This preview shows pages 1–3. Sign up to view the full content.
EMA 405, Spring 2004
Solutions for HW #6
1.1.
Introduction
In general, the deflection of a thin beam with an offcenter load is a combination of bending in
two directions and twist.
The relatively small dimension of the thickness of each web in this
problem makes it challenging for modeling via 3D finite elements, but this approach provides the
maximum amount of information possible if one can obtained converged results.
1.2.
Preliminary Analysis
The approach for an offcenter load acting on a straight beam is described in Section 8.14 of
Roark.
Since the applied load in this problem is directed parallel to a principal axis, we can
consider the behavior to be a linear combination of the same (vertical) force passing through the
flexural center and torque applied about the flexural center.
The magnitude the torque is the
magnitude of the force multiplied by the moment arm, which is 2.5 cm because the pressure
distribution is uniform over a region from 0 cm to 5 cm from a line that is directly above the
flexural center.
Thus,
T
=(20 kN)(0.025 m)=500 Nm.
The location of the crosssection centroid
and the area moment of inertia for bending about the
x
axis (vertical bending when the beam is
aligned with the
z
axis) can be obtained from the information provided in Roark’s Table A.1,
entry 4. for a Tee section with
t
=
t
w
=0.01 m and
b
=
d
=0.1 m.
m
0325
.
0
)
(
2
)
2
(
2
=
+
+
+
=
d
t
tb
d
t
td
bt
y
w
c
(vertical distance from the centroid to the top surface)
4
6
2
3
3
m
10
3542
.
2
)
(
)
(
3
)
(
3
−
×
=
−
+
−
−
−
+
=
c
w
x
y
t
d
A
t
b
d
t
d
b
I
Although this cantilevered beam has a distributed load, it is over a sufficiently small region that
we lose little accuracy by considering the vertical bending as if the entire load were concentrated
at the end of the beam.
The predicted vertical deflection for the beam is then
m
0142
.
0
)
m
10
Pa)(2.3542
10
2
(
3
m)
kN)(1
20
(
3
4
6

11
3
3
=
×
×
≅
=
x
a
EI
L
F
v
y
We can also expect the peak bending stress to occur at the bottom of the vertical web, since that
location is furthest from the centroid.
At
z
=0 (the wall end), we have
MPa
658.4
)
(
)
(
=
−
−
=
−
=
h
y
I
L
F
y
I
M
c
x
a
x
zz
y
σ
1/10
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentTo predict the angle of twist at the end of the beam, we will need the effective polar moment of
inertia,
K
, from Roark’s Table 10.1, entry 23 with
a
=
c
=0.1 m,
b
=
d
=
t
=
t
1
=0.01 m, and
r
=0.
b
d
b
D
cd
K
ab
K
D
K
K
K
4
15
.
0
0105
.
0
3
1
021
.
0
3
1
2
2
3
2
3
1
2
1
+
=
=
−
≅
−
≅
+
+
=
α
Some of the above relations have been simplified based on the parameters for this problem.
The
net
K
is 6.7179
×
10
8
m
4
.
With
G
=
E/
2(1+
ν
)=76.92 GPa, the expected angle of twist is
rad
0968
.
0
=
=
KG
Tl
θ
considering the torque to be applied at the end of the beam,
l
=1 m.
Finally, since the force is
directed vertically, there is no horizontal deflection of the centroid.
1.3.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '04
 Witt

Click to edit the document details