Amir Helil G+4.doc

A DESIGN OF (G+4) LOW COST BUILDING AND ITS COST ESTIMATION (BY USING RIBBED SLAB)

CHAPTER-ONE

ROOF DESIGN

Wind load analysis and design Wind is a moving air which in turn possesses energy and this kinetic energy should be resisted by using appropriate deign for different kinds of structural elements like roofs ,walls The action of wind can be a type of suction or pressure to our structures both externally or internally .How ever these effects are more magnified for structure with more openings and large surface areas .And we focus on sensitive part of the building that is roofs (both hipped and flat roofs) for analysis and design of its parts.

Method of analysis Even though there are two methods for wind load analysis ,namely Quasi static method and dynamic analysis we prefer Quasi static since our structure is assumed to be less susceptible to dynamic excitation and from EBCS-1,1995 section 3.9.3 a building which satisfies the criterion: (For cd <1.2and building height less than 200m) can be analyzed using Quasi static

Roof analysis and design Before going to the direct steps we should classify the location, terrain category our building site. And our building is located at Adigrat and we told that there are buildings around it. And according to

External pressure

External pressure We is obtained from: We =qref * Ce(Ze)*C pe

Where Cpe - is the external pressure coefficient Ce(Ze)- is the exposure coefficient qref - is the reference wind pressure

i. Reference wind pressure

This is determined from; qref = 0.5 *ρ*v2

ref

Where ρ is the air density and for an altitude of 1500m (approximate) above sea level. ρ = 1.00 kg /m3 Vref = the reference wind velocity and is given as Vref = CDIR * CTEM * CALT * Vref,0

Taking the three factors as unity and given Vref,0 =22 m/s, finally we get the reference wind pressure to be qref = 0.5 *(1.00 kg/m3) *(22.0 m/s)2

= 242 pa

MEKELLE UNIVERSITY, FACULITY OF SCIENCE AND TECHNOLOGY DEPARMENT OF CIVIL ENG’G 1


ii. External coefficient According to section 3.8.5(2), the exposure coefficient is given by,

Ce(z) =

Where KT –is the terrain factor Cr(z) – is the roughness coefficient Ct(z) – is the topography coefficient and taken to be unity For the terrain classification, the site is assumed to fall under category iv (the worst case) which is urban area in which at least 15% of the surface is cover with buildings and their average height exceeds 15m. Then, from table 3.2, KT = 0.24 , Z0 = 1m , Zmin = 16mThe actual height of the building, Ze, is equal to 16.7m

Cr(Ze)=KT*ln for Zmin Ze 200m

Cr(Ze) = Cr(Zmin) , for Ze Zmin

Where KT- terrain factor Z0 – roughness length Zmin – minimum height Since 16m <16.4m < 200m, we have to use the 1st one. Then; Cr(Ze) = 0.676 , Ct(Ze) =1.0 Ce(Ze) = 1.59 We have two separated and almost typical adjacent buildings, so the computation, analyze and result of one building can be used for the other one.

iii. External pressure coefficient Our roof is considered as both monopitch and hipped roof. The external pressure coefficient depends on the size of the loaded area, A, and as per the code it is given as: Cpe = Cpe, 1 A 1m2

Cpe =Cpe, 1 + (Cpe, 10 – Cpe, 1) log A 1m2 < A < 10m2

Cpe = Cpe,10 A 10m2

Our roof is modeled and designed as both monopitch (according to appendix A.2.4) and hipped roof (according to appendix A.2.6) of EBCS -1 , 1995. Case-1 Wind direction = 0 5.85m 18.27m M 12.06m 5.188m H L 2.412m F1 G F2 2.412m 6.03m 12.06m 6.03m =0



e = b or 2h which ever is smaller e = min e = 24.12m The pitch angle;

α0 = tan-1 =12.61 Since α0=12.61 applying linear interpolation between +5 +15 the following result is obtained.

Table 1.1 Cpe values for = 0Zone F G H L M

Cpe,10 -1.092 -0.896 -0.372 -1.352 -0.6

-0.256 -0.136 0.008 - -

Cpe,1 -2.12 -1.62 -0.516 -2.0 -0.624

-0.448 -0.328 -0.136 - -

Table 1.2 will gives the Cpe values and their respective wind pressure for the different parts of the roof and the bigger Cpe values is taken for a given roof part.

Table 1.2 wind pressures for =0Zone F1 F2 G H L M

Area(m2) 14.54 12.31 29.09 105.5 14.11 26.839

Cpe -1.092 -1.092 -0.896 -0.372 -1.352 -0.6

-0.256 -0.256 -0.136 +.008 - -

Wind pressure(N/m2

= - 420.18N/m2(suction)= +3.08N/m2(pressure)

= - 520.22N/m2(suction)

Case-2 Wind direction = 90



4.765m 1.085m F2 2.71m 18.27m H G 5.43m =90 7.6m N L F1 M 2.71m 17.61m 5.425m 1.085m

e = min e = 10.85

And pitch angle 90 = tan-1 = 16.2 Applying linear interpolation between = +15 and +30 for = 16.2 the Cpe values are as follows.

Table 1.3 Cpe values at =90Zone F G H L M NCpe,10 -0.788 -0.776 -0.292 -1.4 -0.616 -0.292

0.224 -0.68 -0.244 -1.4 -0.616 -0.292-0.788 0.144 0.166 -1.4 -0.616 -0.2920.224 0.24 0.216 -1.4 -0.616 -0.292

Cpe,1 -1.96 1.26 -0.292 -2.0 1.008 -0.292-1.80 1.436 -0.244 -2.0 1.008 -0.2920.064 0.064 0.168 -2.0 1.008 -0.2920.240 0.240 0.216 -2.0 1.008 -0.292

Table 1.4 wind pressure at =90Zone F1 F2 G H L M NArea(m2) 2.18 2.940 5.892 30.00 8.246 20.615 133.34Cpe -1.56 -0.310 -1.56 -0.292 -1.45 -0.616 -0.292

0.235 0.24 0.235 0.216 -1.45 -0.616 -0.292Wind pressure

= - 600.26N/m2 (suction)= + 92.35 N/m2 (pressure)

= -557.93N/m2 (suction)

Case-3 Wind direction = -90 5.53m 0.32m



0.8m F2

1.6m G2 H2 3.2m

0.8m F2 M =-90 3.8m G1 6.84m H1 I L 3.8m F1 0.76m

0.76m 3.04m 20.32m

e = min e = 7.6m 1 = 12.6m 1 = 90 2 = 16.2m 2 = 180 Table1.5 Cpe values at 90 and 180

Zone F1 G1 H1 I L M F2 G2 H2

Cpe,10 -1.372 -1.876 -0.752 -0.412 -1.352 -0.6 -2.388 -1.26 -0.892 - - - - - - - - -

Cpe,1 -2.048 -2.380 -1.20 -0.792 -2.0 -0.624 -2.76 -1.96 -1.168 - - - - - - - - -

Table 1.6 wind pressureZone F1 G1 H1 I L M F2 G2 H2

Area(m2) 2.888 2.888 23.104 132.202 4.45 9.08 0.256 0.512 17.696Cpe -1.74 -2.15 -0.752 -0.412 -1.58 -0.60 -2.76 -1.96 -0.892

- - - - - - - - -Wind pressure

= - 827.28N/m2(suction) = - 607.95N/m2

(suction)= -1061.993N/m2 (suction)

Case-4

Wind direction =180 =180

2.925m2.925m



3.465m 6.93m 3.465m F2 G2 0.85m

H2 2.34m 1.386m F1 G1 F1

H1 I2 7.66m 6.214m

5.85m e1= min e1 =13.86m e2= min e2 = 5.85m Similar to other cases before 1 = 12.6m 1 = 180 2 = 16.2m 2 = 90 Table 1.7 Cpe valuesZone F1 G1 H1 F2 G2 H2 I1

Cpe,10 -2.452 -1.3 -0.876 -1.292 -1.868 -0.816 -0.708Cpe,1 -2.728 -2.0 -1.2 -2.0 -2.46 -1.208 -1.2 Table1.8 wind pressureZone F1 G1 H1 F2 G2 H2 I1

Area(m2) 8.349 16.68 86.126 8.349 2.486 13.689 44.811Cpe -2.474 -1.3 -0.876 -1.35 -2.23 -0.816 -0.708Wind pressure

= - 951.945N/m2(suction) = - 858.06N/m2(suction) = - 272.43N/m2

(suction)

Internal pressure Internal pressure,Wi is the obtained from Wi = qref *Ce(zi) * Cpi

Where Cpi is the internal pressure coefficient and according to the code for closed buildings with internal partitions and opening windows. Cpi = 0.8 or -0.5 Ce(zi) = Ce(ze) =1.59 qref = 242 pa Therefore the extreme values of the internal pressure are Wi = 242 * 1.59 *(0.8) =307.824 pa (pressure) And Wi = 242 * 1.59 *(-0.5) = -192.39 pa (suction) Net pressure

The net wind pressure is the difference of the pressures on each surface taking due account of their signs. And there will be various net wind pressure combinations out of



which the severe cases are taken. These combinations are determined for purlin design and truss analysis at every step. live load

Since the roof is in accessible except for normal maintenance or other similar things, it follows under category H and accordingly for sloping roof. qk= 0.2 5KN/m2

QK = 1.0KN Dead load

The dead load of the purlin and truss members also determined in the following steps.

- self weight of the roof cover = 130N/m2

- self weight of the ceiling chip board. = 8KN/m3

load combination Combination of actions for persistent and transient design situations is determined according to section 1.9.4.5 of EBCS-1,1995, considering ultimate limit state ,cases to be considered are Case-1 unfavorable = 1.1DL + 1.35(WL (pressure +LL) Case-2 favorable =0.9 DL +1.6 WL (suction)

Truss lay out T2 1.6m T2

1.6m T2

3.8m T3

T1 T1 T1 T1 T1 T1 T1 T1 T1 T4 T5 3.8m

2m 2m 2.5m 2.5m 2m 2m 1.78m 1.78m 2.925m 2.925m

Truss types

TYPE-1(T1) Panel length, s = 1.2979m Pitch angle, =12.6 1.7m Total length of members = 29.5m



1.27m 7.6m TYPE-2(T2) Panel length, s = 1.015m 1.7m Pitch angle, =16.2 Total length of members = 21.6m

1.463m 5.85m TYPE-3(T3)

Panel length, s = 1.2m 1.7m Pitch angle, =13.71 Total length of members =24.52m

1.74m 6.97m TYPE-4(T4)

Panel length, s = 1.625m 1.7m Pitch angle, =10.01 Total length of members =34.82m

1.6m 9.6m

TYPE -5(T5)

Panel length, s = 1.39m Pitch angle, =11.80 1.7m Total length of members =30.92m

1.4m



8.14m

Purlin design

For the purlin section, choose RT-84 as a preliminary one. Dimension and properties are : Dimension nominal size, H = 80mm and B = 40mm Self weight, W = 5.19Kg/m = 50.91 KN/m Properties A = 6.61cm2

Ix = 52.25cm4 , Iy = 17.56cm4

Sx =13.06cm3 , Sy = 8.78cm3

rx = 2.81cm , ry = 1.63cm A) Cross-section classification

For Fe 430 steel, Fy = 275Mpa and

= = 0.92

Internal flange, C = b -3t = 31mm

= = 10.33 26(= 23.92) ………class 1

Web with N.A at mid depth, d = 80-3*3 = 71mm

= = 23.67 72 (=66.24) …………class 4

The section does fall in the class 4 and thus the elastic section modulus is used without reduction for determining the design capacities.

B) Design capacities m0 = material safety factor =1.1

i) Elastic moment capacity

My, Rd = = = 2.195KN.m

Mx, Rd = = = 3.265KN.m

ii) Shear capacity: the design plastic shear resistance is given by

Vpl, Rd =

Where Av is shear area and for RHS it is given by

Av = = = 4.53cm2

Then Vpl, Rd = =65.39KN

C) Loading



i) Dead load, GK1(purlin) = 50.91N/m ,GK2 (roofing) = 130*1.3 =169N/m Total load,GK = 219.91N/m

ii) Live load qK = 0.25*13 = 0.325KN/m QK = 1KN/m ……..located at the center

iii) Net wind load

QK2 = (-827.28-307.824)*1.3 = -1.476KN/m (suction) QK2 = (92.35-(-192.39))*1.3 = 0.37KN/m (pressure)

D) Analysis y

x

= 12.6

Case 1: total design load X-axis, Pdx = 1.1*(219.91*sin) +1.35*(325* sin) = 148.478 Point load, Pdx2 = 1.35*1* sin = 1.35*1* sin12.6 = 294N Y-axis, Pdy =1.1*(219.91*cos) +1.35*(325*cos+370) =1163.76N/m And point load, Pdy2 = 1.35*cos12.6 =1317.5N

y 1317.5N x 294N 1163.76N/m 148.478N/m

z z 2.5m 2.5m

Design bending moments and shear forces at the critical sections are:

My,sd =

= = 0.2997KN.m My,Rd (=2.195KN.m) …ok!

Mx,sd =

=1.732KN.m Mx, Rd = 3.265KN.m………ok!!

Vsd,max = = 2.11KN Vpl,Rd = 65.39KN …ok!



Check for combined action of shear and moment

= = 0.0323 0.5 no moment reduction

Check for combined effect

1.0

= = 0.667 1.0 ……ok!!

Check for deflection According to table 5.1 of EBCS-3, 1995 allowable deflection for roofs generally is

Smax = = =12.5mm

And z = = = 10.0mm

The actual deflection obtained using unfactored load is

x =

= [(5/384)*(0.12*2.54) + (0.22*2.53) /48] * 1/ (210*106*17.56*10-8) =3.60mm Similar, the other components is y = [(5/384)*(0.902*2.54) + (0.976*2.53) /48] * 1/(210*106*52.25*10-8) = 7.44mm Hence, the overall deflection or vertical deflection

= = 8.26mm 10mm

Therefore, the section is safe from deflection.

Case 2: Total design load X-axis: Pdx = 0.9(219.91*sin12.6) = 43.175N/m Y-axis: Pdy = 0.9(219.91*cos12.6) +1.6*(1476) = -2168.45N/m

x y 43.175N/m -2168.45N/m z z

2.5m 2.5m The design moment and shear are

My,sd = =0.034KN.m My,Rd (=2.195KN.m)

Mx,sd = = 1.69KN.m Mx,Rd (=3.265KN.m)



Vsd,max = = 2.71KN Vpl,Rd (=65.39KN)

Check for shear effect on bending

= = 0.041 0.5 …..Ok!!

Therefore no reduction in the design moments is required. Check for combined effect

< 1.0

=1.69/3.265 +0.034/2.195 =0.533 1.0 ……..ok!!

Check for deflection From the above computation, allowable deflections are: Max =12.5mm and 2 =10mm But the actual deflection (vertical) is given as: =y*cos -x*sin where =12.6 Where x = (5/384)*[(0.048*2.54)/ (210*106*17.56*10-8)] = 0.662mm y = (5/384)*[(1.261*2.54)/ (210*106*52.25*10-8)] = 5.845mm After all = 5.845*cos12.6 - 0.662*sin12.6 = 5.56mm 10mm Therefore our section is safe for deflection.

STRESS ANALYSIS OF TRUSS

A) As a sample analysis and design of truss 1 is given. To determine the self weight of the truss we choose RT-64 section as a preliminary section for the truss members. Properties and resistance capacities of the section are as follows:

i) Dimension and properties y Nominal sizes, 60mm by 40mm Wall thickness, t =3mm Self weight, W = 41.59N/m A = 5.41cm2 = 541mm2 x Ix =25.38cm4 Iy = 13.44cm4

Sx =8.46cm3 Sy =6.72cm3

Ry =2.17cm Rx =1.58cm ii) Section capacity in tension For section in axial tension, the design plastic resistance as per EBCS-3, 1995(4.4) is:

Npl,Rd=

Where A - cross sectional area and is equal to 541mm2



Fy = yield stress of the material =275Mpa m1 = material safety factor =1.1 Npl,Rd =541*275/1.1 = 135.25KN

iii) Section capacity in compression Classification for Fe 430 steel

= = 0.92

Internal flange : C=b -3t = 40 -3*3 = 31mm

= =10.33 26 (= 23.92) …class 1

For web where whole section is subjected to compression:

= =18 39 (=35.88) …..Class 1

Then the design compression resistance of the cross section is

Ncomb,Rd =

Where m0 = material partial safety factor for class 1,2 or 3 cross section m0 =1.1 Ncomb,Rd = 541*275/1.1

Loading Dead load

Total length of the truss members = 29.50m -self weight of the truss ,GK1 =41.59N/m *29.50m =1.226KN -self weight of the ceiling ,GK2 = (7.60*2.5*0.008*)*8 KN =1.216KN Total load ,GK = 2.442KN Since we have 7 panel joints 1.226/7 = 0.175KN And 1.216/7 = 0.174KN

0.175KN 0.175KN 0.175KN 0.175KN 0.175KN 1.7m 0.175KN

0.175KN 7.6m

0.174K 0.174KN 0.174KN 0.174KN 0.174KN 0.174KN 0.174KN

Similarly for the 5.85m span the weight of the truss and the ceiling can be sketched.

DESIGN OF TRUSS



In order to get the maximum design actions, we take three different combinations:Case 1 - Dead load +Live load (concentrated)Case 2 - Dead load + Live load (distributed)Case 3 - Dead load + wind load

Analysis

Case-1 P=0.98KN 0.268KN/m =W

2.5m M = P*l/4 + W*l2/8

=0.98*2.5/4 + 0.268*2.52/8 = 0.822KN.m Case-2 0.317KN/m = W1

0.268KN/m = W2

2.5m

M =

= (0.317+0.268)*2.52/8 =0.457KN/m

Case-3 -0.913KN/m

2.5m

M = = (0.913*2.52)/8

=0.713KN.m From the above three cases, case 1 is the critical case which gives maximum moment, Mmax = 0.822KN.m Determination of reaction of purlins

Case-1 DL + LL (concentrated) 0.98KN



0.268KN/m

2.5m

R = 0.825KN, but the reaction on a given truss at panel points is equal to 2R, however at the heel is equal to R.

0.825m 1.65m 1.65m 1.65m 1.65m 1.7m 1.65m 0.825m

5.85m

0.825KN 1.65KN 1.65KN 1.65KN

1.65KN 1.7m 1.65KN 0.825KN

1.7m 5.69m

CHAPTER-TWO

DESIGN OF SLAB

Slabs are horizontal structural elements which transfer service loads to the frame elements and their design should follow procedures. We have slabs of the types: solid slabs and ribbed slabs Ribbed slabs are preferred when we have long spans and they have advantages of the following - Easy to make partion walls -It has a very good sound proof -It provides good architectural view for the ceilings

Geometry

hf ≥ (S/10,50mm)



D bw

bw ≥ 70mm

C/C 1.0m

If center to center spacing is >1.0 the gap between the ribs should be designed as slabs And this center to center spacing should be minimized for the following reasons -to minimize the effect or impact of concentrated loads -to decrease the size of the filling HCB -to increase the bond between the HCB and the ribs considering the above guide lines and from def lection requirement we fix our dimensions

hf =60mm D=300mm C/C =400mm bw =80mm And we assumed HCB of size shown below ` 200mm

40mm

400mm

Design procedures Step-1 Depth determination

d ( )*

Where: fyk=characteristics tensile strength of steel Le=effective span e=accounts for the boundary condition and span ration for slabsOverall depth, D = d + l+ s + c

Step-2 Suitable dimension for the ribbed should be assumed considering deflection requirement and construction suitability.



Step-3 Loading

To calculate the loads on the ribbed the following data are assumed

a) Unit weight - PVC tile ………….. 16kN/m3

- Terrazzo tile........ 23kN/m3

- HCB………………14KN/m3

- cement screed….. 20KN/m3

- RCC……………… 24KN/m3

b) Dimensions - Head room = 2.7m - Partition wall height = 2.7m - Plaster thickness = 1.5cm

Computation of loading:

Lay out of typical floor plan with its select strip

A 5 m B 3.8m C 5m D 5 m E

A D G J 5

B E H K 4



F I L O’ C 3 3’

C’ F’ I’ L’ 2’

B’ E’ H’ K’ N’ 2

G’

A’ D’ J’ M’ 1

A’ B’ C’ D’ E’ F’

4m 5m 4 m 3.56m 5m

For the span – AB

Dead load 2mm of PVC = 0.002*0.4*16 = 0.0128 kN/m 4.8cm of cement screed = 0.048*0.4*20 = 0.384 kN/m RC slab (topping) = 0.06*0.4*24 + 0.08*0.2*24 = 0.96 kN/m HCB with ceiling plaster = 1 kN/m (assumed value) Partition wall = 0.2*1.15*2.7*14*0.4/25 = 0.1391 kN/m Plastering = 0.03*1.15*2.7*20*0.4/25 = 0.0298 kN/m Total dead load = 2.526 kN/m

Live load

Since our building residential type we took live load is 2 kN/m2 from EBCS-2, 1995 Live load = 0.4m*2KN/m2 = 0.8 kN/m

Design load, Pd

Pd = 1.3*DL + 1.6*LL = 1.3*2.526 + 1.6*0.8 = 4.564 kN/m



For the span – BC

Dead load 2mm of PVC = 0.002*0.4*16 = 0.0128 kN/m 4.8cm of cement screed = 0.048*0.4*20 = 0.384 kN/m RC slab (topping) = 0.06*0.4*24 + 0.08*0.2*24 = 0.96 kN/m

HCB with ceiling plaster = 1 kN/m Partition wall = (4.1+1.31)*2.7*0.1*14*0.4/25 = 0.327 kN/m Plastering = (4.1+1.31)*2.7*0.1*20*0.4/25 = 0.0339 kN/m Total dead load (DL) = 2.832 kN/m

Live load Live load (LL) = 0.4*2 = 0.8 kN/m

Design load

Pd = 1.3*DL + 1.6*LL = 1.3*2.832 + 1.6*0.8 = 4.962 kN/m Since the ration of self weight the partitions to the total design load is less than 20% we don’t need to use Reynolds method to determine an equivalent loads.

Table2.1 loading values for ribsspan DL

(KN/m)LL(KN/m)

1.3*DL(KN/m)

1.6*LL(KN/m)

AB 2.526 0.8 3.284 1.28BC 2.832 0.8 3.682 1.28DE 3.268 0.8 4.248 1.28EF 2.384 0.8 3.099 1.28GH 2.932 0.8 3.812 1.28HI 2.384 0.8 3.099 1.28JK 2.961 0.8 3.849 1.28KL 2.793 0.8 3.631 1.28A’B’ 2.472 0.8 3.21 1.28B’C’ 2.384 0.8 3.099 1.28D’E’ 3.256 0.8 4.238 1.28E’F’ 3.118 0.8 4.055 1.28G’H’ 2.472 0.8 3.213 1.28H’I’ 2.528 0.8 3.286 1.28J’K’ 3.333 0.8 4.333 1.28K’L’ 2.528 0.8 3.286 1.28M’N’ 2.641 0.8 3.433 1.28N’O’ 2.930 0.8 3.80 1.28cantilever 3.206 0.8 4.167 1.28

Analysis



Modeling We model the ribs as continuous beams and the direction of ribs is chosen in the shorter direction but some ribs run in the longer direction in contrast to the above theory for construction simplicity. We consider three cases for the application of live loads to know the critical reactions for design and a typical rib is taken to show the behavior of alternate loading along the spans For span – ABC

Case-1 When the live load act on span –AB

1.28

3.6823.284

A B C Case-2 When the live load acts on span –BC

1.28

3.284 3.682

A B CCase-3 When the live load acts on the whole span (fig)

1.28

3.284 3.682

A B C

The analysis of the ribs for the above load cases is done using SAP (v, 7.4) and a typical is taken to show shear and bending diagrams and for the other cases it is tabulated in the following table

Bending shear force diagram for rib ABC in there Cases

Case-1

14.49 6.59



-9.26 11.81

-13.07

8.31 5.36

5m 5m

Case-2 11.29 9.79

-6.06 -15.01

-13.07

4.87 8.97



Case-3

14.89

9.39

-8.86 - 15.41

- 15.05

7.37 7.97

5m 5m

Table 2.2 Sap Results For ribs (moments)

Ribs case 1 case 2 case 3maximum moment

span support span span support span span support span span support ABC 8.31 -13.07 5.36 4.87 -13.07 8.97 7.37 -15.05 7.97 8.97 -15.05DEF 15.04 -4.48 0 10.11 -6.33 0 14.11 -6.33 0 14.11 -6.63GHI 13.67 -4.48 0 8.75 -6.33 0 12.75 -6.33 0 13.67 -6.63JKL 13.4 -5.25 0 8.48 -7.1 0 12.48 -7.1 0 13.4 -7.1A'B'C' 11.65 -4.78 0 6.73 -6.63 0 10.73 -6.63 0 11.65 -6.63D'E'F' 14.84 -4.78 0 9.91 -6.63 0 13.91 -6.63 0 14.84 -6.63G'H'I' 5.06 -4.75 0 2.34 -6.6 0 4.14 -6.6 0 5.06 -6.6J'K'L' 15.17 -4.75 0 10.24 -6.6 0 14.24 -6.6 0 15.17 -6.6M'N'O' 8.6 -14.02 5.31 5.23 -14.01 9.26 7.8 -15.29 8.03 9.26 -15.29



Design for flexure

Material specification C-25 C1 = 0.086 S-300 C2 = 3003.23 Class- work m = 28.78 = 0.0187

For rib ABC

Design for flexure

Mspan = 8.97 KN.m

1-Check for the assumed depth for flexure

d > (M/(k*b))

K = 3.8126 (constant from material property) After substituting for moment in the above equation we get d = 171.49 < 232.0 i.e. safe for the flexure

2- Check for the section either rectangular or T-section

Effective width=min

Effective width, be = min

= 400mm Neutral axis N.A = 0.8*X X = m**d =0.5*(0.086-(0.0862 – 4*8.97*106/ (3000.23*400*2322))) =0.00162Then x=*m*d =0.00162*28.78*232 =10.98And y=0.8*x=0.8*10.98=8.79<60mm (there fore the section is rectangular of width be=400mm

3-reinforcement design

As= *be*d =0.00162*400*232 =150.48mm2



#of bars 10 =150.48/78.54=1.92Use 2 10

M support =15.05KN.m

Similarly:

1) Check for flexure

d where M=15.05 KN.m

Substituting the value we get d=222.14<232mm (section is safe for flexure)

2) Check the section for its classification

The section is obviously rectangular r =0.5*(0.086-(0.0862 – 4*15.05*106/ (3000.23*80*2322))) =0.00165 3) Reinforcement design

As = *b*d 0.0165*80*232 =306.24

#of bars 14 = =1.988

Use 2 14A similar calculation can be applied for the other ribs and tabulated as shown below

Table 2.3 Design summary for ribs

Shear design

Let’s take rib ABC to show the procedure of design for shear From the there loading cases case- 3 (when the

MEKELLE UNIVERSITY, FACULITY OF SCIENCE AND TECHNOLOGY DEPARMENT OF CIVIL ENG’G

Ribs design moments reinforcement

span support span support

ABC 8.97 -15.05 2Ø10 2Ø14

DEF 14.11 -6.63 2Ø14 2Ø10

GHI 13.67 -6.63 2Ø14 2Ø10

JKL 13.4 -7.1 2Ø14 2Ø10

A'B'C' 11.65 -6.63 2Ø14 2Ø10

D'E'F' 14.84 -6.63 2Ø14 2Ø10

G'H'I' 5.06 -6.6 2Ø10 2Ø10

J'K'L' 15.17 -6.6 2Ø14 2Ø10

M'N'O' 9.26 -15.29 2Ø10 2Ø14

24


whole span is loaded with the live load) is critical for shear design (fig above)

The shear is max at

d= = .457m

Shear capacity

Concrete shear capacity

Vc = 0.25*k1*k2*fctd*bw*d bw = rib width =80mmd=effective depth =232mmk1=1.6-d=1.6-0.232=1.368 (d in m)

k2=1+50*r = = 1.0833

Vc = 0.25*1.368*1.083*1.031*80*232=7.087KN

Diagonal shear capacity

Vrd =0.25*fcd*bw*d =0.25*11.33*80*232 =52.57KNBut from theoretical point of view the shear capacity for T-section is 1.1 times than that of the rectangular section For span AB

Vdl = (1.95-0.457)*8.86/1.95 = -6.78 KNVdr= (3.05-0.457)*14.89/3.05=11.78 KN

For span BC

Vdl= (3.11-.457)*15.41/3.11=13.14 KNVdl = (1.89-0.457)*9.39/1.89=7.12 KNTaking the maximum design shear force:For cases where Vc <Vd (reinforcement is needed)That means for Vd =13.14 KN Vs=Vd-Vc = (13.04-7.096) = 6.044 KN S =Asv*fyd*(d-dc)/Vs

=2*28.26*260.87*(232-43)/6.044 =461.1mm

But the maximum spacing, s max =min (d, 800, Asv *fyk/ (.4*bw)) mm =min (232,800,525) mm =232mm

Take 6 c/c 230mm



Similarly for the other ribs the required shear reinforcements are minimum

Load transfer to beam from ribs

On block-1 On axis-3 31.04

Loading Weight of wall on beam = 0.2 * 2.7 *14 = 7.6 kN/m Reaction from ribs = 9.39 / 0.4 = 23.48 kN/m w = 31.08 kN/m On axis-4 83.88

63.88 61.14 64.03

A B C D E

Loading On AB Weight of wall = 7.6 kN/m Reaction from ribs = 75.85 kN/m w = 83.41 kN/mOn BC Weight of wall = 7.6 kN/m Reaction from ribs = 56.32 kN/m w = 63.88 kN/mOn CD Weight of wall = 7.6 kN/m Reaction from ribs = 53.57 kN/m w = 61.14 kN/mOn DE Weight of wall = 7.6 kN/m Reaction from ribs = 56.47 kN/m w = 64.03 kN/m

On axis-5 39.86 37.14 36.985



30.71

A B C D E

Loading On AB Weight of wall = 7.6 kN/m Reaction from ribs = 23.15 kN/m w = 30.71 kN/m On BC Weight of wall = 7.6 kN/m Reaction from ribs = 32.3 kN/m w = 39.86 kN/m On CD Weight of wall = 7.6 kN/m Reaction from ribs = 29.58 kN/m w = 37.14 kN/m On DE Weight of wall = 7.6 kN/m Reaction from ribs = 29.425 kN/m w = 36.985 kN/m

On Block-2

On Axis-3’

30.36

Loading Weight of wall = 7.6 kN/m Reaction from ribs = 22.8 kN/m w = 30.36 kN/m On Axis-2

84.11 65.435 64.8451.91 58.46

A’ B’ C’ D’ E’ F’ Loading

On A’B’ Weight of wall = 7.6 kN/m Reaction from ribs = 76.55 kN/m w = 84.11 kN/m On B’C’ Weight of wall = 7.6 kN/m Reaction from ribs = 57.87 kN/m w = 65.435 kN/m



On C’D’ Weight of wall = 7.6 kN/m Reaction from ribs = 44.35 kN/m w = 51.91 kN/m On D’E’ Weight of wall = 7.6 kN/m Reaction from ribs = 57.275 kN/m w = 64.835 kN/mOn E’F’ Weight of wall = 7.6 kN/m Reaction from ribs = 50.90 kN/m w = 58.46 kN/m

On axis 1-1

33.47 40.26 38.71 33.26

24.735 A’ B’ C’ D’ E’ F’

Loading On A’B’ Weight of wall = 7.6 kN/m Reaction from ribs = 25.15 kN/m w = 33.466 On B’C’ Weight of wall = 7.6 kN/m Reaction from ribs = 32.70 kN/m w = 40.26 kN/mOn C’D’ Weight of wall = 7.6 kN/m Reaction from ribs = 17.175 kN/m w = 24.735 kN/mOn D’E’ Weight of wall = 7.6 kN/m Reaction from ribs = 31.15 kN/m w = 38.710 kN/m On E’F’ Weight of wall = 7.6 kN/m Reaction from ribs = 25.70 kN/m w = 33.26 kN/m

Mass center and stiffness center calculation Table 2.4 Typical Floor 1 Mass Center Calculation (sample)

Items W(KN) Xi(m) Yi(m) Wxi(KN- Wyi(KN-m)



m)S1 152.25 2.5 7.5 380.625 1141.875S2 120.77 6.9 7.5 833.313 905.775S3 154.5 11.3 7.5 1745.85 1158.75S4 154.5 16.3 7.5 2518.35 1158.75S5 154.5 2.5 2.5 386.25 386.25S6 38.66 6.9 4.15 266.754 160.439S7 50.66 11.3 4.15 572.458 210.239S8 50.66 16.3 4.15 825.758 210.239

Beam axis 5 54.14 9.4 10 508.916 541.44 67.68 9.4 5 636.192 338.43 14.4 2.5 0 36 0A 21.6 0 5 0 108B 21.6 5 5 108 108C 14.472 8.8 6.65 127.3536 96.2388D 14.472 13.8 6.65 199.7136 96.2388E 14.472 18.8 6.65 272.0736 96.2388

Column A5 16.13 0 10 0 161.3A4 16.58 0 5 0 82.9A3 16.2 0 0 0 0B5 13.52 5 10 67.6 135.2B4 16.58 5 5 82.9 82.9B3 13.52 5 0 67.6 0C5 13.52 8.8 10 118.976 135.2C4 16.58 8.8 5 145.904 82.9D5 13.52 13.8 10 186.576 135.2D4 16.58 13.8 5 228.804 82.9E5 16.13 18.8 10 303.244 161.3E4 16.58 18.8 5 311.704 82.9

External wall

A5B5 37.8 2.5 10 94.5 378B5C5 28.73 6.9 10 198.237 287.3C5D5 37.8 11.3 10 427.14 378D5E5 37.8 16.3 10 616.14 378A5A4 37.8 0 7.5 0 283.5A4A3 37.8 0 2.5 0 94.5A3B3 37.8 2.5 0 94.5 0B4B3 37.8 5 2.5 189 94.5B4C4 28.73 6.9 5 198.237 143.65C4D4 37.8 11.3 5 427.14 189D4'E4' 37.8 16.3 3.3 616.14 124.74D4D4' 12.85 13.8 4.15 177.33 53.3275E5E4 37.8 18.8 7.5 710.64 283.5E4E4' 12.85 18.8 4.15 241.58 53.3275

Internal wall

A4B4 18.5 2.5 5 46.25 92.5C5C4 28.35 8.8 7.5 249.48 212.625



D5D4 18.9 13.8 7.5 260.82 141.75Partition

wall S2,P1 15.48 6.09 8.63 94.2732 133.5924S2P2 13.89 6.99 7.2 97.0911 100.008

S3 18.9 11.55 7.5 218.295 141.75S4 28.35 16.3 5.9 462.105 167.265S5 18.9 2.5 2.5 47.25 47.25S8 22.68 16.1 5 365.148 113.4

Sum(∑) 1929.886 16762.211 11751.0188

Xm= =8.685597

The remaining mass center calculation can be summarized as follow building-1

Table2.5 total mass center calculation Floor ∑W ∑Wxi ∑Wyi Xm(m) Ym(m)Roof 338.985 3323.75 1991.75 9.8 5.875

Fourth 1929.886 16762.21 11751.02 8.686 6.089Third 1929.886 16762.21 11751.02 8.686 6.089

second 1929.886 16762.21 11751.02 8.686 6.089First 1929.886 16762.21 11751.02 8.686 6.089

Ground 876.78 6855.31 5520.25 7.82 6.29

Similarly the mass center calculation of building-2 can be summarized as follow

Table2.6 total mass center calculation


Ym= = 6.08897

Floor ∑W ∑Wxi ∑Wyi Xm(m) Ym(m)Roof 377.678 4479.26 1559.81 11.86 4.13

Fourth 2259.462 26954 8701.865 11.929 3.851Third 2259.462 26954 8701.865 11.929 3.851

second 2259.462 26954 8701.865 11.929 3.851First 2259.462 26954 8701.865 11.929 3.851

Ground 1024.08 12677.3 4007.37 12.38 3.913

30


CHAPTER THREE

LATERAL LOAD ANALYSIS

The lateral loads to be considered for the design of a building are:- 1. Earth quake load and 2. Wind load NOTE: The analysis for both will be done and the critical load condition will be considered for the Design .

Earth Quake Analysis

According to EBCS-8,1995 static method of analysis is used. The seismic base shear force Fb for each main direction is determined by the equation : Fb=Sd*(T1)*W Where:- Sd*(T1)= Ordinate of the design spectrum T1=fundamental period of vibration of the structure for translational in direction Considered. W= seismic dead load computed. For linear analysis , the design spectrum Sd*(T1) normalized by the acceleration of gravity g is defined by the following expression :- Sd*(T) = ( EBCS-8,1995 Art 1.4.2.2 (4)) Where:- = The ratio of the design bed rock acceleration to the acceleration of gravity g and is given by:- =o*I



o = bed rock acceleration ratio for the site and depend on the Seismic zone (Table 1.1) I = The importance factor (Table 2.4) For zone 4 building location (Mekelle) :- o = 0.1 and use I = 1.0 = 0.1*1.0 =0.1 = The design response factor for the site and is given by :- =1.2*S/(T)2/3 2.5 S= site coefficient for the soil characteristics. =1.0 (for For building up to 80m height the value of T1 may be approximated by the formula:- T1 = C1 *H2/3 (sec) Where :- H = Height of the building (m) = 18.7m C1= 0.075 for reinforced concrete moment resisting frames.

T1 = 0.075 (18.7)3/4 = 0.674 = 1.2*1/(0.62)2/3 The behavior factor to account for energy dissipation capacity is given by :- = o * KD*KR*KW 0.7 (EBCS-8,1995, art.3.3.2.1(1))

Where: o -basic value of the behavior factor , dependent on the structural . type (table 3.2) For frame system, o = 0.2 KD-factor reflecting the ductility class Use KD = 1.0 for DC’H’ KR-factor reflecting the structural regularity in elevation, for regular structures KR= 1.0 Kw-factor reflecting the prevailing failure mode in structural system with walls Kw= 1.0 for frame and frame equivalent dual systems

= 0.2*1.0*1.0*1.0 = 0.2 < 0.7 ………. Ok! Sd*(T) = ** = 0.1*1.56*0.2 = 0.0312

Fb= 0.03*8935.29 kN = 286.1 kN The concentrated force Ft at the top, which is in addition to Fn shall be determined from the equation: Ft = 0.07*T1* Fb (EBCS-8, 1995 art.2.3.3.2.3 (2))

= 0.07*0.62*286.1 = 12.42 kN The remaining portion of the base shear shall be distributed over the height of the structure, including level n according to:-



Fi =

= 876.78*0 +1929.86*(3+6+9+12) + 338.985*15 = 62980.575

From the above equation F1 = 0 kN F2 = 25.16 kN F3 = 50.32 kN F4 = 75.48 kN F5 = 100.63 kN F6 =22.095+12.45 = 34.54 kN

Similarly, the base shear computation for the other floor-2 can be given as :-

Fb = Sb(T1)*W = 0.03*(377.68+4*2259.462+1024.08) = 313.19 KN

The concentrated force, Ft:- Ft = 0.07* T1* Fb

= 0.07*0.62*313.19 = 13.59 kN

Fi =

= 1024.08*0+2259.462*(3+6+12+9) +377.68*15 = 73449.06 From the above equation F1= 0 kN F2= 27.65 kN F3= 55.3 kN F4 =82.95 kN F6 =110.6 kN F5 =23.11+13.59=36.7 kN



Wind load analysis (for block-2)

External pressure

For the majority of the structures the resonant components are small and the wind load can be simplified by considering the background component only. The wind pressure acting on the external surface of structure we shall be obtained from. We = qref *ce(ze)*cpe…….. (EBCS-1, 1995 art.3.5.2 (1)) Where : qref = the reference wind pressure = 242 pa (previous result) Ce(ze) = exposure coefficient Exposure coefficient, ce (ze) , according to 3.8.5 (5) is given by :-

Ce(ze) = cr 2 (z)*ct

2(z)*

From table 3.2, for terrain category IV =0.24 , Zo = 1m and Zmin = 16mFor our case height of the building = 16.7m, hence cr

(z) = cr (Zmin)

= *ln ( )

Substituting the value: cr

(z) = 0.676 cr

(z) = 1.59Cpe = external pressure coefficient (according to ERBCS-1, 1995, table A.1) For A> 10m2

Cpe = Cpe,10

o Wind load in the shorter direction



h =16.7m

e = min =min Use e= 22.92m

For case d < e:-



h = 16.7m < b = 22.92m … the building is consider as one part qref = 242 N/m2 For zone D : We = 242*1.59*0.8 = 0.308 kN/m2

For zone E : We = 242*1.59*-0.3 = -0.115 k N/m2

Total external pressure on the building will be We = 0.308 + 0.115 = 0.423 k N/m2a

o Wind load in the longer direction


zone Cpe valueD +0.8E -0.3A* -1.0B* -0.8

36


e = min =min use e = 10.30m

For the case d > e

h = 16.7m > b = 10.30m < 2*b = 20.6 Hence, the building is considered as two part, comprising: a lower part extending upwards from ground by a height equal to b and an upper with height to h.

cr (z = b = 10.30 < Zmin = 16 ) = 0.24*ln (16/1) = 0.665

cr (z = 16.7 ) = 0.676

ce (z = 10.30) = 0.665 2 *12* = 1.56



ce (z = 16.7) = 0.6762 *12* = 1.59

, hence, we use linear interpolation for Cpe coefficient.

For zone D : We = 0.775*1.59*242 = 0.298 kN/m2, at Ze = 16.7m For zone E : We = -0.3*1.59*242 = -0.115 kN/m2, at Ze = 16.7m At Ze = b = 10.30m For zone D : We = 0.775*1.56*242 = 0.293 kN/m2

For zone E : We = -0.3*1.56*242 = -0.113 kN/m2

The total external pressure on the building will be : At Ze = b = 10.30m We = 0.293 +0.113 = 0.406 kN/m2

At Ze = h = 16.7 We = 0.293\8 + 0.115 = 0.413 kN/m2

Internal pressure

The wind pressure acting on the internal surface of structure Wi shall be obtained from: Wi = qref *Ce(Zi) * Cpi Where: Cpi – internal pressure coefficientFor closed building with internal partition and opening windows the extreme values are : Cpi = 0.8 or Cpi = -0.5…… EBCS-1, 1995, art.A.2.9 (16) Ce(Zi) = Ce(Ze) = 1.59 Wi = 242*1.59*0.8 = 0.3087 kN/m2

Wi = 242*1.59*-0.5 = -0.198 kN/m2

The net wind pressure will be : In the longer direction Wnet = 0.406 + 0.308 – (-0.192) = 0.906 kN/m2 at Ze = b = 10.30m Wnet = 0.413 + 0.308 – (-0.192) = 0.913 kN/m2 at Ze = h = 16.7m


zone Cpe valueA -1.0B -0.8C -0.5D 0.775E -0.8

38


In the shorter direction Wnet = 0.423 + 0.308 – (-0.192) = 0.923 kN/m

Table for values of lateral loadsTable3.1comparison of wind force and earth forcefloor Wind load shorter direction

(kN)Wind load longer direction (kN)

Earth quake (kN)

Ground 156.303 353.291 *368.351st 128.307 289.826 *340.852nd 100.312 226.36 *285.553rd 72.316 162.895 *230.254th 44.198 99.43 *143.3Roof 15.987 35.964 *36.7

From the above result earth quake load is the critical which govern our design Note: - for block-1 we got the same result which is the critical one is the earth quake load.

Distribution of storey shear D-value method

The rigidity center for each floor is determined by:- Xs = (Dyi)*Xi/(Dyi) Ys = (Dxi)*Yi/(Dxi) Where:-Dx and Dy are shear distribution coefficient or D-value for X and Y direction respectively. Moment of inertia for beam and column: For beams

30X40 I = = 90*103 cm4



30X30 I = = 67.5*103 cm4

45X45 I = = 341.72*103 cm4

For column

40X40 I = = 213.33*103 cm4

45X45 I = = 341.72*103 cm4

Stiffness ratio:- Stiffness ratio for a member is given by:

K = , where I = moment of inertia and L = length of a member

Calculation of D-values:

D = a* Kc

Where: a = and = for general case

a= and = for fixed base

Kb and Kc are stiffness ratio for beam and column express in terms of standard K K = 103cmThe stiffness ratio for each beam and column are as shown below for the selected frames











D-value calculation for frame on axis-5

0.4 0.24 0.18 0.43 0.43 0.43

1.14 1.71 1.07

0.4 0.56 0.43

= = 0.62 = 0.502 = 0.402

= 0.414 a = 0.24 a = 0.4 a = 0.38

a = D = 0.15 D = 0.201 D = 0.153

= 0.172D = a* = 0.0713



The center of rigidity is calculated as

Xs =

Ys =



The center rigidity is calculated as

Xs =

Ys = , this values are the same for

floor two and three.



the center of rigidity is calculated as

Xs =

Ys =




Xs =

Ys =




Xs =

Ys =

To allows for accidental torsional eccentricity location of mass center may be in error to either side by : eLi = 0.05 * Li

Where: Li = dimension of building plan perpendicular to the direction of earth quake loading

-when the earth quake is in the y-direction eLi = 0.05 * 18.8 = 0.94m

-when the earth quake is in the x-direction eLi = 0.05 * 10 = 0.5m



Correction factor to account for torsional effect is given by:

Where : , -eccentricities from the center of rigidity for the two extreme positions of the storey shear.

Jx = Dy * X2 and Jy = Dx * Y2

Table3.2 The center of rigidity can be tabulated as follow for block 1

Floor Xc(m) Yc(m)Roof 7.430 6.710

Fourth 7.430 6.710Third 7.430 6.710

second 7.430 6.710First 7.430 6.710

Ground 7.580 6.670 Similarly; Table3.3 the center of rigidity can be tabulated as follow for building 2

Eccentricity calculation for building 1

exi = (Yc-Ym) eli (X-direction) eyi = (Xc-Xm) eli (Y-direction)

The calculated results are tabulated as follow for building Table3.4

Floor ey1 ey2 ex1 ex2


Floor Xc(m) Yc(m)Roof 12.360 3.440

Fourth 12.170 3.500Third 12.170 3.500

second 12.170 3.500First 12.170 3.500

Ground 12.400 3.540

51


Ground -0.120 0.880-

1.180 0.700

Fourth 0.330 1.330-

4.840 -2.960

Third 0.330 1.330-

4.840 -2.960

second 0.330 1.330-

4.840 -2.960

First 0.330 1.330-

4.840 -2.960

Roof 0.350 1.350-

3.310 -1.430

Distribution of story shear in the X & Y direction for ground floor (sample) Base shear=286.1KN (for ground) Table3.5

Dx Dy X Y X2 Y2 Dx*Y2 Dy*X2 X1 X2

1.040 0.440 -7.580 3.330 57.456 11.089 11.532 25.281 1.010 0.928

1.040 0.470 -2.580-

1.670 6.656 2.789 2.900 3.129 0.995 1.036

0.260 0.250 1.220-

6.670 1.488 44.489 11.567 0.372 0.980 1.145 0.250 6.220 38.688 0.000 0.000 9.672 0.240 11.220 125.888 0.000 0.000 30.213

=2.340 1.650 26.000 68.667

Table cont…

Y Y2 Qx(KN)QY

1.000 1.000 128.420 83.3100.925 1.045 131.730 84.0201.023 0.987 36.400 44.4301.027 0.984 48.9001.030 0.982 51.230

The remaining floor distribution of story shear can be tabulated as follow for building-1Table 3.6


FloorBase

shear(KN) Qx(KN) Qy(KN)Roof 34.510 14.970 7.880

11.410 9.060 4.470 5.620 7.090 7.920

Fourth 135.140 59.420 24.320 65.500 33.740 18.230 23.170 31.340 36.570

Third 210.620 91.650 37.510 101.030 52.040 28.120 35.740 48.340 56.710

second 260.090 113.150 46.300 124.720 64.240 34.710 44.120 59.680 70.000

First 286.100 128.420 83.310 131.730 84.020 36.400 44.430 48.900 51.230

52


Similarly, for block-2 storey shear as follow Distribution of story shear can be tabulated as follow for building Table 3.7

FloorBase

shear(KN) Qx(KN) Qy(KN)Roof 36.690 3.380 5.280

18.930 4.840 14.740 4.770 4.730 8.790 8.880

Fourth 147.290 15.040 27.799 77.750 21.270 59.150 19.630 18.780 36.040 37.920

Third 230.240 23.510 43.455 121.530 33.240 92.460 30.690 29.350 56.330 59.280

second 285.540 29.160 53.890 150.720 41.230



114.660 38.060 36.399 69.860 73.520

First 313.190 30.720 42.240 163.370 40.260

122.020 39.820 49.130 75.530 71.890

Sample frames from SAP result as shown below :

AXIS-2 OF BLOCK-2





AXIS-F’ OF BLOCK-2



AXIS-5 OF BLOCK -1

AXIS-A OF BLOCK -1





CHAPTER –FOUR

DESIGN OF BEAM

DESIGN OF BEAM FOR SHEAR

FOR AXIS -4 (for block -1) Design constant:- C-25 S-300 fcd =11.16Mpa fyd = 260.87Mpa fctd = 1.032Mpa The cross section of the beam :-

300mm

600m Vc = 0.25*fctd *k1*k2*bw*d Where:- d = D-cover-s-/2 = 300-25-8-10 = 257mm bw = 600mm k1 = 1.6-d =1.6-.257 = 1.343 k2 = 1 +50r = 1+50(0.5/fyk) = 1.083 Vc = 0.25*1.032*1.343*1.083*600*257 = 60.02(KN) Vrd = 0.25*fcd*bw*d = 0.25*11.16*600*257 = 430.22(KN)



The shear diagram of the beam is:-

218.6 168.9 123.7 169.8 E D C B A

-168.88 -122.6 -168.8-216.65

5m 5m 3.8m 5m

A B C D E

Shear Design Of Span E-D Vd = 123 Vav

Vc

2.5m Vc

Vdr ≈ Vdl = 176.6(KN) > Vc (required shear design )Since Vrd > Vdr ,the section is ok! From –Vc to Vc requires minimum reinforcement.

Smax min



= 2*50.3*300/(0.4*600) = 125.6mm

(from X=1.81m to X=3.186m, use 8 C/C 120mm)

For the remain parts calculated as follow:-

Vav = =

= 118.3(KN) Vs = Vav – Vc = 118.3-60.02 = 58.28(KN Taking 8 :

S =

= 2*50.3*260.87*(257-43)/58.28 = 96. 3

(from X=1.15m to X=1.81m & from X=3.18m to X=4.85 , use 8 C/C 90mm)

For the other portion :- Vs = VdR – Vc

= 176.6 – 60.02 = 116.58(KN ) Taking 10 , S = 75.22mm use 10 C/C 70mm

For span DC: Vd = 168.9

Vc Vav 2.5 m

Vc

VdR ≈ Vdl = 136.3(KN) > Vc (required shear design)

From –Vc to Vc requires minimum reinforcement

Smax min

= = 125.6mm




For the remain parts calculated as follow:-

Vav =

= (60.02 + 136.3)/2 = 98.18(KN) Vs = Vav – Vc

= 98.18 – 60.02 = 38.16

Taking 8 , S =

=

Since 146.14 > 120 (take 120mm) (From X=1.05m to X = 1.61m & from X=3.39m to X= 3.95m ,use 8 C/C 120mm)

For the other portion:- Vs = VdR – Vc

=136.34-60.02 = 76.32(KN) Taking 8 , S = 73.58mm

use 8 C/C 70mm Vd=123.7

For Span CB :

Vc Vav Vc

VdR ≈ Vdl = 92.46(KN) > Vc (required shear design) From –Vc to Vc requires minimum reinforcement

Smax min

= 2*50.3*300/(0.4*600) = 125.6mm

(from X= 0.98m to X= 2.82m, use 8 C/C 120mm) For the remains parts:-



Vav =

= (60.02 + 92.46)/2 = 76.24(KN) Vs = Vav – Vc

= 76.24 – 60.02 = 16.22

Taking 8 , S =

=346.2 > 120mm (Take 120mm) (from X = 0.72m to X = 0.98m & from X= 2.82 to X = 3.07m, use 8 C/C 120mm) For the other portion:- Vs = VdR – Vc

=92.46 – 60.02 = 32.44(KN) Taking 8 , S = 208.3 > 120 (Take 120mm) use 8 C/C 120mm (For all section) For span DC:

Vd = 169.77

Y

Vc Vav X

Vc 2.5 m

VdR ≈ Vdl = 137.14> Vc (required shear design) From –Vc to Vc requires minimum reinforcement

Smax min

= 2*50.3*300/(0.4*600) = 125.6mm


Vav =

= (60.02 + 137.14)/2 = 98.56(KN) Vs = Vav – Vc

= 98.56 – 60.02 = 38.56



Taking 8 , S =

= 2*50.3*260.87*(257-43)/38.56 = 145.65 > 120 (take 120mm) (From X=1.05m to X = 1.62m & from X=3.38m to X= 3.95m ,use 8 C/C 120mm)

For the other portion:- Vs = VdR – Vc

=137.14 – 60.02 = 77.12(KN) Taking 8 , S = 172.82 > 120 use C/C 120mm (For all section)Axis –c’ (block-2)

300mm

600mm

The other beam shear design results can be tabulated as follow for block-1 Table 4.1

Floor Type Axis Span Vc(kN)

Vav(kN) Vdl(kN)

Vdr(kN)

with C/C(mm) Interval(m)

Tie Beam A,B,C,D&E All C/C 160 All Axis

Typical

4.0

ED

60.0 C/C 120 1.81,3.19

118.3 C/C 90 (1.15,1.18),(3.19,3.85)

176.6 C/C 70 (0.48,1.15)

176.6 C/C 70 (3.85,4.52)

DC

60.0 C/C 120 (1.61.3.39)

98.2 C/C 120(1.05,1.61),(3.39,3.95

136.4 C/C 70 (0.48,1.05)

136.4 C/C 70 (3.95,4.52)

CB

60.0 C/C 120

Total section 76.2 C/C 120

92.5 C/C 120

92.5 C/C 120

BA60.0 C/C 120

(3.38,3.95) 98.6 C/C 120



137.1 C/C 70 (0.48,1.05)

137.1 C/C 70 (3.95,4.52)

A,B,C,D&EAll

C/C 120Total section

5.0

AB

38.6 C/C 180(1.20,4.01)

60.8 C/C 180

84.5 C/C 120 (0.46,1.20)

81.2 C/C 120 (4.02,4.54)

BC

38.6 C/C 180

Total section 50.8 C/C 180

64.3 C/C 180

62.5 C/C 180

CD

38.6 C/C 180(0.85,2.97)

60.5 C/C 180

84.1 C/C 120 (0.48,0.85)

81.2 C/C 130 (2.97,3.34)

DE & 3

38.6 C/C 180

(0.46,3.45) 54.5 C/C 180

68.9 C/C 180

72.9 C/C 160 (3.45,4.54)

Grade beam

All All40.5

C/C 160

Total section

The beam shear design result of block-2Table 4.2Floor Type Axis Span Vc(kN)

Vav(kN) Vdl(kN)

Vdr(kN)

with C/C(mm) Interval(m)

Tie beam All All

C/C 180Total Axis

Typical

1' All C/C 160 Total Axis

1 All C/C 120 Total Axis

2 A'A C/C 120 Total span

AB

57.9 C/C 120(0.82,2.98)

97.8 C/C 120

137.7 C/C 120 (0.457,0.82)

137.7 C/C 120 (2.98,3.343)

BC 57.9 C/C 120 Total section



97.8 C/C 120

137.7 C/C 120

137.7 C/C 120

CD C/C 120 Total span

DE C/C 120 Total span

EF

57.9 C/C 120(1.04,3.96)

93.7 C/C 120

129.6 C/C 70 (0.46,1.04)

129.6 C/C 70 (3.96,4.54)

3',A',B',C',D',E',F' C/C 120 All Axis

Grade Beam

All C/C 180 Total section

Shear Diagram:- 61.64

56.87.03

Vdl Vc 4.13

19.971.17m

Vdl = 59.853(KN) Vc = 0.25*fctd *k1*k2*bw*d = 0.25*1.032*(1.6-0.257)*1.083*600*257 =57.88(KN) Vs = Vdl- Vc = 59.853-57.88 = 1.973(KN)

Smax min

= = 125.6m

Taking 8 , S =

=

= 2844.22mm > 120mm (take 120mm) use 8 C/C 120mm (For all section)



Design for torsion

Combined effect of torsion and shear On axis C’ and D’

24.55 kN.m 6.51 kN.m

300 mm 300 mm

600 mm 600 mm

Aeff = (D’-2*d’)*(b-2*d) = (300-2*43)*(600-2*43) = 109.996*103mm2m

heff =

42.8 <

Tc = 1.2*fctd* Aeff * heff = 1.2*1.032*109.996*42.8 =5.83 kN.m

Torsion diagram 24.55 -6.51

1.17 3.83 15.81 -12.91

12.91 12.02

Tc < Tsd = 15.81 we need a reinforcement



Section capacity against diagram compression Trd = 0.8*fcd* Aeff * heff = 0.8*11.33*109.996*42.8 = 42.672 kN.m > Tsd = 15.81 kN.m Vrd = 0.25*fcd*bw*d = 436.772 Vc = 57.88 Trd,cor = t * Trd Vrd,cor = v*Vrd

t = v =

Using the above result we will gett = 0.908 v = 0.419 Therefore, TRd, cor = 38.75 and VRd,cor = 18.3And the torsion and shear resistance of the concrete Tc,cor = tc *Tc Vc,cor = vc*Vc

tc = vc =

By substituting the above result we will have

tc = 0.903 vc = 0.439 Therefore , Tc,cor = 5.265 Vc,cor = 25.41 Tsd < TRd,cor Vsd < VRd,cor Reinforcement Ueff = 2*(b-2*d+D+2*d’) = 2*(600-2*43+300-2*43) = 1456

Asl =

Then use 216 longitudinal barsTransversal bar

S

10 c/c 420 mm and



S

Use 8 c/c 270 or using 6 c/c 150

DESIGN OF BEAM FOR FLEXURE

Axis -4 Section

300 mm

600 mm

From analysis Span moment = 94.72 kN.m , d = 255mm, b = 600mm Km = 49.77, Ks = 4.40

As =

# 20 = 5.2 620 Support moment = -141.49 kN.m (C , D and A) Km = 60.22 (doubly RC)

, Ks = 4.67

, = 1.01

= 0.3, = 1.12

As =

# 20 = 8.33 920

As’ = = 186.43 # 20 = 220



Support moment = -181.62 kN.m (E and D) Km = 68.23

, Ks = 4.55, = 1.026

Ks’ = 1.11, ’ = 1.12

As =

# 20 = 10.59 1120

As’ = = 885.45 mm2m

# 20 = 2.82 320

Axis-4 level-01 at joint-A

Support moment = -60.36 kn.m Km = 41.24 Ks = 4.19

As =

# 20 = 3.47 420

Axis-E on level-02 Support moment 1 = -51.05 kN.m Km =35.89 Ks = 4.11 As = 816.40 mm2

# 20 = 2.6 320 Support moment 2 = -22.53 kN.m Km = 23.84 Ks = 3.982 As = 349.08 mm2

# 20 = 1.74 220 Span moment M = 23.64 kN.m Km = 24.42 , Ks = 3.986 As = 366.65 mm2m # 16 = 1.824 216

The other sample moment results can be tabulated as follow from block -1:-

Tie Beam Table 4.3 Axis Section

positionM(KN-m) Reinforced (As,As’)mm2

calculated(As,As’)mm2Provided()

C,D&E Sup.(4) -23.39 singly 488.14,0 220Span.(5-4) 9.48 singly 185.47,0 216



A,BSup.(3) -26.4 singly 558.61,0 220Sup.(4) -13.06 singly 258.63,0 214Span(5-4) 14.32 singly 284.32,0 214

4,5Sup.(D,E) -14.03 singly 277.21,0 216Sup.(A,B&c) -16.77 singly 337.02,0 216Span(all) 7.48 singly 144.82,0 216

Typical FloorTable 4.4

Axis Section position

M(KN-m) Reinforced (As,As’)mm2calculated

(As,As’)mm2Provided()

C,D&ESup.(4) -58.85 singly 1047.55,0 420

Span(all) 27.03 singly 314

A&BSup.(3) -45.90 singly 780.47,0 416Sup.(4,5) -34.78 singly 0486.96,0 220Span(all) 20.28 singly 0321.17,0 216

4

Sup(E) -175.27 doubly (3178.1,750.2) (624,320)Sup(A,D) -140.37 doubly (2581.7,122.35) (624,320)Sup.(B.C) -83.41 singly 1395.58,0 424Span(all) 94.72 singly 1621.66,0 620

5Sup.(A,B.D) -92.71 singly 1572.82,0 520Sup.(C,E) -70.35 singly 1155.16,0 420Span(all) 46.93 singly 746.86,0 320

3 Sup.( A,B) -54.3 singly 742.82,0 520Span(all) 50.35 singly 1155.16,0 16

Grade BeamTable 4.5

Axis Section position

M(KN-m) Reinforced (As,As’)mm2calculated

(As,As’)mm2Provided()

C,D&ESup.(4,5) -119.59 singly 1240.06,0 420Span(all) 79.20 singly 881.9,0 320

A&BSup.(all) -83.40 singly 841.17,0 320Span(all) 56.82 singly 504.43,0 220

4,5Sup.(all) -66.35 singly 662.85,0 220Span(all 39.8 singly 390.67,0 216

3Sup.(all) -64.24 singly 640.98,0 220Span(all) 26.57 singly 258.37,0 214



CHAPT –FIVE ER

Design of column Before designing columns the members should be classified as sway or non sway depending on their sensitivity to secondary effect.According to EBCS 2, 1995 Article 4.4.4.2:A frame may be classified as non sway if its response to in plane horizontal is sufficiently rigid for it to be acceptability accurate to neglect any additional reactions arising from horizontal displacement of its nodes. If a frame is classified as sway, the effect of the horizontal displacements of its nodes should be taken in to consideration in the design procedure. A frame may be classified as non sway for a given load case if the critical load ratio Nsd/Ncr 0.1Design procedure

1) To design a column in a particular frame first the frame should be classified as sway or non sway

2) To determine the nature of the selected frame we substitute the beams and columns by one substitute frame

3) The value of axial loads on each substitute frame is taken by adding the axial load of each column for the story including self weight (But the sap results already accounts such loads and they are directly taken for design

4) The value of the stiffness coefficients of the substitute fame is given by For beams = 2*kbi



For columns =kci

Where Kbi = stiffness coefficient of beams Kci = stiffness coefficient of columns5) The effective length of the substitute frame is computed for each storey assuming

a sway frame as shown in the following equations For non sway

=

For sway mode Conservatively:

= (1+0.8 αm) 0.5

Where αm is a stiffness coefficient which can be discussed using the following theoretical model k1

k11 k12 α1= (k1+kc)/ α1 (k11+k12 kc

α1= (K2+kc)/ α1 (k21+k22) α1= (α1+ α2)/2 k21 k22

k2 Where k1 and k2 are column stiffness coefficients (=EI/L) K11, k12, k21, k22 are the boundary beam stiffness

Kc is the stiffness coefficient of the column being designed α =1.0 if opposite end elastically rigid α = 0.5 if opposite end free to rotate α = 0 for cantilever beam N.B The above approximate equations for effective length calculation are applicable for values of α1 and α2 not exceeding 10. If a base is designed to resist the column moment, α may be taken as 1.0

6) The dimension of the substitute column is computed to find the moment of inertia of the section, (Ic).

7) The amount of reinforcement required by the substitute column is computed, and the moment of inertia of the reinforcement with respect to the centroid of the concrete section is determined.

For more accurate determination, the first order moment, Mdl at the critical section of the substitute column may be determined using

Mdl = ( α 1+ 3)*H*L (α1+ α2+6) Where H =the total horizontal reaction at the bottom of the storey L=storey height



8) The critical load of a storey may be assumed to be equal to that of the substitute beam-column frame and given as

Ncr = (Π2*EIe)/ (Le2)

Where EIe is the effective stiffness of the substitute column designed Le is effective length In lieu of more accurate determination, the effective stiffness of a column may be taken as EIe= 0.2*EcIc*+EsIs Where :- Ec =1100fcd Es is the modulus of elasticity of steel (in our case =2*10^5)

Ic & Is are the moment of inertia of the concrete and steel bars, respectively of the substitute column, with respect to the centroid of the concrete section We designed four columns but to show the procedure we select two columns How ever, before going to design let’s identify whether they are sway or non sway following the procedure as follows:

Frame on Axis D

0.065 0.13

0.8 0.8 1.51 0.14 0.28

0.8 0.8 1.51 0.14 0.28

0.8 0.14 0.8 1.51 0.28

0.8 1.51 0.14 0.8 0.28



0.8 0.8 1.51 0.43 0.86

1.2 1.07 2.271.0

Substitute frame for frame on axis- D

Table 5.1 level m le s L Mdlfoundation 1 1 1 2.683 483 44.43 0.454 2 44.43ground 10 8.79 9.395 8.755 483 44.43 0.454 3 43.221st 10 10 10 9 483 44.42 0.454 3 66.652nd 10 10 10 9 483 35.74 0.454 3 53.613rd 10 10 10 9 483 23.17 0.454 3 34.764th 10 10 10 9 483 5.62 0.454 3 8.22

Table 5.2

level Nsd(KN V As(mm2 s(109) Ncr(KN Nsd/Ncr Classificationfoundation 1560.9 0.6 0.035 min 1866.31 6 0.7 34554 0.06 non swayground 1459.6 0.561 0.034 min 1866.31 6 0.7 2085 0.88 sway1st 1106.5 0.425 0.053 min 1866.31 6 0.7 1510 0.93 sway2nd 753.67 0.289 0.043 min 1866.31 6 0.7 1510 0.63 sway3rd 405.14 0.156 0.028 min 1866.31 6 0.7 1510 0.34 sway4th 61.13 0.023 0.007 min 1866.31 6 0.7 1457 0.05 non sway

Frame on Axis 4

0.065 0.065 0.086 0.065 0.052

0.71 0.71 3.56 0.270 0.270 0.355 0.270 2.33 0.71 0.71 3.56 0.270 0.270 0.355 0.270 2.33 0.71 0.71 3.56 0.270 0.270 0.355 0.270 2.33 0.71 0.71 3.56

0.270 0.270 0.355 0.270 2.33



0.71 0.71 3.56 0.610 0.610 0.799 0.610 5.26

1.07 1.07 5.35 1.0 E D C B A Axis -4 (substitute frame)

Table 5.3

Level m le s L(m) Mdl(KNm)foundation 1 1 1 2.68 565 131.73 0.855 2 131.73ground 6.11 3.38 4.75 6.57 565 131.73 0.855 3 162.771st 6.11 6.11 6.11 7.28 565 124.72 0.855 3 187.082nd 6.11 6.11 6.11 7.28 565 101.03 0.855 3 151.553rd 6.11 6.11 6.11 7.28 565 65.5 0.855 3 98.254th 6.11 6.11 6.11 7.28 565 16.41 0.855 3 18.09

Table 5.4

Level Nsd(KN V As(mm2 s(109) Ncr(KN Nsd/Ncr Classificationfoundation 1560.9 0.438 0.065 Min 3623 8 0.182 78862 0.02 Non sway ground 1459.6 0.41 0.081 Min 3623 8 0.182 13122 0.14 sway1st 1106.5 0.311 0.093 Min 3623 8 0.182 10687 0.131 sway2nd 753.67 0.212 0.075 Min 3623 8 0.182 10687 0.0895 Non sway 3rd 405.14 0.114 0.049 Min 3623 8 0.182 10687 0.0483 Non sway 4th 61.13 0.017 0.009 Min 3623 8 0.182 7398 0.0101 Non sway

Design of Selected Column For buildings a design method may be adopted which assumes that the compression members are isolated and have a simplified shape for the deformed axis of the column

Total eccentricity The total eccentricity to be used for design of column with constant cross section is given by: etot = ea+ee+e2

Where :-



ee the equivalent constant first order eccentricity There are two cases to compute the equivalent eccentricity

1. for first order eccentricity eo equal at both ends of a column ee= eo

2. for first order moments varying linearly along the length, the equivalent eccentricity is the higher of the following two values

ee = 0.6eo2+0.4eo1

ee = 0.4eo2

eo2 & eo1 are first order eccentricity at the ends eo2 being positive and greater in magnitude than eo1. ea is the additional eccentricity to account for geometric imperfection introduced by increasing the eccentricity of the longitudinal force acting in the most un favorable condition.

ea = Le/300 ≥ 20mm , Le is the effective length of isolated column e2 is the second order eccentricity According to EBCS, 2, 1995 article 4.4.6(2) second order effects in compression members need not be taken in consideration in the following cases

a) for sway frames the greater of -25

- 15/Vd, and Vd =Nsd /fcd*Ac

b) for non sway frames 50-25(M1/M2) where :- M1& M2 are first order moments at the ends and M2 being always positive and greater in magnitude than M1,and M1 being positive if the member bent in a single curvature and negative if bent in bent in double curvature. For sway frames having ratio of Nsd/Ncr <0.25 second order effect is computed using the magnification factor, δs

δs = 1/ (1- Nsd/Ncr) Where :- Nsd =design value of the total axial load Ncr = the critical value for failure in sway mode And Mf = δs*M1 where :- Mf =the amplified sway moment M1= the first order moment at the point of total maximum moment.The total moment used for design of column at the critical section is given by: Mtot= Mf + Ma

As per our EBCS-2, 1995 As, max = 0.08*Ac As, min = 0.008*Ac



For Frame On axis DFor first order momentsTable 5.5Level Mo2 Mo1 0.4*Mo2 0.6*Mo2+0.4Mo1 Mefoundation 4.76 0 1.904 2.856 2.856Ground 12.29 -7.65 4.916 4.314 4.9161st 22.11 -18.9 8.844 5.718 8.8442nd 17.47 -17.1 6.988 3.642 6.9883rd 20.95 -17.7 8.38 5.478 8.384th 12.46 -2.42 4.984 6.508 6.51

For the accidental eccentricity let’s calculate the effective length using the following frame 0.065 0.065

0.71 0.27 0.27 0.71 0.27 0.27

0.71 0.27 0.27 0.71 0.27 0.27

0.71 0.43 0.43 1.07

Table 5.6Level Le/L Le ea Nsd Mafoundation 1 1 1 0.78 1.56 20 1560.9 31.217Ground 2.63 1.65 2.14 1.65 4.95 20 1459.6 29.1921st 2.63 2.63 2.63 1.76 5.28 20 1106.5 22.132nd 2.63 2.63 2.63 1.76 5.28 20 753.67 15.0733rd 2.63 2.63 2.63 1.76 5.28 20 405.14 8.1028



4th 5.46 2.63 4.05 0.92 2.76 20 61.13 1.2226

For Second Order effectTable 5.7

Level Le Li Actual SR Critical SR 2nd order effect (Mtot) xfoundation 1.56 0.116 13.45 50 not considered 34.066Ground 4.95 0.116 42.67 50 not considered 34.1061st 5.28 0.116 45.52 50 not considered 30.9742nd 5.28 0.116 45.52 74.47 not considered 22.0583rd 5.28 0.116 45.52 71.16 not considered 16.484th 2.76 0.116 23.79 54.86 not considered 7.73

For Frame On axis 4Table 5.8 First order momentsLevel Mo2 Mo1 0.4*Mo2 0.6*Mo2+0.4Mo1 Mefoundation 83.33 -6.55 33.332 52.62 52.62Ground 109.04 -4.64 43.62 63.57 63.571st 49.42 -36.8 19.77 14.92 19.772nd 44.81 -20.4 17.924 18.734 18.7343rd 42.55 -3.04 17.02 24.314 24.3144th 8.62 -8.42 3.448 1.804 3.448

0.065 0.065

0.711 0.27 0.27

0.711 0.27 0.27

0.711 0.27 0.27

0.711 0.27 0.27

0.710 0.61 0.61

1.07

D



Axis -D Table 5.9For accidental eccentricityLevel m Le/L Le ea Nsd Ma

foundation 1 1 1 0.778 1.556 20 1560.9 31.217

Ground 2.633 1.46 2.05 1.62 4.86 20 1459.6 29.192

1st 2.633 2.633 2.633 1.76 5.28 20 1106.5 22.13

2nd 2.633 2.633 2.633 0.883 2.649 20 753.67 15.073

3rd 2.633 2.633 2.633 0.883 2.649 20 405.14 8.1028

4th 5.47 2.633 4.05 0.92 2.76 20 61.13 1.2226

Table 5.10For Second Order effect

Level Le Li Actual SR Critical SR 2nd order effect (Mtot)y

foundation 1.556 0.116 13.41 48.03 not considered 83.83Ground 4.86 0.116 41.90 51.06 not considered 92.761st 5.28 0.116 45.52 68.636 not considered 41.92nd 2.649 0.116 22.84 61.37 not considered 33.813rd 2.649 0.116 22.84 51.786 not considered 32.424th 2.76 0.116 23.79 74.405 not considered 4.668

Table 5.12Final Design Summary Of column located at (axis 2, axis D)Level Nsd Mx My V x y s As providedfoundation 1560.9 34.07 83.83 0.874 0.05 0.12 0.31 2121.9 8Ground 1459.6 34.11 92.76 0.817 0.05 0.13 0.293 2005.5 1st 1106.5 30.97 41.9 1.08 0.101 0.137 0.66 2579.8 82nd 753.67 22.06 33.81 0.739 0.072 0.11 0.22 859.95 3rd 405.14 16.48 32.42 0.397 0.054 0.106 0.1 390.88 44th 61.13 7.73 4.668 0.088 0.044 0.002 min 500 4

Similar manner taken for a column of the other buildingTo design a column located at the intersection of frame 2’and frame B’ we follow the same procedure as above Frame On axis 2’Table 5.13Level m Le S L dlfoundation 1 1 1 2.683 626.03 1.28 163.37 2 163.37Ground 3.52 2.974 3.248 5.691 626.03 1.28 163.37 3 234.3271st 1.698 3.521 2.61 5.272 469.52 0.405 150.72 3 262.8162nd 1.698 1.698 1.698 4.607 469.52 0.405 121.53 3 182.2953rd 1.698 1.698 1.698 4.607 469.52 0.405 77.75 3 116.6254th 3.525 1.698 2.612 5.273 469.52 0.405 18.93 3 23.772

Table 5.14

level Nsd(KN) V As(mm2) s(109mm4) Ncr(KN) Nsd/Ncr Classification

foundation 1102.43 0.25 0.06 Min 3135 0.23 43807.32 0.025 Non sway



ground 989.83 0.22 0.08 Min 3135 0.23 9736.68 0.102 sway1st 754.23 0.3 0.22 0.3 2870 0.12 3593.25 0.21 sway2nd 518.21 0.33 0.16 0.125 1760 6 0.07 4700.81 0.11 sway3rd 282.2 0.11 0.1 0.125 1760 6 0.07 4700.81 0.06 Non sway 4th 46.77 0.02 0.02 0.025 1760 6 0.07 3588.34 0.013 Non sway

Frame On axis B’Table 5.15

Level m Le S L dlfoundation 1 1 1 2.683 475.623 0.427 40.26 2 40.26ground 10 5.849 7.92 8.128 475.623 0.427 40.26 3 48.9171st 6.667 10 8.33 8.307 356.762 0.135 41.23 3 70.9392nd 6.667 6.667 6.67 7.550 356.762 0.135 33.24 3 49.863rd 6.667 6.667 6.67 7.550 356.762 0.135 21.27 3 31.9054th 6.923 6.667 6.80 7.611 356.762 0.135 4.84 3 7.165

Table 5.16

level Nsd(KN) V As(mm2) s(109mm4) Ncr(KN) Nsd/Ncr Classification

Foundation 1820.78 0.71 0.03 Min 1810 0.072 14612.55 0.125 swayground 1726.82 0.67 0.04 Min 1810 0.072 1592.6 1.084 sway1st 1309.25 1.75 0.14 1.01 5580 0.1 484.14 2.704 sway2nd 895.57 1.2 0.1 0.467 2580 0.052 584.14 1.53 sway3rd 484.21 0.65 0.06 Min 1020 0.023 583.43 0.83 sway4th 69.5 0.09 0.01 Min 1020 0.023 574.11 0.121 sway

Design of Selected Column

Frame On axis 2’ Table 5.17 First order momentsLevel Mo2 Mo1 0.4*Mo2 0.6*Mo2+0.4Mo1 Mefoundation 36.1 -30.08 14.44 9.628 14.4Ground 6.44 -3.22 2.58 2.576 2.581st 2.43 -0.74 0.972 1.162 1.1622nd 2.08 -1.37 0.832 0.7 0.8323rd 10.96 -3.14 4.384 5.32 5.32

4th 25.88 -14.18 10.352 9.856 10.352

Table 5.18 For accidental eccentricityLevel m Le/L Le ea sd Mafoundation 1 1 1 0.78 1.56 20 1820.78 36.4156Ground 0.964 0.549 0.708 1.25 3.75 20 1726.82 34.53641st 0.68 0.864 0.772 1.27 3.81 20 1309.25 26.1852nd 0.62 0.68 0.65 1.23 3.69 20 895.57 17.9114



3rd 0.448 0.62 0.554 0.71 2.13 20 484.2 9.6844th 0.747 0.488 0.618 0.72 2.16 20 69.5 1.39

Table 5.19 For second Order effectLevel Le Li Actual SR Critical SR 2nd order effect Mtot

foundation 1.56 0.087 17.93 70.83 Not considered 1 50.82Ground 3.75 0.087 43.10 25 considered 1.13 37.461st 3.81 0.087 43.79 25 considered 1.266 27.662nd 3.69 0.087 42.41 25 considered 1.24 18.853rd 2.13 0.087 24.48 57.16 Not considered 1 154th 2.16 0.072 30.00 63.7 Not considered 1 11.74

Frame On axis B’

Table 5.20 First Order momentsLevel Mo2 Mo1 0.4*Mo2 0.6*Mo2+0.4Mo1 Mefoundation 30.28 -21.85 12.112 9.428 12.112Ground 13.51 -11.1 5.404 3.666 5.4041st 14.31 -14.18 5.72 2.908 5.722nd 14.66 -14.35 5.864 3.056 5.8643rd 15.81 -15.36 6.324 3.342 6.3244th 12.85 -7.45 5.14 4.73 5.14

Table 5.21 For accidental eccentricity

Level m Le/L Le ea sd Mafoundation 1 1 1 1.34 2.68 20 1820.78 36.4156Ground 5.615 2.19 3.903 2.03 6.09 20.3 1726.82 35.054451st 2.63 5.615 4.123 2.07 6.21 20.7 1309.25 27.101482nd 1.926 2.63 2.28 1.68 5.04 20 895.57 17.91143rd 1.77 1.926 1.85 1.57 4.71 20 484.2 9.6844th 1.674 1.77 1.72 1.54 4.62 20 69.5 1.39

Table 5.22 For Second Order effectLevel Le Li Actual SR Critical SR 2nd order effect Mtot

foundation 2.68 0.115 23.30 25 Not considered 1 48.532Ground 6.09 0.115 52.96 25 considered 1.13 42.251st 6.21 0.087 71.38 25 considered 1.33 34.712nd 5.04 0.072 70.00 25 considered 1.33 25.713rd 4.71 0.072 65.42 25 considered 1.33 18.24th 4.62 0.072 64.17 47.88 considered 1.14 7.25



Column reinforcement

Using biaxial chart no 10

h

b Design summary for column located at (axis2’, axis B’)Level sd Mx My V x y As(mm2) As provided(mm2)foundation 1820.78 49 51 1.34 0.09 0.125 0.779 4164.22 Ground 1726.82 42.3 37 1.27 0.08 0.093 0.637 3319.91 1st 1309.25 35 28 1.28 0.115 0.092 0.742 2900.36 2nd 895.57 26 19 1.05 0.122 0.075 0.523 1703.61 3rd 484.2 18 15 0.57 0.09 0.059 0.1 325.74 4th 69.5 7 12 0.1 0.045 0.07 0.15 407.17

After designing the above typical columns (namely C2&C3) for demonstrating the steps we group the other columns as C1&C4 and tabulated as follows:-

Table 5.23

C1(x8) C2(x4) C3(x4) C4(x10)

Level As provided

X-section

As provided

X-section

As provided

X-section

As provided

X-section

Foundation 6Ø16 350x350 8Ø20 400x400 14Ø20 300x400 10Ø16 300x300Ground 8Ø16 350x250 10Ø16 400x400 12Ø20 300x400 10Ø20 300x250First 4Ø20 300x200 8Ø20 300x300 10Ø20 300x300 12Ø16 300x250Second 4Ø16 300x200 6Ø14 300x300 10Ø16 300x250 6Ø16 300x250Third 4Ø16 300x200 4Ø14 300x300 4Ø14 300x250 6Ø14 300x250Fourth 4Ø16 200x200 4Ø14 250x250 4Ø14 250x250 6Ø14 250x250

Lateral reinforcement for column

According to EBCS 2, 1995 article 7.2.4.3, (1) and (2)I. The diameter of should not be less than 6mmor one quarter of the diameter of the

diameter of the longitudinal bars II. The center to center spacing of lateral reinforcement shall not exceed:

-12 times the minimum diameter of longitudinal bars =12*16=192mm-Least dimension of column =400mm-300mm

According to EBCS 2, 1995 article 4.5.2. (1) in order to prevent diagonal compression failure in the concrete, the shear resistance Vrd of a section shall not be less than the applied shear force Vsd Where VRd = 0.25*fcd*bw*d bw =the minimum width of the web



VR=0.25*11.33*400*300 =339.9KNTo compare the value for the maximum computed shear forces:Vsy =26.07 KN and Vsx = 33.09 KN taking the resultant force VR = 42.126 KNHence VRd=339.9 KN ≥ 42.126 KN....ok!For members subjected to axial compression the additional shear force VCRT carried by concrete is given by V CRT = 0.1* bw* d*Nsd/Ac …..EBCS-2, 1995 (4.30) Where:- Nsd is the design axial load For the minimum value of Nsd =69.50 KN V CRT = 0.1*400* 300*69.50/400*300 = 5.762 KN VRd= (339.9+5.762) KN and VR =42.126 KN since VR VRd apply min reinforcement. Ø 8 c/c 190mm

Basic anchorage length

The basic anchorage length is the embedment length required to develop the full design strength of a straight reinforcement bar The basic anchorage length, Lb for a bar of diameter Ø is given by Lb=Ø*fyd / (4*fbd)

The design bond strength for deformed bars is given by

fbd = 2*fctd =2*1.0312=2.0624MPa For Ø=20 Lb = 20*260.87/ (4*2.0624) =632.44mm

Required anchorage length The required anchorage length Lb, net= (a* Lb *As, cal/ (As, provided)) ≥ Lb min

a=1 for straight bar anchorage in tension or compression Lb min= 0.6* Lb≥ 10Ø=200mm or Lb min=303.57mm≥200mm Lb, net=1.0*632.44*416.24/4398.22=598.79mm≥200mmUse Lb, net =590mm

Laps and jointsThe length of lap shall not be at least equal to Lo≥a1* lb net ≥ lo min....................EBCS2, 1995 (7.8) where Lo min = 0.3a*a1*Lb≥15Ø = 240mm for Ø16 = 300mm for Ø20 or 200mma1 a function of the percentage of the reinforcement lapped at any one section Assuming 100%lap, a1 =1.4 For Ø16, Lo min = 0.3*1.4*590* = 247.8mm



Lo min = 240mm for Ø16 Lo min =300mm for Ø20 Lo≥ 1.4*590 = 826mm ≥ 300mm Use Lo = 830mm

CHAPTER-SIX

Foundation

Footing Design A) Interior column footing design (column at (4,E))

Y Using comb-4

MyGiven L - -Mx = 83.32 kN.m L’ Mx -My = -0.884 kN.m X -Pd = 1560.67 kN and L’ = B’ = 0.4 m B’

BAssuming an allowable bearing capacity for the soil as = 280 KPa

proportioning

Using unfactored load; Pd =



1.3m

2m 0.7m

Assuming a surface footing, B = L

Solving the equation by trial and error, we get B = 2.19m We use B = L = 2.5m

Assume D = 0.70 and sub soil = 18 kN/m3 Dead weight of sub soil = (2.5*2.5-0.42)*1.3*18 = 142.51 kN Dead weight of footing = 2.5*2.5*0.7*25 = 109.38 kN Pd, tot = 1560.67 + (142.51 + 109.38 ) * 1.3 = 1888.1 kN

Taking maximum value from the above equation we get = 334.34 KPa

For conservative design Let’s check for both punching and wide beam shear failure respectively

i . punching shear

According to EBCS-2, 1995 article 4.7.6 the resistance of footing without punching shear reinforcement is give by: Vrd = 0.25 * fctd * k1 * k2 * u * d Where: K1 = 1+50 2.0

K2 = 1.6 – d 1.0 (d in meter)



d =

e = Assume = 0.002, k1 = 1.0 + 50 * 0.002 = 1.1 L’+3*d L’ U = 4*(B’ + 3*d) = 4* ( 0.4 + 3*d) = 1.6 + 12*d Vrd = 0.25*1043*1.1*1.0*(1.6 + 12*d) B’ Vacting = (2.5*2.5 – (0.4 + 3* d)2)*334.34Then Vrd > Vacting (for safe condition) B’+3*d Hence solving for d d 499 mm

Therefore , D = 499 + 50 + 10 = 550 mm < 600 mm..... ok!.

ii - wide beam shear

According to EBCS-2, 1995, article 4.5.3, the shear force Vc carried by the concrete is given by : Vc = 0.2*fctd*k1*k2*bw*d Using 50mm clear cover and 20 longitudinal reinforcement bar d= 700 – 50 – 10 = 640 mm Vc = 0.25*1043*1.1*1*2.5*0.64 = 458.90 kN Vacting = (B/2 - B’/2 - d)*qu*B = (2.5/2 - 0.4/2 – 0.54) *333.27*2.5 = 342.69 kN Since Vc > Vacting ..... the section is safe !!

Similarly, By comb-1

Given F3 = 1928.028 kN and L’ = B’ = 0.4 m, = 280 KPa Mx = 5.435 kN.m My = 3.164 kN

proportioning

for unfactor , Pd =

Assume the square footing B = L



Solving by trial and error we get B = 2.22 m Use B = L = 2.5 m then section is safe for both load combination

iii - reinforcement

M = 364.18*1*1.05*1.05/2 = 200.75 kN.mUsing table No.1 For design

= 22.14

Ks = 3.965 As = Ks * = 3.965 *

Amin = min * b*d = .002*1000*640 = 1280

Since As < Amin use Amin = 1280

taking 16

Take 16 c/c 150 mm (both direction)

N of reinforcement =

Hence provide 17 16 c/c 150 mm

iv-development length (check)

B) Exterior column footing design (column at (4,E))

Using comb-4 Given My Mx = 94.811 kN.m L L’ Mx My = -4.794 kN.m B’ P = 1067.78 kN and L’ = 0.4 m and B’ = 0.4 m B proportioning We assume an allowable bearing capacity for the soil as = 280 KPa



Assume a surface footing B = L

Solving by trial and error we get B = 1.95 m, use B = 2.5

Assume D = 0.7 m Dead weight of sub soil = (2.5*2.5-0.*0.4)*1.3*18 = 142.05 kN Dead weight of footing = (2.5*2.5*0.7*25) = 109.4 kN Pd,tot = 1067.78 + 1.3*(109.4+142.04) = 1394.65 kN

= 261.40 kN

i - punching shear Vrd = 0.25*fctd*k1*k2*U*d U = 1.6 + 12*d Vrd = 0.25*1043*1.1*1*(1.6+12*d)*d Vacting = (2.5*2.5-(0.4+3*d)(0.4+3*d))*261.40 Vrd > Vacting for safe condition Solving by trial; and error d > 0.499m D = 499 + 50 + 10 = 600 mm < 700 mm ... ok! Hence D = 700 mm ii-wide beam shear

Verso = 0.25*fctd*k1*k2*bw*d, d = 700 – 50 – 10 = 0.640 mm = 0.25*1043*1.1*1*2.5*0.64 = 458.92 kN

Vacating = (B/2 - B’/2 - d)*qu*B = (2.5/2 – 0.4/2 - 0.64 ) * 261.40*2.5 = 267.94 kN Since Vacting < Vresist safe !



Let’s check the dimension for comb-1

Given P = 1258.36 kN B’ = 0.4 and L’ = 0.4 Mx = 7.65 kN.m = 280 KPa My = -2.62

Assume a square footing B = L

by trial and error B = 1.797 m < 2.5 m.... ok!

iii- reinforcement M = 261.4*1*1.05*1.05/2 = 144.96 kN

Km = 18.76 Ks = 3.95 (minimum) As = 894.675 mm2 < Asmin = 1280 mm2 1.05

Therefore take 16 c/c 150 mm (both direction)

N of bar 17 16 c/c 150 mm ult = 261.4 kN

iv - development length

y

C). footing at (axis-1 and axis-A’) Given P = 608.97 kN My Mx = 21.66 kN.m



L L’ Mx

B’

B My = 19.98 kN.m and L’ = 0.35 and B’ = 0.25And we assume an allowable bearing capacity, = 280 KPa

B’= L’

proportioning using unfactored loads, Pd = 604.97/1.4 = 434.98 kN and ex = 0.0459 m and ey = 0.0497 m Assuming a square footing, B = L 1.4m

0.6m

After trial and error we get, B = 1.469 m take B = 1.5 m Assume D = 0.6 m and sub soil = 18 kN/m3

Dead weight of sub soil = (1.5*1.5 – 0.35*0.25)*1.4*18 = 54.495 kN Dead weight of footing = (1.5*1.5*0.6*25) = 33.75 kN Pd,tot = (608.97 + 1.3*(54.495 + 33.75)) = 723.68/8 kN ex = 0.0276 m ey = 0.0299 m ult = max = 395.62 KPa = max min = 261.49 Kpa

i - Check for punching

Vrd = 0.25*fctd*k1*k2*U*d = 1.4 + 12*d Vrd = 0.25*1043*1.1*1*(1.4 + 12*d) Vacting = (2*2 – (0.3 + 3*d0)*(0.4 + 3*d))*248.42 Hence solving for d , d 358.86mm L’

Then Vrd > Vacting ..... for safe condition 0.4+3*d Therefore D =358.86 + 50 + 10 = 418.86 mm B’ = 420 mm < 600 mm....ok !



ii - Check for wide beam shear Vc = 0.25*1043*1.1*1.24*2.0*0.36 = 258.143 kN Vacting = (2/2 – 0.3/2 – 0.36)*248.42*20 = 243.45 kNSince Vacting < Vc ..............the section is safe!

iii - Reiforcement

M = 248.44*0.85*1.0*0.85/2 = 89.74 kN.m 0.4m 0.85m Km = 17.54 ks = minimum 0.3m As = 0.002*1000*540 = 1080 use 10 c/c 80 mm

D). Footing located at (axis-E and axis-3’) Given Mx =31.5 kN.m My = 16.07 kN.m 2.0m Pd =646.3 kN and L’ = 0.4 m and B’ = 0.3 m And we assume an allowable baring capacity for the soil as = 280.0KPa proportion

using un factor loads, pd = 0.035 m ex = 0.035m ey = 0.068 m Assuming a square footing, B = L

After trial and error, we get B = L = 1.52 m, and we use B = l = 2.0 m



Assume, D = 0.6 m and sub soil = 18 kN/m3

Dead weight of sub soil = (2*2-0.4*0.3)*1.4*18 = 97.787 kN Dead weight of footing = 2*20.6*25 = 60 kN Pd,tot = 646.3 + 1.3*(60 + 97.78) = 851.41 kN ex = 0.0188 m ey = 00.369 m

I - Check for punching Vrd = 0.25*fctd*k18k2*U*d U = 2*(0.25+3*d) + 2*(0..35+3*d) = 1.2 + 12*d Vrd = 0.25*1043*1.1*1*(1.2 + 12*d) Vacting = (1.5*1.5-(0.35+3*d)(0.25+3*d))*395.62 Hence solving for d = 0.109 m , D = 169 mm < 600mm , d = 0.54m actually available depth ii - Wide beam shear

Vc = 0.25*1043*1.1*1*1.491*1.5*0.540 = 346.40 kN Vacting = (1.5/2 - 0.25/2 - 0.169)*395.62 = 306.21 kN

iii - reinforcement

M = 394.62*0.625*1*0.625/2 = 77.27 KN.m Km =16.27 Ks = provide minimum reinforcement,As =0.002*540*1000 = 1080

use 10 c/c 80 mm



CHAPTER- SEVEN

DESIGN OF STAIR CASE

1-design of roof slab

-0.2cm pvc -4.8cm cement screed 4.73m S9 -15cm RC slab

-2cm plaster

4.87m

Simple support slab Loading

dead load - pvc water proofing treatment = 0.002*16 = 0.032 kN/m2m -cement screed = 0.048*23 = 1.104 kN/m2

-RC slab = 0.15*24 = 3.6 kN/m2

-ceiling plaster = 0.02*19 = 0.38 kN/m 2 Gk = 5.116 kN/m2

live load

For roof inaccessible except for normal maintenance or other similar things, qk = 0.25 kN/m2

wind load qref = 242 Pa



I-external pressure

Since the slab is at a height of h = 17.40 m Ce(z) = 1.62 For the external pressure coefficient, appendix A.2 of EBCS-1, 1995 is used. The roof is flat with a parapet of height

and

Because the roof is nearly square in shape, the coefficients are computed for one direction of the wind only.

1.22m e=min =4.87m

F2.44m 4.87m

G H I Table 7.1

1.22mF

0.487m 1.95m 2.29m Table 7.2 The coefficients are tabulated as followsZone F G H ICpe -2.08 -1.63 -0.711 -0.2 0.2Wind pressure -815.44 -639.03 -278.74 78.41 78.41 ii-internal pressure Wi = 313.63 Pa Wi = -196.02 Paiii-net wind pressure considering the severe condition Wnet = -815.44 – 313.63 = -1128.8 Pa = 78.41 – (-196.02) = 274.43 Pa

load combination the following three case are consideredcase-1: 1.3*DL + 1.6*LL Pd = 1.3*5.116 + 1.6*0.25 = 7.05 PaCase-2: 1.1*DL + 1.35*(LL + WL(+ve) ) Pd = 1.1*5.116+1.35*(0.25+0.274) = 6.34 Pa


zone Area (m2)F 0.59G 1.19H 9.5I 11.15

95


Case-3: 0.9*LL + 1.6*WL(-ve) Pd = 0.9*5.116 + 1.6*-1.128 = 2.799 paAmong the three case, case-1 is critical and used for analysis and designAnalysis Using table A-1 from EBCS-1, we take S9

a c xs=ys=0xf=0.056 Myf =Mxf = xf * Pd * Lx2

yf=0.056 = 0.056*7.05*4.722

S9 =8.83 kNm/m ux =uy = 0.33

Vxs = Vy = ux*Pd*Lx2

= 0.33*7.05*4.732

b d = 11.00 kN/mBut the load transferred to beam ab and ac

on ac w’ w’ = 0.989*11.00 = 10.879 kN/m self weight

reinforcement M = 8.83 kNm/m , d = 150 – 25 – 6 = 119 m Km = 24.97 , Ks = 3.99

As = but Asmin = min * b * d

= 0.002*1000*119 = 238 mm2 < As

Therefore s = use 8 c/c 160 mm

2-staircase design

We consider it to be simply support

C 1 2 D A…… ….. A

Rise = 200mmThread = 280mm

3 3

B……. …..B



C C1 2

Modeling Axis 1-1

1.68m 1.37m 1.68m

Axis 2

1.68m 1.37m 1.68m

Axis 3-3

1.37m 2.13m 1.37m

Axis 3-3



0.3m

0.167m

Depth determination

From deflection requirement for end span of a = 20

Using 16 bars and a cover of 15mm, overall depth becomes. D = 148.75 + 15 + 6 = 169.75mm Take D = 170mm, d = 149mmLoading

Data’s -Rise = 16.7 cm, thread = 30cm - # of rise = 6 and # of thread = 7 -terrazzo tile floor finishing of = 23kN/m3

-inclination of stair θ =

-live load for category, B, qk = 3 kN/m3

landing load

-Rc slab Gk1 = 0.17*25 = 4.25 kN/m2m -cement screed (2cm thickness) Gk2 = 0.02*23 = 0.46 kN/m2

-Terrazzo tile (3cm thick) Gk3 = 0.03*23 = 0.69 kN/m2

-ceiling plaster Gk4 = 0.02*19 = 0.38 kN/m 2 Total DL, Gk = 5.78 kN/m2

Therefore design loads is, Pd = 1.3*5.78 + 1.6*3 = 12.314 kN/m2

flight load -RC slab Gk1 = 0.17*25*(sec29.10˚) = 4.86 kN/m2

-terrazo tile Gk2 = 0.03*23 = 0.69 kN/m2

-ceiling plaster Gk3 = 0.02*23 = 0.46 kN/m2

-weight of step Gk4 = 0.5*0.167*25 = 2.087 kN/m 2 Total DL, Gk = 8.532 kN/m2

Design load, Pd Pd = 1.3*8.532 + 1.6*3 = 15.892 kN/m2



15.89212.314

43.75 kN.m

33.75

33.75

Reinforcement Fcd = 11.16 b = 1000mm Fyd = 260.87 d = 149 mm M = 43.75 kN.m Km = 44.394 Ks = 4.27

As = > As min

S = , using 12

Take 12 c/c 90 mm

of bars = , use 16 bars

Secondary reinforcement

According to EBCS-1, 1995, provide a secondary reinforcement of As = 0.2*1256.55 = 251.31 mm2m Provide 10 c/c 300, As (available) 261.8 mm2

Axis 1-1

33.795 kN/m



1.68m 1.37m 1.68m Depth determination (axis 1-1)

, Le = 4.73m, a=20

201.03 mm D = 201.03 + 15 + 6 = 222.03 mm Take D = 250 mm d= 250 -15 – 6 = 229 mmLoading Data -rise = 20 cm (# 5) -thread = 28 cm (# 6)

-inclination of stair θ =

-LL = 3 kN/ m3

Similarly from the above loading data I-landing, Pd = 14.914 kN/m2

ii-flight load, Pd = 1.3*11.77 + 1.6*3 = 20.11 KN/ m2

Analysis

48.71 20.11 20.11

95.86 kN.m

67.15

67.15

Reinforcement

Km = 42.76, Ks = 4.23 As = 1770.06 mm2m S = 87.00 mm, 14 Use 14 c/c 80 mm

N of bars = Take 18 bars

Secondary reinforcement

As = 0.2* A axis 2-2s (available) = 0.2*1925 = 385 mm2

10 c/c 200, As available = 392.7 mm2

MEKELLE UNIVERSITY, FACULITY OF SCIENCE AND TECHNOLOGY DEPARMENT OF CIVIL ENG’G100


Axis 2-2

Similar calculation applies for stair case on axis 2-2

Design of beams around the stairs A) For tie beams



Assuming b=250mm D=300-25-10-8 =257mm Msup = -37KN.m Km = ((37/0.25))/0.257 = 47.34 Ks = 4.34 As =Ks*M/d =624.82mm2

16 = 3.108 So use 3 16 Mspan = 20.37KN.m Km =35.123 Ks = 4.10 As = 324.97mm2

16 =1.62 So use 216

B) For intermediate beam at 4th level (30 x40 )

Msup = -49.37KN.m Km = 35.9 Ks =4.1 As = 566.99mm2

16 =2.82 Use 3 16 Mspan =42.89KN.m Km =33.49 Ks = 4.09 As = 491mm2

16 = 2.44 Use 3 16 C) for intermediate beams at other levels (3rd,2nd,1st)

Msup = -71.46KN.m Km = 43.23 Ks = 4.22 As = 844.70mm2

16 = 4.2 Use 416 Mspan = 54.34KN.m Km =37.699 Ks =4.14 As =630.16mm2

16 =3.14 Use 316 D) For beams parallel to the staircase ,use min reinforcement

As,min =0.002*250*0.207 =103.5mm2 Use 2 10 for both support and span

Shear design a)Tie beam Section 250mm



300mm

Vc =0.25*1.343*1.083*1.032*250*257 =24.11KN VRd = 0.25*11.13*250*257 =179.26KN

47.2KN 37.56KN 0.457m

24.11KN

0.457m 24.11KN

Vdl = 37.56 From x = 0 to x =1.35 and x = 0 to x = -1.35 apply minimum reinforcement S = min = use 6 c/c 160mm From x = 0 to x = 1.35 and x = 1.35 to x = 1.978

use 6 c/c 160mm

b)for grade beam and beam on 4th floor level

300mm Vc = 0.25*1.243*1.083*1.032*300*357 = 37.19 kN 400mm Vrd = 298.81 kN

37.19 0.457



0.786

0.786 Vc Vdr145.8

3.627 Vrd = 128.63 kN From x = 0 to x = 1.572 apply minimum reinforcement Use 6 c/c 160mm From x = 1.572 to x = 4.87

S =

Use 6 c/c 90mm

c) for intermediate beams (at 3rd,2nd,1st )

-166.49 Vdr 2.41

Vc Vc = 37.19 kN2.46

163.69 Vrd = 298.81 kN

Vdl = 135.50

From x = 0 to x = 0.55 and from x = 0 to x = -0.55 use minimum reinforcement Use 6 c/c 160mm From x the left portion Apply

S =

Use 6 c/c 80mm

Design of column for the stair case

The following the same procedure, we find it sway except at the foundation level.Table 7.1

level Mo2 Mo1 0.4*Mo2 0.6*Mo2 + 0.4*Mo1 Me(critical)Foundation 19.74 -5.92 7.896 9.472 9.472Ground 34.31 -30.15 13.72 8.526 13.721st 37.43 -36.64 14.97 7.802 14.97



2nd 40.06 -37.90 16.02 8.876 16.023rd 31.4 -19.9 12.56 10.88 12.564th 37.0 -22.6 14.8 13.16 14.80

For accident eccentricity 0.0668 Table 7.2

0.328 0.225

0.328 0.225

0.328 0.225

0.328

1.06

For secondary order effectTable 7.3

level Le Li Actual SR

Critical SR

Mark Ncr(103) Mfactor

Foundation 1.56 0.116 13.45 57.45 Not consider

1.00 230.4 9.742

Ground 4.64 0.087 53.33 25 Consider 1.03 26.10 14.131st 4.34 0.087 49.88 25 Consider 1.02 29.76 15.272nd 4.34 0.087 49.88 25.42 Consider 1.01 29.76 16.183rd 4.34 0.087 49.88 37.12 Consider 1.01 29.76 12.684th 5.11 0.087 58.74 45.74 Consider 1.01 21.50 14.95

Design momentsTable 7.4

level Nsd Mx My x y As (mm2)Foundation 882.8 27.13 - 0.49 0.037 0Ground 720 28.53 - 0.72 0.0946 01st 535.34 25.97 - 0.533 0.086 02nd 249.7 23.17 - 0.347 0.0795 0


level 1 2 m Le ea Nsd MaFoundation 1.0 1.0 1.0 1.56 20 882.8 17.66Ground 1.37 2.1 1.74 4.64 20 720 14.41st 1.37 1.37 1.37 4.34 20 535.34 10.72nd 1.37 1.37 1.37 4.34 20 349.7 6.993rd 1.37 1.37 1.37 4.34 20 160.0 3.24th 3.37 1.37 2.37 5.11 20 107 2.14

105


3rd 160.0 15.88 - 0.159 0.053 04th 107 17.09 18.98 0.106 0.053 0.063

CHAPTER- EIGHT

REINFORCEMENT DETAIL

Note: See in a separate page in AutoCAD drawing



CHAPTER– NINECOST ESTIMATION

Table 1.0 Cost comparison of BL building

Item No Description Unit

pre cast Ribbed slabvolume cost(birr) volume cost(birr)

A.SUB STRUCTURE I. EXCAVATION & EARTH WORK

Bulk excavation m3 263.20 5145.56 263.20 5145.56

Trench excavation m3 32.30 757.76 32.30 757.76

Pit excavation m3 247.25 7079 247.25 7079

backfill m3 307.85 4513.08 307.85 4513.08

cart away m3 542.75 13487.34 542.75 13487.34

hardcore m2 316.30 7894.85 316.30 7894.85 2. CONCRETE WORK 0.00

lean concrete m2 484.60 9939.14 731.09 16237.51 In Footings. m3 32.80 24201.08 96.78 77,169.47

In Foundation Columns. m3 11.00 13393.26 5.30 4,541.73

In grade beams m3 14.30 16327.87 25.19 19,941.66

In 10 cm thick ground floor slab m2 316.30 20008.51 411.31 32,561.36

Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire

a) dia 6mm deformed bar Kg 2779.32 674.62 7,542.25

b) dia 8 mm deformed bar Kg 16081.79 2950.18 32,983.01

c) dia 12 mm deformed bar Kg 0 210.96 2,358.53

d) dia 14 mm deformed bar Kg 12980.23 6.41 71.66

e) dia 16 mm deformed bar Kg 14415.75 4823.99 53,932.21



f) dia 20 mm deformed bar Kg 5880.25 595.74 6,660.37

g) dia 24 mm deformed bar Kg 29415 0.00

3.MASONARY WORK m3 86.10 29412 123.40 44,554.80 Subtotal 233711.79 337432.15

B-SUPER STRUCTURE 1. CONCRETE WORK

a) In elevation columns m3 57.20 76129.12 35.78 31,486.76

b) In suspended floor beams and top tie beam m3 95.60 108352.872 113.92 93,710.59

c) In roof slab . m3 3.27 3421 22.61 18,598.99

d) staircase m3 27.40 23136.85 24.70 20,318.22

e) beam column connections and necks of columns where indicated on the detail drawings. m3 10.08 11502.8

Floor slab m2 1361.60 102256.16 1188.30 94,362.90

precast beam 1 pcs 248 77475.2 0.00 0.00

precast beam2 pcs 112 34988.8 0.00 0.00

precast beam 3 0 0.00 0.00

precast beam 4 0 0.00 0.00

Slab blocketm2 58177.44

Mild steel reinforcement according to structural drawings. Price includes cutting, bending, placing in position and tying wire

a)dia 6 mm plain bar Kg 24309 3738.72 41,798.86 b) dia 8 mm deformed bar Kg 51064.97 5612.35 62,746.02 c) dia 10 mm deformed bar Kg 7612.92 661.06 7,390.63 d) dia 12 mm deformed bar Kg 30632.6 123.01 1,375.25 e) dia 14 mm deformed bar Kg 64897.75 2659.12 29,728.96 f) dia 16 mm deformed bar Kg 55566.75 1470.52 16,440.41 g) dia 20 mm deformed bar Kg 92364 1493.34 16,695.54 h) dia 24 mm deformed bar Kg 0 291.45 3,258.41Sub total 821888.232 Total 1055600.02 898250.17

Summary for comparison item Ribbed Pre cast DifferenceTotal Bars(kg) 25311.46 39529 14217.54

Total Volume(m3) 301.67 251.65 -50.02Total cost (birr) 898250.2 1055600 157349.852


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