(*)τα = Nhomakorabea2
sin 2α + τ xy cos 2α
(**)
(*) 2 + (**) 2
(σ α −
σ x +σ y
2
) + (τ α ) = (
2 2
σ x −σ y
2
2 ) 2 + τ xy
(7 - 6)
In a given problem, σx, σy, τxy are the three constants, σα,, τα are the variables. This equation is an expression for a circle of radius
σ x −α y
2
cos 2α − τ xy sin 2α
(7-1)
τα =
sin 2α + τ xy cos 2α
3. Principle Stresses in Two-dimensional Problems To find the plane for a maximum or a minimum normal stress, let σ x −α y dσ α = −2[ sin 2α + τ xy cos 2α ] = 0 = −2τ α 2 dα 2τ xy tg 2α1 = − σ x −σ y
σ'=
σ x +σ y
(7 - 5)
∴τ max = ±
min
σ1 − σ 2
2
Example 7-1 For the state of stress shown in the figure, (a) find the stresses acting on the inclined plane with θ=-22.5°; (b) find the principle stresses and shown their sense on a properly oriented element; and (c) find the maximum shear stresses with the associated normal stresses and show the results on a properly oriented element. Solution: For original state of stress σx=3 Mpa σy=1 MPa τxy= -2 Mpa (a) From Eq.(7-1)
