AOE 3024: Thin Walled Structures Solutions to Homework # 4

Transcription

1 AOE 34: Thin Walled Structures Solutions to The state of stress at a point in a component is given as σ xx τ xy τ xz 4 4 [σ] = τ yx σ yy τ yz = 4 5 MPa () τ zx τ zy σ zz a) Determine the factor of safety using Tresca (Maximum Shear stress) and Von-Mises failure criteria. Assume the yield stress to be σ y = 5 MPa Factor of safety can be defined as the ratio of the allowable strength to the required strength and is always greater than one. F S = Allowable stresses Required stresses = σ all () First, we calculate the principal stresses. In general, when all three principal stresses are needed the eigenvalue approach is easier (See Homework # for details). However, for problems where τ xz = τ yz = τ zx = τ zy =, one of the principal stresses will be σ 3 = σ zz. Moreover, for a plane stress problem σ 3 = σ zz =. Therefore, Mohr s Circle for plane stress problems can be used. Using the Mohr s Circle σ ave = σ xx + σ yy σ dif = σ xx σ yy = R = (4) + (5) = MPa = 45 MPa (3) (4) (5) MPa = 5 MPa (4) τ xy + σ dif = (4) + ( 5) MPa = 4.33 MPa (5) σ = σ ave + R = (45) + (4.33) = MPa (6) σ = σ ave R = (45) (4.33) = MPa (7) σ 3 = MPa (for plane stress) (8) Using the eigenvalue approach Cauchy s equations can be written in matrix form as follows 4 λ 4 n x 4 5 λ n y = λ n z (9) of 5

2 These equations possess a nontrivial solution when the determinant of the matrix of coefficients of n x, n y, and n z vanishes. This determinant leads to the characteristic equation λ 3 9 λ + 4 λ = ( λ 9 λ + 4 ) λ = () The three roots of the above characteristic equation are the principle stresses: σ = MPa σ = MPa σ 3 =. MPa () I. Tresca (Maximum Shear Stress) Failure Tresca criterion is a method to predict yielding in 3-D state of stress and is defined as: σ max σ min = ±σ y () where σ max = max[σ, σ, σ 3 ] = max[85.33, , ] = MPa (3) σ min = min[σ, σ, σ 3 ] = min[85.33, , ] = MPa (4) The required strength is the overall maximum shear stress and is defined as = τ max 3 D = σ max σ min = MPa = MPa (5) The cross section of a uniaxial test specimen is a principal plane, and the normal stress on that plane is the only nonzero principal stress. Therefore, the allowable strength is the maximum shear stress in a one dimensional tension test and is defined as Then the factor of safety is D σ all = τ max = σ y = 5 = MPa = 5 MPa (6) =.9344 (7) of 5

3 II. Von Mises Failure Von Mises stress is another method to predict yielding in 3-D state of state and can be defined in different, but equivalent, ways (σ σ ) + (σ σ 3 ) + (σ 3 σ ) [(σ xx σ yy ) + (σ xx σ zz ) + (σ yy σ zz ) ] + 3 ( ) τxy + τxz + τyz (8a) (8b) For plane stress problems (τ yz = τ xz = σ zz = ), for which σ 3 = σ zz =, the above simplifies to σ σ σ + σ σxx + σyy σ xx σ yy + 3 τxy Therefore, the required strength is the von Mises stress. Using equation for plane stress (9a) (9b) = (85.33) (85.33)(4.6887) + (4.6887) MPa = MPa () The allowable strength is the yield stress Then the factor of safety is σ all = σ y = 5 MPa () = = () Note that: () the advantage in using von Mises criteria is that one does not need to find the principal stresses in order to calculate σ M ; () the lower factor of safety predicted by the maximum shear stress criterion shows it is slightly conservative with respect to von Mises prediction. b) Your solution should clearly indicate the given state of stress in the σ a -σ b failure diagram. First, note that we will nondimensionalize all stresses by dividing them by the yield stress. Therefore σ a and σ b will be nondimensional quantities σ a = σ σ y σ b = σ σ y (3) (Nondimensionalization is not necessary. criterions.) However, it helps us better understand the different 3 of 5

4 Therefore, let the location of the present state of stress be σ a, σ b σ a = =.344 σ b = =.875 (4) Von Mises σσb= ###### σ b =σ σy /σ y Maximum Shear Stress.5 (.344,.875) σa= σ ###### σ a =σ σy /σ y Actual state of stress Fig. Location of the stress state of the present problem Note that the factor of safety can be also found from the σ a -σ b failure diagram Von Mises σ b = ##### σ b=σ # σ /σ y y Maximum Shear Stress.5 S A C M σa = ##### σ a=σ # σ /σ y y Actual state of stress Fig. General location of the stress state For von Mises Failure : Then the factor of safety is = CM CA 4 of 5 = (5)

5 For Tresca : Then the factor of safety is = CS CA =.9344 (6) 5 of 5