Chapter 1: Design of Roof and Truss (B+G+4 Apartment Building in Ethiopia)

PROJECT BY ABDIRASHID MOHAMED DAHIR EMAIL:[email protected]

Chapter 1: Design of Roof and Truss (B+G+4 Apartment) Writer’s Note: This is a project by Abdirashid Mohamed Dahir who holds BSc in Civil Engineering from Debre Berhan University, Ethiopia. All reference materials can be found at the end of the lecture tailored for civil engineering students. Kindly, don’t hesitate to drop your comments to the email: [email protected] 1. Wind Analysis 1.1 Introduction We consider in our structural project, the design of G+4 apartment building with a basement storey, and located in Debre Berhan City, North Shewa, in the Amhara regional State of Ethiopia. Wind forces are variable loads which act directly on the internal and external surface of structures. The intensity of wind load on a structure is related to the square of the wind velocity and the dimensions of the members that are resisting the wind (frontal area). Wind velocity is dependent on geographical location, the height of the structure, the topography of the area and the roughness of the surrounding terrain. 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 , and walls .The action of wind can be a type of suction or pressure to our structures both externally or internally .However 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. 1.1.1 Method of Analysis Response of structure to variable action of wind could either be background component that involves static deflection or resonant component which is relatively small and involving dynamic vibration. The latter is widely considered for tall and


flexible structures. Here, our analysis is based on Background resonant or Quasistatic gust load. 1.1.2 Why static? It is critical for engineers to investigate dynamic coefficient for gust wind response as stipulated EBCS-1, 1995 section 3.9.3; a building which satisfies the criterion: (For cd <1.2 and building height less than 200m), static analysis can be adopted.

1.2 Analysis 1.2.1 Code Provisions EBCS1, Art 3.5.2 The wind pressure acting on the external surface of a structure shall be: We=qref*Ce(Ze)*Cpe And the internal pressure acting on the internal surfaces becomes, Wi=qref*Ce(Zi)*Cpi

Where:

qref=reference mean wind velocity Ce(Ze)= exposure coefficients for external pressure Ce(Zi)=exposure coefficients for internal pressure Ze,Zi= reference height defined in Appendix A of EBCS1

qref 

 2

Vref

2

Where: ρ= air density in Kg/m3


Vref=refernce wind velocity For Ethiopia Vref= 22m/s Cdir=Ctemp=Calt=1 [All are unity when it comes to Direction, Seasonal Variation in Temporary structures and Altitude] Vref=Cdir*Ctemp*Calt*Vref=22m/s

1.2.2 Flat Roof According to EBCS-1, Art 3.5.2:The wind pressure acting on the external surface of a structure shall be: We=qref*Ce(Ze)*Cpe And the internal pressure acting on the internal surfaces becomes:Wi=qref*Ce(Zi)*Cpi Where:

qref=reference mean wind velocity Ce(Ze)= exposure coefficients for external pressure Ce(Zi)=exposure coefficients for internal pressure Ze,Zi= reference height defined in Appendix A of EBCS1

qref 

 2

Vref

2

Where: ρ= air density in Kg/m3 Vref=reference wind velocity For Ethiopia Vref= 22m/s Cdir=Ctemp=Calt=1


Vref=Cdir*Ctemp*Calt*Vref=22m/s Assuming Debre Berhan altitude the highest in Ethiopia at 2000m Altitude above MSL ρ=0.94 kg/m3. qref 

0.94 * 22 2  227.48 N / m2  0.22748KPa 2

Exposure Coefficient at Reference Height For Terrain Category four in EBCS [Table 3.2 Terrain Categories and Related Parameters] We have assumed that the G+4 building locates in urban areas covered with buildings and their average height exceeds 15m in Debre Berhan town. Kt[Terrain Factor]=0.24, Zo=1m , Zmin=16m

Roughness coefficient Cr(z)=KT x ln(Z/ZO)………Zmin≤Z≤200m KT=0.24…………………….Category IV ZO=1Zmin=16m Cr(17.8)=0.24xln(17.8/1)=0.69 Topograghic coefficient In our case it is assumed that the topography is unaffected Ct=1 for Φ<0.05 Exposure coefficient According to section 3.8.5(2), the exposure coefficient is given by,


  7 * KT 2 Ce(17.8) = C r ( Z ) * Ct ( Z ) * 1    C r ( Z ) * Ct ( Z )  Where KT –is the terrain factor Cr(z) – is the roughness coefficient Ct(z) – is the topography coefficient and taken to be unity

 7 * 0.24  Ce(17.8) = 0.69 2 (17.8) *1(17.8) * 1   1.63  0.69 *1 

Based on EBCS flat roof configuration for wind loads, e=b or 2h, whichever is smaller. Meaning, e=7.72 and e=2(17.8)=35.6m, our e=7.72m

Zone

F

G

H

I


Area 𝑚2

1.49

2.98

23.84

10.73

We assume that we have a flat roof with sharp eaves. Cpe for Zone F Cpe= Cpe,1+ (Cpe,10 - Cpe,1) log A …………. Since 110m2 For closed building with internal partitions and opening windows with extreme values take C pi=0.8 or -0.5……………………EBCS-1, 1995 Therefore the external and internal wind pressure on the flat roof will be calculated as follows We = qref*Ce (Ze)*Cpe = 0.22748*1.63* Cpe = 0.37Cpe

Zone

F

G

H

Cpe

-2.37

-1.62

-0.7

We

-0.877

-0.5994

-0.26


Wi = qref*Ce(Zi)*Cpi Where:

Cpi = 0.8 or -0.5 for closed buildings with partition walls

and openings) Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*0.8 = 0.3KN/m2 Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*(-0.5) = -0.18KN/m2 Summary of wind loads on flat roof Wind pressure Maximum

–ve Maximum +ve pressure

Resultant pressure (we-wi)

pressure external

internal

external

internal

external

Internal

-0.877

-0.18

0

+0.3

0.877 + 0.3

0+18

=-1.18

= +0.18

From this the resultant wind pressures on the roof are Upward

= -1.18KN/m2

Downward = +0.18 KN/m2 1.2.3 Hipped roof @ 17.34m Our roof is considered as 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


1m2 < A < 10m2

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

A  10m2

Cpe = Cpe,10

Our roof is modeled and designed as hipped roof(according to appendix A.2.6)of EBCS -1 ,1995.

Case-1

Wind direction

 = 0

Reference height: ze=h=17.34m DETERMINATION OF EXTERNAL PRESSURE COEFFICIENT (Cpe (z)) External pressure coefficient for hipped roofs. For wind direction θ = 00

For wind direction θ = 900

90

Wind o

.

15

θ = 00

H=17.34

.

11.4

15.06

θ = 900 40.45

12.39 o=

2.28

tan-1(2.28*2/12.39)=200

Reference height: ze=h=17.34m

90=

tan-1(2.28x2/11.4)=220

Reference height: ze=h=17.34m

e=b or 2h whichever is smaller For wind direction

θ = 00

b=cross wind dimension

Crosswind dimension, b (40.45) our e (40.45, 2x17.43=34.68) whichever is smaller.

h=17.34


𝛼=

6.2 = 1.08, 𝑡𝑎𝑛 − 1(1.08) = 47.4ᵒ 5.7

FOR WIND DIRECTION θ=00 Β=90-47.4=42.6ᵒ X=3.47tan42.6ᵒ=3.2m F= (8.67*3.47)-(0.5x3.47x3.2) =24.5 m2 ……………..A>10 m2 G=23.11*3.47=80.2 m2………………………………A>10 m2 H=0.5(29.06+28.91)*2.725=79 m2 I=0.5(17.52+28.91)*2.725=63.26 m2 J=3.47*2.725=9.45*2=18.9 m2 K=0.5(29.06+40.06)*3.47=119.92 m2


L=8.4*3.47=29.15 m2x2=58.3 m2 M=0.5(8.92*5.7) =25.42*2=50.84m2 Roof ɑ0 = 20o Since α0=20 applying linear interpolation between +15 +30 the following result is obtained. Table 1-1 Cpe values for  = 0 Zone

F

G

H

I

J

K

L

M

Cpe,10

-0.43

-0.7

-0.26

-0.47

-0.9

-0.97

-1.4

-0.67

+0.3

+0.36

+0.26

--

--

--

-1.83

+0.5

-0.26

-0.47

-1.4

-1.5

+0.3

+0.36

+0.26

--

--

-0.5

Cpe,1

--2.0 --

-+0.4 --

We = qref*Ce (Ze)*Cpe = 0.22748*1.63* Cpe = 0.37Cpe

Zone

F

G

H

I

J

K

L

M

Area(m2)

24.5

80.2

79

63.26

18.9

119.9

58.3

50.84

2


Cpe

We=0.37C

-0.43

-0.7

-0.26

-0.47

-0.9

-0.97

+0.3

+0.36

+0.26

--

--

--

-0.16

-2.6

-0.1

-0.17

-

-

0.33

0.359

pe

-1.4

-0.67 --

--

-0.52

-0.248

--

--

3 +0.11

+0.13

+0.1

--

--

--

Wi = qref*Ce(Zi)*Cpi Where:


and openings) Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*0.8 = 0.3KN/m2 Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*(-0.5) = -0.18KN/m2 Summary of wind loads on hipped roof Wind pressure Maximum



pressure external

internal

external

internal

external

Internal

-0.52

-0.18

+0.13

+0.3

-0.52- 0.3

+0.13+0.18

=-0.82

= +0.31



= -0.82KN/m2

Downward = +0.31 KN/m2


Case II: For wind direction

θ = 900

E=b or 2h whichever is smaller. E(12.39, 34.68) Therefore, our e=12.39

e=b or 2h whichever is smaller: e=b=12.39m,d=40.45m, e/4=3.1m,e/10=1.24m,e/2=6.2m 𝛼=

6.2 = 1.08, 𝑡𝑎𝑛 − 1(1.08) = 47.4ᵒ 5.7

X=1.24tan47.4ᵒ=1.35m F = (3.1*1.24)-(0.5x1.24x1.35) =3 m2 G = 6.2*1.24=7.69 m2 H = (0.5x5.7x12.39)-(2x3+7.69)=21.62 m2 I = 0.5*9.91*4.46=22 m2


J = (0.5*12.39*6.2)-22=16.41 m2 L = 1.24*6.2=7.7 m2 M=0.5*4.96*4.96=12.3 m2*2=24.6 m2 N=(28.06x6.2)+(0.5x6.2x6.2)=193.2 m2 Table 1-2 Roof ɑ0 = 22o Zone

F

G

H

I

J

Cpe,10

-0.246

-0.66

-0.25

-0.45

-0.86 -1.4

+0.34

+0.43

--

--

-1.76

+0.1

-0.25

-0.45

-1.36

+0.34

+0.43

+0.29

--

--

Cpe,1

+0.29

L

--

M

N

-0.7

-0.25

--

-2.0

+0.08

--

--0.25

--

--

We = qref*Ce (Ze)*Cpe = 0.22748*1.63* Cpe = 0.37Cpe Cpe= Cpe,1+ (Cpe,10 - Cpe,1) log A …………. Since 1
Zone

F

G

H

I

J

Area(m2)

3

7.69

21.62

22

Cpe

-2.7

-0.57

-0.25

-0.45

L

M

N

16.41 7.7

24.6

193.2

-

-0.7

-0.25

-1.4

0.86 +0.34 Wi = qref*Ce(Zi)*Cpi

+0.43

+0.29

--

--

--

--

--


We=0.37C

-0.1

-0.21

-0.1

-0.167

pe

-

-0.52

-0.26

--

--

-0.1

0.32 +0.126

+0.16

+0.1073

Where:

--

--


and openings) Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*0.8 = 0.3KN/m2 Wi = qref*Ce(Zi)*Cpi = 0.22748*1.63*(-0.5) = -0.18KN/m2 Summary of wind loads on hipped roof Wind pressure Maximum



pressure external

internal

external

internal

external

Internal

-0.52

-0.18

+0.16

+0.3

-0.52- 0.3

+0.16+0.18

=-0.82

= +0.34


= -0.82KN/m2

Downward = +0.34 KN/m2


2. Roof covering [EGA sheet design] The roof system of this project is a truss roof system. The type of sheet used is EGA sheet and the trusses and purlins are steel. The roof is classified as hipped roof as per EBCS-2 1995 A.2.6 (page 78).The effect of wind load in both directions (θ=0˚ & θ=90˚) is taken into account and the severest case was considered in the design process. Design in this section includes: 

Design of EGA sheet



Design of purlin



Design of trusses

2.1 EGA SHEET CAPACITY Height of truss=1.0m Spacing of truss=1.2m


Spacing of purlin=1.25m in both wings Inclination of truss= 20 degree on both wings Type of EGA selected= EGA-500, t=0.4mm=0.0004m Load per unit area (GK) = Unit weight of EGA sheet *t =77*0.0004 = 0.0308 kN/m2 Maximum wind load (critical wind load) is WL=-0.82 KN/m2………………………from computation Live load from ……………………………..EBCS-1, 1995 qk=0.25 KN/m2



Distributed live load



Concentrated live load Qk=1KN



Purlin spacing=1.25m

Take EGA-500 from Kaliti Metal on its official website Thickness t=0.4mm, section modulus Sx=3756

Loading

Dead load Weight of EGA=3.14kg/m Coating=0.7kg/m


Total load=3.84kg/m Effective width of EGA=712mm=0.712 3.84𝑥9.81

Load on EGA per meter square=1000𝑥0.712 = 𝟎. 𝟎𝟓𝟑𝑲𝑵/𝒎𝟐 Live load As per EBCS-1, 1995, sec 2.6.3.4.2 Distributed load (qk)=0.25KN/m2 Concentrated load (Qk) =1KN

Main truss Equivalent distributed load for the concentrated load along effective width, P=1.0/0.712=1.404KN/m Mmax=pL/4=1.404*1.25/4=0.439KN/m


For the distributed load Mmax=wL2/8=

𝑤𝑙2 8

=

0.25𝑥1.252 8

=0.049KNm/m

By comparing the two results, the concentrated load governs & therefore it was adopted for design. Components of live load & dead load Perpendicular to the EGA sheet for α=20º DL (Gk) =Gk cos α= 0.053*cos 20=0.05KN/m2 LL (Qk) =Qk cosα=1.404cos20=1.32KN/m Parallel to the EGA sheet for α=20º DL (Gk) =Gk sin α= 0.053*sin20=0.018KN/m2 LL (Qk) =Qk sin α=1.404sin20=0.48KN/m Perpendicular to the EGA sheet for α=22º DL (Gk) =Gk cos α= 0.053*cos 22=0.049KN/m2 LL (Qk) =Qk cosα=1.404cos22=1.3KN/m Parallel to the EGA sheet for α=22º DL (Gk) =Gk sin α= 0.053*sin22=0.02KN/m2 LL (Qk) =Qk sin α=1.404sin22=0.52KN/m

Load combination


As per EBCS-1 1995 section 1.9.4.3 table 1.2 safety factors for favorable and unfavorable conditions are: ‫ץ‬G=0.90 for favorable condition for permanent load. ‫ץ‬G=1.30 for unfavorable “

“

“

“.

‫ץ‬q=1.60 for live and wind loads.

Case-1 unfavorable permanent and live load Pd=1.3Gk+1.6Qk=1.30(0.05)+1.60(1.32)=0.065KN/m2+2.112KN/m, multiplying both by effective width of EGA sheet =0.712m, Pd=0.065*0.712KN/m2+2.112*0.712KN/m Pd=0.0463KN/m+1.5KN Mmax=WL2/8+PL/4=0.0463 (1.25^2)/8 + 1.5(1.25)/4=0.48KNm

Case-2 favorable load condition Pd=0.90Gk+1.60Qk=0.90(0.05)+1.60(-0.82), Qk=wind load=-0.82 KN/m2. Multiplying both by effective width of EGA sheet=0.712m, Pd=0.90*0.05*0.712 + 1.60*-0.82*0.712 Pd=-0.9KN/m Mmax=WL^2/8=0.9*1.25^2/8=0.17KNm Note that W=Pd.


Capacity of EGA-500 Section property of the EGA-500 is taken from Kaliti table: 

Section modulus, Sx=3756mm3



Moment of inertia, Ixx=79972mm4/m



Allowable stress, бall=160Mpa



Modulus of Elasticity=210,000Mpa

Checking for capacity of EGA Actual stress= Mmax/Sx= (0.48KNm/3756*10^-9m3)/1000=127.8Mpa Actual stress=127.8Mpa < allowable stress=160Mpa, OK! Checking for deflection of EGA Allowable deflection, Δall=span/200= 1250/200=6.25mm Actual deflection, Δact. =5WL^4/384EI + PL^3/48EI Δact.=5*.0463*1250^4)/(384*210000*79972)+1.5*1250^3/ =0.0876 + 3.63=3.71mm. Δact. =3.71mm < Δall=6.25mm, OK! 3.DESIGN OF PURLIN Load on purlin Weight of EGA=0.053KN/m2. For 1.25m spacing of purlin,

(48*210000*79972)


The Weight of EGA on purlin per meter= 0.053*1.25= 0.066KN/m, Live load from EGA onto purlin=1.404KN/m

Selection of dimension for purlin As per Kaliti metal table, section index RT-64 with the following properties for rectangular hollow section was selected for trial design. Nominal size=60mmх40mm Wall thickness=3mm Weight of rectangular hollow section per meter=4.24kg/m Weight

of

4.24𝑘𝑔 𝑥10𝑚/𝑠2 𝑚

meter=

1000

purlin

per

4.24Kg/m*10m/s^2/1000=0.0424KN/m

Nominal steel grade, fe-430 [EBCS-3] Tensile strength of the section (fu) =430Mpa, Yield point (fy) =275Mpa Moment of inertia, Ix=253800mm4.

Section modulus, Sx=8460mm3

Moment of inertia, Iy=134,400mm4,

Section modulus, Sy=6720mm3

linear


Size=60mmх40mm Thickness, t=3mm H=60mm, B=40mm, h=54mm, b=34mm Cross section of purlin Total load on purlin DL=imposed load + self weight of purlin=0.066 + 0.0424=0.108KN/m Live load from EGA, LL =1.404KN/m Pd=1.3DL+1.6LL=1.3*0.108 + 1.6*1.404= 2.387KN/m


DESIGN OF PURLIN ON MAIN TRUSS Decomposing the design load (Pd) into perpendicular and parallel direction to the purlin for inclination angle α=20º Pdx= 2.387sin 20 =0.82KN/m [parallel] Pdy= 2.387cos20 =2.24KN/m [perpendicular]

Maximum moments Mmax,x=

𝑤𝑥𝑙2

=

8

𝑤𝑦𝑙2

Mmax,y=

8

=

0.82𝑥1.22 8 2.24𝑥1.22 8

=0.148KNm,

=0.41KN/m,

Wx=Pdx Wy=Pdy

Maximum shear Vx=WxL/2=0.82*1.2/2=0.5KN, Vy=WyL/2=2.24*1.2/2=1.35KN,

Wx=Pdx Wy=Pdy

CHECKING PURLIN CAPACITY Moment resistance As per EBCS-3 1995 section 4.6.1.1 plastic moment resistance of a section is: Mpl,RD=

1.5𝑥 𝑍𝑝𝑙𝑥𝑓𝑦

Mpl,RD,y=

where, Zpl=1.5Zel, fy=minimum yield point,‫=ץ‬safety factor=1.1.

𝛾 1.5𝑥 6720𝑥275

Mpl,RD,x=

106 𝑥1.1

=2.52KNm>Mmax,y=0.41KNM, OK!

1.5𝑥 8460𝑥275 106 𝑥1.1

=3.17KNm>Mmax,x=0.148KNm, OK!


Shear resistance The plastic shear resistance of a section as per EBCS-3 1995, section 4.6.1.2 is, Vpl, RD=Av*(fy/√3)/‫ץ‬, where Av=shear area=A*h/ (b+h), in which no reduction of resistance moment will be provided if design shear force Vsd < 50% of the design shear resistance. RT-64 with a wall thickness of 3.0mm has area=5.41cm2; 60x40mm dimensions and a wall thickness of 3mm Av, y= A*h (b+h) =541*54/ (54+34) =331.98mm2 Av, x=A*b/ (b+h) =541*34/ (34+54) =209.02mm2 Vpl, RD, y=331.98(275/√3)/1.1=47.92KN > Vy=1.35KN, OK! Vpl, RD, x=209.02(275√3)/ 1.1=30.17KN>Vx=0.5KN, OK! Deflection requirement Allowable deflection as per EBCS-3 1995 is, Δall=span of purlin/200 Δall=1200/200=6mm, Actual deflection, Δact. x=5WxL^4/384EI ,

W=Pd.

5𝑥 0.82𝑥12004

Δact. x =384𝑥210000𝑥134400=0.78mm<6mm, OK! Actual deflection, Δact. y=5WyL^4/384EI , 5𝑥 2.24𝑥12004

Δact. Y=384𝑥210000𝑥253800=1.13mm <6mm, OK!

W=Pd.


4. TRUSS DESIGN MAIN TRUSS Selection of trial section A square tube of section index ST-50 from structural steel table of Kaliti metal has been selected for trial design. Section property

Nominal size 50mmх50mm, Area=541mm2 Ix=Iy=194700mm4, Sx=7790mm3 Wall thickness=3mm


Weight of square hollow section per meter=4.25kg/m Weight per linear meter=4.25*10/1000=0.0425KN/m

Main truss

Total

length=(0.65+1.85+2.54)

+0.3*2+1.86*2+0.86*2+1.48*2+1.98*2+2.32*2+1.42*2+2.81*2+6.01*2+1.2*7=51.52m

Weight of truss=51.52*0.0425=2.19KN Number of load carrying joints=18 Load per joint=2.19/18=0.12KN Load transferred from purlin 1. DL=0.108KN/m*1.2=0.13KN+2.19=2.32KN---total truss DL inclusive=2.32KN 

On inner joint=0.13KN



On outer joint=0.065KN

2. LL=1.404*1.2=1.7KN




On inner joint=1.7KN



On outer joint=0.85KN

3. Wind load WL=-0.82*1.2*1.25=-1.23KN 

On inner joint=-1.23KN



On outer joint=-0.615KN

Ceiling load Ceiling material are, chipboard density=8KN/m3, Zigba wood density=6KN/m3.Load of chip wood per 1.2m*1.2m area=1.2*1.2*.008*8=0.0921KN Load of zigba grid per 1.2m*1.2m=6*1.2*0.04*0.05*6=0.0864KN Total ceiling load=0.0921+0.0864=0.1785/ (1.2m*1.2m)=0.124KN/m2 Load

per

lower

joints=0.124*1.2*0.65=0.097KN,

0.124*1.2*1.2=0.178KN

0.124*1.2*1.85=0.27KN,


Note: with the same procedures shown above for the main truss, the inner and outer joint loads and the member forces for trusses 1 was computed and prepared in table format as shown below. Load on Upper inner=load per node (0.12) + inner load (DL=0.13, LL=1.7, WL=-1.23) Load on Upper outer=load per node + outer load (DL=0.065, LL=0.85, WL=-0.615) + ceiling load Load on lower inner=load per node + inner load due to ceiling 1-4 Load on inner and outer joints on main truss

load case

DEAD

LIVE

WIND

Joint

joint type

load(KN)

B,C,D,E,F,G,H, I

upper inner

0.25

A,J

upper outer

0.455

K,T

lower inner

0.39

K,L,M,N,O,P,Q,R,S lower inner

0.39

B,C,D,E,F,G,H,I

upper inner

1.82

A,J

upper outer

1.24

B,C,D,E,F,G,H, I

upper inner

-1.11

A,J

upper outer

-0.225


 Lastly, the Loads are transferred to the Truss at Purlin Position DL=2.32kN ………….Vertical LL=1.7 kN …………………Vertical WL=1.23 kN …………...Perpendicular to the rafter

1-5 Design Loads on Truss at Purlin Position

Design Loads Load Type

Horizontal Vertical

DL(KN)

0

2.32

LL(KN)

0

1.7

WL(KN)

0.42

1.15

 Load Combinations Combination: Pd=1.3*DL+1.6*LL+1.6*WL Horizontal=1.6(0.42)=0.672KN, Pd=0.672kN Vertical=1.3(2.32)+1.6(1.7)+1.6(1.15)=

Pd=7.58 kN

Number of load carrying joints=10 (upper chord joints) Load per joint (vertical)=7.58/10=0.758KN


Fo the above load, we have taken a square tube of section index ST-50 section, consequently, the truss analysis was carried using sap software, and the following result was obtained. Truss model

Reactions


Axial Force diagram

Steel Design Section (Euro code 3-1993)

Section is economical, since there’s no red sign in the truss elements


The result shows Vsd/Vrd ratio, so if the ratio is less than 1, it will be an indication of safe section. So the section is obviously safe!


References 

Asrat Worku, “CALLING

FOR

A

REVISION

OF

THE

CURRENT

ETHIOPIAN SEISMIC CODE –EBCS 8:1995”, Addis Ababa, 2011. 

Computers and Structures, “SAP 2000 (version 16)”, Berkley, California.



Debre Berhan University, “Lecture Notes”, [Unpublished].



European Committee for Standardization, Eurocode 8, “Design of Structures for Earthquake Resistance (EN 1998-1: 2004)”, Brussels, 2004.



Ministry of Works and Urban Development, “Basis of Design of actions on structures”, Ethiopian Building Code Standard (EBCS 8), Addis Ababa, 1995.

Chapter 1: Design of Roof and Truss (B+G+4 Apartment Building in Ethiopia)

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