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8/13/2019 Indrumator Metal - EG
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3. LOADS, LOAD FACTORS, LOAD COMBINATIONS
Loads
Nominal
Load
[KN/m2]
Factor
of
Safety
Factored
Load
[KN/m2]
PermanentLoads
(P)
(dead loads)
Roof weight: ….………………..(corrugated sheet)
Purlin weight: ……………..…...
Technological load: ……………
0.15…0.35
0.10…0.15
0.10…0,20
1.35
1.35
1.35
Variable
Loads
(V)
(environmental
loads)
Snow (CR 1-1-3-2005):
sk = µ i × Ce × Ct × s0,k
where :
* s0,k is the ground snow load (as is
shown in ground snow load map)
Zone on the map s0,k [KN/m2]
A 1,50
B 2,00C 2,50
* Ce is the exposure factor (to account
for wind effects); the value of this
coefficient is:
Ce = 0.80 for full exposure.
* Ct is the thermal factor (to account
for snow melting effects due to the
heat inside the building); the value of
this coefficient is considered 1.00 in
most cases.
* µi is the slope factor;
µ1 = 0.80 whenever 0° < α ≤ 30°
sk 1.50
* Case study:
Loads
Nominal
Load
[KN/m2]
Factor
of
Safety
Factored
Load
[KN/m2]
Permanent
Loads
(P)
(dead loads)
Roof weight: ….………………..
(corrugated sheet)
Purlin weight: ……………..…...
Technological load: ……………
0.15
0.15
0.10
1.35
1.35
1.35
0.203
0.203
0.135
Variable
Loads
(V)
(environmental
loads)
Snow (CR 1-1-3-2005):
s0,k = 2,50 kN/m2
Ce = 0.80 for full exposure.
Ct = 1.00.
µ1 = 0.80
1.60 1.50 2.40
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Load combinations: relevant load combinations for:
a) Ultimate Limit States (U.L.S.) : 1 0,
1 2
1,35 1,50 1,50n m
c
j i i
j i
q P V V ψ = =
= + ⋅ + ⋅ ⋅∑ ∑ [KN/m2]
b) Serviceability Limit States (S.L.S.) : 1 0,
1 2
n mn
j i i
j i
q P V V ψ = =
= + + ⋅∑ ∑ [KN/m2]
where:V 1 is the dominant variable action
ψ 0,i = 0.70 (factor of simultaneity)
* Case study:
The horizontal distributed load on the roof is:
U.L.S. ⇒ Qc = 1,35 (0,15 x 1/cos α + 0,15 + 0,10) + 1,50 × ( 1.60 ) = 2.94 [KN/m
2]
S.L.S. ⇒ Qn = 0,15 x 1/cos α + 0,15 + 0,10 + 1.60 = 2.00 [KN/m
2]
Determine the vertical distributed load acting on the current purlin (d = 2,40 m):
a) U.L.S. : qc= Q
c × d = 2.941 × 2.40 = 7.06 kN/m
b) S.L.S. : qn= Qn
× d = 2.001 × 2.40 = 4.80 KN/m
The vertical load on the purlin must be decomposed into normal (q w) and parallel (q f )
components; the parallel component (in the roof plane) q f is carried by the roof deck if this is
made of corrugated steel. Consequently the purlin needs to be designed for the normal
component (perpendicular to the roof plane) q w only.
* Case study:
- the parallel component:
q f c= q
c × sinα = 7.06 × sin5.71° = 0.70 KN/ml
q f n = q n × sinα = 4.80 × sin5.71° = 0.48 KN/ml
- the normal component (carried by the purlin):
q wc= q
c × cosα = 7.06 × cos5.71° = 7.03 KN/ml
q wn= q
n × cosα = 4.80 × cos5.71° = 4.78 KN/ml
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4. CAL
- for Ul
- for Se
5. CRO
The firs
purlinConseq
depth a
only on
CULUS O
imate Lim
viceability
SS-SECT
t (terminal
ortions hently, ther
d the sam
e depth.
F BENDI
t States ver
Limit Stat
ON SIZI
purlin por
s two croe are four
thickness
G MOME
ification:
s verificati
G
ion and cu
ss-sectionsross-sectio
of flanges
NT AND
on (only n
rent purlin
for sizingns to be si
because th
HEAR F
minal bend
portion are
: midspaned, so that
y belong t
RCE (in e
ing mome
to be calc
section ais necessa
the same
lastic rang
t diagram):
lated; each
d supporty to have t
element, w
e)
of these
section.he same
hich has
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• Choice of the cross-section shape:
Double T - bisymmetric cross-section (most adequate shape for members in bending);
• Choice of the material quality:
For purlin design is possible to use one of the following steel grades: S355, S275 or S235;
the steel grade is chosen such that the deflection requirement should be satisfied, with a
minimum material consumption. For the beginning one should start with S355 grade and if
the actual deflection is greater than the allowable deflection, the steel grade should be
changed in S275 or further on, in S235.
Steps:
1) Determination of the required section modulus:
** 0 M
y
M W
f
γ =
where: M* is the ponderate moment on the purlin
γM0 is a coefficient equal to 1,00
f y is the yield limit of the material
* Case study:
M* =2 0,8 44,41 2 0,4 59,79 2 0,6 18,79 4 0,4 44,98 3 0,6 26,20
7
t t t t t
t
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅=
= 37,22 kNm
W* = 37,22 x 10
6
/275 = 117164 mm
3
= 117,16cm
3
(S275)
2) Determination of the web dimensions : hw tw
hw ≈ 1.15 × w
nec
t
W
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- propose tw (tw = 3, 4, 5, 6, 7, 8, 9, 10, 12 mm).
- calculate hw
100 to 100 mm for hw > 1000 mm;
- round off result: 50 to 50 mm for 500 mm < hw ≤ 1000 mm;
10 to 10 for hw ≤ 500.
- Determine hw and check ifw
w
t
h ∈ (70…150); if not, change tw .
* Case study: propose tw = 3 mm
* 1171641,15 1,15 227,3
3w
w
W h mm
t = = = hw = 230 mm
w
w
t
h= 76,7 (as the result was obtained with the minimum tw it is not possible to change)
3) Determination of the flanges dimensions : bf × tf
Af ≈ i
w
W
h - 0.16 × hw × tw = net area for one flange
where: W i =0 M i
y
M
f
γ
- propose: tf = ( 1.50…2.50 ) × tw ;
- calculate: bf = f
f
t
A ≥ 60 mm
- round up result 10 to 10 mm;
13 S355
-check if: f
t
b' ≤ 14 S275 if not change tf .
15 S235
4) Optimize the material consumption (if possible)
-check if:
wh
b ∈ ( 1/3…1/5 );
2 × Af ≈(0.50…0.60) × A ; (A is the area for the whole cross-section)
Ai ≈ (0.50…0.40) × A .
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* Case
1. Prop
2. Com
- first s
- first s
- curren3. Com
bf1 = A
t
bf2 = A
bf3 = A
6. VER
a)
b)
positio
tudy:
se tf = 6 m
ute the str
an and cur
pport:
t spans:ute the fla
1 f
f
= 600,7/
2 f
f
= 834,9
3 f
f
= 303,8
IFICATI
raw the s
alculate t
of neutral
m
ngth modu
ent suppor
ge areas:
6 = 100,1
6 = 139,2
6 = 50,6 m
NS FOR
etch of the
e geometri
axis; mom
lus:
ts: W1
W2
W3
Af1
Af2
Af3
m
m
HE SIZE
section;
cal charact
nt of inerti
44,98 x 1
59,79 x 1
26,20 x 1
163563,6 /
217418,2 /
95272,7 /
b
b
b
CROSS-
ristics: ar
with resp
6 /275 = 1
6 /275 = 21
6
/275 = 9
230-0,16x
230-0,16x
30-0,16x3
f1 = 110 m
f2 = 140 m
f3 = 60 mm
SECTION
a (A); cent
ct to x-x a
3563,6 m
7418,2 m
272,7 mm
x230 = 60
x230 = 83
230 = 303
oid locatio
is (Ix).
3
3
0,7 mm
,9 mm
,8 mm
n (C.G.)
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6.1. VE
a) CH
1. Esta
• Cas
1. 2.
For the
2. Be
• Cas
3. She
• Cas
RIFICATI
CK FOR
blish the s
e study for
or the we
or flanges
whole secti
ding mom
e study for
r capacity
e study for
ONS OF
ESISTAN
ction class
purlin secti
: hw / tw =
: b’ / tf = 6
on, the cla
nt capacit
purlin secti
of the secti
purlin secti
LTIMATE
E
according t
on 2
230/3 = 77
8,5/6 = 11,
s establish
of the sect
c
M
on 2
Wel
Mc,
on, Vc,Rd :
,c RV
on 2
AV = 3
Vc,
LIMIT S
o SREN 1
→ class 2
→ class
d will be 3
ion, Mc,Rd :
,min
,
el
Rd M
W
γ =
,min = 218,4
d = 60,06
(0
V y
M
A f
γ
=
x 230 = 6
Rd = 109,6
ATES (wit
93-1-1-20
y f
cm3
Nm
)3
0 mm2
kN
factored
6 (table 5.
oads)
)
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4. Resistance check
,
,
2 2
,
0 0
1,0
1,0
3 1,0
Ed
c Rd
Ed
c Rd
x Ed Ed
y M y M
V
V
M
M
f f
σ τ
γ γ
≤
≤
⎛ ⎞ ⎛ ⎞+ ≤⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
• Case study for purlin section 2 :
VEd = 38,27 kN (see diagram)
MEd = 59,79 kNm (see diagram)6
2
,
59,79 10 230260,0 /
2 26431000 2
Ed w x Rd
x
M h N mm
I σ
⋅= = =
3238,27 10
55,47 /230 3 Ed Ed
w w
V N mmh t τ
⋅= = =⋅
( ) ( )
,
,
2 2
2 2,
0 0
38,270, 35 1, 0
109,6
59,790, 996 1, 0
60,06
3 0,945 3 0,20 1,0 1,0
Ed
c Rd
Ed
c Rd
x Ed Ed
y M y M
V
V
M
M
f f
σ τ
γ γ
= = ≤
= = ≤
⎛ ⎞ ⎛ ⎞+ = + = ≤⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
6.2. VERIFICATION OF SERVICIABILITY LIMIT STATE (with nominal loads)
Check for deflection:
f actual =x
2
I10
t × ∑ α × M ≤ f allowable =
200
t ;
Input t ⇒ [m] ;
Ix ⇒ [cm4] ; ⇒ f actual in [cm] ;M ⇒ [KNm] ;
(α ⇒ from tables)
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7. PURLIN CROSS-SECTION VARIATION ON THE LENGTH OF THE BEAM
It is necessary to determine the distances ‘x’ on both sides of the supports, for which
the midspan section resists to the bending moment; for purlin design we choose the greatest
distance ‘x’. Assuming the loads discussed earlier and S275 for the purlin material, we obtain:
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8. PURLIN (ERECTION) SPLICE DESIGN (with factored loads)
It is necessary to design two (erection) splices (connections which provide the purlin
continuity): on the second support and on the current one.
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- Design stress is: M10, (M20) which is resolved in a couple of forces
H =h
M ; h = hw + 2 tf (lever arm = overall depth of the beam)
- Cover plates sizing: Ac.p.
nec
=
0
y
M
H
f γ
a) top flange cover plate ( tc.p.t × bc.p.
t- lc.p.
t):
- propose: bc.p.t
= b – 20 ;
- calculate : tc.p.t
=t
c.p.
nec
c.p.
b
A
(round off to a plate thickness which is manufactured)
- check for cover plate resistance:
σ =t
pc
t
pcbt
H
.... × ≤ f y/γM0 ;
- fillet welds sizing:
- propose weld size aw ≤ 0.70 × tmin ;
tmin = min (tf , tc.p.
t) ;
- weld length : lw =. . . . 0
,2
t t
c p c p y M
w vw d
t b f
a f
γ × ×
× × + 2 aw (round off 5 to 5)
2
,
2
430
233,653 3 0,85 1,25
uvw d
w M
f
f N mm β γ = = =⋅ ⋅ ;
- verify if : lw ≤ 60 × aw .
b) bottom flange cover plate ( tc.p. b
× bc.p. b
- lc.p. b
):
- propose : tc.p. b
= tf ;
- calculate : bc.p. b
= b
pc
nec
pc
t
A
..
.. , round off 10 to 10;
- check for cover plate resistance :
σ =b
pc
b
pcbt
H
.... × ≤ f y/γM0 ;
- fillet welds sizing:
- propose weld size aw ≤ 0.70 × tmin ;
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tmin = min (tf , tc.p.
t) ;
- weld length: lw =. . . . 0
,2
b b
c p c p y M
w vw d
t b f
a f
γ × ×
× × + 2 aw (round off 5 to 5)
- verify if: lw ≤ 60 × aw .
* Case study: H =h
M =
59,79
0,242= 247,1 kN
Ac.p.nec =
R
H =
3247,1 10
275
⋅= 898,4 mm2
a) top cover plate: bc.p.t
= b – 20 = 140 – 20 = 120 mm
tc.p.t
=t
c.p.
nec
c.p.
b
A =
898,4
120
= 7,5 mm tc.p.t
= 8 mm
- check for strength: σ =t
pc
t
pcbt
H
.... ×=
3247,1 10
120 8
⋅⋅
= 257,4 N/mm2 ≤ f y/ γM0 = 275 N/mm
2
- fillet weld sizing: aw = 4 mm
- weld length : lw =. . . . 0
,2
t t
c p c p y M
w vw d
t b f
a f
γ × ×
× × + 2 aw =
8 120 275
2 4 233,65
× ×× ×
+ 2 x 4 = 149,3 mm
lw = 150 mm ≤ 60 x 4 = 240 mm
b) bottom cover plate: tc.p. b
= tf = 6 mm
bc.p. b =
nec
c.p.
b
c.p.
A
t =
898,4
6= 149,7 mm bc.p.
b = 160 mm
- check for strength: σ =. . . .
b b
c p c p
H
t b×=
3247,1 10
160 6
⋅⋅
= 257, N/mm2 ≤ R = 275 N/mm
2
- fillet weld sizing: aw = 4 mm
- weld length : lw =. . . . 0
,2
b b
c p c p y M
w vw d
t b f
a f
γ × ×
× × + 2 aw =
6 160 275
2 4 233,65
× ×
× ×+ 2 x 4 = 149,3 mm
lw = 150 mm ≤ 60 x 4 = 240 mm
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ROOF TRUSS DESIGN
1. STRUCTURAL CONFIGURATION
2.
LOADS, LOAD FACTORS, LOAD COMBINATIONS
Loads are the same as fo the purlins, exept for Permanent Loads, which must take into
account the truss weight as well. The nominal truss weight is assumed to be in between
0.10…0.15 KN/m2. The factor of safety for the truss weight is n = 1.35.
Load Combinations:
a) Ultimate Limit States (U.L.S.) : ⎯ q c = ∑ ni × Pi + ∑ ni × Ci + ng × ∑ ni × Vi [KN/m2]
b) Serviceability Limit States (S.L.S.) : ⎯ q n= ∑ Pi + ∑ Ci + ng × ∑ Vi [KN/m
2]
Determination of the panel point loads:
a) U.L.S.: Qc= ⎯ q
c × Aaff = ⎯ q c× d × t [KN]
b) S.L.S.: Qn = ⎯ q n × Aaff = ⎯ q n× d × t [KN]
where Aaff is the aferent area, detailed in the figure below
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3.2. Determination of the member forces
For instance using the method of sections:
∑ M3 = 0 ⇒ B24 = '33
3
l
M
;
Mi = 0
∑ M4 = 0 ⇒ T35 ='44
4
l
M ;
Yi = 0 ⇒ R - Qc /2- Q
c- Q
c- T35 × sinα - D34 × sinγ = 0
D34 =
γ
α
sin
sin2
35 ×−−−− T QQQ
R ccc
T01 = T13 ='22
2
l
M ;
-Qc
+ T01 × sinα - T13 × sinα + M12 = 0
M12 = Qc
3.3. Truss diagram (with force values resulted from computations)
The sign “?” will be replaced by each student with the force values resulted from calculation.
4. DESIGN OF THE TRUSS MEMBERS
4.1. Choice of the cross-section shape
An adequate cross-section (concerning fabrication) for the truss elements (chords and
web members) is built up of two angles back to back. The two angles are connected to each
other by means of local filler plates spaced as follows:
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- for compression members :
l1 ≤ 15 r 1;
no less than two filler plates along the members between two joints.
- for tension members :
l1 ≤ 80 r 1;
no less than one filler plate along the members between two joints.
In the equations above, r 1 is the radius of gyration for one angle with respect to its 1-1
axis parallel to y-y axis contained in the y-y plane which passes between the two angles.
4.2. Choice of the material quality
The steel grade used for the rolled sections is S235.
5. DESIGN OF TENSION MEMBERS
- bottom chord;
- some diagonal members.
5.1. Cross-section sizing
Areq = y
M
f
N 0γ ⇒ from the tables with rolled angles, select angle:
L…x…x…
to fulfill Aact = 2 A1L ≥ Areq
- extract also from tables the values for: e; r x = ix; r y = iy;
r 1 = i1
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5.2. Check for resistance0/ M yact
Ed
f A
N
γ ⋅ ≤ 1,00
5.3. Check for slenderness λmax = max (λx , λy ) ≤ λallowable = 400
λx = x
x
r
l ; λy =
y
y
r
l
- bottom chord :
lx = li = distance between two joints
ly = L1 = distance between two latteral braced joints (between the diagonal links).
- diagonal members in tension:
lx = 0.80 li
ly = li
li = distance between the two joints of the member (length of the bar).
6. DESIGN OF COMPRESSION MEMBERS- top chord;
- some diagonal members and the vertical members.
6.1. Cross-section sizing
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Areq =1
4,1
M y
Ed
f
N
γ
× ⇒ from the tables with rolled angles, select angle: L…x…x…
to fulfill Aact = 2 A1L ≈ Areq (γM1 = 1,00)
- extract also from tables the values for: e ; r x = ix; r y = iy; r 1 = i1
6.2. Check for buckling 1/ M yact
Ed
f A N
γ χ ×× ≤ 1,00
χ = min y x χ χ ,
)( y x χ =2
)(2
1
y xλ −Φ+Φ
( )[ ]2
)()( 2,012
1 y x y x λ λ α +−⋅+⋅=Φ
xλ =ε ⋅
⋅9,93
1
x
bx
r
l ; yλ =
ε ⋅⋅
9,93
1
y
by
r
l ;
y f
235=ε
- for the top chord:
l bx = li = distance between two joints
l by = L = distance between two lateral braced joints = distance between two purlins
- all other members in compression:
lx = 0.80 li
ly = li
li = distance between the two joints of the member (length of the bar).
The factor α is equal to 0,49 (curve c)
7. DESIGN OF FILLET WELDS (for web members only)
⎯ N = min ( 1.30 NEd , NRd )
NEd → as it results from computation of the
forces in the truss members ;
NRd = Aact f y/γM0 (members in tension); NRd = χmin Aact f y/γM1 (members in
compression).
-propose weld size: a1 ≤ 0.70 × tg
⇒ a1
a1 ≤ 0.85 × tL
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a2 ≤ 0.70 × tg
⇒ a2
a2 ≤ 0.70 × tL
-calculate weld length:
l1 =1
,1
22
a f a
N b
eb
d uw
×+××
×−
rounding up 5 to 5.
l2 =2
,1
22
a f a
N b
e
d uw
×+××
×
where f uw,d =
23 M w
u f
γ β ⋅⋅
- verify if : l1 ≥ 40 mm l2 ≥ 40 mm
l1 ≥ 15 a1 l2 ≥ 15 a2
l1 ≥ b l2 ≥ b
l1 ≤ 60 a1 l2 ≤ 60 a2
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TRANSVERSE FRAME ANALISYS
1. STRUCTURAL LAYOUT
- See the transverse section (cross-section of the building).
2. STRUCTURAL CONFIGURATION AND LOADING
- Single storey sway frame
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• dea
• per
• sno
0.1=α
ga is t
2.r β =
( β r is th
r T is th
cT is th
q = 3 fo
(the Du
r ε = 1.0
structur
LOAD
a) ∑ 1.3
b) ∑ Pi
loads
anent load
(e
γ x pz)
0 is the Im
e ground a
5 r if T <
e Site Stru
e fundame
e Seismic
r transvers
ctility Fact
0 is the Eq
e to the firs
COMBIN
5 × Pi + 1.5
Snoe ×γ
s
;e
γ
cr
ortance fa
celeration
c ; 2.r
β =
ture Reson
tal elastic
one Dump
frame; q
r);
ivalency F
t degree.
TIONS
× Vi + 0.7
Seism+
(n
0.40
q
r ××
β α
tor for nor
according t
(5r c
T T − −
nce Factor
eriod of vi
ng Period (
4 for long
actor betw
∑ 1.5 ×
ominal loa
g (global
al buildin
o seismic r
) 1r
if T ≥
);
ration of t
Dumping
tudinal bra
en effectiv
j (
eγ =0.4
s)
seismic fa
gs;
sk zones (
cT
e building
eriod Map
ced bay.
e structure
= varable
.
tor)
n the map)
;
nd statical
loads)
;
ly indeterm
inate
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DETERMINATION OF THE LOADS AND MOMENT DISTRIBUTION
1. Permanent (P) :
( ) .aff
cc
P APQ ×= [KN]
( ) .aff
nn
P
APQ ×= [KN]
Aaff = t × L / 2
2. Cvasipermanent (C) :
( ) .aff
cc
C AC Q ×= [KN]
( ) .aff nn
C AC Q ×= [KN]
3. Snow (Z) :
.aff zF
c A p Z ××= γ [KN]
.aff ze
n A p Z ××= γ [KN]
4. Wind (W) :
( )whw
g zc p ×××= 80.060.1 [KN/ 2m ] pressure
( ) whw g zc p ×××= 40.060.1, [KN/ 2m ] suction
F w
c
www p pt p p γ ×=→×= [KN/m]; F
c
averagew W W t pW γ ×=→××= 35.1, [KN]
F w
c
www p pt p p γ ×=→×= ,',, [KN/m]; F
c
averagew W W t pW γ ×=→××= ''35.1'' , [KN]
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This frame is indeterminate to the
first degree; it is a sidesway frame
(joint translation is possible).
Calculation of bending moment distribution
h ph pW W R c
w
c
w
cc ××+××++= '8
3
8
3'
8'
2
1
h p M c
w ×=
8''
2
2
h p M c
w ×= ''11 M M M += and ''22 M M M +=
h R
M ×=2
'
5. Seismic Force (S) :
hS
M ×=2
[KNm]
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Results of calculation:
After finishing all the calculations, the results will be centralized in the following table for
both sections 1 –1 and 2 – 2 of the column.
Column
sketchSection Efforts
Permanent Loads
(Pi)
Quasipermanent
Loads (Ci)
Snow
(Z)
Factored Nominal Factored Nominal Factored Nominal
0 1 2 3 4 5 6 7 8
1 1
2 2
1 - 1
M
(kNm)
N
(kN)
T
(kN)
2 - 2
M
(kNm)
N
(kN)
T
(kN)
(The nominal load for snow is considered for the earthquake combination – 0.30 x pz)
Wind
(W)
Eartquake
(S)
Relevant Load Combinations
∑ ni × Pi + ∑ ni × Ci + ng × ∑ ni × Vi
3 5 7 + 9
∑ Pi + ∑ Ci + γ e x Z + S
4 6 8 10
Factored GcS r ×= Mmax
Ncor
Nmax
Mcor
Mmax
Nmin
Mmax
Ncor
Nmax
Mcor
Mmax
Nmin
9 10 11 12 13 14 15 16
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COLUMN DESIGN
1. CONSTRUCTION DETAILS
(Fig. 32)
2. CALCULUS SCHEME
The bracing (of the longitudinal frame) prevents sway in the longitudinal plane of the
hall, or members connecting stanchions to a braced bay;
Effective lengths of stanchion (buckling length) – theoretical is the distance between
two points of contraflexure:
- transverse plane of the hall = plane of the diagram; Ll fx ×= 2 ; if we consider that
in the plane of diagram the stanchions act as cantilevers tied together by the roof
trusses (but in this plane the tops of the stanchions are not otherwise held in position or
restrained in direction), than for two or more spans : Ll fx ×= 50.1 .
- longitudinal plane of the hall = perpendicular to the plane of the diagram;2
Ll fy = ;
3. LOADING
N = …
M = …
T = …
For (maximum) values of axial compression , bending moment and shear force , see frame
analysis.
4. CROSS-SECTION SIZING
- cross-section shape: double T bisymmetric section;
- material quality: steel grade OL 37;
- sizing: by successive tests (propose an initial section and verify it; the proper section
satisfies, in all economical manners, all formulae of verification); for instance:
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5.2. Overall buckling check
R
N
N W
M c
A
N
Ex
xg
x x ≤
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ −××
×+
×1
minϕ
ϕ
y x ϕ ϕ ϕ ,minmin = is the minimum buckling factor;
→= x
fx
xr
lλ from buckling curve A
xϕ →
→= y
fy
yr
lλ from buckling curve B yϕ →
y xr r , are the radius of gyration with respect to x-x axis and y-y axis respectively.
→= yf
fy
yf r
lλ from buckling curve B yϕ →
yf r is the radius of gyration for maximum compressed flange with respect to y-y axis;
xc =0.85 is the equivalent uniform moment factor;
A E
N
x
Ex ××
=2
2
λ
π is the elastic buckling Euler force.
5.3 Check for slenderness
120,maxmax =<= allowable y x λ λ λ λ
5.4.Check for local buckling
- verify the width-to- thickness ratio (plate slenderness)
- for maximum compressed flange:
- 15
'≤
f
f
t
b (for OL 37)
- for web:
[ ]22
3
42
20100
β ψ ψ σ ×++−×
××≤
k
t
h
w
w
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* y I
M
A
N
x
×−−=σ
→−
=⇒σ
σ σ ψ
' table 35 STAS 3k → ;
σ τ β 307.0
k ××=
*' y
I
M
A
N
x
×+−=σ ',σ σ with their signs;
ww t h
T
×=τ
5.5. Anchorage bolts check
(Fig. 37)
Loading:.min,max , corr N M
b
in
R An
T
≤×=σ 210.6.6.180.6.5.
150.6.4.
=→→=→→
=→→
Rgr R Rgr R
Rgr R
b
i
b
i
b
i
[N/
2
mm ]
T is the tensile force in bolts (2 bolts);
⎟ ⎠
⎞⎜⎝
⎛ −×+=
×
××=
−−=
−×××
=
2
2
325.250.1
2
0
2
02
0
2
0
0
ll N M M
M
Rblr
r
N Rbl
T
a
a
b
b
α
α
R b is the strength (compression stress) of the concrete;
B100 B150 B2002/ mm N Rb
5 7 9
4
2
0d
An
×
≈
π
, is the net area of the bolt in the threaded zone;
d d ×≈ 89.00 ;
d = the nominal diameter of the bolt;
n = the number of bolts in tension.
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Base Plate :
Loading:.max,max , corr N M
lb Rlb
M
lb
N bb ,
6
2 ⇒≤
×+
×=σ
2
2
1
l M
b ×= σ
β →→ tablel
l
1
2
2
2 l M b ××= σ β
37
max6OL
R M t ×≥ rounded off to 20, 25, 30.
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A. HO
1.
RIZONT
OADIN
a) Wi
ana
w
p
c
w p
Th
R
L TRANS
:
d pressure
ysis”):
pe
cc ××=
w p×= 50,1
factored l
OF B
VERSE B
on the tran
w
g
ads: p
W 1
pW 2
p
W 3
ACIN
ACINGS
verse clad
p
af
c
w A p ,1×
p
aff
c
w A p ,2×
p
aff
c
w A p ,3×
DESI
ing (accor
[KN
[KN
[K
[K
[KN
GN
ing to “Tr
2m ]
2m ]
]
]
]
nsverse fra
me
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b) Wind friction on the roof:
wewe f f w gcgcc p ××=××= 01,0, [KN/ 2
m ]
01.0= f c wind friction factor;
f w
c
f w p p ,, 50,1 ×= [KN/ 2m ]
The factored loads: f aff
c f w
f A pW ,1,1 ×= [KN]
f
aff
c
f w
f A pW ,2,2 ×= [KN]
f
aff
c
f w
f A pW ,3,3 ×= [KN]
c) Stabilizing force corresponding to restraint the compression top chord of the
afferent trusses to the bracing;
- total force “S” for one truss: max%2 T S ×≈ ;
- Tmax is the maximum compression force in the top chord of the truss (factored load);
- “n” is the number of joints to be laterally restrained (we consider it equal to thenumber of central loaded joints – in a simplifying assumption: n = 3);
- for a maximum of three trusses (considering the load is acting in the same sense):
3
3
0
32
1
S S S
S
×==
= [KN]
Bracing loading for diagonal members design (check in wind pressure):
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f p W W H 111 += [KN]
2222S W W H f p ++= [KN]
3333 S W W H f p ++= [KN]
- where: p p pW W W 321
,, are from wind pressure ( 80.0= pc ).
Bracing loading for bracing chord design (check in tension for wind pressure and for
compresion in wind suction ):
f s W W H 111 += [KN]
2222 S W W H f s ++= [KN]
3333 S W W H f s ++= [KN]
- where: sssW W W 321 ,, are from wind suction ( 40.0= pc ).
paff cws A pW ,11 ×= [KN] p
aff
c
w
s A pW ,22 ×= [KN]
p
aff
c
w
s A pW ,33 ×= [KN]
suction
aff
pressure
aff A A ,1,1 =
suction
aff
pressure
aff A A ,2,2 =
suction
aff
pressure
aff A A ,3,3 =
2. DETERMINATION OF THE STRESSES IN THE MEMBERS OF THETRANSVERSE BRACING
- Stresses in the bracing chord : from bracing truss calculation
.....max = N (tension/compression)
- Stress in the diagonal members: from bracing truss calculation
....= D (compression)
3. DESIGN OF COMPRESSION MEMBERS OF THE BRACING (BRACING
CHORD AND DIAGONALS)
- Choice of the cross-section shape : square hollow section y x r r = ;
- Material quality : Steel grade S235 ;
- Cross-section sizing: propose a section and verify it; repeat this operation until the
verifications for buckling and slenderness are satisfied.
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Example: Bracing chord:
• Sizing for resistance in tension0/ M yact
Ed
f A
N
γ ⋅ ≤ 1,00
• Check for buckling in compression 1/
M yact
Ed
f A
N
γ χ ×× ≤ 1,00
χ = min y x χ χ ,
)( y x χ =2
)(2
1
y xλ −Φ+Φ
( )[ ]
2
)()( 2,012
1 y x y x λ λ α +−⋅+⋅=Φ
xλ =ε ⋅
⋅9,93
1
x
bx
r
l ; yλ =
ε ⋅⋅
9,93
1
y
by
r
l ;
y f
235=ε
The factor α is equal to 0,21 (curve a)
x
x
bx
xr
l χ λ →=
y
y
by
yr l χ λ →=
== bybxll distance between two joints (between two purlins)
Check for slenderness: .250,maxmax =≤= a y x λ λ λ λ
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Example: Diagonal members:
Check for buckling: same as for the bracing chord
x
x
d
xr
l χ λ →
×=
50.0
y
y
d
yr
l χ λ →
×=
70.0
Check for slenderness: .250,maxmax =≤= a y x λ λ λ λ
4. DESIGN OF FILLET WELDS (for all members)
⎯ N = min ( 1.30 NEd , NRd )
NEd → as it results from computation of the
forces in the truss members ;
NRd = Aact f y/γM0 (members in tension);
NRd = χmin Aact f y/γM1 (members in
compression).
-propose weld size: a1 ≤ 0.70 × tg
⇒ a1
a1 ≤ 0.70 × t
a2 ≤ 0.70 × tg
⇒ a2
a2 ≤ 0.70 × t
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-calculate weld length:
l1 =1
,1
22
50,0a
f a
N
d uw
×+×××
rounding up 5 to 5.
l2 =2
,1
2
2
50,0a
f a
N
d uw
×+
××
×
where f uw,d =25,133 2 ⋅⋅
=⋅⋅ w
u
M w
u f f
β γ β
- verify if : l1 ≥ 40 mm l2 ≥ 40 mm
l1 ≥ 15 a1 l2 ≥ 15 a2
l1 ≥ b l2 ≥ b
l1 ≤ 60 a1 l2 ≤ 60 a2