Civil Engineering Reference
In-Depth Information
VRwL
=- =
20 50225 55
.
-
. ()
=
.
k
2
1
11
wL
2
0225
2
.( )
2
11
MRL
=
-
=
20 525
.( )
-
=
450
ft-k
2
1
1
2
V
= 0 at 63.75 ft, which is the location of the maximum moment
0438 75
2
.( .
)
M
max
=
20 5637502255125
.( .
) .( )(
-
.
)
-
=
750 3
. t-k
2
Adjust the moment diagram for the smallest moments of inertia.
M
2
=
450
ft-k
left
1 4553 10
29710
.
×
6
M
2
right
=
(
450
)
=
220 5
. t-k
.
×
6
1 4553 10
29710
.
×
6
M
max
=
(
750 3
. )
=
367 7
.
ft-k
.
×
6
The adjusted moment diagram can be seen in Fig. 3.16. The adjusted moment dia-
gram for the wind loading shown in Fig. 3.15, used in the first part of the example, has
also been included for reference. The reader is encouraged to complete the hand calcu-
lations for that loading condition.
Calculation of the Bending Deflection (By Hand)
Using the moment-area method, the moment diagram can be broken down into simple
sections from which the moment area of the moment diagram can be obtained. The area
of the segment under the curve is equal to
Awa
= ()
3
1/ , with its centroid located at
a
/2
and
w
=uniform load. The area of the semiparabola to the right is equal to
A
= (
2/
)
bh
with its centroid located at (
5/
)
b
from grid line 4. The calculated areas are shown in
Fig. 3.16.
Summing moments about grid line 4:
T
AD
=
[,
15 012 43828
.( .
)
+
1939 58063
.( .
)
+
8544 480 63
.(
.
)
+
2851 74 17
(.
)
-
12270 101 45
,
,
.
+
260 4 112 5
.(
.)
+
5625 108 33
(
.
)]
=
EI
EI
t
t
For simplicity, find the deflection at the maximum moment location, which is close
enough without having to break the moment diagram into additional sections.
63 75
125
.
1 157 751 7
,
,
.
T
=
(,
2 270 101 45
,
.
)
=
f
romsimilartriangles
AC
-
EI
t
Summing moments about the maximum moment location gives:
T
C- ′
=
[
8544 419 375
. (.
)
+
2851 12 92
(
.
)
+
1939 519..
.(
375
)
+
5625 47 08
(
.
)
1
513 0
,
94 98
.
+
260 43188
.( .
]
=
EI
t
EI
t
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