Balance from-4 Flow Beam FB/2

Reinforce for –Ve moment at B=83kn

End portion will be designed as rectangular section

Mulim as calculated earlis is more than MuD is 83kn

So the section is under reinforced,

So 83 x 106=0.87fyAst.d(1-Astfy/bdfck)

or,83 x 106 =141532Ast-29.97Ast2

Or,Ast=686mm2

Use 3-16Φ+1-12Φ at the support top

Negative moment at C =16.27Kn/m

Use 2-12Φ at top 0f section at C.

Transverse Reinforce

% steel at support =(100 x 3 x 113.1)/(250 x 3.92)=0.345%

τcfck 0.346%will be from table=0.36+(0.12/0.25) x 0.096=0.41N/mm2

τuc > τc,Transverse shear Reinforce essestial.

Asu =τu,Su/b1d1(0.87f)+VuSu/2.5d1(0.87fy)

Assume 8Ф stirrap b1=250-(25-4) x 2=208

d1=425-(25-4) x 2=383

6/208 x 383)+(63.34 x
(Asu x 0.87fy/Su)=(τu/b1d1+Vu/2.5 X d1)=(8.43 x 10
1000/383)=106+165=271N/mm

Also Asu =(τve-τc) x bsu/0.87fy

Or,Asu x 0.87fy/su=(τve-τc)b=(0.669-0.41) x 250=65N/mm

We know,(Asu/Su)=271/0.87fy=0.75

Su=(2 x 50.3/0.75)=134N/mm
Hance stirrap 8Ф @130 c/c

This should not excud x1(x1+y1)/4

X1=250-2 x 2=200, y1425-2 x 25=375

(x1+y1)/4=200+375/4=144

Hance 8Ф @130 c/c satisfy the coad,

Beam FB/5 ,FB/6 ,FB/4 ,FB/5 ,FB/6

K=1

K=3.72 N=3.72 O=3.34 P=3,72 Q=3.72 R

K=I/L, 700cos=1146, Table,250 x 425, K=4265, Table=250 x 350,

K =3828, col:Beam:Beam

1:3.72:3.34

DF-MN=(3.72/5.72)=0.65, MN=(3.72/9.44)=0.39, NO=0.39,

ON=0.41, OP=0.41, PO=0.41, PQ=0.41, QR=0.39, QP=0.39, RQ=0.65,

Mome Spa F.EMdue Stirrap moment S.F

nt n todu+lim
FB/ MN/Q 4.5 (47.13 X (47.13 X (47.13 X
5 R 4.52/12)=79.53 4.52/8)=119.3
4.252/2)=100.
15
FB/ NO/Q 4.5 (47.356 X (47.356 X (47.356 X
6 P 4.52/12)=79.91 4.52/8)=119.87 4.52/2)=100.6
3
FB/ OP/2. 2.8 (16.26 X (16.26 X (16.26 X
4 8 282/12)=10.62 282/8)=15.93 282/2)=22.76
(60 X (60 X (60 X
2.8)/8=(21.00/3 2.8)/4=(42.00/5 57/2)=(28.00/
1.62) 7.93) 51.6)

MN, >Span(Span as AB)=(250 x 425)=47.12N/mm

NO, >Span(Span as BC)=(250 x 425)=26.56N/mm

OP, >Span(Span as op)=(250 x 350)=12.50N/mm

Now for slab w.t=0.35 x 0.25 x 25 x 1.5=3.28

Wall w.t=1 x 3.06 x 0.2 x 19 x 0.744 x 1.5=12.98, TOTAL=16.26

Point M N O P

D.F MN NM NO QN OP PQ PQ

D.F 0.65 0.39 0.39 0.41 0.41 0.41 0.41

F.EM -79.53 +79.53 -79.91 +79.91 -31.62 +31.62 -79.91

Dist +51.69 +0.148 +0.148 -19.80 -19.80 +19.80 +19.80

c.o +0.074 +25.84 -9.90 +0.074 +9.90 -9.90 +19.90

Dist -0.048 -6.22 -6.22 -4.03 -4.03
-27.814 +99.303 +95.852 +56.154 -45.55

Due to beam=FB/13=60.00, FE/4=(60 X 2.8)/8=21K/mm

M27.81 N O P Q R27.81

95.85 45.55 +12.4 45.65

F.EM

Beam MN & QR/FB15=250 X 450

+Ve moment 119.3-(27.81 + 99.303/2)=55.74

-ve moment at M & R=27.81

-Ve moment at N & Q=99.303

Beam NO & PQ /FB16=250 X 450

+Ve moment at =119.87-(95.86 x 56.16/2)=43.87

-ve moment at N & Q=95.88

-ve moment at O & P=56.154

Beam OP –FB/15 250X350

+Ve Moment – 57.93-45.55=12.38

-Ve Moment at P80=45.55

Beam- FB 5 450x250 MM

1700 +ve =55.74

155 -Ve at M =27.81

450 -Ve at M =99.303

Bf=(L0/6+6X X DF+bw)

250 =0.7 x 4.5 + 6 x 155 + 250=1700

Using 16Ф rod, d=450–25–8=417mm

Eff. Span=4.5m

D=√99.303 x 106/2.761 x 250=379<417

MuD =55.74

Mulim =.36fck x df x bf x Df x (d-.42 x Df)

=.36 x 20 x 1700 x 155 x (417-65.1)

=1.8972 x 351.9 x 106

=667.62Kn.m>MUD

So it is under reinforce section

AST =.5fck/fy(1-√1-4.6MUD/fck x bf x d2) x bf x d

-Ve moment 56.154 KN.M

1.874x106 = 5024 Ast2

Ast = 5024-4212 = 406m2

Shear force
100.63𝑋1000
S.F Vu = 100.63, there = = 0.965 N/m2
250𝑋417
100𝑋3𝑋113.1
% of poisons ratio = = 0.325
250𝑋417
0.12
ĨC = 0.36+ X 0.075 = 0.396 N/m2
0.25

VuS = Vu- 0.396X250X417 = 100630-41283 = 59347 N

0.87𝑋415𝑋2𝑋50.3𝑋417
Space of 8ɸ stirrup = = 255
59347

Space of 8ɸ stirrup @ 250 C\C

Beam FB14(0P) 250X350

𝑙𝑜
700 bf= + 𝑏𝑤 + 3𝑑𝑦
12
2.8∗.7
= +250+300
12

=163+250+300

=713 ͌700

Calculate beam-

+ve=12.38kn/m

-Ve=45.5kn/m

45.55∗10⌃6
Req d=√ =257
2.76∗250

d=350-25-6 =319>257

effient =2.8m
+ve moment of midsfe =12.38

Xu max =0.48*319=153

mediam=0.36*20*250*153(319-42*153)

=0.2754*106*255 =70.227*106kn>12.38
𝑎𝑠𝑡∗𝑓𝑦
Therefore, 12.38*106=.8780*ast*d(1- )
𝑓𝑎𝑘𝑏𝑑

415𝑎𝑠𝑡
=.87*415*319ast(1- )
20∗250∗319

=115175*29.97ast2

0.413*106=384ast-ast2

Therefore,
3843−3622
Ast=
2

=111m2

Mean ‘m ruf’
Ast 0.85 0.85𝑋250
= ast = = 163m
bd 𝑓𝑦 415

So use – 2-12

-ve moment = 45.55 < Mulim

𝑎𝑠𝑡 𝑓𝑦
45.55X106 =0.87fy ast X d (1- )
𝑓𝑐𝑙 𝑏𝑑

415 𝑎𝑠𝑡
= 0.87X415X319 ast (1- )
20𝑋250𝑋313

=115175 ast – 29.97 ast2

1.52X106=3843ast-ast2
3843−2948
Ast= = 448mm2
2

Use-u12ɸ are= 4X113.1 = 452.4m2

Shear force
 52760
Vu= 52.76KN, ĩu= = 0.662nm2
250319
100𝑋2𝑋113.1
Area of rif = = 0.284
250𝑋319
0.12
Ĩc = 0.36+ X0.034 = 0.376 N/m2
0.25

BEAM FB/16

CALCULATION 250*450

+ve moment = 43.87

-ve moment = 95.88

-ve moment = 56.154

d = 450-25-8=417

= 4.5m

Bf = 1700
95.88∗10
D =√ = 372<417
2.761∗250
+ve moment = 43.83

=0.36*fok*bf*Df(d-0.42 Df )

=0.36 * 20*1700*155 (417-0.42*155)

=1.8972*326.9*106=620.19*106 kn/h>43.87

Under section
415 𝑎𝑠ℎ
Therefour, 43.87*106 = 0.87 fy * ash *d(1- )
𝑏𝑓∗𝑑∗𝑓𝑐𝑙𝑒

415𝑎𝑠ℎ
=.87*415*417 ash (1- )
1700∗417∗20

=141532 ash – 4.41 ash

W 9.948*106 = 32093-ash
32093−31468
Ash= = 313 m2
2

Wlc-3-12 R atbftm

-ve moment 95.88 kn/m

Xu max. =0.48*417=200.16 muti = 0.36* xcle * d * xu max. *(d-

0’’42 xu max. )

= 0.36*20*250*260.16(417-0.42*200.16)

= 0.3607*333 = 119.82 kn/m> 95.88

Undwe requr sec.

𝑎𝑠𝑡𝑓𝑦
Therefour, 95.88*106 =0.87 fy* sh *d (1- )
𝑏𝑑𝑓𝑐𝑘

𝑎𝑠𝑡∗415
=0.87*415*417 ast (1- )
250417∗20

=150588ast -2937 ast2

W 3.2*106=5024 ast- ast2

 5024−3527
Ast = =749 mm2
2

use-4-16 of at top

0.5𝑓𝑐𝑘 4.6∗𝑚𝑢𝑑
ast= [1-√1 − ] log 𝑑
415 𝑓𝑐𝑘 log 𝑑

10
= [1-0.978]1700*417= 376mm2
415

Use_ 2-16 of at top

Hegeli 99.303 at n & q

X max.=0.48, x max=.48*47= 200

Then, = 0.36 fek*d*x max.*(d-0.42x max.)

=0.36*20*250*200(417-0.42*200)

=0.336*106*333=119.88 km/h>99.303 km/h

Sec. b under require

𝑎𝑠𝑡∗𝑡𝑦
So 99.303*106=0.87 fo*ast*d(1- )
𝑙𝑜𝑑 𝑓𝑜𝑘

15055 8 ast-as2

W 3.313*106=5024 ast –art2

5024−3464
Ast = = 780 m2
2

Use 4-16ᶲ

s.f =100’15 kh
100150 100150
Ʈve= = =0.916 kn/m2
𝑏𝑑 250∗417
100∗2∗201
%of stel= = 0.386
250∗417

0.12
Ʈc=0.’36+ *0.136 = 0.425 kn/m2+
.25

Vus= vu-Ʈcbd= 100 sd -.425*250*447

 =100150-44306= 55844kh

.87∗415∗2∗50.3∗417
Spec.981= = 271
55844

Use 8ᶲ @ 270 cle all this.

Beam FB/7 & FB/8 (AG,GM)

A G B

5.0 4.5

LOADING

AG

Self wt = 0.45x0.25x1.5=4.22 kn/m

Wall = 0.80x0.2x2.96x19x1.5= 13.50

Slab 19.82

37.54 Kn/m

GM

Self Wt = 0.4x0.25x25+1.5 = 3.75

Wall = 0.8x0.2x3x19x1.5= 13.73
Slab = 18.63
36.11 Kn/m

MEMBER SPAN F.E.M

SPAN MEM
AG 5.0 37.54X52/12= 78.2
( (37.54X52/8)=117.31

GM 4.536.11X4.52/12=60.94

(36.11X4.52/8) = 91.40

S.F
 (37.54X4.75/2) = 89.16

K = for col = 1145 D.F = AG = 0.67

(36.11X4.25/2) = 76.73

Beam =4556 GA = 0.44

400 x 25 = 3556 GM= 0.34

Col : 450 : 400 = 1 : 4 : 3.1 MG = 0.61

Joint A G M

Member AG GA GM MG

D.f 0.67 0.44 0.34 0.61

F.e.m -78.2 +78.2 -60.94 +60.94

Duf +52.39 -7.6 -5.86 -37.18

C.o -3.8 +26.20 -18.59 -2.39

Dist +2.55 -3.35 -2.59 +1.79

Slab Ag

27.06 93.45 87.98 22.62 117.31-(22.06+93.45)

=59.56 Kn/m2

A G M Gm = 91.40-(87.98-
22.6)

=36.10 Kn/m2

BEAM FB/7 (Am)

+ve moment = 59.56 kn-m

-ve moment at G = 27.06 kn-m

-ve moment at G = 93.45 kn-m

BEAM FB/8(GM)

+ve moment = 36.10 kn-m

-ve moment at G = 87.98 kn-m

-ve moment at M = 22.60 kn-m

BEAM FB/7 450X250 (AG)

1000 Lef = (l0 /12 +lw + 3df) = (5000x0.7)/12

+ 250+3x155

155

450 Eff depth = 450-25-8= 417

750 Eff span = 5000

Tu=19.82x0.75x(0.75/2)(5-0.25)x1/2=13.24

250 Vu=(4.75x37.54/2)=89.16

Vd=Vu+1.6(Tu/6)=89.16+1.6(13.24/6)=92.69

Tue=(92.69/250x417)=0.889 Kn-m2

LAP REINFORCEMENT

Mel=Mu+Me=59.56+21.18=81.37,
Me=Tu(1+D16/1.7)=13.24(1+0.45/1.7)=21.81

Mulim=0.36xfckx(Xumax/d)x(1-0.42xXumax/d)bd2 =
71.72x0.7984x250x4172=3.57

5.97x417= 24891.5 Kn-m>81.37

81.37x106 = 0.87fyxAstxd(1-fyAst/bdfck)

=150558Ast-29.97Ast2

0r, 2.715x106 = 5024 – Ast2

=Ast = (5024 – 3792/2) = 616m2

Use = 2 – 16@ + 2- 12@ at mid span.