## Info

Using #5 hoops with two crossties in the longitudinal and one in the transverse direction, Ash - 4 x 0.31 - 1.24 in.2 in the longitudinal direction and, Ash - 3 x 0.31 - 0.93 in.2 in the transverse direction.

This is larger than 0.92 required by Eq. (21.4). Use #5 hoops and crossties at 4-in. vertical spacing as shown in Fig. 4.40.

Verify Confining Reinforcement for Shear. In the previous step, we determined transverse reinforcement required for confining column concrete and for providing lateral support to column vertical bars. In this step, we check if this reinforcement is adequate to resist shear forces resulting from the probable flexural strengths Mpr at each end of a column.

The positive probable flexural strength of the beam framing to the left face of column at third level is 848 kip-ft. The negative probable strength on the right face is 1055 kip-ft. Assuming that the flexural reinforcement for the beam below the level under consideration is the same, the design strength Ve is given by

Since the factored axial forces are greater than Agf'c /20, the shear strength Vc of concrete may be included in calculating the column shear capacity. For simplicity, we use Vc _ 2^fC'bd, although for members subjected to axial compression (as is the case for the example column), Eq. (11.4) permits higher shear values in concrete.

Therefore, #5 hoops and crossties provided at a spacing of 4 in. for confinement over a length of lo = 34 in. at column ends is also adequate for design shear.

The midlength of the column between the plastic hinging lengths must be provided with hoop reinforcement not exceeding a spacing of 6 times the diameter of the longitudinal bar = 6 x 1.56 = 9.36 in. or 6 in. In our case, the spacing of 6 in. governs. Therefore, provide #5 hoops and crossties at 6 in. for the midlength of the column. See Fig. 4.40 for a schematic layout of reinforcement.

4.2.9.8. Beam Column Joint Example: Special Reinforced Concrete Frame

To ensure that the beam-column joint of special moment-resisting frames have adequate shear strength, first, an analysis of the beam-column panel zone is performed to determine the shear forces generated in the joint. The next step is to check this against allowable shear stress.

The joint analysis is done in the major and the minor directions of the column. The procedure involves the following steps:

• Determination of panel zone design shear force.

• Determination of effective area of the joint.

• Verification of panel zone shear stress.

Determination of panel zone shear force. Consider the free body stress condition of a typical beam-column intersection showing the forces Pu, Vu, ML and MR (Fig. 4.40a). The force Vh, the horizontal panel zone shear force, is to be calculated.

The forces Pu and Vu are the axial force and shear force, respectively, from the column framing into the top of the joint. The moment ML and MR are the beam moments framing into the joint. The joint shear force Vh is calculated by resolving the moments into compression C and tension T forces. The location of C or T is determined by the direction of the moment using basic principles of ultimate strength design. Noting that Tl = CL and Tr = CR, Vuh = TL + TR - Vu.

The moments and the C and T forces from beams that frame into the joint in a direction that is not parallel to the major or minor directions of the column are resolved along the direction that is being investigated.

In the design of special moment-resisting concrete frames, the evaluation of the design shear force is based upon the moment capacities (with reinforcing steel overstrength factor a and no f factors) of the beams framing into the joint. The C and T forces are based upon these moment capacities. The column shear force Vu is calculated from the beam moment capacities as follows:

It should be noted that the points of inflection shown on Fig. 4.40a are taken as midway between actual lateral support points for the columns.

The effects of load reversals, as illustrated in cases 1 and 2 of Fig. 4.40b, are investigated and the design is based upon the maximum of joint shears obtained from the two cases.

Determine the effective area of joint. The joint area that resists the shear forces is assumed always to be rectangular. The dimensions of the rectangle correspond to the major and minor dimensions of the column below the joint, except that if the beam framing into the joint is very narrow, the width of the joint is limited to the depth of the joint plus the width of the beam. The area of the joint is assumed not to exceed the area of the

column below. It should be noted that if the beam frames into the joint eccentrically, the above assumptions may be unconservative. Given. A frame column joint.

Column: 34 in. x 34 in. Floor-to-floor height = 10.23 ft

Beam top reinforcement, six #9 top

Beam bottom reinforcement, eight #8 bottom

Beam moment ML = 1055 kip-ft

Beam moment MR = 186 kip-ft

Beam confined on two faces.

Confining reinforcement through the joint of a frame column is required no matter how low the calculated shear force is. This is to ensure ductile behavior of the joint and to allow it to maintain its load carrying capacity even after possible spalling of concrete outside of transverse reinforcement.

The design shear force is determined by subtracting the column shear force from the tensile force in the top beam reinforcement and the compressive force at the bottom of the beam on the opposite face of the column. The stress in the beam reinforcement is taken as 1.25 fy (Section 21.5.1.1).

Tl due to six #9 = As x 1.25 fy = 6 x 1.0 x 1.25 x 60 = 450 kips

CR due to six #8 = 6 x 0.79 x 1.25 x 60 = 356 kips.

Column horizontal shear force, Vh is obtained by assuming a point of contraflexure at midheight of column and by moment equilibrium condition at the frame joint.

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