Homework # 6: Solutions
Solution to 4.2
Given:
σx = −30 MPa
σy = 90 MPa
τxy = −40 MPa
σz = τxz = τyz = 0
σyp = 250 MPa
(1)
(2)
(3)
The principal stresses can be calculated as the eigenvalues of the stress tensor or from Mohr’s
circle as :
σ1 = 102 MPa
σ2 = 0
σ3 = −42.1 MPa
(4)
The maximum shearing stress can be calculated as:
σ1 − σ3
τmax =
= 72.1 MPa
(5)
2
The octahedral shear stress can be calculated as:
1p
τoct =
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 60.6 MPa
(6)
3
Part a: Based on the maximum shear stress theory, the maximum allowable shear stress is
σyp
. So the factor of safety is given by:
2
n=
σyp
2
τmax
=
125
= 1.73
72.1
(7)
Part b: Based on the maximum energy of distortion criterion (which is identical to the
maximum
octahedral shear stress theory), the maximum allowable octahedral shear stress
√
2σyp
is 3 . So the factor of safety is given by:
√
n=
2σyp
3
= 1.95
(8)
τoct
NOTE: We could have used the simple torsion test results τyp = 145 MPa to calculate the
above results as follows:
Part a: Based on the maximum shear stress theory, the maximum allowable shear stress is
τyp . So the factor of safety is given by:
τyp
145
=
= 2.01
(9)
τmax
72.1
Part b: Based on the maximum
octahedral shear stress theory, the maximum allowable
√
2τyp
√
octahedral shear stress is
. So the factor of safety is given by:
3
n=
n=
√
2τ
√ yp
3
= 1.96
(10)
τoct
Based on the above results its seems that the maximum energy of distortion criterion gives
a result consistent with both the uniaxial tension test as well as simple torsion test.
Solution to 4.3
Given:


0 40
0
[τij ] = 40 50 −60 MPa
0 −60 0
(11)
The principal stresses are:
σ1 = 101 MPa
σ2 = 0
σ3 = −51.3 MPa
(12)
The maximum shearing stress can be calculated as:
τmax =
σ1 − σ3
= 76.3 MPa
2
(13)
The octahedral shear stress can be calculated as:
τoct =
1p
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 63.4 MPa
3
(14)
Part a Based on the maximum shear stress theory, the maximum shear stress at yielding is
σyp
. So the uniaxial tensile strength can be calculated as:
2
σyp = 2τmax = 153 MPa
(15)
Part
b Based on the maximum shear stress theory, the maximum shear stress at yielding is
√
2σyp
. So the uniaxial tensile strength can be calculated as:
3
3τoct
σyp = √ = 135 MPa
2
(16)


120 50 30
[τij ] =  50 80 20 MPa
30 20 10
(17)
σyp = 300 MPa
(18)
Solution to 4.15
Given:
The principal stresses can be calculated as the eigenvalues of the stress tensor as :
σ1 = 162 MPa
σ2 = 46.1 MPa
σ3 = 1.47 MPa
(19)
The maximum shearing stress can be calculated as:
τmax =
σ1 − σ3
= 80.5 MPa
2
(20)
The octahedral shear stress can be calculated as:
τoct =
1p
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 67.8 MPa
3
(21)
Part a: Based on the maximum shear stress theory, the maximum allowable shear stress is
σyp
. So the factor of safety is given by:
2
n=
σyp
2
τmax
= 1.86
(22)
Part b: Based on the maximum energy of distortion criterion (which is identical to the
maximum
octahedral shear stress theory), the maximum allowable octahedral shear stress
√
2σyp
is 3 . So the factor of safety is given by:
√
n=
2σyp
3
τoct
= 2.09
(23)