Single storey building design English version

HCM City, date 18 month 03 year 2016
HO CHI MINH UNIVERSITY OF ARCHITECTURE
Field: Structural Engineering
STRUCTURAL STEEL PART II
Project: Design Single Storey Steel Building.
NAME: NGUYỄN TRÍ THIỆN
Student ID number:
Grade: XD12A2
TUTOR: DR. TRẦN VĂN PHÚC

STRUCTURAL STEEL PART TUTOR: DR.TRẦN VĂN PHÚC
NAME: NGUYỄN TRÍ THIỆN Page 1
SYMBOLS USED IN THIS PROJECT
A: is gross area of cross section.
An: is net cross section area.
Af: is cross section area of flange (chord).
Aw: is cross section area of web.
Abn: is cross section area of bolt.
b: is width.
bf: is flange (chord) width.
bef: is design width.
bs: is the width of stiffener.
h: is height.
hw: is height of web.
hf: is the height of fillet weld.
hfk: is distance between flange’s central axis.
i: is inertia radius of cross section.
ix, iy: is inertia radius of cross section relative to axes x-x, y-y respectively.
If: moment of inertia of flange (chord).
It: torsional constant.
Ix, Iy: moment of inertia about the principal axes.
Inx, Iny: moment of inertia about the principal axes referred to the net area.
L: is length of member.
l0: is design length of compression members.
lx, ly: are design lengths of component in planes perpendicular to axes x-x and y-
y respectively.
lw: is the length of weld seam.
S: is static moment of gross slid portion cross section relative to neutral axis.
t: is thickness.
tf, tw: is thickness of flange (chord), web respectively.
Wnmin: is the minimum moment resistence of net section.
Wx, Wy: are the moment resistance of gross section relative to axes x-x, y-y
respectively.
F, P: is force.
M: is moment or bending moment
Mx, My: are moment or bending moment relative to axis x-x, y-y respectively.
N: is longitudinal force.
V: is lateral force or shear force.
E: is modulus of elasticity.
f: is design yield strength.
fv: is ultimate shear strength.

fc: is design crushing strength.
fub: is ultimate tension strength.
σ: is stress.
σc: is local stress.
σx, σy: are normal stresses which are parallel with axes x-x, y-y respectively.
σct, σc,ct: are normal critical stress and concentrated critical stress respectively.
τ: is shear (tangential) stress.
τct: is ultimate shear stress.
e: is eccentricity of force.
m: is relative eccentricity.
me: is reduced relative eccentricity.
nv: is the number of design section.
βf, βs: are factor for analysis of corner seam through seam metal and through
metal of melting boundary respectively.
γc: is working condition factor.
γb: is working condition factor of bolt.
γg, γp: is load factor (load coefficient).
nc: is load combination factor.
η: is factor of influence of cross section form.
λ: is flexibility (slenderness ratio).
 : is fictitious flexibility.
w : is fictitious flexibility of web.
λx, λy: are design flexibility in planes perpendicular to axes x-x and y-y
respectively.
μ: is length coefficent.
φ: is buckling factor.
φb: is factor of reducing design resistence at bending-and-twisting form of beam
stability loss.
φc: is factor of reducing design resistence at eccentric compression.
Ψ: is intermidiate factor to calculate φb factor.

CHAPTER 1. GIVEN DATA
Name: Nguyễn Trí Thiện Student ID number:
Order number: 87 Grade: XD12A2
Design three span single storey building according to following data
given below:
Span
(m)
Bay
(m)
Level
Crane
Q(T)
Wind
pressure
at 10 m
qo
(daN/m2
)
Length
(m)
Slope
i%
L1 L2 L3
Ground
(m)
Ground
floor
(m)
Rail
(m)
30 33 30 6,5 -0,65 0 12 10 85 162,5 10
 Area type to calculate wind load is B.
 A single model overhead crane is in the center span which is its load
capacity is given above. There is no crane on the side span, and the
dimension of side span is equal L1=L3.
 Roof materia: tole. Using portal frame with I built-up beam, straight
column, and beam with variable section.
 Material using : Steel CCT34, with following properties:
 Design yield strength: f = 2100 daN/cm2
 Ultimate shear strength: fv = 1200 daN/cm2
 Design crushing strength: fc = 3200 daN/cm2
 Ultimate tension strength: fu = 3400 daN/cm2
 Using Shielding metal arc welding method and electrode wire N46 with
fwf = 2000 daN/cm2
 Conection between beam and column: Rigid.
 Conection between column and foundation: Fixed.

CHAPTER 2. FRAME GEOMETRY
CHOSING CRANE.
Accroding to given data: Span L2 = 33m, load capacity Q = 3T, using
catalouge and find out a suitable crane:
 Span: 33-0,75x2=31,5 m
 Gabarit high HK = 875 mm
DEFINE VERTICAL DIMENSION.
Upper column length:
r dcct KhH h H C   
 HK: Gabarit height (the distance between upper surface of the rail and yop
of the crane) – According to catalouge: HK = 875mm.
 C: the safety distance between crane and rafter
 
1 1
100 33000 100 265
200 200
C L mm     
 hdcc: the height of runway beam
 
1 1 1 1
.6500 650 812,5
8 10 8 10
dcth B mm
   
        
   
 Chose hdct = 800mm
 hr: the height of rail 200mm
 So  200 800 875 265 2140r dcc Kt h h H C mH m       
 Chose Ht = 2.2m
Lower column length:
 –d r dcc r mH H h h h  
 Hr – upper surface of the rail level, Hr = 12 m
 hm – ground floor level above ground level 0,65m, hm=0,65m (foundation
are refferd to be placed at the same ground level).
     120– 00 800 200 650 11650d r dcc r nH H h h h mm       
 Chose Hd = 12m
 The length of column:
 2200 12000 14200t dH H H mm    
DEFINE HORIZONTAL DIMENSION:
 The grid of reference axis shall be coincident with axis of column.

 The distance between reference axis and y-y axis of runway beam:
750mm  to match up crane load capacity Q < 75 ton. To prevent
the impact between crane and column,  shall be: 1 ( )tB h a D    
 B1 is the dimension of head crane B1 = 200(mm).
 D is the safety clearence between crane and column D=
60(mm).
 ht is the height of upper column:
1 1 1 1
2200 200 220( )
10 11 10 11
t th H mm
   
         
   
Chose ht =250mm (ht is multiple of 250mm)
 1750( ) ( ) 200 250 60 510tmm B h a D mm         
   
1 1
14200 568 mm 600
25 25
d dh H h mm     
 Distance between rail center point and outer edge of the middle
colum:
 
0.60
0.75 0.45
2 2
h
Z m    
JACK ROOF MONITOR DIMENSION.
 Span of the roof:
1 1
33 3,3 ( )
10 10
cmL L m   
 Height of the roof: Hcm =1.5(m).
HORIZONTAL FRAME MODEL OF CALCULATION.
 Using straight column and beam with variable section.

Horizontal frame model of calculation as follow :

CHAPTER 3. DEFINE LOAD.
DEAD LOAD.
Dead load apply to horizontal frame include in gravity load of purlins,
self weight of frame and runway beam. In this project, self weight of bracing
column and roof are neglected.
Self weight of structure.
It would be automatically calculated by software.
Envelope material.
 Roof structure.
 Roof system: using roof fill insulation, purlins and snag rods. Take
 2
30 d /tc
mg aN m , load factor n = 1,1.
 Normal load apply to roof:
 1 . 30 6,5 195 /tc tc
kh mg g B daN m   
 Design load apply to roof:
 1 1. 1,1 195 214,5 /tt tc
kh khg n g daN m   
 Building lateral side.
 Sidewall girt, sidewall canopy system and tole is assumed to be
 2
30 d /tc
bcg aN m , load factor n = 1,1
 Normal load apply to outside column:
 2 . 30 6,5 195 /tc tc
kh bcg g B daN m   
 Design load apply to outside column:
 2 2. 1,1 195 214,5 /tt tc
kh khg n g daN m   
 Runway beam.
 The height of runway beam Hdct = 0,8m. Preliminary self weight of
runway beam is assumed to be 200 daN/m, load factor n = 1,1. It is
reduced to concentrated load and eccentric moment is located at level of
junction of column and lower surface of rail:
 Normal load:
 200 6,5 1300 dtc tc
dct dctG g B aN   
 1300 0,75 975 d .tc tc
dct dct nM G L aN m   
 Design load:
 . 1,1 1300 1430 dtt tc
dct dctG n G aN   
 1430 0,75 1072,5 d .tt tt
dct dct nM G L aN m   

ROOF LIVE LOAD.
 According to TCXDVN 2737: 1995, standard load of roof live load
should be taken as 30 daN/m2, load factor n = 1,3 (in case live load is
less than 200 daN/m2). It is reduced to distributed load apply to rafter:
 Normal load:  30 6,5 195 /tc tc
htp p B daN m   
 Design load:  . 1,3 195 253,5 /tt tc
pp n p daN m   
CRANE LOAD.
 This following table gives properties of 10 ton crane which is taken from
crane catalouge :
Load
(T)
Span
(m)
Total
mass
Max
wheel
load Pmax
(T)
Min
wheel
load Pmin
(T)
Dimension
H3 B W C2 C1 H2 H1
10 31,5 7,76/8,28 6,60/6,85 1,55/1,56 1200 3500 3000 1230 1830 1640 875
Crane load apply to horizontal frame including vertical impact and
horizontal breaking focre:
Vertical impact.
 The maximum wheel load used for the design of runway beams,

including monorails, their connections and support brackets, shall be
increased by the percentage given below to allow for the vertical impact
or vibration:
_ Monorail cranes (powered) .........................................................25
Cab-operated or radio operated bridge cranes (powered)…...........25
_ Pendant-operated bridge cranes (powered)…..............................10
_ Bridge cranes or monorail cranes with hand-geared bridge, trolley and
hoist…...............................................................................................0
_ Vertical impact shall not be required for the design of frames, support
columns, or the building foundation.
 It should be calculated as follow:
     
 
max max 0.85 1.2 6850 1 0,908 6600 0,446 0,538
19955,48
tt
c iD n n P y
daN
            

     
 
min min 0.85 1.2 1550 1 0,908 1560 0,446 0,538
4582,29
tt
c iD n n P y
daN
            

Where:
 n – is the load factor of crane, n = 1,2.
 nc – is the load combination factor, nc = 0,85 is taken from SNiP
2.01.07-85*
_ If two cranes are taken into account, their loads ought to be multiplied by
the following combination coefficient:
nc= 0,85 – for crane operation mode groups 1K-6K
INFLUENCE LINE FOR REACTION TO DEFINE Dmax, Dmin

 Through rail and runway beam, Dmax and Dmin would be transmited to
bracket support, thus eccentric of column may be e = L1 = 0,75m.
Eccentric moment is given by:
 max max 19955,48 0.75 14966,61 .tt tt
M D e daN m   
 min min 4582,29 0.75 3436,72 .tt tt
M D e daN m   
Breaking force of trolley and lifted load:
 Lateral loads being bi-directional action are applied to the column
throught crane rail to account for such effects as acceleration and
breaking forces of trolley and lifted load, skewing of the travelling crane,
rail misalignment, and not picking up the load up vertically. It is defined
as :
max c kT n nT y 
Where:
 n – overload factor of crane, γp = 1,2
 nc – load combination factor, nc = 0,85 is considered for crane
operation mode groups 1K-6K. Loads and effects SNiP 2.01.07-
85*
 Tk – breaking force of single trolley wheel impact to rail
 
 
0
0,05 10000 8280 77600,05( )
263
2
xc
k
Q G
T daN
n
       
 no – the number of driven wheels
1 0,908 0,446 0,538 2,892iy     
 max 0,85 1,2 263 2,892 775,81tc
c p iT n T y daN       
WIND LOAD:
 According to TCVN 2737-1995 or Loads and effects SNiP 2.01.07-85*,
Upon calculation of internal pressure wi as well as upon calculation of
high buildings up to 40 m and one-storey buildings up to 36 m – in case
the ratio between the height and the span is less than 1,5 – that are located
in areas of A and B types (see item 6.5), the pulsating component of wind
load may be omitted. The standard average component of wind load W at
the height of z over the ground surface is to be calculated under the
following formula:
. . . .o
W W nc k B

Where:
 Wo = 85 daN/m2
: standard wind pressure (see item 6.4)
 B – span 6,5B m
 c – aerodynamic coefficient (see item 6.6)
Coefficient α degrees
h1/l
0 0.5 1 ≥2
ce1
0 0 -0.6 -0.7 -0.8
20 +0.2 -0.4 -0.7 -0.8
40 +0.4 +0.3 -0.2 -0.4
60 +0.8 +0.8 +0.8 +0.8
ce2 ≤60 -0.4 -0.4 -0.5 -0.8
Aerodynamic coefficient
ce1: 1 13,55
0,452
30
h
l
  và 5 42'o
   ce1 = -0,44
 n = 1,2 : confidence factor of service life (50 years).
 k – coefficient of wind pressure change in height (see item 6.5), which
is to be calculated as following formula
2
( ) 1,844
tm
t z g
t
z
k
z
 
  
 

Area type g
tz mt
A 250 0.07
B 300 0.19
C 400 0.14
 According to given data, calculated area type is B so
300; 0,09g
t tz m  . The height is lower than 10m take k = 1.
 The coefficient k considering wind pressure change in height z is set
under following table:
Height z (m) k
0 1
10 1
13,55 1,056
15,05 1,090
15,2 1,093
 According to TCVN 2737-1995 or Loads and effects SNiP 2.01.07-85*
if wind is perpendicular to the end wall of the building so it’s c = -0,7 for
the whole surface of the building.
Outside column
Left wind: c = +0,8
 Level 10m:  . . . . 85 1,2 0,8 1 6,5 530,4 /o
W W nc k B daN m      
 Level 13,55m:
 . . . . 85 1,2 0,8 1,056 6,5 560,10 /o
W W nc k B daN m      
Right wind: c = -0,4
 Level 10m:    . . . . 85 1,2 0,4 1 6,5 265,2 /o
W W nc k B daN m        
 Level 13,55m:
   . . . . 85 1,2 0,4 1,056 6,5 280,05 /o
W W nc k B daN m        
Longitudinal wind: c = -0,7
 Level 10m:    . . . . 85 1,2 0,7 1 6,5 464,1 /o
W W nc k B daN m        
 Level 13,55m:
   . . . . 85 1,2 0,7 1,056 6,5 490,09 /o
W W nc k B daN m        
Left roof’s L1-L2 span
Left wind: c = -0,442
 Level 13,55m:
   . . . . 85 1,2 0,442 1,056 6,5 309,46 /o
W W nc k B daN m        
 Level 15,050m:

   . . . . 85 1,2 0,442 1,090 6,5 319,42 /o
W W nc k B daN m        
 W 314,44 d /tb aN m  
Right wind: c = - 0,4
 Level 13,55m:
   . . . . 85 1,2 0,4 1,056 6,5 280,05 /o
W W nc k B daN m        
 Level 15,050m:
   . . . . 85 1,2 0,4 1,090 6,5 289,07 /o
W W nc k B daN m        
 W 284,56 d /tb aN m  
 Level 13,55m:
   . . . . 85 1,2 0,7 1,056 6,5 490,09 /o
W W nc k B daN m        
 Level 15,050m:
   . . . . 85 1,2 0,7 1,09 6,5 505,87 /o
W W nc k B daN m        
 W 497,98 d /tb aN m  
Right roof’s L1-L2 span
Left wind: c = -0,6
 Level 13,55m:
   . . . . 85 1,2 0,6 1,056 6,5 420,08 /o
W W nc k B daN m        
 Level 15,05m:
   . . . . 85 1,2 0,6 1,090 6,5 433,60 /o
W W nc k B daN m        
 W 426,84 d /tb aN m  
Right wind: c = - 0,5
 Level 13,55m:
   . . . . 85 1,2 0,5 1,056 6,5 350,06 /o
W W nc k B daN m        
 Level 15,05m:
   . . . . 85 1,2 0,5 1,090 6,5 361,34 /o
W W nc k B daN m        
 W 355,7 d /tb aN m  
 Level 13,55m:
   . . . . 85 1,2 0,7 1,056 6,5 490,09 /o
W W nc k B daN m        
 Level 15,050m:
   . . . . 85 1,2 0,7 1,090 6,5 505,87 /o
W W nc k B daN m        
 W 497,98 d /tb aN m  

Center roof
Left wind: c = -0,2
 Level 13,55m:
   . . . . 85 1,2 0,2 1,056 6,5 140,03 /o
W W nc k B daN m        
 Level 15,2m:
   . . . . 85 1,2 0,2 1,093 6,5 144,93 /o
W W nc k B daN m        
 W 142,48 d /tb aN m  
Right wind c = - 0,5
 Level 13,55m:
   . . . . 85 1,2 0,5 1,056 6,5 350,06 /o
W W nc k B daN m        
 Level 15,2m:
   . . . . 85 1,2 0,5 1,093 6,5 362,33 /o
W W nc k B daN m        
 W 356,20 d /tb aN m  
Longitudinal wind c = - 0,7
 Level 13,55m:
   . . . . 85 1,2 0,7 1,056 6,5 490,09 /o
W W nc k B daN m        
 Level 15,2m:
   . . . . 85 1,2 0,7 1,093 6,5 507,26 /o
W W nc k B daN m        
 W 498,68 d /tb aN m  

CHAPTER 4. DETERMINE INTERNAL FORCE,
SHEAR AND MOMENT AT SAP 2000 V18
DEFINE MATERIAL AND SECTION PROPERTIES.
Material.
Properties of CCT34 steel:
 Weight per unit volume 3
7850 /T daN m 
 Modulus of elasticity 10 2
2,1 10 /E daN m 
 Minimum yield stress 7 2
2,2 10 /yf daN m 
 Minimum tensile stress 7 2
3,4 10 /uf daN m 

Section.
a) Column section: (reference book: Structural Steel - Pham Van
Hoi - page 225).
 Wide – flange shape I preliminary sizing
 Brace is located at level (code) +6,35m, ly = 7,0 m.
 Chosing height h and width b:
1 1 1 1
14200 568 947( )
15 25 15 25
h H mm
   
          
   
Chọn  700h mm
 
     
1 1 1 1
7000 233 350
30 20 30 20
0,3 0,5 0,3 0,5 700 210 350
yb l mm
b h mm
    
          
   
       
Chọn  350b mm
 Chosing the thickness of flange tf and web tw:
  
 w
w w
1 1 1 1 21
350 12.5 10
28 35 21 28 35 21
1 1 1 1
700 11.67 5.83
60 120 60 120
; 60 ; 8
f
f f
f
t b mm
t h mm
t t t mm t mm
    
          
   
    
          
   
   


→Chose
 
 w
14
10
ft mm
t mm



 Both kinds of column side and center are defined at the same size as
straight element.

a) Beam section: (reference book: Structural Steel - Pham Van Hoi -
page 225).
 The height of beam is normally bigger than hmin
 min
5 5 210 15000
200 577
24 24 210000 1,1 cos(10 )o
tb
f l l
h mm
E 
 
       
=>Chose  700h mm
 To preventing over all buckling and easier connecting with other
elements, the width is defined as follow:
 
1 1 1 1
700 140 350
2 5 2 5
180 , /10
f
f f
b h mm
b mm b h
    
          
   
  
=>Chose  350b mm
 To preventing vertical buckling of compression flange, the ratio between
width and the thickness of flange is defined as follow:
 
 
6
2100
350 11,07
2,1 10
300
30 10
30 30
f
f
f
f f f
fb E t b mm
Et f
b
b t t mm
      
 
     
Chose  14ft mm
 To preventing shear buckling of web, the thickness of the web is defined
as:
 w
w 6
700 2 14 2100
6,64
3,2 3,2 2,1 10
h f
t mm
E
 
   

Chose  w 10t mm
 Both kinds of beam side and center are defined at the same size as
straight element.

CREATING BUILDING MODEL.
 The length of rafters are too long L1 = L3 = 30 (m), L2 = 33 (m), so to be
easier in transportation, the rafter is devided into 2 separated segment.
Side span’s rafter L1 = L3 = 30 (m), the conection between 2 segment is
loacated at 6 (m) in horizontal direction starting at side column, the rafter
is divided as follow:
 
 
6000
1 4 9 12 6092,56
cos10
9000
2 3 10 11 9138,84
cos10
o
o
D D D D mm
D D D D mm
    
    
 Centre span’s rafter L2 = 33 (m), the conection between 2 segment is
loacated at 6 (m) in horizontal direction starting at side column, the rafter
is divided as follow:
 
 
6000
5 8 6092,56
cos10
10500
6 7 10661,98
cos10
o
o
D D mm
D D mm
  
  

DEFINE LOAD.
 Dead loads include the following:
a) weight of structural parts including weight of bearing and
enclosing structures;
b) weight and pressure of ground (mounds and fillings), rock
pressure.
c) The prestressing forces remained in building structures and
foundations should be adopted in calculations as dead load forces.).
 Crane load include in 6 load cases:
 Dmax vertically impact on column E (Dmax LEFT)
 Dmax vertically impact on column I (Dmax RIGHT)
 T horizontally impact on column E, from right to left direction (T
LEFT)
 T horizontally impact on column E, from left to right direction (-T
LEFT)
 T horizontally impact on column I, from left to right direction (T
RIGHT)
 T horizontally impact on column I, from right to left direction (-T
RIGHT)
 Wind load include in 2 load cases:
 Wind impact on building from left to right direction (LEFT WIND)
 Wind impact on building from right to left direction (RIGHT WIND)
 Live load include in 6 load cases: HT1, HT2, HT3, HT4, HT5, HT6.
Those kinds of loads could be occurred concurrent.

Dead load
Defining load for building model at Sap 2000 to analysis, the Self
Weight Multiplier need to be 1.1
Live load.
 HT1.

 HT2.
 HT3.
 HT4.
 HT5.
 HT6.

Wind load.
 Left wind.
 Right wind.
Crane load.
 Dmax Left.
 Dmax phải.

 T LEFT.
 -T LEFT.
 T RIGHT.
 -T RIGHT.
 Longitudinal wind:

DEAD LIVE1 LIVE2 LIVE3 LIVE4 LIVE5 LIVE6
DMAX
LEFT
DMAX
RIGHT
TLEFT - TLEFT TRIGHT - TRIGHT
LEFT
WIND
RIGHT
WIND
LONGITUDINAL
WIND
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
M(KN.m) -13501.4 -4281.6 -3692.7 691.9 -1116.3 127.5 -1328.7 1971.1 -1622.2 -1309.5 1309.5 1313.6 -1313.6 37250.2 -5522.0 6554.8
N(KN) -10176.9 -2959.1 -800.4 165.6 31.9 -7.4 -57.4 115.2 -47.1 -35.9 35.9 51.4 -51.4 5161.2 3963.5 7205.7
V(KN) -2328.3 -923.6 -646.7 137.9 -85.6 8.1 -145.7 227.9 -167.8 -134.1 134.1 141.6 -141.6 8122.2 -1425.6 -888.9
M(KN.m) 19559.8 8833.1 5490.2 -1266.8 99.7 13.0 739.9 -1265.5 759.9 594.0 -594.0 -696.3 696.3 -24547.8 -12047.6 -27668.0
N(KN) -5105.3 -2959.1 -800.4 165.6 31.9 -7.4 -57.4 115.2 -47.1 -35.9 35.9 51.4 -51.4 5161.2 3963.5 7205.7
V(KN) -2328.3 -923.6 -646.7 137.9 -85.6 8.1 -145.7 227.9 -167.8 -134.1 134.1 141.6 -141.6 537.8 2366.6 5747.5
M(KN.m) -3086.1 3790.0 3651.2 -4520.5 -4157.7 617.5 -1306.9 -3075.6 -3360.6 -2237.4 2237.4 1497.6 -1497.6 9490.0 -11480.0 3843.8
N(KN) -15040.6 -894.9 -3144.7 -3483.8 -917.3 131.1 90.0 -19986.4 -4619.5 15.2 -15.2 -30.7 30.7 9368.7 10457.9 15961.9
V(KN) -459.6 638.9 762.8 -929.5 -705.9 108.1 -139.0 -948.7 -553.0 -323.2 323.2 177.0 -177.0 787.8 -1455.5 560.7
M(KN.m) 2267.8 -3653.4 -5235.9 6307.9 4066.1 -641.7 312.2 7976.7 3082.0 1528.3 -1528.3 -564.3 564.3 312.0 5476.9 -2688.8
N(KN) -13378.8 -894.9 -3144.7 -3483.8 -917.3 131.1 90.0 -19986.4 -4619.5 15.2 -15.2 -30.7 30.7 9368.7 10457.9 15961.9
V(KN) -459.6 638.9 762.8 -929.5 -705.9 108.1 -139.0 -948.7 -553.0 -323.2 323.2 177.0 -177.0 787.8 -1455.5 560.7
M(KN.m) 1195.3 -3653.4 -5235.9 6307.9 4066.1 -641.7 312.2 -6989.9 -354.7 1528.3 -1528.3 -564.3 564.3 312.0 5476.9 -2688.8
N(KN) -11948.8 -894.9 -3144.7 -3483.8 -917.3 131.1 90.0 -30.9 -37.2 15.2 -15.2 -30.7 30.7 9368.7 10457.9 15961.9
V(KN) -459.6 638.9 762.8 -929.5 -705.9 108.1 -139.0 -948.7 -553.0 -323.2 323.2 177.0 -177.0 787.8 -1455.5 560.7
M(KN.m) 2367.2 -5282.6 -7181.1 8678.1 5866.2 -917.3 666.6 -4570.7 1055.5 994.8 -994.8 -1015.6 1015.6 -1697.0 9188.5 -4118.7
N(KN) -11585.0 -894.9 -3144.7 -3483.8 -917.3 131.1 90.0 -30.9 -37.2 15.2 -15.2 -30.7 30.7 9368.7 10457.9 15961.9
V(KN) -459.6 638.9 762.8 -929.5 -705.9 108.1 -139.0 -948.7 -553.0 452.6 -452.6 177.0 -177.0 787.8 -1455.5 560.7
ELEMENT SECTION
INTERNAL
FORCE
LOAD PATTERN
C1
COLUMN
BASE
COLUMN
HEAD
C2
COLUMN
BASE
LOWER
BRACKET
C3
UPPER
BRACKET
COLUMN
HEAD
INTERIAL FORCE AND LOAD COMBINATION.
INTERNAL FORCE TABLE OF COLUMN

DEAD LIVE1 LIVE2 LIVE3 LIVE4 LIVE5 LIVE6
DMAX
LEFT
DMAX
RIGHT
T LEFT - T LEFT T RIGHT - T RIGHT
LEFT
WIND
RIGHT
WIND
LONGITUDINAL
WIND
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
M(KN.m) -19559.75 -8833.08 -5490.22 1266.76 -99.68 -12.99 -739.87 1265.45 -759.94 -593.95 593.95 696.34 -696.34 24547.81 12047.57 27667.99
N(KN) -2824.69 -1213.43 -723.12 153.73 -82.04 7.29 -150.67 238.26 -171.61 -136.95 136.95 145.96 -145.96 1048.7 2749.27 6435.92
V(KN) -4848.31 -2852.54 -732.12 151.1 40.22 -8.14 -42.64 91.96 -30.2 -22.37 22.37 37.04 -37.04 5082.07 3708.38 6598.04
M(KN.m) 3214.44 3781.78 -1075.58 355.66 -342.2 36.1 -482.78 710.92 -577.84 -459.07 459.07 473.01 -473.01 -380.2 -5140.38 -3064.41
N(KN) -2610.4 -1061.33 -723.12 153.73 -82.04 7.29 -150.67 238.26 -171.61 -136.95 136.95 145.96 -145.96 1048.7 2749.27 6435.92
V(KN) -2705.41 -1331.54 -732.12 151.1 40.22 -8.14 -42.64 91.96 -30.2 -22.37 22.37 37.04 -37.04 3186.02 1992.5 3595.26
M(KN.m) 3214.44 3781.78 -1075.58 355.66 -342.2 36.1 -482.78 710.92 -577.84 -459.07 459.07 473.01 -473.01 -380.2 -5140.38 -3064.41
N(KN) -2610.4 -1061.33 -723.12 153.73 -82.04 7.29 -150.67 238.26 -171.61 -136.95 136.95 145.96 -145.96 1048.7 2749.27 6435.92
V(KN) -2705.41 -1331.54 -732.12 151.1 40.22 -8.14 -42.64 91.96 -30.2 -22.37 22.37 37.04 -37.04 3186.02 1992.5 3595.26
M(KN.m) 13147.86 5507.48 5546.37 -1010.99 -705.98 109.73 -97.15 -120.89 -304.68 -256.75 256.75 138.01 -138.01 -16335.27 -11522.44 -15213.23
N(KN) -2288.97 -833.18 -723.12 153.73 -82.04 7.29 -150.67 238.26 -171.61 -136.95 136.95 145.96 -145.96 1048.7 2749.27 6435.92
V(KN) 508.94 949.96 -732.12 151.1 40.22 -8.14 -42.64 91.96 -30.2 -22.37 22.37 37.04 -37.04 341.95 -581.31 -908.92
M(KN.m) 13147.86 5507.48 5546.37 -1010.99 -705.98 109.73 -97.15 -120.89 -304.68 -256.75 256.75 138.01 -138.01 -16335.27 -11522.44 -15213.23
N(KN) -2344.42 -1004.79 -563.83 120.77 -88.38 8.76 -139.24 215.33 -162.23 -129.81 129.81 135.73 -135.73 960.22 2809.94 6488.46
V(KN) 45.6 766.16 -860.82 178.55 23.18 -6.54 -71.63 137.32 -63.58 -49.04 49.04 65.21 -65.21 542.84 -25.39 383.52
M(KN.m) -1801.32 -1422.34 3014.4 -2625.92 -915.62 168.84 550.7 -1362.97 270.43 186.85 -186.85 -451.77 451.77 -3785.33 3257.1 1687.73
N(KN) -2665.86 -1004.79 -791.98 120.77 -88.38 8.76 -139.24 215.33 -162.23 -129.81 129.81 135.73 -135.73 960.22 2809.94 6488.46
V(KN) 3259.95 766.16 1420.68 178.55 23.18 -6.54 -71.63 137.32 -63.58 -49.04 49.04 65.21 -65.21 -3317.88 -3242.65 -4120.65
M(KN.m) -1801.32 -1422.34 3014.4 -2625.92 -915.62 168.84 550.7 -1362.97 270.43 186.85 -186.85 -451.77 451.77 -3785.33 3257.1 1687.73
N(KN) -2665.86 -1004.79 -791.98 120.77 -88.38 8.76 -139.24 215.33 -162.23 -129.81 129.81 135.73 -135.73 960.22 2809.94 6488.46
V(KN) 3259.95 766.16 1420.68 178.55 23.18 -6.54 -71.63 137.32 -63.58 -49.04 49.04 65.21 -65.21 -3317.88 -3242.65 -4120.65
M(KN.m) -27919.36 -6042.23 -10137.97 -3702.54 -1055.38 208.25 982.6 -2191.02 653.83 482.59 -482.59 -844.95 844.95 23981.18 29276.69 35588.23
N(KN) -2880.15 -1004.79 -944.08 120.77 -88.38 8.76 -139.24 215.33 -162.23 -129.81 129.81 135.73 -135.73 960.22 2809.94 6488.46
V(KN) 5402.85 766.16 2941.68 178.55 23.18 -6.54 -71.63 137.32 -63.58 -49.04 49.04 65.21 -65.21 -5891.69 -5387.5 -7123.43
M(KN.m) -30286.59 -759.59 -2956.89 -12380.59 -6921.59 1125.52 316.03 2379.71 -401.65 -512.22 512.22 170.68 -170.68 25678.17 20088.17 39706.95
N(KN) -3363.27 -286.48 103.26 -1117.78 -875.72 127.88 -279.99 -708.81 -725.58 314.9 -314.9 319.01 -319.01 1490.66 1279.75 7095.62
V(KN) -5615.58 -4.11 -134.65 -3222.9 -802.32 111.54 60.76 155.58 -12.15 -52.29 52.29 -11.1 11.1 3278.67 4713.55 7559.7
M(KN.m) -2885.82 -734.81 -2144.95 2467.52 -2083.64 452.95 -50.33 1441.55 -328.41 -196.94 196.94 237.63 -237.63 8498.32 -1858.45 3188.55
N(KN) -3148.98 -286.48 103.26 -965.68 -875.72 127.88 -279.99 -708.81 -725.58 314.9 -314.9 319.01 -319.01 1490.66 1279.75 7095.62
V(KN) -3472.68 -4.11 -134.65 -1701.9 -802.32 111.54 60.76 155.58 -12.15 -52.29 52.29 -11.1 11.1 2419.53 2565.69 4552.69
M(KN.m) -2885.82 -734.81 -2144.95 2467.52 -2083.64 452.95 -50.33 1441.55 -328.41 -196.94 196.94 237.63 -237.63 8498.32 -1858.45 3188.55
N(KN) -3148.98 -286.48 103.26 -965.68 -875.72 127.88 -279.99 -708.81 -725.58 314.9 -314.9 319.01 -319.01 1490.66 1279.75 7095.62
V(KN) -3472.68 -4.11 -134.65 -1701.9 -802.32 111.54 60.76 155.58 -12.15 -52.29 52.29 -11.1 11.1 2419.53 2565.69 4552.69
M(KN.m) 13973.08 -691.45 -724.04 6382.76 6382.76 -724.04 -691.45 -200.24 -200.24 354.81 -354.81 354.81 -354.81 -9100.66 -9100.66 -17088.52
N(KN) -2773.98 -286.48 103.26 -699.51 -875.72 127.88 -279.99 -708.81 -725.58 314.9 -314.9 319.01 -319.01 1490.66 1279.75 7095.62
V(KN) 277.4 -4.11 -134.65 959.85 -802.32 111.54 60.76 155.58 -12.15 -52.29 52.29 -11.1 11.1 916.02 -1193.07 -709.56
ELEMENT SECTION
INTERNAL
FORCE
LOAD PATTERN
B1
START
0m
END
6,03m
B2
START
0m
END
9,04m
B3
START
0m
END
9,04m
B6
START
0m
END
10,55m
B4
START
0m
END
6,03m
B5
START
0m
END
6,03m
INTERNAL FORCE TABLE OF RAFTER

Mmax, Ntư Mmin, Ntư |N|max, Mtư Mmax, Ntư Mmin, Ntư |N|max, Mtư Mmax, Ntư Mmin, Ntư |N|max, Mtư
1,14 1,2,3,5,7 1,2,3,6,7
1,4,6,8,11,1
4, 16
1,2,3,5,7,9
,13, 15
1,2,3,6,7,9,13
1,4,6,8,11,1
4, 16
1,2,3,5,7,9
,13, 15
1,2,3,6,7
M(KN.m) 23748.8 -23920.8 -22677.0 29613.1 -30490.9 -24401.6 29613.1 -30490.9 -22677.0
N(KN) -5015.7 -13962.0 -14001.2 1231.8 -10104.9 -13707.4 1231.8 -10104.9 -14001.2
V(KN) 5794.0 -4129.8 -4036.1 4638.9 -5511.1 -4143.7 4638.9 -5511.1 -4036.1
1,2,3,5,6,7 1, 16 1,2,3,6,7
1,2,3,5,6,7,9
,13
1,4,8,11,1
4, 16
1,2,3,6,7,9,13 1,2,3,5,6,7
1,4,8,11,1
4, 16
1,2,3,6,7
M(KN.m) 34735.6 -8108.2 34635.9 34528.7 -30248.0 34438.9 34735.6 -30248.0 34635.9
N(KN) -8897.8 2100.4 -8929.7 -8607.2 6309.9 -8635.9 -8897.8 6309.9 -8929.7
V(KN) -4121.8 3419.2 -4036.1 -4220.8 3778.4 -4143.7 -4121.8 3778.4 -4036.1
1,14 1, 15 1,8,11
1,2,3,6,14,
16
1,4,5,7,8,1
0, 15
1,2,3,4,5,8,11
1,2,3,6,14,
16
1,4,5,7,8,1
0, 15
1,2,3,4,5,8,1
1
M(KN.m) 6404.0 -14566.1 -3924.3 16167.2 -27186.3 -4953.8 16167.2 -27186.3 -4953.8
N(KN) -5671.9 -4582.8 -35042.2 4239.2 -27482.7 -40638.8 4239.2 -27482.7 -40638.8
V(KN) 328.3 -1915.1 -1085.0 2113.0 -4511.2 -1232.8 2113.0 -4511.2 -1232.8
1,4,5,7 1,2,3,6 1,8,11
1,4,5,7,8,10,
15
1,2,3,6, 16 1,2,3,4,5,8,11
1,4,5,7,8,10
, 15
1,2,3,6, 16
1,2,3,4,5,8,1
1
M(KN.m) 12954.1 -7263.1 8716.3 25369.1 -8729.9 9407.8 25369.1 -8729.9 9407.8
N(KN) -17689.9 -17287.3 -33380.4 -25820.8 -2530.7 -38976.9 -25820.8 -2530.7 -38976.9
V(KN) -2233.9 1050.3 -1085.0 -4511.2 1404.0 -1232.8 -4511.2 1404.0 -1232.8
1,4,5,7 1,2,3,6 1,2,3,4,5
1,4,5,7,9,13,
15
1,2,3,6,8,1
1, 16
1,2,3,4,5,9,12
1,4,5,7,9,13
, 15
1,2,3,6,8,1
1, 16
1,2,3,4,5
M(KN.m) 11881.6 -8335.6 2680.1 15930.8 -17468.8 1704.6 15930.8 -17468.8 2680.1
N(KN) -16259.9 -15857.3 -20389.5 -6422.5 -1142.2 -19606.5 -6422.5 -1142.2 -20389.5
V(KN) -2233.9 1050.3 -693.2 -4023.5 841.0 -1008.2 -4023.5 841.0 -693.2
1,4,5,7 1,2,3,6 1,2,3,4,5
1,4,5,7,9,13,
15
1,2,3,6,8,1
1,14, 16
1,2,3,4,5,9,12
1,4,5,7,9,13
, 15
1,2,3,6,8,1
1,14, 16
1,2,3,4,5
M(KN.m) 17578.1 -11013.8 4447.8 26190.7 -19918.8 4275.6 26190.7 -19918.8 4447.8
N(KN) -15896.1 -15493.6 -20025.8 -6058.8 7653.4 -19242.8 -6058.8 7653.4 -20025.8
V(KN) -2233.9 1050.3 -693.2 -4023.5 851.8 -1008.2 -4023.5 851.8 -693.2
INTERNAL FORCEDESIGN
ELEMENT SECTION
INTERNAL
FORCE
LOAD COMBINATION 1 LOAD COMBINATION 2
C3
START
0m
END
2,55m
C1
START
0m
END
14,2m
C2
START
0m
END
11,65m
LOAD COMBINATION OF COLUMN

Mmax, Ntư Mmin, Ntư |N|max, Mtư Mmax, Ntư Mmin, Ntư |N|max, Mtư Mmax, Ntư Mmin, Ntư |N|max, Mtư
1, 16 1,2,3,5,6,7 1,2,3,5,7
1,4,8,11,14,
16
1,2,3,5,6,7,9
,13
1,2,3,5,7,9,13
1,4,8,11,14,
16
1,2,3,5,6,7 1,2,3,5,7,9,13
M(KN.m) 8108.2 -34735.6 -34722.6 30248.0 -34528.7 -34517.0 30248.0 -34735.6 -34517.0
N(KN) 3611.2 -4986.7 -4994.0 4387.5 -5056.3 -5062.8 4387.5 -4986.7 -5062.8
V(KN) 1749.7 -8443.5 -8435.4 5902.7 -8144.5 -8137.2 5902.7 -8443.5 -8137.2
1,2,4,6 1, 15 1,2,3,5,7 1,2,4,6,8,11
1,3,5,7,9,13,
15, 16
1,2,3,5,7,9,13 1,2,4,6,8,11
1,3,5,7,9,13
, 15, 16
1,2,3,5,7,9,13
M(KN.m) 7388.0 -1925.9 5095.7 8023.6 -6826.1 3961.8 8023.6 -6826.1 3961.8
N(KN) -3510.7 138.9 -4627.6 -3083.0 4510.2 -4711.7 -3083.0 4510.2 -4711.7
V(KN) -3894.0 -712.9 -4771.5 -3672.2 1602.0 -4625.4 -3672.2 1602.0 -4625.4
1,2,4,6 1, 15 1,2,3,5,7 1,2,4,6,8,11
1,3,5,7,9,13,
15, 16
1,2,3,5,7,9,13 1,2,4,6,8,11
1,3,5,7,9,13
, 15, 16
1,2,3,5,7,9,13
M(KN.m) 7388.0 -1925.9 5095.7 8023.6 -6826.1 3961.8 8023.6 -6826.1 3961.8
N(KN) -3510.7 138.9 -4627.6 -3083.0 4510.2 -4711.7 -3083.0 4510.2 -4711.7
V(KN) -3894.0 -712.9 -4771.5 -3672.2 1602.0 -4625.4 -3672.2 1602.0 -4625.4
1,2,3,6 1,14 1, 16 1,2,3,6,8,11
1,4,5,7,9,13,
14, 16
1,2,3,5,7,9,13 1,2,3,6
1,4,5,7,9,13
,14, 16
1,2,3,5,7,9,13
M(KN.m) 24311.4 -3187.4 -2065.4 23317.4 -17276.9 21975.1 24311.4 -17276.9 21975.1
N(KN) -3838.0 -1240.3 4147.0 -3345.4 4090.3 -4184.9 -3838.0 4090.3 -4184.9
V(KN) 718.6 850.9 -400.0 800.6 72.0 642.3 718.6 72.0 642.3
1,2,3,6 1,14 1, 16 1,2,3,6,8,11
1,4,5,7,9,13,
14, 16
1,2,3,5,7,9,13 1,2,3,6
1,4,5,7,9,13
,14, 16
1,2,3,5,7,9,13
M(KN.m) 24311.4 -3187.4 -2065.4 23317.4 -17276.9 21975.1 24311.4 -17276.9 21975.1
N(KN) -3904.3 -1384.2 4144.0 -3437.7 3995.1 -4229.2 -3904.3 3995.1 -4229.2
V(KN) -55.6 588.4 429.1 122.2 880.5 -199.1 -55.6 880.5 -199.1
1,3,6,7 1,2,4,5 1,2,3,5,7
1,3,6,7,9,13,
15, 16
1,2,4,5,8,11,
14
1,2,3,5,7,9,13
1,3,6,7,9,13,
15, 16
1,2,4,5,8,11
,14
1,2,3,5,7,9,13
M(KN.m) 1932.6 -6765.2 -574.2 6659.6 -11070.4 -46.9 6659.6 -11070.4 -46.9
N(KN) -3588.3 -3638.3 -4690.3 4604.3 -2366.2 -4756.0 4604.3 -2366.2 -4756.0
V(KN) 4602.5 4227.8 5398.3 -2274.7 1312.7 5068.6 -2274.7 1312.7 5068.6
1,3,6,7 1,2,4,5 1,2,3,5,7
1,3,6,7,9,13,
15, 16
1,2,4,5,8,11,
14
1,2,3,5,7,9,13
1,3,6,7,9,13,
15, 16
1,2,4,5,8,11
,14
1,2,3,5,7,9,13
M(KN.m) 1932.6 -6765.2 -574.2 6659.6 -11070.4 -46.9 6659.6 -11070.4 -46.9
N(KN) -3588.3 -3638.3 -4690.3 4604.3 -2366.2 -4756.0 4604.3 -2366.2 -4756.0
V(KN) 4602.5 4227.8 5398.3 -2274.7 1312.7 5068.6 -2274.7 1312.7 5068.6
1, 16 1,2,3,4,5 1,2,3,5,7
1,6,7,9,13,
15, 16
1,2,3,4,5,8,1
1
1,2,3,5,7,9,13
1,6,7,9,13,
15, 16
1,2,3,4,5,8,
11
1,2,3,5,7,9,13
M(KN.m) 7668.9 -48857.5 -44172.3 32879.7 -49169.9 -41198.1 32879.7 -49169.9 -41198.1
N(KN) 3608.3 -4796.6 -5056.6 5102.8 -4294.4 -5107.2 5102.8 -4294.4 -5107.2
V(KN) -1720.6 9312.4 9062.2 -6043.3 9089.2 8580.4 -6043.3 9089.2 8580.4
1, 16 1,2,3,4,5 1,2,4,5,7
1,6,7,8,11,14
, 16
1,2,3,4,5,9,1
3
1,2,4,5,7,9,13
1,6,7,8,11,1
4, 16
1,2,3,4,5 1,2,4,5,7,9,13
M(KN.m) 9420.4 -53305.3 -50032.3 32460.2 -51518.5 -48572.9 32460.2 -53305.3 -48572.9
N(KN) 3732.4 -5540.0 -5923.2 3306.1 -6262.4 -6607.4 3306.1 -5540.0 -6607.4
V(KN) 1944.1 -9779.6 -9584.2 4481.1 -9364.1 -9188.2 4481.1 -9779.6 -9188.2
1,14 1,2,3,5,7 1,2,4,5,7
1,4,6,8,11,14
, 16
1,2,3,5,7,9,1
3, 15
1,2,4,5,7,9,13
1,4,6,8,11,1
4, 16
1,2,3,5,7,9,
13, 15
1,2,4,5,7,9,13
M(KN.m) 5612.5 -7899.6 -3287.1 11735.4 -9580.2 -3756.4 11735.4 -9580.2 -3756.4
N(KN) -1658.3 -4487.9 -5556.9 2903.3 -4142.4 -6256.2 2903.3 -4142.4 -6256.2
V(KN) -1053.2 -4353.0 -5920.3 1558.1 -1956.8 -5676.4 1558.1 -1956.8 -5676.4
1,14 1,2,3,5,7 1,2,4,5,7
1,4,6,8,11,14
, 16
1,2,3,5,7,9,1
3, 15
1,2,4,5,7,9,13
1,4,6,8,11,1
4, 16
1,2,3,5,7,9,
13, 15
1,2,4,5,7,9,13
M(KN.m) 5612.5 -7899.6 -3287.1 11735.4 -9580.2 -3756.4 11735.4 -9580.2 -3756.4
N(KN) -1658.3 -4487.9 -5556.9 2903.3 -4142.4 -6256.2 2903.3 -4142.4 -6256.2
V(KN) -1053.2 -4353.0 -5920.3 1558.1 -1956.8 -5676.4 1558.1 -1956.8 -5676.4
1,4,5 1, 16 1,2,4,5,7 1,4,5,8,11
1,2,3,6,7,9,1
3, 15, 16
1,2,4,5,7,9,13 1,4,5,8,11
1,2,3,6,7,9,
13, 15, 16
1,2,4,5,7,9,13
M(KN.m) 26738.6 -3115.4 25355.7 26936.7 -12654.5 23717.9 26936.7 -12654.5 23717.9
N(KN) -4349.2 4321.6 -4915.7 -5113.0 3521.9 -5641.6 -5113.0 3521.9 -5641.6
V(KN) 434.9 -432.2 491.6 606.3 -1405.7 469.2 606.3 -1405.7 469.2
INTERNAL FORCEDESIGN
ELEMENT SECTION
INTERNAL
FORCE
LOAD COMBINATION 1 LOAD COMBINATION 2
B1
START
0m
END
6,03m
B2
START
0m
END
9,04m
B3
START
0m
END
9,04m
B4
START
0m
END
6,03m
B5
START
0m
END
6,03m
B6
START
0m
END
10,55m
LOAD COMBINATION OF RAFTER

CHAPTER 5. PURLINS DESIGN.
TOLE DESIGN.
Properties.
 This product is taken from Ngo Long Cor. catalogue
Sectional drawing:
 Properties:

2
tolef = 2100 daN/cm và
3
ρ = 7850 daN/m
 Table below:
Total
coated
thickness
Single
span
Coating
weight
Rib
height
Moment
of inertia
x-x
Moment
of inertia
y-y
Section
modulus
x-x
Section
modulus
y-y
t (mm)
L
(mm)
P
(daN/m2
)
h (mm)
Jx
(104
mm4
)
Jy
(104
mm4
)
Wx
(103
mm3
)
Wy
(103
mm3
)
0.3 1000 2,65 21 2,117 25000 1,623 500

Define load.
 Self weight of tole is devided into 2 part gx and gy:
 Normal load: tole
tc 2
g = 2,65 (daN/m )
tc tc 2
y(tole)
= g cosα = 2,65×0,995 = 2,64 (daN/m )tole
g
tc tc 2
x(tole)
= p sinα = 2,65×0,1 = 0,27 (daN/m )tole
g
 Design load:
tt tc 2
( ) ( )g = ng =1,1×2,64 = 2,904 (daN/m )y ytole tole
tt tc 2
( ) ( )
g = ng = 1,1×0,27 = 0,3 (daN/m )x xtole tole
 Wind load: the most dangerous load case for tole is negative wind (c = -0.6):
 
gió
gió gió
tc 2
tt tc 2
q = W .c.k = 85×(-0,6)×1,09 = -55,59 (daN/m )
q =nq = 1,2 -55,59 = -66,71 (daN/m )
C

 Live load:
Short-term load tc 2
p = 30 daN/mroof is devided into 2 part px and py:
tc tc 2
y mái
p = p cosα = 30×0,995 = 29,85 (daN/m )
tc tc 2
x mái
p = p sinα = 30×0,1 = 3 (daN/m )
tt tc 2
y y
p = np = 1,3×29,85 = 38,81 (daN/m )
tt tc 2
x xp = np = 1,3×3 = 3,9 (daN/m )
Design tole section.
 Purlins spacing a = 1,1 m
 Model of calculation: width of the section is supposed to be 1m to design:
 These are the most dangerous load case is given below:
a) Load combination 1: Dead load and Wind load:
 Check for allowable stress:
   tt tt
gió ( )q = q + g ×1 = -66,71 + 2,904 1= -69,61(daN/m)tt
y y tole 
( )
q = g 1= 0,3 1=0,3(daN/m)tt tt
x x tole
 
The value  q 0,3 /tt
x
daN m is too small. Thus, it is neglected.
2 2
q a -69,61×1,1
M = = = -10,53 (daN.m)
8 8
tt
y
 Section modulus at least:
 3 3
yc x
c tole
M 10,53×100
W = = = 0,55 (cm ) < W = 1,623 cm
γ f 0,9×2100

 Allowable stress:
 2 2
c
x
M 10,53×100
σ = = =648,8(daN/cm ) < fγ = 0,9 2100=1890 daN/cm
W 1,623

 This section is OK.
 Check for displacement:
   tc tc
gió ( )q = q + g ×1 = -55,59 + 2,64 1= -52,95(daN/m)tc
y y tole 
( )
q = g 1= 0,27 1=0,27(daN/m)tc tc
x x tole
 
The value  q 0,27 /tt
x
daN m is too small. Thus it is neglected.
3 3
6
x
q aΔ 5 5 (52,95/100)×(1,1×100) 1 Δ 1
= = × = < =
a 384 EI 384 2,1×10 ×2,117 484 a 150
tc
y  
  
b) Load combination 2: Dead load and Wind load:
   tt tt
( )q = g + p ×1 = 2,904 38,81 1= 41,71(daN/m)tt
y y tole y  
   tt tt
( )q = g + p ×1 = 0,3 3,9 1= 4,2(daN/m)tt
x x tole x  
 Moment:
2 2
q a 41,71×1,1
M = = = 6,31 (daN.m)
8 8
tt
y
x
2 2
q a 4,2×1,1
M = = = 5,08 (daN.m)
8 8
tt
x
y
 
y 2x
max x y
x y
2
c
MM 6,31 100 5,08 100
σ = σ + σ = + = + =389,8 (daN/cm )
W W 1,623 500
< fγ =1890 daN/cm
 
   tc tc
( )q = g + p ×1 = 3 29,85 1= 32,85(daN/m)tc
y y tole y  
   tc tc
(t )q = g + p ×1 = 0,27 2,64 1= 2,91(daN/m)tc
x x ole x  

       
2222 tc 3 tc 3
y yx x
x
2 23 3
6 6
Δ q aΔ q aΔ 5 5
= + = +
a a a 384 EI 384 EI
32,85/100 1,1 100 2,91/100 1,1 1005 5 1 Δ 1
= + =
384 2,1×10 ×2,117 384 2,1×10 ×25000 781 a 150
y
    
               
        
              
PURLINS DESIGN.
Properties. According to vvvTra co Cor. catalogue
Chose steel C15015
4 3
4 3
4,44 /
353 ; 34,7
39,6 ; 7,17
xg
x x
y y
g daN m
I cm W cm
I cm W cm

 
 
Load impact.
 Purlins self weight:
tc
y(xg) xgg = g cosα = 4,44×0,995 = 4,42 (daN/m)
tc
x(xg) xgg = g sinα = 4,44×0,1 = 0,444 (daN/m)
tt tc
y(xg) yg = ng = 1,1×4,42= 4,86(daN/m)
tt tc
x(xg) xg = ng = 1,1×0,444 = 0,488(daN/m)
 Tole self weight:
tc tc
tole toleq = ag = 1,1×2,65 = 2,92 (daN/m)
tc tc
y(tole)
= q cosα = 2,92×0,995 = 2,91 (daN/m)tole
q

tc tc
x(tole)
= q sinα = 2,92×0,1 = 0,29(daN/m)tole
q
tt tc
( ) ( )= nq = 1,1×2,91 = 3,2 (daN/m)y ytole toleq
tt tc
( ) ( )
= nq = 1,1×0,29 = 0,32 (daN/m)x tole x tole
q
 Wind load: the most dangerous load case for tole is negative wind (c = -0.6).
 
gió
gió gió
tc
tt tc
q = W .c.k.a = 85×(-0,6)×1,09 1,1 = -61,15 (daN/m)
q =nq = 1,2 -61,15 = -73,38 (daN/m)
C 

 Short-term load tc 2
p = 30 daN/mroof is devided into 2 part px and py:
tc tc
y mái
p = p .a.cosα = 30 1,1 0,995 = 32,84(daN/m) 
tc tc
x mái
p = p .a.sinα = 30 1,1 0,1 = 3,3(daN/m) 
tt tc
y y
p = np = 1,3×32,84 = 42,69(daN/m)
tt tc
x xp = np = 1,3×3,3 = 4,29(daN/m)
Purlins design.
 Model of calculation: pinned beam.
 These are the most dangerous load case is given below:
tt tt tt tt
y y(tole) y(xg) gió
q = (q + g ) + q = 4,86 + 3,87 +(-73,38) = -64,65 (daN/m)
tt tt tt
x(tole) x(xg)
q = (q + g ) = 0,32 + 0,488 = 0,81 (daN/m)x
 Moment:
2 2
q -64,65×6,5
M = = = -341,4 (daN.m)
8 8
tt
y
x
B
2 2
q 0,81×6,5
M = = = 4,28 (daN.m)
8 8
tt
x
y
B
 
y 2x
max x y
x y
2
c
MM 341,4 100 4,28 100
σ = σ + σ = + = + =1042,4 (daN/cm )
W W 34,7 7,17
< fγ =1890 daN/cm
 
 This section is OK
tc tc tc tc
q = (q + g ) + q = 2,91 + 4,42 +(-61,15) = -53,82 (daN/m)
tc tc tc
x(tole) x(xg)
q = q + g = 0,29 + 0,444 = 0,734 (daN/m)x

       
2 22 tc 3 tc 32
y xyx
x
2 23 3
6 6
q qΔΔ Δ 5 5
= + = +
384 EI 384 EI
53,82 /100 6,5 100 0,734 /100 6,5 1005 5 1
= +
384 2,1×10 ×353 384 2,1×10 ×39,6 382
Δ 1
=
a 150
y
B B
B B B
     
                 
      
   
   
   
 
   
 Checking for allowable stress:
tt tt tt tt
y y(tole) y(xg)
q = (q + g ) + p = 3,2 + 4,86 +42,69 = 50,75 (daN/m)y
tt tt tt
x(tole) x(xg)
q = (q + g ) + p = 0,32 + 0,488 + 4,29 = 5,098 (daN/m)tt
x x
 Moment:
2 2
q 50,75×6,5
M = = =268,02 (daN.m)
8 8
tt
y
x
B
2 2
q 5,098×6,5
M = = = 26,92 (daN.m)
8 8
tt
x
y
B
 
y 2x
max x y
x y
2
c
MM 268,02 100 26,92 100
σ = σ + σ = + = + =1147,85 (daN/cm )
W W 34,7 7,17
< fγ =1890 daN/cm
 
tc tc tc tc
y y(tole) y(xg)
q = (q + g ) + p = 2,91 + 4,42 + 32,84 = 40,17(daN/m)y
 tc tc tc
x(tole) x(xg)q = q + g + p = 0,29 + 0,444 + 3,3 = 4,034 (daN/m)tc
x x
       
2222 tc 3 tc 3
y yx x
x
2 23 3
6 6
Δ qΔ Δ 5 5 q
= + = +
384 EI 384 EI
40,17 /100 6,5 100 4,034 /100 6,5 1005 5 1
= +
384 2,1×10 ×353 384 2,1×10 ×39,6 384
Δ 1
=
a 150
y
B B
B B B
    
               
      
   
   
   
 
   
This section is OK.

DESIGN CONECTION BETWEEN TOLE AND PURLINS.
 Spacing screw, bđv = 500 mm.
 Chose screw has properties is as follow: d = 5.5 mm; ftb = 1700 (daN/cm2
),
fvb = 1500 (daN/cm2
), fcb = 3950 (daN/cm2
).
 Area force pressure impact on screw:  2
1,1 0,5 0,55đvA ab m   
Load combination 1: Dead load and Wind load:
 Tension force:
     tt tt
gió y(tole)q 0,55 66,71 2,904 35,09tN A g daN       
     
2 2
3,14 0,55
35,09 1700 403,69
4 4
t t bn tb tb
d
N daN N A f f daN
 
       
 This screw is OK.
Load combination 2: Dead load and Wind load:
 Load ( )
tt tt
x tole xg p cause shear and bearing failure
     tt
x(tole) ( ) 0,55 0,3 0,3 0,33tt
x htN A g p daN     
 Shear strength and bearing strength of screw:
   
     
2
min
3,14 0,55
1 0,9 1500 320,57
4
0,55 0,03 3950 0,9 58,66
v b vbvb
cb bcb
N n f A daN
N d t f daN


 
     

       
   min
58.66N daN 
So      min
0,33 58,66N daN N daN  
 This screw is OK.

DESIGN CONECTION BETWEEN PURLINS AND RAFTER:
 Figure is as follow:
tt tt tt tt
q = (g + g ) + q = 2,904 + 4,86 + (-73,38) = -65,62 (daN/m)
tt tt tt
x(tole) x(xg)
q = (g + g ) = 0,3 + 0,488 = 0,788 (daN/m)x
tt tt tt tt
y y(tole) y(xg)
q = (q + g ) + p = 2,094 + 4,86 +42,69 = 49,64 (daN/m)y
tt tt tt
x(tole) x(xg)
q = (g + g ) + p = 0,3 + 0,488 + 4,29 = 5,08 (daN/m)tt
x x
 tt
xq causing shear and moment for fillet weld conected plate purlin and rafter is
neglected.
 tt
yq causing shear and bearing to bolts. Load combination 1 is more dangerous
than 2 one.
 6,5 65,62 426,53tt
yN B q daN   
 Properties of bolt 4.6 ftb = 1700 (daN/cm2
), fvb = 1500 (daN/cm2
), fcb = 3950
(daN/cm2
) - steel CCT34. Chose bolt’s diameter d = 14mm.
 Shear strength and bearing strength of bolt:
   
     
2
min
3,14 1,4
1 0,9 1500 2077,11
4
1,4 0,49 3950 0,9 2438,73
v b vbvb
cb bcb
N n f A daN
N d t f daN


 
     

       
   min
2077,11N daN 
 So      min
426,53 2077,11N daN N daN   . According to geometry
requirement, bolts are located as figure below.

CHAPTER 6. RAFTER DESIGN
DESIGN THE START SECTION OF RAFTER B1.
Section dimention.
 From internal force table, chose N, Mx and My are absolute values of lateral
force and bending moments respectively at the most unfavourable
combination thereof:
M(daN.m) -34735.6
N(daN) -4986.7
VdaN) -8443.5
 This force is at section (4) is caused by these load cases 1, 2, 3, 5, 6,7.
 Section molulus is given as follow :
334735,6 100
1837,9 ( )
0,9 2100
yc
x
c
M
W cm
f 

  

 Preliminary height:
 33
5,5 W 5,5 1837,9 67,37yc
xh cm   
 The smallest rafter height (model of calculation: pinned restraint):
 min 6
5 5 2100 1500
250 67,9
24 24 2,1 10 1,15tb
f l l
h cm
E 
 
        
Limit deflection:
1
250l
 
  
, take γtb = 1,15
 Preliminary thickness of web: assume w minh h
 max
w
3 3 8443,5
0,16
2 2 67,9 1200w v
V
t cm
h f
   


tw is too small if using this formula. tw should be calculated follow shear
buckling of plate girder web:
6
w
w
2,1 10
3,2 3,2 101,19
2100
h E
t f

   
 w
67,9
0,67
101,19
t cm   .
Chose tw = 1 (cm)
 The effective height:
 
w
W 1837,9
1,2 51,44
1
yc
x
kth k cm
t
   
k = 1,2 – coefficient factor.
Chose: h = 70 (cm) > hmin = 67,9 (cm)
Define flange dimention (bf và tf)
 Flange area:
 2W3 3 1873,9
20,08
4 4 70
yc
x
fA cm
h
   
 From local and over all buckling, formula is given below:
 
 
w
6
; 12 24
2,1 10
31,62
2100
1 1 1 1
700 140 350 ; 180
2 5 2 5
f f
f
f
f f
t t t mm
b E
t f
b h mm b mm

   

 
  

               
    
Chose tf = 1,4 (cm), bf = 35 (cm)
Checking section.
a) Properties of section.
 Section area:
2
2 2 1 67,2 2 35 1,4 165,2( )n w f w w f fA A A t h b t cm         
 Moment of inertia relative to axis x-x:
 
3 2 3 3 2 3
w w
w
4
. 35 1,4 68,6 1 67,2
2 2 . 2 35 1,4
12 4 12 12 4 12
140600,73
f f f
x x f f
b t h t h
I I I b t
cm
    
              
  

 Modulus of section:

32 2 140600,73
4017,16 ( )
70
x
x
I
W cm
h

  
 Statical moment of flange area:
 368,6
. . 1,4 35 1680,7
2 2
f
f f f
h
S t b cm    
 Statical moment of ½ area:
 
w w w
x
3
68,6 1 67,2 67,2
1,4 35
2 2 4 2 2 4
2245,18
f
f w f f
h t h h
S S S t b
cm
       
              
      

b) Checking for allowable stress:

 
 
2
2
4986,7 34735,6 100
894,87 /
165,2 4017,16
2100 0.9 1890 /
x
n x
c
N M
daN cm
A W
f daN cm



    
   
 Checking for allowable shear stress:
 2
0,9 1200 1080 /x
c v
x w
VS
f daN cm
I t
     
 2 28443,5 2245,18
134,83( / ) 1080 /
140600,73 1
x
x w
VS
daN cm daN cm
I t


   

 The start rafter section is concurrent impacted by shear and moment. Thus,
allowable stress need to be stratified this formula below:
2 2 2
1 1
3 1,15 1,15 0,9 2100 2173,5 ( / )td c
f daN cm         
Where:
2w
1
x
M h 34735,6 100 67,2
σ = = × = 830,1(daN/cm )
W h 4017,16 70

2
1
8443,5 1680,7
100,93( / )
140600,73 1
f
x w
VS
daN cm
I t


  

2 2 2 2 2 2
td 1 1σ = σ + 3τ = 830,1 + 3×100,93 = 848,3 (daN/cm ) < 2173.5 (daN/cm )
 Because of impacting of purlin so rafter need to be checked by local buckling
limit:
 2
cb
cb w z
F F
σ = = 0,9 2100 1890 /
A t l
c f daN cm   

Where:
F – impact of purlin on rafter.
c) Load combination 1: Dead load and Wind load:
tt tt tt tt
q = (q + g ) + q = 4,86 + 3,87 +(-73,38) = -64,65 (daN/m)
tt tt tt
x(tole) x(xg)
q = (q + g ) = 0,32 + 0,488 = 0,81 (daN/m)x
d) Load combination 2: Dead load and Live load:
tt tt tt tt
y y(tole) y(xg)
q = (q + g ) + p = 3,2 + 4,86 +42,69 = 50,75 (daN/m)y
tt tt tt
x(tole) x(xg)
q = (q + g ) + p = 0,32 + 0,488 + 4,29 = 5,098 (daN/m)tt
x x
 Load combination 2 cause local web buckling in rafter (neglect qx
tt
).
. 50,75 6,5 329,88( )tt
yF q B daN   
lz – is fictitious length of load distribution determined depending on condition
of leaning.
 2 8 2 1,4 10,8z fl b t cm     
   2 2329,88
30,54 / 1890 /
1 10,8
cb
w z
F
daN cm daN cm
t l
    

 Checking for normal stress, shear stress and local buckling of web.
2 2 2 2
1 1 1
- 3 1,15 1,15 0,9 2100 2173,5 ( / )td cb cb c
f daN cm             
 
2
1
2
1
2
830,1( / )
100,93( / )
30,54 /cb
daN cm
daN cm
daN cm



 




 2 2 2 2
2
830,1 30,54 -830,1 30,54 3 100,93 833,79 /
< 2173,5 ( / )
td
td
daN cm
daN cm


     
e) Checking for over all buckling:
 Top flange connect to purlins, with purlins spacing is 1,1 (m).
 So:
0 1100
3,143
350f
l
b
 
 Maximum value of 0
f
l
b
:

6
0,41 0,0032 0,73 0,016
35 35 35 2,1 10
0,41 0,0032 0,73 0,016 20,82
1,4 1,4 68,6 2100
f f fo
f f f fk
b b bl E
b t t h f
    
               
   
         
  
o o
f f
l l
b b
 
   
  
so non checking.
f) Checking for local buckling of flange and web:
 Flange:
6
35 1 2,1 10
12,14 0,5 0,5 15,81
2 1,4 2100
of
f
b E
t f
 
     

 Web:
Slender ratio of web
w
w 6
w
67,2 2100
2,125
1 2,1 10
h f
t E
    

w w2,125 3.2      
web without stiffeners is not buckling by shear.
 Acctually, there is local buckling occur in top flange of rafter but it is doesn’t
mean so it is necglected. According to section 5.6.1.3 TCVN 5575 – 2012,
local buckling web do not need to check.
g) Calculating fillet weld connected between flange and web:
 As rafter bending, flange slip on web and fillet weld is created to prevent
sliping.
 Using shielded metal arc welding, electrode N46: βf = 0.7, βs = 1, fwf = 2000
daN/cm2
; fws = 0,45fu = 0,45 × 3400 = 1530 daN/cm2
.
 So:
 
 
   
2
w
2
ws
2
w wmin
. 0,7 2000 1400 /
. 1 1530 1530 /
. 1400 /
f f
s
f f
f daN cm
f daN cm
f f daN cm


 
   

  
  
 The height of the fillet weld:

 
 
w min
.S 8443,5 1680,7
0,04
2 . . 2 1400 140600,73 0,9
f
f
x c
V
h cm
f I 

  
  
 The height of the fillet weld follow above formula is too small so it should be
calculated by following formula:
 
 
min
min
1.2 1,2 10 12
5
f
f f
h t mm
h h mm
   

 
 Chosse hf = 6 (mm) along longitudinal rafter.
DESIGN SECTION AT THE END OF D1 – BEGIN OF D2.
Section dimension.
M(daN.m) -8023,6
N(daN) -3083,0
V(daN) -3672,2
 This force is at section (5) is caused by these load cases 1, 2, 4, 6, 8, 11.
 Section molulus is given as follow:
38023,6 100
424,53( )
0,9 2100
yc
x
c
M
W cm
f 

  

 Section varies by the height so other values should be constant:
tw = 1 (cm)
tf = 1,4 (cm)
bf = 35 (cm)
 
w
W 424,53
1,2 24,72
1
yc
x
kth k cm
t
   
Chose h = 40 (cm)
 Check tw:
   max
w
6
w
w
3 3 3672,2
1 0,12
2 2 37,2 1200
37,2 2,1 10
37,2 3,2 3,2 101,19
1 2100
w v
V
t cm cm
h f
h E
t f

     

      



Checking section.
 Section area:
2
2 1 37,2 2 35 1,4 135,2( )w f w w f fA A A t h b t cm         
 
3 2 3 3 2 3
w w
w
4
. 35 1,4 38,6 1 37,2
2 2 . 2 35 1,4
12 4 12 12 4 12
40809,93
f f f
x x f f
b t h t h
I I I b t
cm
    
              
  

32 2 40809,93
2040,50 ( )
40
x
x
I
W cm
h

  
 338,6
. . 1,4 35 945,7
2 2
f
f f f
h
S t b cm    
 
w w w
x
3
38,6 1 37,2 37,2
1,4 35
2 2 4 2 2 4
1118,68
f
f x f f
h t h h
S S S t b
cm
       
              
      

b) Checking for allowable stress.

 
 
2
2
3083 8023,6 100
416,02 /
135,2 2040,50
2100 0,9 1890 /
x
n x
c
N M
daN cm
A W
f daN cm



    
   
 2
0,9 1200 1080 /x
c v
x w
VS
f daN cm
I t
     
 2 22672,2 1118,68
73,25( / ) 1080 /
40809,93 1
x
x w
VS
daN cm daN cm
I t


   


2 2 2
1 1
3 1,15 1,15 0,9 2100 2173,5 ( / )td c
f daN cm         
Where:
2w
1
x
M h 8023,6 100 37,2
σ = = × = 365,69 (daN/cm )
W h 2040,50 40

2
1
3672,2 945,7
85,1( / )
40809,93 1
f
x w
VS
daN cm
I t


  

2 2 2 2 2 2
td 1 1
σ = σ + 3τ = 365,69 +3×85,1 = 394,28 (daN/cm ) < 2173,5 (daN/cm )
limit:
 2
cb
cb w z
F F
σ = = 0,9 2100 1890 /
A t l
c
f daN cm   
Where:
c) Load combination 1: Dead load and Wind load:
tt tt tt tt
q = (q + g ) + q = 4,86 + 3,87 +(-73,38) = -64,65 (daN/m)
tt tt tt
x(tole) x(xg)
q = (q + g ) = 0,32 + 0,488 = 0,81 (daN/m)x
d) Load combination 2: Dead load and Live load:
tt tt tt tt
y y(tole) y(xg)
q = (q + g ) + p = 3,2 + 4,86 +42,69 = 50,75 (daN/m)y
tt tt tt
x(tole) x(xg)
q = (q + g ) + p = 0,32 + 0,488 + 4,29 = 5,098 (daN/m)tt
x x
tt
).
. 50,75 6,5 329,88( )tt
of leaning.
 2 8 2 1,4 10,8z fl b t cm     
   2 2329,88
30,54 / 1890 /
1 10,8
cb
w z
F
daN cm daN cm
t l
    


2 2 2 2
1 1 1- 3 1,15 1,15 0,9 2100 2173,5 ( / )td cb cb cf daN cm             
 
2
1
2
1
2
365,69( / )
85,1( / )
30,54 /cb
daN cm
daN cm
daN cm



 




 2 2 2 2
2
365,69 30,54 -365,69 30,54 3 85,1 381,08 /
< 2173,5 ( / )
td
td
daN cm
daN cm


     
e) Checking for over all buckling:
 So:
0 1100
3,143
350f
l
b
 
f
l
b
:
6
0,41 0,0032 0,73 0,016
35 35 35 2,1 10
0,41 0,0032 0,73 0,016 24,96
1,4 1,4 38,6 2100
f f fo
f f f fk
b b bl E
b t t h f
    
               
   
         
  
o o
f f
l l
b b
 
   
  
so non checking.
f) Checking for local buckling of flange and web:
 Flange:
6
17 2,1 10
12,14 0,5 0,5 15,81
1,4 2100
of
f
b E
t f

     
 Web:
w
w 6
w
37,2 2100
1,18
1 2,1 10
h f
t E
    


w w1,18 3,2      
web without stiffeners is not buckling by shear.
g) Calculating fillet weld connected between flange and web:
sliping.
daN/cm2
; fws = 0,45fu = 0,45 × 3400 = 1530 daN/cm2
.
 So:
 
 
   
2
w
2
ws
2
w wmin
. 0,7 2000 1400 /
. 1 1530 1530 /
. 1400 /
f f
s
f f
f daN cm
f daN cm
f f daN cm


 
   

  
  
 
 
w min
.S 3672,2 945,7
0,034
2 . . 2 1400 40809,93 0,9
f
f
x c
V
h cm
f I 

  
  
 
 
min
min
1.2 1,2 10 12
5
f
f f
h t mm
h h mm
   

 
DESIGN SECTION AT THE END OF D2 – SECTION (6).
Section dimension.
M(daN.m) 24311.4
N(daN) -3838.0
V(daN) 718.6
 This force is at section (6) is caused by these load cases 1, 2, 3, 6.

 Section molulus is given as follow:
324311,4 100
1286,32( )
0,9 2100
yc
x
c
M
W cm
f 

  

 Section varies by the height so other values should be constant:
tw = 1 (cm)
tf = 1,4 (cm)
bf = 35 (cm)
 
w
W 1286,32
1,2 43,04
1
yc
x
kth k cm
t
   
 Chose h = 50 (cm)
 Check tw:
   max
w
6
w
w
3 3 718,6
1 0,02
2 2 47,2 1200
47,2 2,1 10
47,2 3,2 3,2 101,19
1 2100
w v
V
t cm cm
h f
h E
t f

     

      


Checking section.
 Section area:
2
2 1 47,2 2 35 1,4 145,2( )w f w w f fA A A t h b t cm         
 
3 2 3 3 2 3
w w
w
4
. 35 1,4 48,6 1 47,2
2 2 . 2 35 1,4
12 4 12 12 4 12
66646,86
f f f
x x f f
b t h t h
I I I b t
cm
    
              
  

32 2 66646,86
2665,87 ( )
50
x
x
I
W cm
h

  
 348,6
. . 1,4 35 1190,70
2 2
f
f f f
h
S t b cm    

 
w w w
x
3
48,6 1 47,2 47,2
1,4 35
2 2 4 2 2 4
1469,18
f
f x f f
h t h h
S S S t b
cm
       
              
     

b) Checking for allowable stress.

 
 
2
2
3838 24311,4 100
938,4 /
145,2 2665,87
2100 0,9 1890 /
x
n x
c
N M
daN cm
A W
f daN cm



    
   
 2
0,9 1200 1080 /x
c v
x w
VS
f daN cm
I t
     
 2 2718,6 469,18
5,06( / ) 1080 /
66646,86 1
x
x w
VS
daN cm daN cm
I t


   

2 2 2
1 1
3 1,15 1,15 0,9 2100 2173,5 ( / )td c
f daN cm         
Where:
2w
1
x
hM 24311,4 100 47,2
σ = = × = 860,88 (daN/cm )
W h 2665,87 50

2
1
718,6 1190,7
12,84( / )
66646,86 1
f
x w
VS
daN cm
I t


  

2 2 2 2
td 1 1
2
σ = σ + 3τ = 860,88 +3×12,84 = 1547,37 (daN/cm )
< 2173,5 (daN/cm )

limit:
 2
cb
cb w z
F F
σ = = 0,9 2100 1890 /
A t l
c f daN cm   

Where:
c) Load combination 1: Dead load and Wind load
tt tt tt tt
q = (q + g ) + q = 4,86 + 3,87 +(-73,38) = -64,65 (daN/m)
tt tt tt
x(tole) x(xg)
q = (q + g ) = 0,32 + 0,488 = 0,81 (daN/m)x
d) Load combination 2: Dead load and Live load
tt tt tt tt
y y(tole) y(xg)
q = (q + g ) + p = 3,2 + 4,86 +42,69 = 50,75 (daN/m)y
tt tt tt
x(tole) x(xg)
q = (q + g ) + p = 0,32 + 0,488 + 4,29 = 5,098 (daN/m)tt
x x
tt
).
. 50,75 6,5 329,88( )tt
of leaning.
 2 8 2 1,4 10,8z fl b t cm     
   2 2329,88
30,54 / 1890 /
1 10,8
cb
w z
F
daN cm daN cm
t l
    

2 2 2 2
1 1 1
- 3 1,15 1,15 0,9 2100 2173,5 ( / )td cb cb c
f daN cm             
 
2
1
2
1
2
860,88( / )
12,84( / )
30,54 /cb
daN cm
daN cm
daN cm



 




 2 2 2 2
2
860,88 30,54 -860,88 30,54 3 12,84 846,32 /
< 2173,5 ( / )
td
td
daN cm
daN cm


     
e) Checking for over all buckling.
 So:
0 1100
3,143
350f
l
b
 

f
l
b
:
6
0,41 0,0032 0,73 0,016
35 35 35 2,1 10
0,41 0,0032 0,73 0,016 23,01
1,4 1,4 48,6 2100
f f fo
f f f fk
b b bl E
b t t h f
    
               
   
         
  
o o
f f
l l
b b
 
   
  
so non checking.
f) Checking for local buckling of flange and web.
 Flange:
6
17 2,1 10
12,14 0,5 0,5 15,81
1,4 2100
of
f
b E
t f

     
 Web:
w
w 6
w
47,2 2100
1,49
1 2,1 10
h f
t E
    

w w1,49 3,2       web without stiffeners is not buckling by shear.
g) Calculating fillet weld connection:
sliping.
daN/cm2
; fws = 0,45fu = 0,45 × 3400 = 1530 daN/cm2
.
 So:
 
 
   
2
w
2
ws
2
w wmin
. 0,7 2000 1400 /
. 1 1530 1530 /
. 1400 /
f f
s
f f
f daN cm
f daN cm
f f daN cm


 
   

  
  

 
 
w min
.S 718,6 1190,7
0,0051
2 . . 2 1400 66646,86 0,9
f
f
x c
V
h cm
f I 

  
  
 
 
min
min
1,2 1,2 10 12
5
f
f f
h t mm
h h mm
    

 

8 9 10
M (daN.m) -53305.25 11735.427 26936.689
N (daN) -5539.99 2903.313 -5113.026
V (daN) -9779.56 1558.077 606.26
Pcb (daN) 329.88 329.88 329.88
h (cm) 80 40 50
hw (cm) 77.2 37.2 47.2
tw (cm) 1 1 1
tf (cm) 1.4 1.4 1.4
bf (cm) 35 35 35
hfk (cm) 78.6 38.6 48.6
bof (cm) 17 17 17
A (cm2
) 175.2 135.2 145.2
Ix (cm4
) 189717.66 40809.93 66646.86
Wx (cm3
) 4742.94 2040.50 2665.87
Sf (cm3
) 1925.70 945.70 1190.70
Sx (cm3
) 2670.68 1118.68 1469.18
σxmax (daN/cm2
) 1155.51 596.60 1045.64
γcf = 1890 (daN/cm2
) OK OK OK
Τmax (daN/cm2
) 137.67 42.71 13.36
γcfv = 1080(daN/cm2
) OK OK OK
σ1 (daN/cm2
) 1084.55 534.87 953.84
τ1 (daN/cm2
) 99.27 36.11 10.83
σtd (σ1, τ1) (daN/cm2
) 1098.09 538.51 954.03
1.15γcf = 2173.5 (daN/cm2
) OK OK OK
σcb (daN/cm2
) 42.29 42.29 42.29
γcf = 1890 (daN/cm2
) OK OK OK
σtd (σ1, τ1, σcb) (daN/cm2
) 1077.84 518.81 933.60
1.15γcf = 2173.5 (daN/cm2
) OK OK OK
3.14 3.14 3.14
19.24 23.85 21.99
Condition OK OK OK
12.14 12.14 12.14
Condition OK OK OK
2.55 1.23 1.56
3.2 3.2 3.2
Non checking Non checking Non checking
hf (design) (mm) 0.39 0.14 0.04
hf (preliminary) (mm) 5 ≤ hf ≤ 12 5 ≤ hf ≤ 12 5 ≤ hf ≤ 12
hf (optimize) (mm) 6 6 6
CHECK FOR FLANGE
STABIITY
CHECK FOR WEB
STABILITY
CONECTION BETWEEN
FLANGE AND WEB
SECTION
INTERIAL FORCE
DIMENSION
SECTION PROPERTIES
CHECK FOR ULTIMATE
STRESS
CHECK FOR OVER ALL
STABILITY
o
f
l
b
o
f
l
b
 
 
  
of
f
b
t
w
w
w
h f
t E
 
w 
 
DESIGN SECTION OF RAFTER D5 AND D6.
 The design process is the same to D1 and D2 rafter, the table below:

CHAPTER 7. COLUMN DESIGN
DESIGN STRAIGHT COLUMN C1.
Design length:
 Length H = 14,2 m
 Side span L1 = L3 = 30m.
a) In the plane frame:
 Factors of design length  for columns with constant section in the frame
plane at stiff fixing of collar beams to columns shall be determined by formula
of Table 17a according to SNiP II – 23 – 81.
 Factor  for extreme column of multi – span frame shall be determined as for
single – span frame column.
 Formulars for determining factor  :
b
c 1
I H 14,2
n = 0,29 0,14
I L 30
  
Where: Ib – the smallest moment of inertia of rafter.
Ic – moment of inertia of column.
 The factor of design lengh is given below:
n + 0,56 0,14 0,56
μ = = 1,6
n + 0,14 0,14 0,14



 Design length ler of columns (posts) with constant section or of separate
portions of stepped columns shall be determined by formula:
 . 1,6 14,2 22,72xl H m   
b) Out of the frame plane:
 Design lengths of columns in the direction along the buildings length (out of
frame planes) shall be adopted equal to distances between points fixed
fromshifting out of the frame plane (column supports, crane and eaves girders;
joints of braces' and collar beams' fixing, etc.). Design lengths may be
determined onthe basis of design schemes which consider actual type of
columns' fixing.
 Setting wide – flange steel bracing system at level code +6,35 m.
 Design length: yl = 7(m)

Section dimension.
combination thereof |N|max:
M(daN.m) -22677.0
N(daN) -14001.2
VdaN) -4036.1
 This force is at base column section is caused by these load cases 1, 2, 3, 6, 7.
 The height of column section:
1 1 1 1
14200 (568 947)mm
15 25 15 25
h H
   
         
   
 Chose h = 700 mm
 The width of flange of column section is chosen follow geometry:
 
     
350
1 1 1 1
7000 350 250
20 30 20 30
0,3 0,5 0,3 0,5 700 210 350
f
f y
f
b mm
b l mm
b h mm
 

    
          
   
       
 Chose bf = 350 mm.
 According to formula of Iasinky x
c
x
N M
+ fγ
φA W
 , area section requirement
shall be conducted as follow:
x x
yc
c x c x
M A MN 1 N 1
A = + = +
fγ φ NW fγ φ ρ N
   
   
   
Preliminary φ = 0,8 and xρ = (0,35 ÷ 0,45)h :
 
 2
14001,2 22677 100
1,25 2,2 2,8 1,25 (2,2 2,8)
2100 0,9 70 14001,2
46,97 57,25
x
yc
c
MN
A
f h N
cm

   
            
 
 The thickness of flange and web column is chosen follow geometry:
 
1 1 1 1 2100
= 35 1,25 1 ( )
28 35 2100 28 35 2100
f f
f
t b cm
   
         
   
 Chose tf = 1,4 (cm)
 
  w
1 1 1 1
70 0,58 1,17
120 60 120 60
8
w
f
t h cm
cm t t
    
         
   
  

 Chose tw = 1 (cm)
Checking section.
 Section area:
2
2 1 67,2 2 35 1,4 165,2 ( )w w f fA t h b t cm       
 Moment of inertia relative to axis x-x, y-y:
 
3 2 3 3 2 3
w w
w
4
. 35 1,4 68,6 1 67,2
2 2 . 2 35 1,4
12 4 12 12 4 12
140600,73
f f f
x x f f
b t h t h
I I I b t
cm
    
              
  

33 3 3
467,2 1 35
2 2 1,4 10009,77 ( )
12 12 12 12
fw w
y f
bh t
I t cm

      
 Inertia radius of cross section relative to axes x-x, y-y:
x
x
I 140600,73
i = = = 29,17 (cm) ;
A 165,2
y
y
I 10009,77
i = = = 7,78 (cm)
A 165,2
3x
x
2I 2×140600,73
W = = = 4017,16(cm )
h 70
 Design flexibility in planes perpendicular to axes x-x and y-y:
x
x
x
l 22,72 100
λ = = = 77,89
i 29,17

x x 6
f 2100
λ = λ =77,89× = 2,46
E 2,1×10
y
y
y
l 7,5×100
λ = = = 96,4;
i 7,78
y y 6
f 2100
λ = λ = 96,4 × = 3,05
E 2,1×10
 max yλ = λ = 96,4 < λ = 165
 Design flexibility [λ] of column is satisfied.
 Relative eccentricity m and reduced relative eccentricity me:

x
x
e A 22677 100 165,2
m = = × = 6,7
W 14001,2 4017,16
M
N


According to Table 73 SNiP II-23-81, with f
w
A 2×1,4×35
= = 1,46 > 1
A 1×67,2
;
x0 λ = 2,46 < 5 and 5 < m = 6,7 < 20, factor of influence of cross section form
η is given as:
1,4 0,2 1,4 0,2 2,46 0,91x      
em = ηm = 0,91×6,7 =6,1
Coefficient e at reduced relative eccentricity 6,1em  and fictitious flexibility
2,46x  : 0,178e 
 Ultimate flexibility compression components:
  180 60 180 60 0,25 165      
where
14001,2
0,25
0,178 165,2 2100 0,9e c
N
Af

 
  
  
b) Checking for strength of eccentric compression.
 According to Section 5.24 SNiP II-23-81: The strength analysis of
eccentrically compressed and compressed-and-bent components according to
formula (49) is not necessary when the value of reduced eccentricity
6,1 20em   , the section weakening does not occur and the values of bending
moments used for strength and stability are equal.
c) Checking for stability in the plane of moment action.
 The stability analysis of eccentric compression and compression-and-bending
components with constant section (with regard to requirements of Sections
5.28 and 5.33 of this Code) in the plane of the moment's action which
coincides with the symmetry plane, shall be conducted by formula:
x c
e
N
σ = fγ
φ A

Factor eφ in formula (51) shall be determined.
 The internal force are calculating is at base section of column Mc = -22677
(daN.m), causing by load case 1, 2, 3, 6, 7.
 2 2
x c
e
N 14001,2
σ = = = 476,14 (daN/cm ) < γ f = 0,9 2100 =1890 daN/cm
φ A 0,178×165,2


 This column is OK.
d) Checking for stability out of the moment action.
 According to Section 5.30 SNiP II-23-81, The stability analysis of eccentric
compression components with constant section out of the moment action plane
at bending thereof in the plane of maximum stiffness (Jx > Jy ) coinciding with
the symmetry plane shall be executed by formula:
y c
y
N
σ = fγ
cφ A

Where:
 yφ is factor calculated according to requirements of Section 5.3 on this Code,
when 2,5 3,05 4,5y   then yφ is given below:
2
2
1,47 13 (0,371 27,3 ) (0,0275 5,53 )
1,47 13 0,001 (0,371 27,3 0,001) 3,05 (0,0275 5,53 0,001) 3,05 0,61
y y y
f f f
E E E
       
           
 c is factor calculated as required by Section 5.31.
 When determining the relative eccentricity mx it is necessary to adopt as the
design moment Mx:
The maximum moment within the middle third of bar length (but not less than
the haft of the maximum moment along the bar length) for bars with hinged
bearing ends which are prevented from shifting perpendiculally to the plane of
moment’s action.
 The internal force are calculating is at base section of column Mc = -22677
(daN.m), causing by load case 1, 2, 3, 6, 7, moment value at head column is
Md = 34635,9 (daN.m).

So:
 1/3
34635,9 22677
max ; ; max 15531,1; ; 17317,9 daN.m
2 2 2 2
d c
x
M M
M M
   
    
  
 Relative eccentricity mx is defined by adopted moment Mx:
x
x
e A 17317,9 100 165,2
m = = × = 5,09
W 14001,2 4017,16
x
x
M
N


 For relative eccentricity 5 < mx = 5,09 < 10, by formula:
   5 102 0,2 0,2 1x xc c m c m   
 c5 is determined at mx = 5:
5
1 x
c
m




α and β are factors adopted according to Table 10:
When mx = 5 the factor 0,65 0,05 0,65 0,05 5 0,9xm      
When
6
2,1 10
96,4 3,14 3,14 99,3
2100
y c
E
f
 

      the factor β = 1.
So 5
1
0,181
1 1 0,9 5x
c
m


  
  
 c10 is determined at mx = 10:
10
1
1 x y
b
c
m 



0,61y  (as calculated)
bφ is factor determined as required by Section 5.15 and Appendix 7 as for a
beam with two or more fixings of compression chord; bφ = 1.0 for closed
sections.
For welded double-T sections composed of three sheets as well as for double-
T sections with chord joints on high strength bolts:
2 23 3
o f w
3
fk f f f
l t at 700 1,4 68,6 1
α = 8 1 + = 8 1 1,81
h b b t 68,6 35 2 35 1,4
       
                 
According to Table 77 formulas for  at 0.1 1,81 40   :
2,25 0,07 2,25 0,07 1,81 2,4      
1φ is given as follow:
2 2 6
1
10009,77 70 2,1 10
2,4 1,49
140600,73 700 2100
y
x o
I h E
I l f
 
   
          

φ1 > 0,85
1
1
0,68 0,21 0,68 0,21 1,49 0,99
b
b

 

 
      
Take φb = 0,99.
10
1 1
0,14
10 0,61
11
0,99
x y
b
c
m 

   


       5 102 0.2 0,2 1 0,181 2 0,2 5,09 0,14 0,2 5,09 1 0,18x xc c m c m            
So:
   2 2
y c
y
N 14001,2
σ = = = 771,89 / γ f = 0,9 2100 =1890 daN/cm
cφ A 0,18 0,61 165,2
daN cm  
 
e) Checking for stability of flange and web.
 For flange:
 When analyzing centric and eccentric compression and compression - and
bending components with the fictitious flexibility  equal to 0.8 to 4, the ratio
of design width of an overhang of chord sheets (flanges) be to thickness t shall
not exceed values determined by formulas of Table 29*:
   
6
2,1 10
0,36 0,1 0,36 0,1 2,46 19,2
2100
ob E
t f

 
        
 Ratio:
0,5 (35 1)
12,14
1,4
o
f
b
t
 
 
 So:
12,14 19,2o o
f
b b
t t
 
    
 For web:
 Checking by: w w
w w
h h
t t
 
  
 
.
 In checking for stability:
   2 2
476,14 / 771,89 /x ydaN cm daN cm    so the strength of this column
is mostly depend on moment’s action in the plane. Thus the slenderness ratio
[hw/tw] is given by Section 7.16 SNiP II-23-81 Code.

14001,2 22677,1 100 67,2
550,4
2 165,2 140600,73 2
x w
x
M hN
A I


      
1
14001,2 22677,1 100 67,2
533,44
2 165,2 140600,73 2
x w
x
M hN
A I


       
 
1
2 2
550,4 533,44 (2 1)
1,97 1 4,35
550,4 2 4
w
w
h E
t
  

    
   
      
     
  
4036,1
1,4 (2 1) 1,4 (2 1) 1,4 (2 1,68 1) 0,36
67,2 1 550,4w w
V
h t

  
 
             
   
   2 22 2
(2 1) (2 1,68 1) 2100000
4,35 4,35 281,66
550,4 2 1,68 1,68 4 0,362 4
2100000
3,8 3,8 120,17
2100
120.17
w
w
w
w
w
w
h E
t
h E
t f
h
t

   
     
    
             
 
    
 
 
  
 
 Check:
w w
w w
h h67,2
= = 67,2 < = 120,17
t 1 t
 
 
 
 In addition:
6
w
w
h 67,2 E 2,1×10
= = 67,2 < 2,3 = 2,3× = 72,73
t 1 f 2100
 Web of this column need not be strengthend by lateral stiffening ribs.
Checking column at |M|max.
combination thereof |M|max:

M(daN.m) 34735.6
N(daN) -8897.8
VdaN) -4121.8
 This force is at top of column section is caused by these load cases 1, 2, 3, 5,
6, 7.
 Using dimension of calculated section.
But:
x
x
e A 34735,6 100 165,2
m = = × = 16,05
W 8897,8 4017,16
M
N


w
A 2×1,4×35
= = 1,46 > 1
A 1×67,2
;
η is given as:
1,4 0,2 1,4 0,2 2,64 0,87x      
em = ηm = 0,87×16,05 = 13,96
Coefficient e at reduced relative eccentricity 13.96em  and fictitious
flexibility 2,64x  : 0,098e 
  180 60 180 60 0,25 162.6      
where
8897.8
0,29
0,098 165,2 2100 0,9e c
N
Af

 
  
  
 max yλ = λ = 96.4< λ = 162.6
a) Checking for strength of eccentric compression.
13.96 20em   , the section weakening does not occur and the values of
bending moments used for strength and stability are equal.
b) Checking for stability in the plane of moment action.
5.28 and 5.33 of this Code) in the plane of the moment's action which
coincides with the symmetry plane, shall be conducted by formula:

x c
e
N
σ = fγ
φ A

Factor eφ in formula (51) shall be determined.
 The internal force are calculating is at top section of column Md = 34735,6
(daN.m), causing by load case 1, 2, 3, 5, 6, 7.
 2 2
x c
e
N 8897,8
σ = = = 549,6 (daN/cm ) < γ f = 0,9 2100 =1890 daN/cm
φ A 0,098×165,2

 This section is OK.
c) Checking for stability out of the moment action.
 According to Section 5.30 SNiP II-23-81, The stability analysis of eccentric
compression components with constant section out of the moment action plane
at bending thereof in the plane of maximum stiffness (Jx > Jy ) coinciding with
the symmetry plane shall be executed by formula:
y c
y
N
σ = fγ
cφ A

Where:
 yφ is factor calculated according to requirements of Section 5.3 on this Code,
when 2,5 3,05 4,5y   then yφ is given below:
2
2
1,47 13 (0,371 27,3 ) (0,0275 5,53 )
1,47 13 0,001 (0,371 27,3 0,001) 3.05 (0,0275 5,53 0,001) 3,05 0,61
y y y
f f f
E E E
       
           
 c is factor calculated as required by Section 5.31.
 When determining the relative eccentricity mx it is necessary to adopt as the
design moment Mx:
The maximum moment within the middle third of bar length (but not less than
the haft of the maximum moment along the bar length) for bars with hinged
bearing ends which are prevented from shifting perpendiculally to the plane of
moment’s action.
 The internal force are calculating is at top section of column Md = 34735,6
(daN.m) , causing by load case 1, 2, 3, 6, 7, 9, 13, moment value at base of
column is Mc = 29613,1 (daN.m).

So:
 1/3
34735,6 29613,1
max ; ; max 33028,1; ; 33028,1 daN.m
2 2 2 2
d c
x
M M
M M
   
    
  
 Relative eccentricity mx is defined by adopted moment Mx:
x
x
e A 34735,1 100 165,2
m = = × =16,05
W 8897,8 4017,16
x
x
M
N


 For relative eccentricity mx = 16.05 > 10, by formula:
1
1 x y
b
c
m 



Where:
bφ is factor determined as required by Section 5.15 and Appendix 7 as for a
beam with two or more fixings of compression chord; bφ = 1.0 for closed
sections.
For welded double-T sections composed of three sheets as well as for double-
T sections with chord joints on high strength bolts:
2 23 3
o f w
3
fk f f f
l t at 700 1,4 68,6 1
α = 8 1 + = 8 1 1,81
h b b t 68,6 35 2 35 1,4
       
                 
According to Table 77 formulas for  at 0.1 1,81 40   :
2,25 0,07 2,25 0,07 1,81 2,4      
1φ is given as follow:

2 2 6
1
10009,77 70 2,1 10
2,4 1,49
140600,73 700 2100
y
x o
I h E
I l f
 
   
          
φ1 > 0,85
1
1
0,68 0,21 0,68 0,21 1,49 0,99
b
b

 

 
      
Take φb = 0,99.
So:
1 1
0 09
16 05 0 61
11
0 99
.
. .
.
x y
b
c
m 

  


So:
   2 2
y c
y
N 8897,8
σ = = = 981,07 / γ f = 0,9 2100 =1890 daN/cm
cφ A 0,09 0,61 165,2
daN cm  
 
d) Checking for stability of flange and web.
 For flange:
 When analyzing centric and eccentric compression and compression - and
bending components with the fictitious flexibility  equal to 0.8 to 4, the ratio
of design width of an overhang of chord sheets (flanges) be to thickness t shall
not exceed values determined by formulas of Table 29*:
   
6
2,1 10
0,36 0,1 0,36 0,1 2,46 19,2
2100
ob E
t f

 
        
 Ratio:
0,5 (35 1)
12,14
1,4
o
f
b
t
 
 
 So:
12,14 19,2o o
f
b b
t t
 
    
 For web:
 Checking by: w w
w w
h h
t t
 
  
 
 In checking for stability:

   2 2
549,6 / 981,07 /x ydaN cm daN cm    so the strength of this column
is mostly depend on moment’s action in the plane. Thus the slenderness ratio
[hw/tw] is given by Section 7.16 SNiP II-23-81 Code:
8897,8 34735,6 100 67,2
883,95
2 165,2 140600,73 2
x w
x
M hN
A I


      
1
8897,8 34735,6 100 67,2
776,23
2 165,2 140600,73 2
x w
x
M hN
A I


       
 
1
2 2
883,95 776,23 (2 1)
1,88 1 4,35
883,95 2 4
w
w
h E
t
  

    
   
      
     
  
4121,8
1,4 (2 1) 1,4 (2 1) 1,4 (2 1,88 1) 0,27
67,2 1 883,95w w
V
h t

  
 
             
   
   2 22 2
(2 1) (2 1,88 1) 2100000
4,35 4,35 77,3
883,95 2 1,88 1,88 4 0,272 4
2100000
3,8 3,8 120,17
2100
77,3
w
w
w
w
w
w
h E
t
h E
t f
h
t

   
     
    
             
 
    
 
 
  
 
 Check:
w w
w w
h h67,2
= = 67,2 < = 77,3
t 1 t
 
 
 
 In addition:
6
w
w
h 67,2 E 2,1×10
= = 67,2 < 2,3 = 2,3× = 72,73
t 1 f 2100
 Web of this column need not be strengthend by lateral stiffening ribs.
DESIGN STRAIGHT COLUMN C2-C3.
Design length.
 Length H = 14,2 m

 Side span L1 = L3 = 30 (m), centre span L2 = 33 (m)
a) In the plane frame:
 Factors of design length  for columns with constant section in the frame
plane at stiff fixing of collar beams to columns shall be determined by formula
of Table 17a according to SNiP II – 23 – 81.
 Factor  for extreme column of multi – span frame shall be determined as for
single – span frame column.
 Formulars for determining factor  :
1 2( )
1
k n n
n
k



Where:
 k – the number of span, k = 3.
 Side span L1 = 30 (m)
b
1
c 1
I H 15
n = 0,29 0,15
I L 30
  
 Centre span L2 = 33 (m)
b
2
c 2
I H 15
n = 0,29 0,13
I L 33
  
 Ib – the smallest moment of inertia of rafter.
 Ic – moment of inertia of column.
 1 2
2 0,15 0,13( )
0,21
1 3 1
k n n
n
k
 
   
 
 The factor of design lengh is given below:
n + 0,56 0,21 0,56
μ = = 1,49
n + 0,14 0,21 0,14



 Design length ler of columns (posts) with constant section or of separate
portions of stepped columns shall be determined by formula:
 . 1,49 14,2 21,16xl H m   
b) Out of the frame plane:
 Design lengths of columns in the direction along the buildings length (out of
frame planes) shall be adopted equal to distances between points fixed
fromshifting out of the frame plane (column supports, crane and eaves girders;
joints of braces' and collar beams' fixing, etc.). Design lengths may be

determined onthe basis of design schemes which consider actual type of
columns' fixing.
 Setting wide – flange steel bracing system at level code +6,35 m.
 Design length: yl = 7(m)
Section dimension.
combination thereof |N|max:
M(daN.m) -4953.8
N(daN) -40638.8
VdaN) -1232.8
 This force is at base column section is caused by these load cases trọng 1, 2, 3,
4, 5, 8, 11.
 The height of column section:
1 1 1 1
14200 (947 568)mm
15 25 15 25
h H
   
         
   
 Chose h = 700 mm
 The width of flange of column section is chosen follow geometry:
 
     
350
1 1 1 1
7000 350 230
20 30 20 30
0.3 0.5 0,3 0,5 700 210 350
f
f y
f
b mm
b l mm
b h mm
 

    
          
   
       
 Chose bf = 350 mm.
 According to formula of Iasinky x
c
x
N M
+ fγ
φA W
 , area section requirement
shall be conducted as follow:
x x
yc
c x c x
M A MN 1 N 1
A = + = +
fγ φ NW fγ φ ρ N
   
   
   
Preliminary φ = 0,8 and xρ = (0,35 ÷ 0,45)h :
 
 2
40638,8 4953,8 100
1,25 2,2 2,8 1,25 (2,2 2,8)
2100 0,9 70 40638,8
35,11 37,36
x
yc
c
MN
A
f h N
cm

   
            
 
 The thickness of flange and web column is chosen follow geometry:

 
1 1 1 1 2100
= 35 1,25 1 ( )
28 35 2100 28 35 2100
f f
f
t b cm
   
         
   
 Chose tf = 1.4 (cm)
 
  w
1 1 1 1
70 0,58 1,17
120 60 120 60
8
w
f
t h cm
cm t t
    
         
   
  
 Chose tw = 1 (cm)
Checking section.
 Section area:
2
2 1 67.2 2 35 1.4 165.2 ( )w w f fA t h b t cm       
 Moment of inertia relative to axis x-x, y-y:
 
3 2 3 3 2 3
w w
w
4
. 35 1.4 68.6 1 67.2
2 2 . 2 35 1.4
12 4 12 12 4 12
140600.73
f f f
x x f f
b t h t h
I I I b t
cm
    
              
  

33 3 3
467.2 1 35
2 2 1.4 10009.17 ( )
12 12 12 12
fw w
y f
bh t
I t cm

      
 Inertia radius of cross section relative to axes x-x, y-y:
x
x
I 140600.73
i = = = 29.17 (cm) ;
A 165.2
y
y
I 10009.77
i = = = 7.78 (cm)
A 165.2
3x
x
2I 2×140600.73
W = = = 4017.16(cm )
h 70
 Design flexibility in planes perpendicular to axes x-x and y-y:
x
x
x
l 21.16 100
λ = = = 72.54
i 29.17


x x 6
f 2100
λ = λ =72.54× =2.29
E 2.1×10
y
y
y
l 7×100
λ = = = 89.97 ;
i 7.78
y y 6
f 2100
λ = λ = 89.97× = 2.85
E 2.1×10
 max yλ = λ = 88.97< λ = 165.54
 Design flexibility [λ] of column is satisfied
x
x
e A 4953.8 100 165.2
m = = × = 0.5
W 40638.8 4017.16
M
N


w
A 2×1,4×35
= = 1,46 > 1
A 1×67,2
;
η is given as:
     1.90 0.1 0.02 6 1.90 0.1 0.5 0.02 (6 0.5) 2.29 1.59xm m            
em = ηm = 1.59 0.5=0.795 
Coefficient e at reduced relative eccentricity 0.795em  and fictitious
flexibility 2.29x  : 0.539e 
  180 60 180 60 0.241 165.54      
where
40638.8
0.241
0.539 165.2 2100 0.9e c
N
Af

 
  
  
b) Checking for strength of eccentric compression.
0.795 20em   , the section weakening does not occur and the values of
bending moments used for strength and stability are equal.
c) Checking for stability in the plane of moment action.

Single storey building design English version

Single storey building design English version

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Single storey building design English version

Similar to Single storey building design English version (20)

Recently uploaded

Recently uploaded (20)

Single storey building design English version