, use 1 D = 2 in 2 2 483. The worm shown is to deliver 65.5 hp steadily at 1750 rpm. It will be integral with the shaft if the shaft size needed permits, and its pitch diameter 3 in. The 12in. pulley receives the power from a horizontal belt in which the tight tension F1 = 2.5F2 . The forces (in kips) on the worm are as shown, with the axial force taken by bearing B. The strength reduction factor for the thread roots may be taken as K f = 1.5 , shear or bending. The shaft is machined from AISI 1045, as rolled. (a) For N = 2.2 (Soderberg criterion) by the octahedral-shear theory, compute the required minimum diameter at the root of the worm thread (a first approximation). (b) What should be the diameter of the shaft 2.5 in. to the left of the centerline of the worm? (c) Select a shaft size D and check it at the pulley A. Problem 483. Solution: For AISI 1045, as rolled s y = 59 ksi su = 96 ksi sn = 0.5su = 48 ksi sn sns 48 1 = = = s y s ys 59 1.229 63,000hp 63,000(65.5) T= = = 2358 in − lb 1750 1750 (F1 − F2 ) 12 = T 2 (2.5F2 − F2 )(6) = 2358 F2 = 262 lb F1 = 2.5F2 = 655 lb Page 23 of 76 SECTION 7 – SHAFT DESIGN FA = F1 + F2 = 655 + 262 = 917 lb Horizontal [∑ M B =0 ] (917 )(6) + (1570)(6.5) = 13Eh Eh = 1208 lb [∑ F h =0 ] 917 + Eh = Bh + 1570 917 + 1208 = Bh + 1570 Bh = 555 lb Shear Diagram M Ah = 0 M Bh = (917 )(6 ) = 5502 in − lb M Ch = (1208)(6.5) = 7852 in − lb Vertical: 3 M ′ = (2540 ) = 3810 in − lb 2 ∑ M E =0 [ ] M ′ + (1170)(6.5) = 13Bv 3810 + (1170)(6.5) = 13Bv Bv = 878 lb Page 24 of 76 SECTION 7 – SHAFT DESIGN [∑ F =0] v Ev + Bv = 1170 Ev + 878 = 1170 Ev = 292 lb Shear Diagram Moment Diagram M Av = 0 M Bv = 0 M Cv = 5707 in − lb M= MA = MB = MC = (M h )2 + (M v )2 (0)2 + (0)2 = 0 (5502)2 + (0)2 = 5502 in − lb (7852)2 + (5707)2 = 9707 in − lb (a) Minimum diameter at the root of the warm thread. K f = K fs = 1.5 M = M C = 9707 in − lb F = 2540 lb smax = 32 M 4F 32(9707 ) 4(2540 ) 310,624 10,160 + = + = + 3 2 π Dr π Dr π Dr3 π Dr2 π Dr3 π Dr2 Page 25 of 76 SECTION 7 – SHAFT DESIGN smin = − 310,624 10,160 + π Dr3 π Dr2 1 (smax + smin ) 2 10,160 sm = π Dr2 1 sa = (smax − smin ) 2 310,624 sa = π Dr3 K f sa s se = n s m + sy SF sm = 1 10,160 1.5 310,624 2632 174,485 se = + 2 3 = D2 + D3 1.229 π Dr 0.85 π Dr r r 16T 16(2358) 12,000 ss = = = π Dr3 π Dr3 Dr3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 9764 1 12,000 ses = +0= 3 3 Dr 1.229 Dr N = 2.2 , Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2632 174,485 2 2 1 + 2 2 2 2 2 3 1 Dr Dr 9764 1 3.635 1 + = + = + 2 0.577(48,000)Dr3 2.2 48,000 Dr3 2.84 Dr3 18.24 Dr By trial and error Dr = 2.023 in 1 say Dr = 2 in 16 (b) D – shaft diameter 2.5 in. to the left of the center line of worm Page 26 of 76 SECTION 7 – SHAFT DESIGN 3 in 16 Figure AF 12 3 r 16 = ≈ 0.1 d 3 2.023 − 2 2 16 D 2.023 = = 1 .2 d 3 2.023 − 2 2 16 K f = K t = 1.65 r= K fs = K ts = 1.34 at 2.5 in to the shaft M h = (917 )(6) + (362)(6.5 − 2.5) = 6950 in − lb M v = (878)(6.5 − 2.5) = 3512 in − lb M= (6950)2 + (3512)2 = 7787 in − lb 10,160 π D2 32 M 32(7787 ) 249,184 sa = = = π D3 π D3 π D3 K f sa s se = n s m + sy SF sm = 1 10,160 1.65 249,184 2632 153,970 + = se = + 2 3 D2 D3 1.229 π D 0.85 π D 9764 ses = D3 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2632 153,970 2 2 1 2 2 2 2 + 2 3 1 9764 1 3.21 1 D + = = D + 3 + 3 2 2.2 48,000 18.24 D D 2.84 D 3 0.577(48,000)D By trial and error D = 1.9432 in 15 say D = 1 in 16 Page 27 of 76 SECTION 7 – SHAFT DESIGN 15 in = 1.9375 in 16 At the pulley A, or 3 in. right of centerline M h = (917 )(3) = 2751 in − lb (c) Selecting D = 1 Mv = 0 M = 2751 in − lb For sled runner keyway K f = 1 .6 K fs = 1.6 sm = 0 32 M 32(2751) sa = = = 3853 psi 3 πD π (1.9375)3 K f sa s se = n s m + sy SF 1 .6 se = 0 + (3853) = 7253 psi 0.85 9764 ses = = 1343 psi (1.9375)3 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 7253 1343 = + N 48,000 0.577(48,000) N = 6.30 > 2.2 , therefore o.k. 484. A propeller shaft as shown is to receive 300 hp at 315 rpm from the right through a flexible coupling. A 16-in. pulley is used to drive an auxiliary, taking 25 hp. The belt pull FB is vertically upward. The remainder of the power is delivered to a propeller that is expected to convert 60% of it into work driving the boat, at which time the boat speed is 1500 fpm. The thrust is to be taken by the right-hand bearing. Let N = 2 ; material cold-worked stainless 410. Use the octahedral shear theory with varying stresses. (a) Determine the shaft size needed assuming no buckling. (b) Compute the equivalent column stress. Is this different enough to call for another shaft size? Compute N by the maximum shear stress theory, from both equations (8.4) and (8.11). Page 28 of 76 SECTION 7 – SHAFT DESIGN Problem 484. Solution: For stainless 410, cold-worked s y = 85 ksi sn = 53 ksi SF = 0.85 Belt drive 63,000hp 63,000(25) TB = = = 5000 in − lb n 315 2T 4(5000) FB = F1 + F2 = 2(F1 − F2 ) = 2 B = = 1250 lb 16 DB Propeller 63,000hp 63,000(300 − 25) TP = = = 55,000 in − lb n 315 Thrust Fvm = ηhp(33,000) F (1500) = (0.60)(300 − 25)(33,000) F = 3630 lb Vertical loading [∑ M E =0 ] (20)(1250) = 60C C = 417 lb [∑ F =0] v A + C = FB A + 417 = 1250 Page 29 of 76 SECTION 7 – SHAFT DESIGN A = 833 lb Shear Diagram M B = (20)(833) = 16,660 in − lb Maximum T at B T = TB + TP = 60,000 in − lb (a) Shaft size assuming no buckling M = 16,660 in − lb F = 3630 lb 4F 4(3630 ) 14,520 = = sm = 2 πD π D2 π D2 32 M 32(16,660 ) 533,120 sa = = = π D3 π D3 π D3 For sled-runner keyway K f = 1 .6 K fs = 1.6 sn sns 53 1 = = = s y s ys 85 1.604 se = K f sa sn sm + sy SF 1 14,520 1.6 533,120 2882 319,430 + = se = + 2 3 D2 D3 1.604 π D 0.85 π D 16 16(60,000 ) 960,000 ss = sms = = = 3 πD π D3 π D3 sas = 0 K fs sas s ses = ns sms + s ys SF 190,510 1 960,000 + 0 = ses = 2 D3 1.604 π D Page 30 of 76 SECTION 7 – SHAFT DESIGN 1 2 1 se ses = + N sn sns N = 2 , Octahedral Shear Theory, sns = 0.577 sn 2 2 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2882 319,430 1 2 2 2 2 + 2 3 1 190,510 1 6.027 6.230 D + = = D + + 3 2 2 53,000 18.39 D D 3 D 3 0.577(53,000 )D By trial and error D = 2.6 in 5 say D = 2 in = 2.625 in 8 2 (b) Equivalent Column Stress 4F s= α π D2 Le = 12 + 60 + 10 = 82 in 1 1 k = D = (2.625) = 0.65625 in 4 4 Le 82 = = 125 > 120 k 0.65625 Use Euler’s equation 2 L sy e 2 85(125) k α= = 2 = 4.486 π 2E π 30 ×103 4F 4(3630 ) s= (4.486) = 3000 psi α= 2 πD π (2.625)2 Since α > 1 , it is different enough to call for another shaft size. ( ) Solving for N by maximum shear theory. 2882 319,430 2882 319,430 se = + = + = 18,078 psi 2 2 3 D D (2.625) (2.625)3 190,510 ses = = 10,533 psi (2.625)3 Equation (8.4) Page 31 of 76 SECTION 7 – SHAFT DESIGN 1 1 2 2 s 2 2 18,078 2 2 τ = ss + = (10,533) + = 13,880 psi 2 2 0.5sn 0.5(53,000) N= = = 1.91 τ 13,880 Equation (8.11) sns = 0.5sn 1 1 2 2 2 18,078 2 10,533 2 2 1 s ss = + = + ( ) N sn sns 53 , 000 0 . 5 53 , 000 N = 1.91 CHECK PROBLEMS 485. A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean diameter is Dm = 10 in and which is keyed (profile) to the left end. Acting on the gear are a radial force G = 1570.8 lb , a driving force Q = 3141.6 lb . The thrust force is taken by the right-hand bearing. Let a = 5 in and L = 15 in ; material, AISI C1040, annealed. Base calculations on the maximum shearing stress theory with variable stress. Compute the indicated design factor N . With the use of a sketch, indicate the exact point of which maximum normal stress occurs. Solution: For AISI C1040, annealed, Figure AF 1 s y = 48 ksi su = 80 ksi sn = 0.5su = 40 ksi sn sns 40 1 = = = s y s ys 48 1.2 FDm (6283.2 )(10 ) T= = = 31,416 in − lb 2 2 16T 16(31,416 ) ss = = = 5926 psi π D3 π (3)3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 1 ses = (5926 ) + 0 = 4940 psi 1 .2 Vertical Page 32 of 76 SECTION 7 – SHAFT DESIGN QDm (3141.6 )(10 ) = = 15,708 in − lb 2 2 ∑ M E =0 [ ] QDm = 5G + 15 AV 2 15,708 = 5(1570.8) + 15 AV AV = 523.6 lb [∑ F =0] v AV + BV = G 523.6 + BV = 1570.8 BV = 1047.2 lb Shear Diagram Moment Diagram M CV = 15,708 in − lb M BV = 7854 in − lb Page 33 of 76 SECTION 7 – SHAFT DESIGN Horizontal [∑ M B =0 ] 15 Ah = 5(6283.2) Ah = 2094.4 lb [∑ F h =0 ] Bh = Ah + F Bh = 2094.4 + 6283.2 Bh = 8377.6 lb Shear Diagram M Ch = 0 M Bh = (15)(2094.4 ) = 31,416 in − lb Maximum Moment 2 M = M B2h + M Bv = (31,416)2 + (7854)2 = 32,383 in − lb Since thrust force is taken by the right-hand bearing sms = 0 32 M 32(32,383) sas = = = 12,217 psi π D3 π (3)3 K f sa s se = n s m + sy SF Assume K f = 1.0 at the bearing B 1 .0 se = 0 + (12,217 ) = 14,373 psi 0.85 Page 34 of 76 SECTION 7 – SHAFT DESIGN Maximum shear theory sns = 0.5sn 1 2 2 2 1 se ses = + N sn 0.5sn 1 2 2 2 1 14,373 4940 = + N 40,000 0.5(40,000) N = 2 .3 Location of maximum normal stress 487. A 2 7/16-in. countershaft in a machine shop transmits 52 hp at 315 rpm. It is made of AISI 1117, as rolled, and supported upon bearing A and B, 59-in. apart. Pulley C receives the power via a horizontal belt, and pulley D delivers it vertically downward, as shown. Calculate N based on the octahedral-shearstress theory considering varying stresses. Problem 487, 488 Solution: For AISI 1117, as rolled s y = 44.3 ksi Page 35 of 76 SECTION 7 – SHAFT DESIGN su = 70.6 ksi sn = 0.5su = 35.3 ksi sn sns 35.3 1 = = = s y s ys 44.3 1.255 SF = 0.85 63,000(52 ) T= = 10,400 in − lb 315 Pulley C 2T 4(10,400) = FC = F1 + F2 = 2(F2 − F1 ) = 2 = 2311 lb D 18 C Pulley D 2T 4(10,400) = FD = F1 + F2 = 2(F2 − F1 ) = 2 = 1664 lb D 25 D Horizontal [∑ M A =0 ] 15(2311) = 59 Bh Bh = 588 lb [∑ F h =0 ] Ah + Bh = 2311 Ah + 588 = 2311 Ah = 1723 lb Shear Diagram M Ch = (1723)(15) = 25,845 in − lb Page 36 of 76 SECTION 7 – SHAFT DESIGN M Dh = (1723)(15) − (588)(26 ) = 10,557 in − lb Vertical [∑ M B =0 ] 18(1664) = 59 Av Av = 508 lb [∑ F =0] v Av + Bv = 1664 508 + Bv = 1664 Bv = 1156 lb Shear Diagram M Cv = (508)(15) = 7620 in − lb M Dv = (1156 )(18) = 20,808 in − lb M C = M C2h + M C2v = (25,845)2 + (7620 )2 M D = M D2 h + M D2 v = (10,557 )2 + (20,808)2 Maximum M at C M = M C = 26,945 in − lb sm = 0 32 M sa = π D3 7 D = 2 in = 2.4375 in 16 Page 37 of 76 = 26,945 in − lb = 23,333 in − lb SECTION 7 – SHAFT DESIGN 32(26,945) = 18,952 psi π (2.4375)3 assume K f = K fs = 1.0 sa = se = K f sa sn sm + sy SF (1.0)(18,952) = 22,300 psi 1 se = (0 ) + 0.85 1.255 16T 16(10,400 ) ss = = = 3658 psi π D 3 π (2.4375)3 sms = s s = 3658 psi sas = 0 ses = K fs sas sns sms + s ys SF 1 ses = (3658) + 0 = 2915 psi 1.255 Octahedral shear theory sns = 0.577 sn 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 22,300 2915 = + N 35,300 0.577(35,300) N = 1.544 489. A shaft for a general-purpose gear-reduction unit supports two gears as shown. The 5.75-in. gear B receives 7 hp at 250 rpm. The 2.25-in. gear A delivers the power, with the forces on the shaft acting as shown; the gear teeth have a o 1 A B pressure angle of φ = 14 ( tan φ = h = h ). Both gears are keyed (profile) to 2 Av Bv the shaft of AISI 1141, cold rolled. (a) If the fillet radius is 1/8 in. at bearing D, where the diameter is 1 3/8 in., compute N based on the octahedral-shear-stress theory (Soderberg line). The shaft diameter at A is 1 11/16 in. What is N here? Page 38 of 76 SECTION 7 – SHAFT DESIGN Problem 489, 490 Solution: For AISI 1141, cold rolled s y = 90 ksi sn = 50 ksi sn sns 50 1 = = = s y s ys 90 1.8 SF = 0.85 63,000(7 ) T= = 1764 in − lb 250 16T sms = πD 3 sas = 0 Gear B: 5.75 Bv = T = 1764 in − lb 2 Bv = 614 lb Bh = Bv tan φ = 614 tan 14.5 = 159 lb Gear A: 2.25 Av = T = 1764 in − lb 2 Av = 1568 lb Ah = Av tan φ = 1568 tan 14.5 = 406 lb Vertical Page 39 of 76 SECTION 7 – SHAFT DESIGN [∑ M D =0 ] 8Cv = 4(1568) − 3(614) Cv = 554 lb [∑ F =0] v Cv + Dv = Av + Bv 554 + Dv = 1568 + 614 Dv = 1628 lb Shear Diagram M Av = (554 )(4 ) = 2216 in − lb M Dv = (614 )(3) = 1842 in − lb Horizontal [∑ M C =0 ] 4(406) + 8Dh = 11(159) Dh = 16 lb [∑ F h =0 ] Ch + Bh = Ah + Dh Ch + 159 = 406 + 16 Ch = 263 lb Shear Diagram Page 40 of 76 SECTION 7 – SHAFT DESIGN M Ah = (263)(4 ) = 1052 in − lb M Dh = (159 )(3) = 477 in − lb M A = M A2h + M A2v = (1052)2 + (2216)2 = 2453 in − lb M D = M D2 h + M D2 v = (477 )2 + (1842)2 = 1903 in − lb (a) At bearing D 1 r = in 8 3 d = 1 in 8 r 0.125 = ≈ 0.10 d 1.375 D 1.375 + 0.25 = ≈ 1 .2 d 1.375 K t ≈ K f = 1 .6 K ts ≈ K fs = 1.34 M = MD sm = 0 32 M 32(1903) sa = = = 7456 psi π d 3 π (1.375)3 K f sa s se = n s m + sy SF se = 0 + sms (1.6)(7456) = 14,035 psi 0.85 16T 16(1764 ) = = = 3456 psi 3 π D π (1.375)3 sas = 0 ses = K fs sas sns sms + s ys SF 1 ses = (3456 ) + 0 = 1920 psi 1 .8 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn Page 41 of 76 SECTION 7 – SHAFT DESIGN 1 2 2 2 1 14,035 3456 = + N 50,000 0.577(50,000) N = 3.28 (b) At A For profile keyway K f = 2.0 , K fs = 1.6 11 in = 1.6875 in 16 M = M A = 2453 in − lb sm = 0 32 M 32(2453) sa = = = 5200 psi 3 πd π (1.6875)3 K f sa s se = n s m + sy SF d =1 se = 0 + sms (2.0)(5200) = 12,235 psi 0.85 16T 16(1764 ) = = = 1870 psi 3 π D π (1.6875)3 sas = 0 ses = K fs sas sns sms + s ys SF 1 ses = (1870 ) + 0 = 1040 psi 1 .8 Octahedral shear theory 1 se ses = + N sn 0.577 sn 2 2 1 2 1 2 2 2 1 12,235 1040 = + N 50,000 0.577(50,000) N = 4.043 THRUST LOADS 491. The high-speed shaft of a worm-gear speed reducer, made of carburized AISI 8620, SOQT 450 F, is subjected to a torque of 21,400 in-lb. Applied to the right Page 42 of 76 SECTION 7 – SHAFT DESIGN end with no bending. The force on the worm has three components: a horizontal force opposing rotation of W = 6180 lb , a vertical radial force S = 1940 lb , and a rightward thrust of F = 6580 lb . The shaft has the following dimensions: a = 6 , 7 9 3 9 b = 4 , c = 10 , d = 4 , e = 2 , f = 13 , g = 11.646 , h = 10.370 , 8 16 4 16 13 D2 = 4 , D3 = 4 , D4 = 3.3469 , D5 = 3.253 , r1 = 0.098 , D1 = 3.740 , 16 3 1 r2 = r3 = , r4 = 0.098 , r5 = , all in inches. The pitch diameter of the worm, 4 16 6.923 in., is the effective diameter for the point of application of the forces. The root diameter, 5.701 in. is used for stress calculations. The left-hand bearing takes the thrust load. Calculate N based on the octahedral-shear-stress theory with varying stresses. (Data courtesy of Cleveland Worm and Gear Company.) Problem 491 Solution: Table AT 11n For AISI 8620, SOQT 450 F s y = 120 ksi su = 167 ksi sn = 0.5su = 83.5 ksi sn sns 83.5 1 = = = s y s ys 120 1.437 SF = 0.85 T = 21,400 in − lb Vertical Page 43 of 76 SECTION 7 – SHAFT DESIGN 6.923 6.923 M ′ = F = 6580 = 22,777 in − lb 2 2 ∑ M A =0 [ ] 22,777 + (11.646)(1940) = (11.646 + 10.370)Gv Gv = 2061 lb [∑ F =0] v S + Av = Gv 1940 + Av = 2061 Av = 121 lb Shear Diagram Moment Diagram M Av = 0 M Bv = −(121)(1.2035) = −146 in − lb M Cv = −(121)(1.2035 + 4.875) = −736 in − lb M Dv = −(121)(1.2035 + 4.875 + 5.5675) = −1409 in − lb at left side M Dv = −1409 + M ′ = −1409 + 22,777 = 21,368 in − lb at right side M Ev = 21,368 − (2061)(4.4325) = 12,233 in − lb M Fv = 12,233 − (2061)(4.5625) = 2830 in − lb M Gv = 2830 − (2061)(1.375) = 0 Horizontal Page 44 of 76 SECTION 7 – SHAFT DESIGN [∑ M A =0 ] (11.646)(6180) = (11.646 + 10.370)Gh Gh = 3269 lb [∑ F h =0 ] Ah + Gv = W Ah + 3269 = 6180 Ah = 2911 lb Shear Diagram Moment Diagram M Ah = 0 M Bh = (2911)(1.2035) = 3500 in − lb M Ch = (2911)(1.2035 + 4.875) = 17,695 in − lb M Dh = 33,900 in − lb M Eh = 33,900 − (3269 )(4.4325) = 19,410 in − lb M Fh = 19,410 − (3269 )(4.5625) = 4495 in − lb M Fh = 4495 − (3269 )(1.375) = 0 Combined M = M h2 + M v2 MA = MB = MC = MD = MD = ME = (0)2 + (0)2 = 0 in − lb (3500)2 + (146)2 = 3503 in − lb (17,695)2 + (736)2 = 17,710 in − lb (33,900)2 + (1409)2 = 33,930 in − lb (left) (33,900)2 + (21,368)2 = 40,073 in − lb (right) (19,410)2 + (12,233)2 = 22,944 in − lb Page 45 of 76 SECTION 7 – SHAFT DESIGN MF = MG = (2830)2 + (4495)2 = 5312 in − lb (0)2 + (0)2 = 0 in − lb Bending stresses (Maximum) At A, s A = 0 32 M B 32(3503) At B, s B = = = 682 psi π D13 π (3.740)3 32 M C 32(17,710 ) = = 1618 psi At C, sC = π D23 π (4.8125)3 32 M D 32(40,073) At D, s D = = = 2203 psi π Dr3 π (5.701)3 32 M E 32(22,944 ) At E, s E = = = 3652 psi π D33 π (4)3 32 M F 32(5312 ) At F, s F = = = 1443 psi 3 π D4 π (3.3469 )3 At G, sG = 0 Shear Stresses: 16T 16(21,400 ) ssA = ssB = = = 2083 psi π D13 π (3.740)3 16T 16(21,400 ) ssC = = = 978 psi π D23 π (4.8125)3 16T 16(21,400 ) ssD = = = 588 psi π Dr3 π (5.701)3 16T 16(21,400 ) ssE = = = 1703 psi π D33 π (4)3 16T 16(21,400 ) ssF = ssG = = = 2907 psi π D43 π (3.3469)3 Tensile stresses: F = 6580 lb 4F 4(6580 ) s′A = s′B = = = 599 psi 2 π D1 π (3.740)2 4F 4(6580 ) sC′ = = = 362 psi 2 π D2 π (4.8125)2 4F 4(6580 ) s′D = = = 258 psi 2 π Dr π (5.701)2 4F 4(6580 ) s′E = = = 524 psi 2 π D3 π (4 )2 Page 46 of 76 SECTION 7 – SHAFT DESIGN s′E = s′F = 4F 4(6580 ) = = 748 psi 2 π D4 π (3.3469 )2 r1 0.098 = = 0.03 D1 3.740 D2 4.8125 = = 1 .3 D1 3.740 Figure AF 12 K f ≈ K t = 2 .3 At B: K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′B = 599 psi se = sa = s B = 682 psi (2.3)(682 ) = 2262 psi 1 se = (599 ) + 0.85 1.437 K fs sas s ses = ns sms + s ys SF sms = ssB = 2083 psi sas = 0 1 ses = (2083) + 0 = 1450 psi 1.437 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 2262 1450 = + N 83,500 0.577(83,500) N = 24.7 r2 0.75 = = 0.16 D2 4.8125 Dr 5.701 = = 1 .2 D2 4.8125 Figure AF 12 K f ≈ K t = 1 .5 At C: K fs ≈ K ts = 1.2 Page 47 of 76 SECTION 7 – SHAFT DESIGN se = K f sa sn sm + sy SF sm = 362 psi sa = 1618 psi (1.5)(1618) = 3107 psi 1 se = (362 ) + 0.85 1.437 K fs sas s ses = ns sms + s ys SF sms = ssC = 978 psi sas = 0 1 ses = (978) + 0 = 681 psi 1.437 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 3107 681 = + N 83,500 0.577(83,500) N = 25.1 At D: Assume K f = 1.5 as in Prob. 483 se = K f sa sn sm + sy SF sm = 258 psi sa = 2203 psi (1.5)(2203) = 4067 psi 1 se = (258) + 0.85 1.437 K fs sas s ses = ns sms + s ys SF sms = s sD = 588 psi sas = 0 1 ses = (588) + 0 = 409 psi 1.437 Octahedral shear theory Page 48 of 76 SECTION 7 – SHAFT DESIGN 1 se ses = + N sn 0.577 sn 2 2 1 2 1 2 2 2 1 4067 409 = + N 83,500 0.577(83,500) N = 20.2 r3 0.75 = = 0.19 D3 4 Dr 5.701 = = 1.43 D3 4 Figure AF 12 K f ≈ K t = 1.45 At E: K fs ≈ K ts = 1.25 K f sa sn sm + sy SF sm = s′E = 524 psi se = sa = s E = 3652 psi (1.45)(3652) = 6595 psi 1 se = (524 ) + 0.85 1.437 K fs sas s ses = ns sms + s ys SF sms = s sE = 1703 psi sas = 0 1 ses = (1703) + 0 = 1185 psi 1.437 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 6595 1185 = + N 83,500 0.577(83,500) N = 12 At F: r4 0.098 = = 0.03 D4 3.3469 Page 49 of 76 SECTION 7 – SHAFT DESIGN D3 4 = = 1 .2 D4 3.3469 Figure AF 12 K f ≈ K t = 2 .3 K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′F = 748 psi se = sa = s F = 1443 psi (2.3)(1443) = 4425 psi 1 se = (748) + 0.85 1.437 K fs sas s ses = ns sms + s ys SF sms = s sF = 2907 psi sas = 0 1 ses = (2907 ) + 0 = 2023 psi 1.437 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 4425 2023 = + N 83,500 0.577(83,500) N = 14.8 Then N = 12 at r3 = 492. 3 in (E) 4 The slow-speed shaft of a speed reducer shown, made of AISI 4140, OQT 1200 F, transmits 100 hp at a speed of 388 rpm. It receives power through a 13.6 in. gear B. The force on this gear has three components: a horizontal tangential driving force Ft = 2390 lb , a vertical radial force S = 870 lb , and a thrust force Q = 598 lb taken by the right-hand bearing. The power is delivered to a belt at F that exerts a downward vertical force of 1620 lb.; sled runner keyways. Use the octahedral shear theory with the Soderberg line and compute N at sections C and D. (Data courtesy of Twin Disc Clutch Company.) Page 50 of 76 SECTION 7 – SHAFT DESIGN Problem 492, 493 Solution: For AISI 4140, OQT 1200 F s y = 83 ksi su = 112 ksi sn = 0.5su = 56 ksi sn sns 56 1 = = = s y s ys 83 1.482 SF = 0.85 63,000(100 ) T= = 16,237 in − lb 388 Vertical 13.6 13.6 M ′ = Q = (598) = 4066.4 in − lb 2 2 ∑ M A =0 [ ] Page 51 of 76 SECTION 7 – SHAFT DESIGN 5 5 3 7 11 13 3 3 3 1 + 1 (870 ) + 1 + 1 + 3 + 1 + 1 + + 2 (1620 ) + 4066.4 8 32 32 16 4 16 8 16 8 5 3 7 3 = 1 + 1 + 3 + 1 Gv 8 32 16 8 Gv = 3573 lb [∑ F =0] v Av + S + F = Gv Av + 870 + 1620 = 3573 Av = 1083 lb Shear Diagram Moment Diagram M Av = 0 3 M Pv = −(1083)1 = −1286 in − lb 16 5 M Bv = −1286 + (− 1083)1 = −3046 in − lb at the left 8 M Bv = −3046 + 4066.4 = 1021 in − lb at the right 3 M Cv = 1021 − (1953) 3 = −5570 in − lb 8 7 M Gv = −5570 − (1953)1 = −7950 in − lb 32 Page 52 of 76 SECTION 7 – SHAFT DESIGN 11 M Dv = −7950 + (1620 )1 = −5773 in − lb 32 13 M Ev = −5773 + (1620 ) = −4457 in − lb 16 3 M Fv = −4457 + (1620 ) 2 = 0 in − lb 4 Horizontal [∑ M A =0 ] 19 13 13 2 (2390 ) + 2 + 4 Gh 32 16 16 Gh = 908 lb [∑ F h =0 ] Ah + Gh = Ft Ah + 908 = 2390 Ah = 1482 lb Shear Diagram M Ah = 0 3 M Ph = (1482 )1 = 1760 in − lb 16 5 M Bh = 1760 + (1482 )1 = 4168 in − lb 8 Page 53 of 76 SECTION 7 – SHAFT DESIGN 3 M Ch = 4168 − (908) 3 = 1104 in − lb 8 7 M Ch = 1104 − (908)1 = 0 in − lb 32 M Dh = 0 in − lb M Eh = 0 in − lb M Fh = 0 in − lb Combined M = M h2 + M v2 M A = 0 in − lb MP = MB = MC = MD = ME = MF = (1760)2 + (1286)2 = 2180 in − lb (4168)2 + (3046)2 = 5163 in − lb (1104)2 + (5570)2 = 5678 in − lb (0)2 + (5773)2 = 5773 in − lb (0)2 + (4457 )2 = 4457 in − lb (0)2 + (0)2 = 0 in − lb 1 in = 0.125 in 8 d = 2.750 in D = 2.953 in r 0.125 = = 0.05 d 2.750 D 2.953 = = 1.10 d 2.750 Figure AF 12 K f 1 ≈ K t = 1 .9 at C: r = K fs1 ≈ K ts = 1.3 For sled runner keyway K f 2 = 1 .6 K fs 2 = 1.6 K f = 0.8 K f 1 K f 2 = 0.8(1.9 )(1.6 ) = 2.4 K fs = 0.8 K fs1 K fs 2 = 0.8(1.3)(1.6 ) = 1.7 Page 54 of 76 SECTION 7 – SHAFT DESIGN se = K f sa sn sm + sy SF 4Q 4(598) = = 101 psi 2 πd π (2.750)2 32 M C 32(5678) sa = = = 2781 psi π d3 π (2.750)3 (2.4)(2781) = 7920 psi 1 se = (101) + 0.85 1.482 K fs sas s ses = ns sms + s ys SF sm = 16T 16(16,237 ) = = 3976 psi π d 3 π (2.750)3 sas = 0 sms = 1 ses = (3976 ) + 0 = 2683 psi 1.482 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 7920 2683 = + N 56,000 0.577(56,000) N =6 1 in = 0.0625 in 16 d = 2.953 in 3 D = 3 in = 3.375 in 8 r 0.0625 = = 0.02 d 2.953 D 3.375 = = 1.14 d 2.953 Figure AF 12 K f ≈ K t = 2 .4 at D: r = K fs ≈ K ts = 1.6 se = K f sa sn sm + sy SF Page 55 of 76 SECTION 7 – SHAFT DESIGN 4Q 4(598) = = 87.3 psi 2 πd π (2.953)2 32 M C 32(5773) sa = = = 2284 psi π d3 π (2.953)3 (2.4)(2284) = 6508 psi 1 se = (87.3) + 0.85 1.482 K fs sas s ses = ns sms + s ys SF sm = 16T 16(16,237 ) = = 3211 psi π d 3 π (2.953)3 sas = 0 sms = 1 ses = (3211) + 0 = 2167 psi 1.482 Octahedral shear theory 1 2 2 2 1 se ses = + N sn 0.577 sn 1 2 2 2 1 6508 2167 = + N 56,000 0.577(56,000) N = 7 .5 TRANSVERSE DEFLECTIONS 494. The forces on the 2-in. steel shaft shown are A = 2 kips , C = 4 kips . Determine the maximum deflection and the shaft’s slope at D. Problems 494-496 Solution: Page 56 of 76 SECTION 7 – SHAFT DESIGN [M B = 0] 2(10) + 25D = 4(15) D = 1.6 kips [Fv = 0] A+C = B + D 2 + 4 = B + 1 .6 B = 4.4 kips Shear Diagram Moment Diagram M 64 M = EI Eπ D 4 M (in − kip ) M 4 4 D 10 EI Page 57 of 76 A 0 B -20 C 16 D 0 0 -135.8 108.6 0 SECTION 7 – SHAFT DESIGN Scale ss = 10 in in M 200 × 10−4 , Scale sM = per in EI EI D4 Slope θ , Scale sθ = 0.2 D 4 rad in y deflection, Scale s y = 2.0 D 4 in in Deflection: At A: y A = 0.625 in D4 Page 58 of 76 SECTION 7 – SHAFT DESIGN At C: yC = 0.375 in D4 Slope: 0.075 rad D4 0.0125 At B: θ = rad D4 0.05625 At D: θ = rad D4 At A: θ = Maximum deflection: 0.625 y = yA = = 0.04 in (2 )4 Shaft’s slope at D 0.05625 θ= = 0.0035 rad (2)4 495. The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine the constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at section C. Solution: (see Problem 494) 0.375 yC = = 0.006 D4 D = 2.812 in 7 say D = 2 in 8 496. The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine a constant shaft diameter that would limit the maximum deflection at section A to 0.003 in. Solution: (see Problem 494) 0.625 yA = = 0.003 D4 D = 3.80 in 7 say D = 3 in 8 497. A steel shaft is loaded as shown and supported in bearings at R1 and R2 . Determine (a) the slopes at the bearings and (b) the maximum deflection. Page 59 of 76 SECTION 7 – SHAFT DESIGN Problem 497 Solution: ∑ M R1 = 0 [ ] (3000 ) 7 + 1 1 − (2100) 7 + 2 1 + 1 = 7 + 2 1 + 2 + 7 R2 8 8 R2 = −444 lb [∑ F = 0] R1 + R2 + 2100 = 3000 R1 − 444 + 2100 = 3000 R1 = 1344 lb Loading Shear Diagram Moment Diagram Page 60 of 76 8 4 8 4 8 SECTION 7 – SHAFT DESIGN M A = 0 in − lb 7 M B = (1344 ) = 1176 in − lb 8 7 1 M C = (1134 ) + 1 = 2688 in − lb 8 8 1 M D = 2688 − (1656 )1 = 825 in − lb 8 M E = 825 − (1656)(1) = −831 in − lb M F = −831 + (444)(1) = −387 in − lb 7 M G = −387 + (444 ) = 0 in − lb 8 M (in − kips ) D(in ) M 10 4 EI ( )( ) A 0 1½ B1 1.18 1½ B2 1.18 2 C 2.69 2 D1 0.83 2 D2 0.83 1¾ E -0.83 1¾ F1 -0.39 1¾ F2 -0.39 1½ G 0 1½ 0 1.58 0.50 1.14 0.35 0.60 -0.60 -0.28 -0.52 0 Scale ss = 2 in in Page 61 of 76 SECTION 7 – SHAFT DESIGN M 2 × 10−4 , Scale sM = per in EI EI D4 Slope θ , Scale sθ = 4 × 10 −4 D 4 rad in y deflection, Scale s y = 8 × 10 −4 D 4 in in (a) Slopes at the bearings ( ) at R1 , θ A = 0.375 4 × 10 −4 = 1.5 × 10 −4 rad at R2 , θ G = 0 rad (b) Maximum deflection at C, yC = 0.1875(8 × 10 −4 ) = 1.5 × 10 −4 in Page 62 of 76 SECTION 7 – SHAFT DESIGN 498. (a) Determine the diameter of the steel shaft shown if the maximum deflection is to be 0.01 in.; C = 1.5 kips , A = 1.58 kips , L = 24 in . (b) What is the slope of the shaft at bearing D? See 479. Problems 498, 505, 506. Solution: Vertical [∑ M D =0 ] 6(1.5) + 8(0.424) = 24 Ev Ev = 0.516 kip [∑ F v =0 ] Dv + 0.424 = 1.5 + Ev Dv + 0.424 = 1.5 + 0.516 Dv = 1.592 kip Shear Diagram M C = 0 ; M D = −6(1.5) = −9 in − kips M B = −9 + 8(0.092) = −8.264 in − kips Page 63 of 76 SECTION 7 – SHAFT DESIGN M A = −8.264 + 11(0.516) = −2.588 in − kips M E = −2.588 + 5(0.516) = 0 M (in − kips ) M D 4 ×10 4 EI ( ) C 0 0 D -9 -61.1 Scale ss = 8 in in M 120 × 10 −4 , Scale sM = per in EI EI D4 Slope θ , Scale sθ = 0.096 D 4 rad in Page 64 of 76 B -8.264 -56.1 A -2.588 -17.6 E 0 0 SECTION 7 – SHAFT DESIGN y deflection, Scale s y = 0.768 D 4 in in Deflections. 0.384 yCv = in D4 0.288 y Bv = in D4 0.168 y Av = in D4 Slope 0.057 θ Dv = rad D4 Horizontal [∑ M D =0 ] 8(0.424 ) + 24 Eh = 19(1.58) Eh = 1.1095 kip [∑ F h =0 ] Dh + Eh + 0.424 = 1.58 Dh + 1.1095 + 0.424 = 1.58 Page 65 of 76 SECTION 7 – SHAFT DESIGN Dh = 0.0465 kip Shear Diagram Moments MC = 0 MD = 0 M B = 8(− 0.0465) = −0.372 in − kip M A = −0.372 + 11(− 0.4705) = −5.5475 in − kips M E = −5.5475 + 5(1.1095) = 0 M (in − kips ) M D 4 ×10 4 EI ( ) C 0 0 D 0 0 Scale ss = 8 in in M 4 × 10−4 , Scale sM = per in EI EI D4 Slope θ , Scale sθ = 0.032 D 4 rad in Page 66 of 76 B -0.372 -2.53 A -5.5475 -37.7 E 0 0 SECTION 7 – SHAFT DESIGN y deflection, Scale s y = 0.256 D 4 in in Deflections. 0.064 yC h = in D4 0.072 y Bh = in D4 0.096 y Ah = in D4 Slope 0.012 θ Dh = rad D4 Resultant deflection: ( y = yh2 + yv2 1 2 ) 1 2 2 [(0.064) + (0.384) ] 2 yC = D 4 [(0.072) + (0.288) ] D 4 [(0.096) + (0.168) ] Slope: Page 67 of 76 = 0.297 D4 = 0.194 D4 1 2 2 2 yA = 0.390 D4 1 2 2 2 yB = = D 4 SECTION 7 – SHAFT DESIGN θ = (θ + θ 2 h 1 2 2 v ) 1 2 2 [(0.012) + (0.057) ] = 2 θD D 4 = 0.05823 rad D4 (a) Diameter D. Maximum deflection = yC = 0.390 = 0.01 in D4 D = 2.50 in (b) slope of the shaft at bearing D 0.05823 0.05823 θD = = = 0.0015 rad D4 (2.5)4 CRITICAL SPEED 499. A small, high-speed turbine has a single disk, weighing 0.85 lb., mounted at the midpoint of a 0.178-in. shaft, whose length between bearings is 6 ½ in. What is the critical speed if the shaft is considered as simply supported? Solution: Table AT 2 3 WL3 ( 0.85)(6.5) y= = = 0.052634 in 4 3EI 6 π (0.178 ) 3 30 × 10 64 ( ) 1 1 1 30 g o (∑ Wy ) 2 30 g o 2 30 386 2 nc = = = 818 rpm = π ∑ Wy 2 π y π 0.052634 500. The bearings on a 1 ½-in. shaft are 30 in. apart. On the shaft are three 300-lb disks, symmetrically placed 7.5 in. apart. What is the critical speed of the shaft? Solution: Page 68 of 76 SECTION 7 – SHAFT DESIGN Table AT 2 Deflection of B. y B = y B1 + y B2 + y B3 y B1 = (300)(22.5)(7.5)([ 30)2 − (22.5)2 − (7.5)2 ] = 0.01273 in 4 6 π (1.5 ) 6(30 × 10 ) (30) 64 (300)(15)(7.5)(30)2 − (15)2 − (7.5)2 = 0.01556 in y B2 = π (1.5)4 6 30 × 106 (30) 64 2 2 2 ( 300)(7.5)(7.5)(30) − (7.5) − (7.5) y B3 = = 0.00990 in 4 6 π (1.5 ) 6 30 × 10 (30) 64 y B = 0.01273 + 0.00990 + 0.01556 = 0.03819 in [ ( ] ) [ ( ] ) Deflection of C. yC = yC1 + yC2 + yC3 yC1 2 2 2 ( 300)(7.5)(30 − 15)( [ 30) − (7.5) − (30 − 15) ] = = 0.01556 in 4 6 π (1.5 ) 6(30 × 10 ) (30) 64 (300)(15)(30 − 15)(30)2 − (15)2 − (30 − 15)2 = 0.02264 in yC2 = 4 6 π (1.5 ) 6 30 ×10 (30) 64 (300)(7.5)(15)(30)2 − (7.5)2 − (15)2 = 0.01556 in yC3 = π (1.5)4 6 30 × 106 (30) 64 yC = 0.01556 + 0.02264 + 0.01556 = 0.05376 in Deflection of D. y D = y D1 + y D2 + y D3 [ ( ) [ ( Page 69 of 76 ] ] ) SECTION 7 – SHAFT DESIGN y D1 = (300)(7.5)(30 − 22.5)([ 30)2 − (7.5)2 − (30 − 22.5)2 ] = 0.00990 in 4 6 π (1.5 ) 6(30 × 10 ) (30) 64 (300)(15)(30 − 22.5)(30)2 − (15)2 − (30 − 22.5)2 = 0.01556 in y D2 = 4 6 π (1.5 ) 6 30 ×10 (30) 64 (300)(7.5)(22.5)(30)2 − (7.5)2 − (22.5)2 = 0.01273 in y D3 = π (1.5)4 6 30 × 106 (30 ) 64 y D = 0.00990 + 0.01556 + 0.01273 = 0.03819 in [ ( ] ) [ ( ] ) 1 2 1 30 g o (∑ Wy ) 30 g o ( y B + yC + y D ) 2 nc = = π ∑ Wy 2 π y B2 + yC2 + y D2 1 30 386(0.03819 + 0.05376 + 0.03819) 2 nc = = 888 rpm π (0.03819)2 + (0.05376)2 + (0.03819)2 501. A fan for an air-conditioning unit has two 50-lb. rotors mounted on a 3-in. steel shaft, each being 22 in. from an end of the shaft which is 80 in. long and simply supported at the ends. Determine (a) the deflection curve of the shaft considering its weight as well as the weight of the rotors, (b) its critical speed. Solution: W1 = 50 lb W3 = 50 lb π 2 W2 = (0.284 ) (3) (80 ) = 160 lb weight of shaft 4 160 w2 = = 2 lb in 80 Deflection of B. y B = y B1 + y B2 + y B3 Page 70 of 76 SECTION 7 – SHAFT DESIGN y B1 = (50)(50)(22)([ 80)2 − (58)2 − (22)2 ] = 0.002844 in 4 6 π (35 ) 6(30 × 10 ) (80) 64 (50)(22)(22)(80)2 − (22)2 − (22)2 = 0.002296 in y B3 = 4 6 π (3) 6 30 × 10 (80) 64 (2)(22)(80)3 − 2(80)(22)2 − (22)3 = 0.006843 in y B2 = π (3)4 6 30 × 106 64 y B = 0.002844 + 0.006843 + 0.002296 = 0.011983 in [ ( ] ) [ ] ( ) Deflection of C. yC = yC1 + yC2 + yC3 yC1 2 2 2 ( 50)(22)(80 − 40)( [ 80) − (22) − (80 − 40) ] = = 0.003317 in 4 6 π (35) 6(30 ×10 ) (80) 64 2 2 2 ( 50)(22 )(40)(80) − (22) − (40) yC3 = = 0.003317 in 4 6 π (3) 6 30 × 10 (80) 64 (2)(40)(80)3 − 2(80)(40)2 − (40)3 = 0.008942 in yC2 = 4 6 π (3) 6 30 × 10 64 yC = 0.003317 + 0.008942 + 0.003317 = 0.015576 in [ ( ] ) [ ] ( By symmetry y D = y B = 0.011983 in (a) Deflection curve Page 71 of 76 ) SECTION 7 – SHAFT DESIGN (b) Critical speed 1 30 g o (∑ Wy ) 2 nc = π ∑ Wy 2 ∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 3.69046 ∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 0.053177 2 2 2 2 1 30 386(3.69046) 2 nc = = 1563 rpm π 0.053177 ASME CODE 502. A cold-rolled transmission shaft, made of annealed AISI C1050, is to transmit a torque of 27 in-kips with a maximum bending moment of 43 in-kips. What should be the diameter according to the Code for a mild shock load? Solution: For AISI C1050, annealed s y = 53 ksi su = 92 ksi 0.3s y = 15.9 ksi 0.18su = 16.56 ksi use τ d = 0.3s y = 15.9 ksi M = 43 in − kips T = 27 in − kips 1 2 2 16 α FD 1 + B 2 2 3 (K sT ) + K m M + D = 8 πτ d 1 − B 4 Reduce to 1 16 2 2 2 3 D = (K sT ) + ( K m M ) πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 2 2 2 3 D = [(1.25)(27,000)] + [(1.75)(43,000)] π (15,900) D = 2.98 in say D = 3 in ( ) ( )[ { Page 72 of 76 ( ) ] } SECTION 7 – SHAFT DESIGN 503. A machinery shaft is to transmit 82 hp at a speed of 1150 rpm with mild shock. The shaft is subjected to a maximum bending moment of 7500 in-lb. and an axial thrust load of 15,000 lb. The material is AISI 3150, OQT 1000 F. (a) What should be the diameter when designed according to the Code? (b) Determine the corresponding conventional factor of safety (static-approach and maximum-shear theory). Solution: For AISI 3150, OQT 1000 F s y = 130 ksi su = 151 ksi 0.3s y = 39 ksi 0.18su = 27.18 ksi use τ d = 0.18su = 27.18 ksi 63000(82 ) T= = 4492 in − lb 1150 M = 7500 in − lb F = 15,000 lb 16 α FD 1 + B 2 2 3 (K sT ) + K m M + (a) D = πτ d 1 − B 4 8 ( ( ) 1 ) 2 2 For mild shock load K m = 1.75 K s = 1.25 B=0 α =1 1 2 ( 16 1)(15,000)D 2 2 3 D = [(1.25)(4492)] + (1.75)(7500) + π (27180) 8 { 1 2 2 D 3 = 0.1874 31.53 + [13.125 + 1.875D ] D = 1.4668 in say D = 1.5 in (b) s = } 32 M 4F 32(7500 ) 4(15,000 ) + = + = 31,124 psi = 31.124 ksi 3 2 πD πD π (1.5)3 π (1.5)2 Page 73 of 76 SECTION 7 – SHAFT DESIGN ss = 16T 16(4492 ) = = 6778.5 psi = 6.7785 ksi π D 3 π (1.5)3 2 2 1 s ss = + N s y s ys Maximum shear theory s ys = 0.5s y 2 2 1 31.124 6.7785 = + N 130 0.5(130) N = 3.83 504. short stub shaft, made of SAE 1035, as rolled, receives 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed midway between the bearings and its pressure angle φ = 20o . See the figure for 471. (a) Neglecting the radial component of the tooth load, determine the shaft diameter for a mild shock load. (b) Considering both tangential and radial components, compute the shaft diameter. (c) Is the difference in the foregoing results enough to change your choice of the shaft size? Solution: Figure for 471. For SAE 1035, as rolled s y = 55 ksi su = 85 ksi 0.3s y = 16.5 ksi 0.18su = 15.3 ksi use τ d = 0.18su = 15.3 ksi Page 74 of 76 SECTION 7 – SHAFT DESIGN Data are the same as 471. From Problem 471. (a) M = 4200 in − lb = 4.2 in − kips T = 6300 in − lb = 6.3 in − kips 16 α FD 1 + B 2 2 ( ) D3 = K T + K M + s m πτ d 1 − B 4 8 Reduce to 1 16 2 2 2 D3 = K T + K M ( ) ( ) s m πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 D3 = [(1.25)(6.3)]2 + [(1.75)(4.2)]2 2 π (15.3) D = 1.5306 in 9 say D = 1 in 16 ( ) ( )[ ( 1 ) 2 2 ] { } (b) M = 4472 in − lb = 4.472 in − kips T = 6300 in − lb = 6.3 in − kips 16 D = [(1.25)(6.3)]2 + [(1.75)(4.472)]2 π (15.3) D = 1.5461 in 9 say D = 1 in 16 { 3 1 2 } (c) Not enough to change the shaft size. 505. Two bearings D and E, a distance D = 24 in . Apart, support a shaft for a punch press on which are an 8-in. gear A, a 44-in. pulley B, and a flywheel C, as indicated (498). Weight of flywheel is 1500 lb.; pulley B receives the power at an angle of 45o to the right of the vertical; gear A delivers it horizontally to the right. The maximum power is 25 hp at 250 rpm is delivered, with heavy shock. For cold-finish AISI 1137, find the diameter by the ASME Code. Solution: Page 75 of 76 SECTION 7 – SHAFT DESIGN Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498. For AISI 1137, cold-finished s y = 93 ksi su = 103 ksi 0.3s y = 27.9 ksi 0.18su = 18.54 ksi use τ d = 0.18s u = 18.54 ksi From Problem 479 M = M B = 14,343 in − lb = 14.343 in − kips T = TA = 12,600 in − lb = 12.6 in − kips For heavy shock load K m = 2.5 K s = 1.75 B=0 16 α FD 1 + B 2 (K sT ) + K m M + 4 πτ d 1 − B 8 1 16 2 2 2 ( ) ( ) D3 = K T + K M s m πτ d 1 − B 4 D3 = ( ) ( )[ ( ] 16 [(1.75)(12.6)]2 + [(2.5)(14.343)]2 π (18.54) D = 2.2613 in 5 say D = 2 in 16 D3 = { 2 1 2 } - end - Page 76 of 76 ) 2 1 2 SECTION 8 – KEYS AND COUPLINGS FLAT AND SQUARE KEYS DESIGN PROBLEMS 521. A 2-in. shaft, of cold-drawn AISI 1137, has a pulley keyed to it. (a) Compute the length of square key and the length of flat key such that a key made of cold-drawn C1020 has the same yield strength as the shaft does in pure torsion. (b) The same as (a), except that the key material is AISI 2317, OQT 1000 F. (c) Would you discard either of these keys? Explain. Solution: For AISI 1137 shaft, Table AT 8, sy = 93 ksi Yield Strength of Shaft sys = 0.6sy = 0.6(93) = 55.8 ksi sysπD 3 (55.8)(π )(2)3 T= = = 87.65 in − kips 16 16 (a) Key Material, cold-drawn, C1020, Table AT 7 sy = 66 ksi sys = 0.6sy = 0.6(66) = 39.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 5.32 in sc tD 66(1 2)(2) Use L = 5.32 in For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 7.09 in sc tD 66(3 8)(2) Use L = 7.09 in (b) Key Material, AISI 2317, QOT 1000 F. Table AT 8 sy = 71 ksi sys = 0.6sy = 0.6(71) = 42.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength. 1 SECTION 8 – KEYS AND COUPLINGS 4T 4(87.65) = = 4.94 in sc tD 71(1 2 )(2 ) Use L = 4.94 in L= For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 6.59 in sc tD 71(3 8)(2 ) Use L = 6.59 in (c) Either of the above is not to be discarded since they are designed based on yield strength with the same factor of safety. 522. A cast-iron pulley transmits 65.5 hp at 1750 rpm. The 1045 as-rolled shaft to which it is to be keyed is 1 ¾ in. in diameter; key material, cold-drawn 1020. Compute the length of flat key and of square key needed. Solution: For shaft: 1045 as-rolled, Table AT 7, sy = 59 ksi For key: Cold-drawn 1020, sy = 66 ksi D = 1 ¾ in = 1.75 in hp = 65.5 hp, n = 1750 rpm 63,000hp 63,000(65.5) = = 2358 in − lb = 2.358 in − kips n 1750 Table AT 19, use b = 3/8 in, t = 1/4 in for D = 1 ¾ in Assume smooth load, N = 1.5 T= For Flat key, b = 3/8 in, t = 1/4 in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.549 in sc tD 39.3(1 4)(1.75) Use L = 0.549 in - answer 2 SECTION 8 – KEYS AND COUPLINGS For Square key, b = 3/8 in, t = 3/8 in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.366 in sc tD 39.3(3 8)(1.75) Use L = 0.366 in - answer 523. A 3 ¼-in. shaft transmits with medium shock 85 hp at 100 rpm. Power is received through a sprocket (annealed nodular iron 60-45-10) keyed to the shaft of cold-rolled AISI 1040 (10% work), with a key of cold-finished B1113. What should be the length of (a) a square key? (b) a flat key? Solution: For sprocket, annealed nodular iron, 60-45-10, Table AT 6, sy = 55 ksi For shaft, cold-rolled AISI 1040 (10% work), Table AT 10, sy = 85 ksi For key, cold-finished B1113, Table AT 7, sy = 72 ksi D = 3 ¼ in = 3.25 in hp = 85 hp n = 100 rpm 63,000hp 63,000(85) = = 53,550 in − lb = 53.55 in − kips n 100 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 ¼ in For medium shock, N = 2.25 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 3.60 in sc tD 24.4(3 4)(3.25) T= 3 SECTION 8 – KEYS AND COUPLINGS Use L = 3.60 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 5.40 in sc tD 24.4(1 2)(3.25) Use L = 5.40 in. - answer 524. A cast-steel gear (SAE 0030), with a pitch diameter of 36 in., is transmitting 75 hp at 210 rpm to a rock crusher, and is keyed to a 3-in. shaft (AISI 1045, as rolled); the key is made of AISI C1020, cold drawn. For a design factor of 4 based on yield strength, what should be the length of (a) a square key, (b) flat key? (c) Would either of these keys be satisfactory? Solution: For cast-steel gear (SAE 0030), Table AT 6, sy = 35 ksi For shaft, AISI 1045, as rolled, Table AT 7, sy = 59 ksi For key, AISI C1020, cold-drawn, Table AT 7, sy = 66 ksi D = 3 in hp = 75 hp n = 210 rpm 63,000hp 63,000(75) = = 22,500 in − lb = 22.5 in − kips n 210 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 in Design factor, N = 4 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 T= 4 SECTION 8 – KEYS AND COUPLINGS 4T 4(22.5) = = 4.57 in sc tD 8.75(3 4)(3) Use L = 4.57 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 4T 4(22.5) L= = = 6.86 in sc tD 8.75(1 2)(3) Use L = 6.86 in. - answer L= 525. An electric motor delivers 50 hp at 1160 rpm to a 1 5/8 in. shaft (AISI 13B45, OQT 1100 F). Keyed to this shaft is a cast-steel (SAE 080, N & T) pulley whose hub is 2 in. long. The loading may be classified as mild shock. Decide upon a key for this pulley (material), investigating both flat and square keys. Solution: hp = 50 hp n = 1160 rpm D = 1 5/8 in = 1.625 in Shaft material – AISI 13B45, OQT 1100 F, Table AT 10, sy = 112 ksi Pulley material – SAE 080, N & T, Table AT 6, sy = 40 ksi L = 2 in N = 2.0 to 2.25 for mild shock From Table AT 19 for D = 1 5/8 in b = 3/8 in, t = ¼ in 63,000hp 63,000(50) T= = = 2,716 in − lb = 2.716 in − kips n 1160 For flat key: b = 3/8 in, t = ¼ in Check for compression: 4T 4(2.716) sc = = = 13.37 ksi LtD (2 )(1 4)(1.625) sy 40 = 2.99 > 2.25 Based on pulley material, N = = sc 13.37 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) 5 SECTION 8 – KEYS AND COUPLINGS N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi Min. s y = (13.37 )(2.25) = 30 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer. For square key: b = t = 3/8 in Check for compression: 4T 4(2.716) sc = = = 8.914 ksi LtD (2 )(3 8)(1.625) sy 40 = 4.49 > 2.25 Based on pulley material, N = = sc 8.914 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi Min. s y = (8.914)(2.25) = 20 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer. CHECK PROBLEMS 526. A cast-steel (SAE 080, N & T) pulley, attached to a 2-in. shaft, is transmitting 40 hp at 200 rpm, and is keyed by a standard square key, 3 in. long, made of SAE 1015, cold drawn; shaft material, C1144, OQT 1000 F. (a) What is the factor of safety of the key? (b) The same as (a) except a flat key is used. Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, C1144, OQT 1000 F, sy = 83 ksi hp = 40 hp N = 200 rpm D = 2 in L = 3 in 63,000hp 63,000(40) = = 12,600 in − lb = 12.6 in − kips n 200 Table AT 19, D = 2 in b = ½ in, t = 3/8 in T= 6 SECTION 8 – KEYS AND COUPLINGS a) Square key, b = ½ in, t = ½ in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 16.8 ksi LtD (3)(1 2)(2) sy (pulley ) 40 N= = = 2.38 < 6.21 sc 16.8 Answer N = 2.38 b) Flat key, b = ½ in, t = 3/8 in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 22.4 ksi LtD (3)(3 8)(2) sy (pulley ) 40 N= = = 1.78 < 6.21 sc 22.4 Answer N = 1.78 527. A cast-steel (SAE 080, N & T) pulley is keyed to a 2 1/2-in. shaft by means of a standard square key, 3 ½-in. long, made of cold-drawn SAE 1015. The shaft is made of cold-drawn AISI 1045. If the shaft is in virtually pure torsion, and turns at 420 rpm, what horsepower could the assembly safely transmit (steady loading)? Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 3 ½ in = 3.5 in D = 2 ½ in Table AT 19, D = 2 ½ in b = 5/8 in, t = 7/16 in Square Key, b = t = 5/8 in N = 1.5 for steady loading (smooth) For shaft: 7 SECTION 8 – KEYS AND COUPLINGS 3 ssπD 3 0.6 syπD 0.6(85)(π )(2.5)3 T= = = = 104.31 in − kips 16 N(16) 1.5(16) Key: By shear: s LbD T= s 2 sys 0.6 sy 0.6(63) ss = = = = 25.2 ksi N N 1. 5 (25.2 )(3.5)(5 8)(2.5) T= = 68.9 in − kips < 104.31 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 sc = = = 26.67 ksi N 1.5 (26.67 )(3.5)(5 8)(2.5) T= = 36.46 in − kips < 104.31 in-kips 4 Use T = 36.46 in − kips = 36,460 in − kips Tn 36,460(420) hp = = = 243 hp 63,000 63,000 528. The same as 527, except that the diameter is 3 in. and the length of the key is 5 in. Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 5 in D = 3 in Table AT 19, D = 3 in b = 3/4 in, t = 1/2 in Square Key, b = t = 3/4 in N = 1.5 for steady loading (smooth) For shaft: 3 s πD 3 0.6 sy πD 0.6(85)(π )(3)3 T= s = = = 180.25 in − kips 16 N(16 ) 1.5(16 ) Key: By shear: s LbD T= s 2 8 SECTION 8 – KEYS AND COUPLINGS 0.6(63) = 25.2 ksi N N 1. 5 (25.2)(5)(3 4)(3) T= = 141.75 in − kips < 180.25 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 = = 26.67 ksi sc = N 1.5 (26.67 )(5)(3 4)(3) T= = 75 in − kips < 180.25 in-kips 4 ss = sys = 0.6 sy = Use T = 75 in − kips = 75,000 in − kips Tn 75,000(420) hp = = = 500 hp 63,000 63,000 MISCELLANEOUS KEYS 529. Two assemblies, one with one feather keys, are shown, with the assumed positions of the normal forces N. Each assembly is transmitting a torque T. Derive an equation for each case giving the axial force needed to slide the hub along the shaft (f = coefficient of friction). Does either have an advantage in this respect? Solution: ND 2 2T N= D a) T = Axial force = F = fN = 2 fT D 2ND = ND 2 T N= D b) T = fT D Assembly (b) is stronger than assembly (a) which has an axial force half that of assembly (b). Axial force = F = fN = 9 SECTION 8 – KEYS AND COUPLINGS 530. A 1 11/16-in. shaft rotating at 200 rpm, carries a cast-iron gear keyed to it by a ¼ x 1 ¼-in. Woodruff key; shaft material is cold-finished SAE 1045. The power is transmitted with mild shock. What horsepower may be safely transmitted by the key, (a) if it is made of cold-drawn SAE 1118? (b) if it is made of SAE 2317, OQT 1000 F? (c) How many keys of each material are needed to give a capacity of 25 hp? Specify a choice. Solution: Only shear is used. D = 1 11/16 in n = 200 rpm Woodruff key = ¼ x 1 ¼ in N = 2 for mild shock Shear force for key 2T F= = ss As D s AD T= s s 2 Table 10.1, ¼ x 1 ¼ in Woodruff Key is Key No. 810 Shear area, As = 0.296 sq. in. (a) Key, cold-drawn 1118, Table AT 7, sy = 75 ksi sys 0.6 sy 0.6(75) = = = 22.5 ksi < 24.06 ksi ss = N N 2 (22.5)(0.296)1 11 16 = 5.62 in − kips = 5620 in − lb T= 2 (5620)(200) Tn hp = = = 17.84 hp 63,000 63,000 (b) Key, SAE 2317, OQT 1000F, Table AT 7, sy = 79 ksi sys 0.6 sy 0.6(79) ss = = = = 23.7 ksi < 24.06 ksi N N 2 (23.7 )(0.296 )1 11 16 = 5.92 in − kips = 5920 in − lb T= 2 (5920)(200) Tn hp = = = 18.79 hp 63,000 63,000 (c) Number of keys for (a) = 25 / 17.84 = 1.4 or 2 keys Number of keys for (b) = 25 / 18.79 = 1.33 or 2 keys Select (b) which is stronger. 531. A 3/16 x 1-in. Woodruff key is used in a 1 3/16-in. shaft (cold-drawn SAE 1045). (a) If the key is made of the same material, will it be weaker or stronger than the shaft in pure torsion? (b) If the key is made of SAE 4130, WQT 1100 F, will it be weaker or stronger? For the purposes here, the weakening of the shaft by the keyway is ignored. 10 SECTION 8 – KEYS AND COUPLINGS Solution: Woodruff key, 3/16 x 1 in. D = 1 3/16 in Shaft: Cold drawn, SAE 1045 (Table AT 8) sy = 85 ksi s AD T= s s 2 Table 10.1, 3/16 x 1 in., Woodruff key is Key no. 608. Shear area = As = 0.178 sq. in. (a) Key material = Shaft Material In yield: For key 3 0.6(85)(0.178)1 ss As D 0.6sy As D 16 = 5.39 in − kips T= = = 2 2 2 For shaft: 3 3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. (b) Key material = SAE 4130, WQT 1100, Table AT 7, sy = 114 ksi In yield: For key 3 0.6(114)(0.178)1 ss As D 0.6 sy As D 16 = 7.23 in − kips T= = = 2 2 2 For shaft: 3 3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. 532. A 2-in. shaft (cold-finished SAE 1137) is connected to a hub by a 3/8-in. radial taper pin made of 4150, OQT 1000 F. (a) What horsepower at 1800 rpm would be transmitted when the pin is about to be sheared off? (b) For this horsepower, what peak torsional stress may be repeated in the shaft? Is the shaft safe from fatigue at this stress? Solution: D = 2 in d = 3/8 in n = 1800 rpm (a) For pin material , 4150, OQT 1000 F, Table AT9, su = 193.5 ksi πd 2 πd 2 = ss F = ss (2 As ) = ss (2) 4 2 πd 2 D 1 FD = ssπd 2D T= = ss 2 2 2 4 s s = sus = 0.75 su = 0.75(193.5 ) = 145.1 ksi 11 SECTION 8 – KEYS AND COUPLINGS 2 1 3 T = (145.1)(π ) (2 ) = 32.05 in − kips = 32,050 in − lb 4 8 (32,050 )(1800) Tn hp = = = 915.7 hp 63,000 63,000 (b) For shaft, cold-finished, SAE 1137, Table AT 8, su = 103 ksi s 103 sns = 0.6 sn′ = 0.6 u = 0.6 = 30.96 ksi 2 2 But, 16T 16(32,050) ss = 3 = = 20,404 psi = 20.4 ksi < 30.96 ksi πD π (2)3 s 30.96 N = ns = = 1.52 > 1.5, therefore safe from fatigue at this stress. ss 20.4 533. A 20-in. lever is keyed to a 1 7/8-in. shaft (cold-finished SAE 1141) by a radial taper pin whose mean diameter is 0.5 in.; pin material, C1095, OQT 800 F. The load on the lever is repeatedly reversed; N = 2 on endurance strength. What is the safe lever load (a) for the shaft, (b) for the pin key (shear only), (c) for the combination? Solution: T = FL where F = safe lever load. L = 20 in D = 1 7/8 in = 1.875 in Shaft Material, cold finished, SAE 1141, Table AT 10. sy = 90 ksi sy sn = 1.8 90 = 50 ksi 1.8 Pin Material, C1095, OQT 800 F, Table AT 9. su = 176 ksi sn = 0.5su = 0.5(176 ) = 88 ksi (a) For the shaft. 0.6sy 0.6(50) ss = = = 15 ksi N 2 s πD 3 15(π )(1.875)3 T= s = = 19.414 in − kips = 19,414 in − lbs 16 16 T = FL 19,414 = F (20 ) F = 970.7 lb (b) For the pin. 0.6 sy 0.6(88) ss = = = 26.4 ksi N 2 sn = 12 SECTION 8 – KEYS AND COUPLINGS 2T d T 4T = = 3 π 2 2 As d (d ) πd 4 s πd 3 26.4(π )(0.5)3 T= s = = 2.592 in − kips = 2592 in − lbs 4 4 T = FL 2592 = F (20 ) F = 129.6 lb (c) For the combination. Use F = 129.6 lb ss = 534. A lever is keyed to a 2 ½-in. shaft made of SAE 1035, as rolled, by a radial taper pin, made of SAE 1020, as rolled. A load of 200 lb. is applied to the lever 22 in. from the center of the shaft. (a) What size pin should be used for N = 3 based on the yield strength in shear? (b) Let the hub diameter be 5 in. and assume that the part of the pin in the hub is uniformly loaded cantilever beam. Compute the bending stress and comment on the bending strength (especially if the loading varies). Solution: Shaft material, SAE 1035, as rolled, Table AT 7, sy = 55 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi F = 200 lb, L = 22 in, N = 3 D = 2 ½ in, Dh = 5 in T = FL = (200)(22) = 4400 in-lb = 4.4 in-kips (a) For the pin: 0.6 sy 0.6(48) ss = = = 9.6 ksi N 3 2T d T 4T ss = = = 3 π 2 2 As d (d ) πd 4 4(4.4) 9.6 = πd 3 d = 0.836 in 7 Use d = in = 0.875 in 8 (b) For the bending stress. As cantilever beam let 1 L = (Dh − D ) 2 1 L = (5 − 2.5) = 1.25 in 2 From Table AT 2. wL2 FL M= = 2 2 Where F is the uniform load. 13 SECTION 8 – KEYS AND COUPLINGS 2T 4T 4(4400) = = = 2347 lb Dm Dh + D 5 + 2.5 F = 1174 lb FL (1174)(1.25) M= = = 734 lb 2 2 Bending stress 32M 32(734) sb = 3 = = 11,160 psi = 11.16 ksi πd π (0.875)3 If the loading varies and factor of safety of 3. sn = Nsb = 3(11.16 ) = 33.48 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi, su = 65 ksu. sn′ = 0.5 su = 0.5(65 ) = 32.5 ksi The bending stress is nearly safe as the load varies. 2F = 535. A sprocket, transmitting 10 hp at 100 rpm, is attached to a 1 7/16 in. shaft as shown in Fig. 10.15, p. 290., Text; E = 3-1/2 in. What should be the minimum shear pin diameter if the computed stress is 85% of the breaking stress mentioned in the Text? Solution: hp = 10 hp, n = 100 rpm, D = 1 7/16 in = 1.4375 in, E = 3 ½ in = 3.5 in 63,000hp 63,000(10) T= = = 6300 in − lb n 100 T 6300 F= = = 1800 lb E 3.5 From text, Page 290. Breaking stress = 50,000 psi s s = 0.85(50 ,000 ) = 42 ,500 psi 4F ss = 2 πd 4(1800 ) 42,500 = πd 2 d = 0.2322 in use ¼ in. 536. A gear is attached to a 2-in. shaft somewhat as shown in Fig. 10-15, p. 290, Text; E = 3 5/16 in.; minimum shear-pin diameter = 3/8 in. with a rated torque of 22 in-kips. (a) For this torque, compute the stress in the shear pin. (b) From the ferrous metals given in the Appendix, select those that would give a resisting torque of about 110% of the rated value. Choose one, specifying its heat treatments or other conditions. Solution: D = 2 in E = 3 5/16 in = 3.3125 in d = 3/8 in T = 22 in-kips (a) Stress in shear-pin T 22 F= = = 6.64 lb E 3.3125 14 SECTION 8 – KEYS AND COUPLINGS 4(6.64) = 60.12 ksi π (3 8)2 (b) Select material. sus = 1.1(60.12 ) = 66.13 ksi From appendix, Table AT 7, select Cold drawn, C1020 with sus = 66 ksi ss = SPLINES 537. A shaft for an automobile transmission has 10 splines with the following dimensions: D = 1.25 in., d = 1.087 in., and L = 1.000 in. (see Table 10.2, p. 287, Text). Determine the safe torque capacity and horsepower at 3600 rpm of this sliding connection. Solution: D = 1.25 in, d = 1.087 in, L = 1.000 in, Nt = 10, n = 3600 rpm T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.25 ) = 0.11875 in But D − d 1.25 − 1.087 h= = = 0.0815 in actual 2 2 1.25 + 1.087 rm = = 0.58425 in 4 T = (1000 )(0.0815)(1.000 )(0.58425 )(10 ) = 476.2 in − lb - ans Tn (476.2 )(3600) hp = = = 27.2 hp - ans 63000 63000 538. The rear axle of an automobile has one end splined. For this fitting there are ten splines, and D = 1.31 in., d = 1.122 in., and L = 1 15/16 in. The minimum shaft diameter is 1 3/16 in. (a) Determine the safe torque capacity of the splined connection, sliding under load. (b) Determine the torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT 1000 F, if one fourth of the splines are in contact. (c) Determine the torsional stress in the shaft corresponding to each of these torques. Solution: D = 1.31 in, d = 1.122 in, L = 1 15/16 = 1.9375 in, Nt = 10 Dr = 1 3/16 in = 1.1875 in T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.31) = 0.12445 in 15 SECTION 8 – KEYS AND COUPLINGS But D − d 1.31 − 1.122 = = 0.094 in (actual) 2 2 1.31 + 1.122 rm = = 0.608 in 4 (a) Safe Torque T = (1000 )(0.094 )(1.9375)(0.608 )(10 ) = 1107.32 in − lb - ans (b) Torque by splines required on the point of yielding with one fourth of splines in contact (Page 288). From Table AT-7, AISI 8640, OQT 1000 F. sy = 150 ksi, ss = sys = 0.6sy = 0.6(150) = 90 ksi s πDL D (90)(π )(1.31)(1.9375) 1.31 T = s = 2 = 58.76 in − kips 8 8 2 T = 58,760 in − lbs (c) Torsional stress in the shaft From safe torque of 1107.32 in-lb 16T 16(1107.32 ) ss = 3 = = 3368 psi - ans πDr π (1.1875)3 From torque at the point of yield 16T 16(58,760) ss = 3 = = 178,711 psi - ans (too high) πDr π (1.1875)3 h= 539. An involute splined connection has 10 splines with a pitch Pd of 12/24 (a) Determine the dimension of this connection. (b) Compute the length of spline to have the same torsional strength as the shaft when one fourth the splines carry the load; minimum shaft diameter is 9/16 in. (no sliding). Check for compression. Solution: Nt = 10, Pd = 12, Dr = 9/16 in = 0.5625 in (a) Dimension D N 10 D = t = = 0.8333 in Pd 12 (b) Length of spline (same torsional strength as the shaft when one fourth the splines carry the load (Page 288). D 3 (0.5625)3 L= r = = 0.2136 in D 0.8333 Check for compression. Failure in compression is not likely (Page 289) and can be checked by using the projected contact area. Projected contact area: Ac = 0.2LD = 0.2(0.2136 )(0.8333) = 0.0356 in 2 based on one-fourth of the teeth being under load. COUPLINGS 540. A flange coupling has the following dimensions (Fig. 10.19, p. 291, Text): d = 5, D = 8 5/8, H = 12 ¼, g = 1 ½, h = 1, L = 7 ¼ in.; number of bolts = 6; 1 ¼ x 1 ¼-in. square key. Materials: key, colddrawn AISI 1113; shaft, cold-rolled, AISI 1045; bolts, SAE grade 5 (§5.8). Using the static 16 SECTION 8 – KEYS AND COUPLINGS approach with N = 3.3 on yield strengths, determine the safe horsepower that this connection may transmit at 630 rpm. Solution: d = 5 in D = 8 5/8 in = 8.625 in H = 12 ¼ in = 12.25 in g = 1 ½ in = 1.5 in h = 1 in L = 7 ¼ in = 7.25 in N = 3.3 nb = 630 rpm Square key = 1 ¼ in x 1 ¼ in Materials: Key: cold-drawn AISI 1113, Table AT 7, sy = 72 ksi, sys = 0.6sy = 0.6(72) = 43.2 ksi Shaft: cold-rolled, AISI 1045, Table AT 8, sy = 85 ksi, sys = 0.6sy = 0.6(85) = 51 ksi Bolt: SAE Grade 5, h = 1 in. sy = 81 ksi, sys = 0.6sy = 0.6(81) = 48.6 ksi No given material for the flange. Bolts in shear: sys 48.6 ss = = = 14.73 ksi N 3. 3 πh 2 F= N b ss 4 FH πh 2 Nb ss H T= = 2 8 2 π (1) (6 )(14.73)(12.25) T= = 425.158 in − kips 8 T = 425,158 in − lbs (425,158)(630) Tn hp = = = 4252 hp 63,000 63,000 Bolts in compression: sy 81 sc = = = 24.55 ksi N 3.3 F = Nb hgs c 17 SECTION 8 – KEYS AND COUPLINGS FH Nb hgsc H = 2 2 (6)(1)(1.5)(24.55)(12.25) T= = 1353.319 in − kips 2 T = 1,353,319 in − lbs (1,353,319)(630) Tn hp = = = 13,533 hp 63,000 63,000 T= Key in shear: sys 43.2 ss = = = 13.09 ksi N 3. 3 s bdL (13.09 )(1.25)(5)(7.25) T= s = = 296.570 in − kips 2 2 T = 296,570 in − lbs (296,570)(630) Tn hp = = = 2966 hp 63,000 63,000 Key in compression: sy 72 sc = = = 21.82 ksi N 3.3 s tdL (21.82)(1.25)(5)(7.25) T= c = = 247.180 in − kips 4 4 T = 247 ,180 in − lbs (247,180 )(630) Tn hp = = = 2472 hp 63,000 63,000 Shaft in shear: sys 51 ss = = = 15.45 ksi N 3.3 πd 3 ss π (5)3 (15.45) T= = = 379.200 in − kips 16 16 T = 379,200 in − lbs (379,200)(630) Tn hp = = = 3792 hp 63,000 63,000 The safest horsepower is the lowest which is 2472 hp. 541. A cast-iron (ASTM 25) jaw clutch with 4 jaws transmits 50 hp at 60 rpm. The inside diameter of the jaws is 3 in. Considering rough handling, choose N = 8 on ultimate strengths. Make reasonable and conservative assumptions and compute (a) the outside diameter of the jaws, (b) the length of jaws h. 18 SECTION 8 – KEYS AND COUPLINGS Solution: For ASTM 25, suc = 97 ksi, in shear sus = 35 ksi (Table AT 6) 63,000hp 63,000(50) T= = = 52,500 in − lbs n 60 T = 52.5 in − kips (a) The outside diameter of the jaws s 35 ss = us = = 4.375 ksi N 8 Assume Dm as the average diameter, t = thickness = Do – Di , Nj = number of jaws Shear area, 1 πD 1 π D + Di Do − Di As = m (t ) = o 2 Nj 2 N j 2 2 1 π D 2 − Di2 π 2 = As = o Do − Di2 2 4 4 32 2T 4T F= = Dm Do + Di F 4T 32 128T ss = = ⋅ = 2 2 π (Do + Di ) Do2 − Di2 As Do + Di π Do − Di 128(52.5) 4.375 = π (Do + 3) Do2 − 9 By trial and error. Do = 7.466 in or D o = 7.5 in (b) The length of jaws h. s 97 sc = uc = = 12.125 ksi N 8 N j h(Do − Di ) Ac = 2 F 2F 2F F sc = = = = Ac N j h(Do − Di ) 4h(Do − Di ) 2h(Do − Di ) ( ( ( sc = ) ) ( ) ) 4T (Do + Di ) 2T = 2 2h(Do − Di ) h Do − Di2 ( ) 19 SECTION 8 – KEYS AND COUPLINGS 2(52.5) h (7.5)2 − (3)2 3 h = 0.1833 in or h = in 16 12.125 = 542. [ ] The universal joint shown is made of AISI 3150, OQT 1000 F; a = 2 7/16 in., D = 9/16; n = 400 rpm. (a) What torque may be transmitted for shear of the pin (N = 5 on ultimate)? (b) Considering the pin as a simply supported beam of length a with the load distributed from a maximum at the outer (triangular), compute the safe transmitted torque (Same N). (c) In order not to have excessive wear on the pin, the average bearing pressure should not excced 3 ksi. Compute this transmitted torque. (d) What is the safe power? Solution: For AISI 3150, OQT 1000 F, Table AT 7, su = 151 ksi, sus = 113 ksi N=5 pb = 3 ksi a = 2 7/16 in = 2.4375 in D = 9/16 in = 0.5625 in n = 400 rpm (a) Torque transmitted for shear of the pin. s 113 ss = us = = 22.6 ksi N 5 Each shear area πD 2 (22.6)(π )(0.5625)2 = F = ss = 5.616 kips 4 4 Fa T = 2 = Fa = (5.616)(2.4375) = 13.687 in − kips 2 T = 13,687 in − lbs (b) Torque transmitted for shear of the pin (simply supported beam) Fa T =M= 3 (5.616)(2.4375) T= = 4.563 in − kips 3 T = 4,563 in − lbs 20 SECTION 8 – KEYS AND COUPLINGS (c) Torque transmitted for shear of the pin (pb = 3 ksi) F F F pb = = = Ab D(a 2) 2Da F 3= 2(0.5625)(2.4375) F = 8.23 ksi Fa T =M= 3 (8.23)(2.4375) T= = 6.687 in − kips 3 T = 6,687 in − lbs (d) Safe power (4,563)(400) Tn hp = = = 28.97 hp 63,000 63,000 544. A diagrammatic representation of a universal joint is shown, two yoke parts, the type being similar to Figs. 10.28 and 12.10, Text. The pin extensions have a diameter D = ¾ in.; a = 11/16 in., material of all parts is 4340, OQT 800. Let N = 4 on ultimate stresses; n = 2400 rpm. Compute the safe torque for (a) shear of pins, (b) the pin extensions in bending, assuming that the load distribution is from zero at the outside pin ends to a maximum at the inside yoke surfaces, (c) an average bearing pressure on pins of 4 ksi. (d) What is the corresponding horsepower capacity? Solution: For AISI 4340, OQT 800 F, Table AT 7, su = 221 ksi, sus = 0.75su = 0.75(221) = 166 ksi N=4 pb = 4 ksi a = 11/16 in = 0.6875 in D = 3/4 in = 0.75 in n = 2400 rpm (a) Torque transmitted for shear of the pin. s 166 ss = us = = 41.5 ksi N 4 Each shear area πD 2 (41.5)(π )(0.75)2 = F = ss = 18.33 kips 4 4 T = M = Fa = (18.33)(0.6875) = 12.602 in − kips T = 12,602 in − lbs 21 SECTION 8 – KEYS AND COUPLINGS (b) Torque transmitted for shear of the pin (simply supported beam) 2Fa T =M= 3 2(18.33)(0.6875) T= = 8.401 in − kips 3 T = 8,401 in − lbs (c) Torque transmitted for shear of the pin (pb = 4 ksi) F F pb = = Ab Da F 4= (0.75)(0.6875) F = 2.063 ksi 2Fa T =M= 3 2(2.063)(0.6875) T= = 0.9455 in − kips 3 T = 945.5 in − lbs (d) Safe power (945.5)(2400) Tn hp = = = 36.02 hp 63,000 63,000 - End - 22 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS LIGHTLY LOADED BEARINGS 551. (a) A 3 x 3 – in. full bearing supports a load of 900 lb., c d D = 0.0015 , n = 400 rpm . The temperature of the SAE 40 oil is maintained at 140 oF. Considering the bearing lightly loaded (Petroff), compute the frictional torque, fhp, and the coefficient of friction. (b) The same as (a) except that the oil is SAE 10W. Solution. (a) T f = µπ DLvips D 2 (cd 2) L = 3 in D = 3 in πDn π (3)(400) vips = = = 20π ips 60 60 c d D = 0.0015 SAE 40 oil, 140 oF, Figure A16. µ = 7.25 µreyns µπ DLvips 7.25 ×10 −6 (π )(3)(3)(20π ) F= = = 17.173 lb (cd 2) (0.0015 2) ( ) D 3 T f = F = (17.173) = 25.76 in − lb 2 2 Fvm fhp = 33,000 π Dn π (3)(400 ) vm = = = 314.16 fpm 12 12 Fvm (17.173)(314.16) = 0.1635 hp fhp = = 33,000 33,000 F 17.173 f = = = 0.0191 W 900 (b) SAE 10W oil, 140 oF, Figure A16. µ = 2.2 µreyns = 2.2 ×10 −6 reyn µπ DLvips 2.2 ×10 −6 (π )(3)(3)(20π ) F= = = 5.211 lb (cd 2) (0.0015 2) ( ) D 3 T f = F = (5.211) = 7.817 in − lb 2 2 Fvm fhp = 33,000 Page 1 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS vm = π Dn π (3)(400 ) = 314.16 fpm 12 12 Fvm (5.211)(314.16) = 0.0496 hp fhp = = 33,000 33,000 F 5.211 f = = = 0.00579 W 900 553. = The average pressure on a 6-in. full bearing is 50 psi, cd = 0.003 in. , L D = 1 . While the average oil temperature is maintained at 160 oF with n = 300 rpm , the frictional force is found to be 13 lb. Compute the coefficient of friction and the average viscosity of the oil. To what grade of oil does this correspond? Solution: W p= LD D = 6 in. L D =1 L = 6 in. W = pLD = (50)(6)(6 ) = 1800 lb F = 13 lb Coefficient of Friction F 13 f = = = 0.0072 W 1800 µπ DLvips F= (cd 2) πDn π (6)(300) vips = = = 30π ips 60 60 µπ DLvips µ (π )(6)(6)(30π ) F= = = 13 lb (cd 2) (0.003 2) µ = 1.8 ×10 −6 reyn = 1.8 µreyns Figure AF 16, 160 oF use SAE 10W or SAE 20W FULL BEARINGS 554. The load on a 4-in. full bearing is 2000 lb., n = 320 rpm ; L D = 1 ; cd D = 0.0011 ; operating temperature = 150 oF; ho = 0.00088 in . (a) Select an oil that will closely accord with the started conditions. For the selected oil determine (b) the frictional loss (ft-lb/min), (c) the hydrodynamic oil flow through the bearing, (d) the amount of end leakage, Page 2 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (e) the temperature rise as the oil passes through the bearing, (f) the maximum pressure. Solution: (a) D = 4 in L D =1 L = 4 in cd = 0.0011D = 0.0011(4) = 0.0044 in ho = 0.00088 in 2h 2(0.00088) ε = 1− o = 1− = 0 .6 cd 0.0044 Table AT 20 ε = 0.6 , L D = 1 Sommerfield Number µns D 2 p cd 320 ns = = 5.333 rps 60 W 2000 p= = = 125 psi LD (4)(4) cd D = 0.0011 S= µ (5.333) 2 1 0.121 = 125 0.0011 µ = 3.4 ×10 −6 reyn = 3.4 µreyns Figure AF-16, 150 oF, use SAE 30 or SAE 20 W Select SAE 30, the nearest µ = 3.9 ×10−6 reyn (b) Table AT 20, L D = 1 , ε = 0.6 r f = 3.22 cr r D 1 = = cr cd 0.0011 1 f = 3.22 0.0011 f = 0.003542 F = f W = (0.003542)(2000 ) = 7.084 lb Page 3 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS vm = π Dn = π (4)(320) = 335.1 fpm 12 12 Frictional loss = Fvm = (7.084 )(335.1)2374 ft − lb min (c) Table AT 20, L D = 1 , ε = 0.6 q = 4.33 rcr ns L D r = = 2.0 in 2 c 0.0044 cr = d = = 0.0022 in 2 2 ns = 5.333 rps L = 4 in q = 4.33rcr ns L = 4.33(2.0 )(0.0022 )(5.333)(4 ) = 0.4064 in 3 sec (d) Table AT 20, L D = 1 , ε = 0.6 qs = 0.680 q qs = 0.680q = 0.680(0.4064 ) = 0.2764 in 3 sec (e) Table AT 20, L D = 1 , ε = 0.6 ρ c∆to = 14.2 p ρ c = 112 , p = 125 psi 14.2 p 14.2(125) ∆to = = = 15.85 o F ρc 112 (f) Table AT 20, L D = 1 , ε = 0.6 p = 0.415 pmax 125 pmax = = 301.2 psi 0.415 555. A 4-in., 360o bearing, with L D = 1.1 (use table and chart values for 1), is to support 5 kips with a minimum film thickness 0.0008 in.; cd = 0.004 in. , n = 600 rpm . Determine (a) the needed absolute viscosity of the oil .(b) Suitable oil if the average film temperature is 160 F, (c) the frictional loss in hp. (d) Adjusting only ho to the optimum value for minimum friction, determine the fhp and compare. (e) This load varies. What could be the magnitude of the maximum impulsive load if the eccentricity ration ε becomes 0.8? Ignore “squeeze” effect. Page 4 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Solution: D = 4 in L = 1.1D = 1.1(4) = 4.4 in W 5000 p= = = 284 psi LD (4.4)(4) ho = 0.0008 in cd = 0.004 in. 2h 2(0.0008) ε = 1− o = 1− = 0 .6 cd 0.004 600 η= = 10 rps 60 (a) Table AT20, L D = 1 , ε = 0.6 S = 0.121 r S = cr 2 µ ns µ n s D = p cd p µ (10) 4 0.121 = 284 0.004 µ = 3.4 ×10−6 reyn 2 2 (b) Figure AF16, 160 F Use SAE 30, µ = 3.2 ×10−6 reyn (c) Table AT 20, L D = 1 , ε = 0.6 r f = 3.22 cr D f = 3.22 cd 4 f = 3.22 0.004 f = 0.00322 F = f W = (0.00322)(5000 lb ) = 16.1 lb π Dn π (4)(600) vm = = = 628.3 fpm 12 12 Page 5 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS fhp = (16.1)(628.3) = 0.3065 hp Fvm = 33,000 33,000 (d) adjusting ho , cd = 0.004 in. Table AT 20, L D = 1 ho cr = 0.30 optimum value for minimum friction r f = 2.46 cr D f = 2.46 cd 4 f = 2.46 0.004 f = 0.00246 F = f W = (0.00246)(5000 lb ) = 12.3 lb π Dn π (4)(600) vm = = = 628.3 fpm 12 12 Fvm (12.3)(628.3) = 0.234 hp < fhp (c ) fhp = = 33,000 33,000 (e) ε = 0.8 , Table AT 20, L D = 1 S = 0.0446 r S = cr 2 µ ns µ n s = p p D cd (3.2 ×10 )(10) 0.0446 = −6 p 2 4 0.004 2 p = 717.5 psi W = pDL = (717.5)(4)(4.4) = 12,628 lb 556. For an 8 x 4 – in. full bearing, cr = 0.0075 in. , n = 2700 rpm , average µ = 4 ×10 −6 reyn . (a) What load may this bearing safely carry if the minimum film thickness is not to be less than that given by Norton, i11.14, Text? (b) Compute the corresponding frictional loss (fhp). (c) Complete calculations for the other quantities in Table AT 20, φ , q , qs , ∆to , pmax . Compute the maximum load for an optimum (load) bearing (d) if cr remains the same, (e) if ho remains the same. Page 6 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Solution: D × L = 8× 4 L D =1 2 cr = 0.0075 in r = D 2 = 4 in µ = 4 ×10 −6 reyn (a) by Norton, ho = 0.00025D = 0.00025(8) = 0.002 in ho 0.002 = = 0.27 cr 0.0075 h Table AT 20, L D = 1 2 , o = 0.27 cr S = 0.172 2 r µns S = cr p 2700 ns = = 45 rps 60 2 ( ) −6 4 4 × 10 (45) S = 0.172 = p 0.0075 p = 298 psi W = pDL = (298)(8)(4) = 9536 lb (b) Table AT 20, L D = 1 2 , ho = 0.27 cr φ = 38.5o r f = 4.954 cr D f = 4.954 cd 4 f = 4.954 0.004 f = 0.0093 F = f W = (0.0093)(9536 lb ) = 88.7 lb π Dn π (8)(2700) vm = = = 5655 fpm 12 12 Fvm (88.7 )(5655) = 15.2 hp fhp = = 33,000 33,000 Page 7 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (c) Table AT 20, L D = 1 2 , ho = 0.27 cr φ = 38.5o q = 5.214 rcr ns L q = 5.214rcr ns L = 5.214(4 )(0.0075)(45)(4 ) = 28.2 in 3 sec qs = 0.824 q qs = 0.824(28.2 ) = 23.2 in 3 sec ρ c∆t = 20.26 p 20.26(298) ∆t = = 54 o F 112 p = 0.3013 pmax 298 pmax = = 989 psi 0.3013 h To solve for maximum load, Table AT 20, L D = 1 2 , o = 0.43 cr r S = cr 2 µns = 0.388 p (d) cr = 0.0075 in 2 ( ) −6 4 4 × 10 (45) S = 0.388 = p 0.0075 p = 132 psi W = pDL = (132)(8)(4) = 4224 lb (e) ho = 0.002 in ho = 0.43 cr 0.002 cr = = 0.00465 in 0.43 2 ( ) −6 4 4 × 10 (45) S = 0.388 = p 0.00465 p = 343.3 psi W = pDL = (343.3)(8)(4) = 10,986 lb Page 8 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 557. A 6 x 6 – in full bearing has a frictional loss of fhp = 11 when the load is 68,500 lb. and n = 1600 rpm ; cr r = 0.001 . (a) Compute the minimum film thickness. Is this in the vicinity of that for an optimum bearing? (b) What is the viscosity of the oil and a proper grade for an operating temperature of 160 F? (c) For the same ho , but for the maximum-load optimum, determine the permissible load and the fhp. Solution: L = 6 in D = 6 in L D =1 r = D 2 = 3 in cr r = 0.001 n = 1600 rpm π Dn π (3)(1600) vm = = = 2513 fpm 12 12 Fvm fhp = 33,000 33,000(11) F= = 144.45 lb 2513 F 144.45 f = = = 0.00211 W 68,500 r 1 (a) f = (0.00211) = 2.11 cr 0.001 r Table AT 20, L D = 1 , f = 2.11 cr Near the vicinity of optimum bearing cr = 0.001r = 0.001(3) = 0.003 in ho = 0.254cr = 0.254(0.003) = 0.0008 in (b) Table AT 20, L D = 1 , r f = 2.11 cr S = 0.0652 2 r µns S = = 0.388 cr p 1600 ns = = 26.67 rps 60 W 68,500 p= = = 1902.8 psi LD (6)(6) Page 9 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 2 1 (µ )(26.67 ) S = 0.0652 = 0.001 1902.8 µ = 4.7 ×10 −6 reyn Figure AF 16, 160 F, use SAE 40. (c) Table AT 20, L D = 1 optimum bearing, maximum load, ho = 0.53 cr ho 0.0008 = = 0.0015 in 0.53 0.53 ho r = 0.53 , S = 0.214 , f = 4.89 cr cr ho the same, cr = r S = cr 2 µns p 2 ( ) −6 3 4.7 × 10 (26.67 ) S = 0.214 = p 0.0015 p = 2343 psi W = pDL = (2343)(6)(6) = 84,348 lb r f = 4.89 cr 3 f = 4.89 0.0015 f = 0.00245 F = f W = 0.00245(84,348) = 206.65 lb vm = 2513 fpm (206.65)(2513) = 15.74 hp Fvm fhp = = 33,000 33,000 558. The maximum load on a 2.25 x 1.6875 in. main bearing of an automobile is 3140 lb. with wide-open throttle at 1000 rpm. If the oil is SAE 20W at 210 F, compute the minimum film thickness for a bearing clearance of (a) 0.0008 in. and (b) 0.0005 in. Which clearance results in the safer operating conditions? Note: Since a load of this order exists for only 20-25o of rotation, the actual ho does not reach this computed minimum (squeeze effect). Solution: D × L = 2.25 × 1.6875 in L 1.6875 = = 0.75 D 2.25 Page 10 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS SAE 20 W at 210 oF µ = 0.96 ×10 −6 reyn W = 3140 lb n = 1000 rpm W 3140 p= = = 827 psi DL (2.25)(1.6875) 1000 ns = = 16.67 rps 60 D r = = 1.125 in 2 µ ns r 2 p cr (a) cr = 0.0008 in S= (0.96 ×10 )(16.67 ) 1.125 S= −6 2 = 0.038 827 0.0008 Table AT 20, L D = 3 4 , S = 0.038 L D 1 ½ ho cr 0.2 0.2 S 0.0446 0.0923 ¾ 0.2 0.0685 L D 1 ½ ho cr 0.1 0.1 S 0.0188 0.0313 ¾ 0.1 0.0251 At L D = 3 4 ho 0.038 − 0.0251 = (0.2 − 0.1) + 0.1 = 0.13 cr 0.0685 − 0.0251 ho = 0.13cr = 0.13(0.0008) = 0.0001 in (b) cr = 0.0005 in (0.96 ×10 )(16.67 ) 1.125 S= −6 827 0.0005 2 = 0.098 Table AT 20, L D = 3 4 , S = 0.098 Page 11 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS L D 1 ½ ho cr 0.2 0.2 S 0.0446 0.0923 ¾ 0.2 0.0685 L D 1 ½ ho cr 0.4 0.4 S 0.121 0.319 ¾ 0.4 0.220 At L D = 3 4 ho 0.098 − 0.0685 = (0.4 − 0.2) + 0.2 = 0.239 cr 0.220 − 0.0685 ho = 0.239cr = 0.239(0.0005) = 0.00012 in use cr = 0.0005 in , ho = 0.00012 in A 360o bearing supports a load of 2500 lb.; D = 5 in. , L = 2.5 in. , cr = 0.003 in. , n = 1800 rpm ; SAE 20 W oil entering at 100 F. (a) Compute the average temperature t av of the oil through the bearing. (An iteration procedure. Assume µ ; compute S and the corresponding ∆to ; then the average oil temperature t av = ti + ∆to 2 . If this t av and the assumed µ do not locate a point in Fig. AF 16 on the line for SAE 20 W oil, try again.) Calculate (b) the minimum film thickness, (c) the fhp, (d) the amount of oil to be supplied and the end leakage. 561. Solution: D = 5 in L = 2.5 in L 2 .5 = = 0 .5 D 5 cr = 0.003 in (a) Table AT 20 ρ c∆to Parameter, , ρ c = 112 p r S = cr 2 µns p Page 12 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W 2500 = = 200 psi DL (5)(2.5) 1800 ns = = 30 rps 60 D r = = 2.5 in 2 cr = 0.003 in Fig. AF 16, SAE 20 W, Table AT 20, L D = 0.5 , ti = 100 o F ρ c∆to ∆to o F Trial µ ( t o F ), reyns S p 3.5 x 10-6 (130 F) 0.365 36.56 65 -6 3.2 x 10 (134 F) 0.333 34.08 61 -6 3.4 x 10 (132 F) 0.354 35.71 64 p= Therefore, use t av = 132 o F , S = 0.354 (b) Table AT 20, L D = 0.5 , S = 0.354 ho = 0.415 cr ho = 0.415(0.003) = 0.00125 in (c) Table AT 20, L D = 0.5 , S = 0.354 r f = 8.777 cr 2 .5 f = 8.777 0.003 f = 0.0105 F = f W = 0.0105(2500 ) = 26.25 lb π Dn π (5)(1800) vm = = = 2356 fpm 12 12 Fvm (26.25)(2356) = 1.874 hp fhp = = 33,000 33,000 (d) Table AT 20, L D = 0.5 , S = 0.354 q = 4.807 rcr ns L q = 4.807 rcr ns L = 4.807(2.5)(0.003)(30 )(2.5) = 2.704 in 3 sec qs = 0.7165 q Page 13 of 63 t av = ti + ∆to 2 o F 132.5 130.5 132.0 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 0.7165(2.704 ) = 1.937 in 3 sec PARTIAL BEARINGS 562. A 2 x 2-in. bearing has a clearance cr = 0.001 in , and ho = 0.0004 in. , n = 2400 rpm , and for the oil, µ = 3 × 10−6 reyn . Determine the load, frictional horsepower, the amount of oil to enter, the end leakage of oil, and the temperature rise of the oil as it passes through for : (a) a full bearing, partial bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o. Solution: D = L = 2 in L D =1 cr = 0.001 in r = D 2 = 1 in n = 2400 rpm ns = 40 rps µ = 3 ×10−6 reyn ho = 0.004 in. ho 0.0004 = = 0. 4 0.001 cr πDn π (2)(2400) vm = = = 1257 fpm 12 12 (a) Full bearing Table AT 20, L D = 1 , ho cr = 0.4 S = 0.121 rf = 3.22 cr q = 4.33 rcr ns L qs = 0.680 q ρ c∆to = 14.2 p p = 0.415 pmax Load W Page 14 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r S = cr 2 µ ns p 2 ( ) −6 1 3 ×10 (40 ) 0.121 = p 0.001 p = 992 psi W = pDL = (992)(2)(2) = 3968 lb fhp: F = fW rf = 3.22 cr 1 f = 3.22 0.001 f = 0.00322 F = f W = (0.00322)(3968) = 12.78 lb (12.78)(1257 ) = 0.4868 hp Fvm fhp = = 33,000 33,000 Oil flow, q q = 4.33 rcr ns L q = 4.33 (0.1)(0.001)(40)(2) q = 0.3464 in3 sec End leakage qs = 0.680 q qs = 0.68(0.3464 ) = 0.2356 in3 sec Temperature rise, ∆to ρ c∆to = 14.2 p (112)∆to = 14.2 992 ∆to = 126 o F (b) 180o Bearing Table AT 21, L D = 1 , ho cr = 0.4 S = 0.128 Page 15 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS rf = 2.28 cr q = 3.25 rcr ns L qs = 0.572 q ρ c∆to = 12.4 p Load W r S = cr 2 µ ns p 2 ( ) −6 1 3 × 10 (40 ) 0.128 = p 0.001 p = 937.5 psi W = pDL = (937.5)(2)(2) = 3750 lb fhp: F = fW rf = 2.28 cr 1 f = 2.28 0.001 f = 0.00228 F = f W = (0.00228)(3750) = 8.55 lb (8.55)(1257) = 0.3257 hp Fvm fhp = = 33,000 33,000 Oil flow, q q = 3.25 rcr ns L q = 3.25 (0.1)(0.001)(40)(2) q = 0.26 in3 sec End leakage qs = 0.572 q qs = 0.572(0.26 ) = 0.1487 in3 sec Temperature rise, ∆to Page 16 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ρ c∆to = 12.4 p (112)∆to = 12.4 937.5 ∆to = 104 o F (c) 12o Bearing Table AT 22, L D = 1 , ho cr = 0.4 S = 0.162 rf = 2.16 cr q = 2.24 rcr ns L qs = 0.384 q ρ c∆to = 15 p Load W r S = cr 2 µ ns p 2 ( ) −6 1 3 × 10 (40 ) 0.162 = p 0.001 p = 741 psi W = pDL = (741)(2 )(2) = 2964 lb fhp: F = fW rf = 2.16 cr 1 f = 2.16 0.001 f = 0.00216 F = f W = (0.00216)(2964) = 6.4 lb Fvm (6.4)(1257 ) = 0.2438 hp fhp = = 33,000 33,000 Oil flow, q Page 17 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS q = 2.24 rcr ns L q = 2.24 (0.1)(0.001)(40)(2) q = 0.1792 in3 sec End leakage qs = 0.384 q q s = 0.384(0.1792 ) = 0.0688 in 3 sec Temperature rise, ∆to ρ c∆to = 15 p (112)∆to = 15 741 ∆to = 99 o F (d) 60o Bearing L D = 1 , ho cr = 0.4 S = 0.450 rf = 3.29 cr q = 1.56 rcr ns L qs = 0.127 q ρ c∆to = 28.2 p Load W r S = cr 2 µ ns p 2 ( ) −6 1 3 × 10 (40 ) 0.450 = p 0.001 p = 267 psi W = pDL = (267 )(2)(2) = 1068 lb fhp: F = fW Page 18 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS rf = 3.29 cr 1 f = 3.29 0.001 f = 0.00329 F = f W = (0.00329)(1068) = 3.514 lb Fvm (3.514)(1257) = 0.1339 hp fhp = = 33,000 33,000 Oil flow, q q = 1.56 rcr ns L q = 1.56 (0.1)(0.001)(40)(2) q = 0.1248 in3 sec End leakage qs = 0.127 q qs = 0.127(0.1248) = 0.0158 in3 sec Temperature rise, ∆to ρ c∆to = 28.2 p (112)∆to = 28.2 267 ∆to = 67 o F 563. A 2 x 2 in. bearing sustains a load of W = 5000 lb. ; cr = 0.001 in. ; n = 2400 rpm ; µ = 3 × 10−6 reyn . Using Figs. AF 17 and AF 18, determine the minimum film thickness and the frictional loss (ft-lb/min.) for (a) a full bearing, and for partial bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o. Solution: L = 2 in D = 2 in W = 5000 lb cr = 0.001 in. n = 2400 rpm ns = 40 rps µ = 3 ×10−6 reyn r = D 2 = 1 in Page 19 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS p= W 5000 = = 1250 psi LD (2)(2) 2 2 ( ) r µns 1 3 × 10 −6 (40 ) S = = = 0.10 1250 cr p 0.001 πDn π (2)(2400) vm = = = 1257 fpm 12 12 Using Fig. AF 17 and AF 18 (a) Full Bearing ho = 0.346 cr r f = 2 .8 cr ho = 0.346(0.001) = 0.000346 in 1 f = 2 .8 0.001 f = 0.0028 F = f W = (0.0028)(5000) = 14 lb Fvm = (14)(1257 ) = 17,600 ft − lb min (b) 180o Bearing ho = 0.344 cr r f = 2 .0 cr ho = 0.344(0.001) = 0.000344 in 1 f = 2 .0 0.001 f = 0.0020 F = f W = (0.0020)(5000) = 10 lb Fvm = (10)(1257 ) = 12,570 ft − lb min (c) 120o Bearing ho = 0.302 cr Page 20 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r f = 1 .7 cr ho = 0.302(0.001) = 0.000302 in 1 f = 1 .7 0.001 f = 0.0017 F = f W = (0.0017 )(5000) = 8.5 lb Fvm = (8.5)(1257 ) = 10,685 ft − lb min (d) 60o Bearing ho = 0.20 cr r f = 1 .4 cr ho = 0.20(0.001) = 0.0002 in 1 f = 1 .4 0.001 f = 0.0014 F = f W = (0.0014)(5000) = 7 lb Fvm = (7 )(1257 ) = 8,800 ft − lb min 564. A 120o partial bearing is to support 4500 lb. with ho = 0.002 in. ; L D = 1 ; D = 4 in. ; cd = 0.010 in. ; n = 3600 rpm . Determine (a) the oil’s viscosity,(b) the frictional loss (ft-lb/min), (c) the eccentricity angle, (d) the needed oil flow, (e) the end leakage, (f) the temperature rise of the oil as it passes through, (g) the maximum pressure. (h) If the clearance given is the average, what approximate class of fit (Table 3.1) is it? (i) What maximum impulsive load would be on the bearing if the eccentricity ratio suddenly went to 0.8? Ignore “squeeze” effect. Solution: W = 4500 lb ho = 0.002 in L D =1 D = 4 in L = 4 in r = D 2 = 2 in cd = 0.010 in. n = 3600 rpm Page 21 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 3600 = 60 rps 60 πDn π (2)(3600) vm = = = 3770 fpm 12 12 W 4500 p= = = 281.25 psi LD (4 )(4) ho 2ho 2(0.002 ) = = = 0 .4 cr cr 0.010 Table AT 22, L D = 1 , ho cr = 0.4 S = 0.162 φ = 35.65o r f = 2.16 cr q = 2.24 rcr ns L qs = 0.384 q ρ c∆to = 15.0 p p = 0.356 pmax ns = r (a) S = cr 2 µns p 2 D µns S = cd p 2 4 µ (60 ) 0.162 = 0.010 281.25 µ = 4.75 ×10−6 reyn (b) r f = 2.16 cr D f = 2.16 cd 4 f = 2.16 0.010 f = 0.0054 F = f W = 0.0054(4500) = 24.30 lb Page 22 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Fvm = (24.30)(3770) = 91,611 ft − lb min (c) φ = 35.65o q 4q = = 2.24 rcr ns L Dcd ns L 4q = 2.24 (4)(0.010)(60)(4) q = 5.4 in3 sec (d) qs = 0.384 q qs = 0.384(5.4 ) = 2.07 in3 sec (e) (f) ρ c∆to = 15.0 p (112)∆to = 15.0 281.25 ∆to = 38 o F (g) p = 0.356 pmax 281.25 pmax = = 790 psi 0.356 (h) cd = 0.010 in , D = 4 in Table 3.1 RC 8, Hole, average = + 0.0025 Shaft, average = - 0.00875 cd = 0.0025 + 0.00875 = 0.01125 ≈ 0.010 in Class of fit = RC 9 (i) ε = 0.80 Table AT 22, , L D = 1 S = 0.162 2 D µ ns S = cd p 2 ( ) −6 4 3 × 10 (60 ) 0.0531 = p 0.010 p = 542 psi Page 23 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W = pDL = (542)(4)(4) = 8672 lb 565. A 120o partial bearing is to support 4500 lb., D = 3 in. , cd = 0.003 in. ; n = 3600 rpm ; SAE 20W entering at 110 F. Calculate (a) the average temperature of the oil as it passes through,(b) the minimum film thickness, (c) the fhp, (d) the quantity of oil to be supplied. HINT: In (a) assume µ and determine the corresponding values of S and ∆to ; then tav = ti + ∆to 2 . If assumed µ and tav do not locate a point in Fig. AF 16 that falls on line for SAE 20W, iterate. Solution: W = 4500 lb D = 3 in L = 3 in L D =1 cd = 0.003 in. 2 D µns S = cd p 3600 ns = = 60 rps 60 W 4500 p= = = 500 psi DL (3)(3) ρ c∆to , (SAE 20W) p (a) Using Table AT22, L D = 1 , ρ c = 112 , ti = 110o F Trial µ t , oF S 3.5 x 10-6 2.0 x 10-6 2.6 x 10-6 2.35 x 10-6 2.4 x 10-6 130 160 145 150 149 0.42 0.24 0.312 0.282 0.288 ∴ Use tav = 149 o F (b) Table AT 22, L D = 1 , S = 0.288 Page 24 of 63 ρ c∆to p 19.8 15.4 17.7 17.2 17.3 ∆to tav = ti + ∆to 2 88 68 79 76 78 154 144 149.5 148 149 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho = 0.513 cr 2ho = 0.513 cd 2ho = 0.513(0.003) ho = 0.00077 in (c) Table At 22, L D = 1 , S = 0.288 r f = 2.974 cr D 3 f = f = 2.974 cr 0.003 f = 0.002974 F = f W = (0.002974)(4500) = 13.383 lb Fvm fhp = 33,000 πDn π (3)(3600) vm = = = 2827 fpm 12 12 Fvm (13.383)(2827 ) = 1.15 hp fhp = = 33,000 33,000 (d) Table At 22, L D = 1 , S = 0.288 q = 2.528 rcr ns L 4q = 2.528 Dcd ns L 4q = 2.528 (3)(0.003)(60)(3) q = 1.024 in3 sec 566. The 6000-lb. reaction on an 8 x 4 –in., 180o partial bearing is centrally applied; n = 1000 rpm ; ho = 0.002 in . For an optimum bearing with minimum friction determine (a) the clearance, (b) the oil’s viscosity, (c) the frictional horsepower. (d) Choose a cd D ratio either smaller or larger than that obtained in (a) and show that the friction loss is greater than that in the optimum bearing. Other data remain the same. Solution: W = 6000 lb Page 25 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS D = 8 in L = 4 in n = 1000 rpm 1000 ns = = 16.67 rps 60 L D =1 2 ho = 0.002 in (a) Table AT 21, L D = 1 2 Optimum value (minimum friction) ho cr = 0.23 0.002 cr = = 0.0087 in 0.23 (b) Table AT 21, L D = 1 2 , ho cr = 0.23 S = 0.126 2 r µns S = cr p W 6000 p= = = 187.5 psi DL (4)(8) D r = = 4 in 2 2 4 µ (16.67 ) S = 0.126 = 0.0087 187.5 µ = 6.70 ×10−6 reyn (c) Table AT 21, L D = 1 2 , ho cr = 0.23 r f = 2.97 cr 4 f = 2.97 0.0087 f = 0.00646 F = f W = (0.00646 )(6000) = 38.76 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12 Page 26 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS fhp = Fvm (38.76)(2094) = 2.46 hp = 33,000 33,000 For (a) cd 2cr 2(0.0087 ) = = = 0.0022 D D 8 cd > 0.0022 D cd = 0.0030 D cd = 0.0030(8) = 0.0240 in cr = 0.0120 in ho 0.002 = = 0.1667 cr 0.012 Table AT 21, L D = 1 2 r f = 1.67 cr 4 f = 1.67 0.0016 f = 0.00668 F = f W = (0.00668)(6000 ) = 40.08 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12 (40.08)(2094) = 2.54 hp > 2.46 hp Fvm fhp = = 33,000 33,000 cd < 0.0022 D cd = 0.0020 D cd = 0.0020(8) = 0.0160 in cr = 0.0080 in ho 0.002 = = 0.25 cr 0.008 Table AT 21, L D = 1 2 r f = 3.26 cr 4 f = 3.26 0.0016 f = 0.00652 Page 27 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.00652 )(6000) = 39.12 lb πDn π (8)(1000) vm = = = 2094 fpm 12 12 Fvm (39.12)(2094) = 2.48 hp > 2.46 hp fhp = = 33,000 33,000 A 120o partial bearing supports 3500 lb. when n = 250 rpm ; D = 5 in. , L = 5 in. ; µ = 3 × 10−6 reyn . What are the clearance and minimum film thickness for an optimum bearing (a) for maximum load, (b) for minimum friction? (c) On the basis of the average clearance in Table 3.1, about what class fit is involved? Would this fit be on the expensive or inexpensive side? (d) Find the fhp for each optimum bearing. 567. Solution: D = 5 in. L = 5 in. L =1 D n = 250 rpm 250 ns = = 4.17 rps 60 µ = 3 ×10−6 reyn W = 3500 lb W 3500 p= = = 140 psi DL (5)(5) (a) Table AT 22, h L = 1 , max. load o = 0.46 D cr S = 0.229 2 r µns S = cr p D r = = 2.5 in 2 2 ( ) 2.5 3.0 × 10−6 (4.17 ) S = 0.229 = 140 cr cr = 0.00156 in ho = 0.46cr = 0.46(0.00156) = 0.00072 in Page 28 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (b) Table AT 22, L h = 1 , min. friction o = 0.40 D cr S = 0.162 2 r µns S = cr p D r = = 2.5 in 2 2 ( ) 2.5 3.0 × 10−6 (4.17 ) S = 0.162 = 140 cr cr = 0.00186 in ho = 0.46cr = 0.40(0.00186) = 0.00074 in (c) cd 1 = 2(0.00156) = 0.00312 in cd 2 = 2(0.00186) = 0.00372 in Use Class RC4, ave. cd = 0.00320 in , expensive side (d) Table AT 22, L h = 1 , max. load o = 0.46 D cr r f = 2.592 cr 2 .5 f = 2.592 0.00156 f = 0.00162 F = f W = (0.00162 )(3500 ) = 5.67 lb πDn π (5)(250) vm = = = 327.25 fpm 12 12 Fvm (5.67)(327.25) = 0.0562 hp fhp = = 33,000 33,000 For minimum friction, ho = 0.40 cr r f = 2.16 cr 2 .5 f = 2.16 0.00186 f = 0.00161 F = f W = (0.00161)(3500) = 5.635 lb Page 29 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS πDn π (5)(250) = 327.25 fpm 12 12 Fvm (5.635)(327.25) = 0.0559 hp fhp = = 33,000 33,000 vm = 570. = A 180o partial bearing is to support 17,000 lb. with p = 200 psi , n = 1500 rpm , ho = 0.003 in , L D = 1 . (a) Determine the clearance for an optimum bearing with minimum friction. (b) Taking this clearance as the average, choose a fit (Table 3.1) that is approximately suitable. (c) Select an oil for an average temperature of 150 F. (d) Compute fhp. Solution: W = 17,000 lb p = 200 psi n = 1500 rpm 1500 ns = = 25 rps 60 L D =1 L=D W p= DL 17,000 200 = D2 D = L = 9.22 in D 9.22 r= = = 4.61 in 2 2 (a) For optimum bearing with minimum friction Table AT 21, L D = 1 , ho cr = 0.44 ho cr = 0.44 0.003 = 0.44 cr cr = 0.00682 in (b) Table 3.1, D = 9.22 in cd = 2cr = 2(0.00682 ) = 0.01364 in Use Class RC7, average cd = 0.01065 in Or use Class RC8, average cd = 0.01575 in (c) Table AT 21, L D = 1 , ho cr = 0.44 S = 0.158 Page 30 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r S = cr 2 µns p 2 4.61 µ (25) 0.158 = 0.00682 200 µ = 2.8 ×10−6 reyn Fig. AF 16, at 150 F Use Either SAE 20W or SAE 30. (d) Table AT 21, L D = 1 , ho cr = 0.44 r f = 2.546 cr 4.61 f = 2.546 0.00682 f = 0.00377 πDn π (9.22)(1500) vm = = = 3621 fpm 12 12 F = f W = (0.00377 )(17,000) = 64.09 lb Fvm (64.09)(3621) = 7.0 hp fhp = = 33,000 33,000 571. The reaction on a 120o partial bearing is 2000 lb. The 3-in journal turns at 1140 rpm; cd = 0.003 in. ; the oil is SAE 20W at an average operating temperature of 150 F. Plot curves for the minimum film thickness and the frictional loss in the bearing against the ratio L D , using L D = 0.25, 0.5, 1, and 2. (Note: This problem may be worked as a class problem with each student being responsible for a particular L D ratio.) Solution: W = 2000 lb D = 3 in. n = 1140 rpm 1140 ns = = 19 rps 60 cd = 0.003 in cr = 0.0015 in For SAE 20W, 150 F µ = 2.75 ×10−6 reyn (a) L = 0.25 D Page 31 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS L = 0.25D = 0.25(3) = 0.75 in W 2000 p= = = 889 psi DL (3)(0.75) L Table AT 22, = 0.25 D D r = = 1.5 in 2 2 2 ( ) r µns 1.5 2.75 ×10 −6 (19 ) S = = = 0.0588 c p 0 . 0015 889 r ho = 0.083 cr ho = 0.083(0.0015) = 0.000125 in r f = 2.193 cr 1 .5 f = 2.193 0.0015 f = 0.002193 F = f W = (0.002193)(2000) = 4.386 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (4.386)(895) = 0.119 hp fhp = = 33,000 33,000 L (b) = 0 .5 D L = 0.5 D = 0.5(3) = 1.5 in W 2000 p= = = 444 psi DL (3)(1.5) L Table AT 22, = 0 .5 D D r = = 1.5 in 2 2 2 ( ) r µns 1.5 2.75 × 10−6 (19 ) S = = = 0.1177 444 cr p 0.0015 ho = 0.2159 cr ho = 0.2159(0.0015) = 0.000324 in r f = 2.35 cr Page 32 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 1 .5 f = 2.35 0.0015 f = 0.00235 F = f W = (0.00235)(2000) = 4.7 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (4.7 )(895) = 0.1275 hp fhp = = 33,000 33,000 L (c) =1 D L = D = 3 in W 2000 p= = = 222 psi DL (3)(3) L Table AT 22, =1 D D r = = 1.5 in 2 2 2 ( ) r µns 1.5 2.75 × 10−6 (19 ) S = = = 0.2354 222 cr p 0.0015 ho = 0.4658 cr ho = 0.4658(0.0015) = 0.000699 in r f = 2.634 cr 1 .5 f = 2.634 0.0015 f = 0.002634 F = f W = (0.002634)(2000 ) = 5.268 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (5.268)(895) = 0.1429 hp fhp = = 33,000 33,000 L (d) =2 D L = 2 D = 2(3) = 6 in W 2000 p= = = 111 psi DL (3)(6) L Table AT 22, =2 D Page 33 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r= D = 1.5 in 2 2 2 ( ) r µns 1.5 2.75 × 10−6 (19 ) S = = = 0.47 111 cr p 0.0015 ho = 0.718 cr ho = 0.718(0.0015) = 0.00108 in r f = 3.8118 cr 1 .5 f = 3.8118 0.0015 f = 0.003812 F = f W = (0.003812)(2000) = 7.624 lb πDn π (3)(1140) vm = = = 895 fpm 12 12 Fvm (7.624)(895) = 0.2068 hp fhp = = 33,000 33,000 L D 0.25 0.5 1.0 2.0 Page 34 of 63 ho , in fhp 0.000125 0.000324 0.000699 0.001080 0.119 0.128 0.143 0.207 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS STEADY-STATE TEMPERATURE A 180o partial bearing is subjected to a load of 12,000 lb.; D × L = 8 × 8 in. , cr r = 0.0015 , ho ≈ 0.0024 in. , n = 500 rpm . The air speed about the bearing is expected to be in excess of 1000 fpm (on moving vehicle) and the effective radiating area is 20 DL . Determine: (a) the eccentricity factor, (b) µreyns, (c) the frictional loss (ft-lb/min), (d) the estimated temperature of oil and bearing ( a self-contained oil-bath unit) for steady-state operation, and a suitable oil.(e) Compute ∆to of the oil passing through the load-carrying area, remark on its reasonableness, and decide upon whether some redesign is desirable. 572. Solution: D = 8 in. L = 8 in. L D =1 W = 12,000 lb D r = = 4 in 2 cr = 0.0015r = 0.0015(4) = 0.0060 in ho 0.0024 = = 0. 4 cr 0.0060 n = 500 rpm 500 ns = = 8.33 rps 60 Table AT 21, ho cr = 0.4 , L D = 1 S = 0.128 r f = 2.28 cr ρ c∆to = 12.4 p W 12,000 p= = = 187.5 psi DL (8)(8) (a) ε = 1 − ho = 1 − 0 .4 = 0 .6 cr r (b) S = cr 2 µns p 2 4 µ (8.33) S = 0.128 = 0.0060 187.5 µ = 6.5 ×10−6 reyn Page 35 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (c) r f = 2.28 cr 4 f = 2.28 0.0060 f = 0.00342 F = f W = (0.00342 )(12,000) = 41.04 lb πDn π (8)(500) vm = = = 1047 fpm 12 12 Fvm (41.04)(1047) = 1.302 hp fhp = = 33,000 33,000 Frictional loss = 43,000 ft-lb/min (d) Q = hcr Ab ∆tb ft-lb/min Q = 43,000 ft − lb min hcr = hc + hr hr = 0.108 ft − lb min − sq.in. − F va0.6 , va ≥ 1000 fpm D 0.4 0.6 ( 1000 ) hc = 0.017 = 0.467 ft − lb min − sq.in. − F (8)0.4 hcr = 0.467 + 0.108 = 0.575 ft − lb min − sq.in. − F Ab = 20 DL = 20(8)(8) = 1280 sq.in. Q = hcr Ab ∆tb 43,000 = (0.575)(1280)(∆tb ) ∆tb = 58.42 F Oil-bath, 1000 fpm ∆toa ≈ (1.2)(1.3)(∆tb ) ∆toa = (1.2)(1.3)(58.42) = 91.1 F assume 100 F ambient temperature tb = 100 + 58.42 F = 158.42 F tb = 100 + 91.1 F = 191.1 F hc = 0.017 (c) ρ c∆to = 12.4 p (112)∆to = 12.4 187.5 ∆to = 20.8 F Solve for to 2 Page 36 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS to1 + to 2 = 2(191.1) = 382.2 F to1 = 382.2 − to 2 to 2 − to1 = 20.8 F to 2 − 362.2 + to 2 = 20.8 to 2 = 201.5 F ≈ 200 F ∴ not reasonable since the oil oxidizes more rapidly above 200 F, a redesign is desireable. 573. A 2 x 2-in. full bearing (ring-oiled) has a clearance ratio cd D = 0.001 . The journal speed is 500 rpm, µ = 3.4 × 10−6 reyn , and ho = 0.0005 in. The ambient temperature is 100 F; Ab = 25DL , and the transmittance is taken as hcr = 2 Btu hr − sq. ft. − F . Calculate (a) the total load for this condition; (b) the frictional loss, (c) the average temperature of the oil for steady-state operation. Is this temperature satisfactory? (d) For the temperature found, what oil do you recommend? For this oil will ho be less or greater than the specified value? (e) Compute the temperature rise of the oil as it passes through the bearing. Is this compatible with other temperatures found? (f) What minimum quantity of oil should the ring deliver to the bearing? Solution: L = 2 in. D = 2 in. cd D = 0.001 cd = (0.001)(2) = 0.0020 in µ = 3.4 ×10−6 reyn ho = 0.0005 in. cr = 0.0010 in ho cr = 0.0005 0.0010 = 0.5 Table AT 20, L D = 1 , ho cr = 0.5 , Full Bearing S = 0.1925 r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p (a) S = 0.1925 Page 37 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 2 r µns S = cr p D r = = 1 in 2 500 ns = = 8.33 rps 60 2 ( ) −6 1 3.4 × 10 (8.330 ) S = 0.1925 = p 0.0010 p = 147 psi W = pDL = (147 )(2)(2) = 588 lb (b) r f = 4.505 cr 1 f = 4.505 0.001 f = 0.004505 F = f W = (0.004505)(588) = 2.649 lb πDn π (2)(500) vm = = = 261.8 fpm 12 12 U f = Fvm = (2.649 )(261.8) = 693.5 ft − lb min (c) Q = hcr Ab ∆tb hcr = 2 Btu hr − sq. ft. − F = 0.18 ft − lb min − sq.in. − F Ab = 25DL = 25(2)(2) = 100 sq.in. Q =Uf (0.18)(100)(∆tb ) = 693.5 ∆tb = 38.53 F ∆toa = 2∆tb = 2(38.53) = 77 F to = 77 + 100 = 177 F , near 160 F ∴ satisfactory. (d) to = 177 F , µ = 3.4 × 10−6 reyn Figure AF 16 Use SAE 40 oil, µ = 3.3 × 10−6 reyn r S = cr 2 µns p Page 38 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 2 ( ) −6 1 3.3 × 10 (8.33) S = = 0.187 147 0.0010 Table AT 20, L D = 1 , S = 0.187 ho cr = 0.4923 ho = 0.4923(0.0010) = 0.00049 in < ho (= 0.0005 in ) ρ c∆to (e) = 19.25 p (112)∆to = 19.25 147 ∆to = 25.3 F ∆to1 + ∆to 2 = 2(177 ) = 354 F ∆t o 2 − ∆t o1 = 25.3 F 2∆to 2 = 354 + 25.3 ∆to 2 = 190 F < 200 F ∴ compatible. q = 4.16 rcr ns L q = 4.16 (1)(0.001)(8.33)(2) q = 0.0693 in3 sec (f) 574. An 8 x 9-in. full bearing (consider L D = 1 for table and chart use only) supports 15 kips with n = 1200 rpm ; cr r = 0.0012 ; construction is medium heavy with a radiating-and-convecting area of about 18 DL ; air flow about the bearing of 80 fpm may be counted on (nearby) pulley; ambient temperature is 90 F. Decide upon a suitable minimum film thickness. (a) Compute the frictional loss and the steady state temperature. Is additional cooling needed for a reasonable temperature? Determine (b) the temperature rise of the oil as it passes through the load-carrying area and the grade of oil to be used if it enters the bearing at 130 F, (c) the quantity of oil needed. Solution: D = 8 in. L = 9 in. W = 15,000 lb. n = 1200 rpm. 1200 ns = = 20 rps 60 cr r = 0.0012 Page 39 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r = D 2 = 4 in cr = 0.0012(4) = 0.0048 in By Norton: ho = 0.00025 D = 0.00025(8) = 0.002 in ho 0.002 = = 0 .4 cr 0.0048 Table AT 20, L D = 1 , ho cr = 0.4 S = 0.121 r f = 3.22 cr q = 4.33 rcr ns L ρ c∆to = 14.2 p (a) r f = 3.22 cr 4 f = 3.22 0.0048 f = 0.003864 F = f W = (0.003864)(15,000) = 57.96 lb πDn π (8)(1200) vm = = = 2513 fpm 12 12 U f = Fvm = (57.96 )(2513) = 145,654 ft − lb min Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F va0.6 ft − lb min − sq.in. − F D 0.4 (80)0.6 = 0.103 ft − lb min − sq.in. − F hc = 0.017 (8)0.4 hcr = hc + hr = 0.103 + 0.108 = 0.211 ft − lb min − sq.in. − F Ab = 18DL = 18(8)(9 ) = 1296 sq.in. Uf =Q hc = 0.017 145,654 = (0.211)(1296)∆tb ∆tb = 533 F , very high, additional cooling is necessary. (b) ρ c∆to p = 14.2 Page 40 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS W 15,000 = = 208 psi DL (8)(9) (112)∆to = 14.2 208 ∆to = 26 F ti = 130 F to = 156 F tave = 12 (130 + 156) = 143 F p= r S = cr 2 µns p 2 4 µ (20 ) S = 0.121 = 0.0048 208 µ = 1.8 ×10−6 reyn Figure AF 16, µ = 1.8 µreyns , 143 F Use SAE 10W q = 4.33 rcr ns L q = 4.33 (4)(0.0048)(20)(9) q = 14.96 in3 sec (c) 575. A 3.5 x 3.5-in., 360o bearing has cr r = 0.0012 ; n = 300 rpm ; desired minimum ho ≈ 0.0007 in . It is desired that the bearing be self-contained (oilring); air-circulation of 80 fpm is expected; heavy construction, so that Ab ≈ 25DL . For the first look at the bearing, assume µ = 2.8 × 10−6 reyn and compute (a) the frictional loss (ft-lb/min), (b) the average temperature of the bearing and oil as obtained for steady-state operation, (c) ∆to as the oil passes through the load-carrying area (noting whether comparative values are reasonable). (d) Select an oil for the steady-state temperature and decide whether there will be any overheating troubles. Solution: D = 3.5 in. L = 3.5 in. cr r = 0.0012 r = D 2 = 1.75 in. cr = (0.0012)(1.75) = 0.0021 in ho ≈ 0.0007 in Page 41 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho cr = 0.0007 0.0021 = 0.333 Table AT 20, 360o Bearing, L D = 1 , ho cr = 0.333 S = 0.0954 r f = 2.71 cr ρ c∆to = 12.12 p 2 r µns (a) S = cr p 300 ns = = 5 rps 60 µ = 2.8 ×10−6 reyn 2 ( ) −6 1.75 2.8 × 10 (5) S = 0.0954 = p 0.0021 p = 102 psi W = pDL = (102)(3.5)(3.5) = 1250 lb r f = 2.71 cr 1.75 f = 2.71 0.0021 f = 0.00325 F = f W = (0.00325)(1250) = 4.0625 lb πDn π (3.5)(300 ) vm = = = 275 fpm 12 12 U f = Fvm = (4.0625)(275) = 1117 ft − lb min (b) Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F va0.6 ft − lb min − sq.in. − F D 0.4 0.6 ( 80 ) hc = 0.017 = 0.143 ft − lb min − sq.in. − F (3.5)0.4 hcr = hc + hr = 0.143 + 0.108 = 0.251 ft − lb min − sq.in. − F Ab = 25DL = 25(3.5)(3.5) = 306.25 sq.in. Uf =Q hc = 0.017 1117 = (0.251)(306.25)∆tb Page 42 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ∆tb = 14.5 F ∆toa = 2∆tb = 2(14.5) = 29 F assume ambient temperature of 100 F tb = 114.5 F to = 129 F (c) ρ c∆to = 12.12 p (112)∆to = 12.12 102 ∆to = 11 F to1 + to 2 = 2(129) = 258 F to 2 − to1 = 11 F 2to 2 = 269 F to 2 = 135 F < 140 F ∴ reasonable (d) to = 129 F , µ = 2.8 × 10−6 reyn use SAE 10W Figure AF 16, to = 126 F ∆toa = 126 − 100 = 26 F ∆toa = 2∆tb 26 ∆tb = = 13 F 2 Q = hcr Ab ∆tb = (0.251)(306.25)(13) = 999 ft − lb min < U f ∴ there is an overheating problem. 576. A 10-in. full journal for a steam-turbine rotor that turns 3600 rpm supports a 20-kip load with p = 200 psi ; cr r = 0.00133 . The oil is to have µ = 2.06 ×10−6 reyn at an average oil temperature of 130 F. Compute (a) the minimum film thickness (comment on its adequacy), (b) the fhp, (c) the altitude angle, the maximum pressure, and the quantity of oil that passes through the load-carrying area (gpm).(d) At what temperature must the oil be introduced in order to have 130 F average? (e) Estimate the amount of heat lost by natural means from the bearing (considered oil bath) with air speed of 300 fpm. If the amount of oil flow computed above is cooled back to the entering temperature, how much heat is removed? Is this total amount of heat enough to care for frictional loss? If not, what can be done (i11.21)? Page 43 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Solution: D = 10 in. n = 3600 rpm 3600 ns = = 60 rps 60 W = 20,000 lb p = 200 psi W p= DL 20,000 200 = 10 L L = 10 in L D =1 D r = = 5 in 2 cr r = 0.00133 cr = 0.00133(5) = 0.00665 in µ = 2.06 ×10−6 reyn tave = 130 F r S = cr 2 µns p 2 ( ) −6 5 2.06 × 10 (60 ) S = = 0.35 200 0.00665 Table AT 20, L D = 1 , S = 0.35 ho cr = 0.647 φ = 65.66o r f = 7.433 cr q = 3.90 rcr ns L p = 0.495 pmax ρ c∆to = 30.8 p qs = 0.446 q (a) ho = 0.647cr = 0.647(0.00665) = 0.00430 in Page 44 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Norton’s recommendation = 0.00025D = 0.00025(10) = 0.00250 in < 0.00430 in ∴ adequate (b) r f = 7.433 cr 5 f = 7.433 0 . 00665 f = 0.0099 F = f W = (0.0099)(20,000) = 198 lb πDn π (10)(3600) vm = = = 9425 fpm 12 12 Fvm (198)(9425) = 56.55 hp fhp = = 33,000 33,000 (c) φ = 65.66o p 200 pmax = = = 404 psi 0.495 0.495 q = 3.90rcr ns L q = 3.90(5)(0.00665)(60)(10) = 77.805 in3 sec q = (77.805 in3 sec )(1 gpm 231 in3 )(60 sec min ) = 0.21 gpm (d) ρ c∆to = 30.8 p (112)∆to = 30.8 200 ∆to = 55 F ∆t tave = ti + o 2 55 130 = ti + 2 ti = 102.5 F (e) Q = hcr Ab ∆tb hr = 0.108 ft − lb min − sq.in. − F va0.6 ft − lb min − sq.in. − F D 0.4 0.6 ( 300 ) hc = 0.017 = 0.207 ft − lb min − sq.in. − F (3.5)0.4 hcr = hc + hr = 0.207 + 0.108 = 0.315 ft − lb min − sq.in. − F Assume hc = 0.017 Page 45 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Ab = 25DL = 25(10 )(10) = 2500 sq.in. ∆toa = 130 − 100 = 30 F ∆toa = 1.3∆tb 30 ∆tb = = 23 F 1 .3 Q = (0.315)(2500)(23) = 18,113 ft − lb min Qr = ρ c(q − qs )∆to in − lb sec Qr = (112 )(1 − 0.446)(77.805)(55)(1 12)(60) = 1,327,602 ft − lb min QT = Q + Qr = 18,113 + 1,327,602 = 1,345,735 ft − lb min U f = Fvm = (198)(9425) = 1,866,150 ft − lb min > QT not enough to care for frictional loss, use pressure feed (i11.21). DESIGN PROBLEMS 578. A 3.5-in. full bearing on an air compressor is to be designed for a load of 1500 lb.; n = 300 rpm ; let L D = 1 . Probably a medium running for would be satisfactory. Design for an average clearance that is decided by considering both Table 3.1 and 11.1. Choose a reasonable ho , say one that gives ho cr ≈ 0.5 . Compute all parameters that are available via the Text after you have decided on details. It is desired that the bearing operate at a reasonable steady-state temperature (perhaps ring-oiled medium construction), without special cooling. Specify the oil to be used and show all calculations to support your conclusions. What could be the magnitude of the maximum impulsive load if the eccentricity ration ε becomes 0.8, “squeeze” effect ignored? Solution: L D =1 D = 3.5 in L = 3.5 in W = 1500 lb n = 300 rpm 300 ns = = 5 rps 60 W 1500 p= = = 122.45 psi DL (3.5)(3.5) Table 3.1, medium running fit, D = 3.5 in RC 5 or RC 6 Use RC 6 Average cd = 0.0052 in Page 46 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Table 11.1, air-compressor General Machine Practice Average cd = 0.0055 in Using cd = 0.0055 in cr = 0.00275 in ho = 0.5cr = 0.5(0.00275) = 0.001375 in Table AT 20, L D = 1 , ho cr = 0.5 ε = 0.5 S = 0.1925 φ = 56.84o r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p p = 0.4995 pmax Specifying oil: Q = hcr Ab ∆tb U f = Fvm r f = 4.505 cr 1.75 f = 4.505 0.00275 f = 0.00708 F = f W = (0.00708)(1500) = 10.62 lb πDn π (3.5)(300 ) vm = = = 275 fpm 12 12 U f = Fvm = (10.62 )(275) = 2921 ft − lb min Q = hcr Ab ∆tb Assume hcr = 0.516 ft − lb min − sq.in. − F Medium construction Ab = 15.5DL = 15.5(3.5)(3.5) = 189.875 sq.in. Oil-ring bearing ∆toa = 2∆tb Q =Uf (0.516)(189.875)(∆tb ) = 2921 Page 47 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ∆tb = 30 F ∆toa = 2∆tb = 2(30) = 60 F assume ambient temperature = 90 F t o = 150 F r S = cr 2 µns p 2 1.75 µ (5) S = 0.1925 = 0.00275 122.45 µ = 11.6 ×10−6 reyn Figure AF 16, 150 F, µ ≈ 11.6 × 10−6 reyn Use SAE 70 oil Maximum load, W with ε = 0.8 Table AT 20, L D = 1 S = 0.0446 r S = cr 2 µns p 2 ( ) −6 1.75 11.6 × 10 (5) S = 0.0446 = p 0.00275 p = 527 psi W = pDL = (527 )(3.5)(3.5) = 6456 580. A 2500-kva generator, driven by a water wheel, operates at 900 rpm. The weight of the rotor and shaft is 15,100 lb. The left-hand, 5 –in, full bearing supports the larger load, R = 8920 lb . The bearing should be above medium-heavy construction (for estimating Ab ). (a) Decide upon an average clearance considering both Table 3.1 and 11.1, and upon a minimum film thickness ( ho cr ≈ 0.5 is on the safer side). (b) Investigate first the possibility of the bearing being a self-contained unit without need of special cooling. Not much air movement about the bearing is expected. Then make final decisions concerning oil-clearance, and film thickness and compute all the parameters given in the text, being sure that everything is reasonable. Solution: n = 900 rpm 900 ns = = 15 rps 60 D = 5 in W = R = 8920 lb Page 48 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (a) Table 3.1, D = 5 in RC 5, average cd = 0.0051 in cr = 0.00255 in ho = 0.5cr = 0.5(0.00255) = 0.00128 in (b) Use L D = 1 L = 5 in D r = = 2.5 in 2 W 8920 p= = = 356.8 psi DL (5)(5) Table AT 20, L D = 1 , ho cr = 0.5 S = 0.1925 r f = 4.505 cr q = 4.16 rcr ns L ρ c∆to = 19.25 p r S = cr 2 µns p 2 2.5 µ (15) S = 0.1925 = 0.00255 356.8 µ = 4.8 ×10−6 reyn r f = 4.505 cr 2 .5 f = 4.505 0.00255 f = 0.00460 F = f W = (0.00460 )(8920) = 41.032 lb πDn π (5)(900) vm = = = 1178 fpm 12 12 U f = Fvm = (41.032 )(1178) = 48,336 ft − lb min Q = hcr Ab ∆tb Medium-Heavy Ab = 20.25DL = 20.25(5)(5) = 506.25 sq.in. Assume hcr = 0.516 ft − lb min − sq.in. − F Q =Uf Page 49 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS (0.516)(506.25)(∆tb ) = 48,336 ∆tb = 185 F , very high Therefore, special cooling is needed. ρ c∆to = 19.25 p (112)∆to = 19.25 356.8 ∆to = 61 F Assume ti = 100 F 61 tave = 100 + ≈ 130 F 2 Figure AF 16, µ = 4.8 µreyns , 130 F Select SAE 30 oil. µ = 6.0 µreyns r S = cr 2 µns p 2 ( ) −6 2.5 6.0 × 10 (15) S = = 0.242 356.8 0.00255 Table AT 20, L D = 1 , S = 0.242 SAE 30 oil at 130 F ho = 0.569 cr φ = 61.17o r f = 5.395 cr q = 4.04 rcr ns L ρ c∆to = 22.75 p p = 0.4734 pmax Oil, SAE 30 cr = 0.00255 in ho = 0.569(0.00255) = 0.00145 in PRESSURE FEED 581. An 8 x 8-in. full bearing supports 5 kips at 600 rpm of the journal; cr = 0.006 in. ; let the average µ = 2.5 × 10−6 reyn . (a) Compute the frictional loss U f . (b) The Page 50 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS oil is supplied under a 40-psi gage pressure with a longitudinal groove at the point of entry. Assuming that other factors, including U f , remain the same and that the heat loss to the surrounding is negligible, determine the average temperature rise of the circulating oil. Solution: L = 5 in D = 5 in W = 5000 lb n = 600 rpm 600 ns = = 10 rps 60 cr = 0.006 in µ = 2.5 ×10−6 reyn L D =1 W 5000 p= = = 78.125 psi DL (8)(8) r S = cr 2 µns p 2 ( ) −6 4 2.5 ×10 (10 ) S = = 0.1422 78.125 0.006 (a) Table AT 20, L D = 1 , S = 0.1422 r f = 3.6 , ε = 0.57 cr 4 f = 3 .6 0.006 f = 0.0054 F = f W = (0.0054)(5000) = 27 lb πDn π (8)(600) vm = = = 1257 fpm 12 12 U f = Fvm = (27 )(1257 ) = 33,940 ft − lb min (b) Longitudinal Groove. c3 p 2π r 2 3 q = 2.5 r i tan −1 1 + 1.5ε in sec 3µ L pi = 40 psi ( ) 3 ( 0.006) (40) −1 2π (4) q = 2.5 tan [1 + 1.5(0.57)2 ]in3 3(2.5 × 10−6 ) Page 51 of 63 8 sec SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS q = 5.41 in3 sec U f = ρ cq∆to (33,940 ft − lb min )(12 in ft )(1min 60sec ) = (12)(5.41)∆to ∆to = 11.2 F 583. A 4-in. 360o bearing, with L D = 1 , supports 2.5 kips with a minimum film of ho = 0.0008 in. , cd = 0.01 in. , n = 600 rpm. The average temperature rise of the oil is to be about 25 F. Compute the pressure at which oil should be pumped into the bearing if (a) all bearing surfaces are smooth, (b) there is a longitudinal groove at the oil-hole inlet. (c) same as (a) except that there is a 360o circumferential groove dividing the bearing into 2-in. lengths. Solution: D = 4 in L = 4 in r = 2 in W = 2500 lb cd = 0.010 in cr = 0.005 in n = 600 rpm 600 ns = = 10 rps 60 ∆to = 25 F W 2500 p= = = 156.25 psi DL (4)(4) ho = 0.00080 in ho 0.0008 = = 0.16 cr 0.005 Table AT 20, L D = 1 , ho cr = 0.16 r f = 1.44 , ε = 0.84 cr 2 f = 1.44 0.005 f = 0.0036 F = f W = (0.0036)(2500) = 9 lb πDn π (4)(600) vm = = = 628 fpm 12 12 U f = Fvm = (9 )(628) = 5652 ft − lb min = 1130 in − lb sec S = 0.0343 Page 52 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS r S = cr 2 µns p 2 2 µ (10 ) S = 0.0343 = 0.005 156.25 µ = 3.35 ×10−6 reyn U f = ρ cq∆to 1130 = (112)(q )(25) q = 0.404 in3 sec (a) Smooth q= cr3 pi −1 2π r 2 3 tan 1 + 1.5ε in sec 3µ L ( 0.404 = ) (0.005)3 ( pi ) tan −1 2π (2) [1 + 1.5(0.84)2 ] in3 3(3.35 × 10−6 ) pi = 12.5 psi (b) Longitudinal groove q= 4 sec 2.5cr3 pi −1 2π r 2 3 tan 1 + 1.5ε in sec 3µ L ( 3 ) 2.5(0.005) ( pi ) −1 2π (2) 2 tan 1 + 1.5(0.84) in3 sec −6 3 3.35 × 10 4 pi = 5 psi (c) Circumferential groove 0.404 = ( ) [ ] 2π rcr3 pi q= 1 + 1.5ε 2 in3 sec 3µL ( ) 3 2π (2)(0.005) ( pi ) 2 0.404 = 1 + 1.5(0.84) in3 sec −6 3 3.35 × 10 (4 ) pi = 5 psi ( ) [ ] BEARING CAPS 584. An 8-in. journal, supported on a 150o partial bearing, is turning at 500 rpm; bearing length = 10.5 in., c d = 0.0035 in ., ho = 0.00106 in . The average temperature of the SAE 20 oil is 170 F. Estimate the frictional loss in a 160o cap for this bearing. Solution: Page 53 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ho = 0.00106 in c d = 0.0035 in c r = 0.00175 in 2 ho hav = cr 1 + 0.741 − in cr 2 0.00106 hav = (0.00175)1 + 0.741 − = 0.00195 in 0.00175 For SAE 20, 170 F µ = 1.7 ×10−6 reyn µAvips F= hav 1 A = θDL 2 D = 8 in L = 10.5 in 160 8π θ = 160o = π= 180 9 1 8π A = (8)(10.5) = 117.3 sq.in. 2 9 500 vips = π Dns = π (8) = 209.5 ips 60 1.7 × 10−6 (117.3)(209.5) = 21.424 lb F= 0.00195 πDn π (8)(500) vm = = = 1047 fpm 12 12 U f = Fvm = (21.424 )(1047 ) = 22,430 ft − lb min = 1130 in − lb sec ( 585. ) A partial 160o bearing has a 160o L = 2 in ., cd = 0.002 in ., ho = 0.0007 in ., n = 500 rpm , and For the cap only, what is the frictional loss? Solution: c d = 0.002 in c r = 0.001 in ho = 0.0007 in ho 0.0007 = = 0 .7 cr 0.001 Page 54 of 63 cap; D = 2 in ., µ = 2.5 ×10−6 reyn . SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS ε = 1− ho = 1 − 0 .7 = 0 .3 cr ( [ ) 2 ] hav = cr 1 + 0.74ε 2 = (0.001) 1 + 0.74(0.3) = 0.001067 in µAvips F= hav πDn π (2)(500) vm = = = 261.8 fpm 12 12 12 vips = (261.8) = 52.36 ips 60 1 160 1 160 A= πDL = (π )(2 )(2 ) = 5.585 sq.in. 2 180 2 180 (2.5 ×10−6 )(5.585)(52.36) = 0.685 lb F= 0.001067 U f = Fvm = (0.685)(261.8) = 179.3 ft − lb min 586. The central reaction on a 120o partial bearing is 10 kips; D = 8 in ., L D = 1 ., cr r = 0.001 . Let n = 400 rpm and µ = 3.4 ×10−6 reyn . The bearing has a 150o cap. (a) For the bearing and the cap, compute the total frictional loss by adding the loss in the cap to that in the bearing. (b) If the bearing were 360o, instead of partial, calculate the frictional loss and compare. Solution: 2 r µns S = cr p 400 ns = = 6.67 rps 60 W 10,000 p= = = 156.25 psi DL (8)(8) 2 ( ) −6 1 3.4 × 10 (6.67 ) S = = 0.145 156.25 0.001 (a) Table AT 22, L D = 1 , S = 0.145 r f = 2.021 cr ε = 0.6367 r f = 2.021 cr 1 f = 2.021 0.001 f = 0.002021 Page 55 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.002021)(10,000) = 20.21 lb πDn π (8)(400) vm = = = 838 fpm 12 12 U f 1 = Fvm = (20.21)(838) = 16,936 ft − lb min CAP: hav = cr (1 + 0.74ε 2 ) cr = 0.001r D r = = 4 in 2 cr = 0.001(4) = 0.004 in ( ) [ 2 ] hav = cr 1 + 0.74ε 2 = (0.004) 1 + 0.74(0.6367 ) = 0.0052 in µAvips F= hav 12 vips = (838) = 167.6 ips 60 1 150 1 150 A= πDL = (π )(8)(8) = 83.78 sq.in. 2 180 2 180 3.4 ×10−6 (83.78)(167.6) F= = 9.18 lb 0.0052 U f 2 = Fvm = (9.18)(838) = 7693 ft − lb min Total Frictional Loss = U f 1 + U f 2 = 16,936 + 7693 = 24,629 ft − lb min ( ) (b) 360o Bearing, L D = 1 , S = 0.145 r f = 3.65 cr ε = 0.5664 BEARING: 1 f = 3.65 0.001 f = 0.00365 F = f W = (0.00365)(10,000) = 36.5 lb πDn π (8)(400) vm = = = 838 fpm 12 12 U f 1 = Fvm = (36.5)(838) = 30,587 ft − lb min CAP: hav = cr (1 + 0.74ε 2 ) [ ] hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5664) = 0.00495 in Page 56 of 63 2 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F= µAvips hav (3.4 ×10 )(83.78)(167.6) = 9.645 lb F= −6 0.00495 U f 2 = Fvm = (9.645)(838) = 8083 ft − lb min Total Frictional Loss = U f 1 + U f 2 = 30,587 + 8083 = 38,670 ft − lb min 587. The central reaction on a 120o partial bearing is a 10 kips; D = 8 in. , L D = 1 , cr r = 0.001 ; n = 1200 rpm . Let µ = 2.5 ×10−6 reyn . The bearing has a 160o cap. (a) Compute ho and fhp for the bearing and for the cap to get the total fhp. (b) Calculate the fhp for a full bearing of the same dimensions and compare. Determine (c) the needed rate of flow into the bearing, (d) the side leakage qs . (e) the temperature rise of the oil in the bearing both by equation (o), i11.13, Text, and by Table AT 22. (f) What is the heat loss from the bearing if the oil temperature is 180 F? Is the natural heat loss enough to cool the bearing? (g) It is desired to pump oil through the bearing with a temperature rise of 12 F. How much oil is required? (h) For the oil temperature in (f), what is a suitable oil to use? Solution: 2 r µns S = cr p 1200 ns = = 20 rps 60 W 10,000 p= = = 156.25 psi DL (8)(8) −6 1 (2.5 ×10 )(20 ) S = = 0.32 156.25 0.001 (a) Table AT 22, L D = 1 , S = 0.32 ε = 0.5417 ho = 0.4583 cr r f = 3.18 cr q = 2.60 rcr ns L 2 Page 57 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 0.305 q ρ c∆to = 17.834 p p = 0.38434 pmax ho = 0.4583cr = 0.4583(0.001)(4) = 0.00183 in BEARING: r f = 3.18 cr 1 f = 3.18 0.001 f = 0.00318 F = f W = (0.00318)(10,000) = 31.8 lb πDn π (8)(1200) vm = = = 2513 fpm 12 12 U f 1 = Fvm = (31.8)(2513) = 79,913 ft − lb min , 2.42 hp CAP: hav = cr (1 + 0.74ε 2 ) cr = 0.001r D r = = 4 in 2 cr = 0.001(4) = 0.004 in [ ] hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5417 ) = 0.00487 in µAvips F= hav 2 12 vips = (2513) = 503 ips 60 1 160 1 160 A= πDL = (π )(8)(8) = 89.36 sq.in. 2 180 2 180 2.5 ×10−6 (89.36)(5036) F= = 23.1 lb 0.00487 U f 2 = Fvm = (23.1)(2513) = 58,050 ft − lb min , 1.76 hp Total Frictional Loss = U f 1 + U f 2 = 79,913 + 58,050 = 137,963 ft − lb min ( ) Uf 137,963 = 4.18 hp 33,000 33,000 (b) Full Bearing, L D = 1 , S = 0.32 fhp = Page 58 of 63 = SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS Table AT 20 ho = 0.6305 cr r f = 6.86 cr ε = 0.3695 ho = 0.6305(0.004 ) = 0.002522 in BEARING: r f = 6.86 cr 1 f = 6.86 0.001 f = 0.00686 F = f W = (0.00686)(10,000) = 68.6 lb U f 1 = Fvm = (68.6 )(2513) = 172,392 ft − lb min , 5.224 hp CAP: hav = cr (1 + 0.74ε 2 ) [ ] hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.3695) = 0.00440 in µAvips F= hav (2.5 ×10 )(89.36)(503) = 25.54 lb F= 2 −6 0.00440 U f 2 = Fvm = (25.54 )(2513) = 64,182 ft − lb min , 1.946 hp Total Frictional Loss = U f 1 + U f 2 = 172,392 + 64,182 = 236,574 ft − lb min fhp = Uf 33,000 = 236,574 = 7.17 hp 33,000 (c) 120o Bearing q = 2.60 rcr ns L q = 2.60 (4)(0.004)(20)(8) q = 6.656 in3 sec q (d) s = 0.305 q qs = 0.305 6.656 Page 59 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS qs = 2.03 in 3 sec (e) Equation(o) U f 1 = ρ cq∆to 12 U f 1 = 79,913 ft − lb min = 79,913 in − lb sec = 15,983 in − lb sec 60 U f 1 = 15,983 = (112 )(6.656 )∆to ∆to = 21.4 F Table 22. ρ c∆to = 17.834 p 112∆to = 17.834 156.25 ∆to = 24.9 F (f) Q = hcr Ab ∆tb assume hcr = 0.516 ft − lb min − sq.in. − F Ab = 25DL = 25(8)(8) = 1600 sq.in. ∆t ∆tb = oa 2 assume ambient = 100 F 180 − 100 ∆tb = = 40 F 2 Q = (0.516 )(1600 )(40 ) = 33,024 ft − lb min < U f 1 Therefore not enough to cool the bearing. (g) Qr + Q = U f 1 + U f 2 Qr + 33,024 = 137,963 Qr = 104,939 ft − lb min Qr = 20,988 in − lb sec Qr = ρ cq∆to 20,988 = (112)q(12) q = 15.62 in3 sec (h) Fig. AF 16, 180 F, µ = 2.5 ×10−6 reyn use SAE 30 oil Page 60 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS IMPERFECT LUBRICATION: 588. A 0.5 x 0.75-in. journal turns at 1140 rpm. What maximum load may be supported and what is the frictional loss if the bearing is (a) SAE Type I, bronze base, sintered bearing, (b) nylon (Zytel) water lubricated, (c) Teflon, with intermittent use, (d) one with carbon graphite inserts. Solution: (a) f = 0.12 πDn π (0.5)(1140) vm = = = 149.23 fpm 12 12 pvm = 50,000 p(149.23) = 50,000 p = 335 psi W = pDL = (335)(0.5)(0.75) = 126 lb F = f W = (0.12)(126) = 15.12 lb U f = Fvm = (15.12 )(149.23) = 2256 ft − lb min (b) f = 0.14 ~ 0.18 , use f = 0.16 pvm = 2500 , water p(149.23) = 2500 p = 16.75 psi W = pDL = (16.75)(0.5)(0.75) = 6.28 lb F = f W = (0.16)(6.28) = 1.005 lb U f = Fvm = (1.005)(149.23) = 150 ft − lb min (c) vm > 100 fpm f = 0.25 pvm = 20,000 , intermittent p(149.23) = 20,000 p = 134 psi W = pDL = (134)(0.5)(0.75) = 50.25 lb F = f W = (0.25)(50.25) = 12.5625 lb U f = Fvm = (12.5625)(149.23) = 1875 ft − lb min (d) pvm = 15,000 p(149.23) = 15,000 p = 100.5 psi W = pDL = (100.5)(0.5)(0.75) = 37.69 lb assume f = 0.20 Page 61 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS F = f W = (0.20)(37.69) = 7.54 lb U f = Fvm = (7.54 )(149.23) = 1125 ft − lb min 590. A bearing to support a load of 150 lb at 800 rpm is needed; D = 1 in. ; semilubricated. Decide upon a material and length of bearing, considering sintered metals, Zytel, Teflon, and graphite inserts. Solution: πDn π (1)(800) vm = = = 209.44 fpm 12 12 assume, L = D = 1 in W 150 p= = = 150 psi DL (1)(1) pvm = (150)(209.44) = 31,416 Use sintered metal, limit pvm = 50,000 THRUST BEARINGS 592. A 4-in. shaft has on it an axial load of 8000 lb., taken by a collar thrust bearing made up of five collars, each with an outside diameter of 6 in. The shaft turns 150 rpm. Compute (a) the average bearing pressure, (b) the approximate work of friction. Solution: (a) p = 4W 4(8000 ) = = 283 psi πDo2 π (6)2 (b) assume f = 0.065 , average F = f W = (0.065)(8000) = 520 lb πDn π (3)(150) vm = = = 117.81 fpm 12 12 U f = nFvm = (5)(520 )(117.81) = 306,306 ft − lb min 593. A 4-in. shaft, turning at 175 rpm, is supported on a step bearing. The bearing area is annular, with a 4-in. outside diameter and a 3/4 –in. inside diameter. Take the allowable average bearing pressure as 180 psi. (a) What axial load may be supported? (b) What is the approximate work of friction? Solution: πDn vm = 12 Page 62 of 63 SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS 1 (4 + 0.75) = 2.375 in 2 πDn π (2.375)(175) vm = = = 108.81 fpm 12 12 assume f = 0.065 , average D= (a) p = 4W π Do2 − Di2 ( ) π 2 3 W = (4) − (180) = 2182 lb 4 4 (b) U f = f Wvm = (0.065)(2182 )(108.81) = 15,433 ft − lb min 2 - end - Page 63 of 63 SECTION 10 - BALL AND ROLLER BEARINGS 601. The radial reaction on a bearing is 1500 lb.; it also carries a thrust of 1000 lb.; shaft rotates 1500 rpm; outer ring stationary; smooth load, 8-hr./day service, say 15,000 hr. (a) Select a deep-groove ball bearing. (b) What is the rated 90 % life of the selected bearing? (c) For b = 1.34 , compute the probability of the selected bearing surviving 15,000 hr. Solution: Fx = 1500 lb Fy = 1000 lb ( ) B10 = (15,000 )(60 )(1500 ) 10−6 = 1350 mr F e= 0.56Cr Fx + Ct Fz C r = 1.2 , outer ring stationary assume Ct = 1.8 F e = 0.56(1.2)(1500) + (1.8 )(1000) = 2808 lb 1 1 B 3 Fr = 10 Fe = (1350) 3 (2808) = 31,034 lb Br (a) Table 12.3 use 321, Fr = 31,800 lb Fs = 32,200 lb To check: Fz 1000 = = 0.03125 Fs 32,000 Table 12.2, Ct = 1.96 , Q = 0.2246 Fz 1000 = = 0.556 > Q C r Fx (1.2)(1500) F e= 0.56Cr Fx + Ct Fz F e = 0.56(1.2)(1500) + (1.96)(1000) = 2968 lb 1 1 B 3 Fr = 10 Fe = (1350 ) 3 (2968 ) = 32,803 lb Br 3.2 % higher than 31,800 lb. Safe. Therefore use Bearing 321, Deep-Groove Ball Bearing. (b) Fr = 31,800 lb F e = 2968 lb 1 B 3 31,800 = 10 (2968 ) 1 mr B10 = 1230 mr ( ) B10 = (HR )(60 )(1500 ) 10 −6 = 1230 HR ≈ 13,700 hr 628 SECTION 10 - BALL AND ROLLER BEARINGS 1 1 b ln B P (c) = B10 1 ln P10 1 1 ln = ln 0.9 P10 B10 = 1230 mr B = 1350 mr 1 1 1.34 ln 1350 P = 1230 1 ln 0.9 P = 0.8875 602. A certain bearing is to carry a radial load of 500 lb. and a thrust of 300 lb. The load imposes light shock; the desired 90 % life is 10 hr./day for 5 years at n = 3000 rpm . (a) Select a deep-groove ball bearing. What is its bore? Consider all bearings that may serve. (b) What is the computed rated 90 % life of the selected bearing? (c) What is the computed probability of the bearing surviving the specified life? (d) If the loads were changed to 400 and 240 lb., respectively, determine the probability of the bearing surviving the specified life, and the 90 % life under the new load. Solution: Fx = 500 lb Fz = 300 lb Assume Cr = 1 Fz 300 = = 0.6 Cr Fx (1.0)(500 ) Fz Table 12.2, >Q Cr Fx (a) F e= 0.56Cr Fx + Ct Fz Cr = 1 Assume Ct = 1.8 F e= 0.56(1)(500) + (1.93)(300) = 820 lb For light shock, service factor ~ 1.1 F e= (1.1)(820 ) = 902 lb 629 SECTION 10 - BALL AND ROLLER BEARINGS 1 1 B 3 Fr = 10 Fe = (1350 )3 (2770) = 30,614 lb Br B10 = (5)(365)(10)(60 )(3000)(10−6 ) = 3285 mr 1 1 B 3 Fr = 10 Fe = (3285)3 (902 ) = 13,409 lb Br Table 12.3, Bearing No. Fr , lb Fs , lb 217 14,400 12,000 312 14,100 10,900 Bore 85 mm 60 mm Select, Bearing No. 312 Fr = 14,100 lb Fs = 10,900 lb (b) Table 12.2 Fz 300 = = 0.0285 Fs 10,900 Ct = 1.99 Q = 0.22 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(500) + (1.99)(300 ) = 877 lb F e= (1.1)(877 ) = 965 lb 1 B 3 Fr = 10 Fe Br 1 B 3 14,100 = 10 (965) 1 B10 = 3119 mr ( ) B10 = (YR )(365)(10 )(60)(3000 ) 10−6 = 3119 YR = 4.75 years 1 1 b ln B P (c) = B10 1 ln P10 use b = 1.125 B10 = 3119 mr 630 SECTION 10 - BALL AND ROLLER BEARINGS B = 3285 mr 1 ln 3285 P = 3119 1 ln 0.9 P = 0.8943 1 1.125 (d) Fx = 400 lb Fz = 240 lb Cr = 1 Fz 240 = = 0.6 Cr Fx (1.0)(400) Table 12.2 Ct = 2.15 Q = 0.21 < 0.6 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(400 ) + (2.15)(240) = 740 lb F e= (1.1)(740) = 814 lb 1 B 3 Fr = 10 Fe Br 1 B 3 14,100 = 10 (814) 1 B10 = 5197 mr 1 1 b ln B P = B10 1 ln P10 1 1 1.125 ln 3285 P = 5197 1 ln 0.9 P = 0.939 Life: B10 = (YR )(365)(10 )(60)(3000 )(10−6 ) = 5197 YR = 8 years 631 SECTION 10 - BALL AND ROLLER BEARINGS 603. The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; n = 300 rpm . An electric motor drives through gears; 8 hr./day, fully utilized. (a) Considering deep-groove ball bearings that may serve, choose one end specify its bore. For the bearing chosen, determine (b) the rated 90 % life and (c) the probability of survival for the design lufe. Solution: Fx = 500 lb Fz = 100 lb Table 12.1, 8 hr./day fully utilized, assume 25,000 hr B10 = (25,000 )(60)(300 )(10−6 ) = 450 mr (a) assume Cr = 1 Fz 100 = = 0.2 Cr Fx (1.0)(500 ) F consider Q > z Cr Fx Fe = Cr Fx = (1.0 )(500) = 500 lb 1 1 B 3 Fr = 10 Fe = (450)3 (500) = 3832 lb Br Table 12.3 Bearing No. Fr , lb Fs , lb 207 4440 3070 306 4850 3340 305 3660 2390 Select 305, Fr = 3660 lb , Fs = 2390 lb Bore (Table 12.4) = 25 mm Fz 100 = = 0.0418 Fs 2390 Table 12.2, 0.22 < Q0.26 F Q> z Cr Fx Fe = Cr Fx = (1.0 )(500) = 500 lb (a) 1 B 3 3660 = 10 (500) 1 B10 = 392 mr Rated Life: B10 = (HR )(60)(300 )(10−6 ) = 392 HR ≈ 22,000 hr 632 SECTION 10 - BALL AND ROLLER BEARINGS 1 1 b ln B P (c) = B10 1 ln P10 b = 1.125 1 1 1.125 ln 450 P = 392 1 ln 0.9 P = 0.884 605. A No. 311, single-row, deep-groove ball bearing is used to carry a radial load of 1500 lb. at a speed of 500 rpm. (a) What is the 90 % life of the bearing in hours? What is the approximate median life? What is the probability of survival if the actual life is desired to be (b) 105 hr., (c) 104 hr.? Solution: Table 12.3, No. 311 Fs = 9400 lb Fr = 12400 lb Fx = 1500 lb assume Cr = 1 Fe = Cr Fx = (1)(1500 ) = 1500 lb 1 B 3 (a) Fr = 10 Fe Br 1 B 3 12400 = 10 (1500 ) 1 B10 = 565 mr ( ) B10 = (HR )(60)(500 ) 10−6 = 565 HR ≈ 18,800 hr For median life = 5( 90 % life) = 5(18,800) = 94,000 hr ( ) ( ) (b) B = 105 (60 )(500) 10 −6 = 3000 mr 633 SECTION 10 - BALL AND ROLLER BEARINGS 1 ln B P = B10 1 ln P10 b = 1.125 1 b 1 1 1.125 ln 3000 P = 565 1 ln 0.9 P = 0.502 (c) 104 hr ( ) ( ) B = 10 4 (60 )(500 ) 10−6 = 300 mr 1 ln B P = B10 1 ln P10 b = 1.125 1 b 1 1 1.125 ln 300 P = 565 1 ln 0.9 P = 0.950 606. The load on an electric-motor bearing is 350 lb., radial; 24 hr. service, n = 1200 rpm ; compressor drive; outer race stationary. (a) Decide upon a deepgroove ball bearing, giving its significant dimensions. Then compute the selected bearing’s 90 % life, and the probable percentage of failures that would occur during the design life. What is the approximate median life of this bearing? (b) The same as (a), except that a 200 series roller bearing is to be selected. Solution: Fx = 350 lb Fe = Cr Fx outer race stationary, Cr = 1 Fe = (1)(350 ) = 350 lb Table 12.1 634 SECTION 10 - BALL AND ROLLER BEARINGS 90 % Life, hrs = 50,000 hrs B = (50,000 )(60 )(1200 ) 10−6 = 3600 mr ( ) 1 1 B 3 (a) Fr = 10 Fe = (3600 )3 (350 ) = 5364 lb Br Table AT 12.3 earing No. Fr , lb Fs , lb 208 5040 3520 209 5660 4010 306 4850 3340 307 5750 4020 Use No. 209 Fr = 5660 lb Table 12.4, Dimension Bore = 45 mm O.D. = 85 mm Width of Races = 19 mm Max. Fillet r = 0.039 mm 90 % Life: 1 B 3 Fr = 10 Fe Br 1 B 3 5660 = 10 (350) 1 B10 = 4229 mr ( ) B10 = (HR )(60 )(1200 ) 10−6 = 4229 HR ≈ 58,740 hr Probability. 1 1 b ln B P = B10 1 ln P10 b = 1.125 1 1 1.125 ln 3600 P = 4229 1 ln 0.9 P = 0.916 % failures = 1 – 0.916 = 0.084 = 8.4 % Median Life = 5(58,740) = 293,700 hrs 635 SECTION 10 - BALL AND ROLLER BEARINGS (b) Table 12.3, Fr = 5364 lb use No. 207, Fr = 5900 lb Bore = 35 mm O.D. = 72 mm Width of Races = 17 mm 90 % life: 1 B 3 Fr = 10 Fe Br 1 B 3 5900 = 10 (350) 1 B10 = 4790 mr ( ) B10 = (HR )(60 )(1200 ) 10−6 = 4790 HR ≈ 66,530 hr Probability. 1 1 b ln B P = B10 1 ln P10 b = 1.125 1 1 1.125 ln 3600 P = 4790 1 ln 0.9 P = 0.926 % failures = 1 – 0.926 = 0.074 = 7.4 % Median Life = 5(66,530) = 332,650 hrs 608. A deep-groove ball bearing on a missile, supporting a radial load of 200 lb., is to have a design life of 20 hr.; with only a 0.5 % probability of failure while n = 4000 rpm . Using a service factor of 1.2 , choose a bearing. ( A 5- or 6- place log table is desirable.) Solution: No need to use log table. Fx = 200 lb assume Cr = 1 Fe = Cr Fx = (1.0 )(200 ) = 200 lb 636 SECTION 10 - BALL AND ROLLER BEARINGS Fe = (1.2 )(200 ) = 240 lb ( ) B10 = (20 )(60)(4000) 10 −6 = 4.8 mr P = 1 − 0.005 = 0.995 1 1 b ln B P = B10 1 ln P10 b = 1.125 1 ln 4.8 0.995 = B10 1 ln 0.9 B10 = 72 mr 1 1.125 1 1 B 3 Fr = 10 Fe = (72 )3 (240) = 998.4 lb Br Table 12.3 Select No. 201, Fr = 1180 lb VARIABLE LOADS 610. A certain bearing is to carry a radial load of 10 kip at a speed of 10 rpm for 20 % of the time, a load of 8 kips at a speed of 50 rpm for 50 % of the time, and a load of 5 kips at 100 rpm during 30 % of the time, with a desired life of 3000 hr.; no thrust. (a) What is the cubic mean load? (b) What ball bearings may be used? What roller bearings? Solution: 1 F13n1 + F23n2 + F33n3 + L 3 (a) Fm = ∑n ∑n = n + n 1 2 + n3 For 1 min. n1 = (0.2 )(10 ) = 2 rev n 2 = (0.5)(50 ) = 25 rev n3 = (0.3)(100 ) = 30 rev ∑ n = 2 + 25 + 30 = 57 rev F1 = 10 kips F2 = 8 kips 637 SECTION 10 - BALL AND ROLLER BEARINGS F3 = 5 kips 1 (10 )3 (2 ) + (8)3 (25) + (5)3 (30) 3 Fm = = 6.88 kips 57 (b) Fx = 6.88 kips = 6880 lb assume Cr = 1 Fe = (1.0 )(6880 ) = 6880 lb 1 min = 57 rev B10 = (3000 )(60 )(57 )(10−6 ) = 10.26 mr 1 3 1 B Fr = 10 Fe = (10.26 )3 (6880) = 14,950 lb Br Table 12.3, Ball Bearing Use Bearing No. 217, Fr = 14,400 lb (c) Table 12.3 (Roller Bearing) Use Bearing No. 213, Fr = 14,900 lb 612. A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm with Fx = 1200 lb and Fz = 600 lb , 55 % of the time at 800 rpm with Fx = 1000 lb and Fz = 500 lb , and 15 % of the time at 1200 rpm with Fx = 800 lb and Fz = 400 lb . Determine (a) the cubic mean load; (b) the 90 % life of this bearing in hours, (c) the average life in hours. Solution: Bearing No. 215, Fr = 11,400 lb , Fs = 9,250 lb Table 12.2, Fz Fs At 30 % of the time, 500 rpm Fz 600 = = 0.065 Fs 9250 Ct = 1.66 Q = 0.266 Fz 600 = = 0.5 > Q Cr Fx (1)(1200 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(1200 ) + (1.66)(600 ) = 1668 lb At 55 % of the time, 800 rpm Fz 500 = = 0.054 Fs 9250 Ct = 1.73 638 SECTION 10 - BALL AND ROLLER BEARINGS Q = 0.257 Fz 500 = = 0.5 > Q Cr Fx (1)(1000 ) Fe 2 = 0.56Cr Fc + Ct Fz = 0.56(1)(1000 ) + (1.73)(500 ) = 1425 lb At 15 % of the time, 1200 rpm Fz 400 = = 0.043 Fs 9250 Ct = 1.84 Q = 0.242 Fz 400 = = 0.5 > Q Cr Fx (1)(800 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(800) + (1.84)(400) = 1184 lb 1 F 3n + F23n2 + F33n3 + L 3 (a) Fm = 1 1 ∑n ∑n = n + n 1 2 + n3 F1 = 1668 lb F2 = 1425 lb F3 = 1184 lb For 1 min. n1 = (0.3)(500 ) = 150 rev n 2 = (0.55)(800) = 440 rev n3 = (0.15)(1200) = 180 rev ∑ n = 150 + 440 + 180 = 770 rev 1 (1668)3 (150 ) + (1425)3 (440) + (1184)3 (180 ) 3 Fm = = 1434 kips 770 (b) Fe = Fm = 1434 lb 1 B 3 Fr = 10 Fe Br 1 B 3 11,400 = 10 (1434 ) 1 B10 = 503 mr For 1 min = 770 rev 639 SECTION 10 - BALL AND ROLLER BEARINGS ( ) B10 = (HR )(60)(770) 10−6 = 503 HR ≈ 11,000 hr (c) Average life = 5(11,000) = 55,000 hrs MANUFACTURER’S CATALOG NEEDED 614. A shaft for the general-purpose gear-reduction unit described in 489 has radial bearing reactions of RC = 613 lb and RD = 1629 lb ; n = 250 rpm . Assume that the unit will be fully utilized for at least 8 hr./day, with the likelihood of the same uses involving minor shock. (a) Select ball bearings for this shaft. (b) Select roller bearings. (c) What is the probability of both bearings C and D surviving for the design life? Solution: 3 Problem 489, D = 1 in = 1.375 in 8 Ref: Design of Machine Members, Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr at C. Fr = RC = 613 lb K t = 1.0 K p = 1. 0 K o = 1.0 Ks = 3 Kr Na Nc N a = 250 rpm N c = 500 rpm K r = 1.5 Ks = 3 (1.5)(250) = 0.90856 500 K a = 1.0 Ha H c K rel Table 12.1, 8 hr/day, fully utilized, Text H a = 25,000 hr Kl = 3 H c = 10,000 hr assume K rel = 1.0 for 90 % reliability Kl = 3 25,000 = 1.3572 10,000 640 SECTION 10 - BALL AND ROLLER BEARINGS Fc = (K a K l )K o K p K s K t Fr Fc = (1.0 )(1.3572)(1.0)(1.0 )(0.90856)(1.0 )(613) = 756 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 207 Bore = 1.3780 in, Fc = 880 lb At D. Fr = RD = 1629 lb Fc = (K a K l )K o K p K s K t Fr Fc = (1.0 )(1.3572)(1.0)(1.0 )(0.90856)(1.0 )(1629) = 2009 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 407 Bore = 1.3780 in, Fc = 2290 lb (b) at C, Fc = 756 lb Table 9.8, Doughtie and Vallance Use No. 207, Bore = 1.3780 in, Fc = 1540 lb at C, Fc = 2009 lb Table 9.8, Doughtie and Vallance Use No. 307, Bore = 1.3780 in, Fc = 2660 lb (c) For probability: (c.1) at C, Bearing No. 207, Two-row spherical bearing, Fc = 880 lb Fc = 880 lb = (1)K l (1)(1)(0.90856)(1)(613) K l = 1.58 Kl = 3 Ha H c K rel 1.58 = 3 25,000 10,000 K rel K rel = 0.634 Table 9-3, Reference Probability = 95.8 % at D, Bearing No. 407, Deep-groove bearing, Fc = 2290 lb Fc = 2290 lb = (1)K l (1)(1)(0.90856)(1)(1627 ) K l = 1.547 Kl = 3 Ha H c K rel 641 SECTION 10 - BALL AND ROLLER BEARINGS 1.547 = 3 25,000 10,000 K rel K rel = 0.675 Table 9-3, Reference Probability = 93.3 % (c.2) at C, Roller Bearing No. 207, Fc = 1540 lb Fc = 1540 lb = (1)K l (1)(1)(0.90856)(1)(613) K l = 2.765 Kl = 3 Ha H c K rel 2.765 = 3 25,000 10,000 K rel K rel = 0.118 Table 9-3, Reference Probability = 98.8 % at D, Roller Bearing No. 407, Fc = 2660 lb Fc = 2660 lb = (1)K l (1)(1)(0.90856 )(1)(1627 ) K l = 1.80 Kl = 3 Ha H c K rel 1.80 = 3 25,000 10,000 K rel K rel = 0.43 Table 9-3, Reference Probability = 95.7 % 615. A shaft similar to that in 478 has the following radial loads on the bearings, left to right: 803 lb, 988 lb, 84 lb, and 307 lb; no thrust. The minimum shaft diameter at the bearings are 1.250 in, 1.125 in, 1.000 in, and 1.0625 in. Assume that the service will not be particularly gentle; intermittently used, with n = 425 rpm . (a) Select ball bearing for this shaft. (b) Select roller bearings. Solution: Ref: Design of Machine Members by Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr K a = 1.0 642 SECTION 10 - BALL AND ROLLER BEARINGS Kl = 3 Ha H c K rel H c = 10,000 hr Table 12.1, Text, H a = 10,000 hr (intermittent) 90 % reliability, K rel = 1.0 Kl = 3 10,000 = 1. 0 10,000 K o = 1.0 K p = 1. 0 K r = 1.5 assumed Ks = 3 Kr Na Nc N a = 425 rpm N c = 500 rpm Ks = 3 (1.5)(425) = 1.0844 500 K t = 1.0 (a) Ball Bearing (a.1) 803 lb, D = 1.250 in Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(803) = 870 lb Table 9-7, Ref. Two-row spherical type, 207 Fc = 880 lb Bore = 1.3780 in (a.2) 988 lb, D = 1.125 in Fc = (1.0 )(1.0)(1.0)(1.0)(1.0844)(1.0 )(988) = 1071 lb Table 9-7, Ref. Two-row spherical type, 306 Fc = 1050 lb Bore = 1.1811 in (a.3) 84 lb, D = 1.000 in Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(84) = 91 lb 643 SECTION 10 - BALL AND ROLLER BEARINGS Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (a.4) 307 lb, D = 1.0625 in Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(307 ) = 333 lb Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (b) Roller Bearing (b.1) 803 lb, D = 1.250 in Fc = 870 lb , Bore = 1.3780 in use No. 207, Fc = 1540 lb (b.2) 988 lb, D = 1.125 in Fc = 1071 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.3) 84 lb, D = 1.000 in Fc = 91 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.4) 307 lb, D = 1.0625 in Fc = 333 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb - end - 644 SECTION 11 – SPUR GEARS INTERMITTENT SERVICE DESIGN PROBLEMS 631. A pair of gears with 20o full-depth teeth are to transmit 10 hp at 1750 rpm of the 3-in. pinion; velocity ratio desired is about 3.8; intermittent service. Use a strength reduction factor of about 1.4, with theload at the tip and teeth commercially cut. Determine the pitch, face width, and tooth numbers if the material is cast iron, class 20. Solution: πD p n p π (3)(1750) vm = = = 1374 fpm < 2000 fpm 12 12 10 hp < 20 hp Commercially cut gears, vm < 2000 fpm 600 + vm Fd = Ft 600 33,000hp 33,000(10 ) Ft = = = 240 lb vm 1374 600 + 1374 Fd = (240 ) = 790 lb 600 Fs = sbY K f Pd For cast-iron, class 20, sn = 0.4 su s = 0.4(20 ) = 8 ksi = 8000 psi K f = 1 .4 10 Pd Table AT 24, Load at tip, 20o F.D. Assume Y = 0.33 assume b = Fs = Fd (8000)(10)(0.33) = 790 (1.4)Pd2 Pd = 4.89 use Pd = 5 N p = Pd D p = (5)(3) = 15 Y = 0.289 Page 1 of 57 SECTION 11 – SPUR GEARS Fs = Fd (8000)(b )(0.33) = 790 (1.4)(5) b = 2.4 in 8 12.5 <b< Pd Pd use 12.5 12.5 b= = = 2.5 in Pd 5 1 b = 2 in 2 Summary of answers: Pd = 5 1 b = 2 in 2 N p = 15 N g = 3.8(15) = 57 633. A pair of gears with 20o full-depth teeth are to transmit 5 hp at 1800 rpm of the pinion; mw = 2.5 ; N p = 18 teeth; commercially cut teeth; intermittent service; K f ≈ 1.45 . (a) Determine the pitch, face width, and tooth numbers if the material is cast iron, class 25. (b) The same as (a) except that the pinion is to be made of phosphor gear bronze (SAE 65, Table AT 3). Solution: Load at tip, Table AT 24, 20o F.D. N p = 18 , Y = 0.308 N g = (2.5)(18) = 45 , Y = 0.399 vm = πD p n p 12 N p 18 Dp = = Pd Pd n p = 1800 rpm π (18)(1800 ) 2700π 12 Pd Pd Commercially cut teeth vm = Page 2 of 57 = SECTION 11 – SPUR GEARS 600 + vm Fd = Ft 600 33,000(5) 550 Pd Ft = = 9π 2700π Pd 2700π 8482.3 600 + 600 + Pd 550 Pd Pd (19.4523Pd ) Fd = = 9π 600 600 14.1372 Fd = 19.4523Pd 1 + Pd Fs = sbY K f Pd 10 Pd (a) Cast iron, class 25, sn = 0.4su = 0.4(25) = 10 ksi = 10,000 psi assume b = Fs = Fd Pinion is weaker (10,000)(10)(0.308) = 19.4523P 1 + 14.1372 d Pd (1.45)(Pd )2 14.1372 21,241 = 19.4523Pd 1 + 2 Pd Pd Pd = 7.16 use Pd = 7 Face width b (10,000 )(b )(0.308) = 19.4523(7 )1 + 14.1372 (1.45)(7 ) 7 b = 1.355 in 8 12.5 <b< Pd Pd 1.143 < b < 1.786 say b = 1.5 in Summary of answers: Pd = 7 1 b = 1 in 2 Page 3 of 57 SECTION 11 – SPUR GEARS N p = 18 N g = 45 (b) Phosphor gear bronze , (SAE 65, Table AT 3) su = 80 ksi sn = 0.4su = 0.4(80) = 32 ksi or sn = 31 ksi (Table AT 3) use sn = 31 ksi Pinion, bronze sY = (31,000)(0.308) = 9548 Gear, cast iron sY = (10,000)(0.399) = 3990 Therefore gear is weaker (10,000)(10)(0.399) = 19.4523P 1 + 14.1372 d Pd (1.45)(Pd )2 14.1372 27,517 = 19.4523Pd 1 + 2 Pd Pd Pd = 8 use Pd = 8 Face width b (10,000 )(b )(0.399) = 19.4523(8)1 + 14.1372 (1.45)(8) 8 b = 1.252 in 8 12.5 <b< Pd Pd 1 < b < 1.5625 1 say b = 1.25 in = 1 in 4 Summary of answers: Pd = 8 1 b = 1 in 4 N p = 18 N g = 45 Page 4 of 57 SECTION 11 – SPUR GEARS 634. It is desired to transmit 120 hp at 1800 rpm of the pinion; intermittent service; with light shock (§13.18); preferably not less than 18, 20o-full-depth teeth on the pinion; K f = 1.5 should be conservative; mw = 1.5 . Decisions must be made concerning the material and quality of cutting the teeth. Since the design is for strength only, it will be convenient to express Ft , Fd , vm , b in terms of Pd and arrange an equation containing s and Pd convenient for iteration. Weak material results in a relatively large pinion with high peripheral speed. A very strong material may be unnecessarily expensive. On a production basesm carefully cut teeth should have a reasonable cost. Specify material, accuracy of cutting, pitch, face width, and tooth numbers. Solution: N 18 Dp = p = Pd Pd 18 (1800 ) πD p n p Pd 8482.3 vm = = = 12 12 Pd 33,000hp 33,000(120 ) Ft = = = 466.85Pd vm 8482.3 Pd 120 hp > 20 hp use Buckingham’s equation 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Assume a material of steel, as rolled AISI 1050, su = 102 ksi For carefully cut teeth, Fig. AF 20, e = 0.001 in (min.) Table AT 25, steel on steel, 20o F.D. C = 1660 π Try b = 10 Pd Fd = 466.85Pd + 8482.3 10 1660 + 466.85 Pd 0.05 Pd Pd 1 2 8482.3 10 + 1660 + 466.85Pd 0.05 Pd Pd Page 5 of 57 SECTION 11 – SPUR GEARS 424 16,600 + 466.85Pd Pd Pd Fd = 466.85Pd + 424 16,600 + + 466.85Pd Pd Pd 1 2 sbY K f Pd Weak pinion Table AT 24, N p = 18 , Y = 0.522 (Load near middle, 20o F.D.) Fs = s = 0.5su = 0.5(102) = 51 ksi = 51,000 psi K f = 1 .5 (51,000) 10 (0.522) Pd (1.5)Pd Fs = = 177,480 Pd2 Fs = N sf Fd (§13.18), light shock, N sf = 1.25 Iteration: Pd 5 4 Use Pd = 5 10 10 b= = = 2 in Pd 5 N p = 18 Ng Np = mw N g = (1.5)(18) = 27 Summary of first computation Material, AISI 1050, as rolled Accurately cut gears Pd = 5 b = 2.0 in N p = 18 N g = 27 Page 6 of 57 Fd 5331 5342 Fs 7099 11093 N sf (≥ 1.25) 1.33 2.08 SECTION 11 – SPUR GEARS CHECK PROBLEMS A pair of carefully cut, full depth, 20o involute gears, made of cast iron, ASTM 30, is transmitting 5 hp at 1150 rpm of the pinion; N p = 24 , N g = 32 , Pd = 8 , 636. 1 b = 1 in . For the teeth, determine (a) the endurance , (b) the dynamic load, (c) 2 the service factor (§13.18). Solution: 5 hp < 20 hp N p = 24 , 20o F.D. Y = 0.337 , Load at tip assume K f = 1.45 average (a) Fs = sbY K f Pd cast-iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi (12,000)(1.5)(0.337 ) = 523 lb Fs = (1.45)(8) 1200 + vm (b) Fd = Ft 1200 For carefully cut gears πD p n p vm = 12 N 24 Dp = p = = 3 in Pd 8 n p = 1150 rpm vm = π (3)(1150) = 903.2 fpm 12 33,000hp 33,000(5) Ft = = = 182.7 lb vm 903.2 1200 + vm 1200 + 903.2 Fd = Ft = (182.7 ) = 320 lb 1200 1200 (c) N sf = 637. Fs 523 = = 1.63 Fd 320 A manufacturer’s catalog for cut-tooth spur gears rates a 25-tooth, cast-iron (ASTM 25) pinion with 5-pitch, 20o full-depth involute teeth at 16.5 hp at 900 Page 7 of 57 SECTION 11 – SPUR GEARS 1 rpm; b = 2 in and mg = 2 ; let K f = 1.5 ; intermittent service; smooth load. (a) 2 What horsepower may these gears transmit? Do you consider the catalog rating too high or too low? (b) The same as (a) except that the teeth are carfully cut. (c) The same as (a) except that the pinion is to be made of phosphor bronze (SAE 65, Table AT 3). Solution: 16.5 hp < 20 hp πD p n p vm = 12 N p 25 Dp = = = 5 in Pd 5 n p = 900 rpm vm = π (5)(900) 12 = 1178 fpm (a) Using commercially cut. 600 + vm Fd = Ft , vm ≤ 2000 fpm 600 Table AT 24, Load at tip, 20o F.D. N p = 25 , Y = 0.340 N g = mg N p = 2(25) = 50 , Y = 0.408 use Y = 0.340 , weaker pinion Cast iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi K f = 1 .5 Pd = 5 1 b = 2 in 2 sbY Fs = K f Pd (12,000)(2.5)(0.340) = 1360 lb Fs = (1.5)(5) Fs = N sf Fd smooth load, N sf = 1 600 + 1178 1360 = (1) Ft 600 Ft = 459 lb Page 8 of 57 SECTION 11 – SPUR GEARS Ft vm (459)(1178) = 16.4 hp = 33,000 33,000 The rating is not too low or too high. hp = (b) Carefully cut, [1000 fpm < vm < 4000 fpm] 1200 + vm Fd = Ft 1200 Fs = N sf Fd 1200 + 1178 1360 = (1) Ft 1200 Ft = 686 lb hp = Ft vm (686 )(1178) = 24.5 hp = 33,000 33,000 (c) Phosphor bronze, (SAE 65, Table AT 3) sn = 24 ksi = 24,000 psi Pinion sY = (24,000 )(0.340) = 8160 Gear sY = (12,000)(0.408) = 4896 Therefore , gear is weaker sbY Fs = K f Pd (12,000)(2.5)(0.408) = 1632 lb Fs = (1.5)(5) Fs = N sf Fd 600 + 1178 1632 = (1) Ft 600 Ft = 551 lb hp = 638. Ft vm (551)(1178) = 19.7 hp = 33,000 33,000 A pair of commercially cut spur gears transmits 10 hp at 1750 rpm of the 25tooth pinion. The teeth are 20o full depth with 6 pitch; material , cast iron, class 9 30; face width is 1 in .; N g = 40 . Allow for stress concentration. (a) Compute 16 Page 9 of 57 SECTION 11 – SPUR GEARS the service factor for the teeth (§13.18). (b) If the drive is for a single-cylinder compressor, would carefully cut teeth be advisable? Show calculations. Solution: 10 hp < 20 hp Dp = vm = Np = Pd πD p n p 25 = 4.167 in 6 12 π (4.167 )(1750) vm = = 1909 fpm 12 For commercially cut gears 600 + vm Fd = Ft 600 33,000hp 33,000(10 ) Ft = = = 173 lb vm 1909 600 + 1909 Fd = (173) = 723.4 lb 600 sbY Fs = K f Pd Assume K f = 1.45 , average, load near tip Table AT 24, 20o F.D. N p = 25 , Y = 0.340 Cast iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi (12,000)1 9 (0.340) Fs = 16 (1.45)(6) = 732.8 lb Fs 732.8 = ≈ 1 .0 Fd 723.4 (b) Carefully cut gears 1200 + vm 1200 + 1909 Fd = Ft = (173) = 448.2 lb 1200 1200 F 732.8 N sf = s = = 1.635 Fd 448.2 (§13.18) Single-cylinder compressore 1.5 < N sf < 1.75 (a) N sf = Page 10 of 57 SECTION 11 – SPUR GEARS 1.5 < 1.635 < 1.75 Therefore advisable. CONTINUOUS SERVICE DESIGN PROBLEMS NOTE: When using Buckingham’s Fd equation and a K f is used, as intended in design for continuous service, use Y for load near middle. 639. The pinion of a pair of steel gears, transmitting 110 hp at 2300 rpm, is to have a 1 diameter of about 4 in .; mg ≈ 2.3 ; 20o full-depth teeth; the drive is to a 3 centrifugal pump, continuous service. (a) Decide upon Pd , b , N p , N g , and the material to be used. Consider the strength with the load near the middle of the profile. (b) The same as (a) except that it is not expected that the maximum loading will occur for more than 107 cycles, can you justify changes in your previous answers? Solution: 110 hp > 20 hp 1 D p = 4 in 3 n p = 2300 rpm 1 πD p n p 3 vm = = = 2609 fpm 12 12 33,000hp 33,000(110 ) Ft = = = 1391 lb vm 2609 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Fig. AF 19, vm = 2609 fpm Maximum permissible error = 0.0012 in Fig. AF 20 Use carefully cut gears Expected errors = 0.001 in Table AT 25, steel on steel, 20o F.D. C = 1660 10 Try b = Pd π 4 (2300) Page 11 of 57 SECTION 11 – SPUR GEARS 10 0.05(2609)1660 + 1391 Pd Fd = 1391 + 1 2 10 0.05(2609) + 1660 + 1391 Pd 16600 130.45 + 1391 Pd Fd = 1391 + 1 16600 2 130.45 + + 1391 Pd Wear Load Fw = D pbQK g Q= 2m g mg + 1 = 2(2.3) = 1.394 2 .3 + 1 with Pd = 5 , Fd = 4478 lb Pd = 4 , Fd = 4919 lb 1 10 Pd = 5 , Fw = 4 (1.394 )K g = 4478 3 5 K g = 371 1 10 Pd = 4 , Fw = 4 (1.394 )K g = 4919 3 4 K g = 326 Table AT 26, 20o F.D. Use sum of BHN = 800. K g = 366 22,109 1 10 Fw = 4 (1.394)(366) = Pd 3 Pd Iteration: Fw ≥ Fd Pd 5 4 8 12.5 <b< Pd Pd 1.6 in < b < 2.5 in use b = 2.5 in Use Pd = 5 , Page 12 of 57 Fd 4478 4919 Fw 4422 5527 SECTION 11 – SPUR GEARS To check for strength sbY Fs = K f Pd 1 Load near middle, N p = Pd D p = (5) 4 = 21.67 3 N g = mg N p = (2.3)(21.67 ) = 49.84 use N g = 50 , N p = 22 D p = 4.4 in 50 = 10 in 5 For s , average BHN = 400 Dg = Use 8630 WQT 800 F, BHN = 375 gear 2330 WQT 600 F, BHN = 429 pinion Sum of BHN = 375 + 429 = 804 ≈ 800 Table AT 24, Load near middle, 20o F.D. N p = 22 Y = 0.559 N g = 50 Y = 0.694 Pinion: s = 250(429) = 107,250 psi sY = (107,250)(0.559) = 59,953 psi s = 250(375) = 93,750 psi sY = (93,750)(0.694) = 65,062 psi Gear: Pinion is weaker sbY (59,953)(2.5) = 17,633 lb > F Fs = = d (1.7)(5) K f Pd Summary of answer: Pd = 5 b = 2.5 in N p = 22 N g = 50 Material: Pinion: Gear: 2330 WQT 600 F, BHN = 429 8630 WQT 800 F, BHN = 375 (b) For 107 cycles Page 13 of 57 SECTION 11 – SPUR GEARS Tables AT 26, K g =< 306 for the less stronger material available, therefore can’t be justify to change the previous answer. ( N p will become less than minimum). Using stronger material is expensive. 640. A pinion with 20o full-depth teeth, transmitting 60 hp at 2400 rpm, is part of a gear reduction for a lobeblower. It is to be about 3.2 in. in diameter; mg ≈ 1.56 . (a) Decide upon a material for the mating gears (and its treatment), Pd , b , N p , and N g . Determine the strength with the load near the middle of the profile. (b) The same as (a), except that the maximum loading will occur for no more than 107 cycles. Solution: D p = 3.2 in n p = 2400 rpm vm = πD p n p = π (3.2)(2400) = 2011 fpm 12 12 33,000hp 33,000(160 ) Ft = = = 2626 lb vm 2011 160 hp > 20 hp 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Fig. AF 19, vm = 2011 fpm Maximum permissible error = 0.0015 in Fig. AF 20 carefully cut gears Expected errors = 0.001 in Table AT 25, steel on steel, 20o F.D. C = 1660 10 Try b = Pd 10 0.05(2011)1660 + 2626 Pd Fd = 2626 + 1 2 10 0.05(2011) + 1660 + 2626 Pd Page 14 of 57 SECTION 11 – SPUR GEARS 16600 100.55 + 2626 Pd Fd = 2626 + 1 16600 2 100.55 + + 2626 Pd Fw = D pbQK g Q= 2(1.56 ) = 1.22 m g + 1 1.56 + 1 2m g = 39.04 K g 10 Fw = (3.2) (1.22)K g = Pd Pd Fw = Fd For Pd = 5 Fd = 5991 lb 39.04 K g Fw = = 5991 5 K g = 767 For Pd = 4 Fd = 6532 lb 39.04 K g Fw = = 6532 4 K g = 651 Table AT 26, Use Steel (600), carburized case hardened, and same 1010 cycles, 20o F.D. K g = 750 Using Pd = 5 , b = 10 = 2 in 5 1 Say b = 2 in 2 To check for strength sn = 250 BHN = 250(600) = 150,000 psi Table AT 24, Load near middle N p = Pd D p = (5)(3.2 ) = 16 Y = 0.503 N g = mg N p = (1.56 )(16 ) = 25 Y = 0.580 Page 15 of 57 SECTION 11 – SPUR GEARS assume K f = 1.7 Fs = (150,000)(2.5)(0.503) = 22,191 lb > F sbY = d (1.7)(5) K f Pd Summary of answer: Pd = 5 b = 2.5 in N p = 16 N g = 25 (b) 107 cycles Table AT 26, K g = 1680 Fw = 39.04(1680 ) 65,587 = Pd Pd Iteration: Fw ≥ Fd Pd 7 8 9 10 11 12 13 Fd 5560 5421 5311 5222 5148 5086 5033 Use Pd = 13 10 b = in 13 N p = D p Pd = (3.2 )(13) = 42 , Y = 0.667 To check for strength sbY (150,000)(10)(0.667) = 3482 lb < F Fs = = d K f Pd (1.7)(13)2 Therefore use Fs ≥ N sf Fd For lobe blower, 1.25 < N sf < 1.5 Assume Y = 0.50 Page 16 of 57 Fw 9370 8198 7287 6559 5962 5466 5045 SECTION 11 – SPUR GEARS (150,000)(10)(0.50) = 441,177 sbY = K f Pd Pd2 (1.7 )(Pd )2 Iteration: Pd Fs 7 9004 8 6894 Fs = Use Pd = 8 10 10 b= = = 1.25 in Pd 8 N p = Pd D p = (8)(3.2 ) = 25.6 Fd 5560 5421 N sf 1.619 1.272>1.25 say 26 N g = mg N p = (1.56 )(25.6 ) = 40 Summary of answer: Pd = 8 1 b = 1 in 4 N p = 26 N g = 40 641. Gears with 20o full-depth teeth are to transmit 100 hp continuously at 5000 rpm with mg = 4 ; pinion D p = 3 in. ; the drive is subjected to minor shocks with frequent starts. First calculations are to be made for carburized pinion teeth of AISI E3310, SOQT 450F, and the gear of cast steel, SAE 0175, WQT. Decide upon Pd , b , N p , and N g . Solution: πD p n p π (3)(5000) vm = = = 3927 fpm 12 12 Fig. AF 19, max. per/ error. e = 0.00075 in Use precision gears, error, e = 0.0005 in Table AT 25, steel on steel, 20o F.D. C = 0.5(1660) = 830 33,000hp 33,000(100 ) Ft = = = 840 lb vm 3927 10 b= Pd 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Page 17 of 57 SECTION 11 – SPUR GEARS 10 0.05(3927 )830 + 840 Pd Fd = 840 + 1 10 2 0.05(3927 ) + 830 + 840 Pd 8300 196.35 + 840 Pd Fd = 840 + 1 8300 2 196.35 + + 840 Pd For minor shock with frequent start 1.25 < N sf < 1.5 sbY K f Pd Pinion: AISI E3310, SOQT 450 F, Rc = C 57.5 , BHN = 600 sn = 250(600) = 150,000 psi Gear: Cast Steel, SAE 0175, WQT sn = 77 ksi = 77,000 psi (Table AT 6) Fs = use s = sn = 77,000 psi sbY (77,000)(10)(0.50) = 226,471 Fs = = K f Pd Pd2 (1.7 )(Pd )2 F Iteration: N sf = s , 1.25 < N sf < 1.5 Fd Pd 5 6 7 8 Fd 2833 2633 2488 2378 Use Pd = 8 Check for wear Fw = Fd Fw = D pbQK g Q= 2m g mg + 1 = 2(4 ) = 1 .6 4 +1 Page 18 of 57 Fs 9059 6291 4622 3539 N sf 3.2 2.4 1.9 1.49 SECTION 11 – SPUR GEARS 10 = 1.25 in 8 Fw = (3)(1.25)(1.6 )K g = 2378 b= K g = 396 K g = 396 < for carburized teeth Therefore Pd = 8 1 b = 1 in 4 N p = D p Pd = (3)(8) = 24 N g = mg N p = (4 )(24 ) = 96 Summary of answer: Pd = 8 1 b = 1 in 4 N p = 24 N g = 96 A 20-tooth (20o F.D.) pinion is to transmit 50 hp at 600 rpm, the service being indefinitely continuous in a conveyor drive; mw = 2.5 . The original pplan is to use a nodular-iron casting, 80-60-03, for each gear. Determine suitable values for the pitch, face width, and diameters. (Warning: compute C .) 642. Solution: N p = 20 Dp = Np Pd = 20 Pd 20 (600 ) πD p n p Pd 3142 = = vm = 12 12 Pd 33,000hp 33,000(50) Ft = = = 525Pd vm 3142 Pd π b= 10 Pd Fd = Ft + 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 Page 19 of 57 SECTION 11 – SPUR GEARS Fd = 525Pd + Fd = 525Pd + 3142 10 C + 525Pd 0.05 Pd Pd 1 2 3142 10 + C + 525Pd 0.05 Pd Pd 157.1 10C + 525Pd Pd Pd 157.1 10C + + 525Pd Pd Pd 1 2 Fs ≥ N sf Fd Conveyor drive, 1 < N sf < 1.25 sbY K f Pd assume Y = 0.50 load near middle K f = 1 .7 For nodular-iron, 80-60-03 sn = 40,000 psi Fs ≥ E = 23 × 106 psi ( ( ) ) 2 k 23 × 106 = k 11.5 × 106 C= = 6 Eg + E p 2 23 ×10 k = 0.111e for 20o full depth kE p E g Fs = ( ) (40,000)(10)(0.50) = 117,647 (1.7 )Pd2 Pd2 Fs Fd Iteration: Use Fig. AF 19 and Fig AF 20. 3142 vm = Pd e C Pd 5 628.4 0.00225 2872 4 785.5 0.002625 3351 N sf = Use Pd = 4 , commercially cut 8 12.5 <b< Pd Pd 2 < b < 3.125 use b = 3.0 in Page 20 of 57 Fd Fs 4765 5005 4706 7353 Fs Fd 0.99 1.46 N sf = SECTION 11 – SPUR GEARS Fw = D pbQK g Table AT 26, K g = 248 20o F.D. Np 20 = 5 in Pd 4 2mw 2(2.5) Q= = = 1.43 m w + 1 2 .5 + 1 Fw = (5)(3.0)(1.43)(248) = 5320 lb > Fd Summary of answer: Pd = 4 , commercially cut gears b = 3 in D p = 5 in Dp = = Dg = mw D p = (2.5)(5) = 12.5 in 643. A 4.8-in. (approximate) pinion with 20o full-depth teeth is to transmit 40 hp at 1000 rpm; indefinitely continuous service with smooth load; mg = 3.5 ; carefully cut teeth to reduce the chance of an explosive spark, the use of a phosphor0gearbronze (SAE 65, Table AT 3) pinion and a cast-iron (class 35) gear is a tentative decision. Decide upon an appropriate Pd and b , using Buckingham’s average dynamic load. Solution: Dd = 4.8 in n p = 1000 rpm vm = πD p n p = π (4.8)(1000) = 1257 fpm 12 12 Fig. AF 19, max. permissible error, e = 0.00225 in Carefully-cut, e = 0.001 in kE g E p C= Eg + E p k = 0.111e for 20o full depth Phosphor-bronze pinion SAE 65 sn = 24,000 psi E p = 16×106 psi Cast iron gear, class 35 sn = 0.4su = 0.4(35,000) = 14,000 psi E g = 14.5 ×106 psi Page 21 of 57 SECTION 11 – SPUR GEARS C= (0.111)(0.001)(14.5 ×106 )(16 ×106 ) = 844 14.5 ×106 + 16 ×106 Assume Pd = 4 e = 0.00125 in C = 1.25(844) = 1055 33,000hp 33,000(40 ) Ft = = = 1050 lb vm 1257 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 10 10 b= = = 2.5 in Pd 4 d 0.05(1257 )[1055(2.5) + 1050] Fd = 1050 + = 2926 lb 1 2 0.05(1257 ) + [1055(2.5) + 1050] sbY Fs = K f Pd assume K f = 1.7 , load near middle, gear, 20o F.D. N p = D p Pd = (4.8)(4 ) = 19 , Y = 0.534 use s = 14,000 psi , gear N g = mg N p = (3.5)(19 ) = 66 , Y = 0.7224 Fs = (14,000)(2.5)(0.7224) = 3718 lb (1.7 )(4) Fs ≥ N sf Fd smooth load, N sf = 1.0 Fs > Fd 3718 lb > 2926 lb Summary Pd = 4 b = 2.5 in 644. The 20o full-depth teeth for a pair of steel gears are to transmit 40 hp at 1200 rpm of the 20-tooth pinion; mg = 3 ; continuous service and indefinite life: The driven machine is an off-and-on reciprocating compressor. (a) Determine the pitch, face width, and steel (with treatment), considering at least three alternatives, including carefully cut teeth. For the gear teeth decided on, what would be the power capacity if only intermittent service (wear not considered) were required? (c) If a limited life of 107 cycles were satisfactory? Solution: Page 22 of 57 SECTION 11 – SPUR GEARS N p = 20 Np 20 Pd Pd n p = 1200 rpm Dp = = 20 (1200 ) πD p n p P 6283 vm = = d = 12 12 Pd π Ft = 33,000hp 33,000(40) = = 210 Pd vm 6283 P d Fd = Ft + 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 For carefully cut, e = 0.001 in Table AT 25, steel on steel, 20o F.D. C = 1660 Fw = Fd Fw = D pbQK g assume b = 10 Pd Fd = 210 Pd + Fd = 210 Pd + 6283 10 1660 + 210 Pd 0.05 Pd Pd 314 16600 + + 210 Pd Pd Pd 2mw 2(3) Q= = = 1 .5 mw + 1 3 + 1 20 10 300 Fw = (1.5)(K g ) = 2 K g Pd Pd Pd Pd = 5 Page 23 of 57 1 2 6283 10 + 1660 + 210 Pd 0.05 Pd Pd 314 16600 + 210 Pd Pd Pd 1 2 SECTION 11 – SPUR GEARS Fd = 3179 lb 300 Fw = 2 K g = 3179 (5) K g = 265 Pd = 6 Fd = 3079 lb 300 Fw = 2 K g = 3079 (6) K g = 369 (a) For indefinite life, Table AT 26 use Sum of BHN = 700, K g = 270 , Pd = 5 10 10 = = 2 in Pd 5 N 20 Dp = p = = 4 in Pd 5 Dg = m g D p = (3)(4 ) = 12 in b= Material combination. (Sum of BHN = 700) Alternatives: (1) 4150 OQT 1200 F, gear, BHN = 331 6152 OQT 1000 F, pinion, BHN = 375 (2) 5150 OQT 1000 F, gear, BHN = 321 8620 OQT 800 F, pinion, BHN = 375 (3) 4150 OQT 1200 F, gear, BHN = 331 C1095 OQT 800 F, pinion, BHN = 363 Using Pinion: C1095, OQT 800 F, BHN = 363 sn = 250(363) = 90,750 psi Gear: 4150, OQT 1200 F, BHN = 331 sn = 250(331) = 82,750 psi Table AT 24, Load near middle, 20o F.D. N p = 20 Y = 0.544 N g = 3(20 ) = 60 , Y = 0.713 Pinion: sY = (90,750)(0.544) = 49,368 Gear: sY = (82,750)(0.713) = 59,000 Therefore, pinion is weaker Page 24 of 57 SECTION 11 – SPUR GEARS K f = 1 .7 Fs = (90,750)(2)(0.544) = 11,616 lb sbY = (1.7)(5) K f Pd Fs = N sf Fd Reciprocating compressor, N sf = 1.4 Fd = Ft + 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 πD p n p π (4)(1200) vm = = = 1257 fpm 12 12 C = 1660 b = 2 in 0.05(1257 )[1660(2) + Ft ] Fd = Ft + 1 0.05(1257 ) + [1660(2) + Ft ]2 62.85(3320 + Ft ) Fd = Ft + 1 62.85 + (3320 + Ft )2 Fs = N sf Fd 11,616 = 1.4 Fd Fd = 8297 lb By trial and error method Ft = 4900 lb Fv (4900)(1257 ) = 186.6 hp hp = t m = 33,000 33,000 (b) 107 cycles, Table AT 26 Use Pd = 4 , K g = 252 Sum of BHN = 500 e = 0.00125 in C = 1.25(1660) 314 1.25(1660)(10) 0.05 + 210 Pd Pd Pd Fd = 210 Pd + 1 314 1.25(1660)(10) 2 + 0.05 + 210 Pd Pd Pd Pd = 4 Fd = 3870 lb Page 25 of 57 SECTION 11 – SPUR GEARS 300 (252) = 4725 > Fd (4)2 Therefore Pd = 4 Fw = 10 10 = = 2.5 in Pd 4 N 20 Dp = p = = 5 in Pd 4 Dg = mg D p = (3)(5) = 15 in b= Material combination. (Sum of BHN = 600) (1) 8630 WQT 1100 F, gear, BHN = 285 9261 OQT 1200 F, pinion, BHN = 311 (2) 6152 OQT 1200 F, gear, BHN = 293 9840 OQT 1000 F, pinion, BHN = 302 (3) 5150 OQT 1200 F, gear, BHN = 269 4150 OQT 1200 F, pinion, BHN = 331 Using Pinion: 4150, OQT 1200 F, BHN = 331 sn = 250(331) = 82,750 psi Gear: 5150, OQT 1200 F, BHN = 269 sn = 250(269 ) = 62,250 psi Table AT 24, Load near middle, 20o F.D. Y = 0.544 N p = 20 N g = 3(20 ) = 60 , Y = 0.713 Pinion: sY = (82,750)(0.544) = 45,016 Gear: sY = (62,750)(0.713) = 47,949 Therefore, pinion is weaker K f = 1 .7 sbY (82,750)(2.5)(0.544) = 16,550 lb = (1.7 )(4) K f Pd 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 πD p n p π (5)(1200) vm = = = 1571 fpm 12 12 C = 1.25(1660) = 2075 b = 2.5 in Fs = Page 26 of 57 SECTION 11 – SPUR GEARS Fd = Ft + 0.05(1571)[2075(2.5) + Ft ] 1 0.05(1571) + [2075(2.5) + Ft ]2 78.55(5187.5 + Ft ) Fd = Ft + 1 78.55 + (5187.5 + Ft )2 Fs = N sf Fd 16,550 = 1.4 Fd Fd = 11,821 lb By trial and error method Ft = 6800 lb Fv (6800)(1571) = 324 hp hp = t m = 33,000 33,000 645. A pair of spur gears, delivering 100 hp to a reciprocating pump at a pinion speed of 600 rpm, is to serve continuously with indefinite life; minimum number of 20o full-depth teeth is 18; mw = 2.5 . Since low weight is highly important, it is decided that the initial design be for carburized case-hardened teeth. (a) Determine a suitable pitch, face width, diameters, and specify the material and its heat treatment. (b) Use the same size teeth as determined in (a), but let the material be flame-hardened 4150, OQT 1100 F. Compute Fs and Fw . If it were decided that the maximum (specified) loading would be imposed only occasionally, would these gears transmit more or less power than the carburized teeth? Explain. Solution: N p = 18 Dp = Np Pd = 18 Pd 18 (600 ) πD p n p Pd 2827 vm = = = 12 12 Pd 33,000hp 33,000(100) Ft = = = 1167 Pd vm 2827 P d π Fd = Ft + 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 Page 27 of 57 SECTION 11 – SPUR GEARS 10 Pd assume first class commercial gears e = 0.002 in Table AT 25, C = 2(1660) = 3320 b= 2827 10 3320 + 1167 Pd 0.05 Pd Pd Fd = 1167 Pd + Fd = 1167 Pd + 1 2 2827 10 + 3320 + 1167 Pd 0.05 Pd Pd 141.35 33200 + 1167 Pd Pd Pd 1 2 141.32 33200 + + 1167 Pd Pd Pd Fw = D pbQK g 2(2.5) = 1.43 m g + 1 2 .5 + 1 Table AT 26, 20o F.D., carburized, indefinite K g = 750 Q= 2m g = 18 10 193,050 Fw = (1.43)(750) = Pd2 Pd Pd Fw ≥ Fd Iteration: Pd 5 6 Will not equal Using precision cut gears C = 0.5(1660) = 830 Page 28 of 57 Fd 8355 9181 Fw 7722 5363 SECTION 11 – SPUR GEARS 141.35 8300 + 1167 Pd Pd Pd Fd = 1167 Pd + 141.32 8300 + + 1167 Pd Pd Pd Pd 5 1 2 Fd 7680 Use Pd = 5 , precision cut To check for strength sbY Fs = K f Pd BHN = 600 sn = 250 BHN = 250(600) = 150,000 psi 10 10 b= = = 2 in Pd 5 Table AT 24m Load near middle, 20o F.D. N p = 18 , Y = 0.522 K f = 1 .7 Fs = (150,000)(2)(0.522) = 18,424 lb > F d (1.7 )(5) BHN = 600 , Rc = 57.5 E3310, SOQT 450 F Pd = 5 , precision cut b = 2 in N 18 D p = p = = 3.6 in Pd 5 Dg = mw D p = (2.5)(3.6 ) = 9 in Material, E3310, SOQT 450 F (b) Flame hardened, 4150 OQT 1100 F BHN ≈ 359 Sum of BHN = 718 K g = 287 Fw = D p bQK g = (3.6 )(2 )(1.43)(287 ) = 2955 lb Page 29 of 57 Fw 7722 SECTION 11 – SPUR GEARS Fs = sbY K f Pd s = 250(359) = 89,750 psi Y = 0.522 (89,750)(2)(0.522) = 11,023 lb Fs = (1.7 )(5) For occasional loading, Fs = Fd For (a) Fd = 7680 lb < 11,023 lb Therefore, these gears would transmit more power than carburized teeth for occasional loading with continuous loading for carburized teeth. CHECK PROBLEMS 646. A 6-ft. ball mill runs at 24.4 rpm, the drive being through 14 1/2o involute spur gears; Pd = 2 , N p = 15 , N g = 176 , b = 5 in ., and hp = 75 . The material of the pinion is SAE 1040, BHN = 180; of the gear, 0.35% C cast steel, BHN = 180. (a) Check for strength and wear and give your decision as to the service to be expected. (b) The foregoing pinion wore out. Actually, the first step was to replace it with one made of SAE 3140, OQT 1000 F. Would you expect this to cure the trouble? (c) The drive in (b) also wore out. The following solution which maintained the same gear diameters, pitch and face, was proposed: 20o full-depth teeth; pinion of SAE 3140 with BHN = 350; gear of SAE 1045 with BHN = 280. Would you predict that these gears will give long service? What are the approximate tempering temperatures to get the specified hardness? Solution: n g = 24.4 rpm vm = πDg ng 12 N 176 Dg = g = = 88 in Pd 2 π (88)(24.4) vm = = 3373 fpm 12 33,000hp 33,000(75) Ft = = = 734 lb vm 3373 Fig. AF 19, max. per. Error = 0.0008 in for vm = 3373 fpm Fig. AF 20, precision cut, Pd = 2 , e = 0.0010 in ≈ 0.0008 in Table AT 25, e = 0.0010 in C = 1600 , 14 1/2o F.D. Page 30 of 57 SECTION 11 – SPUR GEARS Fd = Ft + 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 b = 5 in Fd = 734 + 0.05(3373)[1600(5) + 734] 1 2 0.05(3373) + [1600(5) + 734] = 6354 lb (a) Wear load Fw = D pbQK g Dp = Np = 15 = 7.5 in 2 Pd b = 5 in 2N g 2(176 ) Q= = = 1.843 N p + N g 15 + 176 Sum of BHN = 180 + 180 = 360, 14 1/2o F.D. Table At 26, K g = 46.3 Fw = (7.5)(5)(1.843)(46.3) = 3200 lb Strength: sbY Fs = K f Pd s = 250 BHN = 250(180) = 45,000 psi Pinion is weaker, N p = 15 , 14 1/2o F.D. (involute) Y = 0.415 , Load near middle K f = 1 .7 Fs = (45,000)(5)(0.415) = 27,463 lb (1.7 )(2) Fw < Fd , service is intermittent. (b) Pinion, SAE 3140, OQT 1000 F, BHN = 311 sn = 250 BHN = 250(311) = 77,750 psi sY = (77,750)(0.415) = 32,266 psi Gear, s n = 250 BHN = 250(180) = 45,000 psi N g = 176 , Y = 0.6376 , 14 1/2o F.D. sY = (45,000)(0.6376) = 28,692 psi Strength, Page 31 of 57 SECTION 11 – SPUR GEARS Fs = (45,000)(5)(0.6376) = 42,194 lb sbY = (1.7)(2) K f Pd Wear, Fw = D pbQK g Sum of BHN = 180 + 311 = 491, 14 1/2o F.D. Table AT 26, K g = 92.4 Fw = (7.5)(5)(1.843)(92.4) = 6386 lb Fw ≈ Fd , this will cure the trouble. (c) Pinion, SAE 3140, BHN = 350 sn = 250 BHN = 250(350) = 87,500 psi Gear, SAE 1045, BHN = 280 s n = 250 BHN = 250(280) = 70,000 psi Table AT 25, 20o F.D., e = 0.001 in C = 1660 Fd = 734 + 0.05(3373)[1660(5) + 734] 1 2 0.05(3373) + [1660(5) + 734] = 6512 lb Wear load, sum of BHN = 350 + 280 = 630 Table At 26, K g = 218.8 , 20o F.D. Fw = D pbQK g Fw = (7.5)(5)(1.843)(218.8) = 15,122 lb > Fd Tempering temperatures: Pinion: SAE 3140, BHN = 350 Fig. AF 2, OQT 995 F Gear: SAE 1045, BHN = 280 Table AT 8, WQT 900 F, rod. Diameter = ½ in. 647. A 22-tooth pinion, transmitting 110 hp at 2300 rpm, drives a 45-tooth gear, both steel; 20o full depth; Pd = 5 , b = 1.5 in .; The manufacturing process is expected to result in a maximum effective error of e = 0.0016 in . (a) Compute Buckingham’s average dynamic load. Compute Fs and Fw if the material is (b) case-carburized AISI 8620, DOQT 300 F, (c) AISI 8742, OQT 950 F, (d) induction-hardened AISI 8742. (e) Suppose your company carries a stock of the foregoing materials. For a minimum-service factor of 1.2, which material do you Page 32 of 57 SECTION 11 – SPUR GEARS recommend for (i) intermittent service, (ii) indefinitely continuous service, (iii) cycles of loading not to exceed 107? Solution: N p = 22 N g = 45 Dp = vm = Np = Pd πD p n p 22 = 4.4 in 5 = π (4.4)(2300) = 2649 fpm 12 12 33,000hp 33,000(110 ) Ft = = = 1370 lb vm 2649 (a) Dynamic Load 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 For e = 0.0016 in , Table AT 25, steel, 20o F.D. C = 1.6(1660) = 2656 b = 1.5 in 0.05(2649)[2656(1.5) + 1370] Fd = 1370 + = 4819 lb 1 0.05(2649) + [2656(1.5) + 1370]2 (b) AISI 8620, DOQT 300 F carborized Table AT 26, 20o F.D. K g = 750 , 1010 cycles ≈ indefinite Table AT 11, Rc = C 64 Figure AF 4, BHN = 700 sn = 250(700) = 175,000 psi N p = 22 Table AT 24, Load near middle, 20o F.D. Y = 0.559 K f = 1 .7 Strength sbY (175,000)(1.5)(0.559) = 17,263 lb Fs = = (1.7)(5) K f Pd Wear Fw = D pbQK g Page 33 of 57 SECTION 11 – SPUR GEARS Q= 2N g N p + Ng = 2(45) = 1.3433 22 + 45 Fw = (4.4)(1.5)(1.3433)(750) = 6649 lb (c) AISI 8742, OQT 950 F BHN = 358.5 sum of BHN = 2(358.5) = 717 Table AT 26, 20o F.D. K g = 286 sn = 250(358.5) = 89,625 psi Strength sbY (89,625)(1.5)(0.559) = 8841 lb Fs = = (1.7 )(5) K f Pd Wear Fw = D pbQK g Fw = (4.4)(1.5)(1.3433)(286) = 2536 lb (d) Induction hardened, AISI 8742 Table AT 26, 20o F.D. K g = 555 at 1010 cycles sn = 250(500) = 125,000 psi Strength sbY (125,000)(1.5)(0.559) = 12,331 lb Fs = = (1.7 )(5) K f Pd Wear Fw = D pbQK g Fw = (4.4)(1.5)(1.3433)(555) = 4921 lb (e) (i) intermittent service N sf F d = (1.2 )(4819 ) = 5783 lb use AISI 8742, OQT 950 F FS ≥ N sf F d FS = 8841 lb (ii) indefinitely continuous service Fw ≥ F d Fd = 4819 lb use AISI 8620, DOQT 300 F Page 34 of 57 SECTION 11 – SPUR GEARS Fw = 6649 lb (iii) 107 cycles Use AISI 8742, induction hardened K g = 1190 Fw = (4.4)(1.5)(1.3433)(1190) = 10,550 lb > Fd Two mating steel gears have 16 and 25 teeth, respectively, 20o F.D.; b = 2 in , Pd = 5 ; pinion speed, 2400 rpm. The maximum effective error in the profiles is planned to be 0.0012 in. The drive is for heavy-duty conveyer, continuous service. Compute and specify a reasonable rated horsepower if the gear teeth are: (a) Case carburized AISI 8620, SOQT 300 F, (b) AISI 8742, OQT 950 F and the flame-hardened, (c) AISI 8742, OQT 800 F. 648. Solution: N p = 16 N g = 25 Dp = vm = Np = Pd πD p n p 12 Fd = Ft + 16 = 3.2 in 5 = π (3.2)(2400) 12 0.05vm (Cb + Ft ) = 2011 fpm 1 0.05vm + (Cb + Ft )2 error, e = 0.0012 in Table AT 25, 20o F.D. C = 1.2(1660) = 1992 0.05(2011)[1992(1.5) + Ft ] Fd = Ft + 1 0.05(2011) + [1992(1.5) + Ft ]2 100.55(2988 + Ft ) Fd = Ft + 1 100.55 + (2988 + Ft )2 (a) Case carburized AISI 8620, SOQT 300 F K g = 750 BHN = 700 sn = 250(700) = 175,000 psi N p = 16 , Y = 0.503 , Table AT 24, 20o F.D., Load near middle Page 35 of 57 SECTION 11 – SPUR GEARS Q= 2N g N p + Ng Wear: Fw = D pbQK g = 2(25) = 1.22 16 + 25 Fw = (3.2 )(2)(1.22)(750) = 5856 lb Strength sbY (175,000)(2)(0.503) = 20,712 lb Fs = = (1.7 )(5) K f Pd Use Fw = Fd 5856 = Ft + 100.55(2988 + Ft ) 1 100.55 + (2988 + Ft )2 Ft = 2635 lb (2635)(2011) = 160 hp Fv hp = t m = 33,000 33,000 (b) AISI 8742, OQT 950 F, flame-hardened K g = 555 sn = 250(500) = 125,000 psi Strength sbY (125,000)(2)(0.503) = 14,794 lb Fs = = (1.7)(5) K f Pd Wear: Fw = D pbQK g Fw = (3.2 )(2)(1.22)(555) = 4333 lb Use Fw = Fd 100.55(2988 + Ft ) 4333 = Ft + 1 100.55 + (2988 + Ft )2 Ft = 1590 lb Fv (1590)(2011) = 97 hp hp = t m = 33,000 33,000 (c) AISI 8742, OQT 800 F, BHN = 416.4 (Table AT 9) Sum of BHN = 2(416.4) = 832.8 Table AT 26, K g = 397.5 sn = 250 BHN = 250(416.4) = 104,100 psi Strength (104,100)(2)(0.503) = 12,320 lb sbY Fs = = (1.7)(5) K f Pd Page 36 of 57 SECTION 11 – SPUR GEARS Wear: Fw = D pbQK g Fw = (3.2 )(2)(1.22)(397.5) = 3104 lb Use Fw = Fd 100.55(2988 + Ft ) 3104 = Ft + 1 100.55 + (2988 + Ft )2 Ft = 770 lb Fv (770)(2011) = 47 hp hp = t m = 33,000 33,000 649. 3 The data for a pair of gears are: 20o F.D. teeth, b = 1 in , Pd = 6 , N p = 26 , 4 N g = 60 , n p = 2300 rpm ; as-rolled AISI 1050; carefully cut teeth; N sf = 1.2 . (a) Strength alone considered, find the horsepower that may be transmitted. (b) Determine the required surface hardness in order for Fw = Fd , and specify a treatment that would make the gears long lasting in continuous service. Solution: N 26 Dp = p = = 4.333 in Pd 6 πD p n p π (4.333)(2300) vm = = = 2609 fpm 12 12 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 For carefully cut teeth, Pd = 6 , 20o F.D. C = 1660 0.05(2609)[1660(1.75) + Ft ] Fd = Ft + 1 0.05(2609 ) + [1660(1.75) + Ft ]2 130.45(2905 + Ft ) Fd = Ft + 1 130.45 + (2905 + Ft )2 (a) Strength sbY Fs = K f Pd For as rolled AISI 1050, BHN = 229 sn = 250(229 ) = 57,250 psi or sn = 0.5su = 0.5(102.000) = 51,000 psi use s = 51,000 psi Page 37 of 57 SECTION 11 – SPUR GEARS N p = 26 Table AT 26, 20 o F.D., load near middle Y = 0.588 K f = 1 .7 Fs = (51,000)(1.75)(0.588) = 5145 lb (1.7)(6) Fs = N sf Fd 5145 = 1.2 Fd Fd = 4288 lb Fd = 4288 = Ft + 130.45(2905 + Ft ) 1 130.45 + (2905 + Ft )2 Ft = 1400 lb Fv (1400)(2609) = 110 hp hp = t m = 33,000 33,000 (b) Fw = Fd = 1400 lb Wear Fw = D pbQK g Q= 2N g N p + Ng = 2(60 ) = 1.395 26 + 60 Fw = (4.333)(1.75)(1.395)K g = 4288 lb K g = 405 Table of BHN = 838 BHN1 = 229 BHN 2 = 838 − 229 = 609 Therefore use carburized teeth. 650. Gears with carefully cut, 20o F.D. teeth have Pd = 5 , b = 2 , mg = 5 , N p = 24 . Pinion material is manganeses gear bronze (heading of Table AT 3); gear is cast iron, class 25. Gear speed ng = 200 rpm ; smooth load. They are to transmit 18 hp. (a) Are the teeth strong enough for intermittent service? (b) Does the limiting wear load indicate long life? Suggestion: Compute C for equation (13.7). Solution: N 24 Dp = p = = 4.8 in Pd 5 Page 38 of 57 SECTION 11 – SPUR GEARS n p = mg ng = (5)(200 ) = 1000 rpm vm = πD p n p π (4.8)(1000) = 1257 fpm 12 12 33,000hp 33,000(18) Ft = = = 473 lb vm 1257 Carefully cut: kE g E p C= Eg + E p k = 0.111e for 20o full-depth Fig. AF 20, e = 0.001 in , Pd = 5 C= = (0.111)(0.001)(11.5 ×106 )(16 ×106 ) = 743 Fd = Ft + 11.5 ×106 + 16 ×106 0.05vm (Cb + Ft ) 1 0.05vm + (Cb + Ft )2 0.05(1257 )[743(2) + 473] Fd = 473 + = 1622 lb 1 0.05(1257 ) + [743(2) + 473]2 (a) Fs ≥ N sf Fd N sf = 1.0 , smooth load sbY K f Pd Gear: cast-iron, class 25, 20o F.D. N g = mg N p = (5)(24 ) = 120 Fs = Y = 0.7646 , load near middle sn = 0.4(25) = 10 ksi = 10,000 psi snY = (10,000)(0.7646) = 7646 psi Pinion: manganese gear bronze, 20o F.D. N p = 24 Y = 0.572 , load near middle sn = 17 ksi = 17,000 psi snY = (17,000)(0.572) = 9724 psi Gear is weaker K f = 1 .7 Fs = (10,000)(2)(0.7646) = 1799 lb > N F sf d (1.7 )(5) Page 39 of 57 SECTION 11 – SPUR GEARS Therefore, enough for intermittent service (b) Fw = D pbQK g Q= 2m g mg + 1 = 2(5) = 1.667 5 +1 s 2 sin φ 1 1 + K g = 1.4 E p E g Pinion, manganese gear bronze s = su = 75 ksi φ = 20o (75,000 )2 sin 20 1 1 Kg = + = 205 6 6 1 .4 16 × 10 11.5 × 10 Fw = (4.8)(2)(1.667 )(205) = 3281 lb > Fd Therefore, indicates long life. A 20-tooth pinion. 20o F.D., drives a 100-tooth gear. The pinion is made of SAE 1035, heat treated to Rockwell C15; the gear is cast iron class 35, HT; Pd = 3 , b = 2.5 in .; carefully cut teeth; pinion speed n p = 870 rpm , smooth load. (a) For a continuous service, indefinite life, what is a safe horsepower? (b) For intermittent service (wear unimportant), compute the safe horsepower. 651. Solution: N p = 20 N g = 100 Dp = vm = Np = Pd πD p n p 12 Fd = Ft + 20 = 6.667 in 3 = π (6.667 )(870) 12 0.05vm (Cb + Ft ) 1 = 1519 fpm 0.05vm + (Cb + Ft )2 Carefully cut gears, steel and cast iron, 20o F.D. Pd = 3 , Fig. AF 20, Table AT 25 e = 0.0016 in C = 1.6(1140) = 1824 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Page 40 of 57 SECTION 11 – SPUR GEARS Fd = Ft + Fd = Ft + 0.05(1519)[1824(2.5) + Ft ] 1 0.05(1519) + [1824(2.5) + Ft ]2 75.95(4560 + Ft ) 1 75.95 + (4560 + Ft )2 (a) Continuous ser vice Strength: sbY Fs = K f Pd Pinion: SAE 1035, Rc = C15 Fig. AF 4, BHN = 200 sn = (250)(200) = 50,000 psi N p = 20 Table AT 24, 20o F.D., load near middle Y = 0.544 snY = (50,000)(0.544) = 27,200 psi Gear: Cast iron, class 35 sn = 0.4su = 0.4(35,000) = 14,000 psi N g = 100 Table AT 24, 20o F.D., load near middle Y = 0.755 snY = (14,000)(0.755) = 10,570 psi Gear is weaker K f = 1 .7 Fs = (14,000)(2.5)(0.755) = 5181 lb (1.7)(3) Wear: Fw = D pbQK g Q= 2N g N p + Ng = 2(100 ) = 1.667 20 + 100 s 2 sin φ 1 1 K g = + 1.4 E p E g φ = 20o s = 0.4 BHN − 10 = 0.4(200) − 10 = 70 ksi = 70,000 psi E p = 30×106 psi Page 41 of 57 SECTION 11 – SPUR GEARS E g = 14.5 ×106 psi (70,000 )2 sin 20 1 1 Kg = + = 122 6 6 1 .4 30 ×10 14.5 × 10 Fw = (6.667 )(2.5)(1.667 )(122) = 3390 lb Fw = Fd 3390 = Ft + 75.95(4560 + Ft ) 1 75.95 + (4560 + Ft )2 Ft = 700 lb Fv (700)(1519) = 32 hp hp = t m = 33,000 33,000 (b) Fs = N sf Fd N sf = 1.0 , smooth load 5181 = Ft + 75.95(4560 + Ft ) 1 75.95 + (4560 + Ft )2 Ft = 2000 lb Fv (2000)(1519) = 92 hp hp = t m = 33,000 33,000 652. A pair of steel gears is defined by Pd = 8 , b = 1.5 in , N p = 25 , N g = 75 , e = 0.001 in , 20o F.D. If these gears may transmit continuously and without failure 75 hp at 1140 rpm of the pinion, what horsepower would be satisfactory for n p = 1750 rpm ? Solution: N 25 Dp = p = = 3.125 in Pd 8 πD p n p π (3.125)(1140 ) vm = = = 933 fpm 12 12 33,000hp 33,000(75) Ft = = = 2653 lb vm 933 Table AT 25, e = 0.001 in , 20o F.D. C = 1660 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Page 42 of 57 SECTION 11 – SPUR GEARS Fd = 2653 + 0.05(933)[1660(1.5) + 2653] 1 2 0.05(933) + [1660(1.5) + 2653] = 4680 lb For n p = 1750 rpm vm = πD p n p π (3.125)(1750 ) = 1432 fpm 12 0.05(1432)[1660(1.5) + Ft ] Fd = Ft + = 4680 lb 1 0.05(1432 ) + [1660(1.5) + Ft ]2 Ft = 2260 lb Fv (2260)(1432) = 98 hp hp = t m = 33,000 33,000 654. 12 = A gear manufacturer recommends that the following gears can transmit 25 hp at 600 rpm of the pinion during continuous 24-hr. service, indefinite life, moderate shock: N p = 31 , N g = 70 , b = 3.25 in , Pd = 6 , 20o F.D.; pinion material is SAE 2335 with BHN = 300; gear material is SAE 1040 with BHN = 250. At what horsepower would you rate them? Solution: N 31 D p = p = = 5.167 in Pd 6 πD p n p π (5.167 )(600) vm = = = 812 fpm 12 12 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 Commercial cut, e = 0.001 in , 20o F.D., Pd = 6 Table AT 25 C = (2)(1660) = 3320 0.05(812)[3220(3.25) + Ft ] Fd = Ft + 1 0.05(812) + [3220(3.25) + Ft ]2 40.6(10,465 + Ft ) Fd = Ft + 1 40.6 + (10,465 + Ft )2 Strength: Pinion: SAE 2335, bhn = 300 sn = 250(300) = 75,000 psi N p = 31 , Table AT 24, Load near middle, 20o F.D. Y = 0.6115 Page 43 of 57 SECTION 11 – SPUR GEARS snY = (75,000)(0.6115) = 45,862 psi Gear: SAE 1040, BHN = 250 sn = 250(250 ) = 62,500 psi N g = 70 , Table AT 24, Load near middle, 20o F.D. Y = 0.728 snY = (62,500)(0.728) = 45,500 psi Gear is weaker, K f = 1.7 Fs = (62,500)(3.25)(0.728) = 14,498 lb sbY = (1.7 )(6) K f Pd Wear load: Fw = D pbQK g Q= 2N g N p + Ng = 2(70 ) = 1.386 31 + 70 Sum of BHN = 300+250 = 550 Table AT 26, 20o F.D. K g = 162 Fw = (5.167 )(3.25)(1.386)(162) = 3770 lb Fw = Fd 40.6(10465 + Ft ) 3770 = Ft + 1 40.6 + (10465 + Ft )2 Ft = 675 lb (675)(812) = 16.6 hp Fv hp = t m = 33,000 33,000 For carefully cut, C = 1660 Cb = (1660)(3.25) = 5395 40.6(5395 + Ft ) 3770 = Ft + 1 40.6 + (5395 + Ft )2 Ft = 1500 lb Fv (1500)(812) = 36.9 hp carefully cut hp = t m = 33,000 33,000 use hp = 36.9 hp , carefully cut. Page 44 of 57 SECTION 11 – SPUR GEARS NONMETALLIC GEARS 655. A 4-in. Bakelite pinion meshing with a cast-iron gear, is to be on the shaft of a 12 –hp induction motor that turns at 850 rpm; mw = 5 , 20o F.D. teeth. (a) Determine Pd , b , N p , N g for indefinitely continuous service with a smooth load. Is there serious interference in the gears you have designed? (b) Are the teeth of your design strong enough to taje without damage an occasional 60 % overload? Solution: Pinion as weaker (200 + vm )Ft Fd = v 200 + m 4 πD p n p π (4)(850) vm = = = 890 fpm 12 12 33,000hp 33,000(12 ) Ft = = = 445 lb vm 890 (200 + 890)(445) = 1148 lb Fd = 890 200 + 4 Strength: s = 6000 psi sbY Fs = Pd 10 and Y = 0.33 , 20o F.D. Load at tip. assume b = Pd Fs = Fd (smooth load) (6000)(10)(0.33) = 1148 lb Fs = Pd2 Pd = 4.15 use Pd = 4 10 10 b= = = 2.5 in Pd 4 say b = 3 in N p = D p Pd = (4 )(4 ) = 16 Table At 24, Y = 0.295 , 20o F.D. Load at tip. (6000)(3)(0.295) = 1328 lb > 1148 lb Fs = 4 Check: Page 45 of 57 SECTION 11 – SPUR GEARS Wear: Fw = D pbQK g 2mw 2(5) = = 1.667 mw + 1 5 + 1 Table AT 26, Use K g = 64 , 20o F.D. Q= Fw = (4)(3)(1.667 )(64) = 1280 lb > Fd , o.k. Then Pd = 4 b = 3 in N p = 16 N g = mw N p = (5)(16 ) = 80 N p < 18 , there is interference. (b) Ft = (1.6)(445) = 712 lb Fd > 712 lb , strong enough. 656. A Zytel pinion with molded teeth is to transmit 0.75 hp to a hardened-steel gear; n p = 1750 rpm , D p ≈ 1.25 in . Determine the pitch, face, and number of teeth on the pinion for intermittent service. Solution: πD p n p π (1.25)(1750) vm = = = 573 fpm 12 12 33,000hp 33,000(0.75) Ft = = = 43 lb vm 573 Fd = (VF )Ft For molded teeth, vm < 4000 fpm VF = 1 Fd = (1)(43) = 43 lb Load near middle, assume Y = 0.50 Fs = Fd say s = 4.6 ksi = 4600 psi , §13.27 10 b= Pd sbY (4600 )(10 )(0.50 ) Fs = = = 43 Pd Pd2 Page 46 of 57 SECTION 11 – SPUR GEARS Pd = 23 use Pd = 20 10 = 0.5 in b= 20 N p = D p Pd = (1.25)(20 ) = 25 teeth 657. A 10-in. Textolite pinion, driving a hardened steel gear, transmits power at 400 rpm; Pd = 2.5 , b = 5 in ., 14 1/2o F.D. teeth. Determine the safe horsepower (a) for smooth, continuous, indefinite service, and also (b) for limited-life intermittent service. Solution: D p = 10 in n p = 400 rpm vm = πD p n p = π (10)(400) = 1047 fpm 12 12 (200 + vm )Ft Fd = v 200 + m 4 Strength sbY Fs = Pd s = 6000 psi b = 5 in Pd = 2.5 N p = Pd D p = (2.5)(10 ) = 25 Table At 24, 14 1/2o F.D. Load at tip. Y = 0.305 (6000)(5)(0.305) = 3660 lb Fs = 2 .5 Wear load, Table At 26, 14 1/2o F.D. K g = 46 Fw = D pbQK g Q ≈ 1 .5 Fw = (10)(5)(1.5)(46) = 3450 lb (a) Continuous service Fw = Fd (smooth) Page 47 of 57 SECTION 11 – SPUR GEARS 3450 = (200 + 1047 )Ft 200 + 1047 4 Fs = 1277 lb Fv (1277)(1047) = 40 hp hp = t m = 33,000 33,000 (b) Intermittent service Fs = Fd (200 + 1047 )Ft 3660 = 1047 200 + 4 Fs = 1355 lb Fv (1355)(1047) = 43 hp hp = t m = 33,000 33,000 658. 5 in . 8 (a) What safe horsepower may be transmitted for long-life? (b) For 107 cycles? A 16-pitch Zytel pinion, with 26, 20o F.D. cut teeth, rotates at 600 rpm; b = Solution: πD p n p vm = 12 N 26 Dp = p = = 1.625 in Pd 16 π (1.625)(600) vm = = 255 fpm 12 Cut-teeth, vm < 4000 fpm VF = 1.2 Fd = (VF )Ft Fd = Ft (a) Long life, 5× 108 cycles Pd = 16 s = 2.3 ksi = 2300 psi sbY Fs = Pd For N p = 26 , 20o F.D., Load near middle Y = 0.588 Page 48 of 57 SECTION 11 – SPUR GEARS (2300) 5 (0.588) 8 = 53 lb 16 Fs = 53 = 1.2 Ft Ft = 44 lb (44)(255) = 0.34 hp Fv hp = t m = 33,000 33,000 Fs = (b) 107 cycles. s = 4.2 ksi = 4200 psi , Pd = 16 (4200) 5 (0.588) 8 = 96.5 lb 16 Fs = 96.5 = 1.2 Ft Ft = 80 lb Fv (44)(255) = 0.34 hp hp = t m = 33,000 33,000 Fs = CAST-TOOTH GEARS 659. A pair of cast-iron spur gears, ASTM 20, with cast teeth, transmits 10 hp at 125 rpm of the pinion; mw = 5 , D p ≈ 8 in . Determine Pc , b , N p , N g . Solution: πD p n p π (8)(125) vm = = = 262 fpm 12 12 600 + vm Fd = Ft 600 33,000hp 33,000(10 ) Ft = = = 1260 lb vm 262 600 + 262 Fd = (1260) = 1810 lb 600 Fs = 0.054sbPc say b = 2.5Pc s = 0.4su = 0.4(20) = 8 ksi = 8000 psi Fs = 0.054(8000 )(2.5)Pc2 = 1080 Pc2 Fs = Fd 1080 Pc2 = 1810 Pc = 1.29 in Page 49 of 57 SECTION 11 – SPUR GEARS 1 say Pc = 1 in 4 b = 2.4 Pc = 2.4(1.25) = 3 in πD p π (8) Np = = = 20 Pc 1.25 N g = mw N p = (5)(20 ) = 100 660. Design the cast teeth for a pair of cast-iron spur gears to transmit 35 hp at 50 rpm of the pinion; mw ≈ 2.5 . Decide upon a suitable grade of cast iron and find Pc , b , D p , Dg , and center distance. Solution: Using cast-iron, class 35 sn = 0.4su = 0.4(35) = 14 ksi = 14,000 psi 33,000hp 33,000(35) 1,155,000 Ft = = = vm vm vm 600 + vm Fd = Ft 600 Try N p = 20 Dp = vm = Pc N p π πD p n p 12 = = 20 Pc = 6.366 Pc π π (6.366 Pc )(50) 12 Ft = 1,155,000 13,860 = 83.33Pc Pc Fd = 600 + 83.33Pc 600 = 83.33Pc 13,860 23.1(600 + 83.33Pc ) = P Pc c Fs = 0.054sbPc b = 2.5Pc Fs = 0.054(14,000 )(2.5)Pc2 = 1890 Pc2 Fs = Fd 23.1(600 + 83.33Pc ) Pc Pc = 2.117 in 1890 Pc2 = use Pc = 2 in PN 20(2) Dp = c p = = 12.73 in π Page 50 of 57 π SECTION 11 – SPUR GEARS Dg = mw D p = (2.5)(12.73) = 31.83 in b = 2.5Pc = 2.5(2) = 5 in 1 1 C = (D p + Dg ) = (12.73 + 31.83) = 22.28 in 2 2 Summary: Pc = 2 in D p = 12.73 in Dg = 31.83 in b = 5 in C = 22.28 in 661. A manufacturer’s catalog specifies that a pair of gray cast-iron spur gears with cast teeth will transmit 7.01 hp at a pitch-line speed of 100 fpm; N p = 20 , Pc = 1.5 in ., b = 4 in . Compute the stress and specify the grade of cast iron that should be used. Solution: vm = 100 fpm 33,000hp 33,000(7.01) Ft = = = 2313.3 lb vm 100 600 + vm 600 + 100 Fd = Ft = (2313.3) = 2699 lb 600 600 Fs = 0.054sbPc Fs = Fd 0.054s (4)(1.5) = 2699 s = 8330 psi s ≈ 0.4su 8330 ≈ 0.4su su = 20,825 psi = 21 ksi use Cast Iron, ASTM 25, su = 25 ksi ARMS AND RIMS 662. A 24-in. cast-iron gear transmits 30 hp at 240 rpm; 14 1/2o F.D. teeth, Pd = 4 , b = 2.5 in . The gear is on a 2 ¼ -in. shaft. Determine (a) the hub diameter, rim thickness, and bead, (b) the dimensions of the arms at the hub and at the pitch circle for an elliptical-shaped section, (c) the arm dimensions for a cross shape. Page 51 of 57 SECTION 11 – SPUR GEARS Solution: 1 1 (a) Hub diameter = 2 Ds = 2 2 = 4 in , for cast iron 2 4 7 π Rim thickness = 0.56 Pc = 0.56 = 0.44 in ≈ in 16 4 7 π Bead = 0.56 Pc = 0.56 = 0.44 in ≈ in 16 4 (b) Elliptical section use no. of arms = N a = 4 , D = 24 in < 120 in FL M= Na 1 24 − 4 D − Dh 2 = 9.75 in L= = 2 2 F = Fd M s= Z πh3 Z= 64 0.05vm (Cb + Ft ) Fd = Ft + 1 0.05vm + (Cb + Ft )2 πD p n p π (24)(240) vm = = = 1506 fpm 12 12 assume C = 800 , 14 1/2o F.D. 33,000hp 33,000(30 ) Ft = = = 660 lb vm 1506 0.05(1506)[800(2.5) + 660] Fd = 660 + = 2238 lb 1 0.05(1506 ) + [1800(2.5) + 660]2 M FL 64 FL s= = = 3 Z N a Z πh N a use s = 8000 psi Page 52 of 57 SECTION 11 – SPUR GEARS 64(2238)(9.75) πh3 (4) h = 2.4 in 8000 = At the hub h = 2.4 in h h1 = = 1.2 in 2 At the pitch circle h h′ = = 1.2 in 2 h h1 = = 1.2 in 2 (c) Cross shape 6Z h2 G1 = 0.75G h = 2.4 in G= Z= πh3 , §13.32 64 π (2.4)3 Z= = 0.6786 in 3 64 At the hub 6(0.6786 ) G= = 0.71 in (2.4)2 G1 = 0.75(0.71) = 0.53 in h = 2.4 in h At the pitch circle, h′ = = 1.2 in 2 G = 0.71 in G1 = 0.53 in Page 53 of 57 SECTION 11 – SPUR GEARS INTERNAL GEARS A 20-tooth pinion, with 20o F.D. teeth , drives a 75-tooth internal gear ( Pd = 8 , b = 1.5 in , n p = 1150 rpm ; material, cast iron, class 20). What horsepower may be transmitted continuously (a) if the teeth are commercially hobbed (AGMA equation, §13.15), (b) if they are precision cut? There are minor irregularities in the loading. 664. Solution: (a) Commecially hobbed 1 2 m 50 + v Ft 50 Fw = D pbQK g Fd = Q= 2N g Ng − N p = 2(75) = 2.727 75 − 20 From AT 26, K g = 112 Np 20 = 2.5 in Pd 8 Fw = (2.5)(1.5)(2.727 )(112) = 1145 lb Dp = = For strength: sbY Fs = K f Pd N p = 20 , Y = 0.320 at the tip Let K f = 1.3 , s = 8000 psi (8000)(1.5)(0.320) = 369 lb (1.3)(8) πD p n p π (2.5)(1150 ) = = = 753 fpm Fs = vm 12 Fs = Fd 12 1 50 + (753)2 Fd = 369 = Ft 50 Ft = 238 lb hp = Ft vm (238)(753) = 5.43 hp = 33,000 33,000 (b) Precision cut Page 54 of 57 SECTION 11 – SPUR GEARS 1 78 + vm2 Fd = Ft 78 1 2 Fd = 369 = 78 + (753) Ft 78 Ft = 273 lb hp = 666. Ft vm (273)(753) = 6.23 hp = 33,000 33,000 A planetary gear train is composed of four gears – the sun gear B , two planet gears C , as shown. Gears B and C have 20 teeth each, gear D has 60; Pd = 10 , 1 b = 1 − in ., 20o F.D. teeth, cast iron, class 20, nB = 1750 rpm . (a) Determine the 4 speed of the arm. (b) What horsepower may be transmitted continuously? Note that the dynamic load (AGMA equation , §13.15) depends on the speed of tooth engagement, which is not the absolute pitch line speed of B (pitch-line speed relative to arm for B − C ). Check the speed of tooth engagement of both B − C and C − D . (c) If the designer wishes to increase the power transmitted by using three planet gears, instead of two, what changes must be made in tooth numbers so that the gears can be assembled with the planets 120o apart? Solution: nL = enF + na (1 − e ) − (20)(20) 1 e= =− (20)(60) 3 nF = nB = 1750 rpm nL = 0 = nD 1 1 0 = − (1750 ) + na 1 + 3 3 na = 437.5 rpm (assumed) (a) Speed pf arm = 437.5 rpm (b) For BC vm = πD p nB A 12 Page 55 of 57 Np (nB − n A ) π 20 (1750 − 437.5) D p 10 = = = 687 fpm 12 12 π SECTION 11 – SPUR GEARS For BC vm = − πDg nD A 12 N −π g Dp = (nD − n A ) − π 60 (0 − 437.5) 10 = = 687 fpm 12 12 vm = 687 fpm < 1000 fpm Use commercial teeth 600 + vm Ft 600 For cast-iron, continuous service Fs ≥ N sf Fd Fd = For cast iron, assume N sf = 1 Fs = sbY K f Pd Let K f = 1.2 s = 0.4su = 0.4(20) = 8 ksi = 8000 psi For N = 20 , Y = 0.32 , load at tip, 20o F.D. (8000)1 1 (0.32) Fs = 4 (1.2)(10) = 267 lb Fs = Fd 600 + 687 267 = Ft 600 Ft = 124 lb hp = Ft vm (124)(687) = 2.6 hp = 33,000 33,000 (c) For each planet gear, 2 .6 hp = (1.15) = 1.495 hp 2 For 3 planet gears 1.495 hp = (3) = 3.9 hp 1.15 sbY 600 + vm Fs = = Ft K f Pd 600 Page 56 of 57 SECTION 11 – SPUR GEARS 33,000(3.9 ) = 187 lb 687 assume Y = 0.3 (8000)1 1 (0.30) 600 + 687 4 Fs = = (187) 600 (1.2)(Pd ) Pd = 6.65 Ft = use Pd = 7 N = DPd = (2)(7 ) = 14 teeth - end - Page 57 of 57 SECTION 12 – HELICAL GEARS DESIGN PROBLEMS 701. For continuous duty in a speed reducer, two helical gears are to be rated at 7.4 hp at a pinion speed of 1750 rpm; m w ≈ 2.75 ; the helix angle 15o ; 20o F.D. teeth in the normal plane; let N p = 21 teeth, and keep b < 2 D p . Determine the pitch, face, N g , and the material and heat treatment. Use through-hardened teeth with a maximum of 250 BHM (teeth may be cut after heat treatment). Solution: ψ = 15o φn = 20o πD p n p vm = 12 N p 21 Dp = = Pd Pd n p = 1750 rpm 21 (1750) Pd 9621 vm = = 12 Pd 33,000hp (33,000 )(7.4 ) Ft = = = 25.38 Pd vm 9621 P d π b ≤ 2 Dp 21 42 b = 2 = Pd Pd 0.05vm (Ft + Cb cos 2 ψ )cosψ Fd = Ft + lb 1 2 2 0.05vm + (Ft + Cb cos ψ ) Table AT 25 Assume C = 1660 ψ = 15o Fd = 25.38Pd + 9621 42 25.38Pd + 1660 cos 2 15 cos15 0.05 Pd Pd 9621 42 + 25.38Pd + 1660 cos 2 15 0.05 Pd Pd Page 1 of 14 1 2 lb SECTION 12 – HELICAL GEARS Fd = 25.38 Pd + 465 65050 25.38 Pd + Pd Pd 481 65050 + 25.38 Pd + Pd Pd For continuous service: Fw ≥ Fd bD pQK g Fw = cos 2 ψ 2mg 2(2.75) Q= = = 1.467 mg + 1 2.75 + 1 Table At 26, Bhn = 250 Sum of BHN = 500, φn = 20o K g = 131 42 21 (1.467 )(131) 181,670 Fw = = 2 Pd2 Pd Pd cos 15 Fw ≥ Fd By trial and error method Pd 7 6 Fd 3967 4758 Fw 3708 5046 use Pd = 6 21 21 Dp = = = 3.5 in Pd 6 42 42 b= = = 7 in Pd 6 9621 9621 vm = = = 1604 fpm Pd 6 Fig. AF 19, permissible error = 0.0018 in Fig. AF 20 Use carefully cut gears, Pd = 6 Error = 0.001 in is o.k. For material Strength sbY cosψ Fs = K f Pd Page 2 of 14 1 2 lb SECTION 12 – HELICAL GEARS Np 21 = 23 cos ψ cos3 15 Table AT 24, Load near middle N ep = 23 , φn = 20o FD N ep = 3 = Y = 0.565 assume K f = 2.0 Fs = N sf Fd assume N sf = 2.0 s (7 )(0.565) cos15 = (4758)(2) (2 )(6) s = 29,892 psi s use sn = u 3 su = 3(29,892) = 89,676 psi Use C1050, OQT 1100 F, su = 122 ksi , BHN = 248 < 250 Ans. Pd = 6 b = 7 in N g = mw N p = (2.75)(21) = 58 Material. C1050, OQT 1100 F 703. A pair of helical gears, subjected to heavy shock loading, is to transmit 50 hp at 3 1750 rpm of the pinion.; mg = 4.25 ; ψ = 15o ; minimum D p = 4 in. ; continuous 4 o service, 24 hr/day; 20 F.D. teeth in the normal plane, carefully cut; throughhardened to a maximum BHN = 350. Decide upon the pitch, face width, material and its treatment. Solution: π (4.75)(1750) vm = = 2176 fpm 12 33,000hp (33,000 )(50 ) Ft = = = 758 lb vm (2176) Dynamic load: 0.05vm Ft + Cb cos 2 ψ cosψ Fd = Ft + lb 1 2 2 0.05vm + Ft + Cb cos ψ Fig. AF 19, vm = 2176 fpm Permissible error = 0.0014 in ( ) ( Page 3 of 14 ) SECTION 12 – HELICAL GEARS Use carefully cut gears, e = 0.001 in , Pd = 5 as standard Table AT 25, Steel and steel, 20o FD C = 1660 Fd = 758 + ( ) 0.05(2176) 758 + 1660b cos 2 15 cos15 ( 0.05(2176) + 758 + 1660b cos 15 105.1(758 + 1548.8b ) Fd = 758 + 1 lb 108.8 + (758 + 1548.8b )2 Wear load: bD pQK g Fw = cos 2 ψ 2mg 2(4.25) Q= = = 1.619 mg + 1 4.25 + 1 Table At 26, 20o FD, Sum of BHN =2(350)=700 K g = 270 2 b(4.75)(1.619 )(270 ) = 2225b cos 2 15 2π Fw ≥ Fd , bmin = 2 Pa = = 4.69 in. Pd tanψ By trial and error method Fw = b 5 6 Fd 5203 5811 Fw 11125 13350 use b = 5 in Material: Strength: sbY sbY cosψ Fs = = K f Pdn K f Pd Np N ep = cos3ψ N p = Pd D p = (5)(4.375) = 22 22 = 25 cos3 15 Table AT 24, Load near middle N ep = Page 4 of 14 1 2 ) lb SECTION 12 – HELICAL GEARS N ep = 25 , φn = 20o FD Y = 0.580 assume K f = 1.7 s (5)(0.580) cos15 = 0.32955s (1.7 )(5) Fs = N sf Fd for 24 hr/day service, heavy shock loading N sf = 1.75 Fs = 0.32955s = (1.75)(5203) s = 27,629 psi s use sn = u 3 su = 3(27,629) = 82,887 psi Table AT 9 Use 4150, OQT 1200 F, su = 159ksi , BHN = 331 < 350 Ans. Pd = 5 b = 5 in Material. 4150, OQT 1200 F 705. Design the teeth for two herringbone gears for a single-reduction speed reducer with mw = 3.80 . The capacity is 36 hp at 3000 rpm of the pinion; ψ = 30o ; F.D. teeth with φn = 20o . Since space is at a premium, the initial design is for N p = 15 teeth and carburized teeth of AISI 8620; preferably b < 2 D p . Solution: N 15 Dp = p = Pd Pd b ≈ 2Dp b = 2 Dp = vm = 30 Pd πD p n p 12 15 π (3000) P 11,781 vm = d = 12 Pd Page 5 of 14 SECTION 12 – HELICAL GEARS 33,000hp (33,000 )(36 ) = = 101Pd vm 11,781 Pd Dynamic load 0.05vm Ft + Cb cos 2 ψ cosψ Fd = Ft + lb 1 2 0.05vm + Ft + Cb cos ψ 2 Ft = ( ) ( φn = 20 ) o ψ = 30 o Assume C = 1660 , Table AT 25, 20o FD 11,781 30 101Pd + 1660 cos 2 30 cos 30 0.05 Pd Pd Fd = 101Pd + lb 1 2 11,781 30 + 101Pd + 1660 cos 2 30 0.05 Pd Pd 510 37,350 101Pd + Pd Pd Fd = 101Pd + lb 1 2 589 37,350 + 101Pd + Pd Pd Wear load bD pQK g Fw = cos 2 ψ 2 mg 2(3.80) Q= = = 1.583 mg + 1 3.80 + 1 For AISI 8620, carburized, 20o FD K g = 750 for 1010 cycles 30 15 (1.583)(750) 712,350 Fw = = 2 Pd2 Pd Pd cos 30 By trial and error, Fw ≥ Fd Pd Fd Fw 5 4433 28,494 4 5454 44,522 6 3817 19,788 8 3173 11,130 9 3008 8794 For carefully cut gears, e = 0.001 vmax = 1400 fpm (Fig. AF 9) Page 6 of 14 SECTION 12 – HELICAL GEARS Pd 5 4 6 8 9 11,781 Pd 2356.2 1963.5 1683 1473 1309 fpm vm = use Pd = 9 Fd = 3008 lb Fw = 5794 lb > Fd 30 30 b= = = 3.3 in Pd 9 use b = 3.0 in To check for strength sbY sbY cosψ Fs = = K f Pdn K f Pd Np N ep = cos3ψ N p = 15 15 = 23 cos3 30 Table AT 24, Load near middle N ep = 23 , φn = 20o FD N ep = Y = 0.565 assume K f = 1.7 8620, SOQT 450, su = 167 ksi s sn = u 3 su 167 sn = = = 83.5 2 2 (83,500)(3.0)(0.565)cos 30 = 8011 lb > F (= 3008 lb ) Fs = d (1.7)(9) Designed Data: Pd = 9 b = 3.0 in N p = 15 N g = mw N p = (3.8)(15) = 57 Page 7 of 14 SECTION 12 – HELICAL GEARS N p 15 = = 1.67 in Pd 9 N 57 Dg = g = = 6.33 in Pd 9 Dp = CHECK PROBLEMS 707. The data for a pair of carefully cut gears are: Pdn = 5 , φn = 20o ,ψ = 12o , b = 3.5 in. , N p = 18 , N g = 108 teeth; pinion turns 1750 rpm. Materials: pinion, SAE 4150, OQT to BHN = 350; gear, SAE 3150, OQT to BHN = 300. Operation is with moderate shock for 8 to 10 hr./day. What horsepower may be transmitted continuously? Solution: N Dp = p Pd Pd = Pdn cosψ = (5)cos15 = 4.89 18 Dp = = 3.681 in 4.89 Wear load bD pQK g Fw = cos 2 ψ b = 3.5 in. 2Ng 2(108) Q= = = 1.7143 N p + N g 18 + 108 Table AT 26, φn = 20o Sum of BHN = 350 + 300 = 650 K g = 233 Fw = (3.5)(3.681)(1.7143)(233) = 5379 lb cos 2 12 Strength of gear sbY Fs = lb K f Pdn For gear: SAE 3150, OQT to BHN = 300 su = 151 ksi sn = 0.5su = 0.5(151) = 75.5 ksi Ng 108 N eg = = = 116 3 cos ψ cos3 12 Page 8 of 14 SECTION 12 – HELICAL GEARS Table AT 24, Load near middle, φn = 20o Y = 0.763 snY = (75.5)(0.763) = 57.6 For pinion: SAE 4150, OQT to BHN = 350 su = 0.5BHN = 0.5(350) = 175 ksi sn = 0.5su = 0.5(175) = 87.5 ksi Np 18 N ep = = = 19 3 cos ψ cos3 12 Table AT 24, Load near middle, φn = 20o Y = 0.534 snY = (87.5)(0.534) = 46.7 Therefore use pinion as weak Assume K f = 1.7 Fs = (87,500)(3.5)(0.534) = 19,240 lb (1.7 )(5) For moderate shock, 8 to 10 hr./day Use N sf = 1.5 Fs ≥ N sf Fd 19,240 = 1.5Fd Fd ≤ 12,827 lb Therefore use Fd = Fw = 5379 lb Fd = Ft + 0.05vm (Ft + Cb cos 2 ψ )cosψ lb 1 0.05vm + (Ft + Cb cos 2 ψ )2 Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 πD p n p π (3.681)(1750) vm = = = 1686 fpm 12 12 0.05(1686) Ft + 1660(3.5) cos 2 12 cos12 Fd = Ft + lb 1 2 0.05(1686) + Ft + 1660(3.5) cos 12 2 82.46[Ft + 5559] Fd = Ft + = 5379 lb 1 84.3 + [Ft + 5559]2 Trial and error Ft = 1800 lb Fv (1800)(1686) = 92 hp hp = t m = 33,000 33,000 [ ] [ Page 9 of 14 ] SECTION 12 – HELICAL GEARS 708. Two helical gears are used in a single reduction speed reducer rated at 27.4 hp at a motor speed of 1750 rpm; continuous duty. The rating allows an occasional 100 % momentary overload. The pinion has 33 teeth. Pdn = 10 , b = 2 in. , φn = 20o , ψ = 20o , mw = 2.82 . For both gears, the teeth are carefully cut from SAE 1045 with BHN = 180. Compute (a) the dynamic load, (b) the endurance strength; estimate K f = 1.7 . Also decide whether or not the 100 % overload is damaging. (c) Are these teeth suitable for continuous service? If they are not suitable suggest a cure. (The gears are already cut.) Solution: N Dp = p Pd Pd = Pdn cosψ = (10) cos15 = 9.66 33 Dp = = 3.42 in 9.66 πD p n p π (3.42)(1750) vm = = = 1567 fpm 12 12 33,000hp 33,000(27.4 ) Ft = = = 577 lb vm 1567 (a) Dynamic load 0.05vm (Ft + Cb cos 2 ψ )cosψ Fd = Ft + lb 1 0.05vm + (Ft + Cb cos 2 ψ )2 Fig. AF 20, carefully cut gears, Pdn = 10 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 b = 2 in Fd = 577 + [ ] 0.05(1567 ) 577 + 1660(2)cos 2 15 cos15 [ (b) Endurance strength sbY Fs = lb K f Pdn For SAE 1045, BHN = 180 su = 0.5BHN = 0.5(180) = 90 ksi sn = 0.5su = 0.5(90) = 45 ksi Np 33 N ep = = = 37 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o Page 10 of 14 1 2 ] 0.05(1567 ) + 577 + 1660(2)cos 15 2 = 2578 lb SECTION 12 – HELICAL GEARS Y = 0.645 K f = 1 .7 sbY (45,000)(2)(0.645) = 3415 lb = (1.7 )(10) K f Pdn For 100 % overload Ft = 2(577 ) = 1154 lb Fs = Fd = Ft + 0.05vm (Ft + Cb cos 2 ψ )cosψ 1 2 lb 0.05vm + (Ft + Cb cos ψ ) 2 Fd = 1154 + [ ] 0.05(1567 ) 1154 + 1660(2)cos 2 15 cos15 [ (c) Fw = 1 2 ] 0.05(1567 ) + 1154 + 1660(2) cos 15 Since Fs ≈ Fd , 100 % overload is not damaging 2 = 3475 lb bD pQK g cos 2 ψ b = 2 in. 2mw 2(2.82 ) Q= = = 1.476 mw + 1 2.82 + 1 Table AT 26, φn = 20o Sum of BHN = 2(180) = 360 K g = 62.5 Fw = (2)(3.42)(1.476)(62.5) = 676 lb < F (= 2578 lb ) d cos 2 15 Therefore not suitable for continuous service. Cure: Through hardened teeth For Bhn 2578 Kg = (62.5) = 238 676 min Bhn = 0.5(650) = 325 709. Two helical gears are used in a speed reducer whose input is 100 hp at 1200 rpm, from an internal combustion engine. Both gears are made of SAE 4140, with the pinion heat treated to a BHN 363 – 415, and the gear to 321 – 363; let the teeth be F.D.; 20o pressure angle in the normal plane; carefully cut; helix angle ψ = 15o ; N p = 22 , N g = 68 teeth; Pd = 5 , b = 4 in . Calculate the dynamic load, the endurance strength load, and the limiting wear load for the teeth. Should these gears have long life if they operate continuously? (Data courtesy of the Twin Disc Clutch Co.) Solution: Page 11 of 14 SECTION 12 – HELICAL GEARS N p 22 = = 4.4 in Pd 5 πD p n p π (4.4)(1200) vm = = = 1382 fpm 12 12 33,000hp 33,000(100 ) Ft = = = 2388 lb vm 1382 Dynamic load 0.05vm (Ft + Cb cos 2 ψ )cosψ Fd = Ft + lb 1 2 0.05vm + (Ft + Cb cos ψ )2 Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in Table AT 25, steel and steel, 20o FD C = 1660 b = 4 in Dp = Fd = 2388 + [ ] 0.05(1382) 2388 + 1660(4) cos 2 15 cos15 [ 1 2 ] 0.05(1382 ) + 2388 + 1660(4)cos 15 Endurance strength load sbY cosψ Fs = lb K f Pd 2 = 5930 lb Assume K f = 1.7 Pinion sn = 0.25BHN = 0.25(363) = 90.75 ksi Np 22 N ep = = = 25 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o Y = 0.580 sbY cosψ (90,750)(4 )(0.580) cos15 Fs = = = 23,925 lb (1.7 )(5) K f Pd Gear sn = 0.25BHN = 0.25(321) = 80.25 ksi Np 68 N ep = = = 75 3 cos ψ cos3 15 Table AT 24, Load near middle, φn = 20o Y = 0.735 sbY cosψ (80,250 )(4)(0.735)cos15 Fs = = = 26,811 lb (1.7 )(5) K f Pd use Fs = 23,925 lb Page 12 of 14 SECTION 12 – HELICAL GEARS Limiting Wear Load bD pQK g Fw = cos 2 ψ Table AT 26, φn = 20o Sum of BHN = 684 to 778 use 700 K g = 270 Q= 2Ng 2(68) = = 1.511 N p + N g 22 + 68 Fw = (4)(4.4)(1.511)(270) = 7696 lb cos 2 15 Since Fw (= 7696 lb ) > Fd (= 5930 lb ) these gears have long life if they operate continuously. CROSSED HELICAL 710. Helical gears are to connect two shafts that are at right angles ( N1 = 20 , N 2 = 40 , Pdn = 10 ,ψ 1 = ψ 2 = 45o ). Determine the center distance. Solution: πD cosψ 1 N1 = 1 = Pdn D1 cosψ 1 Pcn 20 = (10)(D1 )cos 45 D1 = 2.83 in N 2 = Pdn D2 cosψ 2 40 = (10)(D2 )cos 45 D2 = 5.66 in C = 12 (D1 + D2 ) = 12 (2.83 + 5.66) = 4.25 in 712. Two shafts that are at right angles are to be connected by helical gears. A tentative design is to use N1 = 20 , N 2 = 60 , Pdn = 10 , and a center distance of 6 in. What must be the helix angles? Solution: Σ = ψ 1 +ψ 2 = 90o N1 D1 = Pdn cosψ 1 N2 D2 = Pdn cosψ 2 1 C = 2 (D1 + D2 ) Page 13 of 14 SECTION 12 – HELICAL GEARS N1 N2 + Pdn cosψ 1 Pdn cosψ 2 20 60 + 2(6 ) = 10 cosψ 1 10 cosψ 2 2 6 12 = + cosψ 1 cosψ 2 1 3 6= + cosψ 1 cosψ 2 By trial and error method 1 3 + 6= cosψ 1 sinψ 1 2C = ψ 1 = 39.5o ψ 2 = 50.5o - end - Page 14 of 14 SECTION 13 – BEVEL GEARS DESIGN PROBLEMS 751. Decide upon the pitch, face, N g , material, and heat treatment of a pair of straight bevel gears to transmit continuously and indefinitely a uniform loading of 5 hp at 900 rpm of the pinion, reasonable operating temperature, high reliability; mg ≈ 1.75 ; D p ≈ 3.333 in . Pinion overhangs, gear is straddle mounted. Solution: ( L = rp2 + rg2 tan γ p = 1 2 ) 1 1 = mg 1.75 γ p = 29.75o L sin γ p = rp L sin 29.75 = 3.333 2 L = 3.358 in 33,000hp Ft = lb vm πD p n p π (3.333)(900) vm = = = 785.4 fpm 12 12 33,000(5) Ft = = 210 lb 785.4 Fd = (VF )N sf K m Ft 1 1 50 + vm2 50 + (785.4 ) 2 VF = = = 1.56 50 50 One gear straddle, one not K m = 1.2 Table 15.2, uniform N sf = 1.0 Fd = (1.56)(1.0)(1.2 )(210) = 393 lb Wear load s2 C Fw = D pbI cd2 l C e K t Cr 2 D p = 3.333 in b = 0.3L = 0.3(3.358) = 1.0 in Temperature factor K t = 1.0 , reasonable operating temperature Life factor for wear Page 1 of 17 SECTION 13 – BEVEL GEARS Cl = 1.0 for indefinite life Reliability factor for wear Cr = 1.25 high reliability Geometry factor for wear, Fig. 15.7 Assume I = 0.080 Elastic coefficient (Table 15.4) Steel on steel , Ce = 2800 Fw = Fd 2 (3.333)(1.0)(0.08) (scd ) 2 1.0 (2800) (1.0)(1.25) 2 = 393 scd = 134,370 psi Table 15.3, use Steel, (300) scd = 135 ksi Strength of bevel gears s bJ K l Fs = d Pd K s K t K r Size factor, assume K s = 0.71 Life factor for strength K l = 1 for indefinite life Temperature factor, K t = 1 good operating condition Reliability factor K r = 1.5 high reliability Geometry factor for strength (Fig. 15.5) Assume J = 0.240 b = 1.0 in sd = design flexural stress Min. BHN = 300 sd = 19 ksi Fs = F d (19,000)(1.0)(0.240) Pd 1 (0.71)(1)(1.5) = 393 Pd = 11 say Pd = 10 10 10 so that b = = = 1.0 in Pd 10 Dg = D p mg = (3.333)(1.75) = 5.833 in Page 2 of 17 SECTION 13 – BEVEL GEARS N g = Pd Dg = (10 )(5.833) = 58.33 say N g = 58 Use Pd = 10 , b = 1.0 in , N g = 58 Material = steel, min. Bhn = 300 752. A pair of steel Zerol bevel gears to transmit 25 hp at 600 rpm of the pinion; mg = 3 ; let N p ≈ 20 teeth; highest reliability; the pinion is overhung, the gear straddle mounted. An electric motor drives a multi-cylinder pump. (a) Decide upon the pitch, face width, diameters, and steel (with treatment) for intermittent service. (b) The same as (a) except that indefinite life is desired. Solution: N 20 Dp = p = Pd Pd 20 (600) πD p n p Pd 1000π vm = = = fpm 12 12 Pd 10 Let b = Pd Dynamic load Fd = (VF )N sf K m Ft π 33,000hp lb vm 33,000(25) Ft = = 262.6 Pd 1000π Pd Ft = 1 1000π 50 + 1 2 Pd 50 + vm 1.121 1.121 VF = = = 1 + 12 = 1 + 50 50 Pd Pd Table 15.2, electric motor drives a multi-cylinder pump Service factor, N sf = 1.25 2 One gear straddle, one not, K m = 1.2 1.121 (1.25)(1.2 )(262.6 )Pd = 394 Pd 1 + 1.121 Fd = 1 + Pd Pd (a) Strength of Bevel Gears s bJ K l Fs = d Pd K s K t K r Size factor, assume K s = 0.71 Page 3 of 17 SECTION 13 – BEVEL GEARS Life factor for strength Intermittent service, use K l = 4.6 Temperature factor, say K t = 1.0 Reliability factor, highest reliability K r = 3.0 Geometry factor for strength N mg = g Np N p = 20 N p = 3(20 ) = 60 Fig. 15.5, J = 0.205 10 b= Pd Design flexural stress, steel Assume sd = 15 ksi Fs = Fd (15,000) 10 (0.205)(4.6) 1.121 = 394 Pd 1 + P d 1.121 66,408 = 394 P d 1 + Pd2 P d Pd = 4.814 say Pd = 5 10 10 b= = = 2.0 in Pd 5 Pd Pd (0.71)(1.0 )(3) Wear load for bevel gears s2 C Fw = D pbI cd2 l C e K t Cr 2 N p 20 = = 4 in Pd 5 K t = 1.0 Dp = Life factor for wear, intermittent service Cl = 1.5 Reliability factor for wear, highest reliability Cr = 1.25 Geometry factor for wear, Fig. 15.7 Page 4 of 17 SECTION 13 – BEVEL GEARS N p = 20 , N g = 60 I = 0.083 Elastic coefficient, steel on steel (Table 15.4) Ce = 2800 Pd = 5 Fw = Fd 2 2 (4)(2)(0.083) scd 2 1.5 = 394(5)1 + 1.121 (2800) (1.0)(1.25) 5 scd = 155,730 psi Table 15.3 Use steel, min. BHN = 360, scd = 160 ksi Pd = 5 b = 2 in D p = 4 in Dg = mg D p = (3)(4 ) = 12 in steel, min. BHN = 360 (b) For indefinite life, K l = 1.0 , life factor for strength Cl = 1.0 , life factor for wear Strength: s bJ K l Fs = d Pd K s K t K r Fs = Fd (15,000) 10 (0.205)(1.0) 1.121 = 394 Pd 1 + P d 1.121 14,437 = 394 Pd 1 + 2 Pd P d Pd = 2.799 say Pd = 3 10 10 b= = = 3.33 in Pd 3 Pd Pd (0.71)(1.0)(3) Wear load Page 5 of 17 SECTION 13 – BEVEL GEARS scd2 Cl Fw = D pbI 2 C e K t Cr 2 N p 20 = = 6.67 in Pd 3 Fw = Fd Dp = 2 2 (6.67 )(3.33)(0.083) scd 2 1.0 = 394(3)1 + 1.121 (2800) (1.0)(1.25) 3 scd = 113,744 psi Table 15.3 Use steel, min. BHN = 240, scd = 115 ksi Pd = 3 b = 3.33 in D p = 6.67 in Dg = mg D p = (3)(6.67 ) = 20 in steel, min. BHN = 240 753. Decide upon the pitch, face, and number of teeth for two spiral-bevel gears for a speed reducer. The input to the pinion is 20 hp at 1750 rpm; mg ≈ 1.9 ; pinion overhung, gear-straddle mounted. It is hoped not to exceed a maximum D p of 4 3/8-in.; steel gears with minimum 245 BHN on pinion and 210 BHN on gear. The gear is motor-driven, subject to miscellaneous drives involving moderate shock; indefinite life against breakage and wear with high reliability. If the gears designed for the foregoing data are to be subjected to intermittent service only, how much power could they be expected to transmit? Solution: (a) vm = πD p n p π (4.375)(1750) = 2000 fpm 12 12 33,000hp 33,000(20 ) Ft = = = 330 lb vm 2000 Dynamic load Fd = (VF )N sf K m Ft One gear straddle, one not K m = 1.2 Table 15.2 Motor-driven, moderate shock Page 6 of 17 = SECTION 13 – BEVEL GEARS N sf = 1.25 1 70 + vm2 2 VF = , spiral 70 1 1 70 + (2000) 2 VF = = 1.254 70 Fd = (1.254)(1.25)(1.2)(330) = 621 lb 1 2 Wear load s2 C Fw = D pbI cd2 l C e K t Cr 2 D p = 4.375 in Temperature Factor, K t = 1.0 Design contact stresses, BHN = 245 , pinion scd = 116 ksi Life factor for wear Cl = 1.0 , indefinite life Reliability factor for wear Cr = 1.25 , high reliability Geometry factor for wear, Fig. 15.8 Assume I = 0.12 Elastic coefficient, steel on steel (Table 15.4) Ce = 2800 (116,000)2 1.0 Fw = (4.375)(b )(0.12) (2800)2 (1.0)(1.25) Fw = Fd 721b = 621 b = 0.8613 in 7 say b = in = 0.875 in 8 Strength of gear s bJ K l Fs = d Pd K s K t K r sd = design flexural stress min. BHN = 210 sd = 15.4 ksi Size factor, assume K s = 0.71 Page 7 of 17 2 = 721b SECTION 13 – BEVEL GEARS Life factor for strength K l = 1 for indefinite life Temperature factor, Kt = 1 Reliability factor K r = 1.5 high reliability Geometry factor Fig. 15.6 Assume J = 0.28 3543 (15,400)(0.875)(0.28) 1 Fs = = P Pd (0.71)(1)(1.5) d Fs = Fd 3543 = 621 Pd Pd = 5.7 say Pd = 6 7 Then, Pd = 6 , b = in , N p = D p Pd = (4.375)(6 ) = 26 8 N g = mw N p = (1.9 )(26 ) = 50 (b) Intermittent service only Strength s bJ K l Fs = d Pd K s K t K r sd = 15,400 psi (Gear) For Pd = 6 , K s = 0.64 For indefinite service, K l = 4.6 K t = 1.0 , K r = 1.5 Geometry factor, Fig. 15.6, N p = 26 , N g = 50 J = 0.292 (15,400)(0.875)(0.292) 4.6 = 3142 lb Fs = (0.71)(1)(1.5) 6 Wear load s2 C Fw = D pbI cd2 l C e K t Cr D p = 4.375 in K t = 1.0 scd = 116 ksi Ce = 2800 Page 8 of 17 2 SECTION 13 – BEVEL GEARS Cl = 1.5 intermittent service Cr = 1.25 Geometry factor for wear, Fig. 15.8 N p = 26 , N g = 50 I = 0.116 (116,000)2 1.5 Fw = (4.375)(0.875)(0.116 ) (2800)2 (1.0)(1.25) 2 = 1098 lb use Fw = Fd Fd = (VF )N sf K m Ft 1098 = (1.254)(125)(1.2)Ft Ft = 584 lb Fv (584)(2000) = 35 hp hp = t m = 33,000 33,000 CHECK PROBLEMS 755. A pair of straight-bevel gears are to transmit a smooth load of 45 hp at 500 rpm 3 of the pinion; mg = 3 . A proposed design is Dg = 15 in. , b = 2 in. , Pd = 4 . 8 Teeth are carburized AISI 8620, SOQT 450 F. The pinion overhangs, the gear is straddle-mounted. Would these gears be expected to perform with high reliability in continuous service? If not would you expect more than 1 failure in 100? Solution: D 15 D p = g = = 5 in mg 3 πD p n p π (5)(500) vm = = = 655 fpm 12 12 33,000hp 33,000(45) Ft = = = 2267 lb vm 655 Dynamic load Fd = (VF )N sf K m Ft 1 1 50 + vm2 50 + (655) 2 VF = = = 1.512 50 50 One gear straddle, one not K m = 1.2 Smooth load, N sf = 1.0 Fd = (1.512)(1.0)(1.2 )(2267 ) = 4113 lb Page 9 of 17 SECTION 13 – BEVEL GEARS Strength of bevel gears s bJ K l Fs = d Pd K s K t K r Size factor, for Pd = 4 , K s = 0.71 Life factor for strength Kl = 1 Temperature factor, Kt = 1 Geometry factor for strength (Fig. 15.5) N p = Pd D p = (4 )(5) = 20 N g = Pd Dg = (4 )(15) = 60 J = 0.205 sd = 30 ksi (55 – 63 Rc) for carburized teeth Fs = (30,000)(2.375)(0.205) 5143 1 (0.71)(1)(K ) = K r r 4 Fs = Fd 5143 = 4113 Kr K r = 1.25 < 1.5 will not perform high reliability. Wear load s2 C Fw = D pbI cd2 l C e K t Cr 2 D p = 5 in b = 2.375 in Table 15.3, scd = 225 ksi Table 15.4, Ce = 2800 Geometry factor for wear, Figure 15.7 N p = 20 , N g = 60 I = 0.083 Kt = 1 life factor for wear Cl = 1 (225,000)2 1 Fw = (5)(2.375)(0.083) (2800)2 (1)(Cr ) Fw = Fd Page 10 of 17 2 = 6364 Cr2 SECTION 13 – BEVEL GEARS 6364 = 4113 Cr2 Cr = 1.244 ≈ 1.25 , high reliability Since K r < 1.5 ,this will not perform high reliability but 1 in 100, K r ≈ 1.12 < 1.25 756. A gear catalog rates a pair of cast-iron, straight-bevel gears at 15.26 hp at 800 rpm of the 16-tooth pinion; mg = 3.5 , b = 3 in. , Pd = 3 ; pinion overhangs, straddle-mounted gear. Assume the cast iron to be class 30. If the load is smooth is this rating satisfactory, judging by the design approach of the Text for good reliability (a) when strength alone is considered, (b) when long continuous service is desired? Solution: N 16 Dp = p = = 5.333 in Pd 3 πD p n p π (5.333)(800) vm = = = 1117 fpm 12 12 33,000hp 33,000(15.26 ) Ft = = = 451 lb vm 1117 Dynamic load Fd = (VF )N sf K m Ft 1 1 50 + vm2 50 + (1117 ) 2 VF = = = 1.668 50 50 One gear straddle, one not K m = 1.2 Smooth load, N sf = 1.0 Fd = (1.668)(1.0)(1.2)(451) = 903 lb (a) Strength s bJ K l Fs = d Pd K s K t K r Pd = 3 , K s = 0.76 Kl = 1 Kt = 1 K r = 1.5 sd = 4.6 ksi , cast-iron class 30 N p = 16 Page 11 of 17 SECTION 13 – BEVEL GEARS N g = mw N p = (3.5)(16 ) = 56 J = 0.184 (4,600)(3)(0.184) 1 Fs = = 742 lb < 903 lb(= Fd ) 3 (0.76)(1)(1.5) with K l = 1.4 for 106 cycles Fs = (1.4)(742) = 1040 lb > 903 lb(= Fd ) Therefore satisfactory for 106 cycles. (b) Continuous service Wear load s2 C Fw = D pbI cd2 l C e K t Cr 2 D p = 5.333 in b = 3 in Table 15.3, scd = 50 ksi , cast-iron class 30 Table 15.4, cast-iron and cast-iron Ce = 2250 Cl = 1 Kt = 1 Cr = 1.25 Geometry factor for wear, Figure 15.7 N p = 16 , N g = 56 I = 0.077 (50,000)2 1 Fw = (5.333)(3)(0.077 ) (2250)2 (1)(1.25) 2 = 389 lb < Fd (= 903 lb ) Therefore, not satisfactory for long continuous service. 757. An 870-rpm motor drives a belt conveyor through bevel gears having 18 and 72 3 teeth; Pd = 6 , b = 1 in . Both gears are straddle-mounted. What horsepower may 4 these gears transmit for an indefinite life with high reliability if both gears are (a) cast-iron, class 40; (b) AISI 5140, OQT 1000 F; (c) AISI 5140, OQT 1000 F, flame hardened (d) AISI 8620, SOQT 450 F? Solution: N 18 D p = p = = 3 in Pd 6 πD p n p π (3)(870) vm = = = 683 fpm 12 12 Dynamic load Page 12 of 17 SECTION 13 – BEVEL GEARS Fd = (VF )N sf K m Ft Both gears straddle mounted K m = 1.0 Table 15.2, N sf = 1.0 1 1 50 + vm2 50 + (683) 2 = = 1.523 50 50 Fd = (1.523)(1.0)(1.0)Ft = 1.523Ft VF = (a) Cast-iron, class 40 Strength s bJ K l Fs = d Pd K s K t K r sd = 7 ksi , cast-iron class 40 3 b = 1 in 4 K l = 1 , indefinite life Pd = 6 K s = 0.64 Kt = 1 K r = 1.5 , high reliability Figure 15.5, N p = 18 , N g = 72 J = 0.204 (7000)(1.75)(0.204) 1 Fs = = 434 lb 6 (0.64)(1)(1.5) Wear: s2 C Fw = D pbI cd2 l C e K t Cr 2 D p = 3 in 3 b = 1 in 4 Table 15.3, scd = 65 ksi , cast-iron class 40 Table 15.4, cast-iron and cast-iron Ce = 2250 Cl = 1 , indefinite life Kt = 1 Cr = 1.25 , high reliability Geometry factor for wear, Figure 15.7 N p = 18 , N g = 72 I = 0.082 Page 13 of 17 SECTION 13 – BEVEL GEARS 2 ( 65,000) 1 Fw = (3)(1.75)(0.082 ) (2250)2 (1)(1.25) 2 = 230 lb Fd = Fw 1.523Ft = 230 Ft = 151 lb Fv (151)(683) = 3 hp hp = t m = 33,000 33,000 (b) AISI 5140, OQT 1000 F, BHN = 300 Strength sd = 19,000 psi Fs = (19,000)(1.75)(0.204) 1 (0.64)(1)(1.5) = 1178 lb 6 Wear: s2 C Fw = D pbI cd2 l C e K t Cr 2 Table 15.3, scd = 135 ksi Table 15.4, steel and steel, Ce = 2800 2 ( 135,000) 1 Fw = (3)(1.75)(0.082) (2800)2 (1)(1.25) 2 = 640 lb Fd = Fw 1.523Ft = 640 Ft = 420 lb Fv (420)(683) = 8.7 hp hp = t m = 33,000 33,000 (c) AISI 5140, OQT 1000 F, Flame Hardened Strength sd = 13.5 ksi Fs = (13,500)(1.75)(0.204) 1 (0.64 )(1)(1.5) = 837 lb 6 Wear: s2 C Fw = D pbI cd2 l C e K t Cr 2 Table 15.3, scd = 190 ksi Table 15.4, steel and steel, Ce = 2800 Page 14 of 17 SECTION 13 – BEVEL GEARS 2 ( 190,000) 1 Fw = (3)(1.75)(0.082) (2800)2 (1)(1.25) 2 = 1269 lb Fd = Fs 1.523Ft = 837 Ft = 550 lb Fv (550)(683) = 11.4 hp hp = t m = 33,000 33,000 (d) AISI 86200, SOQT 450 F, carburized Strength sd = 30 ksi (55 – 63 Rc) Fs = (30,000)(1.75)(0.204) 1 (0.64)(1)(1.5) = 1859 lb 6 Wear: s2 C Fw = D pbI cd2 l C e K t Cr 2 Table 15.3, scd = 225 ksi Table 15.4, steel and steel, Ce = 2800 2 ( 225,000) 1 Fw = (3)(1.75)(0.082) (2800)2 (1)(1.25) 2 = 1779 lb Fd = Fw 1.523Ft = 1779 Ft = 1168 lb Fv (1168)(683) = 24.2 hp hp = t m = 33,000 33,000 758. A pair of straight-bevel gears transmits 15 hp at a pinion speed of 800 rpm; Pd = 5 , N p = 20 , N p = 60 , b = 2 in . Both gears are made of AISI 4140 steel, OQT 800 F. What reliability factor is indicated for these gears for strength and for wear (a) for smooth loads, (b) for light shock load from the power source and heavy shock on the driven machine? Solution: N 20 Dp = p = = 4 in Pd 5 πD p n p π (4)(800) vm = = = 838 fpm 12 12 Page 15 of 17 SECTION 13 – BEVEL GEARS 33,000hp 33,000(15) = = 591 lb vm 838 Fd = (VF )N sf K m Ft Ft = 1 1 50 + vm2 50 + (838) 2 VF = = = 1.579 50 50 assume K m = 1.0 Fd = (1.579 )(N sf )(1.0 )(591) = 933 N sf Strength of bevel gear s bJ K l Fs = d Pd K s K t K r For AISI 4140, OQT 800 F, BHN = 429 sd = 24 ksi assume K l = 1 Kt = 1 Pd = 5 K s = 0.675 Figure 15.5, N p = 20 , N g = 60 J = 0.205 2916 1 (24,000)(2)(0.205) Fs = = K 5 r (0.675)(1)(K r ) Fs = Fd 2916 = 933 N sf Kr 3.1254 Kr = N sf Wear load: s2 C Fw = D pbI cd2 l C e K t Cr 2 BHN = 429 Table 15.3, scd = 190 ksi Table 15.4, steel and steel, Ce = 2800 D p = 4 in b = 2 in Assume Cl = 1.0 , K t = 1.0 Fig. 15.7, N p = 20 , N g = 60 I = 0.083 Page 16 of 17 SECTION 13 – BEVEL GEARS 2 ( 190,000 ) 1 ( )( )( ) Fw = 4 2 0.083 (2800 )2 (1)(Cr ) 2 = 3058 Cr2 Fd = Fw 3058 Cr2 1.810 Cr = N sf 933 N sf = (a) Table 15.2, smooth load N sf = 1.0 For strength, K r = For wear, Cr = 3.1254 3.1254 = = 3.1254 N sf 1 1.810 1.810 = = 1.810 N sf 1 (b) Table 15.2, light shock source, heavy shock driven N sf = 2.0 3.1254 3.1254 = = 1.5627 N sf 2 1.810 1.810 For wear, Cr = = = 1.2799 N sf 2 For strength, K r = - end - Page 17 of 17 SECTION 14 – WORM GEARS DESIGN PROBLEMS 791. (a) Determine a standard circular pitch and face width for a worm gear drive with an input of 2 hp at 1200 rpm of the triple-threaded worm; the 1.58-in. ( Dw ) is steel with a minimum BHN = 250; gear is manganese bronze (Table AT 3); mw = 12 . Consider wear and strength only. Use a φn to match the lead angle λ . (See i16.13, Text.) (b) compute the efficiency. Solution: 1200 + vmg Ft lb a) Fd = 1200 33,000hp Ft = vmg πDg ng vmg = 12 n 1200 ng = w = = 100 rpm mw 12 Dg = mw Dw tan λ tan λ = N t Pc πDw N P N m P (3)(12)Pc Dg = mw Dw t c = t w c = = 11.46 Pc π π πDw π (11.45Pc )(100) vmg = = 300 Pc 12 33,000(2 ) 220 Ft = = 300 Pc Pc 1200 + 300 Pc 220 55(4 + Pc ) Fd = = lb 1200 Pc Pc Wear load Fw = Dg bK w say b = 2 Pc , Dg = 11.46 Pc Fw = Fd (11.46 Pc )(2 Pc )(K w ) = 55(4 + Pc ) Pc 22.92 Pc2 K w = Page 1 of 19 55(4 + Pc ) Pc SECTION 14 – WORM GEARS Dg 11.46 Pc = = 0.60443Pc mw Dw (12)(1.58) tan λ = By trial and error and using Table AT 27 ( φn ≈ λ ) Kw Pc Pc (std) λ 36 50 0.678 0.605 ¾ 5/8 24.4 20.7 Use φn = 20o , λ = 20.7o , Pc = 5 in 8 Fw = Fd (11.46 Pc )(b )(K w ) = 55(4 + Pc ) Pc 5 55 4 + (11.46) 5 (b)(50) = 5 8 8 8 b = 1.1365 in 5 say b = 1 in 32 To check for strength. sYbPcn sYbPc cos λ Fs = = π π For manganese bronze, s = 30,000 psi φn = 20o Y = 0.392 λ = 20.7o 5 Pc = in 8 5 b = 1 in 32 (30,000)(0.392)1 5 5 cos 20.7 32 8 = 2530 lb > Fd Fs = π use Pc = b =1 5 in 8 5 in 32 Page 2 of 19 λmax i16.11 16 25 φn 14 ½ 20 SECTION 14 – WORM GEARS cos φn − f tan λ (b) e = tan λ cos φn tan λ + f φn = 20o λ = 20.7o πDwnw π (1.58)(1200 ) vr = = = 531 fpm > 70 fpm 12 cos λ 12 cos 20.7 0.32 0.32 = = 0.0334 0.36 vr (531)0.36 cos 20 − 0.0334 tan 20.7 e = tan 20.7 = 0.902 = 90.2% cos 20 tan 20.7 + 0.0334 f = 792. A high-efficiency worm-gear speed reducer is desired, to accept 20 hp from a 1750-rpm motor. The diameter Dw of the integral worm has been estimated to be 7 1 in. ; the next computations are to be for a steel worm with a minimum BHN = 8 250; phosphor-bronze gear (Table AT 3); mw = 11 . Probably, the worm should not have less than 4 threads. (a) Considering wear and strength only (i16.13), decide upon a pitch and face width that satisfies these requirements (i16.11, Text); specifying the pressure angle, diameters, and center distance. How does Dw used compare with that from equation (m), i16.11, Text? What addendum and dedendum are recommended by Dudley? Compute a face length for the worm. (b) Compute the efficiency. What do you recommend as the next trial for a “better” reducer? Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 1750 ng = w = = 159.1 rpm mw 11 PN Pm N P (11)(4 ) Dg = c g = c w t = c = 14 Pc π vmg = π π (14 Pc )(159.1) π = 583Pc 12 33,000(20 ) 1132 Ft = = 583Pc Pc Page 3 of 19 SECTION 14 – WORM GEARS 1200 + vmg Ft lb Fd = 1200 1200 + 583Pc 1132 1132(1 + 0.4858 Pc ) = Fd = lb 1200 Pc Pc (a) Wear Fw = Dg bK w b = 2 Pc , Dg = 14 Pc Fw = Fd (14 Pc )(2 Pc )(K w ) = 1132(1 + 0.4858Pc ) Pc 1132(1 + 0.4858 Pc ) 28 Pc2 K w = Pc Table AT 27, steel, min. BHN = 250, and bronze And by trial and error ethod NP 4(Pc ) tan λ = t c = = 0.6791Pc πDw (π )(1.875) By trial and error and using Table AT 27 Kw Pc Pc (std) λ 36 50 60 1.213 1.071 1.000 1¼ 1¼ 1.0 40.33 40.33 34.18 Use φn = 25o , λ = 34.18o , Pc = 1 in Fw = Fd (14 Pc )(b )(K w ) = 1132(1 + 0.4858Pc ) Pc (14)(1)(b )(60) = 1132(1 + 0.4858) 1 b = 2 in To check for strength sYbPcn sYbPc cos λ Fs = = π π For phosphor-bronze, s = sn = 31,000 psi Page 4 of 19 λmax i16.11 16 25 35 φn 14 ½ 20 25 SECTION 14 – WORM GEARS For φn = 25o , Y = 0.470 (31,000)(0.470 )(2.0)(1.0)cos 34.18 = 7674 lb > F , ok Fs = d π use Pc = 1.0 in b = 2.0 in φn = 25o 7 Dw = 1 in 8 7 Dg = mw Dw tan λ = (11)1 tan 34.18 = 14.0 in 8 1 1 7 C = (Dw + Dg ) = 1 + 14 = 7.9375 in 2 2 8 Equation (m) (7.9375)0.875 = 2.785 in > 1.875 in , ok C 0.875 Dw = in = 2 .2 2 .2 Addendum and dedendum (by Dudley) Addendum = a = 0.3183Pcn = 0.3183Pc cos λ = 0.3183(1.0) cos 34.18 = 0.2633 in Whole depth = 0.7 Pcn = 0.7 Pc cos λ = 0.7(1.0)cos 34.18 = 0.5791 in Dedendum = whole depth – addendum = 0.5791 in – 0.2633 in = 0.3158 in N Face length = Pc 4.5 + g 50 N g = mw N p = (11)(4 ) = 44 44 Face length = 1.0 4.5 + = 5.38 in 50 Or [ ] Face length = 2 2a(Dg − 2a ) 1 2 Dg = 14 in a = 0.2633 in 1 Face length = 2{2(0.2633)[14 − 2(0.2633)]} 2 = 5.33 in Use Face length = 5.38 in cos φn − f tan λ (b) e = tan λ cos φn tan λ + f πDwnw π (1.875)(1750 ) vr = = = 1038 fpm > 70 fpm 12 cos λ 12 cos 34.18 Page 5 of 19 SECTION 14 – WORM GEARS f = 0.32 0.32 = = 0.0263 ( 70 < vr < 3000 fpm ) 0.36 vr (1038)0.36 φn = 25o , λ = 34.18o , cos 25 − 0.0263 tan 34.18 e = tan 34.18 = 0.94 = 94% cos 25 tan 34.18 + 0.0263 recommendation for next trial φn = 30o λmax = 45o 793. The input to a worm-gear set is to be 25 hp at 600 rpm of the worm with m w = 20 . The hardened-steel worm is to be the shell type with a diameter approximately as given in i16.11, Text, and a minimum of 4 threads; the gear is to be chilled phosphor bronze (Table AT 3). (a) Considering wear and strength only determine suitable values of the pitch and face width. Let φn be appropriate to the value of λ . (b) Compute the efficiency. (c) Estimate the radiating area of the case and compute the temperature rise of lubricant. Is special cooling needed? Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 600 ng = w = = 30 rpm mw 20 PN Pm N P (20 )(4 ) 80 Pc Dg = c g = c w t = c = π π π π 80 Pc (30) π vmg = = 200 Pc 12 33,000(25) 4125 Ft = = 200 Pc Pc π 1200 + vmg Ft lb Fd = 1200 1200 + 200 Pc 4125 687.5(6 + Pc ) Fd = = lb 1200 Pc Pc shell type: Dw = 2.4 Pc + 1.1 in (4)Pc NP 4 Pc tan λ = t c = = πDw (π )(2.4 Pc + 1.1) π (2.4 Pc + 1.1) Page 6 of 19 SECTION 14 – WORM GEARS (a) Wear load Fw = Dg bK w b = 2 Pc , 80 Pc Dg = π Fw = Fd 687.5(6 + Pc ) 80 Pc (2 Pc )(K w ) = Pc π 687.5(6 + Pc ) 50.93Pc2 K w = Pc Table AT 27, Hardened steel and chilled bronze By trial and error method By trial and error and using Table AT 27 ( φn ≈ λ ) Kw Pc Pc (std) λ 90 125 1.017 0.907 1.0 1.0 20 20 Use φn = 20o , λ = 20o , Pc = 1 in Fw = Fd 1687.5(6 + 1) 80 (b )(125) = 1 π b = 1.512 in 5 say b = 1 in 8 cos φn − f tan λ (b) e = tan λ cos φn tan λ + f φn = 20o λ = 20o Dw = 2.4 Pc + 1.1 = 2.4 + 1.1 = 3.5 in πDwnw π (3.5)(600 ) vr = = = 585 fpm 12 cos λ 12 cos 20 0.32 0.32 f = 0.36 = = 0.0323 ( 70 < vr < 3000 fpm ) vr (585)0.36 cos 20 − 0.0323 tan 20 e = tan 20 = 0.9023 = 90.23% cos 20 tan 20 + 0.0323 Page 7 of 19 λmax i16.11 16 25 φn 14 ½ 20 SECTION 14 – WORM GEARS (c) Radiating area ≈ Amin = 43.2C 1.7 sq. in. 1 (Dw + Dg ) 2 Dw = 3.5 in 80 Pc 80(1) Dg = = = 25.5 in C= π π 1 C = (3.5 + 25.5) = 14.5 in 2 1.7 Amin = 43.2(14.5) = 4072 sq.in. Temperature rise = ∆t Qc = hcr A∆t ft − lb min Q = (1 − e )(hpi ) = (1 − 0.9023)(25) = 2.4425 hp(33,000 ft − lb min − hp ) = 80,600 ft − lb min Figure AF 21, A = 4072 sq.in. = 28.3 sq. ft. hcr = 0.42 ft − lb min − sq.in. − F Q = Qc 80,600 = (0.42)(4072)(∆t ) ∆t = 47 F with t1 = 100 F t2 = 147 F < 150 F Therefore, no special cooling needed. 794. A 50-hp motor turning at 1750 rpm is to deliver its power to a worm-gear reducer, whose velocity ratio is to be 20. The shell-type worm is to be made of high-test cast iron; since a reasonably good efficiency is desired, use at least 4 threads; manganese –bronze gear (Table AT 3). (a) Decide upon Dw and φn , and determine suitable values of the pitch and face width. Compute (b) the efficiency, (c) the temperature rise of the lubricant. Estimate the radiating area of the case. Is special cooling needed? Solution: 33,000hp Ft = vmg πDg ng vmg = 12 n 1750 ng = w = = 87.5 rpm mw 20 PN Pm N P (20 )(4 ) 80 Pc Dg = c g = c w t = c = π Page 8 of 19 π π π SECTION 14 – WORM GEARS 80 Pc (87.5) π vmg = = 583Pc 12 33,000(50 ) 2830 Ft = = 583Pc Pc π 1200 + vmg Ft lb (a) Fd = 1200 1200 + 583Pc 2830 1375(2.06 + Pc ) Fd = = lb 1200 Pc Pc Wear load Fw = Dg bK w b = 2 Pc , 80 Pc Dg = π Fw = Fd 1375(2.06 + Pc ) 80 Pc (2 Pc )(K w ) = Pc π 1375(2.06 + Pc ) 50.93Pc2 K w = Pc NP tan λ = t c πDw Shell-type Dw = 2.4 Pc + 1.1 in 4 Pc tan λ = π (2.4 Pc + 1.1) Table AT 27, high-test cast-iron and manganese bronze By trial and error and using Table AT 27 ( φn ≈ λ ) Kw Pc Pc (std) λ 80 115 1.012 0.885 1.0 7/8 20 19.2 Use λ = 19.2o , φn = 20o , Pc = 7 in 8 7 Dw = 2.4 Pc + 1.1 = 2.4 + 1.1 = 3.2 in 8 Fw = Fd Page 9 of 19 λmax i16.11 16 25 φn 14 ½ 20 SECTION 14 – WORM GEARS 7 1375 2.06 + 80 7 8 (b )(115) = 7 π 8 8 b = 1.80 in 7 say b = 1 in 8 cos φn − f tan λ (b) e = tan λ cos φn tan λ + f λ = 19.2o φn = 20o πDwnw vr = 12 cos λ nw = 1750 rpm Dw = 3.2 in πDwnw π (3.2 )(1750 ) vr = = = 1552 fpm 12 cos λ 12 cos19.2 0.32 0.32 f = 0.36 = = 0.0227 ( 70 < vr < 3000 fpm ) vr (1552)0.36 cos 20 − 0.0227 tan 19.2 e = tan 19.2 = 0.9273 = 92.73% cos 20 tan 19.2 + 0.0227 (c) Q = (1 − e )(hpi ) = (1 − 0.9273)(50 ) = 3.635 hp = 119,955 ft − lb min Qc = hcr A∆t ft − lb min A = Amin = 43.2C1.7 sq.in. 1 C = (Dw + Dg ) 2 Dw = 3.2 in 7 80 80 Pc 8 Dg = = = 22.3 in π π 1 C = (3.2 + 22.35) = 12.75 in 2 1.7 A = 43.2(12.75) = 3272 sq.in. Figure AF 1 3272 A= = 22.7 ft 2 144 hcr = 0.43 ft − lb min − sq.in. − F Page 10 of 19 SECTION 14 – WORM GEARS Q = Qc 119,955 = (0.43)(3272)(∆t ) ∆t = 85 F with t1 = 100 F t2 = 185 F > 150 F Therefore, special cooling is needed. CHECK PROBLEMS 795. A worm-gear speed reducer has a hardened-steel worm and a manganese-bronze gear (Table AT 3); triple-threaded worm with Pc = 1.15278 in. , Dw = 3.136 in. , 1 φn = 25o , b = 2 in. , mw = 12 , nw = 580 rpm . The output is 16 hp. Compute (a) 4 the dynamic load, (b) the endurance strength of the teeth and the indicated service factor on strength, (c) the limiting wear load (is it good for indefinitely continuous service?), (d) the efficiency and input hp, (e) the temperature rise of the oil (estimate case area as Amin , i16.6). (f) Determine the tangential and radial components of the tooth load. (g) Is this drive self-locking? Solution: 33,000hp Ft = vmg πDg ng vmg = 12 nw 580 ng = = = 48.3 rpm mw 20 PN P m N (1.15278)(12 )(3) Dg = c g = c w t = = 13.21 in π vmg = π π (13.21)(48.3) 12 π = 167 fpm 1200 + vmg Ft (a) Fd = 1200 1200 + 167 Fd = Ft 1200 33,000(16 ) Ft = = 3162 lb 167 1200 + 167 Fd = (3162 ) = 3602 lb 1200 Page 11 of 19 SECTION 14 – WORM GEARS (b) Fs = sYbPcn π = sYbPc cos λ π N P (3)(1.15278) tan λ = t c = πDw π (3.136 ) λ = 19.34o For manganese-bronze, s = sn = 30,000 psi For φn = 25o , Y = 0.470 (30,000)(0.470) 2 1 (1.15278)cos19.34 Fs = 4 π = 10,984 lb Service factor F 10,984 = 3.05 N sf = s = 3602 Fd (c) Fw = Dg bK w Dg = 13.21 in b = 2.25 in Table AT 27, hardened-steel worn and manganese bronze gear φn = 25o K w = 100 Fw = (13.21)(2.25)(100) = 2972 lb < Fd (= 3602 lb ) Therefore, not good for indefinitely continuous service cos φn − f tan λ (d) e = tan λ cos φn tan λ + f πDwnw π (1.15278)(580 ) vr = = = 185.5 fpm 12 cos λ 12 cos19.34 0.32 0.32 f = 0.36 = = 0.0488 ( 70 < vr < 3000 fpm ) vr (185.5)0.36 cos 25 − 0.0488 tan 19.34 e = tan 19.34 = 0.85 = 85% cos 25 tan 19.34 + 0.0488 hp 16 hp hpi = o = = 18.82 hp e 0.85 (e) Temperature rise, ∆t Q = (1 − e )(hpi ) = (1 − 0.85)(18.82)(33,000 ) = 93,159 ft − lb min Qc = hcr A∆t ft − lb min A = Amin = 43.2C1.7 sq.in. Page 12 of 19 SECTION 14 – WORM GEARS 1 (Dw + Dg ) 2 1 C = (1.15278 + 13.21) = 7.18 in 2 1.7 A = 43.2(7.18) = 1233 sq.in. Figure AF 1 1233 A= = 8.6 ft 2 144 hcr = 0.47 ft − lb min − sq.in. − F C= Q = Qc 93,159 = (0.47 )(1233)(∆t ) ∆t = 161 F (f) Tangential components on the worm cos φn sin λ + f cos λ cos 25 sin 19.34 + 0.0488 cos19.34 = 3162 Wt = Ft = 1305 lb cos 25 cos19.34 − 0.0488 sin 19.34 cos φn cos λ − f sin λ on the gear Ft = 3162 lb radial components Ft sin φn 3162 sin 25 S= = = 1593 lb cos φn cos λ − f sin λ cos 25 cos19.34 − 0.0488 sin 19.34 (g) λ = 19.34o > 5o , not self-locking 797. A worm-gear speed reducer has a hardened-steel worm and a phosphor-bronze gear. The lead angle of the 5-threaded worm λ = 28o57' , Pc = 1.2812 in. , 1 φn = 25o , b = 2 in. , mw = 8 ; worm speed = 1750 rpm. The gear case is 35 3/8 2 in. high, 22 in. wide, 14 in. deep. Compute (a) the efficiency, (b) the limiting wear load, the strength load, and the corresponding safe input and output horsepowers. (c) The manufacturer rates this reducer at 53-hp input. Is this rating conservative or risky? (d) What is the calculated temperature rise of the oil with no special cooling? (e) The manufacturer specifies that for continuous service power should not exceed 36.5 hp if there is to be no artificial cooling and if ∆t is to be less than 90 F. Make calculations and decide whether the vendor is on the safe side. (Data courtesy of the Cleveland Worm Gear Co.) Solution: Page 13 of 19 SECTION 14 – WORM GEARS cos φn − f tan λ (a) e = tan λ cos φn tan λ + f λ = 28o57' = 28.95o φn = 25o πDwnw vr = 12 cos λ nw = 1750 rpm N P Dg = g c π N g = mw N t = (8)(5) = 40 Dg = (40)(1.2812) = 16.31 in π NP tan λ = t c πDw tan 28.95 = (5)(1.2812) πDw Dw = 3.686 in π (3.686 )(150 ) vr = = 1923 fpm 12 cos 28o57' 0.32 0.32 f = 0.36 = = 0.0210 ( 70 < vr < 3000 fpm ) vr (1923)0.36 cos 25 − 0.0210 tan 28.95 e = tan 28.95 = 0.9475 = 94.75% cos 25 tan 28.95 + 0.0210 (b) Fw = Dg bK w Dg = 16.31 in b = 2.5 in Table AT 27, hardened-steel worn and phosphor bronze gear φn = 25o K w = 100 Fw = (16.31)(2.5)(100) = 4078 lb Fs = sYbPcn π = sYbPc cos λ π For phosphor-bronze, s = sn = 31,000 psi For φn = 25o , Y = 0.470 (31,000)(0.470 )(2.5)(1.2812 )cos 28.95 = 13,000 lb Fs = π Page 14 of 19 SECTION 14 – WORM GEARS For safe input and output 1200 + vmg Ft Fd = 1200 πDg ng vmg = 12 n 1750 ng = w = = 218.75 rpm mw 8 π (16.31)(218.75) vmg = = 934 fpm 12 Fw = Fd 1200 + 934 4078 = Ft 1200 Ft = 2293 lb Fv (2293)(934) = 64.9 hp safe output = hpo = t mg = 33,000 33,000 hp 64.9 safe input = hpi = o = = 68.5 hp e 0.9475 (c) 53-hp input < 68.5 hp. ∴ conservative. (d) Q = (1 − e )(hpi ) = (1 − 0.9475)(68.5)(33,000) = 118,676 ft − lb min Qc = hcr A∆t ft − lb min A = 2[(22)(14) + (22)(35.375)] = 2172.5 sq.in. Figure AF 1 2172.5 = 15 ft 2 A= 144 hcr = 0.45 ft − lb min − sq.in. − F Q = Qc 118,676 = (0.45)(2172.5)(∆t ) ∆t = 121.4 F (e) ∆t ′ = 90 F hpi′ ∆t ′ = hpi ∆t hpi′ 90 = 68.5 124 hpi′ = 50.8 hp Since 36.5 hp < 50.8 hp, therefore on the safe side. Page 15 of 19 SECTION 14 – WORM GEARS HEATING 799. The input to a worm-gear reducer is 50.5 hp at 580 rpm of the 4-threaded worm. The gear case is 22 x 31 x 45 in. in size; φn = 25o , Pc = 1.5 in , Dw = 4.432 in , f = 0.035 , room temperature = 80 F. Compute the steady-state temperature for average cooling. Solution: NP 4(1.5) tan λ = t c = πDw π (4.432 ) λ = 23.3o cos φn − f tan λ e = tan λ cos φn tan λ + f cos 25 − 0.035 tan 23.3 e = tan 23.3 = 0.9025 cos 25 tan 23.3 + 0.035 Q = (1 − e )(hpi ) = (1 − 0.9025)(50.5)(33,000) = 162,484 ft − lb min Qc = hcr A∆t ft − lb min A = 2[(22)(31) + (31)(45)] = 4154 sq.in. Figure AF 1 4154 A= = 28.85 ft 2 144 hcr = 0.42 ft − lb min − sq.in. − F Q = Qc 162,484 = (0.42)(4154 )(∆t ) ∆t = 93 F t1 = 80 F t2 = 173 F 801. A hardened-steel, 4-threaded worm drives a bronze gear; Dw = 1.875 in , Dg ≈ 14 in , Pc = 1.0 in , φn = 25o , area of case ≈ 1500 sq.in. , vr ≈ 1037 fpm ; input = 20 hp at 1750 rpm of the worm; room temperature = 80 F. Compute the steady-state temperature of the lubricant for average ventilation. Solution: NP 4(1.0 ) tan λ = t c = πDw π (1.875) λ = 34.2o f = 0.32 0.32 = = 0.0263 ( 70 < vr < 3000 fpm ) 0.36 vr (1037 )0.36 Page 16 of 19 SECTION 14 – WORM GEARS cos φn − f tan λ e = tan λ cos φn tan λ + f cos 25 − 0.0263 tan 34.2 e = tan 34.2 = 0.94 cos 25 tan 34.2 + 0.0263 Q = (1 − e )(hpi ) = (1 − 0.94)(20)(33,000) = 39,600 ft − lb min Qc = hcr A∆t ft − lb min A = 1500 sq.in. Figure AF 1 1500 = 10.4 ft 2 A= 144 hcr = 0.46 ft − lb min − sq.in. − F Q = Qc 39,600 = (0.46 )(1500)(∆t ) ∆t = 57 F t1 = 80 F t2 = 137 F 802. The input to a 4-threaded worm is measured to be 20.8 hp; Pc = 1.0 in , Dw = 2 in , φn = 25o . The area of the case is closely 1800 sq. in.; ambient temperature = 100 F; oil temperature = 180 F. Operation is at a steady thermal state. Compute the indicated coefficient of friction. Solution: Qc = hcr A∆t ft − lb min Figure AF 1 1800 A= = 12.5 ft 2 144 hcr = 0.46 ft − lb min − sq.in. − F A = 1800 sq.in. ∆t = 180 − 100 = 80 F Qc = hcr A∆t = (0.46)(1800)(80) = 66,240 ft − lb min Q = (1 − e )(hpi )(33,000) ft − lb min Q = Qc (1 − e)(hpi )(33,000) = 66,240 e = 0.9035 cos φn − f tan λ e = tan λ cos φn tan λ + f Page 17 of 19 SECTION 14 – WORM GEARS tan λ = N t Pc 4(1.0 ) = πDw π (2 ) λ = 32.5o cos 25 − f tan 32.5 0.9035 = tan 32.5 cos 25 tan 32.5 + f 0.5217 + 0.9035 f = 0.5774 − 0.4059 f f = 0.0425 FORCE ANALYSIS 804. 1 The input to a 4-threaded worm is 21 hp at 1750 rpm; e = 90% , Dw = 2 in , 4 o Dg = 14 in , N g = 44 , φn = 25 . (a) From the horsepowers in and out, compute the tangential forces on the worm Wt and the gear Ft . (b) Using this value of Ft , compute Wt from equation (k), i16.8, Text. (Check?) (c) Compute the separating force. (d) What is the end thrust on the worm shaft? On the gear shaft? Solution: hpi = 21 hp hpo = (hpi )(e ) = (21)(0.90) = 18.9 hp N Dg mw = g = N t Dw tan λ 44 14 = 4 1 2 tan λ 4 λ = 29.5o cos φn − f tan λ e = tan λ cos φn tan λ + f cos 25 − f tan 29.5 0.90 = tan 29.5 cos 25 tan 29.5 + f 0.4615 + 0.90 f = 0.5128 − 0.32 f f = 0.0420 πDg ng vmg = 12 ng N t = nw N g Page 18 of 19 SECTION 14 – WORM GEARS ng 4 = 1750 44 ng = 159 rpm vmg = π (14)(159) (a) Ft = 12 = 583 fpm 33,000hpo 33,000(18.9) = = 1070 lb vmg 583 33,000hpi vw πD n π (2.25)(1750) vw = w w = = 1031 fpm 12 12 33,000(21) Wt = = 672 lb 1031 Wt = cos φn sin λ + f cos λ cos 25 sin 29.5 + 0.0420 cos 29.5 = 1070 (b) Wt = Ft = 672 lb cos 25 cos 29.5 − 0.0420 sin 29.5 cos φn cos λ − f sin λ Ft sin φn 1070 sin 25 (c) S = = = 589 lb cos φn cos λ − f sin λ cos 25 cos 29.5 − 0.0420 sin 29.5 (d) End thrust Worm shaft = Ft = 1070 lb Gear shaft = Wt = 672 lb - end - Page 19 of 19 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS LEATHER BELTS DESIGN PROBLEMS 841. A belt drive is to be designed for F1 F2 = 3 , while transmitting 60 hp at 2700 rpm of the driver D1 ; mw ≈ 1.85 ; use a medium double belt, cemented joint, a squirrel-cage, compensator-motor drive with mildly jerking loads; center distance is expected to be about twice the diameter of larger pulley. (a) Choose suitable iron-pulley sizes and determine the belt width for a maximum permissible s = 300 psi . (b) How does this width compare with that obtained by the ALBA procedure? (c) Compute the maximum stress in the straight port of the ALBA belt. (d) If the belt in (a) stretches until the tight tension F1 = 525 lb ., what is F1 F2 ? Solution: (a) Table 17.1, Medium Double Ply, Select D1 = 7 in . min. 20 t= in 64 πD n π (7 )(2700 ) vm = 1 1 = = 4948 fpm 12 12 4000 fpm < 4948 fpm < 6000 fpm (F − F2 )vm hp = 1 33,000 F1 − F2 )(4948) ( 60 = 33,000 F1 − F2 = 400 lb F1 = 3F2 3F2 − F2 = 400 lb F2 = 200 lb F1 = 3F2 = 3(200 ) = 600 lb F1 = sbt sd = 300η For cemented joint, η = 1.0 sd = 300 psi 20 F1 = 600 = (300)(b ) 64 b = 6.4 in say b = 6.5 in 757 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (b) ALBA Procedure hp = (hp in., Table 17.1)(bC mC p )(C f1 C f 2 L) Table 17.1, vm = 4948 fpm Medium Double Ply hp in = 12.448 Table 17.2 Squirrel cage, compensator, starting Cm = 0.67 Pulley Size, D1 = 7 in C p = 0.6 Jerky loads, C f = 0.83 hp = 60 = (12.448)(b )(0.67 )(0.6)(0.83) b = 14.5 in say b = 15 in (c) s = F1 = ηbt 1 1 600 = 128 psi 20 (1)(15) 64 1 1 1 (d) 2 Fo2 = F12 + F22 = (600) 2 + (200 )2 Fo = 373.2 lb F1 = 525 lb 1 2 1 2 1 2 2 2(373.2 ) = (525) + F F2 = 247 lb F1 525 = = 2.1255 F2 247 842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The horizontal center distance must be about 8 to 9 ft. for clearance, and operation is continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed recommended as about optimum in the Text? (b) Decide upon a pulley size (iron or steel) and belt thickness, and determine the belt width by the ALBA tables. (c) Compute the stress from the general belt equation assuming that the applicable coefficient of friction is that suggested by the Text. (d) Suppose the belt is installed with an initial tension Fo = 70 lb in . (§17.10), compute F1 F2 and the stress on the tight side if the approximate relationship of the operating tensions 1 1 1 and the initial tensions is F12 + F22 = 2 Fo2 . 758 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Solution: vm = 4000 to 4500 fpm assume vm = 4250 fpm πD n vm = 1 1 12 πD (1750 ) 4250 = 1 12 D1 = 9.26 in say D1 = 10 in 23 in 64 Minimum pulley diameter for vm ≈ 4250 fpm , D1 = 10 in (b) Using Heavy Double Ply Belt, t = Use D1 = 10 in πD n π (10 )(1750) vm = 1 1 = = 4581 fpm 12 12 ALBA Tables hp = (hp in., Table 17.1)(bC mC p )(C f1 C f 2 L) hp in = 13.8 Slip ring motor, Cm = 0.4 Pulley Size, D1 = 10 in C p = 0.7 Table 17.7, 24 hr/day, continuous N sf = 1.8 Assume C f = 0.74 hp = (1.8)(20 ) = (13.8)(b )(0.4)(0.7 )(0.74 ) b = 12.59 in use b = 13 in (c) General belt equation 12 ρvs2 e fθ − 1 F1 − F2 = bt s − 32.2 e fθ 4581 = 76.35 fps 60 ρ = 0.035 lb cu. in. for leather 23 t= in 64 b = 13 in vs = 759 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 33,000(1.8)(20 ) = 260 lb 4581 f = 0.3 on iron or steel D − D1 θ ≈π ± 2 C C = 8 ~ 9 ft use 8.5 ft 1750 D2 = (10 ) = 53 in 330 53 − 10 θ =π − = 2.72 rad 8.5(12 ) fθ = (0.3)(2.72 ) = 0.816 F1 − F2 = e fθ − 1 e 0.816 − 1 = 0.816 = 0.5578 e fθ e 2 23 12(0.035)(76.35) F1 − F2 = 260 = (13) s − (0.5578) 32.2 64 s = 176 psi 1 1 1 (d) F12 + F22 = 2 Fo2 Fo = (70 lb in )(13 in ) = 910 lb F1 − F2 = 260 lb F2 = F1 − 260 lb 1 2 1 1 F1 + ( F1 − 260 )2 = 2(910)2 = 60.33 F1 = 1045 lb F2 = 1045 − 260 = 785 lb 1045 = 224 psi 23 (13) 64 F1 1045 = = 1.331 F2 785 s= F1 = bt 843. A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freon reciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley, D1 = 16 in ; D2 = 53 in , a flywheel; cemented joints;l C = 8 ft . (a) Choose an appropriate belt thickness and determine the belt width by the ALBA tables. (b) Using the design stress of §17.6, compute the coefficient of friction that would be needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial tension was set so that the operating F1 F2 = 2 . Compute the maximum stress in a straight part. (d) The approximate relation of the operating tensions and the 760 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1 2 1 2 2 1 2 o initial tension Fo is F1 + F = 2 F . For the condition in (c), compute Fo . Is it reasonable compared to Taylor’s recommendation? Solution: (a) Table 17.1 πD n π (16)(1140) vm = 1 1 = = 4775 fpm 12 12 Use heavy double-ply belt 23 t= in 64 hp in = 14.1 hp = (hp in., Table 17.1)(bC mC p ) C f1 C f 2 L ( ) line starting electric motor , C m = 0.5 Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor N sf = 1.4 D1 = 16 in , C p = 0.8 assume, C f = 0.74 hp = (1.4)(100 ) = 140 hp hp = 140 = (14.1)(b )(0.5)(0.8)(0.74) b = 33.5 in use b = 34 in (b) §17.6, s d = 400η η = 1.00 for cemented joint. s d = 400 psi 12 ρvs2 e fθ − 1 F1 − F2 = bt s − 32.2 e fθ 4775 vs = = 79.6 fps 60 ρ = 0.035 lb cu. in. for leather 23 t= in 64 b = 34 in 33,000(1.4 )(100) F1 − F2 = = 968 lb 4775 2 12(0.035)(79.6 ) e fθ − 1 23 F1 − F2 = 968 = (34) 400 − fθ 32.2 64 e 761 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS e fθ − 1 = 0.2496 e fθ fθ = 0.28715 D − D1 θ ≈π ± 2 C C = 8 ft 53 − 16 θ =π − = 2.7562 rad 8(12 ) f (2.7562) = 0.28715 f = 0.1042 < 0.3 Therefore satisfactory. (c) F1 − F2 = 968 lb F1 = 2F2 2 F2 − F2 = 968 lb F1 = 2 F2 = 2(968) = 1936 lb F 1936 s= 1 = = 159 psi bt 23 (34) 64 (d) F1 = 1936 lb , F2 = 968 lb 1 1 1 2 Fo2 = F12 + F22 1 2 o 1 1 2 F = (1936) 2 + (968)2 Fo = 1411 lb Fo = 844. 1411 = 41.5 lb in of width is less than Taylor’s recommendation and is reasonable. 34 A 50-hp compensator-started motor running at 865 rpm drives a reciprocating compressor for a 40-ton refrigerating plant, flat leather belt, cemented joints. The diameter of the fiber driving pulley is 13 in., D2 = 70 in ., a cast-iron flywheel; C = 6 ft.11 in. Because of space limitations, the belt is nearly vertical; the surroundings are quite moist. (a) Choose a belt thickness and determine the width by the ALBA tables. (b) Using recommendations in the Text, compute s from the general belt equation. (c) With this value of s , compute F1 and F1 F2 . (d) 1 1 1 Approximately, F12 + F22 = 2 Fo2 , where Fo is the initial tension. For the condition in (c), what should be the initial tension? Compare with Taylor, §17.10. (e) Compute the belt length. (f) The data are from an actual drive. Do you have any recommendations for redesign on a more economical basis? 762 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Solution: (a) vm = πD1n1 = π (13)(865) = 2944 fpm 12 12 Table 17.1, use Heavy Double Ply, Dmin = 9 in for vm = 2944 fpm belts less than 8 in wide 23 t= in 64 ( ) hp = (hp in., Table 17.1)(bC mC p ) C f1 C f 2 L hp in = 9.86 Table 17.2 Cm = 0.67 C p = 0.8 C f = (0.74 )(0.83) = 0.6142 Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating compressor N sf = 1.4 hp = (1.4 )(50) = 70 hp hp = 70 = (9.86)(b)(0.67 )(0.8)(0.6142) b = 21.6 in use b = 25 in (b) General Belt Equation 12 ρvs2 e fθ − 1 F1 − F2 = bt s − 32.2 e fθ b = 25 in 23 t= in 64 ρ = 0.035 lb cu. in. for leather 2944 vs = = 49.1 fps 60 Leather on iron, f = 0.3 D − D1 θ =π − 2 C 70 − 13 = 2.455 rad 6(12) + 11 fθ = (0.3)(2.455) = 0.7365 θ =π − 763 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS e fθ − 1 e 0.7365 − 1 = 0.7365 = 0.5212 e fθ e F1 − F2 = 33,000(1.4 )(50) = 785 lb 2944 2 23 12(0.035)(49.1) F1 − F2 = 785 = (25) s − (0.5212 ) 32.2 64 s = 199 psi Cemented joint, η = 1.0 s = 199 psi (c) F1 = sbt = (199)(25) 23 = 1788 lb 64 F2 = 1788 − 785 = 1003 lb F1 1788 = = 1.783 F2 1003 1 2 o 1 2 1 2 2 (d) 2 F = F1 + F 1 1 1 2Fo2 = (1788) 2 + (1003) 2 Fo = 1367 lb Fo = 1367 = 54.7 lb in 25 Approximately less than Taylor’s recommendation ( = 70 lb/in.) (e) 2 ( D2 − D1 ) L ≈ 2C + 1.57(D2 + D1 ) + L = 2[(6)(12 ) + 11] + 1.57(70 + 13) + 4C (70 − 13)2 = 306 in 4[(6)(12 ) + 11] (f) More economical basis πD n vm = 1 1 12 πD (865) 4500 = 1 12 D1 = 19.87 in use D1 = 20 in 764 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS CHECK PROBLEMS 846. An exhaust fan in a wood shop is driven by a belt from a squirrel-cage motor that runs at 880 rpm, compensator started. A medium double leather belt, 10 in. wide is used; C = 54 in .; D1 = 14 in . (motor), D2 = 54 in ., both iron. (a) What horsepower, by ALBA tables, may this belt transmit? (b) For this power, compute the stress from the general belt equation. (c) For this stress, what is F1 F2 ? (d) If the belt has stretched until s = 200 psi on the tight side, what is F1 F2 ? (e) Compute the belt length. Solution: (a) For medium double leather belt 20 t= in 64 hp = (hp in )(b )C mC pC f Table 17.1 and 17.2 Cm = 0.67 C p = 0.8 C f = 0.74 b = 10 in πD n π (14)(880) vm = 1 1 = = 3225 fpm 12 12 hp in = 6.6625 hp = (6.6625)(10 )(0.67 )(0.8)(0.74) = 26.43 hp 12 ρvs2 e fθ − 1 (b) F1 − F2 = bt s − 32.2 e fθ b = 10 in 20 t= in 64 ρ = 0.035 lb cu. in. 3225 vs = = 53.75 fps 60 D − D1 θ =π − 2 C 54 − 14 θ =π − = 2.4 rad 54 Leather on iron f = 0.3 fθ = (0.3)(2.4 ) = 0.72 765 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS e fθ − 1 e 0.72 − 1 = 0.72 = 0.51325 e fθ e 33,000(26.43) F1 − F2 = = 270 lb 3225 2 20 12(0.035)(53.75) F1 − F2 = 270 = (10 ) s − (0.51325) 32.2 64 s = 206 psi 20 (c) F1 = sbt = (206 )(10) = 644 lb 64 F2 = 644 − 270 = 374 lb F1 644 = = 1.72 F2 374 (d) s = 200 psi 20 F1 = sbt = (200 )(10) = 625 lb 64 F2 = 625 − 270 = 355 lb F1 625 = = 1.76 F2 355 (e) 2 ( D2 − D1 ) L ≈ 2C + 1.57(D2 + D1 ) + 4C (54 − 14)2 = 222 in L = 2(54) + 1.57(54 + 14) + 4(54 ) 847. A motor is driving a centrifugal compressor through a 6-in. heavy, single-ply leather belt in a dusty location. The 8-in motor pulley turns 1750 rpm; D2 = 12 in . (compressor shaft); C = 5 ft . The belt has been designed for a net belt pull of F1 − F2 = 40 lb in of width and F1 F2 = 3 . Compute (a) the horsepower, (b) the stress in tight side. (c) For this stress, what needed value of f is indicated by the general belt equation? (d) Considering the original data,what horsepower is obtained from the ALBA tables? Any remarks? Solution: (a) vm = πD1n1 12 = π (8)(1750) 12 = 3665 fpm b = 6 in F1 − F2 = (40)(6) = 240 lb 766 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS hp = (F1 − F2 )vm = (240)(3665) = 26.65 hp 33,000 33,000 (b) F1 = 3F2 3F2 − F2 = 240 lb F2 = 120 lb F1 = 360 lb F s= 1 bt For heavy single-ply leather belt 13 t= in 64 360 s= = 295 psi 13 (6) 64 12 ρvs2 e fθ − 1 (c) F1 − F2 = bt s − 32.2 e fθ ρ = 0.035 lb cu. in. 3665 vs = = 61.1 fps 60 F1 − F2 = 240 lb 2 12(0.035)(61.1) e fθ − 1 13 F1 − F2 = 240 = (6) 295 − fθ 32.2 64 e e fθ − 1 = 0.7995 e fθ D − D1 θ =π − 2 C 12 − 8 θ =π − = 3.075 rad 5(12) e fθ = 4.9875 fθ = 1.607 f (3.075) = 1.607 f = 0.5226 (d) ALBA Tables (Table 17.1 and 17.2) hp = (hp in )(b )C mC pC f vm = 3665 fpm hp in = 6.965 767 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS b = 10 in Cm = 1.0 (assumed) C p = 0.6 C f = 0.74 hp = (6.965)(6 )(1.0 )(0.6 )(0.74) = 18.6 hp < 26.65 hp 848. A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9in. paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The compensator-started motor turns 1750 rpm; C = 42 in . This is an actual installation. (a) Determine the horsepower from the ALBA tables. (b) Using the general equation, determine the horsepower for this belt. (c) Estimate the service factor from Table 17.7 and apply it to the answer in (b). Does this result in better or worse agreement of (a) and (b)? What is your opinion as to the life of the belt? Solution: πD n π (9)(1750 ) vm = 1 1 = = 4123 fpm 12 12 (a) hp = (hp in )(b )C mC pC f Table 17.1 and 17.2 Medium double leather belt 20 t= in 64 vm = 4123 fpm hp in = 11.15 Cm = 0.67 C p = 0.7 C f = 0.74 b = 10 in hp = (11.15)(10)(0.67 )(0.7 )(0.74) = 38.7 hp 12 ρvs2 e fθ − 1 (b) F1 − F2 = bt s − 32.2 e fθ b = 10 in ρ = 0.035 lb cu. in. s = 400η η = 1.0 cemented joint s = 400 psi D − D1 θ =π − 2 C 768 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 15 − 9 = 2.9987 rad 42 Leather on paper pulleys, f = 0.5 fθ = (0.5)(2.9987 ) = 1.5 θ =π − e fθ − 1 = 0.77687 e fθ 4123 = 68.72 fps vs = 60 2 12(0.035)(68.72) 20 F1 − F2 = (10 ) 400 − (0.77687 ) = 822 lb 32.2 64 hp = (F1 − F2 )vm = (822)(4123) = 102.7 hp 33,000 33,000 (c) Table 17.7 N sf = 1.6 102.7 = 64.2 hp < 102.7 hp 1. 6 Therefore, better agreement hp = Life of belt, not continuous, 60 hp > 38.7 hp . MISCELLANEOUS 849. Let the coefficient of friction be constant. Find the speed at which a leather belt may transmit maximum power if the stress in the belt is (a) 400 psi, (b) 320 psi. (c) How do these speeds compare with those mentioned in §17.9, Text? (d) Would the corresponding speeds for a rubber belt be larger or smaller? (HINT: Try the first derivative of the power with respect to velocity.) Solution: 12 ρvs2 e fθ − 1 F1 − F2 = bt s − 32.2 e fθ (F − F2 )vm hp = 1 33,000 60(F1 − F2 )vs hp = 33,000 hp = 60vs bt 12 ρvs2 e fθ − 1 s − 33,000 32.2 e fθ 60bt e fθ − 1 12 ρvs2 s − vs hp = 33,000 e fθ 32.2 769 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS d (hp ) 60bt e fθ − 1 12 ρvs2 24 ρvs2 s − − = =0 d (vs ) 33,000 e fθ 32.2 32.2 36 ρvs2 s= 32.2 ρ = 0.035 lb cu. in. (a) s = 400 psi 36(0.035)vs2 32.2 vs = 101.105 fps vm = 6066 fpm 400 = (b) s = 320 psi 36(0.035)vs2 32.2 vs = 90.431 fps vm = 5426 fpm 320 = (c) Larger than those mentioned in §17.9 (4000 – 4500 fpm) (d) Rubber belt, ρ = 0.045 lb cu. in. (a) s = 400 psi 36(0.045)vs2 32.2 vs = 89.166 fps vm = 5350 fpm < 6066 fpm 400 = Therefore, speeds for a rubber belt is smaller. 850. A 40-in. pulley transmits power to a 20-in. pulley by means of a medium double leather belt, 20 in. wide; C = 14 ft , let f = 0.3 . (a) What is the speed of the 40-in pulley in order to stress the belt to 300 psi at zero power? (b) What maximum horsepower can be transmitted if the indicated stress in the belt is 300 psi? What is the speed of the belt when this power is transmitted? (See HINT in 849). Solution: 12 ρvs2 e fθ − 1 fθ F1 − F2 = bt s − 32 . 2 e 770 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS hp = 60(F1 − F2 )vs 33,000 hp = 60bt e fθ − 1 12 ρvs2 s − vs 33,000 e fθ 32.2 d (hp ) 60bt e fθ − 1 12 ρvs2 24 ρvs2 s − − = =0 d (vs ) 33,000 e fθ 32.2 32.2 36 ρvs2 for maximum power 32.2 (a) At zero power: s= 12 ρvs2 s= 32.2 s = 300 psi ρ = 0.035 lb cu. in. 12(0.035)vs2 32.2 vs = 151.6575 fps vm = 9100 fpm 300 = Speed, 40 in pulley, n2 = 12vm 12(9100) = = 869 rpm πD2 π (40 ) (b) Maximum power 36 ρvs2 s= 32.2 36(0.035)vs2 300 = 32.2 vs = 87.5595 fps vm = 5254 fpm 60bt e fθ − 1 12 ρvs2 s − vs hp = 33,000 e fθ 32.2 20 in 64 b = 20 in D − D1 θ =π − 2 C 40 − 20 θ =π − = 3.0225 rad 14(12) f = 0.3 t= 771 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS fθ = (0.3)(3.0225) = 0.90675 e fθ − 1 = 0.5962 e fθ 20 60(20 ) 2 64 (0.5962 )300 − 12(0.035)(87.5595) (87.5595) = 118.64 hp = 33,000 32.2 vm = 5254 fpm AUTOMATIC TENSION DEVICES 851. An ammonia compressor is driven by a 100-hp synchronous motor that turns 1200 rpm; 12-in. paper motor pulley; 78-in. compressor pulley, cast-iron; C = 84 in . A tension pulley is placed so that the angle of contact on the motor pulley is 193o and on the compressor pulley, 240o. A 12-in. medium double leather belt with a cemented joint is used. (a) What will be the tension in the tight side of the belt if the stress is 375 psi? (b) What will be the tension in the slack side? (c) What coefficient of friction is required on each pulley as indicated by the general equation? (d) What force must be exerted on the tension pulley to hold the belt tight, and what size do you recommend? Solution: (a) F1 = sbt b = 12 in 20 t= in 64 20 F1 = (375)(12 ) 64 33,000hp vm πD n π (12)(1200) vm = 1 1 = = 3770 fpm 12 12 Table 17.7, N sf = 1.2 (b) F1 − F2 = 33,000(1.2 )(100 ) = 1050 lb 3770 F2 = F1 − 1050 = 1406 − 1050 = 356 lb F1 − F2 = 12 ρvs2 e fθ − 1 fθ (c) F1 − F2 = bt s − 32 . 2 e 3770 vs = = 62.83 fps 60 772 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS ρ = 0.035 lb cu. in. 12(0.035)(62.83) e fθ − 1 20 1050 = (12 ) 375 − e fθ 32.2 64 e fθ − 1 = 0.8655 e fθ fθ = 2.006 Motor pulley π θ = 193o = 193 = 3.3685 rad 180 f (3.3685) = 2.006 f = 0.5955 Compressor Pulley π θ = 2403o = 240 = 4.1888 rad 180 f (4.1888) = 2.006 f = 0.4789 (d) Force: Without tension pulley D − D1 78 − 12 θ1 = π − 2 =π − = 2.356 rad C 84 D − D1 78 − 12 θ2 = π + 2 =π + = 3.9273 rad C 84 773 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS α1 = θ1′ − θ1 − 2 = 3.3685 − 2.356 − π − 2.356 2 = 0.6197 rad = 35.5o 3.9273 − π + 4.1888 − 3.9273 = 0.6544 rad = 37.5o 2 2 Q = F1 (sin α1 + sin α 2 ) = 1406(sin 35.5 + sin 37.5) = 1672 lb of force exerted Size of pulley; For medium double leather belt, vm = 3770 fpm , width = 12 in > 8 in D = 6 + 2 = 8 in α2 = θ2 − π π − θ1 + θ 2′ − θ 2 = 774 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 852. A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted 3 base. In Fig. 17.11, Text, e = 10 in ., h = 19 in . The center of the 11 ½-in. 16 motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley; C = 48 in . (a) With the aid of a graphical layout, find the tensions in the belt for maximum output of the motor if it is compensator started. What should be the width of the medium double leather belt if s = 300 psi ? (c) What coefficient of friction is indicated by the general belt equation? (Data courtesy of Rockwood Mfg. Co.) Solution: (a) R = 1915 lb Graphically b ≈ 26 in a ≈ 9 in [∑ M B =0 ] eR = F1a + F2b (10)(1915) = (F1 )(9) + (F2 )(26) 9 F1 + 26 F2 = 19,150 For compensator started hp = 1.4(rated hp ) = 1.4(40 ) = 56 hp 33,000hp F1 − F2 = vm 775 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS vm = πD1n1 π (11.5)(685) = 2062 fpm 12 33,000(56) F1 − F2 = = 896 lb 2062 F2 = F1 − 896 Substituting 9 F1 + 26(F1 − 896) = 19,150 F1 = 1213 lb F2 = 1213 − 896 = 317 lb 12 = For medium leather belt, t = 20 in 64 F1 = sbt 20 1213 = (300)(b ) 64 b = 13 in 12 ρvs2 e fθ − 1 (c) F1 − F2 = bt s − 32.2 e fθ 2062 vs = = 34.37 fps 60 ρ = 0.035 lb cu. in. 12(0.035)(34.37 ) e fθ − 1 20 896 = (13) 300 − e fθ 32.2 64 e fθ − 1 = 0.775 e fθ fθ = 1.492 D − D1 60 − 11.5 θ =π − 2 =π − = 2.1312 rad C 48 f (2.1312) = 1.492 f = 0.70 853. A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm, 3 and drives a reciprocating compressor; in Fig. 17.11, Text, e = 8 in ., 4 5 h = 17 in . The center of the 12-in. motor pulley is on the same level as the 16 center of the 54-in. compressor pulley; C = 40 in . (a) With the aid of a graphical layout, find the tensions in the belt for maximum output of the motor if it is compensator started. (b) What will be the stress in the belt if it is a heavy double 776 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS leather belt, 11 in. wide? (c) What coefficient of friction is indicated by the general belt equation? (Data courtesy of Rockwood Mfg. Co.) Solution: (a) For compensator-started hp = 1.4(50 ) = 70 hp 33,000hp F1 − F2 = vm πD n π (12 )(1140) vm = 1 1 = = 3581 fpm 12 12 33,000(70) F1 − F2 = = 645 lb 2062 b ≈ 25 in a ≈ 5 in R = 1900 lb eR = F 1a + F2b (8.75)(1900 ) = F 1 (5) + F2 (25) F 1+5F2 = 3325 lb 645 + F 2+5 F2 = 3325 lb F2 = 447 lb F1 = 645 + F2 = 645 + 447 = 1092 lb (b) For heavy double leather belt 23 t= in 64 b = 11 in 777 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1092 = 276 psi 20 (11) 64 12 ρvs2 e fθ − 1 (c) F1 − F2 = bt s − 32.2 e fθ 3581 vs = = 59.68 fps 60 ρ = 0.035 lb cu. in. s= F1 = bt 12(0.035)(59.68) e fθ − 1 23 645 = (11) 276 − e fθ 32.2 64 fθ = 1.241 D − D1 54 − 12 θ =π − 2 =π − = 2.092 rad C 40 f (2.092) = 1.492 f = 0.60 RUBBER BELTS 854. A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor pulley turns 1150 rpm; D2 = 36 in ., fan pulley; C = 23 ft . (a) Design a rubber belt to suit these conditions, using a net belt pull as recommended in §17.15, Text. (b) Actually, a 9-in., 5-ply Goodrich high-flex rubber belt was used. What are the indications for a good life? Solution: (a) θ = π − 36 − 8 D2 − D1 =π − = 3.040 rad = 174o 23(12 ) C Kθ = 0.976 bv N K hp = m p θ 2400 Kθ = 0.976 πD n π (8)(1150) vm = 1 1 = = 2409 fpm 12 12 Np = 5 hp = 20 = b(2409 )(5)(0.976) 2400 b = 4.1 in min. b = 5 in 778 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (b) With b = 9 in is safe for good life. 855. A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650 rpm, to an ore crusher. The center distance between the 33-in. motor pulley and the 108-in. driven pulley is 18 ft. The motor and crusher are so located that the belt must operate at an angle 75o with the horizontal. What is the overload capacity of this belt if the rated capacity is as defined in §17.15, Text? Solution: bv N hp = m p 2400 b = 20 in πD n π (33)(650 ) vm = 1 1 = = 5616 fpm 12 12 N p = 10 hp = (20 )(5616)(10 ) = 468 hp 2400 Overlaod Capacity = 468 − 300 (100%) = 56% 300 V-BELTS NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as well as from data in the Text. 856. A centrifugal pump, running at 340 rpm, consuming 105 hp in 24-hr service, is to be driven by a 125-hp, 1180-rpm, compensator-started motor; C = 43 to 49 in . Determine the details of a multiple V-belt drive for this installation. The B.F. Goodrich Company recommended six C195 V-belts with 14.4-in. and 50-in. sheaves; C ≈ 45.2 in . Solution: Table 17.7 N sf = 1.2 + 0.2 = 1.4 (24 hr/day) Design hp = N sf (transmitted hp) = (1.4 )(125) = 175 hp Fig. 17.4, 175 hp, 1180 rpm Dmin = 13 in , D-section D2 1180 50 = = D1 340 14.4 use D1 = 14.4 in > 13 in D2 = 50 in 779 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS vm = πD1n1 12 = π (14.4 )(1180) 12 = 4449 fpm 103 0.09 v2 v c − Rated hp = a − e m6 m3 K d D1 10 10 vm Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 3.47 D1 K d = 1.14 0.09 103 (4449 )2 4449 = 28.294 hp 137.7 − Rated hp = 18.788 − (0.0848) (1.14)(14.4) 106 103 4449 Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416 103 0.09 v2 v c − Rated hp = a − e m6 m3 K d D1 10 10 vm 0 . 09 103 (4449 )2 4449 = 20.0 hp 38.819 − Rated hp = 8.792 − (0.0416 ) (1.14 )(14.4 ) 10 6 103 4449 Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 50 − 14.4 = = 0.77 C 46 Kθ = 0.88 Table 17.6 (D − D1 )2 L ≈ 2C + 1.57(D2 + D1 ) + 2 4C 2 ( 50 − 14.4) L = 2(46) + 1.57(50 + 14.4 ) + = 200 in 4(46 ) use C195, L = 197.9 in K L = 1.07 Adjusted rated hp = (0.88)(1.07 )(20 ) = 18.83 hp Design hp 175 No. of belts = = = 9.3 belts use 9 belts Adjusted rated hp 18.83 Use 9 , C195 V-belts with 14.4 in and 50 in sheaves 780 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS B + B 2 − 32(D2 − D1 ) C= 16 B = 4 L − 6.28(D2 + D1 ) = 4(197.9) − 6.28(50 + 14.4 ) = 387.2 in 2 C= 857. 387.2 + (387.2 )2 − 32(50 − 14.4)2 16 = 44.9 in A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocating pump at a speed of 330 rpm. The pump is for 12-hr. service and normally requires 44 hp, but it is subjected to peak loads of 175 % of full load; C ≈ 50 in . Determine the details of a multiple V-belt drive for this application. The Dodge Manufacturing Corporation recommended a Dyna-V Drive consisting of six 5V1800 belts with 10.9-in. and 37.5-in. sheaves; C ≈ 50.2 in . Solution: Table 17.7, (12 hr/day) N sf = 1.4 − 0.2 = 1.2 Design hp = (1.2 )(1.75)(50 ) = 105 hp Fig. 17.4, 105 hp, 1160 rpm Dmin = 13 in , D-section D2 1160 46.4 = ≈ D1 330 13.2 use D1 = 13.2 in > 13 in D2 = 46.4 in πD n π (13.2 )(1160) vm = 1 1 = = 4009 fpm 12 12 103 0.09 vm2 vm c Rated hp = a − K D − e 10 6 103 vm d 1 Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D 46.4 Table 17.4, 2 = = 3.5 D1 13.2 K d = 1.14 0.09 2 103 ( 137.7 4009 ) 4009 − Rated hp = 18.788 − (0.0848) = 24.32 hp (1.14)(13.2) 106 103 4009 Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416 Dmin = 9 in 781 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D2 1160 32 = ≈ D1 330 9.1 use D1 = 9.1 in πD n π (9.1)(1160) = 2764 fpm vm = 1 1 = 12 12 0.09 103 (2764 )2 2764 = 10.96 hp 38.819 − Rated hp = 8.792 − (0.0416 ) (1.14 )(9.1) 106 103 2764 Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 32 − 9.1 = = 0.458 C 50 Kθ = 0.935 Table 17.6 2 ( D2 − D1 ) L ≈ 2C + 1.57(D2 + D1 ) + 4C (32 − 9.1)2 = 167 in L = 2(50 ) + 1.57(32 + 9.1) + 4(50 ) use C158, L = 160.9 in K L = 1.02 Adjusted rated hp = (0.935)(1.02 )(10.96 ) = 10.45 hp Design hp 105 No. of belts = = = 10 belts Adjusted rated hp 10.43 B + B 2 − 32(D2 − D1 ) 16 B = 4 L − 6.28(D2 + D1 ) = 4(160.9) − 6.28(32 + 9.1) = 385.5 in 2 C= C= 385.5 + (385.5)2 − 32(32 − 9.1)2 16 Use 10-C158 belts, D1 = 9.1 in D2 = 32 in , C = 46.8 in 858. = 46.8 in A 200-hp, 600-rpm induction motor is to drive a jaw crusher at 125 rpm; starting load is heavy; operating with shock; intermittent service; C = 113 to 123 in . Recommend a multiple V-flat drive for this installation. The B.F. Goodrich Company recommended eight D480 V-belts with a 26-in. sheave and a 120.175in. pulley; C ≈ 116.3 in . Solution: 782 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.7 N sf = 1.6 − 0.2 = 1.4 hp = (1.4 )(200 ) = 280 hp Fig. 17.14, 280 hp, 600 rpm Use Section E But in Table 17.3, section E is not available, use section D Dmin = 13 D2 600 = = 4.8 D1 125 For D1max : D + D2 min C = 1 + D1 2 D + 4.8 D1 113 = 1 + D1 2 D1 = 28 in min C = D2 D2 = 113 in 113 = 23.5 in D1 = 4. 8 1 use D1 ≈ (13 + 23.5) = 18 in 2 D2 = (4.8)(18) = 86.4 in L ≈ 2C + 1.57(D2 + D1 ) + (D2 − D1 )2 4C 2 ( 86.4 − 18) ( ) ( ) L = 2 118 + 1.57 86.4 + 18 + = 410 in 4(118) using D1 = 19 in , D2 = 91.2 in , C = 118 in 2 ( 91.2 − 19) L = 2(118) + 1.57(91.2 + 19) + 4(118) = 420 in Therefore use D420 sections D1 = 19 in , D2 = 91.2 in πD n π (19)(600 ) vm = 1 1 = = 2985 fpm 12 12 103 0.09 v2 v c − Rated hp = a − e m6 m3 K d D1 10 10 vm Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 4.8 D1 783 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS K d = 1.14 0.09 103 (2985)2 2985 = 29.6 hp 137.7 − Rated hp = 18.788 − (0.0848) (1.14 )(19) 106 103 2985 Therefore, Fig. 17.14, section D is used. Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 91.2 − 19 = = 0.612 C 118 Kθ = 0.83 (V-flat) Table 17.6, D420 L = 420.8 in K L = 1.12 Adjusted rated hp = (0.83)(1.12 )(29.6 ) = 27.52 hp Design hp 280 No. of belts = = = 10 belts Adjusted rated hp 27.52 Use10 , D420, D1 = 19 in , D2 = 91.2 in , C = 118 in 859. A 150-hp, 700-rpm, slip-ring induction motor is to drive a ball mill at 195 rpm; heavy starting load; intermittent seasonal service; outdoors. Determine all details for a V-flat drive. The B.F. Goodrich Company recommended eight D270 Vbelts, 17.24-in sheave, 61-in. pully, C ≈ 69.7 in . Solution: Table 17.7, N sf = 1.6 − 0.2 = 1.4 Design hp = (1.4 )(150 ) = 210 hp Fig. 17.4, 210 hp, 700 rpm Dmin = 13 in , D-section 103 0.09 v2 v c − Rated hp = a − e m6 m3 K d D1 10 10 vm d (hp ) For Max. Rated hp, =0 vm d 3 10 v Rated hp = a m3 10 v Let X = m3 10 0.91 c vm vm − − e K d D1 103 103 3 784 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS c X − eX 3 K d D1 v πD1n1 πD1 (700) X = m3 = = 10 12 × 103 12 × 103 12 × 103 X D1 = 700π 700πc hp = aX 0.91 − − eX 3 3 12 × 10 K d d (hp ) = 0.91aX − 0.09 − 3eX 2 = 0 d (X ) 0.91a X 2.09 = 3e Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 hp = aX 0.91 − 2.09 0.91(18.788) v X 2.09 = m3 = 3(0.0848) 10 vm = 7488 fpm πD n vm = 1 1 = 7488 12 πD (700) vm = 1 = 7488 12 D1 = 40.86 in max D1 = 40.86 in 1 ave. D1 = (13 + 40.86 ) = 26.93 in 2 use D1 = 22 in D2 700 79 = ≈ D1 195 22 D1 = 22 in , D2 = 79 in 22 + 79 D + D2 Min. C = 1 + D1 = + 22 = 72.5 in 2 2 Or Min. C = D2 = 79 in 2 ( D2 − D1 ) L ≈ 2C + 1.57(D2 + D1 ) + 4C (79 − 22 )2 = 327 in L = 2(79) + 1.57(79 + 22 ) + 4(79 ) use D330, L = 330.8 in B + B 2 − 32(D2 − D1 ) C= 16 2 785 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS B = 4 L − 6.28(D2 + D1 ) = 4(330.8) − 6.28(79 + 22 ) = 689 in C= 689 + (689)2 − 32(79 − 22 )2 = 81.12 in 16 πD n π (22 )(700 ) = 4032 fpm vm = 1 1 = 12 12 K d = 1.14 0.09 103 (4032)2 4032 = 39.124 hp 137.7 − Rated hp = 18.788 − (0.0848) (1.14)(22) 106 103 4032 Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 79 − 22 = = 0.70 C 81.12 Kθ = 0.84 (V-flat) Table 17.6 D330 K L = 1.07 Adjusted rated hp = (0.84 )(1.07 )(39.124) = 35.165 hp Design hp 210 No. of belts = = = 5.97 belts use 6 belts Adjusted rated hp 35.165 Use 6 , D330 V-belts , D1 = 22 in , D2 = 79 in , C ≈ 81.1 in 860. A 30-hp, 1160-rpm, squirrel-cage motor is to be used to drive a fan. During the summer, the load is 29.3 hp at a fan speed of 280 rpm; during the winter, it is 24 hp at 238 rpm; 44 < C < 50 in .; 20 hr./day operation with no overload. Decide upon the size and number of V-belts, sheave sizes, and belt length. (Data courtesy of The Worthington Corporation.) Solution: Table 17.7 N sf = 1.6 + 0.2 = 1.8 Design hp = (1.8)(30 ) = 54 hp Speed of fan at 30 hp 30 − 24 n2 = (280 − 238) + 238 = 286 rpm 29.3 − 24 at 54 hp, 1160 rpm. Fig. 17.4 use either section C or section D Minimum center distance: C = D2 786 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D1 + D2 + D1 2 D2 1160 = = 4.056 D1 286 use C = 4.056 D1 44 in < C < 50 in , use C = 47 in 47 D1max = = 11.6 in 4.056 use C-section, Dmin = 9 in Let D1 = 10.1 in , D2 = 41 in or C = 2 ( D2 − D1 ) L ≈ 2C + 1.57(D2 + D1 ) + 4C (41 − 10.1)2 = 179.3 in L = 2(47 ) + 1.57(41 + 10.1) + 4(47 ) use C137, L = 175.9 in B + B 2 − 32(D2 − D1 ) C= 16 B = 4 L − 6.28(D2 + D1 ) = 4(175.9) − 6.28(41 + 10.1) = 328.7 in 2 C= 382.7 + (382.7 )2 − 32(41 − 10.1)2 16 C173, satisfies 44 in < C < 50 in 0.91 = 45.2 in ≈ 44 in 3 c vm vm v Rated hp = a m3 − − e K d D1 103 103 10 πD n π (10.1)(1160 ) vm = 1 1 = = 3067 fpm 12 12 Table 17.4 D2 = 4.056 , K d = 1.14 D1 Table 17.3, C-section a = 8.792 , c = 38.819 , e = 0.0416 0.09 2 103 ( 38.819 3067 ) 3067 − Rated hp = 8.792 − (0.0416 ) = 12.838 hp (1.14)(10.1) 106 103 3067 Adjusted rated hp = Kθ K L (rated hp ) Table 17.5, D2 − D1 41 − 10.1 = = 0.68 C 45.2 Kθ = 0.90 787 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.6 L = 175.9 , C173 K L = 1.04 Adjusted rated hp = (0.90 )(1.04 )(12.838) = 12.02 hp Design hp 54 No. of belts = = = 4.5 belts use 5 belts Adjusted rated hp 12.02 Use 5 , C173 V-belts , D1 = 10.1 in , D2 = 41 in POWER CHAINS NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as well as from data in the Text. 861. A roller chain is to be used on a paving machine to transmit 30 hp from the 4cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft speed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to intermittent overloads of 100 %. (a) Determine the pitch and the number of chains required to transmit this power. (b) What is the length of the chain required? How much slack must be allowed in order to have a whole number of pitches? A chain drive with significant slack and subjected to impulsive loading should have an idler sprocket against the slack strand. If it were possible to change the speed ratio slightly, it might be possible to have a chain with no appreciable slack. (c) How much is the bearing pressure between the roller and pin? Solution: (a) design hp = 2(30 ) = 60 hp intermittent D2 n1 1000 ≈ = =2 D1 n2 500 D2 = 2D1 D C = D2 + 1 = 24 in 2 D 2 D1 + 1 = 24 2 D1max = 9.6 in D2 max = 2D1max = 2(9.6) = 19.2 in vm = πD1n1 = π (9.6 )(1000) = 2513 fpm 12 12 Table 17.8, use Chain No. 35, Limiting Speed = 2800 fpm 788 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Minimum number of teeth Assume N1 = 21 N 2 = 2 N1 = 42 [Roller-Bushing Impact] 1.5 100 N ts 0.8 hp = K r P n Chain No. 35 3 P = in 8 N ts = 21 n = 1000 rpm K r = 29 1.5 100(21) 3 hp = 29 1000 8 0.8 = 40.3 hp [Link Plate Fatigue] hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P 3 3− 0.07 8 3 hp = 0.004(21) (1000 ) = 2.91 hp 8 design hp 60 No. of strands = = = 21 rated hp 2.91 3 Use Chain No. 35, P = in , 21 strands 8 Check for diameter and velocity 1.08 D1 = P = 0.9 0.375 = 2.516 in 180 sin 21 180 sin Nt πD n π (2.516)(1000 ) vm = 1 1 = = 659 fpm 12 12 Therefore, we can use higher size, Max. pitch 180 180 = 9.6 sin P = D1 sin = 1.43 in 21 Nt say Chain no. 80 P = 1 in hp = 0.004(21)1.08 (1000)0.9 (1)3−0.07 (1) = 53.7 hp A single-strand is underdesign, two strands will give almost twice over design, Try Chain no. 60 789 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS P= 3 in 4 hp = 0.004(21) 1.08 0.9 (1000) 3 4 3 3− 0.07 4 = 23 hp design hp 60 = = 2.61 or 3 strands rated hp 23 P 0.75 D1 = = = 5.0 in 180 180 sin sin 21 Nt vm = πD1n1 = 12 π (5.0 )(1000) 12 = 1309 fpm The answer is Chain No. 60, with P = ¾ in and 3 chains, limiting velocity is 1800 fpm 2 N + N 2 ( N 2 − N1 ) (b) L ≈ 2C + 1 + pitches 2 40C C= 24 = 32 3 4 N1 = 21 N 2 = 42 L = 2(32) + 21 + 42 (42 − 21)2 + = 95.845 pitches ≈ 96 pitches 2 40(32 ) Amount of slack ( h = 0.433 S 2 − L2 L = C = 24 in 1 2 ) (96 − 95.845) 3 in S = 24 in + [ 2 4 = 24.058in h = 0.433 (24.058)2 − (24)2 1 2 ] = 0.7229 in (c) pb = bearing pressure Table 17.8, Chain No. 60 C = 0.234 in 1 E = in 2 J = 0.094 in 1 A = C (E + 2 J ) = 0.234 + 2(0.094) = 0.160992 in 2 2 FV = 60 hp 33,000 790 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS F (1309) = 60 hp 33,000 F = 1512.6 lb 1512.6 F= = 504.2 lb strand 3 504.2 pb = = 3131 psi 0.160992 862. A conveyor is driven by a 2-hp high-starting-torque electric motor through a flexible coupling to a worm-gear speed reducer, whose mw ≈ 35 , and then via a roller chain to the conveyor shaft that is to turn about 12 rpm; motor rpm is 1750. Operation is smooth, 8 hr./day. (a) Decide upon suitable sprocket sizes, center distance, and chain pitch. Compute (b) the length of chain, (c) the bearing pressure between the roller and pin. The Morse Chain Company recommended 15- and 60-tooth sprockets, 1-in. pitch, C = 24 in ., L = 88 pitches . Solution: Table 17.7 N sf = 1.2 − 0.2 = 1.0 (8 hr/day) design hp = 1.0(2 ) = 2.0 hp 1750 n1 = = 50 rpm 35 n2 = 12 rpm Minimum number of teeth = 12 Use N1 = 12 [Link Plate Fatigue] hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P hp 2.0 P 3−0.07 P ≈ P 3 = = = 1.0 1.08 0.9 1.08 0.9 0.004 N ts n 0.004(12 ) (50) Use Chain No. 80, P = 1.0 in To check for roller-bushing fatigue 1.5 100 N ts 0.8 hp = K r P n K r = 29 100(12 ) hp = 17 1000 1.5 (1)0.8 = 2747 hp > 2 hp (a) N1 = 12 n 50 N 2 = 1 N1 = (12) = 50 teeth 12 n2 791 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS C = D2 + D1 ≈ D2 ≈ PN1 π PN1 π D1 2 = (1.0 )(12) = 3.82 in = (1.0 )(50) = 15.92 in π π 3.82 = 17.83 in 2 use C = 18 in C = 18 pitches chain pitch = 1.0 in, Chain No. 80 C ≈ 15.92 + N + N 2 ( N 2 − N1 ) + (b) L ≈ 2C + 1 2 40C 2 12 + 59 (50 − 12 ) L ≈ 2(18) + + = 69 pitches 2 40(18) use L = 70 pitches 2 (c) pb = bearing pressure Table 17.8, Chain No. 80 C = 0.312 in 5 E = in 8 J = 0.125 in PN ts n1 (1)(12 )(50) vm = = = 50 fpm 12 12 3 A = C (E + 2 J ) = 0.141 + 2(0.05) = 0.04054 in 2 16 FV = 60 hp 33,000 33,000(2 ) F= = 1320 lb 50 F 1320 pb = = = 4835 psi C (E + 2 J ) 5 0.312 + 2(0.125) 8 863. A roller chain is to transmit 5 hp from a gearmotor to a wood-working machine, with moderate shock. The 1-in output shaft of the gearmotor turns n = 500 rpm . The 1 ¼-in. driven shaft turns 250 rpm; C ≈ 16 in . (a) Determine the size of sprockets and pitch of chain that may be used. If a catalog is available, be sure maximum bore of sprocket is sufficient to fit the shafts. (b) Compute the center 792 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS distance and length of chain. (c) What method should be used to supply oil to the chain? (d) If a catalog is available, design also for an inverted tooth chain. Solution: Table 17.7 N sf = 1.2 design hp = 1.2(5) = 6 hp D2 500 = =2 D1 250 D C = D2 + 1 2 D 16 = 2 D1 + 1 2 D1max = 6.4 in D2 max = 2D1max = 2(6.4 ) = 12.8 in vm = πD1n1 π (6.4 )(500) = 12 12 (a) Link Plate Fatigue hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P = 838 fpm Nts = N1 = 21 hp = 0.004(21)1.08 (500)0.9 P 3−0.07 P Max. pitch 180 180 = 6.4 sin P = D1 sin = 0.95 in 21 Nt Try chain no. 60 5 P = in 8 hp = 0.004(21) 1.08 0.9 5 (500) 8 5 3− 0.07 8 = 7.17 hp 5 8 P 0.625 D1 = = = 4.194 in 180 180 sin sin 21 N1 use P = in , Chain No. 60 n 500 N2 = N1 1 = 21 = 42 n 250 2 P 0.625 D2 = = = 8.364 in 180 180 sin sin 42 N2 793 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 5 8 Size of sprocket, N1 = 21 , N2 = 42 , P = in , D1 = 4.194 in , D2 = 8.364 in (b) C = 16 in C= 16 in = 25.6 pitches 5 in 8 N1 + N 2 ( N 2 − N1 ) + 2 40C 2 L ≈ 2C + 21 + 42 (42 − 21)2 + = 83.13 use 84 pitches 2 40(25.6) use L = 84 pitches L ≈ 2(25.6) + (c) Method: πD1n1 π (4.194)(500) vm = 12 = 12 = 549 fpm . Use Type II Lubrication ( vmax = 1300 fpm ) – oil is supplied from a drip lubricator to link plate edges. 864. A roller chain is to transmit 20 hp from a split-phase motor, turning 570 rpm, to a reciprocating pump, turning at 200 rpm; 24 hr./day service. (a) Decide upon the tooth numbers for the sprockets, the pitch and width of chain, and center distance. Consider both single and multiple strands. Compute (b) the chain length, (c) the bearing pressure between the roller and pin, (d) the factor of safety against fatigue failure (Table 17.8), with the chain pull as the force on the chain. (e) If a catalog is available, design also an inverted-tooth chain drive. Solution: Table 17.7 N sf = 1.4 + 0.2 (24 hr/day) design hp = 1.6(20) = 32 hp n 570 (a) 1 = = 2.85 n2 200 D2 n1 ≈ = 2.85 D1 n2 Considering single strand hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P min N ts = 17 hp = 32 = 0.004(17 ) 1.08 (570)0.9 P 3−0.07 P P 3−0.07 P = 1.24 P = 1.07 in use P = 1.0 in 794 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 3− 0.07 (1) hp = 32 = 0.004( N1 ) (570 ) (1) N1 = 21 570 N2 = (21) = 60 200 5 Roller width = in 8 D C = D2 + 1 2 PN1 (1)(21) D1 ≈ = = 6.685 in 1.08 π D2 ≈ π PN 2 π (1)(60 ) = 19.10 in = π C = 19.10 + C= 0.9 6.685 = 22.44 in 2 22.44 = 22.44 pitches 1 Considering multiple strands 5 8 Assume, P = in hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P hp = 0.004(21)1.08 (570)0.9 (0.625)3−0.07 (0.625 ) = 8.07 hp 32 hp No. of strands = = 4.0 8.07 hp Use 4 strands Roller width = 0.4 in D1 ≈ D2 ≈ PN1 π PN2 π 5 (21) 8 = = 4.178 in π 5 (60) 8 = = 11.937 in C = 11.937 + C= π 4.178 = 14.026 in 2 14.026 = 22.44 pitches 5 8 (b) Chain Length For single strands 795 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS N1 + N 2 ( N 2 − N1 ) + 2 40C 2 L ≈ 2C + 21 + 60 (60 − 21)2 + = 87.08 pitches 2 40(22.44) use L = 88 pitches L ≈ 2(22.44) + For multiple strands 2 N1 + N 2 ( N 2 − N1 ) L ≈ 2C + + 2 40C 21 + 60 (60 − 21)2 + = 87.08 pitches 2 40(22.44) use L = 88 pitches L ≈ 2(22.44) + (c) pb = bearing pressure Table 17.8, P = 1 in 5 E = in 8 J = 0.125 in C = 0.312 in 33,000hp F= vm πD n π (6.685)(570 ) vm = 1 1 = = 998 fpm 12 12 33,000(32 ) F= = 1058 lb 998 F 1058 pb = = = 3876 psi C (E + 2 J ) 5 0.312 + 2(0.125) 8 5 8 Table 17.8, P = in 3 E = in 8 J = 0.080 in C = 0.200 in 33,000hp vm πD n π (6.685)(570 ) vm = 1 1 = = 998 fpm 12 12 F= 796 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS F= 33,000(32 ) = 1058 lb 998 1058 = 264.5 lb 4 F 264.5 pb = = = 2472 psi C (E + 2 J ) 3 0.200 + 2(0.080) 8 F= (d) Factor of Safety = Fu , based on fatigue 4F Fu = 14,500 lb , Table 17.8 F 14,500 = 3.43 Factor of Safety = u = 4 F 4(1058) 4-strand, chain no.50 Fu = 6100 lb , Table 17.8 Factor of Safety = 865. Fu 6100 = = 5.77 4F 4(264.5) A 5/8-in. roller chain is used on a hoist to lift a 500-lb. load through 14 ft. in 24 sec. at constant velocity. If the load on the chain is doubled during the speed-up period, compute the factor of safety (a) based on the chain’s ultimate strength, (b) based on its fatigue strength. (c) At the given speed, what is the chain’s rated capacity ( N s = 20 teeth ) in hp? Compare with the power needed at the constant speed. Does it look as though the drive will have a “long” life? Solution: Table 17.8 5 P = in 8 Fu = 6100 lb Fu F F = (500 )(2 ) = 1000 lb 6100 Factor of Safety = = 6. 1 1000 F (b) Factor of Safety = u (fatigue) 4F (a) Factor of Safety = 797 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Factor of Safety = (c) vm = 6100 = 1.5 4(1000) 14 ft 60 sec = 35 fpm 24 sec 1 min N s = 20 5 P = in 8 Rated hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P [Link Plate Fatigue] 5 (20 )n PN s n 8 vm = = = 35 fpm 12 12 n = 33.6 rpm Rated hp = 0.004(20 ) 1.08 (33.6 ) 0.9 5 8 5 3− 0.07 8 = 0.6 hp Hp needed at constant speed Fvm (500)(35) = 0.53 hp < 0.6 hp hp = = 33,000 33,000 Therefore safe for “long” life. WIRE ROPES 866. In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400 ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19, IPS, 1 ¾-in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force but take the static view and compute the factor of safety with and without allowances for the bending load. (b) If N = 1.35 , based on fatigue, what is the expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect of the rope’s weight. A load of 14 kips is gradually applied on the 6-kip cage. How much is the deflection of the cable due to the load and the additional energy absorbed? (d) For educational purposes and for a load of 0.2 Fu , compute the energy that this 400-ft rope can absorb and compare it with that for a 400-ft., 1 ¾-in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the energy per pound of material in each case? Solution: (a) 798 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (1600 1 min fpm) 60 sec = 4.445 fps 2 6 sec v2 − v1 = t Wh = 20 kips For 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft 400 2 wL = 1.6 Dr2 kips = 0.64 Dr kips 1000 Ft − wL − Wh = ma a= m= 20 + 0.64 Dr2 32.2 20 + 0.64 Dr2 (4.445) Ft − 0.64 Dr2 − 20 = 32 . 2 2 Ft = 22.76 + 0.73Dr 3 Dr = 1 in 4 2 3 Ft = 22.76 + 0.731 = 25 kips 4 F − Fb N= u Ft Table AT 28, IPS Fu ≈ 42 Dr2 tons Fu = 42(1.75) = 129 tons = 258 kips with bending load Fb = sb Am EDw sb = Ds EAm Dw Fb = Ds Table At 28, 6 x 19 Wire Rope 2 799 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Dw = 0.067 Dr = 0.067(1.75) = 0.11725 in Ds = 8 ft = 96 in E = 30,000 ksi Am ≈ 0.4 Dr2 Am = 0.4(1.75) = 1.225 sq in (30,000)(1.225)(0.11725) = 45 kips Fb = (96) F − Fb 258 − 45 N= u = = 8.52 Ft 25 without bending load 258 F N= u = = 10.32 Ft 25 2 (b) N = 1.35 on fatigue IPS, su ≈ 260 ksi 2 NFt Dr Ds = ( p su )su (1.75)(96) = 2(1.35)(25) ( p su )(260) p su = 0.0015 Fig. 17.30, 6 x 19 IPS Number of bends to failure = 7 x 105 FL Am Er Am = 1.225 sq in Er ≈ 12,000 ksi (6 x 19 IPS) F = 14 kips L = 400 ft = 4800 in (14)(4800) = 4.57 in δ= (1.225)(12,000) 1 1 U = Fδ = (14 )(4.57 ) = 32 in − kips 2 2 (c) δ = (d) F = 0.2 Fu = 0.2(258) = 51.6 kips FL δ= Am Er (51.6)(4800) = 16.85 in δ= (1.225)(12,000) 800 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1 1 Fδ = (51.6)(16.85) = 434 in − kips 2 2 For 1 ¾ in, as-rolled 1045 steel rod su = 96 ksi U= π 2 Fu = su A = (96) (1.75) = 230.9 kips 4 F = 0.2 Fu = 0.2(230.9) = 46.2 kips FL δ= AE (46.2 )(4800 ) = 3.073 in δ= π 2 (1.75) (30,000) 4 1 1 U = Fδ = (46.2 )(3.073) = 71 in − kips < U of wire rope. 2 2 868. A hoist in a copper mine lifts ore a maximum of 2000 ft. The weight of car, cage, and ore per trip is 10 kips, accelerated in 6 sec. to 2000 fpm; drum diameter is 6 ft. Use a 6 x 19 plow-steel rope. Determine the size (a) for a life of 200,000 cycles and N = 1.3 on the basis of fatigue, (b) for N = 5 by equation (v), §17.25, Text. (c) What is the expected life of the rope found in (b) for N = 1.3 on the basis of fatigue? (d) If a loaded car weighing 7 kips can be moved gradually onto the freely hanging cage, how much would the rope stretch? (e) What total energy is stored in the rope with full load at the bottom of te shaft? Neglect the rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it reasonable? Solution: (2000 1 min fpm) 60 sec = 5.56 fps 2 6 sec v2 − v1 = t For 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft 2000 2 wL = 1.6 Dr2 kips = 3.2 Dr kips 1000 a= 801 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Wh = 10 kips wL + Wh Ft − wL − Wh = a 32.2 a 5.56 Ft = + 1(wL + Wh ) = + 1 3.2 Dr2 + 10 = 1.17267 3.2 Dr2 + 10 32.2 32.2 2 NFt (a) Dr Ds = ( p su )su Fig. 17.30, 200,000 cycles, 6 x 19 p su = 0.0028 PS: su ≈ 225 ksi Ds = 6 ft = 72 in N = 1.3 2(1.3)(1.17267 ) 3.2 Dr2 + 10 Dr (72) = (0.0028)(225) 45.36 Dr = 9.7566 Dr2 + 30.49 Dr2 − 4.64916 Dr + 3.1251 = 0 Dr = 0.815 in 7 say Dr = in 8 ( ( ) ) (b) by N = 5 , Equation (v) F − Fb N= u Ft EDw sb = Ds Dw = 0.067 Dr (30,000 )(0.067 Dr ) = 27.92 D sb = r 72 Fb = sb Am Am = 0.4 Dr2 ( ) Fb = (27.92 Dr ) 0.4 Dr2 = 11.17 Dr3 Fu = 36 Dr2 tons for PS Fu = 72 Dr2 kips Fu − Fb = NFt 72 Dr2 − 11.17 Dr3 = (5)(1.17267 )(3.2 Dr2 + 10 ) 72 Dr2 − 11.17 Dr3 = (5.8634 )(3.2 Dr2 + 10) Dr = 1.216 in 802 ( ) SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1 use Dr = 1 in 4 2 NFt (c) Dr Ds = ( p su )su (1.25)(72) = 2(1.3)(1.17267 )[3.2(1.25) ( p su )(225) 2 + 10 ] p su = 0.00226 Fig. 17.20 Expected Life = 3 x 105 cycles (d) F = 7 kips Er = 12,000 ksi L = 2000 ft = 24,000 in 7 For (a) Dr = in 8 FL δ= Am Er 2 7 Am ≈ 0.4 D = 0.4 = 0.30625 sq in 8 (7)(24,000) = 45.7 in δ= (0.30625)(12,000) 1 For (b) Dr = 1 in 4 FL δ= Am Er 3 r 2 1 Am ≈ 0.4 D = 0.41 = 0.625 sq in 4 (7)(24,000) = 22.4 in δ= (0.625)(12,000) 3 r 1 1 Fδ = (7 )(45.7 ) = 160 in − kips 2 2 1 1 For (b) U = Fδ = (7 )(22.4 ) = 78.4 in − kips 2 2 (f) Limiting pressure, cast-iron sheaves, 6 x19, p = 500 psi . (e) For (a) U = For (a) p su = 0.0028 p = 0.0028(225) = 0.630 kips = 630 psi > 500 psi , not reasonable. For (b) p su = 0.00226 803 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS p = 0.00226(225) = 0.5085 kips = 508.5 psi ≈ 500 psi , reasonable. 869. For a mine hoist, the cage weighs 5900 lb., the cars 2100 lb., and the load of coal in the car 2800 lb.; one car loaded loaded at a time on the hoist. The drum diameter is 5 ft., the maximum depth is 1500 ft. It takes 6 sec. to accelerate the loaded cage to 3285 fpm. Decide on a grade of wire and the kind and size of rope on the basis of (a) a life of 2× 105 cycles and N = 1.3 against fatigue failure, (b) static consideration (but not omitting inertia effect) and N = 5 . (c) Make a final recommendation. (d) If the loaded car can be moved gradually onto the freely hanging cage, how much would the rope stretch? (e) What total energy has the rope absorbed, fully loaded at the bottom of the shaft? Neglect the rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it all right? Solution: Wh = 5900 + 2100 + 2800 = 10,800 lb = 10.8 kips 1 min fpm) 60 sec v2 − v1 a= = = 9.125 fps 2 t 6 sec wL + Wh Ft − wL − Wh = a 32.2 Assume 6 x 19 IPS, w ≈ 1.6 Dr2 lb ft 1500 2 wL = 1.6 Dr2 kips = 2.4 Dr kips 1000 a 9.125 Ft = + 1(wL + Wh ) = + 1 2.4 Dr2 + 10 = 3.08 Dr2 + 13.86 32.2 32.2 5 (a) Fig. 17.30, 2 x 10 cycles p su = 0.0028 2 NFt Dr Ds = ( p su )su Ds = 5 ft = 60 in (3285 ( ) 804 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Ds ≈ 45Dr 60 Dr max = = 1.33 in 45 1 use Dr = 1 in 4 2 1 Ft = 3.081 + 13.86 = 18.67 kips 4 2(1.3)(18.67 ) su = = 231 ksi 1 (0.0028)1 (60 ) 4 1 Use Plow Steel, 6 x 19 Wire Rope, Dr = 1 in . 4 (b) N = sb = Fu − Fb Ft EDw Ds 1 Dw = 0.067 Dr = 0.0671 = 0.08375 in 4 Ds = 60 in E = 30,000 ksi (30,000 )(0.08375) = 41.875 ksi sb = 60 2 1 2 Am = 0.4 Dr = 0.41 = 0.625 in 2 4 Fb = sb Am = (41.875)(0.625) = 26.17 kips N =5 Fu = NFt + Fb = (5)(18.67 ) + 26.17 = 119.52 kips = 59.76 tons Fu 59.76 = = 38.25 2 Dr 1 2 1 4 Table AT 28, Use IPS, 6 x 19, Fu = 42 > 38.25 Dr2 (c) Recommendation: 1 6 x 19, improved plow steel, Dr = 1 in 4 805 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS FL Am Er F = 2100 + 2800 = 4900 lb Er ≈ 12×10 6 psi L = 1500 ft = 18,000 in (4900)(18,000) = 11.76 in δ= (0.625) 12 ×106 (d) δ = ( ) 1 1 Fδ = (4900)(11.76) = 28,800 in − lb 2 2 (f) p su = 0.0028 (e) U = su = 231 ksi p = 0.0028(231,000 ) = 646.8 psi For cast-iron sheave, limiting pressure is 500 psi p = 646.8 psi > 500 psi , not al right. 870. The wire rope of a hoist with a short lift handles a total maximum load of 14 kips each trip. It is estimated that the maximum number of trips per week will be 1000. The rope is 6 x 37, IPS, 1 3/8 in. in diameter, with steel core. (a) On the basis of N = 1 for fatigue, what size drum should be used for a 6-yr. life? (n) Because of space limitations, the actual size used was a 2.5-ft. drum. What is the factor of safety on a static basis? What life can be expected ( N = 1 )? Solution: (a) 365 days 1 wk 1000 trips = 312,857 cycles ≈ 3 × 105 cycles No. of cycles = (6 yr ) 1 yr 7 days 1 wk Figure 17.30, 6 x 37, IPS p su = 0.00225 2 NFt Dr Ds = ( p su )su For IPS, su ≈ 260 ksi Ft = 14 kips N = 1.0 Dr = 1.375 in 2 NFt Dr Ds = ( p su )su (1.375)Ds = 2(1.0)(14) (0.00225)(260) 806 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Ds = 34.8 in (b) Ds = 2 ft = 30 in Static Basis F − Fb N= u Ft Table AT 28, 6 x 37 Dw ≈ 0.048 Dr = 0.048(1.375) = 0.066 in Am ≈ 0.4 Dr2 = 0.4(1.375) = 0.75625 in 2 2 Fu = su Am = (260)(0.75625) = 196.6 kips EDw Am (30,000 )(0.066)(0.75625) Fb = sb Am = = = 49.9 kips Ds 30 F − Fb 196.6 − 49.9 N= u = = 10.5 Ft 10.5 Life: N = 1.0 (fatigue) 2 NFt Dr Ds = ( p su )su (1.375)(30) = 2(1.0)(14) ( p su )(260) p su = 0.0026 Figure 17.30, Life ≈ 2.5 × 105 cycles , 6 x 37. 871. A wire rope passes about a driving sheave making an angle of contact of 540o, as shown. A counterweight of 3220 lb. is suspended from one side and the acceleration is 4 fps2. (a) If f = 0.1 , what load may be noised without slipping on the rope? (b) If the sheave is rubber lined and the rope is dry, what load may be raised without slipping? (c) Neglecting the stress caused by bending about the sheave, find the size of 6 x 19 MPS rope required for N = 6 and for the load found in (a). (d) Compute the diameter of the sheave for indefinite life with say N = 1.1 on fatigue. What changes could be made in the solution to allow the use of a smaller sheave? 807 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Problems 871 – 874. Solution: 4 fps 2 = 2820 lb F2 = (3220 lb )1 − 2 32.2 fps (a) F1 = F2 e fθ θ = 540o = 3π f = 0.10 F1 = (2820 )e (0.10 )(3π ) = 7237 lb (b) For rubber lined, dry rope f = 0.495 F1 = (2820 )e (0.495 )(3π ) = 249,466 lb (c) Ft = F1 = 7237 lb F − (Fb ≈ 0 ) Fu N= u = Ft Ft Fu ≈ 32 Dr2 tons for MPS Fu ≈ 64 Dr2 kips Fu = 64,000 Dr2 lb Fu = NFt 64,000 Dr2 = (6)(7237 ) Dr = 0.824 in use Dr = 0.875 in 808 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 2 NFt ( p su )su Indefinite life, p su = 0.0015 MPS: su ≈ 195 ksi = 195,000 psi (0.875)Ds = 2(1.1)(7237) (0.0015)(195,000) Ds = 62.2 in To reduce the size of sheave, increase the size of rope. (d) Dr Ds = A traction elevator with a total weight of 8 kips has an acceleration of 3 fps2; the 6 cables pass over the upper sheave twice, the lower one once, as shown.. Compute the minimum weight of counterweight to prevent slipping on the driving sheave if it is (a) iron with a greasy rope, (b) iron with a dry rope, (c) rubber lined with a greasy rope. (d) Using MPS and the combination in (a), decide upon a rope and sheave size that will have indefinite life ( N = 1 will do). (e) Compute the factor of safety defined in the Text. (f) If it were decided that 5× 105 bending cycles would be enough life, would there be a significant difference in the results? 872. Solution: 3 fps 2 = 8.745 kips F1 = (8 kips )1 + 2 32.2 fps θ = 3 180o = 3π F F2 = f1θ e Wc = weight of counterweight F2 Wc = = 1.10274 F2 3 1− 32.2 1.10274 F1 Wc = e fθ (a) Iron sheave, greasy rope, f = 0.07 1.10274(8.745) Wc = = 4.986 kips e (0.07 )(3π ) (b) Iron sheave, dry rope, f = 0.12 1.10274(8.745) Wc = = 3.112 kips e (0.12 )(3π ) (c) Rubber lined with a greasy rope, f = 0.205 1.10274(8.745) Wc = = 1.397 kips e (0.205 )(3π ) ( ) 809 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 2 NFt ( p su )su Indefinite life, p su = 0.0015 Ft = F1 = 8.745 kips total 8.745 Ft = = 1.458 kips each rope 6 Ft = 1458 lbs N =1 Table AT 28, 6 x 19 Ds ≈ 45Dr 2(1)(1458) Dr (45Dr ) = (0.0015)(195,000) Dr = 0.47 in 1 Use Dr = in = 0.5 in 2 (d) Dr Ds = Fu − Fb Ft Table AT 28, MPS 2 Fu = 32 Dr2 tons = 64,000 Dr2 lb = 64,000(0.5) lb = 16,000 lb (e) N = Fb = EDw Am Ds E = 30× 106 psi 6 x 19, Dw = 0.067 Dr Ds ≈ 45Dr Am = 0.4 Dr2 = 0.4(0.5) = 0.1 sq. in. 2 (30 ×10 )(0.067)(0.1) = 4467 lb 6 Fb = 45 16,000 − 4467 N= = 7.91 1458 (f) 5 x 105 cycles Fig. 17.30, 6 x 19. p su = 0.0017 2 NFt Dr Ds = ( p su )su 810 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Dr (45Dr ) = 2(1)(1458) (0.0017)(195,000) Dr = 0.44 in since Dr = 0.44 in ≈ 0.47 in as in (d), therefore, no significant difference will result. 873. A 5000-lb. elevator with a traction drive is supported by a 6 wire ropes, each passing over the driving sheave twice, the idler once, as shown. Maximum values are 4500-lb load, 4 fps2 acceleration during stopping. The brake is applied to a drum on the motor shaft, so that the entire decelerating force comes on the cables, whose maximum length will be 120 ft. (a) Using the desirable Ds in terms of Dr , decide on the diameter and type of wire rope. (b) For this rope and N = 1.05 , compute the sheave diameter that would be needed for indefinite life. (c) Compute the factor of safety defined in the Text for the result in (b). (d) Determine the minimum counterweight to prevent slipping with a dry rope on an iron sheave. (e) Compute the probable life of the rope on the sheave found in (a) and recommend a final choice. Solution: (a) Ft = 4500 lb Wh = 5000 lb W + wL Wh + wL − Ft = h a 32.2 assume 6 x 19 w = 1.6 Dr2 lb ft wL = (1.6 Dr2 )(120 ) = 192 Dr2 per rope wL = 6(192 Dr2 ) = 1152 Dr2 5000 + 1152 Dr2 (4 ) 5000 + 1152 D − 4500 = 32 . 2 2 2 1152 Dr + 500 = 621.12 + 143.11Dr Dr = 0.3465 in 3 say Dr = 0.375 in = in 8 2 r 811 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 7 3 Ds ≈ 45Dr = 45 = 16 in 8 8 3 Six – 6 x 19 rope, Dr = in 8 (a) Dr = 0.375 in = 3 in 8 4500 = 750 lb 6 N = 1.05 2 NFt Dr Ds = ( p su )su assume IPS, su = 260 ksi = 260,000 psi Ft = Indefinite life, p su = 0.0015 (0.375)Ds = 2(1.05)(750) (0.0015)(260,000) Ds = 10.77 in Fu − Fb Ft Ft = 750 lb IPS (c) N = 2 3 Fu ≈ 42 D tons = 84,000 D lb = 84,000 lb = 11,813 lb 8 2 r Fb = 2 r EDw Am Ds 6 x 19, Ds = 10.77 in as in (b) 3 Dw = 0.067 Dr = 0.067 = 0.025 in 8 2 3 Am = 0.4 D = 0.4 = 0.05625 sq. in. 8 6 E = 30× 10 psi 2 r (30 ×10 )(0.025)(0.05625) = 3917 lb 6 Fb = 10.77 812 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS N= 11,813 − 3917 = 10.53 750 (c) F1 = Ft = 4500 lb F1 = F2 e fθ For iron sheave, dry rope, f = 0.12 θ = 540o = 3π F2 = F1 4500 = (0.12 )(3π ) = 1452 lb fθ e e a CW 1 + = F2 32.2 4 CW 1 + = 1452 32.2 CW = 1291 lb 874. A traction elevator has a maximum deceleration of 8.05 fps2 when being braked on the downward motion with a total load of 10 kips. There are 5 cables that pass twice over the driving sheave. The counterweight weighs 8 kips. (a) Compute the minimum coefficient of friction needed between ropes and sheaves for no slipping. Is a special sheave surface needed? (b) What size 6 x 19 mild-plowsteel rope should be used for N = 4 , including the bending effect? (Static approach.) (c) What is the estimated life of these ropes ( N = 1 )? Solution: a = 8.05 fps 2 (a) F1 = 10 kips 8.05 F2 = (8 kips )1 − = 6 kips 32.2 θ = 3π F1 = e fθ F2 813 SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 10 = e f (3π ) 6 f = 0.0542 Special sheave surface is needed for this coefficient of friction, §17.21. F − Fb (b) N = u Ft 10 Ft = = 2 kips 5 EDw Am Fb = Ds Table AT 28, 6 x 19, MPS Dw = 0.067 Dr Ds ≈ 45Dr Am ≈ 0.4 Dr2 E = 30× 106 psi (30 ×10 )(0.067 D )(0.4D ) = 17.87 D 6 Fb = r 45Dr 2 r 2 r kips Fu ≈ 32 Dr2 tons = 64 Dr2 kips 64 Dr2 − 17.87 Dr2 2 Dr = 0.4164 in 7 use Dr = in 16 7 (c) Ds ≈ 45Dr = 45 = 20 in 16 2 NFt Dr Ds = ( p su )su Ft = 2 kips each rope MPS, su = 195 ksi N = 1.0 2(1.0 )(2 ) 7 (20) = ( p su )(195) 16 p su = 0.0023 Expected life, Figure 17.30, 3 x 105 bending cycles. - end N =4= 814 SECTION 16 – BRAKES AND CLUTCHES ENERGY TO BRAKES 881. A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4. The drum on which the cable wraps is on the same shaft as the gear, and the torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on the motor shaft. Consider first on which shaft to mount the brake drum; in the process make trial calculations, and try to think of pros and cons. Make a decision and determine the size of a drum that will not have a temperature rise greater than ∆t = 150o F when a 4000-lb. load moves down 200 ft. at a constant speed. Include a calculation for the frp/sq. in. of the drum’s surface. Solution: Consider that brake drum is mounted on motor shaft that has lesser torque. 12,000 ft − lb T f= = 3000 ft − lb = 36,000 in − lb 4 From Table AT 29, Assume f = 0.35 , p = 75 psi , max. vm = 5000 fpm Tf = FD 2 F = fN = N= 2T f D 2T f fD N p= A A = π Db p= 2T f N 2(36,000 ) = = = 75 2 π Db π D bf π D 2b(0.35) D 2b = 873 use D 2b = 873 873 b= 2 D Then, U ft − lb ∆t o F = f Wm c Assume a cast-iron, ρ = 0.253 lb in3 c = 101 Wm = ρ V Page 1 of 97 SECTION 16 – BRAKES AND CLUTCHES D2 D 2t = π t Db + 4 4 U f = (4000 )(200 ) = 800,000 ft − lb V = π Dbt + π ∆t = 150o F Wm = ρ V = Uf c∆t 800,000 0.253V = (150)(101) V = 208.7 in 3 But D2 V = π t Db + 4 873 b= 2 D 873 D 2 V = π t + 4 D For minimum V : dV − 873 D = π t 2 + = 0 dD 2 D 3 D = 2(873) D = 12 in For t : 873 (12 )2 V = 208.7 = π t + 4 12 t = 0.611 in 5 say t = in 8 873 1 = 6.0625 in = 6 in 2 16 (12 ) 5 1 Therefore use D = 12 in , t = in , b = 6 in 8 16 b= For fhp sq. in. = fhp = Fvm 33,000 Page 2 of 97 fhp A SECTION 16 – BRAKES AND CLUTCHES 2(36,000) = 6000 lb D 12 vm = 5000 fpm (max.) (6000)(5000) = 909 hp fhp = 33,000 1 A = π Db = π (12 ) 6 in 2 16 fhp 909 fhp sq. in. = = = 3.98 (peak value) A 228.55 F= 882. 2T f = A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the brake is to be neglected. (a) What total averaging braking torque must be applied? (b) What must be the minimum coefficient of friction between the tires and the road in order for the wheels not to skid if it is assumed that weight is equally distributed among the four wheels (not true)? (c) If the frictional energy is momentarily stored in 50 lb. of cast iron brake drums, what is the average temperature rise of the drums? Solution: (a) Solving for the total braking torque. W 2 U f = − ∆KE = vs − vs22 2g 1 W = 3500 lb vs1 = 60 mph = 88 fps ( ) vs2 = 0 mph = 0 fps g = 32.2 fps 2 3500 Uf = (882 − 02 ) = 421,000 ft − lb 2(32.2) (T ft − lb )ωm = (T f in − lb )n fhp = f 33,000 63,000 2 2 2 vs − vs1 0 − (88) a= 2 = = −14.892 fps 2 2s 2(260) vs − vs1 0 − 88 t= 2 = = 5.91 sec a − 14.892 U − ∆KE 421,000 fhp = = f = = 130 hp (t )(550) 550t 550(5.91) Page 3 of 97 SECTION 16 – BRAKES AND CLUTCHES 1 (88 fps )(60 sec min ) vm 2 n= = = 336 rpm πD 30 π ft 12 T n fhp = f 63,000 63,000(130 ) Tf = = 24,375 in − lb 336 (b) f = F N for each wheel, N = 3500 = 875 lb 4 24,375 = 6094 in − lb 4 2T 2(6094) F= f = = 406 in − lb D 30 F 406 f = = = 0.464 N 875 Tf = (c) ∆t = Uf Wm c U f = 421,000 ft − lb Wm = 50 lb c = 101 ft − lb lb − F for cast-iron 421,000 ∆t = = 83.4o F (50)(101) 884. An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm. It is driven by a 25-hp motor operating at 1750 rpm.The speed reduction from the motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is negligible. (a) How much energy must be absorbed by the brake to stop this crane in a distance of 18 ft.? (b) Determine the constant average braking torque that must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of the cast-iron brake drum, which is 8 in. in diameter, 1 ½ in. thick, with a 3 ¼-in. face, what will be its temperature rise? (d) Compute the average rate at which the energy is absorbed during the first second (fhp). Is it reasonable? Solution: Page 4 of 97 SECTION 16 – BRAKES AND CLUTCHES W 2 ( vs1 − vs22 ) 2g W = 160,000 lb g = 32.2 fps 2 vs1 = 253 fpm = 4.22 fps U f = −∆KE = vs2 = 0 fps Uf = [ ] 160,000 (4.22)2 − 02 = 44,245 ft − lb 2(32.2) fhp (63,000 ) n 2 2 vs − vs1 0 − (4.22)2 a= 2 = = −0.495 fps 2 2s 2(18) vs − vs1 0 − 4.22 t= 2 = = 8.53 sec a − 0.495 U 44,245 fhp = f = = 9.43 hp 550t 550(8.53) fhp (63,000 ) (9.43)(63,000 ) Tf = = = 68 in − lb on the motor shaft. 1 n (1750) 2 (b) T f = (c) ∆t = Uf Wm c V = π Dbt (rim only) on the motor shaft D = 8 in b = 3.25 in t = 0.5 in V = π (8)(3.25)(0.5) = 40.84 in 3 Wm = ρ V ρ = 0.253 lb in3 for cast iron c = 101 ft − lb lb − F for cast-iron Wm = (0.253)(40.84) = 10.33 lb 44,245 ∆t = = 42.4o F (10.33)(101) (d) First second: Page 5 of 97 SECTION 16 – BRAKES AND CLUTCHES vs1 = 4.22 fps a = −0.495 fps 2 vs2 = vs1 + at = 4.22 − 0.495(1) = 3.73 fps U f = −∆KE = fhp = 885. Uf 550t = [ ] 160,000 (4.22)2 − (3.73)2 = 9680 ft − lb 2(32.2) 9680 = 17.6 hp < 25 hp , therefore reasonable. 550(1) The diagrammatic hoist shown with its load weighs 6000 lb. The drum weighs 8000 lb., has a radius of gyration k = 1.8 ft ; D = 4 ft . A brake on the drum shaft brings the hoist to rest in 10 ft. from vs = 8 fps (down). Only the brake frictional energy is significant, and it can be reasonably assumed that the acceleration is constant. (a) From the frictional energy, compute the average braking torque. (b) If the average fhp/sq. in. is limited to 0.15 during the first second, what brake contact area is needed? Problems 885, 886 Solution: 63,000 fhp Tf = n U f = −∆KE1 − ∆KE2 = ( I1 2 W ( ω1 − ω22 ) + 2 vs21 − vs21 2 2g vs1 = 8 fps , vs2 = 0 fps ω1 = 2vs1 = 2(8) = 4 rad s , ω2 = 0 rad s 4 D W k2 I1 = 1 g W1 = 8000 lb k = 1.8 ft W2 = 6000 lb Page 6 of 97 ) SECTION 16 – BRAKES AND CLUTCHES g = 32.2 fps 2 I1 2 W 80000(1.8) ( ω1 − ω22 ) + 2 vs21 − vs21 = (4)2 + 60000 (8)2 = 12,400 ft − lb 2 2g 2(32.2) 2(32.2) 2 2 vs2 − vs1 ( Uf = a= 2 ) 2s s = 10 ft 0 2 − 82 = −3.2 fps 2 2(10 ) vs − vs1 0 − 8 t= 2 = 2.5 sec = a − 3.2 U 12,400 fhp = f = = 9 hp 550t 550(2.5) 60ω n= rpm 2π 1 ω = (4 rad s − 0) = 2 rad s − 0 2 60(2 ) n= = 19.1 rpm 2π 63,000 fhp 63,000(9 ) Tf = = = 29,700 in − lb n 19.1 a= (b) fhp sq. in. = 0.15 (first second) vs2 = vs1 + at = 8 − 3.2(1) = 4.8 fps 2(4.8) = 2.4 rad sec D 4 2 80000(1.8) Uf = (4)2 − (2.4)2 + 60000 (8)2 − (4.8)2 = 6106 ft − lb 2(32.2 ) 2(32.2) U 6106 fhp = f = = 11.10 hp 550t 550(1) fhp 11.10 A= = = 74 in 2 fhp sq.in. 0.15 ω2 = 2vs2 = [ 887. ] [ ] The same as 885, except that a traction drive, arranged as shown, is used; the counterweight weighs 4000 lb. The ropes pass twice about the driving sheave; the brake drum is on this same shaft. Page 7 of 97 SECTION 16 – BRAKES AND CLUTCHES Problem 887. Solution: ( ) WT 2 vs − vs22 2g 1 WT = 4000 lb + 6000 lb = 10,000 lb − ∆KE of pulley is negligible vs1 = 8 fps , vs2 = 0 fps (a) U f = −∆KE = 10,000 2 (8) = 9,940 ft − lb 2(32.2) vs22 − vs21 0 2 − 82 a= = = −3.2 fps 2 2s 2(10) vs − vs1 0 − 8 t= 2 = = 2.5 sec a − 3.2 U 9940 fhp = f = = 7.23 hp 550t 550(2.5) D = 4 ft 2v 2(8) ω1 = s1 = = 4 rad sec D 4 2v 2(0) ω 2 = s2 = = 0 rad sec D 4 1 1 ω = (ω1 + ω2 ) = (4 + 0) = 2 rad sec 2 2 60ω 60(2 ) n= = 19.1 rpm 2π 2π Uf = Page 8 of 97 SECTION 16 – BRAKES AND CLUTCHES Braking torque, T f = 63,000 fhp 63,000(7.23) = = 23,850 in − lb n 19.1 (b) fhp sq. in. = 0.15 (first second) vs1 = 8 fps vs2 − vs1 = at vs2 − 8 = −3.2(1) vs2 = 4.8 fps [ ] 10,000 (8)2 − (4.8)2 = 6360 ft − lb 2(32.2) U 6360 fhp = f = = 11.56 hp 550t 550(1) fhp 11.56 = = 77.1 in 2 Contact area = A = fhp sq.in. 0.15 Uf = SINGLE-SHOE BRAKES 888. For the single-shoe, short-block brake shown (solid lines) derive the expressions for brake torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) In which direction of rotation does the brake have self-actuating properties? If f = 0.25 , for what proportions of e and c would the brake be self-actuating? Problems 888 – 891, 893. Solution: (a) Clockwise rotation (as shown) Page 9 of 97 SECTION 16 – BRAKES AND CLUTCHES FD 2 F = fN Tf = [∑ M ] =0 fN e + Wa = N c N c − fN e = Wa Wa N= c − fe fWa F= c − fe fWaD Tf = 2(c − fe ) H (b) Counter Clockwise Rotation FD 2 F = fN Tf = [∑ M H =0 ] Page 10 of 97 SECTION 16 – BRAKES AND CLUTCHES Wa = fN e + N c Wa N= c + fe fWa F= c + fe fWaD Tf = 2(c + fe ) (c) Clockwise rotation is self-actuating c > fe with f = 0.25 c > 0.25e The same as 888, except that the wheel and brake shoe are grooved, 2θ degrees between the sides of the grooves (as in a sheave, Fig. 17.38, Text). 889. Solution: [∑ F V =0 ] 2 N1 sin θ = N F = 2 f N1 Page 11 of 97 SECTION 16 – BRAKES AND CLUTCHES N fN F =2f = 2 sin θ sin θ (a) Clockwise rotation Wa N= c − fe fWa F= (c − fe )sin θ fWaD Tf = 2(c − fe )sin θ (b) Counter clockwise rotation Wa c + fe fWa F= (c + fe)sin θ fWaD Tf = 2(c + fe ) sin θ N= (c) Clockwise rotation is self-actuating c > fe with f = 0.25 c > 0.25e 890. Consider the single-shoe, short-block brake shown (solid lines) with the drum rotating clockwise; let e be positive measured downward and D = 1.6c . (a) Plot the mechanical advantage MA (ordinate) against f values of 0.1, 0.2, 0.3, 0.4, 0.5 (abscissa) when e c has values 2, 0.5, 0, -0.5, -1. (b) If f may vary from 0.3 to 0.4, which proportions give the more nearly constant brake response? Are proportions good? (c) What proportions are best if braking is needed for both directions of rotation? Solution: Page 12 of 97 SECTION 16 – BRAKES AND CLUTCHES (a) MA = Tf , Clockwise rotation Wa fD MA = 2(c − fe ) D = 1 .6 c 1.6 fc MA = 2(c − fe ) 0 .8 f MA = fe 1 − c Tabulation: f 0.1 0.2 0.3 0.4 0.5 Page 13 of 97 2 0.100 0.267 0.600 1.600 ∞ Values of MA ec 0.5 0 0.084 0.08 0.178 0.16 0.284 0.24 0.400 0.32 0.533 0.40 -0.5 0.076 0.145 0.209 0.267 0.320 -1 0.073 0.133 0.185 0.229 0.267 SECTION 16 – BRAKES AND CLUTCHES Plot: (b) f = 0.3 to 0.4 , e c = −1 , with MA ≈ constant . They are good because 1 > fe except e c = 2 . c (c) e c = 0 is the best if braking is needed for both directions of rotation with MA the same. 891. A single-block brake has the dimensions: cast-iron wheel of D = 15 in ., 1 3 11 a = 32 in ., c = 9 in ., e = 4 in ., width of contact surface = 2 in. The brake 2 8 16 block lined with molded asbestos, subtends 80o, symmetrical about the center line; it is permitted to absorb energy at the rate of 0.4 hp/in.2; n = 200 rpm . Assume that p is constant, that F and N act at K , and compute (a) pvm and the approximate braking torque, (b) the force W to produce this torque, (c) the mechanical advantage, (d) the temperature rise of the 3/8-in.-thick rim, if it absorbs all the energy with operation as specified, in 1 min. (e) How long could this brake be so applied for ∆t = 400 o F ? See 893. Solution: Page 14 of 97 SECTION 16 – BRAKES AND CLUTCHES D = 15 in a = 32.5 in c = 9.375 in e = 4.6875 in b = 2 in (a) Solving for pvm Fvm = fpAvm ft − lb min Fvm = 0.4 hp in 2 A Fvm (0.4 hp )(33,000 ft − lb hp − min ) 13,200 ft − lb min = = A in 2 in 2 Fvm = fpvm A f = 0.35 from Table AT 29, molded asbestos on cast iron Fvm = 13,200 = 0.35 pvm A pvm = 37,700 ft − lb sq in − min Solving for braking torque Fvm = 13,200 ft − lb sq. in. − min A 15 vm = π Dn = π (200 ) = 785 fpm 12 θDb A= 2 π θ = (80) = 1.3963 rad 180 Page 15 of 97 SECTION 16 – BRAKES AND CLUTCHES A= θDb (1.3963)(15)(2) = = 21 sq. in. 2 2 F (785) = 13,200 21 F = 353 lb FD (353)(15) Tf = = = 2650 in − lb 2 2 (b) Solving for W f Wa c − fe F = 353 lb f = 0.35 a = 32.5 in e = 4.6875 in c = 9.375 in F (c − fe ) (353)[9.375 − (0.35)(4.6875)] W= = = 240 lb (0.35)(32.5) fa F= (c) Solving for MA MA = (0.35)(15) fD = = 0.34 2(c − fe ) 2[9.375 − (0.35)(4.6875)] (d) Solving for ∆t ∆t o F = U f , ft − lb Wm c Wm = ρπ Dbt D = 15 in b = 2 in 3 t = in = 0.375 in 8 ρ = 0.253 lb in3 for cast iron Wm = (0.253)(π )(15)(2)(0.375) = 8.942 lb c = 101 ft − lb lb − F for cast iron Page 16 of 97 SECTION 16 – BRAKES AND CLUTCHES U f = 550t ′( fhp ) t ′ = 1 min = 60 sec U f = 550(60 )( fhp ) = 33,000 fhp fhp = Tf n = (2650)(200) = 8.4127 hp 63,000 63,000 U f = 33,000(8.4127 ) = 277,619 ft − lb ∆t = Uf Wm c = 277,619 = 310o F (8.942)(101) (e) Solving for t ′ , (time) with ∆t = 400o F U f = Wm c∆t U f = (8.942 )(101)(400 ) = 361,260 ft − lb 550( fhp )(t ′) = U f 550(8.4127 )(t ′) = 361,260 t ′ = 78 sec = 1.3 min 892. 1 For a single-block brake, as shown, a = 26 in ., c = 7 in ., e = 3.75 in ., 2 1 D = 15 in ., drum contact width b = 3 in . The molded asbestos lining subtends 2 o θ = 60 , symmetrical about the vertical axis; force W = 300 lb .; n = 600 rpm . Assume that p is constant, that F and N act at K , and compute (a) pvm and the braking torque, (b) the energy rate in fhp/in.2 of contact surface. (c) the mechanical advantage, (d) the temperature of the 3/8-in.-thick rim, if it absorbs all the energy with the operation as specified in 1 min. (e) How long could this brake be so applied for ∆t rim = 400o F ? See 894. Page 17 of 97 SECTION 16 – BRAKES AND CLUTCHES Problems 892, 894. Solution: For greater braking torque, T f , use counterclockwise rotation [∑ M ] =0 aW + efN = cN Wa N= c − ef f Wa F= c − ef From Table AT 29, f = 0.35 for molded asbestos W = 300 lb a = 26 in c = 7.5 in e = 3.75 (0.35)(300)(26) = 442 lb F= 7.5 − (3.75)(0.35) A (a) Solving for pvm Fvm = fpAvm π Dn π (15)(600) vm = = = 2536 fpm 12 12 θDb A= 2 π θ = 60 = 1.047 rad 180 (1.047 )(15)(3.5) = 27.5 in 2 A= 2 Fvm = (442)(2536) = (0.35)(27.5) pvm pvm = 116,500 ft − lb sq. in. − min Page 18 of 97 SECTION 16 – BRAKES AND CLUTCHES Solving for the braking torque, FD (442 )(15) Tf = = = 3315 in − lb 2 2 (b) Energy rate, fhp.in2. fhp = Tf n = (3315)(600) = 31.6 hp 63,000 63,000 2 A = 27.5 in 31.6 hp fhp in 2 = = 1.15 hp in 2 27.5 in 2 T 3315 (c) MA = f = = 0.425 Wa (300)(26) U , ft − lb (d) ∆t o F = f Wm c Wm = ρπ Dbt 3 t = in = 0.375 in 8 D = 15 in b = 3.5 in ρ = 0.253 lb in3 for cast iron c = 101 ft − lb lb − F for cast iron Wm = (0.253)(π )(15)(3.5)(0.375) = 15.648 lb For 1 min U f = 33,000(1)( fhp ) = 33,000(1)(31.6 ) = 1,042,800 ft − lb ∆t = 1,042,800 = 660o F (15.648)(101) (e) ∆t rim = 400o F U f = (400 )(15.648)(101) = 632,179 ft − lb ′ = t min Uf 33,000 fhp Page 19 of 97 = 632,179 = 0.61 min 33,000(31.6) SECTION 16 – BRAKES AND CLUTCHES LONG-SHOE BRAKES FIXED SHOES 893. The brake is as described in 891 and is to absorb energy at the same rate but the pressure varies as p = P sin θ . Derive the equations needed and compute (a) the maximum pressure, (b) the moment M F H of F about H , (c) the moment M N H of N about H , (d) the force W , (e) the braking torque, (f) the x and y components of the force at H . Solution: p = P sin θ = P sin φ D r= 2 dN = pbrdφ dF = fpbrdφ Page 20 of 97 SECTION 16 – BRAKES AND CLUTCHES T f = ∫ rdF T f = ∫ fpbr 2 dφ T f = fbr 2 P ∫ sin φdφ T f = fbr 2 P(cos φ1 − cos φ2 ) (a) Solving for P Tf P= 2 fbr (cos φ1 − cos φ2 ) D r= 2 c tan α = r −e c = 9.375 in 15 r = = 7.5 in 2 e = 4.6875 in 9.375 tan α = 7.5 − 4.6875 α = 73.3o θ = 80o θ 80 φ1 = α − = 73.3 − = 33.3o 2 2 θ 80 φ2 = α + = 73.3 + = 113.3o 2 2 f = 0.35 b = 2 in r = 7.5 in Tf P= fbr 2 (cos φ1 − cos φ2 ) Tf Tf P= = psi (0.35)(2)(7.5)2 (cos 33.3 − cos113.3) 48.5 63,000 fhp Tf = n fhp = fhp in 2 ( A) θDb A= 2 π θ = 80 = 1.396 rad 180 ( Page 21 of 97 ) SECTION 16 – BRAKES AND CLUTCHES A= (1.396)(15)(2) = 21 in 2 2 fhp in 2 = 0.4 fhp = (0.4)(21) = 8.4 hp n = 200 rpm 63,000(8.4 ) Tf = = 2646 in − lb 200 T 2646 P= f = = 55 psi = max .P (φ2 > 90o ) 48.5 48.5 (b) M F H = ∫ (r − R cos φ )dF φ2 MF H = ∫ φ1 (r − R cos φ ) fbrP sin φdφ φ2 M F H = fbrP ∫ φ1 (r sin φ − R sin φ cos φ )dφ φ2 MF H R = fbrP − r cos φ − sin 2 φ 2 φ1 R M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 2 ( R = c 2 + (r − e ) = 2 (9.375)2 + (7.5 − 4.6875)2 ) = 9.788 in 9.788 sin 2 113.3 − sin 2 33.3 M F H = (0.35)(2 )(7.5)(55)7.5(cos 33.3 − cos113.3) − 2 M F H = 1900 in − lb ( (c) M N H = ∫ R sin φdN φ2 MN H = ∫ RP sin 2 φbrdφ MN H = brRP ∫ sin 2 φdφ MN H = φ1 φ2 φ1 ( ) brRP φ2 1 − cos 2 φ dφ 2 ∫φ1 φ MN H 2 brRP 1 = φ − sin 2 φ 2 2 φ1 brRP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 φ2 − φ1 = θ = 1.396 rad 2φ2 = 2(113.3) = 226.6o MN H = 2φ1 = 2(33.3) = 66.6o Page 22 of 97 ) SECTION 16 – BRAKES AND CLUTCHES MN H MN H (d) (2)(7.5)(9.788)(55) [2(1.396) − (sin 226.6 − sin 66.6)] = 4 = 8956 in − lb ∑M H =0 Wa + M F H − M N H =0 a = 32.5 in W (32.5) + 1900 − 8956 = 0 W = 217 lb (e) T f = 2646 in − lb (f) ∑F x =0 − H x − W cos α + ∫ dN sin φ + ∫ dF cos φ = 0 φ2 φ2 φ1 φ1 − H x = W cos α − Pbr ∫ sin 2 φdφ − fPbr ∫ sin φ cos φdφ ( brP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − fbrP sin 2 φ2 − sin 2 φ1 4 2 ( 2 )(7.5)(55) − H x = 217 cos 73.3 − [2(1.396) − (sin 226.6 − sin 66.6)] 4 (0.35)(2)(7.5)(55) sin 2 113.3 − sin 2 113.3 − 2 − H x = −931 lb − H x = W cos α − ( ) H x = 931 lb ∑F y =0 − H y + W sin α − ∫ dN cos φ + ∫ dF sin φ = 0 φ2 φ2 φ1 φ1 − H y = ∫ brP sin φ cos φdφ − ∫ fbrP sin 2 φdφ − W sin α ( ) brP fbrP sin 2 φ2 − sin 2 φ1 − [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α 2 4 (2)(7.5)(55) sin 2 113.3 − sin 2 33.3 − Hy = 2 (0.35)(2)(7.5)(55) [2(1.396) − (sin 226.6 − sin 66.6)] − 217 sin 73.3 − 4 − H y = −305 lb − Hy = ( H y = 305 lb Page 23 of 97 ) ) SECTION 16 – BRAKES AND CLUTCHES 894. The brake is as described in 892, but the pressure varies as p = P sin φ . Assume the direction of rotation for which a given W produces the greater T f , derive the equations needed, and compute (a) the maximum pressure, (b) the moment of F about A , (c) The moment of N about A , (d) the braking torque, (e) the x and y components of the force at A . Solution: p = P sin φ dN = pbrdφ dN = Pbr sin φdφ dF = fdN = fPbr sin φdφ Solving for φ1 and φ2 Page 24 of 97 SECTION 16 – BRAKES AND CLUTCHES tan α = c r+e D = 7.5 in 2 7 .5 tan α = 7.5 + 3.75 α = 33.69o r= θ 60 = 3.69o 2 2 θ 60 φ1 = α + = 33.69 + = 63.69o 2 2 M F A = ∫ (R cos φ − r )dF φ1 = α − φ2 MF A = ∫ φ1 = 33.69 − (R cos φ − r ) fPbr sin φdφ φ2 M F A = fPbr ∫ φ1 (R sin φ cos φ − r sin φ )dφ R M F A = fPbr sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2 ( ) R = c 2 + (e + r ) = 2 (7.5)2 + (3.75 + 7.5)2 = 13.52 in 13.52 2 2 M F A = (0.35)P(3.5)(7.5) (sin 63.69 − sin 3.69) + 7.5(cos 63.69 − cos 3.69) 2 M F A = 11.43P MN A = ∫ R sin φdN MN A = ∫ RPbr sin 2 φdφ φ2 φ1 ( ) brPR φ2 1 − cos 2 φ dφ 2 ∫φ1 brPR MN A = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 φ2 − φ1 = θ = 1.047 rad 2φ2 = 2(63.69) = 127.38o MN A = 2φ1 = 2(3.69) = 7.38o (3.5)(7.5)P(13.52) [2(1.047 ) − (sin 127.38 − sin 7.38)] MN A = 4 M N A = 126.68 P (a) ∑M A =0 Wa + M F A − M N W = 300 lb Page 25 of 97 A =0 SECTION 16 – BRAKES AND CLUTCHES a = 26 in (300)(26) + 11.43P − 126.68P = 0 P = 67.68 psi max. p = P sin φ2 = 67.68 sin 63.69 = 60.67 psi (b) M F A = 11.43(67.68) = 774 in − lb (c) M N A = 126.68(67.68) = 8575 in − lb (d) T f = ∫ rdF φ2 T f = ∫ fPbr 2 sin φdφ φ1 T f = fPbr 2 (cos φ1 − cos φ2 ) T f = (0.35)(60.68)(3.5)(7.5) (cos 3.69 − cos 63.69) 2 T f = 2587 in − lb (e) [∑ F x =0 ] − H x − W cos α + ∫ dN sin φ − ∫ dF cos φ = 0 φ2 φ2 φ1 φ1 − H x = W cos α − Pbr ∫ sin 2 φdφ + fPbr ∫ sin φ cos φdφ ( Pbr [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] + fPbr sin 2 φ2 − sin 2 φ1 4 2 (67.68)(3.5)(7.5) [2(1.047 ) − (sin 127.38 − sin 7.38)] − H x = 300 cos 33.69 − 4 (0.35)(67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69 + 2 − H x = −136 lb − H x = W cos α − ( ) H x = 136 lb [∑ F y =0 ] H y + W sin α − ∫ dN cos φ − ∫ dF sin φ = 0 φ2 φ2 φ1 φ1 H y = Pbr ∫ sin φ cos φdφ + fPbr ∫ sin 2 φdφ − W sin α ( ) Pbr fPbr sin 2 φ2 − sin 2 φ1 + [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α 2 4 (67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69 Hy = 2 ( 0.35)(67.68)(3.5)(7.5) + [2(1.047) − (sin 127.38 − sin 7.38)] − 300 sin 33.69 4 Hy = ( Page 26 of 97 ) ) SECTION 16 – BRAKES AND CLUTCHES H y = 766 lb 895. (a) For the brake shown, assume p = P cos α and the direction of rotation for which a given force W results in the greater braking torque, and derive equations for T f in terms of W , f , and the dimensions of the brake. (b) Under what circumstances will the brake be self-acting? (c) Determine the magnitude and location of the resultant forces N and F . Solution: (a) Clockwise rotation has greatest braking torque. p = P cos α dN = pbrdα = Pbr cos αdα dF = fdN = fpbrdα = fPbr cos αdα θ2 MF H = ∫ −θ1 θ2 MF H = ∫ −θ1 (r + c sin α )dF (r + c sin α ) fPbr cos αdα Page 27 of 97 SECTION 16 – BRAKES AND CLUTCHES θ2 MF H = ∫ −θ1 fPbr (r cos α + c sin α cos α )dα θ 2 1 M F H = fPbr r sin α + c sin 2 α 2 −θ1 [ ] 1 M F H = fPbr r [sin (θ 2 ) − sin (− θ1 )] + c sin 2 (θ 2 ) − sin 2 (− θ1 ) 2 1 M F H = fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 2 ( ) θ2 MN H = ∫ cos αdN MN H = ∫ cPbr cos 2 αdα −θ1 θ2 −θ1 cPbr θ 2 (1 + cos 2α )dα 2 ∫−θ1 cPbr MN H = [2 + sin 2α ]θ−2θ1 4 cPbr MN H = [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] 4 ∑MH = 0 MN = H [ ] Wa + M F H − M N H =0 1 cPbr [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] Wa + fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 = 2 4 Wa P= cbr [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − fbr 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 4 2 4Wa P= br c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 ( ) [ [ { )] ( )]} ( T f = ∫ rdF θ2 Tf = ∫ −θ1 fPbr 2 cos αdα T f = fPbr 2 [sin α ]−2θ1 θ T f = fPbr 2 (sin θ 2 + sin θ1 ) 4 fWabr 2 (sin θ 2 + sin θ1 ) br {c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c(sin 2 θ 2 − sin 2 θ1 ) } 4 fWar (sin θ 2 + sin θ1 ) Tf = c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 ) D where r = e = 2 Tf = [ [ Page 28 of 97 ] ] SECTION 16 – BRAKES AND CLUTCHES [ ] (b) c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] > 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 ) 4 fr (sin θ 2 + sin θ1 ) c> 2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) − 2 f (sin 2 θ 2 − sin 2 θ1 ) (c) N = ∫ dN θ2 N = ∫ Pbr cos αdα −θ1 N = Pbr [sin α ]−2θ1 θ N = Pbr (sin θ 2 + sin θ1 ) F = fN F = fPbr (sin θ 2 + sin θ1 ) Solving for the location of F and N . Let A = vertical distance from O . θ2 ( A − r cos α )dF ∑M F Loc . =∫ ∑M F Loc . = ∫ P A cos α − r cos 2 α fbrdα ∑M F Loc . = Pfbr ∫ ∑M F Loc . ∑M −θ1 θ2 −θ1 ( θ2 −θ1 ) (A cos α − r cos α )dα 2 θ2 1 = Pfbr ∫ A cos α − r (1 + cos 2α ) dα −θ1 2 θ2 F Loc . 1 1 = Pfbr A sin α − r α + sin 2α 2 2 −θ1 1 1 = Pfbr [ A(sin θ 2 + sin θ1 )] − r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 2 2 Then ∑ M F Loc . = 0 ∑M F Loc . [A(sin θ 2 + sin θ1 )] − 1 r (θ 2 + θ1 ) + 1 (sin 2θ 2 + sin 2θ1 ) = 0 2 2 1 1 A(sin θ 2 + sin θ1 ) = r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 2 2 1 1 r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 2 2 A= (sin θ 2 + sin θ1 ) r [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] A= 4(sin θ 2 + sin θ1 ) Page 29 of 97 SECTION 16 – BRAKES AND CLUTCHES 896. For the brake shown with θ1 ≠ θ 2 , assume that the direction of rotation is such that a given W results in the greater braking torque and that p = P sin φ . (a) Derive equations in terms of θ1 and θ 2 for the braking torque, for the moment M F H and for M N H . (b) Reduce the foregoing equations for the condition θ1 = θ 2 . (c) Now suppose that θ , taken as θ = θ1 + θ 2 , is small enough that θ sin θ ≈ θ , cos θ ≈ 1 , θ1 = θ 2 = . What are the resulting equations? 2 Solution: (a) Use clockwise rotation p = P sin φ dN = Pbr sin φdφ dF = fdN = fPbr sin φdφ φ1 = 90 − θ1 φ2 = 90 + θ 2 T f = ∫ rdF Page 30 of 97 SECTION 16 – BRAKES AND CLUTCHES φ2 T f = fPbr 2 ∫ sin φdφ φ1 T f = fPbr (cos φ1 − cos φ2 ) 2 T f = fPbr 2 [cos(90 − θ1 ) − cos(90 + θ 2 )] T f = fPbr 2 (sin θ1 + sin θ 2 ) M F H = ∫ (r − c cos φ )dF φ2 M F H = ∫ fbPr (r − c cos φ )sin φdφ φ1 φ2 M F H = fPbr ∫ φ1 (r sin φ − c sin φ cos φ )dφ φ MF H 2 1 = fPbr − r cos φ − c sin 2 φ 2 φ1 1 M F H = fPbr r (cos φ1 − cos φ2 ) − c sin 2 φ2 − sin 2 φ1 2 1 M F H = fPbr r [cos(90 − θ1 ) − cos(90 + θ 2 )] − c sin 2 (90 + θ 2 ) − sin 2 (90 − θ1 ) 2 1 M F H = fPbr r (sin θ1 + sin θ 2 ) − c cos 2 θ 2 − cos 2 θ1 2 1 M F H = fPbr r (sin θ1 + sin θ 2 ) − c 1 − sin 2 θ 2 − 1 − sin 2 θ1 2 1 M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1 2 ( ) [ ( [( ( MN H = ∫ r sinφdN MN H = bPr 2 ∫ sin 2 φdφ ] ) )] ) ( ) φ2 φ1 2 MN H = MN H = MN H = MN H = MN H = Pbr φ2 (1 − cos 2φ )dφ 2 ∫φ1 Pbr 2 [2 − sin 2φ ]φφ12 4 Pbr 2 [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 Pbr 2 {2[(90 + θ 2 ) − (90 − θ1 )] − [sin 2(90 + θ 2 ) − sin 2(90 − θ1 )]} 4 Pbr 2 [2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )] 4 Page 31 of 97 SECTION 16 – BRAKES AND CLUTCHES MN H = Pbr 2 [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] 4 (b) θ1 = θ 2 T f = fbPr 2 (sin θ1 + sin θ 2 ) T f = 2 fbPr 2 sin θ1 1 M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1 2 2 M F H = 2 fPbr sin θ1 ( MN H MN H MN H MN H Pbr 2 = [2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )] 4 bPr 2 = (4θ1 + 2 sin 2θ1 ) 4 bPr 2 = (4θ1 + 4 sin θ1 cos θ1 ) 4 = bPr 2 (θ1 + sin θ1 cos θ1 ) (c) θ = θ1 + θ 2 sin θ ≈ θ cos θ ≈ 1 θ1 = θ 2 = θ 2 T f = 2 fbPr 2 sin θ1 θ θ T f = 2 fbPr 2 sin = 2 fbPr 2 = fbPr 2θ 2 2 MF H = 2 fPbr 2 sin θ1 θ θ M F H = 2 fbPr 2 sin = 2 fbPr 2 = fbPr 2θ 2 2 MN H MN H = bPr 2 (θ1 + sin θ1 cos θ1 ) θ θ = bPr 2 + (1) = bPr 2θ 2 2 Page 32 of 97 ) SECTION 16 – BRAKES AND CLUTCHES 897. The brake shown is lined with woven asbestos; the cast-iron wheel is turning at 60 rpm CC; width of contact surface is 4 in. A force W = 1300 lb . is applied via linkage systemnot shown; θ = 90o . Let p = P sin φ . (a) With the brake lever as a free body, take moments about the pivot J and determine the maximum pressure and compare with permissible values. Compute (b) the braking torque, (c) the frictional energy in fhp. (d) Compute the normal force N , the average pressure on the projected area, and decide if the brake application can safely be continuous. Solution: (a) dF = fdN p = P sin φ dN = pbrdφ = Pbr sin φdφ dF = fPbr sin φdφ M F J = ∫ (R cos φ − r )dF φ2 M F J = fPbr ∫ φ1 Page 33 of 97 (R cos φ − r )sin φdφ SECTION 16 – BRAKES AND CLUTCHES φ2 M F J = fPbr ∫ φ1 (R sin φ cos φ − r sin φ )dφ φ 2 1 M F J = fPbr R sin 2 φ + r cos φ 2 φ1 1 M F J = fPbr R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2 12.5 tan β = 10 β = 51.34o ( φ1 = β − θ = 90 ) θ 2 o 90 = 6.34o 2 θ 90 φ1 = β + = 51.34 + = 96.34o 2 2 b = 4 in r = 10 in for woven asbestos f = 0.4 (Table At 29) φ1 = 51.34 − (12.5)2 + (10)2 R= = 16 in 1 M F J = fPbr R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 ) 2 16 M F J = (0.4 )P(4 )(10 ) sin 2 96.34 − sin 2 6.34 + 10(cos 96.34 − cos 6.34 ) 2 M F J = −51.81P ( ) ( ) M N J = ∫ R sin φdN φ2 M N J = PbrR ∫ sin 2 φdφ φ1 PbrR [1 − cos 2φ ]φφ12 2 PbrR = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 P(4)(10)(16) π = 2(96.34 − 6.34) − (sin 2(96.34) − sin 2(6.34)) 4 180 = 572.9 P MN J = MN J MN J MN J ∑M J = Wa +M F J − M N J = 0 Page 34 of 97 SECTION 16 – BRAKES AND CLUTCHES (1300)(25) + (− 51.81P ) − 572.9P = 0 P = 52 psi max. p = P = 52 psi , φ2 > 90 From Table AT 29, permissible p = 50 psi Therefore pmax ≈ p permissible (b) T f = ∫ rdF φ2 T f = fPbr ∫ sin φdφ φ1 T f = fPbr (cos φ1 − cos φ2 ) T f = (0.4 )(52 )(4 )(10 )(cos 6.34 − cos 96.34 ) = 9188 in − lb Tf n (c) fhp = , n = 60 rpm 63,000 (9188)(60) = 8.75 hp fhp = 63,000 (d) N = ∫ dN φ2 N = Pbr ∫ sin φdφ φ1 N = Pbr (cos φ1 − cos φ2 ) N = (52)(4)(10)(cos 6.34 − cos 96.34) = 2297 lb ave. p = N 2br sin θ = 90 θ 2 o ave. p = 2297 2(4 )(10 )sin 90 2 = 40.6 psi π Dn π pvm = p = (40.6 ) (20 )(60 ) = 12,755 ft − lb sq. in. − min 12 12 since pvm < 28,000 ft − lb sq. in. − min (§18.4) Application is continuous. Page 35 of 97 SECTION 16 – BRAKES AND CLUTCHES PIVOTED-SHOE BRAKES 898. In the brake shown, the shoe is lined with flexible woven asbestos, and pivoted at point K in the lever; face width is 4 in.; θ = 90o . The cast-iron wheel turns 60 rpm CL; let the maximum pressure be the value recommended in Table At 29. On the assumption that K will be closely at the center of pressure, as planned, compute (a) the brake torque, (b) the magnitude of force W , (c) the rate at which frictional energy grows, (d) the time of an application if it is assumed that all this energy is stored in the 1-in. thick rim with ∆t rim = 350 F , (e) the average pressure on projected area. May this brake be applied for a “long time” without damage? (f) What would change for CC rotation? Problem 898. Solution: a = 27 in , b = 4 in , n = 60 rpm CL 2 D sin c= θ 2 θ + sin θ D = 20 in , r = 10 in θ = 90o = 1.571 rad 90 2 = 11.0 in c= 1.571 + sin 90 2(20)sin (a) T f = 2 fPbr 2 sin θ 2 For woven asbestos, Table AT 29, f = 0.4 P = 50 psi 90 2 T f = 2(0.4 )(50 )(4 )(10 ) sin = 11,314 in − lb 2 Page 36 of 97 SECTION 16 – BRAKES AND CLUTCHES (b) θ + sin θ N = Pbr 2 ∑MJ = 0 Wa = 12 N W (15) = 12(2571) W = 2057 lb [ 1.571 + sin 90 = (50 )(4 )(10 ) = 2571 lb 2 ] (c) fhp = Tf n = (11,314)(60) = 10.78 hp 63,000 63,000 rate of frictional energy = 33,000 fhp = 33,000(10.78) = 355,740 ft − lb min (d) Time (min) = ∆t o F = Uf 33,000 fhp U f ft − lb Wm c Wm = ρπ Dbt For cast iron ρ = 0.253 lb in3 c = 101 ft − lb lb − F t = 1 in Wm = (0.253)π (20)(4)(1) = 63.6 lb U f ft − lb ∆t o F = 350 = (63.6 lb )(101 ft − lb lb − F ) U f = 2,248,260 ft − lb Time (min) = 2,248,260 = 6.32 min 33,000(10.78) N (e) Ave. p = 2br sin Page 37 of 97 θ 2 = 2571 90 2(4 )(10 ) sin 2 = 45.45 psi SECTION 16 – BRAKES AND CLUTCHES pπ Dn (45.45)(π )(20 )(60 ) = = 14,280 ft − lb sq. in. − F 12 12 since pvm < 28,000 , this brake may be applied for a long time. pvm = (f) Since the moment arn of F is zero, no change or CC rotation. The pivoted-shoe brake shown is rated at 450 ft-lb. of torque; θ = 90o ; contact width is 6.25 in.; cast-iron wheel turns at 600 rpm; assume a symmetric sinusoidal distribution of pressure. (a) Locate the center of pressure and compute with the location of K. Compute (b) the maximum pressure and compare with allowable value, (c) the value of force W , (d) the reaction at the pin H , (e) the average pressure and pvm , and decide whether or not the application could be continuous at the rated torque. (f) Compute the frictional work from Tω and estimate the time it will take for the rim temperature to reach 450 F (ambient, 100 F). 899. Problem 899. Solution: 2 D sin (a) c = θ 2 θ + sin θ D = 18 in θ = 90o = 1.571 rad 90 2 = 9.9011 in c= 1.571 + sin 90 but location of K = 9.8125 in then, c ≈ location K 2(18)sin (b) T f = 2 fPbr 2 sin θ 2 T f = 450 ft − lb = 5400 in − lb Page 38 of 97 SECTION 16 – BRAKES AND CLUTCHES b = 6.25 in r = 9 in use f = 0.4 (on cast-iron) T f = 2 fPbr 2 sin θ 2 90 2 P = 18.86 psi < allowable (Table AT 9) 5400 = 2(0.4 )P(6.25)(9 ) sin 2 (c) W (20.375) = N (10.375) θ + sin θ 1.571 + sin 90 N = Pbr = (18.86 )(6.25)(9 ) = 1364 lb 2 2 ( 1364 )(10.375) W= = 695 lb 20.375 (d) H = N − W = 1364 − 695 = 669 lb ↓ N (e) Ave. p = 2br sin θ 2 = 1364 90 2(6.25)(9 ) sin 2 = 17.15 psi n = 600 rpm pπ Dn (17.15)(π )(18)(600 ) pvm = = = 48,490 ft − lb sq. in. − F 12 12 since pvm > 28,000 , not continuous 2π (600 rpm ) (f) Frictional work = Tω = (450 ft − lb ) = 28,275 ft − lb per sec 60 sec min U ft − lb ∆t o F = f Wm c Wm = ρπ Dbt For cast iron ρ = 0.253 lb in3 c = 101 ft − lb lb − F π (18) Wm = (0.253)π (18)(6.25)t + 2 t = 154 t 4 ∆t = 450 − 100 = 350 F U f = ∆tWm c = (350 )(154 t )(101) = 5,443,900 t ft − lb Page 39 of 97 SECTION 16 – BRAKES AND CLUTCHES U f = 2,248,260 ft − lb 5,443,900 t = 192.5 t sec 28,275 1 Assume t = in 2 Time = 96 sec Time = TWO-SHOE BRAKES PIVOTED SHOES 900. The double-block brake shown is to be used on a crane; the force W is applied by a spring, and the brake is released by a magnet (not shown); θ = 90o ; contact width = 2.5 in. Assume that the shoes are pivoted at the center of pressure. The maximum pressure is the permissible value of Table AT 29. Compute (a) the braking torque, (b) the force W , (c) the rate of growth of frictional energy at 870 rpm, (d) the time it would take to raise the temperature of the 0.5-in.-thick rim by ∆t = 300 F (usual assumption of energy storage), (e) pvm . (f) Where should the pivot center be for the calculations to apply strictly? Problem 900. Solution: θ 90 2 = 2 = 5.5 in c= π θ + sin θ + sin 90 2 2 D sin Page 40 of 97 2(10)sin SECTION 16 – BRAKES AND CLUTCHES [∑ M ] R1 =0 R2 =0 F1 (5.5 − 0.875) + 12.75W = 6.75 N1 fN1 (4.625) + 12.75W = 6.75 N1 12.75W N1 = 6.25 − 4.625 f [∑ M ] 12.75W = F2 (5.5 − 0.875) + 6.75 N 2 12.75W = 4.625 fN 2 + 6.75 N 2 12.75W N2 = 6.25 + 4.625 f Assume flexible woven asbestos, f = 0.40 , p = 50 psi 12.75W = 2.898W 6.25 − 4.625(0.40) F1 = fN1 = (0.4)(2.898W ) = 1.16W N1 = Page 41 of 97 SECTION 16 – BRAKES AND CLUTCHES 12.75W = 1.574W 6.25 + 4.625(0.40) F2 = fN 2 = (0.4)(1.574W ) = 0.63W N2 = max . T f = T f1 2 fPbr 2 sin θ 2 = F1c 2 90 10 2(0.40 )(50 )(2.5) sin = (1.16W )(5.5) 2 2 W = 277 lb (a) Braking torque = T f1 + T f 2 = (F1 + F2 )c = (1.16 + 0.63)(277 )(5.5) = 2727 in − lb (b) W = 277 lb Tf n (2727 )(870) = 37.66 hp = (c) fhp = 63,000 63,000 (d) Solving for tine: U ft − lb ∆t o F = f Wm c ∆t o F = 300o F c = 101 , ρ = 0.253 for cast iron Wm = ρV V = π Dbt + π D 2t = π (10 )(2.5)(0.5) + 4 Wm = (0.253)(78.54) = 19.87 lb π (10)2 (0.5) 4 = 78.54 in 3 U f = (300 )(19.87 )(101) = 602,061 ft − lb Time = Uf 33,000 fhp = 602,061 = 0.4844 min = 29 sec 33,000(37.66) (e) pvm : π Dn π (10)(870) vm = = = 2278 fpm 12 12 pvm = (50)(2278) = 113,900 (f) c = 5.5 in 901. A pivoted-shoe brake, rated at 900 ft-lb. torque, is shown. There are 180 sq. in. of braking surface; woven asbestos lining; 600 rpm of the wheel; 90o arc of brake contact on each shoe. The effect of spring A is negligible. (a) Is the pin for the Page 42 of 97 SECTION 16 – BRAKES AND CLUTCHES shoe located at the center of pressure? (b) How does the maximum pressure compare with that in Table AT 29? (c) What load W produces the rated torque? (d) At what rate is energy absorbed? Express in horsepower. Is it likely that this brake can operate continuously without overheating? (e) Does the direction of rotation affect the effectiveness of this brake? Problem 901. Solution: θ 90 2 = 2 = 9.9 in (a) c = π θ + sin θ + sin 90 2 13 19 and 16 ≈ 9.9 in , therefore the pin located at the center of pressure 2 2 D sin (b) 4 13 19 16 o α = 11.4 tan α = [∑ M Q =0 ] 4 FA cos α = 8.5W Page 43 of 97 2(18) sin SECTION 16 – BRAKES AND CLUTCHES 4 FA cos 11.4 = 8.5W FA = 2.168W ∑ FV = 0 and ∑ FH = 0 [ ] [ ] Qv = FA sin α + W = (2.168W )sin 11.4 + W = 1.429W Qh = FA cos α = (2.168W ) cos 11.4 = 2.125W [∑ M R1 =0 ] N1 (10.375) = 20.375Qh N1 (10.375) = 20.375(2.125W ) N1 = 4.173W F1 = f N1 For woven asbestos lining, f = 0.40 , p = 50 psi F1 = (0.40)(4.173W ) = 1.67W (either direction) Page 44 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M =0 R2 ] 10.375 N 2 = 20.375FA cos α 20.375 N2 = (2.168W ) cos11.4 = 4.174W 10.375 F2 = (0.40)(4.174W ) = 1.67W (either direction) T f = (F1 + F2 )c (900)(12) = (1.67 + 1.67 )(W )(9.9) W = 326.6 lb T f1 = T f 2 = Fc = 2 fPbr 2 sin but A = θ br Ar br 2 = θ 2 θ (1.67)(326.6)(9.9) = 2(0.4)(P )(180)(9)sin π 90 2 2 P = 9.26 psi < 50 psi (c) W = 326.6 lb (d) fhp = Tf n = (900)(12)(600) = 103 hp 63,000 63,000 π Dn π (18)(600) vm = = = 2827 fpm 12 12 pvm = (9.26)(2827 ) = 26,178 ft − lb sq.in. − F since pvm < 28,000 , it is likely to operate continuously. (e) Since the value of F is independent of rotation, the direction doesn’t affect the effectiveness of this brake. 902. Refer to the diagrammatic representation of the brake of Fig. 18.2, Text, and let 9 9 the dimensions be: a = b = m = t = 4 , c = 14 , D = 15 , h = 9 in ., and the 16 16 o contact width is 4 in.; arc of contact = 90 ; lining is asbestos in resin binder, wheel rotation of 100 rpm CC; applied load W = 2000 lb . (a) Locate the center of pressure for a symmetrical sinusoidal pressure distribution and compare with the actual pin centers. Assume that this relationship is close enough for approximate Page 45 of 97 SECTION 16 – BRAKES AND CLUTCHES results and compute (b) the dimensions k and e if the braking force on each shoe is to be the same, (c) the normal force and the maximum pressure, (d) the braking torque, (e) pvm . Would more-or-less continuous application be reasonable? Figure 18.2 Solution: θ 90 2 = 2 = 8.25 in (a) c = π θ + sin θ + sin 90 2 2 D sin 2(15) sin On Centers: 9 9 + 4 = 9.125 in > c 16 16 9 9 B : a + b = 4 + 4 = 9.125 in > c 16 16 K :t + m = 4 [∑ M RC =0 ] Page 46 of 97 SECTION 16 – BRAKES AND CLUTCHES eRF = (e + c )W e+c RF = W e RC = RF − W cW e+c RC = W − W = e e [∑ M RH =0 ] N1h − F1b = RF a N1h − fN1b = RF a R a N1 = F h − fb fRF a F1 = h − fb fa(e + c )W F1 = e(h − fb ) [∑ M RE =0 ] N 2 h + F2t = RC k Page 47 of 97 SECTION 16 – BRAKES AND CLUTCHES N 2 h + fN 2t = RC k R k N2 = C h + ft fR k F2 = C h − ft fkcW F2 = e(h + ft ) (b) T f 1 = T f 2 F1c = F2c F1 = F2 fa(e + c )W fkcW = e(h − fb ) e(h + ft ) a(e + c ) kc = h − fb h + ft For asbestos in resin binder, f = 0.35 , Table AT 29 9 a = 4 in = 4.5625 in 16 9 b = 4 in = 4.5625 in 16 9 m = 4 in = 4.5625 in 16 9 t = 4 in = 4.5625 in 16 c = 14 in 9 h = 9 in = 9.5625 in 16 4.5625(e + 14) k (14) = 9.5625 − 0.35(4.5625) 9.5625 + 0.35(4.5625) e + 14 = 2.1903k but k + m = e or e = k + 4.5625 then k + 4.5625 + 14 = 2.1903k k = 15.6 in e = 15.6 + 4.5625 = 20.1625 in (c) N = N1 = N 2 = Page 48 of 97 kcW (15.6)(14)(2000) = = 2720 lb e(h + ft ) (20.1625)[9.5625 − 0.35(4.5625)] SECTION 16 – BRAKES AND CLUTCHES (d) T f = T f 1 + T f 2 = f ( N1 + N 2 )c = 0.35(2)(2720)(8.25) = 15,708 in − lb (e) vm = π Dn = π (15)(100) = 393 fpm 12 12 pvm = (64.11)(393) = 25,195 ft − lb sq.in. − F since pvm < 28,000 , continuous application is reasonable. FIXED SHOES 903. A double-block brake has certain dimensions as shown. Shoes are lined with woven asbestos; cast-iron wheel turns 60 rpm; applied force W = 70 lb . For each direction of rotation, compute (a) the braking torque, (b) the rate of generating frictional energy (fhp). (c) If the maximum pressure is to be P = 50 psi (Table AT 29), what contact width should be used? (d) With this width, compute pvm and decide whether or not the applications must be intermittent. Problems 903, 904. Solution: [∑ M =0 4Q = 26W B ] Page 49 of 97 SECTION 16 – BRAKES AND CLUTCHES Q = 6.5W [∑ M R =0 ] 2.25S = 6Q = 6(6.5W ) S = 17.33W RH = S = 17.33W RV = Q = 6.5W e = 10 in R = 12.5 in a = 2.25 + 9 + 12.5 = 23.75 in ∑M ∑M M F1 M N1 H = Sa − M F1 H −M N1 H = 0 (CC) H = Sa + M F1 H −M N1 H = 0 (CL) R = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 2 brRP [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] H = 4 ( H T f1 = 2 fPbr 2 sin T f1 Pbr = 2 fr sin r = 10 in Page 50 of 97 θ 2 θ 2 ) SECTION 16 – BRAKES AND CLUTCHES 2r sin θ 2 2(10 ) sin = 11 in θ = 11 in 2 θ = 66.43o = 1.165 rad f = 0.4 for woven asbestos M F1 H M F1 H R fT f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2 = θ 2 fr sin 2 R T f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2 = 2r sin θ 2 66.73 φ1 = 90 − = 90 − = 56.64o = 0.9886 rad 2 2 2φ1 = 113.28o θ 66.73 φ2 = 90 + = 90 + = 123.36o = 2.1530 rad 2 2 o 2φ2 = 246.72 φ2 − φ1 = θ 12.5 T f1 10(cos 56.64 − cos 123.36) − sin 2 123.36 − sin 2 56.64 2 =T M F1 H = f1 66.73 2(10)sin 2 RT f1 [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] M N1 H = θ 4 2 fr sin 2 12.5T f1 [2(1.165) − (sin 246.72 − sin 113.28)] M N1 H = = 2.96T f1 66.73 8(0.4)(10) sin 2 θ ( CC: ∑ M H = Sa − M F1 H −M N1 (17.33)(70)(23.75) − T f 1 H =0 − 2.960T f1 = 0 T f1 = 7276 in − lb CL: ∑ M H = Sa + M F1 Page 51 of 97 H −M N1 H =0 ) SECTION 16 – BRAKES AND CLUTCHES (17.33)(70)(23.75) + T f 1 − 2.960T f1 = 0 T f1 = 14,700 in − lb e = 10 in d = 12.5 in CC: [∑ M H =0 ] RH a′ − RV d + M F2 CL: [∑ M H =0 ] RH a′ − RV d − M F2 H − M N2 H =0 H − M N2 H =0 T f2 = Tf = M H F1 H 2 Tf 1 Tf M N 2 H = M N1 H 2 = 2.960T f 2 Tf 1 CC: RH a′ − RV d + M F2 H − M N 2 H = 0 M F2 [(17.33)(21.5) − (6.5)(12.5)](70) + T f T f 2 = 10,405 in − lb CL: RH a′ − RV d − M F2 H − M N 2 H − 2.960T f 2 = 0 2 − 2.960T f 2 = 0 =0 [(17.33)(21.5) − (6.5)(12.5)](70) − T f T f 2 = 5150 in − lb (a) Braking Torque = T f1 + T f 2 Page 52 of 97 2 SECTION 16 – BRAKES AND CLUTCHES CC: T f = T f1 + T f 2 = 7276 + 10,405 = 17,681 in − lb CL: T f = T f1 + T f 2 = 14,700 + 5150 = 19,850 in − lb (b) Rate of generating frictional energy T n fhp = f 63,000 (17,681)(60) = 16.84 hp CC: fhp = 63,000 (19,850)(60) = 18.90 hp CL: fhp = 63,000 (c) p = 50 psi T f1 or T f 2 = 2 fPbr 2 sin CC: b = T f1 θ 2 10,405 = 4.73 in 66.73 2 fPr sin 2(0.4)(50)(10) sin 2 2 T f2 14,700 CL: b = = = 6.68 in θ 66.73 2 2 2 fPr sin 2(0.4)(50)(10) sin 2 2 2 θ = 2 (d) pvm πDn π (20)(60) vm = = = 314 fpm 12 12 pvm = (50)(314) = 15,700 < 55,000 pvm = (50)(314) = 15,700 < 28,000 application can be continuous or intermittent. 904. If the brake shown has a torque rating of 7000 lb-in. for counter-clockwise rotation, what braking torque would it exert for clockwise rotation, force W the same? Solution: CC: Sa − M F1 H − M N1 M F1 H = T f1 M N1 H = 2.960T f1 Page 53 of 97 H =0 SECTION 16 – BRAKES AND CLUTCHES S = 17.33W a = 23.75 in (17.33W )(23.75) − T f 1 − 2.96T f1 = 0 T f1 = 103.9W RH a′ − RV d + M F2 H − M N2 H =0 RH = 17.33W RV = 6.53W a′ = 21.5 in d = 12.5 in M F2 H = T f2 M N2 H = 2.960T f 2 (17.33W )(21.5) − (6.5W )(12.5) + T f 2 − 2.960T f 2 = 0 T f 2 = 148.65W T f = T f1 + T f 2 7000 = 103.9W + 148.65W W = 27.7 lb CL: Sa + M F1 H − M N1 H =0 (17.33)(27.7 )(23.75) − T f 1 − 2.96T f1 = 0 T f1 = 5817 in − lb RH a′ − RV d − M F2 H − M N2 H =0 [(17.33)(21.5) − (6.5)(12.5)](27.7) + T f 2 − 2.960T f 2 = 0 T f 2 = 2038 in − lb T f = T f1 + T f 2 = 5817 + 2038 = 7855 in − lb (CL) 905. A double-block brake is shown for which θ = 90o , b = 5 in ., n = 300 rpm , rim thickness = ¾ in., and W = 400 lb . The shoes are lined with asbestos in resin binder. Determine the frictional torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) How much energy is absorbed by the brake? Express in horsepower. (d) Will the brake operate continuously without danger of overheating? How long for a ∆t rim = 300 F ? How does pvm compare with Text values? Page 54 of 97 SECTION 16 – BRAKES AND CLUTCHES Problem 905 Solution: 4 4+4 α = 26.565o tan α = [∑ M R =0 ] (Q cos α )(4) = 16W (Q cos 26.565)(4) = 16(400) Q = 1789 lb RH = Q cos α = 1789 cos 26.565 = 1600 lb RV = Q sin α + W = 1789 sin 26.565 + 400 = 1200 lb Page 55 of 97 SECTION 16 – BRAKES AND CLUTCHES R M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 2 brRP MN H = [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 4 ( T f = 2 fPbr 2 sin MF H θ 2 R T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2 = 2r sin MN H ) = θ 2 RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 8 fr sin θ 2 20 = 10 in 2 4 tan β = 12 β = 18.435o r= θ = 90o = 1.571 rad θ 90 φ1 = 90 − − β = 90 − − 18.435 = 26.565o = 0.464 rad 2 2 2φ1 = 2(26.565) = 53.13o θ 90 φ2 = 90 + − β = 90 + − 18.435 = 116.565o = 2.034 rad 2 2 2φ2 = 2(116.565) = 233.13o R = 4 2 + 12 2 = 12.65 in Asbestos in resin binder f = 0.35 Page 56 of 97 SECTION 16 – BRAKES AND CLUTCHES MF H MN H 12.65 T f 10(cos 26.565 − cos 116.5656) − ( sin 2 116.565 − sin 2 26.565) 2 = 0.6803T = f 90 2(10) sin 2 (12.65)T f [2(2.034 − 0.464) − (sin 233.13 − sin 53.13)] = = 3.03T f 90 8(0.35)(10) sin 2 (a) Clockwise [∑ M H1 =0 ] (Q sin α )(2.5) + (Q cos α )(24) + M F H − M N H = 0 (1789 sin 26.565)(2.5) + (1789 cos 26.565)(24) + 0.6803T f 1 1 1 1 1 T f1 = 17,195 in − lb [∑ M H2 =0 ] 2.5 RV − 24 RH + M N 2 H2 + M F2 H2 =0 2.5(1200 ) − 24(1600 ) + 3.03T f 2 + 0.6803T f 2 = 0 T f 2 = 9541 in − lb Page 57 of 97 − 3.03T f1 = 0 SECTION 16 – BRAKES AND CLUTCHES T f = T f1 + T f 2 = 17,195 + 9541 = 26,736 in − lb (b) Counterclockwise [∑ M H1 =0 ] 24Q cos α + 2.5Q sin α − M F1 H1 − M N1 H1 =0 (24)(1789 cos 26.565) + (2.5)(1789 sin 26.565) − 0.6803T f 1 T f1 = 10,890 in − lb [∑ M H2 =0 ] 2.5 RV − 24 RH − M F2 H2 + M N2 H2 =0 2.5(1200 ) − 24(1600 ) − 0.6803T f 2 + 3.03T f 2 = 0 T f 2 = 15,066 in − lb T f = T f1 + T f 2 = 10,890 + 15,066 = 25,956 in − lb Page 58 of 97 − 3.03T f1 = 0 SECTION 16 – BRAKES AND CLUTCHES (c) CL: fhp = Tf n 63,000 = (26,736)(300) = 127.3 hp = (25,956)(300) = 123.6 hp CC: fhp = Tf n 63,000 (d) vm = πDn 12 = 63,000 63,000 π (20)(300) 12 = 1571 fpm For p : T f = 2 fPbr 2 sin θ 2 = T f1 (CL) 17,195 = 2(0.35)(P )(5)(10 ) sin 2 90 2 P = 69.48 psi pvm = (69.48)(1571) = 109,153 > 28,000 the brake operate continuously with danger of overheating. For time: U ft − lb ∆t o F = f Wm c c = 101 , ρ = 0.253 Wm = ρV V = πDbt + πD 2t 4 2 3 π (20) 3 3 V = π (20 )(5) + = 471.24 in 4 4 4 Wm = ρV = (0.253)(471.24) = 119.22 lb U f = Wm c∆t = (119.22 )(101)(300 ) = 3,612,366 ft − lb Time = Uf 33,000 fhp Uf 3,612,366 CL: Time = = = 0.886 min = 53 sec 33,000 fhp 33,000(123.6) Uf 3,612,366 CC: Time = = = 0.860 min = 52 sec 33,000 fhp 33,000(127.3) pvm > 28,000 , not good for continuous application. Page 59 of 97 SECTION 16 – BRAKES AND CLUTCHES 906. The double-block brake for a crane has the dimensions: a = 14.3 , b = 2.37 , D = 10 , e = 11.05 , g = 7.1 , h = 12 , j = 6.6 , k = 10.55 , m = 3.5 in ., the width of shoes is 4 in., and the subtended angle is θ = 90o ; wocen asbestos lining. Its rated braking torque is 200 ft-lb. The shoes contact the arms in such a manner that they are virtually fixed to the arms. What force W must be exerted by a hydraulic cylinder to develop the rated torque for (a) counterclockwise rotation, (b) clockwise rotation? Is the torque materially affected by the direction of rotation? (c) Compute the maximum pressure and compare with that in Table AT 29. (Data courtesy of Wagner Electric Corporation.) Problem 906. Solution: b 2.37 = e − c 11.05 − 0.83 α = 13.056o tan α = Page 60 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M ] =0 bQ cos α + cQ sin α = eW (2.37)(Q cos13.056) + (0.83)(Q sin 13.056) = 14.3W Q = 5.7286W R RH = Q cos α = 5.7286W cos 13.056 = 5.58W RV = Q sin α − W = 5.7286W sin 13.056 − W = 0.294W k 5.275 = 2j 6.6 o β = 38.63 tan β = θ 90 φ1 = 90 − − β = 90 − − 38.63 = 6.37o = 0.1112 rad 2 2φ1 = 12.74o φ2 = 90 + θ − β = 90 + 2 2φ2 = 192.74o 2 2 90 − 38.63 = 96.37 o = 1.6820 rad 2 2 k 10.55 2 R = + j2 = + (6.6 ) = 8.449 in 2 2 D 10 r= = = 5 in 2 2 Page 61 of 97 SECTION 16 – BRAKES AND CLUTCHES MF H R T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 ) 2 = 2r sin MN H = θ 2 RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] 8 fr sin θ 2 For woven asbestos lining, f = 0.40 8.449 T f 5(cos 6.37 − cos 96.37 ) − ( sin 2 96.37 − sin 2 6.37 ) 2 = 0.1985T MF H = f 90 2(5) sin 2 8.449T f [2(1.682 − 0.1112) − (sin 192.74 − sin 12.74 )] MN H = = 2.6755T f 90 8(0.4 )(5) sin 2 (a) CC: [∑ M H1 =0 ] RH (12 ) − RV (0.25) − M F1 H1 − M N1 H1 =0 (5.58W )(12) − (0.294W )(0.25) − 0.1985T f 1 T f1 = 23.3W Page 62 of 97 − 2.6755T f1 = 0 SECTION 16 – BRAKES AND CLUTCHES [∑ M H2 =0 ] 12Q cos α + 0.25Q sin α + M F2 H2 − M N2 H2 − 3.5W = 0 12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 − 0.1985T f 2 T f 2 = 66.4W T f = T f1 + T f 2 T f = 200 ft − lb = 2400 in − lb 2400 = 23.3W + 66.4W W = 26.8 lb (b) CL: Page 63 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M H1 =0 ] RH (12 ) − RV (0.25) + M F1 H1 − M N1 H1 =0 (5.58W )(12) − (0.294W )(0.25) + 0.1985T f 1 − 2.6755T f1 = 0 T f1 = 27.0W [∑ M H2 =0 ] 12Q cos α + 0.25Q sin α − M F2 H2 − M N2 H2 − 3.5W = 0 12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 + 0.1985T f 2 T f 2 = 57.2W T f = T f1 + T f 2 2400 = 27.0W + 57.2W W = 28.5 lb Since W has different values, torque is materially affected by the direction of rotation. (c) T f = 2 fPbr 2 sin θ 2 For woven asbestos lining, f = 0.40 Use T f = 66.4W = 66.4(26.8) = 1780 in − lb b = 4 in r = 5 in θ = 90o T f = 1780 = 2(0.4 )P(4 )(5) sin 2 P = 31.47 psi Page 64 of 97 90 2 SECTION 16 – BRAKES AND CLUTCHES From Table AT 29, pmax = 50 psi 31.47 psi < 50 psi INTERNAL-SHOE BRAKES 908. Assuming that the distribution of pressure on the internal shoe shown is given by p = P sin φ , show that the moments M N B , M F B , and TF O of N with respect to B and of F with respect to B and to O are ( b = face width) MN B = (Pbar 2 )[θ − (sin 2φ2 − sin 2φ1 ) 2] , [ ( ) ] M F B = fPbr r (cos φ1 − cos φ2 ) − a sin 2 φ2 − sin 2 φ1 2 , TF O = fPbr 2 (cos φ1 − cos φ2 ) . Problems 908 – 910. Solution: p = P sin φ d (M N B ) = kdN dN = P sin φ (brdφ ) = Pbr sin φdφ k = a cos(φ − 90) = a sin φ d (M N MN B ) = (a sin φ )(Pbr sin φ )dφ = Pabr sin φ2 B = Pabr ∫ sin 2 φdφ = φ1 B φdφ Pabr φ2 (1 − cos 2φ )dφ 2 ∫φ1 2 Pabr 1 Pabr (φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 ) φ φ = − sin 2 = 2 2 2 2 φ1 φ MN 2 but φ2 − φ1 = θ Page 65 of 97 SECTION 16 – BRAKES AND CLUTCHES MN B = (sin 2φ2 − sin 2φ1 ) Pabr θ− 2 2 d (M F B ) = edF dF = fdN = fPbr sin φdφ e = r + a sin (φ − 90) = r − a cos φ d (M F B ) = (r − a cos φ )( fPbr sin φdφ ) = fPbr (r sin φ − a sin φ cos φ )dφ [ ] φ2 M F B = fPbr − r cos φ − a sin 2 φ φ1 ( ) a sin 2 φ2 − sin 2 φ1 M F B = fPbr r (cos φ1 − cos φ2 ) − 2 d (TF O ) = rdF = fPbr 2 sin φdφ TF O = fPbr 2 [− cos φ ]φ12 φ TF O = fPbr 2 (cos φ1 − cos φ2 ) The same as 908, except that a pressure distribution of p = P cos α is assumed. 909. [ ( ) ] M N B = Pbr h(2θ + sin 2α 2 + sin 2α1 ) 4 + c sin 2 α 2 − sin 2 α1 2 , [ ( ) M F B = fPbr r (sin α 2 + sin α1 ) + h sin 2 α 2 − sin 2 α1 2 − c(2θ + sin 2α 2 + sin 2α1 ) 4 M F O = fPbr 2 (sin α 2 + sin α1 ) . Solution: k = h cos α + c sin α e = r + h sin α − c cos α dN = pbrdα = Pbr cos αdα dF = fdN = fPbr cos αdα dM N B = kdN = (h cos α + c sin α )(Pbr cos αdα ) ( ) dM N B = Pbr h cos 2 α + c sin α cos α dα MN α2 B = Pbr ∫ −α1 (h cos 2 α + c sin α cos α )dα α2 MN B h(2α + sin 2α ) c sin 2 α = Pbr + 4 2 −α 1 but θ = α1 + α 2 [ ] h[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )] c sin 2 α 2 − sin 2 (− α1 ) M N B = Pbr + 4 2 Page 66 of 97 ] SECTION 16 – BRAKES AND CLUTCHES ( ) h(2θ + sin 2α 2 + sin 2α1 ) c sin 2 α 2 − sin 2 α1 M N B = Pbr + 4 2 dM F B = edF = (r + h sin α − c cos α )( fPbr cos αdα ) ( ) dM F B = fPbr r cos α + h sin α cos α − c cos 2 α dα α2 MF B h sin 2 α c(2α + sin 2α ) = fPbr r sin α + − 2 4 −α1 [ ] h sin 2 α 2 − sin 2 (− α1 ) c[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )] M F B = fPbr r [sin α 2 − sin (− α1 )] + − 2 4 ( ) h sin 2 α 2 − sin 2 α1 c(2θ + sin 2α 2 + sin 2α1 ) M F B = fPbr r (sin α 2 + sin α1 ) + − 2 4 dM F O = rdF = r ( fPbr cos αdα ) = fPbr 2 cos αdα M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )] α M F O = fPbr 2 (sin α 2 + sin α1 ) The same as 909, except that the α is to be measured from OG , a perpendicular to OB ; limits from − α1 to + α 2 . 910. Solution: k = a cos α e = r + a sin α dM N B = kdN = a cos α (Pbr cos αdα ) = Pbar cos 2 αdα = Pbar (1 + cos 2α )dα 2 α MN B MN B Pbar 2α + sin 2α 2 Pbar [2(α 2 + α1 ) + sin 2α 2 − sin 2(− α1 )] = = 2 2 4 −α1 = Pbar (2θ + sin 2α 2 − sin 2α1 ) 4 dM F B = edF = (r + a sin α )( fPbr cos αdα ) = fPbr (r cos α + a sin α cos α )dα [ α2 MF B ] a sin 2 α a sin 2 α 2 − sin 2 (− α1 ) = fPbr r sin α + = fPbr r [ sin − sin ( − ) ] + α α 2 1 2 −α 2 1 ( ) a sin 2 α 2 − sin 2 α1 M F B = fPbr r (sin α 2 + sin α1 ) + 2 2 dM F O = rdF = r ( fPbr cos αdα ) = fPbr cos αdα Page 67 of 97 SECTION 16 – BRAKES AND CLUTCHES M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )] α M F O = fPbr 2 (sin α 2 + sin α1 ) 911. The following dimensions apply to a two-shoe truck brake somewhat as shown: face b = 5 , r = 8 , h = 5.1 , c = 2.6 , w = u = 6.4 in ., θ = 110 o , φ1 = 15o . Lining is asbestos in rubber compound. For a maximum pressure on each shoe of 100 psi, determine the force Q , and the braking torque for (a) clockwise rotation, (b) counterclockwise rotation. See 908. (Data courtesy of Wagner Electric Corporation.) Problems 911, 912. Solution: See 908. (sin 2φ2 − sin 2φ1 ) Pbar θ− MN B = 2 2 ( ) a sin 2 φ2 − sin 2 φ1 M F B = fPbr r (cos φ1 − cos φ2 ) − , 2 2 TF O = fPbr (cos φ1 − cos φ2 ) φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o p = 100 psi b = 5 in r = 8 in a = h2 + c 2 = (5.1)2 + (2.6)2 = 5.7245 in θ = 110o = 1.92 rad For asbestos in rubber compound f = 0.35 Page 68 of 97 SECTION 16 – BRAKES AND CLUTCHES (a) Both sides (clockwise rotation) Q(h + w) + M F B − M N MN = B =0 (100)(5)(5.7245)(8) 1.92 − (sin 250 − sin 30) = 30,224 in − lb 5.7245 sin 2 125 − sin 2 30 M F B = (0.35)(100)(5)(8)8(cos 30 − cos 125) − 2 h = 5.1 in , w = 6.4 in B 2 2 ( Q(5.1 + 6.4) + 14,436 − 30,224 = 0 Q = 1373 lb TF O = (0.35)(100)(5)(8) (cos 15 − cos 125) = 17,242 in − lb 2 T f = 2TF O = 2(17,242 ) = 34,484 in − lb (b) Counterclockwise rotation Q(h + w) − M N B − M F B = 0 Page 69 of 97 ) = 14,436 in − lb SECTION 16 – BRAKES AND CLUTCHES Q(5.1 + 6.4) − 30,224 − 14,436 = 0 Q = 3883 lb T f = 2TF O = 2(17,242 ) = 34,484 in − lb 913. The data are the same as 911, but the shoe arrangement is as shown for this problem. For a maximum pressure on the shoes of 100 psim determine the force Q and TF O for (a) Cl rotation, (b) CC rotation, See 908. Problem 913. Solution: TF O = T f = fPbr 2 (cos φ1 − cos φ2 ) Pbr = Tf fr (cos φ1 − cos φ2 ) MN B = aT f (sin 2φ2 − sin 2φ1 ) = (sin 2φ2 − sin 2φ1 ) Pbar θ − − θ 2 fr (cos φ − cos φ ) 2 2 2 1 2 ( ) a sin 2 φ2 − sin 2 φ1 M F B = fPbr r (cos φ1 − cos φ2 ) − 2 Tf a sin 2 φ2 − sin 2 φ1 ( ) φ φ MF B = r cos − cos − 1 2 2r (cos φ1 − cos φ2 ) 2 From 911: φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o θ = 110o = 1.92 rad ( a = h2 + c 2 = f = 0.35 Page 70 of 97 (5.1)2 + (2.6)2 = 5.7245 in ) SECTION 16 – BRAKES AND CLUTCHES (sin 250 − sin 30) = 1.753T 1.92 − f 2(0.35)(8)(cos 15 − cos 125) 2 Tf 5.7245 sin 2 125 − sin 2 15 ( ) = 8 cos 15 − cos 125 − 28(cos 15 − cos 125) 2 5.7245T f MN B = MF B ( (a) CL rotation: Left Side [∑ M B =0 ] Q(h + w) − M N1 B − M F1 B =0 Q(5.1 + 6.4 ) − 1.753T f1 − 0.43T f1 = 0 T f1 = 5.268Q Right Side: [∑ M B =0 ] Page 71 of 97 ) = 0.43T f SECTION 16 – BRAKES AND CLUTCHES Q(h + w) + M F2 B − M N 2 B =0 Q(5.1 + 6.4 ) + 0.43T f 2 − 1.753T f 2 = 0 T f 2 = 8.6924Q T f max = T f 2 = 8.6924Q T f max = fPbr 2 (cos φ1 − cos φ2 ) 8.6924Q = (0.35)(100 )(5)(8) (cos 15 − cos 125) Q = 1984 lb 2 T f1 = 5.268Q = 5.268(1984 ) = 10,452 in − lb T f 2 = 8.6924Q = 8.6924(1984 ) = 17,246 in − lb Total TF O = T f1 + T f 2 = 10,452 + 17,246 = 27,698 in − lb (b) CC rotation Left Side [∑ M B =0 ] Q(h + w) + M F1 B − M N1 B =0 Q(5.1 + 6.4 ) + 0.43T f1 − 1.753T f1 = 0 T f1 = 8.6924Q Right Side: Page 72 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M B =0 ] Q(h + w) − M F2 B − M N2 B =0 Q(5.1 + 6.4 ) − 0.43T f 2 − 1.753T f 2 = 0 T f 2 = 5.268Q Since values are just interchanged Q = 1984 lb Total TF O = 27,698 in − lb as in (a) 914. A double-shoe internal brake is actuated by an involute cam as shown, where QR is the force on the right shoe at a radius wR and QL is the force on the left shoe at a radius wL . The pressure of each shoe is proportional to the rotation of the shoe about B which is inversely proportional to w ; therefore, the ratio of the maximum pressures is PL PR = wR wL . The dimensions are: face width b = 4 , 9 1 5 5 r = 6 , h = 4 , c = 1 , wL = 9 , wR = 8 in .: for each shoe, θ = 120 o , 16 8 16 16 o φ1 = 30 . The lining is asbestos in rubber compound, Determine the braking torque and forces QR and QL for the maximum permissible pressure for (a) clockwise rotation, (b) counterclockwise rotation. Page 73 of 97 SECTION 16 – BRAKES AND CLUTCHES Problem 914. Solution: (sin 2φ2 − sin 2φ1 ) Pbar θ− MN B = 2 2 ( ) a sin 2 φ2 − sin 2 φ1 M F B = fPbr r (cos φ1 − cos φ2 ) − , 2 2 TF O = fPbr (cos φ1 − cos φ2 ) 2 2 9 1 a = h + c = 4 + 1 = 4.70 in 16 8 5 8 pL wR = = 16 = 0.8926 pR wL 9 5 16 For asbestos in rubber compound, f = 0.35 , p = 75 psi pR = 75 psi pL = 0.8926(75) = 67 psi 2 2 (a) Clockwise rotation Left Side: Page 74 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M BL =0 QL wL − M FL M FL BL ] BL − M NL BL =0 ( ) a sin 2 φ2 − sin 2 φ1 = fPLbr r (cos φ1 − cos φ2 ) − 2 φ1 = 30o 2φ1 = 60o φ2 = θ + φ1 = 120 + 30 = 150o 2φ2 = 300o θ = 120o = 2.094 rad ( 4.70 sin 2 150 − sin 2 30 ( )( )( )( ) ( ) = 0 . 35 67 4 6 6 cos 30 − cos 150 − BL 2 (sin 2φ2 − sin 2φ1 ) P bar θ− M NL BL = L 2 2 (67 )(4)(4.7 )(6) 2.094 − (sin 300 − sin 60) = 11,185 in − lb M NL BL = 2 2 M FL 5 QL 9 − 5849 − 11,185 = 0 16 QL = 1829 lb T( F O )L = fPL br 2 (cos φ1 − cos φ2 ) T( F O )L = (0.35)(67 )(4 )(6 ) (cos 30 − cos 150 ) = 5849 in − lb 2 Right side: Page 75 of 97 ) = 5849 in − lb SECTION 16 – BRAKES AND CLUTCHES [∑ M BR =0 ] QR wR + M FR M FR BR M NR BR = = − M NR BR M FL BL PR PL M NL BL PL PR = BR =0 (5849)(75) = 6547 in − lb = 67 (11,185)(75) = 12,520 in − lb 67 5 QR 8 + 6547 − 12,520 = 0 16 QR = 719 lb T( F O )L PR (5849)(75) T( F O )R = = = 6547 in − lb PL 67 T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb (b) Counterclockwise rotation Left side: Page 76 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M BL =0 ] QL wL + M FL BL − M NL BL =0 5 QL 9 + 5849 − 11,185 = 0 16 QL = 573 lb T( F O )L = 5849 in − lb Right Side: [∑ M BR =0 ] QR wR − M FR BR − M NR BR =0 5 QR 8 − 6547 − 12,520 = 0 16 QR = 2294 lb T( F O )R = 6547 in − lb Page 77 of 97 SECTION 16 – BRAKES AND CLUTCHES T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb BAND BRAKES 915. The steel band for the brake shown is lined with flexible asbestos and it is expected tha the permissible pressure of Table AT 29 is satisfactory; θ = 245o , 1 a = 20 in ., m = 3 in ., D = 18 in ., and face width b = 4 in .; rotation CL. The 2 cast-iron wheel turns 200 rpm. Set up suitable equations, use the average f given and compute (a) the force in each end of the band, (b) the brake torque and fhp. (c) Determine the mechanical advantage for the limit values of f in Table AT 29 and its percentage variation fron that for the average f . (d) Investigate the overheating problem using relevant information given in the Text. Problem 915. Solution: (1) F1 = e fθ F2 [∑ M Fixed point Wa = F2 m Wa (2) F2 = m Page 78 of 97 =0 ] SECTION 16 – BRAKES AND CLUTCHES (3) F = F1 − F2 = fpA FD (4) T f = 2 From Table AT 29, flexible asbestos Ave. f = 0.40 , p = 50 psi (a) For F1 and F2 : θDb A= 2 θ = 245o = 4.276 rad (4.276)(18)(4) = 154 in 2 A= 2 F = F1 − F2 = fpA = (0.40)(50 )(154) = 3080 lb F1 = e fθ = e (0.40 )(4.276 ) = 5.5312 F2 F1 = 5.5312 F2 F = 5.5312 F2 − F2 = 3080 F2 = 680 lb F1 = 5.5312(680) = 3760 lb (b) T f and fhp FD (3080 )(18) = = 27,720 in − lb 2 2 Tf n (27,720)(200) = 88 hp fhp = = 63,000 63,000 (c) For MA T T MA = f = f Wa F2 m FD Tf = 2 F F2 = fθ e −1 FD D e fθ − 1 2 MA = = 2m Fm fθ e −1 D = 18 in m = 3.5 in θ = 4.276 rad Tf = ( Page 79 of 97 ) SECTION 16 – BRAKES AND CLUTCHES Limit values (Table AT 29) f = 0.35 to 0.45 . f = 0.35 [ ] [ ] 18 e (0.35 )(4.276 ) − 1 = 8.914 2(3.5) f = 0.45 MA = 18 e (0.45 )(4.276 ) − 1 MA = = 15.042 2(3.5) with f = 0.40 (average) [ ] 18 e (0.40 )(4.276 ) − 1 = 11.652 2(3.5) Percentage variation from f = 0.40 . f = 0.35 11.652 − 8.914 (100% ) = 23.5% % var = 11.652 f = 0.45 15.042 − 11.652 % var = (100% ) = 29.1% 11.652 MA = (d) Overheating problem fhp 88 = = 0.57 fhp in 2 A 154 Therefore, a problem of overheating is expected as Rasmussen recommends 0.2 to 0.3 fhp per square inch of brake contact area. 916. (a) For the band brake shown, derive the expressions for the braking torque in terms of W , etc., for CL rotation and for CC rotation, and specify the ratio c b for equal effectiveness in both directions of rotation. Are there any proportions of b and c as shown that would result in the brae being self locking? (b) When θ = 270o , a = 16 in ., b = c = 3 in ., and D = 12 in ., it was found that a force W = 50 lb . Produced a frictional torque of 1000 in-lb. Compute the coefficient of friction. Page 80 of 97 SECTION 16 – BRAKES AND CLUTCHES Problem 916. Solution: (a) CL: [∑ M O =0 ] aW = F1b + F2c F1 = F2 e fθ aW = F2e fθ b + F2 c aW F2 = fθ be + c aWe fθ F1 = F2 e fθ = fθ be + c aWe fθ − aW aW (e fθ − 1) F = F1 − F2 = = be fθ + c be fθ + c Page 81 of 97 SECTION 16 – BRAKES AND CLUTCHES Tf = FD WaD e fθ − 1 = 2 2 be fθ + c CC: [∑ M O =0 ] aW = F2b + F1c WaD e fθ − 1 Tf = 2 ce fθ + b No proportions of b and c as shown that would result in the brake being self-locking. (b) W = 50 lb T f = 1000 in − lb D = 12 in a = 16 in b = c = 3 in θ = 270o = 4.7124 rad T f = 1000 = (50)(16)(12) 2 e fθ − 1 3e fθ + 3 e fθ − 1 = 0.625 e fθ + 1 e fθ = e 4.7124 f = 4.333 f = 0.311 917. (a) For the brake shown, assume the proper direction of rotation of the cast-iron wheel for differential acion and derive expressions for the braking torque. (b) Let 3 D = 14 in ., n = 1 in ., m = 4 in ., θ = 235o , and assume the band to be lined with 4 woven asbestos. Is there a chance that this brake will be self-acting? If true, will Page 82 of 97 SECTION 16 – BRAKES AND CLUTCHES it always be for the range of values of f given in Table AT 29? (c) The ratio n m should exceed what value in order for the brake to be self-locking? (d) If the direction of rotation of the wheel is opposite to that taken in (a), what is the braking torque with a force W = 10 lb . at a = 8 in .? (e) Suppose the brake is used as a stop to prevent reverse motion on a hoist. What is the frictional horsepower for the forward motion if the wheel turns 63 rpm? Problems 917, 918. Solution: (a) Assume CL [∑ M O =0 ] Wa + F1n = F2 m F1 = F2 e fθ Wa = F2 m − F2 ne fθ = F2 (m − ne fθ ) Wa F2 = (m − ne fθ ) Page 83 of 97 SECTION 16 – BRAKES AND CLUTCHES F1 = Wae fθ m − ne fθ ( ) Wa (e fθ − 1) F = F1 − F2 = , Braking force. m − ne fθ FD WaD e fθ − 1 , Braking torque. Tf = = 2 2 m − ne fθ (b) D = 14 in 3 n = 1 in 4 m = 4 in θ = 235o = 4.10 rad Table AT 29, woven asbestos f = 0.35 to 0.45 There is a chance of self-acting if ne fθ > m m = 4 in use f = 0.40 ne fθ = 1.75e(0.40 )(4.10 ) = 9.0 > m use f = 0.35 ne fθ = 1.75e(0.35 )(4.10 ) = 7.35 > m use f = 0.45 ne fθ = 1.75e (0.45 )(4.10 ) = 11.07 > m Therefore true for the range of values of f . (c) ne fθ > m , f = 0.40 (average) n 1 > fθ m e n 1 > (0.4 )(4.10 ) m e n > 0 .2 m (d) For CC: Page 84 of 97 SECTION 16 – BRAKES AND CLUTCHES [∑ M O =0 ] Wa + F2 n = F1m Wa = F1m − F2 n F1 = F2 e fθ Wa = F2 e fθ m − F2 n = F2 (me fθ − n ) Wa F2 = me fθ − n Wae fθ F1 = me fθ − n Wa (e fθ − 1) F = F1 − F2 = me fθ − n FD WaD e fθ − 1 Tf = = 2 2 me fθ − n Tf = (10)(8)(14) 2 (e) fhp = 918. e (0.40 )(4.10 ) − 1 4e (0.40 )(4.10 ) − 1.75 = 123.3 in − lb Tf n 63,000 = (123.3)(63) = 0.1233 hp 63,000 A differential band brake similar to that shown and lined with woven asbestos, has the dimensions: D = 18 in ., n = 2 in ., m = 12 in ., θ = 195o . (a) Is there a chance that this brake will be self-acting? (b) If W = 30 lb . and a = 26 in . , compute the maximum braking torque and the corresponding mechanical advantage. (c) What is the ratio of the braking torque for CL rotation to the braking torque for CC rotation? (d) A 1/16-in.-thick steel band, SAE 1020 as rolled, carries the asbestos lining. What should be its width for a factor of safety of 8, based on the ultimate stress? What should be the face width if the average pressure is 50 psi? Solution: Page 85 of 97 SECTION 16 – BRAKES AND CLUTCHES (a) For CL: ne fθ > m θ = 195o = 3.4 rad m = 12 in n = 2 in f = 0 .4 2e 0.4(3.4 ) = 7.8 < m , not self-acting For CC: n > me fθ me fθ < n 12e 0.4(3.4 ) = 46.8 > n , not self-acting Therefore, there is no change that this brake will be self-acting. (b) T f max = T f (CL) fθ 0.4 (3.4 ) −1 Wad e − 1 (30)(26)(18) e Tf = = fθ 0 . 12 − 2e 4(3.4 ) = 4832 in − lb 2 2 m − ne Tf 4832 MA = = = 6.2 Wa (30)(26) (c) T f (CL ) = 4832 in − lb fθ 0.4 (3.4 ) −1 Wad e − 1 (30)(26)(18) e T f (CC ) = = 0.4(3.4 ) = 454 in − lb fθ − 2 2 2 me − n 12e T f (CL ) 4832 Ratio = = = 10.64 T f (CC ) 454 (d) For SAE 1020, as rolled. su = 65 ksi s 65 s= u = = 8.125 ksi = 8125 psi N 8 F s= 1 bt 1 t = in = 0.0625 in 16 Wae fθ max. F1 = (CL) m − ne fθ (30)(26)e 0.4(3.4 ) = 722.3 lb F1 = 12 − 2e 0.4(3.4 ) Page 86 of 97 SECTION 16 – BRAKES AND CLUTCHES 722.3 b(0.0625) b = 1.422 in With p = 50 psi F = fpA θbD A= 2 Wa (e fθ − 1) (30)(26)(e 0.4(3.4 ) − 1) max. F = = = 536.9 lb m − ne fθ 12 − 2e 0.4(3.4 ) fpθbD F= 2 (0.4)(50)(3.4)(b )(18) 536.9 = 2 b = 0.88 in s = 8125 = 919. A differential band brake is to be design to absorb 10 fhp at 250 rpm. (a) Compute the maximum and minimum diameters from both equations (z) and (a), p. 495, Text. Decide on a size. (b) The band is to be lined with woven asbestos. The Rasmussen recommendation (§18.4) will help in deciding on the face width. Also check the permissible pressure in Table AT 29. Choose dimensions of the lever, its location and shape and the corresponding θ . Be sure the brake is not self locking. What is the percentage variation of the mechanical advantage from the minimum value ( f min ) for the f limits in Table AT 29? Solution: 63,000hp 63,000(10 ) Tf = = = 2520 in − lb n 250 (a) Eq. (z) 1 1 1 1 Tf = 5 3 2520 3 = = 7.96 in 5 Tf Dmax = 4 Eq. (a) 3 2520 3 = = 8.57 in 4 Dmin 1 1 1 1 Dmin = (60 fhp )3 = [60(10)]3 = 8.44 in Dmin = (80 fhp )3 = [80(10)]3 = 9.28 in use D = 8.5 in (b) By Rasmussen Energy absorption capacity = 0.2 to 0.3 fhp per sq. in. of brake contact area. Page 87 of 97 SECTION 16 – BRAKES AND CLUTCHES Say 0.25 fhp per sq. in. θ bD A= 2 fhp = 10 hp D = 8.5 in assume θ = 180o = π rad fhp fhp in 2 = A 10 0.25 = π b(8.5) 2 b = 3 in From Table AT 29, f = 0.40 , p per . = 50 psi F fA π (3)(8.5) A= = 40 in 2 2 2T 2(2520) F= f = = 593 lb D 8.5 593 p= = 37.1 psi < 50 psi (OK) 0.4(40) p= For MA : Tf D e fθ − 1 MA = = WA 2 c − be fθ Not self-locking c > be fθ c > e fθ b c > e 0.4π b c > 3 .5 b c say = 4 or c = 4b b ( ( ) ) For f = 0.40 T D e fθ − 1 8.5 e 0.4π − 1 21.96 MA = f = = = fθ 0.4π WA 2 c − be 2 4b − be b For f = 0.35 = f min ( Page 88 of 97 ( ) ( ) ( ) ) SECTION 16 – BRAKES AND CLUTCHES ( ) ( ) ( ) ) D e fθ − 1 8.5 e 0.35π − 1 8.54 = = fθ 0.35π WA 2 c − be 2 4b − be b 21.96 − 8.54 % variation = (100%) = 157% 8.54 MA = Tf = ( DISK CLUTCHES 920. An automobile engine develops its maximum brake torque at 2800 rpm when the bhp = 200. A design value of f = 0.25 is expected to be reasonable for the asbestos facing and it is desired that the mean diameter not exceed 8.5 in.; permissible pressure is 35 psi. Designing for a single plate clutch, Fig. 18.10, Text, determine the outer and inner diameters of the disk. Solution: 1 Dm = (Do + Di ) = 8.5 in 2 ro + ri = 8.5 in 63,000hp 63,000(200 ) Tf = = = 4500 in − lb n 2800 p = 35 psi f N (ro + ri ) Tf = 2 (0.25)(N )(8.5) 4500 = 2 N = 4235 lb N ave. p = 2 π (ro − ri 2 ) 4235 35 = π (ro2 − ri 2 ) ro2 − ri 2 = 38.5 ro = 8.5 − ri (8.5 − ri )2 − ri2 = 38.5 72.25 − 17 ri + ri 2 − ri 2 = 38.5 ri = 1.985 in say ri = 2.0 in ro = 8.5 − 2.0 = 6.5 in Do = 2ro = 2(6.5) = 13 in Di = 2ri = 2(2.0) = 4 in Page 89 of 97 SECTION 16 – BRAKES AND CLUTCHES 921. An automobile engine can develop a maximum brake torque of 2448 in-lb. Which of the following plate clutches, which make up a manufacturer’s standard 7 1 “line,” should be chosen for this car? Facing sizes: (a) Do = 8 , Di = 6 in ., (b) 8 8 1 1 1 Do = 10 , Di = 6 in ., (c) Do = 11 , Di = 6 in . In each case, assume f = 0.3 . 8 16 8 The unit pressures are (a) 34 psi, (b) 30 psi, and (c) 26.2 psi. Solution: f N (ro + ri ) 2 f N (Do + Di ) Tf = 4 π N = pave Do2 − Di2 4 π pave Do2 − Di2 (Do + Di ) Tf = 16 Tf = ( ) ( ) (a) Do = 8.875 in ; Di = 6.125 in , p = 34 psi , f = 0.3 Tf = π pave (Do2 − Di2 )(Do + Di ) 16 π (0.3)(34)(8.875)2 − (6.125)2 (8.875 + 6.125) Tf = = 1239 in − lb 16 [ ] (b) Do = 10 in ; Di = 6.125 in , p = 30 psi , f = 0.3 Tf = π pave (Do2 − Di2 )(Do + Di ) 16 π (0.3)(30)(10)2 − (6.125)2 (10 + 6.125) Tf = = 1780 in − lb 16 [ ] (c) Do = 11.0625 in ; Di = 6.125 in , p = 26.2 psi , f = 0.3 Tf = π pave (Do2 − Di2 )(Do + Di ) 16 π (0.3)(26.2)(11.0625)2 − (6.125)2 (11.0625 + 6.125) Tf = = 2251 in − lb 16 [ use (c) Page 90 of 97 ] SECTION 16 – BRAKES AND CLUTCHES 922. A single-disk clutch for an industrial application, similar to that in Fig. 18.11, Text, except that there are two disks attached to one shaft and one attached to the other. The clutch is rated at 50 hp at 500 rpm. The asbestos-in-resin-binder facing 1 3 has a Do = 8 in . and Di = 4 in . What must be the axial force and average 2 4 pressure? How does this pressure compare with that recommended by Table AT 29? Solution: n = 2 pairs in contact f = 0.35 (Table AT 29) p = 75 psi 63,000hp 63,000(50 ) Tf = = = 6300 in − lb nm 500 Do = 8.5 in Di = 4.75 in nf N (ro + ri ) Tf = 2 nf N (Do + Di ) Tf = 4 (2)(0.35)(N )(8.5 + 4.75) 6300 = 4 N = 2717 lb 4N 4(2717 ) ave. p = = = 69.6 psi < 75 psi 2 2 π (Do − Di ) π (8.5)2 − (4.75)2 [ 923. ] A multiple-disk clutch similar to Fig. 18.11, Text, is rated at 22 hp at 100 rpm. The outside and inside diameters of the disks are 14 and 7 ½ in., respectively. If f = 0.25 , find (a) the axial force required to transmit the rated load, and (b) the unit pressure between the disks. Solution: (a) Fig. 18-11, n = 4 pairs in contact 63,000hp 63,000(22 ) Tf = = = 13,860 in − lb nm 100 nf N (Do + Di ) Tf = 4 Do = 14 in Di = 7.5 in 4(0.25)( N )(14 + 7.5) 13,860 = 4 Page 91 of 97 SECTION 16 – BRAKES AND CLUTCHES N = 2579 lb (b) p = 924. 4N 4(2579 ) = = 23.5 psi 2 2 π Do − Di π (14)2 − (7.5)2 ( ) [ ] A multiple-disk clutch for a machine tool operation has 4 phosphor-bronze driving disks and 5 hardened-steel driven disks. This clutch is rated at 5.8 hp at 100 rpm when operated dry. The outside and inside diameters of the disks are 5 ½ and 4 3/16 in., respectively. (a) If the pressure between the disks is that recommended for metal on metal in Table AT 29, what coefficient of friction is required to transmit the rated power? (b) What power may be transmitted for f and p as recommended in Table AT 29? Solution: Do = 5.5 in Di = 4.1875 in 63,000hp 63,000(5.8) Tf = = = 3654 in − lb nm 100 n = 8 pairs in contact (a) Table AT 29, p = 150 psi , metal to metal nf N (Do + Di ) Tf = 4 π π 2 2 N = p (Do2 − Di2 ) = 150 (5.5) − (4.1875) = 1498 lb 4 4 (8)( f )(1498)(5.5 + 4.1875) T f = 3654 = 4 f = 0.126 [ ] (b) from Table AT 29, p = 150 psi , f = 0.2 (8)(0.2)(1498)(5.5 + 4.1875) = 5805 in − lb Tf = 4 T n (5805)(100) = 9.2 hp hp = f m = 63,000 63,000 925. A multiple-disk clutch with three disks on one shaft and two on the other, similar to that in Fig. 18.11, Text, is rated at 53 hp at 500 rpm. (a) What is the largest value of Di if f and p are given by Table AT 29 for asbestos in resin binder and Do = 10.5 in . (b) For the diameter used of Di = 7 in .,what is the required axial force and the average pressure? Solution: Page 92 of 97 SECTION 16 – BRAKES AND CLUTCHES Table AT 29, asbestos in resin binder, f = 0.3 , p = 75 psi nf N (ro + ri ) Tf = 2 nf N (Do + Di ) Tf = 4 4T f N= nf (Do + Di ) but π N = p (Do2 − Di2 ) 4 π 4T f = nfp Do2 − Di2 (Do + Di ) 4 63,000hp 63,000(53) Tf = = = 6678 in − lb nm 500 n = 4 pairs in contact π 4T f = nfp Do2 − Di2 (Do + Di ) 4 π 2 4(6678) = (4 )(0.3)(75) (10.5) − Di2 (10.5 + Di ) 4 Di = 9.5607 in ( ) ( ) [ ] (b) Di = 7 in nf N (Do + Di ) Tf = 4 ( 4 )(0.3)( N )(10.5 + 7 ) 6678 = 4 N = 1272 lb 4N 4(1272 ) ave. p = = = 26.44 psi 2 2 π (Do − Di ) π (10.5)2 − (7 )2 [ ] MISCELLANEOUS CLUTCHES AND BRAKES 926. For the cone brake shown, find an expression for the braking torque for a given applied force W on the bell crank. Consider the force F ′ , Fig. 18.12, Text, in obtaining the expression. Page 93 of 97 SECTION 16 – BRAKES AND CLUTCHES Problems 926-928. Solution: 927. For the cone brake similar to that shown, certain dimensions are: Dm = 15 in ., 1 c = 2 in ., α = 12o , b = 9 in ., and a = 20 in . The contact surfaces are metal and 2 asbestos. (a) For an applied force W = 80 lb ., what braking torque may be expected of this brake? Consider the resistance F ′ , Fig. 18.12, Text. (b) If the rotating shaft comes to rest from 300 rpm during 100 revolutions, what frictional work has been done? (c) What must be the diameter of the steel pin P , SAE 1020 as rolled, for a factor fo safety of 6 against being sheard off? The diameter 1 of the hub d = 4 in . (d) What is the unit pressure on the face of the brake? 2 Solution: f Dm R f Dm aW = 2(sin α + f cos α ) 2b(sin α + f cos α ) Table AT 29, asbestos on metal, f = 0.40 (0.40)(15)(20)(80) = 890 in − lb Tf = 2(9)(sin 12 + 0.4 cos 12) (a) Tf = 2π (300 ) = 31.42 rad sec 60 ω2 = 0 rad sec θ = 100(2π ) = 628.3 rad 1 θ = (ω1 + ω2 )t 2 1 628.3 = (31.42 + 0 )t 2 (b) ω1 = Page 94 of 97 SECTION 16 – BRAKES AND CLUTCHES t = 40 sec 1 nm = (300 + 0 ) = 150 rpm 2 T n (890)(150) = 2.119 hp fhp = f m = 63,000 63,000 U f = 550( fhp )(t ) = 550(2.119 )(40 ) = 46,618 ft − lb (c) For SAE 1020, as rolled, ssu = 49 ksi s 49 ss = su = = 8.17 ksi N 6 4R ss = πd2 aW (20 )(80 ) R= = = 177.8 lb b 9 4(177.8) ss = 8170 = πd2 d = 0.1665 in 3 say d = in 16 N π Dm c R 177.8 N= = = 297 lb sin α + f cos α sin 12 + 0.4 cos 12 297 p= = 2.52 psi π (15)(2.5) (d) p = 928. A cone clutch for industrial use is to transmit 15 hp at 400 rpm. The mean diameter of the clutch is 10 in. and the face angle α = 10o ; let f = 0.3 for the cast-iron cup and the asbestos lined cone; permissible p = 35 psi . Compute (a) the needed axial force, (b) the face width, (c) the minimum axial force to achieve engagement under load. Solution: 63,000hp 63,000(15) Tf = = = 2362.5 in − lb n 400 f Dm R (a) T f = 2(sin α + f cos α ) Page 95 of 97 SECTION 16 – BRAKES AND CLUTCHES 2362.5 = (0.3)(10)R 2(sin 10 + 0.3 cos 10) R = 739 lb N π Dm c R 739 N= = = 1575 lb sin α + f cos α sin 10 + 0.3 cos 10 1575 35 = π (15)c c = 1.44 in (b) p = (c) max. f = 0.4 (Table AT 29) f Dm R Tf = 2(sin α + f cos α ) (0.4)(10)R 2362.5 = 2(sin 10 + 0.4 cos 10) R = 670 lb , minimum. 929. An “Airflex” clutch, Fig. 18.15, Text, has a 16-in drum with a 5-in. face. This clutch is rated at 110 hp at 100 rpm with an air pressure of 75 psi. What must be the coefficient of friction if the effect of centrifugal force is neglected? (Data courtesy of Federal Fawick Corporation.) Solution: D = 16 in b = 5 in hp = 110 hp rpm = 100 rpm p = 75 psi 63,000hp 63,000(110 ) Tf = = = 69,300 in − lb n 100 FD Tf = 2 F (16 ) 69,300 = 2 F = 8662.5 lb N = p(π Db ) = (75)(π )(16)(5) = 18,850 lb F 8662.5 f = = = 0.46 N 18,850 Page 96 of 97 SECTION 16 – BRAKES AND CLUTCHES 930. The same as 929 except that the diameter is 6 in., the face width is 2 in., and the rated horsepower is 3. Solution: 63,000hp 63,000(3) Tf = = = 1890 in − lb n 100 FD Tf = 2 F (6 ) 69,300 = 2 F = 630 lb N = p(π Db ) = (75)(π )(6)(2 ) = 2827 lb F 630 f = = = 0.22 N 2827 - end - Page 97 of 97 SECTION 17– WELDING DESIGN PROBLEMS 941. A joint welded with a coated rod is to support a steady load of 10 kips; the design is to be for a 3/8-in weld. Compute the length of weld needed for (a) a reinforced butt joint, Fig. 19-2, Text, (b) a lap joint, Fig. 19-3(a), (c) a T-joint, Fig. 19-3(b), (d) fillet welds, parallel loaded, Fig. 19-3(d). Solution: (a) Reinforced butt joint, Fig. 19-2. F = ηs t tL From Table AT 30, using Jennings recommendation, st = 16 ksi F = 10 kips η = 0.90 3 t = in = 0.375 in 8 3 F = 10 = (0.90 )(16 ) L 8 L = 1.85 in (b) Lap joint F = 2ss Lb cos 45 Table AT 30, normal or transverse loading, ss = 16 ksi (Lincoln-Electric) F = 10 kips 3 b = in = 0.375 in 8 3 F = 10 = 2(16 )L cos 45 8 L = 1.18 in (c) T-joint F = 2ss Lb cos 45 Table AT 30, all loading, ss = 14 ksi (Jennings recommendation) F = 10 kips 3 b = in = 0.375 in 8 3 F = 10 = 2(14 )L cos 45 8 Page 1 of 41 SECTION 17– WELDING L = 1.35 in (d) Fillet welds, parallel load F = 2ss Lb cos 45 Table AT 30, ss = 13.6 ksi (AWS Code) F = 10 kips 3 b = in = 0.375 in 8 3 F = 10 = 2(13.6 )L cos 45 8 L = 1.39 in 943. The load F varies from 5 to 10 kips and the arrangement is such that the location is given by m = 3 in and n = 7 in . It is desired that the weld lengths L1 and L2 be such that the line of action of F passes through the centroid G of the weld metal, thereby avoiding eccentric loading. Determine (a) the ratio of L1 L2 , (b) the lengths L1 and L2 of 3/8 in. fillet welds made with E6010 rod, for indefinite life. (c) The same as (b) except that the life expectancy is 105 cycles. Problem 943 Solution: F = 2ss Lb cos 45 F1 = 2 ss L1b cos 45 F2 = 2s s L2b cos 45 (a) Solving for the ratio of L1 L2 [∑ M G =0 ] Page 2 of 41 SECTION 17– WELDING F1m = F2 n F1 n = F2 m F1 L1 7 = = F2 L2 3 L1 : L2 = 7 : 3 (b) Solving for the lengths F1 + F2 = F 3F1 + 3F2 = 3F 7 F2 + 3F2 = 3F F2 = 0.3F F1 = 0.7 F For indefinite life use nc = 2× 10 6 Table AT 30 7 .2 sd = 1 − 0 .5 R 5 kips R= = 0.50 10 kips 7.2 sd = = 9.6 ksi 1 − 0.5(0.5) F = 10 kips F1 = 0.7(10 ) = 7 kips F2 = 0.3(10) = 3 kips F1 = 7 = 2(9.6)L1 (0.375) cos 45 L1 = 1.375 in F2 = 3 = 2(9.6)L2 (0.375) cos 45 L2 = 0.589 in (c) Solving for the lengths, 105 cycles For indefinite life use nc = 105 Table AT 30 12.2 sd = 1 − 0 .5 R 12.5 sd = = 16.67 ksi 1 − 0.5(0.5) Page 3 of 41 SECTION 17– WELDING F1 = 7 = 2(16.67 )L1 (0.375) cos 45 L1 = 0.792 in F2 = 3 = 2(16.67 )L2 (0.375) cos 45 L2 = 0.339 in 944. A bracket somewhat as shown is made of structural steel and supports a repeated ( R = 0 , nc = 2× 10 6 ) load of 2000 lb at a distance a = 10 in from the wall. What should be the length L of a 3/8-in. fillet weld that resists the entire load? Adapt the design shear stress from Table AT 30 (fillet weld). Problem 944, 958 Solution: Fa (L 2 ) I 3 2tL I= 12 FaL 3Fa st = = 2 3 2tL tL 2 12 t = b cos 45 st = 1 1 2 2 F 2 3Fa 2 2 s τ = ss2 + t = + 2 2 2tL 2tL From Table AT 30, R = 0 , nc = 2× 10 6 7.2 7.2 τ= = = 7.2 ksi 1 − 0.5R 1 − 0.5(0) F = 2000 lb = 2 kips a = 10 in Page 4 of 41 SECTION 17– WELDING b= 3 in 8 2 F 3Fa τ = + 2 2tL 2tL 2 2 2 2 1 (7.2) = 12 + 4 L 2(0.375) cos 45 L 14.22 12,800 51.84 = 2 + L L4 51.84 L4 = 14.22 L2 + 12,800 51.84 L4 − 14.22 L2 − 12,800 = 0 L = 3.98 in say L = 4 in 2 945. 3(2)(10) 2(0.375) cos 45 2 A bracket is to be fabricated from flat plates by bending and welding with a shielded rod, E6010. A steady load F = 5000 lb , L = 18 in , h = 4 in , and a = 6 in . (a) Take the design shear stress for a design factor N = 3.75 on the ultimate shear strength, which may be estimated at 80% of su of rod, and find the size of fillet weld ABCDA. Compare the design stress used with values in Table AT 30. (b) Compute the thickness of the SAE 1020, rolled-steel plates of all are the same (cantilever part). Problem 945 Solution: h FL 2 st = I Page 5 of 41 SECTION 17– WELDING I= 2th 3 2at 3 h + + 2(at ) 12 12 2 2 th 3 at 3 ath 2 + + 6 6 2 h = 4 in , a = 6 in , L = 18 in , F = 5 kips I= t (4 ) 6t 3 (6 )t (4 ) I= + + = 58.67t + t 3 6 6 2 (5)(18) 4 180 2 = st = 3 58.67t + t 58.67t + t 3 F 5 0.25 ss = = = 2at + 2th 2(6)t + 2t (4) t 3 s τ = s + t 2 s (a) τ d = us N sus = 0.8su 2 2 2 2 s su = 60 ksi for E6010 N = 3.75 (0.8)(60) = 12.8 ksi τd = 3.75 2 180 0.25 + 3 t 2 58.67t + t 0.0625 8100 163.84 = + 2 2 t 58.67t + t 3 τ 2 = (12.8)2 = ( ( ) 2 ) By trial and error method t = 0.121399 in t = b cos 45 0.121399 = b cos 45 b = 0.1717 in 3 use b = = 0.1875 in 16 Comparing the design stress, from Table AT 30 τ = 13.6 ksi (AISC Building Code) τ = 13.6 ksi > 12.8 ksi (b) Solving for the thickness of the SAE 1020, rolled-steel plates Table 1.1, use N = 2 based on yield strength Page 6 of 41 SECTION 17– WELDING Table AT 7, s y = 48 ksi sd = sy = 48 = 24 ksi 6391 2 N Mc sd = I h 4 c = = = 2 in 2 2 I = 58.67t + t 3 M = FL = (5)(18) = 90 in − kips Mc sd = I (90 in − kips )(2 in ) 24 ksi = 58.67t + t 3 in 4 58.67t + t 3 = 7.5 By trial and error method t = 0.128 in but t ≥ b 3 use t = in 16 946. A bracket is made with two 3/8-in. steel plates A welded with a coated electrode to a vertical I-beam with fillet welds on both sides of the plate, as indicated. It supports a steady vertical load F = 12 kips in a center position; a = 14 in ; h = 8 in . What size and length of weld should be used? Is the stress at G the maximum one? Justify your answer. Problem 946 – 948 Page 7 of 41 SECTION 17– WELDING Solution: a = 14 in h = 8 in F = 12 kips (steady load) use weld size = 3/8 in = plate thickness = 0.375 in τ B > τ G , therefore the stress at G is not the maximum one. L 2 cos θ = = 1 ρ′ 2 By cosine law L 2 = L2 + h 2 1 τ = (ss2 + ss2 + 2ss ss cos θ )2 1 2 F ′ eρ 2 ss1 = Jc L e=a+ 2 Page 8 of 41 1 2 L L2 + h 2 SECTION 17– WELDING 2 3 2tL3 h tL 1 + 2tL = + tLh 2 12 6 2 2 F 2 s s2 = 2tL Jc = Substituting values t = b cos 45 = (0.375) cos 45 = 0.2652 in 1 2 1 2 1 2 2 L + h2 = L + (8) = L + 64 ρ′ = 2 2 2 L cos θ = L2 + 64 e = 14 + 0.5 L 0.2652 L3 1 2 Jc = + (0.2652)L(8) = 0.0442 L3 + 8.1984 L 6 2 12 1 2 (14 + 0.5L ) L + 64 33.94(28 + L ) L2 + 64 2 2 ss1 = = 0.0442 L3 + 8.1984 L L3 + 185.484 L F 12 11.43 2 2 s s2 = = = 2tL 2(0.2652 )L L 2 2 2 τ = ss1 + ss2 + 2ss1 ss2 cos θ From Table AT 30, Use τ d = 14 ksi as recommended by Jennings Solving this by trial and error method τd >τ ) ( ( L, in 8 7 5 4 3.5 3.75 3.875 ) ) cos θ 0.7071 0.6585 0.5300 0.4472 0.4008 0.4244 0.4359 Use L = 3.875 in , τ = 13.92 ksi ≈ 14ksi 7 Or L = 3 in each weld 8 Page 9 of 41 ( ss1 , ksi ss2 , ksi 6.9260 7.6930 10.040 12.053 13.489 12.723 12.377 1.429 1.633 2.286 2.858 3.266 3.048 2.950 τ , ksi 8.00 8.85 11.42 13.57 15.10 14.28 13.92 SECTION 17– WELDING 947. The same as 946, except that the material is aluminum alloy, welded with shielded 1100 wire, and the load is 5 kips. Let the design factor N = 3.4 for the information in Table At 30; but consider other approaches, as available, to a design stress. Solution: With F = 5 kips 5 1 2 (14 + 0.5L ) L + 64 14.14(28 + L ) L2 + 64 2 2 ss1 = = 0.0442 L3 + 8.1984 L L3 + 185.484 L F 5 4.714 2 2 s s2 = = = 2tL 2(0.2652 )L L L cos θ = 2 L + 64 2 2 τ = ss1 + ss22 + 2ss1 ss2 cos θ From Table AT 30, ultimate strength of aluminum alloy welded with 1100 wire, = 12.7 ksi. 12.7 τd = = 3.735 ksi 3 .4 Solving this by trial and error method τd >τ ) ( ( ( ) ) L, in 8 7 6.5 6.75 6.875 cos θ 0.7071 0.6585 0.6306 0.6449 0.6518 ss1 , ksi ss2 , ksi 2.886 3.205 3.397 3.298 3.251 0.589 0.673 0.725 0.698 0.686 τ , ksi 3.326 3.683 3.895 3.786 3.735 Use L = 6.875 in , τ = 3.735 ksi = τ d 7 Or L = 6 in for each weld 8 948. (a) Two ¾-in plates A, arranged as shown, are to be welded with coated electrodes, E6020; a = 12 in ; h = 4 in , and F repeats from 0 to 10 kips. Choose a design stress from Table AT 30 for 2x106 cycles and specify the size and length of weld. (b) The same as (a), except that the design is for 105 cycles. (c) Demonstrate that the stress at G is or is not the maximum. Page 10 of 41 SECTION 17– WELDING Problem 946 – 948 Solution: This problem is the same as 946 except that a = 12 in h = 4 in F = 0 to 10 kips 3 b = in = same as plate thickness 4 t = b cos 45 = (0.75) cos 45 = 0.53 in L L L cos θ = = = 2 2 2 2 2 L +h L +4 L + 16 F ′ eρ 2 ss1 = Jc L e = a + = 12 + 0.5 L 2 1 2 1 2 1 2 2 L + h2 = L + (4 ) = L + 16 ρ′ = 2 2 2 0.53L3 1 2 Jc = + (0.53)L(4) = 0.088L3 + 4.24 L 6 2 F = 10 kips 10 1 2 (12 + 0.5L ) L + 16 14.205(24 + L ) L2 + 16 2 2 ss1 = = 0.088L3 + 4.24 L L3 + 48.18L F 10 4.714 2 2 s s2 = = = 2tL 2(0.53)L L 2 2 2 τ = ss1 + ss2 + 2ss1 ss2 cos θ ) ( ( ) (a) From Table AT 30 , nc = 2× 10 6 Page 11 of 41 ( ) SECTION 17– WELDING 7 .2 ksi 1 − 0 .5 R 0 R= =0 10 7.2 τd = = 7.2 ksi 1 − 0.5(0) Solving for length by trial and error method τd >τ τd = L, in 8 7 6 5.5 5.625 cos θ 0.8944 0.8682 0.8321 0.8087 0.8150 ss1 , ksi ss2 , ksi 4.530 5.219 6.084 6.607 6.469 0.590 0.674 0.786 0.858 0.839 τ , ksi 5.06 5.81 6.87 7.32 7.17 Use L = 5.625 in , τ = 7.17 ksi ≈ τ d 5 Or L = 5 in for each weld 8 (b) From Table AT 30, nc = 105 12.5 τd = ksi 1 − 0 .5 R 0 R= =0 10 12.5 τd = = 12.5 ksi 1 − 0.5(0) Solving for length by trial and error method τd >τ L, in 4 3 cos θ 0.7071 0.6000 ss1 , ksi ss2 , ksi 8.764 11.179 1.179 1.573 Use L = 3 in , τ = 12.188 ksi ≈ τ d Or L = 3 in for each weld (c) Demonstrating that the stress at G is or is not the maximum Page 12 of 41 τ , ksi 9.634 12.188 SECTION 17– WELDING From the figure τ G is not the maximum. 949. One arm of a bracket that is to support a steady load of F = 18 kips without twisting is welded with an E6010 rod, as shown. The plate is 10 in. ( ≈ L2 ) deep. Assume a value of L1 (not less than 5 in.) and compute the size of fillet weld. By sketching vectors (only), compare the stress at C with that of B. Problems 949-951 Solution: Use L1 = 5 in , L2 = 10 in F = 18 kips Solving for the center G. (L1 + L1 + L2 )x = 2 L1 L1 2 Page 13 of 41 SECTION 17– WELDING (5) = 1.25 in L12 = 2 L1 + L2 2(5) + 10 2 x= e = 11.25 in − 1.25 in = 10 in at B, where τ is maximum Feρ1 Jc For J c : ss1 = r12 = (2.5 − 1.25) + (5) r1 = 5.154 in r2 = 1.25 in 2 2 2tL13 tL32 2t (5) t (10 ) 2 2 Jc = + + tL2 r22 + 2tL1r12 = + + t (10 )(1.25) + 2t (5)(5.154 ) = 385.43t 12 12 12 12 2 2 2 ρ1 = (5) + (5 − 1.25) ρ1 = 6.25 in (18)(10)(6.25) = 2.92 ss1 = 385.43t t 3 Page 14 of 41 3 SECTION 17– WELDING F 18 0 .9 = = 2tL1 + tL2 2t (5) + t (10 ) t L − x 5 − 1.25 cos θ1 = 1 = = 0.60 ρ1 6.25 s s2 = τ B2 = τ 2 = ss2 + ss2 + 2ss ss cos θ1 1 2 1 2 2 2 2.92 0.90 2.92 0.90 + + 2 (0.6 ) t t t t 3.534 τ= t From Table AT 30, steady load, use code τ = 13.6 ksi τ2 = τ = 13.6 = 3.534 t t = 0.26 in t = b cos 45 0.26 = b cos 45 b = 0.368 in Therefore use size of fillet weld = b = 0.375 in = 3 in 8 Comparing stress at C with that of B. Stress at B Stress at C τ B > τ C , ρ1 > ρ 2 , ss′ > ss 1 950. 1 The same as 949, except that F makes a 30o angle with the vertical as indicated by the dotted line in the figure. Consider all computed stress to be shear. Solution: Page 15 of 41 SECTION 17– WELDING L1 = 5 in L2 = 10 in F = 18 kips T = (F cos 30)e − (F sin 30)(3) e = 10 in as in 949. T = (18 cos 30)(10) − (18 sin 30)(3) = 128.89 in − kips As in 949, Tρ ss1 = 1 Jc ρ1 = 6.25 in (128.89)(6.25) = 2.09 ss1 = 385.43t t 0 .9 s s2 = (from 949) t at B cos θ1 = 0.60 from 949. θ1 = 53.13 α = 90 − 30 − 53.13 = 6.87 o τ B2 = τ 2 = ss21 + ss22 + 2ss1 ss2 cos α 2 2 2.09 0.90 2.09 0.90 τ = + + 2 cos 6.87 t t t t 2.986 τ= t but τ = 13.6 ksi as in 949 2 Page 16 of 41 SECTION 17– WELDING τ = 13.6 = 2.986 t t = 0.22 in t = b cos 45 0.22 = b cos 45 b = 0.311 in Therefore use size of fillet weld = b = 0.3125 in = 5 in 16 Comparing stress at C with that of B. Stress at B Stress at C τ B >τC 951. The same as 949, except that the load varies from 4 to 18 kips; expected life, 2 x 106 cycles. Solve (a) by using a design stress from Table AT 30 for the given value of R , and (b) by using a design stress for R = −1 , and the Soderberg criterion. Solution: same as 949, but F = 4 to 18 kips F = 18 kips (is the same as in 949) 3.534 then τ = t (a) Table AT 30, nc = 2 × 10 6 cycles 4 R = = 0.222 18 7.2 7.2 τd = = = 8.1 ksi 1 − 0.5 R 1 − 0.5(0.222 ) Page 17 of 41 SECTION 17– WELDING 3.534 t 3.534 8 .1 = t t = 0.4363 in t = b cos 45 0.4363 = b cos 45 b = 0.617 in 5 say in fillet weld. 8 τ =τd = (b) Using a design stress for R = −1 and the Soderberg criterion, nc = 2 × 10 6 cycles 1 sm K f sa = + N sy sn For R = −1 sm = 0 sa = τ d s τd = n KfN Table 19.1, sn = 11.7 ksi , i4.20, use N = 1.4 assume K f = 1.67 11.7 = 5.0 ksi (1.4)(1.67 ) 3.534 τ =τd = 5 = t t = 0.7068 in t = b cos 45 0.7068 = b cos 45 b = 0.9996 in say b = 1 − in fillet weld. τd = 952. A steel plate, welded to a column as shown with E6010 rod, is to support a steady load of F = 5 kips , applied so as to produce no twisting of the plate; m = 24 in , n = 18 in ; the initial design is for a 3/8-in. fillet weld. Compute L . Demonstrate by sketches which stress ss A or ssB is the larger. Page 18 of 41 SECTION 17– WELDING Problems 952-954, 959 Solution: F = 5 kips m = 24 in n = 18 in 3 b = in = 0.375 in 8 t = b cos 45 = (0.375) cos 45 = 0.2652 in maximum τ = ssA > ssD ρ= 1 2 (m − n )2 + L2 cos θ = L (m − n )2 + L2 τ 2 = ss2 + ss2 + 2ss ss cos θ 1 ρ= 1 2 2 1 2 (24 − 18)2 + L2 Page 19 of 41 = 1 2 L + 36 2 SECTION 17– WELDING cos θ = L L2 + 36 Feρ Jc F = 5 kips (steady load) 1 1 1 e = n + (m − n ) = (m + n ) = (24 + 18) = 21 in 2 2 2 ss1 = 2tL3 m−n Jc = + 2tL 12 2 2 2(0.2652 )L3 24 − 18 + 2(0.2652 )L 12 2 J c = 0.0884 L3 + 4.774 L 2 Jc = (5)(21) 1 L2 + 36 594 L2 + 36 2 ss1 = = 0.0884 L3 + 4.774 L L3 + 54 L F 5 9.427 = = s s2 = 2tL 2(0.2652 )L L From Table AT 30, use AISC Building Code τ d = 13.6 ksi Solving for L by trial and error method. τd ≥τ L, in 6 5 4.875 4.75 cos θ 0.7071 0.6402 0.6306 0.6207 ss1 , ksi ss2 , ksi τ , ksi 9.334 11.745 12.113 12.500 1.571 1.885 1.934 1.985 10.50 13.03 13.42 13.82 7 ∴ use L = 4 in = 4.875 in , τ d ≥ τ 8 953. A steel plate, welded as shown with E6010 rod, is to support a load that varies from –5 to 5 kips, without twisting; m = 14 in , n = 8 in ; the initial design is for a 3/8-in. fillet weld; indefinite life. Compute L . Solution: The same as 952, except m = 14 in n = 8 in F = −5 to 5 kips Page 20 of 41 SECTION 17– WELDING 3 in fillet weld 8 t = b cos 45 t = (0.375) cos 45 = 0.2652 in F = 5 kips Then; 1 ρ= (m − n )2 + L2 2 1 ρ= (14 − 8)2 + L2 = 1 L2 + 36 2 2 L cos θ = (m − n )2 + L2 L cos θ = L2 + 36 1 1 1 e = n + (m − n ) = (m + n ) = (14 + 8) = 11 in 2 2 2 b= Jc = 2tL3 m−n + 2tL 12 2 2 2(0.2652 )L3 14 − 8 Jc = + 2(0.2652 )L 12 2 J c = 0.0884 L3 + 4.774 L Feρ ss1 = Jc (5)(11) 1 2 L2 + 36 311 L2 + 36 2 = 0.0884 L3 + 4.774 L L3 + 54 L F 5 9.427 s s2 = = = 2tL 2(0.2652 )L L 2 2 2 τ = ss1 + ss2 + 2ss1 ss2 cos θ ss1 = From Table AT 30, indefinite life ( nc = 2 × 10 6 ) 7 .2 τd = 1 − 0 .5 R −5 R= = −1 5 7.2 τd = = 4.8 ksi 1 − 0.5(− 1) Solving for L by trial and error method. τd ≥τ Page 21 of 41 SECTION 17– WELDING L, in 6 7 8 7.5 7.75 7.25 7.375 cos θ 0.7071 0.7593 0.8000 0.7809 0.7907 0.7704 0.7757 ss1 , ksi ss2 , ksi τ , ksi 4.887 3.977 3.295 3.613 3.448 3.788 3.699 1.571 1.347 1.178 1.257 1.216 1.300 1.278 6.10 5.01 4.30 4.66 4.47 4.86 4.76 3 ∴ use L = 7 in = 7.375 in , τ = 4.76 ≈ τ d 8 3 or L = 7 in for each weld. 8 The same as 953, except that F varies from 0 to 5 kips with a life expectancy of 105 cycles. 954. Solution: Same as 953, but F = 0 to 5 kips nc = 105 From table AT 30, nc = 105 12.5 τd = 1 − 0 .5 R 0 R= =0 5 12.5 τd = = 12.5 ksi 1 − 0.5(0) L cos θ = L2 + 36 311 L2 + 36 ss1 = L3 + 54 L 9.427 s s2 = L 2 2 τ = ss1 + ss22 + 2ss1 ss2 cos θ Solving by trial and error method. τd ≥τ L, in 3 3.5 3.125 Page 22 of 41 cos θ 0.4472 0.5039 0.4619 ss1 , ksi ss2 , ksi τ , ksi 11.038 9.317 10.558 3.142 2.694 3.017 12.76 10.93 12.25 SECTION 17– WELDING ∴ use L = 3.125 in , τ = 12.25 ≈ τ d 1 or L = 3 in for each weld. 8 955. An arm for a machine is to be fabricated by welding, coated welding rod. See figure. The left end is a hollow cylinder with Do = 3 in , and it is keyed to a 2-in. shaft; L = 14 in , steady load F = 600 lb . The arm material is SAE 1020, rolledsteel plate, ½-in. thick. Compute (a) the depth h of the arm at the hub, and (b) the size of the weld. Problem 955 Solution: (a) Solving for the length or depth h For SAE 1020, rolled-steel plate, s y = 48 ksi (Table AT 7) N = 2 (Table 1.1) 48 sd = = 24 ksi 2 h Fe 2 s= I 1 e ≈ L − (Do + Ds ) 2 Do = 3 in Ds = 2 in L = 14 in 1 e = 14 − (3 + 2 ) = 11.5 in 2 3 th I= 12 1 t = in = 0.5 in 2 Page 23 of 41 SECTION 17– WELDING 0.5h 3 12 F = 600 lb steady (600)(11.5) h 2 = 82,800 s= 3 h2 0.5h 12 h = 1.857 in 7 say h = 1 in 8 I= (b) Solving for the size of weld t = b cos 45 h Fe 2 st = I 3 2th I= 12 h Fe 3Fe 2 st = 3 = 2 2th th 12 3(0.6 )(11.5) 5.888 st = = 2 t t (1.875) F 0.6 0.16 st = = = 2th 2t (1.875) t 1 2 2 s τ = ss2 + t 2 From Table AT 30, use Jennings recommendations, τ = 14 ksi 1 0.16 2 5.888 2 2 τ = 14 = + t 2t t = 0.2106 in 0.2106 = b cos 45 Page 24 of 41 SECTION 17– WELDING b = 0.2978 in 5 say b = in fillet weld. 16 956. A pair of gusset plates, 3/8-in. thick, are to be welded with E6010 electrodes, as shown. The load F on the plates varies from 0 to 10 kips (no twisting of plates). For the first approximation, assume that BC = AD = L = 5 in and compute the size of weld. With free hand sketches, compare the resultant stress at each corner A, B, C, and D. Problems 956, 957 Solution: ρ1 = (2)2 + (6)2 Page 25 of 41 = 6.325 in SECTION 17– WELDING ρ2 = (2)2 + (9)2 = 9.220 in F 2tL F = 0 to 10kips F = 10 kips L = 5 in 10 1 s s2 = = 2t (5) t τC =τD ; τ A =τB Feρ ss1 = Jc e = 6 + 2 = 8 in s s2 = 2 2 2tL3 3 2t (5) 3 Jc = + 2tL 6 + = + 2t (5) 6 + = 583.3t 12 2 12 2 at D 2 = 0.3162 ρ1 6.325 Feρ (10 )(8)(6.325) 0.8675 = = = Jc 583.3t t cos θ1 = ss1D 2 3 = 1 τ D = (ss2 + ss2 + 2ss ss cos θ1 )2 1D 2 1D 2 1 0.8675 2 1 2 0.8675 1 2 1.5170 τ D = + + 2 (0.3162) = t t t t t at A cos θ 2 = 2 ρ2 = 2 9.220 θ 2 = 77.47 o α = 180 − 77.47 = 102.53o Page 26 of 41 SECTION 17– WELDING ss1 A = Feρ (10 )(8)(9.220 ) 1.2645 = = Jc 583.3t t 1 τ A = (ss2 + ss2 + 2ss ss cos α )2 1A 2 1A 2 1 1.2645 2 1 2 1.2645 1 2 1.4319 τ A = + + 2 cos 102.53 = t t t t t 1.5170 t From Table AT 30, assume nc = 2 × 10 6 . 7 .2 τd = 1 − 0 .5 R 0 R= =0 10 7.2 τd = = 7.2 ksi 1 − 0.5(0) 1.5170 τ = 7 .2 = t t = 0.2107 in t = b cos 45 0.2107 = b cos 45 b = 0.2980 in 5 say b = in for each weld. 16 τ max = τ D = CHECK PROBLEMS 957. A 3/8-in. gusset plate is welded with an E6010 electrode; ¼-in. fillet weld, as shown. The loading does not twist the plate and the force varies from 0.2 F to F . For a life expectancy of 105 cycles, what is a safe F ? Make clear how you decide upon the point of maximum stress. Solution: Same as 956, except that F is unknown and varies from 0.2 F to F , nc = 105 cycles. Stress vector is the same as shown in 956, τ D = τ C is maximum. At D (prob. 956) ρ1 = 6.325 in F s s2 = 2tL Page 27 of 41 SECTION 17– WELDING t = b cos 45 b = 0.25 in t = (0.25) cos 45 = 0.1768 in L = 5 in F s s2 = = 0.5656 F 2(0.1768)(5) Feρ1 ss1 = Jc e = 6 + 2 = 8 in ρ1 = 6.325 in 2 2 2tL3 3 2(0.1768)(5) 3 Jc = + 2tL 6 + = + 2(0.1768)(5) 6 + = 23.573 in 4 12 2 12 2 Feρ1 F (8)(6.325) ss1 = = = 2.1465 F Jc 23.573 3 τ 2 = ss2 + ss2 + 2ss ss cos θ cos θ = 0.3162 (prob. 956) 1 2 1 2 From Table AT 30, nc = 105 cycles. 12.5 τd = 1 − 0 .5 R 0 .2 F R= = 0 .2 F 12.5 τd = = 13.89 ksi 1 − 0.5(0.2) τ 2 = ss21 + ss22 + 2ss1 ss2 cos θ (13.89)2 = (2.1465 F )2 + (0.5656 F )2 + 2(2.1465F )(0.5656 F )(0.3162) F = 5.82 kips 958. A bracket of the type shown (944) is to support a load of F = 6 kips ; a = 8 in , L = 5 in , weld size is 3/8-in. (a) Determine the stress in the weld. Is this safe value for a steady load? (b) If the welding is shielded and the load varies with R = 0 , is the weld safe for 2 × 10 6 cycles? For 105 cycles? For Q & T alloy and 2 × 10 6 cycles? Solution: Page 28 of 41 SECTION 17– WELDING F = 6 kips a = 8 in L = 5 in 3 b = in = 0.375 in 8 t = b cos 45 = (0.375) cos 45 = 0.2652 in F 6 ss = = = 2.3 ksi 2tL 2(0.2652)(5) L Fa 3Fa 3(6)(8) 2 st = 3 = 2 = = 21.7 ksi 2 2tL tL ( 0.2652)(5) 12 1 2 st 2 2 τ = ss + 2 1 2 21.7 2 2 τ = (2.3) + = 11.1 ksi 2 (a) Steady load, Table AT 30, AISC Building Code τ d = 13.6 ksi Since τ d > 11.1 ksi ∴ a safe value (b) Table AT 30, nc = 2 × 10 6 , R = 0 7.2 7.2 τd = = = 7.2 ksi 1 − 0.5R 1 − 0.5(0) Page 29 of 41 SECTION 17– WELDING Since τ d < 11.1 ksi ∴ not a safe value Table AT 30, nc = 105 , R = 0 12.5 12.5 τd = = = 12.5 ksi 1 − 0.5 R 1 − 0.5(0 ) Since τ d > 11.1 ksi ∴ a safe value Table AT 30, nc = 2 × 10 6 ,Q & T alloy, R = 0 9 9 τd = = = 9 ksi 1 − 0.5R 1 − 0.5(0) Since τ d < 11.1 ksi ∴ not a safe value 959. The 1-in. plate shown (952) is attached with ½-in fillet welds, laid with E6010 rods; L = 4 in , m = 15 in , n = 9 in . What maximum load may be carried if it is (a) static, (b) varies from 0.5 F to F for 2 × 10 6 cycles and for 105 cycles, (c) R = 0 , indefinite life. (d) Considering strengths given in Table 19.1, Text, determine the design factor for parts (b) and (c). Solution: Same as 952, except that m = 15 in , n = 9 in , b = 1 in , L = 4 in 2 t = b cos 45 = (0.5) cos 45 = 0.3536 in τ 2 = ss21 + ss22 + 2ss1 ss2 cos θ ss1 = Feρ Jc 1 (m + n ) = 1 (15 + 9) = 12 in 2 2 1 ρ= (m − n )2 + L2 = 1 (15 − 9)2 + (4)2 = 3.606 in 2 2 e= 2 2 2tL3 2(0.3536 )(4 ) m−n 15 − 9 4 Jc = + 2tL + 2(0.3536 )(4 ) = = 29.23 in 12 12 2 2 ss1 = Feρ F (12 )(3.606 ) = = 1.4804 F Jc 29.23 Page 30 of 41 3 SECTION 17– WELDING F F = = 0.3535F 2tL 2(0.3536)(4) L 4 cos θ = = = 0.5547 2 2 (m − n ) + L (15 − 9)2 + 42 s s2 = 1 τ = (ss2 + ss2 + 2ss ss cos θ )2 1 2 1 2 [ 1 ] τ = (1.4804 F )2 + (0.3535F )2 + 2(1.4804 F )(0.3535 F )(0.5547 ) 2 = 1.7021F (a) Table AT 30, static, AISC Building Code τ d = 13.6 ksi τ = 13.6 = 1.7021F F = 7.99 kips (b) Table AT 30, static, nc = 2 × 10 6 cycles 7 .2 τd = 1 − 0 .5 R 0 .5 F = 0 .5 R= F 7.2 τd = = 9.6 ksi 1 − 0.5(0.5) τ = 9.6 = 1.7021F F = 5.64 kips Table AT 30, static, nc = 105 cycles 12.5 τd = 1 − 0 .5 R 0 .5 F R= = 0 .5 F 12.5 τd = = 16.67 ksi 1 − 0.5(0.5) τ = 16.67 = 1.7021F F = 9.79 kips (c) Table AT 30, R = 0 , indefinite life, nc = 2 × 10 6 cycles 7 .2 τd = 1 − 0 .5 R 7.2 τd = = 7.2 ksi 1 − 0.5(0) Page 31 of 41 SECTION 17– WELDING τ = 7.2 = 1.7021F F = 4.23 kips (d) Design factors nc = 2 × 10 6 cycles R = 0 .5 τ = 9.6 ksi Table 19-1, sn = 40.1 ksi s 40.1 N= n = = 4 .2 τ 9 .6 nc = 105 cycles R = 0 .5 τ = 16.67 ksi Table 19-1, sn = 46.1 ksi s 46.1 N= n = = 2 .8 τ 16.67 Indefinite life ( nc = 2 × 10 6 ), R = 0 τ = 7.2 ksi Table 19-1, sn = 18.1 ksi s 18.1 N= n = = 2 .5 τ 7 .2 960. The plate for a bracket, as shown must be welded to a member in the manner shown; 5/16-in. welds with shielded arc. Compute the safe load for this plate (no twisting) (a) for static loading, (b) for R = 0 and indefinite life, (c) for R = −1 and indefinite life, (d) for R = 0.2 indefinite life. Problem 960 Page 32 of 41 SECTION 17– WELDING Solution: For x (3 + 6)x = (3) 3 + (6) 6 2 2 x = 2.5 in 5 in = 0.3125 in 16 t = b cos 45 = (0.3125) cos 45 = 0.2210 in F F s s2 = = = 0.503F t (3 + 6) (0.2210)(9) Feρ ss1 = Jc e = 10 − 2.5 = 7.5 in b= 2 6 2 Solving for J c ; ρ = + (6 − 2.5)2 = 4.61 in 2 2 6 3 r1 = + − 2.5 = 3.1623 in 2 2 Page 33 of 41 SECTION 17– WELDING 2 2 6 6 r2 = + − 2.5 = 3.0414 in 2 2 t (3) t (6 ) + 3tr12 + + 6tr22 12 12 3 3 ( 0.221)(3) ( 0.221)(6 ) 2 2 Jc = + 3(0.221)(3.1623) + + 6(0.221)(3.0414 ) = 23.37 in 4 12 12 F (7.5)(4.61) = 1.4795 F ss1 = 23.37 ss2 = 0.503F 3 3 Jc = cos θ = 6 − 2.5 ρ [ = 6 − 2.5 = 0.7592 4.61 1 2 ] τ = (1.4795F ) + (0.503F ) + 2(1.4795F )(0.503F )(0.7592) = 1.890 F 2 2 (a) For static loading, Table AT 30, AISC Building Code τ d = 13.6 ksi τ = 13.6 = 1.890 F F = 7.20 kips (b) For R = 0 and indefinite life, ( ≈ nc = 2× 10 6 ) 7.2 7.2 = = 7.2 ksi 1 − 0.5R 1 − 0.5(0) τ = 7.2 = 1.890 F F = 3.8 kips τd = (c) For R = −1 and indefinite life 7.2 7.2 = = 4.8 ksi 1 − 0.5R 1 − 0.5(− 1) τ = 4.8 = 1.890 F F = 2.5 kips τd = (d) For R = 0.2 and indefinite life 7.2 7.2 = = 8.0 ksi 1 − 0.5R 1 − 0.5(0.2) τ = 8.0 = 1.890 F F = 4.2 kips τd = Page 34 of 41 SECTION 17– WELDING 961. A 2-in. round bar is welded to a vertical wall by a 3/8-in. fillet weld as shown in Fig. 19.8, Text. The bar supports a vertical load of 800 lb at a distance of 10 in. from the wall. What is the maximum computed stress in the weld? Would this result be safe for a varying load with R = 0 , shielded weld? Solution: Fig. 19.8, Text, with additional 3 in = 0.375 in 8 D = 2 in F = 800 lb e = 10 in 5.66 M 5.66 Fe 5.66(800 )(10 ) st = = = = 9609 psi = 9.609 ksi π bD 2 π bD 2 π (0.375)(2)2 F F 800 ss = = = = 480 psi = 0.48 ksi π Dt π Db cos 45 π (2)(0.375) cos 45 b= 1 2 2 s τ = ss2 + t 2 1 2 2 9.609 τ = (0.48)2 + = 4.83 ksi 2 Table AT 30, R = 0 Assume nc = 2× 10 6 cycles 7 .2 τd = = 7.2 > 4.83 ksi 1 − 0 .5 R ∴ safe for a varying load, nc = 2× 10 6 cycles. 962. The 14-in. structural-steel disk is welded to the plate by a 7/16-in. fillet weld, 360o, shielded arc. The force F acts on a pin attached to the disk. The pin is short enough that the moment arm to the disk is negligible. Determine a safe force F for (a) static loading, (b) reversed loading, indefinite life, (c) a varying load from 0.3F to F indefinite life. Page 35 of 41 SECTION 17– WELDING Problem 962 Solution: b= 7 in 16 7 t = b cos 45 = cos 45 = 0.3094 in 16 D = 14 in Feρ ss1 = Jc e = 6 sin 60 ρ = 14 in 1 J c = 2π r 3t = π D 3t 4 F (6 sin 60 )(14 ) ss1 = = 0.1091F 1 π (14)3 (0.3094) 4 F F sss = = = 0.0735 F π Dt π (14)(0.3094) τ = ss1 + s s2 = 0.1091F + 0.0735F = 0.1826 F (a) Static loading, Table AT 30, AISC Building Code τ = 13.6 ksi τ = 13.6 = 0.1826 F F = 74.48 kips (b) Reversed loading, indefinite life, Table AT 30 τ = 5 ksi (Jenning recommendation) τ = 5 = 0.1826 F F = 27.38 kips (c) Varying load, R = 0.3 , indefinite life, Table AT30, nc ≈ 2× 10 6 cycles. Page 36 of 41 SECTION 17– WELDING 7.2 7.2 = = 8.47 ksi 1 − 0.5 R 1 − 0.5(0.3) τ = 8.47 = 0.1826 F F = 46.38 kips τd = 963. A bracket is fabricated from ½-in., AISI-1020 rolled plate with 3/8-in. fillet welds on both sides of the plates G, H, and J as shown. The welds are made with E6016 welding rod; a central load at L = 30 in ; h = 11 in . Determine the repeated load that the welding can support. Problem 963. Solution: 3 b = in = 0.375 in 8 t = b cos 45 = 0.375 cos 45 = 0.2652 in L = 30 in h = 11 in ss = F 1 2t (2 ) + 2t (11 − 1) + 2t 2 − 2 h FL 2 st = I Page 37 of 41 = F F = = 0.13966 F 27t 27(0.2652 ) SECTION 17– WELDING t = 0.2652 in 1 3 2 2 − (0.2652) 2 2(2)(0.2652 ) 1 2 11 11 − 1 I= + 2(2)(0.2652) + + 2 2 − (0.2652) 12 12 2 2 2 2 3 2(0.2652)(11 − 1) + = 96.2 in 4 12 11 F (30) 2 = 1.715F st = 96.2 From Table AT 30, using Jennings recommendation, τ = 14 ksi and assume a strength reduction factor = 1.4 14 τ= = 10 ksi 1 .4 3 1 2 2 s τ = ss2 + t 2 1 2 1.715F 2 2 10 = (0.13966 F ) + 2 F = 11.51 kips 964. The bracket shown is made of ½-in. AISI-C1020 rolled plates. The 3/8-in. fillet welds are on both sides of each plate A and B, E7010 welding rod. The entire bracket is normalized after welding; L = 12 in , a = 8 in , and h = 8 in . What is the safe maximum load if it is (a) static, (b) varies from 0.5 F to F for 2× 10 6 cycles and for 105 cycles, (c) R = 0 , indefinite life. (d) Considering the strengths given in Table 19.1, Text, determine the design factor for part (b) and (c). Problem 964 Solution: 3 b = in = 0.375 in 8 t = b cos 45 = 0.375 cos 45 = 0.2652 in L = 12 in Page 38 of 41 SECTION 17– WELDING h = 8 in a = 8 in 1 8 − 2 1 1 1 1 ( ) 8 + 8 − 1 + 4 8 − y = 8 + 4 8 − + 2 2 2 2 2 y = 2.922 in IG 2 3 3 ( ( 0.2652 ) (8) 0.2652 ) (8 − 1) 1 2 = + (0.2652 )(8)(2.922 ) + + (0.2652 )(8 − 1) 2.922 − 12 3 1 1 4(0.2652 ) 8 − 8 − 1 2 + 4(0.2652 ) 8 − (8 − 2.922 ) − 2 + 12 2 2 4 I = 80.3524 in 1 A = 8(0.2652 ) + (8 − 1)(0.2652 ) + 4 8 − (0.2652 ) = 11.934 in 2 2 F F = = 0.0838 F A 11.934 FLc st = I c = 8 − 2.922 = 5.078 in ss = Page 39 of 41 12 2 2 SECTION 17– WELDING st = FLc F (12 )(5.078) = = 0.7584 F I 80.3524 1 2 st 2 2 τ = ss + 2 1 2 2 0.7584 F 2 τ = (0.0838F ) + = 0.3884 F 2 (a) Static loading, Table AT 30, AISC Building Code τ = 13.6 ksi τ = 13.6 = 0.3884 F F = 35 kips (b) Variable, R = 0.5 For nc = 2× 10 6 cycles (Table AT 30) 7.2 7.2 τ= = = 9.6 ksi 1 − 0.5R 1 − 0.5(0.5) τ = 9.6 = 0.3884 F F = 24.7 kips For nc = 105 cycles (Table AT 30) 12.5 12.5 τ= = = 16.67 ksi 1 − 0.5R 1 − 0.5(0.5) τ = 16.67 = 0.3884 F F = 42.9 kips (c) R = 0 , indefinite life (Table AT 30) 7 .2 = 7.2 ksi 1 − 0 .5 R τ = 7.2 = 0.3884 F F = 18.5 kips τ= (d) From Table 19.1 sn = 40.1 ksi , R = 0.5 , nc = 2× 10 6 sn = 46.1 ksi , R = 0.5 , nc = 105 sn = 18.1 ksi , R = 0 , nc ≈ 2× 10 6 (indefinite life) Page 40 of 41 SECTION 17– WELDING Design Factor, N R = 0.5 , nc = 2× 10 6 40.1 N= = 4 .2 9 .6 R = 0.5 , nc = 105 46.1 N= = 2 .8 16.67 R = 0 , nc ≈ 2× 10 6 (indefinite life) 18.1 N= = 2 .5 7 .2 - end - Page 41 of 41 SECTION 18 – MISCELLANEOUS PROBLEMS THIN SHELLS, EXTERNAL PRESSURE 981. A closed cylindrical tank is used for a steam heater. The inner shell, 200 in. outside diameter and 50 ft. long, is subjected to an external pressure of 40 psi. The material is equivalent to SA 30 (ASME Pressure-Vessel Code: min. su = 55 ksi ); assume an elastic limit of s y = su 2 ; let N = 5 . (a) What thickness of shell is needed from a stress standpoint? (b) For this thickness, what must be the maximum length of unsupported section to insure against collapse? (c) Choose a spacing L to give a symmetric arrangement and determine the moment of inertia of the steel stiffening rings. (d) For a similar problem, the Code recommends that t ≥ 0.76 in , L = 50 in , and I = 96 in 4 . How do these values check with those obtained above? (e) Without stiffening rings, what thickness would be needed? Solution: (a) Solving for the thickness of shell, t= pc D 2sy pc = 5 p = 5(40) = 200 psi s 55,000 sy = u = = 27,500 psi 2 2 D = 200 in p D (200)(200 ) t= c = = 0.73 in 2sy 2(27,500 ) say t = 0.75 in (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5 t 2 2.60 E D pc = psi 1 2 L t − 0.45 D D 5 t 2 1 2.60 E 2 L t D + 0.45 = D pc D E = 30× 106 psi pc = 200 psi t = 0.75 in Page 1 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS D = 200 in 5 0.75 2 1 2.60 30 ×10 6 L 0 . 75 2 200 + 0.45 = 200 200 200 L = 72.68 in ( ) (c) Solving for the moment of inertia of the steel stiffening rings. Choosing L = 60 in for 50 ft long shell 0.035D 3 Lpc 0.035(200) (60 )(200 ) = = 112 in 4 E 30 ×106 3 I= (d) For t ≥ 0.76 in - above minimum L = 50 in - below maximum I = 96 in 4 - lighter than above. (e) L = 50 ft = 600 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5 pc = t 2 2.60 E D L t − 0.45 D D 1 2 psi 1 2 L t p − 0 . 45 c 5 D D t 2 = 2.60 E D 1 2 600 t 200 − 0.45 5 200 200 t 2 = 6 200 2 . 60 30 × 10 ( 5 2 137.886t = 600 − 6.364t 5 2 1 2 ) 1 2 137.886t + 6.364t = 600 t = 1.791 in Page 2 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS say t = 1 982. 13 in 16 The same as 981, except that p = 175 psi , D = 4 ft , and the length of the tank is 18 ft. (a) Solving for the thickness of shell, t= pc D 2sy pc = 5 p = 5(175) = 875 psi s 55,000 sy = u = = 27,500 psi 2 2 D = 4 ft = 48 in p D (875)(48) t= c = = 1.53 in 2 s y 2(27,500 ) say t = 1 9 = 1.5625 in 16 (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5 t 2 2.60 E D pc = psi 1 L t 2 − 0.45 D D 5 t 2 1 2.60 E 2 L t D + 0.45 = D pc D E = 30×106 psi pc = 875 psi t = 1.5625 in D = 48 in 5 1.5625 2 1 2.60 30 ×10 L 48 1.5625 2 = + 0.45 48 875 48 L = 822 in ( 6 ) (c) Since length of shell = 18 ft = 216 in < 822 in, there is no need for stiffeners. Page 3 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS (d) L = 216 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5 t 2 2.60 E D pc = psi 1 2 L t − 0.45 D D 1 2 L t p − 0 . 45 c 5 D D t 2 = 2.60 E D 1 2 216 t 875 − 0.45 5 48 48 t 2 = 6 2.60 30 ×10 48 ( ) 5 2 4886.428t = 3937.5 − 56.833t 5 2 1 2 1 2 4886.428t + 56.833t = 3937.5 t = 0.9122 in 15 say t = in 16 19 but minimum t = 1 in 16 19 use t = 1 in 16 Approximate ratio of weight of this shell to the weight of the shell found in (a) = thickness of shell without stiffening rings / thickness of shell with stiffening rings = 0.34375 / 0.09375 = 3.6667 STEEL TUBES, EXTERNAL PRESSURE 983. A closed cylindrical tank, 6 ft in diameter, 10 ft long, is subjected to an internal pressure of 1 psi absolute. The atmospheric pressure on the outside is 14.7 psi. The material is equivalent to SA 30 (ASME Pressure Vessel Code: min su = 55 ksi ); assume an elastic limit of s y = su 2 ; let pc p = 5 . (a) What thickness of shell is needed for the specified design stress? (b) For this thickness, what must be the maximum length of unsupported section to insure against collapse? (c) Choose a symmetric spacing L of stiffening rings, and compute Page 4 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS their moment of inertia and the cross-sectional dimensions h and b if they are rectangular with h = 2b . (d) Suppose that the tank had no stiffening rings. What thickness of shell would be needed? What is the approximate ratio of the weight of the shell found in (a)? Material costs are roughly proportional to the weight. Solution: (a) Solving for the thickness of shell pD t= c 2sy pc = 5 p = 5(14.7 − 1) = 68.5 psi s 55,000 sy = u = = 27,500 psi 2 2 D = 6 ft = 72 in p D (68.5)(72 ) t= c = = 0.08967 in 2 s y 2(27,500 ) say t = 3 = 0.09375 in 32 (b) Solving for the maximum length of unsupported section, use Eq. (20-1) 5 t 2 2.60 E D pc = psi 1 2 L t − 0.45 D D 5 t 2 1 2.60 E 2 L t D + 0.45 = D pc D E = 30×106 psi pc = 68.5 psi t = 0.09375 in D = 72 in 5 0.09375 2 1 2.60 30 ×10 L 72 0.09375 2 = + 0.45 72 68.5 72 L = 6.2 in ( 6 ) (c) For the length of shell = 10 ft = 120 in. use L = 6 .0 in Page 5 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS Moment of inertia of stiffening rings. 0.035D 3 Lpc 0.035(72) (6 )(68.5) I= = = 0.1790 in 4 E 30 ×106 Solving for cross-sectional dimension h = 2b bh 3 8b 4 2b 4 I= = = = 0.1790 in 4 12 12 3 b = 0.72 in 3 say b = in 4 1 3 h = 2 = 1 in 2 4 3 (d) L = 10 ft = 120 in Solving for thickness without stiffening rings By Saunders and Windenburg, Eq. 20-1 5 t 2 2.60 E D pc = psi 1 L t 2 − 0.45 D D 1 2 L t pc − 0.45 5 D D t 2 = 2.60 E D 1 120 t 2 68 . 5 − 0 . 45 5 72 72 t 2 = 6 2.60(30 ×10 ) 72 5 1 25.8865t 2 = 1.667 − 0.053t 2 5 1 25.8865t 2 + 0.053t 2 = 1.667 t = 0.3314 in 11 say t = in = 0.34375 in 32 Page 6 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 984. A long lap-welded steel tube, 8-in. OD, is to withstand an external pressure of 120 psi. with N = 5 . (a) What should be the thickness of the wall of the tube? (b) What is the ratio D t ? Is it within the range of the Stewart equation? (c) Assuming the internal pressure to be negligible relative to the external pressure, calculate the maximum principal stress from equation (8.13), p. 255, text. What design factor is given by this stress compared with s y for AISI C1015 annealed? (d) Compute the stress from the thin shell formula. Solution: (a) Solving for the thickness of the wall Stewart’s formula 3 t pc = 50,200,000 D pc = Np = (5)(120) = 600 psi 3 t 600 = 50, 200,000 8 t = 0.1829 in 3 say t = in = 0.1875 in 16 D 8 = = 42.67 t 0.1875 or D = 42.67t outside the range of the Steward equation ( D < 40t ) (b) ratio (c) Using eq. (8.13) (Lame’s formula) pi ri 2 − po ro2 + ri 2 ro2 ( pi − po ) r 2 ro2 − ri 2 po = 120 psi pi ≈ 0 OD 8 ro = = = 4 in 2 2 ri = 4 − 0.1875 = 3.8125 in r = ri σt = − po ro2 + ro2 (0 − po ) − 2 po ro2 − 2(120)(4 ) = 2 2 = 2 = −2621 psi 2 2 ro − ri ro − ri (4 ) − (3.8125)2 s y of AISI C1015, annealed = 42 ksi 2 σt = Page 7 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS Design factor, N y = sy σt = 42,000 = 16 2621 (d) Solving for the stress from the thin shell formula s= pD (120)(8) = 2560 psi = 2t 2(0.1875) 985. A long lap-welded steel tube, 3 –in. OD, is to withstand an external pressure of 150 psi with N = 5 . Parts (a) – (c) are the same as in 984. Solution: (a) Solving for the thickness of the wall Stewart’s equation 3 t pc = 50,200,000 D pc = Np = (5)(150) = 750 psi 3 t 750 = 50, 200,000 3 t = 0.074 in 5 say t = in = 0.078125 in 64 D 3 = = 38.4 t 0.078125 or D = 38.4t within the range of the Steward equation ( D < 40t ) (b) ratio (c) Using eq. (8.13) (Lame’s formula) po = 150 psi OD 3 ro = = = 1.5 in 2 2 ri = 1.5 − 0.078125 = 1.421875 in − 2 po ro2 − 2(150 )(1.5) = = −2957 psi 2 2 ro − ri (1.5)2 − (1.421875)2 s y of AISI C1015, annealed = 42 ksi 2 σt = Design factor, N y = Page 8 of 25 sy σt = 42,000 = 14.2 2927 SECTION 18 – MISCELLANEOUS PROBLEMS 986. A long lap-welded tube, 3-in. OD, is made of SAE 1015, annealed. Let the shell thickness t = D 40 and N = 5 . (a) What is the corresponding safe external pressure? (b) Compute the maximum principal stress (p. 255, Text), assuming a negligible internal pressure. What design factor is given by this stress compared with s y ? (c) Compare with stress computed from the thin-shell formula. Solution: (a) Solving for safe external pressure, t pc = 50,200,000 D t 1 = D 40 3 3 1 pc = 50,200,000 = 784 psi 40 p 784 p= c = = 157 psi N 5 (b) Solving for maximum principal stress, neglecting internal pressure po = 157 psi OD 3 ro = = = 1.5 in 2 2 D 3 t= = = 0.075 in 40 40 ri = 1.5 − 0.075 = 1.425 in − 2 po ro2 − 2(157 )(1.5) = = −3221 psi 2 2 ro − ri (1.5)2 − (1.425)2 s y of SAE 1015 annealed = 42 ksi 2 σt = Design factor, N y = sy σt = 42,000 = 13.0 3221 (c) Solving for stress from the thin-shell formula s= pD (157 )(3) = = 3140 psi 2t 2(0.075) FLAT PLATES 987. A circular plate 24 in. in diameter and supported but not fixed at the edges, is subjected to a uniformly distributed load of 125 psi. The material is SAE 1020, as Page 9 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS rolled, and N = 2.5 based on the yield strength. Determine the thickness of the plate. Solution: Solving for the thickness of the plate 2 r s = p psi t for SAE 1020,a s rolled, s y = 48 ksi sy 48,000 = 19,200 psi N 2.5 p = 125 psi 24 r= = 12 in 2 2 12 s = 19,200 = 125 t t = 0.968 in say t = 1 in s= 988. = The cylinder head of a compressor is a circular cast-iron plate (ASTM class 20), mounted on a 12-in. cylinder in which the pressure is 250 psi. Assuming the head to be supported but not fixed at the edges, compute its thickness for N = 6 based on ultimate strength. Solution: Solving for the thickness of the head 2 r s = p psi t for cast-iron (ASTM class 20), su = 20 ksi s 20,000 s= u = = 3333 psi N 6 p = 250 psi 12 r= = 6 in 2 2 6 s = 3333 = 250 t t = 1.6432 in 21 say t = 1 in = 1.65625 in 32 Page 10 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 989. A 10x15-in. rectangular opening in the head of a pressure vessel, whose internal pressure is 175 psi, is covered with a flat plate of SAE 1015, annealed. Assuming the plate to be supported at the edges, compute its thickness for N = 6 based on ultimate strength. Solution: Solving for the thickness of the head a 2b 2 p s= 2 2 psi 2t (a + b 2 ) for SAE 1015, annealed, su = 56 ksi s 56,000 s= u = = 9333 psi N 6 a = 10 in b = 15 in p = 175 psi s = 9333 = (10 )2 (15)2 (175) psi 2 2 2t 2 [(10 ) + (15) ] t = 0.8056 in 13 say t = in = 0.8125 in 16 CAMS 990. The force between a 5/8-in. hardened steel roller and a cast-iron (140 BHN) cam is 100 lb.; radius of cam curvature at this point is 1 ¼ in. Compute the contact width. Solution: Solving for the contact width Kcb P= 1 1 N + r1 r2 From Table 20-2, hardened steel and a cast-iron, BHN = 400 Use K c2 = 900 P = 100 lb 15 r1 = = 0.3125 in 28 1 r2 = 1 = 1.25 in 4 N = 1.15 Page 11 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 900b 1 1 1.15 + 0.3125 1.25 b = 0.511 in 100 = A radial cam is to lift a roller follower 3 in. with harmonic motion during a 150o turn of the cam; 1 1.2-in. roller of hardened steel. The reciprocating parts weigh 10 lb., the spring force is 175 lb., the external force during the lift is 250 lb. The cast-iron (225 BHN) cam turns 175 rpm. The cam curvature at the point of maximum acceleration is 1 ½-in. radius. Compute the contact width. 991. Solution: Neglecting frictional forces Q + Fg + Fs + Fre = P cos φ Q = external force during lift = 250 lb Fg = weight of reciprocating parts = 10 lb Fs = spring force = 175 lb Fre = reversed effective force = − qma Fre = −2ma for harmonic motion 2 π L πω cos θ a = &x& = 2 β β L = 3 in 2π (175) ω= = 18.326 rad sec 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2 3 π (18.326) &x& = = 726 in sec 2 2 2.618 P cos φ = Q + Fg + Fs + Fre = 250 + 10 + 175 − 2m&x& 10 lb 10 lb − sec 2 = g (32.2 ft sec 2 )(12 in ft ) 386.4 in assume cosφ ≈ 1 10 P = 435 − 2 (726 ) = 397 lb 386.4 m= Fg = Page 12 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS Solving for the contact width Kcb P= 1 1 N + r1 r2 From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15 2100b 397 = 1 1 1.15 + 1. 5 1. 5 b = 0.29 in 992. The same as 991, except that the motion of the follower is cycloidal. Solution: Fre = −1.1ma for cycloidal motion As a continuation of 991, but 2π 2π Lω 2 a = &x& = sin θ for cycloidal 2 β β L = 3 in 2π (175) ω= = 18.326 rad sec 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2 2π Lω 2 2π (3)(18.326) &x& = = = 924 in sec 2 2 2 β (2.618) 10 P = 435 − 1.1 (924 ) = 409 lb 386.4 Solving for the contact width Kcb P= 1 1 N + r1 r2 From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100 Page 13 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15 2100b 409 = 1 1 1.15 + 1.5 1.5 b = 0.30 in 993. The same as 991, except that the motion of the follower is parabolic. Solution: Fre = −3ma for parabolic motion As a continuation of 991, but 2 2ω for parabolic a = &x& = L β L = 3 in 2π (175) ω= = 18.326 rad sec 60 150 β = 150o = π = 2.618 rad 180 at maximum acceleration 2 2(18.326 ) &x& = (3) = 588 in sec 2 2.618 10 P = 435 − 3 (588) = 389 lb 386.4 Solving for the contact width Kcb P= 1 1 N + r1 r2 From Table 20-2, hardened steel and a cast-iron, BHN = 225 Use K c2 = 2100 1 r1 = 1 in = 1.5 in 2 1 r2 = 1 in = 1.5 in 2 N = 1.15 Page 14 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 2100b 1 1 1.15 + 1. 5 1. 5 b = 0.284 in 389 = FLYWHEELS AND DISK 994. A cast-iron flywheel with a mean diameter of 36 in. changes speed from 400 rpm to 380 rpm while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation, the weight, and the approximate sectional area of the rim? Solution: Solving for coefficient of fluctuation n −n Cf = 1 2 n n1 + n2 n= 2 2(n1 − n2 ) Cf = n1 + n2 n1 = 400 rpm n2 = 380 rpm 2(400 − 380 ) Cf = = 0.0513 400 + 380 Solving for the weight 32.2∆KE w= C f vs2 ∆KE = 8000 ft − lb C f = 0.0513 vs = π Dn 12(60 ) D = 36 in 400 + 380 n= = 390 rpm 2 π (36 )(390) vs = = 61.26 fps 12(60 ) 32.2(8000 ) w= = 1338 lb 2 0.0513(61.26) Page 15 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS Solving for the approximate sectional area of the rim w = ρV assume ρ = 0.254 lb in 3 for cast iron V = π DA D = 36 in w = ρπ DA 1338 = 0.254π (36 ) A A = 36 in 2 995. The energy required to shear a 1-in. round bar is approximately 1000 ft-lb. In use, the shearing machine is expected to make a maximum of 40 cutting strokes a minute. The frictional losses should not exceed 15 % of the motor output. The shaft carrying the flywheel is to average 150 rpm. (a) What motor horsepower is required? (b) Assuming a size of flywheel and choosing appropriate C f , find the mass and sectional dimensions of the rim of a cast-iron flywheel. The width of the rim is to equal the depth and is not to exceed 3 ½ in. It would be safe to assume that all the work of shearing is supplied by the kinetic energy given up by the flywheel. Solution: (a) Solving for the horsepower required (Energy required )(Strokes per min ) (1 − Frictional losses )(33,000 ft − lb hp − min ) (1000)40 hp = = 1.426 hp (1 − 0.15)(33,000 ft − lb hp − min ) hp = (b) Solving for the mass of the rim and size of section 32.2∆KE C f vs2 w = ρV assume ρ = 0.254 lb in 3 for cast iron w= V = π DA w = ρπ DA = 32.2∆KE C f vs2 assume C f = 0.06 (Table 20-3) ∆KE = 1000 ft − lb Page 16 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS vs = π Dn 12(60 ) n = 150 rpm w = ρπ DA = D3 = 32.2∆KE 5π D Cf 24 32.2∆KE 2 2 5π Cf ρπ A 24 using width = depth = 3 ½ in A = (3.5)(3.5) = 12.25 in 2 32.2(1000 ) D3 = 2 5π 0.06 (0.254)(π )(12.25) 24 D = 50.42 in assume D = 51 in 5π ( 51) vs = = 33.38 fps 24 32.2(1000) w= = 482 lb 2 0.06(33.38) w 482 m= = = 15 slugs g 32.2 w 482 A = (depth )(width ) = = = 11.84 in 2 ρπ D 0.254π (51) depth = width = 11.84 in 2 = 3.44 in 1 say depth = width = 3 in 2 996. The same as 995, except that the capacity of the machine is such as to cut 1 ½-in. round brass rod, for which the energy required is about 400 ft-lb./sq. in. of section. Solution: (a) Solving for the horsepower required hp = (Energy required )(Strokes per min ) (1 − Frictional losses )(33,000 ft − lb hp − min ) Page 17 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS π 2 Energy required = 400 ft − lb in 2 (1.5 in ) = 707 ft − lb 4 (707)(40) hp = = 1.01 hp (1 − 0.15)(33,000 ft − lb hp − min ) ( ) (b) Solving for the mass of the rim and size of section 32.2∆KE D3 = 2 5π Cf ρπ A 24 ∆KE = 707 ft − lb C f = 0.06 ρ = 0.254 lb in 3 A = (3.5)(3.5) = 12.25 in 2 32.2(707 ) D3 = 2 5π 0.06 (0.254)(π )(12.25) 24 D = 45 in use D = 45 in 32.2∆KE w= C f vs2 5π D 5π (45) = = 29.45 fps 24 24 32.2(707 ) w= = 438 lb 2 0.06(29.45) w 438 m= = = 13.6 slugs g 32.2 w 438 A = (depth )(width ) = = = 12.20 in 2 ρπ D 0.254π (45) vs = depth = width = 12.20 in 2 = 3.49 in 1 say depth = width = 3 in 2 997. A 75-hp Diesel engine, running at 517 rpm, has a maximum variation of output of energy of 3730 ft-lb. The engine has three 8 x 10 ½ in. cylinders and is directly connected to an a-c generator. (a) What should be the weight and sectional area of the flywheel rim if it has an outside diameter of 48-in.? (b) The actual flywheel and generator have Wk 2 = 6787 lb − ft 2 . Compute the corresponding coefficient of fluctuation and compare. Page 18 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS Solution: (a) Solving for the weight ad sectional areas w= 32.2∆KE lb C f vs2 assume C f = 0.0035 , Table 20-3 ∆KE = 3730 ft − lb π Dm n vs = 12(60) n = 570 rpm Dm = Do − t = 48 − t π (48 − t )(517 ) vs = = 2.25584(48 − t ) 12(60) 32.2(3730) 6,743,418 w= = (0.0035)[2.25584(48 − t )]2 (48 − t )2 w = ρπ DA assume ρ = 0.254 lb in 3 (cast iron), b = 2t w A= , A = bt = 2t 2 ρπ D 0.254π (48 − t )(2t )(t ) = 6,743,418 (48 − t )2 t 2 (48 − t ) = 4,225,387 t = 8.18 in Then w= 6,743,418 = 4253 lb (48 − 8.18)2 A = 2t 2 = 2(8.18)2 = 133.8 in 2 (b) Solving for coefficient of fluctuation Ig = Wk 2 = 6787 lb − ft 2 ( ) I ω12 − ω22 ∆KE = ft − lb 2 Wk 2 (ω12 − ω22 ) ∆KE = ft − lb g 2 Wk 2 (ω1 − ω2 ) (ω1 + ω2 ) ft − lb ∆KE = g 2 (ω1 + ω2 ) = ω 2 Page 19 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 2π (517 ) = 54.14 rad sec 60 ∆KE = 3730 ft − lb 6787 ∆KE = 3730 = (ω1 − ω2 )54.14 32.2 ω1 − ω 2 = 0.327 ω − ω 2 0.327 Cf = 1 = = 0.006 > 0.0035 ω 54.14 ω= 998. A 4-ft flywheel, with an rim 4 in. thick and 3 in. wide, rotates at 400 rpm. If there are 6 arms, what is the approximate stress in the rim? Is this a safe stress? At what maximum speed should this flywheel rotate if it is made of cast iron, class 30? Solution: Solving for the approximate stress, v2 ρ s= s psi 144 g o π Dn π (4 )(400 ) vs = = = 83.78 fps 60 60 ρ = 0.254 lb in 3 (class 30, cast iron) ρ = 0.254 lb in 3 = (0.254)(1728) lb ft 3 = 439 lb ft 3 g o = 32.2 ft s 2 (83.78)2 (439) = 665 psi 144(32.2 ) since v = (83.78)(60) fpm = 5027 s= fpm < 6000 fpm (cast iron) this is a safe stress Solving for maximum speed, max. v = 6000 fpm v = π Dn = π (4 )n = 6000 fpm maximum, n = 477 rpm 999. A hollow steel shaft with Do = 6 in and Di = 3 in rotates at 10,000 rpm. (a) What is the maximum stress in the shaft due to rotation? Will this stress materially affect the strength of the shaft? (b) The same as (a), except that the shaft is solid. Solution: (a) Solving for maximum stress Page 20 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS st = ρω 2 [(3 + µ )r 2 + (1 − µ )ri 2 ] psi 4 go where D 6 ro = o = = 3 in 2 2 D 3 ri = i = = 1.5 in 2 2 for steel, ρ = 0.284 lb in 3 , µ = 0.30 o g o = 386 in sec 10,000 = 1047 rad sec 60 ω = 2π st = ρω 2 4 go [(3 + µ )r 2 o + (1 − µ )ri 2 ] psi 2 ( 0.284)(1047 ) [ (3 + 0.3)(3)2 + (1 − 0.3)(1.5)2 ] = 6306 psi st = 4(386 ) (does not affect the strength of the shaft) (b) Solving for the maximum stress for solid st = ρω 2 (3 + µ )ro 2 2 2 ( 0.284 )(1047 ) (3 + 0.3)(3) = 4(386 ) 4 go 1000. A circular steel disk has an outside diameter Do = 10 in and an inside diameter Di = 2 in . Compute the maximum stress for a speed of (a) 10,000 rpm, (b) 20,000 rpm. (c) What will be the maximum speed without danger of permanent deformation if the material is AISI 3150, OQT at 1000 F? Solution: st = = 2994 psi ρω 2 [(3 + µ )r o 2 4 go where ρ = 0.284 lb in 3 µ = 0.30 D 10 ro = o = = 5 in 2 2 D 2 ri = i = = 1 in 2 2 g o = 386 in sec Page 21 of 25 + (1 − µ )ri 2 ] psi SECTION 18 – MISCELLANEOUS PROBLEMS (a) Solving for maximum stress for a speed of 10,000 rpm 10,000 ω = 2π = 1047 rad sec 60 2 ( 0.284)(1047 ) (3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 16,776 psi st = 4(386 ) [ ] (b) Solving for maximum stress for a speed of 10,000 rpm 20,000 ω = 2π = 2094 rad sec 60 2 ( 0.284)(2094) (3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 67,104 psi st = 4(386) [ ] (c) Solving for maximum speed, ω . For AISI 3150, OQT at 1000 F, s y = 130 ksi st = s y = ρω 2 4go [(3 + µ )r o 2 + (1 − µ )ri 2 ] psi ( 0.284 )ω 2 [ (3 + 0.3)(5)2 + (1 − 0.3)(1)2 ] 130,000 = 4(386) ω = 2914.57 rad sec 60ω 60(2914.57 ) RPM = = = 27,832 rpm 2π 2π 1001. The same as 1000, except that Di = 1 in . Solution: Di = 1 in D 1 ri = i = in = 0.5 in 2 2 use other data as in 1000. (a) Solving for maximum stress for a speed of 10,000 rpm 10,000 ω = 2π = 1047 rad sec 60 2 ( 0.284)(1047 ) (3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 16,670 psi st = 4(386 ) [ ] (b) Solving for maximum stress for a speed of 10,000 rpm Page 22 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS 20,000 = 2094 rad sec 60 2 ( 0.284)(2094) (3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 66,680 psi st = 4(386) ω = 2π [ ] (c) Solving for maximum speed, ω . For AISI 3150, OQT at 1000 F, s y = 130 ksi st = s y = ρω 2 4go 130,000 = [(3 + µ )r o 2 + (1 − µ )ri 2 ] psi (0.284)ω 2 [(3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 ] 4(386 ) ω = 2923.8 rad sec 60ω 60(2923.8) RPM = = = 27,920 rpm 2π 2π 1002. A circular steel disk, with Do = 8 in and Di = 2 in , is shrunk onto a solid steel shaft with an interference of metal i = 0.002 in . (a) At what speed will the pressure in the fit become zero as a result of the rotation? Assume that the shaft is unaffected by centrifugal action. (This effect is relatively small.) (b) Compute the maximum stress in the disk and the pressure at the interface when the speed is 10,000 rpm. Note: The maximum stress in the disk is obtained by adding equations (8.15) of i8.26, Text, and (n of i20.9. The resulting equation together with equation (s) of i8.27 can then be used to obtain pi and σ th ; where σ ts = − pi for a solid shaft. Solution: (a) Solving for speed, pi = 0 From equation 8.15, i8.26, Text. σ + µ h pi σ ts + µ s pi i = Di th − Eh Es Di = 2 in i = 0.002 in Eh = Es = 30 × 10 6 psi µ h = µ s = 0.30 σ ts = − pi = 0 Page 23 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS σ + (0.3)(0 ) 0 + (0.3)(0) i = 0.002 = 2 th − 6 30 ×10 6 30 ×10 σ th = 30,000 psi σ th = σ ti + st From Equation 8-15 pi (ro2 + ri 2 ) − 2 po ro2 σ ti = ro2 − ri 2 but po = 0 pi (ro2 + ri 2 ) ro2 − ri 2 D 8 ro = o = = 4 in 2 2 Di 2 ri = = = 1 in 2 2 pi = 0 σ ti = 0 σ ti = From Equation (n) i20.9 st = ρω 2 4 go [(3 + µ )r o 2 + (1 − µ )ri 2 ] psi ρ = 0.284 lb in 3 µ = 0.30 ro = 4 in ri = 1 in g o = 386 in sec σ th = σ ti + st 30,000 = 0 + (0.284)ω 2 [(3 + 0.3)(4)2 + (1 − 0.3)(1)2 ] 4(386) ω = 1746 rad sec 60ω 60(1746 ) RPM = = = 16,673 rpm 2π 2π (b) Solving for the maximum stress in the disk and the pressure within the interface. σ + µ h pi σ ts + µ s pi i = Di th − Eh Es σ + 0.3 pi (− pi + 0.3 pi ) 0.002 = 2 th − 6 30 ×10 30 ×10 30,000 = σ th + pi Page 24 of 25 SECTION 18 – MISCELLANEOUS PROBLEMS σ th = 30,000 − pi σ th = σ ti + st σ ti = st = 4 2 + 12 17 pi (ro2 + ri 2 ) = = p pi i 2 2 ro2 − ri 2 4 − 1 15 ρω 2 [(3 + µ )r 2 + (1 − µ )ri 2 ] psi 4 go 2π (10,000) ω= = 1047 rad sec 60 (0.284)(1047 )2 (3 + 0.3)(4)2 + (1 − 0.3)(1)2 = 10,788 psi st = 4(386 ) σ th = σ ti + st 17 σ th = 30,000 − pi = pi + 10,788 15 pi = 9000 psi (interface pressure) σ th = 30,000 − pi = 30,000 − 9000 = 21,000 psi (maximum stress) o [ ] - end - Page 25 of 25