Since Vu is greater than fVc /2 = 0.85 x 93/2 = 39.5 kips, and is less than f Vc = 0.85 x 93 = 79 kips, the required shear reinforcement is governed by the minimum specified in Section 11.5.5. Assuming #3 stirrups with four vertical legs, the required spacing s is:

50 bw

The maximum spacing of shear reinforcement, according to Section 11.5.4, is d/2 = 15.25/2 = 7.6 in. or 24 in. Thus, the governing spacing of stirrups is 7.6 in. According to Section, stirrups may be discontinued at sections where Vu < Vc/2. For the example beam, this occurs at 10.2 ft from the face of the column. Provide 18 #3, four-legged stirrups at 7-in. spacing at each end. Place first stirrup 2 in. from the face of support. See Fig. 4.33 for a schematic reinforcement layout. Frame Column Example: Ordinary Reinforced Concrete Moment Frame

Given. Values of axial loads, bending moments, and shear forces obtained from an analysis for column C3 are given in Table 4.5. To keep the calculations simple, the values of bending moments and shear forces due to dead and live loads and the axial load due to wind are assumed negligible.

Required. Design and a schematic reinforcement detail for column C3 using provisions of ACI 318-99.

Solution. Similar to frame beams of OMF, frame columns must satisfy the design provisions of ACI Chapters 1 through 18 and 22. Chapter 22 refers to structural plane concrete and has limited impact on the design.

Figure 4.33. Design example, frame beam; ordinary moment frame. Although by calculations no shear reinforcement is required in the midsection of the beam, it is good practice to provide #3 four-legged stirrups at 15 in. spacing.

Since there are no dimensional limitations specified for frame columns of OMF, the given column dimensions of 26 in. X 26 in. are OK.

From an interaction diagram not shown here, a 26 in. X 26 in. column with 12 #11 vertical bars has been found to be adequate for the ultimate load combinations given in Table 4.5. The reinforcement ratio of (12 X 1.56)/(26 X 26) X 100 = 2.7% is within the maximum and minimum limits of 1 and 8%. Thus the design of column C3 is OK.

Design for Shear. The shear design of a frame column of OMF is no different from that of a nonframe column. The shear strength of column is verified using ACI Eq. (11.4) for members subject to axial compression:

1350,000 2000 x 26 x 26

a/4000 x

26 x 22 1000

= 145 kips

Observe that Nu is the smallest axial force corresponding to the largest shear force Vu = ± 52 kips. (See Table 4.5.)

Since Vu = 52 kips < fVc/2 = 0.85 X 145/2 = 62 kips, column tie requirements must satisfy Section 7.10.5. Using #4 ties, the minimum vertical spacing of ties is given by the smallest of

• 16 times the diameter of vertical bars = 16 X 1.41 = 22.5 in.

• 48 times the diameter of tie bars = 48 X 0.5 = 24 in.

Use #4 ties at 22 in. Observe that at least #4 ties are required for vertical bars of sizes #11, 14, and 18, and for bundled vertical bars. See Fig. 4.34 for column reinforcement.

Figure 4.34. Design example, frame column; ordinary moment frame. Frame Beam Example: Intermediate Reinforced Concrete Moment Frame

Given. A beam 24 in. wide x 26 in. deep as shown in Fig. 4.35. The beam is part of the lateral resisting system that consists of an intermediate reinforced concrete moment frame.

Ultimate design values are as follows: Nominal moments (f = 1) are as follows: Support moment -Mul = 376 kip-ft At supports: -Mnl = 418 kip-ft -Mur = 188 kip-ft -Mnr = 209 kip-ft

Figure 4.35. Frame beam and column example; intermediate moment frame.

Midspan moment +Mu = 157 kip-ft 30

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