Info

Note: Aj = effective cross-sectional area within the joint equal to the joint depth times an effective width. The effective width is the smaller of

• Beam width + twice the smaller distance from beam edge to the column edge equal to 28 + 2 x 3 = 34 in.

Observe that joint shear is a function of effective cross-sectional area Aj of the joint and the square root of the concrete compressive strength .f only. If the net shear exceeds the nominal shear strength fVc (equal to 20-\[fC Aj ,15^f?Aj, or 12^f?Aj, depending upon the confinement provided at the joint), then the designer has no choice but to increase f of concrete and/or the size of columns.

A column face is considered confined by a beam if the beam width is equal to at least 75% of the column width. (No mention is made in ACI 318-02 for the required depth of beam.) When joints are confined on all four sides, transverse reinforcement within the joint required by Section 21.4.4 may be reduced by 50%. Hoop spacing is permitted to a maximum of 6 in. See Fig. 4.41.

4.2.9.9. Special Reinforced Concrete Shear Wall

Given. A shear wall that is part of a lateral load resisting system of a 10-story building located in a high seismic zone that has the following seismic characteristics.

St = maximum considered earthquake, 5% damped, spectral response acceleration at a period of 1 sec = 0.85g.

Figure 4.41. Beam-column joint; special moment frame. Transverse reinforcing in the joint is the same as for the frame column. A 50% reduction is allowed if the joint is confined on all the four faces. Maximum spacing of transverse reinyforcement = 6 in.

Se= horizontal elastic displacement at roof level corresponding to code-level seismic forces = 2.15"

PD =

1600 kips

PL =

300 kips

Vu =

VE= 1350 kips

Mu =

Me= 70,000 kip-ft

Se= horizontal elastic displacement at roof level corresponding to code-level seismic forces = 2.15"

Section modulus = 357216 in.3

16"

366"

Figure 4.42. Design example; partial shear wall elevation and plan.

Ss = maximum considered earthquake, 5% damped, spectral response acceleration at short periods = 1.80g.

Site class = D (as determined by project geotechnical engineer).

Seismic use group, SUG = 1

Seismic design category, SDC = D

Reliability/redundancy factor, p = 1.0

Seismic importance factor, IE = 1.0

Specified compressive strength of concrete f = 5000 psi

Specified yield strength of reinforcement fy = 60 ksi

Figure 4.42 shows a partial elevation and plan of the wall along with the ultimate axial forces and moments due to gravity and lateral loads. The dead load PD includes the self-weight of the wall. PL is the reduced live load. Also shown therein are the section properties of the wall and the horizontal displacement 8e equal to 2.15 inches at the roof level. The displacement is the lateral elastic deflection due to design basis code level earthquake loads. As will be seen presently, this displacement multiplied by the Cd factor is used to determine the requirements for detailing boundary elements.

The wall has been analyzed using the following assumptions:

• Effective section properties of the wall are based on a cracked section.

It should be noted that a computer analysis is almost always necessary to determine the building's response. This is because it is mandated in recent seismic codes to consider variables such as uncracked and cracked concrete section properties and some soil or foundation deformation beneath the structure's base.

Required.

• Calculation of ultimate design loads and moments using ASCE 7-02 load combinations.

• Preliminary sizing of the wall using a rule-of-thumb approach.

• Design of wall for combined axial load and bending moment.

• Determination of boundary element requirements using both stress index and displacement-based methods.

• Design of boundary elements.

• Schematics showing reinforcement layout.

• The design shall be in accordance with ACI 318-02.

Solution.

Load Combinations.

For compression check, E _ pQE + 0.2SdsD

p _ 1 and Ss _ 1.80 as given in the statement of the problem.

SMS _ FaSs, Fa _ 1.0 for site class D with S > 1.25 _ 1.0 Ss _ 1.80

Factored axial load Pu for compression check

Pu _ 1.2(1600) + 1.0(1 x 0 + 0.2 x 1.20 x 1600) + 300 + 0 _ 2604 kips

Factored axial load Pu for tension check

Pu _ 0.9 x 1600 - 1 x 0 - 0.2 x 1.2 x 1600 _ 1056 kips

The two sets of design forces and moments for the example are

4.2.9.9.1. Preliminary Size Determination. Since the length of the wall has been set at 30.5 ft, only the thickness t is adjusted to limit shear stress. The maximum shear stress allowed by Section 21.7.4.4. is , but experience has shown that limiting shear stress between and usually results in an economical wall design. For the example wall, using 4_ 4V5000 _ 283 psi as the limiting shear stress, the required wall thickness equals t _ 1,350,00/(30 x 12 x 283) _ 13.25 in.

However, because of boundary element considerations we will use 16 in. as the wall thickness.

A few thoughts of about preliminary sizing of shear walls. An estimate of wall length and thickness based on a reasonable shear stress using only the base shear may not be adequate for resisting design moments. The resulting area of vertical boundary reinforcement may be too high, quickly leading to unworkable details. Thus it is prudent to verify that the wall thickness determined on the basis of shear stress is also thick enough to allow room for placement of reinforcing steel and concrete.

4.2.9.9.2. Shear Design. Shear design using ACI 318-02 requirements is quite straight forward. Typically, the shear demand is taken directly from the lateral analysis without having to go through load combinations because, most often, horizontal shear resulting from gravity loads is negligible unless, of course, the building is highly irregular with built-in PA effects. For the example wall, Vu = VE = 1350 kips as obtained from a lateral analysis performed by using the ultimate earthquake loads.

Next the required horizontal reinforcement is calculated from the usable shear capacity equation,

where

Vn = nominal shear capacity f = strength reduction factor = 0.6 (see Section 9.3.4) ACV = gross area of wall equal to its length times the thickness. ac = coefficient defining the relative contribution of concrete strength to wall strength, typically taken as equal to 2.0 (Note: Section 21.7.4.1 permits ac = 3.0 for squat walls with hwllw < 1.5, 2.0 for hwllw > 2.0, and a linear variation between 3.0 and 2.0 for intermediate values of hwllw. The controlling ratio for the design of wall pier is based on the larger of overall dimensions of the wall or a segment of the wall. It is permitted to use ac = 2.0 in all cases. pn = ratio of area horizontal reinforcement to gross concreate area perpendicular to it. f = specified compressive strength of concrete, psi fy = specified yield strength of reinforcement, psi

For the example wall, the shear demand Vu = VE = 1350 kips

Assuming #5 @ 15 horizontal reinforcement, each face

Greener Homes for You

Greener Homes for You

Get All The Support And Guidance You Need To Be A Success At Living Green. This Book Is One Of The Most Valuable Resources In The World When It Comes To Great Tips on Buying, Designing and Building an Eco-friendly Home.

Get My Free Ebook


Post a comment