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1.
Calculate
the
maximum
shear
stress
for
two
cases:
(a)
4
xx
yy
zz
xy
MPa
MPa
1
σ
σσ
τ
===
=
(b) same as case a, except that
0
zz
=
.
Why is the maximum shear stress larger in case b?
The three principal stresses for case a) are
123
5,
4,
3
MPa
=
==
so the maximum
shear stress is
max
1
3
1
1
2
MPa
τσ
=−
=
. For case b) the principal stresses are
3,
0
,
so
the
maximum
shear
stress
is
max
1
3
1
2.5
2
MPa
=
. The reason for the smaller shear stress in Case a) is that
it is much closer to hydrostatic pressure where there are no shear stresses at all.
Problem 2.8
A state of hydrostatic stress is given by the following:
[]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
0
0
0
0
0
0
0
0
0
Show that on any surface the force (or stress vector) is always perpendicular to the
surface and that the magnitude of the stress vector is equal to
0
.
Solution:
Lets write stress vector
t
in terms of
[ ]
given on any plane
k
j
i
n
z
y
x
n
n
n
+
+
=
()
k
j
i
z
y
x
z
y
x
z
y
x
n
n
n
n
n
n
t
t
t
+
+
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
0
0
0
0
0
0
0
0
0
0
.
We can rewrite this in vector form as
n
t
0
=
.
That is stress vector on any given plane is
a vector of magnitude
0
and has the same direction with the normal vector of the plane
(perpendicular to the plane).
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 Fall '09
 ACHUTHAN

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