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

e x and e y in terms of σ x and σ y can be substituted into the strain transformation equation,
producing an equation in which the only unknown will be σ y . After solving for σ y , we can
again use Equation (13.24) to compute e x , e y , and e z . These values will then be used to
determine the principal strains, the maximum in-plane shear strain, and the absolute maximum
shear strain in the plate.
SolutioN
(a) Normal Stress σ y
The strain gage is oriented at an angle θ = 150°. Using this angle, write a strain transformation
equation for the gage, where the strain e n is the value measured by the gage:
2 2
230 µε = e cos (150 ° ) + e sin (150 ° ) + γ sin(150 ° )cos(150 ° )
x y xy
Note that the shear strain γ xy is related to the shear stress τ xy by Equation (13.25):
1
γ xy = τxy
G
Since τ xy = 0, the shear strain γ xy must also equal zero; thus, the strain transformation
equation reduces to
−6 2 2
230 µε = 230 × 10 mm/mm = ε cos (150 ° ) + ε sin (150 ° )
Equations (13.24) from the generalized Hooke’s law define the following relationships
between stresses and strains for a plane stress condition (which is observed to apply in
this situation):
1 1
εx = ( σx − νσ y ) and εy = ( σ y −νσ
x )
E
E
Substitute these expressions into the strain transformation equation, expand terms, and
simplify:
−6 2 2
230 × 10 mm/mm = ε cos (150 ° ) + ε sin (150 ° )
Solve for the unknown stress σ y :
x
y
x
1 ( )cos
2
1 (150 ) ( )sin
2
= σ x − νσ y ° + σ y − νσ x (150 ° )
E
E
1 1 [ cos
2 (150 ) sin
2 (150 )] [ sin
2 (150 ) cos
2
= σ x ° − νσ x ° + σ y ° − νσ y (150 ° )]
E
E
σ x
σ
2 2 y 2 2
= [cos (150 ° ) − ν sin (150 ° )] + [sin (150 ° ) − ν cos (150 ° )]
E
E
−6 2 2 2 2
(230 × 10 mm/mm) E − σ [cos (150 ° ) − νsin (150 ° )] = σ [sin (150 ° ) − νcos (150 ° )]
x
−6 2 2
(230 × 10 mm/mm) E − σx[cos (150 ° ) − νsin (150 ° )]
∴ σ y =
2 2
sin (150 ° ) − ν cos (150 ° )
Before computing the normal stress σ y , the value of Poisson’s ratio must be calculated
from the elastic modulus E and the shear modulus G:
G =
E
E 210 GPa
∴ ν = − 1 = − 1 = 0.3125
2(1 + ν) 2G 2(80 GPa)
The normal stress can now be computed:
y
y
σ =
y
−6 2 2
(230 × 10 mm/mm)(210,000 MPa) − (70 MPa)[cos (150 ° ) − (0.3125) sin (150 ° )]
= 81.2 MPa
2 2
sin (150 ° ) − (0.3125) cos (150 ° )
Ans.
569