analysis and design of small reinforced concrete building for earthquake forces

8/13/2019 Analysis and Design of Small Reinforced Concrete Building for Earthquake Forces

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ANALYSIS AND DESIGN OF SMALL REINFORCED

CONCRETE BUILDINGS FOR EARTHQUAKE FORCES

Chapter 1

EARTHQUAKES AND EARTHQUAKE FORCES INBUILDINGS

A severe earthquake often leaves in its wake many collapsed buildings and much

structural damage. But observations of building performance during earthquakes indicate that

properly designed structures are capable of withstanding severe ground motions without

significant damage. Furthermore, with concrete construction this can be accomplished at very

little, if any, added expense. This is particularly true for small buildings.

Earthquakes consist of horizontal and vertical ground vibrations. The horizontal

motion is usually much greater than the vertical, although cases have been recorded where

vertical and horizontal motions have been of approximately the same magnitude. However,

because of the considerably greater stiffness of buildings in the vertical direction, the effect

of vertical ground motion has generally not been considered in design.

The most destructive force is caused by the horizontal components of ground motion.

When the ground beneath a structure moves suddenly to one side, the building tends to

remain in its original position because of its inertia. As a result, the building suffers a

distortion. The dynamic response of a structure to earthquake motions is characterized by a

complex series of vibrations. Analysis of these vibrations has been possible only since theavailability of modem electronic digital computers.

The intensity of earthquake ground motion at a site is usually measured in terms of

the magnitude of the ground acceleration. However, the potential destructiveness of

earthquake ground motions is dependent not only on the magnitude of the ground

acceleration but also on the duration of the large amplitude acceleration pulses. The

frequency characteristics of the ground motion, and the dynamic properties of a structure.

Analysis have shown that a structure subjected to a ground motion such as that

recorded at El Centro, Calif., during the May 18, 1940 Imperial Valley earthquake (see Fig. l)

would suffer considerable inelastic deformations. The design seismic forces specified incurrent are only about one-fourth as much as would be developed in a structurethat responded elastically to an El Centro type excitation. The principal reason for theserelatively low design forces is that many buildings designed to such low forces never

survived earthquakes of moderate intensity such as that recorded at EJ Centro, California in

1940, provided they had adequate capacity to dissipate energy through inelastic deformation

and/or damping. In view of this; code requires that structures be designed elastically to a

certain minimum amount of ductility, the assumption being that they yield when subjected to

earthquakes of moderate to strong intensity. A commonly used measure of ductility, i.e., the

capacity of a structure to undergo inelastic deformation without significant loss of strength or


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capacity, is the

ductility ratio or racier. This is defined as the ratio of the maximum displacement to the

displacement corresponding to first yield and is illustrated in Fig. 2.The seismic forces specified by codes are to be applied as static loads on a structure,

in a manner similar to the design for wind loads. The design lateral force is usually specified

in terms of a total base shear, V, which is distributed along the height of the structure. The

magnitude of the base shear is given as a function of the seismicity of the site, the type of

construction, and the period of vibration of the structure. Most also include otherfactors in determining the base shear, such as an importance factor, depending upon the use

of the building, and a foundation factor to account for soil conditions at the site.


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The analysis of a structure for member forces can be undertaken using well-known methods

of elastic frame analysis. Where deep flexural members are involved, account should be

taken of the contribution of the shear deformation to the total deformation. For multistory

frames and frame shear wall structures, the information given in Reference 6 will be found

useful. The most convenient way of analyzing multistory frames, however, would be to use

any of a number of currently available computer programs. Buildings should be designed to resist lateral forces in any direction because the ground

displacements due to an earthquake can occur in any direction. However, the horizontal

components of ground motion can be replaced by two mutually perpendicular components

acting parallel to the axes of a building. Thus, it is customary to investigate the strength of a

structure in two perpendicular directions. Most codes require that a structure be designed to

withstand the specified lateral seismic forces when assumed to act non concurrently in the

direction of each of the main axes of the structure.

To illustrate the effect of earthquakes on buildings, Fig. 3 shows a house set above the ground

on piers. The bot- toms of these piers must move with the ground and the tendency is to

transmit the motion to the building above. The sudden change from rest to motion causes

inertia forces opposite to the ground movement to act on the superstructure. The purpose of

earthquake-resistant design is to provide for the deformations associated with these inertia

forces.

Causing both walls to deflect equally. If a part of wall be is removed while AD is unchanged,

the center of rigidity will move toward AD; the center of mass will also move in the same

direction, but not as much. The unequal deflections of the two end walls, as shown in Fig.

5(b), give rise to higher stresses in Be and lower stresses in AD.

It is desirable that the height of a building be uniform. Whenever possible, setbacks should be

avoided since, as noted earlier, force concentrations result at and near such regions of

geometric and stiffness discontinuities, In plan, a closed shape, preferably square or nearly

square, is desirable because these shapes lend themselves most easily to symmetrical bracing.

However, unsymmetrically shaped, buildings may be built to withstand earthquakes provided

they are properly designed.

Adjacent buildings or parts of the same building dissimilar in mass or stiffness should be

sufficiently separated to prevent them from pounding one another should they vibrate out of

phase of each other during an earthquake. This can be accomplished with special joints,

sometimes called "crumple sections," that allow adequate free movement of adjacentbuildings. The width of the crumple joint or separation will vary with the height of the

building and will be greater for taller buildings. A gap about 4 to 6 times the maximum lateral

displacement under the code-specified loads is recommended. Stucco or some other easily

crushed material may be used. Fig. 6 shows an arrangement of these special joints for a group

of buildings or for a building with large wings. The separation should be carried down to the

top of the foundation, which may be continuous for an entire group of buildings.

Fire sometimes follows an earthquake, and its ravages can be greater than those of the quake.

The fire resistance of reinforced concrete makes its use doubly desirable in earthquake-

resistant construction.


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Fi3- 6. Separation of adjoining buildings

Fig. 5. Lateral distortion of buildings

The structure shown in Fig. 3, with a very stiff wall supported on piers or columns, tends to

develop large moments and shears, with correspondingly large ductility requirements, at the

column tops. This behavior is typical of the force concentrations that occur at regions of

major stiffness and/or mass and for geometric discontinuity, even under static loading

conditions. Such a configuration should be avoided whenever possible.

A continuous foundation wall is preferable to piers since the force tending to move the

building off its foundation is not concentrated at a few localized areas. The lateral forces,

computed as a specified proportion of the vertical load, are usually assumed to act asconcentrated forces applied at the different floor levels as indicated in Fig.4 (a).


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The lateral forces must be resisted by the walls or by the framing system. Partitions and

outside walls parallel to the direction of the ground motion (crosswalls) possess much greater

strength and stiffness against distortion than those normal to this direction. Moreover,

crosswalls are generally stiffer than frames extending in the same direction and will resist

practically all the lateral forces.

Fig. 4(b) shows the typical distribution and relative magnitude of stresses in a crosswall due

to lateral forces. If the wail thickness is uniform, both shearing and bending stresses will

increase in the lower stories.

All parts of a building should be firmly tied together and so braced that the building will tend

to move as a unit. Floors and crosswalls should be continuous throughout the building and

openings should, if possible, be away from outside comers. Symmetrical arrangement of

crosswalls or approximate coincidence of the center of mass and the center of rigidity (the

point where a lateral force must be placed to produce equal deflections of the crosswalls) is

desirable; otherwise, stresses due to torsion must be taken into consideration.

The Uniform Building Code requires vertical resisting elements that depend ondiaphragm action for the distribution of shear at any level to be designed for a minimum

horizontal torsional moment corresponding to an eccentricity of the horizontal story shear

equal to 5% of the maximum building dimension at that level. Tills provision for accidental

eccentricity applies to dynamically symmetrical buildings. Any known eccentricity of the

mass center from the center of rigidity will have to be provided for in design.

Fig. 5(a) shows exaggerated lateral deflections of a building in which the center of mass, or

point of application of the total lateral force, coincides with the center of rigidity,


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Chapter 2

DESIGN LOADS AND BUILDING CODE REQUIREMENTS

Permissible stresses for building materials, live loads for various classifications ofoccupancy, forces due to wind, and, where applicable, horizontal earthquake forces to be

considered in the design are established in building codes to ensure the safety of buildings.

There is little difference between the basic requirements of the various codes, and a building

properly designed in accordance with anyone of them should safely resist earthquakes if

provision is made for horizontal earthquake forces and if proper attention is given to details,

considering probable inelastic action.

Given below are design data on lateral loads, load factors, and capacity reduction factors as

contained in the latest edition of the Uniform Building Code (UBC-1997), theearthquake-resistant design provisions of which are based essentially on the ASCE/SEI 7-05

Recommendations. UBC-1997 design provisions intended to ensure adequate ductilityin reinforced concrete members and their connections are also discussed briefly. Theseprovisions are based on those in Appendix A of ACI 318-77. 2.1 Design Base Shear

The minimum lateral force for which a structure has to be designed is given in terms of the

design base shear, V, as follows:

(1)

Where

Cv is the seismic coefficient as set forth in Table 208-8

I is an occupancy importance factor-with assigned values equal to or greater than 1.O-

reflecting the greater conservatism in design required for facilities the continued

operation of which is essential after a major earthquake and for public assembly areas.

Given in Table 208-1

R is numerical coefficient representative of the inherent overstreghth and global ductility

capacity of lateral-force-resisting systems, as set forth in Table 208 -11 or Table 208-13

T is the elastic fundamental period of vibration of the structure in the direction under

consideration, to be used in the above equation, may be established by analysis or

experiment, or, in the absence of such data, may be determined from the formula


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(Eq.208-8)

Where

= height (in meter) of the structure considered above the baset = 0.0731 for reinforced concrete moment-resisting frames and eccentrically braced framesAlternatively, the value of Ct for structures with concrete or masonry shear walls may

be taken as 0.0743/ Ac

The value of Ac shall be determined from the following equation

The value of De/hnused in Equation (208-9) shall exceed 0.9

The total design base shear need not exceed the following:

The total design base shear shall not be less than the following:

In addition to seismic Zone 4, the total design base shear also shall not be less than the

following:


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Fig. 7. Seismic risk map of the Philippines -ASEP (Reference 1).


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W

I

s

t

h

e

t

o

ta

l

d

W is dead load including partition loading, except that in storage and warehouse

occupancies, Wshall include both the total-dead load plus 25% of the floor live load.


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2.2 Distribution of Base Shear

The total lateral force or base shear, V, is to be divided into two parts: a concentrated load,F,

applied at the top of the structure, and the balance, (V - F), to be distributed over the entire

height of the building, generally as concentrated loads at the floor levels. The latter is to bedistributed in a "triangular" manner, increasing from' zero at the base to a maximum at the

top.

The top load, is given by= (2)The magnitude of the distributed force, making up the balance of the total

lateral force, i, e., (V-, is given by= (3)


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Where,= that portion the total weight, W., which is located at or assigned to level xor I, respectively

,

= the height (in feet) above the base to levelxor i, respectively

Level n is the uppermost level in the main portion of the structure

At each level x, the force is to be applied over the area of the building inaccordance with the mass distribution at the level. At the top floor, the horizontal

load will generally consist of the sum of (where x=). A typical distribution offorces in a multistory structure having a uniformly distributed mass along its height

is illustrated in Fig. 9

Fig. 9. Typical distribution of code-specified static lateral forces and story

shears in a

building with

uniform mass

distribution

(from

Reference 13).

2.3 Lateral forces

on Parts or

Portions of

Buildings

Although failure of ornamentations, masonry or marble veneers and parapets, or

similar appendages in buildings seldom affects structural integrity, it

represents a serious menace to the safety of the occupants and of


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passersby. Because of this, particular care should be taken to ensure that

portions of buildings, such as parapets, and electrical and mechanical:

fixtures and appliances are adequately anchored with reinforcement, bolts,

or other devices.

UBC76 specifies design lateral forces for parts or portions of buildings and otherstructures in terms of a coefficient, and the weight of the partconsidered, Thus, the design lateral.force, , is to be determined using= (4) is the portion of the structure and values of the coefficient,, are

as given in Table 2. Where in Table 2 is 1.0 or more, the value of J and S neednot exceed 1.0. Note that the design lateral force on cantilevered parapets and walls is

to be applied normal to the plane of such members, while for other parts or

appendages the lateral force may be applied from any direction. In all cases, the intent

is to design the part and its anchorage for the most unfavorable loading condition.

Table 2. Horizontal Force Factor "" for Elements of Structures (from Reference 1)Part or Portion of Buildings Direction of

Force

Values ofExterior b88ring end nonbearing walls, interior

bearing walls and partitions. interior nonbearing

wells and partitions,M850nrv or concrete fences

Normal to flat

surface

Cantilever parapet Normal to flatsurface Exterior and interior ornamentations and

appendages

Any Direction When connected to, part of, or housed within a

building:

(1) Towers, Tanks, towers and tanks plus

contents. chimneys. smokestacks and penthouses.

(2) Storage reeks with the upper storage level at

more than B ft. in height plus contents.

(3) Equipment or machinery not required for life

safety systems or for continued operations of

essential facilities.

(4) Equipment or machinery required for life

safety systems or for continued operation of

essential facilities

Any Direction

When resting on the ground, tanks plus effective

mass of its contents.

Any Direction Suspended ceiling framing systems (Applies to

Seismic Zones Nos. 2, 3, and 4 only.

Any Direction Floor and roofs acting as diaphragms Any Direction

Connection for exterior panels or the elementscomplying with Section 3212 (j) 3C

Any Direction


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Connection for prefabricated structural elements

other than walls, with force applied at center of

gravity of assembly

Any Direction

Footnotes for Table 2:

1 See also Section 2309 (b) for minimum load on deflection criteria for interior

partitions.

2 when located in the upper portion of any building where the /D ratio is the five-to-one or greater the value shall be increased by 20 percent.

3 for storage racks shall be the weight of the racks plus contents. The value of for over two storage support levels in height shall be 0.18 for the levels below the top

two levels. In lieu of the tabulated values steel storage racks may be designed inaccordance with UBC Standard No.27-11.

Where a number of storage rack units are interconnected so that there are a minimum

of four vertical elements in each direction on each column line designed to resist

horizontal forces the design coefficient may be as for a building with Kvalues from

Table No. 23-1, CS= 0.20 for use in the formula V = ZIKCSWand Wequal to the

total deed load plus 50 percent of the rack rated capacity. Where the design and rack

configuration are in accordance with this paragraph the design provisions in in UBC

Standard No.27-11 do not apply.

4 For flexible and flexible mounted equipment and machinery, the appropriate valuesof shall be determined with consideration given to born me dynamic properties oft.he equipment and machinery and to the building or structure in which it is placed but

shall not be less than the listed values. The design of the equipment and machinery

and their anchorage is an integral part of the design and specification of such

equipment and machinery.

5 For essential facilities and life safety systems, me design and detailing of equipment

which must remain in place and be functional following a major earthquake shall

consider drifts in accordance with Section 2312 (k). The product of IS need not

exceed 1.5.

6 Ceiling weight shall include all light fixtures and other equipment which are

laterally supported by the ceiling. For purposes of determining the lateral force, a

ceiling weight of not less than 4 pounds per square foot shall be used.

7 Floor and roots acting as diaphragms shall be designed for a minimum force

resulting from 8 of 0.12 applied to unless a greater force result from thedistribution of lateral forces in accordance with Section 2312 (c).

8 The

shall include 25 percent of the floor five load in storage and warehouse

occupancies.


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2.4 Minimum Horizontal Torsional Moments

As mentioned in Chapter 1, UBC-1997 requires that where the vertical-load-resisting

elements are connected by a floor system that is rigid in its own plane, such as a

monolithically cast reinforced concrete slab, such elements should be designed for the

direct shears plus torsional shears corresponding to a minimum eccentricity of thestory lateral leads not less than 0.05 of the maximum building dimension at that level.

Shears due to such an "accidental torsion," which are directed opposite to the direct

shears, are to be neglected.

2.5 Analysis for Member Forces

Once the lateral loads on the structure have been determined, an analysis for the

corresponding member forces can be made. For small relatively low buildings, an

approximate analysis using the portal method may be carried out. For larger frame so

structures, the use of a frame analysis computer program may be more convenient and

economical.

2.6 load Factors and Loading Combinations to he Used as Bases for Strength

Design

Codes generally require that the strength of a structure and its components be equal to

or greater than the forces corresponding to any of a number of loading combinations

that may reasonably be expected during the life of the structure. The Uniform

Building Code, 1976 edition, calls for de- signs to satisfy the following ultimate load

conditions:

U= } (5)Except that a value of 2.0E shall be used instead of I.40E in Equation 5(d) in

calculating shear in shear walls of buildings without a 100 percent moment-resisting

space frame and located in seismic Zones 2 and 3. Equations 5(c) and 5(d) apply only

to buildings located in Zones 2 and 3. The corresponding equations in ACI 3J 8-77*

are

U = { (6)In Equations 5 and 6.

U = required strength [0 resist the design 10aGS or their related internal moments

and forces

D = dead loads or their related internal forces

L = live loads or their related internal forces

W = wind load or its related internal forces

E = load effects of earthquake or their related internal forces


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Note thatD, L, W, andEare all service loads.

In the process of proportion structural members by the ultimate strength method,

capacity reduction factors, , design to account for variations in material strength, for

the nature of the associated failure mechanism, as well as for the importance of a

member to the structure as a whole, are used either in calculating the effects of thedesign forces or identical for both UBC and ACI 318-77, i, e 0.90 for flexure, 0.85

for shear (diagonal tension), and 0_70 and 0.75 for axial compression with or without

bending-the lower value being for tied members and the higher for spiral reinforced

members.

The word ultimate is no longer used in ACI 318-77. Instead, the word "strength" isused when referring to the ultimate strength design method, as distinguished from the

working stress design method. The adjective "design" is used as in "design moments,to indicate ultimate or load factored moments. Working stress design is refer red toas the ultimate design method.


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Chapter 3

DISTRIBUTION OF LATERAL LOADS

As noted in Sections 1 and 2, the code-specified Lateral loads are applied at the

different floor levels where rigid floor systems acting as diaphragms distribute the

loads to the cross walls and frames. In low buildings where walls possess

considerably greater stiffness than do frames extending in the same direction> the

walls will resist practically all the lateral forces.

If the center of mass coincides with the center of rigidity (the point where' a lateral

force must be applied in order to produce equal deflections of the resisting units), the

supporting walls will deflect equally. Consequently, horizontal forces will be

distributed to the cross walls in inverse pro- - portion to their capacity to deflect, or

flexibility Thus, a very flexible wall will resist only a small portion of the seismic

force, while a stiffer wail will resist a larger portion. In terms of stiffness (i.e. the

lateral force required to produce a unit deflection), which is the reciprocal offlexibility, the lateral forces will be distributed in direct proportion to the relative

stiffness of the resisting elements.

31 Deflection

Fig. 10 indicates two idealized conditions of end support for vertical concrete

elements displaced by lateral loads. Be- cause most concrete piers or cross walls act

as short , deep beams, contributions (0 the displacement due to both flexure and shear

must be considered as indicated in the following equations:For both ends fixed, i.e.; restrained against rotation,

= += + (7)For one end fixed, the other hinged,

= += + (8)Where

= total displacement= displacement due to bending


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Fig. 10. = displacement due to shearP = lateral force on pier

h = height of pier

A = cross-sectional area of pier

I = cross-sectional moment of intertie of pier in direction of bending = modulus of elasticity of concreteG = shear modulus

3.2 Deflection of Wall Consisting of Several Piers

A wall consisting of several piers connected "long their tops is shown schematically

in Fig. 11(b). The lateral deflection at the top of such a wall. , due to a horizontalload,P, call be obtained if the deflections, of the component piers when subjectedseparately to the load, P, are known or calculated first. Identical end-restrain r

conditions are assumed for the piers in both cases. Thus, if (he deflection of a

component pier, i, under the load,P, is denoted by then as stiffness (i.e., the (forcerequired to produce a unit deflection) is given by

= (9)The deflection of a wall consisting of n connected piers each having individual lateralstiffnesses =P/(I = 1, 2 n) can then be obtained from the expression= = (a)or, using Equal (9), (10)

=

=

(b)


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Fig. 11 deflection of piers.

3.3 Relative Stiffness of Connected Walls

Consider the three parallel walls, A, B, and C, shown in the plan in Fig. 12. The walls are

assumed to be tied together at the top by a horizontal floor slab which is stiff enough in its

plane to be considered rigid. Since the centers of rigidity and mass coincide, the walls will

deflect equally and the force, P, will be distributed to the walls in proportion to their relative

stiffness. For any wall, i, the relative stiffness is given by

Relative Stiffness of wall i = =

kii = 1or

=

=

where ki, is the absolute lateral stiffness of wall iand i(1=1,2,3) are the deflections

of the individual walls when subjected separately to the same horizontal load, assuming the

same end restraint conditions prevail when the walls are connected.

Thus, each wall, i, will resist a portion of the total applied load, P, equal to

Vi= ( ) (12)As indicated by Equation (12), the distribution of the applied lateral force among

individual vertical elements depends only on the relative, and not the absolute stiffness of the

connected elements; nor does it depend on the magnitude of the load,P.

The examples in this booklet incorporate concrete with an ultimate strength of fc=

4,000 psi, which has a modulus of elasticity, Ecof about 3.6 x 106psi and a shear modulus,G, of about 1.44 X 106psi.


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Substituting these values of Ecand G in Equations (7) and (8) and using a load P=

103kips, yields the following expressions for the lateral deflection:

For both ends fixed:

(13)For one end fixed, the other end hinged: (14)

3.4 Centers of Mass and Rigidity at Distinct Points

The plan in Fig. 13, consisting of three parallel walls and two transverse walls, has its center

of rigidity and center of mass at two different points. With the load,P, applied at the center

of mass, a torsional moment,Mt, is induced which, in addition to the horizontal load must be

resisted by the walls. This torsional moment is equal to

Mt= P(e) where;

P = horizontal load applied at the center of mass

e = normal distance between the line of force, P, and the center of rigidity

As noted in Section 2.4, codes generally require consideration of a minimum eccentricity of

the specified lateral load with respect to the center of rigidity of each floor.

The distance, xm, from the centerline of wall A to the center of mass is found by taking

statical moments about the centerline of the endwall, using the respective weights of the

walls. The distance, xr from the centerline of wall A to the center of rigidity is found by

taking statical moments about the centerline of the endwall, using the relative lateral

stiffnesses of the component walls in their respective planes as weights.

Thus, if ky is the in-plane (absolute) lateral stiffness (the relative lateral stiffness will serve

the same purpose) of a particular wall in they-direction, then the x-coordinate of the center ofrigidity,xrwith respect to an arbitrary plane of reference such as the axis of wall A in Fig. 13.

is given by

(15a)the summation being taken for all walls parallel to the y- direction. It is assumed here that

walls parallel to thex-direction do not contribute to the lateral resistance in they-direction.

Similarly, they-coordinate of the center of rigidity is given by


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(15b)

where kxis the in-plane lateral stiffness of a particular wall parallel to the x-direction. In the

above equations, x is the distance from the axis of a wall parallel to the y-direction to the

reference plane, whileyis the corresponding distance for walls parallel to thex-direction.

The horizontal shear, Vy, resisted by a particular wall with axis parallel to the y-

direction, due to an applied horizontal load, Py (see Fig. 13) which produces a torsional

moment,Mt= Pye, may be obtained from the expression*

( ) (16a)Similarly, for an applied horizontal force,Px, in thex-direction, (16b)

where:

Jr = rotational stiffness of all walls in a story (corresponding to polar moment of

inertia)

= 2: (kxyJ + kyX2) all walls in story EQUATION (17) = perpendicular distance from the center of rigidity to the axis of a particularwall

kx = in-plane lateral stiffness of a particular wall with axis in the x-direction. For

such a wall it is assumed that ky= 0

ky = lateral stiffness of wall along y-direction. For walls parallel to the y-

axis, kx= 0


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*FootnoteNote that a horizontal force,P, having an eccentricity, e,with respect to the

center of rigidity, can always be replaced by a statically equivalent system consisting of a

force of the same magnitude passing through the center of rigidity and a torsional moment of

magnitude,Pe.

Note that in Equations (16a) and (l6b) the plus sign is always used. This is becausethe load, P, is a reversible load and codes generally require that the effect of torsional

moments be considered only when they tend to increase the shears due to the horizontal load.**

In determining the center of rigidity for a particular story or the distribution of the

total story shear to the different walls in the story, using Equations (15), (16), and (17); only

the relative stiffnesses of the component elements need be considered.

3.5 Example Problem 1: Distribution of Lateral Loads

The plan and elevations of the first story of a three-story (plus basement) building are

shown in Fig. 14. The building, which is of cast-in-place reinforced concrete construction, is

assumed to have been designed for the static loads shown in the figure, with footings

adequate for the static loads plus the increased pressure due to earthquake. For an earthquake

occurring in an east-west direction, determine the design lateral forces at each floor level and

the distribution of shears at the first-story level.

The precision of the following computations is that obtainable by means of a slide

rule. All examples are based on the strength design method and the Uniform Building Code.

**footnoteSee Section 3,5.4 and its footnote for a case when a negative torsional shearhas to be disregarded.

3.5.1 DESIGN BASE SHEAR,

1. Assuming the structure to be located in Zone 4 (Lanao del Norte) : Z= 0.4, with soil type

SD and NV for source type B is 1.0 , therefore Cv= 0.64

2. The structure is Office Warehouse - Miscellaneous Structures and as such is not considered

an essential facility:I= 1.0

3. Arrangement of the resisting elements conforms to the "building frame system ofconstruction. Therefore, (from Table 208-11a):R= 5.5

4. The fundamental period of vibration, T, is given by:

for hn 13.50 m

D = 11 m.


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So, = 11 / 13.50 = 0.81 , less than 0.9

Computing Ae,

Ae = 2430mm (1200mm) + 1010mm (1200mm) + 850mm (1200mm) +1050mm

(1200mm) + 2475mm (3000mm) + 2400mm (3000mm) + 875mm (3000mm)

= 23658000 sq. mm (1/1000000)

= 23.658 sq.m.

Computing Ac,

Ac = 23.658 sq.m. [ 0.2 + 0.812] = 20.44 sq.m.

Computing Ct,

Ct = 0.0743 / 20.44 = 0.02

Therefore,

T = 0.02 ( 13.5)0.75 = 0.12 sec < 0.7 sec

5. Using the loads indicated in Fig. 14, calculate W

Roof Level:

Liveload

Tributary Area 165 sq.m.

Uniform Live Load 1 Kpa

Factored Live Load = 1 Kpa (88%) 0.88 Kpa

r 0.08

R = 0.08(165 -15) 12 %

Deadload

Steel Deck 0.14 KPaReinforced Concrete = 23.6 KN/m3 x .125 m (thick0 2.95 Kpa

Waterproofing 0.03 KPa

parapet 2.98 KPa

1st ,2nd,

3rd Level

Liveload

Tributary Area 165 sq.m.

Uniform Live Load

Storage 12 Kpa12 KPa


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Factored Live Load,L = 12 * 0.25 + 4.57 ( 1/165)= 7.27 Kpa

Factored Live Load,(KN) = 7.27 KPa ( 165) 1,199.43 KN

Deadload

Steel Deck 0.14 KPa

Reinforced Concrete = 23.6 KN/m3 x .125 m (thick0 2.95 Kpa

Partition Load 0.5 KPa

3.59 Kpa

Roof Level = (0.14 + 2.95+0.03)( 165) + ( 2.98 )(11*2+14.5*2) +

(0.88)(165)= 811.98 KN

1st ,2nd, 3rd Level = 1,199.43 (0.25) + (3.59)(165) 892.21 KN

W = 811.98 + 892.21 *3 = 3,488.60 KN

6. Base shear, V 3,488.60 = 3,507.17 KNChecking; 0.11 Ca I W < V < 2.5 Ca I W/R,

0.11 (0.44)(1.0) 3,488.60 = 168.85 < 3,507.17 , true

2.5 (0.44) (1.0) 3,488.60/ 5.5 = 697.72 KN < 3,507.17 KN

Therefore use limiting value of V = 697.72 KN


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Figure 14.


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269.69 KN

219.51 KN

142.68 KN

65.85 KN


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3.5.2 CONCENTRATED LATERAL FORCES AT FLOOR LEVELS

1. 2. 3. , since 4. Forces at each storey:

Froof =697.72 x 811.98 x 13.5 / 28,359.79 269.69 KN

F3rd =697.72 x 892.21 x 10 / 28,359.79 219.51 KN

F2nd =697.72 x 892.21 x 6.5 / 28,359.79 142.68 KN

F1st =697.72 x 892.21 x 3 / 28,359.79 65.85 KN

3.5.3 LOCATION OF CENTER OF RIGIDITY AT FIRST STORY

1. Total load at second floor,Px: The total load at any level is equal to live load appliedat that level plus the shear from all loads applied above that level. Therefore, the total

horizontal shear at the first-story level is equal to:

Px= 269.69 + 199.77 + 129.85 = 599.30 KN

2. Center of mass: As noted in Fig. 14, the center of mass is assumed to be located at thecenter of the building.

3. Center of rigidity: To determine the center of rigidity, the relative stiffness of theresisting elements parallel to the lateral force (i.e., the north and south walls) mustfirst be calculated. The piers in both north and south walls will be assumed fixed at

both ends. The deflections and stiffness of the piers in the two walls are computed by

using Equation (l3), with the associated calculations given in Table 3.


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Table 3. Deflections and Stiffness of Piers thru A to G

Pierwidth,mm height,mm area, mm

2 I, mm

4

f =

h3/12EI

v=

1.2h/GA T 1/T

1-1 A 2430 1200 607500 2.98936E+11 0.0192 0.0002 0.0194 51.53444

B 1010 1200 252500 21464604167 0.2670 0.0006 0.2675 3.737928

C 850 1200 212500 12794270833 0.4479 0.0007 0.4485 2.229422

D 1050 1200 262500 24117187500 0.2376 0.0005 0.2381 4.199139

4-4 E 2475 3000 618750 3.15853E+11 0.2835 0.0006 0.2840 3.520533

F 2400 3000 600000 2.88E+11 0.3109 0.0006 0.3115 3.210476

G 875 3000 218750 13956705729 6.4152 0.0016 6.4168 0.155841

The deflection underP= 103KN and stiffness of the north and south walls are computed

by using Equation (l0b), as follows:

1-1 (North Wall):

Deflection, = 0.0162 mmStiffness, 61.70 KN/mm4-4 (South Wall):

Deflection, = 0.145 mmStiffness, = 6.89 KN/mm

Byusing Equation (11), the relative stiffnesses of the north and south walls are found tobe 0.90 and 0.10, respectively.

The distance , from the centerline of the south wall to the center of rigidity can now befound by taking statical moments about the centerline of the south wall, using the relative

stiffnesses as weights (see Equation (15):

= = 0.10(0) +0.90 (14.75) = 13. 269 mWith the building relatively symmetric in the east-west direction, the center of rigidity will lie

on a line centered between C-C (east) and A-A (west) walls at a distance of 13.394 m (13.269

m + 125mm.) from the exterior face of the 4-4(south wall).

3.5.4 DISTRIBUTlON OF HORIZONTAL SHEARS AMONG EXTERIOR WALLS

OF 1stSTORY

(A)Ground motion in east-west direction (parallel to x- axis)The total horizontal shear carried by each of the exterior walls of the first story due to

the eccentric horizontal load, Px= 599.30 KN (see Fig. 14), will be determined by applying

Equation (16). Thus, for the north and south walls (parallel to the x-axis in Fig. 14)

(16b)


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In the above expression,

Px = 599.30 KN

Pxe = 599.30 x 5,895 = 3,532,881.96 KN-mm

To calculate the total rotational stiffness of the storey about the center of rigidity, Jr the

stiffness of the east and west walls (same for both) will be required. With

width, d height, h area,A

I

due toP= 103KN

Stiffness, ky(east wall = west wall) = 424.84 KN/mmTherefore, (17)=

The total shear carried by the north wall, with

and

= 528.19 KN

For the south wall, kx= 6.89 x 103KN/mm, and

*Note that for the north wall the shear due to the eccentricity is a negative value and hence

must be dropped from the equation.

= 72.12 KN

For ground motion along the east-west direction (parallel to the x-axis), the east and west

walls are assumed not to participate in resisting the direct shears due to the component of the

eccentric load,Pxthrough the center of rigidity. However, the torsional component, Mt= Pe,

will induce forces in these walls. Applying Equation (l6a) with the first term set to zero, and

with ky= 424.84 X 103KN/mm., = 5500 mm., the shear in the east and west walls.


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( ) (B) Ground motion in northsouth direction (parallel to y-axis)

The forces calculated above for the east and west walls due to a design load in the east-west

direction will next be compared to the forces corresponding to ground motion in the north-

south direction. The base shear in this direction is calculated as follows:

Where

Cv= 0.64

I= 1.0

R= 5.5

T, is given by:

for hn 13.50 m

D = 14.5 m.

So, = 14.5 / 13.50 = 01.07 , greater than 0.9 ; use 0.9

Computing Ae,

Ae = 14.5m (3.5m) = 13.25 sq.m.

Computing Ac,

Ac = 13.25 sq.m. [ 0.2 + 0.92] = 13.38 sq.m.

Computing Ct,

Ct = 0.0743 / 13.38 = 0.02

Therefore,

T = 0.02 ( 13.5)0.75 = 0.143 sec < 0.7 sec

W = 3,488.60 KN


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therefore , V 3,488.60 = 2,837.91 KNChecking; 0.11 Ca I W < V < 2.5 Ca I W/R,

0.11 (0.44)(1.0) 3,488.60 = 168.85 < 2,837.91 , true

2.5 (0.44) (1.0) 3,488.60/ 5.5 = 697.72 KN < 2,837.91 KN

Therefore use limiting value of V = 697.72 KN

1. 2.

3. , since 4. Forces at each storey:

Froof =697.72 x 811.98 x 13.5 / 28,359.79 269.69 KN

F3rd =697.72 x 892.21 x 10 / 28,359.79 219.51 KN

F2nd =697.72 x 892.21 x 6.5 / 28,359.79 142.68 KN

F1st =697.72 x 892.21 x 3 / 28,359.79 65.85 KN

Hence, for the first story,Py= 631.87 KN, passing through the center of rigidity (see Fig. 14).

However, in compliance with code requirements (see Section 2.4), aneccentricity equal to

0.05 of the maximum building dimension will be considered. Thus, for this case,

Py= 631.87 KN The horizontal shear in each wall, using Equation (16a). with ky= 424.84 KN/mm.

and = 5500 mm.,


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Results of these calculations are in Table 4.

Table 4. Summary of Load Distribution for Example Problem 1

Summary of Load Distribution

Wall Stiffness, KN/mm Load Parallel to main axis of wall, KN

1-1 61.70 556.89

4-4 6.89 76.04

A-A 424.84 356.90

C-C 424.84 356.90

analysis and design of small reinforced concrete building for earthquake forces

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