Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Since the strain measurements were made on the free surface
of the copper alloy component, we have a plane stress
situation. In such a situation, the third principal stress σ p3
(which is the principal stress in the out-of-plane direction) will
be zero; however, the third principal strain ε p3 (meaning the
principal strain in the out-of-plane direction) will not be zero,
because of the Poisson effect.
The third principal strain in this instance was ε p3 = −554 µε.
This point is plotted on the ε axis, and two additional Mohr’s
circles are constructed. As shown in the sketch, the circle
defined by ε p3 and ε p1 is the largest circle. This result indicates
that the absolute maximum shear strain γ abs max will not occur
in the x–y plane.
(c) Principal and Maximum in-Plane Shear Stress
The generalized Hooke’s law equations are written in terms of
the directions x and y in Equation (13.26); however, these
equations are applicable to any two orthogonal directions. In
this instance, the principal directions will be used. Given the
material properties E = 115 GPa and ν = 0.307, the principal
stresses σ p1 and σ p2 can be computed from the principal strains
ε p1 and ε p2 :
P 3
(–554, 0)
γ
2
(–260, 275) x
(168, 0) (625, 0) (1,082, 0)
P2
P ε
1
R = 818
R = 457
y (990, –275)
σ
σ
E
115,000 MPa
−6 − 6
= ( ε + νε ) =
[(1, 082 × 10 ) + 0.307(168 × 10 )] = 143.9 MPa Ans.
1 − ν
1 − (0.307)
p1 2 p1 p2 2
E
115,000 MPa
−6 − 6
= ( ε + νε ) =
[(168 × 10 ) + 0.307(1,082 × 10 )] = 63.5 MPa Ans.
1 − ν
1 − (0.307)
p2 2 p2 p1 2
Note: The strain measurements reported in microstrain units (µε) must be converted to
dimensionless quantities (i.e., mm/mm) in making this calculation.
Before the maximum in-plane shear stress τ max can be computed, the shear modulus
G for the copper alloy material must be calculated from Equation (13.18):
G =
E 115,000 MPa
=
2(1 + ν)
2(1 + 0.307)
= 44,000 MPa
The maximum in-plane shear stress τ max is calculated from Equation (13.25), which is
solved for the stress:
τ
max
− 6
= G γ = (44,000 MPa)(914 × 10 ) = 40.2 MPa Ans.
max
Alternatively, the maximum in-plane shear stress τ max can be calculated from the principal
stresses:
τ
max
σ
=
− σ
p1 p2
2
143.9 − 63.5
=
= 40.2 MPa
Ans.
2
On the planes of maximum in-plane shear stress, the normal stress is
σ
avg
σ
=
+ σ
p1 p2
2
143.9 + 63.5
=
2
= 103.7 MPa
573