Maximum spacing of stirrups over a length equal to 2h = 2 X 26 = 52 in. from the face of the supports is the smallest of

• 8 X diameter of smallest longitudinal bar 8 X 1 = 8.0 in.

• 24 X diameter of stirrup bar = 24 X 0.375 = 9.0 in.

Observe that the allowable maximum spacing is the same as for the frame beams of SMRFs. However, hoops and crossties with seismic hooks are not required for frame beams of IMFs.

Provide 12 #3 stirrups at each end spaced at 5 in. on centers. Place the first stirrup 2 in. from the face of each column. For the remainder of the beam, the maximum spacing of stirrups is d/2 = 23.5/2 = 11.8 in. Use 11-in. spacing. Figure 4.36 provides a schematic reinforcement layout. Frame Column Example: Intermediate Reinforced Concrete Moment Frame

Given. A 30 in. x 30 in. frame column of an intermediate reinforced concrete moment frame. The column has been designed with 10 #11 longitudinal reinforcement to satisfy the ultimate axial load and moment combinations.

The ultimate design shear force due to earthquake loads E = 35 kips. The smallest axial load, Nu, corresponding to the shear force, = 1040 kips.

Clear height of the column = 11.84 ft.

Figure 4.36. Design example, frame beam; intermediate moment frame.

Figure 4.36. Design example, frame beam; intermediate moment frame.

Required. Seismic design and a schematic reinforcement detail for column C4 using the provisions given in Section 21.10 of ACI 318-99.

Check Limitations on Column Cross-Sectional Dimensions. No limitations are specified in ACI 318-99. Therefore, the given dimensions of 30 in. x 30 in. for the column are OK.

Design for Bending and Axial Loads. The statement of the problem acknowledges that the column has been designed for the governing load combinations with 10 # 11 vertical reinforcement. The reinforcement ratio (equal to 15.6/(30 x 30) x 100 = 1.74%) is within the allowable range of 1% and 8%. OK.

Design for Shear. Similar to that for beams, the shear design of columns in intermediate moment frames is based on providing a threshold of toughness. The design shear in columns may be determined by using either of the two options similar to those given earlier for beams. The first choice is to use the shear associated with development of nominal moment strengths of column at each end of the clear span. The second is to double the earthquake effect E when calculating ultimate design load combinations that include the earthquake effect E.

The ultimate shear force E due to earthquake = 35 kips, as given in the statement of the problem. Using the second option, the design shear force Vu is equal to 2 x E = 2 x 35 = 70 kips. The shear capacity of the column is:

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