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EGE UNIVERSITY
ENGINEERING FACULTY
CIVIL ENGINEERING DEPARTMENT
2013 -2014
STEEL STRUCTURES – 2
PROJECT
1st CHECKING
Student No: Name:
05100000837 Doğan SAMANCIOĞLU
1. PROJECT DATA
Relevant information about steel structure was indicated in Figure1.
1.1.Roof Slope
tan∝= 3/8= 0,375 ; ∝ =20,55
1.2.Length Of Frame Elements
Lower Flange( Ui) ;
U1= U’1=U2 =U’2= U3=U’3=U4=U’4= 16/8 = 2 m
Verticals ( Vi=tan∝*Ui);
V1=V’1= 2*0,375 =0,75 m
V2=V’2=(2+2)*0,375= 1,5 m
V3=V’3=(2+2+2)*0,375= 2,25 m
V4=V’4=(2+2+2+2)*0,375=3 m
Diaganals (Di);D1=D’1=√V 1²+U 2² = √0 ,75²+2²= 2,14 m
D2=D’2=√V 2²+U 3² = √1 ,5²+2² =2,5 m
D3=D’3= √V 3²+U 4²= √2 ,25²+2² =3,01 m
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*Material: St37*Type of purlin: Simple Beam*Type of roof clading: Sandwich Panel*City/Elevation: İzmir / 25 m*Height of the building = 7 m
Upper Flange(Oi);
O1=O’1=√U 1²+V 1²= √2²+0 ,75²=2,14 m
O2=O’2= √(U 1+U 2)²+V 2²- O1= √4²+1 ,5²- 2,14= 2,14 m
O3=O’3= √(U 1+U 2+U 3)²+V 3²- O1-O2= 2,14 m
O4=O’4=√(U 1+U 2+U 3+U 4) ²+V 4² - O1-O2-O3=2,14 m
Calculated values are shown in Figure 2.
Figure 2. Length of the members
2. LOADS
2.1. Snow Load
Snow load is affected like in Figure 3 on the roof.
Figure 3.Loading condition for snow
Pk=m.Pko Pk= snow load calculation value Pko= snow loadm=reduction factor dependent on roof slope
According to TS 498(1980);City: İzmir……I.Bölge pko=75 kg/cm2Elevation: 25 m<200 …..m=1
Pk=1*75= 75 kg/m2Pk vertical= 75*cos20,55= 70,23 kg/m2
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2.2. Wind Load
TS 498 suggests following equations for wind load calculations(Figure 4):
Figure 4. Calculation of wind load
Height of the building 7m<8 m …… q= 50 kg/m2
Wind left; (1,2sin∝ -0,4)q = (1,2sin20,55 – 0,4)*50= 1,06 kg/m2
-0,4q= -0,4*50= -20 kg/m2
There is sucking pressure both side of the roof .
2.3. Roof Clading
In this project Aluminium sandwich panel produced by Atermit Firm was used. Relevant information
is in the Table 1.
Table 1.Bearing capasity
Opening of purlin : 2,14 m
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a=roof slope
q= sucking pressure dependent on height of the building(kg/m2)
There is snow and wind load on the roof and roof clading should be bear 75 kg/m2 .
Choosed clading can bear max 143 kg/m2.
Effect type of cladding load is shown in Figure 5.
Figure 5. Effect of cladding
2.4. Purlin Self Weight
In beginning of the project weight of the purlin was estimated 12 kg/m2 on the horizontal plane.
2.5. Stability Bars
Stability bars will connect to the purlins and weight of the stability bars is used 3 kg/m2 .
2.6. Truss Self Weight
In the beginning of the project weight of the truss was estimated 16 kg/m2 on the horizontal plane.
2.7. Loads Table
First Loads were shown in the table2.
Table2. First Loads
Dead Load
Roof Cladding 5,20 kg/m2Purlin self weight 12 kg/m2Stability Bars 3 kg/m2
total 20,2 kg/m2Truss self weight 16 kg/m2
total 36,2 kg/m2
Snow Load For horizontal plane 75 kg/cm2
Wind LoadFor roof plane Left RightWind-left 1,06 -20Wind-right -20 -1,06
3. PURLIN CALCULATION
Purlin was supported as basic beam. Interval of the purlin was used as 3m . and Interval of the truss beam was used 6 meters(figure6).
3.1. Determination Of Loads And Section Stress For Purlin:
5
Wind load is sucking both side of the roof. Therefore in purlin calculation Wind load was accepted as
0. Figure 6. Intervals
Snow load ;
Pk =75 kg/m2
On the one purlin surface ; 75*3=225 kg/m
Roof cladding; 5.20 kg/m2
On the one purlin surface ; 5,20*2,14=11,128 kg/m
Purlin self weight; 12 kg/m2
On the one purlin surface ; 12*3=36 kg/m
Stability bars;3 kg/m2
On the one purlin surface ;3*3=9 kg/m
Load on the purlin as horizontal:
225+11,128+36+9=281,128 kg/m2
qx=281,128*cos20,55= 263,24 kg/m
qy=281,128*sin20,55= 98,68 kg/m
Direction of the loads is in Figure 7. Figure 7. Directions
Purlin was designed as basic beam:
Mx= qx∗l2
8 = 263,24*6²/8= 1184,58 kgm =118458 kgcm
My=qy∗l2
8 = 98,68*6²/8 = 3444,06 kgm= 44406 kgcm
3.2. Choosing Of Purlin Profil
σ =MW ; σem=1440 kg/cm2
Wx (needed)=Mxσ = 118458/1440= 82,26 cm3
Wy (needed)=Myσ = 44406/1440= 30,84 cm3
Profil I140;
Wx (needed)= 82,26 cm3 > Wx=81,9 cm3
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Profil I140Wx=81,9 cm3Wy=10,7 cm3
Profil I140 is not suitable under these conditions.
Profil I160;
σ Control (N=0)
σ =MxWx +
MyWy ≤ σem
*Tieless :
σ =118458
117 +4440614,8 = 4012,86 ¿ σem=1440 kg/cm2
*Single tie : l/2= 6/2= 3m for My direction
σ =118458
117 +( 98,68∗32 )∗100
814,8
= 1762,56 kg/cm2 ¿ σem=1120 kg/cm2
*Double tie: l/3=6/3=2 m My direction
σ =118458
117 +( 98,68∗22 )∗100
814,8
=1345,84 kg/cm2 ¿σem=1120 kg/cm2
Deflection control:
fx= 5
384 * qx l4
E∗Ix ; fy=
5384 * qy l
4
E∗Iy ; f=√ fx 2+ fy2 < l/300
fx=5
384 *263,24∗10−2∗6004
2100000∗935 =2,26 cm
fy=5
384 *96,68∗10−2∗3004
2100000∗54,7 =0,90 cm
f=√2,262+0,902 = 2,43 > 600/300=2 cm
Profil is not suitable under these conditions even if it has double ties..
7
Profil I160Ix=935 cm4Iy=54,7 cm4Wx=117 cm3Wy=14,8 cm3
Profil I180;
σ =MxWx +
MyWy ≤ σem
Tieless:
σ =118458
161 +4440619,8 = 2978,5 ¿ σem=1440 kg/cm2
Single tie: l/2= 6/2= 3m for My direction
σ =118458
161 +( 98,68∗32 )∗100
819,8
=1296,45 ¿σem=1440 kg/cm2
Deflection control:
fx= 5
384 * qx l4
E∗Ix ; fy=
5384 * qy l
4
E∗Iy ; f=√ fx 2+ fy2 < l/300
fx=5
384 *263,24∗10−2∗6004
2100000∗1450 =1,458 cm
fy=5
384 *98,68∗10−2∗3004
2100000∗81,3 =0,61 cm
f=√1,4582+0,612 = 1,58 cm < 2 cm
Checking shear forces in purlin:
τx=¿ TxA≤ τem=1440
√3=831,38kg/cm2
τy=¿ TyA≤ τem=1440
√3=831,38kg/cm2
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Profil I180Ix=1450 cm4Iy=81,3 cm4Wx=161 cm3Wy=19,8 cm3
Tx= qx*l/2 = 263,24*6/2=789,72kg/cm2 < 831,38kg/cm2
Ty=qy*l/2=98,68*3/2= 148,02 kg/cm2 <831,38kg/cm2
Choosed profil : I180- Single Tie birim boy ağırlığı: 21,9 kg/m
Control Of Choosed Profil with own weight: weight of the unit length: 21,9 kg/m
Load on the purlin as horizontal: snow, roof, purlin, stability, respectively:
225+11,128+ 21,9 +9=267,02 kg/m2
qx=263,24*cos20,55= 246,48 kg/m
qy=263,24*sin20,55= 92,40 kg/m
Purlin was designed as basic beam:
Mx= qx∗l2
8 = 246,48*6²/8= 1109,16 kgm =110916 kgcm
My=qy∗l2
8 = 92,40*6²/8 = 415,8kgm= 41580 kgcm
Profil Control For Purlin:
σ =MW ; σem=1440 kg/cm2
Wx (needed)=Mxσ = 110916/1440= 77,03 cm3 < Wx=161 cm3
Wy (needed)=Myσ = 41580/1440= 28,88 cm3 > Wy=19,8 cm3
Single Tie : l/2= 6/2= 3m for My direction
σ Control (N=0)
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Profil I180Ix=1450 cm4Iy=81,3 cm4Wx=161 cm3Wy=19,8 cm3
σ =110916
161 +( 92,40∗32 )∗100
819,8
=1213,92 ¿σem=1440 kg/cm2
Deplasman:
fx= 5
384 * qx l4
E∗Ix ; fy=
5384 * qy l
4
E∗Iy ; f=√ fx 2+ fy2 < l/300
fx=5
384 *246,68∗10−2∗6004
2100000∗1450 =1,36 cm
fy=5
384 *92,40∗10−2∗3004
2100000∗81,3 =0,57 cm
f=√1,362+0,572 = 1,47 cm < 2 cm
3.3. Last Loads on Purlin:
After purlin calculations loads with real weight of purlin were shown in Table 3.
Table 3. Last Loads after purlin calculationsSnow 225 kg/m : 3m 75 kg/m2Roof 11,128 kg/m :2,14m 5,20 kg/m2Purlin (I180) 21,9 kg/m : 3m 7,3 kg/m2Stability 9 kg/m:3m 3 kg/m2
Total: 90,50 kg/m2
3.4. Control of Ridge Purlin
Loads are used as half of the general purlin loads.(Table 4).
Table 4. Half LoadsSnow 112,5 kg/mRoof 5,564 kg/mPurlin (I180) 10,95 kg/mStability 4,5 kg/m
Total: 133,514 kg/m2
qx=133,514*cos20,55 = 125,02 kg/m
qy=133,514*sin20,55 = 46,87 kg/m
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Ridge purlin was designed as basic beam, too:
Mx= 125,02∗62
8 = 562,59 kgm =56259 kgcm
My=46,87∗62
8 = 210,92 kgm= 21092 kgcm
Choosed profil I180 as tieless :
σ =56259
161 +2109219,8 = 1414,7kg/cm2 < σem=1440 kg/cm2
Deflection Control:
fx=5
384 *125,02∗10−2∗6004
2100000∗1450 =0,69 cm
fy=5
384 *46,87∗10−2∗6004
2100000∗81,3 =4,63 cm
f=√0,692+4,632 = 4,68 cm > 2 cm
A support profil is needed. General for ridge purlin L40.40.4 profil is used.
Single tie:
σ =56259
161 +( 46,87∗32 )∗100
819,8
=615,74kg/cm2 < σem=1440 kg/cm2
Single tie is suitable and L40.40.4 profil can be choosed.4. Tie Bar Dimensions
Purlin with single tie bar;
Z is force in skewed tie bar and qy is the total qy component of all purlin (except ridge purlin between two truss beam;
Z= L
4 cosβ Σqy
In this design there are 8 purlin with 6 meters length between two truss.
Tanβ = 3/2,14 = 1,40 ; β=54,46°
Z= 6
4 cos 54,46 *(8*92,40)=1907,54 kg
Atie= Zσem =
1907,541440 = 1,32 cm2
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ϕ 12 = 1,13 cm2 1907,54/1,13 =1688 kg/cm2 > σem=1440 kg/cm2
ϕ 14 = 1,54 cm2 1907,54/1,54 =1238,7 kg/cm2 ¿ 1440 kg/cm2
***Choosed tie bar: ϕ 14 ***
5. CALCULATION FORCE OF LATTICE BAR
One sample of point loads calculation was mentioned below( for snow). Other loading was calculated by the help of SAP2000(Tablo6) . Used loads in Sap were shown in Table 5.
Table.5. Truss LoadsSnow 75 kg/m2 *3*6 1350 kgRoof 5,20 kg/m2*2,14*6 66,7kgPurlin (I180) 21,9 kg/m*6 132kgStability 3 kg/m2*3*6 54kgWind left 1,06/ -20 kg/m2 13,58/ -256,30Wind right -20/ 1,06 -256,30/ 13,58Truss 16 kg/m2*3*6 288kg
5.1. Snow Loading:
ΣMA=¿ 1350(2+4+6+8+10+12+14)+675*16=By*16 ; By= 86400/16=5400 kgAy=By=5400 kg
Point
O1y=5400-675=4725
0,752
=4725O 1x
;
12
O1x= 12600
O1= √4725²+12600² =13456 (-)
U1= O1x=12600 (+)
Point U2=12600(+)V1=0
Point – - RitterΣM3= 5400*4-675*4-1350*2-O2y*2-O2x*0,75=01,5 O2x = 24300O2x= 16200O2y=4050O2= 13504(-)
3*O2y= 0,75 *O2x (From angle)
Point – - – Ritter
Σ M4=5400*6-675*6-1350*4-1350*2- O3x*1,5-O3y*3=0
30375=2,25* O3xO3x = 13500O3y=3375O3=11490 (-)
3*O2y= 0,75 *O2x (From angle)
Point – - – – Ritter5400*18-675*18-1350*6-1350*4-1350*2- O4x*2,25-O4y*3=0
32400=3* O4xO4x=10800O4y=2700
13
O4=9329 (-)
Point
D1= √2700²+675² = 2783,09 (-)
Point
V2=675 (+)U3=12600-2700=9900(+)
Point
D2=√2700²+1350² =3018,69 (-)
Point V3=1350(+)U4=9900-2700= 7200(+)
Point
D3= √2700²+2025² =3375 (-)
14
Point V4= 2700+2700=4050(+)
The other points are symetric with these points.
Wind Left Loading
Wind Right Loading
Dead Loading
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