Design of Beam gColumn

By

Ass. Prof. Dr. Ehab B. Matar

B C lBeam-Column Nearly all-structural members in building frames or industrial

portal frames are subjected to both bending moments and axial load either in tension or compression formsaxial load either in tension or compression forms

(a) (b) (c)

BEHAVIOR OF BEAM-COLUMN UNDER AXIALBEHAVIOR OF BEAM COLUMN UNDER AXIAL LOAD AND BENDING MOMENT

Behavior of beam column = behavior of beam (considering effect of lateral torsional buckling)

+behavior of columns (considering effect of global

M buckling)M ult

M

M'

M' ult M varies

P = const.

MyM

P

Fig. Relationship for I beam

Modes of failure of Beam- ColumnJohnson [3] has mentioned a number of modes of failure that

characterizes a beam-column behavior which are summarized as follows:follows:

Failure by yielding: in the case of member subjected to bending moment and axial tension force.F il b i t bilit i th l f b di ith t Failure by instability in the plane of bending without twisting for the case of beam-column that is transversely loaded but is stable w.r.t. Lateral torsional buckling.F il b l t l t i l b kli Failure by lateral torsional buckling for beam-column loaded about their strong axis without adequate bracing.

Failure by instability about one of the principle y y p paxes for members subjected to axial compressive force and biaxial bending moment with stiff sections (W-shapes).

Failure by combined twisting and bending for the Failure by combined twisting and bending for the case of thin walled open sections (that are torsionally weak sections) subjected to axial compression force and biaxial bending moment.bending moment.

Design Criteria of Beam-Column

Assessment of Buckling length Effect of P- Effect of P- Design of sections subjected to M,T and

th bj t d t M Nothers subjected to M,N Checking local bucklingg g Preliminary section design

Assessment of Buckling Length

1. Kinds of buckling in beam column: Buckling out of plan (L ): will be Buckling out of plan (Lby): will be

governed by the vertical bracing profileB kli i l (L ) ill b d b Buckling in plan (Lbx): will be governed by the end conditions of the columns and the relative stiffness between different members.

41:10Ex.:- Buckling out of plan

2.5mI 0 23I

1.5m 1 0.7I0.7I

4mI 0.23I

24mVertical wind bracing

Lby=4m

Buckling in PlaneF ll d fi d d di i h b kli l h ld b For well defined end conditions, the buckling length could be estimated as per the following table

Buckling Shape

0.65k values = 0.80 1.00 1.20 2.00 2.10

R i fi d & l i fRotation free & translation fixedRotation fixed & translation fixed

End condition

Rotation free & translation freeRotation fixed & translation free

For Portal Frames: the alignment chart is used to determine the buckling coefficient factor k based on the following assumptions:-assumptions:

All behavior is elastic. The members are prismatic. All columns reach their respective buckling loads

simultaneously. For braced frames (no side sway), the girders are assumed to be For braced frames (no side sway), the girders are assumed to be

in single curvature while for un-braced frames (side sway is permitted), the girders are assumed to be in double curvature.

At a joint the restraining moment provided by the girders is At a joint, the restraining moment provided by the girders is distributed among the columns in proportion to their stiffness.

The girders are elastically restrained at their ends by the l d t th t f b kli th t ti f thcolumns, and at the onset of buckling the rotations of the

girders at its ends are equal and opposite. The girders carry no axial loads.g y

Determine the factor G expressing the relative stiffness at the two ends of the columns as follows:

g

gg

c

cc

LIE

LIE

G

Where EcIc, EgIg are flexural stiffness of the column and girder meeting at a joint while Lc, Lg are the

l d i d l th

g

column and girder lengths It is to be noted based on ECP recommendations

that:that: When the col. End is hinged G=10 When the col End is fixed G=1 0 When the col. End is fixed G=1.0

Alignment Chart

Steps to use alignment chartC l l t th G ffi i t t b th d f th l1. Calculate the G coefficient at both ends of the columns

2. It is to be noted that when the far end of the girder is fixed or hinged a modification factor is to be applied as follows:

The far end is hinged, then multiply EgIg /Lg ratio by 0.5 for the case of side sway is permitted and 1.5 for the case of side sway is preventedis prevented.

The far end is fixed, then multiply EgIg /Lg ratio by 0.67 for case of side sway is permitted and 2.0 for case of side sway is

t dprevented.

3. Decide whether the frame is permitted or prevented from side ec de e e e a e s pe ed o p e e ed o s desway and then use proper alignment chart

upper col. is permitted to sidesway

upper col. is permitted to sidesway

lower& upper col. are lower col. is lower col. is both permitted to sidesway

prevented from sidesway

prevented from sidesway

For truss frame: two options may be adopted1. Treating the column of the frame truss as a column partially

fixed at truss end or2. Solving the truss frame as a portal frame with girder stiffness

50 to 100 times column stiffness50 to 100 times column stiffness.

2a

h

L

50 - 100 Ich'

h'=h+a

Fig.(6.5) Buckling for column-truss framesL

Beam- Column with CraneBeam Column with Crane

ExamplesEx1

Required: Calculate the column b kli l th i l f thbuckling length in plane of the frame if end A is hinged and end D is fixed considering the frameD is fixed considering the frame is permissible to side sway

Solution: B II

1:10

C

For Column AB End A is hinged, then GA=10

B II

7m 1.5 I

C

1.5 I

g , At end B GB = ( ∑Ic/Lc)/( ∑

IG/LG) = (1.5I/7)/(I/26.13) = 5.626m

A D

From the alignment chart with side sway permitted, K = 2.6

Then, Lbx = 2.6x7 = 18.2m

However, on substituting into equation 6.4 for GA and GB K h l f 2 63 t ti f thGB, K approaches a value of 2.63 to satisfy the equality sign.

For Column CD For Column CD End D is fixed, then GD = 1.0

And at end C Gc = ( ∑Ic/Lc)/( ∑ IG/LG) = And at end C Gc = ( ∑Ic/Lc)/( ∑ IG/LG) = (1.5I/7)/(I/26.13) = 5.6

From the alignment chart with side sway permitted K = From the alignment chart with side sway permitted K = 1.72

Then, Lbx = 1.72x7 = 12.04m. Then, Lbx 1.72x7 12.04m. It is worth noting that the fixation at column end has

reduced the buckling length factor K by nearly 33.85% g g y ycompared with hinged case.

Ex2 Given: The shown two story

building frame, knowing that end G and H of the girders is

C I F I H

end G and H of the girders is fixed to the adjacent R.C. shear wall.

Required: Calculate the

3m

B 2 I E 2 I

2 I 2 I

G Shear Wall

Required: Calculate the columns buckling length in plane of the frame if end A is fi d d d D i hi d

7m6m

4m

A

4I

D

3I

fixed and end D is hinged. Solution: For Column AB For Column AB GA = 1.0 (end A is fixed) GB = (4I/4+2I/3)/(2I/6)=5.0 From the alignment chart

for side sway prevented K=0 85K 0.85

Lbx = 0.85 x 4 = 3.4m

For Column DE GD = 10 (end D is hinged) GD = 10 (end D is hinged) GE = (3I/4+2I/3)/(2I/6+2x2I/7) = 1.57 From the alignment chart for side sway

prevented K=0.89prevented K 0.89 Lbx = 0.89 x 4 = 3.56m

Ex.3 Given: The shown truss

frame, knowing that end A is hi d d d B i fi d 3 5mhinged and end B is fixed

Required: Calculate the columns buckling length in plane of the truss frame BA

8m

3.5m

plane of the truss frame. Solution: For Column A

16x2.4=38.4m

9.75m

The buckling length can be found by two different methods:

38 4

9.75m75 Ic

Method 1: assuming the column as hinged – partial fixed to the truss. Then f t bl (5 3) i th

38.4m

from table (5.3) in the Egyptian code K=2.0. The height of the column h’ = 8+3 5/2 = 9 75m8+3.5/2 = 9.75m.

Lbx = 9.75x2.0=19.5m

Method 2: transforming the truss frame into an equivalent gable frame with an assumed large stiffness q g gvalue of the rafter (assumed here 75Ic)

GA = 10 as the column is hinged and GB = (Ic/9.75)/(75Ic/(2x19.3)=0.05 From the alignment chart with side sway permitted

K=1.68 Lbx=1.68x9.75=16.38m The difference in the two methods is referred to the

approximate assumption of modeling the truss stiffness. However in the second method if the truss stiffness isHowever in the second method if the truss stiffness is varied between 50Ic and 100Ic then the K factor will vary between 1.7 and 1.67 respectively.between 1.7 and 1.67 respectively.

For Column B Again the buckling length will be found by two different methods: Method 1: assuming the column as fixed – partial fixed to the

truss. Then from table (5.3) in the Egyptian code K=1.2. Thetruss. Then from table (5.3) in the Egyptian code K 1.2. The height of the column h’ = 8+3.5/2 = 9.75m.

Lbx = 9.75x1.2=11.7mM th d 2 t f i th t f i t i l t bl Method 2: transforming the truss frame into an equivalent gable frame with an assumed large stiffness value of the rafter (assumed here 75Ic)

GA = 1.0 as the column is fixed and GB = (Ic/9.75)/(75Ic/(2x19.3)=0.05 From the alignment chart with side sway permitted K=1 16 From the alignment chart with side sway permitted K=1.16 Lbx=1.16x9.75=11.31m

bPbP

P- EffectcP

b

cPtP tP

f c

b

t

b

f

Sections sub. to M, T Sections sub. To M, NLPLP 23 LPLP 23

EILP

EILP fcb

f 848

3

EILP

EILP ftb

tbf 848

3

AP

ZP

ZLPf cfcb

b

4AP

ZP

ZLPf t

ftb

b 4 AZZ xx

b 4AZZ fxx4

Moment Magnification factor for beamMoment Magnification factor for beam columns prevented from side sway

Assuming that Pe = π2EI/L2 and α = P/Pe y1

W(z)

Py

P

Mo

Lz

Mi

11.1.1 2

0

02

0max LMEIMM z p(y1+ )

L/II

b- Intial B.M.D.

Fig. (6.6) Simplified Analysis of Beam-column

L/2L/II

c- P - effect

1.1 2

0

02

LMEIcm

1m

mcA

mz AMM .0max

Moment Magnification factor for beam gcolumns permitted to side sway

1.. 00max

mm

cMAMM y1

P

IG=infinity

HP

L

1

11 20

2EIcmP

/

L

P( + y1)

4 2

0LMcm H/2

y1

L

H/2

assumed as a qurter sine curveSecondary B.M

4 2PLP

effective length 2L is P

H/2

)2/(

4

3

2

HLLH

EIPL

PP

e

gused in calculating the value of α, Cm

121

2&

3)2/(

32

00

HLEI

HLMEI

LH

164

1 2 HLEILcm

Design of sections subjected to M,T

Failure of these members is governed by yielding

fff y g

N

ptMN

ANf

fff

yxM

netN

MMf

Af

Moreover, the compressive bending stress ynetxnet

M ZZf

shall be checked against the lateral torsional buckling stress.

ExamplesExample 1 Given:-M=20mt, T=15t, St.37, , , ,Beam is restrained laterally at supports Required:- Design of steel beam Required:- Design of steel beam Solution:-

tMN fff

ptx

ptMN

fZM

AT

fff

A;by gmultiplyin then ,

EQptx

x

PAfMZAT ..

x

Assuming A/Zx = 0.07 then: PEQ = 15+20E2x0 07 = 155t PEQ = 15+20E2x0.07 = 155t Areq = PEQ/fpt = 155/1.4 = 110.7cm2

Trying IPE 500 with the following geometrical Trying IPE 500 with the following geometrical properties

Zx = 1930cm3, A = 116cm2, L’ =258.2cm, f 2h=d=50cm, Af=20x1.6=32cm2, rt = 5.17cm

Checking the class of the cross sectionflange is s bjected to p re compression flange is subjected to pure compression

C/tf = (20/2-1.02/2-2.1) /1.6=4.6 O.K. class 1 web is subjected to bending and tension force web is subjected to bending and tension force

therefore < 0.5. Then, a = N/(2tw.Fy)= 3.06cm, , ( y) , , dw/tw = 42.6/1.02 = 41.76

which is smaller than O.K. Class 15.956.63

yFMT y

2/129015 cmtf

ZM

ATff

xMN

2/036.11930

220

/129.0116

cmtEf

cmtf

M

N

the resultant tensile stress is +1.17 t/cm2 < fpt for steel 37, O K

2/91.0&17.1036.1129.0 cmtff MN

O.K. Checking compressive stresses of –0.91 t/cm2, and as the

top flange is braced laterally at supports only which results i t d l th f 500 L 258 2in unsupported length of 500cm > Lumax = 258.2cm therefore, a check for the lateral torsional buckling should be carried out.3 Th ll bl i t i d b th 3- The allowable compressive stress is governed by the lateral torsional buckling stress which is given as the maximum of either:

bfu

ltb CAdL

f ./.

800

ltb

b

cmtxxf

C

/024.10.1.)50(500

328000.1

2

y

b

y

b

t

u

fL

fC

fC

rL

/

84&18871.9617.5

500

2

y

b

ytultb

fcmtxxf

fxCEfrL

f

580/081424.2)71.96(640

.5176.1

./64.0

22

f / 2 f

yltb fcmtxxE

f 58.0/08.14.20.15176.1

64.0

Then fltb = 1.08 t/cm2 and this value < fbc and therefore controls.

Fact=0.91t/cm2<Fbc=1.08t/cm2 => O.k safe

Design of sections subjected to M,N

Failure of these members may be governed by instability bybxca A

fAff

1

my

bcybcxc

cc

Af

Aff

1.. 21

Ey

ca

my

Ex

ca

mx

ffA

ff

cA

1

&1

21

fca = actual compressive stress due to axial compressive force = P/Af h ll bl b kli f h b

EyEx ff

fc = the allowable buckling stress of the member considered as a column

f f = the bending stresses about the x and fbx, fby = the bending stresses about the x and y axes

fbcx, fbcy = the allowable compressive bending stress

Allowable Bending Stresses

The allowable bending stress will depends on:-on:

1. Section ClassL t l t d l th f i2. Lateral un-supported length of compression flange

Local buckling for eccentric Sec.

NThe maximum web to thickness ratio for compact sections are

w

y

w

w

.dthen 0.5for

F/

td

then 0.5 for

663113

699

1- yw F.t

Na2

2- Then determine the value of as follows: yw Ft

The maximum web to thickness ratio for non-compact sections are

y

w

w F/

td

then 1- for

2

190

For rolled section cahdw 2

For welded section Stahd fwyw

wF

)(td

then 1- for

195

For welded section Stad fw 2

Allowable Stresses in Bending

For compact sections bent about major axis64.0 FF ybc bf

138020 ofeither

exccedingnot isLu length dunsupporte lateral max. that thecondition in

Ab

ybcC tf

dw twd

bf

138020

where

CdF

Aor

Fb

by

f

y

f

M1 M2

3.23.005.175.1loadingransversewithout tbeamsfor

*2

11

MM

MMC

tbA

b

fff

For transverse loaded members Cb is taken from table (4.2)For compact section bent about minor axis F =0 72F

g22

MMb

For compact section bent about minor axis Fbc=0.72Fy

concrete slab

composite beam with shear connectorFlange embedded

concrete slab

welded

open web joist

welded or bolted roof purlin

stiffener

bracedunbraced

lateral buckling of floor systemstrut

Types of lateral restraintsTypes of lateral restraints

For Compact sections not satisfying the above limits or Non-Compact sections satisfying the above mentioned limits, Fbc=Fbt=0.58Fy

When the lateral unsupported length Lu exceeds the maximum LuWhen the lateral unsupported length Lu exceeds the maximum Luthen the maximum bending compression Fbc is the min. of either 0.58Fy or max. of either Fltb1 and Fltb2

CF 800

y

btu

bfu

ltb

FCrL

CAdL

F

then84/When

/.1

btu

b

yltb

y

FCrL

FC

F.F

then188/ 84When

5802

wff

fft Atb

btr

6/112/3

yyb

yt

u

ltb

yy

FFCx

FrL

F

FF

58.01017.1

.64.0 5

2

2

y

btu

b

FCrL

Cx

then188/When

1017.1

yb

t

ultb

y

FC

rL

F 58.01200022

continuewhere

7500&750022

yEy

xEx ff

where

factoron modificatimoment

1

m

yx

M

C

loadingansversewithout tr

sway sidefrompreventedframesfor 4.06.02

1

m M

MC

endsat restraint moment and loading transversesway with side from prevented framesfor 85.0

gmC

endsat supportedsimply and loading transversesway with side from prevented framesfor 0.1mC

sway side topermitted framesfor 85.0mC

M2M2

M1

frames prevented from side sway

without transverse loading

frames prevented from side sway

with transverse loading

frames prevented from side sway

with transverse loading

without end moment restraintswith end moment restraintsM1<M2

000022

7500&7500EyEx ff

yx

Preliminary Section DesignBased on the AISC specifications [4] a simplified treatment was carried

out to facilitate the first trial of choosing the beam column cross section.

011 .A.ff

ff

bc

b

c

ca

011

.

ff

cf.ZM

f.AP

E

cam

bcc

bcc

Multiplying the last equation by A.fc

EQcca

m

bc

c Pf.Af

cff

.ZAMP

1

E

cabcf

1

The term (cm/(1-(fca/fE))) may be taken unity & the ratio of A/Z may be

termed B which is given in the steel tables and may roughly be taken equal to

0.07 for most steel sections, then,

bc

cEQ f

f.B.MPP .................................................................(6.23)

h i l b d EQPhil iThus a cross sectional area may be assumed as

c

EQg f

A while a section

modulus can be calculated the same way as before where Z/A14-15

Mff

.AZ.PM

c

bcEQ ..............................................................(6.24)

Thus a trial sectional modulus can be obtained as bc

EQf

MZ

Example 2 Given:- St. 37, shown beam

Case M1 M2 P 1 +20mt +20mt -15t

2 +20mt -20mt -15t

3 +20mt +20mt -60t

4 +80mt +80mt -15t

Required:- Design the beam for different combinations Solution:-

C 2 Case 2 M1= +20mt, M2=-20mt i.e. double curvature, P = -15t

Ass ming that B = A/Z is nearl eq al to 0 07 & Assuming that B = A/Z is nearly equal to 0.07 &the slenderness ratio in the range of 40, and therefore for St.

37 fc will equal to 1.3t/cm2 and while fbc is 1.4t/cm237 fc will equal to 1.3t/cm and while fbc is 1.4t/cm

1453.122007.015.. txExfBMPP cEQ

25.11131

145

4.1

cmf

PA

f

EQreq

bcEQ

If λ = 40, then ix ≥ 500/40 ≥12.5cm

3.1fc

, x

iy ≥ 250/40 i.e. ≥ 6.25 cm, Z > 2000/1 4 i e >1428 6cm3 Zx > 2000/1.4 i.e. >1428.6cm . Try section of I.P.E. 500 may be checked knowing

the following geometrical properties:the following geometrical properties: A = 116 cm2, Zx = 1930 cm3, ix = 20.4cm, iy =

4 31cm r = 5 17cm4.31cm, rt = 5.17cm, Lumax = 258.2cm, h = d = 50cm, Af = 20x1.6 = 32cm2

Checking class of the chosen section; FlangeC/tf = 10/1.6=6.25 yF/.916 = 10.33 O.K class 1

For web; Npw=(50-2x1.6)x1.02x2.4=114.57t >> N=15t. Therefore:

a = N/(2tw.Fy) = 3.06cm,

5707306350 cahd

699an smaller th is which 41.7642.6/1.02dw/tw

57.07.306.322

w cad

O.K. Class 1

4.70)113(

699

yF

2/129015Pf 2

58250&51.24500

/129.011615 cmt

APf

yx

ca

22max /18.1)58(000065.04.1,58

31.44.20

fcmtxfc

yx

2

1

/041220

0.115.0109.0

cmtEMf

Aff

c

ca

/04.11930

cmtZ

fx

bx

As the maximum lateral unsupported length L’=258.2cm for this section,

which is greater than the actual distance between lateral restraints of the

compression flange (250cm), therefore there is no need to check for the lateral

torsional buckling stress Therefore the value of f is controlled by thetorsional buckling stress. Therefore the value of fbc is controlled by the

allowable bending stress, which is 1.4 t/cm2.

0.1. 1 Aff

ff bca

0185004.1129.0

ff bcc

safeO k

0.185.04.118.1

safeO.k

Example 3 12.16mt

5.33

41:10

Given:- 9.5mt

B.M.D

33

0.23I

0.7I0.7I

I

4

21

1:10

1.5m2.5m

4

Vertical wind bracing 24m24m

4m

N.A.

338.7

12 I.P.E. 360

I.P.E. 360

320

360mm

812.7

170

320

the distance between purlins or side girt is restricted to 2m.

Sec. M(mt) N(t)Sec. M(mt) N(t) 1 -16.93(case I) -5.19(case I) 2 0(case I)

-4.72(case II) -11.22(case I) -6.85 (case II)

3 -22.02(case I) -2.66(case I)4 +10.38(case I) -2.13(case I)

Required:-D i th h f i St 52 id i th ff tDesign the shown frame using St. 52 considering the effect of existing knee bracing or not for sections 1 and 3 and compare between the two casescompare between the two cases Solution:-Design of Sec. 1

M = 16.93mt, N = -5.19 t

Considering that λmax will nearly be 80 then fc = 1.24t/cm2.

Calculating the buckling length: Lby = 400cm

For calculating Lbx, the frame is permitted to side sway, GA = 10

(hinged base) and 34122470

8 ../I.

/IG B therefore from the alignment

chart K = 2.5. Thus Lbx = 2.58=20m.

Then for λ = 80, cmi&cmi yx 580

4002580

2000

Hence

75

177512

24129316070195

P

t..

.xE.x..ff

.B.MPP

Q

bc

cEQ

26260

24175 cm..f

PA

c

EQreq

Zx > 16.93E2/2.1 i.e. >806.19cm3.

Case of existing of knee bracing

Knee bracing detail for the rafter

Knee bracing detail for the column

When a knee bracing is used then Lu= the di t b t id i t i 200distance between side girt i.e. 200cm

When there is no knee bracing then Lu= the distance between the point of maximum momentdistance between the point of maximum moment and the point of zero moment i.e. 800cm.

Case 1: no knee bracing Case 1: no knee bracing A trial section of I.P.E.. No. 360 was tried first and

found to be unsafe for low lateral torsional bucklingfound to be unsafe for low lateral torsional bucklingstress, therefore another section of I.P.E. No.500may be checked knowing the following geometrical

tiproperties A = 116cm2, Zx = 1930cm3, ix = 20.4cm, iy =

4 31cm rt = 5 17cm4.31cm, rt = 5.17cm, L’ = 258.2cm, h = d = 50cm, Af = 201.6 = 32cm2

Checking Section class and actual stressesFl C/t 10/1 6 6 25 8 9 O K l 1a- Flange: C/tf=10/1.6=6.25 yF/9.16 =8.9 O.K class 1

b- Web: is subjected to bending moment and compression

force then:force, then:

Npw = (50-21.6) 1.023.6 = 171.85t >> N = 5.19t.

Therefore, a = N/(2tw.Fy)= 0.71cm, 520.cahdw , , ( w y) ,2w ,

dw/tw=42.6/1.02=41.76 which is smaller than 9663113

699 .F)( y

O K Cl 1O.K. Class 1

20450116

195 cm/t..APfca

22 8004980001350120498

8192314

4000498420

2000

cm/t.).(x..f,.

..

&..

cmax

yx

2

1

88029316

01150060

/E.Mf

.A..ff

c

ca

28801930

29316 cm/t.E.ZMf

xbx

As Luact=800cm > Lumax=258cm, a check for lateral torsional buckling is needed as follows

fltb is controlled by the greater value of

21217515080032800800

cm/t...xxC.

d.LA

f bu

fltb

Where: α = 0/16.93 = 0 and Cb = 1.75

or

bltb

y

b

t

u

C.f

fC

..r

L

212000

18874154175

800

b

t

ultb

rL 2

fcm/t.xf 58088075112000 2 yltb f.cm/t.

.f 580880

74154 2

Fbc=1.12 t/cm2.

On checking the stabilit interaction eq ation it as fo nd that the sectionOn checking the stability interaction equation it was found that the section

is safe:

011 .A.ff bca

01840121880

800450

01

....

..

..ff bcc

Case 2: of existing knee bracingSearching for these approximate initial requirements a trial section of

I.P.E. No. 400 may be checked knowing the following geometrical properties: A = 84.5 cm2, Zx = 1160cm3, ix = 16.5cm, iy = 3.95cm, L’ = 189.7cm

Checking class of the chosen section:

a- Flange: C/tf=9/1.35=6.67 yF/.916 =8.9 O.K class 1 b- Web: is subjected to bending moment and compression force, then Npw =

(40-21.35) 0.86 3.6 = 115.48t >> N = 5.19t.

Th f N/(2t F ) 0 84 530hd Therefore, a = N/(2tw.Fy) = 0.84cm, 5302

.cadw ,

dw/tw=33.1/0.86 = 38.49 which is smaller than 5562113

699 .F)(

O.K. 113 F)( y

Class 1

20610584

195 cm/t..

.APfca

22 51021121750021121

27101953

40021121516

2000

/t)/(f

..

&.. yx

1

22

011501190

51021121750021121

.A..ff

cm/t.)./(f,.

c

ca

cmax

24611160

29316 cm/t.E.ZMf

xbx

c

Lumax’=189.7cm < Luact =200cm, therefore there is a need to check for

the lateral torsional buckling stress.

13175030750051751

7509318

697512

2)(C

..

.

13175030750051751 2 .).(x..x..Cb

fltb is controlled by the greater value of

800yb

u

fltb f.cm/t...

x.xxC.

d.LA

f 58075213140200

35118800800 2

Therefore the value of f is controlled by the allowable bending stress forTherefore the value of fbc is controlled by the allowable bending stress for

that type of steel, which is 2.1t/cm2.

O h ki h bili i i i i f d h hOn checking the stability interaction equation it was found that the

section is safe:

4610610

011 .A.ff

ff

bc

b

c

ca

01815012

4615100610 ..

..

..

Therefore on using knee bracing there is a saving on column steel

weights that reach 36.8%.

Design of Sec. 3M = 22.02mt, N = -2.66 t

For the rafter, the effect of the normal force is generally very small and

the rafter may be designed only for bendingthe rafter may be designed only for bending.

Hence:

2123320222 E.MZ 2123312850

cm.x.f

Zbt

req

When knee bracing is used Luact = dist. Between purlins=200cm otherwise

for no knee bracing Luact= distance between the maximum bending moment

and the point of zero moment i.e. 5.33m

Case 1: of no knee bracing

450mmI.P.E. 450 416.19

N.A.

450mm

414 9.4

450

12 I.P.E. 450

416.19

19014.6

Searching for these approximate initial requirements a trial section of built

up section composed of I.P.E. No. 450 + 1/2 I.P.E. No. 450 was tried first and

may be checked knowing the following geometrical properties:

A = 164.084cm2, Zt = 3292.2cm3, Zc = 3059.73cm3, ix = 28.9cm,

iy = 3.91cm, rt = 5.54cm, d = 86.4cm, Af = 19x1.46 = 27.74cm2,

L’ = 200.3cm

cm/t.E.Mfbc272020222

safe K.O.cm/t.cm/t..x.

E.ZMf

cm/t..Z

f

tbt

cbc

22 1279085023292

20222

720733059

.x.Zt 85023292

As Lumax’ = 200.3cm considering that Cb = 1.75 as α = 0 for this section,

< Luact =533cm, therefore there is a need to check for the lateral torsional

buckling stress.

fltb is controlled by the greater value of

28407514865337427800800

cm/t....x

.xC.d.L

Af b

u

fltb

Where: α = 0/22.02 = 0 and Cb =1.75

Or bb CCL 533

ytu

y

b

y

b

t

u

f.r/L

fC

&fC

..r

L

841882196545

533

2

yb

ytultb f.

xCE..f

51761640

f/t.x).(f 58072163632196640 22

yltb f.cm/t..x.xE.

)(.f 580721630151761

640 2

Therefore the lateral torsional buckling fltb = 1.72t/cm2. Therefore the allowable value of fbc is controlled by the lateral torsional buckling stress which is 1.72 t/cm2 which is larger than the actual stress = 0.72 t/cm2 O.K safe.

Case 2 of existing knee bracing

N.A.

360mm 338.7

12 I P E 360

I.P.E. 360

320

170

812.7

2 I.P.E. 360

Searching for these approximate initial requirements a trial section of built

up section composed of I.P.E.. No. 360 + 1/2 I.P.E.. No. 360 was tried and

may be checked knowing the following geometrical properties

A = 118.874cm2, Zt = 2015.73cm3, d = 68cm, 2Af = 17x1.27 = 21.59cm2 , rt = 5.07cm, L’ = 153.35cm

cm/t.E.ZMfbc

28260382000

20222

K.O.cm/t.cm/t..x.

E.ZMf

.Z

tbt

c

22 12291850732015

20222382000

As Lumax’ = 153.35cm considering that Cb = 1.26 as α = -12.16/22.08 =

-0.55 for this section, <Luact=200cm, therefore there is a need to check for the , ,

lateral torsional buckling stress:

fltb is controlled by the greater value of fltb y g

26126168200

5921800800cm/t...

x.xC.

d.LA

f bu

fltb

or

844339075

200fC

.Lu b

212580

075

cm/t.f..f

f.r

yltb

yt

Therefore the value of fbc is controlled by the lateral torsional buckling stress of a value of 2 1 t/cm2 which is greater than the compressive bending stress ofa value of 2.1 t/cm , which is greater than the compressive bending stress of 0.826 t/cm2 . Therefore the chosen section is safe. Although using smaller section may be quite safe but this section satisfies the maximum permissible deflection and any reduction in member stiffness may fall beyond the allowabledeflection and any reduction in member stiffness may fall beyond the allowable deflection limits. However, on using knee bracing there is a saving on rafter steel weights that reach 27.56%

Thank you y&

Good LuckGood Luck

Ass. Prof. Dr. Ehab Boghdadi Matar