06-design of main truss

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Main Truss

Design members 2009

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<15

Y

Y

X

X

/o

Butt Weld

Gusset plate

Diagonalmember

Verticalmember

5 : 1



Main Truss

Design members 2009

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Design of truss members in Bridges

(As welded members)

b

b

tt

b12

23 b

ht

t

UpperChord

LowerChord

Verticalordiagonal

t

th

G.PL

G.PL

10-20cm

fl

w

Gusstplate

flb

Buckling lengths of members: ECP 58, 59 table 4.5

Some notes for table 4.5

• Single triangulated web system is W or N-truss.

• Compression chord effectively braced is for all verticals and

diagonals of Deck Bridge.

• Compression chord un-braced is for all verticals and diagonals of

Pony Bridge.

• Compression chord may be braced or un-braced for verticals and

diagonals of through bridge according to the suggested shape of

the upper bracing.

• For all buckling lengths see page 40



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We have to study how to design truss member subjected to:

1- Compression only 2- Tension only 3- Moment & tension

4- Moment & Comp 5- Zero member 6- M only

(I) Design procedure of upper chord (compression member):

Assume stress for steel 44 f ≈1.2à 1.5 t / cm2

Assume stress for steel 52 f ≈1.5à 1.9 t / cm2

Calculate area = f

F max ------ cm2

= 2 h t + b /

t

Take h =15121512

length

→=

→SPanel ECP 129

Note: Panel length is the length of the member

Take b = (0.75à 1) h for deck or through ECP 129

b = (1à 1.25) h for pony bridge

b /

= b + 2 * (10à 20 cm )

Applying the equation of areaà

get (t) = ---- cmNote: If t ≥ 4 cm, use reduced Fy t≥ 1.2cm

Minimum thickness is 1.2cm for upper and lower chord because this

plate will be used as gusset plate at the position of connection. The

minimum thickness of gusset plate is 12 mm ECP 133.

• If the thickness "t" is very big, we can use box section or use stiffener

10-20 cm

t

t

h

bC=10-20cm

t

b



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(1) Check local buckling:

We must be sure that the section is not slender by applying the non-

compact conditions in the code.

yF t

b 64≤ (Table 2-1 b) code p. 10 ( plate)

yF t

c 21≤ (Table 2-1 c) code p. 11 (as cantilever of L – section)

yF t

h 30≤ (Table 2-1 d) code p. 12 (as cantilever of T – section)

Note: IF yF t

h 30> so we can use "box section"

Then check that yF t

h 64≤

(1) Check Global buckling:

The limit of buckling " λ " ≤ 90 for railway ECP 51

x

inin

r

l=λ = ---- ≤ 90

y

out out

r

l=λ = ---- ≤ 90

To calculate r x , r y :

Calculatet bht

t ht bhht y

/

/

2

)5.0()5.0*2(

+

++=v

Calculate I X = 2 *12

3

th + 2 t h ( y - 0.5 h) 2

+ b /

t ( y - h – 0.5 t)2

= --- cm4

I y = 2 t h (0.5 b + 0.5 t)2

+12

3 / tb

= ---- cm4

Calculate A = b /

t + 2 h t &

A

I r x x = &

A

I r

y y =

Y

t

h

b

X X

Y

Y

b

t

t



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(2) Check compressive stresses

Actual stresses = A

F f I Ld

ca++= = ---- t / cm

2 ≤ F C

Allowable stresses = F C = 1.6 – 8.5 * 10-5 2maxλ for st. 44

F C = 2.1 – 13.5 * 10-5 2

maxλ for st. 52

Important note: There is no check fatigue in the upper chord because

both the maximum and the minimum forces are compression.

II- Design procedure of lower chord (tension member):

Fatigue in tension members must be considered

How to consider fatigue?

)1(max

minmax

T

T

F F sr

−= So we have to get Fsr from ECP 41

The member will be welded, so detail "B"

To calculate number of cycles Table 3.1b ECP40

Member description No F s r

- Chord members (Class I) (L<10m) > 2 * 106

1.12

Double track

(D.T.)

2 * 105

2.80web members

(vl.& diag.)

(Class II) single track (S.T.) 5 * 105

2.00

Double track

(D.T.)

5 * 10 5 2.00web members

(carry load of

X.G. only)

(Class III)

single track (S.T.) > 2 * 106

1.12

Notes:

• For chord member: "L" is length of member and always < 10m



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• If more that 2 tracks, we can use "n" of double tracks

• "n" = 2*105

is not in the table, so we have to calculate it from the

chart ECP 42.

• Class III means that the force in the vertical member calculated

from one X.G. and there is no affect of diagonals. (See marked

members)

Sub-divided W-Truss

Design procedure:

Take "b, h" as the upper chord

b /

= (1/2 2/3 ) * b

Assume

)1(max

minmax

T T

F F sr

−

=

For st 44 Take smaller from F max or 1.6 t / cm2

= Area

T max

For st 52 Take smaller from F max or 2.1 t / cm2

= Area

T max

Get A = -- cm2 = 2 h t + 2 b / t

Get t = -- cm tmin = 12mm t≤ 4 cm or use reduced Fy


There is no check local buckling for

pure tension members.

b

12

23 b

h

t

t

Lower Chord

12

23 bb =y

XX

y

Stiffenerh



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x

inin

r

l=λ = ---- ≤160

y

out out

r

l=λ = ---≤ 160

Where A

I r xin =

A

I r

yout =

And we have to calculate y to get Ix

(3) Check stresses:

Max. Stressà A

T max

= --- ≤ 1.6 t / cm2

or 2.1 t / cm2

Stress rangeà A

F F

A

T T d I Ld −=

− ++minmax = ---- F sr

Note that: For case of railway, Tmax = Fd+L+I .

For case of roadway, Tmax = Fd +0.6 FL+I.

Note that: For lower chord, Fmin = Fd

(4) Check deflection: (roadway)35&(railway)30dl ≤ ECP 129

Note that: for lower chord: d = h + t

(III)- Design procedure of web members (compression members):

Assume f ≈1.2à 1.5 t / cm2

st. 44

1.5à 1.9 st. 52

Get Area = f

forceMax= --- cm

2

= 2 bfl * tfl + b t w

Take "b" as chord members exactly

Get t w from yw F t

b 64≤ & t w ≥ 1.0 cm

t

t

b

Y C Y

x

xFlange

Web

f

flb

W



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Applying the equation of areaà Get bFL * tF L = --- cm2

Assume flt20)(10 −= flb Get bFL& tFL

)20()10(flange.

It is preferable that b > bfl


y fl F t

C 21≤

yw

fl

F t

st b 6422≤

−−


Calculate Iout =2

3

)22

(212

)2( fl fl fl

flw x

t bt b

t bt I −+

−= (Buckling outside

is about the stronger axis)

12*2

fl fl y

bt I =

A = (b – 2 flt ) t w + 2 fl flt b A

I r x x = A

I r y y =

y

inin

r

l=λ = --- ≤ 90 & 90≤−−==

y

out out

r

lλ

(y - y axis with inside plane) & (x-x axis with outside plane)

(3) Check stresses:

2max

510*5.86.1 λ−−≤−−== Area

Force f ca Or 2.1 – 13.5 * 10

-5 2maxλ

(4) Check deflection:

30d

l≤ For web member d = bfl

No check deflection for vertical members



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(IV) Design procedure of web members (as tension members):

Assume −−−=−

=

max

minmax

1

T

T

F F sr

yF 58.0≤ Get area =max

max

F

T --- cm

2

Get b, t w, flt , flb as before. (No check local buckling)

If Area of flange is negative, so use minimum flange which is:

tfl = 10mm bfl = 6φ + tw (assume using M24 or M27)

- Check global buckling ≤λ 160

- Check maximum stress = yF A

I Ld F T

F 58.0

max

max ≤

++=

=

- & stress range (fatigue). = sr I L F

A

F ≤+

Check deflection: 30d

l≤ For web member d = bfl

(V) Design procedure of zero members (web members):

Take yw F t

b 64≤ get t w = -- cm

Take t w = flt = --- 1.0 cm

Get flb by assuming that there is only one

row of bolts each side of the flange taking

bolts M24 , So flb 6d + t w + 2 s

Check local buckling yw

fl

F t

st b 6422≤

−−&

y fl

w fl

F t

st b 212 / 2 / ≤

−−

Check global buckling: y

inin

r

l=λ = --- ≤ 90 & 90≤−−==

y

out out

r

lλ

Check deflection: 30b

l

fl ≤

b

<1.5d

bolt

Gusset

plate

tW

bfl

tf



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(VI) Design procedure of zero members (lower chord):

Take "t" the bigger value of

yF t

h 64≤

yF t

b

t

c 212 / / /

≤≈

Where h, b, b /

are taken as other

chords.

- Check local buckling

- Check global buckling y

inin

r

l=λ ≤ 90 & 90≤=

y

out out

r

lλ

- Check deflection: (roadway)35&(railway)30d

l≤ ECP 129

Note that: for lower chord: d = h + t

(VII) Design vertical or diagonal subjected to tension and

compression at the same time:

Calculate Area from both tension and compression

a) Assume f ≈1.2à 1.5 t / cm2

st. 44

1.5à 1.9 st. 52

Acomp =

f

forcencompressioMax= --- cm

2

b) Aten =1.26.1

forcenMax tensio

or = --- cm

2

Take bigger of Acomp and Aten

b is the same as the truss

Take yw F t

b 64≤ get t w = -- cm

Stiffenerbch

cb =(1

223 )b

t

XX

y

y

1 0 c m



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A2 flanges = A – b*tw Assume bf = 20tf get bf and tf

Checks: 1) As compression member:

Check local buckling yw

fl

F t

st b 6422

≤

−−

& y fl

w fl

F t

st b 212 / 2 /

≤

−−

Check global buckling: y

inin

r

l=λ = --- ≤ 90 & 90≤−−==

y

out out

r

lλ

Check stresses: cF A

≤C

2) As tension member:

Check stresses: yF A

58.0T

≤

Check deflection: 30b

l

fl

≤

Check fatigue: f sr = sr F A

cT ≤

−− )(

i.e. If the forces in the diagonal is as following:

Fd = 40 t(Comp) FL+I(max) = 120t (Comp) FL+I(min) = 50t (tens)

So Fmin = 50-40 = 10t (tens) Fmax = 120+40 = 160t (comp)

We have to design the member as compression member (160t) and check

as tension member (10t).

10/A

60/A

fsr

ft

fc

Check fatigue: f sr = sr F A A A

cT <=

−−=

−− 170)160(10)(



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Design members 2009

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Summary of buckling lengths of members:

Table 4-5 ECP 58 and 59

1- Deck bridge:

• All upper chords: Lin = 0.85L and Lout = 0.85L

• All verticals and diagonals = Lin = 0.7L and Lout = 0.85L

• For lower chord: Lin = 0.85L and Lout = 0.85L or 0.85(2L)

According to the shape of the lower bracing

L1 L3

Lower bracing of deck bridge

• For L1: Lin = 0.85L and Lout = 0.85L

• For L3: Lin = 0.85L and Lout = 0.85(2L)

2- Pony bridge• All verticals and diagonals = Lin = 0.7L and Lout = 1.2L

• For lower chord: Lin = 0.85L and Lout = 0.85L

• All upper chords: Lin = 0.85L and Lout = 2.5 4 δa EI y



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Design members 2009

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45.2 δa EI l yout = a is spacing between X.G.

E = 2100 t / cm2

& a = X.G spacing

I y = y – y inertia of upper chord=δ Flexibility of U- frame (cm / t)

If not given2

22

1

31

23 EI

d

EI

d +=δ

I 1 = I X - X of vertical member I2 = IX – X of X.G

x

x

x xX.G. Verticalmember

Important note: Incase of using U – Frame every X.G, all the verticals

will be subjected to moment due to C/100 and due to wind in addition to

the force from vertical loads which is tension or compression.

d

B

2

d1

y

y

Upper

Chord



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Design members 2009

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3- Through bridge

a) If the upper bracing is as shown:

Upper bracing of through bridge

• All upper chords: Lin = 0.85L and Lout = 0.85L

• All verticals and diagonals = Lin = 0.7L and Lout = 0.85L

• For lower chord: Lin = 0.85L and Lout = 0.85L

b) If the upper bracing is as shown:

For All lower chords: Lin = 0.85L and Lout = 0.85L

Upper chord: For U1 : Lin = 0.85L and Lout = 0.85L

For U2 : Lin = 0.85L and Lout = 0.85(2L)

M.G.Plan of upper bracing

V1V2

V3 V4 V5D1

D2

D3

D4

D5

u1u2



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Design members 2009

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Verticals and diagonals

For V1, D1, V2, D2, V4, D4 the compression chord is

effectively braced Lin = 0.7L and Lout = 0.85L

For V3, D3, the compression chord is not braced

Lin = 0.7L and Lout = 1.2L

Design of members subjected to moment

Design procedure of Verticals of pony bridge:

M Only All verticals(M+N)

(M+T)

Always all verticals of Pony Bridge subjected to moment in addition

to the main force from cases of loading.

Mx of pony = hF

*100

max where h = hmg -2

xgh

Mx of wind = (0.1S*0.5)*2

2h

where "S" is spacing between X.G.

F

100

h

max F

100max

C.L. of X.G. V L

o f p o n

y

b r i d g e

C.L. of X.G. V L

o f p o n

y

b r i d g e

M1M2

The moment is outside the plan.(about x-axis of the vertical member)

Fmax is the maximum compression force in the upper chord. (The force in

the upper chord at mid span of truss)



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Note that: The effect of hF

*100

max and wind is case "B"

Procedures:

(1) If the member is subjected to M + N:

a- Estimate section: "b" of the whole truss or ≈1512 −

S

For web: yw F t

b 64=

For flange: Force =3

N b

M x + (comp)

Assume F max = 1.5 t / cm2

for st. 44 & 2.0 t / cm2

for st.52)

Af =maxF

Force= bfl tfl complete as before

Note that: We can estimate the section as previous in case of web

member subjected to pure compression & decrease the assumed stress.

b- Check member as pure compression member for case "A"

- Local buckling

- Global buckling ≤ 90

- C ca F f <

c- Check member as beam – column (subjected to moment & normal)

for case "B"

2.11 ≤+ AF

f

F

f

bcx

bcx

C

ca Where Cca F& f from previous step "b"

y

I

M f

out bcx

max= [

2

by&IisI Xout = ]

N/2 M

MN

x

b

N/2

b tW

tf

b

Y C

fl

Y

x

x

xM

bx



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Design members 2009

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inbl L −≈actu of member (important)

Because the moment is outside plan, so L.T.B is inside plan.

Calculate L u max then F bcx

bigger

smaller

M

M =α = zero C b = 1.75

If 15.0

1

≤C

ca

F

f A 1 = 1

If 15.0

1

>C

ca

F

f

EX

ca

F

f A

−=

1

85.01

Where2

7500

x

EX F λ

= out is xλ

Note: The moment rotates about the stronger axis

- Check using interaction equation

2.11 ≤+ A

F

f

F

f

bcx

bcx

C

ca

(2) If the member is subjected to M + T:

a - Estimate section as before (web subjected to pure tension) using

ysr F

T

T

F F 58.0

1max

minmax ≤

−=

Or Force in the critical flange =3

T

b

M x + (tens) Afl =

max

max

F

T

b - Check member as pure Tension

- Global buckling ≤ 160

- f t ≤ 0.58 F y

- f sr ≤ F sr



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No check deflection because it is vertical member

c -Check using interaction equation as member subjected to M & T

2.1≤+ btx

btx

t

t

F

f

F

f

f t as before = Area

T max , F t = 0.58 F y

y I

M f

X btx = Where I X = I out & y =

2

b

F btx = 0.58 F y (always because there is no. L.T.B in tension side)

d- Check fatigue:

Fmax – Fmin ≤ Fsr

Fmax = y I

M

A

T

out

*max + M is Mw + hC

*100

max Cmax = CD+L+I

Fmin = y I

M

A

T

out

*min + M is hC

*100

min Cmin = CD

CD

and CD+L+I

are of upper chord at mid span of truss.

DO Not take the effect of wind in studying fatigue

(3) If the member is subjected to M only

a - Estimate section as web zero member (zero member) use b of

truss, t w = 1 cm, b f (min.) = 6φ + 2s + t w (s is size of weld ≈1cm)

t f (min.) = 1 cm.

b - Check as zero member

- Local buckling

- Global buckling < 90

- No check deflection because it is vertical member



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c – Check y I

M

X

X (I X = I out & y =2

b) ≤ F bcx * 1.2 where F bcx is

calculated using inbactu l≈ L & L u max (C b = 1.75).

Design procedure of Vertical of through bridge forming the closed

portal frame:

The member is subjected to

either tension and moment

or compression and

moment.

a- Estimate the section as before (section subjected to moment and

compression or moment and tension).

b- Check the member as pure compression or pure tension (case A)

- To calculate buckling lengths:

Buckling inside: From table of code for web members lin = 0.7l

Buckling outside: The vertical member forms portal frame outside plan,

so lout is calculated using GA and GB.

GA = xg xg

vv

l I

l I

/

/

GB =beambeam

vv

l I

l I

/

/

Lxg is the width of the bridge = Lbeam

Ibeam is the Ix of the horizontal beam.

Note that: Buckling inside and outside of all other verticals and

diagonals is from table of code.

The frame is allowed to sway: Cmx = 0.85

First vertical(M+N) (M+T)

CL of upper beam

CL of XG

V l o f t r u s s

2h/3

h/3

Ru

M



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Design members 2009

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5.0+==bigger

smaller

M

M α Cb = 1.75 + 1.05*0.5+0.3*(0.5)

2= 2.35, taken

2.3

c- Check as beam column (case B)

Design procedure of diagonal of through bridge forming the open

portal frame:

The member is subjected to compression

and moment.

a- Estimate the section as before

using the same section of the upper chord.

b- Check the member as pure compression (case A)

a. To calculate buckling lengths:

Buckling inside: From table of code for web members lin = 0.7l

Buckling outside: The diagonal forms portal frame outside plan, so lout is

calculated using GA and GB.

GA = 10 (hinged base)

GB =beambeam

dldl

l I

l I

/

/

Lbeam is the width of the bridge

Ibeam is the Ix of the horizontal

beam.

Note that: Buckling inside and outside of all other verticals and

diagonals is from table of code.

The frame is allowed to sway: Cmx = 0.85

First diagonal(M+N)

CL of upper beam

V

l o f t r u s s

L(inclined distance)

Ru

R /2u R /2u

M



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Design members 2009

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0==bigger

smaller

M

M α Cb = 1.75

c- Check as beam column (case B)

Summary of all previous:

1. For deck bridge:

- All the members of the truss are subjected to either tension

or compression only (NO moment)

- lin & lout of members are from table in Code ECP 58&59

2. For pony bridge:

- All members of the truss are subjected to either tension or

compression except verticals.

- All verticals are subjected to moment in addition to actual

tension or compression.

M = 0.1*0.4 S * h2 /2 + hC *

100

max where h = hMG -2

XGh

- Buckling length of all members are from ECP 58&59 except

upper chord.

- Buckling length of upper chord is 45.2 δS EI y

3. For through bridge:

• If there is inclined frame, so the section of the first inclined

diagonal is the same as the upper chord.

- All the members subjected to either tension or compression

except vertical or diagonal forming the portal frame.

Vertical: M = Ru /2 * 2/3 h Ru of loaded case

Horizontal: M = Ru /2 * l Ru of loaded case



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Design members 2009

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Lin & lout for all members are from ECP 58&59 except for vertical

or diagonal forming portal frame. We use GA & GB

For vertical: GA = xg xg

vv

l I

l I

/

/

GB = beambeam

vv

l I

l I

/

/

For diagonal: GA = 10 (hinged base) GB =beambeam

dldl

l I

l I

/

/

Very important note;

b is constant for whole of the truss. So before design we must check if

any section is given in the exam, we must take "b" from it

Solved Example 1:

Deck bridge as shown, h of truss = 4m, St.44, single track

U4 U3

U1

V1

d1

V2

d2 V3d

3

V4

d4 V5

L1

L2

L3

R=241t 5.0m40.0m

L4

2U

The forces in the shown members are given in the following table:

Member Maximum Minimum Member Maximum Minimum

V2 -96 -15 D1 +352 +78

U3 -515 -113 D2 -260 -46

L2 +418 +90 D3 +181 -2

L4 +553 +119 D4 -110 0

It is required to design U3, U4, L4, V1 and D3



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Design members 2009

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Solution:

Design (U 3 &, U4 ): They have the same force

h =1512

500

→= 41.7 ≈→ 3.33 40 cm

For deck bridge b ≈ (0.75 →1)* 40

= 30 →40 cm ≈36 cm

Assume f = 1.2à 1.5 for st. 44

A =

2.1

515= 429.2 cm

2Taking b

/ ≈36 + 2*(10à 20) = 56à76 ≈60 cm

So A = 429.2 = (2 * 40 + 60) * t t = 3.07 cm ≈ 3.2 cm

• Check local buckling

Taking weld size ≈ 1 cm

8.2

646.10

2.3

1*236≤=

−=

t

b= 38.2

8.2

214.22.3

12.312

≤=−−

=t

c= 12.5

8.2

302.12

2.3

140≤=

−=

t

h= 17.9

• Check global buckling

)2.3*60()2*2.3*40(

)6.41*2.3*60()2*20*2.3*40(

++

= y = 29.3 cm

]2*12

40*2.3[

3= x I + [3.2*40*(29.3-20)

2*2] + [3.2*60*(29.3-41.6)

2]

= 85322 cm4

]12

60*2.3[

3

= y I + [2*3.2*40*(18+1.6)2] = 155945 cm

4

A = (3.2 * 2 * 40) + (60 * 3.2) = 448 cm2

3.2 cmy

3.2

40

36

X X

Y 2 9 . 3 c m

12

60

Y



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Design members 2009

24/46

488

85322= xr = 13.8 cm &

488

155945= yr = 18.7 cm

Deck bridge (upper) inl = 0.85 l & out l = 0.85 l

908.308.13

500*85.0≤==inλ 907.22

7.18

500*85.0≤==out λ

• Check stress

16.1448

515==ca f t / cm

2 ≤ F C = 1.6 – 8.5*10-5

* 30.82

= 1.52 t / cm2

Safe but waste Try smaller thickness 2.6 cm

Design (L 4 ) Lower tension

h = 40 cm & b = 36 cm

36*)3

2

2

1( / →=b = 18à 24 cm ≈ 20 cm

Chord, L<10m n=over 2,000,000

Detail B Fsr = 1.12 t / cm2

553

1191

12.1max

−=F = 1.43 t / cm

2< 1.6 t / cm

2

1.43 = A

553get A = 386.7 cm

2= (2*20 + 2*40) * t

Get t = 3.22 cm ≈ 3.4 cm

• Check global buckling

)4.3*40*2()4.3*20*2(

)4.23*4.3*40*2()7.1*20*4.3*2(

++

= y = 16.2 cm

A = 408 cm2

]2*12

40*4.3[

3

= x I + [3.4*40*2*(16.2-1.7)2] + [3.4*20*2(16.2- 23.4)

2]

= 100505 cm4

3640

t3.4

Y

Y

X X

20



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Design members 2009

25/46

]2*12

20*4.3[

3

= y I + [3.4*20*2*(2

4.3

2

36+ )

2] + [3.4*40*2(18+1.7)

2]

= 162874 cm4

408

100505= xr = 15.7 cm &

408

162874= yr = 20 cm

Take inl = 0.85 l & out l (from lower bracing)

Assume the lower bracing is as shown below:

Plan of suggested lower bracing

L3 4and L

1601.277.15

500*85.0≤==inλ 1605.42

20

500*85.0*2≤==out λ

• Check stress

6.136.1

408

553.max ≤= → t / cm2

12.106.1408

119553. ≤=−

→ fatiquet /cm

2

3052.114.340

500≤=

+ → Deflection

Design (V 1): Web member comp:

Assume f = 1.1à 1.4 = 1.1 t / cm2

= A

241

A = 219 cm2

b = 36 cm8.2

6436≤

wt = 38.2

t w = 0.94 ≈1.0 cm

A = 219 = (36 * 1) + (2 flb flt ) flb flt = 91.5 cm2

281.0

3.2

36

29.6

Y Y

X

X



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Design members 2009

26/46

Take flb ≈ 10 flt à flt = 3.02 ≈ 3.2 cm flb =2.3

5.91 ≈ 28 cm

1. Check local buckling

Taking size of weld s ≈ 1.0 cm

8.2

219.3

2.3

12 / 12 / 28≤=

−−=

flt

c= 12.5

8.2

646.27

0.1

1*22.3*236≤=

−−=

wt

b= 38

2. Check global buckling

12

6.29*1 3= x I + 2*3.2*28*(18-1.6)

2= 50359 cm

4

12

28*2.3*2

3

= y I = 11708 cm4

A = (1*29.6) + (2*28*3.2) = 209 cm2

209

50359= xr = 15.5 cm &

209

11708= yr = 7.5 cm

In deck bridge inl = 0.70 l & out l = 0.85 l

9037)5.7(

400*70.0≤==

yin

r λ 9022

)5.15(

400*85.0≤==

xout

r λ

3. Check stress

15.1209

241==ca f t / cm

2 ≤ F c = 1.6 – 8.5 * 10-5

* 372

= 1.48 t / cm2

Design (D 3): Web member tension

Fmax= +181t Fmin= -2t

Web member (diagonal), do not carry floor reaction only (class II)

Single track n=500,000 Detail B Fsr = 2 t / cm2

Note that: If double tracks, so n=200,000 Fsr = 2.8 t / cm2



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Design members 2009

27/46

181

21

0.2max −

−=F = 1.98 > 1.6 t / cm

2

So 1.6 = A

181 ∴ A = 113 cm2 , b = 36 cm

8.2

6436≤

wt = 38.2 t w = 0.94 ≈ 1.0 cm

We used local buckling because this member will be subjected to

compression.

A = 113 = (36 * 1) + (2 flb flt ) ∴ flb flt = 38.5 cm2

Taking flb ≈ 10 flt à flt = 1.96 = 2.0 cm & flb =2

5.38 ≈ 20 cm

1 - Check global buckling

12

32*1 3

= x I + 2*20*2*(18-1)2

= 25851 cm4

12

20*2*2

3

= y I = 2667 cm4

& A = 1 * 32 + 2 * 20 * 2 = 112 cm2

112

25851= xr = 15.2 cm &

112

2667= yr = 4.9 cm

inl = 0.70 l & out l = 0.85 l (deck) & 22 54 +=l = 6.4 m

16091)9.4(

640*70.0≤=

==

yin

r λ 16036

)2.15(

640*85.0≤=

==

xout

r λ

2 - Check stress

6.161.1112

181.max ≈= → t / cm2

201.0

2.0

36

32

Y Y

X

X



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Design members 2009

28/46

0.265.1112

)2(181. ≤=−−

→ fatiquet / cm

2

3032

20

640. >= → DeflectionUse bfl = 22cm and recheck

• Since the minimum force in the member is compression, so we

must check this member on "2t"

Local buckling: Taking size of weld s ≈ 1.0 cm

8.2

213.4

2

12 / 12 / 20≤=

−−=

flt

c=12.5

8.2

6430

0.1

1*22*236≤=

−−=

wt

b= 38

• Global buckling 9091 ≈=inλ & 9036 ≤=out λ O.K.

02.0112

2==ca f t / cm

2 ≤ 1.6 – 8.5 * 10-5

* 912= 0.9 O.K.

Solved Example 2:

Pony bridge as shown, h of truss = 5.5m, St.44, single track

5.5 m

10*5.5=55 m

3.9 m

J1

-945t -985t

+ 1 4 4 t

- 7 6 t

-220t-198t

Joint1

5.5 m

5.5 m

It is required to design the upper chord and vertical member at joint J1

Take b = 420 mm δ = 0.12 cm / t



Main Truss

Design members 2009

29/46

Solution:

(1) Upper chord: F = 985 t (comp.)

Assume F C = 1.3 t / cm2

3.1

985=g A = 757.7 cm

2

Choose 451512

550≈

→=h cm

Take b u = 42 + 20 = 62 cm

757.7 = (2 * 45 + 62) t t = 4.98 cm ≈ 5.0 cm

Check local buckling:

8.2

648

5

1*242≤==

−t

b= 38 O.K.

8.2

218.0

5

15)2

4262(

≤=−−

−

= 22.5 O.K.

8.2

308.8

5

145≤=

−=

t

h= 17.9 O.K.

)5*62()2*5*45(

)5.47*5*62()2*5.22*5*45(

++

= y = 32.7 cm

A = 62 * 5 + 45 * 2 * 5 = 760 cm2

12

45*5[

3

= x I +5*45*(32.7-22.5)2]*2 + [62*5*(47.5-32.7)

2] = 191304 cm

4

]1262*5[2*])

25

242(*5*45[

32 ++= y I = 348753 cm4

760

191304= xr = 15.87cm &

760

348753= yr = 21.42 cm

bxbin ll = = 0.85 * 550 = 467.5 cm

4 12.0*550*348753*21005.2== bybout ll = 1172.2 cm

5

45

42 Y

32.7

62

55X X

y



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Design members 2009

30/46

87.15

5.467= xλ = 29.47

42.21

2.1172= yλ = 54.72 < 90

t > 4cm F c = 1.5-7.5*10-5

54.722

= 1.275 t / cm2

760

985=c f = 1.296 t / cm2

> 1.275 t / cm2

Unsafe

Increase dimensions and recheck

(2)Vertical member:

Since this is Pony Bridge, so the vertical is subjected to bending

momentThe maximum force of vertical can be calculated from diagonal.

.)(102)45sin144(sinmaxmax compF F d −=−=−= α

.)(54)45sin76(sinminmin tensF F d +=−−=−= α

For (S.T.), class (II) (do not carry X.G. only) n = 500.000

F sr = 2.0 t / cm2

M add =100

1F max (max. comp. force in upper chord) * h

/ (HM.G – HX.G /2)

h = 5.5 – 0.7/2 = 5.15m

M add (max) =100

1* 985 * 5.15 + 0.1*5.5*0.5*

2

15.5 2

= 53.6 m t

M add (min) =100

1* 220 * 5.15 = 11.33 m t

Since the member is subjected to both tension and compression, so we

have to check the vertical twice.



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Design members 2009

31/46

• Design as M + N:

8.2

6442=

wt à t w = 1.1 cm t w = 1.2 cm

Force in flange =42

5360

2

102+ = 178 cm

2

Assume f = 1.2 t / cm2

f f t b∴ =2.1

178= 157 cm

2

Assume b = 10 t 10 t2

= 157

t = 4 cm & b = 40 cm

I y (inside) = 2 * 4 *12

403

= 42667 cm4

I X (outside) = 1.2 *12

343

+ 2 * 40 * 4 *2)

2

34

2

4( + = 119450 cm

4

A = 2 * 40 * 4 + 34 * 1.2 = 360.8 cm2

8.360

42667

=inr = 10.9cm & 8.360

119450

= yr = 18.2 cm

• Check local buckling:

8.2

647.26

2.1

234<=

−=

w

w

t

h= 38

8.2

216.4

4

12 / 2.12 / 40<=

−−=

f t

C = 12.5

• Check global buckling:

90359.10

550*7.0<==inλ 903.36

2.18

550*2.1<==out λ

• Check stresses:

F C = 1.6 – 8.5 * 10-5

* (36.3)2

= 1.49 t / cm2

8.360

102=ca f = 0.28 t / cm

2< 1.49 t / cm

2

42

Y Y

X

X

4.0

4034*1.2



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Design members 2009

32/46

• Check as beam-column

23.36

7500= EX F = 5.69 t / cm

2

49.1

28.0=

C

ca

F

f = 0.19 > 0.15

69.5

28.01

85.0A 1

−=∴ = 0.89 take A 1 = 1

L U act = L in = 0.7 * 550 = 385 cm α = zero c b = 1.75

8.2

40*20maxu = L = 478 cm or 75.1*

8.2*42

4*40*1380= 3286 cm > 385

bcxF ∴ = 0.58 F

y= 1.6 t / cm

2

1*6.1

2

42*

119450

5360

49.1

28.0+ = 0.78 < 1.2 Waste

Decrease b, try flange 30 * 3 cm (Better to assume f = 1.5 t / cm2

from

the estimation)

• Check for case of M + T

λ < 1608.360

54=t f = 0.15 t / cm

2< 1.6 t / cm

2(case A)

2

42*

119450

5360=btx f = 0.94 t / cm

2

6.1

94.0

6.1

15.0+ = 0.68 < 1.2 (case B)

Check deflection: flb

l=

40

550= 13.75 < 30

Check fatigue: Mfatigue= 15.5*985*100

1= 50.7 mt (without wind)

F max =8.360

54+

2

42*

119450

5070= 0.15+ 0.89 = 1.04 t / cm

2



Main Truss

Design members 2009

33/46

2

42*

119450

1139

8.360

102min +

−=F = - 0.08 t / cm

2

F max – F min = 1.04 – (- 0.08)

= 1.12 t / cm2

< 2 t / cm2

Web member single track,

n = 500000, detail "B"

Solved Example 3:

Through bridge single track, with span 78 m. Spacing between X.G. is 6

m & of depth 0.8 m. Steel used is st 52. Width of bridge is 5.5 m.

It is required to design the marked members.

7.5

Inclined portal frame

6m78m

M

M

4M

M2

1

3

Solution:

D.L.: W S = 0.9 * (0.75 + 0.05 * 78) = 4.185 t / m /

2

6.0

2

185.4+= DLW = 2.39 t / m

/

For Member M 1:65.7tan 1−=α = 510

1 0

0 .

3 5

0 .

5 5

0 .

6 7

1 .

1 6

1 . 1

9 0 .

7 2

1 . 0

3

0 .

9 2

1 2 .

5

6 .

2 5

6 .

2 5

1 2 .

5

1 2 .

5

1 0

6 .

2 5

1 0

1 0

1 0

1 0

6 .

2 5

1 2 .

5

1 2 .

5

1 2 .

5

1 0

1 0

1 0

1 0

1 0

1 0

1 0

1 0

1 0

1 0

0 .

1 5

1 0

1 0

1 0

0 .

0 3

1 0

1 0

0 .

8 4

M

C

minmax

T

M



Main Truss

Design members 2009

34/46

1- Dead load: F D =2

1* 78 * 1.19 * 2.39 = 111 t

2- Live load:

F L = 6.25 * 2 * 0.67 + 12.5 * 3 * 1.16 + 40 * 1.03 + 6.25 * 2 * 0.92 + 3 *

12.5 * 0.84 + 40 * 0.72 + 40 * (0.55 + 0.35 + 0.13) + 10 * 0.03 = 205.2 t

7824

24

+= I = 0.23 take I = 0.25

F L+I = 205.2 * 1.25 = 256.5 t

From Wind: The diagonal is subjected tomoment as a member in the inclined open

portal frame.

W upper (loaded) = 0.1 [0.4 *2

5.7+ 0.4 *

2.45] = 0.25 t / m /

R U (loaded) = 0.25 * 2

78

= 9.75 t

Note that: Loaded case is critical when design the diagonal while un-

loaded case is critical when design upper bracing & the horizontal beam.

22 65.7 +=l = 9.6 m

M =2

75.9* 9.6 = 46.8 m t (case B)

Design of Member M 1:

For the whole truss; assume b =12

600= 50 cm

F D = 111 t & F L+I = 256.5 t & F tot = 367.5 t (compression)

M wind = 46.8 m t & l inclined = 9.6 m

7.5

1.8mFor Upper bracing

0.10.1

2 . 4

5

3 . 7

5

0.8

CL of upper beam

V l o f

t r u s s

9.6

9.75M

4.875 4.875



Main Truss

Design members 2009

35/46

The diagonal will be in the shape of π - section (same shape of the

section of upper chord)


(We do not assume f = 1.8 t / cm2

because the

member is subjected to moment with normal force)

0.1

5.256111+= A = 367.5 cm

2b

/ = 50 + 20 = 70 cm

Assume h = b = 50 cm (2 * 50 + 70) t = 367.5

t = 2.2 cm

Check local buckling =

6.3

307.22

2.2

50>= = 15.8

We can either increase thicknesst

50=15.8 t = 3.2 cm

Or use stiffness as shown and take t = 2.2 cm

Check local buckling:

6.3

2154.4

2.2

10<==

t

c= 11.1

6.3

3054.4

2.2

10 /

<==t

c= 15.8

6.3

647.22

2.2

50<==

t

b= 33.73

6.3

642.18

2.2

40 /

<==t

h= 33.73

Check global buckling:

A = (70 + 2 * 50 + 2 * 10) * 2.2 = 418 cm2

418

)]402.2(*10*2)252.2(*50*22

2.2*0[7.22 ++++

= yv

= 19.16 cm

50

70

50 10

t=22mm

t=22mm

b

b

h h

c

c



Main Truss

Design members 2009

36/46

−++−+== 2

32

3

in )16.192.27(*50*2.212

50*2.22)

2

2.216.19(*2.2*70

12

2.2*70 I I x

+ 2

−+ 2

3

)16.192.42(*2.2*1012

2.2*10= 133720cm

4

++++==

12

10*2.2)

2

2.225(*2.2*50

12

2.2*502

12

70*2.2 32

33

out y I I

+ 22

)52.225(*10*2.2 ++ = 258826 cm4

r x =418

133720= 17.9 cm r y =

418

258826= 24.9 cm

inl = 0.7 * 9.6 = 6.72 m

out l = Since the diagonal is a member in the portal frame, so out l will

be calculated using GA and GB. So we have to know section of

horizontal beam.

- To estimate the beam of the upper bracing, the critical case is the

un-loaded case because this beam carries wind only.

W upper (unloaded) = 0.2 * 0.4 *2

5.7* 2 = 0.6 t / m

/

R U = 0.6 *2

78= 23.4 t M un-loaded =

2

4.23* 9.6 = 112.32 m t

CL of upper beam

V l o f t r u s s

9.6

23.4M

11.7 11.7

7.5m

1.8m

0.8

For Upper bracing

3 . 7

50.2 0.2



Main Truss

Design members 2009

37/46


1.2

11232=∴ X S = 5350 cm

3

Choose H E B 600 or B.U.S Note

2

550L actU = = 275 cm

Or assume upper horizontal beam is HEB 600 (Always if open frame)

GA = 10 (hinged base) GB = 87.05.5 / 171000

6.9 / 258826= K = 1.8

Note: we used I y because it is I out

905.379.17

672<==inλ 903.71

9.24

960*85.1<==out λ

(1) Check stresses:

F C = 2.1 – 13.5 * 10-5

* (71.3)2

= 1.41 t / cm2

418

5.367=ca f = 0.88 t / cm

2< 1.41 t / cm

2O.K. Safe

• For effect of moment due to wind: (case B)

I out = 258826 cm4 rout = 24.9 cm out λ = 79.3

For Π - Section There is no LTB bcxF ∴ = 0.58Fy = 2.1 t / cm2

41.1

88.0=

C

ca

F

f = 0.62 > 0.15 &

2yE3.71

7500)( =out F = 1.47 t / cm

2

47.1

88.01

85.0A 1

−=∴ = 2.12 (too large because f ca is too large)

f bcx = 63.035*258826

4680= t / cm

2

Using interaction equation:

1.2*1.2

63.062.0 + 2 = 1.26 > 1.2 (case B) Use t = 2.4cm

If we want to check for the unloaded case: (NOT CRITICAL)



Main Truss

Design members 2009

38/46

418

111=ca f = 0.27 t / cm

2

41.1

27.0=

C

ca

F

f = 0.19 > 0.15

2yE 3.71

7500)( =out F = 1.47 t / cm

2

41.1

27.01

85.0A

1 −=∴ = 1.05

M un-loaded = 112.32 mt f bcx = 52.135*258826

11232= t / cm

2

Using interaction equation:

05.1*1.2

52.119.0 + = 0.95 < 1.2 (case B) (Not critical)

• Design of Member M 2:

The member is subjected to tension only

F D =2

1* 12 * 1.0 * 2.39 = 14.34 t

F L = 12.5 (1 + 2 * 0.67) + 10 * 0.17

+ 6.25 (0.375 + 0.083) = 33.8 t (tension)

1224

24

+= I = 0.67

F L+I = 33.9 *1.67 = 56.6 t

F total = 14.34 + 56.6 = 70.95 t

b = 50 cm (as in diagonal)

Truss vertical carry floor beam reaction only (class III) Single track

n = Over 2*106

detail B F sr = 1.12 t / cm2

95.70

34.141

12.1max

−=F = 1.4 t / cm

2> 2.1 t / cm

2(0.58 F y)

4.1

95.70=∴ A = 50.7 cm

2

Take t w = 1 cm

50

Y Y

X

X

20

1.0

48*1.0

0 . 1

7

0 .

0 8 3

0 .

6 7

0

. 6 7

1 .

0 0

0 .

3 7 5

6 .

2 5

1 2 . 5

1 2 . 5

6 .

2 5

1 2 . 5

1 0



Main Truss

Design members 2009

39/46

f f t b2∴ = 50.7 – 50 * 1 ≈ zero

b min = 6 ϕ + t w + 2 S

For M27 and size of weld 1cm, bmin = 20cm

So take minimum f b = 20 cm & f t = 1 cm

in I = 2 * 1*12

203

= 1333.3 cm4

12

483

=out I *1 + 2 * 20 * (2)

2

48

2

1+ = 33226 cm

4

A = 20 * 1 * 2 + 48 * 1 = 88 cm2

88

3.1333=inr = 3.9 cm &

88

33226=out r = 19.43 cm

l = 7.5 m inl = 0.7 * 7.5 = 5.25 m

out l = 0.85 * 7.5 = 6.375 m (Supported on upper bracing)

1601359.3

525<==inλ & 16033

43.19

5.637<==out λ

Check stresses:

88

95.70=t f = 0.81 t/cm

2< 2.1 t /cm

2

64.088

34.1495.70=

−=sr f t /cm

2< 1.12 t /cm

2

For verticals: no check deflection

• Design of member M3:

For x = zero

F = tan-1

= 7.5/6 = 51.30

Fmin= 111*cos51.3=69.3t (Tension)

FL+I= 256.5*cos51.3=160t (Tension)

FL+I+D = 69.3+160=229.3 t

Stiffener

c

XX

y

y

1 0 c m

50

25

50 12

1.2



Main Truss

Design members 2009

40/46

We may use stiffener or not

inl = 0.85 * 6 = 510 m & out l = 0.85* 6 = 510 m

Complete as before

• Design of member M4: (zero members)

(1)6.3

6450=

wt t w = 1.5 cm

Take t w = t f = 1.5 cm

b min = 6 ϕ + t w + 2 S Assume M 27

b min = 6 * 2.7 + 1.5+ 2 * 1 = 19.7 cm,

take 20 cm

6.3

215.5

5.1

12 / 5.12 / 20<=

−−=

f t

c= 11.1

Check global buckling

inl = 0.7 * 7.5 = 5.25 m

lout is according to the shape of the upper bracing

Assume out l = 0.85 * 7.5 = 6.375 m

in I = 2 * 1.5 *12

203

= 2000 cm4

12

47

*5.1

3

=out I + 2 * 20 * 1.5 (2

)2

47

2

5.1

+

= 48262 cm4

A = 20 * 1.5 * 2 + 47 * 1.5 = 130.5 cm2

5.130

2000=inr = 3.9 cm

5.130

48262=out r = 19.2 cm

20

50

Y Y

X

X

1.5

47*1.5

S S

Lout

S S

outLout

out

= 0.85*7.5 = 0.85*7.5

= 0.85*7.5 = 1.2*7.5



Main Truss

Design members 2009

41/46

901359.3

525>==inλ

So increase dimensions of flanges (main item in Iin)

Assume flange 300*20 mm

in I = 2 * 2 *12

303

= 9000 cm4

A = 30 * 2 * 2 + 46 * 1.5 = 189 cm2

189

9000=inr = 6.9 cm 9076

9.6

525<==inλ

Solved Example 4:

Through bridge single track, with span 78 m. Spacing between X.G. is 6

m. The XG composed of web plate 800*14 and 2 flange plates 300*30.

Steel used is st 52. Width of bridge is 5.5 m.

It is required to design the marked members.

7.5

6m78m

M1

M2

M4

M3

Solution:

• For Member M1 It is subjected to compression only.

Design of Member M 1:

For the whole truss; Take b =12

600= 50 cm (given for U1)

From previous example:

F D = 111 t & F L+I = 256.5 t & F tot = 367.5 t (compression)

The diagonal will be in the shape of π - section (same section of upper

chord)



Main Truss

Design members 2009

42/46

The section of the upper chord is given, so we use the same section of

diagonal

i.e Section of the diagonal is

Important note: If the section of U1 is not

given, estimate diagonal as for upper chord

i.e. Assume f = 1.8 t / cm2

(No moment)

8.1

5.367= A = 204.2 cm

2

Assume12600=b = 50 cm b

/ = 50 + 20 = 70 cm

Assume h = b = 50 cm (2 * 50 + 70) t = 204.2

t = 1.2 cm

Check local buckling =6.3

307.41

2.1

50>= = 15.8

Use stiffener as shown

6.3

307.41

2.1

50>= = 15.8

t = 204.2 t = 1.07 cm take (t) = 1.2 cm

Check local buckling =6.3

647.41

2.1

50>= = 33.73

Taket

50= 33.73

t = 1.5 cm

For6.3

217.6

5.1

10<==

t

c= 11.1

6.3

307.6

5.1

10 /

<==t

c= 15.8

t=15mm

1050

70

50

h

b c

c

t=15mm

b =70

h =50 =40

=50 =10

=10



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Design members 2009

43/46

6.3

647.26

5.1

40 /

<==t

h= 33.73

Check global buckling:

A = (70 + 2 * 50 + 2 * 10) * 1.5 = 285 cm2

285

)]405.1(*10*2)255.1(*50*22

5.1*0[75.1 ++++

= yv

= 18.6 cm

−++−+== 2

32

3

ni )6.185.26(*50*5.1

12

50*5.12)

2

5.16.18(*5.1*70

12

5.1*70 I I x

+ 2

−+ 2

3

)6.185.41(*5.1*1012

5.1*10= 89824 cm

4

++++==

12

10*5.1)

2

5.125(*5.1*50

12

5.1*502

12

70*5.1 32

33

out y I I

+ 2 2)55.125(*10*5.1 ++ = 172380 cm4

r x =285

89824= 17.75 cm r y =

285

172380= 24.6 cm

(2) Check stresses:

inl = 0.7 * 9.6 = 6.72 m out l = 0.85 * 9.6 = 8.16 m

Note: For out l , the joint is restrained outside with upper bracing ( we do

not use GA and GB as in previous example because the diagonal is not

forming a frame)

9.3775.17

672==inλ 2.33

6.24

816==out λ

F C = 2.1 – 13.5 * 10-5

* (37.9)2

= 1.9 t / cm2



Main Truss

Design members 2009

44/46

285

5.367=ca f = 1.3 t / cm

2< 1.9 t / cm

2O.K. Safe

Not waste because we used minimum thickness for local buckling

• Members M 3 & M 4 is the same as previous

• Member M 2: From previous example, it is subjected to tension

F D = F min = 14.34 t F max = 70.95 t F sr = 1.12 t / cm2

If the section is as in previous example:

Iin = 1333.3 cm4 Iout = 33226 cm4

A = 88 cm2

rin = 3.9cm rout = 19.4cm

1601359.3

750*7.0<==inλ

lout is calculated using GA and GB because

the vertical member is a member in the

closed portal frame.

Ixg = 1.4*12

203

+ 2*3*30* 2)2

80

2

3+ = 369738 cm

4

Lxg = L horizontal beam = 5.5m (Width of truss)

To estimate the horizontal beam: h = 7.5 -2

8.0= 7.1 m

R U (unloaded) = 23.4 t (from previous)

CL of upper beam

V l o f t r u s s

2

23.4M

11.7 11.7

7.5m

1.8m

0.8

For Upper bracing

3 . 7

5

3*7.1

0.2 0.2

50

Y Y

X

X

20

1.0

48*1.0



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Design members 2009

45/46

M =3

2*

2

4.23* 7.1 = 55.38 m t

1.2

5538= X S = 2637 cm

4

Choose H E B 400 or I P E 600 or B U S

Note: We can always assume the section of the horizontal beam as

following:

IPE 600 if the portal frame is vertical and closed

HEB 600 if the portal frame is inclined and open

GA =5.5 / 369738

5.7 / 33226= 0.06 GA =

5.5 / 92080

5.7 / 33226= 0.26 K = 1.1

lout = 1.1*7.5 = 8.25m 1605.424.19

825 <==inλ

f t = 81.088

95.70= t / cm

2< 2.1 t / cm

2

f sr =88

34.1495.70 −= 0.64 t / cm

2< 1.12 t / cm

2

(All the previous values are from previous example)

The vertical is subjected also to moment (closed portal frame)

R U (loaded) = 9.75 t (from previous example)

M =3

2*

2

75.9* 7.1 = 23.1 m t f btx = 74.125*

33226

2310= t / cm

2

Note that: we checked the tension side which is critical. There is no LTB

in the tension side and A1 = 1.0Check moment: case B

2.1≤+btx

btx

t

t

F

f

F

f

2.11.2

74.1

1.2

81.0=+ 1≈ 1.2 O.K.

Check fatigue: The wind is not considered in check fatigue



Main Truss

Design members 2009

f max =88

95.70= 0.81t / cm

2f min =

88

34.14= 0.16 t / cm

2

f sr = f max - f min = 0.81 – 0.16 = 0.65 t / cm2

< 1.12 t / cm2

06-design of main truss

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