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12.1 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.1 

Solution

(215 MPa) cos 25 ( cos 25 ) (70 MPa)sin 25 ( sin 25 ) 0

189.1021 MPa 189.1 MPa (T)

n n

n

 F dA dA dAσ  

σ  

Σ = − ° ° − ° ° =

= =   Ans.

(215 MPa)sin 25 ( cos 25 ) (70 MPa)cos 25 ( sin 25 ) 0

55.5382 MPa 55.5 MPa

t nt 

nt 

 F dA dA dAτ  

τ  

Σ = + ° ° − ° ° =

= − = −   Ans.

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12.2 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.2

Solution

(3, 000 psi) cos 70 ( cos 70 ) (1, 600 psi) sin 70 ( sin 70 ) 0

1,061.9022 psi 1,062 psi (T)

n n

n

 F dA dA dAσ  

σ  

Σ = + ° ° − ° ° =

= =   Ans.

(3, 000 psi) sin 70 ( cos 70 ) (1, 600 psi) cos 70 ( sin 70 ) 0

1,478.4115 psi 1,478 psi

t nt 

nt 

 F dA dA dAτ  

τ  

Σ = + ° ° + ° ° =

= − = −   Ans.

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12.3 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.3

Solution

(190 MPa) cos 40 ( cos 40 ) (320 MPa)sin 40 ( sin 40 ) 0

20.7197 MPa 20.7 MPa (C)

n n

n

 F dA dA dAσ  

σ  

Σ = − ° ° + ° ° =

= − =   Ans.

(190 MPa)sin 40 ( cos 40 ) (320 MPa) cos 40 ( sin 40 ) 0

251.1260 MPa 251 MPa

t nt 

nt 

 F dA dA dAτ  

τ  

Σ = − ° ° − ° ° =

= =   Ans.

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12.4 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.4

Solution

(21.0 ksi) cos 55 ( cos 55 ) (12.5 ksi)sin 55 ( sin 55 ) 0

15.2964 ksi 15.30 ksi (C)

n n

n

 F dA dA dAσ  

σ  

Σ = + ° ° + ° ° =

= − =   Ans.

(21.0 ksi)sin 55 ( cos 55 ) (12.5 ksi) cos 55 ( sin 55 ) 0

3.9937 ksi 3.99 ksi

t nt 

nt 

 F dA dA dAτ  

τ  

Σ = − ° ° + ° ° =

= =   Ans.

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12.5 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.5

Solution

(270 MPa) cos 30 ( cos 30 )

(125 MPa)sin 30 ( cos30 ) (125 MPa) cos 30 ( sin 30 ) 0

310.7532 MPa 311 MPa (T)

n n

n

 F dA dA

dA dA

σ  

σ  

Σ = − ° °

− ° ° − ° ° =

= =   Ans.

(270 MPa) sin30 ( cos 30 )

(125 MPa) cos 30 ( cos 30 ) (125 MPa)sin 30 ( sin 30 ) 0

54.4134 MPa 54.4 MPa

t nt 

nt 

 F dA dA

dA dA

τ  

τ  

Σ = + ° °

− ° ° + ° ° =

= − = −   Ans.

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12.6 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.6

Solution

(2, 300 psi) cos 55 ( cos 55 ) (900 psi) sin 55 ( sin 55 )

(400 psi)sin 55 ( cos55 ) (400 psi)cos55 ( sin 55 ) 0

984.7089 psi 985 psi (T)

n n

n

 F dA dA dA

dA dA

σ  

σ  

Σ = − ° ° − ° °

+ ° ° + ° ° =

= =   Ans.

(2, 300 psi) sin 55 ( cos 55 ) (900 psi) cos 55 ( sin 55 )(400 psi) cos55 ( cos55 ) (400 psi)sin 55 ( sin 55 ) 0

520.9768 psi 521 psi

t nt 

nt 

 F dA dA dAdA dA

τ  

τ  

Σ = + ° ° − ° °

+ ° ° − ° ° =

= − = −   Ans.

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12.7 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.7

Solution

(35 MPa) sin 75 ( sin 75 )

(25 MPa)sin 75 ( cos75 ) (25 MPa)cos75 ( sin 75 ) 0

20.1554 MPa 20.2 MPa (T)

n n

n

 F dA dA

dA dA

σ  

σ  

Σ = − ° °

+ ° ° + ° ° =

= =   Ans.

(35 MPa) cos 75 ( sin 75 )(25 MPa)cos 75 ( cos75 ) (25 MPa)sin 75 ( sin 75 ) 0

30.4006 MPa 30.4 MPa

t nt 

nt 

 F dA dAdA dA

τ  

τ  

Σ = + ° °

− ° ° + ° ° =

= − = −   Ans.

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12.8 The stresses shown in the figure act at a point in a stressed

 body. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.

Fig. P12.8

Solution

(7.4 ksi) cos 25 ( cos 25 ) (14.6 ksi) sin 25 ( sin 25 )

(9.3 ksi)sin 25 ( cos 25 ) (9.3 ksi)cos 25 ( sin 25 ) 0

3.6535 ksi 3.65 ksi (T)

n n

n

 F dA dA dA

dA dA

σ  

σ  

Σ = + ° ° − ° °

− ° ° − ° ° =

= =   Ans.

(7.4 ksi) sin 25 ( cos 25 ) (14.6 ksi) cos 25 ( sin 25 )(9.3 ksi)cos 25 ( cos 25 ) (9.3 ksi)sin 25 ( sin 25 ) 0

14.4044 ksi 14.40 ksi

t nt 

nt 

 F dA dA dAdA dA

τ  

τ  

Σ = + ° ° + ° °

+ ° ° − ° ° =

= − = −   Ans.