Chapter 9 Ansel C.ugural - Solution Manual Mechanics of Materials PDF

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PROBLEM (9.1) A cylinder with external diameter D and internal diameter d is subjected to an axial compressive load P and a torque T (Fig. P9.1). Calculate the maximum principal stress and the maximum shearing stress. Show the results on a properly oriented element. Given: D = 6 in., d = 4 in., P = 20 π kips, T = 15 π kip ⋅ in.

SOLUTION State of stress on an element at the cylinder’s surface is

σx =

σx = Tc τ xy = J

P A

τ =−

− 20π × 103 = −4 ksi π (32 − 2 2 )

15π ×103 (3)

π 2

4

= −1.385 ksi

4

(3 − 2 )

Equation (9.1): 4 4 2 2 σ 1 = 0.433 ksi σ 2 = −4.433 ksi τ max = 2.433 ksi Equation (9.3): 2( −1.385) tan 2θ p = , 2θ p = 34.7 o −4 Equation (4.4a) gives 1 1 σ x ' = ( −4) + ( −4) cos 34.7o − 1.385sin 34.7 o 2 2 = −4.433 ksi Thus θ p " = 17.35o

σ 1,2 = − ± (− )2 + (− 1.385)2 = − 2.433

2.433 ksi

4.433 ksi 17.35o

x

0.433 ksi

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_______________________________________________________________________ PROBLEM (9.2) A hollow outboard motor propeller shaft with an outer diameter D and inner diameter d is simultaneously acted upon by a torque T and an axial thrust P (Fig. P9.1) . What is the maximum shearing stress in the shaft? Given: D = 8 in. d = 4 in. T = 400 kip ⋅ in. P = 250 kips

SOLUTION

A=

π 4

(82 − 42 ) = 37.7 in.2

J=

π 32

(84 − 4 4 ) = 377 in.4

250 P =− = −6.63 ksi A 37.7 Tr 400(4) = 4.24 ksi τ xy = = 377 J Thus

σx = −

τ max = (

σx 2

τ xy σx

) 2 + τ xy2 2

2

= (−3.315) + (4.24) = 5.38 ksi

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_______________________________________________________________________ PROBLEM (9.3) A shaft of diameter d is subjected to an axial compressive load (P) and two torques (T1 , T2 ) as shown in Fig. P9.3. What is the maximum shear stress at point A on the surface of the shaft? Given: d = 120 mm, P = 500 kN, T1 = 4 kN ⋅ m, T2 = 6 kN ⋅ m

SOLUTION

At point A, T = 10 kN and P = 500 kN P 4(500) = − 44.21 MPa σx = − = − π (0.12) 2 A 16T 16(10) = = 29.47 MPa τ xy = 3 π d π (0.12)3

τ xy A

σx

Hence

τ max = (

σx 2

) 2 + τ xy2 2

= (−22.105) + (29.47) = 36.84 MPa = R

2

τ

y

σ avg =

σx 2

R

= −22.105 MPa

tan 2θp " =

C

σ2 29.47 , 22.105

θ p = 26.56o

2θp"

O σavg

σ1

σ

x

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_______________________________________________________________________ PROBLEM (9.4) Determine the largest value of the axial load P that can be carried by the stepped steel shaft shown in Fig. P9.4 for allowable bending and shear shear stresses Given:

σall

τ all = 60

= 100 MPa ,

σ all

and

τ all , respectively.

T = 0.01P N ⋅ m.

MPa,

SOLUTION

ACD = JCD = JAB =

π 4

(582 − 50 2 ) = 216π mm

π 32

π 32

2

(58 4 − 50 4 ) = 0.158π ×10 6 mm (504 ) = 0.195π ×106 mm

4

4

Since J AB > JCD , the critical section is in CD. Thus,

σx =

σx = τ=

TcCD J CD

P ACD

P 4630 P = −6 π 216π ×10

0.01P (29× 10− ) 0.158π × 10−6 1835 P =− 3

τ =−

π

Equation (9.1): 4630P 4630P 2 −1835 P 2 + ( ) σ1 = ) +( π 2π 2π 2315P 2954P 5269P = + =

π

τ max =

π

π

2954 P

π

We have 100 ×10 6 = 60 × 10 6 =

5269P

π

2954P

π

,

,

P = 59.6 kN P = 63.8 kN

Thus, Pall = 59.6 kN

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_______________________________________________________________________ PROBLEM (9.5) Repeat Prob. 9.4 for the case in which the 50-mm-diameter hole in the shaft is eliminated.

SOLUTION

Critical section is in AB.

σx = τ=

4P π d 2AB

σx =

4P 1600 P = 2 π (0.05) π

τ =−

16(0.01P) 1280 P =− 3 π (0.05) π

16T

πd 3AB

Equation (9.1): 800P 800P 2 − 1280P 2 + ( ) σ1 = ) +(

π

π

π

800 P 1509 P 2309 P = + =

π

π

π

We have 100 ×10 6 = 60 × 10 6 =

2309 P

π

1509 P

π

,

,

P = 136 kN P = 125 kN

Thus, Pall = 125 kN

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_______________________________________________________________________ PROBLEM (9.6) Determine the maximum shearing stress and its orientation in the stepped shown in Fig. P9.4. Given: P = 30π kN,

T = 0.01P kN ⋅ m

SOLUTION

See solution of Prob. 9.4: 2954P 4630 P −1835P , τ= , σx = τ max =

π

π

π

Thus,

τ max =

2954(30π ×103 )

= 88.6 MPa

π 1835(30π ×103 ) = −55.1 MPa τ =− π 4630(30π ×103 ) = 138.9 MPa σx = π τ (MPa)

(50.56, 23.4)

σ ' = 69.45 MPa

2θs" σ (MPa)

C R

tan 2θ s " =

88.6

69.45 55.1

θ s " = 25.8o

σ' 69.45 MPa 88.6 MPa o 25.8

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_______________________________________________________________________ PROBLEM (9.7) A rectangular shaft of depth a and width b is subjected to an axial tensile load P and a torque T, as shown in Fig. P9.7. What is the largest permissible value of T if the allowable tensile stress in the shaft is σ all ? Given:

a = 3 in.,

b = 2 in.,

P = 30 kips,

σ all

= 20 ksi

SOLUTION

P 30 = = 5 ksi A 3× 2 From Table 5.1, with a b =1.5 : T T = = 0.3608 T τ= 2 α ab (0.23)(3)(2) 2

σ=

σ1 =

σ

( T in kip ⋅ in.)

σ

+ ( )2 +τ 2 = σ all 2 2

= 2.5 + (2.5) 2 + (0.3608 Tall ) 2 = 20

or Tall = 48 kip ⋅ in.

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_______________________________________________________________________ PROBLEM (9.8) A shaft with an a x b rectangular cross section is acted upon by an axial tensile load P and a torque T (Fig. P9.7). Determine the largest permissible value of P if the allowable tensile stress in the member is limited to σ all . Given:

a = 100 mm,

b = 50 mm,

T = 5 kN ⋅ m,

σ all

= 120 MPa

SOLUTION

By Table 5.1, with a b = 2 : 3

T 5(10 ) = = 81.3 MPa 2 α ab (0.246)(0.1)(0.05) 2 P ( P in MN) = 200 P MPa σ= (0.1)(0.05) Hence

τ=

σ1 =

σ

σ

+ ( )2 +τ 2 = σ all 2 2 2

2

= 100 P + (100 P) + (81.3) = 120 (120 − 100 P) 2 = 10, 000 P 2 + 6,609.7 14, 400 − 24,000 P + 10,000 P 2 = 10, 000P 2 + 6, 609.7 or Pall = 0.3246 MN = 325 kN

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_______________________________________________________________________ PROBLEM (9.9) A turbine shaft of diameter d is under a thrust load P and a torque T (Fig. P9.9). Calculate the largest permissible value of the load P if the allowable normal stress is not to exceed Given:

σ all = 220 MPa

T = 2 kN ⋅ m,

d = 60 mm,

σ all .

SOLUTION

y z

C

A

Maximum normal stress occurs at a point A.

d

We have 16T 16(2 × 10 3) = = 47.16 MPa τ= πd 3 π (0.06)3 4P 4P = = 353.7 P (P in MN) σ= 2 πd π (0.06)2 Hence

σ1 =

σ

σ

+ ( )2 +τ 2 = σ all 2 2

220 = 176.8P + (176.8 P) 2 + (47.16) 2 (220 − 176.8P ) 2 = 31, 258 P 2 + 2, 224 or 48, 400 − 77, 792 P + 31, 258P 2 = 31, 258 P 2 + 2, 224 Solving, P = 0.594 MN = 594 kN

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_______________________________________________________________________ PROBLEM (9.10) A thin-walled cylindrical vessel of radius r and thickness t is acted upon by an internal pressure p as well as an axial compressive load P applied to the vessel through the rigid end plates as shown in Fig. P9.10. What is the magnitude of P needed to produce pure shear in the cylinder wall ?

SOLUTION

We have d=2r. For pure shear, σ1 = −σ 2 pr pr P =− + t 2t 2π rt Solving

σ1 = pr t σ2 =

pr P − 2 t 2π rt

2 P = 3π pr

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_______________________________________________________________________ PROBLEM (9.11) A thin-walled cylindrical tank of radius r and thickness t is subjected to an internal pressure p, axial compression P , and a torque T applied to the tank through the rigid end plates (Fig. P9.11). Calculate the maximum principal stress in the cylinder wall. Given: d = 10 in, t = 0.2 in., p = 400 ksi, P = 5 kips, T = 250 kip ⋅ in.

SOLUTION d = 2r A = 2π rt J = 2π r 3 t

y

σy

Stresses are: P pr σx = − + x A 2t 5 0.4(5) =− + = 4.2 ksi 2π (5)(0.2) 2(0.2) pr 0.4(5) = = 10 ksi σy = t 0.2 Tr T 250 =− = − 7.96 ksi τ xy = − = − 2 J 2π r t 2π (5) 2 (0.2) Maximum principal stress is then 4.2 +10 4.2 −10 2 ) + ( −7.96) 2 = 15.57 ksi + ( σ1 = 2 2

σx

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_______________________________________________________________________ PROBLEM (9.12) A closed-ended cylinder of radius r and wall thickness t is subjected to an internal pressure p and an axial tension P. Determine for the shell wall: (a) The maximum tensile stress. (b) The maximum shearing stresses and the orientation of their planes. Given: r = 10 in., t = ½ in., p = 500 psi, P = 100 kips

SOLUTION

A = 2π rt = 2π (10)(0.5) = 31.42 in.2 500(10) = 10 ksi σt = 0.5

(a) pr σt = t

σa =

P pr + A 2t

3

σa =

1 (b) τ max = (10 − 8.18) = 0.91 ksi 2

100 × 10 + 5 = 8.18 ksi 31.42

1 2

σ ' = (8.15 + 10) = 9.08 ksi 9.08 ksi

0.91 ksi

o 45

9.08 ksi

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_______________________________________________________________________ PROBLEM (9.13) A thin-walled cylinder of radius r and wall thickness t is acted upon by an internal pressure p and a torque T. Calculate the largest shearing stress and its orientation in the wall of this closed-ended vessel. Given: r = 250 mm, t = 10 mm, p = 3 MPa, T = 40 kN ⋅ m

SOLUTION

J = 2π r3 t = 2π (250)3 (10) = 981.7 × 106 mm 4

σt =

pr t

6

σx = τ xy =

τ (MPa)

3 ×10 (250) = 37.5 MPa 2(10) σ t = 2 σ x = 75 MPa

σx =

pr 2t

Tc J

3

σ ' = 56.25 MPa

(37.5, 10.19) 2θs"

O

− − 40 × 10 (250 × 10 ) = −10.19 MPa 6 981.7 × 10−

3

τ=

σ (MPa)

C R

R = (18.75)2 + (10.19) 2 = 21.34

(75, -10.19)

tan 2θ s " =

σ'

18.75 , 10.19

θ s " = 30.7o

τ max = 21.34 MPa 30.7o 21.34 MPa 5 6.25 ΜPa

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_______________________________________________________________________ PROBLEM (9.14) Calculate the normal and shearing stresses on the spiral weld of the steel shaft of diameter d acted upon by an axial load P and a torque T as shown in Fig. P9.14. Given: P = 200 kN, T = 2 kN ⋅ m, d = 50 mm, φ = 60°

SOLUTION

θ = α + 90 = 60 + 90 = 150 o x'

P σx = A x

θ

τ xy =

y'

σx =

200 ×10 3

π

= 102 MPa

(0.05)2

4 16(2 ×10 −3 ) = 81.5 MPa τ= π (0.05)3

16T π d3

Equations(8.4):

σw = σ x ' =

102 102 + cos 300 o + 81.5 cos 300 o = 147.1 MPa 2 2

and

τ w =τ x' y ' = −

102 sin 300 o −81.5 cos 300 o = 44.2 − 40.75 = 3.45 MPa 2

x’

150 o

147.1 MPa

x

3.45 MPa

y’

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_______________________________________________________________________ PROBLEM (9.15) A cantilever steel shaft of diameter d is subjected to an axial load P and a torque T as shown in Fig. P9.14. Determine the permissible value of T. The allowable normal and shear stresses on the spiral weld are σ all and τall , respectively. Given:

σ all

τ all = 12 ksi,

= 20 ksi,

d = 4 in.,

φ=

50o ,

P = 40π kips

SOLUTION

θ = α + 90 = 50 + 90 = 140o x'

θ

P σx = A x

y'

16T τ xy = 3 πd

σx =

40π ×103

π

= 10 ksi

2

(4) 4 − 16T = −0.0796T ksi τ= π (4)3

Equations(8.4): 10 10 + cos 280o − 0.0796T sin 280o 2 2 T = 180 kip ⋅ in. = 5 + 0.8682 + 0.0784T , 10 τ x ' y ' = − 12 = − sin 280o − 0.0796T cos 280o 2 T = 1225 kip ⋅ in. = 4.924 − 0.01382T ,

σ x ' = 20 =

Thus, Tall = 180 kip ⋅ in.

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_______________________________________________________________________ PROBLEM (9.16) A thin-walled cylindrical pressure vessel of radi...