AOE 3024: Thin Walled Structures

Solutions to Homework # 4

The state of stress at a point in a component is given as

[σ] =

σxx τxy τxz

τyx σyy τyz

τzx τzy σzz

=

40 40 0

40 50 0

0 0 0

MPa (1)

a) Determine the factor of safety using Tresca (Maximum Shear stress) and Von-Mises

failure criteria. Assume the yield stress to be σy = 250 MPa

Factor of safety can be defined as the ratio of the allowable strength to the required strength

FS =Allowable stresses

Required stresses=

σall

σreq

(2)

and is always greater than one.

First, we calculate the principal stresses.In general, when all three principal stresses are needed the eigenvalue approach is easier (See Home-

work # 2 for details). However, for problems where τxz = τyz = τzx = τzy = 0, one of the principal

stresses will be σ3 = σzz. Moreover, for a plane stress problem σ3 = σzz = 0. Therefore, Mohr’s

Circle for plane stress problems can be used.

Using the Mohr’s Circle

σave =σxx + σyy

2=

(40) + (50)

2MPa = 45 MPa (3)

σdif =σxx − σyy

2=

(40)− (50)

2MPa = −5 MPa (4)

R =√

τ 2xy + σ2

dif =√

(40)2 + (−5)2 MPa = 40.3113 MPa (5)

σ1 = σave + R = (45) + (40.3113) = 85.3113 MPa (6)

σ2 = σave −R = (45)− (40.3113) = 4.6887 MPa (7)

σ3 = 0 MPa (for plane stress) (8)

Using the eigenvalue approach

Cauchy’s equations can be written in matrix form as follows 40− λ 40 0

40 50− λ 0

0 0 0− λ

nx

ny

nz

=

0

0

0

(9)

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Homework # 4

These equations possess a nontrivial solution when the determinant of the matrix of coefficients of

nx, ny, and nz vanishes. This determinant leads to the characteristic equation

λ3 − 90 λ2 + 400 λ =(λ2 − 90 λ + 400

)λ = 0 (10)

The three roots of the above characteristic equation are the principle stresses:

σ1 = 85.3113 MPa σ2 = 4.6887 MPa σ3 = 0.0 MPa (11)

I. Tresca (Maximum Shear Stress) Failure Criterion

Tresca criterion is a method to predict yielding in 3-D state of stress and is defined as:

σmax − σmin = ±σy (12)

where

σmax = max[σ1, σ2, σ3] = max[85.3113, 4.68871, 0] = 85.3113 MPa (13)

σmin = min[σ1, σ2, σ3] = min[85.3113, 4.68871, 0] = 0 MPa (14)

The required strength is the overall maximum shear stress and is defined as

σreq = τmax

∣∣∣3−D

=σmax − σmin

2=

85.3113− 0

2MPa = 42.6556 MPa (15)

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

σall = τmax

∣∣∣1−D

=σy

2=

250

2MPa = 125 MPa (16)

Then the factor of safety is

FS =σall

σreq

=125

42.6556= 2.93044 (17)

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Homework # 4

II. Von Mises Failure Criterion

Von Mises stress is another method to predict yielding in 3-D state of state and can be defined

in different, but equivalent, ways

σM =

√(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2

2(18a)

σM =

√1

2[(σxx − σyy)2 + (σxx − σzz)2 + (σyy − σzz)2] + 3

(τ 2xy + τ 2

xz + τ 2yz

)(18b)

For plane stress problems (τyz = τxz = σzz = 0), for which σ3 = σzz = 0, the above simplifies to

σM =√

σ21 − σ1σ2 + σ2

2 (19a)

σM =√

σ2xx + σ2

yy − σxxσyy + 3 τ 2xy (19b)

Therefore, the required strength is the von Mises stress. Using equation for plane stress

σreq = σM =√

(85.3113)2 − (85.3113)(4.68871) + (4.68871)2 MPa = 83.0662 MPa (20)

The allowable strength is the yield stress

σall = σy = 250 MPa (21)

Then the factor of safety is

FS =σall

σreq

=250

83.0662= 3.00965 (22)

Note that: (1) the advantage in using von Mises criteria is that one does not need

to find the principal stresses in order to calculate σM; (2) 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 σa-σb failure

diagram.

First, note that we will nondimensionalize all stresses by dividing them by the yield stress.

Therefore σa and σb will be nondimensional quantities

σa =σ1

σy

σb =σ2

σy

(23)

(Nondimensionalization is not necessary. However, it helps us better understand the different

criterions.)

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Homework # 4

Therefore, let the location of the present state of stress be σa1 , σb1

σa1 =85.3113

250= 0.34124 σb1 =

4.68871

250= 0.01875 (24)

-1 -0.5 0.5 1σa=

σ1#######σy

-1

-0.5

0.5

1

σb=σ2#######σy

Actual state of stress

Von Mises Criterion

Maximum Shear Stress

Criterion

(0.34124, 0.01875)

σb=σ2/σy

σa=σ1/σy

Fig. 1 Location of the stress state of the present problem

Note that the factor of safety can be also found from the σa-σb failure diagram

-1 -0.5 0.5 1 σ a = σ 1

#### ## # σ y

-1

-0.5

0.5

1

σ b = σ 2 #### ## # σ y

Actual state of stress

Von Mises Criterion

Maximum Shear Stress

Criterion

σb=σ2/σy

σa=σ1/σy

C A

S M

Fig. 2 General location of the stress state

For von Mises Failure Criterion: Then the factor of safety is

FS =σall

σreq

=CM

CA= 3.00965 (25)

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Homework # 4

For Tresca Criterion: Then the factor of safety is

FS =σall

σreq

=CS

CA= 2.93044 (26)

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Homework # 4