Reinforced Concrete Design by Computer - R. Hulse and W.H. Mosley

8/10/2019 Reinforced Concrete Design by Computer - R. Hulse and W.H. Mosley

1/297


2/297


3/297


4/297


5/297


6/297


7/297


8/297


9/297


10/297


11/297


12/297


13/297


14/297


15/297


16/297


17/297


18/297


19/297


20/297


21/297


22/297


23/297


24/297


25/297


26/297


27/297

2

Programs for the Analysis of the

Structure

One &the first stages in reinforced concrete design is to carry out the structural

analysis to determine the moments, shears and axial forces that have to be

resisted. As this usually involves a considerable amount of numerical calcu-

lation it is best i f it can be processed with the aid of the computer.

This chapter describes several analytical programs, starting with a simple

case of the one span beam, then a continuous beam, and so to further

developments of these programs in order to calculate the envelopes of shear

and bending moments with the option to redistribute the moments and also to

carry out the analysis for beams with haunches. The programs have been

developed in a structured form with the use of standard subroutines that can be

utilised in successive programs so that the logic should be more readily

understood and applied.

2.1. Limit State Analysis of a Single Span Beam

The program calculates the shear forces, bending moments and deflections at

2O&intervals along a one span simply supported beam, and sorts for the

maximum moment and deflections. Any of the limits states can be considered

by inputting the relevant partial factors of safety. Many of the features of this

program are similar to those used in the longer programs for the analysis of

continuous beams which also include the calculations of the shear force and

bending moment envelopes.

The load distribution on the beam can be either concentrated, uniform or

triangular, as specified in figure 2.1. The input data for each load on a span

consists of

(a) the type of load distribution, identified by a code number 1, 2, 3 or 4

17


28/297

18

REINFORCED CONCRETE DESIGN BY COMPUTER

A 2 Uniformly

distributed

A 3 Triangular

left

(b) the weight, W, of the load in kiloNewtons

(c) the distance,

A,

in metres from the spans eft-hand support to the start of

the load

(d) the distance, C, in metres that the load cavern

(e) whether the load is dead or imposed, identified by an input D or I.

21.1 AnaIytIcaI procedure

The shear forces and bending moments are calculated at 20th increments

along the span for each load and progressively accumulated for the total

loading, so providing the final distribution. Further explanation of the

procedure is given in the program description of section 2.1.3.

The deflection is calculated based on the second moment of area of the

untracked concrete section. The method of analysis uses the flexibility

equations and a process of numerical integration applying Simpsons rule.

At any point along the beam

deflection =

I

mm1

-dx

EI


29/297

PROGRAMS FOR THE ANALYSIS OF THE STRUCTURE

19

where m, is the bending moment along the span due to the applied loads and ml

is the corresponding bending moment due to a unit load applied at the point

where the deflection is being calculated. Applying Simpsons rule to the

numerical integration to achieve greater accuracy gives

jydr = s/3 [(sum of end ordinates) + 4 (sum of even ordinates)

(2.2)

+ 2 (sum of remaining odd ordinates)]

where s is the length of the increment, equal to span/20 in this program.

In this numerical integration, greater accuracy is attained if any disconti-

nuities in the bending moment diagram occur at an odd-numbered section.

Also in applying Simpsons rule the number of intervals must always be an even

number.

The actual procedure is probably best explained by working through a

numerical example by hand calculation. In the beam of figure 2.2 the deflection

due to the loads shown is required at section number 4 along the span.

For this hand calculation the numerical integration has been carried out over

10 increments at 2 metres spacing. The bending moment diagrams m, due to

the span loads and m, due to a unit load at section 4 are shown.

The numerical values of the bending moments and their products at each

section are listed in table 2.1.


30/297

20


So applying Simpsons rule the deflection of the beam at section 4 is given

by

EI x y4 = j[(O + 0) + 4(39.2 + 352.8 + 345.0 + 163.8 + 19.2)

+ 2 (156.8 + 403.2 + 259.2 + 76.8)]

= 3648 kN m3

therefore

y4 = 3648/H = 3648070

= 21.5 mm

This compares with a result of 21.1 mm using the computer program with

an integration over 20 intervals.

2.1.2 Flow diagram for the single span beam analysis program

lNPT DATA

Title. T$

Partial safety factors. GK. OK

Elastic modulus, EC

Beams span. I.

2nd moment of area. I2

Characteristic loads:

No. of loads. NL

For each load:

Load type ~ 1.2, 3 or 4. T1 (J)

Load weight, WC(J)

start distance. A(J)

Cover distance. C(J)

Dead or imposed, TS(J)

I

I

Calculate design loads

-50

1

GOT0 SUBROUTINE

1

Calculate span shear

and moments. Sort

for maximm moment


31/297


Table 2.1 Values of moments and their products

21

Section 1 2 3 4 5 6 7 8 9 10 11

f% 0 28.0 56.0 84.0 112.0 115.0 108.0 91.0 64.0 32.0 0

ml 0 1.4 2.8 4.2 3.6 3.0 2.4 1.8 1.2 0.6 0

m,ms 0 39.2 156.8 352.8 403.2 345.0 259.2 163.8 76.8 19.2 0

2.1.3

Lkscription of the single span beam analysis program

Lines

30

40-330

35&370

380

390

4c+710

4oi3u

4010-4040

4050+380

407&4140

4150-4260

The variable arrays are dimensioned for the shears, VS, the

moments, MS, the deflections, DS, and Simpsons ordinates, S.

Input of data, commencing with the title and then the partial

factors of safety on lines 70-110. The beam information of

elastic modulus, the span and second moment of area are

input on lines 12&180, then the data for the characteristic

loading is input on lines 19&330.

The design loads WD are calculated as a product of the

characteristic loads, WC, and the partial factors of safety,

GK or QK, depending on whether the load is dead or

imposed.

The program transfers to the subroutine on lines 4Ol%4450

in order to calculate the shears and moments along the

beams span.

The program transfers to the subroutine on lines 6-260

for calculating the deflections along the span.

Printout of data and results. An example of the printout of

data is shown in section 2.1.6. Although the moments and

deflections are calculated at 21 sections along the span, only

the results for 11 alternate sections are. printed.

Subroutine to calculate span sheets and moments.

Variables that are used in accumulative totals are initialised to

zero.

Within a FOR-NEXT loop the reactions, shearsand moments are

calculated for each design load and accumulated for the complete

loading arrangement.

The reactions Rl and R2 are calculated for each load according to

the type of load distribution, and the total reactions RA and RB

are accumulated on line 4140.

The span shears, VS(K), are calculated at 11 sections along the

beam.


32/297

22


42704UXl The span moments, MS(K), are calculated at 21 sections along the

beam. As with the shears, the moment calculation is carried out in

three parts according to whether the section being considered is to

the left or right of the load, or within the load distribution.

441LL4440 The moments along the span are sorted to determine the maxi-

mum moment, MM, and its distance, SM, from the left-hand

support.

-260 Subroutine to calculate deflections. The span deflections, DS(K),

are calculated at 21 sections along the beam based on the numeri-

cal integration method and Simpsons rule described in section

2.1.1, using the span moments already calculated in the previous

subroutine.

2.1.4 Listing of the single span beam analysis program

10 REM ANALYSIS OF R SINGLE SPAN BER

211REM

.~~*~*ff,.~ff*t.fff+,,,,*~~*~*


33/297


34/297

24


2.1.5 Lid of variables in the single span beam program

A ( ) Distance from the left-hand support to the start of the load

C ( ) Distance of load cover

DK Sum of integration product in deflection calculation

DM

Maximum deflection

DS ( ) Deflection at intervals along the beam


35/297


Elastic modulus

25

Partial factor of safety for dead load

Second moment of area of the beams section

Span of beam

Bending moment

Bending moment at intervals along the beam

Maximum moment

Moment due to unit load

Number of loads on the span

Partial factor of safety for imposed load

Support reactiotis

Support reactions calculated for each load

Distance from left support to load centroid

Simpsons rule factors

Distance from left support of maximum deflection

Distance from left support of maximum moment

Distance from right support to load centroid

Code for type of load distribution

Load type, dead or imposed

Shear force

Shear force at intervals along the span

Weight of characteristic load

Weight of design load

Weight of part of load to interval along the beam

Distance of section intervals along the beam

EC

GK

I2

L

MK

MS ( )

MM

MX

NL

QK

RA, RB

RL, RR

: 0

SD

SM

T .

Tl

T$ ( )

VK

vs ( )

WCC )

WD( 1

Wl

z, Zl

and 22

2.1.6 One span beam analysk example

Section +*;

2Om


36/297

26 REINFORCED CONCRETE DESIGN BY COMPUTER

BEMS BPnN - 20 m.tr..

CHWacCTERR1STIC LOIIDBCLN, AND PoaITI0Nhmtr.r~

twMBER a LOms ON SPW - 2

This program analys,es a continuous beam which may have cantilever moments

acting at the supports at either end. It caters for uniformly distributed loads

but by inputting the load cover as zero it will also deal with concentrated loads.

Using the fixed end moments tabulated in table 2.2 the program would be

readily modified to include loads with a triangular distribution; in addition the

subroutine for the span moments would also require altering so as to be similar

to that for the one span beam of the previous section.

The sign convention for the results is that sagging moments are positive

while hogging moments are negative.

2.2.1 Am&tid procedure

The analysis of a continuous beam to determine the support moments can be

carried out by setting up the slope deflection equations as a series of simul-

taneous equations and solving these equations for the rotation at each support.


37/297


27


38/297

28


The

support moments can then

be

derived from the rotations.

For a continuous beam of N spans there are N + 1 equations of the form

4 XK, R, + 2 k, RZ = MCL - MF,u

2 k, R, + 4 SK, R2 + 2 k2 R3 = M~nl - Mm2

2k,, Rim,4 XK, Ri + 2 ki R,+, = M,,(,,) - Mm

(2.3)

2kvR.w+4W++,R,y+, =MFRN-Men

In these equations

ki = Ii/L+ the stiffness of the beam in the ith span

XKi = the sum of the stiffness of the two spans meeting at the ith support

Ri = the rotation of the beams at the ith support

M

CL =

the cantilever moment at the first support

Men = the cantilever moment at the last support

Mni = the tixed end moment at the left-hand support of the ith span

MFni = the tixed end moment at the right-hand support of the ith span

The slope deflection equation can be rewritten in the form

D, Rx + E, RZ = Fl

E, R, + Dz Rz + EZ Rs = Fz

Ei-, Rim, + Dj Ri + Ej R,+l = F,

EN RN +

DN+I RN+I

=

F~+~

which corresponds to the form they have been set up in the program subroutines

starting at lines 4000 and 6000 of the program listing.

The solution of the simultaneous equations to provide the rotation Ri at each

support is achieved by a process of forward elimination and backward substi-

tution similar to the Gauss elimination method. The support moments at the

ends of each span are calculated from the equations

M,j = ki (4 Rj + 2 R,+l) + Mmj

(2.4)

MRi = ki (2 Ri + 4 Ri+l) - Mpn;

The shears and moments at 20th intervals along each span are calculated

using a method similar to that for the one span beam program of the previous

section, but in this case the effect of the end support moments must also be

included.

This program does not include the subroutine for calculating the span


39/297


22.2 Flow &?gmm for the continuous heam program

29


40/297

30


deflections as in the single span beam program, but if required it could be

readily introduced using the same routine procedures of numerical integration

applied to the moments along each span which are stored in the array MS(I,K).

The effect of full fixity at the end supports could be provided by arranging to

modify the program so that the left-hand stiffness coefficients D(1) or

D(NS + 1) in the subroutine starting at line 4000 are set to a very large

number.

This continuous beam program could also be used for the analysis of a

continuous slab at the ultimate limit state for the single loading condition of all

spansbeing fully loaded, with 1.4 GK + 1.6 QK as specified in clause 3.5.2.3

of BS 8110.

Lines

30

4M60

170-320

33&370

42W450

460

470

480

490

500

510-560

570-950

The variable arrays are dimensioned.

Input of the beam data consisting essentially of the length and

second moment of area for each span.

Input of the data for the loads on each span.

Input of the data for any cantilever moments, which in this

instance are keyed in as positive when they cause hogging at the

relevant support.

Calculation of the beam stiffness, In, for each span.

Transfer to the subroutine for setting up the coefficients for the

left-hand side of the slope deflection equations.

Transfer to the subroutine to calculate the fixed end moments at

the ends of each span.

Transfer to the subroutine for setting up the coefficients for the

right-hand side of the slope deflection equations.

Transfer to the subroutine for solving the slope deflection equations

and calculating the moments at the ends of each span.

Transfer to the subroutine for calculating the shearsand bending

moments along each span.

Sort for maximum span moments, MM(I), and their position,

.SM(I).

Printout of the data and results. The result consistsof a tabulated

list of the shearsand moments at 11 section along each span. The

maximum saggingmoment and its location along the span are also

printed out.

4OOU4080 Subroutine to set up the coefficients for the left-hand side of the

slope deflection equations. These coefficients are a function of the

member stiffnesses only.

500&5130

Subroutine to calculate fixed end moments at the end of each

span. These moments are stored in the arrays FL(I) and FR(1);


41/297

PROGRAMS FOR THE ANALYSIS OF THE STRLKTURE

31

they are calculated according to the equations in table 2.2 which

are set out on lines 5070 and 5080.

6WO60

Subroutine to set up the coefficients for the right-hand side of the

slope deflection equations, these coefficients being a function of

the fixed end moments and any cantilever moments at the end

supports.

7OGG7160

Subroutine to solve the slope deflection equation and calculate the

support moments at the ends of each span. The processof forward

elimination is carried out between lines 7010 and 7050, while the

back substitution to determine the member end rotations, R(I), is

between lines 7060 and 7100.

The support moments at the end of each span, ML(I) and

MR(I), are calculated on lines 71 &7160. The sign convention for

these moments is that clockwise moments are negative.

80004320

Subroutine to calculate span shearsand moments. This subroutine

is very similar to that used in the single. span beam program,

except that it caters only for uniformly distributed loads; also it is

necessary to include the effect of the span end moments on lines

8020-8050.

2.2.4 Lkting of the

continuous

beam andysk program

16 RE

ANALYSIS OF A CONTINOS BEI\H

ZB REM

fff*ffftt~.f~**~...~.~**.~.**

311DIM s,1B,ll~,Ms~1B,*1~

4y REM l ~f..~**.*.*.tft*f*f..~*.*.~ ~*.~~ ***.~~ Input Of data

58 DISP ENTER TIT LE

66 INPUT TS

78 ~~ pj l *.~~~*~f*~f~~ ff+lf l********.*.~~~** Beam information

88 DISP ENTER NO. OF SPANS

98 INPUT NS

IOk7

110

128

I ,@


42/297

32



43/297


33


44/297

2.2.5 List of variables for tbe conthuouv beam program

Many of the variables used in the one span beam program are used again to

define the same term in this program, and so they are not listed again in this

section. In some instances a variable such as W(I) has been changed to a two-

dimensional array, W(I,J), to take account of the number of spans. For these

array variables, in general I refers to the span number whilst J is the load

number on that span. Similarly for a variable such as MS(I,K), the span

number is I and K is the interval along the span, which is from 1 to 21 for a span

moment or 1 to 11 for a shear.


45/297


35

CL

Cantilever moment at the first support

Cantilever moment at the last support

$) and E(1)

Coefficients for the left-hand side of the slope deflection

equations

F(I) Coefficients for the right-hand side of the slope deflection

equations

FL(I) and FR(I) Fixed end moments at the ends of each span

WI)

Stiffness of the beam in each span = 12(1)/L(I)

NS

Number of spans

R(I)

Rotation of the beam at each support


46/297

36


2.3 Envelope Program

In order to design a continuous beam it is necessary to construct the envelopes

of maximum shear forces and bending moments along the beam. These

envelopes are used to calculate the areas of reinforcement at the key sections

and to set out the curtailment of the reinforcing bars. In constructing the

envelopes it is required to consider all the critical combinations of dead and

imposed load, usually at the ultimate limit state.

2.3.1 Andytkal precedure for the envelope program

Consider the continuous beam shown in figure 2.7, which has cantilever spans

at both ends and the number of internal spans NS = 4. The critical loading

306 kN

100 kN

306 kN

llltit

Gm 1 bm , Gm

I

I


47/297


37

&

-pattern

&.

1

2

3

L

5

G

7

patterns which determine the shear force and bending moment envelopes at

the ultimate lim it state are shown in figure 2.7.

Load patterns number 1 and 2 cause the maximum sagging moments in the

odd and even numbered spans respectively, and also with possible maximum

hogging moments in adjacent spans. The following three loading patterns, 3,4

and 5, provide the maximum design hogging moments at the successive internal

supports, as specified by the requirements of some Codes of Practice.

The last two loading patterns, 6 and 7, need be considered only when there

are cantilever spans. These patterns induce the maximum shear force at the end

supports, but they will also cause the maximum end moments in the end spans

of a substitute frame described in section 2.4.


48/297

38


2.3.2 Flow diagram for the envelope pro.@xm


49/297


39

Thus the number of critical loading patterns for a continuous beam of NS

internal spansare

NS + 1 with no cantilevers present

NS + 2 with one cantilever span

NS + 3 with two cantilever spans

A modification of this program to conform with the more simplified require-

ments of BS 8110, clause 3.2.1.2.2, is described at the end of this section.

2.3.3 Lkscription of the envelope program

This program is a further development of the continuous beam program of

section 2.2. The major changes from the previous program are as follows.

(1) Input of the partial factors of safety GK and QK for the dead and imposed

loads, lines 7&110.

(2) Input of the characteristic weight of the loads and specifying if the loads are

dead or imposed, lines 30&310 and 36G370.

(3) Input of the cantilever moments as dead and imposed components caused

by the characteristic loading, lines W80. Moments causing hogging at

the support are keyed in aspositive.

(4) Initialising the variables VE(I,K), MN(I,K), and MP(1.K) for the shear,

hogging and sagging moment envelopes, lines 49GS60.

(5) Calculation of the number of critical load patterns, NP, lines 62W550.

(6) Repeating the analysis for each critical load pattern, LP, from lines 680 to

750.

(7) Adding a subroutine to calculate the design loads on each span for each

loading pattern, lines 9OGG-9440.

(8) Adding a further subroutine to sort for the shear force and bending

moment envelopes, lines lOiXM-10180.

Much of the remaining program remains the same, using identical sub-

routines for setting up and solving the slope deflection equations and for

calculating the shears and moments along each span of the continuous beam.

2.3.4 Listing of the envelope program

1s REM SHEARAND HOENT ENVELOPES OR A CONTINOS BEAM

28 REM

~f..f.**f~**.*ff**f~~***~*~**~~.~~.**.~.....~**~

38 0, S(1B,11),E(ls,ll),nSo.~~,~~,*~~

48 RE *~,ff..fff*f***~f*~*.~...,....***...~. Ipt of data

5&i ISP ENTER TITLE

611 NPVT TS

76 REH l ft*f*ff***f*~*f*f,f.**. ... Partial factors Of safety

88 DISP ENTER PARTIAL FRCTOR OF SAFETY FOR DEAD LOAD,GK

98 INPUT GK

188 DISP ENTER PARTIAL FACTOR OF SAFETY FOR IPOSED LOAD,QK


50/297

40


33Y.

348


51/297


52/297

42



53/297


43

7866 R,NS+l)-F,D


54/297

44


FOR K=l TO II

IF ABS tE,I,K,,


55/297


45

CL(l) and CR(l) The cantilever moments caused by the characteristic dead

loads at the first and last supports respectively

CL(2) and CR(2) The cantilever moments caused by the characteristic im-

posed loads at the first and last supports

GG Partial factors of safety applied to the dead loads for each

load pattern

GK Maximum partial factor of safety for the dead loading

LP Load pattern number

MN(W)

Moments along the hogging moment envelope

MV,K)

Moments along the sagging moment envelope

NP Number of critical loading patterns

QG

Partial factors of safety applied to the imposed loads for

each load pattern

QK

Maximum partial factor of safety for the imposed loading

WD(LJ)

The design loads

WLK)

Shears along the shear force envelope


56/297


SPm4NO. 1

SECTION SHEAR

HODGINO

NO.

nOnENT

1 131.10 -.oo

2 LOO.Jo 0.00

: (19.909.50 0.00.00

s

8.70 0.00

d

-27.23 0.00

7

-57.83 0.00

B

-ea. 43 0.00

; 119.03 -.20

: se.7.833 0.00.00

b 27.23 0.00

7

-8.70 0.00

e

-39.30 0.00

9

-6V.

90 0.00

10 -Loo.SO 0.00

II -151.10 -.oo

nnxnltm emw naENT - L&e.06 ld4.m

BnwING

lWYlENT

0.00

0.00

0.00

0.00

7.67

11.75

7.67

0.00

0.00

0.00

0.00

Dead Load

gk=25kN/m

Imposed Load qk=10 kN/m

m

I -A

Figure .9 Sheor ndmomentwelopes xample


57/297

PROGRAMS FOR THE ANALYSIS OF THE STRUCI-URE

Modification

of

the envelope program to conform with BS 81 O

41

The Code of Practice BS 8110 allows the maximum design moments at the

supports to be calculated from one loading arrangement with all spans oaded

with the ultimate load of (1.4 GK + 1.6 QK). In this case there are only three

critical loading arrangements and the program can be readily modified for the

option by adding the following program lines.

2.4 Substitute frame analysis

With some relatively minor modifications and additions, the previous program

for determining the shear force and bending moment envelopes of a continuous

beam may be applied to the analysis of a substitute frame. Briefly these changes

and additions are as follows.

(1) Input the column dimensions.

(2) Initialise to zero the column moments.

(3) Calculate the column stiffnesses and distribution factors.

(4) Modify the left-hand side of the slope deflection equations in the sub-

routine starting at line 4ooO n order to include the effects of the column

stiffnesses.

(5) Calculate the column moments by means of a subroutine at line llOC+J

which is called on line 725.

(6) Print out the column data.

(7) Print out the column moments.

These changes and additions have been inserted into the program without

altering the previous line numbering, and they are presented in sequence

shown in the listing of the revised and additional program lines.

2.4.1 Listing of the amended and additional lioes for the substitute frame

analysis program


58/297


59/297


24.2 List of variables for tbe substitute frame analysis program

49

DA(I), DBU)

WI), WI)

KNI), WI)

MA(I), MB(I)

MD

TA,TB

Distribution factors for the column above and below

Second moments of area of the column section above and

below

Column stiffness above and below

Column moments above and below the beam

Sum of the column moments at a support which is equal to the

difference in the end moments of the beams meeting at that

support

Temporary values of the column moments

2.5 Moment Redistribution

All the previous programs in this chapter have carried out an elastic analysis.

Howevtx, at the ultimate limi t state reinforced concrete no longer behaves

elastically but acts more as a plastic material with the formation of plastic

hinges at the most highly stressed sections. To allow for this plastic behaviour

the Code of Practice allows for a redistribution of the moments calculated from

an elastic analysis. Up to a 30 per cent reduction of the elastic moments is

permitted, but this is restricted to 10 per cent for structures over four storeys in

height. There is also a restriction on the maximum depth of the neutral axis, x,

at a section such that

x sl (pb - 0.4)d

where d is the sections effective depth and Pb is

moment at the section after redistribution

Pb =

moment at the section before redistribution

This restriction will in general rule out any redistribution of the moments in

columns which are primarily axially loaded members, with a relatively large

depth of neutral axis.

It is also important that after redistribution there must still be equilibrium

between the external loads and the internal member forces.

25.1 Andytic~I procedure for the redistribution program

For simplicity and ease of understanding the redistribution is applied only to

the continuous beam envelope program and not to the substitute frame

program.

Most redistribution is usually applied to reduce the maximum support

moments where the continuous beam acts as a rectangular section and where


60/297

50


there is also the possibility of congestion of steel reinforcing bars. With

reference to figure 2.1, the maximum hogging moment at the Nth support is

caused by the (N + 1)th loading pattern.

The effect of moment redistribution to the hogging moments at a support is

shown in figure 2.10. It can be seen that the codes restriction that the moment

at any section must not be less than 70 per cent of the elastic moment prevents

the movement of the point of contraflexure at point c. This ensures that there

should always be an adequate curtailment of reinforcement to resist the elastic

moments at the serviceability limit state.

Elastic moments

Redistributed moments

__----.---- 70% of elastic moments

Figure *.lll Moment redishibuzion

When redistributing the moments in a substitute frame the column moments,

as already explained, should not be changed from those calculated in the elastic

analysis. Hence forthe support hogging moments shown in figure 2.11 the

difference in beam moments, length ef, represents the moment induced in the

columns and this value should not be changed by redistribution of the moments

for any of the loading patterns. Also redistribution must never be applied to

any cantilever moments at the end supports.


61/297

PROGRAMS FOR THE ANALYSIS OF THE STRlJClVRE

25.2 Flow diagram for the moment rdstributhn program

A

51


62/297

52


25.3 Listing of the amended and additional lines of the moment redistribution

Program


63/297


53

Foil 1 =1 TO 21 STE P 2

PRINT SING ,918 ; ,R+1),Z;R,I, K+l),Z);RN I,K);RP,I,K)


64/297

54


2.5.4 List of variables for the envelope redistribution ~ro& ram

ML(LP,I)

MR(LP.1)

R$

RW)

RF

RL(LP.1)

RR(LP,I)

RN(W)

WW)

WLK)

RX

Moments at the end of each span, I, and for each load pattern, LP

Moment redistribution to be applied -Y or N?

Percentage redistribution at each support

Fractional change of moment due to redistribution

Moments at the end of each span after redistribution

Negative (hogging) and positive (sagging) envelope moments after

redistribution

Shear force envelope after redistribution

Variable for temporary storage of moments at the end of each span

2.6 Continuous Beam with a Varying Cross-section

In many reinforced concrete structures, for reasons of strength, stiffness or

perhaps aesthetic requirements, the beams are constructed with haunches at

the supports. To allow for this the continuous beam program hasbeen modified

so that it will cater for the three types of haunches shown in figure 2.13. The

prismatic haunch, type 1, could be used with the analysis of flat slabswith drop

panels. The program could be readily further modified to deal with a beam of

any variable section by either the program calculating the second moments of

area or by direct input of the members second moments of area at the discrete

intervals along the span.

=+ETD2

LL

k

L2

L

(a) Prismati: haunches

(b) Straight haunches

(c) Parabolic haunches

Figure2.13

ypes of haunches for a memberf varying ection.


65/297


55

The beams are considered to be of constant unit breadth but, by a small

modification to the subroutine which calculates the second moments of area,

beams of varying breadth could also be included.

2.61 AnaIyticdprocedure

For a member of constant cross-section the slope deflection equations for the

rotations at the supports of a continuous beam take the form shown in equation

(2.3). In the equation the stiffness at each end of a member is 4EIIL while the

carry over effect due to a unit rotation is 2EIIL.

With a member of non-constant section the slope deflection equations have

to be modified to allow for the stiffness of kA x El/L and ks x EIIL at each

end of the member and a carry over effect of kc x H/L. In these stiffnesses

the coefficients kA, kB and kc depend on the type and size of the haunches,

while EIIL is based on the second moment of area of the member at the centre

of span between the haunches.

The stiffness coefficients kA, ks and kc can be determined from the flexi-

bility matrix for a member AB such that

The terms in the flexibility matrix are derived in most standard textbooks of

structural analysis by applying the method of virtual work, so that

and

fl2 = fil =

(2.6)

The moment diagrams for ml and mZ are shown in figure 2.14.

The stiffness at end A of a member is the moment MA required to rotate it

through unit angle, so setting 13

= 1 and tlB = 0 and cross-multiplying the

flexibility matrix 2.5 gives


66/297

56


kA = MA =

kB =

(2.7)

The carry over effect moment is

In the computer program the stiffness coefficients for each span are calcu-

lated by the method of numerical integration and applying Simpsons rule to

give a better accuracy, in a manner similar to that used to calculate the

deflections in the one span beam program.

It is also necessary to calculate the tixed end moments for each span using a

numerical integration procedure. The tixed end moments are given by

M

FA = kA 8~ + kc es (2.8)

MF~ = ks 0~ + kc On

where MFA, Mm are the fured end moments at ends A and B of span AB

and OA, OS are the free rotations with ends A and B released of moment

restraint.

Applying the principles of virtual work the free rotations are


67/297


51

when m, and m, are the moments in the bending moment diagrams of figure

2.14

and

m, are the moments of the free bending moment diagram due to the

applied loads on the span.

The fixed end moments for each span are also calculated by applying numerical

integration with Simpsons rule.

Within the program for each span I, the fixed end moments MFA, MFB and

the free rotations AA, eB are represented by the variables FL(I), FR(I), DL

and Dri respectively.


68/297

58


2.6.2 Flow diagram for continuous beam with varying section program


69/297


59

2.6.3 Descri@m of tbe contim~ous beam wit6 vrvying cross-section program

Lines

30

40-290

3w50

520

530

560

570

600

610 -

620

630

640

650

660-1180

4WO80

5cK&s120

6OOCMO60

7000-7160

The variable arrays are dimensioned.

Input of the structure data, consisting of the span lengths, and

the types of haunches with their lengths and depths.

Input of the data for the loads on each span.

Initialising variables to zero.

Transfer to subroutine for setting the ordinates for Simpsons

rule and for the integrals of the unit force moment diagrams.

Transfer to subroutine for calculating second moments of areas

along each span.

Transfer to subroutine to calculate member stiffness coef-

ficients, KA(I), KB(I) and KC(I).

Transfer to subroutine for setting up the coefficients of the left-

hand side of the slope deflection equations.

Transfer to subroutine to calculate the free bending moments

on each span with no support restraint.

Transfer to subroutine to calculate the fixed end moments.

Transfer to subroutine for setting up coefficients of the right-

hand side of the slope deflection equations.

Transfer to subroutine for solving slope deflection equations

and calculating the support moments.

Transfer to subroutine to calculate the bending moments along

each span.

Print out of data and the results, consisting of the member

stiffness factors and the bending moments at intervals along

each span together with the maximum sagging moment and its

location.

Subroutine to set up the coefficients for the left-hand side of the

slope deflection equations. This subroutine is very similar to

that in the continuous beam program, except that the stiffness

factors, KA(1) and KB(I), replace the value 4, and KC(I)

replaces the value 2.

Subroutine to calculate the fixed end moments. The fixed end

moments, FL(I) and FR(I), are calculated using equation (2.8)

and (2.9).

Subroutine to set up the coefficients for the right-hand side of

the slope deflection equations. This subroutine is also very

similar to that in the continuous beam program, except in this

case the cantilever moments have not been included.

Subroutine to solve the slope deflection equations and calculate

the support moments. This subroutine is again very similar to

the one in the continuous beam program, except the stiffness


70/297


800&8120

14OOLL14170

15cO&15330

1600&16130

17cOO-17170

coefficients, KA(I), KB(I), replace the value 4 and KC(I) the

value 2 in calculating the end moments on lines 7130 and 7140.

Subroutine to calculate the span moments. The span moments

at intervals along the span are calculated by adding together on

line 8040 the moment due to the support moments and the free

span bending moments which are calculated in the final sub-

routine starting on line 17ooO.

Subroutine to set up the ordinates for Simpsons rule and the

integrals of the moment diagrams due to unit forces. The

ordinates for Simpsons rule, lines 14010-14060, are as defined

in section 2.1.1. The integrals of the moment diagrams and their

products, lines 14070-14160, are according to equations (2.6)

and (2.9) used in calculating the stiffness factors and the fixed

end moments.

Subroutine to calculate the second moments of areas at inter-

vals along each span. These values are calculated at the 21

intervals along each span for each type of haunch.

Subroutine to calculate the member stiffnesses. Equations (2.6)

and (2.7) are used to calculate the stiffness coefficients KA(I),

KB(I) and KC(I) for each span. The products of the Simpsons

ordinates, S(K), and the relative second moments of area,

Il(I)/I2(K), are calculated on lines 1601~16030 and stored in

the variable P(I,K).

Subroutine to calculate the ordinates of the free bending

moment diagram for each span with no moment restraint at the

supports. These values are used in calculating the fixed end

moment in the subroutine starting at line 5000, and also in

calculating the final moments along each span in the subroutine

starting at line 8CCil.


71/297

PROGRAMS FOR THE ANALYSIS OF THE STRUCKJRE

DISP ENTER SPAN LENGT - metres w

INPUT ,I)

DISP ENTER LENGTH Ll - metres

INPUT l,I,

DXP ENTER LENGTH r.* - metres

INPUT L?(I,

L-llSP -ENTER DEPTH D - metres *I

INPUT DC(I,

DISP ENTER DEPTH Dl - metres

INPUT Dl(f,

DfSP ENTER DEPTH DZ - metres

INPUT DZ,I,


72/297


73/297

PROGRAMS FOR THE ANALYSIS OF THE STRUDURE

63

4#9@ RE I.tfff*t*.~.~.*.*t,*....,**.,*~~,~.*******~**..~.

NEXT K

KA=KA(*, c KB=KB,I, P KC=KC(I,

FL (I, =KA*D-KCDR

FR,I,=KBDR-KCDL


74/297


X=0

FOR K=l TO 21

Ml K)= l-X)-Z

MZ K)=X-2

Pl3 K)=X* l-X)

PM K)=l-X

ns ,K)-X

x=i+i/a

NEXT K

Lx=Dc

12 K) .DX^3/12

x=x+w*o

NEXT K


75/297


65

2.6.5 Lirt of variables

Again most of the variables used in the program listing are consistent with

those in the orevious programs, and only the principle new variables are

defined here.

DW)

W )

DC(I)

DX

Fl, F2, F3

WI)

11(I)

12(K)

IWO, KB(I),

KC(I)

Ml(K) to

M-W)

Depth of the haunch at the left-hand side of each span

Depth of the haunch at the right-hand side of each span

Depth of the beam between haunches for each span

Depth of beam at each section along a span

Flexibility factor coefficients as used in equations (2.6)

Type of haunch-prismatic, straight OTparabolic

Second moment of area of the section between the haunches

for each span

Second moment of area at each section along a span

Stiffness factors as defined in equations (2.7)

Ordinates of the moment integrals of equations (2.6) and

(2.9)


76/297

66


Loading : 25 kN/m

WAN NO. 2

--_-_---_

Pa. OF LOADS ON SPlaN - 1

Lom No. 1

LOALl UEIBHT - 150 kN


77/297

PROGRAMS FOR THE ANALYSIS OF THE STRUCTURE 67

DISTANCE TO LOm ST(VIT - 0 I.W..

DIBTIWCE a Lam COVER - 6 m&r..

STIFFNESS .=acTms


78/297

3

Beam Analysis and Design

Once the ultimate bending and torsional moments and shear forces have been

determined, either by continuous beam or substitute frame analysis, the next

most likely step in the design of any reinforced concrete structure is the design

of tbe beam elements.

Any such beam element must have adequate strength at the ultimate limit

state to resist the design forces, and must also satisfy serviceability require-

ments of deflection, cracking etc. under serviceability conditions. Programs for

serviceability conditions are given in chapter 7 and this chapter is concerned

only with analysis and design at the ultimate limit state. This chapter therefore

includes the following programs

(i) The analysis of rectangular and flanged beam sections.

(ii) The design of rectangular and flanged beam sections for bending

(iii) Design of shear links.

(iv) Design for torsion.

Whereas the first of the above programs is concerned with analysis, the

others form part of the overall design procedure for concrete beam design. The

inter-relation of these programs is shown in the flow chart in figure 3.1 which

shows a generalised approach to reinforced concrete beam design. The pro-

grams in this chapter are indicated by the heavy black lines.

Although these programs are presented as stand alone programs, the inter-

relation of both the bending design and shear design programs with the analysis

programs given in chapter 2 should be noted.

The output from the analysis programs include shear force and bending

moments at discrete intervals along each beam span. Using the computer file

storage system, these moments and shears can be stored and the bending and

shear design programs can be modified to read this data and carry out the

section design at critical locations along each span. The linking of the programs

in these two chapters will lead to the development of a powerful design aid, and

means of doing this are discussed further in chapter 8.

68


79/297

BEAM ANALYSIS AND DESIGN 69

3.1 Section Analysis for Bending Resistance

The program in this section calculates the ultimate moment of resistance of any

rectangular or flanged beam containing a single level of compression and/or

tension reinforcement. The analytical procedure is based on the use of the

rectangular concrete stess block, the parameters of which can be altered within

the program to suit the requirements of different Codes of Practice. In this


80/297

70


program, stress block parameters have been assigned to give an average stress

of 0.45f,, and a stress block depth of 0.9 x.

Figure 3.2 shows the assumed stress and strain conditions in a concrete

beam at ultimate load conditions. The beam section can be analysed using

fundamental principles of compatibility of strain, equilibrium and material

stress/strain relationships. The steps in the method can be summarised as

(i) Assume a value of the neutral axis depth, x.

(ii) From the geometry of the strain diagram calculate the strain in the

compression and tension reinforcement, E.~ and E.~, for this assumed

value of x.

(iii) Using the design stress/strain graph for reinforcement, figure 3.3,

-

determine the stress in the compression and tension reinforcement, fs

andfst.

For a strain in the tensile reinforcement of l ., the values of stress are

given by

fst =

200.103 Est

+ts El

fst = fyhn

%f El

For a strain in the compressive reinforcement of l IC the values of

stress are given by identical expressions

d

Section Strains

FIgwe 3.2 Se&m subjm 10 bending

Stress


81/297


~igum 3.3 S~res risnain graph for reinforcemem (remion and com pression)

(iv) Calculate the compressive force, C, above the neutral axis and the

tensile force, T, in the tension steel given by

C = F,, + F,,

=f(bkx) +A,f,

T = F,,

= 4 fst

(v)

If T and Care not within agreement to an acceptable level of accuracy,

then repeat from step (i) until an acceptable level of accuracy is

reached.

(vi) Calculate the ult imate moment of resistance, Mu, by taking moments

about the tension steel

M,=(fbkx) d-; +A,f,(d-d)

( 1

(b) Fhged sections

The procedure for a flanged section is identical to that described above,

except that the possibility of the stress block extending into the web must be

considered.

Stress

block

in theflange. The section can be treated as a rectangular beam

of breadth equal to the breadth of the flange, bf, as shown in figure 3.4. The

analytical equations are identical to those given above.


82/297

72


+

hf

l P

AS

m

Pf;di3:

d

AS

&

F-

St

Section Stress

Figure 3.4 Fhnged seckm - s,res s block in rheflnnge

Stress block in fhheweb. In this case the equations have to be modif ied to

allow for the area of web in compression above the neutral axis. With

reference to figure 3.5, the modified equations are given below and can be

used in steps (iv) -(vi) above.

C=f(b,h,+b,(k*-h,))+A:f,,

T = As fst

+fb,(kx-h,) d-$

+A:fsc(d-4

bf

f

(3.6)

d

Section Stress

Fiyre 3.5 Fkmgczd sec tion - stres s block in he web


83/297

BEAM ANALYSIS AND DESIGN

Concrete grade FU

Steel strength FY

Beam type B

73


84/297

74


3.1.3 Description of the section analysis program

Lines

N-300

Data input. The program will analyse either a rectangular or

flanged section and the appropriate input data is requested after

the response to the question displayed on line 120. In lines 10 and

20 the concrete stress block factors, K8 and K9, are equated to

0.45 and 0.9. These factors give an average concrete stress, , of

0.45 fcu and a stress block depth of 0.9 x.

310 (-500) An iteration procedure is used to determine the neutral axis depth

370

410/420

440

4501460

470

490/500

530

560

57&730

of the section (steps (i) to (v)).

The iteration commences with the neutral axis depth, x, equal to

the depth of the compression reinforcement, d, or in the case of

a singly reinforced beam with x equal to 11100of the effective

depth, d.

The program is routed to the subroutine to calculate strains and

stresses n both tension and compression reinforcement (steps (ii)

and (iii)).

If the stress block lies within the web of a flanged beam then the

area of concrete in compression is modified to include the web

area.

The compressive force, C, and tensile force, T, are calculated

(equations (3.1), (3.2), (3.4) and (3.5), step (iv)),

The current value of (T-C) is compared with the corresponding

value (T9 - C9) from the previous iteration. For small values of

neutral axis depth the term (T - C) will be positive. As the

neutral axis depth increases, the compressive force will increase as

the tensile force decreases, and hence the term (T - c) will

eventually become negative. Hence if (T - C) and (T9 - C9)

are both.positive, the iteration is repeated in increments of d/100

until (T - C) is found to be negative.

The value of x at which T and C are equal must lie somewhere

between the current negative value of (T - C) and the positive

value from the previous iteration. The iteration procedure is

repeated, commencing with the value of x from the previous

iteration and increments of d/loo0 to obtain a more accurate

estimate of the neutral axis depth (step(v)). Figure 3.7 illustrates

this iterative technique.

The ultimate moment of resistance of a rectangular section, or a

flanged section with the stress block in the flange, is calculated

(equation (3.3) and step (vi)).

The ultimate moment of resistance of a flanged section with the

stressblock in the web iscalculated (equation (3.6) and step (vi)).

Data, ultimate moment of resistance and xld ratio are printed.


85/297


Small values of x Large values of x

il

T>C

TcC

+

xc=

4OtXb4050 Calculation of the strain and stress in the compression reinforce-

406&tloO

ment (steps (ii) and (iii)).

Calculation of the strain and stress in the tension reinforcement

(steps (ii) and (iii)).


86/297

76


276 INPUT AC


87/297


77

3.1.5 List of variables for the section analysk program

3.1.5.1 Data variables

B : F for flanged section, R for rectangular section

3.1.5.2 Program variables

C : Compressive force in concrete and steel above the neutral axis

El : Steel yield strain tl

E3 : Strain in compression steel

E4 : Strain in tension steel

F3 : Stress in compression steel

F4 : Stress in tension steel

G : Iteration variable

KS : Concrete stress block factor

K9 : Neutral axis depth factor

T : Tensile force in the steel below the neutral axis

U : increment for iteration of X

3.1.6. I

Determine the ultimate moment

of

resistanceof the beam section given in

figure 3.8.

fc. =

30 N/mm2,

f, =

460 N/mm2

Input:

TITLE

: ANALYSIS EXAMPLE

CONCRETE GRADE, fcu - N/sq.mm

: 30

CHARACTERISTIC STRENGTH OF

REINFORCEMENT, fy - Nlsq.mm

: 460

FLANGED OR RECTANGULAR BEAM? :R

BREADTH OF BEAM, b-mm

: 280

AREA OF TENSION STEEL, AS - sqmm

: 2410

EFFECTWE DEPTH, d-mm :510

AREA OF COMPRESSION STEEL,

AS(C) - sqmm : 628


88/297


1280..-

Figure 3.8

DEPTH TO COMPRESSION STEEL,

d(C) -mm

Output:

2L10rnm2

: 50

MOMENT OF RESISTANCE OF SECTION= 411.98 kNm

X/D RATIO

= 0.41

3.1.6.2 Determine the ultimate moment

of resistance of

the flanged beam in

figure 3.9. fC

= 25 N/mmz, f, = 460 N/mm*

TiTLE

: ANALYSIS EXAMPLE


: 25


REINFORCEMENT, fy - Nlsqmm

: 460

FLANGED OR RECTANGULAR BEAM?

:F

800

1 300 /

Figure 3.9


89/297


DEPTH, hf, AND BREADTH, h, OF

FLANGE - mm

: 150,800

BREADTH OF WEB, hw - mm

: 300

AREA OF TENSION STEEL, AS - sqmm

: 1470

EFFECTIVE DEPTH, d-mm : 420

AREA OF COMPRESSION STEEL AS (C)

:o

79

Output:

MOMENT OF RESISTANCE OF SECTION= 227.94 kNm

XID RATIO

= 0.17

3.1.7 Further developments

Modify the program to permit the analysis of beams containing reinforcement

at different levels.

3.2 Design of Bending Reinforcements

This program will design the areas of longitudinal tension reinforcement for a

rectangular or flanged beam section. If required the area of longitudinal

compression reinforcement is also determined. The program is based on the

development of the design expressions for the following five situations

(i) Rectangular section with tension steel only.

(ii) Rectangular section with tension plus compression steel.

(iii) Flanged section with the concrete stress block within the flange and no

compression steel.

(iv) Flanged section with the concrete stress block extending below the

flange, and no compression steel.

(v) Flanged section with tension plus compression steel.

3.2.1

Design procedure

For each of the five design situations the expressions for the required steel

areas can he developed as follows.

(i) Rectangular section with tension steel only

Referring to figure 3.2, the compressive force, FCC,n the concrete above

the neutral axis is given by

F,, = stress x area of action

=fkxb


90/297


The moment of resistance, by taking moments about the centroid of the

steel area, is

M = F, x lever arm

=fkxb(d- kx/2)

The solution of this quadratic equation provides a value for x, the depth of

the neutral axis as

x = d/k - [(d/k) ~ 2Mlf b kz]

The force in the tension steel, F,,, is

Fst =

f,

A,ly, = 0.87 f, A,

and for equilibrium

Fs, = Fee

therefore

A, = f kx b/0.87 f,

(3.7)

(3.8)

(ii) Rectangular section with tension plus compression steel

In order that the beam should be designed so that a bending failure will

ccunmence with a gradual yielding of the tension steel, the depth of neutral

axis calculation from equation (3.7) should not exceed the neutral axis

depth xba,, at the balanced condition. This balanced state is taken at a

neutral axis depth equal to half the effective depth, which ensures that the

tension steel has yielded at collapse.

However, if moment redistribution has been carried out, the neutral

axis is further limited to

where

+,a~ = 0% - 0.4) d (3.9)

moment at section after redistribution

I%=

moment at section before redistribution

(5 1)

This lim itation applies where redistribution exceeds 10 per cent (that is,

Pb s 0.9).

The moment of resistance of the section in terms of the concrete

strength, obtained by taking moments about the tension steel, is


91/297


M

ha, = f b kxh;i, (d - kx,aJ2)

If

M

applied Mbal

Xl

(3.10)

then compression steel is required and the depth of neutral axis, x, is

restricted to xbal in order to ensure a gradual tension-type failure and not a

sudden compressive failure.

Referring to figure 3.2 and taking moments about the tension steel

M = F,, (d d) + F,, (d - kx,,,/Z)

where

F,, = fs, A : and

Fee = f b kXbz.1

so-that solving for the area of compression steel, A:

A, = M - f b k%a, (d - kXtJ2)

P

fx (d - d)

(3.11)

A: = (M - Mb&& (d - d)

(3.12)

The stress, fsC, in the compression steel is derived from the steels stress/

strain diagram using the equations of section 3.1.1,

For the equil ibrium of the tensile and compressive forces on the section

(3.13)

(iii ) Flanged section with the concrete stress block within the flange: no com-

pression steel

Referring to figure 3.5, with the depth of the stress block equal to the

flange depth, ht, the force in the concrete is

Fx-fb,h,

and the moment of resistance of the section in this case is obtained by

taking moments about the tension steel

Mr = F,, (d - h,/2)

=fbthr(d - h,/2) (3.14)


92/297

82


If

M

app,ied < Mr

then the stress block is within the flange. In this case the depth of the

neutral axis, x, can be calculated from equation (3.7) with the flange

breadth, bt, replacing the beam breadth, b. The force in the concrete, F,,

is

Fcc=fbrh

For equilibrium of the tensile and compressive forces

Fs, = Fc,

therefore the area of tension steel is given by

A, =

f

bf kx/0.87fy (3.15)

(iv)eFlangedsection with the concrete stress block extending below theflange: no

compression steel

When Mapplied

> Mr the neutral axis must l ie within the web. In this case,

taking moments about the tension steel for the compressive forces de-

veloped by the concrete gives

M = F,, (d - hr/2) + F,w [d - (kr + h,)/2] (3.16)

where F,,, the compressive force in the Range, is

F,,=f brht

and F,, the compressive force in the web, is

Fw=fbw(h - h,)

so that equation.(3.16) is a quadratic which can be solved for x, giving a

solution of the standard form

x = - B ? (B - 4 AC)?2A

(3.17)

where

and

A = k212, B = - kd

c = M - f h hf Cd- W) +

h d _ ,,z /2

fbw

f f

For equilibrium of the tensile and compressive forceson the

section

Fs, = F,, + Few


93/297


83

therefore

A = f br h, + f bw (kx h,)

s

0.87

f,

(v) Flangedsection with tension&s compression steel

When the depth of the neutral axis equalsx,., the force in the concrete,

F,, is

Fee = Fc, + Few

= f br hr + f bw &al - h,)

and the moment of resistance of the concrete, taking moments about the

tension steel is

MM = Fc, d - W) + Fav d - (&,a, + Q/21

so that if

then compression steel is required and the depth of neutral axis, x, is

limited to xbal for a tensile-type failure.

The area of compression steel required is given by

A: = W - Mdfsc Cd - 0

(3.19)

where the steel stress,

fsc,

s derived using the equations of sections 3.1.1.

For equilibrium of the section, the tensile and compressive forces

developed by the steel and concrete must balance. So that

Fst = Fee + Fc, + Few

Therefore the area of tension steel required can be calculated from

A

I

= ~A: x + f bf hf + f bw ha, - W

0.87

fy

(3.20)

If the depth of the flange, hf, is greater than+,,,, then the section should be

designed as a rectangular section with a breadth of bf.


94/297


32.2 Flow diagram for the design of bending4inforcementprogram

Lines

3w190

200

25&280

2W310

320

33c-340

Title, T

Concrete grade FU

Stee l strength FY

Beam type BS

Beam breadth B

lNPT DATA

lange breadth BF

.%mge depth HF

Applied ultimate moment M

% Redistribu tion RD

SECTION

MOMENT. MB

(1) TENSlON STE EL ONLY:

3Hk3-90

CALCULATE TENSION

STEEL

(2) TENS ON ST EE L PLUS COM-

/

W&410 PREWON STEEL: NEUTRAL AXIS

DEPTH = BALANCED DERH (XB)

420 GOT0 SUBROUTINE

43w50 CALCULATE COMPRESSION

AND TENSION STEEL AREAS

460 FLANGED SECTION

F


95/297


SECTlON MOMENT, MB

CALCULATE MOMENT FOR

STRESS BLOCK DEPTH

= FLANGE DEPTH. MF

470

4X

490

S,,

510

(3)STRESSBLOCKWlTHlN

5X--5X1

FLANGE:CALCLATE

TENSIONSTEELAREA

-..-~_

J

sw1n BEI.OWFLANGE:CALCLATE

TENSION STEEL AREA I

(S)TENSlON +COMPRESS,ON

h2w430 STEEL:NETRALAXlSDEPTH,

=BALANCEDDEPTH

MO ~?i%OS+RO~

CALCULATESTEEL

COMPRESSWE STRESS,

hxk-810 PRINTDATAANDSTEELAREAS

32.3 lkscrtption ofthe designofbendingreinforcementp~am

Lines

30480

290-310

Input of data.

Variables are assigned, including values for K8 and K9 which

speci fy the depth and magnitude of the concrete stress block. The


96/297


320

340

350

360-390

4wo

46&510

520-550

560-610

62&660

67&800

steel yield strain, El, is calculated and the depth of neutral axis,

XB, at the balanced condition, allowing for redistribution if

applicable. The percentage redistribution assumes a decrease in

moment.

If the section is flanged the program goes to line 460 to commence

the design for a flanged beam.

For a rectangular section the moment or resistance, MB, of the

concrete at the balanced condit ion is calculated using equation

(3.10).

If the applied moment, M, is greater than MB then compression

steel is required and the program jumps to line 400.

The design of the area of tension reinforcement for a singly

reinforced section is carried out. The depth of neutral axis, x, and

the area of tension steel are calculated using equations (3.7) and

(3.8) respectively.

The design of compression and tension steel is carried out in this

part of the program. The depth of neutral axis, X, is set equal to

the balanced depth, XB, and the stress in the compression steel is

determined from the subroutine at line 4000 which uses the

equations of section 3.1.1. The areas of steel are calculated from

equations (3.12) and (3.13).

These lines commence the design of reinforcement for a flanged

section. If the depth of the flange HF is greater than the depth of

the stress block at the balanced condition, then the section is

designed as a rectangular section and the program is directed back

to 330 from line 470.

The moments of resistance of the concrete section, MB at the

balanced condition and MF for the flange alone, are calculated

using equations (3.16) and (3.14). If the applied moment, M,

exceeds MB, then the program goes to the design of a section with

tension and compression steel at line 620. If the applied moment is

less than MB but exceeds MF, then the program goes to line 560

for a design with the concrete stress block extending below the

flange.

With M less than MF the stress block lies within the flange and the

tension steel is designed using similar equations to that for a

rectangular section of width BF, the flange width.

This part of the program is for the design of a section with the

stress block extending below the flange.

The design of a flanged section with tension plus compression steel

is carried out in this part of the program. The depth of neutral axis,

X, is set equal to the balanced depth, XB, and the steel compres-

sive stress is obtained from the subroutine at line 4000. The areas

of reinforcement are calculated using equations (3.19) and (3.20).

Printout of data and steel reinforcement areas, AS and AC.


97/297

Subroutines


87

4WO40 Subroutine to calculate strains and stresses in the compressive

reinforcement.

3.2.4 Listing the design of bending reinforcement program


98/297


99/297


89

32.5 List of variables for the design of bending reinforcement program


B : F for flanged section, R for rectangular section


Bl

El

E3

F3

K8

K9

MB

MF

RD

XB

Equals, B, the breadth of the beam web

Steel yield strain t,

Strain in compression steel

Stress n compression steel

Concrete stressblock factor

Neutral axis depth factor

Moment of resistance of the concrete section at the balanced condition

Moment of resistance of the concrete flange

&centage redistribution at section

Depth of neutral axis at the balanced condition

3.2.6 Beam section design examples

3.2.6.1 Determine the area

of

reinforcement required if the beam section in

figure 3.11 carries an ultimate moment of 145 kN m. No moment

redistribution has been carried out. fCu = 25 N/mm, f, = 460 N/mm.


100/297

90


Input:

TITLE

: BEAM DESIGN

CONCRETE GRADE, fcu- N/sq.mm

: 25


REINFORCEMENT, fy - N/sq.mm

: 460

FLANGED OR RECTANGULAR BEAM

:R

BREADTH OF BEAM, b-mm

: 230

EFFECTIVE DEPTH OF BEAM, d - mm

: 490

DEPTH TO COMPRESSION STEEL, dl -mm

: 50

APPLIED ULTIMATE MOMENT kNm

: 14.5

% REDISTRIBUTION AT SECTION

:o

Output:

AREA OF TENSION STEEL = 854.72 sqmm

3.2.6.2 Determine the required steel areas for the T beam in figure 3.12 which

is subject to an ultimate moment of 160 kN m. No redistribution has

been carried out. fcu =

25 Nlmm2, f, = 460 N/mm2

Input:

TITLE : FLANGED BEAM

CONCRETE GRADE, fcu - N/sq.mm : 25

CHARACIXRISTIC STRENGTH OF

REINFORCEMENT, fy - N/sq.mm

:460

600

r

1 250 .

Figure 3.12


101/297


FLANGED OR RECTANGULAR BEAM

BREADTH OF BEAM, b - mm

EFFECTIVE DEPTH OF BEAM, d-mm

DEPTH TO COMPRESSION STEEL, dl-

BREADTH OF FLANGE, bf - mm

DEPTH OF FLANGE, hf - mm

APPLIED ULTIMATE MOMENT - kNm

% REDISTRIBUTION AT SECTION

Output:

AREA OF TENSION STEEL = 789.17 sqmm

F

250

530

50

600

150

160

0

3.27 Further developments

(a) Write an additional subroutine to select a suitable number and size of

reinforcing bars to satisfy the calculated areas of reinforcement. Consider

the-following alternatives

(i) Automatic selection

of

bars

from

all available bar sires. Write the

available sizes n a DATA list and for each size calculate the required

number of bars. PRINT answers for all sizes to give a selection of

results.

(ii) Oser selection

of

sizes. Permit the user to INPUT as data his chosen

number and sizes of bars. The program can check whether the chosen

bars are adequate.

(iii) User specification

of

sizes. Permit the user to specify a limited number

of acceptable bar sizes. The program should calculate the numbers of

each size required in combination to give the steel area nearest to the

required value.

(b) Once the above bar selection subroutine is written, add additional routines

tp calculate bar spacingsand to see f the chosen combination of reinforce-

ment satisfies requirements for spacing. See BS 8110, clause 3.12.11.

(c) Clause 3.12.5 of BS 8110 specifies minimum areas of tension and compres-

sion reinforcement and clause 3.12.6 specifies maximum areas. Add a

subroutine to check whether the limitations are met and to print suitable

warning messagesf they are not met.

3.3 Design of Shear Links

In any concrete beam, other than those of minor importance, shear reinforce-

ment, in the form of links, will be provided in addition to the longitudinal


102/297

92


bending reinforcement. The size and spacing of these links can be determined

using this program which calculates the link area/spacing ratio for any given

beam section and ultimate shear force.

3.3.1

Designpmcf?dwe

The design of shear links for a beam section is based on the equation

AS

WV - 4

-=

S

0.87f,

(3.21)

where A,, is the sum of the cross-sectional area of the legs of a link and sv is the

link spacing. The shear stress v at the beams cross-section is calculated from

and this value of v should not exceed a maximum value given by

0.8Vfc, or 5

N/mm2, whichever is the lesser.

The ultimate shear stress, v~, that can be resisted by the beam without shear

reinforcement is a function of the grade of concrete, the beams effective depth

and the percentage area of the longitudinal tension reinforcement at the

section considered. This tension reinforcement must extend at least an effec-

tive depth beyond the section or be adequately anchored at a support section.

The value of Y= s obtained from the formula

..=0.79(100 2 )"'(Y)'" -;t;;

where the steel ratio, 100 Aslbd, should not be taken as greater than 3 or less

than 0.15, and the effective depth d should not exceed 400 mm. -yrn is taken as

1.25

For characteristic concrete strengths in excess of 25 N/mm the value of V~

obtained from equation (3.22) can be multip lied by &/25), but in using this

multiplying figure

fc.

should not exceed 40 N/mm*.

Where the average shear stress is less than (v~ + 0.4) N/mm*, minimum

shear reinforcement should be provided according to the formula

0.4 b

0.87 f,

(3.23)

For members of minor structural importance, where Y < 0.5 v,, minimum

links can be omitted; but the program in this section calculates minimum link

requirements in al l cases where the average shear stress is less than (v~ + 0.4)

N/mm. The characteristic strength of the shear links should not be taken as

greater than 460 N/mm.


103/297


3.3.2 Flow diagram for the design of shear links program

93

3.3.3 Description of the design of shear links program

Lines

3c-170

190

Input of data.

Calculation of the average concrete shear stress, v = Vlbd.


104/297

94

200

230

25&26a

270

28&370


The program calculates the maximum shear stress, 0.8Vf=,. If the

value of Y exceeds this maximum stress, then the program displays

MAXIMUM SHEAR STRESS EXCEEDED and goes to the

end of the program.

The program goes to a subroutine at line 4CKKl to determine the

ult imate concrete shear stress, v,, using equation (3.22).

Minimum values for A,Js, are calculated.

Values of

A&,

are determined when the average shear stress

exceeds (v. + 0.4) N/mm.

Data and

A,,/s,

are printed.


105/297


3.3.5 List of variables for the design of shear links program


Reserved variables only used.


D9 : Effective depth

M : A string to display that A,,/s, is the minimum required

U : Equals concrete grade, FU

VC : Design concrete shear stress, v=

Vl : Maximum shear stress

3.3.6 Shear links example

Determine the required A&, ratio for a beam section 300 mm wide and

effective depth 550 mm subject to a shear force of 196 kN. The area of

longitudinal steel is 982 mm2, fen = 25 N/mm*, f, = 250 N/mm*.

Input:

TITLE



SHEAR LINKS, fyv - N/sq.mm

: SHEAR LINK EXAMPLE

: 25

: 250


106/297

96


BEAMS BREADTH, b - mm

: 300

BEAMS EFFECTIVE DEPTH, d - mm

: 550

AREA OF LONGITUDINAL TENSION

STEEL, As - sqmm

: 982

ULTIMATE SHEAR FORCE, V - kN : 196

Output:

SHEAR LINKS-A&, = 0.900

3.3.7 Further developments

(a) Write an additional subroutine to select a suitable size and spacing of shear

links to satisfy the calculated A,,/s, ratio. The links can be either automati-

cally selected from available sizes written into a DATA list; or user-

selected with the user selecting the size and the program calculating the

spacing.

(b)-Further develop this subroutine to reject those combinations of link sizes

and spacings which fail to meet the requirement that links must be spaced

no more than 0.75 d apart.

3.4 Design for Torsion

The design of a reinforced concrete beam to resist torsional moments is based

on a number of design formulae given in BS 8110. Torsional stresses are

resisted by the addition of torsional reinforcement in the form of closed links

and also additional longitudinal reinforcement. The required areas of both

types of reinforcement can be calculated with the aid of this program.

3.4.1 Design procedure

The design procedure for torsional design may be summarised as follows.

(i)

Calculate, by an appropriate method of structural analysis the torsional

moment, T, for which the section is to be designed.

(ii)

Calculate the torsional shear stress, Y,, according to the formula

2T

Vt =

h2min (Lx - hmiJ3)

where

h,in = the smaller dimension of the beam section

h

mar

= the larger dimension of the beam section.


107/297


97

(iii) If vt > O.O67Vf=, (with an upper limit of 0.40 N/mm), then torsional

reinforcement is required.

(iv) If vt + v > 0.80dff,, (with an upper limi t of 5.00 N/mm), then the

design is inadmissable. v is the shear stress due to the shear forces on the

section.

(v) In the case of small sections (y, < 550 mm) the design is inadmissible. if

Yt > 0.80VfC x Ji

550

where

y, = the larger dimension of a reinforcing link

(measured centre to centre).

(vi) Calculate the required torsional reinforcement, in the form of closed

links according to the formula

4,

T

->

0.8 XI YI (0.87 f,)

(3.25)

&

where

x, = the smaller dimension of a reinforcing link

f, = the characteristic strength of the links.

(vii) Calculate the link spacing which must be lim ited to the least ofx,, y,/2 or

200 mm.

(viii) Calculate the add tmnal longitudinal reinforcement required according

to the formula

69

4 - - (XI + Y,)

fY

(3.26)

S

where

f,, = characteristic strength of the longitudinal

reinforcement.

In the case of Ranged sections the section should be divided into com-

ponent rectangles and each component designed to carry a torsional

moment given by

(3.27)


108/297

98


3.4.2 Flow diagram for the torsion design pqram

Concrete grade FU

Number of com,mnents N

Hmax, min. H ( )

Number of shear component N1

Breadth and effect ive depth

Steel men$hs P( and F

Link dimensions Y1 ( )

andX,( )


109/297


3.4.3 Description of the torsion design program

Lines

1@350

38&500

51lK610

620-790

SO@820

All the data is entered at the begin

Reinforced Concrete Design by Computer - R. Hulse and W.H. Mosley

Documents

Transcript of Reinforced Concrete Design by Computer - R. Hulse and W.H. Mosley