SOLUTION Use Appendix D

3

Substitute given data:

6 3 4

(21.29 lb)(24 in.) 510.96 lb in.

  ^L   

A B

C

2.5 ft 9.5 ft 16 in.

2.65 in.

20 in.

PROBLEM 11.55

A 160-lb diver jumps from a height of 20 in. onto end C of a diving board having the uniform cross section shown.

Assuming that the diver’s legs remain rigid and using 1.8 106

E  psi, determine (a) the maximum deflection at point C, (b) the maximum normal stress in the board, (c) the equivalent static load.

SOLUTION

3 4

1 (16)(2.65) 24.813 in 129.5 ft 114 in.

Over portion AB:

2 2 2 2

Over portion BC:

2 2 3

1 35.799 2

2 4.4694 89.388 0

m m

B

PROBLEM 11.56

A block of weight W is dropped from a height h onto the horizontal beam AB and hits it at point D. (a) Show that the maximum deflection y_m at point D can be expressed as

where y represents the deflection at D caused by a static load W _st applied at that point and where the quantity in parenthesis is referred to as the impact factor. (b) Compute the impact factor for the beam and the impact of Prob. 11.52.

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

Work of falling weight: WorkW h y(  _m)

Strain energy: 1 1 ²

2 ^m 2 ^m

U  Py  ky where k is the spring constant for a load applied at point D.

Equating work and energy,

2

PROBLEM 11.56 (Continued)

For Problem 11.52, W mg (2)(9.81) 19.62 N 

4

9 3 4 9 4

3

200 10 Pa 16 3.217 10 mm 3.217 10 m 4 2

0.6 m 40 mm 40 10 m

E I

L h

 _



 

        

   

Using Appendix D, Case 1, ³

st 3 y WL

 EI

3 3

9 9

3 3

(19.62)(0.6)

2.196 10 m (3)(200 10 )(3.217 10 )

2 (2)(40 10 )

36.44 2.196 10

st

st

y h y









  

 

  



(b) Impact factor.  1 1 36.44 Impact factor 7.12 

B

PROBLEM 11.57

A block of weight W is dropped from a height h onto the horizontal beam AB and hits point D. (a) Denoting by y the exact value of _m the maximum deflection at D and by y the value obtained by _m neglecting the effect of this deflection on the change in potential energy of the block, show that the absolute value of the relative error is (y_m y_m)/ ,y_m never exceeding y_m/2 .h (b) Check the result obtained in part a by solving part a of Prob. 11.52 without taking y_m into account when determining the change in potential energy of the load, and comparing the answer obtained in this way with the exact answer to that problem.

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

1 1 2

2 ^{m m} 2 ^m U  P y  ky where k is the spring constant for a load at point D.

Work of falling weight: exact: Work ( ) approximate : Work

W h ym

Wh

 

 Equating work and energy, ²

2

1 ( ) (1) exact

2

1 (2) approximate

2

Relative error: 2

( )

PROBLEM 11.57 (Continued)

Approximate solution:

9 8.936 10 175.65 10 m Relative error: 15.63 13.25

15.63

A

L

a b

D B

P

PROBLEM 11.58

Using the method of work and energy, determine the deflection at point D caused by the load P.

SOLUTION

Reactions: _A Pb, _B Pa

EIL EIL

    

D B

a L

A

P

PROBLEM 11.59

Using the method of work and energy, determine the deflection at point D caused by the load P.

SOLUTION

See solution of Prob. 11.28 for beam of length L with overhang of length a, load at end of overhang.

2 2

( )

 P a6 

U a L

EI

1 1 2 2

: ( )

2 _D  2 _D  P a6 

Py U Py a L

EI ²( )

 3  

D

y Pa a L

EI 

M₀

A B

D

L

a b

PROBLEM 11.60

Using the method of work and energy, determine the slope at point D caused by the couple M ₀.

SOLUTION

Reactions: _A M⁰ _B M⁰

M₀

A D

B

a L

PROBLEM 11.61

Using the method of work and energy, determine the slope at point D caused by the couple M ₀.

SOLUTION

Reactions: _A  M⁰  _B  M⁰ 

R R

L L

Over portion AB:  M x⁰

M L

2 2

0 2

0 2 2 2 0





^L 



^L

AB

M M

U dx x dx

EI EIL

02

 M L6 EI Over portion BD: M  M ₀

02

 2

BD

U M a EI Total:

02( 3 ) 6

 

 

AB BD

U U U

M L a EI

0 0

1

2M  U

0

_D  2U M

0( 3 )

_D  M L3 ^ a

EI 

A B

L/2

2EI EI

L/2 C

P

PROBLEM 11.62

Using the method of work and energy, determine the deflection at point C caused by the load P.

SOLUTION

Bending moment: M  Pv

2EI

EI EI

P

A C B

a a a a

PROBLEM 11.63

Using the method of work and energy, determine the deflection at point C caused by the load P.

SOLUTION

Symmetric beam and loading:

1

A B 2 R R  P

From to ,A C 1

A 2

M R x Px

2 2

0

2 2 2 2 2

0

2 4

8 16

a a

AC a

a a

a

M M

U dx dx

EI EI

P P

x dx x dx

EI EI

 

 

 

 

2 3 2 2 3

3 3 3

24 48 (2 ) 16

P a P P a

a a

EI  EI   EI

By symmetry, 3 ^{2 3}

CB AB 16 U U P a

  EI

Total: 3 ^{2 3}

AB BC 8 U U U P a

   EI

1 2

2 ^{C C}

P U U

    P 3 ³

C 4 Pa

  EI  

2EI

Using the method of work and energy, determine the slope at point B caused by the couple M₀.

SOLUTION

0 0

2EI

Using the method of work and energy, determine the slope at point D caused by the couple M ₀.

SOLUTION

Reactions: _A M⁰ , _B M⁰

R R

L L

   

Bending moment diagram.

0

C

PROBLEM 11.66

The 20-mm-diameter steel rod BC is attached to the lever AB and to the fixed support C. The uniform steel lever is 10 mm thick and 30 mm deep. Using the method of work and energy, determine the deflection of point A when L600 mm. Use E200 GPa and G77.2 GPa.

SOLUTION

Member AB. (Bending)

3 3 4

(450)(0.500) 225 N m

B

(450) (0.500)

0.9375 J (6)(200 10 )(22.5 10 )

Member BC. (Torsion)

4 3 4 9 4

(225) (0.600)

12.5242 J 2 (2)(77.2 10 )(15.708 10 )

60 in.

PROBLEM 11.67

Torques of the same magnitude T are applied to the steel shafts AB and CD. Using the method of work and energy, determine the length L of the hollow portion of shaft CD for which the angle of twist C is equal to 1.25 times the angle of twist at A.

SOLUTION

T is the same for each shaft.

1 1

1.25 and

2 2

2.3873 0.039789

2 2 (25.138) 2 2

2.3873 0.018416

2 2

2.3873 0.018416 (1.25)(2.3873)

2 2 2

T T L T

G  G  G

E T

PROBLEM 11.68

Two steel shafts, each of 0.75-in. diameter, are connected by the gears shown. Knowing that G11.2 10 psi ⁶ and that shaft DF is fixed at F, determine the angle through which end A rotates when a 750-lb in. torque is applied at A. (Neglect the strain energy due to the bending of the shafts.)

SOLUTION

Work-energy equation:

1

Portion AB of shaft ABC:

750 lb in.

0.75 31.063 10 in

2 2 2 2

PROBLEM 11.68 (Continued)

Portion EF of shaft DEF:

4

3 4

1000 lb in.

8 in. 31.063 10 in

2 2

EF E

EF EF

T T

L J d ^

  

       



2 2

6 3

(1000) (8)

11.497 lb in.

2 (2)(11.2 10 )(31.063 10 )

EF EF EF

EF

U T L

GJ ^

   

 

Total: 20.389 U U _ABU_BC U_DE U_EF  in. lb

2 (2)(20.389) 3

54.4 10 rad

A 750

A

U

  T    ^ 3.12_A   

T_B

PROBLEM 11.69

The diameter steel rod CD is welded to the 20-mm-diameter steel shaft AB as shown. End C of rod CD is touching the rigid surface shown when a couple T_B is applied to a disk attached to shaft AB. Knowing that the bearings are self aligning and exert no couples on the shaft, determine the angle of rotation of the disk when 400 T_B N m. Use E  200 GPa and G  77.2 GPa.

(Consider the strain energy due to both bending and twisting in shaft AB and to bending in arm CD.)

SOLUTION

 _AB 0: _{CD C}  _{B C}  ^B

CD

M r F T F T

r

3

400 1333.3 N, 300 10^

Bending of shaft ADB:

0: 0

PROBLEM 11.69 (Continued)

4

3 4 9 4

7.854 10 mm 7.854 10 m 4 2

  

       

I d

(270 10 ) m^3

 

LAB

2 3 2 3 2

9 9 9

(1333.3) (70 10 ) (200 10 )

0.137 J (6)(200 10 )(7.854 10 )(270 10 )

 

 

 

 

  

U

Torsion: Only portion DB carries torque. J 2J 15.708 10 m  ^{9 4}

2 2 3

9 9

(400) (200 10 )

13.194 J 2 (2)(77.2 10 )(15.708 10 )





   

 

T LB DB

U GJ

Total: U 5.0930.137 13.194 18.424 J 1

2T_{B B} U

3

2 (2)(18.424) 400 92.1 10 rad





 

 

B B

U T

_B 5.28 

L

T T'

A

B

t

x ds

PROBLEM 11.70

The thin-walled hollow cylindrical member AB has a noncircular cross section of nonuniform thickness. Using the expression given in Eq. (3.50) of Sec. 3.10 and the expression for the strain-energy density given in Eq. (11.17), show that the angle of twist of member AB is

4 2

TL

 G

 A

ds t

where ds is the length of a small element of the wall cross section and A is the area enclosed by the center line of the wall cross section.

SOLUTION From Eq. (3.53),

2 T

 t

 A Strain energy density:

2 2

2 2

2 8

  T u G Gt Ꮽ

0

U 



L ut ds dx

2 0 8 2





^L _G^T_Ꮽ ^ds_t ^dx₈^{T L}_G²_Ꮽ^{2 ds}_t

Work of torque: 1 ² ₂

2  8

  T L

T GᏭ ds

t 4TL²

GᏭ ds t 

3l

4

l

P A B

C D

PROBLEM 11.71

Each member of the truss shown has a uniform cross-sectional area A. Using the method of work and energy, determine the horizontal deflection of the point of application of the load P.

SOLUTION

Members AB and BD are Zero force members.

Joint A: 4 5

0: 0

5 4

F_x  F_AD  P F_AD   P

3 3

0: 0

5 4

F_y  F_AC  F_AD  F_AC  P

Member F L F²L

AB 0 l 0

BD 0 3

4l 0

AD 5

 P 4 5

4l 125 ² 64 P l

CD P l P²l

AC 3

4P 3

4l 27 ² 64P l

 27 2

8 P l

Joint D: 4 5

0: 0

F_x  5 4 PF_CD 

CD  F P

2 1 2

2 2

 F L  

U F L

EA EA 27 2

 16 P l

EA Work of 1

P2P U

2 27

  U  8 Pl

P EA   3.38 Pl 

EA 

P

3l

4

l

A B

C D

PROBLEM 11.72

Each member of the truss shown has a uniform cross-sectional area A. Using the method of work and energy, determine the horizontal deflection of the point of application of the load P.

SOLUTION

Members AB, AC and CD are zero force members.

Joint B: 4 5

0: 0

5 4

F_x  P F_BC  F_BC  P

3 3

0: 0

5 4

F_y  F_BD  F_BC  F_BD   P

2 1 2

2 2

 F L  

U F L

EA EA 19 2

 16 P l EA Work of 1

 2   P P U

2 19

  U  8 Pl

P EA

Member F L F²L

AB 0 l 0

AC 0 3

4l 0

CD 0 l 0

BC 5

4P 5

4l 125 ²

64 P l

BD 3

 P 4 3

4l 27 ²

64P l

∑ 19 ²

8 P l

  2.38Pl  EA 

6 ft 6 ft 20 kips

2.5 ft A D

C B

PROBLEM 11.73

Each member of the truss shown is made of steel and has a uniform cross-sectional area of 5 in². Using E  29 10 psi, ⁶ determine the vertical deflection of joint B caused by the application of the 20-kip load.

SOLUTION

2 2

10 kips 6 ft 72 in.

Equilibrium of joint A.

0: 2.5 10 0 26 kips Strain energy:

2 1 2

0.64966 kip in.

(2)(29 10 )(5)

U   

 Vertical deflection of point B.

1 2

2 (2)(0.64966) P B U

1.2 m

60 kN

PROBLEM 11.74

Each member of the truss shown is made of steel. The cross-sectional area of member BC is 800 mm and for all other members the cross-² sectional area is 400 mm². Using E200 GPa, determine the deflection of point D caused by the 60-kN load.

SOLUTION

Entire truss: M_A 0: 2.4R_D(0.5)(60) 0 R_D12.5 kN 12.3633 10

30.908 J (2)(200 10 ) Work-energy:

3

1.030 mm

   

6 ft 6 ft

15 kips

PROBLEM 11.75

Each member of the truss shown is made of steel and has a cross-sectional area of 5 in². Using E29 10 psi, ⁶ determine the vertical deflection of point C caused by the 15-kip load.

SOLUTION

Members BD and AE are zero force members.

For entire truss, M_A 0: 2.5R_E (12)(15) 0 R_E 72 kips

39 kips

x 36 kips

x

Strain energy:

2 1 2

2.7488 kip in.

(2)(29 10 )(5)

12 kN

480 mm 480 mm

PROBLEM 11.76

The steel rod BC has a 24-mm diameter and the steel cable ABDCA has a 12-mm diameter. Using E  200 GPa, determine the deflection of joint D caused by the 12-kN load.

SOLUTION

Owing to symmetry, _{AB BD DC CA}

200 10 113.097 10 452.39 10

D ^{ }  ^{ } _   ^{ }_ ^  ^

     _D1.111 mm 

Joint D: Joint B:

L

B A

M₀

P

PROBLEM 11.77

Using the information in Appendix D, compute the work of the loads as they are applied to the beam (a) if the load P is applied first, (b) if the couple M is applied first.

SOLUTION

From Appendix D, Case 1,

3 AP 3 y PL

 EI ²

AP 2 PL

   EI From Appendix D, Case 3,

0 2 AM 2 y M L

 EI _AM M L⁰

  EI

(a) First P, then M₀.

1 2 3

0

1 1

2 ^{AP AM} 2 ^AM

U A A A

Py Py M 

  

  

2 2

2 3 0 0

6 2 2

PM L M L U P L

EI EI EI

   

(b) First M₀, then P.

4 5 6

0 0

1 1

2 ^{AP AP} 2 ^AM

U A A A

Py M  M 

  

  

2 2

2 3 0 0

6 2 2

M PL M L U P L

EI EI EI

   



B

A C

M₀

L/2 L/2

P

PROBLEM 11.78

Using the information in Appendix D, compute the work of the loads as they are applied to the beam (a) if the load P is applied first, (b) if the couple M is applied first.

SOLUTION

Appendix D, Cases 1 and 3,

A B

C

L/2 L/2

P

P

PROBLEM 11.79

For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec. 11.2A and show that it is equal to the work obtained in part a.

SOLUTION

(a) Label the forces P and _B P _C. Using Appendix D, Case 1,

3 3

PROBLEM 11.79 (Continued)

C

PROBLEM 11.80

For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec.11.2A and show that it is equal to the work obtained in part a.

SOLUTION

(a) Label the applied couples M and _A M Apply _B. M at point A _A during this second phase.

From Appendix D, applied,

2

Total strain energy:

2 2

PROBLEM 11.80 (Continued)

(b) Bending moment diagram.

Over portion AB: 0

2 x L

   

 

 

0

2 2

/2 0 0

0 2 4

L AB

M M

M M L

U dx

EI EI

 







Over portion BC:

2

L x L

   

 

 



0

2 2

0 0

/2

2 (2 )

2

L

BC L

M M

M M L

U dx

EI EI

 





 ^

Total strain energy:

 U U_AB U_BC

 5 ⁰²

4 U M L

 EI 

D E

For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec. 11.2A and show that it is equal to the work obtained in part a.

SOLUTION

(a) Label the forces P and _D P _E. Using Appendix D, Case 5,

   

Likewise,

3 3

3 7

and

256 768

D D

512 768 512

D DD D DE E EE

PROBLEM 11.81 (Continued)

(b) Reactions: R_A R_B  P Over portion AD: 0 :

4

x L M Px

    

 

 

2 2 2 3 2 3

/4 /4 2

0 0

1 1

2 2 2 3 4 384

L L

AD

M P P L P L

U dx x dx

EI EI EI EI









      

Over portion DE: ²

 

2 ^{2 2 2} 1 ^{2 3}

4 2 2 16 2 64

L DE

PL M P L L P L

M U

EI EI EI

     

Over portion EB: By symmetry, 1 ^{2 3}

EB AD 384 U U P L

  EI

Total: 1 1 1 ^{2 3}

384 64 384

AB DE EB

U U U U P L

EI

 

      

1 2 3

48 U P L

 EI 

L

B A

M₀ M₀

PROBLEM 11.82

For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec. 11.2A and show that it is equal to the work obtained in part a.

SOLUTION

(a) Label the couples M and _A M . _B Using Appendix D, Case 7.

_AA  M L3^A

EI _BA  M L6^A

EI _BB  M L3^B

EI _AB  M L6^B EI Apply M_A first, then M . _B

1 2 3

  

U A A A

2 2

1 1

2 2

1 1 1

6 6 6

  

  

  

A AA A AB B BB

A A B B

U M M M

M L M M L M L

EI EI EI

With M_A M_B M ₀

02

1

 2M L

U EI

02

 M L2 U EI  (b) Bending moment: M  M ₀

2 2

0

0 2 2





^LM ^{M L}

U dx

EI EI

02

 M L2 U EI 

L/2 L/2 B A

D

w

PROBLEM 11.83

For the prismatic beam shown, determine the deflection of point D.

SOLUTION

Add force Q at point D.

D B

L/2 L/2

A

P

PROBLEM 11.84

For the prismatic beam shown, determine the deflection of point D.

SOLUTION

Add force Q at point D.

A B w

E D

L/2 L/2 L/2

PROBLEM 11.85

For the prismatic beam shown, determine the deflection of point D.

SOLUTION

Add force Q at point D.

Reactions: 1 2 , Over portion DE,

2

768 768

  ^ ^  ^     

^AD ^DE ^EB

D

U U U wL wL

Q Q Q EI EI 

L/2 L/2 B A

D

w

PROBLEM 11.86

For the prismatic beam shown, determine the slope at point D.

SOLUTION

Add couple M at point D. ₀

D B

L/2 L/2

A

P

PROBLEM 11.87

For the prismatic beam shown, determine the slope at point D.

SOLUTION

Add couple M at point D. ₀

Over portion AD:

0

A B w

E D

L/2 L/2 L/2

PROBLEM 11.88

For the prismatic beam shown, determine the slope at point D.

SOLUTION

Add couple M₀ at point D.

384 384

  ^  ^  ^    

A B D

L

a b

P

PROBLEM 11.89

For the prismatic beam shown, determine the slope at point A.

SOLUTION

Add couple M at point A. _A

Reactions: _A  Pb  M^A, _B  Pa  M^A

R R

L L L L

2 2 2

0 0 0

1 1

2 2 2

LM a b

U dx M dx M dv

EI EI EI













0 0

1 1

  ^  ^  ^





^a 



^b 

A

A A A

U M M

M dx M dv

M EI M EI M

Over portion AD (0 x a),  ^A ^A  ^A1  ,  _A  1

x Pbx M x

M M R x M

L L M L

Over portion DB (0 v b),    ,  



B A

A

Pav M v M v

M R v

L L M L

Set 0M_A  A  1



₀^a ^1 ^  1



₀^b

Pbx x dx Pav v dv

EI L L EI L L

2 3 3

2

1 1 1

2 3 3

 

    

P bLa ba ab

EIL

2 2

2(3 2 2 )

_A  6Pab  

La a b

EIL 

L/2 L/2

A B

C

M₀

PROBLEM 11.90

For the prismatic beam shown, determine the slope at point B.

SOLUTION

Add couple M at point B as shown. _B Strain energy.

2

S8 3 13 5 ft

1.5 kips 1.5 kips

B C

A

5 ft

PROBLEM 11.91

For the beam and loading shown, determine the deflection of point B. Use E  29  10⁶ psi.

SOLUTION

Add force Q at point B. 7975 kip ft EI  

1 218.75

0 143.75 27.43 10 ft

_B    7975   ^

EI 0.329 in._B   

S8 3 13 5 ft

1.5 kips 1.5 kips

B C

A

5 ft

PROBLEM 11.92

For the beam and loading shown, determine the deflection of point A. Use E  29  10⁶ psi.

SOLUTION

Add force Q at point A.

7975 kip ft

 

1 656.29

62.5 593.75 82.29 10 ft

_A    7975   ^

EI _A 0.987 in. 

8 kN

PROBLEM 11.93

For the beam and loading shown, determine the deflection at point B. Use E200 GPa.

SOLUTION

Add force Q at point B.

A

PROBLEM 11.94

For the beam and loading shown, determine the deflection at point B. Use E200 GPa.

SOLUTION

2 2

PROBLEM 11.94 (Continued)

Data: 0.6 m,a b0.9 m, L a b  1.5 m

3 3

3 6 4

6 4

9 6 2

5 10 N/m 4 10 N

1 (40)(80) 1.70667 10 mm 12

1.70667 10 m

(200 10 )(1.70667 10 ) 341,333 N m w

P I

EI





 

 

  

 

    

3 4 3 3 3 3

3

5 10 (1.5) (0.6)(1.5) (0.6) (4 10 )(0.9) 0 (2)(341,333) 4 3 12 (3)(341,333) 7.25 10 m

B



 

 

     

 

 



7.25 mm

B   

C B

PROBLEM 11.95

For the beam and loading shown, determine the slope at end A. Use 200 GPa.

E 

SOLUTION

Add couple M at point A. _A

Reactions: 80 80

4.8 4.8

(80) (2.4) (16.6667) (2.4)

2 3 (16.6667)(2.4) 76.8

3

{153.6 76.8}

32,600

A  EI  

 _A  7.07 10 rad ^3 

B

PROBLEM 11.96

For the beam and loading shown, determine the deflection at point D. Use E200 GPa.

SOLUTION

Units: Forces in kN, lengths in m.

Reactions:

0: 3.2 2.6 (0.6)(90) 0 16.875 0.8125

     

  

MB A Q

A Q

0: 3.2 0.6 (2.6)(90) 0 73.125 0.1875

MA B Q

B Q

    

  

Strain energy:

3.2 2 0

1

U 2 M dx

 EI



Deflection at point D: (formula)

3.2

(16.875 0.8125 )

0.8125 ; Set 90 kN.

PROBLEM 11.96 (Continued)

Over portion DE: (0.6 m x 2.6 m)

(16.875 0.8125 ) ( 0.6) 16.875 0.1875 0.6

0.1875 0.6; Set 90 kN.

(73.125 0.1875 ) 0.1875 Set 90 kN.

90

16.875 1.215 kN m 3

 v  

Deflection at point D: (calculated)

5.265 32.4 1.215 38.88 3

3.797 10 m

D B

For the beam and loading shown, determine the slope at end A. Use 29 10 psi.6

E 

SOLUTION

Units: Forces in kips; lengths in ft.

3 4

D B

For the beam and loading shown, determine the deflection at point C.

Use E29 10 psi. ⁶

SOLUTION

Units: Forces in kips; lengths in ft.

3 4

l

l D B

C

1 2 1l

2

A

A P

PROBLEM 11.99

For the truss and loading shown, determine the horizontal and vertical deflection of joint C.

SOLUTION

Add horizontal force Q at point C.

From geometry, 5

BC CD 2

L  L   Equilibrium of joint C.

0: 2 ( ) 0

F_x  5 F_BC F_CD Q 

0: 1 ( ) 0

F_y  5 F_BC F_CD P 

Solving simultaneously,

5 5 5 5

2 4 2 4

    

BC CD

F P Q F P Q

Strain energy:

2

2

i i i

U F L

  EA

Deflections. Horizontal: 1 _{i i i}

C

i

U F L F

x Q E A Q

 

  

 

Vertical: 1 _{i i i}

C

i

U F L F

y P E A P

 

  

 

PROBLEM 11.99 (Continued)

In the table, Q is set equal to zero in the last two columns.

F i L _i A_i F_i/P F_i/Q F L F^{i i i}

A A

PROBLEM 11.100

For the truss and loading shown, determine the horizontal and vertical deflection of joint C.

SOLUTION

Add horizontal force Q at point C.

From geometry, 5

BC CD 2 L L  l

Equilibrium of joint C.

2 2

Solving simultaneously,

5 5 5 5

2 4 2 4

   

BC CD

F P Q F P Q

Equilibrium of joint D.

2 5 5

Strain energy:

2

Deflections. Horizontal: 1 _{i i i}

C

PROBLEM 11.100 (Continued)

In the table, Q is set equal to zero in the last two columns.

F_i L _i A _i  F P_i/  F Q_i/ 

2.5 ft 3 in²

4 in²

6 in²

6 ft

80 kips 48 kips B

C

D 2.5 ft

PROBLEM 11.101

Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E  29  10⁶ psi, determine the deflection indicated.

Vertical deflection of joint C.

SOLUTION

Joint C: 12 12

0: 0

13 13

F_x   F_BC  F_CD Q  13

  12

BC CD

F F Q (1)

5 5

0: 0

13 13

F_y  F_BC  F_CD P  13P

BC CD 5

F F  (2)

Solving (1) and (2) simultaneously,

13 13

10 24

 

FBC P Q

13 13

10 24

  

FCD P Q

Joint D: 5

0: 0

F_y  13F_CD F_BD 

5 1 5

13 2 24

   

BD CD

F F P Q

PROBLEM 11.101 (Continued)

Lengths of members: 78 in.

78 in.

60 in.







BC CD BC

L L L

2

 2F L

U EA

  ^   ^  1 ^

  

P

U FL F FL F

P EA P E A P

Member F L (in.) 

 F P

(in )2

A FL F

A P





BC 13 13

10P 24Q 78 13

10 4 32.955P13.73125Q

CD 13 13

10 24

 P Q 78 13

 10 6 21.97P9.15417Q

BD 1 5

2P 24Q 60 1

2 3 5.00P2.08333Q

 59.975P2.49375Q

Further data: 29 10 psi⁶ 29,000 ksi E  



80 kips 48 kips



 P Q

(59.975)(80) ( 2.49375)(48)

0.1613 in.

29,000

P

    

 

2.5 ft 3 in²

4 in²

6 in²

6 ft

80 kips 48 kips B

C

D 2.5 ft

PROBLEM 11.102

Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E  29  10⁶ psi, determine the deflection indicated.

Horizontal deflection of joint C.

SOLUTION

Joint C:

12 12

0: 0

13 13

x BC CD

F F F Q

     

13

  12

BC CD

F F Q (1)

5 5

0: 0

13 13

F_y  F_BC  F_CD P  13

  5

BC CD

F F P (2)

Solving (1) and (2) simultaneously,

13 13

10 24

 

FBC P Q

13 13

10 24

  

FCD P Q

Joint D:

0: 5 0

F_y  13F_CD F_B 

5 1 5

13 2 24

   

BD CD

F F P Q

PROBLEM 11.102 (Continued)

Lengths of members: 78 in.

78 in. 29,000 ksi E  



80 kips 48 kips



 P Q

( 2.49325)(80) (10.40365)(48)

0.01034 in.

29,000

Q

 

  

 

2.5 m

PROBLEM 11.103

Each member of the truss shown is made of steel and has a cross-sectional area of 500 mm . Using ² E200 GPa, determine the vertical deflection of joint B.

SOLUTION

Find the length of each member as shown.

Add vertical force Q at joint B.

2 1

Solving simultaneously, 6.25 0.8333 kN 1.75 0.8333 kN

PROBLEM 11.103 (Continued)

Member F(10 N)³  F Q/ L(m)

with 0Q ( / )

F F Q L  (10 N m)³ 

AB 6.25 0.8333Q 0.8333 2.0 10.4167

AD 1.05 0.5Q 0.5 2.4 1.26

BD 1.75 0.8333Q 0.8333 2.0 2.9167

BC 6.0 0 1.5 0

CD 3.6 0 2.5 0

 14.593

 ₉ ³ ₆

6

1 ( / )

14.593 10 (200 10 )(500 10 ) 145.9 10 m

B F F Q L

EA





 

 

 

 

  



 _B 0.1459 mm 

2.5 m 1.6 m

1.2 m

1.2 m

4.8 kN

C D

B

A

PROBLEM 11.104

Each member of the truss shown is made of steel and has a cross-sectional area of 500 mm . Using ² E200 GPa,determine the horizontal deflection of joint B.

SOLUTION

Find the length of each member as shown.

Add horizontal force Q at joint B.

2 1

B 2

U F L F

F L

Q Q EA EA Q

      

   

Joint C: 4

0: 4.8 0 6.0 kN

y 5 CB CB

F F F

    

Joint B: 3

0: 0 3.6 kN

x 5 CB CD CD

F F F F

     

4 4

0: 3.6 0

5 5

x AB BD

F F F Q

     

3 3

0: 4.8 0

5 5

y AB BD

F F F

    

Solving simultaneously, F_AB 6.25 0.625 kN Q

1.75 0.625 F_BD    QkN

Joint D: 3

0: 0

y 5 BD AD

F F F

   

3 1.05 0.375

AD 5 BD

F   F   Q

PROBLEM 11.104 (Continued)

Member F(10 N)³  F Q/ L(m)

( / ) F F Q L 

(10 N m)3 

AB 6.25 0.625Q 0.625 2.0 7.8125

AD 1.05 0.375Q 0.375 2.4 0.9450

BD 1.75 0.625Q 0.625 2.0 2.1875

BC 6.0 0 1.5 0

CD 3.6 0 2.5 0

 4.680

 ₉ ³ ₆

6

1 ( / )

4.680 10 (200 10 )(500 10 ) 46.8 10 m

B F F Q L

EA





 

 

 

 

  



 _B 0.0468 mm 

L

L C

B

A

60

P

PROBLEM 11.105

A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the vertical deflection of point A, (b) the horizontal deflection of point A.

SOLUTION

Add horizontal force Q at point A.

Over AB:

(a) Vertical deflection of point A.

BC (b) Horizontal deflection of point A.

3 3

A

R

B

P

PROBLEM 11.106

For the uniform rod and loading shown, and using Castigliano’s theorem, determine the deflection of point B.

SOLUTION

Use polar coordinate .

Strain energy: ²

0 2

By Castigliano’s theorem, U P

B

R A

P

PROBLEM 11.107

For the beam and loading shown, and using Castigliano’s theorem, determine (a) the horizontal deflection of point B, (b) the vertical deflection of point B.

SOLUTION

Add horizontal force Q at point B.

Use polar coordinate .

/2 2

Bending moment.

0: 0

C l

l B

A

P

PROBLEM 11.108

Two rods AB and BC of the same flexural rigidity EI are welded together at B. For the loading shown, determine (a) the deflection of point C, (b) the slope of member BC at point C.

SOLUTION

Add horizontal force Q and couple M at C. _C

PROBLEM 11.108 (Continued)

(a) Deflection at C. _C U^AB U^BC

Q Q



 

 

  2 ³

C 3 Pl

 EI 

 

(b) Slope at C. _C ^{AB BC}

A C

U U

M M



 

 

  ²

C 6 Pl

 EI

 

P

L

L

A D

B C

PROBLEM 11.109

Three rods, each of the same flexural rigidity EI, are welded to form the frame ABCD. For the loading shown, determine the deflection of point D.

SOLUTION

Add dummy force Q at point D as shown.

PROBLEM 11.109 (Continued)

Member CD:   



M Qy M y

Q

Set Q  0 M  0

2

0 0 0

2

 

  

 



^{L CD}



^L

CD

M dy U M M

U dy

EI Q EI Q

3 3 5 3

3 2 0 6

D

PL PL PL

EI EI EI

    

 

P

L

L

A D

B C

PROBLEM 11.110

Three rods, each of the same flexural rigidity EI, are welded to form the frame ABCD. For the loading shown, determine the angle formed by the frame at point D.

SOLUTION

Add couple M at point D. ₀ By Castigliano’s theorem,

0

PROBLEM 11.110 (Continued)

Member CD: ₀

0

M 1 M M

M



  

Set M₀  0

2

0 0 0 0

0 2 0

L L

M

M U M M

U dy dx

EI M EI M







_^ 



_^ 

2

0 0

D 6

PL

   EI  ²

D 6 PL

  EI 

A B C

P

L/2 L/2

PROBLEM 11.111

Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION

Remove support B and add reaction R as a load. _B

A B

L

M₀

PROBLEM 11.112

Determine the reaction at the roller support and draw the bending-moment diagram for the beam and loading shown.

SOLUTION

Remove support B and add reaction R_B as a load.

2 0 2





^LM

U dv

EI

0

1 0

 

  





^L 

B

B B

U M

y M dv

R EI R

0

   

B 

B

M R v M M v

R

0 0

1 ^L( )

B B

y R vM v dv

 EI



2 0

0 0

 ^R_EI^B



^{Lv dv}  ^M_EI



^{Lv dv}

3 2

0 3 0

3 2 0 2

 R L^B  M L  _B  M 

EI EI R L 

0 0 0 0

3 1

2 2

    

A B

M R M M M M 

A

Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION

Remove support A and add reaction R as a load. _A

Bending moment diagram drawn to scale for 1 3 . a L By singularity functions,

3 0

3 ( ) /2

M  M b L a x L M L a   

L/2 L/2 B A

C

w

PROBLEM 11.114

Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION

Remove support A and add reaction R_A as a load.

Portion AC:

2

PROBLEM 11.114 (Continued)

Bending moments. Over AC: 7

128

M wLx

2 2

7 0.02734

C 256

M  wL  wL 

Over CB: 7 1 ²

128 2 2

 

   

M wL v L wv

2

2 2

7 1 9

128 2 2 128

       

B

M wL w L wL

0.07031 2

MB   wL 

7 7

128 0 128

  _m  _m 

dM wL wv v L

dv

7 7 1 7 2

128 128 2 2 128

   

     

m

M wL L L w L

2 2

945 0.02884

32,768wL wL

  

or 7 ²

/2 /2

128    

M wLx w x L 

D B

P

PROBLEM 11.115

Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION

Remove support A and add reaction R as a load. _A

Portion DB:

3 ^A 3

PROBLEM 11.115 (Continued)

Bending moments: 14

3 81

D A

M R    L  PL 0.1728M_D PL 

2 4

3 27

B A

M R L P  L  PL 0.1481M_B   PL 

By singularity functions, 14 ¹

27 /3

M  Px P x L    

B C w

A

L/2 L

PROBLEM 11.116

For the uniform beam and loading shown, determine the reaction at each support.

SOLUTION

Remove support A and add reaction R_A as a load.

P l

D C

 

E

B

PROBLEM 11.117

Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

In document BeerMOM_ISM_C11.pdf (Page 56-170)