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

(b) Principal and Maximum in-Plane Shear Strains
From Equation (13.10), the principal strains can be calculated as
ε
p1, p2
εx + εy ⎛εx − εy⎞
γ xy
= ± ⎜ ⎟ + ⎛ 2 ⎝ 2 ⎠ ⎝ ⎜
⎞
⎟
2 ⎠
260 + 990
= ±
2
= 625 ± 457
2 2
2 2
⎛ 260 − 990 ⎞ 550
⎜ ⎟ + ⎛−
⎞
⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
= 1, 082 µε , 168 µε Ans.
and from Equation (13.11), the maximum in-plane shear strain is
2 2
γ max ⎛ εx − εy⎞
γ xy
=±
2 ⎝
⎜
2 ⎠
⎟ + ⎛ ⎝ ⎜ ⎞
2 ⎠
⎟
2 2
⎛ 260 − 990⎞
550
=±
⎝
⎜
2 ⎠
⎟ + ⎛ − ⎞
⎝
⎜
2 ⎠
⎟
= 457 µ rad
∴ γ = 914 µ rad
Ans.
max
The in-plane principal directions can be determined from Equation (13.9):
tan2θ
p
x
xy
y
−550
=
260 − 990
550
= − −730
∴ 2θ
= 37.0° and thus θ = 18.5°
p
=
ε
γ
− ε
p
Note: ε
x
− ε < 0
y
Since ε x − ε y < 0, θ p is the angle between the x direction and the ε p2 direction.
The strain rosette is bonded to the surface of the copper alloy component; therefore,
we have a plane stress condition. Consequently, the out-of-plane normal strain ε z will not
be zero. The third principal strain ε p3 can be computed from Equation (13.15):
ε
= 0.307
ε = − ν
( )
(260 990) 554
1 − ν ε + ε = − + = − µε Ans.
1 − 0.307
p3
z x y
The absolute maximum shear strain will be the largest value obtained from three possibilities
(see Table 13.2):
γ = ε − ε or γ = ε − ε or γ = ε −ε
absmax p1 p2 absmax p1 p3 absmax p2 p3
In this instance, the absolute maximum shear strain will be
γ = ε − ε = 1, 082 − ( − 554) = 1,636 rad
absmax p1 p3 µ
To better understand how γ abs max is determined in this instance, it is helpful to sketch
Mohr’s circle for strain. Strains in the x–y plane are represented by the solid circle with
its center at C = 625 µε and radius R = 457 µ. The principal strains in the x–y plane are
ε p1 = 1,082 µε and ε p2 = 168 µε.
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