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Chapter 6Chapter 6

Plate girder

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5.1 Introduction

5.3 Fiber stress of plate girder

5.4 Width of compression flange

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5.1 Introduction5.1 Introduction

A plate girders consist of a vertical plate called web, and two flanges each consisting of horizontal plate. Sufficient weld must be used to insure that the bottom flange, top flange and the web plate acts as one unit. For spans less than 15 m the rolled beams or plated beams are used. But above that (15 m) and till spans to 30 -35 m, the plate girders are economic. The weight of the plate girder is greater than that of truss of the same span but the fabrication costs and maintenance are small.

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1. Economic web depth of girders

For simply supported main girders of railway bridges the height of web could not be less than 1/10 of the bridge span. And for roadway bridges the height of web could not be less than 1/12 of the bridge span. While for continuous and cantilever girders the web height could not be less than (1/10 – 1/14) of the bridge span. For stringers and cross-girders the height could not be less than (1/12 – 1/10) respectively.

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1.                  Thickness of the web

Girder without long stiffener (with or without transverse stiffener)

120

d t)2(

190

Fd t)F58.0(Ffif

145

fd t(1)

w

y

wybcbcbc

w

6

120

d

120

d

110

d

120

d

100

d

105

d

Steel

Grade

Minimum web thickness(tw)

t 40mm t>40mm

st37

st44

st52

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Girder with long stiffener at (d/5 – d/4)

tw 60% of values in clause 7.3.2 (clause 7.3.3)

Nothing given for stiffeners at d/2

In the web of a plate girder there are :-

a.a.    In a vertical plan there are

normal stresses due to B.M. and

shear stress due to S.F.

 

b. b.   On a horizontal plan we have only shear stresses.

Unit shear stresses at section b-b

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byallweb

x qorqAt

S 35.0Q

*I

Q = q D

x

Dact

where, Ix = cross section moment of inertia of the whole

section about axis x-x.

Sx = gross statical moment of shaded area about axis x-x.

Aweb = gross area of the web, qb = shear buckling stress

c. The principal stresses in the web occur in inclined plans.

d. The buckling of the web in a diagonal direction due to principal compression stresses should be considered

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a.   e.    We provide for this buckling by using a lower shear stress in the web (increase t) and by adding vertical and horizontal stiffeners

For panel under Q and M

If qact > 0.6 qb

The all bending stress shall be limited to

Fyq

q

b

act *36.0

8.0 Fb

or assume the flanges alone resist total bending without reducing Fb

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Check shear stress

= d1/ d

)( stiffeners without or webf 34.5K

1 for /434.5K

1 for /34.54K

q

2q

2q

d1

d

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If ;

) tincreasenot if(qq

F35.0qq

buckling webno F

k45/td

wbact

yallb

y

qww

12) tincreaseor and/ stiffeners e transvers therearrangenot if(qq

)F35.0(90.0

q20.1

)F35.0)(625.05.1(q20.180.0

)F

k45/td (i.e; F35.0qq80.0

for

k

F

57

t/d

Calculate

buckling Check web F

k45/td

wbact

yq

bq

yqbq

y

qwwyballq

q

ywq

y

qww

If ;

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5.3 Fiber stress of plate girder

Each flange of a plate consists of one or more flange plates. Theses parts are connected together and the web by a sufficient weld, to guarantee that the girders act as a solid beam. The fiber stresses at any point may be computed from the beam formula:

bFI

yM = f

byallgrossweb

x qorqAt

S 35.0Q

*I

Q = q D

x

Dact

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Approximate method (flange area method)

This method is used to obtain a reasonable cross section before we check with the exact method. The flange material is grouped quite closed together and nearly the moment of inertia of the flanges is equal to 85 % of the total moment of inertia of the whole section. If we assumed that the two flanges acts as the upper and lower chord of truss, i.e the flange stress is nearly constant.

Hence, the required flange area:

bflange Fh

M = A

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h = effective depth = distance between the assumed centroids

of the two flanges

= height of web – (5-10) cm

= 97 % hweb

Fb = 0.58 Fy

Fb = 1.4 for St 37.

Fb = 1.6 for St 44.

Fb = 2.1 for St 52.

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 a.  A part of the web is considered to belong the flange can be calculated as:

Gross moment of inertia of the web = 12

hA =

12

hh)(t =

12

ht = I

2web

2w

3w

webgross

     b.  If we assumed that an area of web = Aweb/6, is placed in each

flange, the moment of inertia is:

Gross moment of inertia of the web= 12

hA=

2

h

6

A2 = I

2web

2web

webgross

   C.   Hence, 1/6 gross area of the web acts with the flange (welded section only):

Gross moment of inertia of the web= 12

hA=

2

h

6

A2 = I

2web

2web

webgross

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d.  Finally, the required flange area can be obtained by:

allflange fh

M = A

6

A-A = A web

flangeplates flange

The number of flange plates should be minimized

as many as you can but not less than one flange plate can be used.

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  e. Flange plates thickness may be chosen as following:

Plate thickness = t = 10 – 12 – 14 mm,for smaller spans bridges.

.Plate thickness = t = 14 – 26 mm, for bigger spans bridges.

 f. Check the value of to ensure that the assumed value h is

correct, otherwise repeat the calculations.

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5.4 Width of compression flange

a. a.   The compression flange is liable to buckle perpendicular to the plan of web. In the plan of web it will prevented from buckling by the stiffness of the web.

   b.   The width of compression flange (b) must be chosen in such away that buckling may be prevented.

 

c.  C.   In a deck bridge the buckling length of the compression flanges is limited by on upper wind bracing and is equal to

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d. In a through bridge where an upper wind bracing is not possible, the upper flanges must be laterally supported by bracket plates bolted to the cross girders.

As, these brackets are elastic supports, the buckling length of the compression flange is not the distance between the brackets but it could be calculated as in clause 4.3.4, code 2001.

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a.a.I.E2.50 = L 4yb

E = 2100 t/cm2

Iy = moment of inertia of plate girder compression flange =

IY flange only

a = spacing between U-frames = (a Lu 2 a)

 e.  There are cross girders and stiffeners forming U-frames provide lateral restrained. Hence, the effective buckling length is according to clause 4.3.2.3 (Table 4.9):-

2

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1

31

EI*2

B*d

EI*3

d =

22

d1 = dw – Hx.G.

d2 = dw – Hx.G./2

I1 = moment of inertia of bracket.

I2 = IX = moment of inertia of X-G. about the axis of

bending

B = the distance between centers of Main Girders.

yf F

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t

C (Table 2.1c)

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