Mechanics of Materials Solutions Chapter12 Probs1 8

The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.1

Fig. P12.1 Solution Σ Fn = σ  n dA − (215 σ  n

189.10 1021 21 MPa MPa = = 189.

Σ Ft = τ  nt dA + ( 215 τ  nt 

MPa) co cos 25 25°(dA cos 25 25° ) − (70 MPa) sin 25 25° (dA sin 25 25° ) = 0

= −55.5382

189. 189.1 1 MPa (T)

Ans.

MPa) sin 25 25°(dA cos 25 25° ) − (70 MPa) cos 25 25° (dA sin 25 25° ) = 0

MP MPa =

− 55.5

MPa

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.2

Fig. P12.2 Solution

000 Σ Fn = σ  n dA + (3, 00 σ  n

,061.902 022 2 = 1,061.9

psi psi = 1,062 psi psi (T)

Σ Ft = τ  nt dA + (3, 00 000 τ  nt 

psi) co cos 70 70°(dA cos 70 70° ) − (1, 60 600 psi) si sin 70 70° (dA sin 70 70° ) = 0

,478.4115 = −1,478

Ans.

psi) si sin 70 70°(dA cos 70 70°) + (1, 60 600 psi) co cos 70 70° (dA sin 70 70° ) = 0

psi =

,478 − 1,478

psi

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.3

Fig. P12.3 Solution Σ Fn = σ  n dA − (190 σ  n

= −20.7197

MP MPa = 20.7 MP MPa (C (C)

Σ Ft = τ  nt dA − (190 τ  nt 

=

MP MPa) co cos 40 40° (dA cos 40 40° ) + (320 MP MPa) sin 40 40° (dA sin 40 40° ) = 0 Ans.

MP MPa) sin 40 40°(dA cos 40 40°) − (320 MP MPa) co cos 40 40° (dA sin 40 40° ) = 0

251.1260 MPa = 251 MPa

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.4

Fig. P12.4 Solution Σ Fn = σ  n dA + ( 21.0 σ  n

15.296 964 4 = −15.2

ksi ksi = 15.3 15.30 0 ksi ksi (C)

Σ Ft = τ  nt dA − ( 21.0 τ  nt 

= 3.9937

ks ksi) co cos 55 55°(dA cos 55 55° ) + (12.5 ksi) sin 55 55° (dA sin 55 55° ) = 0 Ans.

ks ksi) sin 55 55°(dA cos 55 55°) + (12.5 ksi) co cos 55 55° (dA sin 55 55° ) = 0

ks ksi = 3.99 ks ksi

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.5

Fig. P12.5 Solution Σ Fn = σ  n dA − (270 −(125 σ  n

= 310.7 0.7532

MPa) sin 30 30°(dA cos 30°) − (125 MPa) co cos 30 30° (dA sin 30 30° ) = 0 MPa = 311 MPa (T)

Σ Ft = τ  nt dA + ( 270 −(125 τ  nt 

= −54.4134

MPa) co cos 30 30° (dA cos 30 30° )

Ans.

MPa) si sin 30 30°(dA cos 30 30°)

MPa) co cos 30 30° (dA cos 30 30° ) + (125 MPa) sin 30 30° (dA sin 30 30° ) = 0 MPa =

− 54.4

MPa

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.6

Fig. P12.6 Solution

300 Σ Fn = σ  n dA − (2, 30 + ( 400 σ  n

.7089 = 984.70

psi) sin 55 55°(dA cos 55°) + (400 psi) cos 55°(dA sin 55 55° ) = 0 ps psi = 985 985 ps psi (T)

300 Σ Ft = τ  nt dA + ( 2, 30 + ( 400 τ  nt 

= −520.9768

psi) co cos 55 55°(dA cos 55 55° ) − (900 psi) si sin 55 55° (dA sin 55 55° )

Ans.

psi) si sin 55 55°(dA cos 55 55°) − (900 psi) co cos 55 55° (dA sin 55 55° )

ps psi) co cos 55°(dA cos 55° ) − (400 ps psi) sin 55 55° (dA sin 55 55° ) = 0

psi =

− 521 psi

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.7

Fig. P12.7 Solution Σ Fn = σ  n dA − (35 + ( 25 σ  n

=

MPa) sin 75 75°(dA cos 75° ) + (25 MPa) cos 75° (dA sin 75 75° ) = 0

20.15 .1554 MPa = 20.2 MPa (T)

Σ Ft = τ  nt dA + (35 −( 25 τ  nt 

MP MPa) si sin 75 75°(dA sin 75 75°)

= −30.4006

Ans.

MP MPa) co cos 75 75°(dA sin 75 75°)

MPa) cos 75 75°(dA cos 75° ) + (25 MP MPa) sin 75 75° (dA sin 75 75° ) = 0 MPa =

− 30.4

MPa

Ans.

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The stresses shown in the figure act at a point in a stressed  body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. 12.8

Fig. P12.8 Solution Σ Fn = σ  n dA + (7.4 −(9.3 σ  n

ksi) sin 25 25°(dA cos 25 25°) − (9.3 ksi) cos 25 25° (dA sin 25 25° ) = 0

.6535 = 3.65

ks ksi = 3.65 ks ksi (T)

Σ Ft = τ  nt dA + (7.4 + (9.3 τ  nt 

= −14.4044

ksi) co cos 25 25°(dA cos 25 25° ) − (14.6 ksi) si sin 25 2 5° (dA sin 25 2 5° )

Ans.

ksi) si sin 25 2 5°( dA cos 25 25°) + (14.6 ksi) co cos 25 25° (dA sin 25 2 5° )

ksi) cos 25 25°(dA cos 25 25°) − (9.3 ksi) sin 25 25°(dA sin 25 25° ) = 0 ks ksi =

− 14.40

ks ksi

Ans.