UFC 4-023-03
25 January 2005
4-2.2
Strength Reduction Factor Φ for Reinforced Concrete Tie Forces
The strength reduction factor Φ for properly anchored, embedded, or spliced
steel reinforcement in tension is 0.75 (based on Section 9.3.2.6 of ACI 318-02 for strut-
and-tie models).
4-2.3
Proportioning of Ties.
Reinforcement that is provided for other purposes, such as flexure or shear,
may be regarded as forming part or whole of the required ties.
4-2.4
Continuity and Anchorage of Ties.
Splices in longitudinal steel reinforcement used to provide the design tie
strength must be lapped, welded or mechanically joined with Type 1 or Type 2
mechanical splices, per ACI 318-02. Locate splices away from joints or regions of high
stress and should be staggered.
Use seismic hooks, as defined in Chapter 21 of ACI 318-02, and seismic
development lengths, as specified in Section 21.5.4 of ACI 318-02, to anchor ties to
other ties. At re-entrant corners or at substantial changes in construction, take care to
insure that the ties are adequately developed.
4-2.5
Internal Ties.
Distribute the internal ties are distributed at each floor and roof level in two
directions approximately at right angles. As shown in Figure 3-1, they must be straight
and made continuous from one edge of the floor or roof to the far edge of the floor or
roof, using lap splices, welds or mechanical splices. The internal ties must be anchored
to peripheral ties at each end (unless continuing as horizontal ties to columns or walls).
They may, in whole or in part, be spread evenly in the slabs or may be grouped at or in
beams, walls or other appropriate positions. Spacings must not be greater than 1.5 lr,
where lr is the greater of the distances between the centers of the columns, frames, or
walls supporting any two adjacent floor spaces in the direction of the tie under
consideration . In walls, they must be within 0.5 m (1.6 ft) of the top or bottom of the
floor slabs.
In SI units and in each direction, internal ties must have a required tie
strength (in kN/m width) equal to the greater of:
a)
(1.0D + 1.0L) lr Ft
(kN/m)
7.5
5
or
b)
1.0 Ft
(kN/m)
= Dead Load (kN/m2)
where:
D
4-2
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