REFERENCES

1. BS 5400 Part 2: 1978, Steel, Concrete and Composite Bridges- Specification for loads.

2. BS 5400 Part 4: 1990 Steel, Concrete and Composite Bridges- Code of Practice for

Design of Concrete Bridges.

3. Jayasinghe M.T.R., Lecture Notes given forM. Eng. Degree Course in Structural

Engineering Design.

4. Clark L.A., 1981 ,Concrete Bridge design to BS 5400, Construction Press London and

New York.

5. Hurst M.K., Nanyang, Pre-Stresses Concrete Design, Technological Institute Singapore

6. User Manual, SAP 2000 V14 Integrated Solution for Structural Analysis and Design.

7. :; \' ',(.

8.

9.

1 ng1tj~i !5orgicdc rJ~ti<b'.L' !• II U.

10. Gee, A.F., "Bridge winners and losers", The Structural Engineer, Vol: 65 A, No 4, pp

141-145, 1987.

11. Burgoyne, C.J ., Jayasinghe, M. T.R., Rationalization of section design philosophy for

prismatic pre-stressed concrete beams.

12. Swann, R.A ( 1972), "A future survey of concrete box spine-beam bridges", Technical

Report 469, Cement and Concrete Association, London, 1972.

13. Podonly, W. & Muller, J.M., Construction and design of pre-stressed concrete segmental

bridges, John Wiley & Sons, New York, 1982.

36

APPENDIXES

APPENDIX 1- Manual calculation for box girder

APPENDIX 2- Computer output of the box girder

APPENDIX 3- Computer output of the doubleT beam

APPENDIX 4- Comparison of the forces bending moment and stresses

37

CONTENTS OF MANUAL CALCULATION

A. Notations and their descriptions

B. Load calculations

C. Design of pre-stressed concrete Box Girder

1. Magnal Diagram

2. Calculation of number of ducts

3. Profile of individual ducts

4. Analysis of the Box Girder

5. Losses

6. Check for ultimate limit state

7. Design of End Blok and transverse reinforcement.

~efere nee Description

s 540

199(

I able

::) 540

199(

324

; 5401.)

199l

3.2.4

3.2 <

)ble,

5400

1990

3.2.2

1ble 2

Data

Density of Concrete, Pc = 24 kN/m3

Density of Screed Concrete, P.vc = 24 kN/m3

Density of Asphalt Concrete, Pac = 23.6 kN/m3

Pre stress Loss ratio, R = 0.8

Span of the bridge, I 30 m

Effective span of the bridge, /e = 29 m

Carriageway width, = 7.4 m

Number of lanes = 2

0-4 Width of the foot walk = 1500 mm

Characteristic Concrete cube strength of the beam, feu = 50 N/mm2

20 Concrete strength at (initial) transfer( at the age of 7 < fc; s · = 36 N/mm2

Characteristic strenght of high yield bars fyv = 460 N/mm2

Modulus of Elasticity of grade 50 concrete Ec = 34000 N/mm2

Characteristic strenght of mild steel bars fy = 250 N/mm2

1-4 Permissible stresses for class 2 member

a (2' Allowable tensile stress at service, larrin = 0.36xvfcu N/mm3

Characteristic Concrete cube strength of the beam, feu = 50 N/mm2

= o.36xv5o

farrin = 2.55 N/mm2

4 Allowable tensile stress at transfer, farrint = 1 N/mm2

(1) Allowable compressive stress at service, fatmX = Design load in Bending

(a) = 0.4xfcu

= 0.4x50 N/mm2

farmx = 20 N/mm2

4 Allowable compressive stress at transfer, !, _ Triangular or near Triangular DIIBX t - rfic::trin11tinn nf c:tr<=>c::c::

= 0.5xfci buts 0.4xfcu

:b) = 0.5x36

farmxt = 18 N/mm2

s 0.4x50

s 20

Output

farrin

2.55

N/mm2

fanint

1

N/mm2

fatmX

20

N/mm2

farm.xt

18

N/mm2

Reference I Description I Output

Details of Precast section

Thckness of bottom flange tb = 200 mm

Thckness of top flange II = 250 mm

Thckness of one web I,. = 350 mm

Width of the cantilever overhang we = 1000 mm

Width of bottom flange wb = 3200 mm

Width of top flange WI = 4400 mm

Overall depth of beam (Pre cast section) hp = 1667 mm

Number of web N = 2

Cross section of the brigde ~ 1.500M --1 __r_!l.225M E5 ''"~ f ,.,~ I r=}-. M I

_L 'l~r o.2sor0 ~ r~ o.200M

0.350M 2.500M ------j

~-200M 0.200M 1 0.200..;-r- 3.200M

Figure 1

Details of the Precast Section after grouting

I 4.400M I r 0.250r~

-J ~ Q.200M ~

::Jf290M r

-r[ 0.350r1=

0.3 50 M +----il------ 2.5001'1

~ Q.200M 0.200!'1 1 1-

I

Q.200M J 3.2001'1

Figure 2

.. ___ .___t_:_ ________________ __:_ __________ ..J_ __ ____.

---- -Reference 1 Description I Output

Cross sectional Area, AP = 2.740E+06 mm2

Total Depth, hp = 1.667E+03 mm

Total width of the top flange wt = 4.400E+03 mm

Total width of the bottom flange wb = 3.200E+03 mm Height to the top fiber from the neutral axis. Ypt = 7.150E+02 mm

Height to the bottom fiber from the neutral axis, Ypb = 9.520E+02 mm

Second Moment of Inertia along x axis, Jxp = 1.016E+12 mm4

Sectional Modulus of the top fiber, zpt = 1.421E+09 mm3

Sectional Modulus of the bottom fiber zpb = 1.067E+09 mm3

Weight of the beam per unit length, w - A xp xi

I wpgl pgl - c c

wpgl = 65.76 kN/m 65.76

kN/m Details of the Composite Sections

1.Edge beam

I 5.200M

. O.OF-1 -

~.350M ~ ---------.......

t I _(r 0.200M

Q.J50M 2.500M

__r__9.200M 0.200M ~ 1-

I I ~

. :)UM L_Q.132M

c--- f250M

f.- 1.000M ~1.667M 0.200;-r-

3.200M I l

Figure 3

Cross sectional Area, Ac = 3.45E+06 mm2

Total Depth, hcl = 1764 mm

Total width of the top flange, welt = 5200 mm

Total width of the bottom flange wclb = 3200 mm

Height to the top fiber from the neutral axis, Yc1t = 665 mm

Height to the bottom fiber from the neutral axis, Yclb = 1099 mm

Moment of Inertia, /xcl = 1.31E+12 mm4

Seactional Modulus of the top fiber, zeit = 1.97E+09 mm3

Sectional Modulus of the bottom fiber, zclb = 1.19E+09 mm3

I Weigl

Weight of the beam per unit length, welgl =A P x p c x 1 kN/m 82.80

Weigl = 82.80 kN/m I kN/m

Refer, :mce

) 5400

1990

I 6.2

-4

Description

Load CaJcuJations

Dead Loads

Dead Loads due to grecast beam

Weight of the beam per unit length, wpgl = 65.76 kN/m

Dead Loads due to comgosite beams

1.Edge beam

Weight of the beam per unit length, Weigl = 82.80 kN/m

Cross sectional Area of the Screed concrete, Asci = 7.15E+05 mm2

Weight of the screed concrete per unit length of bea w sci = AscxPscx1

wscl = 17.16 kN/m

Super imposed dead loads

Super imposed dead loads on edge beam

Cross sectional Area of the Asphalt concrete, Aocl = 1.85E+05 mm2

Weight of the Asphalt concrete, woe! = Aocl X Poe X 1

woe! = 4.37 kN/m

Weight of the hand railings per unit length of beam, whr = 0.585 kN/m

Cross sectonal area of the footwalk, Afw = 3.38E+05 mm2

Weight of the footwalk, wfw = AfwxPscx1

= 8.10 kN/m

Weight of the kerb per unit length of beam, W.v = 0.85 kN/m

HA Loading

HA Uniformly Distributed Load (UDL)

Carriage way width = 7.4 m

Number of notional lanes = 2

For loaded length up to 30 m HA UDL = 30 kN/mllane

HA UDL per unit width of the beam = 30x2x29 7.4x29

= 8.11 kN/m2

Output

wpgl

65.76

kN/m

Weigl

82.80

kN/m

wscl

17.16

kN/m

woe)

4.37

kN/m

whr

0.585

kN/m

wfw

8.10

kN/m

Akr

0.85

kN/m

--~efer' ence

's 54

19'

d 6

S54

199

7 1

s 54

199

d6.

00-4

'()

{ 2

)0-4

D

1 (a)

10-4

I .;

1

Description Output

Effective width for edge beam = 3700 mm

HA UDL on the edge beam = 8.11x3.7 HAUDL

= 30.00 kN/m 30.00

kN/m

HA Knife Edge Load (KEL)

KEL Load for per notional lane = 120 kN/Iane

KEL for a meter width of road = 120x2 7.4

= 32.43 kN/m

Effective width for edge beam = 3700 mm KEL

HA KEL on the edge beam = 120.00 kN 120.00

kN

Pedestrian Load

Pedestrian load per beam = 5 kN/m2

Pedestrian load per beam per unit length of bear = 5x1.5 I

= 7.5 kN/m 7.5

kN/m

HB Loading

Case I 300 300 300 300

~ 0 ~ :1.8: tR1 ... 1.8 ... 6.0 i R2

Figure 4

Where, 0 - Bridge Centre

300 a ... b

v ...

~ - .~ l ... _

Figure 5

From the theory ;

q1 = Pab I (a+b)

= 300*13.2*~30-13.2}

30 2218

= 2217.6 kNm kNm --- ------- - -

~eference

From Figure 5;

From Figure 6;

Description

qo1 = (q1 * Ll2)1b

= 1980 kNm

300

Q02=CI29

L

Figure 6

q2 = qo2 = =

(P * L/2 * L/2) I L

2250 kNm

......

~ a ~ ? ~ b • " > :

From the theory;

From Figure 7 ;

L

Figure 7

q3 =

=

=

qo3 =

=

Pab I (a+b)

300*21*(30-21!

30

1890 kNm

(q3 * U2)1b

1350 kNm

300 a ~ b .,

~ L

..... Figure 8

From the theory;

q4 = Pab I (a+b)

= 300*22.8*(30-22.8)

30

= 1641.6 kNm

Output

1980

kNm

2250

kNm

1890

kNm

1350

kNm

1642

kNm

----- -

teference

From Figure 8;

Moment @ mid span 0 ;

Similarlly;

Case II

1R1

300

Moment @ mid span 0 ;

Case Ill

~ .1.8

Description

qo4 = (q4 * U2)/b

= 1080 kNm

mq1 = qo1 + qo2 + qo3 + qo4

= 6660 kNm

300 300 300

i 0 u .. 3.0 ~~ 3.0 ~ 1.8 ...

Figure 9

mq2 = qo'1 + qo'2 + qo'3 + qo'4

qo4

= 6660 kNm

300 300 300 300

LAo u

i R2

1R1 ... 1.8~ 1.5~~ 4.5 .. 1.8 ... j R2

Moment @ mid span 0 ;

Similarfly;

Moment @ mid span A ;

Section Check

1.Edge beam

Figure 10

mq3 = qo"1 + qo"2 -t qo"3 ; qo"4

= 6660 kNm

mq4 = qo"11 +qo"22· qo"33 . +qo"44

= 6750 kNm

Moment at mid span due to self weight of the beam, M gi = Wc1g 1 x 12

8

Moment at mid span due to screed concrete,

M gl = 65. 76x292

8

= 6913.02 kNm

Mg2 = wscl xP 8

Output

1080

kNm

6660

kNm

6660

kNm

6660

kNm

6750

kNm

Mg2

6913.02

kNm

_____ ....._ ___________________________ .....J...-----1

··-·

Reference Description Output

Mcz = 17.16x292

8 Mcz

= 1803.95 kNm 1803.95

kNm

Moment at mid span due to Asphalt concrete Macl = Wacl X /2

8

= 4.37x292

8 Macl

= 458.98 kNm 458.98

kNm

Moment at mid span due to hand railings M = whr x/2 hr

8

= 0.585x292

8 Mhr

= 61.50 kNm 61.50

kNm

Moment at mid span due to footwalk Mfw = wfw x/2

8

= 8.10x292

8 Mfw

= 851.51 kNm 851.51

kNm

Moment at mid span due to kerbs M~cr = wkr x/2

8

= 0.85x292

8 Mkr

= 89.36 kNm 89.36

kNm

Moments due to HA MHAUDLCI

Moment at mid span due to HA UDL, M HAUDLCI = 3153.75 kNm 3153.75

kNm

Moment at mid span due to HA KEL, M HAKELCI = 870.00 kNm MHAKELCI

870.00

Moment at mid span due to pedestrian load, M PL = 788.44 kNm kNm

MPL

788.44

kNm - .

-·---·-

Reference Description Output

Load Amount (KN/m) Moment at mid span

(KNm)

"01/) Unit weight of pre cast section, wpg1 65.76 6913.02 ro-o G> ro

Unit weight of screed concrete, W5c 17.16 1803.95 o.3 "0

Unit weight of Asphalt concrete, Wac Q) 4.37 458.98 If)

0 If)

a.."' E ro Unit weight of handrails, Whr 0.585 61.50 - 0 L.

...J Unit weight of kerbs, Wkr 0.85 89.36 Q)"' a..ro -:::s Q) Unit weight of foot walk, Wrw 8.10 851.51 U)Q

Uniformly distributed pedestrian live load 7.50 788.44

HA UDL on the beam 30.00 3153.75 If)

HA KEL on the beam u 120.00 870.00 ro 0

...J HB Loading - Case I 6660.00 Q) > ::::; HB Loading- Case II 6660.00

HB Loading- Case Ill(@ mid span) 6660.00

HB Loading -Case Ill(@ point "A") 6750.00

Table 1

Table 1 Total moment due to HA loading = 4023.75 kNm '

Maximum moment due to HB loading = 6750.00 kNm

If compare the moments due to HA and HB, moment due to HB is the higher value than the

moment due to HA loading. But for the design purpose should use moment due to HA loading.

By using Table 1

M gl = 6913.02 kNm Mgl

6913.02

Mg2 = 1803.95 kNm kNm

( M.., + M" + M1• + M,,J Mg2

Mq = x120 1803.95 MHA UDL + MHAKEL

kNm

+MPL xl.OO Mq

= 7370.55 kNm 7370.55

kNm

- L_ _____ . . -- ---------- -

~eference Description

Design of prestressed concrete beam

Design criterian:Stress in concrete should not exceed the allowable values during the life time of

structure.

Sign convention:

Axial compressive force positive

Distances measured upwards from the neutral axis positive

Compresive stress positive

Transfer condition stresses

Top fiber (fur tensi on)

(p) (-PxeJ (Mg1J A + Z pi + z pi 2:: -fa mini

( :) -( ~:: J + ( ~:,' J ~-/ .. ,., .......... ·········· .............. (!)

Bottom fiber (fur compressio n)

(p) (PxeJ (-Mg1J A + Z ph + Z ph < fa maxi

(:)+ ( ~:e J-( ~~I J < fomnr········ ·········· .............. (2)

Service condition stresses

Top fiber (for compressio n)

( R:P)+( -R;:xe ]+( M,,;#M,2 J+( ::) < fomn

( R: p) -( R X:,x e J +( M •';#M,, ]+( ::) < / •• ~·········· .......... ·········· .... (J)

Bottom fiber (fur temi on)

(RxP)+(RxPxeJ+(-(Mg1 +Mg2 )J+(-MqJ>- . A Z z z famm

ph ph ph

( R ~ p)+( R~:xe J-( M,,2+"M,, J-( ~ J > _ / •• , •.................................. (4)

.

Output

•< ____

teference Description Output

M gl + M g2 - R X M gl

Mq = 1.87E+08

R X famin t + famax - z cit

= 1.87E+08 < zp, 1.42E+09

Section is ok at the top fiber

M gl + M g2 - R X M gl

Mq = 2.96E+08

R X famaxt + famin- z clh

= 2.96E+08 <Zpb 1.07E+09

Section is ok at the bottom fiber

--~-

·---- -!terence Description Output

Magnel digram at the mid span of the bridge (1) =>

e ~ ( z~l ]+ zpl X:anint +;I·········· ••••······ .............. (S)

(2) =>

e ~ -( z~h J + z ph x;armx I+ M;l .................... ·········· .... (6)

(3) =>

Mq x(____e_)

e~(Z~~ J- ZP~x:;rrm +((Mg~:~g2)J+ Rx:ct .................................. (?) [ z l (4)=>

M X ( ______e_)

e~-(z~h J- ZP~x:;ni• +((Mg~:~g2)J+ qRx!ch .................................. (8) [ z l 1.Edge Beam

farrin = 2.55 N/mm2

fanint = 1 N/mm2

hrmx = 20 N/mm2

hrraxt = 18 N/mm2

R = 0.80 AP = 2.74E+06 mm2

zpl = 1.42E+09 mm3

zph = 1.07E+09 mm3

zeit = 1.97E+09 mm3

zclb = 1.19E+09 mm3

Mgi = 6913.02 kNm

Mg2 = 1803.95 kNm

' Mq = 7370.55 kNm

e/(mm) -665 0 1099 2500

1/P1 -1.12E-07 -4.92E-08 5.51E-08 1.88E-07

1/P2 -1.05E-08 1.49E-08 5.70E-08 1.11E-07

11P3 6.58E-08 2.88E-08 -3.23E-08 -1.10E-07

11P4 -1.75E-08 2.47E-08 9.45E-08 1.83E-07 .

---

~terence

I -665 em ax

1/P -1.5E-07

lemin 1099

1/P -1.5E-07

3.E-07 1

2.E-071

~:-.:

2.E-07

l.E-07

-1000

i(~ -2.E-07

-2.E-07

-665

1.6E-07

1099

1.6E-07

z ::::::: a. -....

Description

Chosen1/P 4.00E-08 4E-08

e 0 1000

-+-1/Pl

~ --a--1/P2

__._1/P3

·""""*'·-1/P4

Eccentricity,e/(m"!), ------,-- ·-r--- --- , --~- E max

I •

2000 2500 3000 . ·!!!- Emrn

- -6 - Chosenl/P

Chosen e

Prestressing force for the section,P = 2s,ooo,ooo N

Eccentricity,e = 500 mm

Feasible Tendon Profile Zone

Bendingmomentat a point x from one support in a simply supportedbeamof length~ due to a uniformly

distributerlload w, M • ..u

wxlxx wxx2

IM =-----xudl 2 2

! Bendingmomentat a point x from one supportin a simply supportedbeamof length I due to a point load load P, Mxpr

Pxx(l-x) Mrpl= [

j1.Edge Beam

l, = 29 m

Weigl = 65.76 kN/m

wscl = 17.16 kN/m

wac! = 4.37 kN/m

whr = 0.585 kN/m

wfo' = 8.10 kN/m

WAr = 0.85 kN/m

HA UDL on the edge beam = 30.00 kN/m

HA KEL on the edge beam = 120.00 kN

udl due to pedestrian load = 7.5 kN/m

Output

p

25000000

N

---Ference Description Output

Length along the Beam /(m),X 0 3.625 7.25 10.875 14.5

Moment due to self wt of the beam,Mg1 /(Nmm) 0 3.0E+09 5.2E+09 6.5E+09 6.9E+09

Moment due to Screed Concrete,Mg2/(Nmm) 0 7.9E+08 1.4E+09 1.7E+09 1.8E+09

Total Moment due to dead load/(Nmm) 0 3.8E+09 6.5E+09 8.2E+09 8.7E+09

Moment due to Asphalt Concrete, /(Nmm) 0 2.0E+08 3.4E+08 4.3E+08 4.6E+08

Moment due to hand raii/(Nmm) 0 2.7E+07 4.6E+07 5.8E+07 6.1E+07

Moment due to footwalki(Nmm) 0 3.7E+08 6.4E+08 8.0E+08 8.5E+08

Moment due to pedestrian load/I(Nmm) 0 3.4E+08 5.9E+08 7.4E+08 7.9E+08

Moment due to kerb/(Nmm) 0 3.9E+07 6.7E+07. 8.4E+07 8.9E+07

Total Moment due to super imposed load/(Nmm) 0 9.5E+08 1.6E+09 2.0E+09 2.2E+09

Moment due to live loads,HAUDL/(Nmm) 0 1.4E+09 2.4E+09 3.0E+09 3.2E+09

Moment due to live loads,HA KEL/(Nmm) 0 3.8E+08 6.5E+08 8.2E+08 8.7E+08

Total Moment due to live loads 0 1.8E+09 3.0E+09 3.8E+09 4.0E+09

Eccentricity,e1/(mm) 575 696 783 835 852

Eccentricity,e2/(mm) 379 500 586 638 655

Eccentricity,e3/(mm) -902 -613 -406 -281 -240

Eccentricity,e4/(mm) -525 -212 12 147 192

Emin/(mm) -902 -613 -406 -281 -240

Emax/(mm) 575 696 783 835 852

i ----l

Cable Zone

-1000~ --+- Emin/(mm)

-Emax/(mm)

E -500 ~-E

::::::::: 2 4 6 8 lU 16 cu ChianagiaT'ong the beam/(m) ~ 0 ·;::; ·.: .... c cu

500 L u u LLI - - -- -1000-

. -"

i ·---

~eference

:s 5896

1980

-able 6

.cl. 20

s 5400

199l'

16.i '

'--

I

i

Calculation of number of ducts

1.Edge beam

Prestressing force,P

Type of strand

Nominal tensile strength

Nominal steel area

Description

Specified characteristic breaking load

= 25,000,000 N

= BS 5896-3 super strand-1770-15. 7 -relax 1

= 1770

= 150

= 265,500

N/mm2

mm2

N

Output

Maximum prestress force allowed for tendons = 70%xCharacteristic strength

Number of tendons needed

External diametre of duct

Internal diametre of duct

Number of strands per duct

Number of ducts

:n

he h

BJ I le

X

= 185,850 N

= Prestressing force

Maximum prestress force allowed for tendons

= 25.000.000

185850

= 135 Nos.

= 60

=50

=7

= 135 7

= 20

t y

0 ho

/A

mm

mm

Nos.

Nos.

Number of, ducts

20

Nos.

ference

>40(

90

! 36

e 371

Description Output

Assumed equation for parabolic portion Y=aX 2 +hX+c

I At A; X = O,Y = ho

X = 0 dy = 0 'dx

At 8; X = l dy = _!__ 'dx n

So; b = 0

c = ho

i dy = 2ax dx

At 8; dy = _!__ dx n 1 - = 2ax n

1 a =

2nl

Assumed equation for straight portion; Y=mX+c

At 8; dy 1 X=l-=m=-

'dx n

X= lc,Y =he Straight y = (X - le) + he

n

Curve xz

Y=--+ho 2nl

At 8; X= l,and,Y = h

n= (2le -!)

2(he-ho)

. . (he- ho) 2 Equation of Parabolic curve Y = ( ) X + ho

l 2/e -I Equation of Straight line

y = 2(he-hoXX-le) +he 2/e-1

Minimum cover to the ducts at the end of parabolic section = 50 mm

Minimum spacing between the centrelines of ducts at the = 140 mm end of the parabolic section

------.-------------------------------------,.-----, 1ference Description

Profiles of individual ducts

:; 5400 I Cover to ducts

tart 4 I Minimum cover to ducts

3.8.2 3

Clear distance between ducts

Maximum size of coarse aggregate, hagg

hagg +5mm

. 5400 I vertical internal dimension of the duct

art 4 I Horizontal internal dimension of the duct

i.8 3 3 I Clear distance between ducts

Height to centorid of a duct from bottom fibre at level n

Cross sectional area of a duct at level n

=50

= 19

= 24

=50

=50

=50

= Yn

= AJn

Height to the bottom fiber from the neutral axis (Composite = Y pb -

Output

Cover

mm 50

mm

mm

mm Clear

mm spacing

mm 50

mm mm

i :·

:.. t.'

Centroid of all ducts from the bottom fibre, y = (Adl x Yt + Ad2 x Y2 + ····· + Adn x Y I)

Eccentricity of all ducts in the section considered,e

At mid span

Chainage

Eccentricity obtained from the Magnel Diagram

External diametre of the duct

Number of ducts

Strands/duct

Height to the bottom fiber from the neutral axis,

--t

(Adl + Ad2 + ···· + Adn)

= y ph - y

=0

= 500

=50

= 20

=7

Ypb = 952

m

mm

mm

= 25,000,000 N

tference Description Output

Duct Cross sectional

Duds No. of duds No of Strands area of a position

ducU(mm2)

y1 100 12 84 1963

y2 300 2 14 1963

y3 500 2 14 1963 -y4 700 2 14 1963

y5 900 2 14 1963

Total 20

Centrad of all ducts from bottom fiber = 300 mm

Resultant Eccentricity of all tendons,e = 652 mm

Profile5 ~

Projile4~ Profile 3 -------

Profile 2 ~--------~ r--!---

Profile~ -

lo lp Is

Zone I Zone2 Zone3

Length of Eccentricity at

Length of Eccentricity at end

Length of Zone Lenth up to mid Number of start of zone 1 of zone 2 (mid Number of ducts

Zone 1/(m) /(mm)

Zone 2/(m) span) /(mm)

3/(m) span /(m) strands

Profile 1 1 852 8 852 5.5 14.5 12 84

Profile 2 1 542 8 652 5.5 14.5 2 14

Profile 3 1 232 8 452 5.5 14.5 2 14

Profile 4 1 -78 8 252 5.5 14.5 2 14

Profile 5 1 -388 8 52 5.5 14.5 2 14

Total 20 140

tferen ce Description Output

Chainage Eccentricity/(mm)

I (m) Profile 1 Profile 2 Profile 3 Profile 4 Profile 5 Resultant

0 852 542 232 -78 -388 542

1 852 564 276 -12 -300 564

2 852 585 317 50 -218 585

3 852 603 353 104 -146 603

4 852 618 383 149 -86 618

5 852 630 408 186 -36 630

6 852 640 427 215 3 640

7 852 647 441 236 30 647

8 852 651 449 248 47 651

9 852 652 452 252 52 652

! 10 852 652 452 252 52 652

I 11 852 652 452 252 52 652 I 12 852 652 452 252 52 652

13 852 652 452 252 52 652

14 852 652 452 252 52 652

14.5 852 652 452 252 52 652 100 410 720 1030 1340

I

'

-1000 Eccentricity of cables along the beam • • • Emin

~ ~ .. - •- Emax -800 ...... .. --600

.. .. ---ts- Profile 1 ... E

.. .. ... ... -)'( ·Profile 2 E-400 ..... -~ 6 Cha~a9~in~~ • • "1£ • "14• - --------Profile 3 Cl)-200

~ 0 ~ --1-Profile 4 (.)

~~ -ProfileS :s 200 c:: G) 400 - ..... "' .... - ..... ~-Resultant (.) ~

JJ 600) 1·- ·; - ': .:.-:'-."':-:"*·-----#-~-- "'' . +- --~~ -~ ...---*·--~ 800 - • • -.A. • • - • • • - _Ill -- - - - - - - - - - - - -~

1000

i - ------

terence Description Output

--Analysis of the beam

-------·- ~~- -----~--------- -- ----- --·- ------ -----·----- ··------------- ---------- ~--~-- 1----------

Post tensioning sequence

feu =50 N/mm2

Stage 1 Stage 2 Stage 3 Stage4 Stage 5

Age 14 days 1 month

5400-4 Strength 36 50

990

Jle 20

When cables are prestressed,all cables are not stresses at once.Differents cable sets are chosen for stressing procedure.A post tensioning seaquence is introduced and cables are stressed taking each set of cables at once.At each stage after tensioning stresses are checked at top and bottom fiber.

Number of cables in each set of cables Total no of Cable cables profile Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6

Profile 1 8 4 12

Profile 2 2 2

Profile 3 2 2

Profile 4 2 2

Profile 5 2 2

Total 20

Cables are stressed according to the the sequence chosen in the above table

The sectional properties at diifferent sections of the beam changes according to the tendon profiles at each section of the beam. Therefore the section properties has to be found at each section of the beam before grouting of the beam.when tendons are stressed at stage one( cable set 1) at transfer the section will have all the ducts without grout. But when the stressing is done at stage two in cable set two,the the ducts which consists of the cables that are stresses at stage one wil be grouted,thus section properties will be changed. Therefore section properties at each stage of stressing has to be calculated as well.

Calculation of sectional l!rol!erties

L J 0

- Cl 0

- C2 Y5 0

Y4

Y3 0 G--

Y2

!fl -------------

tference I Description I Output

Cross sectional Area of the pre cast section AP' = A - A - A - - A . d! d2 •••• dn

before groutmg P

Height to the centriod of precast section after grouting y p

Height to the centriod of precast section before grouting Y' -

Centroidof the precast section before grouting cl

Centroidof the precast section after grouting c2

Y' = (Ap X Yp- Adl X Y1- Ad2 X Y2····.Adn X Yn) , - AP

Y' = (A p X y p - L A di X y i)

Ap

LAd; xy; = Adl xyl +Ad2 xy2 + ..... +An xyn

I xp - Moment of inertia of precat section after grouting

, fxp - Moment of inertia of precat section before grouting

From parallel axis theorem

Ixp'= Ixp + (Y'- Y)2 x Ap -{(Idl + Adl x (i''- ~)2 + /~2 + Ad2 x (i''- y2)2·····}

····· + 1dn + Ad2 x (Y'- Yn)

Lfdi = Jdl +fd2 + ... +fdn

LAd; X (Y'- y;)2 = Adl X (Y'- YI}2 + A2 X (Y'- Y2)2 + .. + Adn X (Y'- Yn)2

, - -fxp = fxp +(Y' -Y) 2

X Ap- Lfdi- 'LAd;

moment of inertia of a duct of diametre d I d

with respect to x axis

Area of a duct of diametre d Ad

= mf4 -64

mf2 = -~

4

L----------------------------------------------~----~

eference Description

1.Edge beam

At mid span (Chainage 14.5 m)

Cross sectional Ducts

Duct I No.ofducts INoofStrands Externaldiametre, areaofa ITotaiAreaofthel Ad;XY; position at one level,n of duct/(mm) 2 ducts /(mm2)

ductl(mm) Ad

y1 100 12 84 60 I 2827 I 33929 I 3,392,920

y2 300 2 14 60 I 2"827 I 5655 I 1,696,460

y3 500 2 14 60 I 2827 I 5655 I 2,827,433

y4 700 2 14 60 I 2827 I 5655 I 3,958,407

y5 900 2 14 60 I 2827 I 5655 I 5,089,380

Total I 20 I 140 I I 56549 I 16964600

IAd;XY;

I

Ixp = I xp + ( y I - y) 2 X A p - L I di - L Adi X (Y I - y i) 2

yl = (A p X y p - I A di X y i) I

Cross sectional Area of the pre cast section after grouting

yp

AP

Adl + Ad2 + ···· + Adn

Height to the centriod of precast section before grouting

A~

IAd;XY;

Y'

y~_y

Second moment of area of a circle with diametre d around

= Ypb

= 952

= 2.74E+06

= 56549

AP

mm

mm2

= A P - Ad! - Ad2 - ···· - Adn

= 2.68E+06 mm2

= 16,964,600 mm3

= 966 mm

= 14 mm

its centre = 7ld4

64

External diametre of a duct = 60

Second moment of area of a duct around its centre = 636,173 mm4

Output

Y' 966

mm

·----eference Description Output

Duct No. of ducts External Id nxld (Y'- y;)z Adi X (Y'- y;) Ducts position at one level,n diametre of

ducU(mm) /(mm4)

y1 100 12 60 636,173 7634070 749505 25430111423

y2 300 2 60 636,173 1272345 443209 2506289580

y3 500 2 60 636,173 1272345 216913 1226616598

y4 700 2 60 636,173 1272345 70618 399332958

y5 900 2 60 636,173 1272345 4322 24438660

Total 20 12723450 29586789218

IIdi L(Adi x(Y'- Y;)2)

_LJdi = 12,723,450 mm4

L(Ad; x(Y'- Y;)2) = 3.0.E+10 mm4

Jxp = 1.0E+12 mm4

, = Ixp + (Y'- .YY X Ap- L Idi- L Adi X (Y'- Y;)2 Jxp

,,

9.87E+11 , Jxp = 9.87E+11 mm4

mm4

, Ypb

Total depth of the precast section = 1667 mm 966 I Y' mm Ypb =

= 966 mm

Ypt I

Ypt1 = 701 mm 701

, mm , I

zp, =~ Ypt zp,~

= 1.41E+09 mm3 1.41E+09

, mm3

' Jxp zpb =--,

Ypb zpb I

= 1.02E+09 mm3 1.02E+09

mm3

tference Description Output

Similarly,

when stressing is done at stage 1 Mid Span

At qarter At edge of span beam

14.5 7.25 0

Cross sectonal area of the precast section before A~ 2.68E+06 2.68E+06 2.68E+06 grouting/(mm2

)

, 9.87E+11 9.87E+11 9.96E+11

Second moment of area before grouting/(mm4) Jxp

Heght to the top fiber from the neutral axis/(mm) Ypt I 701 701 704

Heght to the bottom fiber from the neutral axis/(mm) 966 966 963

Ypb I

Sectional modulus at the top fiber of the section before 1.41E+09 1.41E+09 1.42E+09

grouting/(mm3)

- zptl

Sectional modulus at the bottom fiber of the section before 1.02E+09 1.02E+09 1.03E+09

grouting/(mm3) zpb

I

Moment due to self weight before grouting/(kNm) Mgl 6770 5078 0

when stressing is done at stage 2 Mid Span

At qarter ·At edge of span beam

14.5 7.25 0

Cross sectonal area of the precast section before

A~ 2.72E+06 2.72E+06 2.72E+06

grouting/(mm2)

Second moment of area before grouting/(mm4)

, 1.01E+12 1.01E+12 1.01E+12

]xp

Heght to the top fiber from the neutral axis/(mm) 711 711 712

Ypt1

Heght to the bottom fiber from the neutral axis/(mm) y 1 956 956 955

pb

Sectional modulus at the top fiber of the section before 1.42E+09 1.42E+09 1.42E+09 grouting/(mm3

) zpll

Sectional modulus at the bottom fiber of the section before 1.05E+09 1.05E+09 1.05E+09 grouting/(mm3

) zpb I

, Moment due to self weight before grouting/(kNm) Mel 6870 5153 0

' i l

I

-------.-------------------------------------------------------------------------------.-------, eference Description

Prestressing force along a cable changes from point to pont because of friction present Therefore the prestressing force along the cable is calculated as follows

Friction in the duct due to unintentional variation from the specified profile

: 5400-4

1990 I Prestressing force at distance x from the jack ~ = Poe -Kx Equation 31

67 3 3j where

Kx ~ 0.2,e-Kx = 1-Kx

P. Pre stressing force in the tendon at the 0

- jacking end

K - constant depending on the type of duct

Friction in the duct due to curvature of the tendon

5400-4

990 I Prestressing force at distance x from the jack

3.7.3.4

Prestressing force alonQ the profile 1 I I

Start Zone Chainage length

Zone 1 0 1

Zone2 1 8

~ne3 9 5.5

-px

px = Poe rP,

where

Equation 32

-px

J.LX ~ 0.2, e rps =I- J.LX rps rps

(Kx + px) ~ 0.2, rps

-(Kx+JLX)

e rps = 1- (Kx + JlX)

End

rps

J.l - Coefficient of friction

rps - Radius of curvature,R

Chanage

1

9

14.5

Output

eferem.e

540u--

199t

" .., J,l

i400

390

733 & 734

i

4j t

I

I

I

I

Description Output

Zone 1 is a staright section px = Poe-Kx

Equation 31 , where

Kx :S: 0.2,e-Kx = 1- Kx

Po = 1300950 N

Start Chainage X :0 m

-K = 0.0033

Kx = 0.000 < 0.2

Therefore,

1-kx = 1.000

p = X

Prestressing force at the beam edge px = 1300950 N 1,300,950

N

End Chanage X = 1 m

Kx = 0.0033 < 0.2 ok

eKx = 1-Kx

= 0.9967 p = X

px = 1296657 N 1,296,657

N

Zone 2 has a curvature

-( J.IX +Kx)

~=Poe rps

Equation 31 and 32, where

(Kx+ ~) :s:; 0.2, e -(Kx+JIX)

rp, = 1-(Kx+ ,ux) rps rps

Start Chainage = 1 m

Po = 1296657 N

for steel moving on steel J.L = 0.3

rps =R

Radius of curvature at the end of zone 2 = I -R

1 -R = 2.70E-06 mm

eferenc e Description Output

Therefore,

R = 370.37 m

At the end of zone 2, Chainage = 9 m

X

f..K -+Kx = 0.0065 < 0.2 ok rps -(Kx + JiX)

- ( J.iX e rps -1- Kx+-) p = rps

X

1,288,255

px = 1 ,288,255 N N

i

Zone 3 is a straight section px = Poe-Kx I

where

Kx ~ 0.2,e-Kr = 1-Kx

I Po = 1,288,255 N

1 At the end of zone 3, Chianage = 14.5 m i

X = 5.50 m

Kx = 0.01815 < 0.2 ok

P= X 1264873 N

Chianage/(_mj_ 0 1 9 14.5

t-'restressmg rorce or 1,300,950 1,296,657 1,288,255 1,264,873 the orofile 2HN}

t· Beam edg_e Quarter span Midspan

! Chianage/(m) 0.0 2.0 7.25 12.0 14.5 t -0 Profile 1 1300950 1292364 1269825 1249432 1238700 Q) ...--. oZ ._ ........

Profile 2 1300950 1295607 1290093 1275501 1264873 o--- ..--OlQ)

1289803 -~ li= Profile 3 1300950 1295560 1275134 1264509 en o en ._ Q)O. Profile 4 1300950 1295045 1286584 1271054 1260463 -= Q) en..c Q) ......

Profile 5 1300950 1294524 1283324 1266923 1256366 ._ a_

_j_

,-Reference I Description

At mid span

Chosen cables for tension in 1

Profile I Duct Number of cables Prestressing

1

T t 1 t tensioned from each . o a orce

position force 1n one t 1 11 Force X y

name y/(mm) profile in stage 1 cable a one eve

Profile 1 I 100 I 8 1,238,700 I 9909596.341 990959634

Profile 2 I 300 I 2 1,264,873 I 2529745.4 I 758923620

Profile 3 I 500 I 2 1,264,509 12529017.4511264508723

Profile 4 I 700 I 0 1,260,463 J 0

0 I 0

Profile 5 I 900 I 2 1,256,366 12512732.6712261459407

Total I 17481092 15275851385

Total prestressing force at stage one= 17,481,092 N

Centroid of forces from the bottom of the beam = L Force x Y Totalforce

= 5275851385

17481091.9

yf = 302 mm

Eccentricity = Y'-Y f

Y' =966 mm

Eccentricity of force = 664 mm

Calculation of stresses

Stage 1

Pre-cats section before grouting at transfer condition-Mid span

Neutral Axis Level

PIA

8

8

-Pxe/Zpt' Mg l/Zpt'

~ ~ ~~

Pxe/Zpb' -Mgl/Zpb'

6.514 O.OOOE+OO 4.80

Stress at top most fibre = _!_ _ P x e + M gl , , , Ap zpt zpt

I Output

·.Reference Description

P = 17,481,092 N

Eccentricity of force = 664 mm

, A , 1 ( l,x x' J Mg, = xp X X---

p c 2 2

A~ = 2.68E+06 mm2

Pc = 24 kN/m3

[e = 29 m

At mid span x = 14.5 ,

Mg, = 6770

zpr' = 1.41E+09

zpb ' = 1.02E+09

Stress at top most fibre = 3.08

Allowable tensile stress at transfer, = -1

m

kNm

mm3

mm3

N/mm2

N/mm2

Stress at the bottom most fibre= P P x e M gl --+ ---1 I 1

AP z pb Z pb

Stress at the bottom most fibre = 11.26

Allowable compressive stress at transfer, = 18

N/mm2

N/mm2

P Pxe Mgl Stress at tendon level ----, + , - --,

, Z pe

AP Zpe Zpe

- J xp

e

,

= 1.49E+09 mm3

Stress at tendon level = 9. 77 N/mm2

Output

3.08

N/mm2

11.26

N/mm2

10

N/mm2

··~---------------------j_ _ _j

-·-·--~eference Description Output

Similarly

Stage 1 Midspan Quarter Span Beam Edge

Prestressing force,P/(N) 17481091.9 17,885,037 18,213,300 ,

Moment due to self weight before grouting/(kNm} Mal 6770 5078 0

Eccentricity of the force,e/(mm) 664 660 553

Sectional modulus at the top fiber of the section before 1.41E+09 1.41E+09 1.42E+09

grouting/(mm3) zpt,

Sectional modulus at the bottom fiber of the section before - 1.02E+09 1.02E+09 1.03E+09

grouting/(mm3) zpb '

, 9.87E+11 9.87E+11 9.96E+11 Second moment of area before grouting/(mm4

) /xp

Sectional modulus at the centroid of force before 1.49E+09 1.50E+09 1.80E+09

grouting/(mm3)

Cross sectonal area of the precast section before A~

2.68E+06 2.68E+06 2.68E+06

grouting/(mm2 )

Stress at the top most fibre/(N/mm2) 3.08 1.88 -0.33

Stress at bottom most fiber/(N/mm2)

11.25 13.25 16.53

Stress at tendon leveV(N/mm2) 9.77 11.16 12.39

Stage 2 Midspan Quarter Span Beam Edge

Prestressing force,P/(N) 7475724.28 7,652;467 7,805,700

Moment due to self weight before grouting!(kNm} M _,' 6870 5153 0

Eccentricity of the force of stage 2,e/(mm) 654 650 545

Sectional modulus at the top fiber of the section before 1.42E+09 1.42E+09 1.42E+09

grouting/(mm3) zpt,

Sectional modulus at the bottom fiber of the section before 1.05E+09 1.05E+09 1.05E+09

grouting/(mm3) zpb I

, 9.87E+11 9.87E+11 9.96E+11 Second moment of area before grouting/(mm4

) /xp

Sectional modulus at the centroid of force before 1.51E+09 1.52E+09 1.83E+09

grouting/(mm3)

Cross sectonal area of the precast section before A~ 2.72E+06 2.72E+06 2.72E+06

grouting/(mm2 )

Stress at the top most fibre/(N/mm2) -0.70 -0.70 -0.14

Stress at bottom most fiber/(N/mm2)

7.38 7.53 6.90

Stress at tendon level of the cables in stage 2/(N/mm2)

5.98 6.09 5.20

Eccentricity of the force of stage 1 ,e/(mm) 654 650 545

Stress at tendon level of the cables in stage1/(N/mm2) 5.99 6.09 5.20

r------.--------------------------------------------------------------~----~ Description .,Reference

BS 5400

Part 4

1990

ci.6J.2. &

cl.6. 7.2.3

Short term prestress losses

A. loss of Prestress due to elastc defonnaion of concrete

Strain in concrete, = 8 c

(J"c &c=E

_c

O" c - Stress of concrete

& c - Strain in concrete

Ec - Modulus of Elasticity of concrete

Strain in concrete = &s

/l(J" s & =--­

s E s

& s - Strain in concrete

llO" s - Loss of prestress in steel

Es - Modulus of Elasticity of steel

At the tendon level Strain in steel = Strain in concrete

&s =Ec

/l(J"s = (J"c

Es E c

Loss of force in the steel, llP = !lO" x A s s

Cross sectional area of steel = As

= (J"c x-As XEs

Ec

.......... ()

Since the tensioning of the steel is done gradually during post tensioning, the stress in tendons are taken as the half of the stress in the steel for calculation of prestress loss

Loss of prestressing force= O.S x O"c x As xEs Ec

Output

r--------,------------------------------------------------------------------------------~------~ .~eference

3S 5400-4

1990

~.6.7.2.3.

I Table3

;r.4.3.2.2.

Figure2

Stage 1

Chainage/(m)

14.5 (mid span)

7.25 (quarter span)

0 (beam edge)

Average stress along the cable/Nmm, u

c

Description

Stress in concete at tendon level

/(N/mm2)

9.77

11.16

12.39

11

Cross sectional area of steel = Cross sectional area of one tendon X A number of tendons stresses

s

Cross sectional area of one tendon = 150 mm2

Number of tendons needed = 98

As = 14700 mm2

Characteristic Concrete cube strength at transfer = 36

Modulus of Elasticity of concrete E c = 29.8

= 29,800

Modulus of Elasticity of steel Es = 200

= 200,000

N/mm2

kN/mm2

N/mm2

KN/mm2

N/mm2

Loss of prestressing force = 0.5 x Uc x As xEs Ec

Loss of pre stress

= 547,823 N

Loss of stress due to direct force loss = AP A' p

= 547,823

2.68E+06

= 0.20

loss of stress due to loss of moment at the top fiber = AP x e zp,'

N/mm2

after 7 days

Output

AP

547,823

N

'-.• Reference Description Output

AP = 547,823 N

e = 664 mm

zpl, = 1.41E+09 mm3

zpb ' mm3 = 1.02E+09

loss of stress due to loss of moment at the top fiber= -0.26 N/mm2 -0.05

Loss of stress due to loss of moment at the bottom fiber = 0

M x e zpb '

= 0.36 N/mm2 0.56

Stresses after stage 1 stressing -Mid span,

Stress at the top most fibre/(N/mm2) = 3.08

Stress at bottom most fiber/(N/mm2) = 11.25

Stress at tendon leveii(N/mm2) = 9. 77

Stresses after the losses,

Resultant stresses at top fiber = 2.62 N/mm2

Resultant stresses at botttom fiber = 1 0.69 N/mm2

Similarly,

Mid Span Average

Loss of stress Loss of stress

Loss of stress due Cables stress along M due to direct

due to loss of to loss of moment

considered the A. moment at the cable/N/mm2 mm2 N

force loss top fiber

at the bottom fiber

Stage 1 St1 cbls 11.11 14700 547823 0.20 -0.26 0.36

St2 cbls 5.76 6300 121668 0.04 -0.06 0.08 Stage 2

St1 cbls 5.76 14700 283968 0.10 -0.13 0.18

Total prestress loss 953458

Quarter Span Average

Loss of stress Loss of stress

Loss of stress due Cables stress along As M due to direct

due to loss of to loss of moment

considered the moment at the cable/N/mm2 mm2 N

force loss top fiber

at the bottom fiber

Stage 1 St1 cbls 11.11 14700 547823 0.20 -0.26 0.35

St2 cbls 5.76 6300 121668 0.04 -0.06 0.08 Stage 2

283968 St1 cbls 5.76 14700 0.10 -0.13 0.18

Total prestress loss 953458 --

.---.Reference I Description I Output

BS 5400-4

1990

Cl6 7.2.6

Beam Edse Average

Loss of stress Loss of stress

Cables I stress along A. I1P due to direct due to loss of I Loss of stress due

to loss of moment considered the moment at the t th b 11 fibe

cable/N/mm2, mm2 force loss topfiber a e o om r N

Stage 1 St1 cbls 11.11 14700 547823 0.20 -0.21

St2 cbls 5.76 6300 121668 0.04 -0.05 Stage 2

St1 cbls 5.76 14700 283968 I 0.10 I -0.11

Total prestress loss 953458

B. loss of prestress due to slip during anchorage

Stage 1

At mid span

if,Anchorage slip = 8

loss of prestressing force = ~ x E x A I s s

Modulus of Elasticity of steel E.

A.

I

= 200,000

= 14700

= 30

Assume Slip of the cable = 6

8 = 12

N/mm2

mm2

m

mm per 15m

mm

loss of prestressing force = 1, 176,000 N

loss of direct stress = M A' p

A' = 2.68E+06 mm2

p

loss of direct stress = 0.44

loss of stress at the top most fiber = M x e zpt,

N/mm2

e = 664 mm

zpt' Mxe zpt,

= 1.41E+09

= -0.55

Mxe loss of stress at the bottom most fiber = ---

Zph'

mm3

N/mm2

I I I

0.29

0.06

0.15

1,176,000

N

Reference Description Output .. z pb ' = 1.02E+09 mm3

Mxe = 0.76 N/mm2

zph,

Stresses after the elastic deformation,

Stress at top most fibre = 2.62 N/mm2

Stress at bottom most fibre = 10.69 N/mm2

. Stresses after the losses,

Resultant stresses at top fiber = 1.62 N/mm2

Resultant stresses at botttom fiber = 9.48 N/mm2

Mid Span Loss of stress

Loss or stress Loss of stress due

Cables due to loss of considered A llP due to direct

moment at the to loss of moment

mms:z N force loss top fiber

at the bottom fiber

Stage 1 St1 cbls 14700 1176000 0.44 -0.55 0.76

St2 cbls 6300 504000 0.19 -0.23 0.31 Stage 2

St1 cbls 14700 0 0.00 0.00 0.00

Total prestress loss 1680000

Quarter Span Loss of stress

LOSS 01 Suess Loss of stress due

Cables due to loss of considered A M due to direct

moment at the to loss of moment

(Jr mmi N force loss too fiber

at the bottom fiber

Stage 1 St1 cbls 14700 1176000 0.44 -0.55 0.76

St2 cbls 6300 504000 0.19 -0.23 0.31 Stage 2

St1 cbls 14700 0 0.00 0.00 0.00

Total prestress loss 1680000

Beam Ed e Loss of stress

Loss of stress Loss of stress due

Cables As due to direct

due to loss of to loss of moment

considered M moment at the a mm2 N force loss

top fiber at the bottom fiber

Stage 1 St1 cbls 14700 1176000 0.44 -0.46 0.63

St2 cbls 6300 504000 0.19 -0.19 0.26 Stage2

0 St1 cbls 14700 0.00 0.00 0.00

Total prestress loss 1680000

-- . .

.--.. Reference

BS 5400-4

1990

cl.6.7.2.5 1

BS 5400-4!1

1990

T.able 20

Description

C.Loss of prestress due to creep of concrete

Stage 1 at mid span

Loss of prestress of the tendon = Creep coefficient X Modulus of elasticity of the tendon X stress at the tendon level

After 14 days of concreting,

Stress at tendon level = 9. 77

Strength of concrete =. 36

N/mm2

N/mm2

After 14 days of concreting, Creep coefficient = 0.000036 X 40/fci

Es

= 0.00004

= 200,000 N/mm2

<40

N/mm2

perN/mm2

After 14 days of concreting,Loss of prestress of the tendon = 0.5X(Creep coefficient X Modulus of elasticity of the tendon X stress at the tendon level)

Maximum stress in the section = 11.25 <fa/3 ok Stress at tendon level = 9. 77

Loss of prestress of the tendon = 39 N/mm2

A, = 14700

Loss of prestressing force/(N) = 574343 N

Loss of stress

Direct loss = M A' p

= 574.343 2.68E+06

= 0.21

Loss of stress due to loss of moment at the top fiber = AP x e zp,~

N/mm2

AP = 574,343 N

e=664 mm z I 3

pr = 1.41E+09 mm

Z ph

1

= 1.02E+09 mm3

Loss of stress due to loss of moment at the top fiber = -0.27 N/mm2

Output

574,343

N

Reference Description Output

Loss of stress due to Joss of moment at the bottom fiber = Mxe

zpb '

= 0.37 N/mm2

Stresses after the elastic deformation and anchorage slip mid span after 14 days,

Stress at top most fibre = 1.62 N/mm2

Stress at bottom most fibre := 9.48 N/mm2

Resultant stresses after 14 days creep of concrete

Resultant stresses at top fiber = 1.14 N/mm2

I Resultant stresses at botttom fiber = 8.90 Nlmm2

Similarly,

Mid Span Stress at M Loss of stress

Loss of stress Loss of stress due

Cables cable A, N due to direct due to loss of

to loss of moment considered moment at the

levei/N/mm2 mm2 force loss

top fiber at the bottom fiber

Stage 1 St1 cbls 9.768 14700 574343 0.21 -0.27 0.37

St2 cbls 5.983 6300 135685 0.05 -0.06 0.08 Stage 2

St1 cbls 5.985 14700 316736 0.12 -0.15 0.20

Total prestress loss 1026763

Quarter Span Stress at Loss of stress

Loss of stress Loss of stress due

Cables cable A, N due to direct due to loss of

to loss of moment considered moment at the

leveVN/mm2 mm2 force loss

top fiber at the bottom fiber

Stage 1 St1 cbls 11.164 14700 656426 0.24 -0.31 0.42

St2 cbls 6.086 6300 138036 0.05 -0.06 0.09 Stage 2

St1 cbls 6.088 14700 322192 0.12 -0.15 0.20

Total prestress Joss 1116654

Beam Ed! e Stress at Loss of stress

Loss ot stress Loss of stress due

Cables cable A, N due to direct due to loss of

to loss of moment considered moment at the

leveVN/mm2 mm2 force loss

top fiber at the bottom fiber

Stage 1 St1 cbls 12.385 14700 728247 0.27 -0.28 0.39

St2 cbls 5.196 6300 117854 0.04 -0.05 0.06 Stage 2

274993 St1 cbls 5.196 14700 0.10 -0.11 0.14

Total prestress loss 1121095

-·~·~

r----------,-----------------------------------------------------------------------------------------~---------, ·.Reference Description

D.Loss of prestress due to shrinkage of the concrete

Shrinkage per unit length, = c_,h

BS 5400

Part4

ci.6.7.2.4,After 14 days

Table29

Prestress loss due to shrinkage= &sh x E, x A.

Prestress loss due to shrinkage = 0.5 x & sh x E s x As

Total area of steel A. E.

= 21000

= 200000

= 0.0002

mm2

kN/mm2

For normal exposure, Esh

Prestress loss due to shrinkage = 420000 N

Loss of stress Direct force = ~ p

A p

= 420000

2.74E+06

= 0.15

Loss of stress due to loss of moment at the top fiber = .6. P x e zpl

N/mm2

e = 652 mm

zp, = 1.42E+09

Loss of stress due to loss of moment at the top fiber = -0.19

Loss of stress due to loss of moment at the bottom fiber = Ll P X e

z ph

mm3

N/mm2

Z ph = 1.07E+09 mm3

Loss of stress due to loss of moment at the bottom fiber = 0.26 N/mm2

Resultant stresses after 14 days of creep of concrete

Resultant stresses at top fiber = 1.14

Resultant stresses at bottom fiber = 8.90

N/mm2

N/mm2

Resultant stresses after 14 days of the cable set one before stressing cable set 2

Resultant stresses at top fiber = 0. 79

Resultant stresses at bottom fiber = 8.49

N/mm2

N/mm2

Output

420000

N

·Reference I Description

Loss of stress Prestress loss due I d t d" ct ~~:st~~~::~; Loss of stress due

Age I A., . ue o 1re to shnnkage/(N) ~

1 /(mm2

)

Mid Span 1 month 21000 420,000

Quarter Span 1month 21000 420,000

Beam Edge 1 month 21000 420,000

· •After stressing cable set 2 at stage 2 after one month

At mid span

Stress at top fiber due to stresses in cable set 2

Stress at bottom fiber due to stresses in cable set 2

Loss due to elastic deformation

Cable set 1

Loss of direct stress in cable set 1 Loss of stress in cable set 1 due to moment loss at the top fibre Loss of stress in cable set 1 due to moment loss at the bottom fibre

Cable set 2 loss due to elastic deformation

Loss of direct stress in cable set 2 Loss of stress in cable set 2 due to moment loss at the top fibre Loss of stress in cable set 2 due to moment loss at the

bottom fibre

Anchorage loss

Cable set 1 anchorage loss

Loss of direct loss in cable set 1 Loss of stress in cable set 1 due to moment loss at the top fibre Loss of stress in cable set 1 due to moment loss at the bottom fibre

orce oss

I 0.15

I 0.15

I 0.15

= -0.70

= 7.38

= 0.10 = -0.131

= 0.176

= 0.04 = -0.06

= 0.08

= 0.00 = 0.00

= 0.00

t t th to loss of moment momen a e t the bott fibe

top fiber a om r

I -0.19

I -0.19

I -0.16

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

I 0.26

I 0.25

I 0.21

Output

-----~------------------------------------------------------------------------------~------~

Reference Description Output

Cable set 2 anchorage loss

Loss of direct stress in cable set 2 = 0.19 N/mm2

Loss of stress in cable set 2 due to moment loss at the = -0.23 N/mm2

top fibre Loss of stress in cable set 2 due to moment loss at the = 0.31 N/mm2

bottom fibre

0

Total losses

Loss of direct stress = 0.33 N/mm2

Loss of stress at top most fibre = -0.420 N/mm2

Loss of stres at bottom most fibre = 0.565 N/mm2

Resultant stresses at top fiber = -0.67 N/mm2 I

Resultant stresses at botttom fiber = 14.97 N/mm2

Loss of prestress due to creep of concrete after 30 days

"

Cable set 1

Loss of direct stress in cable set 1 = 0.12 N/mm2

Loss of stress in cable set 1 due to moment loss at the = -0.15 N/mm2

top fibre Loss of stress in cable set 1 due to moment loss at the = 0.20 N/mm2

bottom fibre

Resultant stresses at top fiber = -0.40 N/mm2

Resultant stresses at botttom fiber = 14.66 N/mm2

Cable set 2 loss due to elastic deformation

Loss of direct stress in cable set 2 = 0.05 N/mm2

Loss of stress in cable set 2 due to moment loss at the top fibre = -0.06 N/mm2

Loss of stress in cable set 2 due to moment loss at the bottom fibre = 0.08 N/mm2

I_ Resultant stresses at top fiber = -0.29 N/mm2

Resultant stresses at botttom fiber = 14.52 N/mm2

i!.

Reference Description

E. loss of prestress due to relaxation of steel

BS 5400-41 For relaxation class 1, Maximum relaxation after 1000 h for an initial load of 70% of the breaking load, is 8 % of prestressing force

1990

cl.6.7.2.2 Type of strand = BS 5896-3 super strand-1770-15.7-relax 1

BS 5896 1980

,~_h•- a In all cables

At mid span

Loss of stress

Relaxation class = class 1

Prestresing force at the jacking end = 26,019,000 N

Percentage prestress loss = 8 %

Loss of Prestressing force = 2081520 N

As = 21000 mm2

Loss of stress in cable = 99.12

Direct force = ~ A p

= 2081520 2.74E+06

= 0.76

N/mm2

N/mm2

!l.P x e Loss of stress due to Joss of moment at the top fiber =

z pt

e = 652 mm

Z P1 = 1.42E+09 mm3

Loss of stress due to Joss of moment at the top fiber = -0.96 N/mm2

Loss of stress due to Joss of moment at the bottom fiber _ !l.P x e

zph

Z ph = 1.07E+09 mm3

Loss of stress due to Joss of moment at the bottom fiber = 1.27

Resultant stresses after 30 days

Resultant stresses at top fiber

Resultant stresses at botttom fiber

= -0.29

= 14.52

N/mm2

N/mm2

N/mm2

After 2 months

Output

---· Reference Description Output

Resultant stresses after 1 month

Resultant stresses at top fiber = 1.42 N/mm2

Resultant stresses at botttom fiber = 12.49 N/mm2

Similarly,

"

' Loss of stress Loss of stress Loss of stress

M due to direct due to toss of due to loss of

/(N) force loss moment at the moment at the top fiber bottom fiber

I

Mid Span 2081520 0.76 -0.96 1.27

Quarter Span 2081520 0.76 -0.95 1.26

Beam Edge 2081520 0.76 -0.79 1.08

Total prestress force loss = 4411399 N

Total prestressing force = 24956816 N

Percentage prestress loss = 17.68 %

Smilarly,After stressing cable set 2 at stage 2 after one month streesses at quarter and mid span can be summarised as follows,

MidSpan Quarter span Beam edge

Stresses after stage 1 stressing 3.08 1.88 -0.33

11.25 13.25 16.53

Loss due to elastic defonnation after stage 1 stressing

Top fibre -0.26 -0.26 -0.21

Bottom fibre 0.36 0.35 0.29

Direct 0.20 0.20 0.20

Resultant

Top 2.62 1.42 0.09

Bottom 10.69 12.69 16.04

Loss of slip stage 1

Top -0.55 -0.55 -0.46

Bottom 0.76 0.76 0.63

Direct 0.44 0.44 0.44 -- ---

--·-· ··Reference Description Output

Resultant

Top 1.62 0.43 -0.81

I Bottom 9.48 11.49 14.97

I I

Creep loss after stage 1stressing

-0.27 -0.31 -0.28 I Top I I Bottom 0.37 0.42 0.39 I !

Direct 0.21 0.24 0.27

Resultant

Top 1.14 -0.12 -0.25

Bottom 8.90 10.82 14.31

Loss of prestress due to shrinkage of the concrete

Top -0.19 -0.19 -0.16

Bottom 0.26 0.25 0.21

Direct 0.15 0.15 0.15

Resultant

Top I 0.79

I 0.23

I 0.06

I Bottom 8.49 10.41 13.94

After stressing cable set 2 at stage 2 after one month

Top -0.70 -0.70 -0.14

Bottom 7.38 7.53 6.90

Resultant stresses after stressing cable set 2

Top 0.09 -0.47 -0.08

Bottom 15.87 17.95 20.84 -- -- --- -· -~-------·-·

nee Description Output

Loss due to elastic deformation

Cable set 1

Top -0.13 -0.13 -0.11

Bottom 0.18 0.18 0.15

Direct 0.10 0.10 0.15

Resultant

Top -0.15 -0.24 0.18

Bottom 15.59 17.67 20.55

Cable set2

Top -0.06 -0.06 -0.05

Bottom 0.08 0.08 0.06

Direct 0.04 0.04 0.04

Resultant

Top -0.05 -0.14 0.09

Bottom 15.47 17.55 20.44

Anchorage loss

Cable set 1 anchorage loss

Top 0.00. 0.00 0.00

Bottom 0.00 0.00 0.00

Direct 0.00 0.00 0.00

Top -0.05 -0.14 0.09

Bottom 15.47 17.55 20.44

Resultant

Cable set 2 anchorage loss

Top -0.23 -0.23 -0.19

Bottom 0.31 0.31 0.26

Direct 0.19 0.19 0.19

r-·

.. Reference

Resultant

Top

Bottom

0.37

14.97

Description

0.28

17.05

Loss of prestress due to creep of concrete after 30 days

cable set 1

Top

Bottom

Direct

Resultant

Top

Bottom

cable set 2

Top

Bottom

Direct

Resultant

Top

Bottom

-0.15

0.20

0.12

0.11

14.66

-0.06

0.08

0.05

0.00

14.53

Loss of prestress due to relaxation of steel

Top

Bottom

Direct

Resultant

Top

Bottom

-0.96

1.27

0.76

1.71

12.49

-0.15

0.20

0.12

0.01

16.73

-0.06

0.09

0.05

-0.10

16.60

-0.95

1.26

0.76

1.61

14.57

Output

-0.29

20.00

-0.11

0.14

0.10

-0.09

19.75

-0.05

0.06

0.04

0.00

19.65

-0.79

1.08

0.76

-1.55

17.81

r-------,-------------------------------------------------------------------------.-------, ·Reference Description

Immediately after placing the concrete

Mid span

Prestressing force after initial losses, P = 20,545,417 N

Moment at mid span due to self weight of the beam, Mgt = 6913 kNm

Moment at mid span due to screed concrete, Mg2 = 1804 kNm

Sectional Modulus of the top fiber,

Sectional Modulus of the bottom fiber,

Cross sectional Area,

Resultant Eccenticity of cables,

PIA Pxe/Zpt

8 c±>

Neutral Axis

c±> CB

PIA Pxe/Zpb

Z pi = 1.42E+09 mm3

zpb = 1.07E+09 mm3

AP = 2. 7 4E+06 mm2

e = 652 mm

Mgl/Zpt Mg2/Zpt

(±) [r 8 ~ Mgl!Zpb Mg2/Zpb

P Pxe Mg, Mg2 ----+--+--Stress at top most fibre = A z z z

pi pi pi

= 2.98 N/mm2

p Pxe Mg, Mg2 stress at bottom most fibre = A + -

2 - z-z

ph ph ph

= 10.80 N/mm2

Output

2.98

N/mm2

10.80

N/mm2

---------L---------------------------------------------------------------------------------------------------L--------~

,.----··-

Reference Description Output

Similarly,

Quarter Beam Edge

Span

Initial prestressing force 25,537,504 26,019,000

! Total pre stress losses 4419206 4351826

Prestressing force after initiallosses,P/N 21,118,298 21,667,174 Moment due to self weight of the Mg, 5184.77 0

beam/(N/mm) .

Moment due to screed concrete/(N/mm) Mgz 1352.96 0

Sectional Modulus of the top fiber, zp, mm3 1.42E+09 1.42E+09

Sectional Modulus of the bottom fiber, zpb mm3 1.07E+09 1.07E+09

Cross sectional Area, AP mm2 2.74E+06 2.74E+06

Resultant Eccentricity of all tendons, e mm 648 542

Stress at top most fibre N/mm2 -0.67 1.54

Allowable stress at bottom most fibre N/mm2 12.16 16.77

I

loss of prestress due to differential shrinkage

After placing screed on the beam,

__[_ t.Q6QM 5.200M

__[_Q.132M

Prestress l rm A ~w, -r250M 0

Designers p.290M .J cr.1isr1 a .......... - - 7 Handbook 0.400rv P.W.Ables eP

1.667M

Q 952M

To find the neutral axis of added concrete,

Taking moments along the axis through point A

- (60x5200x30+0.5x72x5200x84-X = 2x400x250x125-2x0.5x400x40x263)

(60x5200+0.5x72x5200+2x400x250+2x0. 5x400x40)

= -5.77 mm

ea +ep = 1670-5.77-ypb ......... (1)

= 712 mm

e a I a --= m -- .......... (2)

eP Ixp .

ence

'•

400-4

90

t3.4

At mid span

(1) And (2)

Description l I Output

I ' _ 9.40E+09 mm4 ar -

IxJ = 1.02E+12 mm4

I

nil - Modular ratio=1 I

I) _ Moment of Inertia of added concrete

I x~ _ Moment of inertia of precast concrete

e / ~= e ;

I m __ a

P: I xp i j

i = 0.01

eP = 706

ea = 7

AP = 2.74E+06

lxp = 1.02E+12

Au = 7.15E+05

Ia = 9.40E+09

Ea,Ep = 34000

mm

mm

mm2

mm4

mm2

mm4

N/mm2

F - Force exerted by differential Shrinkage

F 2 . 2 ' F F xeP Fxe

11& = + ++ a

APxEP AaxEa lxpxEP /axEa

11& = 0.43x200x1 0-6

F= 11& 2 2 e e ---+ ++ p __ a_

1 1

AP xEP Aa xEa lxp xEP Ia xE,

= 1.30E+06 N

Bending moment due to this force= 917586487 Nmm

Loss of direct stress = 0.47

Bending stress at the top fibre = 0.65

Bending stress at the bottom fibre= -0.86

N/mm2

N/mm2

N/mm2

. jstresses at mid span after placing screed

N/mm2

N/mm2 l Stress at the top fiber = 2.98

. Stress at the bottom fiber= 10.80

-renee Description Output

- M Stress at the top fibre of the composite section due to Mq - z q

ct

r.·fu = 3.14 N/mm2

Stress at the top level of the pre cast section = 2.70 N/mm2

Stress at the bottom fibre of the composite section due to = Mq

Mq zch

= -5.19 N/mm2

Stress at the top most fiber of composite section = 1.86 N/mm2

Resultant stress on the precast section top fibre = 4.56 N/mm2

Stress at the bottom most fiber of pre cast section = 9.47 N/mm2

Resultant stress on the precast section bottom fibre = 4.28 N/mm2

Similarly

Quarter span Beam Edge

Moment due to live loads 4.64E+09 O.OOE+OO

Stress at the top fibre of the composite section due to Mq 2.36 0.00

Mq

zct Stress at the top level of the pre cast section 2.02 0.00

Stress at the bottom fibre of the composite section due to -3.89 0.00

Mq Mq

zcb Resultant stress on the precast section top fibre 1.35 1.54

Resultant stress on the precast section bottom fibre 8.27 16.77 ----·-- - - -- -

r· Reference Description

Claculation of the balance losses after 2nd month­

Loss of prestress due to shrinkage of the concrete

Loss of prestress due to shrinkage of the concrete = Rest 50% of the shrinkage loss

BS 5400 I Loss of stress Direct force = I!J>

Part4

cl6.7.2.4

Table29

Ac

="420000

3.45E+06

= 0.12

tiP x e Loss of stress due to loss of moment at the top fiber =

zclt

N/mm2

e = 799 mm

zc!, = 1.97E+09 mm3

Loss of stress due to loss of moment at the top fiber = ..0.17

Stress at top of precast section = ..0.15

tiP x e Loss of stress due to loss of moment at the bottom fiber = ----

Z clb

N/mm2

N/mm2

Zc!b = 1.19E+09 mm3

Loss of stress due to loss of moment at the bottom fiber = 0.28

Resultant stresses after live loading

Loss due to creep

Resultant stresses at top fiber= 9.47

Resultant stresses at botttom fiber = 4.28

Resultant stresses at top fiber = 9.20

Resultant stresses at botttom fiber = 3.87

In cabe set one at mid span

N/mm2

N/mm2

N/mm2

N/mm2

N/mm2

Loss of pre stress = Rest 50% of the initial loss

N

N/mm2

N/mm2

Loss of prestress = 1026763

Loss of direct stress at the top fibre = 0.30

Loss of strees at the top fiber due to moment = -0.42

Loss of strees at the bottom fiber due to moment = 0.69 --·· N/mm

2

Output

r---

Reference Description Output

Resultant stresses

Resultant stresses at top fiber = 9.20 N/mm2

Resultant stresses at botttom fiber = 3.87 N/mm2

Resultant stresses after creep loss

Resultant stresses at top fiber after all the losses = 8.49 N/mm2

-

Resultant stresses at botttom fiber after all the losses = 4.26 N/mm2 Mid Span

Stress resultants after rest of the losses for the quarter span and beam edge

Resultant streses after placing the screed

Quarter span Beam edge

Resultant stresses at top fiber 1.35 1.54

Resultant stresses at botttom fiber 8.27 16.77

Quarter span Beam edge

Losses due to diferential shrinkage

Loss of prestress 1300240 1300240' N I

Loss of direct stress at the top fibre 0.47 0.47 N/mm2

Loss of strees at the top fiber de to moment 0.65 0.65 N/mm2

Loss of strees at the bottom fiber de to moment -0.86 -0.86 N/mm2

Resultant stresses at top fiber 0.23 0.42 N/mm2

Resultant stresses at botttom fiber 6.93 15.44 N/mm2

Quarter span Beam edge

Losses due to shrinkage

Loss of prestress 420000 420000 N

Loss of direct stress at the top fibre 0.12 0.12 N/mm2

Loss of strees at the top fiber due to moment -0.17 -0.17 N/mm2

Loss of strees at the top fiber of the pre cast section due -0.15 -0.15

to moment N/mm2

Loss of strees at the bottom fiber due to moment 0.28 0.28 N/mm2

Resultant stresses at top fiber after final shrinka_g_e losses -0.04 0.15 N/mm2

Resultant stresses at botttom fiber after final shrinkage 6.53 15.03.

- losses N/mm2 ~-

,----· Reference Description Output

Losses due to creep

Loss of p_!estress 1116654 1121095 N

Loss of direct stress at the top fibre 0.32 0.32 N/mm2

Loss of strees at the top fiber due to moment -0.45 -0.46 N/mm2

Loss of strees at the top fiber of the pre cast section due

to moment -0.39 -0.39

Loss of strees at the bottom fiber due to moment ~ 0.75 0.75 N/mm2

Resultant stresses at top fiber -0.04 0.15 N/mm2

Resultant stresses at botttom fiber 6.53 15.03 N/mm2

Resultant stresses at top fiber after all the losses 0.74 -0.63 N/mm2

Resultant stresses at botttom fiber after all the losses 5.46 13.96 N/mm2

I

Allowable tensile stress at service, famn = -2.55 N/mm2

; Allowable compressive stress at service, farrax = 20 N/mm2

Resultant prestress force after all the losses

Initial prestressing Prestress Final Percentage

force/(N) Loss/(N) prestressing prestress

force/(N) Loss/(%)

Mid Span 24,956,816 8,908,745 16,048,072 36

Quarter span 25,537,504 9,088,525 16,448,979 36

Beam edge 26,019,000 9,097,408 16,921,592 35

--

--·.Reference Description Output

Check for ultimate limit state

BS 5400-4 -"' E

Cl.6.3.3. ~ . ~ -... rJ) rJ)

~ - "

1pu

Ym ___ /! 0.8fpu

Ym I I

I

I I I

I I I

I

200kNif2

... 0.005

... &I &2 Strain

fpu Characteristic strength of pre-stressing tendons

r m Partial safety factor for strength

fpu = 1770 N/mm2

:16.3.3.3.1e Ym = 1.15

fpu = 1539 N/mm2

Ym 0.8/pu = 1231 N/mm2

Ym

E, = 200,000 N/mm2

&I = 1231 200000

= 0.0062

&2 = 0.005 + 1539 200000

= 0.0127 -----

Reference

Mid Span

Description

Prestressing force after all the losses R x P = 16,048,072 N

Area of steel = 21000

Stress in steel = 764

Corresponding strain in steel = 0.0038

mm2

N/mm2

Check the ultimate capacity immediately after laying the screed

Height to the top fiber from the neutral axis, Y pi = 715 mm

Height to the bottom fiber from the neutral axis, = 952 mm

Second Moment of Inertia along x axis, = 1.016E+12 mm4

Cross sectional Area, = 2.74E+06 mm2

Eccentricity of cables, e = 652 mm

Stressatthetendonlevel = RxP +RxPxe2 (Mxl +Mg2)xt Ap IX I

= 6.98

Ec = 34000

Strain in concrete at tendon level = 6.98 34000

= 0.0002

Strian in prestressing steel due to strain in concrete = 0.0002

Nlmm2

Nlmm2

X

x - Depth to the neutral axis from the top fibre

d - Effective depth of tendons

d = Eccentricity + Y pi

= 1367 mm

Assume X = 0.6 d

Depth to the neutral axis,x = 820

Output

·.

-·-· •• Reference Description Output

0.0035

820.2 v /o 547

.

0.00233333

. Total strain in steel = 0.0064

This strain value is within c 1andc2 i

Therefore linear ineterpolating witin this region in the above graphthe corresponding stress can be obtained 0.2xfu

(E-E1)x( P)

0.8xfpu Corresponding stress = Ym ( )

(c2 -&1) Ym

= 1241 N/mm2

Force in steel = 26,058,076 N

I

:L6 3.3.3.1b Stress in concrete = 0.4x feu

Sectional area under compression = Section above the neutral axis

= 1,670,000 mm2

Force in concrete = 33,400,000 N

Ultimate Moment carrying capacity = Force in concrete x ( d- ~) 2

d = 1367 mm

X= 820.2 mm

Ultimate Moment carrying capacity = 31960 kNm

.

_, ___

.. Rete renee Description Output

Maximum moment when screed is laid = (M gl + M g2) x r fJ x r Ft.

3S 540 0-2 rf3 = 1.1

:15. 1 YFL = 1.15

:1 5.1 < 1 Mgl +Mg2 = 8716965000 Nmm

Maximum moment when screed is laid = 11 027 kNm ok .

Check the ultimate capacity with full live load present

Cross sectional Area, = 3450000 mm2

Total Depth, = 1764 mm

Total width of the top flange, = 5200 mm

Height to the top fiber from the neutral axis, = 665 mm

Height to the bottom fiber from the neutral axis, = 1099 mm

Moment of Inertia, = 1.31E+12 mm4

Ultimate moment= 21997 kNm

Assume X = 0.25 -d

d =1464 mm

Depth to the neutral axis,x = 366 mm

0.0035

366 7 /o 1398

0.0134

Total strain in steel = 0.0174 mm

This strain value is within e1ande2

Reference Description Output

Therefore linear ineterpolating witin this region in the above graph the corresponding stress can be obtained

0.2xf u (&-&J)x( P )

0.8xfpu Corresponding stress stress = Ym

(&2-&1) ( )

Ym 0

= 1760 N

Force in steel = 36,967,391 N

Stress in concrete = 0.4xfcu

Sectional area under compression = Section above the neutral axis

= 1,716,000 mm

Force in concrete = 34,320,000 N

I

Ultimate Moment carrying capaCity = Force in concrete x ( d-!.) 2

"

d = 1464 mm

x=366 mm

UHimate Moment carrying capacity = 43964 kNm ok

Oeasign for shear at ultimate limit state

Calculation of shear forces For a beam of udl w/m and length I, the shear force at a distance x fro the support can be derived as follows

f t Rl Rz

J X

tsF - -- -- -- --- -· -- ---····------ -

r-----"

. Reference Description Output

R1 -wxx-SF = 0

SF=R1 -wxx

R-R-wxl I- 2---

2

i i I. I = 29 m

Weigl c = 65.76 kN/m wscl = 17.16 kN/m

wacl = 4.37 kN/m

whr = 0.585 kN/m

WJW = 8.10 kN/m

W.v = 0.85 kN/m !

HA UDL on the edge beam = 30.00 kN/m

HA KEL on the edge beam = 120.00 KN

udl due to pedestrian load = 7.5 kN/m

Length along the Beam /(m),X 0 7.25 14.5 25 29

Shear Force due to self wt of the beam,SF1 /(N) 953520 476760 0 -690480 -953520

Shear force due to Screed Concrete,SF2/(N) 248820 124410 0 -180180 -248820

Total shear force due to dead load/(N) 1202340 601170 0 -870660 -1202340

Shear force due to Asphalt Concrete, /(N) 63307 31654 0 -45843 -63307

Shear force due to hand raii/(N) 8483 4241 0 -6143 -8483

Shear force due to footwalki(N) 117450 58725 0 -85050 -117450

Shear force due to kerb/(N) 12325 6163 0 -8925 -12325

Total sher force due to super imposed loas/(N) 189240 94620 0 -137036 -189240

Shear force due to Live loads,HAUDL /(N) 435000 217500 0 -315000 -435000

Shear force due to Live loads,HA KEL /(N) 60000 60000 -60000 -60000 -60000

Shear force due to pedestrian load/I(N) 108750 54375 0 -78750 -108750

Total shear force due to live loads/(N) 603750 331875 -60000 -453750 -603750

·.Reference Description Output

Check for Maximum shear stress

BS 5400-4 Section uncracked in flexture

cl.6.3.4.2 The ultimate shear resistance vco

vco = 0.67b~(f/ + fcpJ;) Eq(28)

J; = 0.24x fJ: = 1.70 N/mm2

f. RP RPey cp = YFLX(-+--)

Act I

RP = 16,048,072 N i Ac = 3.45E+06 mm

cl.4.2.3 YFL = 1.15

at centroid y =0

/cp = 5.35 N/mm2

BS 5400-4 h = 1764 mm

Cl.6.3.4.5 h = (350-2/3X60)X2

= 620 mm vco

vco = 2,533,945 N/mm2 2,533,945

N/mm2

BS 5400-4 Section cracked in flexture

cl.6.3.4.3 The ultimate shear resistance

Vcr = 0.037bd.fl: +Me, V Eq(29) M

Mer = (0.37fl: + /P,)x I I y

RP RPey fP, = YFL x(-+--) . Acl I

:1.4.2.3 YFL = 1.15

RP = 16,048,072 N

Ac = 3.45E+06 mm2

. /xcl = 1.31E+12 mm4

-·

-·· ·.Reference Description Output

y = Ycth

= 1099 mm

At mid span e = 652 mm

fpt = 15.45 Nmm

feu =50 N/mm2

Me: =(0.37.Jf: + fp1 )x//y

= 2.15E+10 Nmm

Mid Span Quarter Span Beam Edge

YFL 1.15 1.15 1.15 I RP 16,048,072 16,448,979 16,921,592 I

A 3.45E+06 3.45E+06 3.45E+06

Jxcl 1.31E+12 1.31E+12 1.31E+12

y 1099 1099 1099 e 652 648 542 fpt

15.45 15.77 14.50

Mer 2.15E+10 2.19E+10 2.04E+10

BS 5400-2

Table 1 Load Fcators-Uitimate limit state

3S 5400-4

d 4.2.3. Type of load Nominal shear force/(N) Ultimate load/(N) v YFL YJ3

Mid Span Quarter Span Beam Edge Mid Span Quarter Span Beam Edge

Dead Load 1.15 1.1 0 601170 1202340 0 760480 1520960

Super lmpos- 1.75 1.1 0 94620 189240 0 182143 364286

Live Loads(H 1.5 1.1 -60000 331875 603750 -99000 547594 996188

Total Ultimate shear force -99,000 1,490,217 2,881,434

:s 5400-4 feu =50 N/mm2

16 3.4.5 b = 620 mm

d = 749 mm

Allowable shear force for grade 50 concrete v IIIIX = 23,219,000 N -·

---·-Reference Description Output

At mid quarter and beam edge the shear force is less than allowable shear force force for grade 50 concrete.

Type of load Nominal Moment/(Nmm) Ultimate Momeni/(Nmm) M rFI. rf3

Mid Span Quarter Span Beam Edge Mid Span Quarter Span Beam Edge

I

Dead Load 1.15 1.1 8.72E+09 6.54E+09 . 0.00 1.10E+10 8.27E+09 O.OOE+OO

Super ImposE 1.75 1.1 2.16E+09 1.62E+09 0.00 4.16E+09 3.12E+09 O.OOE+OO

Live Loads(H 1.5 1.1 4.02E+09 3.02E+09 0.00 6.64E+09 4.98E+09 O.OOE+OO

Total Ultimate Moments 2.18E+10 1.64E+10 O.OOE+OO

Mer= 2.15E+10

At mid span v = -99000 N

At mid span M = 2.18E+10 Nmm

At mid span d = 749 mm

b = 620 mm

. Vc, = 0.037bdfJ: +Mer V M-

= 2.19E+05 . Midspan Quarter Span Beam Edge

Chianage/(m) 14.5 7.25 0

b /(mm) 620 620 620

d /(mm) 749 745 639

Mcri(Nmm) 2.15E+10 2.19E+10 2.04E+10

M /(Nmm) 2.18E+10 1.64E+10 O.OOE+OO

v /(N) -99000 1490217 2881434

Sin(e) 0 0.005 0.022

Prestressing force/(N) 24,956,816 25,537,504 26,019,000

Vertical component of 0 131,676 572,280

prestressing force

Resultant ultimate -99,000 1,358,541 2,309,154

shear force

V,, /(N) 219,115 2,114,664 0 - vco /(N) 2,533,945 2,533,945 2,533,945 vc is lesser of vcr and v" v c /(N) 219,115 2,114,664 0

If Vis greater than Vc shear reinforcement are needed. - ---· ------·

eference ·• R'

BS

cl.6

-·-

400-4

1.4.4

Shear reinforcement

----

Description Output

Mid span Quarter Span Beam Edge

not needed not needed needed

Asv = v + 0.4bdt- vc sv- 0.87 fyvdt

f yv - Characteristic strength of reinforcement

A.v - Total cross sectional area of the leg of the links

s. - link spacing along the length of the beam

1l A =-¢ 2 xn sv 4

¢ - diametre of the links

n - number of legs

Assumed diametre of the links = 16 mm

n =4 I

b = 620 mm.

Asv = 804 mm2

/yv = 460 N/mm2

d, = 749 mm

At beam span v = 2881434

vc =0

Asv X (0.87 fyvdt) s =

v v + 0.4bdt - vc s.

= 79 mm 79

mm

Maximim spacing = 0.75d1 or 4b

0.75d, = 562 mm

4b = 2480 mm

-- ----- -~ ----- ---- ---·- -- - -

...----. Reference

BS 5400-4 'At mid span provide minimum reinforcement

cl.6.3.4.4

BS 5400-4 'Check for Deflection

1990

Description

A~ ( 0.8:/,.) ~ 0.4 N I mm'

Sv Asv(0.87Jyv) s = v

0.4b

= 1298 mm

>maximum spacing

provide spacing of 400 mm

ci.A21 The deflection of a beam is given by,

2

a= K I 1 I e-rb

a - Deflection

I e - effective span of the member

K, - coefficient depending on the shape of the bending moment diagram

1 - curvature at mid span

rb

1 d2 -=____1:'_ rb dx2

From simple bending formula

Therefore

1 M -=---r6 EJ

Ec - Youngs modulus of concete

I - Second moment of area of the section

a=K/2 M 1 -e E J

Output

1298

mm

·.Reference Description Output

BS 5400-4 Deflection of the beam due to dead load

1990 KI = 0.104

Table 34 I, = 29 m

At mid span M=Mg1

= 6913020000 Nmm

Ec = 34000 Nmm2

I = lxp mm4

= 1.02E+12 mm4

zM downward a=Kl, --EJ a

a = 17.50 mm downward 17.50

mm

Deflection due to prestress

Prestressing force afetr all the losses = 16,048,072 N

Eccentricity = 652 mm

Bending Moment = 1.0463E+1 0 Nmm

K 1 = 0.125 "

I = 1.02E+12 mm4

Creep coefficient = 0.000036 perN/mm2

t/J = creep coefficient X elastic modulus

= 1.224

E elf

E =

1+¢

= 15,288 N/mm2

Deflection due to prestress = 70.82 mm upward

Resultant deflection = 53.31 mm upward

BS8110-3

1997 Allowable deflection for flange or rectangular beams = Span 250

CL3 4.6.3

= 116 mm

Resultant deflection ok -·-- -- -- ------ ---·-

·. Reference Description Output

Deflection of the beam during service condition

Deflection due to screed

Bending moment due to screed concrete = Mgz - = 1803945000 Nmm

Kl = 0.104

I = 1.02E+12 mm4

Creep coefficient = 0.000036 per N/mm2

(; = creep coefficient X elastic modulus

= 1.224

E efT E

= 1 + ¢

= 15,288 N/mm2

2M a=Kle -- ,,

EJ

= 10.16 mm downwards

Resultant deflection = 43.16 mm upwards

Deflection due to footwalk,hand rails, wearing surface,kerb and pedestrian load(Super imposed loads)

Bending moment due to footwalk = 851512500 Nmm

Bending moment due to hand rails = 61498125 Nmm

Bending moment due to wearing surface = 458975750 Nmm

Bending moment due to pedestrian load = 788437500

Bending moment due to kerb = 89356250 Nmm

Bending moment due to footwalk,hand rails, wearing surface and pedestrian load = 2249780125 Nmm

..----Reference Description Output

K 1 = 0.104

I = Ixcl mm4

I = 1.31E+12 mm4

Creep coefficient = 0.000036 perN/mm2

¢ = creep coefficient X elastic modulus

-= 1.224

Eeff E

= 1+¢

= 15,288 N/mm2

2M a=Kl-I e £I

c

= 9.83 mm downwards

Resultant deflection = 33.32 mm upwards

Deflection due to HA UDL(SLS) '·

BS 5400-2 Bending moment due to HA udl = 3153750000 Nmm

Table 1 Load factor (HA alone) = 1.2

K, = 0.104

I = Ixcl mm4

= 1.31E+12 mm4

Creep coefficient = 0.000036 perN/mm2

¢ = creep coefficient X elastic modulus

= 1.224

Eeff E

= 1+¢

= 15288 N/mm2

12M a=K,. --

EJ

= 13.78 mm downwards

~

~-

Reference Description Output

Resultant deflection = -19.54 mm downwards

Deflection due to HA KEL(SLS)

BS 5400-2 Bending moment due to HA KEL = 870000000 Nmm

Table 1 Load factor (HA alone) = 1.2

K~· = 0.083

I = Jxcl mm4

j = 1.31E+12 mm4

Creep coefficient = 0.000036 perN/mm2

rjJ = creep coefficient X elastic modulus

= 1.224

E eff E

= 1+¢

= 15288 N/mm2

2M a=K1/e --

EJ '·

= 3.66 mm downwards

a

Resultant deflection = -15.88 mm downwards -15.88

mm

B.S8110-3

1997 Allowable deflection for flange or rectangular beams = Span

cl.34.6.3 250

= 116 mm

Resultant deflection = ok

-·

r-·

.~eference I n-u-· - I I

BS 5400-4,Design of End Block

1990

cl.6.7.5

3S 5400-4

1990

·able 30

~

Tensile stress

J;m 1------

0 0.2yo

I I

0.5y0 2y0 Distance from loaded face

Fig: Transverse stress distribution along block centreline

---1 0.125M t- J;m - Maximum transverse tensile stress

Olt@] L_t_

2yp0f] f

Design of Spiral

Size of the end block= 175 x 175 mm

Yo - Half the side of end block

Y po - Half the side of loaded area

J>* - Loadinthetendon

Fhst - Bursting tensile force

2y0 = 175 mm

2ypo = 125 mm

2Ypo = 0.71

2yo

F;,SI : 0.11 pk

pk = 1m X Jacking end fOrce

Jacking end force = 1300950

Ym = 1.15

N

Pk = 1496092.5 N

2yo

Output

J>k

1496093

N

.~eference Description

Fhst = 164570

Design strength of reinforcement = 0.87 JY Characteristic strength of mild steel fY = 250

Allowable strength of mild steel bars 0.87 JY = 218

Area of steel required = F;,s, 0.87xfY

N

N/mm2

Nlmm2

= 757 mm2

Using 10 mm diameter bar, number of required= 4.817 turns in the spiral

Say,= 5

Pitch of a spiral = (2Yo- 0.2yo) number of turns of the spirals

Concrete Bridge Design

LA.Ciark p 146

= 31.5 mm

Reinforcement to resist spalling force

Spalling force = 0.04Pk

= 59844

Area of reinforcement required= 0.04P*

0.87 Ivy

Characteristic strength of Tor steel f yv = 460

N

Area of reinforcement required = 149.5 mm2

Diameter of the bar = 10

Number of bars required = 1.90

Say,= 2

END BLOCK

Considering the blocks in the bottom flange ·

h5 h4

Output

Fbst

164570

N

.--.• Reference

BS 5400-4

- 1990

Te.ble 30

Description

Number of plates = 12

Width of the equivalent area of all the plates = 125 mm

Area of one plate = 15625 mm2

Length of the equivalent plate = 1500 mm

h1 = 100 mm

Centroid of the equivalent plate from the bottom = 100 mm

In horizontal direction 2 Y po = 1500 2y

0 3200

Fbst

pk

= 0.47

= 0.18

In vertical direction 2Y po = 125

Selecting maximum from above two cases

2y0 200

= 0.625

Fbst = 0.23 pk

Fbst = 0.23 pk

Total prestressing force pk = Y m X Jacking end fOrce

Jacking end force= 15611400 N

Ym = 1.15

Pk = 17953110 N

Total force on the plate Fbst = 4129215

Design strength of reinforcement = 0.87 JY Characteristic strength of Tor steel

Allowable strength of Tor steel bars

/y = 460

0.87/y = 400

Fbst Area of steel required = 0.87 x fY

= 10318

N

N/mm2

N/mm2

mm2

mm.

mm

Diameter of the bar = 20

Reuired number of bars with two legs = 16.42

Say,= 17

Legth of the beam reinforcement are provided = 1700

Spacing = 106

·say, 100 ----~--------------------~----------

Output

·Reference Description Output

'• Considering the blocks in the web

Number of plates = 5

Width of the equivalent area of all the plates = 125 mm

Area of one plate = 15625 mm2

Total area of the equvalent plate = 78125

Length of the equivalent plate = 625 mm

Centroid of the equivalent plate from the bottom = 625 mm

BS 5400-4 In horizontal direction 2Ypo = 125

1990 2yo 350

Table 30 = 0.36

Fbsr = 0.213 pk

In vertical direction 2yp0 = 625

2yo 1670

= 0.37 ~

Fbsr = 0.208 pk

Selecting maximum from above two cases

Fbst = 0.213 pk

Total prestressing force pk = y m X Jacking end furce

Jacking end force= 6504750 N

Ym = 1.15

pk = 7480463 N

Total force on the plate Fbst = 1592270 N

Design strength of reinforcement = 0.87 [y

Characteristic strength of Tor steel [y = 460 N/mm2

Allowable strength of Tor steel bars 0.87 fY = 400 N/mm2

Area of steel required = Fbst

0.87xfy

= 3979 mm2

Diameter f the bars = 16

Required number of bars with two legs = 9.89

Say,= 10 .

Leath of the beam reinforcement are orovided ::; 1700

Reference Description Output .. Spacing = 189 mm

'·· Say, 100 mm

I Concrete , Bridge

Checking of Interface shear stresses Design to 5400by V,S

vh =--L.A. Clerk Jbe

pp 108 v 1o Horizontal interface shear stress 109

be width of the interface

v_ " Vertical shear stres at the point considered

S- First moment of area about the neujral axis of the one ~ side interface

I - Second moment of area the transformed composite section

__J.e6el'l 5.2001'1 I

-~"x 0.250JLJ·

4.4t01'1

o~rl'l -

.4001'1

NA of composite

secti.m

Neutral axis of the Screed X

(60x5200x30+0.5x72x5200x84-2x250x400x125-

X = 2x0.5x40x400x265) (60x5200+0.5x72x5200+2x250x400+2x0.5x40x400)

= -5.81 mm

Contact width of the beam be = 4400 mm

Distance between the neutral axes of slab and = 563 mm composite beam .

•• Reference

·.

s 5400-4

·1990 .7.423

Description

Area of the slab = 715200 mm2

First moment of area S = 402793600

Second Moment of area of the composite sectiO!J = 1.31 E+12

Ultimate shear force

Horizontal interface shear stress

Similarly,

vu = 99,000

vus vh = Ib.

VIF 0.007

mm3

mm4

N

N/mm2

Quarter Span Beam Edge

Contact width of the bearn/(mm) b.

Distance between the neutral axes of slab and composite beam/(mm)

Area of the slab/(mm2)

First moment of area/(mm3) S

Second Moment of area of the composite section/(rd

Ultimate shear force[{Nl Vu vh

Longitudinal Shear

4,400

563

715,200

402793600

1.31E+12

1,490,217

0.104

Longitudinal shear should not exceed the lesser of following

(a) k,fcuLs

(b) v, Ls + 0. 7 A.fy

Where,

4,400

563

715,200

402793600

1.31E+12

2,881,434

0.202

v1 _ Ultimate longitudinal shear stress (Table 31)

k, _ Constant depending on the concrete bond

A - Area of fully anchored reinorcement crossing shear e plane

L s _ Length of the shear plane

f Y _ Characteristic strenght of reinforcement

feu _ Characteristic cube strength of concrete

v; _ Longitudinal shear force

Maximum longitudinal shear stress applied = 0.202

Applied shear force = 887

N/mm2

N/mm

Output

~eference

BS 5400-4

1990

Table 31

BS 5400-4

1990

Table 31

Eqn (a)

Description

~= krf,-uL,.

fcu= 50

Surface type 2 k1 = 0.09

Ls= 4400

klfcuLf= 19800

N/mm2

mm

N/mm

Eqn (b) ~ = v1Ls + 0.7 Aefy

for Surface type 2 N/mm2

Assume there are no other reinforcement

VI = 0.5

Ls = 4400

Ae = 0

v1Ls + 0.7 Aefy = 2200 N//mm

Therefore, ultimate shear force 2200N/mm is greater than the applied shear force 893 N/mm.

Therefore interface shear reinforcement is not necessary.

~S 5400-41According to the BS 5400 provide 0.15% of contact area of reinforcement acorss the surface.

1990

:;t."7 .4.2.3 0.15 % of surface area = 660 mm2/m

Diameter of the bar = 12 mm

No. bars needed = 6 /m

Check for shear at support

Ultimate shear at support occurs due to HB Load????????

Ultimate shear force at the support = 21881,434 N

Prestressing force at the support (force along the cable)= 161921

1592 N

ngle of the cable at the beam edge to the vertical = 1.05727 rad

= 61 degrees

Vertical component of the Prestressing force at= 8 312 790 N the support I I

Resultant vertical force at the beam edge = -5,431,356 N

3 5400-4 Shrinkage Reinforcement and Temperature reinforcement

1990 To prevent cracking due to shrinkage and thermal movement,reinforcement should be provided in the

5 direction of any restraint to such movements.

,8.9

As> k,(Ac -0.5Aco,)

£. 0 OOfi fnr nmriA 4F\O d"""l

Output

Reference

:; 5400-4 990 5.3.2.3

Design of transverce reinforcement Consider load combinations

Calculation

Combination1 = Dead loads + Superimposed dead loads Combination2 = Combination1 + HB load on mid of lanes Combination3 = Combination1 + HA UDL + HA KEL mid Combination4 = Combination1 + HA UDL + HA KEL edge CombinationS = Combination1 + HB load on lane

Maximum transverce ultimate bending moment of top flange at mid of the beam using grillage analysis

-Combination

Distance(m) 0 1.2 2.6 4

Combination3 -5.14 16 22.4 18.5 Combination4 -5.44 -48.3 38.11 -82.9 CombinationS -0.08 -17.4 9.25 -15.68 Max bending moment -5.44 -48.3 38.11 -82.9

Ultimate Bending Moment

60

.E 40

E 20 z ::.::: ~ 0 c CD

~~J~ 1 / 2 3 \. 4 E

0 :IE Distance(m) m c :a c -60 Gl m -80

-100

Design of top reinforcement of top flange Ultimate bending moment, Mu = 82.9 KNm/m

Assume,Serviceble bending moment Ms = Mu/1.5 = 55.3 KNm/m

Assume 16 mm diameter Tor steel can be used h=

Effective depth for cover of 50mm, d =

M

= b = feu= [y =

= bd2fcu

250 250-50-8

192 1000 50

460

0.045

Single reinforcement is enough

mm

mm mm N/mm2

N/mm2

< 0.15

5.2 -5.18 -2.98 -0.77 -s.18 I

y

Mu = 0.87J;,AsZ ----(1)

Z 1.1/, As (2) = (I y )d ---- . fcubd

Output

6

Reference '•

BS 5400-4 1\990

cl 5.8.4

BS 5400-4 1990 ~ 4.3.2.2 fable 3

3S ~00-4

1~90

:15.8.8.2

-

Calculation

from equations (1) and (2)

z2 -dZ + 1.1Mu = 0 0.87 fcub

z 2 - 192 z + 2096 = 0

Z = 180 mm < 0.95d

If Z < 0.95d, therefore Z = Z

Z = 180 mm

M A=--~

s 0.87Zfy

= 1151 mm2

l OOAS = 0.599 bd

Which is greater than the minimum of 0.15% of bd

Therefore

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Checking of crack width for top flange

A = s

= = =

=

=

1151 mm2

5.7 10

1000 10

100 mm

2011 mm2

Assume reinforcement provided T 16 @ 100 mm

Modulus of elasticity of steel, Modulus of elasticity of concrete,

Stress and strain distribution of section

N/A h II _,JJ~--

0 0 I I

Step- 1

where,

E

E = s

E = c

200 KN/mm2

28 KN/mm2

f.

; = af/J[ ~l + :l/J -1] E

a =-s = 14.29 Ec A

A. =-s = 0 0105 'I' bd .

28 2 E =-KN/mm

c 2

x = 80.17 mm

• X

Output

----Reference ..

) 6400-4 990 5.8.4.2

Step- 2

Step- 3

Step- 4

Step- 5

Step- 6

Calculation

Z=d-x 3

= 165.3 mm

hb=~~ = 8.342 N/mm2 < 0.45fcu

Satisfied

Ms fs=AZ

s

= 166.2 Nfl!lm2 < 0.87fy Satisfied

&I= .fs [~] Es d-x

= 0.001262

&=Is s E

s

= 0.000831

[3.8b,h(a'-dc)][( Mq) _9 ] & 2 = 1-- xlO

&sAs(h- de) Mg

Moment due to pennanent load, Moment due to live load,

M= g

M= q

48 35

KNm KNm

Step -7

c5

Step- 8 Design crack width

Secondary reinforcement

£2 = 0.0002 > 0

Therefore, &m =&! -&2

= 0.0011

Therefore section is cracked

A coo. _ _U ac,- 68.6

>0

mm

3ac,£m

= l+2(acr -Cco%-dJ

= 0.22 mm < 0.25mm Crack width satisfied

Therefore, provide T 16@ 100 mm

Minimum area of secondary reinforcement = 0.12 %ofbd

For grad~ of 460, reinforcement = 0.12x1000x192 100 •

= ?~n mm2/m

Output

T 16 @ 100

..

3S 5400-4

1r.9o :15.3.2 3

s 5400-4 1990 5.8 4

I Use 10 mm diameter Tor steel

No. of bar required = 2.93

No. of bar provided = 4

Spacing of bar = 250 mm

Area of reinforcement provided = 314 mm2/m

Therefore, provide T 10@ 250 mm

Ultimate bending moment, Mu = 38.11 KNm/m

Assume,Serviceble bending moment Ms = Mu/1.5 = 25.4 KNm/m

Assume 12 mm diameter Tor steel can be used

h= 250 mm

Effective depth for cover of 50mm, d= 250-50-6

from equations (1) and (2)

= 194 mm

b = 1000 mm

feu= 50 N/mm2

!y = 460 N/mm2

M 0.020 < 0.15 ---

hd2 fcu Single reinforcement is enough

Mu = 0.87 /yA.Z ----(1)

Z 1.1 +A = {1- Jy s)d fcubd

----(2)

zz -dZ + 1.1Mu =0 0.87 fcub

Z 2 -I94Z + 964 = o Z = 189 mm > 0.95d

If Z > 0.95d, therefore Z = 0.95d Z = 184 mm

M A=--

• 0.87Zfy

= 517 mm2

IOOAS = 0.266

bd Which is greater than the minimum of 0.15% of bd

Therefore A = s 517 mm2

No. of bar required = 4.6

No. of bar provided = 8 Spacing of bar = 1000

8 . ~?t:: ........

T 10

@ 250

Reference

as· 5400-4

1990 cl ~.8.4.2

..... ~

Calculation

Area of reinforcement provided = 905 mm2

Checking of crack width for top flange

Assume reinforcement provided T 12 @ 125 mm

&m =&1-&2 = 0.0008 > 0

Therefore section is cracked •

3acr&m Design crack width =

l+2(acr -cco%-dJ

= 0.17 mm <0.25mm

Crack width satisfied

Therefore, provide T 12 @ 125 mm

Crack width is calculated using above precedure in the top of top flange

Seconda~ reinforcement Minimum area of secondary reinforcement = 0.12 %ofbd

For grade of 460, reinforcement = 0.12x1000x194 100

= 233 mm2/m

Use 10 mm diameter Tor steel

No. of bar required = 2.96 No. of bar provided = 4 Spacing of bar = 250 mm

Area of reinforcement provided = 314 mm2/m

Therefore, provide T 10@ 250 mm

Maximum transverce ultimate bending moment of bottom flange upto 1. ?Om from edge of the beam using grillage analvsis

Combination Distance(m)

0 1.6 3.2 Combination3 -112.87 36.48 -112.45 Combination4 -116.4 38.34 -119.31 CombinationS -108.99 36.23 -106.67 Max bending moment -116.4 38.34 -119.31

,.. ,... "· "" ~ ...... '-. ~ ~ l)

Output

T 12 @ 125

T 10 @ 250

r I

--Reference

) 5400-4 }90

5.3.2.3

5400-4 ~0

.8_4

·- ·~-

Calculation Output

Ultimate Bending Moment

60

'E 40 ... 20 / ~ "E z 0

~ - -20 0.5 1 1.5 2 2.5 3 3.5 c: Gl E -40 0

Distance(m) ::E -60 Cl c: -80 'i5 c: -100 Gl -Ol

-120 j

-140

Design of toQ reinforcement of bottom flange Ultimate bending moment, Mu = 119.31 KNm/m

Assume,Serviceble bending moment Ms = Mu/1.5 = 79.5 KNm/m

Assume 16 mm diameter Tor steel can be used

h= 250 mm Effective depth for cover of 50mm, d= 250-50-8

= 192 mm b = 1000 mm

feu= 50 N/mm2

fy = 460 N/mm2 '·

M 0.065 < 0.15 =

bd2fcu Single reinforcement is enough

Mu = 0.87 f;,A.Z ----(1)

z I. I}; A, (2) = (1- y )d ----fcubd

from equations (1) and (2)

Z 2 -dz + I.IMu =0

0.87fcub

z 2 - 192 z + 3017 = 0

z = 175 mm <0.95d

If Z < 0.95d, therefore Z=Z z = 175 mm

A= M s 0.87Z/"y

= 1704 mm2

lOOA. = 0.887 bd

Wbich is greater than the minimum of 0.15% of bd .

A

Therefore A,= 1704

No. of bar required = 8.5 No. of bar provided = 13.33 Spacing of bar = 1000

13.33 = 75 mm

Area of reinforcement provided = 2681 mm2

Assume reinforcement provided T 16 @ 75 mm

&m =&1-&2 = 0.0013 > 0

Therefore section is cracked

Design crack width 3acr&m

= 1+2(acr-Cco%-dJ

= 0.24 mm < 0.25mm Crack width satisfied

Therefore, provide T 16 @ 75 mm

Crack width is calculated using above precedure in top of top flange

BS 5400-4 Minimum area of secondary reinforcement = 0.12 % ofbd 1990 cl5.8.4.2 For grade of 460, reinforcement = 0 .12x1 000x192

100

= 230 mm2/m

Use 10 mm diameter Tor steel

No. of bar required = 2.93 No. of bar provided = 4 Spacing of bar = 250 mm

Area of reinforcement provided = 314 mm2/m

Therefore, provide T 10@ 250 mm

:;ign of bottom reinforcement of bottom flange ugto 1.70m from edge Ultimate bending moment, Mu = 38.34 KNm/m

Assume,Serviceble bending moment Ms = Mu/1.5

= 25.6 KNm/m

Assume 12 mm diameter Tor steel can be used

h = 250 mm Effective depth for cover of 50mm, d = 250-50-6

=

IT @

I

16 75

10 250

Reference

BS 5400-4 1990

cl 5.3.2.3

BS 5400-4 1990

cl5.8.4

-- I

from equations (1} and {2)

Calculation b = 1000 mm

fc..u = 50 N/mm2

fy = 460 N/mm2

M 0.020 < 0.15 ---

bd2fc.u

Single reinforcement is enough

Mu = 0.87 f;,A.Z --- -(1)

Z 1.1/, A. (2) = (I - Y )d - - - -fcubd

z2 -dZ + l.lMu =0 0.87/cub

z 2 - 194 z + 970 = 0

Z = 189 mm > 0.95d

If Z > 0.95d, therefore Z = 0.95d

Z = 184 mm

M A=--

s 0.87Zf;,

= 520 mm2

1 OOA. = 0.268 bd

Which is greater than the minimum of 0.15% of bd

Therefore

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Checking of crack width for top flange

Assume reinforcement provided

A = s 520 mm2

= 4.6

= 8

= 1000 8

= 125 mm

= 905 mm2

T 12 @ 125 mm

&m =&, -&2

= 0.0008 > 0

Therefore section is cracked

Design crack width 3acr&m

= 1+2(acr -Cco%-dJ

= 0.17 mm < 0.25mm

Crack width satisfied •

Output

,. .

Reference I Calculation l I Output

BS 5400-4 1990

cl 5.8.4.2

Therefore, provide T 12 @ 125 mm

Crack width is calculated using above precedure in top of top flange

Seconda!Y reinforcement Minimum area of secondary reinforcement = 0.12 % ofbd

For grade of 460, reinforcement = 0.12x1000x194 100

= 233 mm2/m

Use 10 mm diameter Tor steel

No. of bar required = 2.96 No. of bar provided = 4 Spacing of bar = 250 mm

Area of reinforcement provided = 314 mm2/m

Therefore, provide T 10@ 250 mm

Maximum transverce ultimate bending moment of bottom flange of interior slab of the beam using grillage analysis

.€ E z

X:: ;::: c Cl)

E 0

:::E CJ c :s c Cl)

a:a

-1

-2

-3

Combination I Distance(m) 0 1.6 3.2

Combination3 1.33 -2.01 0.25 Combination4 1.23 -2.11 2.12 CombinationS 2.57 -2.02 0.61 Max bending moment 2.57 -2.11 2.12

Ultimate Bending Moment

Design of top and bottom reinforcement of bottom flange

I

3

Ultimate bending moment, Mu = 2.57 KNrnlm

Assume,Serviceble bending moment Ms = Mu/1.5 = 1.7 KNm/m

Assume 12 mm d~ameter Tor steel can be used h = 200 mm

Effective depth for cover of 50mm. . d = . 200-50-6

3.5

T 12 @ 125

T 10 @ 250

Reterence Ca\cu\at\on

BS 5400-4 1990

cl5.3.2.3

IS 5400-4 1990 15.84

= 144 mm b = 1000 mm f..u = 50 N/mm2

~= 460 N/mm2

M 0.002 < 0.15 ---

bd2 fcu Single reinforcement is enough

Mu = 0.87 ~AsZ ----(1)

Z 1.11, A =(1- ~s)d

fcubd ----(2)

from equations (1) and (2)

z2 -dZ+ l.IMu =0 0.87 fcub

Z 2 -144Z +65 = o Z = 142 mm > 0.95d

If Z > 0.95d, therefore Z = 0.95d

Z = 137 mm

M As = 0.87ZJ;,

= 47

IOOAS = 0.033 bd

mm2

Which is not greater than the minimum of 0.15% of bd

Therefore

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Checking of crack width for top flanae

Assume reinforcement provided

A = s 216 mm2

= 1.9

= 8 = 1000

8 = 125 mm

= 905 mm2

T 12 @ 125 mm

Em =El -Ez

= -0.0029 < 0

Therefore section is uncracked

Therefore, provide T 12 @ 125 mm

Crack width is calculated using above precedure in top of top flange

Secondary reinforcement

S00-4 I Minimum area of secondarv reinforcement = 0 1? OL. nf brl

Output

T 12 @ 125

.-Reference I Calculation I Output I 1990

cl 5.8.4.2 For grade of 460, reinforcement

Use 10 mm diameter Tor steel

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

= 0.12x1000x192 100

= 173 mm2/m

= 2.20

= 4

= 250 mm

= 314 mm2/m

Therefore, provide T 10 @ 250 mm

Maximum transverce ultimate bending moment of web upto 3.00m from edge of the beam using grillage analysis

Combination Distance(m)

0 0.68 1.32 Combination3 23.57 -103.36 -117.61 Combination4 11.2 -112 -120.89 CombinationS 26.86 -98.12 -113.1 Max bending moment 26.86 -112 -120.89

Ultimate Bending Moment

40

:§ 20~ E

-2~ z "b.Z 0.4 0.6 0.8 1 ~

~ c -40 I CD

""'-E Distance(m) 0 -60

::E IJ) -80 c :c -100 c CD m -120

-140

Design of reinforcement of web Ultimate bending moment, Mu =

Assume,Serviceble bending moment Ms = =

Assume 16 mm diameter Tor steel can be used h=

Effective depth for cover of 50mm, d =

M

= b = feu= fy =

= bd2fcu

120.9

Mu/1.5 80.6

350 350-50-8

292 1000 50

460

0.028 .

SinniA rAinfnrr<>mi:>nt ic: onnoonh

KNrn/m

KNrn/m

mm

mm mm N/mm2

N/mm2

< 0.15

1.2 1.4

T 10 @ 250

.---Reference

BS 5400-4 1990

cl 5.3.2.3

BS 5400-4 1990

cl 5.8.4

s 5400-4 990 5.8.4.2

from equations (1) and (2}

Calculation

Mu = 0.87 fyAsZ

Z I.l~"A = (1- Jy s )d

z2 -dZ + 1.1Mu = 0 0.87 f:-ub

Z 2 -292Z +3057 = o

fcubd

----(1)

----(2)

Z = 281 mm > 0.95d

If Z > 0.95d, therefore Z = 0.95d Z = 277 mm

M As= 0.87Zfy

= 1089 mm2

IOOAS = 0.373

bd Which is greater than the minimum of 0.15% of bd

Therefore

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Checking of crack width for web

Assume reinforcement provided

A = 1089 mm2 s

= 5.4 = 8 = 1000

8

= 125 mm

= 1609 mm2

T 16 @ 125 mm

8 m =&1-&2

= 0.0010 >0

Therefore section is cracked

Design crack width =

= Crack width satisfied

3ac,em 1 + 2(acr -ccom)/

/(h-dc) 0.22 mm < 0.25mm

Therefore, provide T 16@ 125 mm

Crack width is calculated using above precedure in top of top flange

Secondary reinforcement Minimum area of secondary reinforcement = 0.12 % ofbd

For grade of 460. reinforcement = n 1?Y1nnnv?Q?

Output

T 16 @ 125

I Reference

Use 10 mm diameter Tor steel

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Calculation 100

= 350 mm2/m

= = =

=

4.46 5

200 mm

393 mm2/m

Therefore, provide T 10@ 200 mm

Maximum transverce ultimate bending moment of interior web of the beam using grillage analysis

--Distance(m)

Combination 0 0.68 1.32

Combination3 -9 2.12 -1.65 Combination4 -4.1 2.14 -0.1 CombinationS 2.8 2.2 3.1 Max bending moment -9 2.2 3.1

Ultimate Bending Moment

4

.E 2 E z ~

0

-r::: Gl -2

~ 0.6 0.8 1.2 0.2 0

E 0

::!!: -4 Distance(m)

Cl r::: -6 '6 r::: Gl -8 Ill

-10

Design of reinforcement of web Ultimate bending moment, Mu = 9 KNm/m

Assume,Serviceble bending moment Ms = Mu/1.5

= 6.0 KNm/m

Assume 12 mm diameter Tor steel can be used h= 350 mm

Effective depth for cover of 50mm, d= 350-50-6

= 294 mm

b = 1000 mm

feu= 50 N/mm2

fy = 460 N/mm2

M 0.002 < 0.15 =

bd2 fcu Single reinforcement is enough

~

1.4

.

Output

T 10 @ 200

Reference

BS 5400-4

11990 cl 5 3.2.3

3S 5400-4 1990 :15.84

) 5400-4 ~30

5.8.4.2

--

I I

from equations (1) and (2)

Calculation Mu = 0.87 fYA.,Z

Z I. If. A

= (I y ·' )d

z2-dZ+ I.lMu =0 0.87 f."Ub

Z 2 - 294Z + 228 = o

f:.ubd

----(1)

----(2)

Z = 293 mm > 0.95d

If Z > 0.95d, therefore Z = 0.95d Z = 279 mm

M As = 0.87 Zf;,

= 81 mm

100As = bd

0.027

Which is not greater than the minimum of 0.15% of bd

2

Therefore, A = 441 mm2

No. of bar required No. of bar provided Spacing of bar

Area of reinforcement provided

Checking of crack width for top flange

Assume reinforcement provided

s

= 3.9

= 6.67

= 1000 6.67

= 150 mm

= 755 mm2

T 12 @ 150 mm

8 m =&I -&2 = -0.0030 < 0

Therefore section is uncracked

Therefore, provide T 12 @ 150 mm

Crack width is calculated using above precedure in top of top flange

Seconda!Y reinforcement Minimum area of secondary reinforcement = 0.12 %ofbd

For grade of 460, reinforcement = 0.12x1 000x294 100

= 353 mm2/m

Use 10 mm diameter Tor steel

No. of bar required = 4.49 .

Output

Reference Calculation Output

No. of bar provided = 5 Spacing of bar = 200 mm

Area of reinforcement provided = 393 mm2/m

Therefore, provide T 1 0 @ 200 mm T 10

@ 200

-

.

I

,,

-----

l~Jll!D xog ~lp JO lfld lflO ~dWO:J

ZXIGN3ddV

.,.._, I '~ ,frlent forces -frames 'Station Outputcase case Type Step Type V2 T M2 M3

m Text Text Text KN KN-m KN-m KN-m

0.00 3 COMBl Combination -1191.699 -8.027E-14 2.224E-07 4026.564

0.00 3 COMBl Combination -1191.699 -8.027E-14 2.224E-07 4026.564

0.50 2 COMBl Combination -1142.132 -8.027E-14 2.054E-07 4608.8551

1.00 1 COMBl Combination -1092.564 -8.027E-14 1.885E-07 5166.412.

1.5 COMBl Combination -1042.997 -8.027E-14 1.716E-07 5699.H

1.999 COMBl Combination -993.43 -8.027E-14 1.547E-07 6207.3231

2.498 COMBl Combination -943.863 -8.027E-14 1.378E-07 6690.678:

2.997 COMB1 Combination -894.296 -8.027E-14 1.209E-07 7149.298:

2.997 COMB! Combination -894.296 -8.027E-14 1.209E-07 7149.298:

3 COMB! Combination -893.998 -8.027E-14 1.208E-07 7151.981:

0 COMB2 Combination Max -1175.509 220.606 2.225E-07 5561.953:

0.003 COMB2 Combination Max -1175.178 220.5727 2.224E-07 5566.932~

0.003 COMB2 Combination Max -1175.178 220.5727 2.224E-07 5566.932~

0.502 COMB2 Combination Max -1121.369 214.803 2.054E-07 6348.324~

1.001 COMB2 Combination Max -1067.56 209.0334 1.885E-07 7104.982:1

1.5 COMB2 Combination Max -1013.752 203.2638 1.716E-07 7836.905~

1.999 COMB2 Combination Max -959.943 197.4942 1.547E-07 8544.095~

2.498 COMB2 Combination Max -906.134 191.7246 1.378E-07 9226.551

2.997 COMB2 Combination Max -852.326 185.9549 1.209E-07 9884.2725

2.997 COMB2 Combination Max -852.326 185.9549 1.209E-07 9884.2725

3 COMB2 Combination Max -852.01 185.9188 1.208E-07 9887.909

0 COMB2 Combination Min -1664.485 -220.606 2.225E-07 4022.9894

0.003 COMB2 Combination Min -1664.106 -220.5727 2.224E-07 4026.5649

0.003 COMB2 Combination Min -1664.106 -220.5727 2.224E-07 4026.5649

0.502 COMB2 Combination Min -1599.818 -214.803 2.054E-07 4608.8556

1.001 COMB2 Combination Min -1535.53 -209.0334 1.885E-07 5166.4123

1.5 COMB2 Combination Min -1471.243 -203.2638 1.716E-07 5699.235

1.999 COMB2 Combination Min -1406.955 -197.4942 1.547E-07 6207.3236

2.498 COMB2 Combination Min -1342.667 -191.7246 1.378E-07 6690.6781

2.997 COMB2 Combination Min -1278.38 -185.9549 1.209E-07 7149.2987

2.997 COMB2 Combination Min -1278.38 -185.9549 1.209E-07 7149.2987,

3 COMB2 Combination Min -1277.986 -185.9188 1.208E-07 7151.9811

0 COMB3 Combination Max -1150.057 599.316 2.225E-07 6686.6294

0.003 COMB3 Combination Max -1149.639 599.316 2.224E-07 6692.8686

0.003 COMB3 Combination Max -1149.639 599.316 2.224E-07 6692.8686 -0.502 COMB3 Combination Max -1090.092 585.8432 2.054E-07 7598.5113

1.001 COMB3 Combination Max -1030.544 572.3703 1.885E-07 8479.42

1.5 COMB3 Combination Max -970.997 558.8974 1.716E-07 9335.5946

1.999 COMB3 Combination Max -911.45 545.4246 1.547E-07 10167.0352

2.498 COMB3 Combination Max -851.903 531.9517 1.378E-07 10973.7418 -2.997 COMB3 Combination Max -792.356 518.4788 1.209E-07 11755.7143

2-:997 COMB3 Combination Max -792.356 518.4788 1.209E-07 11755.7143

3 COMB3 Combination Max -792.058 518.3168 1.208E-07 11759.8011 t-·

0 COMB3 Combination Min -2079.877 -599.316 2.225E-07 4022.9894

0.003 COMB3 Combination Min -2079.579 -599.316 2.224E-07 4026.5649

0.003 COMB3 Combination Min -2079.579 -599.316 2.224E-07 4026.5649

0.502 COMB3 Combination Min -2010.052 -585.8432 2.054E-07 4608.8556

1.001 COMB3 Combination Min -1940.524 -572.3703 1.88SE-07 5166.4123 r··- 1.5 COMB3 Combination Min -1870.997 -558.8974 1.716E-07 5699.235

1.999 COMB3 Combination Min -1801.47 -545.4246 1.547E-07 6207.3236 -·--2.498 COMB3 Combination Min -1731.943 -531.9517 1.378E-07 6690.6781

2.997 COMB3 Combination Min -1662.416 -518.4788 1.209E-07 7149.2987

2.997 COMB3 Combination Min -1662.416 -518.4788 1.209E-07 7149.2987

3 COMB3 Combination Min -1661.878 -518.3168 1.208E-07 7151.9811

0 COMB1 Combination -893.998 -8.027E-14 1.208E-07 7151.9811

0.003 COMB1 Combination -893.7 -8.027E-14 1.207E-07 7154.6627

'"'"··) I ent Forces •:frames

Station OutputCase case Type Step Type V2 T M2 M3

m Text Text Text KN KN-m KN-m KN-m

2.498 COMB1 Combination 844.132 -8.027E-14 -4.721E-07 7588.251

2.997 COMB1 Combination 893.7 -8.027E-14 -0.000000489 7154.662

2.997 COMB1 Combination 893.7 -8.027E-14 -0.000000489 7154.662

3 COMB1 Combination 893.998 -8.027E-14 -4.891E-07 7151.981

0 COMB2 Combination Max 900.486 159.5566 -3.875E-07 12977.867

0.003 COMB2 Combination Max 900.871 159.5844 -3.876E-07 12975.5021

0.003 COMB2 Combination Max 900.871 159.5844 -3.876E-07 12975.5021

0.502 COMB2 Combination Max 963.662 163.9693 -4.045E-07 12523.368~

1.001 COMB2 Combination Max 1026.453 168.3542 -4.214E-07 12046.500~

1.5 COMB2 Combination Max 1089.244 172.7391 -4.383E-07 11544.898:

1.999 COMB2 Combination Max 1152.034 177.1241 -4.552E-07 11018.562:

2.498 COMB2 Combination Max 1214.825 181.509 -4.721E-07 10467.4921

2.997 COMB2 Combination Max 1277.616 185.8939 -0.000000489 9891.688~

2.997 COMB2 Combination Max 1277.616 185.8939 -0.000000489 9891.688~

3 COMB2 Combination Max 1277.986 185.9188 -4.891E-07 9887.90~

0 COMB2 Combination Min 519.51 -159.5566 -3.875E-07 9386.975~ r--·

0.003 COMB2 Combination Min 519.835 -159.5844 -3.876E-07 9385.186~

0.003 COMB2 Combination Min 519.835 -159.5844 -3.876E-07 9385.186S

0.502 COMB2 Combination Min 575.141 -163.9693 -4.045E-07 9075.2679

1.001 COMB2 Combination Min 630.447 -168.3542 -4.214E-07 8740.6149

1.5 COMB2 Combination Min 685.753 -172.7391 -4.383E-07 8381.2279

1.999 COMB2 Combination Min 741.058 -177.1241 -4.552E-07 7997.1069

2.498 COMB2 Combination Min 796.364 -181.509 -4.721E-07 7588.2518

2.997 COMB2 Combination Min 851.67 -185.8939 -0.000000489 7154.6627

2.997 COMB2 Combination Min 851.67 -185.8939 -0.000000489 7154.6627

3 COMB2 Combination Min 852.01 -185.9188 -4.891E-07 7151.9811

0 COMB3 Combination Max 1243.878 437.3176 -3.875E-07 15326.7952

0.003 COMB3 Combination Max 1244.416 437.4796 -3.876E..()7 15324.1064

0.003 COMB3 Combination Max 1244.416 437.4796 -3.876E-07 15324.1064

0.502 COMB3 Combination Max 1313.944 450.9525 -4.045E-07 14792.6315

1.001 COMB3 Combination Max 1383.471 464.4254 -4.214E..()7 14236.4225 I-·

1.5 COMB3 Combination Max 1452.998 477.8982 -4.383E-07 13655.4795

1.999 COMB3 Combination Max 1522.525 491.3711 -4.552E-07 13049.8025

2.498 COMB3 Combination Max 1592.052 504.844 -4.721E-07 12419.3914

2.997 COMB3 Combination Max 1661.58 518.3168 -0.000000489 11764.2463

2,997 COMB3 Combination Max 1661.58 518.3168 -0.000000489 11764.2463

3 COMB3 Combination Max 1661.878 518.3168 -4.891E..()7 11759.8011

0 COMB3 Combination Min 422.088 -437.3176 -3.875E-07 9386.9752 -··

0.003 COMB3 Combination Min 422.386 -437.4796 -3.876E-07 9385.1868

0.003 COMB3 Combination Min 422.386 -437.4796 -3.876E-07 9385.1868

0.502 COMB3 Combination Min 483.929 -450.9525 -4.045E-07 9075.2679

1.001 COMB3 Combination Min 545.471 -464.4254 -4.214E-07 8740.6149

1.5 COMB3 Combination Min 607.Q13 -477.8982 -4.383E-07 8381.2279

1.999 COMB3 Combination Min 668.555 -491.3711 -4.552E-07 7997.1069

2.498 COMB3 Combination Min 730.097 -504.844 -4.721E-07 7588.2518

2.997 COMB3 Combination Min 791.64 -518.3168 -0.000000489 7154.6627

2.997 COMB3 Combination Min 791.64 -518.3168 ..().000000489 7154.6627

3 COMB3 Combination Min 792.058 -518.3168 -4.891E..()7 7151.9811

0 COMB1 Combination 893.998 -8.027E-14 -4.891E..()7 7151.9811

0.003 COMB1 Combination 894.296 -8.027E-14 -4.892E..()7 7149.2987

0.003 COMB1 Combination 894.296 -8.027E-14 -4.892E..()7 7149.2987

0.502 COMBl Combination 943.863 -8.027E-14 -5.061E..()7 6690.6781

1.001 COMBl Combination 993.43 -8.027E-14 -5.231E..()7 6207.3236

1.5 COMBl Combination 1042.997 -8.027E-14 -0.00000054 5699.235

1.999 COMBl Combination 1092.564 -8.027E-14 -5.569E..()7 5166.4123

2.498 COMBl Combination 1142.132 -8.027E-14 -5.738E-OJ 4608.8556 -

-~ -r ~·· , - - me ___ - ----- ,_ -,-----,-- ... , 1t Forces -F1

Station Output Case Case Type Step Type V2 T M2 M3 m Text Text Text KN KN-m KN-m KN-m

2.99 7 COMB! Combination 1191.699 -8.027E-14 -5.907E-07 4026.5649

2.99 7 COMB! Combination 1191.699 -8.027E-14 -5.907E-07 4026.5649 3 COMB! Combination 1191.997 -8.027E-14 -5.908E-07 4022.9894 0 COMB2 Combination Max 1277.986 185.9188 -4.891E-07 9887.909

0.003 COMB2 Combination Max 1278.38 185.9549 -4.892E-07 9884.2725

0.003 COMB2 Combination Max 1278.38 185.9549 -4.892E-07 9884.2725 0.502 COMB2 Combination Max 1342.667 191.7246 -5.061E-07 9226.551 1.001 COMB2 Combination Max 1406.955 197.4942 -5.231E-07 8544.0954

1.5 COMB2 Combination Max 1471.243 203.2638 -0.00000054 7836.9058

1.999 COMB2 Combination Max 1535.53 209.0334 -5.569E-07 7104.9821

2.498 COMB2 Combination Max 1599.818 214.803 -5.738E-Q7 6348.3245

2.997 COMB2 Combination Max 1664.106 220.5727 -5.907E-Q7 5566.9328

2.997 COMB2 Combination Max 1664.106 220.5727 -5.907E-D7 5566.9328

3 COMB2 Combination Max 1664.485 220.606 -5.908E-Q7 5561.9533

0 COMB2 Combination Min 852.01 -185.9188 -4.891E-07 7151.9811

0.003 COMB2 Combination Min 852.326 -185.9549 -4.892E-07 7149.2987

0.003 COMB2 Combination Min 852.326 -185.9549 -4.892E-07 7149.2987

0.502 COMB2 Combination Min 906.134 -191.7246 -5.061E-07 6690.6781

1.001 COMB2 Combination Min 959.943 -197.4942 -5.231E-07 6207.3236

1.5 COMB2 Combination Min 1013.752 -203.2638 -0.00000054 5699.235

1.999 COMB2 Combination Min 1067.56 -209.0334 -5.569E-07 5166.4123

2.498 COMB2 Combination Min 1121.369 -214.803 -5.738E-07 4608.8556

2.997 COMB2 Combination Min 1175.178 -220.5727 -5.907E-D7 4026.5649

2.997 COMB2 Combination Min 1175.178 -220.5727 -5.907E-07 4026.5649

3 COMB2 Combination Min 1175.509 -220.606 -5.908E-07 4022.9894

0 COMB3 Combination Max 1661.878 518.3168 --4.891E-07 11759.8011

0.003 COMB3 Combination Max 1662.416 518.4788 -4.892E-07 11755.7143

0.003 COMB3 Combination Max 1662.416 518.4788 -4.892E-07 11755.7143

0.502 COMB3 Combination Max 1731.943 531.9517 -5.061E-07 10973.7418 ' 1.001 COMB3 Combination Max 1801.47 545.4246 -5.231E-07 10167.0352

1.5 COMB3 Combination Max 1870.997 558.8974 -0.00000054 9335.5946

1.999 COMB3 Combination Max 1940.524 572.3703 -5.569E-07 8479.42

2.498 COMB3 Combination Max 2010.052 585.8432 -5.738E-Q7 7598.5113

2.997 COMB3 Combination Max 2079.579 599.316 -5.907E-Q7 6692.8686

2.997 COMB3 Combination Max 2079.579 599.316 -5.907E-07 6692.8686

3 COMB3 Combination Max 2079.877 599.316 -5.908E-07 6686.6294

0 COMB3 Combination Min 792.058 -518.3168 -4.891E-07 7151.9811

0.003 COMB3 Combination Min 792.356 -518.4788 -4.892E-07 7149.2987

0.003 COMB3 Combination Min 792.356 -518.4788 -4.892E-07 7149.2987

0.502 COMB3 Combination Min 851.903 -531.9517 -5.061E-07 6690.6781

1.001 COMB3 Combination Min 911.45 -545.4246 -5.231E-07 6207.3236

1.5 COMB3 Combination Min 970.997 -558.8974 -0.00000054 5699.235

1.999 COMB3 Combination Min 1030.544 -572.3703 -5.569E-D7 5166.4123

2.498 COMB3 Combination Min 1090.092 -585.8432 -5.738E-D7 4608.8556

2.997 COMB3 Combination Min 1149.639 -599.316 -5.907E-D7 4026.5649

2.997 COMB3 Combination Min 1149.639 -599.316 -5.907E-07 4026.5649

3 COMB3 Combination Min 1150.057 -599.316 -5.908E-D7 4022.9894

0 COMB1 Combination 1191.997 -8.027E-14 -5.908E-07 4022.9894

0.003 COMB1 Combination 1192.295 -8.027E-14 -5.909E-07 4019.4129

0.003 COMB1 Combination 1192.295 -8.027E-14 -5.909E-07 4019.4129

0.502 COMB1 Combination 1241.862 -8.027E-14 -6.078E-07 3412.0908 1.001 COMB1 Combination 1291.429 -8.027E-14 -6.247E-D7 2780.0346

1.5 COMB! Combination 1340.996 -8.027E-14 -6.416E-07 2123.2444

1.999 COMB1 Combination 1390.564 -8.027E-14 -6.585E-07 1441.7201 2.498 COMB1 Combination 1440.131 -8-027E-14 -6.754E-07 735.4619 2.997 ~OMB1 Combination ·1489.698 -8.027E-14 -6.923E-07 4.4695

-----

~a ~ ;)Jqnoa ;}tp JO Jlld Jno J;)Jndwo;)

£XIGNHddV

8ridge0bj Distance OutputCase Case Type Step Type V2 T M3

Text m Text Text Text KN KN-m KN-m

BOBJl 12 COMB3 Combination Min -832.284 -356.3189 10958.532:

BOBJl 12 COMB3 Combination Max -832.284 -356.3189 10958.532J

BOBJl 12 COMB3 Combination Min -16.524 356.3189 17618.172J

BOBJl 15 COMB3 Combination Max 407.88 299.7 18075.137f

BOBJl 15 COMB3 Combination Min -407.88 -299.7 11415.137€

BOBJl 15 COMB3 Combination Max -407.88 -299.7 11415.137€

BOBJl 15 COMB3 Combination Min 407.88 299.7 18075.137«:

BOBJl 18 COMB3 Combination Max 832.284 356.3189 17618.1721

BOBJl 18 COMB3 Combination Min 16.524 -356.3189 10958.5321

BOBJl 18 COMB3 Combination Max 16.524 -356.3189 10958.5321

BOBJl 18 COMB3 Combination Min 832.284 356.3189 17618.1721

80811 21 COM83 Combination Max 1256.687 437.3188 15528.5356 80811 21 COMB3 Combination Min 434.897 -437.3188 9588.7156 BOBJl 21 COMB3 Combination Max 434.897 -437.3188 9588.7156 BOBJl 21 COMB3 Combination Min 1256.687 437.3188 15528.5356 808J1 24 COMB3 Combination Max 1681.091 518.3188 11913.5081 BOBJ1 24 COMB3 Combination Min 811.271 -518.3188 7305.6881 BOBJl 24 COMB3 Combination Max 811.271 -518.3188 7305.6881 BOBJl 24 COMB3 Combination Min 1681.091 518.3188 11913.5081 BOBJl 27 COMB3 Combination Max 2105.495 599.3187 6773.0895 BOBJ1 27 COMB3 Combination Min 1175.675 -599.3187 4109.4495 BOBJ1 27 COMB3 Combination Max 1175.675 -599.3187 4109.4495

IBOBJl 27 COMB3 Combination Min 2105.495 599.3187 6773.0895 IBOBJ1 30 COMB3 Combination Max 2529.999 680.3996 0 BOBJl 30 COMB3 Combination Min 1522.018 -680.3996 0

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OUT PUT DATA FOR DOUBLET BEAM

COMB! Moment about Horizontal axis

Distance p V2 V3 T M2 M3

m KN KN KN KN-m KN-m KN-m

0 -2.95E-07 -1522.018 6.84E-09 1.75E-14 6.39E-08 -1.33E-06

3 -2.95E-07 -1217.615 6.84E-09 1.75E-14 4.34E-08 4109.4495

3 -2.95E-07 -1217.615 6.84E-09 1.75E-14 4.34E-08 4109.4495

6 -2.95E-07 -913.211 6.84E-09 1.75E-14 2.29E-08 7305.6881

6 -2.95E-07 -913.211 6.84E-09 1.75E-14 2.29E-08 7305.6881

9 -2.95E-07 -608.807 6.84E-09 1.75E,14 2.37E-09 9588.7156

9 -2.95E-07 -608.807 6.84E-09 1.75E-14 2.37E-09 9588.7156

12 -2.95E-07 -304.404 6.84E-09 1.75E-14 -1.82E-08 10958.5321 12 -2.95E-07 -304.404 6.84E-09 1.75E-14 -1.82E-08 10958.5321

15 -2.95E-07 -3.02E-08 6.84E-09 1.75E-14 -3.87E-08 11415.1376

15 -2.95E-07 -3.01E-08 6.84E-09 1.75E-14 -3.87E-08 11415.1376

18 -2.95E-07 304.404 6.84E-09 1.75E-14 -5.92E-08 10958.5321

18 -2.95E-07 304.404 6.84E-09 1.75E-14 -5.92E-08 10958.5321

21 -2.95E-07 608.807 6.84E-09 1.75E-14 -7.97E-08 9588.7156 21 -2.95E-07 608.807 6.84E-09 1.75E-14 -7.97E-08 9588.7156 24 -2.95E-07 913.211 6.84E-09 1.75E-14 -l.OOE-07 7305.6881

24 -2.95E-07 913.211 6.84E-09 1.75E-14 -l.OOE-07 7305.6881

27 -2.95E-07 1217.615 6.84E-09 1.75E-14 -1.21E-07 4109.4495 27 -2.95E-07 1217.615 6.84E-09 1.75E-14 -1.21E-07 4109.4495

30 -2.95E-07 1522.018 6.84E-09 1.75E-14 -1.41E-07 -4.23E-07

COMB3

Distance Item Type p V2 V3 T M2 M3

m KN KN KN KN-m KN-m KN-m

0 Max -2.95£-07 -1522.018 1.54£-04 680.3996 -0.0046 -1.33£-0(

0 Min -4.56£-07 -2530.018 -1.54£-04 -680.3996 0.0046 -2.06£-0(

3 Max -2.95E-07 -1175.675 1.54£-04 599.3187 -0.0042 6773.0895 3 Min -4.56£-07 -2105.495 -1.54£-04 -599.3187 0.0042 4109.4495 3 Max -2.95E-07 -1175.675 1.54£-04 599.3187 -0.0042 6773.0895 3 Min -4.56E-07 -2105.495 -1.54E-Q4 -599.3187 0.0042 4109.4495 6 Max -2.95E-Q7 -811.271 1.54£-04 518.3188 -0.0037 11913.508 6 Min -4.56£-07 -1681.091 -1.54£-04 -518.3188 0.0037 7305.6881 6 Max -2.95£-07 -811.271 1.54£-04 518.3188 -0.0037 11913.508 6 Min -4.56£-07 -1681.091 -1.54£-04 -518.3188 0.0037 7305.6881 9 Max -2.95E-07 -434.897 1.54£-04 437.3188 -0.0032 15528.536 9 Min -4.56E-07 -1256.687 -1.54£-04 -437.3188 0.0032 9588.7156 9 Max -2.95E-07 -434.897 1.54£-04 437.3188 -0.0032 15528.536 9 Min -4.56E-07 -1256.687 -1.54E-04 -437.3188 0.0032 9588.7156

12 Max -2.95E-07 -16.524 1.54E-Q4 356.3189 -0.0028 17618.172 12 Min -4.56E-07 -832.284 -1.54E-Q4 -356.3189 0.0028 10958.532 12 Max -2.95E-07 -16.524 1.54E-Q4 356.3189 -0.0028 17618.172

12 Min -4.56E-Q7 -832.284 -1.54E-D4 -356.3189 0.0028 10958.532 15 Max -2.95E-07 407.88 1.54£-04 299.7 -0.0023 18075.138 1

15 Min -4.56E-07 -407.88 -1.54£-04 -299.7 0.0023 11415.138 15 Max -2.95E-07 407.88 1.54£-04 299.7 -0.0023 18075.138 15 Min -4.56E-07 -407.88 -1.54£-04 -299.7 0.0023 11415.138

18 Max -2.95£-07 832.284 1.54£-04 356.3189 -0.0018 17618.172 18 Min -4.56£-07 16.524 -1.54E-04 -356.3189 0.0018 10958.532

18 Max -2.95£-07 832.284 1.54£-04 356.3189 -o.0018 17618.172 18 Min -4.56E-07 16.524 -1.54E-Q4 -356.3189 0.0018 10958.532 21 Max -2.95E-07 1256.687 1.54£-04 437.3188 -0.0014 15528.536 21 Min -4.56E-07 434.897 -1.54£-04 -437.3188 0.0014 9588.7156 21 Max -2.95£-07 1256.687 1.54£-04 437.3188 -0.0014 15528.536 21 Min -4.56£-07 434.897 -1.54E-04 -437.3188 0.0014 9588.7156 24 Max -2.95£-07 1681.091 1.54£-04 518.3188 -9.24£-04 11913.508 24 Min -4.56E-07 811.271 -1.54E-04 -518.3188 9.24£-04 7305.6881 --· 24 Max -2.95E-07 1681.091 1.54E-04 518.3188 -9.24E-Q4 11913.508 24 Min -4.56£-07 811.271 -1.54E-04 -518.3188 9.24E-04 7305.6881 27 Max -2.95£-07 2105.495 1.54£-04 599.3187 -4.62£-04 6773.0895 27 Min -4.56£-07 1175.675 -1.54£-04 -599.3187 4.62£-04 4109.4495 27 Max -2.95E-07 2105.495 1.54£-04 599.3187 -4.62£-04 6773.0895 27 Min -4.56E-07 1175.675 -1.54E-04 -599.3187 4.62E-04 4109.4495 30 Max -2.95E-07 2529.999 1.54£-04 680.3996 -2.18E-Q7 -4.23£-07

30 Min -4.56E-07 1522.018 -1.54£-04 -680.3996 -1.41£-07 -6.54£-07 "--·-

COMB2

Distance Item Type p V2 V3 T M2 M3

m KN KN KN KN-m KN-m KN-m

0 Max -2.95E-07 -1522.018 5.41£-05 263.6248 -0.0016 -1.33£-01 0 Min -3.99E-07 -2092.018 -5.41£-05 -263.6248 0.0016 -1.80£-01

3 Max -2.95E-07 -1201.127 5.41£-05 220.6068 -0.0015 5648.413L 3 Min -3.99E-07 -1690.103 -5.41£-05 -220.6068 0.0015 4109.449~

3 Max -2.95E-07 -1201.127 5.41£-05 220.6068 -0.0015 5648.413~

3 Min -3.99£-07 -1690.103 -5.41£-05 -220.6068 0.0015 4109.449~

6 Max -2.95£-07 -871.223 5.41£-05 185.9194 -0.0013 10041.61€ 6 Min -3.99E-07 -1297.199 -5.41£-05 -185.9194 0.0013 7305.6881 6 Max -2.95£-07 -871.223 5.41E-Q5 185.9194 -0.0013 10041.616 6 Min -3.99£-07 -1297.199 -5.41£-05 -185.9194 0.0013 7305.6881 9 Max -2.95£-07 -532.319 5.41£-05 159.5569 -0.0011 13179.608 9 Min -3.99£-07 -913.295 -5.41£-05 -159.5569 0.0011 9588.7156 9 Max -2.95£-07 -532.319 5.41£-05 159.5569 -0.0011 13179.608 9 Min -3.99£-07 -913.295 -5.41£-05 -159.5569 0.0011 9588.7156

12 Max -2.95E-Q7 -184.416 5.41E-Q5 141.5194 -9.74£-04 15062.388 12 Min -3.99E-Q7 -538.392 -5.41£-05 -141.5194 9.74£-04 10958.532 12 Max -2.95£-07 -184.416 5.41£-05 141.5194 -9.74£-04 15062.388 12 Min -3.99£-07 -538.392 -5.41£-05 -141.5194 9.74E-04 10958.532 15 Max -2.95£-07 172.488 5.41E-Q5 131.8069 -8.12E-04 15689.958 15 Min -3.99£-07 -172.488 -5.41£-05 -131.8069 8.11£-04 11415.138 15 Max -2.95£-07 172.488 5.41E-05 131.8069 -8.12£-04 15689.9581

15 Min -3.99£-07 -172.488 -5.41£-05 -131.8069 8.11£-04 11415.138 18 Max -2.95£-07 538.392 5.41£-05 141.5194 -6.49£-04 15062.388 18 Min -3.99E-Q7 184.416 -5.41E-05 -141.5194 6:49E-04 10958.532 18 Max -2.95E-07 538.392 5.41E-05 141.5194 -6.49E-04 15062.388 18 Min -3.99E-07 184.416 -5.41E-05 -141.5194 6.49E-04 10958.532 21 Max -2.95E-07 913.295 5.41£-05 159.5569 -4.87E-04 13179.608

21 Min -3.99E-07 532.319 -5.41E-05 -159.5569 4.87E-04 9588.7156 21 Max -2.95E-Q7 913.295 5.41E-05 159.5569 -4.87E-04 13179.608

21 Min -3.99£-07 532.319 -5.41E-05 -159.5569 4.87E-04 9588.7156 24 Max -2.95E-07 1297.199 5.41E-05 185.9194 -3.25E-04 10041.616 24 Min -3.99E-07 871.223 -5.41E-Q5 -185.9194 3.25E-04 7305.6881 24 Max -2.95E-07 1297.199 5.41£-05 185.9194 -3.25E-04 10041.616 24 Min -3.99E-07 871.223 -5.41E-Q5 -185.9194 3.25E-04 7305.6881 27 Max -2.95E-07 1690.103 5.41E-05 220.6068 -1.62E-04 5648.4134 27 Min -3.99E-07 1201.127 -5.41E-05 -220.6068 1.62E-Q4 4109.4495 27 Max -2.95E-07 1690.103 5.41E-05 220.6068 -1.62E-04 5648.4134 27 Min -3.99E-07 1201.127 -5.41E-05 -220.6068 1.62E-04 4109.4495 30 Max -2.95E-07 2091.973 5.41E-05 263.6248 -1.91E-07 -4.23E-Q7

30 Min -3.99E-07 1522.018 -5.41E-05 -263.6248 -1.41E-07 -5.71E-07

PXICINHddV

....

Distance Item Type V2

m KN

0 Max -1490 0 Min -2060 3 Max -1176 3 Min -1664 3 Max -1176 3 Min -1664 6 Max -852 6 Min -1278 6 Max -852 6 Min -1278 9 Max -520 9 Min -900 9 Max -520 9 Min -900 12 Max -178 12 Min -532 12 Max -178 12 Min -532 15 Max 172 15 Min -172 15 Max 172 15 Min -172 18 Max 532 18 Min 178 18 Max 532 18 Min 178 21 Max 900 21 Min 520 21 Max 900 21 Min 520 24 Max 1278 24 Min 852 24 Max 1278 24 Min 852 27 Max 1664 27 Min 1176 27 Max 1664 27 Min 1176 30 Max 2060 30 Min 1490

Shear Force ,Bending Moment Torsion BOX DOUBLET

COM2 COM3 COM2

T M3 V2 T M3 V2 T M3 V2

KN-m KN-m KN KN-m KN-m KN KN-m KN-m KN

264 0 -1490 680 0 -1522 264 0 -1522 -264 0 -2498 -680 0 -2092 -264 0 -2530 221 5562 -1150 599 6687 -1201 221 5648 -1176

-221 4023 -2080 -599 4023 -1690 -221 4109 -2105 221 5562 -1150 599 6687 -1201 221 5648 -1176

-221 4023 -2080 -599 4023 -1690 -221 4109 -2105 186 9888 -792 518 11760 -871 186 10042 -811

-186 7152 -1662 -518 7152 -1297 -186 7306 -1681 186 9888 -792 518 11760 -871 186 10042 -811

-186 7152 -1662 -518 7152 -1297 -186 7306 -1681 160 12978 -422 437 15327 -532 160 13180 -435

-160 9387 -1244 -437 9387 -913 -160 9589 -1257 160 12978 -422 437 15327 -532 160 13180 -435

-160 9387 -1244 -437 9387 -913 -160 9589 -1257 142 14832 -10 356 17388 -184 142 15062 -17

-142 10728 -826 -356 10728 -538 -142 10959 -832 142 14832 -10 356 17388 -184 142 15062 -17

-142 10728 -826 -356 10728 -538 -142 10959 -832 132 15450 408 300 17835 172 132 15690 408

-132 11175 -408 -300 .11175 -172 -132 11415 -408 132 15450 408 300 17835 172 132 15690 408

-132 11175 -408 -300 11175 -172 -132 11415 -408 142 14832 826 356 17388 538 142 15062 832

-142 10728 10 -356 10728 184 -142 10959 17 142 14832 826 356 17388 538 142 15062 832

-142 10728 10 -356 10728 184 -142 10959 17 160 12978 1244 437 15327 913 160 13180 1257

-160 9387 422 -437 9387 532 -160 9589 435 160 12978 1244 437 15327 913 160 13180 1257

-160 9387 422 -437 9387 532 -160 9589 435 186 9888 1662 518 11760 1297 186 10042 1681

-186 7152 792 -518 7152 871 -186 7306 811 186 9888 1662 518 11760 1297 186 10042 1681

-186 7152 792 -518 7152 871 -186 7306 811 221 5562 2080 599 6687 1690 221 5648 2105

-221 4023 1150 -599 4023 1201 -221 4109 1176 221 5562 2080 599 6687 1690 221 5648 2105

-221 4023 1150 -599 4023 1201 -221 4109 1176 264 0 2498 680 0 2092 ·, 264 /I .·, IQ 2530

-264 0 1490 -680 0 1522 ·-264 - .· _. 0. /i522 - -- -- :____ _______ -· __ ____:___ ------- -- - ..

COM3

T M3

KN-m KN-m

680 (

-680 (

599 6773 -599 4109 599 6773

-599 4109 518 11914

-518 7306 518 11914

-518 7306 437 15529

-437 9589 437 15529

-437 9589 356 17618i

-356 10959 356 17618

-356 10959 300 18075

-300 11415 300 18075

-300 11415 356 17618

-356 10959 356 17618

-356 10959 437 15529

-437 9589 437 15529

-437 9589 518 11914

-518 7306 518 11914

-518 7306 599 6773

-599 4109 599 6773

-599 4109 680 0

-680 0 -