Transcript Chapter 6

Chapter 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 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)
(1) t w 
d f bc
145
d
(2) t w 
120
if f bc  Fbc ( 0.58Fy )  t w 
d Fy
190
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Steel
Grade
Minimum web thickness(tw)
t  40mm
st37
st44
st52
d
120
d
110
d
100
t>40mm
d
120
d
120
d
105
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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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QD  S x
QD
q act =

 qall  0.35 y
Ix *t
Aweb
or qb 
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

0.36q act 
Fb  0.8 
 * Fy
qb 

or assume the flanges alone resist total bending without
reducing Fb
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Check shear stress
 = d1/ d
K q  4  5.34 /  2 
K q  5.34  4 /  2 
K q  5.34
for   1
for   1
for web withoutstiffeners(  )
d
d1
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If ;
d w /t w  45
kq
Fy
no web buckling
 q b  q all  0.35Fy
q act  q b (if not increaset w )
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If ;
d w /t w  45
kq
Fy
Check webbuckling
Calculat e
d / tw
q 
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Fy
kq
for
 q  0.80  q all  q b  0.35 Fy (i.e; d w /t w  45
kq
Fy
)
0.80   q  1.20  q b  (1.5  0.625 q )(0.35Fy )
0.90
 q  1.20  q b 
(0.35Fy )
q
q act  q b (if not rearranget he t ransverse st iffenersand/ or increase t w )
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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:
M y
f=
 Fb
I
QD  S x
QD
q act =

 qall 0.35 y or qb 
Ix *t
Aweb gross
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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:
M
A flange =
h   Fb
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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 = I gross web
t w h 3 (t w h)  h 2 A web h 2
=
=
=
12
12
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b. If we assumed that an area of web = Aweb/6, is placed in each
flange, the moment of inertia is:
2
A webh 2
 A web   h 
  =
Gross moment of inertia of the web= I gross web = 2  
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 6  2
C. Hence, 1/6 gross area of the web acts with the flange (welded
section only):
2
A webh 2
 A web   h 
Gross moment of inertia of the web= I gross web = 2   6    2  = 12

  
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d. Finally, the required flange area can be obtained by:
M
A flange =
h   f all
 A web 
A flange plates = A flange - 

 6 
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
bridges.
.Plate thickness = t = 14 – 26 mm,
bridges.
smaller
spans
for bigger spans
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
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2001.
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):L b = 2.50 4 E . I y . a .   a
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)
=
3
d1
3 * EI1

2
d2 * B
2 * EI2
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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.
C
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
tf
Fy
(Table 2.1c)
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