Analysis Between a Beam Supported Structure and a Flat Plate Structure

Analysis Between a Beam Supported Structure and a Flat Plate Structure

INTRODUCTION

General:

When a slab is supported directly on columns, without beams and girders, it is called a flat plate

slab. Although thicker and more heavily reinforced than slabs in beam and girder construction,

flat plate slabs are advantageous because they offer no obstruction to passage of light (as beam

construction does); savings in story height and in the simpler form work involved; less danger of

collapse due to overload; and better fire protection with a sprinkler system because the spray is

not obstructed by beams.

Objective of the Study:

The objectives of the study were:

To analyze and design a six-storied beam supported building system.

To analyze and design the same building for flat plate slab system.

To compare the concrete and steel requirement of the two types of building.

Methodology

i. Analysis Phase

Requires extensive FEM analysis

Correction of analysis results for BNBC since in the software we used UBC* 94 code

Development of shear force & moment envelope to determining the critical sections & values of

critical shear and moment for design

ii. Design & Estimation Phase

Slab, Beam & Column Design & Estimation through manual calculation

ORGANIZATION of the Thesis Works:

The thesis comprises of the following five chapters:

Chapter- 1: Includes a brief introduction, objectives of the study andORGANIZATIONS of the

thesis paper.

Chapter- 2: Includes compilation of the relevant literature that has been reviewed for the

study.

Chapter- 3: Includes a detailed description of the analysis.

Chapter- 4: Includes the comparison of the results of analysis.

Chapter- 5: Includes conclusions and recommendations for further study.

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CHAPTER II LITERATURE REVIEW

Introduction

This chapter elaborated a detailed literature review that was required for the through

understanding and proper conducting of this work.

Wind Load

The minimum deign wind load on buildings and components shall be determined based on the

velocity of the wind, the shape and size of the building and the terrain exposure condition of the

site. The design wind load shall include the effects of the sustained wind velocity component and

the fluctuating component due to gusts. For slender buildings, the design wind load shall also

include additional loadings effects due to wind induced vibrations of the building.

Terrain exposure

A terrain exposure category that adequately reflects the surface roughness characteristics of the

ground shall be determined for the building site, taking intoACCOUNT the variations in ground

roughness arising from existing natural topography, vegetation and man made constructions. The

exposure category is divided into three types-

1. Exposure A: Urban and sub-urban areas, industrial areas, wooded areas, hilly or other terrain

covering at least 20 percent of the area with obstructions of 6 m or more in height and extending

from the site at least 500 m or10 times the height of the structure whichever is greater.

2. Exposure B: Open terrain with scattered obstructions having heights generally less Than 10 m

extending 800 m or more from the site in any full quadrant. This category includes air fields, open

park lands, sparsely built-up outskirts of towns, flat open country and grasslands.

3. 3. Exposure C: Flat and unobstructed open terrain, coastal areas and riversides facing Large

bodies of water, over 1.5 km or more in width, it extends inland from the

shoreline 400m or 10times the height of structure, whichever is greater.

Wind pressure on building:

Wind is one of the significant forces of nature that must be considered in the design of buildings.

Structural load applied by high winds is readily appreciated, even if the method of determining

them is not so easily understood. Other effects that can be caused even by moderate breezes

are commonly overlooked, however, because very often there is no obvious link between wind

and the behavior of a building.

Rain leakage around flashings and through joints in curtain walls may be due to a pressure

gradient across the wall and the functioning of ventilating and heating systems may be affected

by pressure distributions where ducts and openings are located.

Thus it is only the structural engineer who must consider wind action but the architect and

mechanical engineer as well. The latter are often concerned with the maximum pressures that

can reasonably by expected to occur during the useful life of the structure.


Conversion from wind speed to wind pressure:

Wind pressures exerted on a structure depend on the speed of the wind as well as the interaction

between the airflow and the structure. The wind speed to be used in computing the design

pressure depends on the particular component of the building being designed. For structural

purposes the maximum value is required and will vary with the geographical location.

Meteorological records of wind speed are analyzed to yield the most probable maximum that will

be equaled or exceeded, on the average, once during a given period of a time comparable to the

life of a structure.

Sustained Wind pressure

Qz = Cc * Ci* C z* Vb2

Where, Qz = Sustained wind pressure at height z, KN / m2

Ci = Structural importance coefficient

Cc = Velocity – to – pressure conversion

Cz = Combined height & exposure coefficient

Vb = basic wind speed in Km / hr

Design wind pressure:

Pz = Cg * Cp* Qz

Where,

Pz = design wind pressure at height z, KN / M2

Cg = gust coefficient

Cp = pressure coefficient for structures or components

Qz = sustained wind pressure

Pressure coefficients

Pressure coefficients used in practice have usually been obtained experimentally by testing

models of different types of structures in wind tunnels. Commonly used coefficients refer to the

average pressure or suction over a surface. Tangential forces are considered insignificant, so

that the forces referred to act at right angles to the surfaces in question.

Variables affecting pressure distributions

Building shape:

Pressure on certain parts of a structure is rather sensitive to changes in the shape of the building.

The suctions on the windward roof slope, for instance, very considerably with the slope of the

roof, the ratio of height to width, and the ratio of width to length of the building. Suctions on the

leeward wall, on the other hand, are not greatly affected by such variables.

Sometimes shape details have an unexpectedly large effect on the wind pressure distribution.

Parapet walls, large chimneys, silos and spires may have considerable influence and often the

only way to assess such effects is to test a scale model in a wind tunnel.

Openings:

The size and location of opening such as windows and doors determine the internal pressure that

must be considered in the calculation of net forces of walls and roofs. Internal pressure tend to

take on the values appropriate to the exterior of the wall in which in which the opening

predominate. If they are small and uniformly distributed, values of ± 2 are recommended, the

more unfavorable of the two to be considered in each case.

Wind direction:

The orientation of a building to the wind has a market effect on pressure distribution, particularly

on suction maxima, which occur over a small area near the leading edges of roofs.

Increase of wind speed with height:

Since the wind speed and consequently the velocity pressure increase with height above the

ground, a height factor is applied to the basic pressure in the design of building.

Shielding:

Other buildings, trees and similar large objects in the immediate vicinity have a bearing on

pressure distribution. The shielding provided is usually difficult to estimate and model tests

provide the most convenient means of determining design values. The assignment of reduction

for shielding is completed by the fact that conditions could change during the life of the structure.

Shielding does not always has a beneficial effect, and in some cases suction coefficients should

be increased because of the proximity of a neighboring building.

Wind pressures on various part of building

Roofs:

The roof is usually the critical area in the wind design of low building, particularly residential

structures. Where it is made up of light- weight components particular attention must be paid to

anchorage details because of the suction condition prevailing over most, if not all, of it. A good

example of such precautions is the time –honored custom of weighting roofs in alpine areas with

large stones.

Critical angle, windward slope:

For every sloped roof there is a certain slope angle at which the suction coefficients over the

windward slope reaches a numerical maximum.

Steep roofs:

As the roof slope increases beyond the critical angle the average pressure coefficient decreases

numerically to zero; it the increases in a positive direction, indicating pressure, to maximum of +

0.8 or so for a slope angle of 90 degrees.

Leeward slope:

The effect of slope and building dimension ratios are much less pronounced of suctions on the

leeward slope and for general purposes could probably be disregarded.

Local suctions:

Local suctions are more serious for wind at an angle (usually about 45 degree) to the side of the

building.

Walls:

For tall, slender structures the design of the walls and the frame, with regard to overturning

moment, are likely to be critical. The trend toward high-rise buildings and

curtain wall construction may lead to greater problems in limiting sway and specifying the

strength of fastenings for the wall panels.

Earthquake Load

Minimum design earthquake forces for buildings, structures or components of buildings or

structures, can be calculated either by the Equivalent static force method of by the Dynamic

response method. We will calculate earthquake load by equivalent static force method.

Seismic zoning map

The seismic zoning map of Bangladesh is provided by BNBC. Based on the severity of the

probable intensity of seismic ground motion and damages, Bangladesh has been divided into

three seismic zones.

These are –

1. Zone-1

2. Zone-2

3. Zone-3

Selection of lateral force method

Seismic lateral forces on primary framing systems shall be determined by using either the

equivalent static force method of the dynamic response method with the restriction given bellow-

a) The equivalent static force method

1. All structures, regular or irregular, in seismic zone 1 and in structure importance

2. Regular structures less than 75 m in height with lateral force resistance provided by structural

systems listed in BNBC except case 4 below.

3. Irregular structures not mire than 20 m in height.

4. A tower like building or structure having a flexible upper portion supported on a rigid lower portion

where:

Both portion of the structure considered separately can be classified as regular structures,

The average story stiffness of the lower portion is at least tin times the average storey stiffness of

the upper portion.

The period of the entire structure is not greater than 1.1 times the period of the upper portion

considered as a separate structure fixed at he base.

b) The Dynamic response method shall be used for structures of the following types-

1. Structures 75 m or more in height except as permitted by case a (1).

1. Structures having stiffness, weight or geometric vertical irregularity of type 1, 2& 3 is defined

in the BNBC table or structures not described.

2. Structure over 20 m in height in seismic zone 3 not having the same structural

system throughout their height except as permitted by BNBC.

1. Structures, regular or irregular, located on soil profile type S4 as described, which have a period

greater than .1 second. The analysis shall include the effect of the soil at the site.

Seismic dead load

Seismic dead load, W, is the total dead load of a building or a structure, including permanent

partitions and applicable portions of other loads listed below:

1) In storage and warehouse occupancies, a minimum of 25 percent of the floor live

load shall be applicable.

2) Where an allowance for partition load is include in the floor design in accordance

with BNBC all such loads but not less than .6 KN / m2 shall be applicable.

3) Total weight of permanent equipment shall be included

Equivalent static force method

Design base shear:

V = (ZIC / R) * W

Where,

Z = Seismic zone coefficient

I = Structure importance coefficient

R = Response modification coefficient for structural systems

W = Total seismic dead load

C = Numeric coefficient given by the relation-

C = (1.25 S) / T2/3

Where,

S = Site coefficient for soil characteristics

T = Fundamental period of vibration in seconds

Where, T = Ct (hn) 3/4

Ct = 0.083 for steel moment resisting frames

= 0.073 for reinforced concrete moment resisting frames and

Eccentric braced steel frames.

= 0.049 for all other structural systems.

hn = Height in meters above the base to level n

Vertical distribution of lateral force

In the absence of a more rigorous procedure, the total lateral force, which is the base shear V,

shall be distributed along the height of the structure in accordance with the following equation:

V = Ft + ∑ Fi

Where,

Fi = Lateral force applied at storey level I

Ft = Concentrated lateral force considered at the top of the building in

Addition to the force Fn

Where,

Ft = 0.070 TV ≤ 0.25V When T > 0.70 second

Ft = 0.0 When T ≤ 0.70 second

The remaining portion of the base shear (V-Ft) shall be distributed over the height of the building,

including level – n, according to the relation:

Fi = (V- Ft) * W l * hi) / ∑Wi * hi

Direct Design Method

Assumption

Moments in two way slabs can be found using the semi empirical direct design method, subjected

to the following restrictions:

1. There must be a minimum of three continuous spans in etch direction.

2. The panels must be rectangular, with the ratio of the longer to the shorter spans

within a panel not greater than 2.

1. The successive span lengths in each direction must not differ by more than one-third

of the longer span.

1. Columns may be offset a maximum of 10 percent of the span in the direction of the

offset from either axis between center lines of successive columns.

5. Loads must be due to gravity only and the live load must not exceed two times the

dead load.

6. If beams are used on the column lines, the relative stiffness of the beams in the two

perpendicular directions, given by the ratio: α1l2² / α2l12, must be between 0.2 and 5.0

Static moment

For purpose of calculating the total static moment Mo in a panel, the clear span ln in the direction

of moments is used. The clear span is defined to extend from face to face of the columns,

capitals, brackets, or walls but is not to be less than 0.65 l1. The total factored moment in a

span, for a strip bounded laterally by the centerline of the panel on each side of the centerline of

supports is-

Mo = wu l2 ln² / 8

Reciprocal Method

A simple, approximate design method developed by Bresler has been satisfactorily verified by

comparison with results of extensive tests and accurate calculation. The column interaction

surface can alternatively be plotted as a function of the axial load Pn and eccentricities ex = Mny /

Pn and ey = Mnx / Pn, The surface S1 of fig-a can be transformed into an equivalent failure surface

S2, as shown in fig – b, where ex and ey are plotted against 1 / Pn rather than Pn. Thus, ex = ey = 0

corresponds to the inverse of the capacity of the column if it were concentrically loaded, Po and

this is plotted as point C. For ey = 0 and any given value of ex, there is a load Pnyo that would

result in failure. The reciprocal of this load is plotted as point A. Similarly , for ex = 0 and any

given value of ey, there is a certain load Pnxo that would cause failure, the reciprocal of which is

point B. The values of Pnxo and Pnyo are easily established, for known eccentricities of loading

applied to a given column, using the methods already established for uniaxial bending or using

design charts for uniaxial bending. An oblique plane S2 is defined by the three points A, B, C. This

plane is used as an approximation of the actual failure surface.

Figure 2.1: Interaction surfaces for the reciprocal load method.

The vertical ordinate 1 / Pn, exact to the true failure surface will always be conservatively

estimated by the distance 1 / Pn, approx to the oblique plane ABC because of the concave

upward eggshell shape of the true failure surface. In other words,

1 / Pn, approx is always greater than 1 / Pn, exact which means that Pn, approx is always less

than Pn, exact. Breslers reciprocal load equation derives from the geometry of the approximating

plane, it can be shown that-

1 / Pn = 1 / Pnxo + 1 / Pnyo + 1 / Po

Where, Pnxo = nominal load when only eccentrically ey is present (ex = 0)

Pnyo = nominal load when only eccentrically ex is present (ey = 0)

Po = nominal load for concentrically loaded column.

The above equation has been found to be acceptably accurate for design purposes provided ρn ≥

0.10ρo. It is not reliable where biaxial bending is prevalent and accompanied by an axial force

smaller than ρo / 10.

Analysis and Design Basis:

This thesis is prepared properly based on Bangladesh National Building Code. Every part of this

thesis is properly maintained the recommendation of this code. Here some features are

described below-

Bangladesh country paper for WCDR 7

Does your country have building codes of practice and standards in place?

Which takes into account seismic risk?

The National Building Code was formulated and published in 1993. Bangladesh does not have

any separate code for the design or construction of earthquake resistant structure. However, a

new seismic zoning map and detailed seismic design provisions were incorporated into the

National Building Code in 1993 that replaces the code prepared in 1979. The Bangladesh

Earthquake Society has recently published a Bengali translation of the Guidelines for Earthquake

Resistant Non-Engineered Construction, written by the International Association of Earthquake

Engineering 3. The enforcement of the standards presented in the National Building Code

requires close monitoring by concerned agencies. The shortage of trained staff to monitor new

construction impedes the effectiveness of the building standards.

Analysis Software:

There is much finite element software for analyzing structure. ETABS is one of them. Every

analysis is this thesis is done by using ETABS 8 package. In the following paragraph we will

discuss some of its features-

Introduction

ETABS is a sophisticated, yet easy to use, special purpose analysis and design program

developed specifically for building systems. ETABS version 8 features an intuitive and powerful

graphical interface coupled with unmatched modeling, analytical and design procedures, all

integrated using a common database. Although quick and easy for simple structures, ETABS can

also handle the largest and most complex building models, including a wide range of nonlinear

behaviors, making it the tool of choice for structural engineers in the building industry.

History and advantages of ETABS

Dating back more than 30 years to the original development of TABS, the predecessor of ETABS,

it was clearly recognized that buildings constituted a very special class of structures. Early

releases of ETABS pro-vided input, output and numerical solution techniques that took into

consideration the characteristics unique to building type structures, providing a tool that offered

significant savings in time and increased accuracy over general purpose programs. As computers

and computer interfaces evolved, ETABS added computationally complex analytical options such

as dynamic nonlinear behavior, and powerful CAD-like drawing tools in a graphical and object-

based interface. Although ETABS Version 8 looks radically different from its predecessors of 30

years ago, its mission remains the same: to provide the profession with the most efficient and

comprehensive software for the analysis and design of buildings. To that end, the current release

follows the same philosophical approach put forward by the original programs, namely: Most

buildings are of straightforward geometry with horizontal beams and vertical columns. Although

any building configuration is possible with ETABS, in most cases, a simple grid system defined by

horizontal floors and vertical column lines can establish building geometry with minimal effort.

Many of the floor levels in buildings are similar. This commonality can be used numerically to

reduce computational effort. The input and output conventions used correspond to common

building technology. With ETABS, the models are defined logically floor-by-floor, column, bay-by-

bay and wall-by-wall and not as a stream of non-descript nodes and elements as in general

purpose programs. Thus the structural definition is simple, concise and meaningful. In most

buildings, the dimensions of the members are large in relation to the bay widths and story height.

Those dimensions have a significant effect on the stiffness of the frame ETABS corrects for such

effects in the formulation of the member stiff-ness, unlike most general-purpose programs that

work on center-line-to-center-line dimensions. The results produced by the programs should be in

a form directly usable by the engineer. General purpose computer programs produce result in a

general form that may need additional processing before they are useable in structural design.

An integrated approach

ETABS is a completely integrated system. Embedded beneath the simple, intuitive user interface

are very powerful numerical methods, design procedures and international design codes, all

working from a single comprehensive database. This integration means that you create only one

model of the floor systems and the vertical and lateral framing system to analyze and design the

entire building. Everything you need is integrated into one versatile analysis and design package

with one Windows-based graphical user interface. No external modules are maintained, and no

data is transferred between programs or modules. The effects on one part of the structure from

changes in another part are instantaneous and automatic. The integrated modules include:

Drafting module for model generation.

Seismic and wind load generation module.

Gravity load distribution module for the distribution of vertical loads to columns and beams when

plate bending floor elements are not provided as a part of the floor system.

Finite element-based linear static and dynamic analysis module.

Finite element-based nonlinear static and dynamic analysis module.

Output display and report generation module.

Steel frame design module (column, beam and brace).

Concrete frame design module (column and beam).

Composite beam design module

Steel joist design module

Shear wall design module.

Modeling features

The ETABS building is idealized as an assemblage of area, line and point object. Those objects

are used to represent wall, floor, column, beam, brace and link/spring physical members. The

basic frame geometry is defined with reference to a simple three dimensional grid system. With

relatively simple modeling techniques, very complex farming situations may be considered. The

buildings may be unsymmetrical and non-rectangular in plan. Torsional behavior of the floors and

inter story compatibility of the floors are accurately reflected in the result. The solution enforces

complete three dimensional displacement compatibility, making it possible to capture tubular

effects associated with the behavior of tall structures having relatively closely spaced columns.

Semi-rigid floor diaphragms may be modeled to capture the effects of in plane floor deformations.

Floor objects may span between adjacent levels to create sloped floors (ramps), which can be

useful for modeling parking garage structures. Modeling of partial diaphragms, such as in

mezzanines, setbacks, atriums and floor openings, is possible without the use of artificial

(“dummy”) floors and column lines. It is also possible to model situations with multiple

independent diaphragms at each level, allowing the modeling of buildings consisting of several

towers rising from a common base. The column, beam and brace elements may be non-

prismatic, and they may have partial fixity at their end connections. They also may have uniform,

partial uniform and trapezoidal load patterns, and they may have temperature loads. The effects

of the finite dimensions of the beams and columns on the stiffness of a frame system are

included using end offsets that can be automatically calculated. The floors and walls can be

modeled as membrane elements with in-plane stiffness only, plate bending elements with out-of-

plane stiffness only or full shell-type elements, which combine both in-plane and out-of-plane

stiffness. Floor and wall objects may have uniform load patterns in-plane or out-of-plane, and

they may have temperature loads. The column, beam, brace, floor and wall objects are all

compatible with one another.

Analysis features

Static analysis for user specified vertical and lateral floor or story loads are possible. If floor

elements with plate bending capability are modeled, vertical uniform loads on the floor are

transferred to the beams and columns through bending of the floor elements. Otherwise, vertical

uniform loads on the floor are automatically converted to span loads on adjoining beams, or point

loads on adjacent columns, thereby automating the tedious task of transferring floor tributary

loads to the floor beams without explicit modeling of the secondary framing. The program can

automatically generate lateral wind and seismic load patterns to meet the requirements of various

building codes. Three-dimensional mode shapes and frequencies, modal participation factors,

direction factors and participating mass percentage are evaluated using eigenvector. P-Delta

effects may be included with static or dynamic analysis. Response spectrum analysis, linear time

history analysis, nonlinear time history analysis, and static nonlinear (pushover) analysis are all

possible. The static nonlinear capabilities also allow you to perform incremental construction

analysis so that forces that arise as a result of the construction sequence are included. Result

from the various static load conditions may be combined with each other or with the result from

the dynamic response spectrum or time history analysis. Output may be viewed graphically,

displayed in tabular output, sent to a printer, exported to a database file, or saved in an ASCII file.

Types of output include reactions and member forces, mode shapes and participation factors,

static and dynamic story displacements and story shears, inter story drifts and joint

displacements, time history traces, and more.

Shell element internal forces

The shell element internal forces, like stresses, act throughout the element. They are present at

every point on the mid surface of the shell element. ETABS reports values for the shell internal

forces at the element nodes. It is important to note that the internal forces are reported as forces

and moments per unit of in-plane length. The basic shell element forces and moments are

identified as F11, F22, F12, M11, M22, M12, V13 and V23. You might expect that there would also be an

F21 and M21, but F21 is always equal to F12 and M21 is always equal to M12, so it is not actually

necessary to report F21and M21.

Conclusion

Materials problem is a great problem in our country especially the shortage of constriction raw

materials in our country. This thesis is based on the previously discussed topics.

This thesis may result an effective solution of this problem.

ANALYTICAL STUDY

General:

The analysis is made by using ETABS finite element package. Analysis was made for two

different types of building systems. One is beam column slab system and another is flat plate slab

system. Total ten loads combination was considered for design of different elements of the

building. The whole analysis and design was performed based on ACI and BNBC code.

The Building Geometry

The building geometries are as follows:

Option I:

All the floors have 16 columns. All the slabs of the structure are beam supported. Story height is

10 ft. column and beam size is different. The layout is shown in figure 3.2

Option II:

All the floors have 16 columns. All the slabs are directly supported on column (Flat Plate

structure). Column size is different. The layout is shown in figure 3.1

The Loads Considered

Dead Load, D. L = 50 psf (for wall)

Floor Finish, F. F = 30 psf

Live Load, L. L = 40 psf

Load Combination

COMB 1 = 1.4 D.L

COMB 2 = 1.4 D.L + 1.7 L.L

COMB 3 = 0.75 (1.4 D.L + 1.7 L.L + 1.7 WLX)

COMB 4 = 0.75 (1.4 D.L + 1.7 L.L. – 1.7 WLX)

COMB 5 = 0.75 (1.4 D.L + 1.7 L.L + 1.7 WLY)

COMB 6 = 0.75 (1.4 D.L + 1.7 L.L – 1.7 WLY)

COMB 7 = 0.75 (1.4 D.L + 1.7 L.L + 1.87 ELX)

COMB 8 = 0.75 (1.4 D.L + 1.7 L.L – 1.87 ELX)

COMB 9 = 0.75 (1.4 D.L + 1.7 L.L + 1.87 ELY)

COMB 10 = 0.75 (1.4 D.L + 1.7 L.L – 1.87 ELY)

Legends:

D.L = Dead load

L.L = Live load

WLX = Wind load in X direction

WLY = Wind load in Y direction

EQX = Earthquake load in X direction

EQY = Earthquake load in Y direction

Figure 3.1: Option II (Typical floor plan of the flat plate structure).

Wind Load Calculation

This load is a function of the wind speed which in turn is depended on the location of the building,

the exposure of the location, gusting effect, importance of the building and the geometry of the

building. Wind load calculations were done by UBC – 94 codes by ETABS. The wind speed was

adjusted to convert it to BNBC code. In this study, wind load was calculated by the diaphragms

method.

Input data

Windward Coefficient, Cq = 1.4

Leeward Coefficient, Cq ≈ 0

Wind Speed, V = 210 km / hr (131.25 mph)

Exposure Type = B, Importance Factor = 1.0

Along X axis, Wind Direction Angle = 00

Along Y axis, Wind Direction Angle = 900

Velocity adjustment

The table below shows the reaction due to wind load in X and Y direction.

Table 3.1: ETABS output value for reaction forces:

Story Load Fx Fy

BASE WLX - 7.56 - 0.27

BASE WLY 0.27 - 8.11

BASE WLX - 10.27 - 0.20

BASE WLY 0.20 - 8.03

BASE WLX - 10.68 - 0.20

BASE WLY - 0.20 - 8.03

BASE WLX - 8.11 - 0.27

BASE WLY - 0.27 - 8.11

BASE WLX - 8.03 - 0.20

BASE WLY - 0.20 - 10.68

BASE WLX - 10.63 - 0.15

Table 3.1: ETABS output value for reaction forces (continued…).

Story Load Fx Fy

BASE WLY - 0.15 - 10.63

BASE WLX - 10.33 - 0.15

BASE WLY 0.15 - 10.63

BASE WLX -7.64 - 0.20

BASE WLY 0.20 - 10.68

BASE WLX - 7.64 0.20

BASE WLY 0.20 - 10.27

BASE WLX - 7.56 0.27

BASE WLY 0.27 - 7.56

BASE WLX - 10.33 0.15

BASE WLY 0.15 - 10.33

BASE WLX - 10.27 0.20

BASE WLY 0.20 -7.64

BASE WLX - 10.63 0.15

BASE WLY - 0.15 - 10.33

BASE WLX - 10.68 0.20

BASE WLY - 0.20 - 7.64

BASE WLX - 8.03 0.20

BASE WLY - 0.20 - 10.27

BASE WLX - 8.11 0.27

BASE WLY - 0.27 -7.56

Summation WLX - 146.48 0

Summation WLY 0 - 146.48

From Hand Calculation:

Table 3.2: Wind force X / Y direction.

height Vb2 qz Pz Pz A F

H(m) Cc Ci cz 210 (KN/m2) Cg Cp (KN/m2) (psf) (sft) (Kips)

F-1 3.05 4.7E-5 1 0.37 44100 0.76 1.38 1.4 1.48 30.92 250 7.73

F-2 6.1 4.7E-5 1 0.41 44100 0.86 1.38 1.4 1.68 35.08 500 17.54

F-3 9.15 4.7E-5 1 0.50 44100 1.03 1.38 1.4 2.01 41.97 500 20.99

F-4 12.2 4.7E-5 1 0.57 44100 1.17 1.38 1.4 2.28 47.77 500 23.89

F-5 15.24 4.7E-5 1 0.63 44100 1.30 1.38 1.4 2.52 52.74 500 26.37

F-6 18.3 4.7E-5 1 0.68 44100 1.41 1.38 1.4 2.74 57.29 500 28.65

Summation of F 125.16

So the adjusted wind speed in both direction,

V = 131.25 * √ (125.16 / 146.48) = 121.32 mph

Table 3.3: The table below shows the reaction due to wind load in X and Y direction after

correction of wind speed.

Story Load Fx Fy

BASE WLX - 6.44 - 0.25

BASE WLY 0.25 - 6.94

BASE WLX - 8.75 - 0.14

BASE WLY 0.2 - 6.83

BASE WLX - 9.15 - 0.14

Table 3.3: The table below shows the reaction due to wind load in X and Y direction after

correction of wind speed (continued…).

Design of beam supported slab

Design code: ACI.

Design method: USD

Design procedure: Direct design method

Slab system: Beam supported slab.

Material properties: fy = 60 ksi; fc’= 3.5 ksi

Unit wt. of concrete, wc = 150 psf

Unit wt. of brick, wb = 120 psf.

Loads: Live load = 40 psf,

Floor finish = 30 psf

Partition wall load = 50 psf

Figure 3.2: Option I (Typical floor plan of the beam supported structure)

Calculation of slab thickness:

Let, slab thickness h = 6 inch.

Beam size = 10 in * 16 in.

Now, 4hf = 24 in, hw = 10 in.

For the edge beam: I = 10 * 1.50 * 16³ * 1 / 12 = 5120 in4

For the interior beam: I = 10 * 2 * 16³ * 1 / 12 = 6827 in4

For the slab strips:

For the 7.92 ft edge width: I = 7.92 * 12 * 6³ * 1/12 = 1711 in4

For the 17.5 ft width: I = 17.50 * 12 * 6³ * 1 / 12 = 3780 in4

Thus for the edge beam α1 = 5120 / 1711 = 2.99

For the interior beam α2 = 6827 / 3780 = 1.81

Average value, αm = 2.40

For 15’*15’ slab panel: β = 1

For 15’*20’ slab panel: β = (20 – 10 / 12) / (15 – 10 / 12) = 19.17 / 14.17 = 1.35

For 20’*20’ slab panel: β = 1

Now, slab thickness h = [12 * 19.17{0.8+ (60000 / 200000)}] / [36 + 9 * 1] = 5.62 in.

The 3.50 in limitation clearly does not control in this case. 6 in. depth is ok.

Factored load, W = 285 psf.

For 15 ft * 15 ft panel:

Slab – beam strip centered on the Interior column line.

Mo = 0.285 * 17.50 *14.17² * 1 / 8 = 137 ft – kip.

Interior negative moment: 137 * 0.70 = 96 ft-kip

Positive moment: 137 * 0.57 = 78 ft – kip

Exterior negative moment: 137 * 0.16 = 22 ft – kip.

The torsional constant, C = {1 – (0.63 * 10 / 10)16 * 10³ / 3 + {1 – (0.63*6 / 10)10 * 6³ / 3

= 3681 in4

Now, l2 / l1 = 1; α1l2 / l1 = 1.81; βt = 3681 / (2*3780) = 0.50

Exterior negative moment 90%, positive moment 75%, Interior negative moment 75% is taken by

column strip.

The table below is showing slab strip moment at different locations for 15’*15’ panel.

Table 3.5: Slab-beam strip centered on the Interior column line.

Slab-beam strip Column strip slab moment ft-

kip

Middle strip slab moment ft-kip

Exterior negative moment 3 2.2

Positive moment 9 20

Interior negative moment 11 24

Slab-beam strip at the edge of the building:-

Mo = 0.285 * 7.92 * 14.17² * 1 / 8 = 62 ft-kips

Interior negative: 62 * 0.70 = 43 ft-kips

Positive: 62 * 0.57 = 35 ft-kips

Exterior negative: 62 * 0.16 = 10 ft-kips

Now, l2 / l1= 1; α112 / l1 = 2.99 * 1 = 2.99; βt = 3681 / (2 * 1711) = 1.10

Positive moment 75%, Exterior negative moment 90%, Interior negative moment 75% is taken by

column strip.

Table 3.6: Slab-beam strip at the edge of the building: (slab, 15′ * 15′).

Exterior slab-beam strip Column strip slab moment ft- Middle strip moment

kip ft-kip

Exterior negative moment 1.50 1

Positive moment 4 9

Interior negative moment 5 11

For 15 ft * 20 ft Panel:

Slab-beam strip at the edge of the building.

Mo = 0.285 * 7.92 * 19.17² * 1 / 8 = 113 ft-kips.

Interior negative moment: 113 * 0.65 = 73.5 ft-kips

Positive moment: 113 * 0.35 = 39.5 ft-kip

Exterior negative moment: 137 * 0.16 = 22 ft-kip.

Now, l2 / l1 = 15 / 20 = 0.75; α1l2 / l1 = 2.99 * 0.75 = 2.24

Negative moment 83%, positive moment 83%, is taken by column strip.

Table 3.7: Slab-beam strips at the edge of the building (slab 15 ft * 20 ft).

Exterior Slab-beam strip (20

ft span)

Column strip slab moment ft-

kips

Middle strip slab moment

ft-kips

Negative moment 9 12.50

Positive moment 5 7

Slab-beam strip centered on the Interior column line

Mo = 0.285 * 17.50 * 19.17² * 1 / 8 = 250 ft – kips

Negative: 250 * 0.65 = 163 ft – kips

Positive: 250 * 0.35 = 88 ft – kips

Now, l2 / l1= 15 / 20 = 0.75; α1l2 / l1 = 0.75 * 1.81 = 1.40;

Positive moment 83% and negative moment 83% is taken by column strip.

Table 3.8: Slab-beam strips centered on the Interior column line.

Interior slab-beam strip (20 ft

span)

Column strip slab moment

ft-kips

Middle strip moment

ft-kips

Negative moment 20 28

Positive moment 11 15

For 15 ft * 15 ft panel:

The table below is showing reinforcement requirement for different strip such as column strip,

middle strip for 15’*15’ slab.

Table 3.9: Design of slab reinforcement.

Description Location Mu

ft-

kip

Strip

width

b

inch

Effective

depth d

inch

Mu*12/

b

Steel

density

ρ

Required

steel area

As

in²

Exterior half

column strip

Exterior:

negative

positive

Interior:

negative

1.5

4

5

50

50

50

5

5

5

0.36

0.96

1.2

0.0022

0.0022

0.0022

0.55

0.55

0.55

Middle strip Exterior:

negative

positive

Interior:

negative

2.1

19

23

90

90

90

5

5

5

0.28

2.53

3.1

0.0022

0.0022

0.0024

0.99

0.99

1.08

Interior half

column strip

Exterior:

negative

positive

Interior:

negative

1.5

4.5

5.5

45

45

45

5

5

5

0.4

1.2

1.5

0.0022

0.0022

0.0022

0.5

0.5

0.5

Exterior half

column strip

Exterior:

negative

positive

Interior:

negative

1.5

4

5

50

50

50

4.5

4.5

4.5

0.36

0.96

1.2

0.0024

0.0024

0.0024

0.54

0.54

0.54

Table 3.9: Design of slab reinforcement (continued…).


middle strip for 15 ft * 20 ft slab.



ft-

kip

Strip

width

b

inch

Effective

depth of

slab d

inch

Mu*12/b Steel

density

ρ

Required

steel area

As in²

15’span two

half column

strip

Exterior:

negative

positive

Interior:

negative

3

9

11

75

75

75

5

5

5

0.48

1.44

1.76

0.0022

0.0022

0.0022

0.83

0.83

0.83

Middle strip Exterior:

negative

positive

Interior

negative

2.2

20

24

150

150

150

5

5

5

0.2

1.6

1.92

0.0022

0.0022

0.0022

1.65

1.65

1.65

Table 3.10: Design of slab reinforcement (continued…).


middle strip for 20 ft * 20 ft slab.



ft-

kip

Strip

width

b

inch

Effective

depth of

slab d

inch

Mu*12/b

Ft-kip/ft

Steel

density ρ

Required

steel area

As

in²

Two half

column strip

Negative

Positive

20

11

120

120

5

5

2

1.1

0.0022

0.0022

1.32

1.32

Middle strip Negative

Positive

28

15

120

120

5

5

2.8

1.5

0.0022

0.0022

1.32

1.32

Two half

column strip

Negative

Positive

20

11

120

120

4.5

4.5

2

1.1

0.0024

0.0024

1.3

1.3

Middle strip Negative

Positive

28

15

120

120

4.5

4.5

2.8

1.5

0.0026

0.0024

1.4

1.3

Minimum steel As min = 0.0018 * 12 * 6 = 0.13 in²

ρ min = 0.13 / (5 * 12) = 0.0022

Or ρ min = 0.13 / (4.5 * 12) = 0.0024

Maximum spacing = 2h = 2 * 6 = 12 in.

The figure below is showing the reinforcement arrangement in slab

Figure 3.3: Reinforcement arrangement in slab (Beam Supported Structure).

Design of flat plate slab:

Thickness = ln / 30 = (20 * 12-16) / 30 = 7.5”

Dead Load

Self wt of slab: 7.50 * 12.5

Wall Load: 50 psf

Floor Finish: 30 psf

Total Dead Load: 173.75 psf

LL = 40 psf

Factored Load Wu = 1.4DL + 1.7LL = 311.25 psf

For Interior slab 20 ft * 20 ft:

Mo = (1 / 8) Wul2ln² = (1 / 8) * 311.25 * 20 * {20 – (16 / 12)}² = 271.2 kip-ft

Negative moment = 271.2 * 0.65 = 176.28 kip – ft

Positive moment = 271.2 * 0.35 = 94.92 kip – ft

Column strip negative: 0.75 * 176.28 = 132.21

Column strip positive: 0.60 * 94.92 = 56.95

Middle strip negative: 176.28 – 132.21 = 44.07

Middle strip positive: 94.92 – 56.95 = 37.97

For Corner slab 15 ft*15 ft:

Mo = (1 / 8) Wl2ln² = (1 / 8) * 311.25 * 15 * {15 – (16 / 12)}² = 109 kip – ft.

Interior negative: 109 * 0.70 = 76.3

Exterior negative: 109 * 0.26 = 28.34

Exterior positive: 109 * 0.52 = 56.68

Exterior negative column strip = 28.34

Exterior negative middle strip = 0

Interior negative column strip = 0.75 * 76.3 = 57.23

Interior negative middle strip = 76.30 – 57.23 = 19.1

Interior positive column strip = 0.60 * 56.68 = 34

Interior positive middle strip = 56.68 – 34 = 22.67

ρ max = 0.85² * (3.5 / 60) * (0.003 / 0.003+0.004) = 0.018

As min = 0.0018 * 7.5 * 12 = 0.162 in²

20 ft direction ρmin = 0.162 / (6 * 12) = 0.00225

15 ft direction ρmin = 0.162 / (6.50 *12) = 0.00208

d² = Mo / [0.90 * 0.018 * 60000 *12{1 – (0.59 * 0.018 * 60 / 3.50)}]

d = √ (Mo / 9540.50)

d = √ (26.44*12000/9540.50)

d = 5.80 in.

For exterior middle slab 20 ft*15 ft:

Long direction:

Mo = [311.25 * 15 * (20-16 / 12) ²] / 8 = 203.35 ft-kip

Negative moment = 0.65 * 203.35 = 132.2 ft – kip

Positive moment = 0.35 * 203.35 = 71.2 ft – kip

Column strip negative moment: 0.75 * 132.20 = 99.15 ft – kip

Column strip positive moment: 0.60 * 71.20 = 42.7 ft – kip

Middle strip negative moment: 132.20 – 99.15 = 33.05 ft – kip

Middle strip positive moment: 71.20 – 42.70 = 28.5 ft – kip

Short direction:

Mo = [311.25 * 20 * (15 – 16 / 12)²] / 8 = 145.34 ft – kip

Interior negative moment = 0.70 * 145.34 = 101.74 ft – kip

Interior positive moment = 0.52*145.34 = 75.58 ft – kip

Exterior negative moment = 0.26 * 145.34 = 37.8 ft – kip

Exterior negative column strip: 37.80 * 1 = 37.8 ft – kip

Exterior negative middle strip: 0

Interior negative column strip: 101.74 * 0.75 = 76.31 ft-kip

Interior negative middle strip: 101.74 – 76.31 = 25.43 ft – kip

Positive column strip: 0.60 * 75.58 = 45.35 ft – kip

Positive middle strip: 0.40 * 75.58 = 30.23 ft – kip

The table below is showing reinforcement requirement for 15 ft *15 ft panel

Table 3.13: Design of slab reinforcement in 15 ft *15 ft panel

Strip type Moment type Strip width b in. Mu

Ft-kipEffective depth of the slab d inMu*12/b

The table below is showing reinforcement requirement for 20 ft * 20 ft panel




Ft-kip/ft

Steel density ρRequired steel area As=ρbd in²

The table below is showing reinforcement requirement for 15 ft * 20 ft panel




Ft-kip/ft

Steel density ρRequired steel area As=ρbd in²

The figures below is showing the reinforcement arrangement in slab in flat plate structure

Figure 3.4: Reinforcement arrangement in slab in flat plate structure (long direction).

Figure 3.5: Reinforcement arrangement in slab in flat plate structure (short direction).

Punching Shear Check:

For Column- C6, C7, C11, C13:

Maximum spacing = 7.50 * 20 = 15 in

Vu = 0.31125 [17.50*17.5 - (22 * 22 / 144)] = 94.27 kip

D / 2 = 6 / 2 = 3 in

bo = 4 (16+6) = 88

Since, βc < 2;

φVc = φ 4√fc’ * bod = 0.75 * 4√3500 * 0.08 * 6.50 = 101.55 kip

For Column- C1, C4, C10, C16:

Vu = 0.31125 [8.20 * 8.20 - (19 * 19 / 144)] = 20.20 kip

bo = 19+19 = 38

φVc = φ 4√fc’ * bod = 0.75 * 4√3500 * 38 * 6.50 = 43.84 kip

For Column – C2, C3, C5, C8, C9, C12, C14, C15:

Vu = 0.31125 [17.50 * 8.20 - (19 * 22 / 144)] = 44 kip

bo = 19 * 20 + 22 = 60

φVc = φ 4√fc’ * bod = 69.22 kip]

Figure 3.6: Distribution of total static moment M0 to critical section for positive and negative

bending.

Beam design

For beam design required data such as moment values, shears are taken from ETABS analysis

report shown in table – and designed various beams for various floors of the beam supported

structure.

Design of the beam: B 1, B3, B4, B6, B7, B9, B10, B11 ( at 6th story )

From load combination:

Maximum moment:

End section:

Negative moment = 42.61 k-ft

Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0 .75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.50 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 42.61 * 12 / 0.90 * 0.0187 * 60 *10 (1- 0.59*0.0187*60/3.50)

d = 7.90, Clear cover = 2”, Total depth = 7.90+2 = 9.90 say, d = 10”

Provided Beam size = 10” * 10”, d = 10”- 20”= 8”

Main steel calculation:

As = Mu / φfy (d – a / 2) = 42.61 * 12 / 0.9 * 60 (8 – 1 / 2)

= 1.262 in², a = Asfy / 0.85fc bw, a = 1.262 * 60 / 0.85 * 3.5 * 10 = 2.545 in.

As = 42.61 * 12 / 0.9 * 60 (8 – 2.545 / 2) = 1.407 in²

a = 1.407 * 60 / 0.85 * 3.50 * 10 = 2.837 in.

As = 42.61 * 12 / 0.9 * 60 (8 – 2.837 / 2) = 1.438 in²

Use – 2 # 7 +1 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2) = 30.30 * 12 / 0.9 * 60 (8 – 1 / 2)

= 0.897 in², a = Asfy / 0.85fc bw, a = 0.897 * 60 / 0.85 * 3.5 *10 = 1.809 in

As = 30.30 * 12 / 0.9 * 60 (8 – 1.809 / 2) = 0.949 in²

a = 0.949 * 60 / 0.85 * 3.5 * 10 = 1.91 in.

As = 30.30 * 12 / 0.90 * 60 (8 – 1.91 / 2) = 0.955 in² use – 3 # 5 bars

Shear reinforcement design:

V= Vu – φVc

= 16.21 – (2 * 0.85√3500 * 10 * 8) / 1000 = 8.164 kip

4 √fc bw d = (4√3500 * 10 * 8) / 1000 = 16.09 kip

Vs < 4√fc bw d So, ok

Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 8 / 2 = 4”

3) Smax = 24”

4) S = φAvfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 8 / (8.164 * 1000) = 10.99”

Use stirrups # 3 bar @ 4” c/c

Design of the beam: B2, B5, B8, B12 ( at 6th story )


Maximum moment:

End section:

Negative moment = 80.00 k – ft

Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / фρfy bd (1 – 0.59 ρfy / f’c)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 80.0 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60 / 3.50)

d = 10.82, Clear cover = 2”, Total depth = 10.82 + 2 = 12.82. Say d = 15



As = Mu / φfy (d – a / 2)

= 80.0 *12 / 0.9 * 60 (13 – 1 / 2)

= 1.422 in², a = Asfy / 0.85fc bw, a = 1.422 * 60 / 0.85 * 3.5 * 10 = 2.868 in

As = 80.0 * 12 / 0.90 * 60 (8 – 2.868 / 2) = 1.537 in²

a = 1.537 * 60 / 0.85 * 3.50 * 10 = 3.099 in.

As = 80.0 * 12 / 0.90 * 60 (8 – 3.099 / 2) = 1.552 in², use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 57.29 * 12 / 0.90 * 60 (13 – 1 / 2)

= 1.018 in², a = Asfy / 0.85fc bw, a = 1.018 * 60 / 0.85 * 3.5 *10 = 2.053 in

As = 57.29 * 12 / 0.90 * 60 (13 – 2.053 / 2) = 1.063 in²

a = 1.063 * 60 / 0.85 * 3.50 * 10 = 2.143 in.

As = 57.29 * 12 / 0.90 * 60 (13 – 2.143 / 2) = 1.067 in² use – 4 # 5 bars


Vs = Vu - φVc

= 24.14 – (2 * 0.85√3500 * 10 *13) / 1000 = 11.06 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11* 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (11.06 * 1000) =13.1”

Use stirrups # 3 bar @ 6½” c/c

Design of the beam: B14, B17, B 20, B23 ( at 6th story )


Maximum moment at end section


Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / (фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.50 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 132.38 * 12 / 0.90 * 0.0187 * 60 * 10 (1-0.59 * 0.0187 * 60 / 3.50)

d = 13.92, Clear cover = 2”, Total depth = 13.92 + 2 = 15.92.92. Say, d = 18”



As = Mu / φfy (d – a / 2)

= 132.38 * 12 / 0.90 * 60 (16 – 1 / 2)

= 1.897 in², a = Asfy / 0.85fc bw, a = 1.897 * 60 / 0.85 * 3.5 * 10 = 3.825 in

As = 132.38 * 12 / 0.90 * 6 (16 – 3.825 / 2) = 2.088 in²

a = 2.088 * 60 / 0.85 * 3.50 * 10 = 4.211 in.

As = 132.38 * 12 / 0.90 * 60 (16 – 4.211 / 2) = 2.117 in² use – 2 # 8 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 103.96 * 12 / 0.90 * 60 (16 – 1 / 2)

= 1.49 in a = Asfy / 0.85fc, bw a = 1.49 * 60 / 0.85 * 3.5 *10 = 3.00 in.

As = 103.96 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.593 in²

a = 1.593 * 60 / 0.85 * 3.50 * 10 = 3.217 in.

As = 103.96 *12 / 0.90 * 60 (16 – 3.217 / 2) = 1.605 in². Use – 2 # 6 +2 # 5 bars


Vs = Vu - φVc

= 40.84 – (2 * 0.85√3500 * 10 * 16) / 1000 = 24.78 kip

4 √fc bw d = (4√3500 * 10 * 16) / 10 00 = 32.183 kip

Vs < 4√fc bw d. So, ok

Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φAvfy d / Vs = 0.85 * 2 * 0.11* 60000 *16 / (24.78 * 1000) = 7.24 ”


Design of the beam: B13, B15, B16, B18, B19, B21, B22, B24 (at 6th story)




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1 – 0.59 ρfy / fc)

ρ max = 0 .75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 64.59 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.018 * 60 / 3.50)

d = 9.72, Clear cover = 2”, Total depth = 9.72+2= 11.72. Say, d = 12”

Provided Beam size = 12” * 10”, d = 12”- 2” = 10”


As = Mu / φfy (d – a / 2)

= 64.59 * 12 / 0.90 * 60 (10 – 1 / 2) = 1.51 in², a = Asfy / 0.85fc bw

a = 1.51 * 60 / 0.85 * 3.50 * 10 = 3.045 in

As = 64.59 * 12 / 0.90 * 60 (10 – 3.045 / 2) = 1.693 in²

a = 1.693 * 60 / 0.85 * 3.5 * 10 = 3.414 in.

As = 64.59 *12 / 0.90 * 60 (10 – 3.414 / 2) = 1.73 in² use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 51.98 * 12 / 0.90 * 60 (10 – 1 / 2)

= 1.215 in², a = Asfy / 0.85fc, bw a = 1.215 * 60 / 0.85 * 3.5 * 10 = 2.450 in

As = 51.98*12 / 0.90 * 60 (10 – 2.45 / 2) = 1.316 in²

a = 1.316 * 60 / 0.85 * 3.50 * 10 = 2.609 in.

As = 51.98 * 12 / 0.90 * 60 (10 – 2.609 / 2) = 1.328 in² use – 2 # 6 +2 # 5 bars


Vs = Vu - φVc

= 25.40 – (2 * 0.85√3500 * 10 * 10) / 1000 = 15.34 kip

4 √fc bw d = (4√3500 * 10 * 10) / 1000 = 23.66 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 10 / 2 = 5”

3) Smax = 24”

4) S = φAvfy d / Vs = 0.85 * 2 * 0.1 1* 60000 * 10 / (15.34 * 1000) = 7.31 ”


Design of the beam: B1, B3, B4, B6, B7, B9, B10, B11 ( at 5th story )




Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb ,

ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 42.61 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60/3.50)

d = 7.90, Clear cover = 2”, Total depth = 7.90 + 2 = 9.90. Say, d = 10”



As = Mu / φfy (d – a / 2)

= 42.61 * 12 / 0.90 * 60 (8 – 1 / 2) = 1.262 in², a = Asfy / 0.85fc bw,

a = 1.262 * 60 / 0.85 * 3.50 * 10 = 2.545 in

As = 42.61 * 12 / 0.90 * 60 (8 – 2.545 / 2)

As = 1.407 in²

a = 1.407 * 60 / 0.85 * 3.50 * 10 = 2.837 in.

As = 42.61 * 12 / 0.90 * 60 (8 – 2.837 / 2) = 1.438 in². Use – 2 # 7 +1 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 30.30 * 12 / 0.90 * 60 (8 – 1 / 2) = 0.897 in², a = Asfy / 0.85fc bw,

a = 0.897 * 60 / 0.85 * 3.50 * 10 =1.809 in.

As = 30.30 * 12 / 0.9 * 60 (8 – 1.809 / 2) = 0.949 in²

a = 0.949 * 60 / 0.85 * 3.50 * 10 = 1.91 in.

As = 30.30 * 12 / 0.90 * 60 (8 – 1.91 / 2) = 0.955 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 16.21 – (2 * 0.85√3500 * 10 * 8) / 1000 = 8.164 kip

4φ √fc bw d = (4 * 0.85√3500 *10 * 8) / 1000 = 16.09 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2*.11*60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 8 / 2 = 4”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 8 / (8.164 * 1000) = 10.99”




Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb , ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.50 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 80.0 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60 / 3.50)

d = 10.82, Clear cover = 2”, Total depth = 10.82 + 2 = 12.82. Say, d = 15



As = Mu / φfy (d – a / 2)

= 80.0 * 12 / 0.90 * 60 (13 – 1 / 2)

= 1.422 in², a = Asfy / 0.85fc bw,

a = 1.422 * 60 / 0.85 * 3.50 * 10 = 2.868 in

As = 80.0 * 12 / 0.90 * 60 (8 – 2.868 / 2) = 1.537 in²

a = 1.537 * 60 / 0.85 * 3.50 * 10 = 3.099 in.

As = 80.0 * 12 / 0.90 * 60 (8 – 3.099 / 2) = 1.552 in². Use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 57.29 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.018 in², a = Asfy / 0.85fc bw,

a = 1.018 * 60 / 0.85 * 3.5*10 = 2.053 in.

As = 57.29 * 12 / 0.90 * 60 (13- 2.053 / 2) = 1.063 in²

a = 1.063 * 60 / 0.85 * 3.50 * 10 = 2.143 in.

As = 57.29 * 12 / 0.90 * 60 (13 – 2.143 / 2) = 1.067 in². Use – 4 # 5 bars


Vs = Vu - φVc

= 24.14 – (2 * 0.85√3500 * 10 * 13) / 1000 = 11.06 kip

4φ √fc bw d = (4 * 0.85√3500 * 10 * 13 / 1000 = 30.763 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (11.06 * 1000) = 13.1”






Mu = ф ρfy bd² (1 – 0.59 ρfy / f’c)

d² = Mu / (ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 132.38 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60 / 3.50)

= 13.92, Clear cover = 2”, Total depth = 13.92 + 2 = 15.92. Say, d = 18”



As = Mu / φfy (d – a / 2)

= 132.38 * 12 / 0.90 * 60 (16 – 1 / 2)

= 1.897 in², a = Asfy / 0.85fc bw, a = 1.897 * 60 / 0.85 * 3.5 * 10 = 3.825 in.

As = 132.38 * 12 / 0.90 * 60 (16 – 3.825 / 2) = 2.088 in²

a = 2.088 * 60 / 0.85 * 3.50 * 10 = 4.211 in.

As = 132.38 * 12 / 0.90 * 60 (16 – 4.211 / 2) = 2.117 in². Use – 2 # 7 +3 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 103.96 * 12 / 0.90 * 60 (16 – 1 / 2)

= 1.49 in², a = Asfy / 0.85fc bw, a = 1.49 * 60 / 0.85 * 3.5*10 = 3.00 in.

As = 103.96 * 12 / 0.90 * 60 (16 -3.00 / 2) = 1.593 in²

a = 1.593 * 60 / 0.85 * 3.50 * 10 = 3.217 in.

As = 103.96 * 12 / 0.90 * 60 (16 – 3.217 / 2) = 1.605 in². Use – 4 # 6 bars


Vs = Vu - φVc

= 40.84 – (2 * 0.85√3500 * 10 * 16) / 1000 = 24.78 kip

4 √fc bw d = (4√3500 * 10 * 16) / 1000 = 32.183 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φ Avfy d / Vs= 0.85 * 2 * 0.11 * 60000 * 16 / (24.78 * 1000) = 7.24”






Mu = ф ρfy bd² (1- 0.59 ρfy / f’c)

d² = Mu / (фρfy bd (1- 0.59 ρf / f’c)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 64.59 * 12 / 0.90 * 0.0187 * 60 * 10 (1-0.59 * 0.0187 * 60 / 3.50)

d = 9.72, Clear cover = 2”, Total depth = 9.72 + 2= 11.72. Say, d = 12”



As = Mu / φfy (d – a / 2)

= 64.59 * 12 / 0.90 * 60 (10 – 1 / 2) = 1.51 in², a = Asfy / 0.85f’c bw

a = 1.51 * 60 / 0.85 * 3.50 * 10 = 3.045

As = 64.59 * 12 / 0.90 * 60 (10 – 3.045 / 2) = 1.693 in²

a = 1.693 * 60 / 0.85 * 3.5 * 10 = 3.414 in.

As = 64.59 * 12 / 0.90 * 60 (10 – 3.414 / 2) = 1.73 in². Use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 51.98 * 12 / 0.90 * 60 (10 – 1 / 2) = 1.215 in², a = Asfy / 0.85f’c bw

a = 1.215 * 60 / 0.85 * 3.50 * 10 = 2.450 in.

As = 51.98 * 12 / 0.90 * 60 (10 – 2.45 / 2) = 1.316 in²

a = 1.316 * 60 / 0.85 * 3.50 * 10 = 2.609 in.

As = 51.98 * 12 / 0.90 * 60 (10 – 2.609 / 2) = 1.328 in². Use – 2 # 6 +2 # 5 bars


Vs = Vu - φVc

= 25.40 – (2 * 0.85√3500 * 10 * 10) / 1000 = 15.34 kip

4 √fc bw d = (4√3500 * 10 * 10) / 1000 = 23.66 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 10 / 2 = 5”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 10 / (15.34*1000) = 7.31 ”


Design of the beam: B1, B3,B4, B6, B7, B 9, B10, B11 ( at 4th story )


Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu/ фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0 .75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 87.25 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 87.25 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.551 in², a = Asfy / 0.85fc bw

a = 1.551 * 60 / 0.85 * 3.50 * 10 = 3.128 in.

As = 87.25 * 12/0.90 * 60 (13 – 3.128 / 2) = 1.695 in²

a = 1.695 * 60 / 0.85 * 3.50 * 10= 3.418 in.

As = 87.25 * 12 / 0.90 * 60 (10 – 3.418 / 2) = 1.717 in². Use – 2 # 7 + 2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 35.0 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.622 in², a = Asfy / 0.85f’c bw

a = 0.622 * 60 / 0.85 * 3.50 * 10 = 1.254 in.

As = 35.00 * 12 / 0.90 * 60 (13 – 1.254 / 2) = 0.628 in²

a = 0.628 * 60 / 0.85 * 3.50 * 10 = 1.266 in.

As = 35.0 * 12 / 0.90 * 60 (13 – 1.266/2) = 0.63 in². Use – 2 # 6 bars


Vs = Vu - φVc

= 22.23 – (2 * 0.85√3500 * 10 * 13) / 1000 = 9.15 kip

4 √fc bw d = (4√3500 * 10 * 13) /1000 = 30.76 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11* 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Sma = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (9.15 * 1000) = 15.94”




Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 108.93 * 12 / 0.90 * 0.0187 * 60 * 10(1- 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 108.93 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.936 in², a = Asfy / 0.85fc bw,

a = 1.936 * 60 / 0.85 * 3.50 * 10 = 3.904 in.

As = 108.93 * 12 / 0.90 * 60 (13 – 3.904 / 2) = 1.867 in²

a = 1.897 * 60 / 0.85 * 3.50 * 10 = 3.765 in.

As = 108.93 * 12 / 0.90 * 60 (13 – 3.765 / 2) = 2.177 in². Use – 2 # 8 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 59.56 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.058 in², a = Asfy / 0.85fc bw,

a = 1.058 * 60 / 0.85 * 3.50 * 10 = 2.133 in.

As = 59.56 * 12 / 0.90 * 60 (13 – 2.133 / 2) = 1.108 in²

a = 1.108 * 60 / 0.85 * 3.5 * 10 = 2.234 in.

As = 59.56 * 12 / 0.90 * 60 (13 – 2.234 / 2) = 1.113in². Use – 4 # 5 bars


Vs = Vu - φVc

= 26.90 – (2 * 0.85√3500 * 10 * 13) / 1000 = 13.825 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip


Stirrups Spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (13.825 * 1000) = 10.55”


Design of the beam: B14, B17, B20, B23 (at 4th story)


Maximum Moment:

End section:


Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 164.32 *12 / 0.90 * 0.0187 * 60 * 12 (1 – 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 164.32 * 1 / 0.90 * 60 (16 – 1 / 2) = 2.355 in², a = Asfy / 0.85fc bw,

a = 2.355 * 60 / 0.85 * 3.50 * 12 = 3.957 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 3.597 / 2) = 2.604 in²

a = 2.604 * 60 / 0.85 * 3.50 * 10 = 4.376 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 4.376 / 2) = 2.643 in². Use – 2 # 9 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 103.66 * 12 / 0.90 * 60 (16 – 1 / 2) = 1.49 in², a = Asfy / 0.85fc bw,

a = 1.49 * 60 / 0.85 * 3.5 * 10 = 3.00 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.59 in²

a = 1.59 * 60 / 0.85 * 3.50 * 10 = 3.206 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.206 / 2) = 1.60 in². Use – 2 # 7 +2 # 5 bars


Vs = Vu - φVc

= 40.84 – (2 * 0.85√3500 * 10 * 16) / 1000 = 24.78 kip

4 √fc bw d = (4√3500 * 10 * 16) / 1000 = 32.183 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φ Avfy d / Vs= 0.85 * 2 * 0.11 * 60000 * 16 / (24.78 * 1000) = 7.24”






Mu = ф ρfy bd² (1- 0.59 ρfy / f’c)

d² = Mu / (фρfy bd (1- 0.59 ρf / f’c)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 64.59 * 12 / 0.90 * 0.0187 * 60 * 10 (1-0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 64.59 * 12 / 0.90 * 60 (10 – 1 / 2) = 1.51 in², a = Asfy / 0.85f’c bw

a = 1.51 * 60 / 0.85 * 3.50 * 10 = 3.045

As = 64.59 * 12 / 0.90 * 60 (10 – 3.045 / 2) = 1.693 in²

a = 1.693 * 60 / 0.85 * 3.5 * 10 = 3.414 in.

As = 64.59 * 12 / 0.90 * 60 (10 – 3.414 / 2) = 1.73 in². Use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 51.98 * 12 / 0.90 * 60 (10 – 1 / 2) = 1.215 in², a = Asfy / 0.85f’c bw

a = 1.215 * 60 / 0.85 * 3.50 * 10 = 2.450 in.

As = 51.98 * 12 / 0.90 * 60 (10 – 2.45 / 2) = 1.316 in²

a = 1.316 * 60 / 0.85 * 3.50 * 10 = 2.609 in.

As = 51.98 * 12 / 0.90 * 60 (10 – 2.609 / 2) = 1.328 in². Use – 2 # 6 +2 # 5 bars


Vs = Vu - φVc

= 25.40 – (2 * 0.85√3500 * 10 * 10) / 1000 = 15.34 kip

4 √fc bw d = (4√3500 * 10 * 10) / 1000 = 23.66 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 10 / 2 = 5”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 10 / (15.34*1000) = 7.31 ”


Design of the beam: B1, B3,B4, B6, B7, B 9, B10, B11 ( at 4th story )


Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu/ фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0 .75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 87.25 * 12 / 0.90 * 0.0187 * 60 * 10 (1- 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 87.25 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.551 in², a = Asfy / 0.85fc bw

a = 1.551 * 60 / 0.85 * 3.50 * 10 = 3.128 in.

As = 87.25 * 12/0.90 * 60 (13 – 3.128 / 2) = 1.695 in²

a = 1.695 * 60 / 0.85 * 3.50 * 10= 3.418 in.

As = 87.25 * 12 / 0.90 * 60 (10 – 3.418 / 2) = 1.717 in². Use – 2 # 7 + 2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 35.0 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.622 in², a = Asfy / 0.85f’c bw

a = 0.622 * 60 / 0.85 * 3.50 * 10 = 1.254 in.

As = 35.00 * 12 / 0.90 * 60 (13 – 1.254 / 2) = 0.628 in²

a = 0.628 * 60 / 0.85 * 3.50 * 10 = 1.266 in.

As = 35.0 * 12 / 0.90 * 60 (13 – 1.266/2) = 0.63 in². Use – 2 # 6 bars


Vs = Vu - φVc

= 22.23 – (2 * 0.85√3500 * 10 * 13) / 1000 = 9.15 kip

4 √fc bw d = (4√3500 * 10 * 13) /1000 = 30.76 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11* 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Sma = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (9.15 * 1000) = 15.94”




Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 108.93 * 12 / 0.90 * 0.0187 * 60 * 10(1- 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 108.93 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.936 in², a = Asfy / 0.85fc bw,

a = 1.936 * 60 / 0.85 * 3.50 * 10 = 3.904 in.

As = 108.93 * 12 / 0.90 * 60 (13 – 3.904 / 2) = 1.867 in²

a = 1.897 * 60 / 0.85 * 3.50 * 10 = 3.765 in.

As = 108.93 * 12 / 0.90 * 60 (13 – 3.765 / 2) = 2.177 in². Use – 2 # 8 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 59.56 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.058 in², a = Asfy / 0.85fc bw,

a = 1.058 * 60 / 0.85 * 3.50 * 10 = 2.133 in.

As = 59.56 * 12 / 0.90 * 60 (13 – 2.133 / 2) = 1.108 in²

a = 1.108 * 60 / 0.85 * 3.5 * 10 = 2.234 in.

As = 59.56 * 12 / 0.90 * 60 (13 – 2.234 / 2) = 1.113in². Use – 4 # 5 bars


Vs = Vu - φVc

= 26.90 – (2 * 0.85√3500 * 10 * 13) / 1000 = 13.825 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip


Stirrups Spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (13.825 * 1000) = 10.55”


Design of the beam: B14, B17, B20, B23 (at 4th story)


Maximum Moment:

End section:


Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 164.32 *12 / 0.90 * 0.0187 * 60 * 12 (1 – 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 164.32 * 1 / 0.90 * 60 (16 – 1 / 2) = 2.355 in², a = Asfy / 0.85fc bw,

a = 2.355 * 60 / 0.85 * 3.50 * 12 = 3.957 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 3.597 / 2) = 2.604 in²

a = 2.604 * 60 / 0.85 * 3.50 * 10 = 4.376 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 4.376 / 2) = 2.643 in². Use – 2 # 9 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 103.66 * 12 / 0.90 * 60 (16 – 1 / 2) = 1.49 in², a = Asfy / 0.85fc bw,

a = 1.49 * 60 / 0.85 * 3.5 * 10 = 3.00 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.59 in²

a = 1.59 * 60 / 0.85 * 3.50 * 10 = 3.206 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.206 / 2) = 1.60 in². Use – 2 # 7 +2 # 5 bars


Vs = Vu - φVc

= 43.60 – (2 * 0.85√3500 * 10 * 16) / 1000 = 24.28 kip

4 √fc bw d = (4√3500 * 12 * 16) / 1000 = 45.435 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φAvfy d / Vs = 0.85 * 2 * .11 * 60000 * 16 / (24.28 * 1000) = 7.39 ”


Design of the beam: B13, B15, B16, B18, B19, B21, B22, B 24 (at 4th story)




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1- 0.59 ρf / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 115.36 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 13.0, Clear cover = 2”, Total depth = 13 + 2 = 15.0. Say, d = 15”



As = Mu / φfy (d – a / 2)

= 115.36 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.05 in², a = Asfy / 0.85fc bw.

a = 2.05 * 60 / 0.85 * 3.50 * 10 = 4.134 in

As = 115.36 * 12 / 0.90 * 60 (13 – 4.134 / 2) = 2.344 in²

a = 2.344 * 60 / 0.85 * 3.5 * 10 = 4.727 in.

As = 115.36 * 12 / 0.90 * 60 (13 – 4.727 / 2) = 2.41 in². Use – 2 # 8 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 48.69 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.865 in², a = Asfy / 0.85fc bw,

a = 0.865 * 60 / 0.85 * 3.50 * 10 = 1.744 in

As = 48.69 * 12 / 0.90 * 60 (13-1.744/2) = 0.892 in²

a = 0.892 * 60 / 0.85 * 3.50 * 10 = 1.798 in.

As = 48.69 * 12 / 0.90 * 60 (13 – 1.798 / 2) = 0.894 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 32.18 – (2 * 0.85√3500 * 10 * 13) / 1000 = 19.10 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * .11 * 60000 * 13 / (19.10 * 1000) = 7½ ”


Design of the beam: B1, B3, B4, B6, B7, B9, B10, B11 ( at 3rd story )


Maximum moment:

End section:


Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.50 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 87.25 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 11.29, Clear cover = 2”, Total depth = 11.29+2 = 13.29. Say, d = 15”



As = Mu / φfy (d – a / 2)

= 87.25 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.551 in², a = Asfy / 0.85fc bw,

a = 1.551 * 60 / 0.85 * 3.50 * 10 = 3.128 in.

As = 87.25 * 12 / 0.90 * 060 (13 – 3.128 / 2) = 1.695 in²

a = 1.695 * 60 / 0.85 * 3.5 * 10 = 3.418 in.

As = 87.25 * 12 / 0.90 * 60 (10 – 3.418 / 2) = 1.717 in². Use – 2 # 7 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 35.0 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.622 in², a = Asfy / 0.85f’c bw

a = 0.622 * 60 / 0.85 * 3.50 * 10 = 1.254 in.

As = 35.00 * 12 / 0.90 * 60 (13 – 1.254 / 2) = 0.628 in²

a = 0.628 * 60 / 0.85 * 3.50 * 10 = 1.266 in.

As = 35.0 * 2 / 0.90 * 60 (13 – 1.266 / 2) = 0.63 in². Use – 2 # 6 bars


Vs = Vu - φVc

= 22.23 – (2 * 0.85√3500 * 10 * 13) / 1000 = 9.15 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.76 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (9.15 * 1000) =15.94


Design of the beam: B2, B5, B8, B12 ( at 3rd story )




Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 108.93 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 12.63, Clear cover = 2”, Total depth = 12.63+2= 14.63. Say, d = 15



As = Mu / φfy (d – a / 2)

= 108.93 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.936 in², a = Asfy / 0.85fc bw.

a = 1.936 * 60 / 0.85 * 3.50 * 10 = 3.904 in.

As = 108.93 * 12 / 0.90 * 60 (13 – 3.904 / 2) = 1.867 in²

a = 1.897 * 60 / 0.85 * 3.5 * 10 = 3.765 in

As = 108.93 * 12 / 0.90 * 60 (13-3.765 / 2) = 2.177 in². Use – 2 # 8 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 59.56 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.058 in², a = Asfy / 0.85fc bw

a = 1.058 * 60 / 0.85 * 3.5 * 10 = 2.133 in.

As = 59.56 * 12 / 0.90 * 60 (13 – 2.133 / 2) = 1.108 in²

a = 1.108 * 60 / 0.85 * 3.50 * 10 = 2.234 in.

As = 59.56 * 10.90 * 6 (13 – 2.234 / 2) = 1.113in². Use – 4 # 5 bars


Vs = Vu - φVc

= 26.90 – (2 * 0.85√3500 * 10 * 13) / 1000 = 13.825 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip

Vs < 4√fc bw d. So, ok

Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (13.825 * 1000) = 10.55”


Design of the beam: B14, B17, B 20, B23 ( at 3rd story )




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 164.32 * 12 / 0.90 * 0.0187 * 60 * 12 (1-0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 164.32 * 12 / 0.90 * 60 (16 – 1 / 2) = 2.355 in², a = Asfy / 0.85fc bw

a = 2.355 * 60 / 0.85 * 3.50 * 12 = 3.957 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 3.597 / 2) = 2.604 in²

a = 2.604 * 60 / 0.85 * 3.5 * 10 = 4.376 in.

As = 164.32 * 12 / 0.90 * 60 (16 – 4.376 / 2) = 2.643 in². Use – 2 # 9 +5 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 103.66 * 12 / 0.90 * 60 (16 – 1 / 2) = 1.49 in², a = Asfy / 0.85fc bw

a = 1.49 * 60 / 0.85 * 3.5 * 10 = 3.00 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.59 in²

a = 1.59 * 60 / 0.85 * 3.50 * 10 = 3.206 in.

As = 103.66 * 12 / 0.90 * 60 (16 – 3.206 / 2) = 1.60 in². Use – 2 # 7 +2 # 5 bars


Vs = Vu - φVc

= 43.60 – (2 * 0.85√3500 * 10 * 16) / 1000 = 24.28 kip

4 √fc bw d = (4√3500 * 12 * 16) / 1000 = 45.435 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 16 / (24.28 * 1000) = 7.39 in


Design of the beam: B13, B15, B16, B18, B19, B21, B22, B24 (at 3rd story)




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (фρfy bd (1- 0.59 ρfy / f’c)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000+60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 115.36 * 12 / 0.90 * 0.0187 * 60 *10 (1- 0.59 * 0.0187 * 60 / 3.50)

d = 13.0, Clear cover = 2”, Total depth = 13+2 = 15.0. Say, d = 15”



As = Mu / φfy (d – a / 2)

= 115.36 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.05 in², a = Asfy / 0.85fc bw.

a = 2.05 * 60 / 0.85*3.5*10 = 4.134 in.

As = 115.36 * 12 / 0.90 * 60 (13 – 4.134 / 2) = 2.344 in.²

a = 2.344 * 60 / 0.85 * 3.5 * 10 = 4.727 in.

As = 115.36 * 12 / 0.90 * 60 (13 – 4.727 / 2) = 2.41 in². Use – 2 # 8 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 48.69 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.865 in², a = Asfy / 0.85fc bw,

a = 0.865 * 60 / 0.85 * 3.50 * 10 = 1.744 in.

As = 48.69 * 12 / 0.90 * 60 (13 – 1.744 / 2) = 0.892 in²

a = 0.892 * 60 / 0.85 * 3.5 * 10 = 1.798 in

As = 48.69 * 12/0.90 * 60 (13 – 1.798 / 2) = 0.894 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 32.18 – (2 * 0.85√3500 * 10 * 13) / 1000 = 19.10 kip

4 √fc bw d = (4√3500 *10 * 13) / 1000 = 30.763kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * .11 * 60000 * 13 / (19.10 * 1000) = 7½”


Design of the beam: B1, B3, B4, B6, B7, B9, B10, B11 ( at 2nd story )




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 89.00 * 12 / 0.90 * 0.0187 * 60 * 12 (1 – 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 89.00 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.582 in², a = Asfy / 0.85fc bw

a = 1.582 * 60 / 0.85 * 3.50 * 12 = 2.658 in.

As = 89.0 * 12 / 0.90 * 60 (13 – 2.658 / 2) = 1.694 in²

a = 1.694 * 60 / 0.85 * 3.50 * 10 = 2.847 in.

As = 89.0 * 12 / 0.90 * 60 (13 – 2.847 / 2) = 1.708 in². Use – 4 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 42.26 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.751 in², a = Asfy / 0.85fc bw

a = 0.751 * 60 / 0.85 * 3.50 * 10 = 1.514 in.

As = 42.26 * 12 / 0.90 * 60 (13 – 1.514 / 2) = 0 .767 in²

a = 0.767 * 60 / 0.85 * 3.50 * 10 = 1.546 in.

As = 42.26 * 12 / 0.90 * 60 (13 – 1.546 / 2) = 0.768 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 23.17 – (2 * 0.85√3500 * 10 * 13 / 1000 = 10.10 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.76 kip


Stirrup spacing:

1) Smax = Avfy/ 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (10.10 * 1000) = 14.44”


Design of the beam: B2, B5, B8, B12 ( at 2nd story )




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187.

d² = 114.44 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50).




As = Mu / φfy (d – a / 2)

= 114.44 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.034 in², a = Asfy / 0.85fc bw

a = 2.034 * 60 / 0.85 * 3.50 * 10 = 4.102 in.

As = 114.44 * 12 / 0.90 * 60 (13 – 4.102 / 2) = 2.322 in²

a = 2.322 * 60 / 0.85 * 3.50 * 10 = 4.683 in.

As = 114.44 * 12 / 0.90 * 60 (13 – 4.683 / 2) = 2.386 in². Use – 2 # 9 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 60.52 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.075 in², a = Asfy / 0.85fc bw

a = 1.075 * 60 / 0.85 * 3.50 * 10 = 2.168 in.

As = 60.52 * 12 / 0.90 * 60 (13 – 2.168 / 2) = 1.128 in²

a = 1.128 * 60 / 0.85 * 3.50 * 10 = 2.274 in.

As = 60.52 * 12 / 0.90 * 60 (13 – 2.274 / 2) = 1.133 in². Use – 4 # 5 bars


Vs = Vu – φVc

= 24.47 – (2 * 0.85√3500 * 10 * 13) / 1000 = 17.412 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip


Stirrups Spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (17.412 * 1000) = 8.37”


Design of the beam: B14, B17, B20, B23 ( at 2nd story )




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 169.91 * 12 / 0.90 * 0.0187 * 60 * 12 (1-0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 169.91 * 12 / 0.90 * 60 (16 – 1 / 2) = 2.43 in², a = Asfy / 0.85fc bw

a = 2.43 * 60 / 0.85 * 3.50 * 12 = 4.08 in.

As = 169.91*12 / 0.90 * 60 (16 – 4.08 / 2) = 2.70 in²

a = 2.70 * 60 / 0.85 * 3.50 * 12 = 4.54 in.

As = 169.91 * 12 / 0.90 * 60 (16 – 4.54 / 2 = 2.75 in². Use – 2 # 9 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 104.10 *12 / 0.90 * 60 (16 – 1 / 2) = 1.49 in², a = Asfy / 0.85fc bw

a = 1.49 * 60 / 0.85 * 3.50 * 10 = 3.00 in.

As = 104.10 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.59 in²

a = 1.59 * 60 / 0.85 * 3.50 * 10 = 3.206 in.

As = 104.10 * 12 / 0.90 * 60 (16 – 3.206 / 2) = 1.60 in². Use – 2 # 7 +2 # 5 bars


Vs = Vu - φVc

= 44.19 – (2 * 0.85√3500 * 12 * 16) / 1000 = 24.88 kip

4 √fc bw d = (4√3500 * 12 * 16) / 1000 = 45.435 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φAvfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 16 / (24.88 * 1000) =7.21 ”


Design of the beam: B13, B15, B16, B18, B19, B21, B22, B24 (at 2nd story)




Mu = ф ρfy bd² (1- 0.59 ρfy / fc)

d² = Mu / (фρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 114.84 *12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 12.97, Clear cover = 2”, Total depth = 12.97+2= 14.97

Say, d = 15”



As = Mu / φfy (d – a / 2)

= 114.84 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.041 in², a = Asfy / 0.85fc bw.

a = 2.041*60 / 0.85*3.5*10 = 4.116 in.

As = 114.84 * 12 / 0.90 * 60 (13 – 4.116 / 2) = 2.332 in²

a = 2.332 * 60 / 0.85 * 3.5 * 10 = 4.703 in.

As = 114.84 * 12 / 0.90 * 60 (13 – 4.703 / 2) = 2.396 in². Use – 2 # 8 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 51.04 * 12 / 0.90 * 60 (13 – 1 / 2) = 0 .907 in², a = Asfy / 0.85fc bw,

a = 0.907 * 60 / 0.85 * 3.5 * 10 = 1.829 in.

As = 51.04 * 12 / 0.90 * 60 (13 – 1.829 / 2) = 0.938 in²

a = 0.938 * 60 / 0.85 * 3.50 *10 = 1.891 in.

As = 51.04 * 12 / 0.90 * 60 (13 – 1.891 / 2) = 0.941 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 32.71- (2 * 0.85√3500 * 10 * 13) / 1000 = 18.92 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (18.92 * 1000) = 7.71”


Design of the beam: B1, B4, B6, B7, B9, B10, B11 ( at 1st story )




Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1- 0.59 ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 f’c / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 89.00 * 12 / 0.90 * 0.0187 * 60 * 12 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 10.42, Clear cover = 2”, Total depth = 10.42+2 =12.42. Say, d = 15”

Provided Beam size = 15” * 12”, d = 15”- 2” = 13”


As = Mu / φfy (d – a / 2)

= 89.00 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.582 in², a = Asfy / 0.85fc bw.

a = 1.582 * 60 / 0.85 * 3.5 * 12 = 2.658 in.

As = 89.0 * 12 / 0.90 * 60 (13 – 2.658 / 2) = 1.694 in²

a = 1.694 * 60 / 0.85 * 3.50 * 10 = 2.847 in.

As = 89.0 * 12 / 0.90 * 60 (13 – 2.847 / 2) = 1.708 in². Use – 4 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 42.26 * 12 / 0.90 * 60 (13 – 1 / 2) = 0.751 in², a = Asfy / 0.85fc bw.

a = 0.751 * 60 / 0.85 * 3.5 * 10 = 1.514 in.

As = 42.26 * 12 / 0.90 * 60 (13 – 1.514 / 2) = 0.767 in²

a = 0.767 * 60 / 0.85 * 3.50 * 10 = 1.546 in

As = 42.26 * 12 / 0.90 * 60 (13 – 1.546 / 2) = 0.768 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 23.17 – (2 * 0.85√3500 * 10 * 13) / 1000 = 10.10 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.76 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw= (2 * 0.11 * 60000 ) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φAvfy d / Vs = 0.85 * 0.11 * 60000 * 13 / (10.10 * 1000) =14.44”


Design of the beam: B2, B5, B8, B12 ( at 1st story )




Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / ф ρfy bd (1 – 0.59 ρfy / fc)

ρ max = 0.75 ρb , ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρmax = 0.75 * 0.02494 = 0.0187

d² = 114.44 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)

d = 12.95, Clear cover = 2”, Total depth = 12.95+2 = 14.95. Say, d = 15 “



As = Mu / φfy (d – a / 2)

= 114.44 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.034 in², a = Asfy / 0.85fc bw

a = 2.034 * 60 / 0.85 * 3.5 * 10 = 4.102 in.

As = 114.44 * 12 / 0.90 * 60 (13 – 4.102 / 2) = 2.322 in²

a = 2.322 * 60 / 0.85 * 3.5 * 10 = 4.683 in

As = 114.44 * 12 / 0.90 * 60 (13 – 4.683 / 2) = 2.386 in². Use – 2 # 9 +2 # 5 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 60.52 * 12 / 0.90 * 60 (13 – 1 / 2) = 1.075 in², a = Asfy / 0.85fc bw

a = 1.075 * 60 / 0.85 * 3.5 * 10 = 2.168 in.

As = 60.52 * 12 / 0.90 * 60 (13 – 2.168 / 2) = 1.128 in²

a = 1.128 * 60 / 0.85 * 3.50 * 10 = 2.274 in

As = 60.52 * 12 / 0.90 * 60 (13 – 2.274 / 2) = 1.133 in². Use – 4 # 5 bars


Vs = Vu – φVc

= 24.47 – (2 * 0.85√3500 * 10 * 13) / 1000 = 17.412 kip

4 √fc bw d = (4√3500 * 10 * 13) / 1000 = 30.763 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d / Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (17.412 * 1000) = 8.37”


Design of the beam: B14, B17, B20, B23 ( at 1st story )


Maximum moment at mid section


Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / (фρfy bd (1 – 0.59 ρfy / fc)

ρ max = 0.75 ρb , ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 169.91 * 12 / 0.90 * 0.0187 * 60 * 12 (1- 0.59 * 0.0187 * 60 / 3.50)

d =14.40, Clear cover = 2”, Total depth = 14.40+2= 16.40. Say, d = 18”



As = Mu / φfy (d – a / 2)

= 169.91 * 12 / 0.90 * 60 (16 -1 /2) = 2.43 in², a = Asfy / 0.85fc bw.

a = 2.43 * 60 / 0.85 * 3.5 * 12 = 4.08

As = 169.91 * 12 / 0.90 * 60 (16 – 4.08 / 2) = 2.70 in²

a = 2.70 * 60 / 0.85 * 3.5 * 12 = 4.54 in.

As = 169.91 * 12 / 0.90 * 60 (16 – 4.54 / 2) = 2.75 in². Use – 2 # 9 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 104.10 * 12 / 0.90 * 60 (16 – 1 / 2) = 1.49 in², a = Asfy / 0.85f’c bw

a = 1.49 * 60 / 0.85 * 3.5 * 10 = 3.0 in

As = 104.10 * 12 / 0.90 * 60 (16 – 3.00 / 2) = 1.59 in²

a = 1.59 * 60 / 0.85 * 3.50 *10 = 3.206 in.

As = 104.10 * 12 / 0.90 * 60 (16 – 3.206 / 2) = 1.60 in². Use – 2 # 7 +2 # 5 bars


Vs = Vu - φVc

= 44.19 – (2 * 0.85√3500 * 12 * 16) / 1000 = 24.88 kip

4 √fc bw d = (4√3500 * 12 * 16) / 1000 = 45.435 kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0 .11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 16 / 2 = 8”

3) Smax = 24”

4) S = φ Avfyd / Vs = 0.85 * 2 * 0.11 * 60000 * 16 / (24.88 * 1000) = 7.21 ”


Design of the beam: B13, B15, B16, B18, B19, B21, B22, B24 (at 1st story)




Mu = ф ρfy bd² (1 – 0.59 ρfy / fc)

d² = Mu / (ф ρfy bd (1 – 0.59ρfy / fc)

ρ max = 0.75 ρb, ρb = 0.85 β1 fc / fy * 87000 / (87000 + 60000)

= 0.85 * 0.85 * 3.5 / 60 * 87000 / (87000 + 60000) = 0.02494

ρ max = 0.75 * 0.02494 = 0.0187

d² = 114.84 * 12 / 0.90 * 0.0187 * 60 * 10 (1 – 0.59 * 0.0187 * 60 / 3.50)




As = Mu / φfy (d – a / 2)

= 114.84 * 12 / 0.90 * 60 (13 – 1 / 2) = 2.041 in², a = Asfy / 0.85fc bw

a = 2.041 * 60 / 0.85 * 3.5 * 10 = 4.116 in.

As = 114.84 * 12 / 0.90 * 60 (13 – 4.116 / 2) = 2.332 in²

a = 2.332 * 60 / 0.85 * 3.5 * 10= 4.703 in.

As = 114.84 * 12 / 0.90 * 60 (13 – 4.703 / 2) = 2.396 in². Use – 2 # 8 +2 # 6 bars


Mid section:

As = Mu / φfy (d – a / 2)

= 51.04 * 12 / 0.90 * 60 (13 – 1 / 2) = 0 .907 in², a = Asfy / 0.85fc bw

a = 0.907 * 60 / 0.85 * 3.5 * 10 = 1.829 in.

As = 51.04 * 12 / 0.90 * 60 (13 – 1.829 / 2) = 0.938 in²

a = 0.938 * 60 / 0.85 * 3.50 * 10 = 1.891 in.

As = 51.04 * 12 / 0.90 * 60 (13 – 1.891 / 2) = 0.941 in². Use – 3 # 5 bars


Vs = Vu - φVc

= 32.71 – (2 * 0.85√3500 * 10 * 13) / 1000 = 18.92 kip

4 √fc bwd = (4√3500 *10 * 13) / 1000 = 30.763kip


Stirrup spacing:

1) Smax = Avfy / 50 bw = (2 * 0.11 * 60000) / 50 * 10 = 26.40”

2) Smax = d / 2 = 13 / 2 = 6½”

3) Smax = 24”

4) S = φ Avfy d/ Vs = 0.85 * 2 * 0.11 * 60000 * 13 / (18.92 * 1000) = 7.71 ”


The table below is showing cross section of different floor beams at their end sections and mid

section.

Column design for beam supported structure.

Design of the Column: C1, C4, C10, C16


Pu = 329.81 Kips = 329.81*1.20 = 396 Kips

Now, Pu = α ф Ag (0 .85 f’c (1- ρg) + ρg fy)

Or 396 = 0.80 * 0.65 * Ag (0.85 * 3.5 (1- 0.025) + 0.025 * 60)

Or 396 = 2.288 Ag Or Ag = 173.076 in2 Or Ag = 10” * 18”

Provided Column size = 10” * 18”, Ag = 180 in2


Pu = α ф (0 .85 f’c (Ag - A) + Asfy)

Or 396 = 0.80 * 0.65 (0.85 * 3.5 (180 – As) + As * 60)

Or 396 = 278.46 – 1.547As + 31.20 As. Or As = 3.963 in2

Use 4 # 8 bars +2 # 6 bars, As = 4.04 in2

Tie design:

Spacing: S = 16 D = 16 * 6 / 8 = 12” c/c, S = 48 d = 48 * 3 / 8 = 18” c/c

At least lateral dimension, S = 10” c/c. Use # 3 bar ties @ 10” c/c.

Y – Axis:

γ = 13 / 18 = 0.72, ex = Mx / P = 56.093 * 12 / 329.81= 2.04 in

ey = My / P = 1.78 * 12/329.81 = 0.06 in, ex / h = 2.04 / 18 = 0.113

Reinforcement Ratio: ρg =As / A g = 4.04 / 180 = 0.022

From Graph: Pn yo / f′c Ag = 0.91

Or Pn yo = 0.91 * 3.50 * 180 = 573.30 kips

Po / f′c Ag = 1.11 Or Po = 1.11 * 3.50 * 180 = 699.30 kips

X – Axis:

γ = 5 / 10 = 0.5, ex = Mx / P = 56.093 * 12/329.81= 2.04 in

ey = My / P = 1.78*12 / 329.81 = 0.06 in, ey / h = 0.06 / 10 = 0.006

Reinforcement ratio: ρg =As / A g = 4.04 / 180 = 0.022

From Graph: Pn xo / f′c Ag = 1.02

Or Pn xo = 1.02 * 3.50 * 180 = 642.60 kips

Po /f′c Ag = 1.11 Or Po = 1.11 * 3.50 * 180 = 699.30 kips

Here – 1 / Pn = 1 / Pnxo + 1 / Pnyo – 1 / Po

Or 1 / Pn = 1 / 642.60 + 1 / 573.30 – 1 / 699.30. Or Pn = 534.64 kips

Now, Pu = ф Pn. Or Pu = 0.65 * 534.64

Or Pu = 347.516 kips > 329.81 kips. So design is ok.

Design of the Column : C2, C3, C5, C 8, C9, C12, C14, C15


Pu = 532.65 Kips = 532.65 * 1.20 = 639.18 Kips


Or 639.18 = 0.80 * 0.65 * Ag (0.85 * 3.5 (1- 0.025) + 0.025 * 60)

Or 639.18 = 2.288 Ag or Ag = 279.36 in2 or Ag= 12” *24”



Pu = α ф (0 .85 f′c (Ag - As) + Asfy)

Or 639.18 = 0.80 * 0.65 (0.85 * 3.5 (288 – As) + As * 60)

Or 639.18 = 445.54 – 1.547 As + 31.20 As

Or As = 6.53 in2 Use 4 # 8 bars + 8 # 6 bars As = 6.68 in2

Tie design:


At least lateral dimension: S = 12” c/c, Use # 3 bar ties @ 12” c/c.

Y – Axis:

γ = 19 / 24 = 0.79, ex = Mx / P = 54.898 * 12 / 532.65 = 1.24 in

ey = My / P = 1.534 * 12 / 532.65 = 0.035 in, ex / h = 1.24 / 24 = 0.052

Reinforcement ratio: ρg = As / A g = 6.68 / 288 = 0.023


Or Pn yo = 0.98 * 3.50 * 288 = 987.84 kips

Po / f′c Ag = 1.13 Or Po = 1.13 * 3.50 * 288 = 1139.04 kips

X – Axis: γ = 7 / 12 = 0.58, ex = Mx / P = 54.898 * 12 / 532.65 = 1.24 in

ey = My / P = 1.534 * 12 / 532.65 = 0.035 in, ey / h = 0.035 / 12 = 0.0029

Reinforcement Ratio: ρg = As / A g = 6.68 / 288 = 0.023


Or Pn xo = 1.13 * 3.50 * 288 = 1139.04 kips

Po / f′c Ag = 1.0 Or Po = 1.05 * 3.50 * 288 = 1058.40 kips


Or 1 / Pn = 1 / 1058.40 + 1 / 987.84 – 1 / 1139.04 or Pn = 926.61 kips

Now, Pu = ф Pn or Pu = 0.65 * 926.61

Or Pu = 602.30 kips > 532.65 kips, so design is ok.

Design of the Column : C6, C7, C11, C13


Pu = 865.56 Kips = 865.56 * 1.20 = 1038.67 Kips


Or 1038.67 = 0.80*0.65 * Ag (0.85 * 3.5 (1- 0.025) + 0.025 * 60)

Or 1038.67 = 2.288 Ag or Ag = 454 in2 or Ag = 18” * 26”



Pu = α ф (0 .85 f’c (Ag - As) + Asfy)

Or 1038.67 = 0.80 * 0.65 (0.85 * 3.5 (468 – As) + As * 60)

Or 1038.67 = 723.99– 1.547As + 31.20 As

Or As = 10.61 in2 use 4 # 10 bars + 6 # 9 bars As = 11.08 in2

Tie design:

Spacing: S = 16 D = 16 * 9/8 = 18” c/c, S = 48 d = 48 * 3/8 = 18” c/c

At least lateral dimension: S = 18” c/c, use # 3 bar ties @ 18” c/c.

Y – Axis:

γ = 21 / 26 = 0.80, ex = Mx / P = 0.971 * 12 / 865.56 = 0.013 in

ey = My / P = 61.64 * 12 / 865.56 = 0.85 in, ex / h = 0.013 / 26 = 0.0005

Reinforcement ratio: ρg = A s / A g = 11.08 / 468 = 0.024


Or Pn yo = 1.15 * 3.50 * 468 = 1883.70 kips

Po / f′c Ag = 1.15 or Po = 1.15 * 3.50 * 468 = 1883.70 kips

X – Axis:

γ = 13 / 18 = 0.72, ex = Mx / P = 0.013 in, ey = My / P = 0.85 in

ey / h = 0.85 / 18 = 0.047

Reinforcement ratio: ρg = As / A g = 0.024


Or Pn xo = 1.00 * 3.50 * 468 = 1638 kips

Po / f′cAg = 1.15 or Po = 1.15 * 3.50 * 468 = 1883.70 kips


Or 1/Pn = 1 / 1638 + 1 / 1883.70 – 1 / 1883.7 or Pn = 1638 kips

Now, Pu = ф Pn, or Pu = 0.65 * 163

The table below is showing size of different columns of the beam supported structure.

Table 3.16: Cross -section of the Column of the beam supported structure

Column design for flat plate structure



Pu = 232.26 Kips = 232.26 * 2.0 = 464.52 Kips


Or 464.52 = 0.80 * 0.65 * Ag (0.85 * 3.50 (1- 0.03) + 0.03 * 60)

Or 464.52 = 2.44 Ag Or Ag = 190.40 in2. Or Ag = 13.79” *13.79”

Provided Column size = 15” *15”, Ag = 225 in


Pu = α ф (0 .85 fc (Ag - As) + Asfy)

Or 464.52 = 0.80 * 0.65 (0.85 * 3.50 (225 – As) + As * 60)

Or 464.52 = 348.70 – 1.547As + 31.20 As. Or As = 3.92 in2

Use 4 # 9 bars, As = 4.00 in2

Tie design:

Spacing: S = 16 D = 16 * 9/8 = 18” c/c, S = 48 d = 48 * 3 / 8 = 18” c/c

At least lateral dimension: S = 15” c/c, use # 3 bar ties @ 15” c/c.

Y – Axis:

γ = 10 / 15 = 0.67, ex = Mx / P = 94.34 * 12/232.26 = 4.87 in

ey = My / P = 3.293 * 12/232.26 = 0.17 in, ex / h = 4.87 / 15 = 0.32

Reinforcement ratio:

ρg = As / A g = 4.00 / 225 = 0.018

From Graph,

Pn yo / f′c Ag = 0.53

Or Pn yo = 0.53 * 3.50 * 225 = 417.37 kips, Po/ f′c Ag = 1.09

Or Po = 1.09 * 3.50 * 225 = 858.37 kips

X – Axis:


ey / h = 0.17 / 15 = 0.011

Reinforcement Ratio: ρg = As / A g = 0.018

From Graph:

Pn xo / f′c Ag = 1.06 Or Pn xo = 1.06 * 3.50 * 225 = 834.75 kips

Po / f′c Ag =1.09 Or Po = 1.09 * 3.50 * 225 = 858.37 kips

Here – 1/ Pn = 1 / Pnxo + 1 / Pnyo – 1 / Po

Or 1/Pn = 1 / 834.75+ 1 / 417.37– 1 / 858.37 Or Pn = 411.70 kips

Now, Pu = ф Pn. Or Pu = 0.65 * 411.70


Design of the Column : C2, C3, C5, C8, C9, C12, C14, C15


Pu = 469.38 Kips = 469.38 * 1.20 = 563.25 Kips


Or 563.25 = 0.80 * 0.65 * Ag (0.85 * 3.50 (1- 0.025) + 0.025 * 60)

Or 563.25 = 2.288 Ag Or Ag = 246.18 in2

Or Ag =12” *24”

Provided Column size =12” *24”

Ag = 288 in2


Pu = α ф(0 .85 f’c (Ag - As) + Asfy)

Or 563.25 = 0.80 * 0.65 (0.85 * 3.50 (288 – As) + As * 60)

Or 563.25 = 445.54 – 1.547As + 31.20 As Or As = 3.97 in2

Use 4 # 8 bars + 2 # 6 bars, As = 4.04 in2

Tie design:


At least lateral dimension, S = 12” c/c

Use # 3 bar ties @ 12” c/c.

Y – Axis:

γ = 19 / 24 = 0.79, ex = Mx / P = 94.668 * 12 / 469.38 = 2.42 in

ey = My / P = 2.533 * 12 / 469.38 = 0.065 in, ex / h = 2.42 / 24 = 0.10

Reinforcement ratio:

ρg = As / A g = 4.04 / 288 = 0.014


Or Pn yo = 0.86 * 3.50 * 288 = 866.88 kips

Po / f′c Ag = 1.04 Or Po = 1.04 * 3.50 * 288 = 1048.32 kips

X – Axis:


ey / h = 0.85 / 18 = 0.047

Reinforcement ratio: ρg =As / A g = 0.014


Or Pn xo = 1.04 * 3.50 * 288 = 1048.32 kips

Po / f′c Ag = 1.04 Or Po = 1.04 * 3.50 * 288 = 1048.32 kips


Or 1 / Pn = 1 / 1048.32+ 1 / 866.88– 1 / 1048.32

Or Pn = 869.56 kips

Now, Pu = ф Pn. Or Pu = 0.65 * 869.56




Pu = 1075.60 Kips = 1075.60 * 1.40 = 1505.84 Kips


Or 1505.84 = 0.80 * 0.65 * Ag (0.85 * 3.50 (1- 0.03) + 0.03 * 60)

Or 1505.84 = 2.44 Ag Or Ag = 617.15 in Or Ag = 22” * 30”

Provided Column size = 22” * 30”

Ag = 660 in2


Pu = α ф (0 .85 f’c (Ag - As) + Asfy)

Or 1505.84 = 0.80 * 0.65 (0.85 * 3.50 (660 – As) + As * 60)

Or 1505.84 =1021.02 – 1.547As + 31.20 As

Or As = 16.35 in2

Use 4 # 10 bars + 12 # 9 bars, As = 17.08 in2

Tie design:


At least lateral dimension, S = 22” c/c, Use # 3 bar ties @ 18” c/c.

Y – Axis:

γ = 25 / 30 = 0.83, ex = Mx / P = 4.06 * 12 / 1075.60 = 0.045 in

ey = My / P = 135.407 * 12 / 1075.60 = 1.51 in

ex / h = 0.045 / 30 = 0.0015

Reinforcement ratio: ρg = As / A g = 17.08 / 660 = 0.026


Or Pn yo = 1.108 * 3.50 * 660 = 2559.48 kips

Po / f′c Ag = 1.20 Or Po = 1.20 * 3.50 * 660 = 2772 kips

X – Axis:


ey / h = 1.51 / 22 = 0.0686

Reinforcement ratio: ρg = As / A g = 0.026


Or Pn xo = 1.09 * 3.50 * 660 = 2517.90 kips

Po/ f′c Ag = 1.20

Or Po = 1.20 * 3.50 * 660 = 2772 kips


Or 1 / Pn = 1 / 2517.90 + 1 / 2559.48 – 1 / 2772

Or Pn = 2341.31kips

Now, Pu = ф Pn. Or Pu = 0.65 * 2341.31


The table below is showing size of different columns of the flat plate structure

Table 3.17: Cross section of the column elements for flat plate structure.

COMPARATIVE STUDY:

General:

The buildings are analyzed for determine the best condition which can be constructed

economically. The behavior of each condition is observed carefully. Then suitable condition is

selected and suitable dimensions as well as steel area are also determined. The chapter gives a

comparative of the material, dimensions etc required for both the options I building and potion II

building. It is obvious that the amount of material required for option I building will be higher than

the option II building. But the option I building got beam free floor height. Due to this beam free

space, looks nice and the floor will be heavily reinforced and it will be durable.

Comparison Between the Two Types of Structure:

The comparison is made in terms of element dimension, volume of concrete and steel

requirement, their cost and economic view.

Dimension of different parts of beam supported structure and flat pate structure:

Table 4.1: Dimension of different portion of the flat plate structure and beam-supported

structure.

Element Option I building

(Beam supported structure)

Option II building

(Flat plate structure)

Column no There are 16 columns at each floor. So the total nos

of columns of this six storied building is 16.

There are 16 columns at each floor. So the total nos

of columns of this six storied building is 16.

Table 4.1: Dimension of different portion of the flat plate structure and beam-supported structure

(continued…).



Option II building


Column size All the columns are:

1. C1,C4,C10,C16:

Size = 18” * 10”

2. C2,C3,C5,C8,C9,C12,C14,C15

Size = 24” * 12”

3. C6,C7,C11,C13:

Size = 26” * 18”

All the columns are:

1. C1,C4,C10,C16:

Size = 15” * 15”

2. C2,C3,C5,C8,C9,C12,C14,C15

Size = 24” * 12”

3. C6,C7,C11,C13:

Size = 30” * 22”

Column

height

All floor columns height are 10 ft All floor columns height are 10 ft

Beams Slab with beams Slab without beams

Slab

thickness

All floor slab thickness are 6″ All floor slab thickness are 7.5″

Required concrete volume for beam supported structure and flat plate structure:

Table 4.2: Comparative concrete volumes.



Option II building


Column Column: C1, C4, C10, C16= 4 * 6 * 0’- 10” * 1’- 6” *

10’- 0” = 333.12 cft.Column: C2, C3, C5, C8, C9,

Column: C1, C4, C10, C16= 4 * 6 * 1’- 3” * 1’- 3” * 10’- 0” =

375.00 cftColumn: C2, C3, C5, C8, C9, C12, C14, C15= 8 * 6 *

C12, C14, C15= 8 * 6 * 1’- 0” * 2’- 0 * 10’- 0” =

960.00 cft

Column: C6, C7, C11, C13

= 4 * 6 * 1’- 6” * 2’- 2” * 10’- 0” =

779.76 cft

Total R.C.C = 2072.88 cft.

1’- 0” * 2’- 0 * 10’- 0” = 960.00 cft


= 4 * 6 * 1’- 10” * 2’- 6” * 10’- 0” = 1099.80 cft

Total R.C.C = 2434.80 cft

Table 4.2: Comparative concrete volumes (continued…)



Option II building


Beams Beam: B1, B3, B4, B6, B7, B9, B10, B11= 1 * 10” * 10” * 13’ – 7” * 8=

75.46 cftBeam: B2, B5, B8, B12

= 1 * 15” * 10” * 18’ – 0” * 4

= 75.00 cft

Beam: B14, B17, B20, B23

= (1 * 18” * 10” * 17’-10” * 2) + (1 * 18”10” * 18’-5” * 2)

= 90.83 cft

Beam: B13, B15, B16, B18, B19, B21, B22, B24

= 1 * 12” * 10” * 13’-5” * 8

= 89.46 cft

Total R.C.C = 75.46 + 75.00 + 90.83 + 89.46 = 330.75

cft

None

Table 4.2: Comparative concrete volumes (continued…).



Option II building


Beams (5th + 6th) Floor R.C.C= 2 * 330.75 = 661.50 cftR.C.C Beam at 4 th Story: Beam:

B1, B3, B4, B6, B7, B9, B10, B11

= 1 * 15” * 10” * 13’ – 7” * 8

= 113.17 cft

Beam: B2, B5, B8, B12

=1 * 15” * 10” * 18’ – 0” * 4

None

= 75.00 cft

Beam: B14, B17, B20, B23

= (1 * 18” * 10” * 17’- 10” * 2) + (1 * 18” *10” * 18’- 5” * 2)

= 90.83 cft

Beam: B13, B15, B16, B18, B19, B21, B22, B24

= 1 * 15” * 10” * 13’-5” * 8

= 113.17 cft

Total R.C.C = 113.17 + 75.00 + 90.83 + 113.17 = 392.17 cft

(3rd + 4TH) Floor R.C.C = 2 * 392.17 = 784.34 cft

R.C.C Beam at 2 nd Story:

Beam: B1, B3, B4, B6, B7, B9, B10, B11

= 1 * 15” * 12” * 13’- 7” * 8

= 135.80 cft

Beam: B2, B5, B8, B12

= 1 * 15” * 10” * 18’- 0” * 4

= 75.00 cft

Beam: B14, B17, B20, B23

Table 4.2: Comparative concrete volumes (continued..).



Option II building


Beam = (1 * 18” * 12”*17’- 10” * 2) + (1 * 18” * 12” * 18’-5” *

2)= 108.99 cftBeam: B13, B15, B16, B18, B19, B21, B22,

B24= 1 * 15” * 10” * 13’-5” * 8

= 113.17 cft

Total R.C.C = 135.80 + 75.00 + 108.99 +

113.17 = 432.96 cft

(1st + 2nd) Floor R.C.C = 2 * 432.96 =

865.92 cft

Total R.C.C = 135.80 + 75.00 + 108.99 +

113.17 = 432.96 cft

(1st + 2nd) Floor R.C.C = 2 * 432.96 =

865.92 cft

Total R.C.C of beam: 661.50 + 784.34 +

None

865.92 = 2311.76 cft.

Slab Concrete in slab for one story:1 * 50.83 * 50.83 * (6 /

12)= 1292 cftConcrete in slab for six stories:

1 * 1292*6 = 7752 cft

Concrete in slab for one story:1 * 51.33 * 51.33 * 7.5

/ 12= 1647 cft.Concrete in slab for six stories:

1 * 1647 * 6 = 9882 cft

Required steel volume for beam supported structure and flat plate structure.

Table 4.3: Comparative steel volumes.



Option II building


Column Column: C1, C4, C10, C164 # 9 bars = 4 * 4 * 65’- 6”=

1048 rft# 3 bars @ 15” c/c

= 49 * 4 * 4’- 4”= 849.26 rft

Column: C2, C3, C5, C8, C9, C12, C14, C15

4 # 8 bars = 4 * 8 * 65’- 0”

= 2080 rft

2 # 6 bars = 2 * 8 * 64’ – 0”

= 1024 rft

# 3 bars @ 12” c/c = 61 * 8 * (5’- 4” + 1’-

1”) = 3131 rft


4 # 10 Bars = 4 * 4 * 66’- 4”

= 1061.28 rft

12 # 9 bars = 12 * 4 * 65’- 6”

= 3144 rft

# 3 bars @ 18” c/c = 41 * 4 * (8’- 0”+ 5’-

10”+ 4’- 5”) = 2992.84 rft

# 8 Bars = 2080 rft = 2444 kg

# 6 bars = 4608 rft = 3464 kg

# 9 Bars = 4192 rft = 6180.64 kg

Column: C1, C4, C10, C164 # 8Bars = 4 * 4 * 65’- 0”=

1040 rft2 # 6 Bars = 2 * 4 * 64’- 0”

= 512 rft

# 3 bars @ 10” c/c

= 4 * (73 * 4’- 0” + 73 * 0’- 11”)

= 1435.48 rft

Column: C2, C3, C5, C8, C9, C12, C14, c15

4 # 8 Bars = 4 * 8 * 65’- 0”

= 2080 rft

8 # 6 Bars = 8 * 8 * 64’- 0”

= 512 rft

= 4096 rft

# 3 Bars @ 12” c/c

= 61 * 8 * (5’- 4”+3’- 3”+1’- 11”) = 5123.67

rft

Table 4.3: Comparative steel volumes (continued..).



Option II building


Column # 10 Bars = 1061.28 Rft= 2043 kg# 3 Bars = 6973.10 Rft =

1309.58 kgTotal M.S Rod = 11977.00 kgTotal M.S Rod

=11977.00 kg

= 11.98 ton

Column: C6, C7, C11, C134 # 10 Bars= 4 * 4 * 66’-

4” = 1061.28 rft6 # 9 bars

= 6 * 4 * 65’- 6”= 1572 rft

# 3 bars @ 18” c/c

= 41* 4 * (6’- 8”+2’- 3”+ 4’- 2”) =

2145.55 rft

# 8 bars = 3120 rft = 3666 kg

# 6 bars = 4608 rft = 3464 kg

# 9 Bars = 1572 rft = 2318 kg

# 10 bars = 1061.28 rft = 2043 kg

# 3 bars = 8704.22 rft = 1635 kg

Total M.S Rod = 13126.00 kg

Total R.C.C = 2072.88 cft

Total M.S Rod =13126 kg

=13.126 ton

Beam M.S Rod of beam at 6 th Story: Beam: B1, B2, B3, B4, B5, B6, B7,

B8, B9, B10, B11, B12Top Bars: 2 # 5

=1 * 51.17’ * 2 * 2 = 204.68 ft * 0.481 = 98.45 kg

Bottom Bars: 3# 5

= 1 * 51.17’ * 3 * 4 = 614.04 ft * 0.481 = 295.35

kg

Mid = 1 * 10’ * 2 * 4 = 80.00 ft * 0.481 = 38.48 kg

None

Table 4.3: Comparative steel volumes (continued…).



Option II building


Beam Extra top End: 2 # 7= 1 * 5.92’ * 2 * 8 = 94.72ft *

0.911 = 86.29 kgExtra top mid: 2 # 7 & 1 # 5= 1 *

None

9.5’ * 2 * 8 = 152.00 ft * 0.911 = 138.47 kg

= 1* 9.5’ * 8 = 76.00 ft * 0.481

= 36.56 kg

Stirrups #3Bars@ 6.50” C/C

= 85 * 3.75 * 4 = 1275 ft * 0.188 =

239.70 kg

Total Rod = 933.30 kg

Beam: B13, B14, B15, B16, B17, B18

Top Bars: 2#5

= 1 * 51.17’ * 2 * 2 = 204.68 ft * 0.481 =

98.45 kg

Bottom Bars: 2 # 6 &2 # 5

= 1 * 51.17’ * 2 * 2 = 204.68 ft * 0.752 =

153.92 kg

= 1* 51.17’ * 2 * 2 = 204.68 ft * 0.481 =

98.45 kg

Extra top End: 2 # 7

= 5.92’ * 2 * 2 * 2 = 47.36 ft * 0.911 =

43.14 kg

Extra top mid: 2 # 8

= 1 * 9.5’ * 2 * 2 * 2 = 760.00 ft * 1.176 =

89.34 kg




Option II building


Beam Total = 483.30 * 2 = 966.60 kgStirrups # 3 bars @

6.50” C / C= 85 * 3.75 * 4 = 1275 ft * 0.188 =

239.70 kgTotal = 1206.30 kg

Total Rod at 6th Floor = 933.30 +

1206.30 = 2139.60 kg

M.S Rod of Beam at 1 st floor:

Beam: B1, B2, B3, B4, B5, B6, B7, B8, B9,

None

B10, B11

Top Bars: 2 # 5

= 2 * 51.17’ * 4 = 409.36 ft * 0.481 =

196.90 kg

Bottom bars: 3 # 5

= 1 * 51.17’ * 3 * 4 = 614.04 ft * 0.481 =

295.35 kg

Extra top end: 4 # 6

= 1 * 5.92’ * 4 * 8 = 189.44 ft * 0.75 =

142.46 kg

Extra top mid: 2 # 9

= 1 * 9.5’ * 2 * 8 = 152.00 ft * 1.536 =

233.47 kg

Stirrups # 3 bars @ 6.50” C/C

= 85 * 3.75 * 4 = 1275 ft * 0.188 =

239.70 kg

Beam: B13, B14, B15, B16, B17, B18, B19, B20,

B21, B22, B23, B24

Top bars: 2 # 5




Option II building


Beam = 2 * 51.17’ * 4 = 409.36 ft * 0.481 = 196.90

kgBottom bars: 2 # 7 & 2 # 5= 2 * 20’ * 8 = 320.00

ft * 0.911 = 291.52 kg= 2 * 51.17’ * 8 = 818.72 ft *

0.481 = 393.80 kg

Extra top end: 2 # 8 & 2 # 6

= 2 * 5.92’ * 2 * 2 * 2 = 94.72 ft * 1.176 =

111.39 kg

= 2 * 5.92’ * 8 = 94.72 = 0.752

= 71.22 kg

Extra top mid: 2 # 9, 2 # 6 & 1 # 5

= 2 * 9.5’ * 2 * 2 * 2 = 152.00 ft * 1.536 =

None

233.47 kg

= 2 * 9.5’ * 8 = 152.00 ft * 0.752 =

114.30 kg

= 1*15’* 8= 120.00 ft *0.481

= 57.72

Stirrups # 3 bars @ 6.50” C/C

= 85 * 3.75 * 8 = 2550.00 ft * 0.188 =

479.40 kg

Total = 3077.43 kg

6th Story = 2139.60 kg

1st Story = 3077.43 kg

Total = 5217.03 kg




Option II building


Beam 5th to 2nd story = 5217.03/2= 2608.51 kgTotal Rod =

2139.60+2608.51 * 4+3077.43 = 15651.07 kg

None

Slab 15 ft*15 ft slab:4.375 * (6+9+5) = 87.5 ft15.42 *

(6+9+5) = 308 ft5.14 * (6+10+5) = 108 ft

4.375 * (6+9+5) = 87.5 ft

15.42 * (6+9+5) = 308 ft

5.14 * (6+11+5) = 113 ft

= 1012 ft

15 ft*20 ft slab:

4.16 * (8+15) = 96 ft

5.14 * (8+15) = 118 ft

15.42 * (8+15) = 355 ft

5 * (6+13+5) * 2 = 240 ft

20 * (6+9+5) = 400 ft

= 1209 ft

20 ft*20 ft slab:

6.81 * (12+12) = 327 ft

1S: 6 * 15.7’ = 94 ft (+ve)5 * (4.1’+ 16/12)= 27.2 ft

(Ex. – ve)11 * (4.1’+ 8/12)

= 52.43 ft (In. – ve)

= 94 + 27.2 + 52.43 = 173.63 ft.

1L: 7 * 15.7’ = 109.9 ft (+ve)

6 * (4.1’+ 16 / 12)

= 32.6 ft (Ex. – ve)

12 * (4.1’+ 8 / 12)

= 57.2 ft (In. – ve)

= 109.9 + 32.6 + 57.2

= 199.7 ft.

2S: 7 * 15.7’ = 109.9 ft (+ve)

7 * (3’+ 16/12)

= 30.33 ft (Ex. – ve)

7 * (3’+ 8/12)

20 * (12+12) = 480 ft

6.81 * (12+13) * 2 = 341 ft

20*(12+12) = 480 ft

= 1628 ft

= 25.6 ft (In. – ve)

= 109.9 + 30.33 + 25.6

= 165.83 ft.




Option II building


Slab Total steel in slab = 1012 + 1209 + 1628 = 3849 ft

= 3849 * (0.11 / 144)= 2.94 cft= 1440.6 lb= 653 kg.

= 0.653 tonTotal steel in slab for six stories: 1 *

0.653 * 6 = 3.92 ton

2L: 165.83 ft.For panel 1: 173.63 * 2 + 199.70 * 2 +

165.83 * 2 = 1078.32 ft.3: 9 * 20’ = 180 ft (+ve)11 *

2 * (5.6’ + 8/12)

= 137.87 ft (-ve)

11 * 2 * (3.74’+ 8/12)

= 96.82

= 180 + 137.87 + 96.82

= 414.69 ft.

4: 7 * 20’ = 140 ft (+ve)

= 206.73 ft.

5: 9 * 15.7 = 141.3 ft (+ve)

7 * (5.6’+ 16/12)

= 48.53 ft (Ex. – ve)

8 * (5.6’+ 8/12)

= 50.13 ft (In. – ve)

7 * (3.74’+ 8/12)

= 30.85 ft (In. – ve)

=141.3+48.53 +50.13 + 30.85

= 270.81 ft.

6: 11 * 15.7 = 172.7 ft (+ve)

11 * (4.1’+ 16/12)

= 59.77 ft (Ex. – ve)

= 172.7 + 59.77 = 284.9 ft.

For Panel 2: 414.69 * 2 + 206.73 +

270.81 * 2 + 284.9 = 1862.63 ft.




Option II building


Slab 7S: 11 * 20 = 220 ft (+ve)14 * 2 * (5.6’+8/12)= 175.5

ft (-ve)13 * 2 * (3.74’+ 8/12)

= 114.57 ft (-ve)

= 175.5 + 114.57 = 510.1 ft.

7L: 12 * 20 = 240 ft (+ve)

14 * 2 * (5.6’+ 8/12)

= 175.5 ft (-ve)

13 * 2 * (3.74’+ 8/12)

= 114.57 ft (-ve)

= 240 +175.5 +114.57= 530.1 ft.

8S: 9 * 20 = 180 ft (+ve)

9 * 2 * (4.1’+ 8 / 12)

= 85.8 ft (-ve)

= 180 + 85.80 = 265.8 ft.

8L: 265.8 ft.

For panel 3: 510.10 * 2 + 265.8 +

530.10 * 2 + 265.8 = 2612 ft

Total steel: 1078.32 * 4 + 1862.63 * 4 +

2612 = 14376 ft

= 19.67 cft = 4.38 ton.

Total steel in slab for six stories

= 1 * 4.38 * 6 = 26.28 ton

Summary of the Comparative Study:

This cost analyses, shown in Table 4.4, are completed according to “schedule of rate for civil

works”, 12thedition, PWD and as per considerations made in Chapter IV, Art 4.3.

Table 4.4: Cost analysis for volume of concrete of beam supported structure.

Sl. no. Short Unit Total Rate Amount (Tk.)

description (Tk.)

Option I:

Beam supported structure

i. Ground

Floor

201 406576.77

cft 2022.77

ii. 1st Floor cft 2022.77 205 414667.85

iii. 2nd Floor cft 2022.77 209 422758.93

iv. 3rd Floor cft 2022.77 213 430850.01

v. 4th Floor cft 2022.77 217 438941.09

vi) 5th Floor cft 2022.77 221 447032.17

Total costing for concrete works = Tk.25,60,826.82

Table 4.5: Cost analysis for volume of concrete of flat plate structure.

Sl. no. Short

description

Unit Total Rate

(Tk.)

Amount (Tk.)

Option II: Flat plate

structure

i. Ground Floor cft 2052.80 201 412612.80

ii. 1st Floor cft 2052.80 205 420824.00

iii. 2nd Floor cft 2052.80 209 429035.20

iv. 3rd Floor cft 2052.80 213 437246.40

v. 4th Floor cft 2052.80 217 445457.60

vi) 5th Floor cft 2052.80 221 453668.80


The cost analyses of the both structures are summarized in table below.

Table 4.6: Cost analysis for volume of 60 grade deformed bar (steel) of beam supported

structure.

Sl. no. Short Unit Total Rate Amount (Tk.)

description (Tk.)

Option I: Beam supported

structure

i. Ground Floor Kg 5450 86.9

9

474095.50

ii. 1st Floor Kg 5450 87.2

3

475403.50

iii. 2nd Floor Kg 5450 87.4

7

476711.50

iv. 3rd Floor Kg 5450 87.7

1

478019.50

v. 4th Floor Kg 5450 87.9

5

479327.50

vi. 5th Floor Kg 5450 88.1

9

480635.50


Table 4.7: Cost analysis for volume of 60 grade deformed bar (steel) for flat plate

structure.

Sl. no. Short

description

Unit Total Rate

(Tk.)

Amount (Tk.)

Option II: Flat plate

structure

i. Ground Floor Kg 6050 86.9

9

526289.50

ii. 1st Floor Kg 6050 87.2

3

527741.50

iii. 2nd Floor Kg 6050 87.4

7

529193.50

iv. 3rd Floor Kg 6050 87.7

1

530645.50

v. 4th Floor Kg 6050 87.9

5

532097.50

vi. 5th Floor Kg 6050 88.1

9

533549.50


Grand Total (I+II) = Tk.11,203,381.62

The table below is showing total concrete requirement, steel requirement and cost difference for

flat plate structure and beam-supported structure.

Table 4.8: Summary of cost analyses for both structures

Type of

structure

Total

volume of

concrete

works

(cft)

Total

volume of

steel

works

(kg)

Total costing

of concrete

works

(Tk)

Total costing

of steel works

(Tk)

Remarks

Beam

supported

structure

12136.64 32700.00 25,60,826.82 28,64,193.00 About 6.5

% more

cost

required for

flat plate

structure

Flat plate

structure

12316.80 36300.00 25,98,844.80 31,79,517.00

CONCLUSION AND RECOMMENDATION

Recommendations for Further Study:

For further study in this field, the following recommendations are put forward:

i) The study needed use of conventional finite element software and manual calculation for both

analysis and design of whole structures to give a comprehensive conclusion.

ii) Instead of one residential square building of about 3.5 katahs it requires other geometrically

shaped and larger areas residential building for accurate comparison.

Conclusions:

From the comparative study of beam supported structure and flat plate structure, we gathered

knowledge that:

a) From the Finite Element Analysis result it can be said that the internal forces in the flat plate

structure is higher than that of beam supported structure.

b) The construction of flat plate structure requires more construction material which results in

more cost. Such as, concrete requirement increased in flat plate structure about 1.5% and steel

requirement about 11% and finally increased cost of 6.5% than beam supported structure.

c) In case of flat plate structure interior space of building looks nice, due to absence of beams

offset. Flat plate slab is thicker and more heavily reinforced than slabs with beams and girders.

d) It is apparent from the cost comparison that the difference between two estimates is very

insignificant and moreover this difference is for only frame of the building. The cost per unit area

for finishing items will remain same for all cases. In compare to the enormous benefit that can be

gained for aesthetic view and also for light provision, the cost increase in this case is very

insignificant.

References:

1. ACI Code, 1995, USA

2. ETABS -Version 8.2.7

3. BNBC (1993), “Bangladesh National Building Code” 1st edition, city Art Press,

Bangladesh.

4. “Design of concrete structure”- 13th edition by Arthur H. Nilson, David Darwin, Charles W.

Dolan.

1. “Schedule of Rate for Civil works”, 11th edition, Public works department,

Government of the Peoples Republic of Bangladesh, 21st November. 2008.

Analysis Between a Beam Supported Structure and a Flat Plate Structure

Documents

design wind load

building site

building system

analysis phase

results of analysis

types of building

flat plate slab system

terrain exposure condition