Civil Engineering Reference
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against lateral end deflections by the other columns on that level. For a compression
member which has
u
r
35
(ACI Equation 10-19)
P u
c A g
f
the maximum moment will occur at a point between the member ends and will exceed the
maximum end moment by more than 5%. For such a case, the maximum moment must be
determined by magnifying the end moments with the previously presented ACI Equations
10-15 and 10-16. For these equations
d must be determined for the load combination
under consideration. The effective length factor k will equal 1.0 unless an analysis shows
that a lesser value can be used.
For this case the design moment is to be determined by the following expression:
M c
ns M 2 ns
s M 2 s
The reader should understand that it is possible for sway frames to become unstable
under gravity loads alone. Whichever of the three methods for determining magnifying
moments described in this section is used, a gravity load instability check must be made,
as described in ACI Section 10.13.6. These checks follow.
1. If
s M s is determined with a second-order analysis, the ratio of the second-order
lateral deflection to the first-order lateral deflection (for 1.4 D
1.7 L plus the lat-
eral load applied to the structure) shall not be larger than 2.5.
2. If
s M s is computed with ACI Equation 10-17, the value of Q , the stability index,
computed with 1.4 D
1.7 L must not exceed 0.60. This value is equivalent to
s
2.5.
3. If
P c
corresponding to the factored dead and live loads must be positive and may not
exceed 2.5. (If
s M s is determined with ACI Equation 10-18 the
s obtained using
P u and
s is
2.5 it will be necessary to stiffen the frame.)
Example 11.4 illustrates the design of a slender column subject to sway.
EXAMPLE 11.4
Select reinforcing bars using the moment magnification method for the 18 in. 18 in. unbraced
column shown in Figure 11.8 if u 17.5 ft, k 1.3, f y 60 ksi, and
c
f
4 ksi. A first-order
analysis has resulted in the following axial loads and moments:
P u 600 k for loads not causing sway
P u 125 k due to wind
M 1 ns 50 ft-k
M 2 ns 130 ft-k
M u 120 ft-k due to wind
P u 16,000 k for all columns on floor
P c 48,000 k for all columns on floor
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