SYSTEMATIC DESIGN
OF
REINFORCED CONCRETE STRUCTURES
By Microcomputer
Ma1
Mb1
Mc1
10 REM A three dim array
Ma2
Mb2
Mc2
20 DIM M(3,3,3)
Ma3
Mb3
Mc3
30 FOR a = 1 to 5
Ma4
Mb4
Mc4
40 FOR b = 1 to 5
Ma5
Mb5
Mc5
50 FOR c = 1 to 5
Matrix Algebra
Computer Output
Advance Basica
By:Bienvenido C. David – Civil Engineer
The cover page of the book is
our UC – BCF Alumni building
located in Harrison Road. The
cover design suggest that a
portion of our alumni building
have been designed using a
micro computer by Matrix
Method of Structural Analysis
and the software language
used is Advanced Basica.
In a construction project a field engineer and his men are busy excavating the last footing of a
building. The depth of the footing as per plan is 12 Inches. They are half way when
unexpectedly they encountered a big boulder. Removing the boulder is costly and time
consuming it’s on site decision no time to lose the only alternative the field engineer is to
revise the dimensions of the footing. The field engineer had only his programmable
calculator. His problem what dimensions the footing be with the depth reduce to 6 inches?
He has no tables or charts and worst of all he forgot his theory of “Reinforced Concrete”
Time is running out in four hours’ time pouring of concrete will start. Since design is both
safety and economy there should be logical reasons for the new dimensions. Luckily the field
engineer remembers when he was reviewing for his Civil board exams he made a program
called “Square foot” and save it in his programmable calculator (Square foot is a program
used to design a square footing using the ultimate strength theory. Given the depth of the
footing it solves the dimensions, size of rebar and spacing and vice versa). Break time and the
engineer got his programmable calculator retrieved the file “Square footing” and in five
seconds his programmable calculator gives all the information (Dimensions rebar size and
spacing). Without the program the outcome would have been different. The program
“Square footing” saves the engineer that day. The structure had been completed ahead
of schedule as a reward for his judgment and performance the boss promoted the engineer to
Team Leader with a cash bonus in return the good Samaritan engineer donated one
third of his bonus to the BCF-UC alumni for the two incoming projects Speech
lab and E-Review center the other one third he donated it to the recent typhoon
victims in his home country and the remaining third he reserve it for his second
honeymoon.
Promoted to team leader the engineer went home happily and that night having
missed his wife for a long time he started singing “Wash now my love” and in
return he heard his wife sings “What I am washing for it’s not for you” and
moments later the light was turned off.
About the author
The author finished his Bachelor of Science in Civil Engineering in the
University Of Cordilleras year 1969 and passed the Civil Engineering Board
Year 1969. Year 1970 he was employed by Monark International as Instrument
man/Surveyor the work includes the construction of a 6 kilometers highway
with a 1 kilometer airport construction in Sison, Pangasinan. Late 1971
He worked in the Pantabangan Dam Multipurpose project, a special project of
the National Irrigation Administration first as Assistant Construction Engineer
and later promoted to Construction Engineer year 1972 to 1977. In this project
he held positions such as Field Inspector, Field Reporting Engineer and Cost
Engineer.In 1977 when the project was finished he worked as a Feasibility study
Engineer under the office of his uncle then the city administrator of Baguio City
where he designed the Civil Infrastructure component of the proposed project.
In 1980
He ventured into Civil Engineering practice in his hometown Baguio City year
1980 to 1987. During his early years of practice the conventional method of
Designing Reinforced Concrete Structures is inadequate enough to meet the
schedules of his project
This prompted him to write his book “Systematic Design of Reinforced
Concrete Structures by Micro computer”
In 1988 He migrated to Australia and worked in the Work Cover Authority of
New South Wales from 1988 to 2000 and different private telecommunication
companies namely SIM PLUS, RSL, VIRGIN MOBILE AND OPTUS from year 2001
to 2012. He gained extensive experience in computer hardware and software in
his two previous jobs in Australia. He retired in year 2012 at age 66.
Dedication
This book is dedicated to my daughter Erikka David, to
my fellow UC _BCF alumni, the ultimate success of the EReview centre of UC BCF and the success of the UC BCF
CE full membership as per Washington accord.
Inspiration
May this book serves an inspiration for the
undergraduates students of UC –BCF- CE that someday
they may become TOP GUNS, BOARD TOP NOTCHERS,
LEGENDS AND AUTHORS in the UC – BCF CORPS OF CIVIL
ENGINEERS, UNIVERSITY OF CORDILLERAS
EXCERPTS FROM THE SECOND EDETION
The book is primarily intended for the busy field Civil Engineer/Architect who
has no time opening up books, tables or recalling his theory of Reinforced
Concrete never less the book would be a valuable reference for undergraduate
students in Civil Engineering and Architecture whose interest the transition from
classical to digital approach.
Experienced based from my Civil Engineering practice Designing a typical six
storey reinforced concrete structure would normally takes a number of months
using the conventional method but the combinations of all the programs
presented in my book just takes an average of thirty minutes to do the same
job.
EXCERPTS FROM THE FIRST EDETION
Sometimes Civil or particularly Structural Engineers demands that we meet
deadlines from our clients. Without the use of computers team of Engineers
would be needed to do the structural design of buildings. It is the application of
the microcomputer in the fields of Reinforced Concrete Design that I made this
book.
SYSTEMATIC DESIGN OF REINFORCED
CONCRETE STRUCTURES
BY MICRO COMPUTER
By
Bienvenido C. David
B.S.C E. Class 1969 – University Of Cordilleras
P.R.C. NO: 10170
Formerly Construction Engineer (Dam Aspect) Pantabangan Dam, National
Irrigation Administration – 1971 to 1977
Formerly feasibility Study Engineer - Low Cost Housing Development Project
Baguio City, Philippines – 1979 to 1980
Formerly Civil Engineering Practitioner – Baguio City, Philippines 1980 – 1987
Formerly held various positions Work cover Authority of New South Wales,
Australia 1988 – 2000.
Formerly held various positions RSL/SIM PLUS COMMUNICATION COMPANY
Australia 2001 – 2010
Formerly Customer Support Consultant OPTUS COMMUNICATION PTY LTD
Australia 2010 – 2012
PREFACE TO THE SECOND EDETION
The second edition of this book is completely rewritten using the latest
Microsoft office software
So it really looks a technical book. Like the first edition the second edition has
the same objective the application of the micro computer in the fields of
Reinforced Concrete Design. Among the new features of the second edition
includes detail analysis and formula derivations for computer applications. I
have added two classical solutions under chapter eight design of trapezoidal
footing and chapter nine design of cantilever retaining wall to illustrate clearly
in a simple and concise manner how the classical solution is being cast into the
digital solution.
The book consist of 275 pages nine chapters , twenty one computer programs
and two classical solutions.
Chapter one includes three mathematical programs namely the General Cubic
equation (program1), The Newton’s method of approximation (program 2) and
the Simultaneous equations in many unknowns (program no 3). The three
mathematical programs are usually included as a subroutine programs within
the main programs such as program no 16 analysis of columns and
program no 20 design of trapezoidal footing.
As an aid to the practicing Civil Engineer I have included three programs of
structural analysis, program no 4 the three moment equation, program no 5 the
slope deflection and program no 6 the moment distribution method. In program
no 4 and program no 5 the applied mathematical program no 3 is included as a
sub routine program within the main program indicating the importance of
chapter 1
Chapter three is a summary treatment of Reinforced concrete theory and design
useful as a refresher course for practicing Civil Engineers and Architects.
Chapter 4 is all about design of beams and consist program no 7 Design of
Single Reinforced Concrete Beam, program no 8 design of Double Reinforced
Concrete Beam and program no 9 Design of Reinforced Concrete Tee Beam.
Chapter 5 treats one way slab design and two way slab design and under this
chapter program no 10 One Way Slab Design and program no 11 Two Way Slab
Design.
Chapter 6 covers design of web reinforcements using the more accurate AC.I.
code program no 12 design of diagonal tension, deflections of beams and slabs
program no 13 and program no 14 bending of steel reinforcements and
inflection points. In program no 12 and program no 13 the mathematical
program no 3 Newton’s method of approximation is included as a sub – routine
program within the main program.
Chapter 7 covers design of Reinforced Concrete column and made up of two
programs namely program no 15 design of column at balance load and program
no 16 analysis of column in two modes of failure. In program no 16 the applied
mathematical program no 1 is included as a subroutine program within the
main program to solve the depth of stress rectangular block a clearly indicates
the importance of chapter 1
Chapter 8 is all about footings and consists of Square footing program no 17,
Rectangular footing program no 18 and Combined footing program no 19 and
program no 20 trapezoidal footing. I have included under program no 20 a
classical solution (analytical method) to illustrate clearly in a simple and concise
manner how the classical solution is cast into the digital solution.
Chapter 9 Covers Cantilever retaining wall design program no 21. I have
included as well a classical solution which precedes the computer solution.
The programs are based entirely on the Ultimate Strength Theory (USD) except
program no 13 Deflection which is WSD FORMAT (Serviceability theory).
Each program has its own program record showing program description,
drawing and other information to enable the users to understand what the
computer do.
With the emergence of many software languages such as Fortran, Sol, Sql, Pl1,
Pl2 and Turbo Basic. I finally decided to use Advance Basica for Basica is simple,
easy to learn it’s adaptability to a wide variety of programmable calculators
available on the market.
Although the program steps appear simple it is assumed that the reader have a
basic knowledge of computer programming.
The book is primarily intended for the busy field Civil Engineer/Architect who
has no time opening up books, tables or recalling his theory of Reinforced
Concrete never less the book would be a valuable reference for undergraduate
students in Civil Engineering and Architecture whose interest the transition
from classical to digital approach.
Experienced based from my Civil Engineering practice Designing a typical six
storey reinforced concrete structure would normally takes a number of months
using the conventional method but the combinations of all the programs
presented in my book just takes an average of thirty minutes to do the same
job.
The support of my three brothers was a vital factor in the completion of this
book to Engineer Carlos David , a Mechanical Engineer for his financial backing
owing my first generation computer, to Engineer Caesar David, Electrical
Engineer for teaching me the fundamentals of computer programming and to
Engineer Danilo David, Electronic Engineer basic computer programming and
computer housekeeping.
My close associations with competent Engineers & architects during my practice
was exciting and gratifying their valuable suggestions and critical comments
have been greatly acknowledged.
but most of all I am indebted to my UC BCF mentors during my five years as a
student particularly Engineer Avelino Cruz my instructor in Theory of Structures,
Engineer Daniel Cervantes my instructor in Timber Design, Steel Design and
Concrete Design, Engineer Conrado Foronda my instructor in Foundation Of
Structures and Engineer Cecilio Tuanquin my instructor in Graphics 111. The
knowledge I gained from their expertise prompted me to write this book.
The extensive computer experienced I gained from my two previous
employments first Work Cover Authority of New South Wales for 12 years and
second RSL/SIM-PLUS/VIRGIN MOBILE/OPTUS TELECOMMUNICATION
companies in Australia for 12 years was instrumental in rewriting the copy of
my original book.
Finally I want to thank my daughter Erikka David, a graduating Medical
Science/Electrical Engineering student of the university of Sydney for her time
and effort checking the output of my computer programs.
The beautiful clients I met during my Civil Engineering practice in my hometown
Baguio City makes the writing of the original edition of my book a favorable past
time.
This book is dedicated to my daughter Erikka David, to my fellow UC _BCF
alumni, the ultimate success of the E-Review centre of UC BCF and the success
of the UC BCF CE full membership as per Washington accord.
May this book serves an inspiration for the undergraduates students of UC –
BCF- CE that someday they may become TOP GUNS, BOARD TOP NOTCHERS,
LEGENDS AND AUTHORS in the UC – BCF CORPS OF CIVIL ENGINEERS,
UNIVERSITY OF CORDILLERAS.
Bienvenido C. David – January 15, 2015 - Sydney -Australia
PREFACE TO THE ORIGINAL
TYPEWRITTEN EDETION
Three or four years ago I was designing structural
members of various parts of a building , thru
experienced the time involved by using a calculator or
slide rule was to prohibitive that it took me several days
or weeks to design a three or four story reinforced
concrete building.
Sometimes Civil or particularly Structural Engineers
demands that we meet deadlines to submit a structural
plan of the proposed building and to accomplish the
time schedule team of Engineers would be needed to do
the manual calculations.
With the advent of microcomputers, programmable
calculators and mainframe computers there is a change in the technique of structural
computations which maybe called revolutionary and the term is usually referred as
systematized computations.
It is due to these revolutionary changes that this manual was developed. Systematic design of
concrete structure is different from the classical method of approach encountered in different
textbooks in Reinforced Concrete Design. Here the microcomputer plays a major role , no
longer had the Structural Engineer spent his time cracking down his slide rule or calculator,
opening up books and tables but by inserting the correct module desired data can be obtained
in seconds with accuracy far greater than manual calculations.
The arrangement sof the programs follows closely the basic steps in a complete story. It is
applicable to three and four story structure and other type of reinforced concrete members
faced by Civil Engineers from day to day. The design programs is entirely based using the 1977
and 1983 A.C.I. code with the Ultimate Strength Design Theory as the alternative procedure
used.. The program was written in plain basic language and is applicable to a wide variety of
programmable calculators , microcomputers and with some peripherals may be attached to a
main frame computer system.
Each program record has its own drawing and program description to enable the user
understands the program steps involved. Some of the constants appearing were derived by the
author for computer use . Formula derivations are beyond the scope this book . However the
reader is referred to any textbooks in Reinforced Concrete Design for reference..
The manual consists of the following programs. Computer program no 1 and 2 pertains to one
way and two way slab design. Programs no 3,4 and 5 deals with compression beams,
continuous beams, and tee beam. program no 6 deals with columns with bending moment and
axial load combinations followed by program no 7,8, and 9,10 deals with square footing
,trapezoidal footing, combined footing and rectangular footing.. program no 11 is a cantilever
retaining wall design followed by program no 12 design of dia gonal tension followed by
program no 13 calculation of deflection. I have developed two mathematical programs for
design applications namely the General cubic equation program and the Newton’s method of
Approximation for finding roots of equations. These two programs are usually included as sub
routines with in the main program.
I have not included matrix method of structural analysis because I am convinced that there are
many excellent textbooks devoted to this subject. Although it appears quite simple the reader
must have a basic knowledge of computer programming , principle of reinforced concrete
design and a micro computer for practice.
I would like to acknowledge the following textbooks for reference Design of Reinforced
Concrete Structures By George Winter and Nilson, The Theory and practice of Reinforced
Concrete by Clarence W Dunham, Concrete Fundamentals by Phil Moss Ferguson, Foundation
Analysis and Design by Joseph Bowles and Reinforced Concrete Design by William Todd.
I would like to thanks my brother Caesar David, a fifth year electrical engineering graduating
student of Saint Louis University for checking much of the program steps involved and also for
Engineer Carlos David, a mechanical Engineer form lending his personal programmable
calculator.
I hope that by reading this manual the reader is encouraged to continue his studies in this
direction.
Bienvenido C. David
August 30, 1984 Trancoville, Baguio City, Philippines
v
–
APPLIED MATHEMATICS APPLICATION TO
CONCRETE DESIGN
INTRODUCTION
Page 1
THE GENERAL CUBIC EQUATION INTRODUCTION
AND EXAMPLES
Page 2 – 3
PROGRAM RECORD
Page 4 - 7
NEWTON’S METHOD OF APPROXIMATION
INTRODUCTION, ANALYSIS AND EXAMPLES
Page 8 - 9
PROGRAM RECORD
PAGE 10 – 13
SIMULTANEOUS EQUATIONS IN MANY UNKNOWNS
INTRODUCTION, ANALYSIS AND EXAMPLES
PAGE 11 – 17
PROGRAM RECORD
PAGE 18 – 23
vi
STRUCTURAL ANALYSIS
INTRODUCTION
THE THREE MOMENT EQUATION INTRODUCTION
PROGRAM RECORD
THE SLOPE DEFLECTION METHOD INTRODUCTION
PROGRAM RECORD
PAGE 24
PAGE 25 – 26
PAGE 27 – 39
PAGE 40 – 42
PAGE 43 – 55
THE MOMENT DISTRIBUTION METHOD
INTRODUCTION
PRORAM RECORD
PAGE 56 – 57
PAGE 58 – 69
MECHANICS AND BEHAVIOUR OF
REINFORCED CONCRETE
PAGE70 77
vii
DESIGN OF REINFORCED CONCRETE BEAMS
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
Page 78 – 81
SINGLE REINFORCED CONCRETE BEAM
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
Page 82 – 85
PROGRAM RECORD
Page 86 – 91
DOUBLE REINFORCED CONCRETE BEAM
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
page 92 - 95
PROGRAM RECORD
Page 95 – 101
viii
DESIGN OF TEE BEAM
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
page 102 - 105
PROGRAM RECORD
Page 106 – 114
DESIGN OF SLABS
INTRODUCTION, ANALYSIS AND FORMULAS
Page 115 – 116
ONE WAY SLAB DESIGN
INTRODUCTION, ANALYSIS AND FORMULAS
PROGRAM RECORD
Page 116 – 117
Page 117 – 122
TWO WAY SLAB DESIGN
INTRODUCTION, ANALYSIS AND FORMULAS
Page 122 124
PROGRAM RECORD
Page 125 – 129
ix
WEB REINFORCEMENTS, DEFLECTIONS AND
INFLECTION POINTS
WEB REINFORCEMENTS
INTRODUCTION, ANALYSIS AND FORMULAS
Page 130 – 131
PROGRAM RECORD
Page 132 – 140
DEFLECTIONS
INTRODUCTION, ANALYSIS AND FORMULAS
Page 141 - 143
PROGRAM RECORD
Page 144 – 151
INFLECTION POINTS –Bending of positive and
Negative Reinforcements
INTRODUCTION, ANALYSIS AND FORMULAS
Page 152 - 154
PROGRAM RECORD
Page 154 – 159
x
COLUMNS
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
Page 160 – 165
Design of Column at Balance Load
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
Page 166 – 169
PROGRAM RECORD
Page 169 – 174
Analysis of Column In two Modes of Failure
INTRODUCTION, ANALYSIS AND FORMULA DERIVATIONS FOR
COMPUTER APPLICATIONS
Page 175 – 180
PROGRAM RECORD
Page 180 – 187
xi
FOOTINGS
INTRODUCTION
Page 188 – 190
Design of Square Footing
INTRODUCTION, ANALYSIS, DESIGN STEPS
Page 191 – 193
PROGRAM RECORD
Page 194 – 199
Rectangular Footing Design
INTRODUCTION, ANALYSIS, DESIGN STEPS
Page 200 - 201
PROGRAM RECORD
Page 202 – 207
Combined Footing
INTRODUCTION, ANALYSIS, DESIGN STEPS
Page 208 - 209
PROGRAM RECORD
Page 210 – 216
Design of Trapezoidal Footing
INTRODUCTION, ANALYSIS AND FORMULA DERIVATION FOR
COMPUTER APPLICATION
Page 217 – 219
CLASSICAL SOLUTION (ANALYTICAL METHOD)
Page 220 – 227
DESIGN STEPS FOR COMPUTER APPLICATIONS
Page 228 - 229
PROGRAM RECORD
Page 229 – 239
xii
DESIGN OF RETAINING WALLS
INTRODUCTION
Page 240 – 241
CANTILEVER RETAINING WALL DESIGN
INTRODUCTION, ANALYSIS AND FORMULA DERIVATION
FOR COMPUTER APPLICATION
Page 242 – 249
CLASICAL SOLUTION (ANALYTICAL METHOD)
Page 249 – 257
PROGRAM RECORD
Page 257 – 274
xiii
Page1
APPLIED MATHEMATICS
In this chapter, the three applied mathematics is discussed and explain well.
Included in this chapter are the general cubic equation, the Newton’s method of
approximation and the solution of linear equations in many unknowns. A
general program for each applied mathematics is formulated. In the general
cubic equation program three roots are evaluated and the real positive root is
the only one selected by the computer for design application. In the Newton’s
method of approximation, two general programs were formulated. A program
incorporating 15 trial cycles and the general cycle program. In the simultaneous
equations of many unknowns, a direct computer solution and the general
formulation of struct - math solver three is presented. In the formulation of
struct math solver three, a knowledge of invertion matrix is a pre requisites to
understand the program derivation.
Page2
2
APPLICATION TO CONCRETE DESIGN
INTRODUCTION
The general cubic equation is used to solve roots of a third degree equation. It was first
published in 1545 by H. Cardan in his famous traits called Ars Magna. The clue to this method is
supposed to have been discovered independent by Tartaglia and an earlier writer perhaps
Scipione Del Fierro, all mathematicians just mentioned were Italians.
In the generalized form
X3 + BX2 + CX + D = 0 the following solutions apply when B, C & D are
any complex numbers. In the above equation we substitute X = Y – b/3 and we obtained
Y3 + py + q = 0 equation 3 where p = c – b2/3 and q = d – bc/3 + 2b3/27
in
equation 3 we put
y=z-
then z3 – p3/27z3 + q = 0 or z6 + qz3 – p3/27 = 0 we note that the above
equation is in quadratic form in z3 then by the quadratic formula we get z3 = -q/2
+ R and z3 = -q/2 – R where R = p3/27 + q2/4 here R maybe an imaginary
number, in such a case, R is understood to represent any one of the two square
root of R.
General cubic equation appears in finding the point of zero shear using the
accurate A.C.I. code, in column design case three for solving the depth of stress
rectangular block, in trapezoidal footing for finding the point of inflection
(bending of bars) and in locating point of maximum deflection of beams and end
moments in continuous beams.
The following equations are examples how the general cubic equation program
is included as a subprogram in the main program.
3
Example 1
Y = .5X3 – 20.652 -150X + 273.43 = 0 Solve for X. This equation is a part of deflect
program solving mathematically we get X = 12.5 by computer it takes just 10
seconds to solve the roots.
Example 2
Consider the equation M = 1.322a3 – 10.577a2 + 162.889 – 2588 = 0 solving for a
= 11.73 by computer this takes around 4 seconds to solve the positive root a. This
is actually a sub program of column design code name anal col.
Example 3
Consider the equation M = -17.9366X3 + 618.5X2 – 3067(X – 0.23) = 0 Solving for
X X is equals to 2.828. This is actually a sub program for TRAP FOOT design of
trapezoidal footing. The value of X here is the point of inflection. Points by which
bars are bent for positive and negative bending. By these examples we note that
the general cubic equation is essential or necessary for design, hence the general
cubic equation program is included as a sub – routine program if their need arise.
*** For further reference see any textbook on algebra.
4
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Application of STRUCT MATH/Solver 1 to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: General Cubic Equation program CODE NAME: STRUCT
MATH/Solver 1
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 33
LIBRARY MODULE; Floppy Disk PROGRAM NAME: Cubic
PTR NO:
3046165 at Bagiuo City 01/09/1982
The general cubic equation has the form X3 + BX2 + CX + D = 0 We
let P = C – B2/3
Let Q = D - + 2B3/27 Let R = P3/27 + Q2/4 X1 = Z1 – P/3Z1 – B/3
then
X2 = Z2 – P/3Z2 – B/3 X3 = Z3 – P/3Z3
Z3 = - Q/2 + R1/2 Here Z1, Z2 &
Z3 are roots of cubic equation.*** Note for a more elaborate
discussion of the general cubic equation see reference textbook
REFERENCE TEXTBOOK:
College Algebra by William
Hart
CHAPTER 16 pages 254 – 257
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
5
My first generation home computer
STRUCT MATH/SOLVER 1 Is a computer program that solves the roots of the general cubic
equation. It solves three roots Z1, Z2 AND Z3. This program is written in Advance Basic
language. Struct math solver 1 is included as a sub – program in the shear and diagonal
program to find the root of the cubic equation in the design and analysis of rectangular
column and to find the point of inflection in the bending of steel bars. The program is written
in Advance Basic and can be feed to a wide variety of programmable calculators and micro
computers.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
6
COMPUTER INSTRUCTION CODE
5) Call clear
10) “This is general cubic equation solving roots of a cubic equation”
15) Print “Format is A1X3 + B1X2 + C1X + D = 0
20) REM given coefficients of cubic equations and constant C as input
statements we solve three roots of cubic equations
30) INPUT “A1, B1, C1, D1”:A1, B1, C1, D1
35) B =B1/A1
-8
40) C= C1/A1
123.214
45) D = D1/A1
-1957.64
50) P = C – b^2/3
101.8767
55) Q = D – B*C/3 + 2*B^3/27
60) R = P^3/27 + Q^2/4
70) IF R<0 THEN 150
80) Z = -Q/2 + R^.5
90) IF Z<0 THEN 120
100) ZA = Z^.33333
110) GOTO 230
120) ZB = ABS (Z) ^.3333
130) ZA = -ZB
140) GOTO 230
1666.9795
733.8627
7
150) O = ATN (ABS(R) ^.5/ (-Q/2))/3
160) PI = 2.094395102
170) ZC = ((-Q/20^2 –R) ^.5
180) ZA = (ZC) ^.33333
11.91143
190) IF (-Q/2)>0 THEN 220
200) ZA = -ZA
210) XA = COS (O + 2*PI) *(ZA – P/ (3*ZA)) – B/3)
220) X = COS (O+2*PI)*(ZA – P/ (3*ZA)) – B/3
230) XB = COS (O)*(ZA – P)/(3*ZA)) – B/3
11.7274
240) PRINT “First root XA =”; XA
250) PRINT “Second root X=”; X
260) PRINT ‘Third root XB=”; XB
270) PRINT” Programmed by Bienvenido C. David a Civil Structural Engineer on
Jan 25, 1983 at Baguio City”
280) END
8
Of Approximation
APPLICATION TO
Concrete Design
INTRODUCTION
Although the General Cubic Equation can be used for solving the roots of a third degree
equation. In actual case especially design application we are only interested in the real positive
integer root. The Newton method of approximation is sometimes preferable to the General
Cubic equation in relation to concrete design application. The program is usually included as a
sub -routine program in the main program. In the succeeding programs, we see how this
applied mathematical program is included in the main program.
In college algebra the solution of the general polynomial equation in the form
AXn-1 + BXn-1 + CX + D or the general form AXn + BXn-1 + CXn-2 + D = 0 where N is a
positive integer. Either we solved the roots by synthetic division, the general cubic, and the
quartic or assuming a value of X then substituting this value in the said function, such a case
solving the roots would be tedious, time consuming especially when the exponent of X is in the
order greater than two or three. We now introduce a new method known as “The Newton’s
Method of Approximation”. This method is to find to any desired degree the root of an
equation which can’t be solved by elementary methods. To derive the method, let us consider
n
n-1
the figure below. Let Yn = F(X) as a function of X = AX + BX
Y axis
Y = f(X)
O
X2
X1
X axis
+ CXn-2 + XDn-3
9
From the figure the root of the equation X coordinate of a point at which the curve crosses the
X axis. Let the first approximation to the root be X = X1 as shown. The point B where the
ordinate AB intersects the curve has the ordinates X =X1 Y = Y1 == F(X1). The tangent at line B
will intersect the X axis at C, whose coordinate X2 maybe a better approximation to the desired
root that is X1. To find X2 knowing X1, note that BA = f(X1), CA = X1 – X2
and
= f’(X1) thus f(X1)/X1 – X2 = f’(X1) which yields X2 = X – f(X1)/f’(X1) if we have one
approximation X1 to a root of f(X) = 0, equation 3 gives us another approximation, X2 to that
root... From X2 another approximation X3 is obtained in the same way buy using X3, is
obtained X3 = X2 – f(X2)/f’(X2) and the process can be repeated as many times we wish.
For design application, we shall consider a value of n-3 substituting this value in the general
polynomial equation we have the general cubic equation : Yn = AX3 + BX2 + CX + D to solve the
three roots we put Yn = 0 differentiating Yn we have f’y = 3AX2 + 2BX2 + CX + D thus if X1 is the
or X1 = (AX3) + BX2 + CX + D)/(3AX2 + 2BX +
C). This process can be continued until the desired root is obtained.
From the previous discussion, the process of obtaining the roots is
simply a cyclic process which can be easily access to basic
programming. We simply treat the following mathematical expression a
subroutines.
first root then X2 = X1 –f(y)/f’(y)
1) Y(X) = AX3 + BX2 + CX + D
2) Y’ = 3AX2 + 2BX + C
3) X2 = X1 – F(Y)/F’(Y) in using the input data direct from the
keyboard we simply put coefficients of X3, X2, X and constant D as
numerical datas using the read and data statement. The known
coefficients A, B, C & D must be arranged in their sequential order.
Newton’s method of approximation is used as a sub –routine program
in retaining wall design, dia shear program and case three of column
design. This example is a part of design program.
0.0563556X3 + -0.069626x2 + .2639169x – 162.98 = 0 this value of X is a
point by which minimum steel reinforcement ratio from code
governs.
10
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Application of STRUCT MATH/Solver 2 to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Newton’s Method Of Approximation CODE NAME: STRUCT
MATH/Solver 2
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 37
LIBRARY MODULE; Floppy Disk PROGRAM NAME: Newton
PTR
NO: 3046165 at Bagiuo City 01/09/1982
The Newton’s Method Of Approximation has the form of
X2 = X1 –F(X1)/F’(X1) here X1 is the first trial root and X2 is the second
trial root. For the General Cubic Equation: AX3 BX2 + CX + D = 0
F’ = 3AX2 + 2BX + C here F = AX3 BX2 + CX + D here A, B, C & D are
known. For a more elaborate explanation of the Newton’s Method Of
Approximation. Ps refer to reference textbook as noted.
REFERENCE TEXTBOOK:
Differential & Integral Calculus
By Clyde e Love & Earl D.
Rainville
CHAPTER 17 pages 226 – 230
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
11
My first generation home computer
STRUCT MATH/SOLVER 2 Is a computer program that determines the real positive root of the
general cubic equation computers. Upon input of the first trial computer solves second trial
root the process continues until the desired root is obtained. We can have many input cycles
T. This program is usually included as a sub routine program in the determination of point of
zero shear using the more accurate A.C.I. code and the inflection points of bars trapezoidal
footing and the analysis of column case three of column design. The program is written in
Basic Language and applicable to a wide variety of programmable calculators and computers.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
12
COMPUTER INSTRUCTION CODE
10) CALL CLEAR
20) PRNT “This is Newton’s Method of Approximation” a program finding the real root
(positive root of a cubic equation)”
25) PRINT “This program was developed by Bienvenido C. David on January 1981”
30) PRINT “ Try a value of X as first trial see if value of trial root is not changing then trial root
approaches real root so that function Y approaches 0”
35) PRINT”Format is AX3 + BX2 + CX + D = 0
40) PRINT “ Input Coefficients of A, B, C, D program line no. 60 as input data’s if ready then
run line no 60”
50 PRINT “Put value of T as no. of trials input statement line no. 70”
55 STOP
60 INPUT” Coefficients of “A, B, C, D”: A, B, C, D
70 INPUT” How many trials T”: T
80 INPUT “Value of X as first trial “: X
90 P = 1
100 Print “I am now performing trial no =”; P
110 PRINT “Assume trial root is =”; X
120 REM Format is Y = AX3 + BX2 + CX + D
130 Y = A*X^3 + B*X^2 + C*X + D
140 IF Y = 0 THEN 150 ELSE 170
150 PRINT “Real root is =; X
160 STOP
13
170 T = T-1
180 IF T=0 THEN 190 ELSE 200
190 STOP
200 R = X
210 GOSUB 280
220 X = S
230 P = P + 1
240 PRINT “I am now performing trial no=”; P
250 PRINT”Trial root is =’; X
260 GOTO 130
270 STOP
280 REM This is a sub routine
290 M = A*R^3 + B*R^2 + C*R + D
300 N =3*A*R^2 + 2*B*R + C
310 S = R – (M/N)
320 RETURN
330 END
14
EQUATIONS IN MANY UNKNOWNS
APPLICATION
Structural Analysis 1, & 2
Reinforced concrete beams & frames
INTRODUCTION
In elementary algebra, we met how to solve systems of Linear simultaneous
equations in many unknowns. We either solve these by elimination of addition
and subtraction or by determinants.
Let us consider the following equations.
A1Ma + B1Mb = C1
(1)
A2Ma + B2Mb = C2
(2) HERE A1, A2, B1, B2, C1 & C2
are coefficients of unknowns Ma & Mb and C1 & C2 are constants solving Ma in
terms of Mb in (1) we have Ma = (C1 – B1MB)/A1. Substituting the value of Ma in
equation (2) a linear equation in Mb results which can be easily solved.
Substituting the numerical value of Mb in (3) yields the value of Ma. We call this
method solving by substitution. In (2) Multiplying both sides by A1/A2 we get
B2(A1/A2 (Mb) = A1/A2 (C2) Equation 4 subtracting equation (1) and (4) we have
Mb (B1 – B2A1 /A2) = C1 - A1/A2 (C2) (5) By inspection Mb can be solved and
substituting in equation (3) the value of Ma can be evaluated.. The second method
the determinant method, to understand this procedure considers these two
simultaneous equations.
a1x + b1y = k1
(1)
a2x + b2y = k2 (2) solving the generalized
equations multiplying equation (1) by b2 gives a1b2x + b1b2y = k1b2 multipling
equation (2) by b1 gives A2b1x + b1b2y = k2b1 subtracting we get
x(a1b2 – a2b1) = k1b2 - k2b1 from which
15
(3)
(4)
k1b2 - k2b1
X
a1k2 – a2k1
a1b2 – a2b1
Y
a1b2 – a2b1
we can therefore solve any simultaneous equations by memorizing (3) and (4)
The numerators and denominators of these generalized equations are called
determinants . Note that the same determinants appears in the denominators of
both equations (3) and (4), hence we can rewrite equation (3) and (4) as x =
And y =
we now need a simple way of memorizing the formula D, Dx and Dy
We can achieve this by writing the numerical coefficients which makes up the
determinants in matrix form. We first write the simultaneous equations in the
systematic form of equation (1) & (2) directly. Below we write the matrices for the
determinants each determinant matrix contains as many vertical columns and
horizontal rows as there are unknowns. The denominator matrix D simply
reproduces the numerical coefficients of the unknowns in the same matrix
position as in the original equation.
Column 1
column 2
Substitution column
a1x
+
b1y
=
k1
(1)
a2x
+
b2y
=
k2
(2)
k1
b1
k2
b2
a1
b1
a2
b2
x
(5)
a1
k1
a2
k2
a1
b1
a2
b2
Y
(6)
16
The numerator matrix for the determinant Dx replaces the coefficients of X(a1 and
a2) with the numerical constants k1 & k2 .
To evaluate the three determinants so as to arrive at the same results as eq(3)
and (4) we use the following procedure. Using the denominator determinant D as
an example we require the product a1b1 minus the product a2b1. This can be
accomplished in the D matrix by multiplying coefficients along the diagonals,
always moving diagonally to the right. To obtained the required solution
(a1b2 - a2b1) we always subtract the upward diagonal from the downward
diagonal.
a1
b1
Second operation – Subtract diagonal reverse sign of the
= a1b2 – a2b1
a2
b2
First operation additive diagonal . Checking against the
numerator determinants in eq (3) and (4), we find the same rule applies to the
evaluation of the matrices for Dx & Dy.
For structures consisting of simple spans and two to three storeys heights,
applying the slope deflection method would yield an average of 4 or 5
simultaneous equations. Applying the procedure previously discussed poses no
problem, we either solved the unknown by slide rule or scientific calculator.
However in actual design practice (as usually in the case for high rise buildings)
applying the slope deflection method would yield as many as 100 simultaneous
equations or more. Using the standard procedure would be tedious, laborious and
the time involve would be prohibitive, transforming this to computer language
would require a very large amount of memory (RAM) and typing the programs
would be difficult hence for practical purposes not feasible for computer
application . We therefore keep this method for some time and introduced a new
17
method not encountered in elementary mathematics we call this the matrix
reduction method for solving simultaneous equations in many unknowns. ****
Detailed discussion of this method is not discussed here. The reader is referred to
“Elementary Structural Analysis by Noris & Wilbur” 4rth edition or William H Tall
analysis of structures a computer approach.
The advantage of this method , it can be access in BASIC language and second the
method permits the solutions of many simultaneous equations in many
unknowns.
For our computer input we simply arrange the coefficients of the unknowns in
their sequential order from left to right beginning equation 1 as data statement
and put this numerical values program line no 360.
Arrange constants from top to bottom beginning equation one put numerical
values program line no 230
Put no of equations INPUT statement program line no 150
COMPUTER OUTPUT
Upon input of numerical datas into keyboard, computer prints on the screen
both the arrays and numerical values of unknowns in their sequential order. An
average microcomputer can solve 20 unknowns in just 1 minute.
To show you an example Consider the equations
1) 12.8 Mb + 2.4Mc = -1555.2
2) 2.4 Mb + 16.8 Mc = -1495.2
in this example data 360 would be 12.8, 2.4,
16.8
Here data 230 would be -1555.2, -1495.2
Input no of equations = 2 program line no 150
type run program line no 140
Mc = -73.62
Answers Mb = -107.7 Kip foot
18
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: STRUCT MATH/Solver 3 Application to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Simultaneous Equations CODE NAME: STRUCT MATH/Solver 3
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 72
LIBRARY MODULE; Floppy Disk PROGRAM NAME: Simul Bas
PTR
NO: 3046165 at Bagiuo City 01/09/1982
Struct math has the generalized form of
AX + BX + CZ = C Equation1 In the above equations we put constants as
A1X + B1Y + C1Z = C1 Equation 2 numerical data statements and coefficients of X,Y,Z
A2X + B2Y + C2Y = C2 Equation 3 & Q in their respective orders from left to right.
A3X + B3Y + C3Z = C3 Equation 4 Put no of equations N & press enter.
REFERENCE TEXTBOOK:
Analysis of Structures by
William H Tall a computer
approach
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
19
My first generation home computer
STRUCT MATH/SOLVER 3 is a computer program that solves the unknowns in the said
equations. Any no. of equations can be solved. Upon input of numerical datas, computer
solves the unknowns in sequence printing on the screen numerical values of the unknowns.
This program is actually incorporated as a sub program in the slope deflection method for
solving joint rotations and translations. The program is written in Basic Language and
applicable to a wide variety of programmable calculators and computers.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
20
COMPUTER INSTRUCTION CODE
5 CALL CLEAR
10 PRINT “This is a computer program for solving simultaneous equations linear
algebraic equations in many unknowns”
15 PRINT” Be sure you know matrix method of structural analysis otherwise I
can’t solve the unknowns”
17 PRINT” Transform slope equations into array form and by matrix algebra for
computer use”
18 PRINT” If ready then run line no 110”
20 STOP
110 CALL CLEAR
120 PRINT” Put coefficients in data order starting on line 460. Coefficients must
be arrange from their sequential order left to right”
125) PRINT “List value of constants starting from top to bottom as data
statement program line no 230, put no of equations input statement line no 50
then run line 140”
130 STOP
140 CALL CLEAR
150 INPUT “No of equations”: N
160 OPTION BASE 1
170 DIM X (20, 20)
21
180 DIM Y (20)
190 DIM X (20)
200 FOR E = 1 TO N
210 READ Z9E)
220 NEXT E
230 DATA
Values of constants
240 GOSUB 400
250 GOSUB 600
260 MAT = DELTA
270 FOR E = 1 TO N
280 GOSUB 400
290 FOR AB = 1 TO N
300 X (AB, E) =Z (AB)
310 NEXT AB
320 GOSUB 600
330 F = DELTA/MAT
340 PRINT” CHR$(64 + E);”=”; F
350 NEXT E
360 END
400 RESTORE 460
410 FOR Q =1 TO N
420 FOR P=1 TO N
22
430 READ X (Q, P)
440 NEXT P
450 NEXT Q
460 DATA
Note value of Coefficients
470 RETURN
480 STOP
590 REM A SUB ROUTINE
600 A = 1
610 FOR P = 0 TO N-3
620 for s = 1 N – P
630 IFX (1, 1) = 0 THEN 780
640 Y(S) = X(S, 1)
650 NEXT S
660 A = Y (1)*A
670 FOR R =1 TO N-P-1
680 FOR Q = 1 TO N-P-1
690 X (Q, R) = X (Q+1, R+1)-X (1, R+1)*Y (Q+1)/Y (1)
700 NEXT Q
710 NEXT R
720 NEXT P
730 DELTA = A*(X (1, 1)*X (2, 2)-X (2, 1)*X (1, 2))
*** Note for array type program line no: 740 to 760
23
740 PRINT”X (1, 1); X (1, 2)
750 PRINT X (2, 1); X (2, 2)
760 PRINT” TAB 95);”DELTA=”DELTA
770 RETURN
780 FOR D =1 TO N-P
790 IF X (D, 1) <>0 THEN 810
800 NEXT D
810 FOR B=1 TO N-P
820 W = X (1, B)
830 X (1, B) = X (D, B)
840 X (D, B) = W
850 NEXT B
860 IF INT (D/2)><D/2 THEN 880
880 GOTO 620
890 END
900 PRINT” This program was developed by Bienvenido C. David on March
1982”
24
In this chapter three programs of “STRUCTURAL ANALYSIS” are presented, the
three moment equation, the slope deflection and the moment distribution
method. In the moment distribution method two methods were employed, the
seven cycle distribution and the general moment distribution program. The use
of “Struct – math solver three is used as a sub – routine program in slope
deflection and three moment equation respectively. Any of the three programs
presented can be used to analyze a continuous reinforced beams and frames. If
précised moments are desired, the three moment equation will be use full. If
joint rotations and translation is of prime importance, the slope deflection
program will provide accurate numerical results. If modified stiffness is used the
moment distribution method alternative three will be use full. In any case, the
“General moment distribution program” is use full to any desired degree of
accuracy. Employing one method, the other two will be a cross checked.
25
INTRODUCTION
The three moment equation can be used to analyze statically indeterminate
beams. The three moment equation has the generalized form of
M1L1 + 2M2 (L1 + L2) + M3L2 + 6A1 1/L1 + 6A2
2 /L2 =
6EI (h1/L1 + h3/L2)
Where the points are on the same level in the deflection beam, the height h1 and
h3 becomes zero and so thus the right hand term of the above equation.
Below are the following legends used
M1 = Moment at support 1 Span one
M2 = Moment at support 2 Span two
M3 = Moment at support 3 Span three
L1 = length of Span one in feet
L2 = Length of Span two in feet
h1 = Deflection in inches if support yields
h3 =Deflection in inches if support yields at support three.
I = Moment of inertia of beams.
E = Young’s modulus of elasticity
For type of loadings on span table 8 – 1 page 276 of “Strength of Materials by
Ferdinand Singer” list values of 6A /L and 6A /L
In our case we shall deal with a concentrated load and uniform loads see attached
figure as shown in the “PROGRAM RECORD”.
A)
B)
C)
D)
For concentrated load: 6A /L = Pa/L(L2 – a2)
6A /L = Pb/L (L2 – b2)
For uniform loads = 6A /L = wl3 /4 and 6A /L = wl3/4
here a is the distance of P from R1 and b is the distance of P from R2 where
26
E) R1 and R2 are reactions of simply supported span.
P = Concentrated load in pounds or in kips.
L = Span length in feet
w = Uniform loads in pounds or in Kips per foot.
*** Note For a more detail discussion of this subject refers to Ferdinand Singer
pages 270 – 277 “Strength of Materials 2nd edition.
From the figure and applying the generalized form of the three moment equation
we have
X1(Ma) + 2Y1 (Mb) + Z1 (Mc) = C1 Equation 1
X2(Mb) +2Y2 (Mc) + Z2 (Md) = C2 Equation 2
here X, Y, Z is coefficients of moments where the numbers 1, 2, 3are coefficients
subscripts... Also we note that Ma = 0 and Md = W4 (L4)2 + P4 (Y4)
COMPUTER OUTPUT
Upon input of numerical data’s computer first solved coefficients X, Y, C and
constants C1 and C2. it prints on the screen equation one and two. With values of
coefficients as numerical data’s, it solves the unknowns in the said equations thus
a solutions of simultaneous equations is included as a sub – routine program in
the main program. In the third run with end moments known. Computer solves
shear reactions at support.
27
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: ANALYSIS ONE: Application to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Analysis of continues beam by the three moment equation
CODE NAME: ANALYSIS ONE
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 217
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Analysis one
PTR NO: 3046165 at Bagiuo City 01/09/1982
P2
Loading Diagram
W1/ft
W2/FT
L1
L2
Shear Diagram
P3
L3
P4
L4
28
Moment Diagram
REFERENCE TEXTBOOK:
Statically Indeterminate
Structures By Chu Kia
Wang
CHAPTER 6
PAGES 121 - 127
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
ANALYSIS ONE: IS a computer program that analyzes a Statically Indeterminate Beam or
frame by the three moment equation. Three spans with overhang uniform loads and
additional superimposed loads (concentrated loads located at any distance X is included). The
program is written in Basic Language and applicable to a wide variety of programmable
calculators and computers.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
Branch out
Branch out
Main program
IF False
SGN = SIGNUM NOTATION
IF THEN ELSE STATEMENT
29
COMPUTER INSTRUCTION CODE
10 CALL CLEAR
20 PRINT “This is computer program no. 4 in basic language Analysis of statically
indeterminate beams and frames by the three moment equation.”
40 PRINT” If all data’s are in their consistent units then run line no 60”
50 STOP
60 DATA W1, L1
3, 12
70 DATA W2, L2
2, 24
75 DATA X1, L1
0, 12
80 DATA P1, X1, L1
0, 0, 12
90 DATA W2, L2
2, 24
100 DATA X2, L2
12, 24
110 DATA P2, X2, L2
20, 12, 24 Computer output only
120 DATA X3, L3
4, 12
130 DATA W3, L3
0, 12
140 DATA P3, X3, L3
18, 4, 12
for debugging purposes only to check
150 INPUT “Relative stiffness factors “: K1, K2, K3, K4:K1, K2, K3, K4
160 INPUT “Span length in feet L1, L2, L3, L4”:L1, L2, L3, L4
170 INPUT” Uniform and concentrated loads W4, P4”: W4, P4
12, 24, 12, 3
0, 6
180 INPUT” Distance of concentrated loads from support X”: X 3
190 X1 = 2*(L1/K1 + L2/K2)
200 X2 = L2/K2
3, 10, 2, 0
30
210 RESTORE 60
220 GOSUB 930
230 T1 = T
240 GT =T1/K1
250 RESTORE 70
260 GOSUB 830
270 A1L1 = A
280 RESTORE
290 GOSUB 980
300 Q1 = Q
310 R1 = 6*Q1*A1L1/ (K1*L1)
320 U1 = G1 + R1
340 REM For span two
350 RESTORE 90
360 GOSUB 930
370 T2 = T
380 G2 =T2/K2
390 RESTORE 100
400 GOSUB 880
410 B2L2 = B
420 RESTORE 110
430 GOSUB 980
31
440 Q2 = Q
450 R2 = 6*Q2*B2L2/ (K2*L2)
460 U2 = G2 + R2
470 C1 = - (U1 + U2)
480 PRINT TAB (2); X1;”MB”; TAB (7); X2;”MC”; TAB (15); =”C1
490 REM For second equation span two
500 RESTORE 90
510 GOSUB 930
520 T3 = T
530 G3 = T3/K2
540 REM For concentrated load i.e. centroid
550 RESTORE 100
560 GOSUB 830
570 A2L2 = A
580 RESTORE 110
590 GOSUB 980
600 Q3 = Q
610 R3 = 6*Q3*A2L2/ (K2*L2)
620 U3 = G3 + R3
630 REM For span three uniform load w/Ft
640 RESTORE 130
650 GOSUB 930
32
660 T4 = T
670 G4 = T4/K3
680 REM For concentrated load span three
690 restore 120
700 GOSUB 880
710 B3L3 = B
720 RESTORE 140
730 GOSUB 980
740 Q4 = Q
750 R4 = 6*Q4*B3L3/ (K3*L3)
760 U4 = R4 + G4
770 C = (W4*L4^2/2 + P4*J)*L3/K3
780 C 2 = - (U4 + C +U3)
790 X3 = L2/K2
800 X4 = 2*(L2/K2 + L3/K3)
810 PRINT TAB (2); X3;”MB”; TAB (7); X4;”MC”; TAB (15);”=”; C2
820 STOP
830 REM this is sub routine 1
840 READ X, L
850 A = (X + L)/3
860 RETURN
870 STOP
33
880 REM this is sub routine no 2
890 READ X, L
900 B = -1/3*X + 2/3*L
910 RETURN
920 STOP
930 REM this is sub routine no 3
940 READ W, L
950 T = 1/4*W*L^3
960 RETURN
970 STOP
980 REM this is sub routine no 4
990 READ P, X, L
1000 Q = 1/2*(-P*X^2 + P*X*L)
1010 RETURN
1020 STOP
3000 REM With three moment equations known solved the equations. Start at
3010 line number
3010
CALL CLEAR
3020 PRINT” Put coefficients in data order form starting line 3130, then run line
no 3040”
3030 STOP
34
3040 CALL CLEAR
3050 INPUT” No of equations”N: N
3060 OPTION BASE 1
3070 DIM X (20, 20)
3080 DIM Y (20)
3090 DIM Z (20)
3100 for E = 1 TO N
3110 READ Z (E)
3120 NEXT E
3130 DATA
* Note put coefficients of Ma, Mb
3140 GOSUB 3300
3150 GOSUB 3500
3160 MAT = DELTA
3170 FOR E = 1 TO N
3180 GOSUB 3300
3190 FOR AB = 1 TO N
3200 X (AB, E) = Z (AB)
3210 NEXT AB
3220 GOSUB 3500
3230 F = DELTA/MAT
3240 PRINT CHRS$(64 + E);”=”; F
3250 NEXT E
35
3260 END
3300 RESTORE 3360
3310 FOR Q = 1 TO N
3320 FOR P = 1 TO N
3330 READ X (Q, P)
3340 NEXT P
3350 NEXT Q
3360 DATA
***Put constants of equation 1 & 2
3380 RETURN
3500 REM this is a sub routine
3505 A = 1
3510 FOR P = 0 TO N – 3
3520 FOR S=1 TO N-P
3530 IX X (1, 1) = 0 THEN 680
3540 Y(S) = X(S, 1)
3550 NEXT S
3560 A = Y (1)*A
3570 FOR R =1 TO N-P-1
3580 FOR Q =1 TO N-P-1
3590 X (Q, R) = X (Q + 1, R + 1) – X (1, R+1)*Y (Q+1)/Y*(1)
3600 NEXT Q
3610 NEXT R
36
3620 NEXT P
3630 DELTA = A*(X (1, 1)*X (2, 2)*X (1, 2))
*** Note for array
3640 PRINT X (1, 1); X (1, 2)
3650 PRINT X (2, 1); X (2, 2)
3660 PRINT TAB (5);”DELTA =”DELTA
3670 RETURN
3680 FOR D = 1 TO N-P
3690 IF X (D, 1)><0 THEN 3710
3700 NEXT D
3710 FOR B = 1 TO N-P
3720 W = X (1, B)
3730 X (1, B) = X (D, B)
3740 X (D, B) = W
3750 NEXT B
3760 IF INT (D/2)><D/2 THEN 3780
3770 A = -1*A
3780 GOTO 3520
3790 STOP
*Note with moments at ends already solved reactions at support
4010 CALL CLEAR
37
4020 PRINT “This is a sub routine program for three moment equation
determination of shear reaction at supports with end moments known”
4025 PRINT ‘Input the following values as data statements in program line no
4030”
4030 DATA MA1, MB1, W1, L1, P1, A1 MA2, MB2, W2, L2, P2, A2, MA3, MB3, W3,
L3, P3, A3
4035 DATA W4, A4, P4
4040 REM this is a one dimension array
4045 REM let us consider three spans
4050 dim ma (3)
4055 DIM MB (3)
4060 DIM W (3)
4070 DIM L (3)
4080 DIM P (3)
4090 DIM A (3)
4100 FOR X =1 TO 3
4110 READ MA(X), MB(X), W(X), L(X),P(X),A(X)
4120 RESTORE 4030
4130 R1(X) = MA(X) + MB(X) +W(X)*L(X) ^2/2 + P(X)*L(X) –A(X))
4140 RA(X) = R1(X)/L(X)
4150 PRINT RA(X)
4160 NEXT X
38
4170 RA1 =RA (1)
4180 RB1 = RA (2)
4190 RC1 = RA (3)
4200 REM For shear reactions at R2
4205 DIM MA1 (3)
4210 DIM MB1 (3)
4220 DIM W1 (3)
4230 DIM PI (3)
4225 DIM L1 (3)
4240 DIM A1 (3)
4250 FOR Y=1 TO 3
4260 READ NA1(Y), MB1(Y), W1(Y), L1(Y), P1(Y), A1(Y)
4280 R2(Y) = MA1(Y) + MB1(Y) + W1(Y)*L1(Y) ^2/2 + P1(Y)*A1(Y)
4290 RB(Y) = R2(Y)/ (L1(Y))
4300 PRINT RB(Y)
4310 NEXT Y
4320 RA2 = R2 (1)
4330 RB2 = R2 (2)
4340 RC2 = R2 (3)
4350 QA = RA1
4360 PRINT TAB (3);”Shear reaction at support=”; QA;”KIPS”
39
4370 PRINT TAB (6);”Shear reaction at support B=”; QB;”Kips”
4365 QB = RA2 + RB1
4390 QC = RB2 + RC1
4400 PRINT TAB (10);”Shear reaction at support C=”; QC;”Kips”
4410 READ W4, A4, P4
4420 DATA *** Note for data put value of w4, A4 & P4
4430 RD1 = W4 + A4 + P4
4435 QD = RD1 + RC2
4440 PRINTTAB (15);”Shear reaction at support D=”; RD;”Kips”
4450 PRINT Programmed by Bienvenido C. David a Civil Engineer on May 5, 1981
at Baguio City”
4460 END
40
.
41
40
INTRODUCTION
In the early’s 1950 two methods in Structural Engineering were developed namely the Matrix method of
“Structural Analysis” and ‘Dynamics of Structures”. The coverage of these new methods is so extensive
that a number of excellent books are devoted to this subject.
Systematic structural analysis, “Finite element method” and matrix method of “Structural Analysis” has
all the same meaning. According to its basic principle, because of a structure instead of being a
continuation of differential elements is idealized as a composition of a number of finite pieces. This idea
enables the step by step building of the force displacement relationship of a structure from those basic
elements of which the structure is composed.
Many complicated problems in various fields are thus solved by this new computational technique
including truss beams, rigid frames, plates and shells, composite structures and pressure vessels and
torsion in members with irregulars section.
During the past decades, the rapid development of computers and the growing demand for better
method of analysis for complex and lightweight structures led to the development of “Matrix method of
Structural Analysis”.
It is true that classical methods of “Structural Analysis” such as the method of consistent deformation,
Slope Deflection method, castigliano’s theorem, which have only limited use in the
Past because of operational difficulties, have now regained their strength because of the
invention of the digital computer. Indeed solving a set of 100 simultaneous equations with a
modern computer would hardly take a minute and the solution of simultaneous equations is
equivalent to inverting a matrix.
The slope deflection method can be used to analyze all types of statically indeterminate beams
or rigid frames. The assumption is that all joints are considered rigid, i.e. the angles between
the members at the joints are considered not to change in value as loads are applied.
41
By derivation the general analysis “Slope Deflection Equation” is
Mab = Mfab +
(-2
A
-
B)
Mba = Mfba +
(-2
B
-
A)
Here
Equation 1
Equation 2
is the relative stiffness ratio and represented by the symbol K ,
A
&
B
are end joint rotations in radians at support A, B respectively.
Mfba = Fixed end moments of member BA
Mfab = Fixed end moments of member AB
For uniform loads Mfab = + wl2/12 and equals to - Mfba = - wl2/12
For concentrated loads Mfab = + Pab2/L2 = Mfba – Pba2/L2
Substituting in the value of K =I/L in equation one
We get Mab = Mfab + KAB (-2
Mba = Mfba + Kab (-2
A
B
-
-
B)
A)
Equation one
Equation two
SUGGESTED STEPS
a) Determine the fixed end moments at the end of e ach span using
the formulas mentioned.
b) Express all end moments in terms of the fixed end moments and
the joint rotations by using the slope deflections.
c) Established simultaneous equations with rotations at the supports
as unknowns by applying the conditions that the sum of the end
42
moments acting on the ends of the two members meeting at the
support should be zero.
d) Solve for the rotations at the supports and substitute back into
the “Slope deflection Equations” and comute end moments.
e) Determine all reactions, draw shear and moment diagram and
sketch the curve.
f) Referring from figure in the program record the moment at Joint
A is zero Mab = 0
g) At Joint B Mba + Mbc = 0
h) At joint C Mcb + Mcd + 0
i) At Joint D Mdc = (W4x(L4)2) /2 + P4xy4
COMPUTER OUTPUT
Upon input of numerical data’s computer first solved fixed end
moments. Plot on the screen slope deflection equations, a break
statement is given to enable the user to substitute numerical data’s
into slope deflection equations.
Computer solves joint rotations of end members in succession
substitute back known joint rotations in radians in slope deflection
equations and solves corresponding end moments. With the end
moments as input data it solves required reactions at support A, B,C &
D.
43
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: ANALYSIS TWO: Application to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Slope Deflection Method CODE NAME: ANALYSIS TWO
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 169
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Analysis Two
PTR NO: 3046392 at Bagiuo City 01/09/1982
P2
Loading Diagram
W1/ft
W2/FT
L1
L2
Shear Diagram
P3
L3
P4
L4
44
Moment Diagram
REFERENCE TEXTBOOK:
Statically Indeterminate
Structures By Chu Kia
Wang
CHAPTER 7
PAGES 137 - 146
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
ANALYSIS Two: IS a computer program that solves joint rotations , displacements, end
moments and reactions of three spans with overhang statically indeterminate beams and
frames variable uniform and concentrated loads at any location Y is included (refer to figure
program record 5). The program is written in Basic Language and applicable to a wide variety
of programmable calculators and computers with peripheral attachments can be integrated
to E Review center.
BASIC COMPUTER SYMBOLS
+ ADDITION
- SUBTRACTION
MULTIPLICATION *
^ RAISED TO THE POWER
SQR SQUARE ROOT OF THE NUMBER
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
VALUE
If true
Main program
Branch out
If False
Branch out
SGN = SIGNUM NOTATION
IF THEN ELSE STATEMENT
45
COMPUTER
INSTRUCTION
CODE
10 CALL CLEAR
20 REM this is Slope deflection method of determining moment
coefficients of indeterminate beams as analysis of structures
30 PRINT “Be sure you know matrix method of structural analysis or
matrix algebra otherwise I can’t perform the mathematical operation”
40 PRINT “ L1, L2 ,L3, L4 as span length on feet DL1, DL2, DL3, DL4, LL1,
LL2,LL3, LL4 as uniform loads respectively”
50 PRINT “ Y1, Y2, Y2, Y3, Y4 as distance of concentrated load from
support respectively and let I1, I2, I3 & I4 as moment of inertia constant
for uniform cross section”
60 PRINT “PD1, PD2, PD3, & PD4 as concentrated dead load and PL1,
PL2, PL3 and PL4 as concentrated live load in pounds and pounds per
foot”
70 PRINT “If all data’s are in their consistent units then run line no 80”
75 STOP
80 CALL CLEAR
90 REM first determine moment equation in slope deflection form and
put final results into array format
100 INPUT” DL1, DL2, DL3, DL4, LL1, LL2, LL3, LL4”:DL1, DL2, DL3, DL4,
LL1, LL2, LL3, LL4
110 INPUT “PD1, PD2, PD3, PD4, PL1, PL2, PL3, PL4”:PD1, PD2, PD3,
PD4, PL1, PL2, PL3, PL4
46
120 W1 =1.4*DL1 + 1.7*LL1
2299.1
130 W2 = 1.4*DL2 + 1.7*LL2
1999.5
140 W3 = 1.4*DL3 + 1.7*LL3
150 W4 = 1.4*DL4 + 1.7*LL4
160 P1 = 1.4*PD1 + 1.7*PL1
170 P2 = 1.4*PD2 + 1.7*PL2
180 P3 = 1.4*PD3 + 1.7*PL3
190 P4 = 1.4*PD4 + 1.7*PL4
200 REM Let MFAB, MFBA, MFBC, and MFCB & MFDC & MFCD as end
moments respectively
210 INPUT “Span length in feet L1, L2, L3, L4”:L1, L2, L3, L4
220 INPUT “Distance of concentrated loads from end of supports in feet
YI, Y2, Y3, Y4”:Y1, Y2, Y3, Y4
230 MFAB = (W1*L1^2/12 + P1*Y1*(L1-Y1) ^2/L1^2)/1000 36
240 MFBA = - (W1*L1^2/12 + P1*(L1-Y1)*Y1^2/L1^2)/1000 -36
250 MFBC = (W2*L2^2/12 + P2*Y2*(L2-Y2)*Y2^2
260 MFCB = - (W2*L2^2/12 + P2* (L2-Y2)*Y2^2/L2^2)/1000 -156
270 MFCD = (W3*L3^2/12 + P3*Y3*(L3-Y3) ^2/L3^2)/ 1000
32
280 MFDC = - (W3*L3^2/12 + P3*(L3-Y3)*Y3^2/L3^2)/1000 = -16
47
290 PRINT “ Assume dimension of beam constant tru out then let
relative stiffness factors as K1, K2, K3, K4 then type continue to resume
running”
300 BRAK
310 INPUT” Moment of inertia I1, I2, I3, I4”:I1, I2, I3, I4
320 K1 = I1*12/L1
330 K2 =I2*12/L2
340 K3 = I3*12/L3
350 K4 = I4*12/L4
360 X1 =-2*K1
370 X2 =-K1
380 X3 = --K1
390 X4 = -2*(K1 + K2)
400 X5 = -K2
410 X6 =-2*(K2 + K3)
420 X7 = -K2
430 X8 = -K3
440 X9 = -2*K3
450 X10 = -K3
460 C1 = -MFAB
48
470 C2 = - (MFBA + MFBC)
480 C3 = - (MFBC + MFCD)
490 C4 = - (P4*Y4 + W4*L4^2/2) + MFDC
500 REM Establish slope deflection equations
510 PRINT TAB (2);
X1;”A”;TAB(8);X2;”B”;TAB(15);”C”;TAB(22);”D”;TAB(27);”=”;C1
***Note resulting equation is -6A -3B = -36
Equation 1
520) PRINT TAB (2); X3;”A”; TAB98); X4;”B”; TAB (15); X5;”C”; TAB (22)
;”D”; TAB (27);”=”; C2
*** Note resulting equation is -3XA -16B -5C = -120 Equation 2
530 PRINT TAB(8);X7;”B”;TAB(15);X6;”C”;TAB(22);X8;”D”;TAB(27);”=”C3
*** Note resulting equation is -5B -14C -2D = 124
Equation 3
540 PRINT TAB (15); X10;”C”; TAB (22); X9;”D”TAB (27);”=”; C3
*** Note resulting equation is -2C -4D = 2 Equation 4
550 PRINT” Can you solved the above equations or would you like I will
solve it for you if so then run line no. 70”
560 STOP
570 PRINT “This is computer program no 3 solutions of simultaneous
equations in many unknowns a sub program of slope deflection”
580 PRINT” Be sure you know matrix method of structural analysis or
matrix algebra otherwise I can’t solve the above equations”
49
585 PRINT “ Arrange coefficients of unknowns in their sequential order
from left to right as data statement program line no 70”
600 PRINT “Arrange value of constants top to bottom as data
statements program line no 690”
605 PRINT “ Put no of equations program line no 650 as input
statement if all data’s are secured then run line number 640”
620 STOP
640 CALL CLEAR
650 INPUT” NUMBER OF EQUATIONS”: N
660 OPTION BASE 1
670 DIM X (20, 20)
680 DIM Y (20)
690 DIM Z (20)
700 FOR E = 1 TO N
710 READ Z (E)
720 NEXT E
730 DATA C1, C2, C3, C4
740 GOSUB 900
750 GOSUB 1100
760 MAT = DELTA
770 FOR E= 1 TO N
50
780 GOSUB 900
790 FOR AB = 1 TO N
800 X (AB, E) =Z (AB)
810 NEXT AB
820 GOSUB 1100
830 F = DELTA/MAT
840 PRINT CHR$(64 + E);”=; F
850 NEXT E
860 END
900 RESTORE 960
910 FOR Q=1 TO N
920 FOR P=1 TO N
930 READ X (Q, P)
940 NEXT P
950 NEXT Q
960 DATA X1,X2,0,0,X3,X5,0,0,X7,X6,X8,0,0X10,X9 Example only
980 RETURN
990 REM this is a sub routine
1100 A = 1
1010FOR P=0 TO N-3
51
1020 FOR S=1 TO N-P
1030 IF X (1, 1) = 0 THEN 1080
1040 Y(S) = X(S, 1)
1050 NEXT S
1060 A = Y (1)*A
1070 FOR R =1 TO N-P-1
1080 FOR Q =1 TO N-P-1
1090 X (Q, R) = X (Q + 1, R + 1) – X (1, R + 1)*Y (Q + 1)/Y (1)
1100 NEXT Q
1110 NEXT R
11120 NEXT P
1130 DELTA = A*(X (1, 1)*X (2, 2)-X (2, 1)*X (1, 2))
1140 PRINT X (1, 1); X (1, 2)
1150 PRINT X (2, 1); X (2, 2)
1160 PRINT TAB (5);”DELTA=”DELTA
1170 RETURN
1180 FOR D=1 TO N- P
1190 IF X (D, 1) <>0 THEN 1310
1200 W = X (1, B)
1230 X (1, B) = X (D, B)
52
1240 X (D, B) = W
1250 NEXT B
1260 IF INT (D/2) <>D/2 THEN 1280
1270 A = -1*A
1280 GOTO 1120
1290 STOP
1300 REM with known moments and joint rotations solved substitute
Numerical values back slope deflection equations
1310 INPUT” Known joint rotations in radians A, B, C, D”: A, B, C, D
1320 INPUT” Relative stiffness factors K1, K2, K3, and K4”:K1, K2, K3,
and K4
1330 INPUT “Fixed end moments in foot kips”: MFBA
1335 INPUT “Fix end moments
MFAB,MFBA,MFBC,MFCB,MFCD,MFDC,”:MFAB,MFBA,MFBC,MFCD,MF
DC,MFCB
1340 MAB = MFAB +K1*(-2*A-B)
1350 PRINT “Moment at joint A first span MAB =”MAB;”Ft Kips”
1360 MBA = MFBA + K1 (-2*B-A)
1370 PRINT “Moment at joint B first span MBA =”; MBA;”Ft Kips”
1380 MBC = MFBC + K2*(-2*B-C)
1390 PRINT “Moment at joint C=”; MBC;”Ft Kips”
53
1400 MCB = MFCB + K2*(-2*C-B)
1410 PRINT “Moment at joint C second span =”; MCB;”Ft Kips”
1420 MCD = MFCD + K3*(-2*C-D)
1430 PRINT “Moment at joint C third span MCD”=”; MCD;”Ft Kips”
1440 MDC = MFDC + K3*(-2*D-C)
1450 PRINT” Moment at joint D third span MDC” =”; MDC;”Ft kips”
1460 PRINT “Copy end moments then type continue to resume
running”
1470 BREAK
1480 REM Computations of shear reactions at supports
1490 REM Let RA, RB, RC, RD Total vertical shear reactions at support
1500 INPUT “Uniform loads in pounds per Ft W1, W2, W3, W4”:W1,
W2, W3, W4
1510 INPUT” Concentrated loads in pounds
P1,P2,P3,P4”:P1,P2,P3,P4”:P1,P2,P3,P4
1520 INPUT” Span length in feet L1,L2,L3,L4”:L1,L2,L3,L4
1530 INPUT “Distances of concentrated loads from support
Y1,Y2,Y3,Y4”:Y1,Y2,Y3,Y4
1540 RA1 = (W1*L1^2/2 + P1*(L1-Y1) – MBA)/(1000*L1)
in kips
54
1550 INPUT “Fixed end moments
MAB,MBA,MBC,MCB,MCD,MDC”MAB,MBA,MBC,MCB,MCD,MDC
1560 RA1 = (W1*L1^2/(2*1000) + P1*(L1-Y1)/1000 – MBA)/L1
1570 RB1 = (MBA + P1*Y1/1000 + W1*L1^2/2000)/L1
1580 RB2 = (MBC – MCB + W2*L2^2/2000 + P2*(L2–Y2)/1000)/L2
1590 RB = RB1 + RB2
1600 PRINT “ Total vertical shear reaction at support A=”;RA1;”Kips”
1610 PRINT “Total vertical shear reaction at support B=”;RB;”Kips”
1620 RC1 = (MCB – MBC + P2*Y2/1000 + W2*L2^2/2000)/L2
1630 RC2 = (MCD – MDC + W3*L3/2000 + P3*(L3-Y3)/1000)/L3
1640 RC = RC1 + RC2
1650 PRINT “Total vertical reaction at support C=”RC;”Kips”
1660 REM Summation of forces vertical equals zero
1670 RD1 = (MDC - MCD + W3*L3^2/2000 + P3*Y3)/1000)/L3
1680 RD2 = P4/1000 + W4*Y4
1690 RD = RD1 + RD2
1700 PRINT “ Total vertical shear reaction at support D=”;RD;”Kips”
1710 PRINT “ Draw shear and moment diagram by hand and determine
point of maximum positive moment”
55
1720 PRINT” programmed by Bienvenido C. David a Civil/Structural
Engineer on December 16, 1982”
1730 END
56
56
INTRODUCTION
In 1932 professor Hardy Cross of University of Illinois developed the method of
“Moment Distribution Method” to solve problems inn beams and frame analysis which
involves many unknowns’ joint displacements and rotations. For the next three decades,
moment distribution provides the standard means in engineering offices for the analysis of
indeterminate frames.
The moment distribution method can be regarded as an iterative solution of the slope
deflection equations starting with fixed end moments for each member; these are modified in a
series of cycles, each converging on the precise final result, to account for translation and
rotations of the joints. The resulting series can be terminated whenever one reaches the degree
of accuracy required.
SUMMARY STEPS
a) Assume that all supports are fixed or locked and compute fixed end moments for each
span considered separate from every other span.
57
b) Unlock each support and distribute the unbalanced moment at each one to each
adjacent span by the equation DF = K/SUM K
c) Here DF is the distribution factor K is the stiffness factor for that beam and SUM K is
the sum of the stiffness factors for adjacent beams.
d) After distributing the unbalance moment to each adjacent span, carry over half this
amount, with the same sign to the other end of each span. This completes one cycle of
distribution. If there are N cycles’s steps b & c must be repeated because of the new
unbalanced moment caused by the carry over moments. Such repetitions are made until
the carry over moments become zero. The process maybe stopped when any
distribution is completed. The accuracy of the final results ending on the no. of cycles.
Referring to program no 6 record from the figure
Let Ja = Moment acting on member AB
let Jb = Moment acting on member BA
let Jc = Moment acting on member BC
Let Jd = Moment acting on member CB
Let Je = Moment acting on member CD
Let Jf = Moment acting on member DC
Let Jg = Moment acting on member DE
At any cycle the accumulative moment is JA = JA + AC + AB
Jb = Jb + Bc + bb
Jc = Jc + cc + Cb
Jd = Jd + + Dc + Db
Je = Je + Ec + Eb
Jf = Jf + Fc + Fb
Jg = Fde
For constant cross sections: K: = K1 = K2
58
COMPUTER OUTPUT
Upon input of numerical data’s computer first solves fixed end moments refer figure shown
on program no 6. The user put N = No of cycles and fixed end moments as numerical
input data’s, computer determines and prints on the screen unbalance, carryover moments
and accumulated end moments of last cycles. With end moments known, it solves shear
reactions at supports.
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: ANALYSIS Three: Application to Concrete Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Moment Distribution Method CODE NAME: ANALYSIS Three
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 197
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Analysis Three
PTR NO: 3046392 at Bagiuo City 01/09/1982
Loading Diagram
W1/ft
W2/FT
P2
P3
P4
59
L1
L2
L3
L4
Shear Diagram
Moment Diagram
REFERENCE TEXTBOOK:
Statically Indeterminate
Structures By Chu Kia
Wang
CHAPTER 8
PAGES 216 - 221
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
60
ANALYSIS Three: Is a computer program that determines the moments and shear reactions of
the figure as shown above. The program is good for three spans with a cantilever portion,
uniform loads, and concentrated loads located at any distance X from support is variable.
Span length and cross section variable see computer output for results (refer to figure
program record 6). The program is written in Basic Language and applicable to a wide variety
of programmable calculators and computers with peripheral attachments can be integrated
to E Review center.
BASIC COMPUTER SYMBOLS
+ ADDITION
- SUBTRACTION
MULTIPLICATION *
^ RAISED TO THE POWER
SQR SQUARE ROOT OF THE NUMBER
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
VALUE
If true
Main program
Branch out
If False
COMPUTER
SGN = SIGNUM NOTATION
IF THEN ELSE STATEMENT
Branch out
INSTRUCTION
CODE
COMPUTER INSTRUCTION CODE
10 PRINT “This is computer program no 6 “Analysis of continuous beams and frames by the
moment distribution method”
20 PRINT” First let us consider three spans only with overhang. See drawing figure and program
record for reference”
25 PRINT “All units of force in Kips, Kips per foot, and linear dimensions in feet if all units are in
their respective units then run line no. 30”
30 REM First compute fixed end moments
40 DATA P1, A1, L1
50 DATA P2, A2, L2
60 DATA P3, A3, L3
65 DATA P4, A4, W4, L4
70 REM First for uniform loads a one dimensional array
680 DIM W (3)
61
90 DIM L (3)
80 INPUT “P1, L1, P2, L2, L3”:P1, L1, P2, L2, P3, L3
90 FW1 = 1/12*W1*L1^2
100 FW2 = 1/12*W2*L2^2
120 PW3 = 1/12*W3*L3^2
180 REM For concentrated loads
190 RESTORE 40
200 GOSUB 520
210 FP1 = Q1
220 RESTORE 50
230 GOSUB 620
240 FP2 = Q1
250 RESTORE 60
260 GOSUB 520
270 FP3 = Q1
280 RESTORE 40
290 GOSUB 570
300 FB1 = Q2
310 RESTORE 50
320 GOSUB 570
330 FB2 = Q2
340 RESTORE 60
350 GOSUB 570
360 FB3 = M2
62
370 FAB = FW1 + FP1
380 FBA = - (FW1 + FB1)
390 FBC = FW2 + FP2
400 FCB = - (FW2 + FB2)
410 FCD = FW3 + FP3
420 FDC + - (FW3 + FP3)
430 READ P4, A4, W4, L4
440 RESTORE 65
445 FDE = P4*A4 + W4*L4^2/2
450 PRINT TAB (2);”FAB=”; FAB;”Foot Kips”
460 PRINT TAB (12);”FBC=”; FBC;”Foot Kips”
470 PRINT TAB (16);”FCB=”; FCB;”Foot Kips”
480 PRINT TAB (18);”FCD=”;”Foot Kips”
490 PRINT TAB (20);”FDC=”; FDC;”Foot Kips”
500 PRINT TAB (25);”FDE=”; FDE;”Foot Kips”
455 PRINT TAB (8);”FBA=”; FBA” Foot Kips”
520 REM A sub routine one
530 READ Z, H, N
*** Note Z, H, N stands for P, A, & L previously
540 Q1 = Z*H*(N – H)/N^2
550 RETURN
560 STOP
570 REM A sub routine no two
580 READ Z, H, N
590 Q2 = Z*H^2*(N – H)/N^2
63
600 STOP
610 PRINT “With fixed end moments computed input values for moment distribution method”
Start at program line no 1010
1010 CALL CLEAR
1020 PRINT “This is computer program no 6 moment distribution method in basic language
good for moment tables and accumulated moments at end of last cycle”
1030 PRINT” I can only tabulate all the carry over and balance moments however since my
visual display is limited to 14 program lines I can only display 5 to 6 cycles”
1040 PRINT”Input all fixed end moments previously computed in foot kips FAB, FBC, FCD, FCB,
FDC, FDE”
1050 PRINT” For figure and program description see program record no 6 if all data’s are
secured then run line no 1070”
1060 STOP
1070 REM Let the following symbols stands for carry over moments at AC, BC, CC, DC, EC, FC,
GC
1080 REM Let the following legends for balance moments at AB, BB, CB, DB, EB, FB, GB
1090 REM LET K1, K2, K3 & K4 as relative stiffness factors hand calculated
1100 INPUT” Fixed end moments” FAB, FBA, FBC, FCB, FDC, FCD FDE”: FAB, FBA, FBC, FCB,
FDC, FCD, FDE
1110 INPUT” Relative stiffness factors K1, K2, K3, K4”:K1, K2, K3, K4
1120 INPUT” How many cycles do you like N”: N
1140 LET X = 1
1145 PRINT “I am now performing cycle no =”; X
1150 REM Lock joint A
1160 BC = FBA
64
1165 CC = FBC
1170 REM Lock joint C
1180 DC = FCB
1190 EC = FCD
1200 LOCK Joint D
1210 FC = FDC
1220 GC = FDE
1230 REM Release Joint A
1240 A = AC
1250 GOSUB 600
1260 AB = C
1270 REM Release Joint B
1280 UB + BC + CC
1290 GOSUB 690
1300 BB = MBA
1310 CB = MBC
1320 REM Release Joint C
1330 UC = DC + EC
1340 GOSUB 790
1360 DB = MCB
1360 EB = MCD
1370 REM Release joint D
1372 IF X = 1 THEN 1374 ELSE 1378
1374 UD = FC + GC
65
1375 GOSUB 890
1376 Y = MDC
1377 GOTO 1380
1378 Y = -FC
1380 FB = Y
1420 JA = JA + AC + AB
1430 JB = JB + BC + + BB
1435 JC = JC + CC + CB
1440 JD = JD + DC + DB
1445 JE = JE + JE + EC + EB
1448 JF + JF + FC + FB
1450 JG = FDE
1460 IF N = 0 THEN 1461 ELSE 1470
1461 PRINT TAB (7); JA”Foot Kips”
1463 PRINT TAB (12); JB;”Foot kips”
1465 PRINT TAB (15); JC;”Foot Kips”
1466 PRINT TAB918); JD;”Foot Kips”
1467 PRINT TAB (21); JE;”Foot Kips”
1468 PRINT TAB (24); JF;”Foot Kips”
1464 PRINT TAB (27); JG;”Foot Kips”
1469 STOP
1470 Rem lock Joint A
1480 FAB = 1/2*BB
1490 FBA = 1/2*AB
66
1500 FBC + 1/2*DB
1510 REM Locke Joint C
1520 FCB = 1/2*CB
1530 FCD = 1/2*FB
1540 REM Lock Joint D
1550 FDC = 1/2*EB
1560 FDE = FDE
1570 X = X + 1
1580 PRINT “I am performing cycle no=”; X
1590 GOTO 1150
1595 STOP
1600 REM This is a sub routine no 1 determination of unbalance moment at joint A. First joint A
is held lock (i.e. degree of freedom is zero) then release (degree of freedom is set free)
1620 IF X>0 THEN 1630 ELSE 1640
1630 B = ABS (A)
1640 GOTO 1660
1650 B = -A
1660 C = B
1670 RETURN
1680 STOP
1690 REM This is sub routine no two. Determination and distribution of unbalance moment at
joint B (first joint B is held lock then release (degree of freedom is set free)
1700 IF UB>0 THEN 1710 ELSE 1730
1710 M = ABS (UB)
67
1720 GOTO 1740
1730 M = -UB
1740 MC = M
1750 MBA = K1/ (K1 + K2)*MC
1760 MBC = K2/ (K1 + K2)*MC
1770 RETURN
1780 STOP
1790 REM This is a sub routine no three. Determination and distribution of unbalanced
moment at joint C (first joint C is held lock degree of freedom is zero then release degree of
freedom is set free)
1800 IF UC>0 THEN 1810 ELSE 1830
1810 M = ABS (UC)
1820 GOTO 1840
1830 M = -UC
1840 MC = M
1850 MCB = K2/ (K2 + K3)*MC
1860 MCD = K3/ (K2 + K3)*MC
1870 RETURN
1880 STOP
1890 REM This is sub routine no four cyclic determination of unbalance moment at joint D (First
joint D is held lock degree of freedom is zero then release degree of freedom is set free)
1900 IF UD>0 THEN 1910 ELSE 1930
1910 M = ABS (UD)
1920 GOTO 1940
1930 M + -UD
68
1940 MC = M
1850 MDC = MC
1960 RETURN
1970 STOP
*** Start at program line no 2380 this time
2380 REM With end moments already computed determine shear reactions at supports
2390 REM Let RA, RB, RC, RD Total vertical reactions at support A, B, C, D respectively
2400 INPUT” Uniform loads in pounds per foot W1, W2, W3, W4”:W1, W2, W3, W4
2410 INPUT” Concentrated loads in pounds P1, P2, P3, P4”:P1, P2, P3, P4
2420 INPUT” Span length in feet L1, L2, L3, L4”:L1, L2, L3, L4
2430 INPUT” Distance of concentrated loads from support Y1, Y2, Y3, Y4”:Y1, Y2, Y3, Y4
2440 RA1 = (W1*L1^2/2 + P1*(L1 - Y1) – MBA)/1000*L1)
in kips
2450 INPUT” Fixed end moments
MAB,MBA,MBC,,MCB,MCD,MDC”:MAB,MBA,MBC,MCB,MCD,MDC
2460 RA1 = (W1*LI^2/ (2*1000) + P1*(L1 –Y1)/1000 – MBA)/L1 Kips
2480 RB2 = (MBC – MCB) + W2*L2^2/2000 + P2 (L2 – Y2)/1000)/L2 Kips
2470 RB1 = (MBA + P1*Y1/1000 + W1*L1^2/2000)/L1 in kips
2490 RB = RB1 + RB2 Kips
2500 PRINT” Vertical shear reaction at support B=”; RB;” Kips
2445 PRINT” Vertical shear reaction at support A=”; RA1;”Kips
2510 RC1 = (MCB – MBC + P2*Y2/1000 + W2*L2^2/2000)/L2 Kips
2520 RC2 = (MCD – MDC + W3*L3/2000 + P3*(L3 – Y3)/1000)/L3 Kips
69
2530 RC = RC1 + RC2 Kips
2540 PRINT “Total vertical reaction at support C=”; RC;”Kips”
2550 REM Summation of forces vertical equals zero
2560 RD1 = (MDC – MC + W3*L3^2/2000 + P3*Y3)/1000/L3 Kips
2570 RD = RD1 + RD2
in kips
2580 PRINT” Total vertical shear reaction at support D=”; RD;”Kips”
2590 PRINT” Draw shear and moment diagram by hand and determine maximum positive
moment”
2600 PRINT” Programmed by Bienvenido C. David a civil/structural engineer on Jan 20, 1983 at
Baguio City”
2700 END
*** Call clear Resequence 10, 10 for correct numbering
70
70
In this chapter, elaborate discussion of concrete, the aggregates and steel reinforcement is given in
detail. The absolute volume computation by empirical formula and the equivalent numerical example is
given. A table for overload factors used in design is included for the convenience of the designer. The
analysis of:”Reinforced Concrete” is discussed in detail, at the end of the chapter; a summary
treatment of both W.S.D. and U.S.D. format is included. This chapter will serve as an introduction to the
succeeding chapters in the book For Engineers who have a limited time, a review of
chapter three will serve as a refresher course in “Reinforced Concrete”.
INTRODUCTION
The structural Engineering profession has grown tremendously in the last twenty
years. The accelerated development of the digital computer and their expanding
application to structural engineering profession have a tremendous impact on
handling the analytical and design task encountered. With the availability of the
digital computer, the practitioner is able to use classical theories whenever
appropriate, even though large amount of simultaneous equations maybe
involved.
71
Today the “Structural Engineering” is involved in many Engineering projects; in the Civil
Engineering field, he assists the Transportation Engineer, Hydraulic Engineer and Sanitary
Engineer by providing the structures needed to implement their projects. In building
construction, he is one of the principal collaborators of the Architect. In the mechanical,
Electrical and mechanical field, he is responsible for the designing the heavy machineries
required or facilities required for their projects. He may shift his entire activity into naval
architecture and become a specialist in the design of ship structures. In aerospace engineering,
he maybe involve in providing special structures for launching space vehicles. Where design of a
structure; such as a large dam or large industrial facility, he may provide the leadership for
undertaking the project.
“Structural mechanics” is the main tool used in Structural Engineering”. Subjects such as
“Engineering mechanics”, “Strength of Materials”, Theory of Structures” are only
a part of such wide field activity in the structural science. Recent development s includes shells
and plates, finite pieces and the law of continuum mechanics.
Before the advent of the digital computer, the classical method of “Structural Analysis” is
usually employed at that time, the available calculating machine is the slide rule followed by
the more precise scientific calculator.
However in the late 1970 and earl’s 1980; a new development in the field of computation
appeared worldwide, known today as the digital computer, thus with this new innovation, new
methods of “Structural Analysis” were developed; one such development is known as
“Systematic Structural Analysis” or simply the “Finite Element method”. This
new method employs “Matrix algebra as its main mathematical tools.
72
Difference between Classical and Systematic Analysis
Classical analysis uses the standard method of manipulating the desired unknowns in a given
structure. Geometry and algebra are the analytical used in this method, whereas in the
systematic analysis Matrix algebra is the main tools used. In erecting a structure there are three
phases involved. The planning stage, the design phase and the construction stage. In our case
our concerned will be part two the design phase particularly “Design of Reinforced Structures”.
In analogous to “Structural Analysis”, there are two methods employed in the design phase, the
standard procedure and the systematic procedure.
The standard procedure is analogous to classical Structural analysis where algebra and the
geometry of the structure is the main analytical used. Systematic procedure is analogous to
Systematic Structural Analysis”, however there is a difference between the two; while in
systematic structural analysis one is forced to use arrays while in systematic design procedure
logical steps is the method employed. As a rule then we can put it this way “matrix algebra” for
Systematic Structural Analysis” and Boolean algebra for “Systematic Design procedure”
From the above discussion, we can summarize it in three steps
1 Mastery of the subject defined (i.e. Reinforced Concrete
2 Knowledge of any computer’s language (i.e. Basic, Fortran and Sol or Sql)
3, Transforming the standard method into computer language.
The last step is probably the most difficult to accomplish. Sometimes this requires formula
derivations, evaluation of units and evaluating numerical constants. The need for such
transformation is essential for computer application, one step three is accomplished and feed
into the computer, the resulting results is the same as the standard procedure.
73
WHYS IS SYSTEMATIC STRUCTURAL ANALYSIS IMPORTANT
The answer to this question is time element. Computer is a modern approach to “Structural
Analysis”; rather than bogging yourself with slide rule, opening up charts and tables we can
concentrate on Engineering design. Second unlike human beings, computers seldom commit
errors thus the numerical values obtained are précised and the third reason is speed,
sometimes Civil or particularly “Structural Engineers demands that we meet deadlines,
without these machines ; teams of Engineers would be needed to do the manual
computations thus lengthening the time between design and final design.
SUMMARY TREATMENT
Design Behavior and Philosophy
CONCRTE: Is an artificial stone that is cast in place in a plastic condition. Its essential
ingredients are cement and water, which reacts with each other chemically to form another
material having useful strength.
AGRGGREGATES: One of the principal composition of concrete, the other being cement. The
fine aggregate is composed of sand and the course aggregate is gravel or crushed stone.
CEMENT: Cement is the principal composition of concrete, briefly it’s made by mixing and
then burning to incipient fusion, the two materials, one composed principally of lime, the
other being clay or argillaceous material containing silica, alumina and iron.
ADMIXTURE: Substances added to concrete to improve its workability, accelerates its set,
hardens its surface, increase its waterproofing qualities.
74
REINFORCEMENT: Steel bars for reinforcement in concrete are made from billet steel and rail
steel. The three grades of billet steel are structural, intermediate and hard. Manufacturers give
different specifications for different brands of steel.
ABSOLUTE VOLUME AND DESIGN MIXTURE: It has been custom to express the relative
quantities of the concrete ingredients by bulk volume, in this order: cement, sand, stone.
For example a 1:2:4 mix signifies that, for every cubic foot of cement, 2 cubic foot of sand and
4 cubic foot of gravel are to be used. Important conversions as follows.
1 Cubic Ft. = 7.48 Galloons
1 Cu Yard = 27 Cubic Ft.
1 Cubic foot of water weighs 62.4 Lbs
1 Gallons of water weighs 8.35 Lbs
A sack of Portland cement is assumed to have a volume of 1 Cu.Foot and weighs 94 Lbs
Specific gravity of cement is 3.10 and that of aggregates is 2.65
COMPRESSIVE STRENGTH: Depending on the mix especially the water cement ratio and the
time and quality of curing, compressive strength of concrete can be attained up to 14000 P.S.I.
or more. an ordinary aggregate is usually in the 3000 p.s.i. to 5000 p.s.i.
TENSILE STRENGTH: The tensile strength of concrete is relatively low, about 10 – 20 % of the
compressive strength of lightweight concrete, but not at all cases, has a lower strength that
ordinary weight of concrete.
A.C.I. CODES SPECIFICATIONS, ANALYSIS AND FORMULA DERIVATIONS
ULTIMATE STRENGTH: From investigation and experimental results, it has been found out that
at or near ultimate , stress are no longer proportional to strain, thus the general working stress
theory based on elastic analysis is no longer applicable.
75
Test conducted at Lehigh university and University of Illinois indicates that at higher loads axial
and bending moments adhere closely to the ultimate strength theory which is the main
alternative procedure now used in “structural design practice.”
TYPICAL FORMULAS USED IN ULTIMATE STRENGTH DESIGN
1. Cu = 0.85abfc’
4. a = 1.18wd
2. Tu = Asfy
5. c = 1.18wd
6. Mu = AsFy(d – )
3. w =
For two way slab design only
7. Mu =
bd2Fc’w(1 – 0.59w)
8. Mu = AsFyd(1 – 0.59w)
9. As = bdFc’ -
(bdFc’)2 – 2bFc’Mu/
Fy
10. 10 Cb =
(d)
11 Pb = 0.85B1(
12 Wb = ,85B1 (87000/(87000 + Fy)
P
)(87000/(87000 + Fy)
(1 ±
)
1 – 2.622Mu/(bd2Fc’)
for two slab design
76
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
d2
–
For Computer application
STRENGTH REDUCTION FACTORS IN A.C.I CODE 1977 & 1983
KIND OF STRENGTH
Flexural with or without axial tention
Axial Tention
Axial compression, with or without flexure
Members with spiral reinforcements
Other reinforced members
Except that for low values of axial load,
For members in which Fy does not exceed 60,000 p.s.i., with symmetrical
reinforcement, maybe taken as 0.90 as n decreases from 0.10 fc’Ag to
zero.
For others reinforced members,
from .10fc’Ag or Pnb,
whichever is smaller, to zero
Shear and torsion
Bearing on concrete
Flexure in plain concrete
REDUCTION
FACTOR
.90
,90
.80
.70
.70
.85
.70
.65
77
Factored - load combinations for determining required strength in A.C.I. code
Dead load D plus Live load L
Dead + Live + wind Load W when additive
Same as item 2 when gravity counteracts wind – loads effects
In structures designed for earthquake loads or forces E, replace W by 1.1 E
in items 2, and 3.
When lateral earth pressure H acts in addition to gravity forces when
effects are additive
Same as item 5 when gravity counteracts earth pressure effects
1.4D + 1.7L
.75(1.4D + 1.7L +
1.7W)
.9D + 1.3W
1.4D + 1.7L + 1.7H
.9D + 1.7H
Reference textbooks
Design of Reinforced Concrete Structures 9th Edition By
Reinforced Concrete Fundamentals By
The Theory and practice of Reinforced concrete 4rth
edition By
Simplified Design of reinforced Concrete By
Foundation Engineering By
Foundation Analysis and design 3rd Edition
A.C.I. Code of practice 1977 & 1983
George Winter & Arthur Nielsen
Phil M Ferguson
Clarence w Dunham
Parker
Peck And Hanson
Joseph Bowles
78
INTRODUCTION:
In this chapter three types of “Reinforced concrete beams will be discussed, namely the design
of “Single Reinforced Beam”, design of “Double Reinforced Beam and the third one “Design
of Tee Beam”.
The author discussed in details the analysis for each type of beam. Formula derivations have
been derived for computer applications. Design steps and A.C.I code provisions; legends have
been included to understand the program steps involve. Basic theory and analysis shown in the
figure below is a guide for deriving formulas to be used in the programs presented in the
succeeding topics.
LEGENDS
DESCRIPTION OF LEGENDS
Fc’
Fy
As
b
d
Cylinder strength of concrete at ultimate in Kips per square inch
P
Pb
Pm
Steel reinforcement ratio
Ultimate yield strength of steel in kips per square inch
Area of tention steel in square inches
Width of beam in inches
Depth of beam distance from centroid tention steel up to outer concrete fiber stress.
Balanced steel reinforcement ratio from code
Maximum steel reinforcement ratio from code = 0.75pb
79
Pn
Es
ec
Es
B1
Minimum steel reinforcement ratio from code = 200/fy
Modulus of elasticity of steel in pounds per square inch, given
Strain of concrete at ultimate = 0.003 at failure
Strain of steel = 6fs/Es = fs/29,000,000 at failure
Concrete stress block parameter
Moment workmanship factor usually equals to 0.90
Pd
Md
Pn
Mn
Design moment = Pn
Design moment = Mn
Nominal axial load`
Nominal axial moment
The above notations conform to the latest 1977 A.C.I. code and will be used thru
out this book. The analysis of a “Reinforced Concrete Beam” can be visualized
well based from the figure below.
ec
⃝
d
⃝
⃝
As
⃝
c
d h
⃝
⃝
.85fc’
es’
a
N.A
es
Cc
(d - a/2)
(d – c)
Ts
b
Figure 1 Cross section
Figure 2 Strain diagram Figure 3 Stress block diagram
From the strain diagram as shown above and by similarity of triangles, the distance c for balance
eccentricity, let the symbol cb = distance of N.A. from the outer fiber concrete then for simultaneous
failure of steel and concrete, substitute the strain of concrete = 0.003 and strain of steel equals
es = fs/29x106 in the strain diagram then Cb = (ec/ (es + ec) d
Cb = (.003)/(fs/Es + 0 .003) d = (0.003/ (fy/29x106 + 0.003) d
Cb = 87/ (fy + 87) d Equation (1)
But Cb is also equals to (Pb) (fy)/B1fc’ (d) Equation (2)
80
Equating (1) and (2) we get (87)/ (fy + 87) d = (Pb) (fy)/B1fc’ (d) solving for Pb
we get
Pb = (B1)(fc’)/fy(87/(fy + 87)) Equation (4) This equation is known as the balance steel
reinforcement ratio from code , the factor B1is a Para meter that relates the depth of the of
equivalent rectangular stress block to the depth of the actual neutral axis.. The factor B1is
equals to 0.85 for fc’ 4000 psi and decreases by 0.005 for every 1000 psi above 4000 psi to a
minimum of 0.65. The code specifies the steel maximum steel ratio shall not exceed 0.75pb
and a minimum of pn = 200/fy
= 0 .85fc’ab = Asfy or solving for As =
From the figure applying
or solving the depth of stress rectangular block a =
Eq (3)
Eq (4)
in terms of steel reinforcement ratio p As = p (b) (d) Eq (5)
Taking moments about centroid compression concrete block
MN = Asfy (d – a/2)
Eq (8) from Eq (5) substitute the value of As in (8) we get
Md = (p) (b) (d) (fy) (d –
Md = (p) (b) (d) (fy) (1 –
) simplifying further we have
)
Equation 3, 4, 5 can be used to solve any required unknowns a, As, or p
There are many techniques in designing a beam. one method is to assume the
depth of stress rectangular block a then compute the area of steel then checked
the assumed value of a by recomputing the new depth of stress rectangular
block a. . If the computed depth of stress rectangular block a does not coincide
with the first assumption then a second trial a will be performed, the process goes
81
on until the correct value of a is obtained. Design experience indicates that a
value of a from 1/3 to 1/2 of depth is a good assumption.
The second method is thru the use of the steel reinforcement ratio from graph,
the mathematical expression Mu/ d2 is evaluated first and from the graph the
equivalent steel reinforcement ratio is selected. The above techniques are useful
if there are only few beams to design. In the second method, if the evaluated
mathematical expression Mu/ d2 is too small, it is difficult to read the steel
±reinforcement ratio p from graph, besides the two methods discussed is not
applicable for a micro computer solution. we therefore derived a formula for
finding the depth of stress rectangular block in terms of material strength
specifications, external moment Mu and dimension of beam in inches. In a
similar manner, it is more convenient to express a direct formula for area of steel
As in terms of material strength specifications, external moment Mu and beam
dimension b & d. Likewise for steel reinforcement ratio. The reasons for such
derivations are the need for computer application. In the following programs, we
2
often encountered the mathematical expression a = d ± d –
As =
and
this is just one example how the author use this mathematical
formula to solve area of steel at any section along the beam by a micro computer
solution. The other formula is useful both for beam design and slab design, the
mathematical expression for p is equals to p =
( 1 ± 2.622(Mu)/bd2) and
for the direct formula for As using the refined but less accurate formula for As
bdFc’ ±
As =
(bdFc’)2 – 2bFc’Mu/
fy
In our general computer program , we use the standard sequence for solving the area of
steel at the supports and midspan, however by using the author’s derived formula for As and
steel reinforcement ratio p, we can treat the mathematical expression as a sub routine program
within the main program.
82
INTRODUCTION
Most common type of structures has a cross section composed of single reinforced beams. This means
that the top section of the beam above neutral axis is in compression and the lower section below the
●●neutral axis is in tention where the placement of steel bars is usually located. Code specifications as
listed in chapter 3 will be used. In addition design of diagonal tention (stirrups) using the less accurate
A.C.I code is integrated in this design.
ANALYSIS AND FORMULAS DERIVATIONS FOR COMPUTER APPLICATIONS
c
.85Fc’
ec
d
d
●
●
b
Cross section
es
Pressure diagram
(d-c)
Strain diagram
From the figure a single reinforced concrete beam, it is evident that AsFy = Cc
where Cc = 0.85Fc’ (a) (b) or solving for As =
Eq (1) Taking moments
about centroid of compressive concrete block we get Mu = AsFy (d – a/2) Eq (2)
here is the moment reduction factor which is equals to 0.90
substituting the values of As in (2) we have
Mu = (.90) (Fy) (0.85Fc’b (a) (d – a/2) or simplifying further we get
83
Mu =
= 0.85(Fc’) (b) (ad –a2/2) or
= ad – a2/2
= a2 – 2ad adding d2 to both sides of the equations. We have
=
2
2
2
+ d = a -2ad + d
or d2 -
= (a – d)2
Extracting square roots of both sides we get and transposing we get the formula
for solving the depth of stress rectangular block a =
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
d2
–
For Computer application
In this equation we take the smaller value of a. The following formula is useful for
computer application given Mu as bending moment, d as depth of beam and Fc’
and Fy as material strength specifications **** Note author use this mathematical
expression as sun routine programs for solving required depth of stress
rectangular block at any section.
Following the same analysis we get the following formula for the area of steel As
As =
(b)(d) ±
(0.7225Fc’2b2d2)/Fy2 – 1.888Fc’(b)(Mu)/Fy2))
The above formula is for direct computer application . A more refined but less
conservative formula in lieu of the above formula can be used is given as
84
9. As = bdFc’ -
(bdFc’)2 – 2bFc’Mu/
Fy
IMPORTANT DESIGN STEPS
U.S.D. ALTERNATIVE
1 PB = .85B1
(
) where B1 -.85 for Fc’ = 4000 psi and decreases 0.05
for every 1000 psi above 4000.00 but not less than 0.65
2) Limit of Pb from code = .75Pb Pmax = 0.75Pb where Pb is the balance steel
reinforcement ratio from code. Given the area of steel as and dimension of beam
b, d & h a- depth of stress rectangular block is a =
reinforcement ratio from code p a =
or in terms of steel
nominal flexural strength is
Mn = AsFy(d – a/2) in terms of steel reinforcement ratio p
Mn = p(Fy)(b)(d)2( 1 - .59p
) with a value of
= .90 as workmanship factor for
moment we let Md =Design moment then
Md =
p(Fy)(b)(d)2( 1 - .59p
) the code limits Pmin = 200/Fy
Note for computer application use author derives formula a depth of stress
rectangular block.
a =±
d2 –
where M is in inch kips Fc’ = K.P.S.I
85
Note the quantity d2 –
must not be less than zero take lower
positive real root of a
LEGENDS
LEGENDS
Fy
Fc
CA
CB
CC
CU
N
LC
LL
DL
1.4
1.7
DESCRIPTION
Yield strength of steel at ultimate in Kips Per Square Inch
Cylinder strength of concrete at ultimate
coefficients of moments at support a
Coefficient of moment at support b
Coefficient of moment at support c
Largest coefficient of moment for solving depth of beam
Percentage of steel ratio i.e. designer’s choice
Clear distance of beam in feet.
Live loads in pounds
Dead load in pounds
Dead load factor at ultimate
Live load factor at ultimate
Diagonal factor taken as 0.85
Note for more detail discussion please refer “Design of Reinforced Concrete
Structures” by Clarence W Dunham or George Winter
Other reference books listed below
Reinforced Concrete Fundamentals By
The Theory and practice of Reinforced concrete
4rth edition By
Simplified Design of reinforced Concrete By
Phil M Ferguson
Clarence w Dunham
Parker
86
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design of Single Reinforced Concrete Beam
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Single Reinforced Concrete Beam CODE NAME: CON BEAM
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 169
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Con Beam
PTR
NO: 3236392 at Bagiuo City 01/10/1984
CA = 1/9WL2, CB = 1/14WL2, CC = 1/16WL2
If .C.I. code is applicable w/ Ft in pounds
LC
Distance c from N.A
eu
.85fc’
●
●
●
● ●
d
h
●
a
N.A
● ● ●
B
EXTERIOR SUPPORT
B
MIDSPAN
es
B
INTERIOR SUPPORT
AsFy
STRESS & STRAIN DIAGRAM
87
REFERENCE TEXTBOOK:
Design of Concrete
Structures By George
Winter 9th edition
CHAPTER 4
PAGES 112 - 117
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
CON – BEAM: Is a computer program that determines the size, steel reinforcement and
diagonal tension bars of a single reinforced concrete beam. With given live loads and assume
width of beam, computer first determines height of beam, dead load of section is automatically
computed, computes areas of reinforcement steel at interior support, midspan & exterior
support all in conformity with A.C.I. code 1977 & 1983 specifications. Checks whether section is
adequate for shear. Design stirrups required and spacing at any desired locations X distance
from column face. The program is written in Advance Basic and can be feed to a wide variety of
programmable calculators and micro computers.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
88
COMPUTER INSTRUCTION CODE
PROGRAM STATEMENT
PROGRAM
LINE NO
10
20
25
30
35
40
50
60
65
***
70
75
80
85
90
95
100
CALL CLEAR
PRINT “ This is design of continues beam by
U.S.D. method in English units”
PRINT” For legends and drawings refer to
program record and program description”
PRINT” This program was developed by
Bienvenido C. David a civil/structural engineer”
PRINT” All material strength specifications in
Kips per square Inch, Live loads in pounds per
foot, Linear dimension in feet”
PRINT” If all data’s are in their respective units
then run line no 60"
STOP
INPUT” Width of beam in inches”;B
INPUT” Percentage of steel ratio N”:N
Note here N depends on the designer choice
INPUT” Live load in Kips per SQUARE ft”:LL
INPUT” Cylinder strength of concrete at
ultimate”:FC
INPUT” Yield strength of steel at ultimate FY”:FY
INPUT” Length of beam in feet”:LC
INPUT” Coefficient of negative moment at
support A”:CA
INPUT’Coeffecients of positive moment at
center span”:CB
INPUT”Coeffecients of negative moment at
support C”:CC
NUMERICAL
OUTPUT *** For
debugging
purposes only
sample only
16
1
4310
3
40
20
.07143
0.111111
0.0625
89
105
110
120
130
140
150
160
170
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
335
340
INPUT”Coeffecients of maximum shear reaction
at support CR for VU max”:CR support CR for VU
max”:CR
INPUT”Coeffecients of maximum moment for
solving depth of beam CU”:CU
IF FC≥4 THEN 130 ELSE 150
BA = 0.85
GOTO 160
BA = 0.85 – (FC – 4)/1*(.05)
350
355
360
370
375
PB = 0.5418758*FC/FY*87/(87 + FY)*BA
PC = N*PB
PM = 0.2/FY
IF PC<=PM THEN 220
IF PC>PM THEN 240
P = PM
GOTO 250
P = PC
X = 1 – 0.59*P*(FY/FC)
Y = 0.90*P*FY*B
G = X*Y
E = 83.33*G/(CU*LC^2)
F = -1.456*B
C = -(3.64*B + 1.7*LL)
Q = (F^2 -4*E*C)^.5
D = (-F + Q)/(2*E)
H = D + 2.5
PRINT” Height of beam in inches =”;H;”Inches”
REM Compute areas of steel at supports and
midspan
DL = 1.456*B*H
LL = 1.7*LL
W = DL + LL
MA = 3/250*CA*W*LC^2
IF (D^2-2.61*MA/(FC*B))<0 THEN 385 ELSE 395
385
PRINT” Depth of stress rectangular block is not
0.11111
BA = .85 sample
only
.0236694
.0236694
.005
0.005
.0236643
0.8138408
13.630637
11.093169
20.79
-23.296
-7,385.24
784.2343
19.211087
21.91087
510.4356
7,327
7837.436
2687.174
90
390
395
400
405
410
415
420
425
430
435
440
445
450
460
465
470
475
480
495
500
510
520
530
540
545
550
555
possible a is imaginary review given data”
STOP
AA = D – (D^2-2.61*MA/(FC*B)).5
AE = MA/(.90*FY*(D-AA/2))
PRINT” Area of steel at exterior
support=”AE;”Square Inches”
REM For positive steel at center span
MB = 3/250*CB*W*LC^2
IF (D^2 – 2.61*MB/(FC*B))<0 THEN 425 ELSE
435
PRINT” Depth of stress rectangular block at
center is not possible review given data”
STOP
AB = D –(D^2-2.61*MB/(FC*B))^.5
ASP = MB/(.90*FY*(D-AB/2))
PRINT” Area of steel at midspan=”ASP;”Sq Inch”
REM For steel at interior support
MC = 3/250*CC*W*LC^2
IF (D^2-2.61*MC/(FC*B))<0 THEN 470 ELSE 480
PRINT” Depth of stress rectangular block at
interior support is imaginary not possible review
given data”
STOP
AC = D-(D^2-2.61*MC/(FC*B))^.5
ASI = MC/(.90*FY*(D-AC/2))
PRINT” Area of steel at interior
support=”;ASI;”Sq.Inch”
PRINT” Select from tables size and bar no then
type continue and press enter to resume
running”
BREAK
REM Design of diagonal tention using the less
accurate A.C.I. code
VU = CR*W*LC/2000
VCA = 0.053754*FC^.5*B*D
IF VU<=VCA THEN 555 ELSE 580
PRINT” Section adequate for diagonal tention
4.223156
4.314843
4179.548
7.183078
7.339031
2351.231
3.633238
3.7121
79.299
34.83
91
560
565
580
590
600
610
620
630
635
640
645
650
660
670
680
690
700
no stirrups needed in this region just provide
extra diagonal tention for unforeseen loads or
as directed by field Engineer”
PRINT” Programmed by Bienvenido C. David A
Civil/Structural engineer on February 16, 1981”
STOP
580 REM Stirrups needed in this region let us
use U shape stirrups
VCD = VU – VCA
J = 1000*VU/W
JA = 0.5*CR*VCA*LC/VY
Z = J - JA
ZA = 0.187*FY*D/VCD
Rem Let’s take an increment of every 1 feet
distance from support
PRINT” Point of zero excess shear from support
location where stirrups are needed is
=”;Z;”feet”
REM For every 1 feet interval
FOR S = 1 TO Z STEP 1
Z1 = ZA*Z/(Z-S)
PRINT”Recqired spacing at 1 feet interval is
=”;Z1;”Feet”
NEXT S
PRINT”Programed by Bienvenido C. David A
Civil/Structural engineer on February 1982 at
Baguio City”
END
44.469
10
4.39
5.61
3.93
92
INTRODUCTION
The name compression beam is analogous to double reinforced beam.
When the capacity of a given cross section the compressive force of concrete is less
than the maximum external moment, then steel reinforcement is needed in that
part of the beam in which the stresses are compressive to supplement the
concrete in resisting compressive stresses. Beams containing reinforcement
for both tension and compression are known as double reinforced
beam.
Double reinforced beam arises when computed dimensions of single
reinforced beam is not possible for architectural reasons, thus the only
alternative is to analyze the given beam as double reinforced beam.
In recent years, due to the development of high strength materials, compression
beam is seldom used. however there are still situations in which use of double
reinforced beam is the only alternative.
Our analysis will be based entirely on the U.S.D. method
ANALYSIS AND FORMULA DERIVATIONS
b
●
eu
● ●
d’
As’
Ast
●
● ●
Neutral Axis
d
es
d – d’
d
Astfy
Asc
CROSS SECTION STRAIN DIAGRAM CROSS SECTION WITH COMPRESSION
BEAM
h
93
b
h
(d-d’)
a
.85fc’
(d – a/2)
(As – As’)
PRESSURE DIAGRAM
Let us consider the figure below. In the analysis of compression beam if Pmax = 0.75Pbor less
than Pb then equations of single reinforced beam is valid, conversely if P is greater than Pmax
then somewhat difference analysis is required. The following nomenclature will be used for
double reinforced concrete beam.
d’ = Covering of top bars reckoned from centroid compression steel up to outer fiber of
concrete.
d = Depth of beam (I.E distance of centroid tention steel up to outer fiber of concrete).
d – d’ = Lever arm of couple Ts and Cs’
b = Width of the beam given
h = height of the beam
fc’ = Cylinder strength of concrete in Kips per square inch.
fy = Yield strength of steel at ultimate in kips per square inch.
As’ =Area of compression steel in square inches
a = Depth of stress rectangular block
By the superposition method, the total resisting moment is equals to the sum of the individual
moments namely Mn -1 moment provided by the couple consisting of the force in compression
steel As’ and the force equal to the area of tention steel, then from the figure
94
Mn – 1 = As’(fy)(d – d’) We call this equation (1)
The second part Mn – 2 = is the contribution of the remaining tention steel (As – As’) acting
with the compression concrete with a lever arm of (d – a/2) so that Mn – 2 = (As –As’)(fy)(d –
a/2). In analogous to single reinforced concrete beam the depth of stress rectangular block is
given by As
– As’
or a =
we call this equation (3) but by definition As =
pbd and As’ = p’bd
substituting in (3) we get a =
The total resisting moment then would be
Mn = Mn – 1 + Mn – 2 = (As – As’)(fy)(d – a/2) + As’(fy)(d – d’) introducing a factor as
reduction factor then the design moment would be
Md =
–
=
–
+
–
Equation (4)
Equation 4 is the design moment for double reinforced concrete beam. For allowable steel
reinforcement ratio from code see program name comp beam.
In our computer program, upon input of numerical data’s. Computer first analyze given section.
A visual display into the monitor screen the word “ This is double reinforced” or “Single
reinforced beam”. A break command temporarily stops the computer from running to let the
designer sketch the section. Upon entry of the command CONTINUE computer resumes
running, evaluates required area of steel at midspan and supports respectively all in accordance
with the 1977 and 1983 A.C.I code.
STEPS FOR COMPUTER APPLICATION
a=
or in terms of steel reinforcement ratio a =
The limiting value of
lim
= .85B1(
)( )(
) + P’
where P’ = As’/bd for balanced steel ratio B = B + P’Fs’/Fy
Maximum steel ratio permitted by code is max = 0.75 PB + P’Fs/Fy and the nominal
moment is Md = (0.85Fc’ab(d – a/2)) + As’Fs’(d – d’) here = .90
95
Fy Yield stress of steel at ultimate in K.S.I
Fc = Cylinder strength of concrete in K.S.I.
b = Dimension of beam in inches CA = Coefficient of moment at exterior support
CB = Coefficient of moment at midspan
CC = Coefficient of moment at interior support
N = Percentage of steel ratio B1 is taken as 0.85 for 4000 psi and decreases for every 1000 psi
above 4000 psi.
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design of Double Reinforced Concrete Beam
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Double Reinforced Concrete Beam CODE NAME: COMP BEAM
MACHINE LANGUAGE : IBM BASICA
COMPUTER: I.B.M Home
Computer
Program steps: 120
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Comp Beam
PTR
NO: 3235345 at Bagiuo City 02/11/1983
96
DRAWING / FIGURE
CODE NAME: COMP - BEAM
W in LBS per foot
A
B
C
A
B
C
● ●
h
●
● ●
ASc
●
●
h
ASt
●
●
SECTION A – A
●
h
●
b
ASc
● ●
SECTION B – B
●
●
SECTION C - C
a
eu
es’
h
(d-d’)
.85fc’
(d – a/2)
es
(As – As’)
PRESSURE DIAGRAM
STRAIN DIAGRAM
REFERENCE TEXTBOOK: Dean Peabody Jr.Reinforced Concrete Design
97
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
COMP – BEAM: Is a computer PROGRAM that designs and analyze whether a given beam is
singly reinforced or or double reinforced. Computer first checked capacity of given section
compares moment capacity to actual moment and designs required area of steel at midspan ,
exterior support , interior support in square inch, using the less accurate A.C.i code. It
determines whether section is adequate for dia gonal tension or not all following the provisions
of the 1977 or 1988 A.C.I code. Evaluates spacing of stirrups The program is written in Advance
Basic and can be feed to a wide variety of programmable calculators and micro computers as
well as E- Review Center.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
98
COMPUTER INSTRUCTION CODE
PROGRAM
LINE NO
10
20
30
35
40
45
50
60
70
75
80
85
90
100
120
130
140
150
160
170
180
190
200
210
220
230
240
PROGRAM STATEMENT
COMPUTER OUTPUT
FOR DEBUGGING ONLY
CALL CLEAR
PRINT” This is a program for double reinforced beam
design by U.S.D. method in English units”
PRINT” This programmed was developed by
Bienvenido C. David a Civil?Structural engineer”
PRINT” For drawing and program record please see
attached sheets”
PRINT” All material strength specifications in Kips per
Square Inch, Linear dimensions in inches, Live loads
in pounds per foot”
PRINT” If all datas are in their consistent units then
run line 60”
STOP
REM A program for compression beam
INPUT” Material strength FC,FY”:FC,FY
INPUT” Live loads in pounds per foot”:LL
INPUT” Width and height of beam B,H”:B,H
INPUT” Shear react ion factor at support CR”:CR
INPUT”Coeffecients of negative moments at face of
exterior support, midspan, interior support
CA,CB,CC”:CA,CB,CC
INPUT “Clear distance of beam in feet LC”:LC
REM Determine if given section is single reinforced
or double reinforced
D = H-4
IF FC=<4 THEN 150 ELSE 170
BA = 0.85
GOTO 180
BA = 0.85–((FC-4)/1*).05
PB = 0.541875*FC/FY*87*BA/(87+FY)
AS = PB*B*(H-4)
A = AS*FY/(0.85*FC*FY*B)
MD = 0.90*AS*FY*(D-A/2)
DL = 1.0466*B*H
W = 1.4*DL + 1.7*LL
MA = 3/250*CA*W*LC^2
3.5, 40
3000
12, 18
1.15
.0625,.0714285,0.111111
20
14
.85
0.0276084
4.63
5.1968
1900.4853
226 POUNDS PER FEET
741.4 Pounds per foot
222 kips
99
250
260
270
280
290
300
310
320
330
340
350
360
370
380
385
390
395
400
405
410
415
420
425
430
440
445
450
455
460
470
475
480
485
490
495
MB = 3/250*CB*W*LC^2
MC = 3/250*CC*W*LC^2
IF A>MB THEN 300
IF MB>MC THEN 330
GOTO 350
IF MA>MC THEN 310
MU = MA
GOTO 360
MU MB
GOTO 360
MU = MC
IF MU<=MD THEN 380
IF MU.MD THEN 810
REM Assume dimension okay for design no revision
necessary. Determine area of steel bars at support
and midspan.
REM Let AE area of steel at exterior support
IF (d^2-2.61*MA/(FC*B))<0 THEN 395 ELSE 405
PRINT” Depth of stress rectangular block results in
imaginary not possible review given data”
STOP
A1 = D-SQR(D^2-2.61*MA/(FC*B))
AE = MA/(0.90*FY*(D-A/2))
PRINT”Recquired area of steel at exterior support
t=”:AE”Square inch”
REM Let ASP Required area of steel at mid span
IF (D^2-2.61*MB/(FC*B))>0 THEN 430 ELSE 450
PRINT” Depth of stress rectangular block is imaginary
not possible review given data”
STOP
A2 = D- (D^2-2.61*MB/(FC*B))^.5
ASP = MB/(0.90*FY*(D-A2/2)
PRINT”Recquired area of steel at
midspan=”;ASP;”Square Inches”
REM Let S1 area of steel at interior support
IF (D^2-2.61*MC/(FC*B))<0 THEN 475 ELSE 485
PRINT” Depth of stress rectangular block is imaginary
not possible review given data”
STOP
A3 =D – SQR(D^2-2.61*MC/(FC*B))
ASI = MC/(0.90*FY*(D-A3/2))
PRINT”Recquired area of steel at interior support
254 Kips
395.37 Kips
0.5016932
0.4485 Square Inch
0.4485
0.575555
0.5145449 Square Inch
0.9059719
.810695
100
500
505
510
512
513
515
520
525
528
530
540
550
560
570
580
590
600
610
620
630
640
650
660
810
820
840
850
860
870
880
885
890
=”;ASI;”Square Inches”
PRINT” Select from tables appropriate bar size no.
and then type continue to resume running and press
enter”
BREAK
REM Design of diagonal tension
VU = CR8W/2000*LC
VCA = 0.053754*FC^.5*B*D
IFVCA=>VU THEN 530
PRINT” Section safe for shear stress just provide
stirrups for unforeseen loads or as directed by Field
Engineer”
PRINT” Programmed by Bienvenido C. David A
Civil/Structural Engineer”
STOP
REM Design of diagonal tension needed in this region
VCD = VU – VCA
J = 1000*VU/W
JA = 0.5*CR*VCA*LC/VU
ZA = .187*FY*D/VCD
PRINT” Point of zero excess shear is =”;Z;”Feet
distance from support stirrups needed in this region”
REM Let us consider 2 Ft. interval
FOX = 1 TO Z STEP 2
Z1 = ZA*Z/(Z-X)
PRINT”Recquired spacing at 2 Ft. interval is
=”;Z1;”Inches”
NEXT X
PRINT” Detail and sketch section”
PRINT “Programmed by Bienvenido C. David a
Civil/Structural Engineer in Baguio City on April
29,1983”
STOP
REM This is a double reinforced beam
M1 = MU-MD
AS1 = M1/(0.90*FY*(D-2.5))
AT = AS1 + AS
P1 = AS1/(B*D)
PC = 0.85*BA*FC/FY*2.5/D*87/(87- FY) + P1
P2 = AT/(B*D)
REM Compare P2 and PC
PRINT”Recquired area of steel at interior support
=”;AT;”SQUARE Inches”
2.875
16.8948 Kips
939.765
4.5893719
9.21937
0.0273176
0.0482143
0.020896
101
900
910
930
935
940
950
970
980
990
1000
1020
1030
1040
1060
1070
1080
1090
1100
1110
1120
1130
1140
1150
1160
1170
PRINT” Compare P2 and PC allowable steel ratio from
code
PRINT” Area of steel at exterior support can be taken
as equals to area of steel at interior support”
PRINT” Area of steel at exterior support
=”;AT;”Square Inches”
PRINT” Select from tables 950appropriate bar size
and no. then type continue to resume running and
press enter”
BREAK
REM Design of diagonal tention
VU = CR*W*LC/2000
VCA = 0.053574*B*D*FC^.5
IF VCA = > VU THEN 1020
IF VCA<VU THEN 1060
PRINT” This is a double reinforced concrete beam
with a section adequate for diagonal tention just
provide stirrups as directed by Field Engineer”
PRINT” Programmed by Bienvenido C. David a
Civil/Structural Engineer on April 30, 1983
STOP
REM Design of diagonal tention
VCD = VU – VCA
J = 1000*VU/W
JA = 0.5*CR*VCA*LC/VU
Z = J – JA
PRINT” Point of zero excess shear is at a distance
=”;Z;”Feet from support stirrups needed in this
region”
REM Let us consider a 1 Ft interval distance
ZA = 0.187*FY*D/VCD
FOR X = 1 TO Z
Z1 = ZA*Z/(Z – X)
NEXT X
END
102
INTRODUCTION: When a beam and slab are poured at once, resulting in monolithic
construction, a portion of the slab on each side of the beam maybe considered as the flange
of a Tee – Beam. The portion of the beam below the flange serves as the web, sometimes
called the stem.
Flange bending stresses is not uniform from bean to beam, being largest over the web and
tending to drop off with distances from the web. Tee beams are analyzed in much the same
way as rectangular beams. There are two analysis employed.
a) First analysis: The neutral axis lies within the flange, if this happens then the formulas
for rectangular beams is valid program no 7 & program no 8.
b) Second case: The neutral axis lies below the flange, when this happens, this becomes a
Tee –Beam and the analysis is different from the rectangular analysis.
To be sure whether it is a rectangular or Tee – Beam, it is best to locate the neutral axis of the
section. When the location of the N.A. is less than the flange thickness, then this is a
rectangular beam therefore formulas for rectangular beam is applicable (program no 7 & 8).
On the other hand if the N.A. is greater than the flange thickness different analysis is required.
In deriving formulas for Tee –Beam, it is best to consider the figure below and the following
symbols will be used.
ANALYSIS AND FORMULA DERIVATIONS FOR COMPUTER APPLICATION
a = Depth of stress rectangular block
p =As/bd = Steel reinforcement ratio from code
hf = Flange thickness b = Width of beam if rectangular bw Width of beam web portion
103
d = Effective depth of beam = h – 2.5 or the distance of centroid tention steel to outer fiber of
concrete.
Asf = Area of steel required in the flange section
Fy = Ultimate yield strength of steel in K.S.I
Fc = Cylinder strength of concrete in K.S.I
b
d
hf
cb
SLAB
SLAB
SLAB
N.A.
● Ast●
bw
FIGURE 1
hf
FIGURE 2
b
.85Fc’
eu
d
c
N.A.
● Ast●
(d – c)
d – a/2
es
AsFy
bw
SECTION
STRAIN DIAGRAM
PRESSURE DIAGRAM
In the figure below is a cross section of tee Beam with the neutral axis located below the flange as
before a =
Asf =
=
then from the figure, it is evident that
Equation (1)
here the symbol Bw is the beam web. If we let
the symbol Mn-1 = The nominal moment carried by the flange neglecting the web portion then
the equivalent resisting nominal moment will be Mn -1 = (Asf)(fy)(d – hf/2) we call it equation (2)
If As is the total area of the tensile steel, then the remaining steel area (As – Asf) at stress fy is
balanced by the rectangular portion of the beam (section bw by d) with the rectangular portion
as free body shown below and applying
104
.85fc’
Cc =0.85Fc’abw
d
As
● ●
(d – a/2)
Ts = (As – Asf)fy
bw
0.85fc’abw = (As – Asf)fy or solving for a =
(Fy)
thus an additional moment
say
M(n-2) = (as – Asf)fy (d – a/2) the total moment then by the method of superposition say
Mn will be the sum of moments Mn -1 plus Mn – 2 = Mn -1 + Mn- 2
Mn = Asf(fy)(d – hf/2) + (As Asf)(fy)(d – a/2) introducing a factor of , the design moment then
would be Md =
reinforcements see design steps program name Trap
Foot.
The criteria for overhanging width are given below A.C.I. code recommendations.
1. For symmetrical T Beams the effective width b shall not exceed one fourth the span
length of the beam.. The overhanging slab width on the either side of the beam web
shall not exceed eight times the thickness of the slab or one half the clear distance to
the next beam.
2. For beams having a slab on one side only, the effective overhanging slab width shall not
exceed one – twelfth the span length of the beam, nor six times the slab thickness , nor
one half the clear distance to the next beam.
3 For isolated beams in which the flange is used only for the purpose of providing additional
compressive area, the flange thickness shall not be less than one – half the width of the
web, and the total flange width shall not be more than four times the web width.
105
In our general computer program , computer analyze given section , determine whether
given section analyzed as a rectangular or tee beam if so it prints on the T.V monitor
screen” This is a rectangular or Tee Beam”. A break command statement causes the
computer to stop running temporarily, the designer can then sketch the actual section .
After the command continue , computer performs the remaining program lines and
prints on the monitor T.V screen areas of steel in midspan and supports respectively all
in conformity with 1077 A.C.I. code.
*** Note: if the depth of stress rectangular block is equals to or less than the flange thickness
then a rectangular beam analysis is considered. If a is greater than the flange thickness
then a Tee beam analysis is required.
Pwb = Pb + Pf where Pb is the balanced steel ratio of rectangular portion of the beam and Pf
is the unbalance steel reinforcement ratio of the flange. The recommends that that the
steel ratio used shall not exceed Pmax = 0.75(Pb + Pf) here
Pb = 0.85(B1)(Fc/Fy)
here Pf = Asf/Bw where Asf = Area of steel at flange
Bw = Area of web Spacing of bars shall not exceed 5 times the thickness of the
flange nor in any case exceed 18”
Again the depth of stress rectangular block is *** Note given d, b and Mu depth
of stress rectangular block is
a=d ±
d2
–
For Computer application where b = web
thickness if rectangular analysis prevails = to Bw if Tee beam prevails
b = effective flange width.
LEGENDS
FC
Fy
DESCRIPTION
Cylinder strength of concrete at ultimate (K.S.I)
Yield Strength of steel at ultimate (K.S.I)
106
Hf
Cb
Ca
Cc
N
CR
LC
LD
CU
Flange thickness in inches
Coefficient of positive moment at midspan no unit
Coefficient of negative moment at exterior A
Coefficients of negative moment at interior support C
Percentage of steel ratio
Coefficient of shear reaction at support
Length of beam in feet
Centerline spacing in Feet
Coefficient of largest maximum moment for solving section of beam.
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design of Tee Reinforced Concrete Beam
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1977 ACI
Code)
TITLE: Design Of Reinforced Concrete Tee Beam CODE NAME: TEE
BEAM
MACHINE LANGUAGE : TI BASIC
COMPUTER: AMSTRAD
Computer
Program steps: 195
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Tee Beam
PTR
NO: 3235323 at Bagiuo City 02/11/1983
DRAWING FIGURE
CODE NAME: TEE BEAM
REFERENCE TEXTBOOKS: Design of Concrete Structures By George Winter CHAPTER 9 Pages
124 to 131, The Theory and Practice of Reinforced Concrete by Clarence W. Dunham
107
b
N.A.
●●
eu
.85Fc’
d
C
B1C
es
AsFy
bw
SECTION
STRAIN DIAGRAM
STRESS DIAGRAM
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
TEE – BEAM: Is a computer program that analyzes whether a given beam is a rectangular or
Tee Beam. Determines effective flange width and print on the screen value of depth of stress
rectangular block. Computes area of required steel reinforcements in conformity with the 1977
A.C.I code. Checks given cross section using the less accurate A.C.I code the required diagonal
108
tention. Plots spacing of stirrups all in accordance with the 1977 A.C.I. code. The program is
written in advance basica software language and runs on programmable calculators, main
frame computer and E review centre. Design to be used in a time sharing systems in a
conversational mode.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
VALUE
If true
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODES
LINE
NO
10
15
20
25
30
45
50
55
60
65
STATEMENT
NUMERICAL OUTPUT
(SAMPLE ONLY) ***For
debugging only
CALL CLEAR
PRINT” This is design of Tee Beam one flange only
by U.S.D in English units“
PRINT” All material specifications in Kips per
square inch , Linear dimensions in feet, slab
thickness in inches”
PRINT” This program was developed by Bienvenido
C. David A Civil?Structural engineer on March 1983
in Baguio City”
PRINT” If all data’s are in their consistent units then
run line no 50”
STOP
REM First determine if beam is rectangular or Tee
beam
INPUT” Cylinder strength of concrete at
3
ultimate”:FC
INPUT” Yield Strength of steel at ultimate”:FY”
60
INPUT” Thickness of flange in inches”:HF
3
109
70
75
80
85
90
95
100
110
120
140
150
160
170
180
190
200
210
135
220
230
240
250
260
270
280
290
300
310
320
*****
330
340
350
400
410
INPUT” Clear distance of beam in feet”:LC
INPUT’ Centerline spacing in feet”:LD
INPUT” Live load in pounds per foot”:LL
INPUT”Coeffecient of negative moment at exterior
support”:CA
INPUT”Coeffecient of positive moment at center
span”:CB
INPUT’Coeffecient of negative moment at interior
support”:CC
INPUT”Coeffecient of largest moment for
determining section of beam”:CU
INPUT” Percentage of steel ratio N designers
choice”:N
REM Find or analyze given section
IF FC>= THEN 150 ELSE 170
BA = 0.85
GOTO 180
BA = 0.85 – (FC-4)/1*0.05
PB = 0.541857*FC/FY*87*BA/(87+FY)
PC = N*PB
PM = 0.2/FY
IF PC<PM THEN 230
**** Note insert as input statement
INPUT”Dimention of beam in inches”:BW
IF PC < PM THEN 250
P = PM
GOTO 260
P = PC
X = 1- 0.59*P*FY/FC
Y = 0.90*P*FY*BW
G = X*Y
E = 83.33*G/(CU*LC^2)
F = -1.456*BW
C = -(3.63*BW + 1.7*LL)
Q = (F^2 – 4*E*C)^.5
T he quantity (F^2 – 4*E*C) must be greater than 0
D= (- F + Q)/(2*E)
H = D + 2.5
REM With section known determine if rectangular
section or tee beam
DL = 6.25*LD*HF + 1.0466*BW*H
W = 1.4*DL + 1.7*LL
24
3.91
1132
0.0625
0.0714285
0.111111
0.111111
0.60
0.013629
0.008177
0.003333
11
0.008177
0.9035144
4.8571
4.3884597
5.71448
-16.016
-1964.33
212.505
19.99 say 20
22.5
332.346 LBS/FT.
2389 POUNDS PER FOOT
110
420
430
440
450
460
470
480
490
500
510
520
530
540
550
560
***
570
580
590
595
600
605
610
615
625
630
635
640
645
650
655
665
670
680
690
MU = 3 /250*CU*LC^2*W
B1 = 1/12*LC*12 + BW
B2 = 6*HF + BW
B3 = 1/2*LD*12
IF B1<B2 THEN 490
IF B2<B3 THEN 520
GOTO 540
IF B1<B3 THEN 500
B = B1
GOTO 550
B = B2
GOTO 550
B = B3
R = 2.61*MU/(FC*B)
A = D – (D^2 – R)^.5
Note the quantity (D^2 – R) must be greater 0
IF A<=HF THEN 590
IF A>HF THEN 1070
PRINT” Since A is less than or equal to HF then
rectangular beam analysis is valid neutral axis
within flange program no 7 & no 8”
PRINT” Detail section of beam then type continue
to resume running and press enter”
BREAK
MA = 3/250*CA*W*LC^2
IF (D^2 – 2.61*MA/(FC*BW))<0 THEN 615 ELSE 630
PRINT” Depth of stress rectangular block is
imaginary not possible review given data”
STOP
AA = D – SQR(D^2 – 2.61*MA/(FC*BW))
AE = MA/(0.90*FY*(D – AA/2))
PRINT” Area of steel at exterior
support=”;AE”Square Inch”
MB = 3/250*CB*W*LC^2
IF (D^2 – 2.61*MB/(FC*BW))<0 THEN 655 ELSE 670
PRINT “Depth of stress rectangular block not
possible review given data”
STOP
AB = D – (D^2 – 2.61*MB/(FC*BW))^.5
ASP = MB/(0.90*FY*(D – AB/2))
PRINT”Recquired area of steel at midspan=”;ASP”
Square Inches”
1833 KIPS
35”
29”
23.46 SAY 24
B =23.46, R = 67.97
1.778
HF = 3
1032.048 Inch kips
2.15696
1.01000 Square Inch
1179.48
2.4868
1.1645086
1.1645086
111
695
700
705
710
720
725
730
735
740
745
750
770
775
780
785
790
800
810
820
830
850
860
870
880
890
900
905
910
MC = 3/250*CC*W*LC^2
IF (D62 – 2.61*MC/(FC*BW))<0 THEN 705 ELSE 720
PRINT” Depth of stress rectangular block is
imaginary not possible review given data”
STOP
AC = D – (D^2 – 2.61*MC/(FC*BW))^.5
ASI = MC/(.90*FY**(D – AC/2))
PRINT” Select from table appropriate bar size No
then type continue to resume running detail and
sketch section and reinforcements”
BREAK
Vu = 1.15*w*lc/2000
VCA = 0.053754*FC^.5*BW*D
PRINT” IF VCA >= VU THEN 770 ELSE 850
PRINT” Section adequate for diagonal tention
provide stirrups for unforeseen loads or as directed
by Field Engineer”
PRINT” Shear reaction at support =”;VU;”Kips
PRINT” Negative moment at exterior support
=”;MA”Inch Kips”
PRINT” Positive moment at midspan =”;MB”Inch
Kips”
PRINT” Negative moment at interior support
=”;MC;”Inch Kips”
PRINT” Copy end moments for column design type
continue then press enter to resume running”
BREAK
PRINT” Programmed by Bienvenido C. David a
Civil/Structural engineer on July 1983 in Baguio
City”
STOP
REM Stirrups needed in this region VU is greater
than VCA
VCD = VU – VCA
J = 1000*VU/W
JA = 0.5*VCA*LC/VU
REM Determine spacing of stirrups at every 1
Ft.interval
Z = J – JA
PRINT” Point of zero excess shear is at a distance
=”;Z;”Feet from support stirrups needed in this
region”
ZA = 0.187*FY*D/VCD
1834 Kips
4.032940
1.8888 Sq Inch
32.968 Kips
20.483 Kips
12.485 Kips
13.799 Feet
7.455
6.344
17.9735
112
915
920
930
935
****
940
945
950
960
970
975
980
985
990
995
1000
1010
1070
1075
1080
1090
1100
1110
1120
1130
1140
1145
1150
1155
FOR SX = 1 TO Z
Z1 = ZA*Z/(Z – SX)
PRINT” Spacing required at every 1 Ft
interval=”;Z1;”Inches”
NEXT SX
Note for answers example only spacing of stirrups
are
PRINT” Height of beam in inches=”;H
PRINT” Area of steel at exterior
support=”;AE;”Square Inches”
PRINT”Recquired area of steel at
midspan=”;ASP;”Square Inches”
PRINT”Recquired area of steel at interior
support=”;ASI”Square Inches”
PRINT” Total maximum shear reaction at support is
VU=”;VU;”Kips”
PRINT” Copy negative moment equals to first at
exterior support=”MA;”Inch Kips”
PRINT” Copy positive moment at midspan equals to
=”MB;”Inch Kips”
PRINT “Copy end moment at face of interior
support equals to =”MC;”Inch Kips”
PRINT” Sketch and detail rectangular beam then
type continue press enter to resume running”
BREAK
PRINT” Programmed by Bienvenido C. David a
Civil/Engineer on the year 1983 in Baguio City”
STOP
REM Neutral axis within web a tee beam analysis
ASF = 0.85*FC*(B – BW)*HF/FY
M1 = 0.90*ASF*FY*(D – HF/2)
MA = 3/250*CA*W*LC^2
M2 = MA – M1
IF (D^2 – 2.61*M2/(FC*BW))<0 THEN 1120 ELSE
1140
PRINT “Depth of stress rectangular block is
imaginary not possible review given data”
STOP
A2 = D – SQR(D^2 – 2.61*M2/(FC*BW))
AS2 = M2/(0.90*FY*(D – A2/2))
AT1 = AS2 + ASF
PRINT”Recquired area of steel at exterior
support=”AT1;”Square Inches”
21.33,26.48,34.098,48.6446
& 84.838 in inches
`
1032.048
1179.48
1834
1.588 Inch
3306 Inch Kips
4012.416
706.416
0.7035 Inch
0.32944 Square Inches
1.9174
113
1160
1165
1170
1175
1180
1185
1190
1200
1205
1210
1215
1215
1225
1230
1240
1250
1255
1260
1265
1270
1275
1280
*****
1253
1290
1295
1300
1310
1320
1325
1330
1335
1340
1350
P1 = AT1/(BW*D)
PF = ASF/(BW*D)
PU = 2.7225*FC/FY*87/(87 + FY)
PW = 0.75*(PF + PU)
PRINT” Compare P1 and PW maximum allowable
stress from code”
MB = 3/250*CB*LC^2
M3 = MB – M1
IF (D^2 – 2.61*M3/(FC*BW))<0 THEN 1205 ELSE
1215
PRINT” Depth of stress rectangular block is
imaginary not possible review given data”
STOP
A3 = D – (D^2 – 2.61*M3/(FC*BW))^.5
AS3 = M3/(0.90*FY*(D – A3/2))
AT2 = ASF + AS3
PRINT”Recquired area of steel at
midspan=”;AT2;”Square inches”
P2 = AT2/(BW*D)
REM Compare P2 and PW
M4 = MC – M1
IF (D^2 – 2.61*M4/(FC*BW))<0 THEN 1265 ELSE
1275
PRINT “Depth of stress rectangular block is
imaginary not possible review given data”
STOP
A4 = D – (D^2 – 2.61*M4/(FC*BW))^.5
AS4 = M4/(0.90*FY*(D – A4/2))
Note insert 1253 program line value of MC
MC = 3/250*CC*W*LC^2
AT4 = AS4 + ASF
PRINT “Area of steel at interior
support=”;AT4;”Square inches”
P3 = AT4/(BW*D)
REM Compare P3 and PW allowable steel
reinforcement ratio from code
REM Design of diagonal tention
VU = CR*W*LC/2000
VCA = 0.053754*FC^.5*BW*D
IF VCA>=VU THEN 1350
IF VCA<VU THEN 1460
PRINT” Section is a Tee beam section is adequate
for shear and diagonal tention provide stirrups as
0.00435
0.0036036
0.02138
0.0187377
4585.61
1279
1.283
0.60086 Square Inch
2.18966
2.1896
0.004969
3820 Inch Kips
3.96
1.85768
7126
3.44568
0.00781936
111
41
114
1355
1360
1370
1380
1390
1400
1460
1470
1480
1490
1500
1505
1510
1515
1520
1530
1540
1550
1560
1570
1575
1580
1590
1595
1600
1610
1620
1630
directed by Field Engineer”
PRINT” Negative moment at exterior support for
column design is=”MC;”Inch Kips”
PRINT” Copy moment at interior support for
column design =”;MA;”Inch kips”
PRINT” Copy shear reaction at end for column
design=”;VU;”Kips”
PRINT” Detail and sketch steel reinforcements and
section then type CONTINUE to resume running
and press ENTER”`
PRINT” Programmed by Bienvenido C. David a
Civil/Structural engineer on the year 1983 in Baguio
City”
STOP
REM Design of diagonal tention
VCD = VU – VCA
J = 1000*VU/W
JA = 0.5*CR*VCA*LC/VU
Z = J – JA
PRINT” Point of zero excess shear is located at a
distance =”;Z;”Feet from support stirrups needed in
this region”
REM Determine spacing at 1 Feet interval
ZA = 0.187*FY*D/VCD
Z1 = ZA*Z/(Z – 1)
Z2 = ZA*Z/(Z – 2)
Z3 = ZA*Z/(Z – 3)
Z4 = ZA*Z/(Z – 4)
Z5 = ZA*Z/(Z – 5)
PRINT ZA
PRINT”Z1 =”;Z1;”Inches”
PRINT Z2=”;Z2;”Inches”
PRINT”Z3=”Z3;”Inches”
PRINT” Z4=”Z4;”Inches”
PRINT”Z5=”;Z5;”Inches”
PRINT”MC=”;MC;”MA=”;MA;”VU=”;VU
PRINT “Programmed by Bienvenido C. David a
Civil/Structural engineer on the year 1983 in Baguio
City”
END
70
11.95
4.432
7.7518
6.42
7.404
8.746
10.68
13.71
19.16
115
INTRODUCTION: Design of Slab is similar to that of a rectangular beam.
In general there are four kinds of slabs in a typical building.
1 One way solid slab and beam.
2. Two way solid slab and beam.
3 Floor joist, sometimes called ribbed floor.
4. Flat slab or girdles floors.
In this chapter we will be dealing with one way slab and two way slab the two most common slabs built
by the practicing Civil engineers from day to day practice.. The ultimate method of structural theory will
be used thru out in the computer solutions. One way slab is used in foot bridges, small bridges
supported by stringers. The reinforcements in one direction runs in one direction only, from beam to
beam. The slab is uniform in thickness and there is no filler material. The number of beams in a panel
depends on the column spacing, as the name implies the bending moments are assumed to be resisted
by the beam action between two or more lines of supports, so that all of the main bars will be running
parallel to each other and perpendicular to those supports (For figure computer program name “Slab
one”)
In one way slab design, the minimum thickness is governed by the A.C.I. CODE COEFFECIENTS AS WELL
AS TEMPERATURE BARS. A unit strip 1 Ft. (12”) wide is cut perpendicular or at right angles to the
supporting bars, The A.C.I. code coefficients is normally employed for determining the moments at
supports and midspan. To determine the required area of steel per foot of width the area of one bar
times the average number of bars in a foot strip (12 divided by the spacing in inches). As an aid for
design engineers a table for various sizes of bars are included in English as well as in metric units).
In our computer program, two methods are employed the standard method, the sub –routine
method. A third method which is advantageous for many slabs is a combination of one
dimensional array in three elements with the RESTORE COMMAND of advance
BASICA. The standard formula for finding the depth of stress rectangular block as previously derived
(See chapter 4) as shown below is
116
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
d2
–
For Computer application
and the required area of steel As =
for area of steel. These two mathematical
expressions were used by the author a s a sub routine to solved areas of steel reinforcements at the
end supports and midspan. (See program listing).
One way slab is essentially a rectangular beam of comparatively large ratio of width to slab,. The
design of one way slab is similar to the design of a rectangular beam. Referring from the figure program
no 10. a strip of 12” wide is cut at right angles to the supporting beam, the bending moment being
computed for a 12: wide strip.
As per A.C.I. code minimum thickness of slab is
a)
b)
c)
d)
For simply supported ------------------------- L/20
For one end continuous ---------------------- L/24
Both end continuous -------------------------- l/28
Cantilever -----------------------------------------L/10
TEMPERATURE AND SHRINKAGE PROVISIONS
1 ) where grade 40 or 50 deformed bars were used p = 0.002
2 ) Where grade 60 deformed bars or welded wire fabric p = 0.0018
3) Slabs where reinforcements with yield strength exceeding 60,000 p.s.i. yield strength measured at
yield strain of o.35% used
p=
less than 0.0014
4) Weight of concrete is taken as 150 Lbs/Cu.Foot
In no case shall the steel reinforcement ratio be
117
5) Covering of steel is 1” top cover.
6) Dead and Live load factor is 1.4 and 1,7
7) Moment factor is 0.90
8) Shear reaction is 1.115
9) For computer application :
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
d2
–
For Computer application
and the required area of steel As =
*** Where A.C.I.
for area of steel
code is coefficients are applicable
Exterior support ------------------ L/24
Midspan -----------------------------L/14
Interior support -------------------L/9
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design of One Way Slab
118
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: One Way Slab Design CODE NAME: One Slab
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 74
LIBRARY MODULE: Floppy Disk PROGRAM NAME: One Slab
PTR
NO: 3236355 at Baguio City 01/10/1983
CODE NAME:
One Slab
a
One way slab●
b
1 Foot strip
TRANSVERSE SECTION
PLAN
● ●
● ● ●
●
TYPICAL SECTION
REFERENCE TEXTBOOKS:
Concrete Fundamentals by
The Theory and Practice of
Design of Concrete
Phil M. Ferguson
Reinforced Concrete by
Structures by George Winter
Clarence w. Dunham.
CHAPTER 5 Pages 202 - 210
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
119
My first generation home computer
ONE SLAB: Is a computer program that designs depth and required area of steel
reinforcements at midspan, interior support and exterior support. Computer first solves
required depth of slab using the maximum steel reinforcement ratio from code. It then solves
required area of steel reinforcements at midspan and interior supports in accordance with the
A.C.I. code provisions. Computer also solves temperature bars and verify if section is safe for
diagonal tension . The computer runs in Advance BASICA language and programmable
calculators. It can easily be integrated into the E _Review center of UC BCF .
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE NO
10
20
25
30
STATEMENT
CALL CLEAR
PRINT” This is one way slab design by U.S.D. method
in English units”
PRINT” Material strength in Kips Per Square Inch, Live
and Dead loads in Pounds per square Foot, Thickness
of slab in Inches and span length if feet”
PRINT” If all data’s are in their consistent units then
NUMERICAL OUTPUT
***(SAMPLE ONLY FOR
DEBUGGING PURPOSES)
120
50
60
65
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
run line no. 60”
STOP
INPUT”LL’CA,CB,CC,FC,FY”:LL.CA,CB,CC,FY
INPUT”SPAN LENGTH LC”:LC
H =3/7*LC
DL = 12.5*H
W = 1.4*DL + 1.7*LL
PC = 0.541875*FC/FY*87/(87 + FY)
X = 1 – 0.59*PC*FY/FC
Y = 10.8*PC*FY*X
MA = CA*3/250*W*LC^2
MB = CB*3/250*W*LC^2
MC = CC*3/250*W*LC^2
IF MA>MB THEN 190
IF MB>MC THEN 220
GOTO 240
IF MA>MC THEN 200
MU = MA
GOTO 250
MU = MB
GOTO 250
MU = MC
DA = (MU/Y)^.5
DC = H-1
IF DA<=DC THEN 280 ELSE 300
D = DC
GOTO 310
D = DA
H1 = D+1
PRINT” Overall depth of slab =”;H1;”inches”
Print” Detail and sketch section then type continue
to resume running”
BREAK
M= MU
GOSUB 610
AS1 = AS
PRINT” Area of steel at interior support=”ASI;”Square
Inches”
M=MB
GOSUB 610
CA = 0.111, CB = 0.0714, CC
=0.04166, LL = 100 P.S.F,FC
= 4K.S.I , FY = 60 K.S.I.
LC = 15 Feet
6.42
80.25
282.224
0.021
0.81415
11.08
84.72
54.36
31.80
84.72
2.765
5.42
5.42
6.42
84.72
121
410
420
430
440
450
460
470
475
480
485
490
495
500
505
510
520
530
540
550
570
580
590
600
610
620
630
640
650
660
ASP =AS
PRINT “Area of steel at midspan=”;ASP;”Square
Inches”
M=MC
GOSUB 610
ASE=AS
PRINT” Area of steel at exterior
support=”ASE;”Square Inches”
PRINT” Select from tables appropriate size and no. of
bars”
PRINT” Detail and sketch steel reinforcements”
ASB = 0.0216*H
PRINT” Area of temperature bars=”ASB;”Square
Inches”
STOP
PRINT” After detailing bars type continue to resume
running”
BREAK
REM Check shear and diagonal tension
VU = 1/12000*(6.9*W*LC-W*D)
VN=758.88*FC^.5*D
IF VN>=VU THEN 540 ELSE 580
PRINT” Section adequate for shear and diagonal
tension just provide stirrups as directed by the Field
Engineer”
PRINT” Programmed by Bienvenido C. David, a
Civil/Structural Engineer on Jan 10, 1983 at Baguio
City”
STOP
PRINT” Section not adequate for shear and diagonal
tension provide stirrups entire region spacing as
directed by Field Engineer or Design Engineer”
PRINT” Programmed by Bienvenido C. David, a
Civil/Structural Engineer on Jan 10, 1983 at Baguio
City”
STOP
REM This is a sub Routine
IF (D^2-0.2175*M/FC)>0 THEN 630 ELSE 660
PRINT” Depth of stress rectangular block is imaginary
not possible review given data”
STOP
RETURN
A = (D-(D^2-0.2175*M/FC)^.5)
24.7698
0.4431
122
670
680
690
700
S = 0.90*FY*(D-A/2)
AS=M/S
RETURN
END
ANSWERS FOR THIS PROBLEM
a = .4431 a = .2798 a = .1619 M = 84.72
M = 54.36
AS =0.3022
AS = 0.190
AS = 0.11029
S = 280.316
S = 285
S = 288.3087
M = 31.8
INTRODUCTION
TWO WAY SLABS: Two way slabs are used when a floor panel is square or nearly
so, having beams or walls on four sides. The tensile reinforcements are used in four sides both
in the long direction and short direction. These bars in two directions transfer the loads to the
four supporting beams or walls (for figure see program no 10 code name “Slab two”). They are
usually used in large continues monolithic floor systems. They maybe solid slabs, two way ribs,
or joist with some sort of block filler between them and with a thin concrete slab poured
monolithically on top of them or “waffle construction “ having two way ribbed with a
monolithic top slab and air spaces between the ribs.
The analysis is a combination of empirical and analytical methods. Consider the figure
below, from the geometry of the figure, it can be seen that at the point of intersection of the
short beam and long beam the two have the same deflection. Thus for a uniform load W/Lb Ft.
123
4
the deflection in the short direction is Dshort = 5wal /384EI. Similarly the deflection in the long
4
direction is DLong = 5wbl /384EI equating we get the relationship Wa/Wb = Lb4/La4, From these
equations one sees that the larger share of the load w is carried in the short direction, the ratio
of the two portions of the total load being inversely proportional to the fourth power of the
ratio of the spans.
The precise determinations of moments in two way slabs with various conditions of
continuity at the supported edges are mathematically formidable and not suited for design
practice.. For this reason various simplified methods have been adopted for determining
moments, shears and reactions of such slabs.
column strip
middle strip
short direction
long direction
In the moment coefficient methods , the moment in the short direction is designated by
Ma = Caw(la)2 and the moment in the long direction is Mb= Cbw(lb)2 Here the symbol Ca
and Cb are coefficients for positive , negative and shear reaction dead and live
load respectively both in the short and long direction . For the column strip. it is
assumed that the moment is equals to one third of the moment in the middle
strip. This applies to both long and short direction. The point of inflection based
from the A.C.I. code coefficients is located at a distance of one sixth of the span
length in short and long direction. Referring to chapter three of the book
In our general computer program , the author uses the mathematical expression
10. 10 Cb =
(d)
11 Pb = 0.85B1(
12 Wb = ,85B1 (87000/(87000 + Fy)
P
)(87000/(87000 + Fy)
(1 ±
)
1 – 2.622Mu/(bd2Fc’)
124
Here the mathematical expression under the radical sign must not be less than zero.
A.C.I. CODE PROVISIONS FOR MIMINUM THICKNESS OF SLAB
a) For simply supported ----------------------------------L/20
b) One end continues -------------------------------------L/24
c) Both end continues ------------------------------------L/28
D) Cantilever ------------------------------------------------L/10
TEMPERATURE & SHRINKAGE PROVISIONS
1) Where grade 40 or 50 deformed bars were used use p = .0002
2) Where grade 60 deformed bars or welded wire fabric p = 0.0018
3) 3) Slabs where reinforcements with yield strength measured at yield strain o0f 0.35%
is used p
4)
5)
6)
7)
8)
=
In no case shall the steel reinforcement ratio be less than
0.0014
Covering of steel is 1” top cover
Dead and live load factors 1.4 and 1.7
Moment factor is o.90
Shear reaction factor is 1.116
For computer application author use formula for a refer chapter three page 76 of my book
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
d2
–
For Computer application
where A.C.I code coefficients is applicable
Exterior support -------------------------------1/24
Midspan ------------------------------------------1/14
Interior support ----------------------------------1/9
In the above formula Fc’ = cylinder strength of concrete in kips per square inch.
d = depth of slab and B = Width of slab per foot strip = 12 inches
125
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Two Way Slab Design
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Two Way Slab Design CODE NAME: Slab Two
MACHINE LANGUAGE : BASICA
COMPUTER: AMSTRAD 640
Program steps: 61
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Slab Two
PTR
NO: 3236281 at Baguio City 05/09/1982
lb
lb/4
la/4
lb/2
lb/4
la
Column strip
middle strip
column strip
la/2
La/4
PLAN LONG DIRECTION
Mb/3
SHORT DIRECTION
Ma/3 Variations of design moments across width of critical sections for simply
supported two way slabs
REFERENCE TEXTBOOK:
Concrete Fundamentals
Phi Ferguson
126
REFERENCE TEXTBOOK
Pages 214 to 224
Design Of Concrete
Structures by George Winter
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
SLAB TWO: Is a computer program that designs and determines area of required steel per foot
of width for a two way slab using the A.C.I. code coefficients in English units, Computes shear
reaction of slab and determines bending of bars both long and in short direction. It checks steel
reinforcement ratio from code limitations . The program is written in advance basica and
designed to be used in a time sharing system in a conversational mode. Can be run to a variety
of –programmable calculators, main frame computers can be easily integrated into the E
Review centre of the UC _BCF –CE.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
127
LINE
NO
10
20
30
40
50
60
75
80
85
90
100
110
120
125
130
140
150
145
170
175
178
180
190
200
210
220
230
240
STATEMENT
CALL CLEAR
PRINT” THIS IS DESIGN OF TWO WAY SLAB BY U.S.D.
METHOD IN ENGLISH UNITS”
INPUT” Live loads in pounds per square foot”:LL
INPUT” Cylinder strength of concrete in Kips Per Square
Inch”:FC
INPUT” Ultimate strength of steel at ultimate in Kips per
square inch”:FY
INPUT”Dimention of larger side of slab in feet B”:B
PRINT” Select from tables coefficients of moments then type
continue and press enter to resume running”
BREAK
INPUT” Coefficient of negative moment in short
direction”:CA
INPUT” Coefficient of negative moment in long direction”:CB
INPUT” Coefficient of positive dead load moment in short
direction”:CD
INPUT” Coefficient of positive live load moment in short
direction”:CB
INPUT”Coeffecient of positive live load moment in long
direction”:CG
INPUT”Coeffecient of positive dead load moment in long
direction”:CF
INPUT”Coeffecients of shear reaction in long direction”:CH
INPUT”Coeffecients of shear reaction in short direction”:CI
D=H-1
H=0.13333*(A+B)
DL=12.5*1.4*H
W=1.7*LL+DL
MNA= 3/250*CB*W*A^2
MNB = 3/250*CB*W*B^2
MDA = 3/250*CD*DL*A^2
MLA = 3/250*1.7*CE*LL*A^2
MTD = MDA+MLA
MDB = 3/250*CF*DL*B^2
MLB = 3/250*CG*LL*1.7*B^2
MTB = MDB+MLB
NUMERICAL OUTPUT
***(SAMPLE ONLY
FOR DEBUGGING
PURPOSES)
137
3
60
25
.071
.029
.039
.048
.020
.016
.71
.29
4.99 SAY 5
5.99 SAY 6
104.825
337.72
115
73.45
19.623
53.66
73.283
12.5136
34.935
47.448
128
250
260
270
280
Z = 0.847*FC/FY
Q= 1 – (1-2.622*MTA/(12*D^2*FC))^.5
PAT =Z*Q
APS=12*PAT8D
285
PRINT “Area of steel bottom reinforcement in short
direction =”APS;”Square Inches per foot of width”
R = 1-(1-.2185*MNA/(D^2*FC))^.5
PNA = Z*R
ANS=PNA*12*D
290
300
310
315
320
330
340
345
350
360
370
385
390
395
400
410
415
420
425
430
440
445
450
PRINT” Area of negative moment steel top bars in short
direction =”;ANS;”Square Inches per foot of width”
S =1-(1-2.185*MTB/((D-2.50^2*FC))^.5
PTB = Z*S
ANB = 12*PTB*(D-.5)
PRINT “Area of steel bottom reinforcement in long
direction=”;ANB;”Square Inch per foot of width”
U=1-(1-.2185*MNB/(D^2*FC))^.5
PNB=Z*U
BNS=12*PNB*D
PRINT” Area of steel top reinforcement in long direction
=”;BNS;”Square Inches” per foot of width”
WB=CH8A8W/2
WA=CI*W*B/2
VC =645*D8FC^.5
X =1/6*A
PRINT” Inflection point in short direction is located at a
distance =”X;”feet”
Y=1/6*B
PRINT” Inflection point in long direction is located at a
distance =”;Y”Feet”
PRINT “Select from tables appropriate no. and size of bars
detail and sketch slab then type continue to resume
running”
BREAK
PRINT” Load acting on the long direction beam per foot
=”;WB”Pounds per foot”
PRINT” Load acting on short beam per foot =”;WA;” Pounds
per foot”
0.04235
0.113633
0.0048
.288 Square Inch
per ft.
.18455
0.0078
0.468 Square Inch
per Ft.
.0893
0.003782
.204 Square Inch
per foot of width
.113425
0.0048
0.2882 Square
Inches per foot of
width”
2397
1224
5585
3.33
4.16666
129
460
470
480
490
PRINT” Bend up every third bar from the bottom to provide
negative steel at discontinuous edge”
PRINT” Safe shear capacity of section in pounds=”;VC
PRINT” Programmed by Bienvenido C. David on September
1982
END
130
,
WEB REINFORCEMENTS
INTRODUCTION: Web reinforcements or diagonal tension commonly known as stirrups are
needed if the allowable shear carried by the concrete is less than the actual shear due to
superimposed loads. i.e.
factor loads,
Vu = Vn where Vu the total shear force applied to the section
due to
Vn Normal shear strength equals to the sum of the contribution of concrete and web steel
0.85. In our previous programs refer
. The strength reduction factor
chapter 4 programs no. 7, 8 and 9 I have included spacing of stirrups using the less accurate
code. Program no 12 incorporates design of web reinforcements using the more accurate
A.C.I. code.
The formula using the more accurate A.C.I. code is VC = 1.9Fc’ + 2500Pw (
But Vc = 3.5Fc’bwd
bw Web width for Tee section or beam width for rectangular section. d
= Effective depth to tensile steel in inches.
pw Longitudinal tensile steel ratio
or As/bd for tee or rectangular
section. Vu = Shear force at section at factored loads, LBS or Kips
Mu = Moment at section at factored loads, Inch
strength of concrete in K.S.I.
)
Lbs or Inch Kips and Fc’ specified compressive
131
In the above equation using the more accurate code we notice that at any section X distance from
support Vc is a function of X. To find the point of zero excess shear we equate Vc = Vu a cubical
3
2
equation in the third degree results in the form A1x + B1x + C1X + D1 = 0
*** Note for computer use. author use the “General cubic equation program no1” to
solve the value of X and checked it by the :”Newton’s approximation program no2”
The quantity Vud/Mu is not to be taken greater than 1. As an alternative the A.C.I. code
allows the use of simpler, less accurate A.C.I. CODE Vc = 2Fc’bwd which is adequate for
most design purposes.
Av = Required area of vertical stirrups given by the formula Av = .50bws/Fy
where S is the longitudinal spacing of web reinforcement in inches
Fy Yield point of steel at
ultimate in K.S.I.
Av = Total area of web reinforcement steel in square inch.
This provision holds unless Vu is less than one – half of the design strength Vc provided by
the concrete.
● ● ●
● ● ●
● ● ●
● ● ● ● ● ● ● ●
PLAN
STIRRUPS (ELEVATION)
Theoretical spacing
Practical spacing
Graph showing practical and theoretical spacing of web
reinforcements
132
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design of Diagonal Tension using the more accurate A.C.I.
code in English Units
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1983 ACI
Code
TITLE: Design of Diagonal Tension CODE NAME: Dia - shear
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 174
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Dia Shear
PTR
NO: 4236357 at Baguio City 09/11/1981
CODE NAME: Dia Shear
Ma
w Lbs per foot
Mb
top cover of steel 2.5”
● ● ●
shaded portion represents points of zero excess shear
stirrups needed in this region
● ● ●
b
PLAN
Lc/2
Lc/2
stirrups a shown
Lc
ELEVATION
h
133
REFERENCE TEXTBOOKS:
Concrete Fundamentals by
The Theory and Practice of
Design of Concrete
Phil M. Ferguson
Reinforced Concrete by
Structures by George Winter
Clarence w. Dunham.
CHAPTER 4 Pages 131 - 137
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
DIA – SHERA: : Is a computer program that DESIGNS THE RECQUIRED STIRRUPS AND SPACING
USING THE MORE ACCURATE A.C.I code in English units using the USD format. Computer solves
point of zero excess shear using the “General mathematical program cubic” and rechecked it
using the “Newton’s method of approximation program “. Determines maximum and minimum
spacing of web reinforcements as per A.C.I code requirements. It display on the monitor screen
whether given section is adequate for diagonal tension or not. For a given five distances from
column face, It evaluates required spacing in between stirrups The computer runs in Advance
BASICA language and programmable calculators. It can easily be integrated into the E _Review
center of UC BCF .
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
134
COMPUTER INSTRUCTION CODE
LINE
NO
5
10
15
20
25
30
35
45
50
60
65
70
75
80
85
90
100
110
120
STATEMENT
CALL CLEAR
PRINT” This is design of diagonal tension using the more
accurate A.C.I code”
PRINT” For drawing and legend please see program
record and description”
PRINT” This programmed was developed by Bienvenido
C. David a Civil Structural engineer on November 1981
at Baguio City”
PRINT” All units of linear dimensions in inches, clear
span of beam in feet, material specifications in Kips per
square Inch, clear span of beam in feet.”
PRINT” Shear diagonal factor is taken as 0.85”
REM Design of diagonal tension using the accurate A.C.I
code is VC =(1.98*FC^.5 + 2500*PW*VU/MU*D)*B*D
PRINT” If all data’s are in their consistent units then run
line no 60”
STOP
CALL CLEAR
REM Design of diagonal tension using the more accurate
A.C.I code
INPUT”DB,BW,CV,W1,LC,FC”:DB,BW,CV,W1,LC,FC
REM Determine if section is adequate for diagonal
tension or not
W = W1/1000
VU=CV*W*(LC/2)
VC=94.077*BW*DB*FC^.5/1000
IF VU<=VC THEN 120
IF VU>VC THEN 140
PRINT” Section adequate for diagonal tension provide
stirrups for unforeseen loads or as directed by the File
NUMERICAL
OUTPUT
***(SAMPLE
ONLY FOR
DEBUGGING
PURPOSES)
7.9 Kips
79 Kips
57.35 Kips
135
130
140
150
160
170
180
190
200
210
220
225
230
235
240
250
260
270
280
290
295
300
305
310
315
320
Civil Engineer spacing as per field conditions
STOP
REM Let us determine point of zero excess shear let L the
distance of zero excess shear from column face in feet.
PW = AS/(BW*DB)
Y1 =177.0805*PW*BW*DB^2
Y2 = 51.068*BW*DB*FC^.5
A1 =500*CV*W1
B1 = 0.58*Y2-750*CV*W1*LC
C1 = 250*CV*W1*LC^2+Y1*CV-0.5*Y2*LC
D1 = 1/2*(-Y1)CV*LC
PRINT” This is a cubic equation a third degree”
PRINT TAB(1):A1;”L Cube”;TAB(10);B1” L
Square”;TAB(15);C1”L”;TAB(22)D1;TAB(25);”=0”
*** Note when computer reach line no 225 it prints on
the computer screen A1L3 + B1L2 + C1L + D1 = 0
In our example this is 3950L3 -102932.84L2 +517.068.99L
– 384122 = 0
PRINT” There are two methods of solving this third
degree equation by Newton’s method or by the General
cubic equation”
PRINT” Put value of struct math = 1 to solve equation by
the General cubic equation program as input statement”
PRINT” Put value of struct- math =2 to solve the third
degree equation by Newton’s method”
INPUT” Value of structmath”:Structmath
IF STRUCTMATH=1 THEN 280
IF STRUCTMATH=2 THEN 590
REM This is the General Cubic equation A1X3 + B1X2 +
C1X + D1 = 0
B1=B1/A1
C =C1/A1
D =D1/A1
P =C-B2/3
Q = D-B*C/3+2*B^3/27
R =P^3/27+Q^2/4
IF R<0 THEN 390
0.0280113
38,412..2
31134.321
3950
-102,932.84
517068.99
-384,122.00
-26.06
230.95
-97.2475
136
325
330
340
350
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
515
520
530
535
540
550
560
570
580
590
600
610
Z =-Q/2+R^.5
IF Z<0 THEN 360
ZA = Z^0.3333
GOTO 470
ZB = ABS(Z)^.33333
ZA = -ZB
GOTO 470
O =ATN(ABS(R)^.5/(-Q/2))/3
PI =2.094395102
ZC = ((-Q/2)^2-R)^.5
ZA =(ZC)^.333333
IF (-Q/2)>0 THEN 460
ZA - -ZA
XA = COS(O+2*PI)*(ZA-P/(3*ZA))-(B/3)
X = COS(O+PI)*ZA-P/(3*ZA))-(B/3)
XB = COS(O)*(ZA-P/(3*ZA))-(B/3)
IF XA>0 THEN 510
IF X>0 THEN 530
IF XB>0 THEN 550
L1=XA
PRINT “Point of zero excess shear is =”;L1”Feet from
support”
GOTO 950
L1 = X
PRINT” Point of zero excess shear is =”L1;”Feet from
support”
GOTO 950
L1 = XA
PRINT” Point of zero excess shear is L1 is=”;L1;”Feet
from support”
GOTO 950
STOP
REM This is finding the real positive root by Newton’s
method of approximation
REM Format is A1X3 + B1X2 + C1X + D1 = 0
CALL CLEAR
137
620
630
650
660
670
680
690
700
710
720
730
740
750
760
770
780
790
800
810
820
830
840
850
860
870
880
890
900
910
920
930
940
950
PRINT” This is Newton’s method of approximations
finding the real positive root of a cubic equation”
PRINT” Try a value of X as first trial if ready then run line
660”
STOP
INPUT” Coefficients of A,B,C,D”:A,B,C,D
INPUT” Value of X as first trial”:X
INPUT” How many trials T”:T
P =1
PRINT”I am performing trial no=”;P
PRINT” Assume trial root is=”X
REM Format is Y =AX3 + BX2 + CX + D =0
Y = A*X^3+B*X^2+C*X+D
IF Y=0 THEN 750 ELSE 770
PRINT” Real root is=”;X
GOTO 940
T = T-1
IF T=0 THEN 780 ELSE 800
STOP
R-X
GOSUB 880
X=S
P=P+1
PRINT” I am performing trial no=”;P
PRINT” Trial root is=”;X
GOTO 730
STOP
REM This is a sub routine
M =A*R^3+B*R^2+C*R+D
N=3*A*R^2+2*B*R+C
S = R-(M/N)
RETURN
STOP
L1=X
PRINT’ Point of zero excess shear is located at a distance
L1=”X;”Feet from support”
138
955
960
1590
1600
1610
1630
1640
1650
1660
1670
1680
1690
1700
1710
1720
1730
1740
1750
1760
1770
1780
1785
1781
1786
1790
1795
1800
1805
1810
1815
PRINT” Copy important data’s for next input statement
then type continue to resume running”
BREAK
INPUT”DB,LC,W1,FC,FY,PW,BW”:DB,LC,W1,FC,FY,PW,BW
VD =W/2*(LC-DB/6)
MD –W/24*(LC*DB-DB^2/12)
X3 =VD*DB/(12*MD)
IF X3<1 THEN 1650 ELSE 1670
X4=X3
GOTO 1680
X4-1
VC1 =BW*DB*(51.07055*FC^.5+2125*PW*X4)/1000
VC2 =0.0940777*BW*DB*FC^.5
IF VC1<VC2 THEN 1710 ELSE 1730
VC=VC1
GOTO 1750
VC =VC2
VU = W/2*LC
S1=0.85*0.22*FY*DB/(VU-VC)
GOSUB 1910
PRINT’ Maximum spacing of stirrups as per A.C.I
code”=;SN
REM To find spacing at any distance say X1,X2,X3,X4 &
X5 we dim Xs in five elements with TS as a pointer
PRINT” Copy or put numerical values of X1,X2,X3,X4 &
X5 as data statement program line no 1795 then type
continue to resume running”
REM A one dimensional array in XS
SO = 0.187*FY*DB/(CV*(W1*LC/20.0537587*BW*DB*FC^.5))
DATA Note put numerical values of X1,X2,X3,X4 & X5 as
data statement
DIM XS(5)
FOR TS=1 TO 5
READ XS(TS)
REM Let SP spacing required at any distance X
SP(TS) = (SO-L)/(L-XS(TS))
64.5 Kips
131.554
0.8988744
49.969
57.355
79 Kips
5.66 Inches
3.599 Inches
139
1820
1825
1830
1840
1850
1860
1870
1880
1890
1900
1905
1910
1920
1930
1950
1960
1970
1980
1990
2000
2010
2020
2030
2040
2050
2060
2070
2080
2090
2100
2120
2130
2140
PRINT SP(TS)
NEXT TS
PRINT” Spacing at X1 is=”;SP(1);”Inches”
PRINT “Spacing at X2 is=”;SP(2)”Inches”
PRINT”Sapcing at X3 is =”;SP(3);”Inches”
PPRINT” Spacing at X4 is=”;SP(4);”Inches”
PPRINT” Spacing at X5 is=”;SP(5);”Inches”
PRINT” programmed by Bienvenido C. David. a
Civil/Structural Engineer on November 1981 in his
hometown Baguio City, Philippines”
PRINT” Designed by Bienvenido C. David”
STOP
REM This is a sub – routine
REM Determination of maximum spacing of stirrups per
A.C.I code of 1983
VU = CV*W*LC/2
REM Using the less accurate A.C.I code
V1 = 0.0537587*BW*DB*FC^.5
V2 =VU-V1
V3 = 0.1075174*BW*BD*FC^.5
IF V2<V3 THEN 1990 ELSE 2010
S3 =DB/2
GOTO 2020
S3=DB
S4=S3
S2 =4.4*FY/BW
IFS1<S2 THEN 2070
IF S2<S4 THEN 2100
GOTO 2120
IF S1<S4 THEN 2080
SM=S1
GOTO 2130
SM=S2
SM=S4
SN=SM
RETURN
79 Kips
32.774 Kips
46.225 Kips
65.549
11 Inches
22 Inches
11 Inches
11 Inches
140
2150
END
141
INTRODUCTION
DEFLECTION: Deflection of Reinforced concrete must be checked prior to final detailing of
members. There are two reasons for these a) a structure must be safe and b) serviceable.
a) A structure is safe if it is able to resist without distress and with some margin to spare,
all forces which foressebly will act on it during its lifetime.
b) Serviceability implies, among other things that deflections and other distortions under
load will be unobjectionable small. For example excessive beam and slab deflections can
lead to objectionable cracking of partitions, poor drainage and misalignment of sensitive
machineries. It becomes important, therefore to be able to predict deflections with
reasonable accuracy, so that members can be dimensioned to ensure both adequate
strength and appropriate small deflection.
We shall deal with deflections which occur immediately upon application of loading, the so
called instantaneous deflection. Instantaneous deflections can be calculated on
elastic behavior of flexural members. From the theory of Structural Mechanics, elastic
deflections can be expressed in the form Delastic =
where EI is the flexural rigidity
and F (loads, span) is a function of the particular load and span arrangement. Thus for a
3
uniform load in a simple span the deflection is equals to DEF = 5wl /384. For various types of
loadings and joint conditions, refer to any textbooks on Theory of Structures.
Using the transformed section of uncracked section and Ec is the modulus of concrete we have
deliu= F/EcIut here Iut = the moment of inertia of uncracked transform section and Ec is the
modulus of concrete. Investigations and experimental results from AC.I studies gives the
effective moment of Ie =
Iut + ( 1-
)3 )Ict
142
Here Ict is the moment of inertia of the cracked, transformed section Ie = effective moment of
inertia Mcr = cracking moment. For design purposes Mcr =
here Ig is the moment of
inertia of the gross concrete section neglecting reinforcement and fr is the modulus of rupture
for normal concrete taken as 7.5Fc’. Yt is the distance from the neutral axis to the tension
fiber stress.
For continuous beam the A.C.I. calls for the use of the average value.
LONG TIME DEFLECTIONS: These types of deflection are caused shrinkages and creep.
Additional longtime deflection can be computed by the formula Y = Deltaxx instantaneous
deflection where coefficients depends on the duration of sustained loads. Based from
experimental results the A.C.I. codes gives Delta = 2 -
– 0.6 here As’ = Area of
compression steel
and As = Area of steel reinforcement tension.
MAXIMUM ALLOWABLE COMPUTED DEFLECTIONS
TYPE OF MEMBER
DEFLECTION TO BE CONSIDERED
Roofs not supporting or attached
to non structural elements likely to
be damage by large deflections
Floors not supporting or attached
to non structural elements likely to
be damaged by large deflections
Roof or floor construction
supporting or attached to non
structural elements likely to be
damaged by long deflections
Immediate deflections due to live load L
DEFLECTION
LIMITATIONS
1/180
Immediate deflections due to live load l
1/360
That part of the total deflection which
occurs after attachments of the non
structural elements’ the sum of the long
time deflections due to all sustained loads
and the immediate deflections due to any
additional live load.
1/480
MINIMUM THICKNESS OF BEAMS ON ONE WAY SLABS
UNLESS DEFLECTIONS ARE COMPUTED
Member
Minimum thickness, h
143
Cantilever simply supported
One end continues
Both end continues
Members not supporting or attached to partitions or other constructions likely
to be damaged by large deflections
Solids one way slab
Beams or ribbed one way slab
L/20
L/24
L/28
L/10
L/16
L/18.5
L/21
L/8
With uniformly loaded loads and variable moments, the deflection at any point is given as
Delta =
(X3 – (2l +
)X2 +
we differentiate above equation
AX3 + BX2 + CX + D = 0
+ L3 -
-
To locate the point of deflection
and set to zero and arrived at the following generalized form
Here A,B,C & D are coefficients of unknown X while D is a
constant.
By computer author used Newton’s method Of Approximation to solve for X in 15 trials
(Program no 2) Also the maximum positive moment is given as Mp = L/2 +
here Ma & Mb are variable end moments and W and L is uniform distributed load per ft. and
L span length in feet.
COMPUTER OUTPUT
Upon input of numerical data’s, into the keyboard, computer first determines maximum
positive
moment then prints on the monitor screen equation of the deflection curve. It computes
theoretical deflection by the transformed section method (service load design). With given
longtime multiplier factor as input data it computes actual deflection based from code and
prints on the screen whether given cross section and steel reinforcements is adequate for
deflection.
144
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Deflection of beams by the working stress theory in
English Units
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1977 ACI
Code*** Deflection calculations by service load design (WSD. theory)
TITLE: Deflection of beams and slabs CODE NAME: Deflect
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 201
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Deflect
PTR NO: 4696357 at Baguio City 10/11/1982
CODE NAME: Deflect
Longitudinal
B
Ld
Section
H
N.A
Mb
bw
SECTION
Ma
reference line
Moment Diagram
Mc
145
REFERENCE TEXTBOOKS:
Design of Concrete
Structures by George Winter
CHAPTER 4 Pages 161 - 169
Concrete Fundamentals by
The Theory and Practice of
Phil M. Ferguson
Reinforced Concrete by
Simplified design of
Clarence w. Dunham.
Reinforced Concrete by
Parker
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
DEFLECT : Is a computer program that calculates the maximum deflection of a given beam with
variable end moments. Load is uniformly distributed. The program, computer solves location of
maximum positive moment then solves and locate point of maximum theoretical deflection.
Additional longtime deflection is also solved by computer and compares whether given section
is adequate or not. It uses the service load design method and the transformed section
method. The program is written in Advance BASICA language and can be use in
programmable calculators . It can easily be integrated into the E _Review center of UC BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
If false
IF THEN ELSE STATEMENT
146
computer instruction code
LINE
NO
STATEMENT
5
Call clear
10
This is calculation of beam deflection by the
U.S.D. method
PRINT” Are you ready first determine point of
maximum deflection using the 1977 A.C.I code
revised”
PRINT’ Dimension of beam in Inches, material
strength in Kips per square Inch, uniform loads
in pounds per foot.”
PRINT” Let CA moment coefficient at exterior
support and Cc moment coefficient at support
C. If all data’s are in their respective units
then run line no 70”
STOP
INPUT”CA,CB,LC”CA.CB,LC
INPUT”DL,LL”:DL,LL
REM This is Newton’s method of
Approximation solving roots for a cubic
equation good for 15 trials only
3
2
REM Format is AX + BX + CX + D = 0
REM Input the following values as given
A,B,C,D and X1 where X1 is a trial root
X = LC/3
GOSUB 1150
X2=Z
Y2=A^X2^3+B*X2^2+C*X2+D
IF Y2=0 THEN 180 ELSE 200
PRINT” Real root is=”;X2
STOP
X=X2
GOSUB 1150
X3=Z
Y3=A*X3^3+B*X3^2+C*X3+D
IF Y3=0 THEN 250 ELSE 270
PRINT” Real positive root is=”;X3
30
40
50
60
70
80
90
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
NUMERICAL OUTPUT ***(SAMPLE
ONLY FOR DEBUGGING PURPOSES)
0.0725,0.0725,25
1100,2200
147
260
270
280
290
300
310
320
330
340
350
360
370
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
610
620
630
640
650
660
670
580
STOP
X=X3
GOSUB 1150
X4=Z
Y4=A*X4^3+B*X4^2+C*X4+D
IF Y4=0 THEN 320 ELSE 340
PRINT “Real positive root is=”;X4
STOP
X=X4
GOSUB 1150
X5=Z
IF Y5=0 THEN 390 ELSE 410
PRINT” Real positive root is=”;X5
STOP
X=X5
GOSUB 1150
X6=Z
Y6=A*X6^3+B*X6^2+C*X6+D
IF Y6=0 THEN 460 ELSE 480
PRINT” Real positive root is=”;X6
STOP
X=X6
GOSUB 1150
X7=Z
Y7= A*X7^3+B*X7^2+C*X7 +D
IF Y7=0 THEN 530 ELSE 550
PRINT” Real positive root is=”X7
STOP
X=X7
GOSUB 1150
X8=Z
Y8=A*X8^3+B*X8^2+C*X8+D
IF Y8=0 THEN 600 ELSE 620
PRINT” Real positive root is X8=”;X8
STOP
X=X8
GOSUB 1150
X9=Z
Y9=A*X9^3+B*X9^2+C*X9+D
IF Y9=0 THEN 670 ELSE 690
PRINT” Real positive root is X9=”X9
STOP
148
690
700
710
720
730
740
750
760
770
780
790
800
810
820
830
840
850
860
870
880
890
900
910
920
930
940
950
960
970
980
990
1000
1010
1020
1030
1040
1050
1055
1060
1070
1080
1090
X=X9
GOSUB 1150
X10=Z
Y10= A*X10^3+B*X10^2+C*X10+D
IF Y10 = 0 THEN 740 ELSE 760
PRINT” Real positive root is=”;X10
STOP
X-X10
GOSUB 1150
X11=Z
Y11= A*X11^3+B*X11^2+C*X11+D
IF Y11 =0 THEN 810 ELSE 830
PRINT” Real positive root is X11=”;X11
STOP
X=X11
GOSUB 1150
X12=Z
Y12=A*X12^3+B*X12^2+C*X12+D
IF Y12=0 THEN 880 ELSE 900
PRINT” Real root is=”;X12
STOP
X=X12
GOSUB 1150
X13=Z
Y13=A*X13^3+B*X13^2+C*X13+D
IF Y13=0 THEN 950 ELSE 970
PRINT” Real root is=”;X13
STOP
X=X13
GOSUB 1150
X14=Z
Y14=A*X14^3+B*X14^2+C*X14+D
IF Y14=0 THYEN 1020 ELSE 1040
PRINT” Real positive root is =”;X14
STOP
X=X14
Y14=A*X14^3+B*X14^2+C*X14+D
GOSUB 1150
X15=Z
Y15= A*X15^3+B*X15^2+C*X15+D
PRINT” Real positive root is=”;X15
PRINT”Y15=”Y15;”At the end of 15 cycle”
149
1100
1110
1115
1150
1155
1160
1170
1180
1190
1157
1159
2030
2040
2050
2060
2070
2080
2100
2110
2120
2130
2140
2145
2150
2160
2170
2180
2200
2210
2220
2230
2240
2250
2260
2270
STOP
REM A sub routine for Newton’s Method”
REM with X=Lc/3 as first trial
W=1.4*DL+1.7*LL
W1=W/1000
A=1/6*W1
B= -(1/4*W1*LC+1/2*MA/LC-1/2*MB/LC)
C=MA
D=W1*LC^3/24-MA*LC/3-MB*LC/6
MA=CA*W1*LC^2
MB=CB*W1*LC^2
REM Actual sub routine
Y=A*X^3+B*X^2+C*X+D
Q=3*A*X^2+2*B*X+C
Z=X-(Y/Q)
RETURN
STOP
PRINT” This is actual computation of
deflection”
REM For symbols see program record and
drawing
INPUT”BF,BW,FC,FY”:BF,BW,FC,FY
INPUT”H,HF,X,DLP,LLP”:H,HF,X,DLP,LLP
INPUT”ASB.AST”:ASB,AST
INPUT”ASN:ASN
AT1=BW*(H-HF)+BF*HF
M1=(H-HF)*BW*((H-HF)/2+HF)+BF*HF*HF/2
Y1=M1/AT1
REM Compute moment of inertia for
uncracked section
IHC=1/12*(BF-BW)*HF^3+(BF-BW)*HF*(Y1HF/2)^2
IVC=1/12*BW*H^3+BW*H(H/2-Y1)^2
IGU=IHC+IVC
REM Compute moment of inertia of cracked
section i.e positive bending
EC=1802.498*FC^.5
ES=29000
N=INT(ES/EC)
IHU= 1/12*BF*HF^3+BF*HF*(Y2-HF/2)^2
INSERT THE FOLLOWING PROGRAM LINES
5280
3.3. IN KIPS
-20.625
-150
273.43
150 Foot kips
150 Foot kips
Example BF=75,BW=14,FC=2.5 &
FY=40
H=24.5,HF=5,X=12.5,DLP=20,LLP=80
ASB=3.58,AST=5.56
ASN=1.58 Sq Inch
648
4964.25
7.65 Inches
8756.24
24384.50
33139.503
2849.995
10
1864.
150
2262
2265
2268
2280
2290
2300
2500
AT2=BF*HF+N*ASB
M2=BF*HF*HF/2+N*ASB*(H-2.5)
Y2=M2/AT2
IHT=N*ASB*(H-2.5-Y2)^2
IGC=IHU+IHT
REM Let MCR Cracking moment & FR as
modulus of rupture and YT as extreme fiber
stress from N.A. For cracked section
FR=237.17*FC^.5
MCR=FR*IGU/(H-Y1)*1/12000
MC=(W1*LC^2/8-(MA+MB)/2-(MA-MB)/
(2*W*LC^2)
IE=(MCR/MC)^3*IGU+(1-MCR/MC^3)*IGC
IF IE < IGU THEN 2360 ELSE 2380
IE1-IE
GOTO 2385
IE1-IGC
IE2=IE1
REM Determine effective moment of inertia of
negative region
Y3=H/2
IRU=1/12*BW*H^3
REM Let IRC as moment of inertia of cracked
section i.e negative bending region
REM Let Y4 extreme fiber stress of cracked
section from N.A
REM Solve Y4
Q=N*AST*(H-2.5)+(N-1)*ASB*2.5
R=N*AST+(N-1) *ASB
S=SQR(4*R^2+8*BW*Q)
Y4= (S-2*R)/(2*BW)
I4=N*AST*(H-2.5-Y4)^2+(N-1)*ASB*(Y42.5)^2+1/12*BW*Y4^3 + +BW*Y4*(Y4-Y4/2)^2
REM Let IF as average moment of inertia to be
used for negative bending region let MCR2
cracked moment for negative bending region
MCR2=FR*IRU/H/2*1/12000
2510
2520
2530
2535
2540
I2=(MCR2/MA)^3*IRU+(1-(MCR2/MA)^3)*I4
IAV=1/2*(IE+I2)
REM Find multiplier for a long time deflection
FP=2-1.2*ASB/AST
IF FP2<=0.6 THEN 2250 ELSE 2570
2310
2320
2330
2340
2350
2360
2370
2380
2385
2390
2400
2410
2420
2430
2440
2445
2450
2460
2470
2480
2490
410.8
1725.00
4.19 FROM TOP
11343.67865
13207.907
374.99 PSI
61.53
108.00 Ft kips
16892.017
33,139
16892.017
16892.617
12.25
17,157
1303.75
87.82
420.558449
8.746
13377
43.7678 Foot kips
13470
15569.13
1.22733
151
2550
2560
2570
2580
2590
2600
2595
2610
2615
2620
2630
2640
2650
2660
2670
2680
2690
2673
2675
2680
2685
2690
2700
FP1-FP
GOTO 2580
FP1=0.6
FN=2-1.2*ASN/AST
FAV=1/2*(FN+FP1)
BREAK
PRINT” Review given numerical results then
type continue to resume running”
REM Let YA as actual beam deflection
REM We let D1,D2,D3,D4,& D5, D6 as
preliminary variable to arrived at actual beam
deflection
D1=W1*X/24
D2=X^3-2*LC*X^2+4*MA*X^2/(W1*LC)4*MB*X^2/(W1*LC)
D4=1728*D1*(D2+D3)/(EC*IAV)
rem let pd&pl as percent dead load and live
load respectively
D5=(D4*DL*FAV)/(W1*1000)
D7=PL*D4*LL/(W1*1000)
D8=D5+D6+D7
PRINT” Total deflection of member is equals
to the sum of immediate and sustained dead
and live loads=”D8;”Inches”
DC=LC*12/480
PRINT”Allowable deflection from code is
=”;DC;”Inches”
PRINT’ Compare actual deflection and
deflection as per code requirements”
PRINT” Revise section and make another run if
necessary or revise material strength
specifications”
PRINT” This was programmed by Bienvenido
C. David on November 1982 in his hometown
Baguio City
END
1.65
1.43165
1.7185
-5859.38
0.2014
0.1225
0.077
0.309
0.309
0.650
152
INTRODUCTION: For simply supported span the end moments is zero ant the
2
maximum positive moments for uniformly load w /per ft. is given by 1/8 wl . For continues
beams and frames the end moments is not equals to zero and usually negative at the faces of
2,
2
2
the interior and exterior support (1/10wl 1/12wl , & 1/9 wl ). The end moments varying
upon the type of loadings thus for a uniform load of w Lbs/Ft Please refer figure below as
shown. Approximate moment diagram only.
Ma
Mb
Mc
W Lbs per Ft.
Lc
Va
Shear Diagram
Mb (maximum moment)
Vc
Moment Diagram
Ma
Mc
● ● ●
●
Neg steel
● ● ● ●
●
●
●
●
●
Pos steel
153
BAR CUT OOF AND BEND POINTS
1) Every bar should be continued at least a distance equal to the effective depth of the
beam or 12 d bar diameter (whichever is the larger) beyond the point at which it is
theoretically no longer required to resist the stress.
2) At least one third (1/3) of the positive moment steel (one fourth in continuous span)
must be continued uninterrupted along the same face of the beam a distance at least
6” into the support.
3) At least one third (1/3) of the total reinforcement provided for negative moment at
the support must be extended beyond the extreme position of the point of inflection a
distance not less than one sixteen (1/16) the clear span or 12db,whichever is
greater.
BAR CUT OFF RECQUIREMENTS
If the positive bars are cut off, they must project past the point of theoretical maximum
moment, as well as d, or 12db beyond the cutoff point from the positive moment diagram.
The remaining positive bars must extend Ld past the theoretical point of cutoff bars and must
extend at least 6” into the face of the support.
Development length of deformed bars in Compression
a) Basic development length Ld -----------------0.02dbfy’’/fc’ but not less than 0.003dbld.
b) Modification factors to be applied to ld reinforcements in excess of that required by
analysis same as tension.
Development length of deformed bars and deformed wire in tension
A) basic development length ld
No 11 bars and smaller -----------------------------------0.045Abfy /fc’ but not to exceed
0.0004dbfy
No 14 bars --------------------------------------------------------------------------------------.0.85fy/fc’
154
No 18 bars ------------------------------------------------------- 0.11fy/fc’
Deformed wire----------------------------------------------------------- 0.03dbfy/fc’
B) Modification factors to be applied to ld
Top reinforcements (horizontal reinforcement so placed that more than 12” of concrete is
cast in the member below the bar) -------------------------------------------- 1.4
Reinforcement with fy greater than 60,000 psi (2-60000)/fy
Lightweight aggregate concrete when fc’ is specified 6.7.fc’/fct
when fct is not specified
all lightweight concrete ---------------------1.33
Sand – Lightweight concrete ---------------1.18
Linear interpolation maybe used when partial sand replacement is used.
Reinforcement spaced laterally at least 6 inches on centers with at least 3 inch clear cover
from face of member to edge bar ---------------0.8
Reinforcement in flexural members in excess of that required by analysis ----As required/As
provide
Reinforcement enclosed within spiral reinforcement not less than 1/4” in diameter and not
more than 4” pitch ----------------------- 0.75
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Bending of steel reinforcements & inflection points
155
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1983 ACI
Code
TITLE: Inflection points CODE NAME: Bendi-flect
MACHINE LANGUAGE : BASICA
COMPUTER: Tandy Computer
Program steps: 75
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Bendi-flect
CODE NAME: Bendi flect
Ma
A
w Lbs per foot
Mb
2.5”
b
●
●
●
h
A
●
Lc
●
●
Negative steel
●
●
●
●
●
●
●
positive steel
●
●
● ●
●
stirrups
●
SECTION AA
point of inflection bending of pos & neg steel
●
●
ELEVATION
REFERENCE TEXTBOOKS:
Design of Concrete
Structures by George Winter
Concrete Fundamentals by
Phil M. Ferguson
Simplified Design of Reinforced
Concrete by Parker
The Theory and Practice of
Reinforced Concrete by
Clarence w. Dunham.
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
Concrete Design by Dean
Peabody Jr.
156
My first generation home computer
BEN-DIFLECT : Is a computer program that calculates the point of inflection of a continuous
reinforced concrete beams by the U.S.D. format method in English units. It plots on the monitor
screen bending of bars and determines whether the no. of bars at any section is adequate or
not. The program is written in Advance BASICA language and can be use in programmable
calculators . It can easily be integrated into the E _Review center of UC BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE NO
10
15
20
25
STATEMENT
CALL CLEAR
PRINT “This is computer program no. 14. Finding
point of inflection i.e. bending of bars from exterior
column face and interior faces of column
respectively”
Print” This program was developed by Bienvenido
C. David, a Civil/structural Engineer on July 1983 in
his hometown at Baguio City, Philippines”
PRINT” For legends and drawings and other data’s
SAMPLE ONLY FOR DEBUGGING PURPOSES
157
80
refer to program record”.
PRINT” Material unit’s specifications in Kips per
square inch, Linear dimension in inches, live and
dead load in pounds per foot & distance of column
to column in Ft.”
PRINT” If all data’s are in their respective units then
run line no 60”
STOP
CALL CLEAR
REM Let AS1,AS2,AS3 & AS4 required steel areas
Respectively at a distance Y1,Y2,Y3 & Y4 from
exterior column face
REM Let CA,CB & CC as coefficients of maximum
moment at exterior face, midspan and interior
faces of column
INPUT”FC,FY,CA,CB,CC”:FC,FY,CA,CB,CC
85
INPUT”BW,H,HF,X,BF”:BW,H,HF,X,BF
90
100
110
120
130
140
INPUT”DL,LL”:DL,LL
W1=1.4*DL+1.7*LL
W1=W/1000
MA=3/250*CA8W*LC^2
MC=3/250*CC*W*LC^2
REM Let XA & XB as distance from maximum
moment center line
XA=LC^2/4-(MA+MC)/W1
XB=(MA-MC)/(W1*LC)
XC=XB^2
XD=SQR(XA+XC)
REM with X0 & X10 as point of inflection and X as
point of maximum deflection then
X8-LC-X
X9= X-XD/12
X10=X8-X9
PRINT “Point of inflection from exterior column
face =”X10;”Inches”
REM Determine required area of steel at Y1,Y2,Y3 &
Y4
REM At any point Y distance from column face
MX=W1/Y*(LC-Y)+(MA-MB)/LC*Y
30
35
50
60
70
75
145
150
160
170
180
190
200
220
230
240
250
FC=3.5, FY=50,
CA=1/14(0.0714285),
CB=1/16(0.0625) &
CC=1/9(0.11111)
BW=6”,H=14”,HF=4”,X=12.5”
& BF=18”
DL=275, LL=350, LC=25 Ft.
980 Lbs/ft.
.980 Kips per ft.
524.9735 Inch Kips
816.64625 Inch Kips
145.36 Inches =12.114038 Ft.
12.5 Inches
0.385962 Inches
0.385962
0.385962 Inches
158
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
450
580
585
590
600
605
610
615
620
625
630
635
640
650
660
670
680
Y=Y1
GOSUB 470
AS1=AS
PRINT” Area of steel at Y1 distance from column
face=”;AS1;”Square Inches”
Y-Y2
GOSUB 450
AS2=AS
PRINT” Area of steel at Y2 distance from exterior
column face=”;AS2;”Square Inches”
Y=Y3
GOSUB 450
AS3=AS
PRINT “ Area of steel at Y3 distance from exterior
column face=”;AS3;”Square inches”
Y=Y4
GOSUB 450
AS4=AS
PRINT” Area of steel at Y4 distance from exterior
column face=”;AS4”Square Inches”
PRINT” Programmed by Bienvenido C. David on July
1983 at Baguio City”
PRINT” Designed by Bienvenido C. David a
Civil/Structural Engineer on July 1983 at Baguio
City, Philippines”
REM This is a sub routine one
M1=W1/2*Y*(LC-Y)*12+(MC-MA)*Y/LC-MC
M=ABS(M1)
D=H-2.5
IFD^2-2.61*M/(FC*BW))<0 THEN 605 ELSE 615
PRINT “Dept of stress rectangular block a is
imaginary not possible review given data”
STOP
A= D-SQR(D^2-2.61*M/(FC*BW))
AST=M/(0.90*FY*(D-A/2))
P=AST/(BW*D)
PN=.2/FY
PM=0.541875*FC*87/FY/(87+FY)
IF P>=PN THEN 650 ELSE 680
PRINT” Actual steel reinforcement ratio governs”
P1=P
GOTO 700
PRINT” Minimum steel reinforcement ration from
-734.41
11.5”
159
690
700
710
720
730
740
750
760
770
780
code governs”
P1=PN
IF P1<=PM THEN 710 ELSE 730
P2=P1
GOTO 750
PRINT” Maximum steel reinforcement ratio from
code governs”
P2=PM
P3=P2
AS=P3*BW*D
RETURN
END
160
In This chapter, the working stress method of design is compared to the ultimate
method of design
giving the designer a guide where to apply the two design formats. Discussions
of rectangular columns in four cases are all explained . The first case concentric compression, the
second case the balanced eccentric compression, the third case the eccentric compression
and the fourth case the eccentric tension. Formula derivations applicable to different cases were
derived by the author for computer application. An interesting feature of this chapter is the use of the
PROGRAM DES-COL PROGRAM NO 15 & and PROGRAM ANAL- COL PROGRAM
NO 16, how the two programs are used in designing a rectangular column subject to combined
bending and axial loading. Interaction diagram is also presented in the succeeding pages showing the
relationship of the three modes of failures.
A column is any member which carries an axial compression load with or without the presence of axial
bending . Members such as strut (usually found in trusses either in buildings, highways and railways
bridges) is usually classified as a column). To be more particular, our discussion on columns will be
applied to buildings and other vertical structures.
In general , columns are classified in accordance with the reinforcement that is
used.
a) Tied columns in which reinforcements consists of longitudinal bars and separate lateral ties.
b) Spirally reinforced columns in which closely spaced spirals encloses a circular concrete with
longitudinal tie bars.
c) Composite columns, having a structural steel or cast iron column thoroughly encased in
concrete reinforced with both longitudinal and spiral reinforcement.
d) Combination column Consisting of structural steel encased in concrete at least 2 1/2” thick.
161
ANALYSIS
As mentioned earlier, test results conducted at Lehigh University and University of Illinois
indicates that at higher loads the strain diagram of steel and concrete does not coincide closely to the
strain based on the W.S.D. method. Due to this findings, the U.S.D. method (Ultimate Strength Design)
was formulated. The A.C.I. code of 1963 edition permits the use of either the W.S.D. or the U.S.D.
method, the choice is left to the designer. However in the 1977 or 1983 A.C.I. code, the U.S.D.
method is the principal analysis used. Textbooks appearing on the market are based entirely on this
concept.
In less developed countries , a question of prime importance to the Structural Engineer is where to
apply the U.S.D. method.
U.S.D. method emphasize a strict quality control of materials, strict supervision and quality
workmanship. As an example a structure was constructed based on the U.S.D. format, material
specifications is Fc’ = 3000 P.S.I. and Fy = 50000 P.S.I the original workmanship factor is taken as 0.90 as
usual batch of concrete will be by weight. The site is located at 30 kilometers from the nearest
laboratory. During construction a sample of the material is brought for testing. The owner demands
that the structure be finished in one day , the aggregates were taken from nearby roadside. There is no
adequate supply of water, thus concrete curing is neglected . Workmen from nearby locality are not
skilled. proportioning of concrete is done by volumetric measurement. Slump test were never
conducted thus workability of concrete is a guess technique. After seven days , the Engineer brought
the samples with the following results in tabulated form.
DESIGNED
SPECIFICATIONS FOR
CONCRETE
3000 P.S.I
ACTUAL USED DURING
CONSTRUCTIONS AS
PER LABORATORY
RESULTS
2500 P.S.I
YEID STRENGTH OF
STEEL SPEFIFICATIONS
50000 P.S.I
YIELD STRENGTH OF
STEEL AS PER
LABORATORY RESULTS
40000 P.S.I
When the structure was finished , the clearance of steel reinforcement deviates from the standard 2.5”
to 2”. Using author’s derived formula we solved the required area of steel based from the design
material specifications.
As = (bdfc – (bdfc)2 – 2bdfcMu/ ))/fy
here Mu = 223 Inch Kips
fc= 3KSi and Fy = 50 KSI substituting in the above formulas.
= 0.90
b = 12’ d= 16 “
162
As = ((12 x 16 x 3) – (12 x 16 x 3)2 –(2 x12 x3 x112/0.90))/50 which when simplified gives As =
0.1586 In2
Based from the test results with b=12” and d=16.5” fc’ – 2.5 K.S.I. and fy=40 K.S.I the original moment
factor then is found to be doriginal/dactual 0.90 =
(.90) = .872
Substituting the above values in the author’s derived formula we have a new value of As
As = ((12 x 16.5 x 2.5 ) – (12 x 16.5 x 2.5)2 –(2 x 12 x 2.5 x 112.5/0.872))/40 =0.197004 In2
Since this is quite larger than the previous calculations, there is a possibility that the
structure will fail.
In contrast to the U.S.D. method , the W.S.D. method stipulates that the stress use in design
computation be reduced by a factor of 66% of that of the allowable compressive stress of concrete and
60 to 70 % reduction factor for the yield strain of steel . If fc’= 3000 P.S.I, the equivalent design stress
will be 0.45 x 3000 = 1575 P.S.I and for steel Fy =0.36 x 50000 = 18000 P.S.I. and for the test results
fc =0.45 x 2500 = 1 125 P.S.I and for steel fy 0.36 x 40000 = 14400 P.S.I.
From these calculations , it is evident that whether the steel used does not meet the required material
yield strength and on the other hand the required compressive strength is not attained , nor quality
workmanship is observed still the W.S.D. method has a factor of safety far greater than the u.s.d.
method . From the above examples , we can therefore conclude where quality workmanship is not a
standard practice. The W.S.D. method is safer to use than the US.D format.
The working stress theory is beyond the scope of this book, the designer
is referred to “The
Theory and Practice of reinforced Concrete” by Clarence W. Dunham or Simplified Design of
reinforced Concrete” by Parker
The U.S.D. method of column analysis will be used in the computer solution both combined bending
and axial loading.
For combined loading and axial loading, the A.C.I. code provides reduction factor. For tied column
0.90 and for spirally reinforced member
. The reduction factor is
similar to 0.90 for rectangular beams and 0.85 for shear and diagonal tention. With these factors and
using the the 1977 or 1983 A.C.I. code notations. The strength design then would be Pd = Pn =
.70Pn and
Md= Mn where the symbols Pn and Mn are nominal axial and nominal moment respectively.
163
RECTANGULAR COLUMN ANALYSIS
In the figure below is a section of column, the equivalent stress and strain distribution. We let Ast = as
tensile area of tension steel and Asc area of compression steel. If the steel is placed symmetrically then
Ast = Asc = At/2
2.5”
●
Asc
●
d
●
c
N.A
●
ec
●
es
Cs
h
Cc
●
(d-c)
b
Ast
.85Fc’
Ts
es
FIG A
FIG B
SECTION
FIG C
STRAIN DIAGRAM
STRESS DIAGRAM
A TYPICAL STRUCTURAL SECTION
A
E
I
M
3rd floor
B
F
C
G
K
O
H
L
P
D
W lbs.Ft
w Lbs/ft
J
↶
E
l2
1st floor
J
FIG 2.4
B
2nd floor
L1
G
↷
⇩
164
Pn =Nominal axial load at ultimate in Kips
Mn = Nominal bending moment in foot kips
d = Distance of centroid tensile area from outer face of column.
Ts = tensile force of tension steel in Kips
Cs = Compressive force of steel compressive region in Kips
a= Depth of stress rectangular block at ultimate in inches
e’ =Lever arm of Pn based from centroid tension steel in inches
ec= Strain of concrete at ultimate = 0.003 at failure
es’ = Strain of compression steel = Fy/29000
es = Strain of tension steel at ultimate = Fy/29000
c = Distance from neutral axis to extreme fiber of concrete = a/0.85
Let us consider the structural frame shown in fig 2.4 for the continuous beam B-F—J a free
body diagram is shown below. In the free body diagram since w1 = w2 and l1= l2 it follows
that MFB= MFG the unbalance moment then is equals to Mu=MFB – MFG = 0. When the
unbalance moment is equals to zero, this is known as concentric compression. Again for frame
G-K-O the loading is not symmetrical Mgk is greater or lesser than MKO either MGK OR MKO is
greater or lesser or vice versa, an unbalanced moment is produced at column J-K-L. The
unbalanced moment can be verified from the free body diagram of JK as shown on the figure.
2.4
MJK= MGK – MOK in kips. If an unbalance moment occurs, the analysis required is different
from that of concentric compression, an eccentricity is introduced. The 1977 A.C.I. code
notation denotes this eccentricity by e and can be computed by the formula e =
the
165
subscript u means the ultimate loads and this kind of loading is usually referred to as eccentric loading.
Eccentric loading has two meaning eccentric tension and eccentric compression which will be
discussed later.
Concentric compression results when the eccentricity ratio is zero (0). The point of application of Pn
must coincide thru a point known as plastic centroid. If we let
of tension steel then from the figure below.
As
.85fc’
●
Pno
●
P.C.
●
as the distance of Pn from the centroid
●
Cs
h
h
N.A
Ts
b
At
FORCE DIAGRAM
The axial force Pn would be the sum of the two materials concrete and steel neglecting the area
displaced by the steel bars then summation of forces horizontal equals zero ∑ we have
Pn = .85fc’(b)(h) + Asc(Fy) + Ast (Fy)
(1) thus for a cross section (example only) b = 14”,h=18”
, fc’=3000p.s.i. and Fy=50000k.s.i. we have Pno = 0.85(3000)(14)(18) + 2(50000) + 2(50000) = 842600
Pounds.Also taking moments about centroid tension steel (i.e. same figure) we have
=
For symmetrical steel reinforcements, the centroid of cross
section coincides with the plastic centroid.
Concentric compression occurs when the moment (exterior and interior faces of column are
equals to each other) that is Mu1 = Mu2 so that the net moment is equals to zero therefore
e = (Mu1 – Mu2)/Pu = 0 For continues reinforced concrete frames and beams, the moments in
the column face is not equals to zero and for corner columns the magnitude of the moment is
greater than the moment in the interior columns. Due to the unbalance moment a different
analysis is formulated. We call this eccentric compression. Analysis of eccentric
compression and tension, eccentric tension and eccentric compression are discussed under
program no 15 & 16.
166
166
ANALYSIS & FORMULA DERIVATIONS FOR COMPUTER APPLICATION
Program no 15 code name Des-Col is all about eccentric compression, eccentric tension.
Eccentric compression occurs when the actual eccentricity ea is equals to the balanced eccentricity eb
of the actual cross section of the column. Interior columns usually have an eccentricity less or equals to
the balanced eccentricity eb in comparison to corner columns. For eccentric compression a different
analysis is required. Consider the figure below as shown.
Pn
●
Ast
.85fc’
● ●
es’
eu
Cs
y
Cc
ec
d
N.A.
Asc
●
C
d-c
● ●
Ts
es
b
A Section
B Force Diagram
C Strain Diagram
because of the eccentricity e due to unbalance moment Mu, the point of application Pn lies at a distance
e reckoned from the N.A (neutral axis) of the section. The equivalent strain diagram is shown on figure
C. The neutral axis shifts from the original concentric compression by a distance say Y. The new N.A can
be solved by the formula
C=
assuming that compression steel yields then from the stress
diagram summation of forces horizontal equals zero ∑ we get Cc – Ts + Cc = Pn (Equation 3).
As the eccentricity e increases so does Y( the deviation from original concentric compression) increases
causing the area of stress rectangular block decreases and respectively
Cc decreases, the right hand
term of equation 3 Pn approaches (Cc – Ts) as a limit. There will be a point along the cross
section of the column where the tension steel, compression steel and compression concrete block Cc
167
will fail simultaneously. We call this the balance point and the eccentricity at that point is designated
by eb. It is customary to find the nominal load at balance point we designate the load by PnB. Since
concrete failure commence at ec and es’ and es commerce at e = fy/29000 substituting these values
in the strain diagram and putting eu = ec = 0.003 diagram figure C we get
eu)(d) =
cb = eu/(Ey +
(d) with Es as the modulus of elasticity of steel = 29x106 we get Cb =
(d) (Equation 5) Equation 5 is the formula to locate eccentricity at balance load,
given material strength specifications and column section.
From the stress diagram figure B summation of forces horizontal ∑ equals to zero Cs and Ts
cancels each other so that a =
(Equation 6) but ab = B1(Cb) (equation 7) with B1 =
0.85 and ab depth of stress rectangular block at balance load. Equating (6) and (7) and
substituting the value of Cb from equation 5 in equation 7 we get the nominal balance load
PNB = 0.72(FC’) (B) (D) (
= 0.72(FC’) (B) (D) (
) (Equation 8)
Equation 8 is known as balanced load. For computer application
LEGENDS
Fc = Cylinder strength of concrete at ultimate (K.S.I)
Fy = Yield strength of steel at ultimate (K.S.I)
H = Larger dimension of column in inches
B = Smaller dimension of column in inches
EB = Balance eccentricity of section in inches
EA = Actual eccentricity of loading in inches
Ec = Yield strain of concrete at ultimate (0,003)
Ey = Yield strain of steel at ultimate = Fy/29000
168
ES = Yield strain of steel at ultimate = 29000
ES’ = Yield strain of compression steel at ultimate
d = Depth of column in inches
d’= Depth of compression & tension steel (lever arm)
MATHEMATICAL STEPS
a) Compute PU & MU
b) Compute EA = MU/PU
C) Referring to program record (figure) summation forces horizontal equals to zero Pnb = 0
d) Cc = A (0.085) (Fc’) (A) (B)
*** Note taking moments at centroid compression steel
Pnb(E1) = Pnb(D-AB/2) + ASC(Fy)(D-5) Equation 1
*** Put PU = Pnb Equation two
also E1 = EB + (h-5)/2 Equation 3
with equation ,1,2, and 3 known solve unknowns with given data’s. Note if compression steel
yields i.e. must not exceeds Fy/29000
COMPUTER OUTPUT
PUTTING = 1 Computer evaluates cross sectional area of column with side S.
Putting T = 2 Computer evaluates with given area of steel dimension B and dimension H, a
rectangular section results.. Putting T = 3 with given small dimension B computer solves
dimension H and area of steel.
Putting T = 4 with given larger dimension H computer solves small dimension B and required
area of steel
program line no 10 -90 is a guide. When computer encounters a value of 2 it goes to line 240
computes sides of column S then jumps to program line 540 by a goto statement performs line
169
540 to 560. A gosub statement at 660 computer takes the value of D and perform sub –
program 680-800 which checks whether compression steel yields or not.
*** Note there are instances for architectural reasons dimensions at balance load is not
possible. Revising any numerical output obtained may yield different design capacity of
section. Co0mputer program no 16 code name Anal Col will be useful in analyzing this type of
problem.
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170. PTR NO: 345678 at Baguio City 1983
DESCRIPTION: Design of columns at balance load. Simultaneous failure
of concrete, compression steel and tension steel
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1983 ACI
Code
TITLE: Design of columns at balance load CODE NAME: Des - Col
MACHINE LANGUAGE : TI BASICA
COMPUTER: T I - 99
Program steps: 83
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Des - Col
CODE NAME: Des - Col
CORNER COLUMN
INTERIOR COLUMN
170
Pn
●
Ast
.85fc’
● ●
es’
Cs
y
Cc a
Cb
ec
d
( Cb–2.5)
N.A.
Asc
●
ec
d-c
● ●
Ts
es
b
A Section
B Force Diagram
C Strain Diagram
COLUMN SECTION AT BALANCE LOAD PROGRAM NO 15
REFERENCE TEXTBOOKS:
Concrete Fundamentals by
The Theory and Practice of
Design of Concrete
Phil M. Ferguson
Reinforced Concrete by
Structures by George Winter
Clarence w. Dunham.
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
Concrete Design by Dean
Peabody Jr.
Simplified Design of Reinforced
Concrete by Parker
Strength of Materials by
Ferdinand Singer
My first generation home computer
171
DES - COL : Is a computer program that that designs a rectangular column at balance load
(simultaneous crushing of concrete, yielding of compression steel & tension steel
simultaneously). Given axial load and bending moment in one direction, computer solves
require unknowns for a specified given data’s. Factor for column load is given as .70 and live
load and dead load factor are 1.4 & 1.7 respectively. For less developed countries or where
workmanship factor is a must the factor 0.70 and dead and live load factors may be reduced.
The program is written in Advance BASICA language and can be use in programmable
calculators . It can easily be integrated into the E _Review center of UC BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
VALUE
If true
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE NO
10
20
25
26
30
STATEMENT
CALL CLEAR
PRINT” This is design of column at balance load,
simultaneous failure of concrete, tension steel and
compression steel”
PRINT” This programmed was developed by
Bienvenido C. David, a Civil/Structural engineer on
Nov 1983 in Baguio City”
PRINT “ All material strength specifications in kips
per square inch , dimension of column in inches,
axial and bending moment in kips and foot kips”
PRINT” If all data’s are in their respective units then
SAMPLE ONLY
FOR DEBUGGING
PURPOSES
172
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
run line number 100”
PRINT” Put T=1 For square section computer solves
area of steel and side S of square column”
PRINT” Put T=2 For rectangular section given area of
steel computer solves dimension B and dimension H
larger side of column section”
PRINT” Put T=3 For rectangular section given
smaller side of column computer solves other
dimension H of column and area of steel AS”
PRINT” Put T=4 For rectangular section given large
dimension H of column computer solves small
dimension B and area of steel AS”
Print” For legends and drawing refer to program
record”
INPUT”VALUE OF T”:T
PU = 1.4*PD + 1.7*PL
MU = 1.4*MD + 1.7*ML
EA = 12*MU/PU
EY = FY/29000
X = 0.003/(EY + 0.003)
PNB = PU
Y = PNB/(0.7225*FC*X)
IF T=1 THEN 240
IF T =2 THEN 300
IF T = 3 THEN 440
IF T = 4 THEN 500
J = Y + 1.5625
D = SQR(J)-1.25
S = D+2.5
270
280
290
300
310
320
330
PRIN” Side of column S =”;S;”Inches”
B=S
GOTO 540
INPUT” Cross sectional area of steel”:AS
ABC = AS/2
Z = ABC*FY
PNB = Z/PNB + 0.50 -0.425*X
35
40
45
60
70
336.8 Kips
273.65 Kips
9.75 Inches
0.001724
.635055
336.8 Kips
209.726
211.2885
13.285
15.785 SAY 16
Inches
4
2 Square Inch
100
0.52701
173
340
350
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
EB = EA
Q = EB + Z*2.5/PNB – 1.25
D = Q/R
H = D + 2.5
PRINT” Larger dimension of column =”;H;”Inches”
B =Y/D
PRINT” Smaller dimension of column B=”;B;”Inches”
D=D
GOSUB 680
STOP
INPUT” Given smaller dimension b=”:B
D = Y/B
H = D = 2.5
PRINT “Larger dimension H=”;H;”Inches”
GOTO 540
STOP
INPUT’ Given larger dimension H”=:H
D = H-2.5
B = Y/D
530
540
550
560
570
580
590
600
610
620
630
PRINT” Smaller dimension of column B=”;B”Inch”
CB = X*D
AB = 0.85*CB
PNB = 0.85*FC*AB*B
L =PNB*(D-AB/2)
F = 1/2*FY*(D-2.5)
K = 1/2*(D-2.5)
EB = EA
U = PNB*(EB + K)
AS =(U-L)/F
PRINT” Total area of steel AST=”AST;”Square
Inches”
PRINT” Select from tables appropriate no and bar
size”
D=D
GOSUB 680
640
650
660
9.75
9.2422
17.537048
20 Inches
11.95 say 12 “
12”
17.477 Inches
19.97 SAY 20”
20
17.5 Inches
11.98 say 20
Inches
8.4367
7.17119 Inches
336.76 Kips
3266.4036
269.625
5.3925
9.75
5099.3883
6.798
174
670
680
690
700
710
715
720
730
740
750
760
770
780
790
800
STOP
REM This is a sub routine one
W = 0.003/(FY/29000 + 0.003)
CB = W*D
ES = 0.003*(CB-2.5)/CB
EY = FY/29000
IF ES>EY THEN 730 ELSE 770
PRINT” Compression steel yields”
PRINT “Programmed by Bienvenido C. David , a
Civil/Structural Engineer on November 1983 in
Baguio City”
RETURN
STOP
PRINT” Compression steel does not yield”
PRINT “Programmed by Bienvenido C. David , a
Civil/Structural Engineer on November 1983 in
Baguio City”
RETURN
END
.635055
8.4367
0.0021111
0.001724
175
Program 16 code name (Anal – Col) which means analysis of column in two modes of failure. The
first case tension failure and the second case compression failure. When the moment such as corner
columns is extremely large that is the actual eccentricity
ea is greater than the balanced ecentricy eb of
the section then failure is iniated by yielding of tension steel followed by the crushing of concrete.
ANALYSIS AND FORMULA DERIVATION FOR COMPUTER APPLICATION – First Case – Tension failure.
eu
Pn
●
Ast
.85fc’
● ●
es’
Cs
y
Cc
ec
d
N.A.
Asc
●
C
d-c
● ●
Ts
es
b
A Section
B Force Diagram
C Strain Diagram
FIGURE 10.5
Referring to the figure above (Figure A & B) as e (eccentricity) becomes large, the deviation of Y
increases so that Cc approaches zero (0) as a limit. If that happens Pn approaches (Cs – Ts) as a limit.
From the above mathematical expression, it is evident that for large eccentricity the nominal capacity
of section is very small. This clearly indicates why large eccentricity is very dangerous in corner columns
such as columns A,B,C& M as shown in figure 2.4
176
Test results indicate that failure is initiated by yielding of tension steel followed by the shifting of N.A
(neutral axis) towards the compression side of concrete (as verified by the figure C stress strain
diagram).
Assuming that at failure, the compression steel is also yielding so that Ast = ASC again based from the
stress diagram figure B applying summation of forces horizontal ∑ equals zero we have PNB
=0.85fc’ab = Cc. Taking moments about centroid tension steel with e’ as level arm of PN from
centroid tension steel we get PN(e’) = CC(d -
) + CS(d – d’)
Substituting the value of Cc = PN =
0.85fc’ab and substituting the value of a from equation (6) we get a quadratic equation in PN in the
form
PN2/2(.85)(fc’)(b) + (e’ – d)PN – AS’fy(d – d’) = 0 Simplifying further
=
and introducing
p=
and the parameter u = fy/0,85fc’b and dividing thru by d, we obtained the nominal
moment of section for large eccentricity.
PN = 0.85fc’bd – p – ( -1) + (1 – ) + 2p(u – (1 –
)+
) Equation (9)
CASE 1 FOR COMPUTER APPLICATION
ANALYSIS AND FORMULA DERIVATION FOR COMPUTER APPLICATION – Second Case – compression
failure.
When the actual eccentricity ea is less than the balanced eccentricity eb of the section then analysis two
prevails compression failure. Interior columns are likely to have small eccentricity comparable to
corner columns. Referring to figure 10.5 stress and strain diagram Summation of forces horizontal ∑ =
zero(0)
CC + CS = TS = PN Equation (10) By inspection from the figure 10.5 as e is less than eb or in other
way as e approaches zero (0) as a limit which is in fact the N.A (neutral axis) for concentric
compression, likewise Cc as a limit approaches Ccc becomes large so that the value of PN = CC + CS
– TS then from the strain diagram , with given column section by similar triangles the distance C is =
eu/(fs/Es + eu))d) and the tension steel stress fs =
but c =
taking moments
about centroid tension steel figure 10.5 A we have Pn(e’) = Cc (d – a/2) + Cs(d-2.5) Equation
(13)
but e’ = eb +
177
Expressing C in terms of a (depth of stress rectangular block) and fs in terms of
)a3 - (
equation results in the form (
a a third degree
)a2 –
- AscFy
-87Ast)a –87 (0.85d)Ast = 0
A general cubic equation results in a third degree. This formula is
valid only for case three of column design and very useful for computer application. Eccentric
compression usually occurs in interior columns where the unbalanced moment Mu is less comparable
to the unbalanced moment at the corner columns.
3
If we let A the coefficients of a which is
-(
(
)
and B the coefficients of a
) and C the coefficients of a
- AscFy -87Ast) then
2
which is equals to
which is equals to
Aa3 + Ba2 + Ca + D = 0
Case two of column
design compression
failure
this becomes the general cubic equation (program no 1 Struct math solver 1 or program no 2
– struct math solver -Newton’s method of approximation
This clearly shows the importance of the two mathematical programs I & 2 that I
developed in relation to the application of computer in “Reinforced Concrete
Design” Author uses the General cubic program no 1 as a sub routine within the
main program.
Shown below is an equivalent inter action diagram between programs no 15 &
program no 16.
Compression failure
program no 16 Anal Col
ec =
eb =
Balanced Load
program no 16 Anal Col Tension failure
program no 15 Des Col
et =
178
LEGENDS AND DESIGN STEPS
B: Smaller Side of column
Asc Area of compression steel in square inches
Ast Area of tension steel in square inches
fc’ Cylinder strength of concrete in Kips per square inch
fy Yield strength of steel at ultimate
in kips per square inch
d Depth distance of tension steel to outer fiber of concrete
Cb Distance of extreme concrete fiber to N.A.
a Depth of stress rectangular block
e’ Distance of Ast from Pn
ey Strain at ultimate
ec Strain of concrete at ultimate (0.003)
es’ Strain of compression steel at ultimate
Pnb Axial force at column at balance load
Mnb Nominal moment at balance load
EA Actual eccentricity of applied loads
EB Eccentricity of given section at balance load
DESIGN STEPS
FIRST CASE - TENTION FAILURE –EA is greater than EB
1) Cs = Acs(Fy)
8) Pn = - X
[(X2 – 4(Z)(M)]1/2
179
2) Et = e +(H-5)/2
9) a =
3) dc = h – 5
10 c =
4) y = (h = 2.5)
5) X = (et – y)
6) Z =1/(1.7(Fc’)(B)
Note Resulting equation is ZPn2 + XPn + M = 0 for step no 8
DESIGN STEPS
SECOND CASE - COMPRESSION FAILURE – EA is LESS than EB
1) Compute for e = Mu/Pu = (Mu/Pu)
2) Compute for y =
3 Compute for f = y + e
4) Compute for d = h – 2.5
5) Compute for g = h – 5
6) Compute for x = 0.85(Fc’)(b)/2(f)
7) Compute for z = 0.85(Fc’)(b) 8) Compute for N = Acs(Fy) + Ast(87) –
9) Compute for M = -0.85(87)(Ast)(d) = 73.95(Ast)(d)
10 ) Resulting equation is depth of stress rectangular block a cubic equation in the
format as Xa3 + Za2 = Na + M = 0
11) B = Z/X
12) C = N/X Resulting equation is: A3 + BA2 = Ca + D = 0
12) C = N/X
13) D = M/X
14) Solve for Cc = 0.85(Fc’)(A)(B)
180
15 Solved for Cs = ABC(Fy)
–
16) Solved for Ts =
18) Compute for Pd = (Pn)
where
17) Pn = cc + Cs + Ts
= .70
**** Note compare Pd and Pu revise section if necessary and re run program no
16
COMPUTER OUTPUT
Upon input of numerical data’s , computer evaluates eccentricity of section
compares actual eccentricity and solves capacity of column in Kips. In second case
of failure it prints on the monitor screen equation of the stress rectangular block
a and solves the depth a using the general cubic equation (program no 1 – Struct
math solver 1). Checks whether compression steel yields or not.
*** The program was written in a time sharing system environment and can be
easily integrated into the E _Review Centre of UC – BCF.
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170. PTR NO: 345678 at Baguio City 1984
DESCRIPTION: Design of columns in two modes of failure. Eccentric
tension and eccentric compression
181
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1983 ACI
Code
TITLE: Design of columns in Two modes of failure CODE NAME: Anal Col
MACHINE LANGUAGE : TI BASICA
COMPUTER: T I - 99
Program steps: 116
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Anal - Col
CODE NAME: Anal - Col
Pn
●
Ast
.85fc’
● ●
es’
Cs
y
Cc a
Cb
ec
d
( Cb–2.5)
N.A.
Asc
●
ec
d-c
● ●
Ts
es
b
A Section
B Force Diagram
C Strain Diagram
COLUMN SECTION IN TWO MODES OF FAILURES PROGRAM NO 16
REFERENCE TEXTBOOKS:
Concrete Fundamentals by
The Theory and Practice of
Design of Concrete
Phil M. Ferguson
Reinforced Concrete by
Structures by George Winter
Clarence w. Dunham.
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
182
Concrete Design by Dean
Peabody Jr.
Simplified Design of Reinforced
Concrete by Parker
Strength of Materials by
Ferdinand Singer
My first generation home computer
ANAL - COL : Is a computer program that analyzes a rectangular in two modes of failure. this
program is a cross checked for PROG NO 15 – DES COL. Given dimension of column, area of
compression steel and tension steel in English units. Computer computes safe capacity of given
section either in tension case 1 or compression case two with a factor reduction of 0.70. The
user selects whether the assumed section is safe for design or not. He may revise the section or
area of steel reinforcement and repeat the run. The program is based on the Ultimate Theory
of Strength Design and conforms to the latest 1977 A.C.I code of practice. For architectural
reasons dimensions at balance load is not applicable so the user/designer revise his dimensions
and checks the safe capacity using this program no 16. Program is written in Advance BASICA
language and can be use in programmable calculators . It can easily be integrated into the E
_Review center of UC BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
VALUE
If true
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
183
LINE NO
5
10
15
20
30
40
50
80
85
90
95
100
110
120
130
140
STATEMENT
SAMPLE ONLY FOR
DE-BUGGING
PURPOSES
CALL CLEAR
PRINT “This is computer program no 16
analysis of column”
PRINT ‘ See computer program record for
drawing and figure as well as program
description”
PRINT” If all data’s are in their respective
units then run line no 85”
PRINT” Material strength specifications in
Kips Per Square Inch, Dimensions of column
in inches, area of steel reinforcements in
square Inches”
Print” This programmed was developed by
Bienvenido C. David a Civil/Structural
Engineer in his hometown Baguio City on the
11 th of November 1983”
PRINT “This program is good only as shown
on the drawing”
STOP
REM let us determine which case the given
section of column will fail (First case or
second case)
INPUT”H,B,EA”:H,B,EA
H=20,B=12,EA=15,6
Note first case
EA=6” Second case
EA=15”
INPUT”ABC,AST,FC,FY”:ABC,AST,FC,FY
2,2,3.5,50
D = H-2.5
17.5 Inches
EY =FY/29000
CB =(0.003/(EY+0.003))*D
A =0.85*FC*CB
PNB =0.085*FC*A*B
184
150
160
170
180
190
200
205
210
220
230
240
250
260
270
275
280
285
290
300
310
320
340
350
360
370
380
390
ET = (PNB*(D-A/2)+ABC*FY*(D-2.5))/PNB
EB =ET-(H/2-2.5)
IF EA=EB THEN 200
IF EA>EB THEN 290
IF EA<EB THEN 500
PRINT” This is case one it is actually program
no 15 code name Des-Col (Simultaneous
failure of steel and concrete at balance load
ESC =0.003*(A-2.125)/A
EY = FY/29000
IF ESC=>EY THEN 230 ELSE 250
PN = ABC*FY+0.85*FC*A*B-AST*FY
GOTO 260
PN = 87/A*ABC*(A-2.125)+0.85*FC*A*BAST*FY
PD = 0.70*PN
PRINT” Safe design capacity of column
is=”;PD;”Kips”
PRINT “This is first case of column design
failure at balance load”
PRINT “This programmed was developed by
Bienvenido C. David a Civil/Structural
Engineer in his hometown Baguio City on the
11 th of November 1983”
STOP
PRINT” This is second case of column design
failure of assumed section by yielding of
tension steel”
ETS =EA+(H-5)/2
Y = H-2.5
G = ETS-Y
K =1/(1.7*FC*B)
M =-CS*(H-5)
N =G^2-4*K*M
L =N^.5
PNT =(-G+L)/(2*K)
AT = PNT/(0.85*FC*B)
185
400
410
420
430
440
450
460
470
480
485
490
500
510
520
530
540
550
560
570
580
590
600
610
620
630
640
650
660
670
680
690
CT =AT/0.85
EST =0.003*(CT-2.5)/CT
EY = FY/29000
IF EST=>EY THEN 440 ELSE 460
PN =CS+PNT-AST*FY
GOTO 470
PN =87/AT*ABC*(AT-2.125)+0.85*FC*AT*BAST*FY
PD =0.70*PN
PRINT “ This is second case of column design
safe capacity of column is =”;PD;”Kips”
PRINT “This programmed was developed by
Bienvenido C. David a Civil/Structural
Engineer in his hometown Baguio City on the
11 th of November 1983”
STOP
V = (H-5)/2
U =V+EA
D =H-2.5
W =H-5
I = 0.85*FC*B/(2*U)
J = 0.85*FC*B-0.85*FC*B*D/U
T = ABC*FY+AST*87-ABC*FY*W/U
S =-73.95*AST*D
E =J/I
F = T/I
C =S/I
P =F-E^2/3
Q = C-E*F/3+2*E^3/27
R =P^3/27+Q^2/4
IF R<0 THEN 720
Z =-Q/2=R^.5
IF Z<0 THEN 690
ZA =Z^0.3333
GOTO 800
ZB =ABS(Z)^.3333
7.5”
13.5”
17.5”
15”
1.32222
-10.5777
162.888
-2588.25
-8
123.193
-1957
101.859
-1666.91
733789.9877
-1690.07
11.88
0
186
700
710
720
730
740
750
760
770
780
790
800
810
820
830
840
850
860
865
869
870
880
890
894
897
900
1000
1010
1025
1028
ZA =-ZB
GOTO 800
O =ATN(ABS(R)^.5/(-Q/3))/3
PI =2.09439102
ZC =((-Q/2)^2-R)^.5
ZA =ZC^.3333
IF (-Q/2)>0 THEN 790
ZA =-ZA
XA = COS(O+2*PI)*(ZA-P/(3*ZA))-E/3
X = COS(O+PI)*(ZA-P/(3*ZA))-E/3
XB = COS(O)*(ZA-P/(3*ZA))-E/3
IF XA>0 THEN 840
IF X>0 THEN 870
IF XB>0 THEN 900
AC =XA
GOSUB 1040
PDXA = PDC
PRINT” Safe design capacity of column
is=”;PDC;”Kips”
PRINT “ Programmed by Bienvenido C.
David, a Civil Structural Engineer in his
hometown Baguio City on November 1983”
AC = X
GOSUB 1040
PX =PDC
PRINT “ Safe design capacity of column
is=”;PDC;”Kips”
PRINT “ Programmed by Bienvenido C.
David, a Civil Structural Engineer in his
hometown Baguio City on November 1983”
AC = XB
GOSUB 1040
PDXB =PDC
PRINT “Safe capacity of column
is=”;PDC;”Kips”
PRINT “ Programmed by Bienvenido C.
David, a Civil Structural Engineer in his
-11.88205
187
1030
1040
1050
1060
1070
1080
1090
1100
1110
1130
1140
1150
1160
hometown Baguio City on November 1983”
STOP
REM A sub routine
ES = 0.003*(AC-2.125)/AC
EY =FY/29000
IF ES=> EY THEN 1080 ELSE 1130
PNC =ABC*FY+0,85*FC*AC*B87*(AST)*(0.85*D-AC)/AC
PDC =0.70*PNC
RETURN
STOP
PMC = 87*ABC*(AC2.125)/AC+0.85*FC*AC*B-AST*87*(0.85*DAC)/AC
PDC = 0.70*PMC
RETURN
END
188
INTRODUCTION
A building or a bridge is generally considered two have two main portions, the
superstructure and the substructure, the latter is known as foundation. What is a
foundation?
Foundation is a very important in Civil Engineering Structures, author based from experienced
observes a good foundation design, for example is seldom appreciated. Owners who are willing
to spend large sums of money for external beautification are at times reluctant to spend even
small sums for proper soil investigation. They fail to realize that however beautiful the
exteriors, the structure will not perform its functions properly unless there is a good
foundation support. Even more regrettable is the fact that there are Structural Engineers and
Architects who do not appreciate the integral relationship between the superstructure and the
foundation of a building and therefore reduce construction cost by economizing the
foundations; This false economy often proves expensive in the long run.
Let us site three examples
1) A newly built first class hotel in one of the Southeast Asian countries had to cut three stories
from the building during construction because of the possible danger of foundation failure.
189
2) An office building costing 5 million dollars showed sudden excessive settlement at the
completion of the construction. The cost of underpinning and of repairing the structural
damage was more than 500000 dollars, an amount twice the original cost of the foundation.
3) An 18 storey office building used precast concrete piles for its foundation. During
construction, the contractor using ordinary driving equipment claimed that he had driven a
large number of piles to a depth in excess of 200 Ft. A trained soil Engineer would not have
permitted continuation of the piling without an investigation. Subsequent load test proved that
most of the piles could not carry the design load and that most of them were probably broken,
as a result the construction was delayed for six months.
These few cases show the lack of proper soil investigation, of knowledge of soil mechanics, of
proper control during construction or a combination of any of these factors can spell failures
for a project. from these case histories, it is evident that foundation designs play an important
role in the “Structural Engineering “profession.
The foundation supports the superstructures, but it may consist various parts of its own. There
are many kinds of foundations, raft, mat, caissons or structures supported by piles.
Chapter eight is all about the most common type of footings encountered by the “practicing
Civil Engineer or Architect met in everyday practice namely the square footing, the rectangular
footing, the combined footing and the trapezoidal footing. The combinations of the two
mathematical programs “General cubic equation and Newton’s method of approximation are
included within the main program as sub routine programs within the main program.
The method of attack used in footing design is similar to that in beams, however since footing
rest on a medium such as soil or rock, a combination of “Soil Mechanics, “Rock Mechanics”
and “Structural Design” is essential to understand the analysis involved. These topics are
sometimes included in textbooks on “Soil Mechanics”,” Rock Mechanics” and others in
“Foundation Engineering”.
The ultimate strength theory will be used thru out with both English units and metric
units.
There are six critical stresses developed in footing design.
1) Dowels into the footing
2) Strength of soil beneath footing soil pressure q
190
3) Shear strength
4) Bearing (compression) from column face on top of footing
5) Reinforcement provided.
6) Development length of bars.
,
191
Square footing or simply known as Spread Footing from the world alone means to distribute the
concentrated load over a large area which has a lower intensity of pressure, sometimes this is called an
isolated footing and used generally to distribute the load from the base of the column. The depth is
usually controlled by diagonal tension. Let us consider the figure, the perimeter in shear for a square
column is 4(w + d) and the depth to the centre of the steel area is d. The soil pressure on the base of
the footing is q =
. Summing forces vertical ∑ on the diagonal tension. The footing weights
cancels
which gives Pu = (w + d)
2
q + 4d’ (w + d)vc = 0 rearranging terms we get
d2(vc + ) + d(vc + )w = (BL – W2)/4
Equation (1)
to facilitate the design of footings, various charts are published by the A.C.I. code. The quantity
(A2)/(A1) should not be less than 2 equation 1 is the general equation for solving the depth
of footing a quadratic equation and very useful in computer application.
L
B
A1
2
d
2
1
1
A2
ART. 11.10.1A
Bearing pressure on top of footing
192
However a direct formula derived by the author can be used to solve the depth of
footing directly it may be used also for a sub – routine program in lieu of the standard
mathematical expression used by the author. (See design steps or program no 17 code
name “Square Foot’).
●
●
●
●
Column reinforcement with ties
d
● ● ● ● ● ● ●
B
2” or 2.5” covering
Square Footing with column section
(3.4v1 + Pu)d12 + s(3.4v1 + 2Pu)D1 = Pu(A – a2)
here the legends
v1 = Nominal shearing stress caused by the column load shear.
Pu = Ultimate column load
d1 = Effective depth of footing with respect to v1
S = Side of column
A = Area of footing in plan. The above mathematical expression is helpful for computer
application . rather than using the design steps outlined, we treat the mathematical expression
directly to solve depth of footing.
DESIGN STEPS
1) Compute footing dimensions
2) Convert Qa to ultimate
3) Find effective depth d (Diagonal punching shear usually governs)
4) Compute required steel for bending treat a unit strip as a cantilever beam.
193
5) Checked bearing pressure on top of footing
6) Compute development length and area of dowels
7) Sketch and design footing
MATHEMATICAL STEPS
Find PU = 1.4*D + 1.7*L
Find footing size S =
l’ = (S – W)/2
Compute for Mu = Qul2/(2)(12) Inch Kips
Compute for Qu = Pu/S2
Compute for Vc = 15.48(
)
Solve for A = (Vc + Qu/4)
Solve for B = (Vc + Qu/2)/w
Solve for C = (S2 – W2)Qu/4
Solved for d = Ad2 + Bd – C = 0 a quadratic equation
Solve for As = 10.2*Fc’/Fy( 12d±
144d2 – 0.2178Mu/Fc’
)
LEGENDS
Fc = Kips per square inch
Fy = Kips per square inch
Vc = shear strength of concrete
Mu = Inch Kips
P = Concentrated load in Kips
d = In inches
L = Live loads in Kips
As = Square inch per foot.
W = Width of column
194
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170. PTR NO: 345678 at Baguio City 1984
DESCRIPTION: Design of Square Footing by U.S.D. format
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE) 1983 ACI
Code
TITLE: Design of Square Footing by U.S.D. format CODE NAME:
Square - Foot
MACHINE LANGUAGE : IBM BASICA
COMPUTER: IBM
Program steps: 88
LIBRARY MODULE: Floppy Disk PROGRAM NAME: Square - Foot
CODE NAME: Sqr - Ft
L
(w + d)
Y
●
●
●
●
Column reinforcement with ties
d
B diagonal punch shear
PLAN
V=2
● ● ● ● ● ● ●
B
2” or 2.5” covering
ELEVATION Square Footing with column section
Vc = 4
Code article 11.10.1a wide beam shear
195
REFERENCE TEXTBOOKS:
Foundation & Structures By
Foundation Of Structures By
Foundation Analysis &
Gregory Chebotariof
Clarence w. Dunham.
design By Joseph Bowles
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
Soil Mechanics In
Engineering Practice By R.
Peck & Hansen
Strength of Materials by
Ferdinand Singer
My first generation home computer
SQUARE FOOT : Is a computer program that designs dimension of square footing. It calculates
size of square footing with side S given axial load in kips (concentric loading and material
strength specifications). Computer computes required area of steel reinforcements per foot of
width and checks actual steel reinforcement ratio from code requirement 1977. It checks
allowable soil pressure and bearing pressure on top of footing and the required development
length of steel, area of dowels included in accordance with the A.C.I. code specifications
.Program is written in Advance BASICA language and can be use in programmable calculators
. It can easily be integrated into the E _Review center of UC BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
196
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE NO
5
10
STATEMENT
75
CALL CLEAR
PRINT” This is design of Square Footing by U.S.D.
method in English units”
PRINT “ Consult drawing and program record for
legends and symbols”
PRINT “ Be sure all given data’s are in their respective
units dead load in Kips , material strength
specifications in Kips per square inch dimension of
column in inches”
PRINT “ If all data’s are in their respective units then
run line no 60”
STOP
CALL CLEAR
REM This program was developed by Bienvenido C.
David a Civil/Structural Engineer in his hometown
Baguio City year 1984
INPUT”LL,DL,FC,FY,W1,QA”: LL,DL,FC,FY,W1,QA
80
85
90
100
110
120
W =W1/12
S = ((DL+LL)/QA)^.5
PRINT “Side of footing=”S;”Feet”
PU = 1.4*DL+1.7*LL
QU =PU/S^2
VC =15.48*FC^.5
15
20
25
50
60
70
SAMPLE ONLY FOR
DE-BUGGING
PURPOSES
Example only
LL=100,DL=71,
FC=3,FY=50,W1=14,QA=4
1.1666 Ft.
6.53 Ft
6.53 ft.
269.4 Kips
6.3180112
26.8113 Ksf
197
130
140
150
155
160
170
180
190
200
220
225
230
235
245
250
255
260
270
280
290
300
310
315
320
330
340
350
360
379
380
390
400
410
420
430
440
A = VC+QU/4
B =(VC+QU/2)*W
C = (S^2-W^2)*QU/4
E =(B^2+4*A*C)^.5
D =(-B+E)/(2*A)
D1 =12*D
PRINT” Depth of footing=”;D1;”Inches”
REM Determine required steel reinforcement area per
foot of width
LC = (S-W)/2
MU = 6*QU*LC^2
IF (144*D^2-0.2178*MU/FC)<0 THEN 230 ELSE 245
PRINT “ Value of F is negative not possible review
given data”
STOP
F = SQR(144D^2-0.2178*MU/FC)
G =10.2*FC/FY
H = 12*D-F
AS = G*H
PI = 0.006944*AS*D
GOSUB 540
AS4 =P4*D1
PRINT” Area of steel per foot of width=”;AS4;”Square
Inches”
PRINT” Select from tables no and bar diameter sketch
and draw footing then type continue to resume
running”
BREAK
INPUT” Diameter of selected bar in inches”:DB
L1 = 12*LC-3
GOSUB 680
LD = LD
PRINT”Recquired development length as per A.C.I.
code requirement is =”;LD;”Inches”
REM Check bearing pressure on top of footing
A1 = 144*W^2
A2 = (12*W+48*D)^2
K = (A2/A1)^.5
IF K>=2 THEN 420 ELSE 440
T=K
GOTO 450
T=2
28.3908
34.96
65.207
92.88011
1.0199626
12.239
12.239 Inches
2.6817
272.81642 Inch Kips
11.40239
0.612
0.837161
0.51234 Square inch
0.003628
29.1804
29.1804
29.1804
195.9779
3963.6191
4.497
198
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
610
620
630
640
650
660
670
680
690
700
710
720
725
730
740
750
760
F1 = 0.595*FC*T
F2 = PU/(12*W)^2
IF F1>=F2 THEN 480 ELSE 510
PRINT” Actual stress on top of footing is less than
A.C.I. code limitations design okay”
PRINT” This programmed was developed by
Bienvenido C. David, a Civil/Structural Engineer in his
hometown Baguio City on the 10th of October year
1984”
STOP
PRINT “Actual bearing stress is greater than A.C.I.
code”
PRINT” This programmed was developed by
Bienvenido C. David, a Civil/Structural Engineer in his
hometown Baguio City on the 10th of October year
1984”
STOP
REM This is sub – routine one Checking of steel
reinforcement ratio within A.C.I. code limitation
PN = 0.2’FY
PM = 0.6375*FC/FY*87/(97+FY)
IF P1=> PN THEN 580 ELSE 600
P2 – P1
GOTO 610
P2 = PN
IF P2 =< PM THEN 620 ELSE 640
P3 = P2
GOTO 650
P3 = PM
P4 = PM
RETURN
STOP
This is a sub – routine two required development
length as per A.C.I. code
L2 = 0.0993463*FY*DB^2/FC^.5
L3 = 0.4*DB*FY
IF L1>L2 THEN 730
IF L2>L3 THEN 760
GOTO 780
IF L1>L3 THEN 740
L4 = L1
GOTO 790
L4 = L2
3.57 **Here T = 2
1.374 Ksi
0.004
0.0226377
0.004
0.004
11.20 Inches
12.5 Inches
199
770
780
790
800
810
GOTO 790
L4 = L3
LD = L4
RETURN
PRINT” Here LD is the required development length
(i.e. the largest of the three quantities)
200
Rectangular footing design is similar to square footing design except that wide beam
shear probably controls the depth of the footing. The reinforcement in long direction is again
uniformly distributed over the pertinent shorter width. In locating the bars in the short direction, one
has to consider that the support provided to the footing by the column is concentrated to the middle.
Depth will be controlled by shear except that wide beam action will probably control if the
length exceeds the width by a ratio greater than one.
Percentage of steel required in short direction to be placed in zone width B
E=
where E =
percentage of total steel required in short direction to be placed in zone of which of width B
and S is the ratio of long side to short side L/B. Below section and plan
Dowels
Steel bars in short direction
d
● ●
● ● ●
● ●
Steel bars in long direction
L
d
B
S
PLAN AND SECTION
L’
201
SUGGESTED DESIGN STEPS
1) Find footing size B & L
2) Find depth D for shear, check wide beam first for a strip of 1 Mtr. wide at a distance d from
column.
3) Find steel As required in long direction.
4) check diagonal tension approximate
5) Find steel As in short direction allow placing short side steel on top of longitudinal steel
6) Check if actual steel reinforcement ratio is within A.C.I. code allowable
7 Check bearing and design of dowels
8) Check development length required as per A.C.I. code
9) Design and sketch section
10) Revise section if necessary
LEGENDS
Fc = Cylinder strength of concrete at ultimate in K.S.I. VCD = 4
Fy = Yield point of steel at ultimate K.S.I.
Qu = Ultimate soil pressure F1 = Allowable bearing pressure on top of footing as per A.C.I.
code
d = Depth of footing in meters
B + Smaller dimension of footing in meters
L = larger dimension of footing in meters
Pm = Maximum steel reinforcement ratio from code
Pn = Minimum steel reinforcement ratio from code
F2 = Actual bearing pressure on top of footing.
202
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Rectangular Footing Design by the U.S.D. method
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Rectangular Footing Design CODE NAME: RECT FOOT
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 125
LIBRARY MODULE Floppy Disk PROGRAM NAME: Rect Foot
PTR
NO: 337066 Feb 1984 at Baguio City 01/09/1984
⇩ Pu
Dowels
L
w
L’
d
Ast
● ● ●
Ast
● ● ●
B
w
w
w
Ast
● ● ● ● ● ● E = 2/(S+1)
B
Longitudinal steel bars uniformly spaced
(L-B)/2
B
(L-B)/2
Code article 15.4.4
SHOWN ABOVE SECTION AND PLAN BOTH LONG & SHORT DIRECTION
203
REFERENCE TEXTBOOKS
Foundation Analysis & Design
By Joseph Bowles
Foundation Of Structures By
Clarence w. Dunham
Gregory Chebotariof
Foundations
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
RECT FOOT: Is a computer program that design and sizes RECTANGULAR FOOTINGS. The
U.S.D. method is used in this program using the international system of units in metric.
Computer solves dimension of footing given other dimension solves depth of footing then
check whether the depth of footing is okay for diagonal punch shear. It solves area of steel in
both long and short direction per meter width and checks whether actual steel
reinforcements ratio is in conformity with current A.C.I code of 1977 & 1983.Checks actual
bearing pressure on top of footings, design area of dowels and the required development
length of steel as per A.C.I. code requirements. The program is written in Advance Basica and
can be feed to a wide variety of programmable calculators and micro computers. The
program can be easily integrated into the E- REVIEW CENTRE OF UC –BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
A.B.S ABSOLUTE
If true
GO SUB = GO SUB ROUTINE
VALUE
branch out
Branch out
Main program
SGN = SIGNUM NOTATION
IF THEN ELSE STATEMENT
204
If false
COMPUTER INSTRUCTION CODE
LINE NO
10
15
20
25
30
35
40
45
50
55
60
65
67
70
75
80
STATEMENT
CALL CLEAR
PRINT “This is computer program no 18design of rectangular
footing in metric units by U.S.D. format.”
PRINT “ This program was developed by Bienvenido c. David
in his hometown Baguio City on the year February 1984
PRINT “ Material specifications in KG per Cm2”
PRINT” Dead and live loads in Kn”
PRINT” Linear dimensions of footings in Meters”
PRINT” Problem to solve other dimension and area of steel”
PRINT” For drawing and other data’s refer to program
record”
PRINT” If all data’s are in their respective units then run line
no 60”
STOP
REM First find other dimension given width of footing
INPUT”DL,LL,FC,FY,W1,W,QA,VCD,VCW,PN,PM”:
DL,LL,FC,FY,W1,W,QA,VCD,VCW,PN,PM
REM *** Notes data’s for PN, PM,VCW,VCD taken from
tables by Joseph Bowles “Foundation Analysis & Design”
second edition. By formula VCD = 4
L =(DL+LL)/(W*QA)
PRINT” Other dimension of footing =”;L;”Meters”
QU = (1.4*DL+1.7*LL)/(L*W)
SAMPLE ONLY
FOR DEBUGGING
PURPOSES
Example only
DL=1110, LL=122,FC = 211,FY
=4219,W1 =
0.45,W = 2.2, QA =
240, VCD
=1283.7KN/M2,
VCW =642 Kpa,PN
=0.002,PM = 0.016
4.037 Meters
370.59 K.P.A
205
85
90
95
97
98
99
100
110
115
120
130
140
150
160
170
180
190
200
210
220
230
240
250
300
310
320
330
340
350
360
370
380
390
400
410
420
430
435
450
540
REM Find footing depth
E1 = L/W
IF E1>1 THEN 100
IF E1<1 THEN 98
PRINT” Length exceeds width by a ratio of greater than 1
wide beam action controls”
STOP
L1 = L/2-W1/2
D = L1*QU/(QU+VCW)
PRINT” Depth of footing=;D;”Meters round up value to
positive integer if necessary”
REM Check diagonal tension
P =(W1+D)*4
PS = V*VCD*D
PU = 1.4*DL+1.7*LL
IF PS<PU THEN 180
IF PS=>PU THEN 200
PRINT” Punching shear greater than code requirements”
STOP
REM Solve steel reinforcements in long direction As
MU = 1/2*QU*L1^2
X = FY/(1.7*FC)
Y =-D
Z = MU/(0.90*98.07*FY)
R = (Y^2-4*X*Z)^.5
U = (-Y-R)/(2*X)
P = AS/(1*D)
PN = 14/FY
PM = 0.016 From table
IF P=>PN THEN 350 ELSE 370
P1 = P
GOTO 380
P1 = PN
IF P1<= THAN PM THEN 390 ELSE 410
P2 = P1
GOTO 420
P2 = PM
P3 = P2
ASL = P3*1*D
PRINT” Area of steel per foot of meter=;ASL;”Long direction”
REM Design of steel reinforcements in short direction
INPUT”FC,FY,WW1,D,DB”:FC,FY,W,W1,D,DB
1.835
1.7935
0.65638 Meters
4.42552
3728.92105
3291
596.027 kn –Mtr.
11.76
-0.65638
0.00160058
0.5962746
0.65638
0.0038933
0.002
0.0038933
0.002
0.0038933
0.016
0.002555
Example only=
206
322,FY = 4219
KG/Cm2,W1 =0.45
m,D = 0.65638, QU
= 375.59,L = 4.037,
DB = 20 MM(2CM)
536
537
560
570
580
590
600
610
620
630
640
650
660
670
675
680
685
690
700
710
720
730
740
750
760
770
780
790
795
800
810
815
PRINT” Select from table appropriate bar size no then type
continue to resume running”
BREAK
L2 =(W-W1)/2
MU = QU/2*L2^2
X = FY/(1.7*FC)
Y = -D+DB+DB/2
Z = MU/(0.90*FY*98.07)
R = (Y^2-4*X*Z)^.5
U = (-Y-R)/(2*X)
P = U/(1*D)
GOSUB 940
P4 = P3
AS =1*D*P4
AST = 10000*AS*L
Print” required area of steel in short direction
=”;AST;”Square Centimeters per meter width”
PRINT “ Select from tables appropriate bar size then type
continue to resume running for computation for bearing
pressure on top of footing”
BREAK
REM Checked bearing pressure on top of footing
A1 = W1^2
A2 = (W1+4*D)^2
C = (A2/A1)^.5
IF C=> 2 THEN 750
IF C<2 THEN 770
G=2
GOTO 780
G=C
F1 = 58.35765*FC*G here G = 2
F2 = (1.4*DL+1.7*LL)/W1^2
IF F1>F2 THEN 800 ELSE 815
PRINT “ Allowable pressure on top of footing as per code
requirement greater than actual bearing pressure”
GGOTO 820
PRINT” Actual bearing pressure greater than code review
0.875
141.866
11.76
0.62638
0.000380969
0.611906
0.0006153
0.0009375
0.002
0.0013127
13.127
0.2025
9.4588
6.83448
24626
162.53.8
207
820
825
830
840
860
870
880
890
900
910
920
940
950
960
970
980
990
1000
1010
1020
1030
1040
1045
1050
1060
1070
1080
1100
1110
1120
1130
1140
1150
1160
1170
1180
1190
material specifications”
AG = 50*W1^2
PRINT” Recquired area of dowels =”;AG;””Square Cm”
REM Compute required development length *** Note see
page 212 of Bowles about code requirements as per A.C.I.
code
LA = 20
GOSUB 1050
LD = LF
“PRINT” Required development length =”;LD;”Centimeters”
PRINT” programmed by Bienvenido C. David, A
civil/Structural Engineer on Feb 1984 in his hometown
Baguio City”
PRINT” Designed by Bienvenido c. David”
STOP
REM A sub routine for computation of development length
IF P=>PN THEN 950 ELSE 970
P1 = P
GOTO 980
P1 = PN
IF P1<=PM THEN 990 ELSE 1010
P2 – P1
GOTO 1020
P2 = PM
P3 = P2
RETURN
STOP
REM Sub routine two
LB =0.0004278FY*DB
LC = 0.00755*FY*DB/FC^.5
IF LA>LB THEN 1110
IF LB>LC THEN 1140
GOTO 1160
IF LA>LC THEN 1120
LU = LA
GOTO 1170
LU = LB
GOTO 1170
LU = LC
LF = LU
RETURN
END
10.125
43.86 CMS.
0.000937553
0.002
0.002
0.016
0.002
36.030
43.8 Cms
208
208
Combined footing is a special case of spread footing. Combined footing are used if columns are so
close to the property line that single column footings cannot be made without projecting that line
and if some adjacent columns are so close to each other that their footing would merge. In designing a
combined footing the resultants of column loads must coincide to the centroid of the footing area to
prevent eccentricity. Depth is usually based on either wide beam shear or diagonal punch
tension. First criteria, determine d based on wide beam shear with d obtained checked diagonal
tension for three conditions:
1) 3 side zone column one
2) 4 side zone column two
P1
P2
3) 3 side zone column two
⇩
⇩
SECTION
d
● ●
● ● ●
● ●
● ● ●
●
●
● ●
soil pressure Lbs/Ft.
● ● ●
●
● ●
● ● ●
●
● ●
soil pressure Lbs/Ft
B
F1
F2
Lc
PLAN
209
SUMMARY STEPS
1) Determine column loads appropriate for considerations of settlements. These consist of dead
loads plus only a portion of live load specified for design of column.
2) Using the resultant of the loads in step one, select the plan dimension of the footing to obtain a
uniform soil pressure that does not exceed the pressure appropriate for this condition of loading.
3) Using the column load specified in the building code (without load factors) and the plan dimensions
determined in step two, calculate the corresponding soil pressure. If the maximum soil pressure
under this loading exceeds the value considered appropriate for this condition of loading , the
width of the footing must be increased whereas the position of the centroid must remain unchanged.
4) Compute the soil pressure beneath the footing corresponding to the column loads multiply
by appropriate load factors.
5) Draw shear and moment diagrams for the footing when it is subjected to the maximum of
step four.
6) Using step five as the basis for design, determine the depth of the footing and the necessary
amount of reinforcing steel at appropriate locations.
MATHEMATICAL STEPS
FOR THE FORMULATION OF PROGRAM NO 19
*** Please refer to chapter 3 or chapter 4 for formula derivations
1) Depth of stress rectangular block is given as a =
144d2 – 2.61M/Fc/12
where d is the depth of footing, M is the moment in Inch Kips and Fc is the cylinder strength
of concrete at ultimate.
2) development bond i.e. for compression bars
ld = 0,02fydbfc’ or 0.0003fydb or 8 “
ld = 0,24 fydbfc’
or 0.044fydb or 200 mm where Ab area of bar in In2 or mm2
fy Yield strength of steel. PSI or MPa
db bar diameter in or mm
fc 28 day compressive strength of concrete, P.S.I. or MPa
vu =
210
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design Of Combined Footing By U.S.D. format
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Design Of Combined Footing CODE NAME: FOOT COMB
MACHINE LANGUAGE : T.I. BASIC
COMPUTER: T.I. 99/4A Texas
Instruments
Program steps: 133
LIBRARY MODULE: Floppy Disk PROGRAM NAME: FOOT COMB
PTR NO: 3046269 at Baguio City 01/09/1984
P1
⇩
P2
⇩
L
SECTION
d
● ●
● ●
● ● ●
● ● ●
●
●
● ●
SOIL PRESSURE
● ● ●
● ● ●
P2
Lc
PLAN
F1
● ●
● ●
SOIL PRESSURE
P1
B
●
●
F2
211
SHEAR DIAGRAM
MOMENT DIAGRAM
REFERENCE TEXTBOOKS:
Foundation Design By
Soil Mechanics In
Foundation Analysis &
Gregory Chebotariof
Engineering Practice By
Design By Joseph Bowles
Charles B. Peck
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
FOOT COMB Is a computer program that sizes and designs a Combined Footing by US.D.
analysis in English units. Computer finds the dimensions L & B, determines shear and moment
212
equation and plots on the monitor screen shear & moment equation. Computer solves the
depth of footing for both wide beam and checks the depth d obtained for diagonal tension.
Draw negative steel between column one and two. Designs steel reinforcement in short
direction in accordance with the A.C.I. code of 1977 and checks steel reinforcement ratio
within A.C.I. code allowable. Checks bearing pressure on top of footing and the required
development length in feet all in close conformity with A.C.I. code of 1977. The program is
written in advance basica language and can be feed to programmable calculators, personal
computers. Can be easily incorporated to the E. REVIEW CENTRE OF UC – BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE
NO
5
10
15
20
25
30
50
STATEMENT
CALL CLEAR
PRINT” This is design of Combined Footing by U.S.D.
method in English units using the 1983 A.C.I. code”
PRINT “ This program was developed by Bienvenido C.
David a Civil/Structural Engineer on September 1984
in Baguio City”
PRINT” For legends and other data’s please see
drawing and computer program record”
PRINT “All units of length in feet, material strength
specifications in kips per square inch, dimensions of
column in inches, dead load and live load in kips and
soil allowable pressure in kips per square foot”
PRINT” If all data’s are in their consistent units then
run line no 60”
STOP
SAMPLE ONLY FOR DEBUGGING PURPOSES
213
60
65
70
REM Find footing dimensions step one
INPUT”W1,W2”:W1,W2
INPUT”DL1,DL2,LL1,LL2QA,LC”: DL1,DL2,LL1,LL2QA,LC
75
80
85
90
95
100
105
130
140
141
145
150
155
160
165
180
190
200
210
220
225
230
240
250
260
270
280
290
300
310
320
330
340
360
365
INPUT’FC.FY:FC,FY
P1 =1.4*DL1+1.7*LL1
P2 =1.4*DL2+1.7*LL2
P = DL1+DL2+LL1+LL2
UR =(P1+P2)/F
QU = QA*UR
P3 = P1+P2
X = P2*LC/P3
L = (X+W1/24)*2
REM Note for value of L round to integer no
PRINT “Length of footing in Feet=”;L;”Feet”
B = P3/(L*QU)
PRINT “W width of footing =”;B;”Feet”
Y = L/B
REM Determine moment equation by integration
QB = QU*B
V = QB*X
VM = P1
VCW = 7.74125*FC^.5
D = VM/(VCW*B+QB)
PRINT “Depth of footing =”;d;”Feet”
REM Check D obtained for diagonal tension
C = (24*D+3*W1)/12
A = (W1+6*D)*(W1+12*D)/144
VD = P1-A*QU
VCD = 15.48251*FC^.5
VC = VD/(C+D)
REM Check shear stresses at column 2
A2 = (W2+12*D)^2/144
V2 = P2-A2*QU
C2 = 4*(W2+12*D)/3
V3 = V2/(C2*D)
PRINT “Actual shear stress at column one=”;VC;”KSF”
PRINT “Actual shear stress at column two=”;V3;”KSF”
PRINT “Allowable shear stress from code=”;VCD;”KSF”
370
PRINT” Compare actual shear stress from code stress”
12” 15”
Example only DL1=60K,
DL2=110K,LL1=60K,LL2=90K,
QA =2,LC=15 Ft
3 Ksi , 60 ksi
186 Kips
307 Kips
320 Kips
1.54
3.08
493Kips
9.341 Ft
19.682 Ft
8.13 Ft
2.417
25.05
13.4K.S.F
1.39 Ft
1.39 Ft
5.78 Ft
4.05 Square Ft
26.8 KS.F
21.63 K.S.F
6.96
285.6 K
10.56
19.4 K.S.F
VC =21.63,V3=19.40 &VCD
=26.80
214
400
410
STOP
REM Design of negative steel between column one
and two
420
430
INPUT”P2,VM,FC,FY”: P2,VM,FC,FY
INPUT”QB,B,L,D,W1,W2”:QB,B,L,D,W1,W2
440
450
460
470
480
490
500
X =VN/QB
M1 = 12*(QB/2*X^2-VM*(X-W1/24)
MU = ABS(M1)
M – MU
GOSUB 710
ASL = B*AST
PRINT” Total no of bars required in long
direction=”ASL;”Square Inch”
REM Design of steel in long direction
L1 =(W1+9*D)/12
QC = VM/(L1*B)
LU = = (B-W1/12)/2
M2 = QC/2*LU^2*12
M = M2
GOSUB 710
ASW = L1*AST
PRINT” Total area of steel reinforcements in short
direction=”;AST;”Square Inches”
PRINT” Select from tables appropriate size and bar no
then type continue and press enter to resume
running”
BREAK
REM Design of steel in short direction column two
L2 = (W2+18*D)/12
L3 = (B-W2/12)/2
QD = P2/(L2*B)
M3 = 6*QD*L3^2
M = M3
GOSUB 710
ASS = L2*AST
PRINT” Total area of steel reinforcements in square
inches at column two=”ASS;”Square Inches”
PRINT””Select from tables appropriate bar size and
no. then type continue and press enter to resume
510
520
530
540
550
560
570
580
590
595
600
605
610
620
630
640
650
660
670
680
685
307,186,3,60
QB
=25.05,B=8.13Ft,L=19.682Ft,
D=1.39’Ft,W1=12
Inch,W2=15Inch
7.425 Ft
-7170.468 Inch kips
7170.468
1.043
2.04 Ft
11.38 Ft
3.565 Ft
867.78588
8.47 Square Inch
3.335
3.46
11.56
830.35 Inch Kips
215
690
700
705
710
715
720
725
730
735
740
750
755
760
770
780
790
800
810
820
890
900
905
910
***
910
920
930
940
950
960
970
980
990
1000
1010
1020
1025
1030
1040
1045
running”
BREAK
STOP
REM A sub routine one Computation of steel areas
IF (144*D^2-2.61*M/(FC*12))<0 THEN 715 ELSE 730
PRINT “Depth of stress rectangular block is imaginary
not possible review given data”
RETURN
STOP
A = 12*D-SQR(144*D^2-2.61*M/(FC*12))
AS = M/(0.90*FY*(12*D-A/2))
P = AS/(144*D)
PN = 0.2/FY
IF P>=PN THEN 770
IF P<PN THEN 790
AS1 = AS
GOSUB 800
AS1 = 144*D*PN
AST = AS1
RETURN
STOP
REM Computation of development length
LA = LU*12-3
PRINT” Select bar diameter in inches”:DB
LB = 1.2649*3.1416/4*DB^2*FY/FC^.5
Note simplifying further line no 910
LB = 0.99345246*DB^2/FC^.5
LC = 0.4*DB*FY
IF LA>LB THEN 960
IF LB>LC THEN 990
GOTO 1010
IF LA>LC THEN 970
LD = LA
GOTO 1020
LD = LB
GOTO 1020
LD = LC
LE = LD
PRINT “Development length=”;LE;”Inches”
FC1 = 0.85*0.7*FC
PC1 = W1^2*FC1
PRINT “Actual load on top of footing=”;PC1;”Kips”
39.79 Inches
0.785 Inch
26.29 Inches
21 Inches
1.785
257
216
1050
1055
1060
1070
1075
1090
1100
1120
1140
1150
1160
1170
1180
1190
1195
1200
PRINT” Allowable load from code=”;FC1;’Kips”
PRINT” Compare PC1 and FC1”
AG1 = 0.005*W1^2
PRINT “Area of dowels required=”AG1;”Square Inch”
A2 = W2^2
A3 = (W2+48*D)^2
A4 = (A3/A2)^.5
IF A4=>2 THEN 1140 ELSE 1160
G=2
GOTO 1170
G = A4
FC2 = 0.85*0.70*G*FC
PC2 = W2^2*FC2
PRINT’ Allowable load on top of footing at column two
as per code requirement=”FC2;”KSI”
PRINT” Compare value of PC2 and FC2 make revision if
necessary”
END
0.72 Square Inch
225
6678
5.448
3.57 K.S.I
803.25 Kips
3.57
217
In this program I have attached a classical solution (analytical) to illustrate
clearly how the classical solution is cast into the digital solution.
Trapezoidal footing is in fact a combined footing where one side is bigger than the other,
as in the case of combined footing, the centroid of column loads must coin side with the
centroid of the trapezoidal pressure diagram. The second condition, the total bearing pressure
of the trapezoidal diagram must equal to the resultant column loads. With these conditions
two equations are formed with unknowns B1 and B2. Please see figure below
M = AX3 + BX2 + CX + D
R
B1
C1
C2
L
B2
218
Here B1
and B2 are footing dimensions from both ends. From the figure below with L as
- 1)
footing length outside to outside then B1 =
and B2
= ( ) - B1
Equation (2) here
Equation (1)
is the center of gravity of base from the outer
edge of footing larger end a shown on the figure above.
= (2B1 + B2)/ (B1 + B2) Equation (3) Equating this to C.G of column loads and solving
equation 1 and 2 simultaneously. The numerical values B1 and B2 are evaluated.
Since width varies from end to end. It is necessary to compute the required area of steel
reinforcements per meter width. This problem can be analyzed by the analytical solution
presented. In the computer programmed the author used the substitution method to solve B1
and B2 rather than using the simultaneous equation program code name simul bas as a sub
routine program within the main program. In areas of steel determination, author use a one
dimensional array to solve steel reinforcements in ten elements. See program listing.
Bearing pressure is important in footing design. Bearing pressure is obtained by soil testing or
from past construction records, building codes. An approximate empirical formula for bearing
capacity factors is given by the formula.
qultimate = Cnc + 0.5By1N1 + Y2DfNq for strip footing where Nc,By1, Nq are soil bearing
capacity factors whose values depends on the angle of internal friction. C = cohesion of soil
below footing level Y1 is the effective unit weight of soil below soli level. B = footing width
and Df is the depth of footing below lowest adjacent soil surface and qult. = Ultimate gross
bearing capacity or soil bearing pressure (developed by Terzhagi). For cohesive soil factor for
shallow strip footing qult.= CNc = 5.2C. When bearing capacity equations is used in design, it
is convenient to apply a factor of safety between 2.5 and 3 to the value of q ultimate. The q
design factor therefore is qdesign = qultimate/3 or qdesign = qultimate/2.5
For square footing: qultimate = 1.2CNc + .4Y1BNY + Y2DfNQ where a1 and a2 are shape
factors related to length to width (L/B) ratio.
219
Listed below are tables for different values of L/B
L/B
1
2
3
4
6
Strip
a1
1.2
1.12
1.07
1.05
1.03
1.00
a2
0.42
0.45
0.46
0.47
0.48
0.50
For circular footing with radius R
Qultimate = 1.2CNc + 0.6Y1RNY + Y2DfNq
The following table gives allowable bearing pressure
Value in short tons per foot square.
KINDS OF SOIL
Quick sand and
alluvial soli
Soft clay
Wet clay and soft
wet sand
Clay and sand in
alternative layers
Firm and dry loam or
hard clay, dry clay or
fine sand
Sand confined
Compact coarse sand
or stiff gravel
Sand and gravel well
compacted
Good hard pan or
hard shale
Rock
MINIMUM
1/2
MAXIMUM
1
USUAL
1/2
3/4
1
3
4
2
2
1
4
2
2
4
3
1
3
4
6
3
4
5
10
6`
5
10
8`
5
25
15
220
ANALYTICAL SOLUTION
Design a trapezoidal footing given data as shown below. Area for columns = 0.46m2 (side)
D1 = 1,200
Property Line
P1
P2
2100 KN
L1 = 816 Kn
⇩
5.48 mtrs
L2 = 660 Kn
fy= 4,219 Kg/cm2
fc’ =211
1476KN
⇩
D2 = 900
qa 190 Kpa
.46 mtrs
Step 1 Find qultimate
Ultimate ratio:
q ult =
5.94 mtrs
= 1.52 and qultimate =190(1.52) = 289.5 Kps
program line no 5 to line no 90
Step 2 Find end dimensions (i.e Resultants of column loads must coincide with resultant load
area). Taking moments at centre line column 1 we get and let as C.g of loads resultants
5449.2( ) = 5.489[1.4(900) + 1.7(600)] Solving for
=
= 2.395
and X’ = 2.395 + 0.46/2 = 2.625 from property line and from the area of the trapezoid
A=
(5.94) but required area of footing based on ultimate load and ultimate soil
pressure is A =
=
in two unknowns dimensions a and b
= 18.823
Equating A we have two equations
(5.94) = 18.23 Simplifying further we get
= 6.338 equation (1) Another condition is to prevent eccentricity, resultant of
column loads must coincide resultant of trapezoidal area. The
2.395.
was previously solved to be
221
The C.G of a trapezoid is given as
=
= 1.326
= 2.625 or
=
Equation (2) b = 6.338 - 2.065 substituting this
= 1.326 from which a = 6.338(1.326 –
value to equation (2) we have
0.326)
a = 2.065 Meters and back substitution we get b = 6.338 – 2.065 = 4.273 Meters
program line no 100 to 190
Step 3 Draw shear and moment See figure below (pressure diagram).
y = 1237 – 107.5X
1273
597.82
5.94 Mtrs.
PRESSUE DIAGRAM
From pressure diagram big end is = PBig end = 4.273(289.5) = 11,2373.03 KPA/M
and Psmall end = 2.065(289.5) = 597.82 KPA/M so the slope of the pressure diagram
is
=S=
–
= 107.5 From the figure above, the
shear at any section X – X is qdx therefore dv = ydx but Y = (1237 – 107.5X)
hence substituting the value of Y We get dv = (1237 – 107.5X)dx and the total
shear by integration is therefore V =
V = - 53.75X2 + 1237X + C
The constant C can be computed by the fact that at X =
0.23 m C = 0 also at X = 0.23 + dx C = -3067 also at column two X = 5.71 C = -
222
3067 with constant C known the shear at any section between faces of column 1 and 2 is
2
therefore equals to V = 1237X – 53.75X – 3067 = 0
In a similar manner the moment at any section can be found by integrating the shear
53.75X2 + 1237X + C)dx The total moment then is found
by integration M = - 53.75 2dx + 1237
+C
equation dm = (-
and simplifying further yields to M = -107.5X3/6 – 1237X2/2 – C’X’ = 0
The constant of integration can be found immediately, by inspection at X = 0.23m X’ = 0 so
3
/6 – 1237X2/2 – 3067(X
– 0.23) valid only up to the faces between column s only.
that the moment equation r at any section is M = - 107.5X
The point of maximum moment is found to be 2.828 by differentiating
M with respect to X and equating the derivative to zero. (i.e.
BELOW IS A graph of SHEAR AND MOMENT equations.
V = 1237X – 53.75X2 – 3067
= 0)
223
M = - 107.5X3/6 + 1237X2/2 – 3067(X – 0.23)
STEP 4 Find depth for wide beam shear at small end check diagonal tension at large end.
Reasoning
Vb/Vs =
= 1.2
b/a = 4.27/2.06 = 2.07 From A.C.I. code
maximum shear occurs at a distance d from column two .Putting X = 5.48 – d in the shear
equation
V = 1237X – 53.75X2 – 3067 we get V = 1237(5.48 – d) – 53.75(5.48 – d)2 – 3067
V = 2095.5 – 647.5d – 53.75d2 Net shear at section
The shear carried by concrete is at that section is equals to Vc = Wd(d)(Vc)
5.94
2.06 From the figure by proportion
=
4.27
is W = 2.065 +
Y/2
from which the width is
(D + 0.46) simplifying
further we get 2.065 + 0.37d + 0.17 = 2.24 + 0.372d
224
From tables Vc = 642 KPA substituting this value in the safe resisting shear of concrete at
that section we get Vc = (2.24 + 0.372d)d(642) Equating this to the actual shear at that section
we get the folloeing equation (2.24 + 0.372d)d(642) = 2095.5 – 647.5d – 53.75d2 Simplifying
further we get 292.6d2 + 2.085d – 2095.5 = 0 *** A quadratic equation can be solve easily by
completing squares or by the quadratic formula. Solving for d we get d = 0.89 Meters.
In determining the area of steel reinforcements in the long direction since width varies across
the length take 10 distances from the column face say 0.6, 1.2, 1.8, 2.4, 2.28 (max) , 3 , 3.6,
4.8 & 5.94.At any section X-X along the longitudinal section
W = Width = B –
X
(X) = 4.27 – 0.37205X = 4.27 –
W = 4.27 - 0 .37205X
at any section
V = Shear at any section is equals to = 1237X – 53.75X2 – 3067
3
2
M = Moment at any section is equals to = 107.5X /6 – 1237X /2 – 3067(X –
0.23) and As = area of steel reinforcements at any section is equals to
As =
.
below in table form the required area of steel at different distances
along the longitudinal section of the footing.
X
0
V , KN
0
M KN –M
0
0.6
1.2
1.8
2.4
2.28(Max)
3.0
3.6
4.8
5.94
-2344.6
-1660.6
-1015.4
-408.9
0.0
159.0
688.1
1630.3
0.00
-916.1
-2115.8
-2916.6
-3342.0
-3428.7
-3415.0
-3159.0
-1752.4
0.00
W M
4.27
4.05
3.83
3.6
3.38
3.22
3.16
2.94
2.49
2.07
As cm2/M
0
REMARKS
The maximum steel
permitted by the code is
153.3 CM2mtr, The minimum
steel is =32.48cm2/mtr
8.2
20.4
30.2
37.2
42.0
41.2
40.6
26.1
0.00
To solve for the point of inflection equate moment equation to zero and solve for X
225
M=
107.5X3/6 – 1237X2/2 – 3067(X – 0.23) = 0
Author used “Newton’s method of approximation program – Program no 2 –Struct math
solver 2) to solve for the real positive root of X)
The following computer program shows how a one dimensional array of advance basica can
be put to advantage in computing the required area of steel reinforcements as previously
calculated by the analytical method.
This line no is the same as the first
Let us begin with line no 140
column of the table on page 224
140 DATA Note example only 0.6, 1.2, 1.8, 2.4, 2.8, 3.0, 3.6, 4.84, 5.94
150 M (Q) = A*X (Q) ^3 + B*X (Q) ^2 + C*X (Q) + D
160 V (Q) = E*X (Q) ^2 + F*X (Q) + G
170 W (Q) = H – (H – J)/LC*X (Q)
180 D1 (Q) = V (Q)/ (W (Q)*VC
190 M1 (Q) = ABS (M (Q))
195 MU (Q) = M1 (Q)/W (Q)
200 R = FY/ (1.7*FC)
205 P = -88.263*FY*R
210 N = 88.263*FY This line no is the same as the fifth column of table page 224
215 Z (Q) = MU (Q)/P
220 K (Q) = N*D1 (Q)/P
225 I (Q) = Z (Q) + 0.25*K (Q) ^2
230 AS1 (Q) = Z (Q) ^.5 - .5*K (Q)
240 AS2 (Q) = Z (Q) ^.5 + .5*K (Q)
250 PRINT AS2 (Q)
226
260 NEXT Q
this line no is the same as the third column of table page 224
270 PRINT M(1);M(2);M(3);M(4);M(5);M(6);M(7);M(8);M(9);M(10)
275 PRINT V(1);V(2);V(3);V(4);V(5);V(6);V(7);V(8);V(9);V(10)
280 PRINT W(1);W(2);W(3);W(4);W(5);W(6);W(7);W(8);W(9);W(10)
This line no is the same as the second column of table page 224
This line no is the same as the first column of table page 224
The following program is how the program no 2 “Code name Newton can be used to locate
the point of inflection (bending of steel reinforcements “We start at line no: 285)
PRINT “Locate point of inflection. Let us use “Newton’s method of approximation “
290 PRINT” This is Newton’s method of approximation to solve for the real positive root”
295 PRINT” Try a value of X as first trial root X = LC/2 say 2.97 or 3
300 PRINT “If all coefficients are known then run line no 310”
305 STOP
310 CALL CLEAR
360 INPUT”Coeffecients of A, B, C, D”: A, B, C, D
705.41
370 INPUT “How many trials T: T
380 INPUT ‘Value of X as first trial”: X
3:00
390 P = 1
400 PRINT “I am now performing trial no=”; P
410 PRINT “Assume trial root is=”; X
420 REM Format is AX3 + BX2 + CX + D
430 Y = A*X^3 + B*X^2 + C*X + D
A = -17.936, B = 618.5, C = -3067, D =
227
440 IF Y = 0 THEN 450 ELSE 470
450 PRINT” Real root is=”; X
460 STOP
470 T = T-1
480 IF T= 0 THEN 490 ELSE 500
490 STOP
500 R = X
510 GOSUB 580
520 X = S
530 P = P+1
540 PRINT”I am now performing trial no=”; P
550 PRINT “Trial root is=”; X
560 GOTO 430
570 STOP
580 REM This is a sub routine
590 M = A*R^3 + B*R^2 + C*R + D
600 N = 3*A*R^2 + 2*B*R + C
610 S = R-(M/N)
620 RETURN
Negative steel reinforcements
630 END
● ●
● ●
● ●
● ●
● ● ● ● ● ● ●
● ●
● ●
Positive steel reinforcements
LONGITUDINAL SECTION
228
DESIGN STEPS FOR COMPUTER APPLICATION
1) The footing geometry necessary for a trapezoidal shaped footing is a =
and
=(
}
SUGGESTED MATHEMATICAL STEPS
1) Find q ultimate.
2) Find end dimensions
3) If time permits draw shear and moment diagrams.
4) Determine pressure diagram at Big End and Small End.
5) Design flexural steel, since width varies one should check As for several locations.
6) Checked steel in short direction same as rectangular footing using appropriate zone
w + 0.75d
7) Check bearing pressure and development length of steel reinforcements.
8) Detail and sketch footing
9) *** Note author use a one dimensional array to solve for M, V, D & As for different
locations.
LEGENDS
DL1 = Dead load acting on footing 1
DL2 = Dead load acting on footing two
LL1 = Live load acting on footing 1
LL2 = Live load acting on footing 2
UR = Ultimate ratio factor
229
QU = Ultimate allowable soil pressure
A = Dimension of footing small end
B = Dimension of footing big end
D1 = Depth of footing column 1
D2 = depth of footing column 2
ASS1 …….2….3….4 Subscripts for required area of steel at every locations along footing length.
As =
Vc = 0.17
in metric conversions
wide beam shear.
Vcd = (0.17 + 0.16/B)
Diagonal tension
Moment equation for Trapezoidal Footing is
M = A1X3 + B1X2 + C1X + D1 = 0 Equation (1)
V = A1X2 + B1X + C1
Equation (2)
To locate point of inflection solve for X for M= 0
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Design Of trapezoidal Footing By The U.S.D. format
230
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Design of Trapezoidal Footing CODE NAME: TRAP – FOOT
COMPUTER
PROGRAM STEPS: 225
LIBRARY MODULE: Floppy Disk PROGRAM NAME: TRAP - FOOT
PTR NO: 4046175 at Baguio City 11/11/1984
P1
p2
s
w/2
b
F1
F2
a
SECTION
2( + c/2)<s
1/3<x’<1/2
F1
F2
Rectangular footing is too short to reach
column two
PLAN
Negative steel reinforcements
● ●
● ●
● ●
● ● ● ●
● ● ● ●
● ●
●● ●
LONGITUDINAL SECTION
Positive steel reinforcements
REFERENCE TEXTBOOK:
Foundation Analysis and
Foundation Of Structures By
Clarence W. Dunham
Foundations By Gregory
Chebotarioff, Soli Mechanics in
231
Design By Joseph Bowles
Engineering Practice By Peck &
Hanson
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
My first generation home computer
TRAP FOOT Is a computer program that designs a Trapezoidal Footing using the U.S.D. format
of analysis in Metric Units. Computer evaluates dimensions of footing a & b and find d (depth
of footing) based on wide beam shear, it then checks the calculated depth for dia gonal
tension all in conformity with the 1977 A.C.I. code in metric units. Computer determines
shear and moment equations plots on the monitor screen equation of shear and moment
diagram. Designs steel reinforcement areas in long direction at every ten locations. Computer
solves the point of inflection (point of zero moment) by the general cubic equation and
checks it by the Newton’s method of approximation. Computer solves required development
length for bond and allowable pressure on top of footing. The program is written in Advance
Basic and can be feed to a wide variety of programmable calculators and micro computers.
Can be easily integrated into the E REVIEW CENTRE OF UC –BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
A.B.S ABSOLUTE
If true
GO SUB = GO SUB ROUTINE
VALUE
Branch out
Main program
SGN = SIGNUM NOTATION
branch out
IF THEN ELSE STATEMENT
232
If false
LINE
NO
5
10
15
20
25
30
40
50
55
60
65
70
75
80
85
90
100
110
120
130
STATEMENT
CALL CLEAR
PRINT” This is design of Trapezoidal Footing by U.S.D. method in
Metric units”
PRINT” This programmed was developed by Bienvenido C. David in
his hometown Baguio City o the 11th of November 1984.”
PRINT” For drawing and program description see attached
program record no 20”
PRINT “ All material strength specifications in Kg/Cm2, dimensions
of footing big and small end in meters, dead and live load small
and big end in KN allowable soil pressure in KPA and linear
dimensions in meters.”
PRINT “If all data’s are in their respective units then run line no 50”
STOP
REM Step 1 Find Q ultimate
INPUT”DL1,DL2,LL1,LL2,FC.QA.LC”: DL1,DL2,LL1,LL2,FC,QA,LC
140
150
INPUT”W1,W2”:W1,W2
PU1 = 1.4*(DL1 + DL2)
PU2 = 1.7*(LL1 + LL2)
P3 = DL1 + DL2 + LL1 + LL2
P4 = 1.4*DL2 + 1.7* LL2
UR = (PU1 + PU2)/P3
QU = QA*UR
PU3 = PU1 + PU2
X1 = P4*LC/PU3
X2 = X1 + W2/2
REM X2 Distance in meters from centerline of column 1 to centroid
of column loads
LC1 = LC +W1/2 + W2/2
A1 = PU3/QU
160
Y = A1*2/LC1
SAMPLE
ONLY FOR
DE-BUGGING
PURPOSES
DL1 = 1200, DL2
= 900, LL1 =
816, LL2 =660,
FC = 211, QA =
190, LC = 5.48
0.46,0.46
2940
2509
3576
2382
1.52
289.5 KSF`
5449.20
2.395
2.625
5.94
18.823 Sq
meters
6.3377 MTERES
233
170
180
190
200
210
220
230
240
Z = 3*X2/LC1
A = Y*(Z – 1)
B = Y-A
PRINT” Width of footing at small end in meters=”;A;”Meters”
PRINT” Width of footing at big end in meters=”;B;”Meters”
REM Determine depth of footing in meters
REM Determine shear equation by integration
QA = A*QU
250
QB = B*QU
260
270
280
290
295
300
305
QS = (QB – QA)/LC1
P1 = 1.4*DL1 + 1.7*LL1
R = QB*LC-QS/2*LC^2-P1
U = QS*LC-QB
C = B-A
INPUT “Value of wide beam shear VCW”:VCW
REM Value of VCW from Bowles table 8.2 page 213 is VCW =
2
units in Kn/m2
E= C*VCW/LC1 + QS/2
F = A*VCW+C*W2*VCW/LC1-U
G = -R
IF (F^2-4*E*G)<0 THEN 340 ELSE 360
PRINT” Value of D is imaginary not possible review given data”
STOP
H = -F+(F^2-4*E*G)^.5
D1 = H/(2*E)
J = -F-(F^2-4*E*G)^.5
D2 = J/(2*E)
PRINT” Depth of footing at column one =”;D1;”Meters”
PRINT “Depth of footing at column two=”;D2;”Meters”
PRINT” Width of footing in meters=”;B1;”at X distance from outer
face of column”
550
X = X2
GOSUB 860
AS3 = 10000*U1
AS4 = 1000*U2
B3 = W
PRINT” Area of steel at X2 distance =”;AS3;”CM Sq”
PRINT “Area of steel at X2 distance =”;AS4;”CM Sq”
PRINT” Width of footing at X2=”;B3;”Meters”
307
310
320
330
340
350
360
370
380
390
400
410
540
550
560
570
580
590
600
610
680
1.325 Meters
2.0597 Meters
4.278 Meters
2.06587 Meters
4.27 meters
597.82
kpa/meter
1237.03
KPA/Meter
107.5 KPA
3067 Kips
2097
107.5 KPA
2.2183
642 KPA
293.505
2080.25
-2097.626
525.53
0.895 Meters
-4686.03
7.98 Meters
0.895 Meters
7.98 Meters
1.2
234
690
700
710
720
730
740
750
760
770
780
790
800
810
805
815
820
830
835
840
836
838
840
845
850
870
865
880
890
900
910
920
930
940
950
960
970
X = X3
GOSUB 860
AS5 = 10000*U1
AS6 =1000*U2
B4 = W
PRINT” Area of steel bars at a distance X3=”;AS5;”Cm sq”
PRINT” Area at a distance X3=”;AS6;”Cm Sq”
PRINT” Width at a distance X3=”;B4;”Meters”
X = X4
GOSUB 860
AS7 = 10000*U1
AS8 = 10000*U2
B5 = W
PRINT “Area of steel bars at a distance X4=”;AS7;”Cm Sq”
PRINT” Area of steel bars at a distance X4=”;AS8;”Cm Sq”
PRINT “Width at a distance X4=”;B5;”Meters”
STOP
REM This is sub routine
PRINT” Below is a moment equation between column one and
two”
QS1 = -QS/6
QB1 = QB/2
REM Let M1 moment equation between column one and two and
MU as absolute moment
PRINT
TAB(1);”M1=’;TAB(5);QS1;”Xcube”;TAB(100);QB1;”Xsquare”;TAB(1
8);P1;”X-W1/2)”
*** Note resulting equation is M1 = -17.9366X3 + 618.5X2 – 3067(X
– 0.23)
M1 = QS1*X^3 + QB1*X^2 – P1*(X-W1/2)
MU = ABS(M10
W = B-(B-A)/LC1*X
K = FY/(1.7*FC)
F = -D
H = MU/(0.90*98.07*FY)
R = (F^2-4*K*H)^.5
U1 = (-F+R)/(2*K)
U2 = (-F-R)/(2*K)
RETURN
STOP
REM Design of steel in short direction
WL = W1 + 0.75*D
2.4
17.916
618.5
916.1
4.05
11.7619
0.895
0.00246011
0.8278
0.0732
0..002856
1.13
235
980
990
1000
1010
1020
1030
WS = B-WL*(B-A)/LC1
BS = (B+WS)/2
LS = (BS-W1)/2
X = LS
GOSUB 1310
ASS1 = 10000*U1
1040
ASS2 = 10000*U2
1050
1060
1070
1080
1090
1100
1110
1120
1130
1140
PRINT “Area of steel in square cm=”;ASS1
PRINT” Area of steel in square cm=”;ASS2;”Cm2”
REM Design of steel in short direction at column two
WL2 = W2 + 1.5*D
LS2 = (A+WL2*(B-A))/LC1
BS2 = (A+LS2)/2
LS3 = (BS2 – W2)/2
X = LS3
GOSUB 1310
AL1 = 10000*U1
1150
AL2 = 10000*U2
1160
1170
***
965
PRINT” Area of steel in Sq Cm=”AL1;”Per meter width”
PRINT “Area of steel in Sq Cm=”;AL2;”Per meter width”
Insert program line no 965
INPUT”W1,W2,D”:LC1,W1,W2,D
975
INPUT”QS,FC,FY,QU”:QS,FC,FY,QU
1180
1190
1200
1210
1220
1230
1240
1250
1260
1270
REM Determine point of inflectioni.e. bending of bars
A1 = -QS/6
GOSUB 1410
LX1 = X
LX2 = XA
LX3 = XB
PRINT” First root is=”;LX1
PRINT” Second root is =”;LX2
PRINT” Third root is =”;LX3
PRINT” Small real positive root represents point of zero moment
(I’e point of inflection”
3.856
4.067
1.8035
5.13522
CM2/Mtr
755.79
CM2/Mtr
5.13522
1.8025
1.019896
1.539798
0.539899
0.539899
1.26807
Cm2/Mtr
759.66
Cm2/Mtr
W1 =.46,W2 =
.46,LC1 = 5.94,D
= 0.895
QS =107.5,FC
=211,
FY=4219,QU =
289.5
-17.916
236
1275
1290
1300
1310
1320
1330
1340
1350
1360
PRINT” Programmed by Bienvenido C. David Civil/Structural
Engineer on the 11th of November year 1984 in his hometown
Baguio City”
PRINT” Designed By Bienvenido C. David A Civil/Structural
Engineer”
END
REM This is a sub routine no two
MU = QU/2*X^2
K = FY/(1.7*FC)
F = -D
H = MU/(.90*98.07*Fy)
R = (F^2-4*K*H)^.5
U1 = (-F+R)/(2*K)
1370
1380
1390
1400
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500
1510
1520
1530
1540
1550
1560
1570
1580
1590
1600
1610
1520
1630
1640
U2 = (-F-R)/(2*K)
RETURN
STOP
REM This is sub routine no three General cubic equation program
B1 = QB/2
C1 = -P1
D1 = P1*W1/2
B2 = B1/A1
C2 = C1/A1
D2 = D1/A1
P3 = C2-B2^2/3
Q3 = D2-B2*C2/3+2*B2^3/27
R3 = P3^3/27+Q3^2/4
IF R3<0 THEN 1580
Z3 = - Q3/2+R3^.5
IF Z3<0 THEN 1550
ZA = Z3^0.3333
GOTO 1660
ZB = ABS(Z3)^0.33333
ZA = -ZB
GOTO 1660
O = ATN(ABS(R3)^.5/(-Q3/2))/3
PI = 2.094395102
ZC = ((-Q3/2)^2-R3)^.5
ZA = ZC^0.3333
IF (-Q3/2)>0 THEN 1650
ZA = -ZA
XA = COS(O+2*PI)*(ZA-P3/(3*ZA))-B2/3
1280
169.932 KN Mtr
11.7619
0.895
0.0004563
0.88292
0.000513522
CM2/Mtr
0.075579
618.515
3067
705.41
-34.5217
171.181
-39.3717
-226.029
-1116.6
-115,990
YES
YES
8.23422
-823422
0.203339
653.979
8.6800
7.602
237
1650
1660
1670
1680
X = COS(O+PI)*(ZA-P3/(3*ZA))-B2/3
XB = COS(O)* (ZA-P3/(3*ZA))-B2/3
RETURN
END
DETERMINATION OF STEEL REINFORCEMENT AREAS AT DIFFERENT LOCATIONS
Same program listing on page 225
Let us begin with line no 140
140 DATA Note example only 0.6, 1.2, 1.8, 2.4, 2.8, 3.0, 3.6, 4.84, 5.94
150 M (Q) = A*X (Q) ^3 + B*X (Q) ^2 + C*X (Q) + D
160 V (Q) = E*X (Q) ^2 + F*X (Q) + G
170 W (Q) = H – (H – J)/LC*X (Q)
180 D1 (Q) = V (Q)/ (W (Q)*VC
190 M1 (Q) = ABS (M (Q))
195 MU (Q) = M1 (Q)/W (Q)
200 R = FY/ (1.7*FC)
205 P = -88.263*FY*R
210 N = 88.263*FY
215 Z (Q) = MU (Q)/P
220 K (Q) = N*D1 (Q)/P
225 I (Q) = Z (Q) + 0.25*K (Q) ^2
230 AS1 (Q) = Z (Q) ^.5 - .5*K (Q)
240 AS2 (Q) = Z (Q) ^.5 + .5*K (Q)
250 PRINT AS2 (Q)
260 NEXT Q
238
270 PRINT M(1);M(2);M(3);M(4);M(5);M(6);M(7);M(8);M(9);M(10)
275 PRINT V(1);V(2);V(3);V(4);V(5);V(6);V(7);V(8);V(9);V(10)
280 PRINT W(1);W(2);W(3);W(4);W(5);W(6);W(7);W(8);W(9);W(10)
The following program is how the program no 2 “Code name Newton can be used to locate
the point of inflection (bending of steel reinforcements “We start at line no: 285)”
The program is a cross check for program no 1 “General cubic equation”
PRINT “Locate point of inflection. Let us use “Newton’s method of approximation “
290 PRINT” This is Newton’s method of approximation to solve for the real positive root”
295 PRINT” Try a value of X as first trial root X = LC/2 say 2.97 or 3
300 PRINT “If all coefficients are known then run line no 310”
305 STOP
310 CALL CLEAR
360 INPUT”Coeffecients of A, B, C, D”: A, B, C, D
705.41
370 INPUT “How many trials T: T
380 INPUT ‘Value of X as first trial”: X
3:00
390 P = 1
400 PRINT “I am now performing trial no=”; P
410 PRINT “Assume trial root is=”; X
420 REM Format is AX3 + BX2 + CX + D
430 Y = A*X^3 + B*X^2 + C*X + D
440 IF Y = 0 THEN 450 ELSE 470
A = -17.936, B = 618.5, C = -3067, D =
239
450 PRINT” Real root is=”; X
460 STOP
470 T = T-1
480 IF T= 0 THEN 490 ELSE 500
490 STOP
500 R = X
510 GOSUB 580
520 X = S
530 P = P+1
540 PRINT”I am now performing trial no=”; P
550 PRINT “Trial root is=”; X
560 GOTO 430
570 STOP
580 REM This is a sub routine
590 M = A*R^3 + B*R^2 + C*R + D
600 N = 3*A*R^2 + 2*B*R + C
610 S = R-(M/N)
620 RETURN
630 END
240
This chapter is about Designing a Cantilever
Retaining Wall. The author discusses in
details the forces acting on retaining walls. The different criteria in retaining wall design.
An illustrative example explains in detail step by step the procedure in designing a Cantilever
retaining Wall. The author explains how the two MATHEMATICAL PROGRAMS (Struct
math Solver 1 and Structmath Solver 2) were used by the author as a sub
routine program within a main program.
Retaining wall is a wall whose purpose is to resist the trust of a bank of earth or other
materials. Sometimes this is unanimous to a concrete dam. There are three types of
Retaining walls. The gravity, the Cantilever and Counter forth retaining wall. In a
241
similar manner if the wall is used to confined a water as in spillway in dam construction
it may be called intermediate Training Wall (ITW) or Right Training Wall (RTW).
a) Gravity retaining walls depends mostly upon their own weight for stability usually low in
height. They are expensive because of their inefficient use of materials.
b) Cantilever retaining wall is a reinforced concrete wall that utilizes the weight of the
soil itself to provide the desired weight. Stem, toe and heel are each designed as
cantilever slab.
c) Counter forth retaining wall is similar to a cantilever retaining wall, except that it
is used where the cantilever is too long or for very high pressures behind the wall and has
counterforths which tie the wall together.
In this chapter I have included a classical solution (ANALYTICAL METHOD) as well to
illustrate clearly how the classical solution is cast into the digital solution. Program no. 21 is
all about “Design of Cantilever Retaining Wall”
242
DESIGN, ANALYSIS AND FORMULA DERIVATIONS FOR COMPUTER
APPLICATIONS
Design and analysis of cantilever retaining wall is quite complicated because it combines the
science of soil mechanics, hydraulics and structural mechanics. In ancient times, this is usually
done by trial and error method. However in the beginning of the 20 th century with the
acceptance of Foundation Engineering, scientific discipline plays an important role in the
analysis of retaining wall and lately with the emergence of the micro computers and the
matrix method of structural analysis accurate design of retaining walls have improved to
a mark degree.
Below is a typical section of a Cantilever retaining wall and the forces acting on it.
243
T
Fr = Rtan + c’B + Pp
Active Soil Pressure Pah
h
F = Fr (1.5)
Tb
Slab thickness
B
Pp =1/2(y) (Hp)2(Kp)
There are three criteria for retaining wall design
1) Sliding stability
2) Overturning stability
3) Uplift stability
By sliding stability, we mean the entire structure must be in equilibrium (i.e. the
summation of forces horizontal is equals to zero. Then by definition
F sliding =
The safety factor against sliding should be at least 1.5 for cohessionless backfill and
about 2 for cohesive backfill.
244
Overturning stability means that the structure must be safe against overturning with respect
to the toe. Then by definition Foverturning
=
The
usual safety of factor against overturning with respect to toe is 1.5 for cohessionless
backfill and 2 suggested for cohesive soil.
Uplift stability means, there should be no negative pressures at either toe or heel usually
known as the principle of the middle third which will be defined later.
The pressure acting on the stem of a retaining wall is analogous to a concrete dam except that
a factor Ka is introduced. Ka is usually known as Rankin’s coefficient of active soil pressure
which will be discussed later. Let us analyze the figure below.
Stem portion
M=
(W1)(h)3
y
h
dy
q
SECTION
PRESSURE DIAGRAM
MOMENT DIAGRAM
Figure 20.2
Let y = Distance from top of wall
Let w1 = Weight of active soil in pounds per cubic Foot
Let LF = Load factor used (This is similar to 1.4 and 1.7 used in beams design)
Let H = Height of retaining wall
245
Then from the pressure diagram it is evident that the pressure at any instant is
p = Ka(LF)(w1)
in Kips . The total pressure then at any point from top of wall is
equals to the sum ∑ of all infinisitimal elements summation areas pdy. The total pressure
then would
dy =
P = Ka(LF)(w1)
(Ka)(w1)y2 = =
(Ka)(w1)h2 Equation (1)
In a similar manner, the moment acting at any point is equals to M = P(Y)(dy) substituting the
values of P from Equation (1) we get DM =
then by integration M =
Y = h so that M = =
(Ka)(w1)y2dy the total moment
(Ka)(w1)
2
dy =
Ka (w1)(y)3
but
Ka (w1)(h)3 In inch Kips
From figure 20.2 the driving forces causing horizontal sliding will be the horizontal component
Pah while the resisting horizontal force
will be the frictional force FR which is equals to FR =
R tan
+ c’B + Pp here is equivalent to the base soil factor in degrees. B is the overall
base length and Pp is the passive soil pressure preventing sliding of structure here Kp is known
as the Rankin coefficient of passive pressure.
The overturning moment can be found by taking moments about the heel. It is best to divide
the section into individual elements for easy computation shown in figure 20.1. The forces
causing overturning moment will be the horizontal component of P = Pah acting at a lever arm
equals to 1/3(h) from top of wall.
Now we defined Ka known as the Rankin’s coefficient of active earth pressure which is equals
to Ka =
It may also be expressed by Trigonometric transformation K a = tan2 (45 -
Ka can also be computed in terms of B and
with respect to the top of wall and
here B is the angle of inclination, soil surface
246
Ka
(COSB)(COSB - COS2B - COS2
COSB + COS2B - COS2
with Ka known then Pah is equals to
(Ka)(w1)(LF)(h)3 in Kips Equation (4)
Equation 4 is known as the active soil pressure, it is called active because it causes the
structure to slide and overturn. In a similar manner, the coefficients Kp known as the passive
Kp =
(COSB) (COSB + COS2B - COS2
with Kp evaluated
COSB - COS2B - COS2
Pp is equals to Pp = (Kp)(w)(h2)
here h is the reckoned depth of soil from top cover of
concrete. The thickness of stem is usually governed by shear at the junction.
The thickness of stem is usually obtained by shear. The shear force of concrete is equals to
Vc = vc(b)(d) equating this to the shear at junction. A linear equation in t is solved. A covering
of 2” is added as covering of steel..
By uplift stability, this means there is no negative pressure at toe or heel. The principle of the
middle third will be applied . For a rectangular pressure diagram as shown below in Figure A
20.2 the resultant is located at the center which is equals to B/2. . For a triangular pressure
diagram as shown in figure B 20.2, the resultant is located at a distance B/3 from hill. It is
evident that for rectangular pressure distribution the pressure at any point is equal and
uniform thru out while in the second case a maximum pressure occurs at the heel and a zero
pressure at the toe.
The third case, if the resultant lies between B/2 and B/3 from heel then a trapezoidal
pressure diagram shown in figure C 20.2. However if the resultant is located at a distance less
than B/3 from heel or less than B/3 from toe, then the resulting pressure diagram would be
shown in Figure D 20.2.
247
To satisfy uplift stability, it is therefore necessary that the resultant of all forces must be
located between the distance B/3 to B/2. If we let the symbol
the heel edge, then taking moments about the heel edge
R( ) = Summation Mr - (Pah)(h/3) from which
as the location of R from
is readily solved.
here R = Weight of the section plus the vertical component of Pa Introducing the letter e as
eccentricity reckoned from resultant pressure diagram then is equals to
e=
-
then
for a value of e for positive pressure at either toe or heel
e must be within the range =
-
or equals to L/6
In short the eccentricity must not be less than
FIGURE 20.2
FIGURE A
R
B/2
=
FIGURE B
FIGURE
R
R
B/3
B
C
FIGURE D
R
B/6
B
B/6
B
B
with e solved the pressure acting at the heel and toe can be solved by combined axial and
flexure formula from which Qheel =
(1 +
) and Toe = (1 -
)
here V is the
summation of vertical loads and B is the overall length of base. Like the footing the bearing
capacity can be obtained from soil test, building codes. It can also be computed by the
empirical formula
qultimate = CNcdcIc + qNqdqiq +
Bousinique equation.
BNydyIy this formula is known as
248
Once the stability of section is confirmed the next move is to determine the required
area of steel reinforcements in base and at the stem. The analysis is similar to that in
footing, however in the stem portion since width varies linearly with h it is necessary to
determine the required area of steel reinforcements at different locations, often the minimum
steel reinforcement ratio from code governs. Sometimes it is customary to find where the
cutting point of steel reinforcements for a given moment, this is usually the case where a
supply of steel bars is limited. Shown below is a graphical representation both for Pn(limit)
and given area of steel bars.
INTERACTION DIAGRAM CUTTING POINTS OF STEEL REINFORCEMENTS
SECTION
CUTTING POINTS Y
Y1
MINIMUM VALUE OF Y
PROG LINES 2240 - 2820
A1Y3 + B1Y2 + C1Y + D = 0
M = A1X3 + C1X + D1X
Y2
Y
PROG LINES 1610 - 2830
Y3
Mmax
B
equation of straight line
Mmax
M = AS (.90)(FY)(d
- )
given As three values
From the figure the point of intersection Y is the distance by which minimum steel
reinforcement ratio from code governs. We note that the “General Cubic Equation” is used
two times. The above discussion can be visualized by a detailed classical solution (Analytical
solution).
249
In our general computer program, if either of the three criteria is not satisfied the computer
displays into the monitor screen the word “Revise dimensions or material strength
specifications failure of retaining wall by sliding or overturning, uplift stability eminent”.
Computer stops running and the designer may revise his dimensions.
We note that in locating the cutting point of steel reinforcement bars, the third degree
equation is encountered twice, so the general cubic equation program no 1 (Struct
math solver 1) is included as a sub routine program within the main program
“Ret Wall program no. 21”. Instead using a one dimension array author prefers
the combination of RESTORE and GOSUB statement. (See program listings).
ANALYTICAL SOLUTION
Design a Cantilever Retaining Wall for the condition shown. Use Rankin Ka even though
wall is high.
B = 10
Backfill soil
X
= 34
y = 115 P.C.F no water
1Ft
height of wall = 26 Ft
h’
Fc’ = 3 K.S.I Fy = 60 K.S.I.
5 Ft.
Batter on front face of wall = 1:48
3 Ft.
9.5 Ft.
top thickness = 16” Load factor is taken as 1.8 Weight of concrete is 150 Pounds per Cubic
Foot. Estimate 3.5” from CGS to soil interface to allow approximately 3” covering of clear cover.
Base soil
= 32
and c = 0.40
250
1) Step one Establish stem dimensions
Compute Rankin’s active earth coefficients Ka = (CosB) [CosB – Cos2B – Cos2 ]/ [CosB + Cos2B –
Cos2 ] = COS 10 (COS 10
(COS 10 )2 – (COS 34 )2)/ (COS 10 + (COS 10 )2 – (COS 34 )2)
= .294 Find pressure acting on wall (the horizontal component of pressure. Let that Pressure be
Pa = ((y)(h)2(y)
here y is the unit backfill weight of soil in pounds per cubic foot or .115
Kips. per Cubic Foot. Substituting values we get Pa =
(0.115) (26)2 (0.294) = 11.43
Kips/Foot *** A strip of 1 foot is considered.
Pa- horizontal = Pa (COS B) = 11.43(COS 10 = 11.25 Kips per Foot from code wide beam shear is
given as vc= 2 fc’ = 2(0.85) (3000) = 0.09311 in K.S.I. WITH A LOAD FACTOR OF 1.8
Pa- horizontal = 1.8(11.25) The shear carried by concrete at junction is Vc = (vc) (t)
(12) equating Vc to Pa- horizontal we get Pah = Vc 0.09311(12) (t) = 1.8(11.25)
solving for t
=
= 18.14” Allow covering of 3.5” then T at junction is 18.14” + 3.5”
= 21.6” with a slope batter of 1/4” per foot then thickness at top of wall is t = t junction – h (0.25)
= 21. 6 – 26(0.25)
Ttop = 15.1 Inches use t = 16” to maintain even dimensions let us use t =
16 inches + 26((0.25) = 22.5” use 23”
Step 2 Compute overturning and sliding stability of wall
From figure it is evident that H’ = H + 2.42 + 9.5(Tan 10 ) = 30.1 Feet
Pah’ = 1/2(y)(H’)2(Ka) = 0.50(30.1)2(0.115)(0.294) – 15.1
From these data’s and wall dimensions we can set up the following in table form
PART
Weight, Kips
1
2
3
4
Pav’ =
0.5(26+27.65) x 0.115(9.5) = 29.3
1.33 x 26 x 0.15 x 1 =5.2
0.5 x 0.59 x 26 x 0.15 = 1.2
2.42 x 14.42 x 0.15 x 1 = 5.2
Pah’(sin10 ) = 15.1 sin 10 = 2.6 Kips
/Foot
Arm, Foot
9.67
4.28
3.39
7.21
14.42
Mr’ ft kips
283.3
22.1
4.0
37.5
37.5
251
Total Sum
Sum weight = 43.15
Sum
Overturning
M=
384.4 Foot
Kips
Taking moments at toe the overturning moment is MO
mo = P’AH
= 151
Ft Kips/foot here
The safety factor is sum
=
= 1/3 x h =
= 151 Ft kips/foot
= 2.54 greater than 1.5 Okay. Here 1.5 is the
recommended factor of safety against overturning which is equals to 1.5
Determine sliding stability of wall. The factor of safety against sliding will be based on using
3 feet of depth of soil at the toe.
Kp = tan2(45 +
) = 3.255
The friction coefficient factor Fr = R tan ’ + cB’ + Pp here
the limiting value of c’ is 0.50c to 0.75c here c = 0.40 given see datas problem with R = 43.5
substituting values we get Fr = 43.5 tan 32 + 0.67(0.4)(14.42) = 31 Kips From Bowles
2
reference text book page 438 “Foundation Analysis and Design ” Pp = 1/2YH Kp + 2cHKp
see drawing here H is the depth of soil from surface of toe up to soil surface with Kp = 3.255 &
1.804 respectively substituting values in the above equation we get
Pp = 0.50(0.112)(3)2(3.25) + 2(0.4)(3)(1.804) = 6 kips
Summation SFr = Pp + Fr = 31 + 6 = 37 Kips The resulting F =
= 37/15.1 = 2.45
greater than 1.5 (i.e recommended o kay)
Now locate resultant = Summation weight on base and the eccentricity. Taking moments at
toe we have
e = B/2 -
=
=
= 5.37 ‘ Compute eccentricity e
14.42/2 – 5.37 = 1.84 Feet less than L/6 recommended for eccentricity okay.
252
Step 3 Compute bearing capacity of soil. For datas see page 134 reference textbook
“Foundation Analysis and Design by Joseph Bowles”. The actual soil bearing pressure using
bousiniques equation is qultimate = cNcdcic + qNqdqiq +
YBNydyiy B’ = 14.42 – 2(1.84) =
10.7 From tables Nc =35.5, Nq = 23.2 , Ny = 20.8, Ic = 0.42 , Iq = 0.44, Iy = 0.309. dc= 1.19, dq
=1.13 and dy= 1.00 substituting values in the above equation we have
qult = 0.4(35.5)(0.42)1.19 + 5(0.112)(23.2)(1.13)(0.44) + 0.5(0.112)(1.0)(10.7)(20.8)(0.309) =
7.1 - 6.5 - + 3.9 = 17.5 qa = 17.5/3 = 5.8 Ksf
Compute actual soil pressure. Actual soil pressure is given as q =
=
(1
= 3.02
(1
(1
)
) = 5.3 ksf maximum at toe q = 0.7 ksf
max at heel.
Step 4 Compute base slab shear and bending moments toe and heel.
For toe at stem face x = 3 feet. slope pressure is q = 5.3 – 0.36 – 0..32X. Neglecting
soil over toe and integrating the pressure diagram we get the shear at any section
V = 4.94X – 0.32X2/2 = 13.4 Kips Integrating the shear diagram we have the moment at any
section M = 4.94
dx - .32/2
2
dx = 4.94/2X2 – 0.32/6X3 = 20.8 Foot
kips
For heel at approximately CG of tension steel X = 9.5 +
= 9.79 Feet for moment , use 9.5
feet for shear.
Use average height of soil on heel for downward pressure includes Pav = 2.6 kips, the
pressure q is = 3.45 - .70 -.32X. The shear at any section X is V = 2.75X –
PavX
Mheel = 107.2 Foot kips
X2 +
253
Step 5. Check base slab shear using largest V, with LF = 1.8 and d = 2.417 – 0.29 = 2.13 feet
Actual shear stress V =
=
= 0.84 less than 0.093 okay *** Note
we could reduce the base slab shear stress by about 1 to 1.5”
Step 6 Compute steel reinforcement ratio, using authors derived
formula ***see chapter three
P=
(1 ±
1 – 2.622Mu/(bd2Fc’)
)
with Mu = 107.2(12)(1.8) = 2315.52 Inch Kips and b = 12” ( 1 foot strip length) and
d = 25.5 inches evaluating the terms separately 2.622 x 2315.52/(12(25.5)2(3) ) =
0.2582691 and
0.847fc’/fy =
= 0.042235 Substituting values and taking the smaller
negative sign we get p = 0.042235 x (1 – 0.861237) = 0.042235(0.138763) =
0.0058606
but from code minimum is p = 200’/fy = 200/60000 = 0.0033 hence actual steel
reinforcement ratio governs use p = 0.00586 Then area of steel Asheel = p(b)(d) =
0.00586(12)(25.5) = 1.79 in2/ft
For steel reinforcement in toe, let us use authors derived formula ***
Refer to chapter 4
As =
(b)(d) ±
(0.7225Fc’2b2d2)/Fy2 – 1.888Fc’(b)(Mu)/Fy2))
with Mu = 20.8 Using a load factor of 1.8 Mutoe = 20.8(12)(1.8) = 449.28 Inch kips with d =
25.5” and b = 12” substituting values and evaluating the mathematical expressions one by
one. .7225fc’b2d2/fy2 = 0.7225(3)2(12)2(25.5)2/(60)2 = 169.1300
254
0.85fc’bd/fy =
= 13.005
1.88888fc’bMu/fy2 =
=
8.48636
Substituting numerical values in the above equation we get
13.05
13.05
= 13.05
12.674 taking the least value we further get As = 13.005 - 12.674 = o.331 in2/ft.
Check actual steel reinforcement from code allowable
p = As/bd =
= 0.0018 less than P minimum from code Therefore p
minimum from code governs = 0.0033
Required area of steel at toe = Astoe = minimum (b)(d) = 0.0033(12)(25.5) = 1.02 in2/feet
***Note author by computer use the mathematical expression below *** See
chapter 4
*** Note given d, b and Mu depth of stress rectangular block is
a=d ±
As =
d2
–
Mu/( fy(d – a/2))
For Computer application
to solve areas of steel reinforcements as sub routines
program within the main program both heel and stem portions (See program listings)
Below is the load, shear and moment diagram
255
3 Ft
9.5 Ft
q = 2.429.15) = 0.36 STEM Q = 2.42(1.5)+26.8(.115) = 3.45 KSF
pressure diagram heel and toe
5.3 k.s.f.
1
TOP OF WALL
Z
.32
TOP OF WALL
TOP OF WALL
16”
1.4 Ft kips
.2h
17.4”
22 Ft kips
.5h
19.5”
.8h
90 Ft kips
175.5 Ft kips
Pressure diagram
o.7 k.s.f.
Moment diagram
h
21.6”
h
23”
d = 19.5”
Section
STEM PORTION
From the pressure diagram profile the pressure at any point is q = ka(w)(z)/1000 in Kips. The
total pressure then is found by integrating the pressure diagram dv =qdz and V =
V = Ka(w)/1000 dz from which V = Ka(w)(z)2/2000 Likewise integrating the
shear diagram . The moment at any section is M = vdz = Ka(w)/2000 2dz
M=
(z)3 Applying a load factor of 1.8 and w = 115 Pounds per cubic foot
the moment then at any section is M = 0.010143Z3 from the figure the section at
any height h is d = 0.26923(h) + 12.5
256
Step 7 Compute steel reinforcements stem portions from the above figure at
section 0.5h, 0.8h and h using a load factor of 1.8
Using authors derived formula we get the following information’s in table form.
Point
M,ft. Kips
0.5h
0.8h
H
22.00
90
175.8
Wall thickness
inches
19.5
21.6
23.0
D inches
16
18.1
19.5
For longitudinal shrinkage and temperature steel Use
As Square
Inch
0.64
1.18
2.26
P(used)
0.003
0.005
0.010
(12)(0.002) =
0.43 in2/Ft. in top half
Use 2 no 5 bars at one on each face = Use (19.5 + 23)/2(12)(0.002)=
051in2
*** Note author by computer use the General Cubic equation
program to solve the cutting points of steel reinforcements stem
portions. (See program listings)
257
SECTION DETAILS
Programmer/Designer/ Structural Engineer : Bienvenido C. David Date: Jan
19, 1970 PRC NO: 10170
DESCRIPTION: Cantilever Retaining Wall Design By U.S.D. format
SUBJECT: Reinforced Concrete Design (USD ALTERNATIVE 1983 ACI
Code)
TITLE: Design Of Cantilever retaining Wall
258
CODE NAME: Ret - Wall
MACHINE LANGUAGE : IBM BASIC COMPUTER: IBM PERSONAL
Program steps: 310
LIBRARY MODULE: Floppy Disk PROGRAM NAME: RET - WALL
PTR NO: 40589345 at Baguio City 01/12/1987
DRAWING FIGURE
REFERENCE TEXTBOOKS:
Foundation Design By
Soil Mechanics In
Foundation Analysis &
Gregory Chebotariof
Engineering Practice By
Design By Joseph Bowles
Charles B. Peck
PROGRAM DISCLAIMER: Any use of the programs to solve problems other than those
displayed is the role responsibility of the user as to whether the output is correct or correctly
interpreted.
259
My first generation home computer
RET WALL.Is a program that sizes and designs the stem, heel and toe of a cantilever retaining
wall by using the Ultimate Strength Design theory. The program determines the stability of
the section i.e. sliding, overturning and uplift pressure at heel and toe. Determines the
required areas of steel reinforcements at base slab, heel and toe. Prints on the monitor
screen required equation locate cutting points of steel with given area of steel
reinforcements and vice versa. The program is written in advance basica language and can be
feed to a wide variety of programmable calculators, personal computers. Can be easily
incorporated to the E. REVIEW CENTRE OF UC – BCF.
BASIC COMPUTER SYMBOLS
+ ADDITION
^ RAISED TO THE POWER
- SUBTRACTION
SQR SQUARE ROOT OF THE NUMBER
MULTIPLICATION *
GOTO = JUMP LINE NUMBER
\ DIVISION
GO SUB = GO SUB ROUTINE
A.B.S ABSOLUTE
If true
VALUE
SGN = SIGNUM NOTATION
branch out
Branch out
Main program
IF THEN ELSE STATEMENT
If false
COMPUTER INSTRUCTION CODE
LINE
NO
10
15
STATEMENT
CALL CLEAR
PRINT” THIS IS COMPUTER PROGRAM NO 21
“Design of Cantilever Retaining Wall by U.S.D
method” in English units
SAMPLE ONLY FOR
DE-BUGGING
PURPOSES
260
20
25
30
35
40
45
50
55
60
70
80
85
PRINT” This program was developed by
Bienvenido C. David, a Civil/Structural Engineer
on May 1, 1984 in Baguio City”
PRINT “ For drawing and legend refer to
program record see attached sheets”
PRINT” All units i.e. material strength
specifications in Kips per square inch, Height of
retaining wall in feet, Linear dimensions of heel
and toe i.e length in feet, thickness of stem in
inches at top”
PRINT “Length of base from toe up to stem in
feet, length of base from heel up to back face
of stem in feet, weight of backfill soil in pounds
per cubic foot, weight of base soil in pounds per
cubic foot”
PRINT” Coefficients of base soil factor in
degrees passive soil pressure, slope batter front
face of wall in decimal i.e X:Y where X is run
and Y is rise”
PRINT” Angle of inclination of backfill soil from
top of wall in degrees, coefficient factor of base
soil in degrees, passive soil pressure in degrees”
PRINT “Weight of concrete is taken as 150
pounds per cubic foot, refer to tables by Joseph
E. Bowles for determining soil bearing pressure
at base i.e. Nc,Ny,Ic,Iq, Iy,Dc,Dq, & Dy as input
statement”
PRINT” For Shear and moment use 0.90 for
moment and 0.85 for shear”
PRINT” If all data’s are in their respective units
then run line no 80”
STOP
REM Step one Compute dimension and wall
stability
INPUT”H,W1,W2,B1,H1,B2,B3”
Example only H = 26
H,W1,W2,B1,H1,B2,B3
feet, W1 = 115
Lbs,W2 = 112 Lbs, B1
261
90
INPUT”FY,LF,SB,C,TS”:FY,LF,SB.C,TS
440
450
460
WD = 150/1000*TS*L4
XD = L4/2
MD = WD*XD
470
480
495
WT = WA + WB + WC + WD
MT = MA + MB + MC + MD
REM Compute overturning moment take
moment at toe
H4 = H3 + TS
P2H = 0.50*W1/1000*H4^2*COS(K3)*K4
P2V = 0.5*W1/1000*H4^2*KA*SIN(K3)
MAH = P2H*H4//3
FM = MT/MAH
IF FM<1.5 THEN 560
IF FM>=1.5 THEN 580
PRINT” Assumed dimension of slab not
okay failure will be initiated by
overturning of structure either revise
dimension L1, L2 or thickness of slab or
material strength specifications”
STOP
REM Determine sliding stability of wall
PRINT” Overturning stability of wall okay”
KB = COS(K3)*(COS(K3) + K1)/(COS(K3)K1)
KP1 = KB^.5
REM Take C as 0.625
K5 = 0.0174533*B3
500
510
515
520
530
540
550
560
570
580
585
590
600
610
615
= 10 ,H1 = 5 Ft, B2 =
34 and B3 = 32
FY = 60,LF =1.8,SB
=0.25, C= 0.40 & TS =
2.417
5.18537 Kips
7.1525
37. 088358 in foot
kips
40.46162 kips
342.517 foot kips
30. 092 Feet
15.075
2.661589
151.212
2.265
3.2946
1.8151076
Average value
.5585150
262
620
630
640
650
660
670
680
700
710
720
95
100
105
120
130
135
138
230
FR = WT*TAN(K5) + 0.625*C*L4
REM Let PP passive force supporting slab
PP = 0.5*W2*KB*(H1-TS)^2/1000 +
2*C*(H1-TS)*KP1
SFH = PP + FR
REM P2H = Force causing horizontal
motion
REM Let FU actual factor of safety for
sliding must not exceed 1.5 as per code
requirement.
FU = SFH/P2H
IF FU<1.5 THEN 720
IF FU>=1.5 THEN 740
PRINT “Failure of retaining wall by sliding
revise either dimension of slab L1,L2,TS or
material strength specification”
*** REM Have to start at this point line no
95
INPUT”Nc,Nq,Ny,Ic,Iq,Dc,Dq,Dy”:
Nc,Nq,Ny,Ic,Iq,Dc,Dq,Dy
28.8588
4.981696
33.85692
2.2448
720
Nc = 35.5, Nq =
23.2, Ny = 20.8, Ic
=0.42, Iq = 0.44,
Dc = 1.19, Dq =
1.13,Dy = 1.00, Iy
= 0.30796
9.5,3
INPUT”L1,L2”:l1,L2
K3 = B1*3.141516/180
K3 = 0.0174533*B1
K4 = 0.0174533*B2
.5585056.0593
IF (COS(K3)^2 – COS(K4)^2<0 THEN 137
ELSE 138
STOP
KA = COS(K3)*(COS(K3)–K1)/(COS(K3)+K1) 0.2943727
263
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
730
740
745
750
760
770
780
790
PA = 0.5*W1*H^2*KA/1000
PAH = PA*COS(K30
VC = 0.0537587*FC^.5
T1 = PAH*LF/(12*VC)+3.5
T2 = T1-H*SB
REM Compute overturning and sliding
stability of wall
HO = L1*TAN(K3)
H3 = HO+H
WA = 0.5*(H+H3)*L1/1000*W1
AT =L1*H+0.5*L1*HO
X1 = (H*L1*L1/2+0.5*L1*H0*2/3*L1)AT
XA =X1+L2+T1/12
MA = WA*XA
WB = T2/12*H*150/1000
XB = T2/24+L2+SB*H/12
MB =WB*XB
WC = 0.5*(T1-T2)*150*H/12000
XC = 0.6667*(T1-T2)/12+L2
MC = WC*XC
L4 = L1+L2+T1/12
REM Our next line number would be 730
STOP
PRINT” Sliding stability of retaining wall
section okay no revision necessary”
REM Checked uplift pressure at toe or
base
X2 = (MT-MAH)/WT
E1 = XD-X2
E2 = L4/6
IF E1>E2 THEN 800
IF E1<E2 THEN 820
11.42778
11.254169
0.09311
21.63
15.13
1.675
27.675106
29.31 Kips
254.956
4.7994
9.6019
281.4316
4.924685
4.17 Feet
20.477 Feet
1.05625 Inch Kips
3.361129
3.5501926
14.3025
4.728 Feet
2.4245 Feet
2.38375 Feet
264
800
810
820
830
835
840
860
870
880
885
890
900
910
915
920
925
PRINT” Assumed dimensions not okay for
eccentricity uplift pressure at base or heel
eminent either revise section or material
strength specifications”
STOP
PRINT “Section okay for eccentricity
resultant lies within the middle third no
revision necessary”
REM Compute bearing capacity of soil at
base
L5 = L4-2*E1
9.4535
REM Let Q1 Actual soil pressure at base
Q1 = H1*W2/1000
.56 Kips per
square foot
REM Let Qu Ultimate soil pressure and Qa
= Ultimate soil pressure for cohessionless
base soil
QU
16.94716
=C*NC*DC*IC+Q1*NQ*DQ*IQ+0.5*W2*L5
*NY*DY*IY/1000
REM Note values of the above variables
can be found from the textbook
“Foundation Analysis and Design by
Joseph H Bowles”
QA = QU/3
5.6490
REM Let QT and QH actual pressure at
heel and toe respectively
QT = WT/L4*(1+6*E1/L4)
5.70634
PRINT “Actual soil pressure at toe is
5.70634
=”;QT;”Kips Sq Ft”
QH = WT/L4*(1-6*E1/L4)
-0.0483613
PRINT” Actual soil pressure at heel is
-0.0483613
265
930
940
945
950
1020
1030
1040
1050
1070
1080
1090
1100
=”;QH;”KSF”
REM Compute base slab shear and
bending moments for toe up to stem face
determine shear and moment equation
by integration if time permits draw shear
and moment diagram by hand
PRINT” Sketch and detail retaining wall
section copy values of
QT,QH,L4,L1,W1,L2,P2V,H4,TS,LF,T1,T2,
FC and FY FOR INPUT STATEMENTS
STARTING PROGRAM LINE NO STARTING
1020 for design of steel reinforcements
slab portions and stem
PRINT” If finished then type continue to
resume running press enter to run”
REM FIRST COMPUTE THICKNESS d STEP
ONE
INPUT”QT,QH,TS,L4,L1”QT,QH,TS,L4,L1,L2
INPUT”FC,FY,H,H4,P2V”:FC,FY,H,H4,P2V
INPUT”W1,LF”:W1,LF
QS = (QT-QH)/L4
V =(QT-150/1000*TS)*L2*QS/2*L2^2
VT = ABS(V)
D = (TS-3.5/12)*12
VA = VT/(12*D)
Example only QT =
5.3,QH
=.7,L4=14.42,L1=9.
5,L2 = 3,P2V
=2.6,H =26,H4 =
30.07,FC=3 &
FY=60,LF=1.8,TS=2
.42,KA=0.294,T1=2
3”,T2=16”
115,1.8
0.3190013
21.26
21.26
25.54
0.04364
266
1110
1120
1130
1140
1150
1160
1170
1180
1190
1200
1205
1220
1225
1230
1240
1250
1260
1270
1275
1280
1290
1295
1297
1300
VC = 0.053758*FC^.5
IF VA>VC THEN 1140
IF VA<=VC THEN 1160
PRINT “Actual shear stress exceeds
allowable from code revise D”
STOP
M1 = (QT-150/1000*TS)*L2^2/2QS/6*L2^3
MT = ABS(M1)
M = MT
GOSUB 1310
AT= AST
PRINT” Area of steel at toe=”;AT;”Square
Inches Per Foot of width”
X= L1+3.5/12
QR = 150/1000*TS+(H+H4)/2*W1/1000
M2 = (QR-QH)/2*X^2-1/6*QS*X^3+P2V*X
MH = ABS(M2)
M-MH
GOSUB 1310
AH =AST
PRINT”Recquired area of steel
reinforcements at heel=”;AH;”Square Inch
per foot”
STOP
REM A sub routine required area of steel
at slab toe and hill
REM This is a sub routine determination
of steel reinforcements at slabs toe and
heel
REM This is sub routine one
IF (D^2-2.61**M*LF*12/(FC*12))<0 THEN
0.0931088
20.781
20.781
1.02 Square Inch
9.79 Feet
3.5874
113
25.454
267
1305
1310
1315
1320
1330
1340
1350
1360
1370
1380
1390
1400
1410
1420
1430
1305 ELSE 1315
PRINT “Depth of stress rectangular block a
is imaginary not possible review given
data”
STOP
A = D – SQR(D^2-2.61*M*LF*12/(12*FC))
AS = M*LF*12/(0.90*FY*(D-A/2))
P =AS/(12*D)
PN = 0.2/FY
IF P>PN THEN 1370
IF P>=PN THEN 1390
AS1=AS
GOTO 1400
AS1=12*PN*D
AST=AS1
RETURN
`
STOP
**** Note For stem portion there are
three alternatives. First determine the
cutting point by which minimum steel
reinforcement ratio from code governs
then as a guide we can find the area of
steel reinforcements for a value of Y
distance from top of wall. Third for a
given area of steel reinforcements
determines the point of cutting at what
distance from top of wall should a portion
of steel reinforcements be cut. A general
cubic equation results, a third degree
equation.
PRINT “This is a continuation of cantilever
retaining wall design finding the cutting
268
1435
1440
1450
1460
1470
1480
1475
1490
1500
1510
1520
1530
1540
1550
1560
1570
1575
1580
1590
points by which minimum steel
reinforcements from code governs”
PRINT” With dimensions already
computed input this numerical values
following program lines”
INPUT”TB,TT,H”:TB,TT,H
INPUT”FC:FY”:FC,FY
INPUT”KA,W1,LF,”:KA,W1,LF
DT =TT-3.5
DC =DB-DT
DB =TB-3.5
PN=0.2/FY
U =PN*FY*/(1.78*FC)
V =DC/H
J = 10.8*FY*PN
MU=1/500*KA*W1*LF
A1 =MU/J
B1 = -(V^2-U*V^2)
C1 = 2*U*DT**V
D1 = -(DT^2+2*V*DT)
REM A1,B1,C1 and D1 are coefficients of
the General Cubic Equation AY13 + B1Y2 +
C1Y + D1 = 0 To solve for Y we restore
general cubic equation program no 1
PRINT” TAB(1);A1;”Y CUBE”;TAB(10);B1;”
Y Square”;TAB(15);C1;”Y”;TAB(22);”=0”
***Note computer prints on the monitor
screen the third degree equation AY13 +
B1Y2 + C1Y + D1 = 0
PRINT “Copy coefficients of Y cube, Y
square and y and constant D1 for data
statement then type continue to resume
23”,16”,26’
3,60 IN k.s.i
0.294, 115,1.8
12.5”
7”
19.5”
0.0033333
0.039215
0.2692
2.159784
0.121716 Inch Kips
-0.069626
0.2639169
-162.98
269
1600
1610
1620
1630
1640
1645
1650
1655
running”
BREAK
DATA Put value of A1,B1,C1 & D1
GOSUB 2400
PRINT “Location of minimum steel
reinforcements is valid at a distance
Y=”;”Feet from top of wall”1640
REM To find required area of steel
reinforcements for a given value of y
INPUT “Thickness of top T1,T2,:T1,T2
REM Let the factor FR1,FR2,,FR3,,FR4 as
multiplication factor of H
INPUT”VALUE OF
FR1,FR2,FR3,FR4”:FR1,FR2,FR3,FR4
1660 Z = FR1*H
1670 GOSUB 2860
1675 PRINT”Recquired area of steel at first
FR=”;AS1;”Square Inches per foot of
width”
1680 Z =FR2*H
1685 GOSUB 2860
1690 PRINT”Recquired area of steel at second
FR2=”;AS1;”Square Inch per foot of
width”
1700 Z=FR3*H
1705 GOSUB 2860
1710 PRINT”Recquired area of steel a5t a value
of FR4=”;AS1;”Square inch per foot of
wall”
1720 Z=FR4*H
1730 GOSUB 2860
23”,16”
Example only FR1
=0, FR2 = 0.5, FR3
=0.7,FR4 =.9
270
1740 PRINT”Recquired area of steel at a value
of FR4=”ASI;”Square inch per foot”
1750 REM This is third alternative to find the
value of Y(cutting point at a distance from
the top)at a given AS a cubic equation
results
1755 REM Let AS3,AS4 & AS5 as given area of
steel reinforcements
1760 REM Let us DIM AS4 in three elements
1770 PRINT “Put numerical values of AS as data
statement”
1780 DIM AS4(3)
1790 FOR JX=1 TO 3
1800 READ AS4(JX)
1810 DATA
Put numerical
values in
sequential order
1820 VX =1/500*KA*W1*LF
0.121716
1830 LX(JX) = FY*AS4(JX)/(20.4*FC)
1.1568
1890 UX(JX)=0.90*FY**AS4(JX)
63.72
2000 DT =TT-3.5
12.5 Inches
2010 DB = TB-3.5
19.5 Inches
2020 REM Let DA= Depth of stem at any
section from top
2030 DA = DB-DT
7 Inches
2040 A1X=VX
2050 B1X=0
2060 C1X(JX)=-DA*UX(JX)/H
2070 D1X(JX)= UX(JX)*LX(JX)-UX(JX)*DT
722..78888
2080 PRINT D1X(JX)
2090 NEXT JX
2100 PRINT”A1X=”;VX
271
2110
2120
2130
2140
2150
2160
2170
2180
2190
2200
2210
2220
PRINT”A2X=VX
PRINT”A3X=”;VX
PRINT”B1X=”;B1X
PRINT”B2X=”;B1X
PRINT”B3X=”;B1X
PRINT”C1X=”;C1X(1)
PRINT”C2X=”;C1X(2)
PRINT C3X=”;C3X(3)
PRINT”D1X=”;D1X(1)
PRINT ‘D2X=”;D1X(2)
PRINT “D3X=”;D1X(3)
PRINT” Above data’s are coefficients of
cubic equation. Copy coefficients for sub
routine no one then continue to resume
running”
2230 BREAK
2240 REM Format is A1X3 + B1X2 + C1X + D1 – 0
2250 DATA
2260 DATA
2270 DATA
2280 RESTORE 2250
2290 GOSUB 2400
2300 PRINT “Location of y for first area of steel
Note put
Coefficients of
AX1,B1X,C1X &
D1X
Note put
coefficients
A2X,B2X,C2X &
D2X
Note put
coefficients
A3X,B3X,C3X &
D3X
272
2310
2320
2330
2340
2350
2360
2370
2380
2390
2400
2410
2420
2440
2450
2460
2470
2480
2490
2500
2510
2520
2530
is located at a distance =”Y;”Feet from top
of wall”
RESTORE 2260
GOSUB 2400
PRINT “Location Y for second steel area is
located at a distance Y=”;”Feet from top”
RESTORE 2270
GOSUB 2400
PRINT “Location Y for third area of steel is
located at a distance Y=;Y;”Feet from top
of wall”
PRINT Programmed by Bienvenido C.
David a Civil/Structural Engineer on May3,
1984 in Baguio City”
PRINT”Designed by Bienvenido C. David
STOP
REM A sub routine
REM This is the general cubic equation
program no 1 finding the roots of a third
degree equation format is A1X3 + B1X2 +
C1X + D1 – 0
READ A1,B1,C1,D1
P = C- B^2/3
Q = D - B*C/3+2*B^3/27
R = P^3/27+Q^2/4
IF R<0 THEN 2530
Z = -Q/2+R^0.5
IF Z<0 THEN 2520
ZA = Z^.33333
GOTO 2630
ZB = ABS(Z)^.33333
ZA = -ZB
273
2540
2550
2560
2570
2580
2590
2600
2610
2620
2630
2710
2720
2730
2740
2743
2745
2748
2755
2760
2770
2773
2775
2778
2785
2790
2800
2810
2820
2830
GOTO 2630
O =ATN(ABS(R)^.5/(-Q/2))/3
PI = 2.094395102
ZC = ((-Q/2)^2-R)^.5
ZA = (ZC)^.33333
IF (-Q/2)>0 THEN 2620
ZA = -ZA
XA = COS(O+2*PI)*(ZA-P/(3*ZA))-(B/3)
X = COS(O+PI)*(ZA-P/(3*ZA))-(B/3)
XB = COS(O)* (ZA-P/(3*ZA))-(B/3)
IF XA>0 THEN 2740
IF X>0 THEN 2770
IF XB>0 THEN 2800
Y = XA
DA = DT +Y/H*DC
AS =12*PN*DA
PRINT”Recquired area of steel
AS=”AS:;”Square inches per foot length of
wall”
RETURN
STOP
Y=X
DA = DT + Y/H*DC
AS = 12*PN*DA
PRINT”Recquired area of steel
AS=”;AS;”Square Inches per foot of wall”
RETURN
STOP
Y = XB
DA = DT+ Y/H*DC
AS = 12*PN*DA
PRINT” Required area of steel
274
2840
2850
2860
2870
2880
2885
2887
2890
2895
2900
2905
2910
2920
2930
2940
2950
2960
2970
2980
2990
3000
3010
AS=”;AS;’Square inches per foot of wall”
RETURN
STOP
REM This is a sub routine no three
REM Determination of steel
reinforcements for a value of y
M = 1/6000*KA*W1*LF*Z^3
D = Z/H*(T1-T2) + T2
D1 = (D-3.5)
IF (D1^2-2.61*M*12/(12*FC))<0 THEN
2895 ELSE 2905
PRINT “Depth of stress rectangular block
is imaginary not possible review given
data of sub routine no two”
STOP
A = D1- SQR(D1^2-2.61*12/(12*FC))
AS = 12*M/(0.90*FY*(D1-A/2))
P = AS/(12*D1)
PN = 0.2/FY
IF P>=PN THEN 2960
IF P<PN THEN 2980
AS1 = AS
GOTO 2990
AS1 = 12*PN*D1
AS1 = AS1
RETURN
END
275