Local Equilibrium
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Transcript of Local Equilibrium
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similarly 0yF =
0xy yy
x y
+ =
Equations of equilibrium
(2-D stresses)
0
0
yxxx
yx yy
x y
x y
+ =
+ =
General 3-D Equations
of equilibrium (with
body forces)
0
0
0
yxxx zxx
xy yy zy
y
yzxz zz z
fx y z
fx y z
fx y z
+ + + =
+ + + =
+ + + =
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Stresses from axial loading of a bar through the centroid
of the cross sectional area
PP
A
x
constantxxP
A = =
all other stresses are zero
Equations of equilibrium are all satisfied (with zero body force)
0
0
0
yxxx zx
xy yy zy
yzxz zz
x y z
x y z
x y z
+ + =
+ + =
+ + =
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Stresses from pure bending of a beam
MMx
y
z
xx
yy
zI
=
all other stresses are zero
Equations of equilibrium are all satisfied (with zero body force)
0
0
0
yxxx zx
xy yy zy
yzxz zz
x y z
x y z
x y z
+ + =
+ + =
+ + =
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Stresses due to bending (engineering beam theory)
xy
z
( )xx
yy
x z
I
=
all other stresses are neglected
( ) ( )( )xz yy
V x Q z
I t z =
0
0
0
yxxx zx
xy yy zy
yzxz zz
x y z
x y z
x y z
+ + =
+ + =
+ + =
Are the equilibriumequations satisfied?
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Consider the equilibrium equations for a rectangular beam where t(z) = t
= constant. Since there are no loads in the x and y directions it is
reasonable to assume that yy = 0, xy = 0, yz=0, but the thirdequilibrium equation shows us that we cannot also set zz = 0 and stillsatisfy that equilibrium equation
( ) ( ) ( )
( ) ( )
0 0
0 0 0 0
0 0
yy yy
zz
yy
M x z V x Q zx I z I t
V x Q z
x I t z
+ + =
+ + =
+ + =
The first equilibrium equation is satisfied since( )
( )dM xV x
dx=
and ( ) / 2
0 0
h
z
dQ z dz tdz zt
dz dz
= =
t
zh/2
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From the third equilibrium equation
( )
2 2
22
3
/
8 2
6
4
zz
yy
dV dx h z
z I
q x hz
th
=
=
dx
( )V x ( ) dV
V x dxdx
+
( )q x
( )dV
q xdx
=
( ) ( ), / 2 /zz x z h q x t = =
Integrating, we find
( ) ( ) 2 3
36,
4 3zz q x h z zx z C
th = +
and using we find
3
12
hC=
( ) ( ) 2 3 3
3
6,
4 3 12zz
q x h z z hx z
th
= +
force/length
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( ) ( ) 2 3 3
3
6,
4 3 12zz
q x h z z hx z
th
= +
( )
( )
22
max max
max
6zz
xx max
q L h
L
=
O(1) very small
for long,
slenderbeams
0-q(x)/t
z
h/2
-h/2
zz0
