indrumator metal - eg

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



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



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



• 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



- 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 .



* 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.)



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)

)



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)



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:



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.



- 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 ;



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


,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


,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



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



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:



- 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



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



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



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



TRANSVERSE FRAME ANALISYS

1. STRUCTURAL LAYOUT

- See the transverse section (cross-section of the building).

2. STRUCTURAL CONFIGURATION AND LOADING

- Single storey sway frame



• 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



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]



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]



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



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:



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



* 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.



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.



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



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):



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.



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



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



-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

indrumator metal - eg

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