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

Then solve for γ xy :
2 2
380 = ( − 420)cos (45 ° ) + (240)sin (45 ° ) + γ sin(45 ° )cos(45 ° )
380 + (420)(0.5) − (240)(0.5)
∴ γ xy =
= 940 µ rad
0.5
Since the strain rosette is bonded to the surface of the aluminum component, we have a
plane stress condition. Use the generalized Hooke’s law equation (13.26) and the material
properties E = 10,000 ksi and ν = 0.33 to compute the normal stresses σ x and σ y from the
normal strains ε x and ε y :
σ
σ
E
10,000 ksi
−6 −6
= ( ε + νε ) = [( − 420 × 10 ) + 0.33(240 × 10 )] =−3.82 ksi
1 −
2 2
ν
1 − (0.33)
E
10,000 ksi
−6 −6
= ( ε + νε ) = [(240 × 10 ) + 0.33( − 420 × 10 )] = 1.138 ksi
2 2
1 − ν
1 − (0.33)
x x y
y y x
Note: The strain measurements reported in microstrain units (µε) must be converted to
dimensionless quantities (i.e., in./in.) in making this calculation.
Before the shear stress τ xy can be computed, the shear modulus G for the aluminum
material must be calculated from Equation (13.18):
G =
E 10,000 ksi
=
= 3,760 ksi
2(1 + ν)
2(1 + 0.33)
Now we solve Equation (13.25) for the shear stress:
τ
xy
−6
= G γ = (3,760 ksi)(940 × 10 ) = 3.53 ksi
xy
Finally, the normal stress acting in the direction of θ = 45° can be calculated with a stress
transformation equation, such as Equation (12.5):
2 2
σ = σ cos θ + σ sin θ + 2τ sinθcosθ
n x y xy
2 2
= ( − 3.82 ksi) cos 45 °+ (1.138 ksi)sin 45°+ 2(3.53 ksi)sin45° cos45°
= 2.19 ksi (T)
Ans.
xy
ExAmpLE 13.8
y
60°
σ
y
x
σ
x
A thin steel [E = 210 GPa; G = 80 GPa] plate is subjected to biaxial stress. The normal
stress in the x direction is known to be σ x = 70 MPa. The strain gage measures a normal
strain of 230 µε in the indicated direction on the free surface of the plate.
(a) Determine the magnitude of σ y that acts on the plate.
(b) Determine the principal strains and the maximum in-plane shear strain in the plate.
Show the principal strain deformations and the maximum in-plane shear strain distortion
in a sketch.
(c) Determine the magnitude of the absolute maximum shear strain in the plate.
Plan the Solution
To begin this solution, we will write a strain transformation equation for the strain gage
oriented as shown. The equation will express the strain ε n measured by the gage in terms
of the strains in the x and y directions. Since there is no shear stress τ xy acting on the plate,
the shear strain γ xy will be zero and the strain transformation equation will be reduced to
terms involving only ε x and ε y . Equations (13.24) from the generalized Hooke’s law for
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