Solution: 12.9 The stresses shown in the figure act at a point in a stressed body

The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown. 12.9

Fig. P12.9 Solution

The given stress values are: σ x = 4, 200 psi, σ y = 1, 800 psi,

τ xy =

0 psi,

θ = + 50 °

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ =

(4,200 ,200 ps psi)co )cos2 (50°) + (1,80 ,800 ps psi)si )sin 2 (50°) + 2(0 ps psi)si )sin(50°)co )cos(50°)

=

2,791. ,791.6 6222 ps psi = 2,790 ,790 ps psi (T) (T)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) 200 = −[( 4, 20

p si si) − (1, 80 800 p si si)] si sin(50°) co cos(50°) + (0 psi)[cos2 (50°) − sin 2 (50°)]

= −1,181.7693

psi =

− 1,182

psi

Ans.

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The stresses shown in the figure figure act at a point point in in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown. 12.10

Fig. P12.10 Solution

The given stress values are: σ x = −90 MPa, σ y = −140 MPa,

τ xy =

0 MPa,

θ = + 65°

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ = ( −90

MPa ) co cos2 (65°) + (−140 MPa ) si sin 2 (65°) + 2(0 MPa ) si sin(65° ) co cos(65° )

131.06 0697 97 = −131.

MPa MPa = 131. 131.1 1 MPa MPa (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[( −90

MPa ) − (−140 MPa )] sin(65°) co cos(65°) + (0 MPa )[ )[cos2 (65°) − sin 2 (65°)]

= −19.1511 MPa = − 19.15

MP MPa

Ans.









The given stress values are: σ x = −5.5 ksi, σ y = 18.7 ksi,

τ xy =

0 ksi,

θ = − 20 °

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ = ( −5.5

ksi) co cos2 (−20°) + (18.7 ksi) si sin 2 (−20°) + 2(0 ksi) si sin(−20°) co cos( −20°)

6691 ksi = = −2.669

2.67 ks ksi (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[( −5.5

ksi) − (18.7 ksi)] sin( −20°) co cos(−20°) + (0 ksi)[cos 2 (−20°) − sin 2 (−20°)]

= −7.7777

ksi =

− 7.78

ksi

Ans.







The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses stresses at this point on the inclined plane shown. 12.12


The given stress values are: 17, 700 psi, σ y = −12,500 psi, σ x = 17,700

τ xy =

0 psi,

θ = − 60°


(17, 70 700 psi) co cos2 (−60°) + (−12, 50 500 psi) si sin2 (−60° ) + 2(0 psi) si sin(−60°) co cos(− 60° )

,950.000 0000 = −4,950

psi psi = 4,950 ,950 psi psi (C) (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) 700 = −[(17, 70

psi) − (−12, 50 500 psi)] sin(−60°) co cos(− 60°) + (0 psi)[cos2 (−60°) − sin 2 (−60°)]

13,076.9 .983 836 6 = 13,076

psi psi = 13,080 13,080 psi psi

Ans.







The stresses shown in the figure act at a point in a stressed body. Determine the normal normal and shear stresses at this point on the inclined plane shown. 12.13


The given stress values are: σ x = −8 ksi, σ y = 6 ksi,

τ xy = 10

ksi,

θ =

75°

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ cos = ( −8) co

2

10.062 622 2 = 10.0

(75°) + (6 ks k si) si sin 2 (75°) + 2(10 ks ksi) si sin(75°) co cos(75°) ksi ksi = 10.0 10.06 6 ksi ksi (T) (T)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[( −8

ksi) − (6 ksi)] sin(75 °) co cos(75°) + (10 ksi)[cos 2 (75°) − sin 2 (75°)]

= −5.1603

ksi =

−5.16

ksi

Ans.







The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses stresses at this point on the inclined plane shown. 12.14


The given stress values are: σ x = 82 MPa, σ y = 48 MPa,

τ xy = − 26

MPa,

θ = − 25°


(82 MPa ) co cos2 (−25°) + (48 MPa ) sin 2 (−25° ) + 2(−26 MPa ) sin(−25° ) co cos(− 25° )

5.8445 = 95.8

MPa MPa = 95. 95.8 MPa (T)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[(82

MPa ) − (48 MPa )] sin(−25°) co cos(−25°) + (−26 MPa )[cos2 (−25°) − sin 2 (−25°)]

= −3.6897

MPa =

− 3.69

MPa

Ans.









12.15 The

stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.


The given stress values are: σ x = 108 MPa, σ y = − 14 MPa,

τ xy = − 72

MPa,

θ =

50°


(108 MP MPa ) co cos2 (50°) + (−14 MP MPa ) si sin 2 (50° ) + 2(− 72 MP MPa )s ) sin(50° ) co cos(50° )

4987 = −34.498

MPa MPa = 34.5 MPa MPa (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[(108

MP MPa ) − (−14 MP MPa )] )] sin(50°) co cos(50°) + (−72 MP MPa )[cos2 (50° ) − sin 2 (50° )]

= −47.5706

MP MPa =

− 47.6

MP MPa

Ans.









The given stress values are: σ x = −2,150 psi, σ y = 860 psi,

τ xy = − 1, 460

psi,

θ = − 40°

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ = ( −2,150

psi) co cos2 (−40°) + (860 psi) si sin 2 (−40°) + 2(−1, 46 460 psi) si sin(−40°) co cos(− 40° )

31.4788 psi = = 531.

531 psi (T) (T)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[( −2,150

psi) − (860 psi)] sin(−40°) co cos(−40°) + (−1, 46 460 psi)[cos2 (−40°) − sin 2 (−40°)]

,735.6620 = −1,73

psi =

,736 − 1,73

psi

Ans.







12.17 The

stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.


The given stress values are: σ x = 18 MPa, σ y = −42 MPa,

τ xy =

30 MPa,

θ =

° 68.1986


(18 MPa)co a)cos2 (68.1986° ) + (− 42 MPa)si a)sin2 (68.1986° )

2(30 +2(30

MPa) MPa) sin(6 sin(68. 8.198 1986 6° ) cos( cos(68 68.1 .1986 986° )

13.034 345 5 = −13.0

MPa MPa = 13.0 13.03 3 MPa MPa (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[(18 + (30

MPa MPa)) − (−42 MPa)]s a)]siin(68 (68.198 1986°)cos )cos(68 (68.198 1986° )

MPa)[ a)[cos2 (68.1986°) − sin 2 (68.1986° )]

= −42.4138

MPa =

− 42.4

MP MPa

Ans.









The given stress values are: σ x = 24 MPa, σ y = 80 MPa,

τ xy = − 32

MPa,

° θ = − 33.6901


(24 MP MPa ) co cos2 (−33.6901° ) + (80 MP MPa )s ) sin2 (− 33.6901° )

+2( −32 =

MP MPa)si a)sin(−33.6901° )co )cos(− 33.6901° )

70. 70.7693 MPa = 70. 70.8 MP MPa (T)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[(24 (24 + (−32

MP MPa) − (80 MP MPa)]s a)]siin(−33.6901° )cos )cos(− 33.6901° ) MPa )[cos2 (−33.6901°) − sin 2 (−33.6901° )]

= −38.1538

MPa =

− 38.2

MP MPa

Ans.







12.19

The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear shear stresses at this point on the inclined plane shown.


The given stress values are: σ x = −3,800 psi, σ y = − 2,500 psi,

τ xy =

8,200 psi,

° θ = − 59.0362

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ 800 = ( −3, 80

ps psi) co cos2 (−59.0362°) + (−2, 50 500 ps psi) si sin2 (− 59.0362° )

,200 +2(8,200

psi)s psi)siin(−59.036 0362° )cos )cos(− 59.036 0362° )

10,079. 9.41 4185 85 = −10,07

psi psi = 10,080 10,080 psi psi (C) (C)

Ans.

The shear stress transformation equation [Eq. (12-4)] gives τ nt : 2 2 τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos θ − sin θ ) = −[( −3,80 ,800

200 + (8, 20

psi) − (−2,50 ,500 psi)]si ]sin(−59.0362°)co )cos(−59.0362° )

psi)[cos2 (−59.0362° ) − sin 2 (−59.0362° )]

432.3424 = −4, 43

psi =

430 − 4, 43

psi

Ans.











The given stress values are: σ x = −3.8 ksi, σ y = 9.4 ksi,

τ xy =

5.7 ksi,

θ =

° 38.6598

The normal stress transformation equation [Eq. (12-3)] gives σ n: 2 2 σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ )cos = ( −3.8)co 2(5.7 +2(5.7 =

2

(38.6598°) + (9.4 ksi)si )sin 2 (38.6598° )

ksi)sin ksi)sin(3 (38. 8.659 6598 8°) cos( cos(38 38.6 .6598 598° )

6.9 6.9122 ksi = 6.91 ksi (T)

The shear stress transformation equation [Eq. (12-4)] gives τ nt :

Ans.







The stresses shown in Fig. P12.21a P12.21a act at a point on the free surface of a stressed body. Determine the normal stresses σ n and σ t and the shear stress τ nt at this point if they act on the rotated stress element shown in Fig. P12.21b P12.21b. 12.21

(a)

(b) Fig. P12.21

Solution

The given stress values are: σ x = 50 MPa, σ y = −15 MPa,

τ xy = − 40

MPa,

θ = − 36°


(50 MPa ) co cos2 (−36°) + (−15 MPa ) sin 2 (−36°) + 2(−40 MPa ) sin(− 36° ) co cos(−36° )

=

65. 65.5853 853 MPa = 65.6 MP MPa (T)

Ans.







The stresses shown in Fig. P12.22a P12.22a act at a point on the free surface of a stressed body. Determine the normal stresses σ n and σ t and the shear stress τ nt at this point if they act on the rotated stress element shown in Fig. P12.22b P12.22b. 12.22

(a)

(b) Fig. P12.22

Solution

The given stress values are: σ x = 1, 200 psi, σ y = 700 psi,

τ xy =

400 psi,

θ =

20°


(1,200 ,200 ps psi)co )cos2 (20°) + (700 ps psi)si )sin 2 (20°) + 2(400 ps psi)si )sin(20° )co )cos(20° )

,398.62 6262 62 = 1,398.

ps psi = 1,399 ,399 psi psi (T) (T)

Ans.







The stresses shown in Fig. P12.23 act at a point on the free surface of a machine component. Determine the normal stresses σ x and σ y and the shear stress τ xy at the point.

12.23


Redefine the axes, calling the rotated axes x axes x and and y y.. The angle from the rotated element to the unrotated element is now a positive value (since it is counterclockwise). Thus, the given stress values can be expressed as: σ x = 35 MPa, σ y = −27 MPa, τ xy = − 50 MPa, θ = 30°







The stresses shown in Fig. P12.24 act at a point on the free surface of a machine component. Determine the normal stresses σ x and σ y and the shear stress τ xy at the point. 12.24


Redefine the axes, calling the rotated axes x axes x and and y y.. The angle from the rotated element to the unrotated element is now a negative value (since it is clockwise) clockwise) . Thus, the given stress values can be expressed as: σ x = 18.2 ksi, σ y = 2.8 ksi, τ xy = 5.0 ksi, θ = − 24 ° The normal stress transformation equation [Eq. (12-3)] gives σ n, which is actually the normal stress in the horizontal direction (i.e., the original direction) on the unrotated element:

Solution: 12.9 The stresses shown in the figure act at a point in a stressed body

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