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CEE 415
1
Example 7.1:
We are to check deflections and bar spacing (crack width control)
for an interior span of a multispan floor beam.
The floor beams frame between
columns (18 x 18 in.), so the total span is 24 ft. and the clear span is 22.5 ft.
Assume the total distributed dead load is 1.4 k/ft. and the distributed live load is
1.0 k/ft., and 20% of the live load is sustained for periods of two years.
Use
f
c
'
=
4000 psi,
f
y
= 60 ksi.
Sketches of the span and sections at the end and midspan are given at the end of
this problem.
Case 1:
Assume the beam does not support partitions
From Table 9.5(a), read h(min) = ℓ/18.5 for an exterior span, which will govern
for this multispan floor beam.
Using ℓ = 24 ft., h(min) = 15.6 in.
For this
beam, h = 18 in. > h(min), so no deflection checks are required.
Case 2:
Assume the beam is supporting partitions or other nonstructural elements
that are
likely
to be damaged by large deflections.
For this case we must check
deflections as governed by row 3 of Table 9.5(b).
There we find that the
permitted total deflection after the attachment of partitions must be kept less
than
ℓ
/480.
For a span length of 24 ft., this allowable deflection is 0.60 in.
Assume that 85 percent of the dead load is acting when the partitions are
installed.
We will assume that the midspan deflection due to dead load for this interior
span can be estimated using the deflection formula for Case 1 in Table 92,
which is:
34
2.6 10
w
EI
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View Full DocumentCEE 415
2
For the live load, use the deflection coefficient from row 4 of Table 92:
34
4.8 10
w
EI
For
E
, we will use
E
c
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 Spring '11
 Wight

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