C
PROBLEM 11.1
Determine the modulus of resilience for each of the following metals: ( ) Stainless steel
AISI 302 (annealed): 190 GPa 260 MPa ( ) Stainless steel
AISI 302 (cold-rolled): 190 GPa 520 MPa ( ) Malleable cast iron: 165 GPa 230 MPa
Y Y Y a E b E c E σ σ σ = = = = = = SOLUTION (a) E=190 10 Pa,× 9 σY =260 10 Pa× 6 2 6 2 3 3 9 (260 10 ) 177.9 10 N m/m 2 (2)(190 10 ) Y Y u E σ × = = = × ⋅ × 3 177.9 kJ/m Y u = (b) 190 10 Pa,9 520 10 Pa6 Y E= × σ = × 2 6 2 3 3 9 (520 10 ) 712 10 N m/m 2 (2)(190 10 ) Y Y u E σ × = = = × ⋅ × 3 712 kJ/m Y u = (c) E=165 10 Pa,× 9 σY =230 10 Pa× 6 2 6 2 3 3 9 (230 10 ) 160.3 10 N m/m 2 (2)(165 10 ) Y Y u E σ × = = = × ⋅ × 3 160.3 kJ/m Y u =
PROBLEM 11.2
Determine the modulus of resilience for each of the following alloys:
6 6 6
( ) Titanium: 16.5 10 psi 120 ksi
( ) Magnesium: 6.5 10 psi 29 ksi
( ) Cupronickel (annealed): 20 10 psi 16 ksi
Y Y Y a E b E c E σ σ σ = × = = × = = × = SOLUTION
(a) 16.5 10 psi,6 120 10 psi3
Y E= × σ = × 2 3 2 6 (120 10 ) 2 (2)(16.5 10 ) Y Y u E σ × = = × 3 436 in lb/in Y u = ⋅ (b) 6.5 10 psi,6 29 10 psi3 Y E= × σ = × 2 3 2 6 (29 10 ) 2 (2)(6.5 10 ) Y Y u E σ × = = × 3 64.7 in lb/in Y u = ⋅ (c) 20 10 psi,6 16 10 psi3 Y E= × σ = × 2 3 2 6 (16 10 ) 2 (2)(20 10 ) Y Y u E σ × = = × 3 6.40 in lb/in Y u = ⋅
PROBLEM 11.3
Determine the modulus of resilience for each of the following grades of structural steel: ( ) ASTM A709 Grade 50: 50 ksi
( ) ASTM A913 Grade 65: 65 ksi ( ) ASTM A709 Grade 100: 100 ksi
Y Y Y a b c σ σ σ = = = SOLUTION
Structural steel: E=29 10 psi× 6 for all three steels given.
(a) 50 ksi 50 10 psi3
Y σ = = × 2 3 2 6 (50 10 ) 2 (2)(29 10 ) Y Y u E σ × = = × 3 43.1 in lb/in Y u = ⋅ (b) 65 ksi 65 10 psi3 Y σ = = × 2 6 2 6 (65 10 ) 2 (2)(29 10 ) Y Y u E σ × = = × 3 72.8 in lb/in Y u = ⋅ (c) 100 ksi 100 10 psi3 Y σ = = × 2 3 2 6 (100 10 ) 2 (2)(29 10 ) Y Y u E σ × = = × 3 172.4 in lb/in Y u = ⋅
PROBLEM 11.4
Determine the modulus of resilience for each of the following aluminum alloys:
( ) 1100-H14: 70 GPa 55 MPa ( ) 2014-T6: 72 GPa 220 MPa ( ) 6061-T6: 69 GPa 150 MPa σ σ σ = = = = = = Y Y Y a E b E c E SOLUTION Aluminum alloys: (a) 70 10 Pa9 55 10 Pa6 Y E= × σ = × 2 6 2 3 3 9 (55 10 ) 21.6 10 N m/m 2 (2)(70 10 ) Y Y u E σ × = = = × ⋅ × 3 21.6 kJ/m Y u = (b) 72 10 Pa9 220 10 Pa6 Y E= × σ = × 2 6 2 3 3 9 (220 10 ) 336 10 N m/m 2 (2)(72 10 ) Y Y u E σ × = = = × ⋅ × 3 336 kJ/m Y u = (c) 69 10 Pa9 150 10 Pa6 Y E= × σ = × 2 6 2 3 3 9 (150 10 ) 163.0 10 N m/m 2 (2)(69 10 ) Y Y u E σ × = = = × ⋅ × 3 163.0 kJ/m Y u =
PROBLEM 11.5
The stress-strain diagram shown has been drawn from data obtained during a tensile test of an aluminum alloy. Using E=72 GPa, determine (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.
SOLUTION (a) σY =EεY 2 2 9 2 3 3 1 1 (72 10 )(0.006) 2 2 2 1296 10 N m/m Y Y Y u E E σ ε = = = × = × ⋅ 1296 kJ/m3 Y u =
(b) Modulus of toughness total area under the stress-strain curve= The average ordinate of the stress-strain curve is
6 2
500 MPa 500 10 N/m .= ×
The area under the curve is A=(500 10 )(0.18) 90 10 N/m .× 6 = × 6 2
PROBLEM 11.6
The stress-strain diagram shown has been drawn from data obtained during a tensile test of a specimen of structural steel. Using
6
29 10
E= × psi, determine (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.
SOLUTION (a) σY =EεY 2 2 6 2 1 1(29 10 )(0.002) 2 2 2 Y Y Y u E E σ ε = = = × 3 58.0 in lb/in Y u = ⋅
(b) Modulus of toughness total area under the stress-strain curve=
2 1 3 (57)(0.25 0.002) 14.14 kips/in 14.14 in kip/in A = − = = ⋅ 2 2 3 (28)(0.25 0.021) 3.21 kips/in 2 3.21 in kip/in A =1 − = = ⋅ 2 3 3 2 (20)(0.25 0.075) 2.33 kips/in 3 2.33 in kip/in A = − = = ⋅ modulus of toughness =uY +A1+A2+ A3
PROBLEM 11.7
The load-deformation diagram shown has been drawn from data obtained during a tensile test of a 0.875-in.-diameter rod of an aluminum alloy. Knowing that the deformation was measured using a 15-in. gage length, determine (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.
SOLUTION
Volume of stressed material involved in the measurement:
2 2 3 4 (0.875) (15) 9.0198 in 4 V πd L π = = =
(a) Modulus of resilience.
30 kips, 0.104 in. Y Y P = δ = 1 1 (30)(0.104) 1.56 in kip 1560 in lb 2 2 Y Y Y U = Pδ − = ⋅ = ⋅ modulus of resilience 1560 9.0198 Y Y U u V = = = 173.0 in lb/in3 Y u = ⋅ (b) Modulus of toughness. 1 2 3 1 2 3 (30)(1.85 0.104) 52.38 kip in 52380 in lb 1 (5)(1.85 0.104) 4.365 kip in 4365 in lb 2 2 (4)(1.85 0.104) 4.656 kip in 4656 in lb 3 62961 in lb Y A A A U U A A A = − = ⋅ = ⋅ = − = ⋅ = ⋅ = − = ⋅ = ⋅ = + + + = ⋅ 62961 9.0198 U V = modulus of toughness 3 6980 in lb/in = ⋅
PROBLEM 11.8
The load-deformation diagram shown has been drawn from data obtained during a tensile test of structural steel. Knowing that the cross-sectional area of the specimen is 250 mm2and that the deformation was measured using a 500-mm gage length, determine (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.
SOLUTION
Assuming that yielding occurs at P=62.5 kN and δ =0.6 mm,
3 3 1 (62.5 10 )(0.6 10 ) 18.75 N m 18.75 J 2 Y U = × × − = ⋅ =
Volume of stressed material: V=AL=(250)(500) 125 10 mm= × 3 3 =125 10 m× −6 3 3 6 18.75 150 10 125 10 Y Y U u V − = = = × × 3 150 kJ/m Y u = 6 3 3 3 1 (62.5 10 )(96 10 ) 6 10 N m 6 10 J A = × × − = × ⋅ = × 3 3 3 3 2 1 (28 10 )(96 8.6) 10 1.22 10 N m 1.22 10 J 2 A = × − × − = × ⋅ = × 6 3 3 3 3 2(15 10 )(61 10 ) 0.61 10 N m 0.61 10 J 3 A = × × − = × ⋅ = × Total energy: U U= Y +A1+A2+A3 =7.85 10 J× 3 3 6 3 6 7.85 10 63 10 J/m 125 10 U V − × = = × × modulus of toughness =63 MJ/m3
PROBLEM 11.9
UsingE=29 10× 6psi, determine (a) the strain energy of the steel rod ABC when 8
P= kips, (b) the corresponding strain energy density in portions AB and BC of the rod. SOLUTION 3 8 kips, 29 10 ksi P= E= × 2 2, , , 4 2 P A d V AL u A E π σ σ = = = = U =uV
Portion d(in.) L(in.) A(in2) V(in3) σ(ksi) u(in kip/in )⋅ 3 U(in kip)⋅
AB 0.625 24 0.3608 7.363 26.08 11.72 10× −3 86.32 10× −3 BC 0.75 36 0.4418 15.904 18.11 5.65 10× −3 89.92 10× −3 Σ 176.24 10× −3 (a) U =176.2 10 in kip× −3 ⋅ U =176.2 in lb⋅ (b) In :AB u=11.72 10 in kip/in× −3 ⋅ 3 11.72 in lb/in3 AB u = ⋅ 3 3 In :BC u=5.65 10 in kip/in× − ⋅ 5.65 in lb/in3 BC u = ⋅
PROBLEM 11.10
Using E=200GPa, determine (a) the strain energy of the steel rod ABC when P=25kN, (b) the corresponding strain-energy density in portions AB and BC of the rod.
SOLUTION 2 2 6 2 2 2 6 2 (20) 314.16 mm 4 314.16 10 m (16) 201.06 mm 4 201.06 10 m AB BC A A π π − − = = = × = = = × 3 2 3 2 3 2 9 6 9 6 25 10 N 2 (25 10 ) (1.2) (25 10 ) (0.8) (2)(200 10 )(314.16 10 ) (2)(200 10 )(201.06 10 ) P P L U EA − − = × = × × = + × × × × (a) U =5.968 6.213 12.18 N m+ = ⋅ U =12.18 J (b) 3 6 6 25 10 79.58 10 Pa 314.16 10 AB AB P A σ = = × − = × × 2 6 2 3 9 (79.58 10 ) 15.83 10 2 (2)(200 10 ) AB AB u E σ × = = = × × 3 15.83 kJ/m AB u = 3 6 6 25 10 124.28 10 Pa 201.16 10 BC AB P A σ = = × − = × × 2 6 2 3 9 (124.28 10 ) 38.6 10 2 (2)(200 10 ) BC BC u E σ × = = = × × 3 38.6 kJ/m BC u =
PROBLEM 11.11
A 30-in. length of aluminum pipe of cross-sectional area
2
1.85 in is welded to a fixed support A and to a rigid cap B. The steel rod EF, of 0.75-in. diameter, is welded to cap B. Knowing that the modulus of elasticity is 29 10× 6psi for the steel and
6
10.6 10× psi for the aluminum, determine (a) the total strain energy of the system when P= kips, (b) the corresponding 8 strain-energy density of the pipe CD and in the rod EF.
SOLUTION
Member EF carries a force P =8000 lb in tension while member CD carries 8000 lb in compression. Area of member EF: 2 (0.75)2 0.4418 in2
4 4
A=πd =π = (a) Strain energy.
: CD 2 ( 8000) (30)26 48.95 in lb 2 (2)(10.6 10 )(1.85) CD P L U EA − = = = ⋅ × : EF 2 2 6 (8000) (48) 119.89 in lb 2 (2)(29 10 )(0.4418) EF P L U EA = = = ⋅ × Total: U =UCD +UEF =168.8 in lb⋅ U =168.8 in lb.⋅
(b) Strain energy density. : CD 8000 4324 psi, 1.85 σ = − = − 2 ( 4324)26 2 (2)(10.6 10 ) u E σ − = = × 3 0.882 in lb/in u= ⋅ : EF 8000 18108 psi, 0.4418 σ = = 2 (18108)26 2 (2)(29 10 ) u E σ = = × 3 5.65 in lb/in u= ⋅
PROBLEM 11.12
Rod AB is made of a steel for which the yield strength is σY =450MPa and E=200GPa; rod BC is made of an aluminum alloy for which σY=280MPa and E 73= GPa. Determine the maximum strain energy
that can be acquired by the composite rod ABC without causing any permanent deformations. SOLUTION 2 2 6 2 2 2 6 2 (10) 78.54 mm 4 78.54 10 m (14) 153.94 mm 4 153.94 10 m AB BC A A π π − − = = = × = = = × all Y
P =σ A for each portion. : AB 6 6 all 3 (450 10 )(78.54 10 ) 35.343 10 N P = × × − = × : BC 6 6 all 3 (280 10 )(153.94 10 ) 43.103 10 N P = × × − = ×
Use the smaller value. P=35.343 10 N× 3 2 2 3 2 9 6 3 2 9 6 (35.343 10 ) (1.2) 2 2 (2)(200 10 )(78.54 10 ) (35.343 10 ) (1.6) (2)(73 10 )(153.94 10 ) BC AB AB AB BC BC P L P L U E A E A − − × = + = × × × + × × 136.6 J U =
PROBLEM 11.13
A single 6-mm-diameter steel pin B is used to connect the steel strip
DE to two aluminum strips, each of 20-mm width and 5-mm
thickness. The modulus of elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that for the pin at B the allowable shearing stress is τall =85 MPa, determine, for the loading shown, the maximum strain energy that can be acquired by the assembled strips. SOLUTION 2 2 2 pin 6 2 6 all (6) 28.274 mm 4 4 28.274 10 m 85 10 Pa A πd π τ − = = = = × = × Double shear: 6 6 3 2 (2)(28.274 10 )(85 10 ) 4.8066 10 N P= Aτ= × − × = ×
For strips AB, DB, BE, 2 6 2
3 (20)(5) 100 mm 100 10 m 1 2.4033 10 N 2 AB DB A F F P − = = = × = = = × 2 3 9 6 (2.4033 10 )(0.5) 0.2063 J 2 (2)(70 10 )(100 10 )− × = = = = × × AB AB AB DB a AB F L U U E A 2 3 2 9 6 (4.8066 10 ) (1.25 0.5) 0.4332 J 2 (2)(200 10 )(100 10 )− × − = = = × × BE BE BE s BE F L U E A Total: U U= AB+UDB+UBE =0.846 J U =0.846 J
2 4 Y Y U ALu A πd = =
PROBLEM 11.14
Rod BC is made of a steel for which the yield strength is σY =300 MPa and the modulus of elasticity is E=200 GPa. Knowing that a strain energy of 10 J must be acquired by the rod when the axial load P is applied, determine the diameter of the rod for which the factor of safety with respect to permanent deformation is six.
SOLUTION
For factor of safety of six on the energy,
2 6 2 9 3 3 (6)(10) 60 J (300 10 ) 2 (2)(200 10 ) 225 10 J/m Y Y Y U u E σ = = × = = × = × Y Y U = ALu 3 6 2 60 (1.8)(225 10 ) 148.148 10 m Y Y U A Lu − = = × = × 2 4 A= π d 6 3 4 (4)(148.148 10 ) 13.73 10 m A d π π − − × = = = × 13.73 mm d=
PROBLEM 11.15
The assembly ABC is made of a steel for which E = 200 GPa and σY =320 MPa. Knowing that a strain energy of 5 J must be acquired by the assembly as the axial load P is applied, determine the factor of safety with respect to permanent deformation when (a) x =300 mm, (b) x = 600 mm. SOLUTION 6 9 2 2 2 6 2 2 2 2 6 2 min
320 MPa 320 10 Pa, 200 GPa 200 10 Pa
(12) 113.097 mm 113.097 10 m 4 4 (18) 254.47 mm 254.47 10 m 4 4 Y AB AB BC BC AB E A d A d A A σ π π π π − − = = × = = × = = = = × = = = = × =
Force at yielding or allowable axial force.
6 6 3 min (320 10 )(113.097 10 ) 36.191 10 N Y Y P = P =σ A = × × − = × (a) 2 2 2 3 2 9 6 6 3 300 mm: 0.300 m, 0.600 m 2 2 2 (36.191 10 ) 0.300 0.600 (2)(200 10 ) 113.097 10 254.97 10 (3.2745 10 )(2652.6 2353.2) 16.392 J − − − = = = = + = + = + × = + × × × = × + = AB BC AB BC AB BC Y AB BC AB BC AB BC x L L P L P L P L L U U U EA EA E A A Applied energy: U =5 J Factor of safety: 16.392 5 = Y U U F S. .= 3.28 (b) 3 9 6 6 3 600 mm: 0.600 m, 0.300 m 36.191 10 0.600 0.300 (2)(200 10 ) 113.097 10 254.97 10 (3.2745 10 )(5305.2 1176.6) 21.225 J − − − = = = × = + × × × = × + = AB BC Y x L L U Factor of safety: UY = 21.225 F S. .= 4.25
PROBLEM 11.16
Using E=10.6 10× 6psi, determine by approximate means the maximum strain energy that can be acquired by the aluminum rod shown if the allowable normal stress is σall=22ksi.
SOLUTION
2 2
min 4(1.5) 1.7671 in
A =π =
all
all all min
22000 psi 38877 lb P A σ σ = = = 2 2 2 2 2 2 2 2 4 P dx P dx P dx U EA E πd πE d =
=
=
Use Simpson’s rule to compute the integral. h=1.5 in.
Section d(in.) 1/ (in )d2 −2 multiplier m d(1/ ) (in )2 −2
1 1.50 0.4444 1 0.4444 2 2.10 0.22675 4 0.9070 3 2.55 0.15379 2 0.3076 4 2.85 0.12311 4 0.4924 5 3.00 0.11111 1 0.1111 Σ 2.2625 1 2 2 1 1.5 (2.2625) 1.13125 in 3Σ 3 B A dx h m d d -æ ö÷ ç = çç ÷÷÷= = è ø
ò
2 6 (2)(38877) (1.13125) (10.6 10 ) U π = × U =102.7 in lb⋅ PROBLEM 11.17
Show by integration that the strain energy of the tapered rod AB is
2 min 4 P L U EA =
where Amin is the cross-sectional area at end B.
SOLUTION Radius: 2 min cx r A c L π = = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 L L L L L L c A r x L P dx P L dx U EA E c x P L x E c π π π π = = = = = −
2 2 min 1 1 2 2 P L EA L L = − + 2 min 4 P L U EA = PROBLEM 11.18
In the truss shown, all members are made of the same material and have the uniform cross-sectional area indicated. Determine the strain energy of the truss when the load P is applied.
SOLUTION
Equilibrium of joint C. Solving the force triangle,
0.577 tan 60 1.155 sin 60 BC CD P F P P F P = = ° = = ° Member CB: 2 (0.577 )2 2 0.1667 2 2 BC BC BC F L P l P l U EA EA EA = = = Member CD: 2 (1.155 ) (2 )2 2 1.3333 2 2 CD CD CD F L P l P l U EA EA EA = = =
Total strain energy: U =UBC +UCD
2
1.500P l
U
EA
PROBLEM 11.19
In the truss shown, all members are made of the same material and have the uniform cross-sectional area indicated. Determine the strain energy of the truss when the load P is applied.
SOLUTION 2 2 1 5 2 2 BC CD L = L = l + l = l Joint C. (equilibrium) 2 2 0 : 0 5 5 x BC CD CD BC F F F F F = − − = = − 0 : 15 15 0 5 5 2 2 y BC CD BC CD F F F P F P F P = − − = = = − Strain energy. 2 2 2 2 2 1 2 2 1 5 5 5 5 2 2 2 2 2 BC BC CD CD F L U F L F L EA EA P l P l EA = = + = + 2 1.398P l U EA =
PROBLEM 11.20
In the truss shown, all members are made of the same material and have the uniform cross-sectional areas indicated. Determine the strain energy of the truss when the load P is applied.
SOLUTION Geometry. 11 tan 26.57 2 /2 2 , 1.118 sin 26.57 BD BC CD l l l l l l β = − = ° = = = = ° Equilibrium of joint C.
From the force triangle, /2 1.118 sin 26.57 BC CD P F = F = = P ° Equilibrium of joint D. 0: (1.118 ) cos 26.57 0 = − ° = = − x BD BD F F P F P Strain energy.
[
]
2 2 2 2 2 2 2 1 (1.118 ) (1.118 ) ( ) (2 ) (1.118 ) (1.118 ) 2 2 1.3974 1 1.3974 3.795 2 2 i i BC BD CD F L U U U U AE P l P l P l E A A A P l P l AE AE = = + + − = + + = + + =
2 1.898P l U AE = PROBLEM 11.21
In the truss shown, all members are made of the same material and have the uniform cross-sectional area indicated. Determine the strain energy of the truss when the load P is applied. SOLUTION Equilibrium of joint C. + 0: 3 0 2 2 3 = − − = = −
Fy FCD P FCD P 1 1 0: 0 2 3 = − − = =
Fx FBC FCD FBC P Equilibrium of joint D. 3 0: 0 2 = + = =
Fy FBD FCD FBD P Strain energy. 2 2 1 1 2 2 F L F L U EA E A =
=
2 1 P l P l2 Member F L A F2L/A BC 1 3P l A 1 2 3P l A / CD 2 3P − 2l A 8 2 3P l A / BD P 3l A 3 2/ P l A Σ 4.732P2l/APROBLEM 11.22
Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E =29 10 psi,× 6 determine the strain energy of the truss for the loading shown.
SOLUTION 2 2 2 2 7.5 4 8.5 ft 102 in. 7.5 ft 90 in. 3 in , 4 in 29000 ksi BC CD BC CD L L A A E = + = = = = = = = Equilibrium at joint C. 4 0: 24 0 51 kips 8.5 = − = =
Fy FBC FBC 7.5 0: (51 kips) 20 kips 0 8.5 25 kips = − − + = = −
x CD CD F F F Strain energy. 2 2 2 2 2 2 2 2 (51) (102) (25) (90) (2)(29000)(3) (2)(29000)(4) 1.5247 0.2425 i i BC BC CD CD i BC CD F L F L F L U EA EA EA = = + = + = +
1.767 kip in U = ⋅ PROBLEM 11.23
Each member of the truss shown is made of aluminum and has the cross-sectional area shown. Using E =72 GPa, determine the strain energy of the truss for loading shown.
SOLUTION Lengths of members: 2 2 1/ 2 2 2 1/ 2 (3.2 2.4 ) 4 m (1 2.4 ) 2.6 m BC CD L L = + = = + = E =72 GPa =72 10 Pa× 9 Forces in kN. Equilibrium of truss. 0: (30)(2.4) (80)(3.2) (2.2) 0 = − + =
MB Dy 83.636 kN y D = ↑ 0: 80 0 = − − =
Fy Dy By 83.636−By −80 =0 By =3.636 kN ↓ Member forces. 4 m (3.636 kN) 4 6.061 kN 2.4 m 2.4 2.6 m 2.6 (83.636 kN) 90.606 kN 2.4 m 2.4 BC y CD y F B F D = = = = − = − = − Strain energy. 2i i BC CD F L U U U AE = + =
2 2 3 2 3 2 9 3 9 3 (6.061 10 ) (4) (90.606 10 ) (2.6) 2 2 (2)(72 10 )(2 10 ) (2)(72 10 )(2.5 10 ) BC BC CD CD BC CD F L F L U EA EA − − × × = + = + × × × ×PROBLEM 11.24
Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.
SOLUTION 2 0: ( ) 0 2 1 2 = − − = = −
K v M M wv M wv 2 2 2 0 0 2 2 5 4 0 0 1 1 2 2 2 8 8 5 L L L L M U dv wv dv EI EI w w v v dv EI EI = = = =
2 5 40 w L EI = 2 5 40 w L U EI = PROBLEM 11.25
Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.
SOLUTION 0 : ( ) 0 2 2 = − + = =
B A A L wL M R L wL R Bending moment: 1 2 ( 2) 2 2 A w M = R x− wx = Lx−x 2 2 2 2 0 0 2 2 2 3 4 5 2 2 3 4 0 0 2 5 ( ) 2 8 2 ( 2 ) 8 8 3 4 5 1 1 1 8 3 2 5 L L L L M w U dx Lx x dx EI EI w w L x Lx x L x Lx x dx EI EI w L EI = = − = − + = − + = − +
2 5 240 w L U EI = PROBLEM 11.26
Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.
SOLUTION 0 0 0: 0 B A A M M R L M R L Σ = − − = = ↓ 0 0 0: 0 A B B M M R L M R L Σ = − = = ↑ A to D: ΣMJ = 0: M x0 M 0 L + = 0 M x M L = − 2 2 3 2 2 0 0 2 2 0 2 2 0 6 a a AD M M a M dx U x dx EI EIL EIL =
=
= D to B: 0: 0 K M v M M L Σ = − + 0 M v M L = 2 2 3 2 2 0 0 2 2 0 2 2 0 6 b b DB M M b M dv U v dv EI EIL EIL =
=
= Total: U U= AD+UDB 02 3 3 2 ( ) 6 M a b U EIL + = PROBLEM 11.27
Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.
SOLUTION
Symmetric beam and loading: RA =RB
0: 2 0
y A B A B
F R R P R R P
Σ = + − = = =
Over portion AD, M =R x PxA =
2 2 2 3 2 3 2 0 0 0 2 2 2 3 6 a a a AD M P P x P a U dx x dx EI EI EI EI =
=
= =Over portion DE, 2 2( 2 )
2 DE P a L a M Pa U EI − = =
Over portion EB, By symmetry, 2 3 6 EB AD P a U U EI = = Total: U U= AD+UDE +UEB 2 2(3 4 ) 6 P a U L a EI = −
PROBLEM 11.28
Using E=200 GPa, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)
SOLUTION
Over portion AC, 1
2 M = Px 2 2 /2 /2 2 0 0 /2 2 3 2 3 0 2 8 8 3 192 L L AC L M P U dx x dx EI EI P x P L EI EI = = = =
By symmetry, 2 3 192 CB AC P L U U EI = = Total: 2 3 96 AC CB P L U U U EI = + = Data: 3 9 6 4 6 4 180 10 N, 4.8 m, 200 10 Pa 178 10 mm 178 10 m P L E I − = × = = × = × = × 3 2 3 9 6 (180 10 ) (4.8) 1048 N m (96)(200 10 )(178 10 ) U − × = = ⋅ × × U =1048J PROBLEM 11.29
Using E = 200 GPa, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)
SOLUTION
Over portion AD, M =Px
2 2 0 0 2 3 2 3 0 1 ( ) 2 2 2 3 6 a a AD a M U dx Px dx EI EI P x P a EI EI = = = =
Over portion DE, M =Pa
2 2 3 ( ) 2 2 DE Pa a P a U EI EI = = By symmetry, 2 3 6 EB AD P a U U EI = = 2 3 5 6 AD DE EB P a U U U U EI = + + = Data: 3 9 6 4 6 4 80 10 N, 1.6 m, 200 10 Pa 163 10 mm 163 10 m P a E I − = × = = × = × = × 3 2 3 9 6 5 (80 10 ) (1.6) 670 N m 6 (200 10 )(163 10 ) U = × − = ⋅ × × U =670 J
PROBLEM 11.30
Using E=29 10 psi,× 6 determine the strain energy due to bending for
the steel beam and loading shown. (Ignore the effect of shearing stresses.)
SOLUTION
Equilibrium.
0: (4 kip)(24 in.) (96 in.) 0 1 kip B M C C = − = = ↓
Bending moment diagram.
Over AB: (0≤ ≤x 24 in.)
4 kip in M = − x ⋅ Over BC: (0≤ ≤v 96 in.) 1 kip in M = − v ⋅ Strain energy. 2 2 24 96 0 2 0 2 CB AB AB CB M M U U U dx dv EI EI = + =
+
Data: W6 9,× I =16.4 in ,4 E = 29 10 ksi× 3 24 2 96 2 3 3 0 0 3 1 1 16 1 ( 4 ) ( ) (24) (96) 2 2 3 3 184320 184320 0.38755 in kip (29 10 )(16.4) U x dx v dv EI EI EI = − + − = + = = = ⋅ ×
388 in lb U = ⋅ PROBLEM 11.31
Using E=29 10 psi,× 6 determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.) SOLUTION Over A to B: 2 2 2 3 2 0 2 2 0 6 a a AB M Px M dx P P a U x dx EI EI EI = − =
=
= Over B to C: 2 2 2 constant 2 2 BC M Pa M b P a b U EI EI − = = = = By symmetry, 2 3 6 CD AB P a U U EI = = Total: 2 2(2 3 ) 6 AB BC CD P a a b U U U U EI + = + + = Data: 3 3 4 2 10 lb, 15 in., 60 in. 1 (1.5)(3) 3.375 in 12 P a b I = × = = = = 3 2 2 6 (2 10 ) (15) [(2) (15) (3)(60)] (6) (29 10 ) (3.375) U = × + × U =322 in lb⋅ PROBLEM 11.32
Assuming that the prismatic beam AB has a rectangular cross section, show that for the given loading the maximum value of the strain-energy density in the beam is
max 15 U u
V
=
where U is the strain energy of the beam and V is its volume.
SOLUTION 2 0: ( ) 0 2 1 2 K v M M wv M wv Σ = − − = = − 2 2 2 0 0 2 2 5 4 0 0 2 5 1 1 2 2 2 8 8 5 40 L L L L M U dv wv dv EI EI w w x v dv EI EI w L EI = = = = =
2 max max max 1 2 M c M wL I σ = = 2 4 2 2 2 2 1 2 4 2 max max 4 max 2 2 2 2 2 8 2 w L c M c w L c u E EI EI EI σ = = = =(
)
( )
3 1 12 2 2 max 2 1 1 15 15 5 5 d L bd U LI Lbd V u = c = = = max 15 U u V = PROBLEM 11.33
The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft. The steel drill pipe has an outer diameter of 8 in. and a uniform wall thickness of 0.5 in. Knowing that the top of the drill pipe rotates through two complete revolutions before the drill bit at B starts to operate and using G=11.2 10 psi,× 6 determine the maximum strain energy acquired by the drill pipe.
SOLUTION
(
)
3 4 4 4 2 2 2 (2)(2 ) 4 rad 5000 ft 60 10 in. 4 in. 3.5 in. 2 166.406 in 2 2 2 2 o o i o o i L d c c c t J c c TL GJ T GJ L T L GJ L GJ U GJ L GJ L ϕ π π π ϕ ϕ ϕ ϕ = = = = × = = = − = = − = = = = = = 6 2 3 (11.2 10 )(166.406)(4 ) (2)(60 10 ) U = × π × 6 2.45 10 in lb U = × ⋅ PROBLEM 11.34
Rod AC is made of aluminum and is subjected to a torque T applied at C. Knowing that G = 73 GPa and that portion BC of the rod is hollow and has an inner diameter of 16 mm, determine the strain energy of the rod for a maximum shearing stress of 120 MPa.
SOLUTION 12 mm 2 8 mm 2 o o i i d c d c = = = =
(
)
4 4 3 4 9 4 4 4 4 4 3 4 9 4 (12) 32.572 10 mm 2 2 32.572 10 m (12 8 ) 26.138 10 mm 2 2 26.138 10 m AB o BC o i J c J c c π π π π − − = = = × = × = − = − = × = × 9 6 min all all 3 min 2 2 3 9 9 2 2 3 9 9 (26.138 10 )(120 10 ) 261.38 N m 12 10 (261.38) (400 10 ) 5.747 J 2 (2)(73 10 )(32.572 10 ) (261.38) (500 10 ) 8.951 J 2 (2)(73 10 )(26.138 10 ) AB AB AB BC BC BC J Tc T J c T L U GJ T L U GJ τ τ − − − − − − × × = = = = ⋅ × × = = = × × × = = = × × Total: U U= AB+UBC U =14.70 J PROBLEM 11.35
Show by integration that the strain energy in the tapered rod AB is
2 min 7 78 T L U GJ =
where Jmin is the polar moment of inertia of the rod at end B.
SOLUTION 4 4 4 4 min 4 2 2 2 2 4 4 4 2 4 2 4 min , 2 2 2 2 2 2 2 L L L L L L cx r L c J r x J c L T dx T dx U GJ c G x L T L dx GJ x π π π π = = = = = = =
2 2 4 3 min 1 2 3 L L T L GJ x = − 2 4 3 3 min 1 1 2 3(2 ) 3 T L U GJ L L = − + 2 min 7 48 T L U GJ = PROBLEM 11.36
The state of stress shown occurs in a machine component made of a grade of steel for which σY =65 ksi. Using the maximum-distortion-energy criterion, determine the factor of safety associated with the yield strength when (a) σy = +16 ksi, (b) σy = −16 ksi.
SOLUTION ave 1 (0 8) 2 4 ksi 8 0 2 2 4 ksi 14 ksi x z xz σ σ σ τ = + = − = − = = 2 2 2 2 ave ave 2 4 14 14.56 ksi 18.56 10.56 x z xz a b c y R R R σ σ τ σ σ σ σ σ σ − = + = + = = + = = − = − = 2 2 2 2 ( ) ( ) ( ) 2 . . Y a b b c c a F Sσ σ −σ + σ −σ + σ −σ = (a) 2 2 2 2 16 ksi 65 (18.56 10.56) ( 10.56 16) (16 18.56) 2 . . c y F S σ =σ = + + − − + − = 2 8450 847.97 705.43 6.55 ( . .)F S + + = F S. . 2.33= (b) σc=σy= −16 ksi 2 2 2 2 65 (18.56 10.56) ( 10.56 16) ( 16 18.56) 2 . . F S + + − + + − − = 2 8450 847.97 29.59 1194.39 ( . .)F S + + = F S. . 2.02=
PROBLEM 11.37
The state of stress shown occurs in a machine component made of a grade of steel for which σY =65 ksi. Using the maximum-distortion-energy criterion, determine the range of values of σy for which the factor of safety associated with the yield strength is equal to or larger than 2.2. SOLUTION ave 1 (0 8) 2 4 ksi 8 0 2 2 4 ksi 14 ksi x z xz σ σ σ τ = + = − = − = = 2 2 2 2 ave ave 2 4 14 14.56 ksi 18.56 ksi 10.56 ksi x z xz a b c y R R R σ σ τ σ σ σ σ σ σ − = + = + = = + = = − = − = 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) 2 . . 65 (18.56 10.56) ( 10.56 ) ( 18.56) 2 2.2 847.97 (111.51 21.12 ) ( 37.12 344.47) 1745.87 2 16 441.92 0 Y a b b c c a y y y y y y y y F S σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ − + − + − = + + − − + − = + + + + − + = − − = 2 16 16 (4)(2)(441.92) (2)(2) 4 15.39 19.39 ksi, 11.39 ksi y σ σ ± + = = ± = −
PROBLEM 11.38
The state of stress shown occurs in a machine component made of a brass for which σY =160 MPa. Using the maximum-distortion-energy criterion, determine the range of values of σz for which yield does not occur. SOLUTION ave 1(100 20) 2 60 MPa 100 20 2 2 40 MPa 75 MPa x y xy σ σ σ τ = + = − − = = = 2 2 2 2 ave ave 2 40 75 85 MPa 145 MPa 25 MPa x y xy a b c z R R R σ σ τ σ σ σ σ σ σ − = + = + = = + = = − = − = 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) 2 (145 25) ( 25 ) ( 145) (2)(160) 28900 (625 50 ) ( 290 21025) 51200 2 240 650 0 a b b c c a Y z z z z z z z z σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ − + − + − = + + − − + − = + + + + − + = − − = 2 240 240 (4)(2)(650) 60 62.65 (2)(2) 122.65 MPa, 2.65 MPa z z σ σ ± + = = ± = − 2.65 MPa <σz < 122.65 MPa −
PROBLEM 11.39
The state of stress shown occurs in a machine component made of a brass for which σY =160 MPa. Using the maximum-distortion-energy criterion, determine whether yield occurs when (a) σz = +45 MPa, (b) σz = −45 MPa. SOLUTION ave 1(100 20) 2 60 MPa 100 20 2 2 40 MPa 75 MPa x y xy σ σ σ τ = + = − − = = = 2 2 2 2 ave ave 2 40 75 85 MPa 145 MPa 25 MPa x y xy a b c z R R R σ σ τ σ σ σ σ σ σ − = + = + = = + = = − = − = ? 2 2 2 2 (σa−σb) +(σb−σc) +(σc−σa) <2σY (a) σc=σz = +45 MPa ? 2 2 2 2 (145 25)+ + − −( 25 45) +(45 145)− < 2(160) =51200 28900 4900 10000 43800 51200+ + = < (No yield.) (b) σc=σz = −45 MPa ? 2 2 2 (145 25)+ + − +( 25 45) + − −( 45 145) <51200 28900 400 36100 65400+ + = >51200 (Yield occurs.)
PROBLEM 11.40
Determine the strain energy of the prismatic beam AB, taking into account the effect of both normal and shearing stresses.
SOLUTION Reactions: 0 , 0 A B M M R R L L = ↓ = ↑ Shear: V M0 L = − Bending moment: M M0v L = For bending, 2 2 2 0 1 0 2 0 3 2 0 0 2 2 2 6 6 LM M L U dv v dv EI EIL M L M L EI EIL = = = =
For shear, 2 2 2 2 2 2 2 2 4 0 2 2 2 2 2 4 2 2 4 0 2 0 2 2 2 0 2 4 2 3 5 0 2 2 2 4 3 1 1 2 2 9 9 1 1 2 2 8 8 ( ) 9 1 2 8 9 2 1 9 3 5 8 xy xy L c L c c c c c V y c d A c M V y y y u G GA c G bd L c c M b y y U udv u b dy dx dy dx Gb d L c c M y y y dx Gbd L c c τ τ − − − = − = = = − = − + = = = − + = − + =
2 0 2 2 0 0 4 2 2 3 5 8 L M L c c c dx Gbd L − +
2 2 2 0 0 0 2 2 2 9 16 6 3 15 5 5 8 M M c M c L GbdL Gbd L Gbd L = = Total: 2 2 0 0 1 2 3 6 5 M L M U U U EI GbdL = + = + with 1 3 12 I = bd 2 2 0 0 3 2 3 5 M L M U GbdL Ebd = + 02 2 3 2 2 3 1 10 M L E d U G Ebd L = + ⋅ ⋅ PROBLEM 11.41*
A vibration isolation support is made by bonding a rod A, of radius R1, and a tube B, of
inner radius R2, to a hollow rubber cylinder.
Denoting by G the modulus of rigidity of the rubber, determine the strain energy of the hollow rubber cylinder for the loading shown.
SOLUTION 0: (2 ) 0 2 x F rL Q Q rL τ π τ π Σ = − + = = 2 1 2 2 2 2 2 2 2 2 2 2 2 2 0 2 2 8 2 8 8 L R R Q u G r L G Q dV Q r dr U u dV dx GL r GL r τ π π π π = = =
=
=
( )
2 2 1 1 2 2 2 0 2 0 4 4 L R L R R R Q dr Q dx lnr dx r GL GL π π =
=
2 2 1 4 R Q U ln GL R π = PROBLEM 11.42
The cylindrical block E has a speed v0 =16 ft/s when it strikes squarely the yoke BD that is attached to the 7
8-in.-diameter rods AB and CD. Knowing
that the rods are made of a steel for which σY =50 ksi and
E=29 10 psi,× 6 determine the weight of block E for which the factor of
safety is five with respect to permanent deformation of the rods.
SOLUTION
At the onset of yielding, the force in each rod is .
Y F=σ A
Corresponding strain energy:
2 2 2 2 2 2 2 2 0 0 2 2 2 2 1 1 ( . .) ( . .) 2 2 AB AB Y Y AB AB Y CD AB Y m AB CD m F L A L AL U EA EA E AL U U E AL U U U E W U mv F S v F S g σ σ σ σ = = = = = = + = = = Solving for W, 2 2 2 0 0 2 2 ( . .) ( . .) m Y gU g AL W v F S v F S E σ = =
Data: g=32.17 ft/ sec2 =386 in/ sec ,2 50 10 psi,3
Y σ = × 2 2 7 0.60132 in2 4 4 8 A=πd =π = 6 29 10 psi E= × 3.5 ft 42 in. L= = F S. . 5= 0 16 ft/sec 192 in/sec v = = 3 2 2 6 (2)(386)(50 10 ) (0.60132)(42) (192) (5)(29 10 ) W = × × W =9.12 lb
PROBLEM 11.43
The 18-lb cylindrical block E has a horizontal velocity v0 when it strikes squarely the yoke BD that is attached to the 7
8-in.-diameter rods AB
and CD. Knowing that the rods are made of a steel for which σY = 50 ksi and E =29 10 psi,× 6 determine the maximum allowable speed v0 if the
rods are not to be permanently deformed.
SOLUTION
At the onset of yielding, the force in each rod is .
Y F =σ A
Corresponding strain energy:
2 2 2 2 2 2 2 2 2 2 2 AB AB Y Y AB AB CD CD Y CD CD F L A L AL U EA EA E F L AL U EA E σ σ σ = = = = = Total: Y2 m AB CD AL U U U E σ = + = 2 2 0 0 1 1 2 2 m W U mv v g = = Solving for 2 0, v 2 2 0 2gUm 2g YAL v W EW σ = = 2 0 2g YAL v EW σ = Data: 2 2 3 2 2 2 6 32.17 ft/s 386 in/s 50 10 psi 7 0.60132 in , 29 10 psi 4 4 8 3.5 ft 42 in. 18 lb Y g A d E L W σ π π = = = × = = = = × = = = 3 2 0 6 (2)(386)(50 10 ) (0.60132)(42) 305.6 in/sec (29 10 )(18) v = × = × 0 25.5 ft/sec v =
PROBLEM 11.44
Collar D is released from rest in the position shown and is stopped by a small plate attached at end C of the vertical rod ABC. Determine the mass of the collar for which the maximum normal stress in portion BC is 125 MPa. SOLUTION Portion BC: 6 2 2 6 2 125 10 Pa (9) 63.617 mm 63.617 10 m 4 7952 N m BC m m BC A P A σ π σ − = × = = = × = =
Corresponding strain energy:
2 2 9 6 2 2 6 2 2 2 9 6 (7952) (2.5) 17.750 J 2 (2)(70 10 )(63.617 10 ) (12) 113.907 mm 113.907 10 m 4 (7952) (4) 10.574 J 2 (2)(105 10 )(113.907 10 ) 28.324 J m BC BC BC BC AB m AB AB AB AB m BC AB P L U E A A P L U E A U U U π − − − = = = × × = = = × = = = × × = + = Corresponding elongation Δ m: 3 1 2 2 (2)(28.324) 7.12 10 m 7952 m m m m m m P U U P − Δ = Δ = = = × Falling distance: h=0.6 7.12 10+ × −3=0.60712 m Work of weight =Um Wh mgh U= = m 28.324 (9.81)(0.60712) m U m gh = = m=4.76 kg
PROBLEM 11.45
Solve Prob. 11.46, assuming that both portions of rod ABC are made of aluminum.
PROBLEM 11.44 Collar D is released from rest in the position shown and
is stopped by a small plate attached at end C of the vertical rod ABC. Determine the mass of the collar for which the maximum normal stress in portion BC is 125 MPa. SOLUTION Portion BC: 6 2 2 6 2 125 10 Pa (9) 63.617 mm 63.617 10 m 4 7952 N m BC m m BC A P A σ π σ − = × = = = × = =
Corresponding strain energy:
2 2 9 6 2 2 6 2 2 2 9 6 (7952) (2.5) 17.750 J 2 (2)(70 10 )(63.617 10 ) (12) 113.907 mm 113.907 10 m 4 (7952) (4) 15.861 J 2 (2)(70 10 )(113.907 10 ) m BC BC BC AB m AB AB AB P L U EA A P L U EA π − − − = = = × × = = = × = = = × × Total: Um=UBC +UAB=33.611 J Corresponding elongation Δ m: 3 1 2 2 (2)(33.611) 8.45 10 m 7952 m m m m m m P U U P − Δ = Δ = = = × Falling distance: h=0.6+ Δ =m 0.60845 m Work of weight =Um. Wh mgh U= = m 33.611 (9.81)(0.60845) m U m gh = = m=5.63 kg
PROBLEM 11.46
The 48-kg collar G is released from rest in the position shown and is stopped by plate BDF that is attached to the 20-mm-diameter steel rod CD and to the 15-mm-diameter steel rods AB and EF. Knowing that for the grade of steel used
all 180 MPa
σ = and E = 200 GPa, determine the largest allowable distance h.
SOLUTION
Let Δ be the maximum elongation. m
6 9 6 9 180 MPa 180 10 Pa 2.5 m 200 10 Pa (180 10 )(2.5) 0.00225 m 200 10 AB CD EF m AB CD EF m L L L E E E L E σ σ σ σ σ σ Δ = = = = = = = × = = × × Δ = = × For each rod,
2 ( / )2 2 2 2 2 m m m F L EA L L EA U EA EA L Δ Δ = = = Rod CD: (20)2 314.16 mm2 314.16 10 m6 2 4 CD A = π = = × − 9 6 2 (200 10 )(314.16 10 )(0.00225) 63.617 J (2)(2.5) CD U = × × − =
Rods AB and EF: (15)2 176.71 mm2 176.71 10 m6 2
4 AB EF A = A = π = = × − 9 6 2 (200 10 )(176.71 10 )(0.00225) 35.674 J (2)(2.5) AB EF U =U = × × − =
Total strain energy: Um =UAB +UCD +UEF =134.97 J Work of falling collar:
( ) (48)(9.81)( ) m m m U = mg h+ Δ = h+ Δ Equating, (48)(9.81)(h+ Δm) =134.97 h+ Δ =m 0.28662 m 0.28662 0.00225 0.285 m h= − = h = 285 mm
PROBLEM 11.47
Solve Prob. 11.46, assuming that the 20-mm-diameter steel rod CD is replaced by a 20-mm-diameter rod made of an aluminum alloy for which
all 150 MPa
σ = and E = 75 GPa.
PROBLEM 11.46 The 48-kg collar G is released from rest in the position
shown and is stopped by plate BDF that is attached to the 20-mm-diameter steel rod CD and to the 15-mm-diameter steel rods AB and EF. Knowing that for the grade of steel used σall =180 MPa and E = 200 GPa, determine the largest
allowable distance h.
SOLUTION
Let Δ be the maximum elongation. m L =2.5 m
AB CD EF m AB CD EF L L L E E E σ σ σ Δ = = =
If rod CD yields, σCD =150 10 Pa,× 6 ECD =75 10 Pa× 9
6 9 (150 10 )(2.5) 0.005 m 75 10 m × Δ = = ×
If rods AB and EF yield, 6
9 180 10 Pa 200 10 Pa AB EF AB EF E E σ =σ = × = = × 6 9 (180 10 )(2.5) 0.00225 m 200 10 m × Δ = = ×
The smaller value of Δ governs. m Δ =m 0.00225 m
For each rod, 2 ( / )2 2
2 2 2 m m F L EA L L EA U EA EA L Δ Δ = = = Rod CD: (20)2 314.16 mm2 314.16 10 6 4 CD A = π = = × − 9 6 2 (75 10 )(314.16 10 )(0.00225) 23.857 J (2)(2.5) CD U = × × − =
Rods AB and EF: (15)2 176.71 mm2 176.71 10 m6 2
4
AB EF
A = A = π = = × −
9 6 2
PROBLEM 11.47 (Continued)
Work of falling collar:
( ) (48)(9.81)( ) m m m U = mg h+ Δ = h+ Δ Equating, (48)(9.81)(h+ Δm) =95.205 0.20218 m 0.20218 0.00225 0.19993 m m h+ Δ = h = − = 200 mm h =
PROBLEM 11.48
The steel beam AB is struck squarely at its midpoint C by a 45-kg block moving horizontally with a speed v0 =2 m/s. Using
200 GPa,
E = determine (a) the equivalent static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of the midpoint C of the beam.
SOLUTION
From Appendix C, for W150 × 13.5,
6 4 6 4 3 3 6 3 6.83 10 mm 6.83 10 m 91.1 10 mm 91.1 10 m x x I S − = × = × = × = × Kinetic energy: 2 2 0 1 1 (45)(2) 90 J 2 2 T = m v = =
From Appendix D, Case 4:
3 max , 48 4 m PL PL y M EI = =
Principle of work and energy.
2 3 1 2 96 m m m P L U P y T EI = = =
(a) Equivalent static load.
9 6 3 3 3 96 (96)(200 10 )(6.83 10 )(90) 20.907 10 N (3.0) m EIT P L − × × = = = × 21.0 kN m P =
(b) Maximum normal stress.
3 6 max 6 (20.907 10 )(3.0) 172.1 10 Pa 4 (4)(91.1 10 ) m m M P L S S σ = = = × − = × × 172.1 MPa m σ = (c) Maximum deflection. 3 3 2 (2)(90) 8.61 10 m 20.907 10 m m U y P − = = = × × ym =8.61 mm
PROBLEM 11.49
Solve Prob. 11.48, assuming that the W150 × 13.5 rolled-steel beam is rotated by 90° about its longitudinal axis so that its web is vertical.
PROBLEM 11.48 The steel beam AB is struck squarely at its
midpoint C by a 45-kg block moving horizontally with a speed
0 2 m/s.
v = Using E = 200 GPa, determine (a) the equivalent static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of the midpoint C of the beam.
SOLUTION
From Appendix C, for W150 × 13.5,
6 4 6 4 3 3 6 3 0.916 10 mm 0.916 10 m 18.2 10 mm 18.2 10 m y y I S − − = × = × = × = × Kinetic energy: 2 2 0 1 1 (45)(2) 90 J 2 2 T = m v = =
From Appendix D, Case 4:
3 max , 48 4 m PL PL y M EI = =
Principle of work and energy.
2 3 1 2 96 m m m P L U P y T EI = = =
(a) Equivalent static load.
9 6 3 3 3 96 (96)(200 10 )(0.916 10 )(90) 7.657 10 N (3.0) m EIT P L − × × = = = × 7.66 kN m P =
(b) Maximum normal stress.
3 6 max 6 (7.657 10 )(3.0) 316 10 Pa 4 (4)(18.2 10 ) m m M P L S S σ = = = × − = × × 316 MPa m σ = (c) Maximum deflection. 3 3 2 (2)(90) 23.5 10 m 7.657 10 m m U y P − = = = × × ym = 23.5 mm
PROBLEM 11.50
A 25-lb block C moving horizontally with a velocity v0 hits the post AB squarely as shown. Using E = 29 10 psi,× 6 determine the largest speed v0 for which the maximum normal stress in the pipe does not
exceed 18 ksi.
SOLUTION
W5 × 16: 21.4 in ,4 8.55 in3
x x
I = S =
Maximum stress: σm =18 ksi
Maximum bending moment:
3 (18 ksi)(8.55 in ) 153.9 kip in m m x M =σ S = = ⋅ Equivalent force: P Lm = Mm 153.9 kip in 1.71 kips 1710 lb 90 in. m m M P L ⋅ = = = = From Appendix D, 3 3 6 1710)(90) 0.66956 in. 3 (3)(29 10 )(21.4) 1 1 (1710)(0.66956) 572.48 in lb 2 2 47.706 ft lb m m m m m P L y EI U P y ( = = = × = = = ⋅ = ⋅ Kinetic energy: 2 0 1 2 W T v g = 2 2 0 0 25 0.3882 ft lb (2)(32.2) T = v = v ⋅ Equating, T =Um 2 0 0.3882v = 47.706 Maximum speed. v0 =11.09 ft/s
PROBLEM 11.51
Solve Prob. 11.50, assuming that the post AB has rotated 90° about its longitudinal axis.
PROBLEM 11.50 A 25-lb block C moving horizontally with a
velocity v0 hits the post AB squarely as shown. Using E =29 10 psi,× 6 determine the largest speed v0 for which the maximum normal stress in
the pipe does not exceed 18 ksi.
SOLUTION
W5 × 16: 7.51 in ,4 3.00 in3
y y
I = S =
Maximum stress: σm =18 ksi
Maximum bending moment:
3 (18 ksi)(3.00 in ) 54.0 kip in m m y M =σ S = = ⋅ Equivalent force: P Lm = Mm 54.0 kip in 0.600 kips 600 lb 90 in. m m M P L ⋅ = = = = From Appendix D, 3 3 6 600)(90) 0.66945 in. 3 (3)(29 10 )(7.51) 1 1(600)(0.66945) 200.83 in lb 2 2 16.736 ft lb m m m m m P L y EI U P y ( = = = × = = = ⋅ = ⋅ Kinetic energy: 2 0 1 2 W T v g = 2 2 0 0 25 0.3882 ft lb (2)(32.2) T = v = v ⋅ Equating, T =Um 2 0 0.3882v =16.736 Maximum speed. v0 = 6.57 ft/s
PROBLEM 11.52
The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E=200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.
SOLUTION 4 4 3 4 9 4 3 16 3.2170 10 mm 4 2 4 2 3.2170 10 m 8 mm 8 10 m 0.6 m 2 AB d I d c L π π − − = = = × = × = = = × = Appendix D, Case 1: 3 9 9 3 3 3 3 2 3 2 3 3 (3)(200 10 )(3.217 10 ) 8.9361 10 (0.6) 1 1 (8.9361 10 ) 4.4681 10 2 2 m AB m m m AB m m m AB m m m m m P L y M P L EI EI P y y L U P y y y − = = × × = = = × = = × = ×
Work of dropped weight: ( ) (2)(9.81)(0.040 ) 0.7848 19.62 m m m mg h y y y + = + = +
Equating work and energy,
3 2 0.7848 19.62+ ym =4.4681 10× ym 2 4.3911 10 3 175.645 10 6 0 m m y − × − y − × − = (a) 1
{
4.3911 10 3 (4.3911 10 )3 2 (4)(175.645 10 )6}
2 m y = × − + × − + × − 3 15.629 10 m− = × ym=15.63 mm 3 3 (8.9361 10 )(15.629 10 ) 139.66 N m P = × × − = (b) Mm= −P Lm AB = −(139.66)(0.6) |Mm| 83.8 N m= ⋅ (c) 3 6 9 | | (83.8)(8 10 ) 208 10 Pa 3.2170 10 m m M c I σ = = × −− = × × σm=208 MPa PROBLEM 11.53
The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E=200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.
SOLUTION 4 4 3 4 9 4 3 16 3.2170 10 mm 4 2 4 2 3.2170 10 m 8 mm 8 10 m 0.6 m 2 d I d c a π π − − = = = × = × = = = × = Over AB: M = −P xm Mm= −P am 2 2 2 3 0 3 2 9 9 6 2 2 6 (0.6) (6)(200 10 )(3.2170 10 ) 55.953 10 a m m AB m m P x P a U dx EI EI P P − − = = = × × = ×
By symmetry of bending moment diagram,
6 2 6 2 6 2 3 3 2 55.953 10 111.906 10 1 111.906 10 2 4.4681 10 1 2.2340 10 2 BC AB m m AB BC m m m m m m m m m m m U U P U U U P P y U P P y U P y y − − − = = × = + = × = = × = × = = ×
Work of dropped weight: ( ) (2)(9.81)(0.040 ) 0.7848 19.62 m m m mg h y y y + = + = +
Equating work and energy,
3 2
0.7848 19.62+ ym=2.2340 10× ym 2 8.7825 10 3 351.298 10 6 0
m m
PROBLEM 11.53 (Continued) (a) 1
{
8.7825 10 3 (8.7282 10 )3 2 (4)(351.298 10 )6}
2 m y = × − + × − + × − 3 23.636 10 m 23.6 mm− = × = ym=23.6 mm 3 6 (4.4681 10 )(23.636 10 ) 105.61 N m P = × × − = (b) Mm= −(105.61)(0.6)= −64.4 N m⋅ |Mm| 64.4 N m= ⋅ (c) 3 6 9 | | (64.4)(8 10 ) 157.6 10 Pa 3.2170 10 m m M c I σ = = × −− = × × σm=157.6 MPa PROBLEM 11.54
The 45-lb block D is dropped from a height h =0.6 ft onto the steel beam AB. Knowing that E =29 10 psi,× 6 determine (a) the maximum deflection at point E, (b) the maximum normal stress in the beam.
SOLUTION
S5 × 10 rolled steel shape:
4 4 12.3 in 4.90 in 2 ft 24 in. 4 ft 48 in. 6 ft 72 in. 0.6 ft 7.2 in. x x I S a b L h = = = = = = = = = = , , m m m A B m E P b P a P ab R R M M L L L = ↑ = ↑ = =
From Appendix D, Case 5,
2 2 2 2 6 6 (24) (48) 17.2245 10 3 (3)(29 10 )(12.3)(72) 58057 m m E m m E P a b P y P EIL P y − = = = × × =
Total strain energy: 1 29028 2
2
m m m E
U = P y = y
Work of falling weight: W h( + yE) =45(7.2+ yE) =324+45yE
Equating work and energy: 324+45yE = 29028yE2
2 1.5502 10 3 11.1615 10 3 0 E E y − × − y − × − = (a) 1 1.5502 10 3 (1.5502 10 )3 2 (4)( 11.1615 10 )3 2 E y = × − + × − − − × − Deflection at E. yE = 0.1061 in. (b) 3 3 3 3 3 (58057)(0.1064) 6.179 10 lb (6.179 10 )(24)(48) 98.86 10 lb in 72 98.86 10 20.2 10 psi 4.90 m m m m x P M M S σ = = × × = = × ⋅ × = = = ×