24.02.2017 Views

Structural Concrete - Hassoun

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

836 Chapter 20 Seismic Design of Reinforced <strong>Concrete</strong> Structures<br />

where<br />

s 0 = longitudinal spacing of transverse reinforcement within the length l 0<br />

h x = maximum horizontal spacing of hoop or crosstie legs on all faces of the column<br />

The remaining member length should be reinforced with the spiral or hoop transverse reinforcement<br />

spaced as follows: { }<br />

6 × smallest longitudinal diameter bar<br />

s ≤<br />

(20.48)<br />

6in.<br />

Transverse reinforcement should be designed to resist the design shear force. Design shear force<br />

for flexural members of special moment frames can be determined using the following equation:<br />

V u = M pr t<br />

+ M prb<br />

(20.49)<br />

l c<br />

where index t is for top and index b is for bottom of the column and l c is the length of the column.<br />

20.5.1.2.4 Summary: Design of the Special Moment-Resisting Frame Members Subjected<br />

to Bending and Axial Force<br />

Step 1.<br />

Step 2.<br />

Step 3.<br />

Step 4.<br />

Step 5.<br />

Determine seismic design category, base shear, lateral seismic force, and seismic shear<br />

according to 20.2 and 20.3.<br />

Calculate the member forces and using the different load combinations determine the<br />

values of member forces that govern the design. Design the reinforcement.<br />

Check whether the frame member is a flexural member or whether the member is subjected<br />

to the bending and axial force, and check general requirements for the special<br />

moment-frame member according to Section 20.5.1.2.1.<br />

Check the special requirements for the longitudinal reinforcement according to Section<br />

20.5.1.2.2.<br />

Design the transverse reinforcement for confinement and shear resistant using Section<br />

20.5.1.2.3.<br />

Example 20.7<br />

Design the edge column on the second floor of a building from Example 20.6.<br />

Given:<br />

First Floor:<br />

Second Floor:<br />

P u = 1022 K<br />

M nc = 580 kip ⋅ ft<br />

P u = 935 K<br />

M nc = 528 kip ⋅ ft<br />

M nb = 723 kip ⋅ ft

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!