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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 preexisting 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 preaccident 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. 72 SOLUTION (7.2)
Known: A 5ton 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 preexisting 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. Preaccident 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. 75 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
MddleT—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. 76 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 ) 225
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. 78 SOLUTION (7.5) Known: A rescue car, of 1400 kg mass, attempted to jerk a stuck vehicle back onto a
road using a Sm 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 79 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
710 Hence, (7554 Nm)/(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 ropestretch would make a further signiﬁcant addition. Thus the possibility of storing considerably more than 7554 Nm 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) 711 _ 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 21h = 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(0167) — 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 Cl).
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 Cl).
Thus, U = (1/4)1tr4tp(o2 2
= (1/4) 1t(0.150 m)4 (0.025 m) (7700 £53 1%9) = 7554 Nm
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 bodynotched 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 717 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 720 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 inc...
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 Spring '09
 Wang
 Energy, Force, Stress

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