where

Fpx = the diaphragm design force F = the design force applied to level i Wi = the weight tributary to level i WPx = the weight tributary to the diaphragm at level x

Fpx need not exceed 0.4 SDSIw but shall not be less than 0.2 SDSIWpx

• Observe that the force Fpx computed from this equation is typically larger than the force Fx determined by

where

Cvx = vertical distribution factor V = total design lateral force or shear at the base of the structure (kip or kN)

Wi and Wx = the position of the total gravity load of the structure (W) located or assigned to level i or x hi and hx = the height (ft or m) from the base to level i or x k = an exponent related to the structure period as follows: For structures having a period of 0.5 sec or less, k = 1 For structures having a period of 2.5 sec or more, k = 2 For structures having a period between 0.5 and 2.5 seconds, k shall be 2 or shall be determined by linear interpolation between 1 and 2.

This is in recognition of the fact that higher-mode participation can result in larger forces at individual diaphragm levels than predicted by the preceding equation for Fpx

• Perform a three-dimensional lateral load analysis of the building by applying Fpx at the floor and roof levels. Include effects of torsion but ignore the effects if they reduce shear in the vertical elements of the lateral-load-resisting system.

• Determine the net shear in the vertical elements of the lateral-load-resisting system due to Fpx. This is equal to the difference in shears resisted by the vertical elements immediately above and below the level of the diaphragm being designed. Conceptually the shear forces may be considered as reactions to the inertial forces of the diaphragm.

• Determine a set of equivalent loads at the diaphragm level that is in equilibrium with the shear forces by using both force and moment equilibrium conditions. The equivalent loads may be derived as a combination of primary action due to Fpx and a secondary action due to torsional effects, as will be explained shortly in the numerical examples.

• Using the equivalent loads, determine shear and bending moment at critical sections of the diaphragm.

• Compute the shear per unit length to check the shear capacity of the diaphragm. Provide collectors, also referred to as drag beams, to carry the shear that is in excess of force transferred directly into the vertical elements. Use the special seismic load combinations with Qo QE for the design of collectors, if design is by strength design method. See AISC 7-02, Sect 9.5.2.6.3.1. Note that the published capacities of metal deck diaphragms are in working stress design (WSD) format.

• For reinforced concrete diaphragms, use the following equation to calculate the ultimate shear capacity of the diaphragm

Note that the strength reduction factor for shear, f, in diaphragm designs must not exceed the value used for the shear design of vertical elements of lateral-force-resisting systems.

• Check perimeter beams and their connections to columns for diaphragm chord forces.

• Extend chords into the diaphragm at reentrant corners, if any, to develop the forces calculated at the critical sections.

Example 1: Composite Diaphragm.

Given. A typical floor plan of a multistory steel building as shown in Fig. 7.49a1. The lateral-load-resisting system consists of a combination of rigid frames and braced frames. The floor and roof framing consist of a 3-in.-deep 20-gauge composite metal deck with a 3 |-in.-thick lightweight concrete topping.

By comparing the average interstory drift of the floor below the diaphragm to the deflection of the diaphragm itself, the engineer has determined that for analysis purposes, the composite floor and roof system may be considered as rigid (see AISC 7-02, Sect. 9.5.2.3.1).

The values of Fpx at various floor levels and roof have been determined from the ASCE 7-02 Eq. (9.5.2.6.4.4).

Using these forces, a three-dimensional, lateral-load analysis of the building has been performed to determine the shears in various vertical elements of the lateral-load-resisting vu = f(vc + v)

185 k

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