Solution Manual Mechanics of Material by Ansel C.Ugural Chapter 8 PDF

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Solution Manual Mechanics of Material by Ansel C.Ugural Chapter 8...

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CHAPTER 8 PROBLEM (8.1) A 60-mm by 40-mm plate of 5-mm thickness is subjected to uniformly distributed biaxial tensile forces (Fig. P8.1). What normal and shear stresses exist along diagonal AC? As was done in the derivations given in Sec. 8.2, use an approach based upon the equilibrium equations applied to the wedge-shaped half ABC of the plate.

SOLUTION

∠ ACB = tan − 1

50 ΔA cos56.3o

y

y’ C

B 100 ΔA sin56.3o

56.3o

σx =

30(103 ) = 100 MPa 0.06(0.005)

σy =

10(10 ) = 50 MPa 0.04(0.005)

τ x ' y 'ΔA σ x 'Δ A

A

∑F

x'

x

x’

3 = 56.3 o 2

3

= 0 : σ x ' ΔA − 50ΔA cos 2 56.3 o − 100 ΔA sin 2 56.3 o = 0

σ x ' = 15.39 + 69.21 = 84.6 MPa

∑F

y'

= 0 : τ x' y' ΔA + 50ΔA sin 56.3o cos 56.3o − 100ΔA sin 56.3o cos 56.3o = 0

τ x ' y ' = −23.08 + 46.16 = 23.1 MPa

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_______________________________________________________________________ PROBLEM (8.2) Redo Prob. 8.1 using Eqs. (8.4). SOLUTION

From solution of Prob. 8.1: σ x = 100 MPa σ y = 50 MPa 50 MPa

y’

y B 100 MPa

A

C

56.3o

θ

θ = − (90 − 56.3) = − 33.7 o

τx 'y ' σ x' x

33.7o

x’

1 1 2 2 = 75 + 9.6 = 84.6 MPa 1 τ x ' y ' = − (100 − 50) sin(−67.4o ) = 23.1 MPa 2

σ x ' = (100 + 50) + (100 − 50) cos(− 67.4 o )

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_______________________________________________________________________ PROBLEMS (8.3 through 8.5) The states of stress at three points in a loaded body are represented in Figs. P8.3 through P8.5. Determine for each point the normal and shearing stresses acting on the indicated inclined plane. As was done in the derivations given in Sec. 8.2, use an approach based upon the equilibrium equations applied to the wedge shaped element shown in each figure.

SOLUTION (8.3)

y

τ x ' y ' ΔA

y’

σ x ' ΔA

50 ΔA cos15o 60 ΔA cos15o

x’ 15o x o 60 ΔA sin15

15o

80 ΔA sin15o

∑F

x'

∑F

y'

= 0:

σ x'Δ A + 50 cos2 15o − 80Δ A sin 2 15o

σ x'

− 2(60ΔA sin 15o cos15o ) = 0 = −46.65 + 5.36 + 30 = −11.3 MPa

= 0 : τ x' y' Δ A − 50ΔA sin 15 o cos15 o

− 80ΔA sin 15 o cos15 o − 60ΔA cos 2 15 o + 60ΔA sin 2 15o = 0 τ x 'y ' = 12.5 + 20 + 55.98 − 4.02 = 84.5 MPa

SOLUTION (8.4)

x’

σ x 'Δ A o

τ x ' y ' ΔA y’

10 ΔA sin30 30o 10 ΔA cos30o

25 ΔA cos30

∑F

x'

o

o 2 = 0 : σ x' Δ A − 25Δ A cos 30

+ 2(10Δ A sin 30o cos 30o ) = 0 σ x ' = 18.75 − 8.66 = 10.09 ksi Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

∑F

y'

= 0 : τ x' y' ΔA + 25ΔA cos 30 o sin 30 o

+ 10ΔA cos 2 30o − 10ΔA sin 2 30o = 0 τ x ' y ' = − 10.83 − 7.5 + 2.5 = − 15.83 ksi

SOLUTION (8.5) o

40 ΔA cos22.5

y’

10 ΔA cos22.5o o

10 ΔA sin22.5

22.5

o

70 ΔA sin22.5

o

τ x ' y 'ΔA

σ x 'Δ A x’

∑F

x'

o 2 = 0 : σ x' Δ A + 40Δ A cos 22.5

− 70ΔA sin 2 22.5o − 2(10ΔA sin 22.5o cos 22.5o ) = 0 σ x ' = −34.14 +10.25 + 7.07 = −16.8 MPa

∑F

y'

= 0 : − τ x' y' ΔA − 40ΔA sin 22.5o cos 22.5o

−70ΔA sin 22.5 o cos 22.5 o −10 ΔAcos 2 22.5 o + 10ΔA sin 2 22.5o = 0 τ x ' y ' = 14.14 + 24.75 + 8.54 − 1.46 = 46 MPa

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_______________________________________________________________________ PROBLEM (8.6 through 8.8) Redo Probs. 8.3 through 8.5 using Eqs. (8.4). SOLUTION (8.6)

y

τ x ' y'

y’

σ x' 50 MPa 60 MPa

x’ θ x

θ =15o

80 MPa

1 1 2 2 = 15 − 56.3 + 30 = −11.3 MPa 1 τ x ' y ' = − ( −50 − 80) sin 30o + 60 cos 30o 2 = 32.5 + 51.96 = 84.5 MPa

σ x ' = (− 50 + 80) + (− 50 − 80) cos 30 o + 60 sin 30 o

SOLUTION (8.7)

θ = 30 + 90 = 120o

σ x' x’

θ y’

τx 'y '

10 ksi

30o 25 ksi

1 1 2 2 = 12.5 + 6.25 − 8.66 = 10.09 ksi 1 τ x ' y ' = − (0 − 25) sin 240o + 10 cos 240o 2 = −10.83 − 5 = − 15.83 ksi

σ x ' = (0 + 25) + (0 − 25) cos 240o + 10 sin 240o

Thus, τ x ' y ' acts in opposite direction to that shown in the figure above. Continued on next slide

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SOLUTION (8.8)

σx' x’

τ x ' y' 70 MPa

θ 22.5

y’

o

10 MPa

x

θ =112.5o

40 MPa

1 1 2 2 = 15 − 38.89 + 7.07 = −16.8 MPa 1 τ x ' y ' = − (70 + 40) sin 225o − 10 cos 225o 2 = 38.89 + 7.07 = 46 MPa

σ x ' = (70 − 40) + (70 + 40) cos 225 o − 10 sin 225 o

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_______________________________________________________________________ PROBLEM (8.9) The state of stress at a point A of the airplane bracket under a service pull P is represented by an element as illustrated shown in Fig. P8:9. Given: σ x = σ y = 0, τ xy = 15 ksi (pure shear) Determine the stresses on all sides of an element rotated through an angle on the element.

θ

= 22.5o and show their sense

SOLUTION

σ x =σ y = 0

τ xy = 15 ksi

θ = 22.5o

Equations (4.4): σ x ' = 15 sin 45 o = 10.61 ksi

τ x ' y ' = 15 cos 45o = 10.61 ksi σ y ' = −15 sin 45 o = −10.61 ksi y’ 10.61 ksi 10.61 ksi

x’ 22.5o

x 10.61 ksi

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_______________________________________________________________________ PROBLEM (8.10) At a point in a loaded member, the stresses are as shown in Fig. P8.10. The normal stress at the point on the indicated plane is 7 ksi (tension). What is the magnitude of the shearing stress τ ?

SOLUTION

y

τ 30o

θ

x’

6 ksi

x y’ 6 +0 6 −0 o o + cos 300 − τ sin 300 2 2 7 = 3 + 1.5 + 0.866τ

σ x' = 7 = or

τ = 2.89 ksi

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_______________________________________________________________________ PROBLEM (8.11) A triangular plate is subjected to stresses as shown in Fig. P8.11. Determine σ x ,

σy

, and

τ xy

and sketch the results on a properly oriented element.

SOLUTION 50 MPa

y

200 MPa

y’

x’ 45o

x

τ x'y ' = 0 = −

σ x −σ y 2

sin 90o +τ xy cos 90o

or

σ x =σ y σ x ' = −200 = σ x + 0 + τ xy sin 90 o o

σ y' = −50 = σ x − 0 − τ xy sin 90

(1) (2)

Subtract (1) from (2): τ xy = −75 MPa Add (1) and (2): σ x = σ y = −125 MPa 125 MPa 125 MPa 75 MPa

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_______________________________________________________________________ PROBLEM (8.12) Calculate the normal and shear stresses acting on the plane indicated in Fig. P8.10 for

τ

= 4 ksi.

SOLUTION

y 4 ksi

4 ksi 30

o

θ

x’

593 psi 6 ksi

x

7.96 ksi

y’ 6+0 6−0 cos 300 o − 4 sin 300 o + 2 2 = 3 + 1.5 + 3.464 = 7.96 ksi 6 −0 sin 300o − 4 cos 300 o τ x'y ' = − 2 = 2.598 − 2 = 593 psi

σ x' =

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_______________________________________________________________________ PROBLEM (8.13) At a critical point A in the loaded member in Fig. P8.13, the stresses on the inclined plane are σ = 28 MPa and τ = 10 MPa, and the normal stress on the y plane is zero. Calculate the normal and shear stresses on the x plane through the point. Show the results on a properly oriented element.

SOLUTION

σ x’

θ = 120o

τ

θ 30o

x

y’

σ x + 0 σx − 0

cos 240o +τ xy sin 240o + 2 2 σ x = 112 + 3.464τ xy

σ = 28 =

σ x −0

sin 240 o + τ xy cos 240 o 2 σ x = −23.094 + 1.155τ xy

τ = −10 = −

(1)

Solving (1) and (2): σ x = −90.6 MPa

(2)

τ xy = −58.5 MPa

y 58.5 MPa

A

90.6 MPa

x

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_______________________________________________________________________ PROBLEM (8.14) At a particular point in a loaded machine part (use Fig. P8.9), the stresses are as follows:

σ x = 6 ksi, σ y

= -4 ksi , and τ x y = 0. Determine the normal and shear stresses on the plane

whose normals are at angles of -30o and 120o with the x axis. Sketch the results on a properly oriented elements.

SOLUTION

y’ x At θ = −30o :

30

o

x’ 4.33 ksi

3.5 ksi

6− 4 6+ 4 o cos( −60 ) + 0 = 1 + 2.5 = 3.5 ksi + 2 2 6+ 4 o τx 'y ' = − sin(− 60 ) + 0 = 4.33 ksi 2

σ x' =

1.5 ksi

x’

4.33 ksi

At θ = 120 o : 120o

y’

x

6− 4 6+ 4 o + cos 240 = −1.5 ksi 2 2 6+ 4 o τx 'y ' = − sin 240 = 4.33 ksi 2

σ x' =

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_______________________________________________________________________ PROBLEM (8.15) At a point in a loaded structure, the normal and shear stresses have the magnitudes and directions acting on the inclined element depicted in Fig. P8.15. Calculate the stresses σ x ,

τ xy

σy ,

on an element whose sides are parallel to the xy axes.

SOLUTION

Transform from θ = 30o to θ = 0 . For convenience in computations, Let σ x = −200 M Pa, σ y = −60 MPa, τ xy = 35 MPa and θ = −30 o Then 1 1 σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ 2 2 1 1 = ( −200 − 60) + (−200 + 60) cos(−60o ) + 35sin(−60o ) 2 2 = − 195.3 MPa 1 2 1 = − ( −200 + 60) sin(−60o ) + 35 cos(− 60o ) 2 = −43.1 MPa

τ x ' y ' = − ( σ x − σ y ) sin 2 θ + τ xy cos 2θ

So

σ y ' = σ x + σ y − σx ' = − 200 − 60 + 195.3 = − 64.7 MPa For θ = 0o :

y

64.7 MPa 43.1 MPa

195.3 MPa

x

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and

_______________________________________________________________________ PROBLEM (8.16) At a point in a loaded steel bracket the stresses are σ x = 50 MPa, σ y = -15 MPa,

τ xy = -30 MPa , as shown in Fig. P8.16. Determine the stresses acting on an element oriented at an angle θ = 40o from the x axis. Sketch the result on a properly oriented element.

and

SOLUTION

1 1 2 2 1 1 = (50 −15) + (50 +15) cos80 o − 30 sin 80 o = −6.45 M Pa 2 2

σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ

1 2 1 = − (50 + 15) sin 80o − 30 cos80o = − 37.2 MPa 2

τ x ' y ' = − ( σ x − σ y ) sin 2 θ + τ xy cos 2θ

Hence,

σ y ' = σ x + σ y − σx ' = 50 − 15 + 6.45 = 41.4 MPa

41.6 MPa

y’ 29.8 MPa

6.4 MPa x’

θ = 40o x

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_______________________________________________________________________ PROBLEM (8.17) Solve Prob. 8.16 for σ x = 20 ksi, σ y = 5 ksi, τ x y = 4 ksi, and θ = 25o , shown in Fig. P8.17.

SOLUTION

1 1 2 2 1 1 = (20 − 5) + (20 − 5) cos 50o + 4 sin 50o = 15.38 ksi 2 2

σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ

1 2 1 = − (20 − 5)sin 50o + 4 cos50o = −3.17 ksi 2

τ x ' y ' = − ( σ x − σ y ) sin 2 θ + τ xy cos 2θ

Therefore,

σ y ' = σ x + σ y − σx ' = 20 + 5 − 20.4 = 4.6 ksi 4.6 ksi

y’ 3.17 ksi

15.38 ksi

x’

θ = 25o x

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_______________________________________________________________________ PROBLEM (8.18) Two triangular plates are weld together and subjected to biaxial stresses with σ x = -5 MPa and

σy

= 15 MPa, as depicted in Fig. P8.18. Determine:

(a) The normal stress (b) The shear stress

σw

τw

acting perpendicular to the weld.

acting parallel to the weld.

SOLUTION

We have

x’

τx 'y '

θ

o

θ = 25 + 90 = 115 σ x = −5 MPa σ y = 15 MPa τ xy = 0

σx '

5 MPa 25

o

y’

x

15 MPa

1 1 (σ x + σ y ) + (σ x − σ y ) cos 2θ 2 2 1 1 = ( −5 +15) + ( −5 −15) cos 230o = 11.43 MPa 2 2 Thus, σ w = σ x' = 11.43 MPa

(a) σ x' =

1 (b) τ x ' y ' = − ( σ x − σ y ) sin 2θ 2 1 = − (−5 − 15)sin 230 o = − 7.66 MPa 2 τw So τ w = τ x ' y ' = − 7.66 MPa

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_______________________________________________________________________ PROBLEM (8.19) The stresses at a point in the enclosure plate of a boiler are as shown in the element of Fig. P8.19. Calculate ...