BeerMOM_ISM_C11.pdf

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C

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PROBLEM 11.1

Determine the modulus of resilience for each of the following grades of structural steel:

( ) ASTM A709 Grade 50: 50 ksi

Y Y Y a b c       SOLUTION

Structural steel: _E__{29 10 psi}_ 6 _{for all three steels given.}

(a) _{50 ksi 50 10 psi}3

Y     2 3 2 6 (50 10 ) 2 (2)(29 10 ) Y Y u E      3 43.1 in. lb/in Y u    (b) _Y 65 ksi 65 10 psi  3 2 6 2 6 (65 10 ) 2 (2)(29 10 ) Y Y u E      3 72.8 in. lb/in Y u    (c) _{100 ksi 100 10 psi}3 Y     2 3 2 6 (100 10 ) 2 (2)(29 10 ) Y Y u E      3 172.4 in. lb/in Y u   

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PROBLEM 11.2

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          Y Y Y a E b E c E SOLUTION Aluminum alloys: (a) E70 10 Pa 9 _Y 55 10 Pa 6 2 6 2 3 3 9 (55 10 ) 21.6 10 N m/m 2 (2)(70 10 ) Y Y u E         3 21.6 kJ/m Y u   (b) _{72 10 Pa}9 _{220 10 Pa}6 Y E     2 6 2 3 3 9 (220 10 ) 336 10 N m/m 2 (2)(72 10 ) Y Y u E         3 336 kJ/m Y u   (c) E69 10 Pa 9 _Y 150 10 Pa 6 2 6 2 3 3 9 (150 10 ) 163.0 10 N m/m 2 (2)(69 10 ) Y Y u E         3 163.0 kJ/m Y u  

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PROBLEM 11.3

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          SOLUTION (a) E190 10 Pa, 9 _Y 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         3 177.9 kJ/m Y u   (b) _{190 10 Pa,}9 _{520 10 Pa}6 Y E     2 6 2 3 3 9 (520 10 ) 712 10 N m/m 2 (2)(190 10 ) Y Y u E         3 712 kJ/m Y u   (c) E165 10 Pa, 9 _Y 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         3 160.3 kJ/m Y u  

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PROBLEM 11.4

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             SOLUTION

(a) _{16.5 10 psi,}6 _{120 10 psi}3

Y E     2 3 2 6 (120 10 ) 2 (2)(16.5 10 ) Y Y u E      3 436 in. lb/in Y u    (b) E6.5 10 psi, 6 _Y 29 10 psi 3 2 3 2 6 (29 10 ) 2 (2)(6.5 10 ) Y Y u E      3 64.7 in. lb/in Y u    (c) _{20 10 psi,}6 _{16 10 psi}3 Y E     2 3 2 6 (16 10 ) 2 (2)(20 10 ) Y Y u E      3 6.40 in. lb/in Y u   

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0.002 0.021 0.2 0.25 100 (ksi) 80 60 40 20 0 ␴ ⑀

PROBLEM 11.5

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) _Y E_Y 2 2 6 2 1 1 (29 10 )(0.002) 2 2 2 Y Y Y u E E       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      2 3 3 2_{(20)(0.25 0.075) 2.33 kips/in} 3 2.33 in. kip/in A      modulus of toughness u_Y A₁A₂ A₃

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(MPa) 600 450 300 ⑀ 150 0.006 0.14 0.18 ␴

_{PROBLEM 11.6}

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) _Y E_Y 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  _        _{1296 kJ/m}3 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) 90.0 10 N/m .}_ 6 _ _ 6 2

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400 300 200 100 2.8 50 P (kN) (mm) d P P' 400 mm d

PROBLEM 11.7

The load-deformation diagram shown has been drawn from data obtained during a tensile test of a specimen of an aluminum alloy. Knowing that the cross-sectional area of the specimen was 600 mm2 and that the deformation was measured using a 400-mm gage length, determine by approximate means (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.

SOLUTION 6 2 600 10 m    _  P P A 400 mm      L Draw curve:  (a) Modules of resilience:

(shaded area) 1 (500 MPa)(0.007) 2  Y u 3 1.750 MJ/m  Y u 

(b) Modules of toughness: (total area under   curve)

3 1 1.750 MJ/m (500 MPa)(0.125 0.007) (6.33 500)(0.125 0.007) 2       3 3 3 3 1.750 MJ/m 59 MJ/m 10.46 MJ/m 71.2 MJ/m     

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0.025 0.36 3.2 4 15 20 P (kips) (in.) d 10 5 P' d 18 in. P

PROBLEM 11.8

The load-deformation diagram shown has been drawn from data obtained during a tensile test of a 5

8-in.-diameter rod of structural steel. Knowing that the deformation was measured using an 18-in. gage length, determine by approximate means (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.

SOLUTION 5 -in.-diameter rod: 3 2 2 5 0.3068 in 4 8    _{ }    A 2 12.5 kips 40 ksi 0.3068 in P A     3 0.025 in. 1.389 10 18 in. L        40 ksi_Y  Draw - curve:  (a) Mod. of resilience:

(shaded area) 3 3 1 1 (40 10 psi)(1.389 10 ) 2 2 Y Y Y u        3 28.0 in. lb/in Y u   

(b) Mod. of toughness: (total area under -  curve)

2 1

28 in. lb/in (40 ksi)(0.02 0.0014) (40 65)(0.1778 0.02) (62.5)(0.222 0.1778) 2

        

2

28 744 9860 2760 13,390 in. lb/in

     

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P B C 2 ft 3 ft A in. 3 4 in. 5 8

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  _      U uV

Portion d(in.) L(in.) A(in2) V(in3) (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

 _{176.24 10}_ 3

(a) _U __{176.2 10 in. kip}_ 3 _ _176.2_U _ _{in. lb}_{ }

(b) _{In :}_AB _u__{11.72 10 in. kip/in}_ 3 _ 3 _{11.72 in. lb/in}3

AB

u   

3 3

In :BC u5.65 10 in. kip/in   _{5.65 in. lb/in}3

BC

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20-mm diameter 1.2 m 0.8 m 2 m 16-mm diameter P B A C

PROBLEM 11.10

Using 200E GPa, 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             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) 5.968 6.213 12.18 U    N m 12.18 U  J  (b) 3 6 6 25 10 79.58 10 Pa 314.16 10 AB AB P A     _    2 6 2 3 9 (79.58 10 ) 15.83 10 2 (2)(200 10 ) AB AB u E        3 15.83 kJ/m AB u   3 6 6 25 10 124.28 10 Pa 201.16 10 BC AB P A     _    2 6 2 3 9 (124.28 10 ) 38.6 10 2 (2)(200 10 ) BC BC u E        3 38.6 kJ/m BC u  

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30 in. D B A E F P C 48 in.

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

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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d   (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 (8000) (48)₆2 119.89 in. lb 2 (2)(29 10 )(0.4418) EF P L U EA     

Total: U U_CD U_EF 168.8 in. lb 168.8 U  in. lb. 

(b) Strain energy density. : CD 8000 4324 psi, 1.85     2 ( 4324)2₆ 2 (2)(10.6 10 ) u E      3 0.882 in. lb/in u   : EF 8000 18,108 psi, 0.4418    2 (18,108)2₆ 2 (2)(29 10 ) u E     3 5.65 in. lb/in u  

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1.25 m 0.5 m 5 mm 20 mm B A D C E P

PROBLEM 11.12

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 _all 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 d            Double shear: 6 6 3 2 (2)(28.274 10 )(85 10 ) 4.8066 10 N P A       

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: 0.846 JU U _ABU_DBU_BE  0.846 JU  

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10-mm diameter a 6 m 6-mm diameter P B A C

PROBLEM 11.13

Rods AB and BC are made of a steel for which the yield strength is Y  300 MPa and the modulus of elasticity is E  200 GPa.

Determine the maximum strain energy that can be acquired by the assembly without causing any permanent deformation when the length a of rod AB is (a) 2 m, (b) 4 m.

SOLUTION 2 2 6 2 (10) 78.54 mm 78.54 10 m 4       AB A 2 2 6 2 (6) 28.274 mm 28.274 10 m 4       BC A 6 6 min (300 10 )(28.274 10 )    _Y    P A 3 8.4822 10 N   2 2 

_

P L U EA (a) a  2 m L   a 6 2 4 m 3 2 3 2 9 6 9 6 (8.4822 10 ) (2) (8.4822 10 ) (4) (2)(200 10 )(78.54 10 ) (2)(200 10 )(28.274 10 )         U 4.5803 25.4466 30.0 N m 30.0 J       (b) a  4 m L   a 6 4 2 m 3 2 3 2 9 6 9 6 (8.4822 10 ) (4) (8.4822 10 ) (2) (2)(200 10 )(78.54 10 ) (2)(200 10 )(28.274 10 )         U 9.1606 12.7233 21.9 N m 21.9 J      

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1.8 m

B C

P

PROBLEM 11.14

Rod BC is made of a steel for which the yield strength is _Y 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          Y Y U  ALu 3 6 2 60 (1.8)(225 10 ) 148.148 10 m Y Y U A Lu       2 4 A  d 6 3 4 (4)(148.148 10 ) 13.73 10 m A d          13.73 mm d 

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P A x _{900 mm} 18-mm diameter 12-mm diameter B C

PROBLEM 11.15

The assembly ABC is made of a steel for which E  200 GPa and 320 _Y  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) 600 x  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                        

Force at yielding or allowable axial force.

6 6 3 min (320 10 )(113.097 10 ) 36.191 10 N Y Y P  P  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: 5 U  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: 21.225 5  Y U U . .F S  4.25 

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L B 2c c A P

PROBLEM 11.16

Show by integration that the strain energy of the tapered rod AB is

2 min 1 4 P L U EA 

where A_min is the cross-sectional area at end B.

SOLUTION Radius: 2 min cx r A c L    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            _ _  



2 2 min 1 1 2 2 P L EA L L    _  _   2 min 4 P L U EA  

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4 @ 1.5 in. 6 in. A B 1.5 in.2.10 in. 2.55 in.2.85 in. 3 in. P

PROBLEM 11.17

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 _all 22ksi.

SOLUTION

2 2

min ₄(1.5) 1.7671 in A  

all

all all min

22,000 psi 38,877 lb P A      2 2 2 2 2 2 2 2 4 P dx P dx P dx U EA E  d E 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  2.2625 1 2 2 1 1.5_(2.2625) _{1.13125 in} 3 3 B A dx h _m d  d -æ _ö÷ ç = ç_ç_è ÷_÷_÷_ø= =

ò

2 6 (2)(38877) (1.13125) (10.6 10 ) U    102.7 U  in. lb 

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l l D B C 1 2 l 1 2 A A P

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 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  

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C 308 B l A A D P

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 3 2 0: 0 2 3 y CD CD F F P F P        1 1 0: 0 2 3 x BC CD BC F F F F P       2 ₁ 2 2 2 

_

F L 

_

F L U EA E A 2 1 3 2    _ _   P l U E A 2 1.5  P l U EA  Member F L A F2L/A BC 1 3P l A 2 1 / 3P l A CD 2 3P  2l A 8 2/ 3P l A  ₃_{P l A}2_/

(22)

C 30° B l A A A D P

PROBLEM 11.20

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 y CD CD F   F P  F   P



1 1 0: 0 2 3 x BC CD BC F  F  F  F  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 4.732 2 P l U E A    _ _   2 2.37P l U EA   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  4.732P2l/A

(23)

D B C l 3 4 l 2A A 2A P

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 3 3 0: 0 5 5 F_x  F_CD  F_CB   CB CD F F 4 0: 2 0 5 y CD F P F       5 8    CB CD F F P 3 0: 0 5 F_x  F_BD  F_CD  3 0 5 F_BD  F_CD  3 5 3 5 8 8     BD F P P 2 2 2 2 1 2 2 1 179 2 384 179 768        _ _    F L F L U EA E A P l E A P l EA 2 0.233  P l U EA  Member F L A _{F L A}2 _/ CB 5 8P  5 6l 2A 2 125 / 768P l A CD 5 8P  5 6l 2A 2 125 / 768P l A BD 3 8P l A 2 9 / 64P l A  179_{P l A}2_/

(24)

30 kN 80 kN 2500 mm2 2000 mm2 C D B 2.2 m 1 m 2.4 m

PROBLEM 11.22

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.



MB 0: (30)(2.4)(80)(3.2)Dy(2.2)  0 83.636 kN y D  _ 0: 80 0    



Fy Dy By 83.636B_y 80 0 B_y 3.636 kN_ Member forces. 4 m 4 (3.636 kN) 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              0.510 J59.290 J U 59.8 J_

(25)

D 2.5 ft 6 ft 3 in2 2 in2 5 in2 B C 2.5 ft 40 kips 24 kips

PROBLEM 11.23

Each member of the truss shown is made of aluminum and has the cross-sectional area shown. Using E  10.5  106_{psi, determine the}

strain energy of the truss for the loading shown.

SOLUTION 2 2 6 2.5 6.5 ft 78 in.      BC CD L L Joint C: 6 6 0: 24 0 6.5 6.5 F_x   F_BC  F_CD   (1) 2.5 2.5 0: 40 0 6.5 6.5 F_y   F_BC  F_CD   (2)

Solving (1) and (2) simultaneously,

65 F_BC  kips F_CD  39 kips Joint D 2.5 0: 0 15 kips 6.5 F_y  F_BD F_CD  F_BD 2 ₁ 2 2 2  F L  F L U EA E A

Member F(10 lb)3 L(in.) A(in )2 F L A2 / (10 lb /in.)9 2

BC 65 78 3 109.85 BD 15 60 2 6.75 CD 39 78 5 23.73  140.33 9 6 140.33 10

6682 lb in. 6.68 kip in. (2)(10.5 10 )



    



(26)

w

B A

L

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  

(27)

A

a L

B D

P

PROBLEM 11.25

SOLUTION 0: 0 M_D  aPLR_B      B aP aP R L L

Over portion AD:

  M Px 2 2 3 2 3 2 2 0 0 0 1 2 2 2 3 6 

_



_

  a a a AD M P x P a U dx P x dx EI EI EI EI Over portion DB:  aP M v L 2 2 2 2 2 0 0 1 2 2 

_

L 

_

L DB M a P U dv v dv EI EI L 2 2 2 2 3 2 2 2 2 0 2 0 3 6 2 2 

_

  L L P a P a v P a L v dv EI EIL EIL Total: U U_AD U _DB 2 2( ) 6  P a  U a L EI 

(28)

D E B A a a L P P

PROBLEM 11.26

SOLUTION

Symmetric beam and loading: R_A R_B

0: 2 0

y A B A B

F R R P R R P

      

Over portion AD, M R x Px_A 

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U_DEU_EB 2 2(3 4 ) 6 P a U L a EI   

(29)

M0 A B D L a b

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 0 0 0: 0 B A A M M R L M R L        0 0 0: 0 A B B M M R L M R L       A to D: M_J _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     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 U_DB 02 3 3 2 ( ) 6 M a b U EIL   

(30)

S8 3 18.4 6 ft 3 ft 8 kips D B A

PROBLEM 11.28

Using _E__{29 10 psi,}_ 6 _{determine the strain energy due to}

bending for the steel beam and loading shown. (Neglect the effect of shearing stresses.)

SOLUTION 0: 0 M_D  R L_A aP    A aP R L

Over portion AD: M  aPx

L 2 2 0 0 1 2 2     _ _  



L



L AD M aP U dx x dx EI EI L 2 2 2 2 0 2  P a

_

Lx dx EIL 2 2 6  P a L EI Over portion DB: M  Pv 2 2 3 2 2 0 0 1 2 2 6 

_

a 

_

a  BD M P a U dv P x dx EI EI EI Total: 2 2 ( ) 6  _AD  _DB  P a  U U U a L EI Data: 6 4

8000 lb, 6 ft 72 in., 3 ft 36 in., 29 10 psi

57.5 in P L a E I         2 2 6 (8000) (36) (72 36) (6)(29 10 )(57.5)    U 895 in. lb U   

(31)

B A C D D 60 in. 15 in. 15 in. 1.5 in. 3 in.

2 kips 2 kips

PROBLEM 11.29

Using E29 10 psi, 6 determine the strain energy due to bending for the steel beam and loading shown. (Neglect 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     322 U  in. lb 

(32)

B C 180 kN A 2.4 m 2.4 m 4.8 m W360 64

PROBLEM 11.30

Using E200 GPa, determine the strain energy due to bending for the steel beam and loading shown. (Neglect 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   _     1048JU  

(33)

B D E 80 kN A 80 kN W310 74 1.6 m 1.6 m 1.6 m 4.8 m

PROBLEM 11.31

Using 200 E  GPa, determine the strain energy due to bending for the steel beam and loading shown. (Neglect 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   _     670 U  J 

(34)

w

B A

L

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 45 8 U u V 

where U is the strain energy of the beam and V is its volume.

SOLUTION 1 0: ( ) 0 2 2 B A A L M  R L wL  R  wL 2 2 1 1 ( ) 2 2 A M R x wL  w Lx x 2 2 2 3 4 0 0 2 2 3 4 5 0 2 5 2 5 ( 2 ) 2 8 2 8 3 4 5 1 1 1 8 3 2 5 240 L L L M w U dx L x Lx x dx EI EI w L x Lx x EI w L EI w L EI        _   _      _   _   



2 2 max 2 max max 1 1 2 2 2 8 8 L L M w L wL M c wL c I I   _{ }     _{ }         





 

2 2 4 2 max max ₂ 3 1 12 2 2 max ₂ 2 128 8 8 15 ₁₅ 8 8 45 45 d w L c u E EI L bd U LI u c Lbd V        max 45 8 U u V  

(35)

30 mm A B C 0.9 m 0.75 m T_A 5 300 N · m TB 5 400 N · m 46 mm

PROBLEM 11.33

In the assembly shown, torques TA and TB are exerted on disks A and B,

respectively. Knowing that both shafts are solid and made of aluminum (G  73 GPa), determine the total strain energy acquired by the assembly.

SOLUTION

Over portion AB: 300 T_AB T_A  N m

4 4 30 _{79.52 10 mm}3 4 _{79.52 10 m}9 4 2 2 2       _ _       AB J c 0.9 L_AB  m 2 2 9 9 (300) (0.9) 2 (2)(73 10 )(79.52 10 )     AB AB AB AB T L U GJ 6.977 J  Over portion BC: 300 400 700 N m, 0.75 m        BC A B BC T T T L 4 3 4 9 4 46 439.57 10 mm 439.57 10 m 2 2     _ _       BC J 2 2 9 9 (700) (0.75) 5.726 J 2 (2)(73 10 )(439.57 10 )      BC BC BC BC T L U GJ Total: 6.977U U_AB U_BC  5.726 12.70 J 

(36)

2.5 in. 25 kip · in. 36 in. B A

PROBLEM 11.34

The design specifications for the steel shaft AB require that the shaft acquire a strain energy of 400 in.  lb as the 25-kip  in. torque is applied. Using G  11.2  106_{psi, determine (a) the largest inner diameter of the}

shaft that can be used, (b) the corresponding maximum shearing stress in the shaft.

SOLUTION

3 400 in. lb

25 kip in. 25 10 lb in.

U T        48 L in. 2 2  T L U GJ 2 3 2 4 6 (25 10 ) (48) 3.3482 in 2 (2)(11.2 10 )(400)      T L J GU But





4 4 4 4 0 0 2 2 2 32        _ _ _ _            i i d d J d d (a) 4 4 0 4 4 32 J 32 2.5 (3.3482) 4.9580 in        i d d 1.492 in.  i d  (b) 0 3 3 (25 10 )(1.25) 2.5112 9.33 10 psi       Tc J 9.33 ksi   

(37)

L B 2c c A T

PROBLEM 11.35

Show by integration that the strain energy in the tapered rod AB is

2 min 7 48 T L U GJ 

where J_min 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                 



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  

(38)

z σz 75 MPa y x 100 MPa 20 MPa

PROBLEM 11.36

The state of stress shown occurs in a machine component made of a brass for which _Y 160 MPa. Using the maximum-distortion-energy criterion, determine the range of values of _z 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         _    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              _ _               2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) 2 (145 25) ( 25 ) ( 145) (2)(160) 28,900 (625 50 ) ( 290 21,025) 51,200 2 240 650 0 a b b c c a Y z z z z z z z z                                       2 240 240 (4)(2)(650) 60 62.65 (2)(2) 122.65 MPa, 2.65 MPa z z          2.65 MPa <_z < 122.65 MPa  

(39)

z σz 75 MPa y x 100 MPa 20 MPa

PROBLEM 11.37

The state of stress shown occurs in a machine component made of a brass for which _Y 160 MPa. Using the maximum-distortion-energy criterion, determine whether yield occurs when (a) 45 _z   MPa, (b) 45 _z   MPa. SOLUTION ave 1 (100 20) 2 60 MPa 100 20 2 2 40 MPa 75 MPa x y xy         _    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                              ? 2 2 2 2 (_a _b) (_b _c) (_c_a) 2_Y (a) 45 _c _z   MPa ? 2 2 2 2 (145 25)   ( 25 45) (45 145)  2(160) 51,200 28,900 4900 10,000 43,800 51,200    (No yield.)  (b) 45 _c _z   MPa ? 2 2 2 (145 25)   ( 25 45)   ( 45 145) 51,200 28,900 400 36,100 65,400   51,200 (Yield occurs.) 

(40)

z x 8 ksi 14 ksi y σy

PROBLEM 11.38

The state of stress shown occurs in a machine component made of a grade of steel for which _Y 65 ksi. Using the maximum-distortion-energy criterion, determine the range of values of _y 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         _    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   _           _ _               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                       _{ } _           _{ } _             2 16 16 (4)(2)(441.92) (2)(2) 4 15.39 19.39 ksi, 11.39 ksi y y          11.39 ksi _y 19.39 ksi    

(41)

z x 8 ksi 14 ksi y σy

PROBLEM 11.39

The state of stress shown occurs in a machine component made of a grade of steel for which _Y 65 ksi. Using the maximum-distortion-energy criterion, determine the factor of safety associated with the yield strength when (a) 16 _y  ksi, (b) 16 _y  ksi.

SOLUTION ave 1 (0 8) 2 4 ksi 8 0 2 2 4 ksi 14 ksi x z xz         _    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   _           _ _               2 2 2 2 ( ) ( ) ( ) 2 . . Y a b b c c a _{F S}          _{ } _   (a) 2 2 2 2 16 ksi 65 (18.56 10.56) ( 10.56 16) (16 18.56) 2 . . c y F S            _{ } _   2 8450 847.97 705.43 6.55 ( . .)F S    . . 2.33F S   (b) 16 _c_y  ksi 2 2 2 2 65 (18.56 10.56) ( 10.56 16) ( 16 18.56) 2 . . F S          _{ } _   8450 847.97 29.59 1194.39   . . 2.02F S  

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B b d A L M0

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 M0_v L  For bending, 2 2 2 0 1 ₀ 2 ₀ 3 2 0 0 2 2 2 6 6 L_M _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 u dv ub dy dx dy dx Gb d L c c M y y y dx Gb d L c c       _  _          _  _  _   _          _   _      _   _   



 

  2 0 2 2 0 0 4 2 2 3 5 8 L _M L c c c dx Gb d L  _ _     



2 2 2 0 0 0 2 2 2 9 16 6 3 15 5 5 8 M M c M c L Gb dL Gb d L Gb d L  _{ } _     Total: 2 2 0 0 1 2 3 6 5 M L M U U U EI Gb dL     with 1 3 12 I  b d 2 2 0 0 3 2 3 5 M L M U Gb dL Eb d   02 2 3 2 2 3 1 10 M L Ed U Eb d GL             

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Q B A L (b) A B R2 R1 (a) A Q

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           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        









 

 

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   

 





2 2 1 4 R Q U ln GL R   

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1.2 m A V₀ B D

PROBLEM 11.42

A 5-kg collar D moves along the uniform rod AB and has a speed

0 6

v m/s when it strikes a small plate attached to end A of the rod. Using E  200 GPa and knowing that the allowable stress in the rod is 250 MPa, determine the smallest diameter that can be used for the rod.

SOLUTION 2 2 0 2 2 max 9 6 2 2 6 2 max 1 1 (5)(6) 90 J 2 2 ( ) 2 2 2 (2)(200 10 )(90) 480 10 m (250 10 ) (1.2) m m m m U mv P L A L U EA EA EU A L               6 2 4 (4)(480 10 ) _{24.7 10 m}3 4 A d A d            24.7 d mm 

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3.5 ft A E B D C v0

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 _Y  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  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         Total: 2 Y m AB CD AL U U U E     2 2 0 0 1 1 2 2 m W U mv v g   Solving for v ₀2, 2 2 0 2gU_m 2g _YAL v W EW    2 0 2g _YAL v EW   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         _{ }            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  

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3.5 ft A E B D C v0

PROBLEM 11.44

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 _Y 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 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                 _ _ _{ } _     Solving for W, 2 2 2 0 0 2 2 ( . .) ( . .) m Y gU g AL W v F S v F S E   

Data: _g__{32.17 ft/ sec}2 __{386 in./ sec ,}2 _{50 10 psi,}3

Y    2 2 7 _{0.60132 in}2 4 4 8 Ad  _{ }    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    9.12 W  lb 

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2 m 1.5 m 40-mm diameter 30-mm diameter h B D A C m

PROBLEM 11.45

The 35-kg collar D is released from rest in the position shown and is stopped by a plate attached at end C of the vertical rod ABC. Knowing that

E  200 GPa for both portions of the rod, determine the distance h for which the maximum stress in the rod is 250 MPa.

SOLUTION

Portion BC has smaller cross section, thus

2 all (250 MPa) (0.030 m)₄ 176.7 kN      m BC P A

Maximum strain energy (for P_m 176.7 kN)

2 2 2 2 2 (176.7 kN) 2 m 1.5 m 2 2 2(200 GPa) _{(0.040 m)} _{(0.030 m)} 4 4          _  _    



m i m



i m i i P L P L U A E E A 3 78.06 10 [1591.6 2122.1] 289.9 J     m U Max. deflection: 1 2Pm m U m 1 (176.7 kN) 289.9 J 2  m 3 3.28 10 m 3.28 mm  _m   Work of weight U _m (  _m)  _m W h U (mg h)(  _m) U _m 2 (35 kg)(9.81 m/s )(h _m)  289.9 J 0.8443 h  _m m 844.3 mm 844.3h    _m 844.3 3.28 h 841 mm 

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2 m 1.5 m 40-mm diameter 30-mm diameter h B D A C m

PROBLEM 11.46

The 15-kg collar D is released from rest in the position shown and is stopped by a plate attached at end C of the vertical rod ABC. Knowing that

E  200 GPa for both portions of the rod, determine (a) the maximum deflection of end C, (b) the equivalent static load, (c) the maximum stress that occurs in the rod.

SOLUTION 200 GPa  E (a) 2 2 2 m 1.5 m 200 GPa _{(0.04 m)} _{(0.03 m)} 4 4           _  _    



m i m



i m m i i P L P L P A E E A 9 6 18.57 10 53.85 10 m m m m P P        Strain energy: 1 1_{(53.85 10 )}6 2 2 2      m m m m U P

Weight falls distance of h _m Work of weight  strain energy



6 2



1 ( ) 53.85 10 2  _m   _m W h (1) (35 kg)(9.81 m/s) 343.4 N    W mg For h 0.5 m Eq. (1): 343.4(0.5 ) 26.93 106 2 m m      Solve quadratic:  _m 2.531 10 m 3 2.53 mm  (b) P_m 53.85 10 6 _m (53.85 10 )(2.531 10 m) 6  3 136.3 kNP_m   (c) 2 136.3 kN 192.82 MPa (0.030 m) 4   m  _  m BC P A 192.8 MPam  

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2.5 m B F h D G A C E

_{PROBLEM 11.47}

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

  and E  200 GPa, determine the largest allowable distance h.

SOLUTION

Let  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                        For each rod,

2 ₍ _{/ )}2 2 2 2 2 m m m F L EA L L EA U EA EA L      Rod CD: (20)2 314.16 mm2 314.16 10 m6 2 4 CD A       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 mm}2 _{176.71 10 m}6 2

4 AB EF A  A       9 6 2 (200 10 )(176.71 10 )(0.00225) _{35.674 J} (2)(2.5) AB EF U U     

Total strain energy: 134.97 U_m U_AB U_CD U_EF  J

Work of falling collar:

( ) (48)(9.81)( )

m m m

U  mg h   h 

Equating, (48)(9.81)(h _m) 134.97 h  _m 0.28662 m

(50)

7.5 ft A B v0 W5 16 C

PROBLEM 11.48

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 in}3

x x

I  S 

Maximum stress: 18 _m  ksi

Maximum bending moment:

3

(18 ksi)(8.55 in ) 153.9 kip in.

m m x M  S    Equivalent force: P L_m M_m 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: 1 ₀2 2 W T v g  2 2 0 0 25 0.3882 ft lb (2)(32.2) T  v  v  Equating, T U_m 2 0 0.3882v  47.706 Maximum speed. v₀ 11.09 ft/s

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7.5 ft A B v0 W5 16 C

PROBLEM 11.49

Solve Prob. 11.48, assuming that the post AB has rotated 90 about its longitudinal axis.

PROBLEM 11.48 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 in}3

y y

I  S 

Maximum stress: 18 _m  ksi

Maximum bending moment:

3

(18 ksi)(3.00 in ) 54.0 kip in.

m m y M  S    Equivalent force: P L_m  M_m 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: 1 ₀2 2 W T v g  2 2 0 0 25 0.3882 ft lb (2)(32.2) T  v  v  Equating, T U_m 2 0 0.3882v 16.736 Maximum speed. v₀  6.57 ft/s

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100 mm B 0.9 m 0.9 m v0 80 mm 100 mm t = 10 mm C A

PROBLEM 11.50

An aluminum tube having the cross section shown is struck squarely in its midsection by a 6-kg block moving horizontally with a speed of 2 m/s. Using E  70 GPa, determine (a) the equivalent static load, (b) the maximum stress in the beam, (c) the maximum deflection at the midpoint C of the beam.

SOLUTION Kinetic energy: 1 ₀2 1(6 kg)(2 m/s)2 12 J 2 2    T mv Moment of inertia. Aluminium 70 GPa E  3 3 6 4 6 4 1 [80 100 60 80 ] 12 4.1067 10 mm 4.1067 10 m         aa I From Appendix D: 3 48  m m P L y EI 2 3 1 2 96   m m m m P L U P y EI (a) : 2(1.8 m)3 ₆ ₄ 12 J 7535.5 N 96(70 GPa)(4.1067 10 m ) m m m P U T _  P   7.54 kN  m P  (b) 1 1(7535.5 N)(1.8 m) 3391 N m 4 4     m m M P L 6 4 (3391 N m)(0.050 m) 41.28 MPa 4.1067 10 m m m M c I        41.3 MPa m    (c) 3 3 3 6 4 (7535.5 N)(1.8 m) 3.184 10 m 48 48(70 GPa)(4.1067 10 m )        m m P L y EI 3.18 mm  m y 

(53)

100 mm B 0.9 m 0.9 m v₀ 80 mm 100 mm t = 10 mm C A

PROBLEM 11.51

Solve Prob. 11.50, assuming that the tube has been replaced by a solid aluminum bar with the same outside dimensions as the tube.

PROBLEM 11.50 An aluminum tube having the cross

section shown is struck squarely in its midsection by a 6-kg block moving horizontally with a speed of 2 m/s. Using

E  70 GPa, determine (a) the equivalent static load, (b) the maximum stress in the beam, (c) the maximum deflection at the midpoint C of the beam.

SOLUTION

For solid aluminum bar,

3 6 4 1 80 100 6.667 10 m 12      a I

Follow solution of Prob. 10.50: T 12 J

2 ₁ 2 3 ; 48 2 96  m   m m m m m P L P L y U P y EI EI (a) 2 3 6 4 (1.8 m) : 12 J 96(70 GPa)(6.667 10 m )    m m P U T P_m 9601 N 9.60 kN  m P  (b) 1 1(9601 N)(1.8 m) 4320.5 N m 4 4     m m M P L 6 4 (4320.5 N m)(0.05 m) 32.40 MPa 6.667 10 m m m M c I        32.4 MPa m    (c) 3 3 3 6 4 (9601 N)(1.8 m) 2.500 10 m 48 48(70 GPa)(6.667 10 m )        m m P L y EI 2.50 mm  m y 

(54)

B D 0.6 m 2 kg 40 mm A

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        _{ }  _ _                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 y_m 4.4681 10 y_m 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   15.63 y_m mm  3 3 (8.9361 10 )(15.629 10 ) 139.66 N m P      (b) (139.66)(0.6)M_m  P L_{m AB}  |M_m| 83.8  N m  (c) 3 6 9 | | (83.8)(8 10 ) 208 10 Pa 3.2170 10 m m M c I     _    208 m  MPa 