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EMA 405, Spring 2004
Solutions for HW #2
1.1.
Introduction
Here we consider a simply supported beam with a linearly varying transverse body force.
From the notes, we know that the second derivative of moment is equivalent to the transverse
force per unit length, so we would expect transverse displacement (still two more integrals) to be
a fifthorder polynomial.
The problem therefore helps us see the behavior of beam elements in
an application where they do not reproduce the exact solution at all locations along the beam.
1.2
Preliminary Analysis
Although we can use tabulated information to determine the deflection of each end of the
beam, it is straightforward to use basic methods and integrating the beam equation.
Summing
the moments at the right side, gives the reaction force at the left side
()
∫
=
+
−
L
l
dx
x
q
L
F
0
0
Writing
with
q
>0 meaning an upward force (here,
q’
=2
×
10
()
x
q
x
q
′
=
4
N/m
2
), we have
()
0
6
2
V
L
q
F
l
−
=
′
=
where
V
is the shear that satisfies
x
q
dx
M
d
dx
dV
′
−
=
−
=
2
2
.
Integrating once and using
V
(0) gives the distribution of shear.
()
−
′
=
2
6
2
2
x
L
q
x
V
Integrating again and using the fact that neither end has an applied moment gives the distribution
of moment.
()
()
( )
x
L
x
L
x
x
q
x
M
2
3
2
2
.)
3333
(
6
−
−
→
−
′
=
Nm.
We can then integrate the beam equation,
M
dx
v
d
EI
=
2
2
, to get the transverse displacement.
The two constants of integration are determined by the boundary conditions associated with the
simply supported ends,
v
(0)=
v
(
L
)=0.
1/16
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()
+
−
′
=
=
+
−
′
=
360
7
12
24
therefore
and,
360
7
36
120
4
2
2
4
4
3
2
5
L
x
L
x
EI
q
x
dx
dv
x
L
x
L
x
EI
q
x
v
θ
At the two ends, angle of deflection evaluates to
(0)=5.732
×
10
3
rad and
(5 m)=6.551
×
10
3
rad.
For beam elements, the workequivalent load of a transverse body force amounts to a force
and a moment on each node.
For a single beam element, the workequivalent forces will be
counteracted by the reaction forces from the supports, so we only need the workequivalent
moments.
Using information from the nodes and from Cook, Chapter 2,
()()
()()
20
30
2
3
0
2
4
3
0
3
0
2
4
3
2
0
L
q
L
x
L
x
dx
q
x
q
x
dxN
M
L
q
L
x
L
x
x
dx
q
x
q
x
dxN
M
L
L
J
L
L
I
J
I
′
−
=
+
−
′
=
=
′
=
+
−
′
=
=
∫
∫
∫
∫
For the given magnitude of
q’
, the workequivalent moments are
M
I
= 8.333
×
10
4
Nm and
M
J
=
1.25
×
10
5
Nm.
1.3
Computation Description
The ANSYS computation can be run with a single BEAM3 element with the material
properties and real constants specified in the problem.
Vertical displacement is constrained to be
zero at both ends, representing the simple supports, and horizontal displacement is constrained at
one end so that rigidbody translation is avoided.
The linearly varying transverse body force is
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 Spring '04
 Witt

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