Tie rods (continuous rods that run through the slab to the opposite column line) are typically used to counteract large shear forces associated with gravity loads on rigid frame structures. When using tie rods with large clear span rigid frames, consideration should be given to elongation of the tie rods and to the impact of these elongations on the frame analysis and design. In addition, significant amounts of sagging or bowing should be removed before tie rods are encased or covered, since the tie rod will tend to straighten when tensioned. Tie rods and hair pin bars should be placed as close to the top surface of the concrete slab as concrete cover requirements allow.
4.0 DESIGN EXAMPLES 4.1 Example: Base Plate for Concentric Axial Compressive Load (No concrete confinement) A W12×96 column bears on a 24-in. × 24-in. concrete pedestal. The minimum concrete compressive strength is fc′ = 3 ksi, and the base plate yield stress is Fy = 36 ksi. Determine the base plate plan dimensions and thickness for the given required strength, using the assumption that A2 = A1 (Case I). 1. The required strength due to axial loads. LRFD
ASD
Pu = 700 kips
Pa = 430 kips
2. Calculate the required base plate area. LRFD
A1( req ) = =
ASD
Pu φ0.85 f c′
700 kips (0.65)(0.85)(3 ksi)
= 422 in.2
Figure 3.5.4. Typical detail using hairpin bars.
A1( req ) = =
ΩPa 0.85 f c′
430 kips × 2.50 (0.85)(3 ksi)
= 422 in.2
Note: Throughout these examples a resistance factor for bearing on concrete of φ = 0.65 has been applied, per ACI 318-02. This resistance factor is more liberal than the resistance factor of φ = 0.60 presented in the 2005 AISC Specification. Although it was intended that the AISC provision would match the ACI provision, this deviation was overlooked. As both documents are consensus standards endorsed by the building code, and ACI 318-02 has been adopted by reference into the 2005 AISC Specification for Structural Steel Buildings, the authors consider a φ factor of 0.65 appropriate for use in design. However, ACI 318 is written using strength design only and does not publish an equivalent Ω factor. Therefore, an Ω = 2.50 has been used in the ASD calculations presented here to remain consistent with the value published in the AISC Specification. 3. Optimize the base plate dimensions, N and B. ∆=
0.95d − 0.8b f
2 0.95 (12.7 in.) − 0.8 (12.2 in.) = 2 = 1.15 in.
N ≈ A1( req ) + ∆ Figure. 3.5.5 Alternate hairpin detail.
≈ 422 in.2 + 1.15 in. ≈ 21.7 in.
DESIGN GUIDE 1, 2ND EDITION / BASE PLATE AND ANCHOR ROD DESIGN / 31