Civil Engineering Reference
In-Depth Information
3.8.3
Example: Design of the Support Beams for the Standard Pan Joist Floor along
a Typical N-S Interior Column Line (Building #1)
(1)
Data: › = 4000 psi (normal weight concrete, carbonate aggregate)
f
y
= 60,000 psi
Floors: LL = 60 psf
DL = 130 psf (assumed total for joists and beams + partitions + ceiling & misc.)
Required fire resistance rating = 1 hour (2 hours for Alternate (2)).
Preliminary member sizes
Columns interior = 18
18 in.
exterior = 16
16 in.
width of interior beams = 36 in. 2
depth - 2
19.5 = 39.0 in.
The most economical solution for a pan joist floor is making the depth of the supporting beams equal to
the depth of the joists. In other words, the soffits of the beams and joists should be on a common plane.
This reduces formwork costs sufficiently to override the savings in materials that may be accomplished
by using a deeper beam. See Chapter 9 for a discussion on design considerations for economical
formwork. The beams are often made about twice as wide as they are deep. Overall joist floor depth =
16 in. + 3.5 in. = 19.5 in. Check deflection control for the 19.5 in. beam depth. From Table 3-1:
h = 19.5 in. > /18.5 = (28.58
12)/18.5 = 18.5 in. O.K.
where ˜
n
(end span) = 30 - 0.67 - 0.75 = 28.58 ft (governs)
˜
n
(interior span) = 30 - 1.50 = 28.50 ft.
(2)
Determine factored shears and moments from the gravity loads using the approximate coefficients
(see Figs. 2-3, 2-4, and 2-7).
Check live load reduction. For interior beams:
K
LL
A
T
= 2(30
30) = 1800 sq ft > 400 sq ft
L = 60(0.25 + 15/
) = 60(0.604)* = 36.2 psf > 50% L
o
1800
1
1000
DL = 130
30
= 3.9 klf
1
1000
LL = 36.2
30
= 1.09 klf
w
u
= [1.20(130) + 1.6(36.2) = 214 psf]
30 ft = 6.4 klf
* For members supporting one floor only, maximum reduction = 0.5 (see Table 2-5).
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