ch07 - SOLUTION (7.1) Known: A 950—lb gin pole is...

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Unformatted text preview: SOLUTION (7.1) Known: A 950—lb gin pole is supported by a stranded steel cable with known cross section and modulus of elasticity. As the gin pole is descending at a constant velocity, an accident causes the top of the cable, 70 ft above the gin pole, to suddenly stop. Find: Estimate the maximum elongation and maximum tensile stress developed in the cable. Schematic and Given Data: Steel cable 2 A = 0.110 in. E=12x106psi Assumptions: The mass of the cable is negligible. 1. 2. Neglect any stress concentrations. 3. Ignore damping due to internal friction within the cable. 4. The cable responds elastically to the impact. 5. The cable is attached at the top of the gin pole (L = 70 feet). Analysis: Solution 1 — Using Egs. 17.1a) 1. 5:5st 1+ 1+ Vz gést 2. 5m: (0.60 in.) = (0.60 in.)(2.07) = 1.24 in. 1+’\/1 + in/slz [386 in/SZHO .60 in) 3. Impact factor = amax/Sst = 124/060 = 2.07 4_ cm = Fm/A = (950 1b)(2.07)/0. 11 111.2 = 17.9 ksi I 7—1 olution 2 — Usin E . 7. a 1. 5 = 5st wags“ = (0.60 in. (6)2/386(0.60) = 0.237 in. But this is due to the accident only. Adding the pre-existing deflection of 0.60 in. gives a total Smax = 0.837 in. I . 2 sin/s)2 2. U E. 7.4 ,F =W4/l—=9504/———(——-———————=3751b mg q( a) e g5g ( ) (386 in/s2)(060 in.) 3. cm=W+Fe=2§0—fl7—5—= 12.0 ksi A 0.11 I Solution 3 - Usin Basic Ener Relationshi s 1. The rope stretch due to impact involves converting the kinetic energy of the elevator mass to elastic energy in the cable: i 2 = 1 2 2mv 51(5 6 __13_A_E_(0.11)(12x10)_ . where k— 5 — L —————————————840 - 1,5711b/1n. 2. Using Eq. (7.3b), 5 = ZU— where U = 1/2 mv2 k “"2 l I _ mv _ (950/386)(6) _ . 8_,\/—k _ —————1571 —O.2371n. 3. Pre—accident equilibrium deflection of 0.60 in. is added to give 5max = 0.837 in. Hence, Gmax = 12.0 ksi (as in Solution 2) I Solution 4 - Usin Differential E uation of Mass M ti n 1, )3va = 0: kx + m = 0 (neglecting damping, and also the gravity force which is in equilibrium with the pre-accident cable force; and where x is measured from the equilibrium point.) 2. The solution is x = x0 sin Vk/m t where Xmax = x0 ilk/m. This is known to be 6 in./s. 1,571 lb/in. 3. Therefore, 6 = x0,\/————————_————2 :x0= 0.237 in. 9501b/386 1n/s 4. Adding 0.60 gives 511m = 0.837 in., and omax = 12.0 ksi as in Solution 2. I Comment: Comparison of Solution 1 with Solutions 2, 3, and 4 reveals that the static deflection is not significant compared to the total deflection. 7-2 SOLUTION (7.2) Known: A 5-ton elevator is supported by a stranded steel cable with known cross section and modulus of elasticity. As the elevator is descending at a constant velocity, an accident causes the top of the cable, 70 ft above the elevator, to suddenly stop. Find: Estimate the maximum elongation and maximum tensile stress developed in the cable. Schematic and Given Data: Steel cable A = 2.5 in. 2 E=12x106psi Assumptions: 1 . The mass of the cable is negligible. 2. Neglect any stress concentrations. 3. Ignore damping due to internal friction within the cable. 4. The cable responds elastically to the impact. Analysis: Solution 1 - Using Egs. (7.1a) 1+1/1+—V2—) gfist where Sgt = i = —————————(10’000 1b) (840 m') = 0.28 in. AB (25 in.2)(12 x 106 psi) 1. 5:5SI 1+ 1 +_fl)fl/s__ ) = (0.28 in.)(8.76) = 2.45 in. 2. = .2 ' - Smax (0 81m) (386 in./sZX0.28 in.) 3. Impact factor = Sum/851 = 245/028 = 8.75 4. Gum = Fmax/A = (5 tons)(2000 lb/ton)(8.75)/2.5 in.2 = 35.0 ksi I Solution 2 — Using Eg. 17.3a2 1. 5 = Est v 2/g5st = (0.28 in.)1/ (80)2/386(0.28) = 2.155 in. But this is due to the accident only. Adding the pre-existing deflection of 0.28 in. gives a total Smax = 2.435 in. I . 80 in./s) 2. 13.7.4 ,F=W1/l2—=10,000 4/————(——_————=7,901 Usmg q ( a) e gsst ( ) (386 in./s2)(0.28 in.) 6 5 b W+F 10000+76950 . . = ———e— = ~—’——-—————’——— = 4. 3 (Sm.1X A 2.5 3 8 k51 I S lution 3 - Usin Basic Ener Relationshi s 1. The rope stretch due to impact involves converting the kinetic energy of the elevator mass to elastic energy in the cable: %mv2 = éekSZ 2. 2 6 where k = g = % = LXEIMOIE = 35,7141b/in. 2. Using Eq. (7.3b), 6 = Z—kll where U = 1/2 my2 ___ mvz ____ (10,000/386)(80) z . 8 2‘] k 4/———————35,714 2.1551n. 3. Pre-accident equilibrium deflection of 0.28 in. is added to give 5am = 2.435 in. Hence, omax = 34.8 ksi (as in Solution 2) I Solution 4 — Usin Differential E nation of Mass Motion 1. Eva = 0: kx + m = O (neglecting damping, and also the gravity force which is in equilibrium with the pre—accident cable force; and where x is measured from the equilibrium point.) 2. The solution is x = x0 sin ilk/m t where km“ = x0 ilk/m. This is known to be 80 in./s. ' 35 714 1b/in. . 3. Theefor,8 11L: ——’———————: =2.155 . I e 0 S X" ‘V 10,000 lb/386 in/s2 X" m 4. Adding 0.28 gives fimax = 2.435 in., and O'max = 34.8 ksi as in Solution 2. I 74 SOLUTION (7.3) Known: A tensile impact bar is fractured in service. A new bar is made exactly like the old one except that the middle third is enlarged to twice the diameter of the ends. Find: Compare the impact capacities of the new and old bars Schematic and Given Data: Assumptions: 1. Neglect stress concentrations. 2. The static load capacities of the old and the new bars are the same. 3. The mass of the bars are negligible. 4. Ignore damping due to internal friction. Analysis: Neglecting stress concentration, the load capacities of the old and the new bars are the same. 7-5 Solution 1. l. The corresponding deflections are: = PIE?) 5 = EL + PL + LL. = 2 LL. AE AE 4AE AB 4 AB 2. The new bar deflects only 3/4 as much as the old bar. Energy oc (load capacity) (deflection). Hence, the new bar can absorb 3/4 times the energy of the old bar. I Solution 2. l. The elastic capacity for the old bar is determined directly from the Eq. (7.5a), where O' = Sy: 82 V UO = y 1d 2B 2. For the new bar, the energy absorbed is equal to the energy absorbed by the middle section plus the energy absorbed by the two end sections. The end sections are weakest and can be loaded to a stress of Sy. Their volume is (2W3) where V is the volume of the old rod. The energy capacity of the two end sections is g (8%)(V) 2E Uends = = Z 3 Uold The middle section has four times the cross sectional area as an end section; hence, U _ (Sy/4)2(4V/3) _ (33W) _ 1 U Mddle-T—W—‘E Old The total energy capacity is Uends plus Umidcne: Unew = + Uold = % Uold Hence, the new bar can absorb 3/4 times the energy of the old bar. Comment: The stress concentrations at the changes in section of the stepped bar would further reduce its impact capacity and would tend to promote brittle fracture at a step. 7-6 SOLUTION (7.4) Known: A vertical rod is subjected to an axial impact by a 100 lb weight dropped from a height of 2 ft. The rod is made of steel, with Sy = 45 ksi and E = 30 X 106 psr. Find: The length of the member to avoid yielding for a diameter of (a) l in., (b) 1.5 in, and (c) 1 in. for half of its length, and 1.5 in for the other half. Schematic and Given Data: Assumptions: 1. The mass of the members are negligible. 2. Neglect any stress concentrations. Analysis: Neglecting stress concentration, the load capacities of the old and the new bars are the same. ' Area = 0.785 in2 Area = 1.77 in2 7 1 d d: 1.0 in E 7—7 (3) Diameter is 1.0 inches 1. For a rod with uniform axial stress, 6 = 23E, or 6 V : 2U2E z <2><100)(24)<3g x 10 > z 71 my 0 45,000 V 71 - 2. L=—~=—=90.5 1n. 1: I A /4 . (b) Diameter is 1.5 inches 1. V = 71 in.3 2. L = ill—>92— : 40 in. I (n)(1.5>2 (c) Diameter of 1.0 inches and 1.5 inches 1. Let Vs = small end volume. (Large end volume = (1.5)2 V3) 2. Small end stress = Sy = 45,000 psi 45,000 (1.5)2 3. Large end stress = V563, Vs(45,000)2 2E (2x30 x 106) 4. Small end energy = Us = 2 62 (2.25VS)(4:’3(5)0) 5. Large end energy = UL = L L = 25 (2)00 x 10 ) 2 6. U=U5+UL=2400in. 1b. =w 1+4.) (2)(30 x 10 ) 2-25 7. Therefore, Vs = 49.1 in.3 8. L3 = 34734— = 62.6 in. /4 9. Therefore, L = 125.2 in. I Comment: Note physically why, if the "excess diameter" were machined off, the rod could be shortened from 125 inches to 90 inches. * This assumes a deflection which is negligible in comparison with the 24 in. drop. 7-8 SOLUTION (7.5) Known: A rescue car, of 1400 kg mass, attempted to jerk a stuck vehicle back onto a road using a S-m steel tow cable of stiffness k = 5000 N/mm. The rescue car reached a speed of 4 km/hr at the instant the cable became taut. Find: Estimate the maximum impact force and resulting elongation that develops in the cable. Schematic and Given Data: Rope k=5000N/ “Hm/hf L=5m mm m=1400kg "/’/////////l””' Assumptions: The cable is attached rigidly to the masses of the cars. Ignore the mass of the rope. Neglect any stress concentrations. Ignore damping due to internal friction within the rope. The rope responds to the impact elastically. MAWNH Analysis: 1. From Eq. (7.4a), Fe = W's/VZ/gSSt where isst = % W = gravity force = mg = (1400 kg)(9.8l III/$2) = 13,734 N Therefore, 551: 13’734 N = 2.75 m; V = 4km = 1.11% 5000 N/mm hr _ (1.11m/s) _ 2. Fe—(13,734N)/\/——.—-—(9'81m/SZXO00275 m) —(13,734)(6.76)N 3. Maximum impact force = 6.76 times the gravity weight of the car, or 92.9 kN I - ___ 92.9kN = , 4. Cable elongation SkN/mm 18.6 m I 7-9 SOLUTION (7.6) Known: A rescue car attempted to jerk a stuck vehicle back onto a road using a 12 m elastic cable of stiffness 2.4 N/mm. The rescue car was able to reach 12 km/hr at the point of becoming taut. Find: Estimate the impact force developed and the resulting cable elongation. Determine the energy stored in the cable. State the warning that you suggest be provided with elastic cables sold for this purpose. Schematic and Given Data: Assumptions: 1. The cable is attached rigidly to the masses of the cars. 2. Ignore the mass of the rope. 3. Neglect any stress concentrations. 4. Ignore damping due to internal friction within the rope. 5. The rope responds to the impact elastically. 6. The stuck vehicle does not move significantly until the rescue car has just come to a stop. Analysis: 1. From Eq. (7.4a), Fe = WV v2/g Sgt where 5st = W/k W = gravity force = mg = (1400)(9.81) = 13,734 N Sgt — —————13734 N = 5723 mm _ 2.4 N/rnm v = lzhlfm = 3.33 m/s Fe = (13,734 N)4 / fl_— = (13,734)(0.44) = 6043 N (9.81)(5.723) Maximum impact force is 6043 N I Cable elongation = = 2518 m z 2.5 m I 2. Energy = (average force)(deflection) = (60"? N) (2.5 m) = 7554 Nm I Gravity force on a 100 kg mass is (100 kg)(9.81 III/$2) = 981 N 7-10 Hence, (7554 N-m)/(981 N) = 7.7 m is the equivalent free fall. Comment: Note that in Problem 7.5, the added traction force with half the weight on the driving wheels and a coefficient of friction (on ice) of 0.1 would be (13734/2)(O.1) = 687 N. This would add negligibly in Problem 7.5, but would be significant in Problem 7.6. Also, in Problem 7.6, the extra energy of adding a 687 N force throughout the distance of rope-stretch would make a further significant addition. Thus the possibility of storing considerably more than 7554 N-m of energy exists. Warrnup should include ( 1) attach ends securely to appropriate structures on each car, (2) keep all personnel well away from the rope when in use, and (3) examine rope and do not use it if weakened by damage. I SOLUTION (7.7) Known: A 6000 lb tow truck, attempted to jerk a wrecked vehicle back onto a road using a 15 foot length of steel tow cable 1 in. in diameter (E = 12 x 106 psi of the cable). The tow truck reached a speed of 3 mph at the instant the cable became taut, but the wrecked vehicle does not move. The cable breaks in the middle and the two 7.5 foot halves are connected in parallel for a second attempt. Find: (a) Estimate the maximum impact force applied to the wrecked car and the stress produced in the cable. (b) Estimate the impact force and the cable stress developed in the second attempt if the wrecked vehicle still remains fixed. Schematic and Given Data: Rope d=1.0in. E =12 x 10 psi L =15 feet Assumptions: 1. The cable is attached rigidly to the masses of the cars. 2. Ignore the mass of the rope. 3. Neglect any stress concentrations. 4. Ignore damping due to internal friction within the rope. 5. The rope responds to the impact elastically. Analysis: (a) L = 15 feet 1. From Eq. (7.4a), Fe: Wa/VZ/gSg where v = 3 mph = 52.8 in./sec 2. 55t=i=w=01145m AB (.785)(12x106) 7-11 _ Z / (52.8) __ __ 3. Fe — (6000 —————(386)(0.1 145) — (6000)(7.94) — 47,600 lb I 4. G=§£=W=606ksi I A 0.785 111.2 (b) L = 7.5 feet (two parallel 1 in. cables) 1. 53: is decreased by 4, therefore Fe is increased by 2. Thus, the impact force is 95,200 lb. I 2. Both Fe and A are doubled; hence, Ge remains at 60.6 ksi. I Comment: Note that in (a) and (b), the same volume of equally stressed material absorbs the same amount of energy. SOLUTION (7.8) Known: A 100 lb weight falls from a known height onto an aluminum beam of known geometry and material properties. The beam is supported by two springs. Find: Determine the maximum beam stress and deflection. Schematic and Given Data: 1.0 in x 1.0 in aluminum bar 100 lb/in. 100 lb/in. 30 in. 60 in. Assumptions: 1. The masses of the beam and spring can be neglected. 2. The beam and springs respond elastically. 3. The impact load is applied uniformly at the center of the beam. Analysis: 1. The static deflection for the beam only, supporting springs only, and total system are respectively: = PL3 55‘(bem) 48EI where E = 10.4 Mpsi (Appendix 01) I = bh3/12 = 1.0(1.0)3/12 = 0.0833 in.4 7—12 100(60)3 55t(beam) = —-———-——————— = 0.519 in. 48(10.4x106)(0.0833) 55t(springs) - P 100 - 0.50 in. ‘ 5E = 2000) ‘ 235.com) = 0.519 + 0.50 = 1.019 in. 2. From Eq. (7.1) or (7.2) the impact factor is / 2(12) 1 1 21-h = 1 1 : . +1/ +6“ + +1_019 596 3. Hence, the total impact deflection is (1.019)(5.96) = 6.07 in. The deflection of the beam is (0.519)(5.96) = 3.09 in. 4. The extreme—fiber beam stress is estimated from Fe = (100)(5.96) = 596 lb as =M=§£ h : =0.0833= 1 -3 0' Z 42 w ereZ I/c (1.0/2) 0. 67 1n. __ 596(60) _ . (S — 4(0-167) — 53.5 k51 I Comments: 1. From Appendix C—2, Aluminum alloys §__,,gksiz figksi) 1 100—0 12 4.5 2024—T4 65 43 7055—0 34 14.3 7075—T6 86 78 Among the above alloys, only 7075—T6 will not yield with the stress of 53.5 ksi. 2. If the weight of the beam is considered in the analysis, the impact factor from 2nh 55‘ where Burr, Mechanical Analysis and Design, Elsevier, 1982, is 1 + 1 + n is the correction factor. Also from Burr, for impact at the mid—length of a simply supported beam, 11 = m where m = total mass of the beam, m = mass of falling object. mb = Vp where p = 0.1 lb/in.3 for aluminum (Appendix C-l). mb = (1)(1)(60)(0.1) = 6 lb. 1 = 0.972 Thus’ " = 1 + (17/35)(6/100) 7—13 The impact factor is 1 +1/ 1 + W = 5.89 which gives 6 = 52.9 ksi. Neglecting the weight of the beam, we found that 0' = 53.5 ksi. Thus, the effect of including the weight of the aluminum beam in the analysis is small. SOLUTION (7.9) Known: The vertical drive shaft is 20 mm in diameter, 650 mm long, and made of steel. The motor is equivalent to a steel flywheel 300 mm in diameter and 25 mm thick. When the vertical shaft is rotating at 3000 rpm, the propeller strikes a heavy obstruction, bring it to a Virtually instantaneous stop. Find: Calculate the elastic torsional shear stress in the vertical shaft. Schematic and Given Data: 300 mm dia. 20 mm dia. 3000 rpm Assumptions: 1. The weight of the shaft may be neglected. 2. The shaft acts as a torsional spring and responds elastically to the impact. 3. The short horizontal propeller shaft and the bevel gears have negligible flexibility. Analysis: 1. For "flywheel", U = (1/2) 1(02 where I = (1/2) mr2 and m = mztp where p = 7700 kg/m3 (Appendix C-l). Thus, U = (1/4)1tr4tp(o2 2 = (1/4) 1t(0.150 m)4 (0.025 m) (7700 £53 1%9) = 7554 N-m 2. From Eq. (7.6): r; = UTE}. where G = 79 GPa (Appendix C—l) T _ 2 / (7554 N m)(79 x 109 N/mz) _ (0.010)21t(0.650) T=3.4X109Pa=3.4GPa I SOLUTION (7.10) Known: Given a tensile bar with known dimensions. Find: Estimate the ratio of impact energy that can be absorbed with and without the notch which reduces the diameter to 24 mm. Schematic and Given Data: Assumption: The tensile bar material exhibits brittle behavior. Analysis: 1. Points a, b, and c are as noted in the above diagram. 2. Let S = allowable stress, or material strength. Let 6n = stress in main body-notched case. Let cu = stress in main body- unnotched case. 3. At critical point a, 0' = S, for unnotched case. At critical point b, 6 = S, for the notched case. At critical point c, cu = S/ 1.55 = .658 =§ 24mm =016S 0“ 4(30mm ' U = ‘é—gz [Eq. (7.5a)] Un _ g§__ (.168)2 Therefore, Uu — 63 - (.653? U H —“ = . ence, Uu 0 06 I SOLUTION (7.11) Known: A platform is suspended by long steel rods. The steel rod geometry is modified to obtain greater energy absorbing capacity. Find: (a) Determine the smallest effective threaded section area A that would provide maximum energy absorbing capacity for a new design. (b) Using this value of A, determine the increase in energy absorbing capacity that would be provided by the new rod design. Schematic and Given Data: KThread: Area = 600 m K = 3.9 Area = 800 mm2 (a) Original design (b) New design Assumption: Under impact loading, the rod material exhibits brittle behavior. 7—16 Analysis: Allowable stress here = S; then stress here =(3—t°’§)l% = 0.19238 Allowable stress at these two places = S; then stress here = 8/13 = 0.7698 (a) Original design (b) New design 1. To balance stress at thread and fillet, A = 800 = 1600 mm2 I 2 . 2. g? = g; = = 16: An increase of 1500% (the new design has 16 times the capacity of the original). I 7-17 SOLUTION (7.12) Known: A bolt has been fractured. A proposed redesign of the bolt involves drilling an axial hole in the unthreaded portion, and incorporating a larger fillet radius under the bolt head. Find: (a) Determine the theoretical optimum diameter of the drilled hole. (b) Using this hole size, calculate the factor by which the modifications increase the energy absorbing capacity of the bolt. Schematic and Given Data: Hole dia., d Fracture ‘ : location Assume that hole 5 drilled to this depth does not significantly change the K = 3.8 factor of the thread. (negligible) (3) Original design (b) Modified design Assumption: The bolt material exhibits brittle behavior. 7—18 Analysis: Fillet, K = 1.5 C: S , Thread, K = 3.8 m S (3) Original design (b) Modified design 1 . Drill hole to make 0' at fillet = 0' at thread P _ P — (3.8) _ 75—— (1.5) 800 Z(362 — d2) Hence, (1 = 29.9 m. I _ 02v . g: GM 2. U— 2E [Eq. (7.5a)] . Ua Cgva In Fig. P7.12(a), stress in the large center volume of presumably uniformly ‘ ' =—S— 800 =0207s s '11 =i=0667s stressedmaterialis Ga 3‘8(n(18)2) . . irniary,6t> 1'5 . The large center volume in (a) is 2179 (36)2(250) The large center volume in (b) is % (362 - 29.92)(250) Hence fi=w= 0.310 ’ Va 36?- g = (0.6673)2 _ Ua (0.2078)2 The answer is by a factor of 3.22 I 0.310 = 3.22 7—19 SOLUTION (7.13) Known: A tensile impact bar has a small transverse hole. Find: Determine the factor by which the hole reduces the impact energy absorbing capacity of the bar. Schematic and Given Data: 0.1 in.—dia.-hole Assumption: Under impact loading, the rod material exhibits brittle behavior. Analysis: 1. Drilling the "very long" rod will not significantly reduce the volume of uniformly stressed material, but it will reduce the stress, 0, to which this material can be subjected. 2. (S (for rod without hole) = T55 = g = 0.58, where K = 2 at the fillet. o (for rod with hole) = % (*mgesv where Kh is the stress concentration at hole 2 2.7 (Fig. 4.37) E — 0 1) =_S____._.(4 ' = ch 27 E 0.323 s 4 3. From Eq. (7.5a), U = , _ (0.5 S)2 _ (0.323 S)2 U ‘ 2E V’ Uh ‘ 2E V 7-20 2 U — (0‘5 S) — 2.4, or Uh: _ _ ____ _ L Uh (0.323 S)2 2.4 4. The hole reduces the impact energy capacity of the bar by a factor of 2.4. I SOLUTION (7.14D) Known: The bolt shown in Figure P7.12(a) is to be redesigned. Find: Redesign the bolt in Figure P7.12(a) to increase the energy absorbing capacity by a factor of three or more. Schematic and Given Data: Fracture 5 location Assume that hole ' 5 drilled to this depth does not significantly change the K = 3.8 factor of the thread. (negligible) (:1) Original design (b) Modified design Assumption: The bolt material exhibits brittle behavior. Decisions: 1 . Increase the radius under the bolt head changing the stress concentration from K = 2.2 to K: 1.5. 2. Drill an axial hole in the unthreaded portion of the bolt to make the stress at the fillet equal to the stress at the thread. 7-21 Analysis: Fillet, K = 1.5 G: S / Stress, S (21) Original design (b) Modified design 1. Drill hole to make 6 at fillet = O at thread J’— = ___.£____ 800 (3.8) £862 _ d2) (1.5) 4 Hence, (1 = 29.9 m. E = GfiVb _ 62V . 2. U ———2E [Eq. (7.5a)] . Ua Gama In Fig. P7.12(a), stress in the large center volume of presumably uniformly ‘ ‘ =—S— 800 =0207s S' '11 =_S—=0667s stressedmaterlalisca 3.8(n(18)2) . . imiary,0'b 1-5 . The large center volume in (a) is g (36)2(250) The large center volume in (b) is % (362 - 29.92)(250) Hence, X9 = M = 0.310 Va ER : (0.6673)2 Ua W' = The answer is by a factor of 3.22 I 7-22 Comment: 1. The energy absorbing capacity is increased by a factor of 3.22 because of the design decisions. 2. Clearly this answer is only one of several possible solutions. SOLUTION (7.15D) Known: A plain tensile impact bar has a one-inch diameter and stress concentration of 1.5 at each end. Find: Redesign the impact bar to reduce the impact energy absorbing capacity of the bar by a factor of 2 or more. Schematic and Given Data: Decisions: Drill a 0.1 in-diameter home transverse to the axis. Assumption: 1. Under impact loading, the rod material exhibits brittle behavior. 2. The bar is "very long". Analysis: 1. cs (for rod without hole) = %= i = 0.678, where K = 1.5 at the fillet. 1.5 where K}, is the stress concentration at hole 2 2.7 (Fig. 4.37) E- oh=2—$7—(—4%—0J)=O.323 3 7—23 2. From Eq. (7.5a), U = 9-2—25— , 2 U = (0.67 S) 323 sf V 2B 0 V’ Uh=( 2B 3. The ratio of impact energy absorbing capacities is _U_ __ (0.67 sf U — =4.30,0rUh= Uh (0323 S)2 13 Comments: 1. The hole reduces the impact energy capacity of the bar by a factor of 4.3. 2. Drilling the "very long" rod will not significantly reduce the volume of uniformly stressed material, but it will reduce the stress, 0, to which this material can be subjected. 3. The decision to drill a 0.1 in—diameter hole produces a design that clearly reduces the impact energy absorbing capacity by a factor of 2 or more. 4. This design is only one of several possible redesigns. I 7-24 ...
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This note was uploaded on 03/27/2009 for the course MECHENG EN.530.215 taught by Professor Wang during the Spring '09 term at Johns Hopkins.

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ch07 - SOLUTION (7.1) Known: A 950—lb gin pole is...

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