TEST BANK FOR Elasticity By J. R. Barber (Solution Manual)
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1.1. Show that
(i)
∂xi
∂xj
= δij and (ii) R = √xixi ,
where R = |R| is the distance from the origin. Hence find ∂R/∂xj in index
notation. Confirm your result by finding ∂R/∂x in x, y, z notation.
For an orthogonal co¨ordinate system,
∂x
∂y
= 0
(this is what is meant by orthogonality) and
∂x
∂x
= 1 .
In index notation, these results can be combined as
∂xi
∂xj
= δij .
The distance from the origin is
R =
q
x2
1 + x2
2 + x2
3 = √xixi .
Combining these results, we have
∂R
∂xj
=
∂
∂xj
√xixi =
1
2√xixi
∂xi
∂xj
xi + xi
∂xi
∂xj
!
=
2xiδij
2√xixi
=
xj √xixi
.
In x, y, z notation, we would have R = √x2 + y2 + z2 and hence
∂R
∂x
=
(2x)
2√x2 + y2 + z2 =
x
R
,
which agrees.
1.2. Prove that the partial derivatives ∂2f/∂x2; ∂2f/∂x∂y; ∂2f/∂y2 of the
scalar function f(x, y) transform into the rotated co¨ordinate system x′, y′
by rules similar to equations (1.15–1.17).
We first note from equation (1.43) that
∂
∂x′ = cos θ
∂
∂x
+ sin θ
∂
∂y
and by a similar argument
∂
∂y′ = ∇.j′
= i.j′ ∂
∂x
+ j.j′ ∂
∂y
= −sin θ
∂
∂x
+ cos θ
∂
∂y
.
We then have
∂2f
∂x′2 =
cos θ
∂
∂x
+ sin θ
∂
∂y
!
cos θ
∂f
∂x
+ sin θ
∂f
∂y
!
= cos2 θ
∂2f
∂x2 + sin2 θ
∂2f
∂y2 + 2 sin θ cos θ
∂2f
∂x∂y
∂2f
∂x′∂y′ =
−sin θ
∂
∂x
+ cos θ
∂
∂y
!
cos θ
∂f
∂x
+ sin θ
∂f
∂y
!
= (cos2 θ − sin2 θ)
∂2f
∂x∂y
+ sin θ cos θ
∂2f
∂y2 −
∂2f
∂x2
!
∂2f
∂y′2 =
−sin θ
∂
∂x
+ cos θ
∂
∂y
!
−sin θ
∂
∂x
+ cos θ
∂
∂y
!
= cos2 θ
∂2f
∂y2 + sin2 θ
∂2f
∂x2 − 2 sin θ cos θ
∂2f
∂x∂y
and these equations are clearly of the same form as (1.15–1.17).
1.3. Show that the direction cosines defined in (1.19) satisfy the identity
lij lik = δjk .
Hence or otherwise, show that the product σijσij is invariant under co¨ordinate
transformation.
For a given value of j, lij defines the components in x′
i co¨ordinates of a unit vector
in the direction of the xj -axis. It follows that
lij lik ,
is the dot product between two unit vectors defined in the x′
i-system. One of these
vectors represents the xj -axis and the other the xk-axis. This dot product is unity if
the axes are identical and zero if they are not, since the three axes are orthogonal.
Hence
lij lik = δjk .
Now consider
σ′
ij = lipljqσpq ,
from equation (1.22). We can write another version of the same quantity using different
dummy indices as
σ′
ij = lirljsσrs .
We need to do this because otherwise when we take the product the same index
would appear more than twice which leads to an ambiguity in terms of the summation
convention.
Taking the product of these quantities, including the implied summations, we then
have
σ′
ijσ′
ij = lipljqlirljsσpqσrs
and using the identity we proved above, this gives
σ′
ijσ′
ij = δprδqsσpqσrs = σpqσpq ,
showing that the product is invariant under co¨ordinate transformation.
1.4. By restricting the indices i, j etc. to the values 1,2 only, show that
the two-dimensional stress transformation relations (1.15–1.17) can be obtained
from (1.22) using the two-dimensional direction cosines (1.20).
From (1.20), we have
l11 = cos θ ; l12 = sin θ ; l21 = −sin θ ; l22 = cos θ
and from (1.22),
σ′
ij = lipljqσpq .
Thus,
σ′
11 = l11l11σ11 + l12l11σ21 + l11l12σ12 + l12l12σ22
= σ11 cos2 θ + 2σ12 sin θ cos θ + σ22 sin2 θ
σ′
12 = l11l21σ11 + l12l21σ21 + l11l22σ12 + l12l22σ22
= −σ11 sin θ cos θ − σ21 sin2 θ + σ12 cos2 θ + σ22 sin θ cos θ
σ′
22 = l21l21σ11 + l22l21σ21 + l21l22σ12 + l22l22σ22
= σ11 sin2 θ − 2σ12 sin θ cos θ + σ22 cos2 θ
and these are identical to (1.15–1.17).
1.5. Use the index notation to develop concise expressions for the three
stress invariants I1, I2, I3 and the equivalent tensile stress σE.
In index notation,
I1 = σ11 + σ22 + σ33 = σii ,
see equation (1.3).
I2 = σ11σ22 + σ22σ33 + σ33σ11 − σ2
12 − σ2
23 − σ2
31 .
This is clearly a product of stresses. Two simple product terms in index notation are
I2
1 = σiiσjj = σ2
11 + σ2
22 + σ2
33 + 2σ11σ22 + 2σ22σ33 + 2σ33σ11
and
σijσij = σ2
11 + σ2
22 + σ2
33 + 2σ2
12 + 2σ2
23 + 2σ2
31 .
Comparing these expressions, we see that I2 can be written
I2 =
σiiσjj − σijσij
2
.
The third stress invariant
I3 = σ11σ22σ33 − σ11σ2
23 − σ22σ2
31 − σ33σ2
12 + 2σ12σ23σ31
is the determinant of the stress matrix σij . It can be written
I3 =
1
6
ǫijkǫpqrσipσjqσkr .
From the above expressions,
I2
1 − 3I2 = σiiσjj −
3(σiiσjj − σijσij)
2
=
(3σijσij − σiiσjj)
2
.
It follows that
σE =
s
(3σijσij − σiiσjj)
2
,
from equation (1.31).
1.6. Choosing a local co¨ordinate system x1, x2, x3 aligned with the three
principal axes, determine the tractions on the octahedral plane defined by
the unit vector
n =
(
1
√3
,
1
√3
,
1
√3
)T
which makes equal angles with all three principal axes, if the principal
stresses are σ1, σ2, σ3. Hence show that the magnitude of the resultant
shear stress on this plane is √2σE/3, where σE is given by equation (1.31).
The traction on the octahedral plane is given by equation (1.21) as
ti =
σ1 0 0
0 σ2 0
0 0 σ3
1/√3
1/√3
1/√3
=
σ1/√3
σ2/√3
σ3/√3
or
ti =
σi √3
.
To find the resultant shear stress, we first find the normal compont of the vector t as
t · n = tini =
σ1 + σ2 + σ3
3
= ¯σ ,
from equation (1.75).
The vector corresponding to this component is
(t · n)n = ¯σni
and the shear stress on the octahedral plane is a vector defined by
ti − ¯σni =
(
σ1 − ¯σ
√3
,
σ2 − ¯σ
√3
,
σ3 − ¯σ
√3
)T
.
To find the magnitude τoct of this resultant, we take the square root of the sum the
squares of the compontents, obtaining
τoct =
s
(σ1 − ¯σ)2 + (σ2 − ¯σ)2 + (σ3 − ¯σ)2
3
=
s
(2σ1 − σ2 − σ3)2 + (2σ2 − σ3 − σ1)2 + (2σ3 − σ1 − σ2)2
27
.
Expanding and simplifying, we obtain
τoct =
1
3
q
2 (σ2
1 + σ2
2 + σ2
3 − σ1σ2 − σ2σ3 − σ3σ1) .
Comparing this with equation (1.31), we see that
τoct =
√2σE
3
.
[Solved] TEST BANK FOR Elasticity By J. R. Barber (Solution Manual)
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