1.
Design of Bridge
g g
Main Girders
According to ECP
According to ECP
Assis. Prof. Dr. Ehab B. Matar

2.
Design Phases of Main Girders
1. Structural analysis for the main system determining the max.
and min. straining actions at critical sections
2. Design of Web Plate
g
3. Design of Flange Plate
4. Design of Stiffeners (end bearing and intermediate)
5 Design of Connections between web plate and flange plate
5. Design of Connections between web plate and flange plate
6. Design of Splices
7. Check fatigue for all details

3.
Bridges Main systems

4.
1- Structural Analysis
 Structural analysis is carried out using either influence lines,
grillage analysis or finite elements depending on complexity of
grillage analysis or finite elements depending on complexity of
structures and its importance.
 Max. and Minimum shear, moments, reactions…. are determined
at critical sections
at critical sections.
Stress range
Stress range
Stress range

5.
2- Design of Web Plate
 This includes the following tasks:
1. Determining Web height
2 Determining Web thickness
2. Determining Web thickness
3. Checking shear Buckling

6.
2.1 Web height
 Overall depth, h:
 Lo/18 ≤ h ≤ Lo/12 (highway two lanes)
L /10 ≤ h ≤ L /7 ( il i l t k)
 Lo/10 ≤ h ≤ Lo/7 (railway single track)
 For double tracks railway or four lanes roadway, increase
the above limits by 60-85%
the above limits by 60 85%
 If the maximum moment is known; the web height can be
approximately calculated as
3
3 7
.
5
3
.
5
2
3
or
b
b F
M
F
Mk
d
h 


 d= plate girder depth (cm)
 M= maximum bending moment (t.cm)
k ratio of height to thickness of eb plate 100
b
b
 k= ratio of height to thickness of web plate  100
 Fb= allowable bending stress ≈ 0.58 Fy

7.
2.2 Web Thickness
 web thickness, tw:
 tw ≥8mm
y
w F
t
d
In
830
case
any
* 
y
w F
t
d 190
used
are
stiffeners
sverse
when tran
* 
y
F
t
d 320
used
stiff.
e
transvers
and
d/5)
(at
al
longitudin
When
* 
y
w F
t

8.
2 3 Shear Buckling
2.3 Shear Buckling

9.
 No need to check shear buckling resistance if :
 No need to check shear buckling resistance if :
 for un-stiffened web
 for stiffened web
y
w
w F
t
d /
105
/ 
q
w
w
F
k
t
d /
45
/ 
y
w
w
F
1 0
f
)
/
34
5
(
4
shear
for
factor
buckling
K
2
q


where
/d
d
0
.
1
for
)
(4/
5.34
1.0
for
)
/
34
.
5
(
4
2
2












where /d
d1


where
d
d1

10.
Shear Buckling Resistance
 If the above mentioned limits are exceeded the shear buckling
resistance should be checked as follows:
y
w
F
t
d
Calculate
*  
q
q
q
K
0.35F
q
then
8
.
0
For
*
57
Calculate
y
b





y
q
q
q
F
9
.
0
35
.
0
)
0.625
-
(1.5
q
then
2
.
1
0.8
For
* b
y

 


y
q
q F
35
.
0
*
9
.
0
q
then
2
.
1
For
* b

 


11.
 It should be noted that longitudinal stiffener at mid depth would be
ff ti f ti bj t d t h th th t
more effective for sections subjected to pure shear than that
positioned at d/5 for pure bending.
 For continuous plate girder, the web panel over an interior
support will be subjected to simultaneous action of bending
support will be subjected to simultaneous action of bending
moment and shearing force. Therefore:
 If the actual shear stress qact ≤0.6qb then the allowable bending
stresses in the girder flanges will not be reduced and should
stresses in the girder flanges will not be reduced and should
not exceed 0.58Fy.
 If the actual shear stress qact >0.6qb then two alternatives may
be followed. The first is to reduce the allowable bending
g
stress for flange plates according to the following interaction
equation.
y
act
b F
q
F 












 36
.
0
8
.
0
 The second alternative is to design the girder flanges to
y
b
b
q








g g g
resist the whole acting bending moment without any
participation of web for resisting bending moment without
reducing the allowable bending stress.

12.
Example
Design a continuous two spans welded plate girder as shown below for a
roadway bridge. The cross girders are arranged each 2.4m. The deck slab is 20cm
thick and the asphalt cover is 10cm. The steel used for the design of all elements is
St. 37. Moreover, design a field splice located at 16.8m from the end support. The
St. 37. Moreover, design a field splice located at 16.8m from the end support. The
bending moments and shearing forces at the critical locations are tabulated as
follows:
Sec. Md.l (t.m) Qd.l (t) Mll+I (t.m) Qll+I (t)
1 0 +38.2 0 +50.8
2 +171 0 +267 +4 43
2 +171 0 +267 +4.43
3 -305.3 -63.6 -296 -61.65
4 +96 -33.07 +96.69,-98.63 -34.92,+3.36
Solution
 Proportioning of plate girder
 cm
cm
F
Mk
d
b
190
186
4
.
1
*
2
100
*
100
*
)
3
.
305
296
(
*
3
2
3 3
3 



 where k is
assumed 100
 Design of web plate
 t
t
d 56
1
190
190
/ 


 cm
t
F
t
d w
y
w 56
.
1
122
/ 



 By using vertical stiffeners arranged @2.4m (distance between
cross girders), then,
 26
.
1
240
1



d

190
d
 85
.
7
)
26
.
1
/(
4
34
.
5
)
/
4
(
34
.
5 2
2




 
q
k
 37
.
81
4
.
2
85
.
7
45
45 


y
q
w F
k
t
d
190
 thcikness
large
a
is
which
34
.
2
37
.
81
190
cm
tw 


 Assume tw = 16mm and check shear buckling resistance.
 2
.
1
8
.
0
15
.
1
85
.
7
4
.
2
57
6
.
1
190
57





 q
q
y
w
q
k
F
t
d


q
 2
/
66
.
0
)
4
.
2
35
.
0
)(
15
.
1
625
.
0
5
.
1
(
)
35
.
0
)(
625
.
0
5
.
1
(
cm
t
q
x
x
F
q
b
y
q
b






 
 safe
/
41
.
0
6
.
1
190
25
.
125 2




 b
w
q
cm
t
x
A
Q
q


13.
3- Design of Flange Plate
 This includes the following tasks:
1. Determining Flange cross section (using Flange Area
Method)
Method)
2. Determining Flange width and thickness
3. Checking Bending Stresses

14.
3.1 Flange Area Method
MY
 Plate Girder



3
2
d
d
I
MY

c
tf
b










12
2
2
2
3
2
3
2
w
f
d
t
d
A
I
c
d










6
2
12
2
2
3
2
w
f
w
f
A
A
d
d
t
d
A
I
Welded Plate Girder








6
/ w
f
A
A
d
Y
I









w
f
b
A
A
d
M
F




 6
f

15.
A
d
F
M
A w
b
f
6



A
d
F
M
A
d
F
M
A w
w
f
8
4
3
*
6
@4
arranged
bolts
assume
BUS
riveted
or
bolted
For






d
C
b
t
b
A
where
d
F
d
F
f
f
b
b
%
30
20
2
;
*
8
4
6




Section class Part Stress Profile
C tf C tf
D b
3.2 Local Buckling limits
Section class Part Stress Profile
Compact Flange Uniform Comp. Rolled Sec.
dw tw dw tw
Rolled section Welded section t
h
t
y
f F
t
C /
9
.
16
/ 
p
section
g p
B.U.S
Non-compact
section
Flange Uniform Comp. Rolled Sec.
y
f
y
f F
t
C /
3
.
15
/ 
y
f F
t
C /
23
/ 
section
B.U.S
y
f F
t
C /
21
/ 

16.
Remember
 Whenever the width / thickness ratio of compression
flange exceeds the aforementioned limits, the flange should
be treated as a slender section and the effective flange
be treated as a slender section and the effective flange
area should be calculated to account for local buckling.
 The thickness of flange plates as well as web
 The thickness of flange plates as well as web
plates should be reduced by 1.0mm in design calculations
due to the effect of undercut in welding unless special
ti i th ldi t h i t k
precautions in the welding technique are taken.

17.
Note that:
 When plates with
unequal thickness or
 Change of flange plate
width or thick.
q
width butt welded
together, the thicker or
g
wider plate should be
tapered with a slope
2−4mm
1
4
4
1
not exceeding one to
four as shown
2mm
R
=
6
0
c
m
R

18.
3.3 Allowable Compressive Stresses
The allowable compressive
stresses in the compression
flange depend on whether the
flange is braced laterally or un-
flange is braced laterally or un-
braced. The lateral unsupported
length of the compression flange
Lu is calculated as follows:
For deck railway bridges with
For deck railway bridges with
open timber floor where there
exist upper wind bracing or not,
and with the existence of cross
girders that are rigidly connected
girders that are rigidly connected
to the main girder compression
flange, then, Lu = distance
between x-girders.
For deck railway bridges with
For deck railway bridges with
ballasted floor or for roadway
bridges where the compression
flange is supported by
continuous reinforced concrete
continuous reinforced concrete
or steel deck, where the frictional
or connection of the deck to the
flange is capable to resist a
lateral force of 2% of the flange
lateral force of 2% of the flange
force at maximum bending
moment, then, Lu = 0.

19.
For continuous
deck roadway or
dec oad ay o
railway bridges
where the
compression flange
at interior supports
at interior supports
is located in the
bottom side of the
girder away from the
g y
deck slab, then,
Lu = the distance
between the centers
f i t ti f th
of intersection of the
lower wind bracing
with the
compression flange
compression flange
Lu = if there is no
lower wind bracing
or transverse bracing
g
then Lu is taken as
the distance from the
point of maximum
bending moment to
bending moment to
the point of contra-
flexure.

20.
•Lateral unsupported length for through bridge (+M
a e a u suppo ed e g o oug b dge (
region)
 The lateral unsupported length of the
compression flange is taken as the full
4
*
5
.
2 a
EI
Lu y
 
compression flange is taken as the full
girder length if the compression flange is
unrestrained against lateral bending. If
the cross girders and the stiffeners
forming U-frames, then,
Wh
2
2
2
1
3
1
2
3 EI
B
d
EI
d



 Where
 E= The Young's modulus of steel (t/cm2)
 Iy= moment of inertia of the chord
member about the Y-Y axis (cm4) as
shown
d2
d1
I1
shown
 a = distance between the U- frames
(distance between x- girders)
  = the flexibility of the U- frame
d1 di t f th t id f th
I2
I1
 d1 = distance from the centroid of the
compression flange to the nearest face of
the cross girder of the U-frame.
 d2 = distance from the centroid of the
compression flange to the centroidal axis
B
compression flange to the centroidal axis
of the cross girder
 I1 = Moment of inertia of the vertical
member forming the arm of the U-frame
about the axis of bending.
 I2 = Moment of inertia of the cross girder
 B = distance between centers of
consecutive main girders connected by
U-frame.

21.
Continue Example
7.5m
2.5m
1.25
1.0m 1.0m
1.25
2.5m
Continue Example Bridge Cross section
24m 24m
1 2 4 3
16.8
 Flange area Method
 For sec. 3, moment and shear acting together, therefore, checking
%
60
62
.
0
66
.
0
41
.
0



b
q
q
, then, a reduction in the allowable bending stress
should be carried out as follows
 Assuming the compression flange is braced laterally by the lower
wind bracing which has joints @240cm (distance between cross girders),
then Lu=240cm.
 Checking whether lateral torsional buckling controls the
allowable compression stresses or not as follows:
f x
b 60
20
20
24m 24m
should be carried out as follows

 
 
 
  2
/
38
.
1
4
.
2
66
.
0
/
41
.
0
36
.
0
8
.
0
/
36
.
0
8
.
0
cm
t
F
F
q
q
F
b
y
b
b





 Calculating the flange area at sec.3 as follows:
 2
66
.
178
6
6
.
1
190
190
38
1
2
3
.
601
6
cm
x
E
A
d
M
A w
f 




 u
y
f
L
cm
x
F
b


 6
.
774
4
.
2
60
20
20

u
b
y
f
b
y
f
L
cm
C
F
d
A
x
x
x
x
x
C
F
d
A


















86
.
580
.
.
1380
3
.
601
78
.
323
3
.
0
3
.
601
78
.
323
05
.
1
75
.
1
(
4
.
2
196
60
3
1380
.
.
1380 2
6
190
38
.
1
6 x
d
Fb
f
 Assuming bf=0.3d=0.3x190=57cm→60cm
 Then tf=178.66/60=2.98cm→3cm

y
f F
t
c 21
73
.
9
30
2
/
16
300




 Therefore Fbc =1.38t/cm2
& Ft=1.4 t/cm2
 /
38
.
1
543
,
43
2
3
.
601 2
cm
t
E
F 


 O.K safe in both tension and
compression.
 Checking fatigue stress by considering only 60% of the live loads
are acting i.e. 0.6x274.32=164.6tm.
2
6
164 E
 However, although the flange satisfied the limit of compact
section ( y
F
/
3
.
15 ) but the web is non-compact, therefore, the whole section
will be considered non-compact.
 Calculating the flange area at sec.2 as follows:
 2
114
6
.
1
190
2
438 x
E
A
M
A w
 Stress range = 2
/
378
.
0
543
,
43
2
6
.
164
cm
t
E

 For ADTT >2500 truck, No. of stress cycles =2E6 and for welded
plate girder Class B' is taken therefore the allowable fatigue stress range
Fsr=0.85t/cm2
. Therefore, the section is safe.
 Checking Actual stresses at section 2
     2
3
3
4
2
50
6
1
190 x
x 

 2
114
6
190
4
.
1
6
cm
x
d
F
A w
b
f 




 Assuming bf=50cm
 Then tf=114/50=2.28cm→2.4cm

y
f F
t
c 21
08
.
10
24
2
/
16
250





     
3
4
2
194
,
32
,
714
,
135
,
3
2
.
1
95
4
.
2
50
12
4
.
2
50
2
12
6
.
1
190
cm
Z
cm
I
x
x
x
x
x
I
gross
gross













 Assuming the compression flange is braced laterally by the
existence of R.C. deck directly rested over compression flange therefore,
Lu=0. Therefore, there is no need to check Fltb.
 Therefore Fb =1 4 t/cm2
& Ft=1 4 t/cm2
y
f
 Checking Actual stresses at section 3
     
3
4
2
3
3
543
,
43
,
213
,
267
,
4
5
.
1
95
3
60
12
3
60
2
12
6
.
1
190
cm
Z
cm
I
x
x
x
x
x
I
gross
gross













 Therefore Fbc 1.4 t/cm & Ft 1.4 t/cm
 /
36
.
1
194
,
32
2
438 2
cm
t
E
F 


 O.K safe in both tension and
compression.
 Checking fatigue stress by considering only 60% of the live loads
is acting. The live load is ranged at this section between +250.5 tm and -
73.971 tm. Then,

22.
Example for through bridge
 A railway through plate girder
y g p g
bridge of 27 m simple span has
two main girders 9 m apart. The
bridge is for double track and has
bridge is for double track and has
an open timber floor. The cross
girders are arranged every 2.25 m.
The bridge is provided by stringer
 The bridge is provided by stringer
bracing, while braking force
bracing is not arranged. The
f
material of construction is steel 44;
the weight of the timber floor for
each track is 0.6 t/m’.
 Calculate Lu considering, for main
girder tw=20mm, dw=3000mm,
M =1060mt Q =160t while
Mmax=1060mt, Qmax=160t, while
for XG Pl#1000x20/2Pl#300*40mm

23.
Solution
 For the main girder, tw=20mm,
dw=3000mm
 Af=M/(dFb)-Aw/6
Af M/(dFb) Aw/6
 Af=1060*100/(300*1.6)-
2*300/6=120.8cm2
 Bf=0.2*300=60cm
 tf>120.8/60>2.01cm
 c/t <21/sqrt(F )t =2 4cm
d2
d1
I1
 c/tf<21/sqrt(Fy)tf=2.4cm
 Z=(50x2x27.4+60x2.4x1.2)/(60x2.4+50
x2)=11.94cm
I2
B
 Iy=2.4x60^3/12+50X2^3/12=43233cm4
 I2=2X100^3/12+2x(30x4^3/12+30x4x5
2^2)=815947cm4
2 2)=815947cm
 d1= (302.4-100-11.94)=190.46cm
 d2= 190.46+100/2+4)=244.46cm
 B=930cm

24.
 Upon calculating the
I1, the section is
composed of part of
composed of part of
the web plate (25
twxtw), vertical stiffener
assumed 250x20 mm
assumed 250x20 mm
(satisfying local
buckling) and the
b k t i t b
bracket is to be
defined according to
max. distance allowed
by train edges.
 Calculate the centriod
 X=(25x2x1+40x2x22+
 X=(25x2x1+40x2x22+
50x2x43+25x2x56.5)/(
2x(25+40+50+25)=31.
9
9cm
 I1=25x2x30.9^2+40^3x
2/12+40x2*9.91^2+50
EI
B
d
EI
d


2
3 2
2
2
1
3
1

x2x11^2+2x25^3/12+2
x25x24.6^2=111226c
m4 ok
m
a
cm
e
a
EI
L
cm
e
y
u 










25
.
2
379
2
6
.
2
*
225
*
43233
*
2100
*
5
.
2
5
.
2
2
6
.
2
815947
*
2100
*
2
930
*
2
^
46
.
244
111226
*
2100
*
3
3
^
46
.
190
4
4 


25.
3- Design of Stiffeners
 This includes the following tasks:
D i f ti l tiff (Wh th d b i
1. Design of vertical stiffeners (Whether end bearing or
longitudinal stiffeners)
2. Design of horizontal stiffeners
g

26.
Importance of stiffeners
 Reducing slenderness ratio of web plate
 Increase the shear capacity of web plate
p y p
Location of stiffeners
 Vertical stiffeners are located at location of X-
girders and at supports
 Horizontal stiffeners at a distance of d/5 from
compression flange and at d/2

27.
A- Vertical Stiffeners
Compression Flange
x
Tension Flange
The outstanding length of vertical stiffeners
1- stiffeners in pairs x  dw/30 + 5 cm
2- single stiffeners x dw/30 + 10 cm

28.
 Vertical intermediate stiffeners composed of single or
i f t i l ti hil E d b i tiff
pair of symmetrical sections while End bearing stiffeners
should be in pairs fastened on each side of the web
 End bearing stiffeners should be well ground or
machined to fit tightly against the top and bottom flange
angles and should never be crimped.
 Vertical stiffeners act as a compression members with
 Vertical stiffeners act as a compression members with
buckling length equal to 0.8dw
 For the sake of design of the vertical stiffeners it will be
considered as a column of cross section consisting of
considered as a column of cross section consisting of
the area of angles and a certain length of the web ( 25
tw for intermediate vertical stiffeners and 12 tw for end
bearing stiffeners)
bearing stiffeners).
6tw
6tw
25tw
End Bearing
Stiffener
Stiffener
Intermediate

29.
• End Bearing Stiffeners
 Design Steps:
 1- Define the design force d.F (Max. Reaction at support)
 2- Choosing stiffeners in pairs x  dw/30 + 5 cm
 3- Required area of End bearing stiffeners
A d F/1 t/ 2 A + 12t 2
 Areq = d.F/1 t/cm2  Astiff + 12tw
2
 Astiff = (Areq - 12tw
2)
 4- Choose stiff. sections to avoid local buckling such that
f
t
x 21

 5- Calculate the slenderness ratio of the proposed column as follows
 calculate moment of inertia and cross sectional area Ix & A &
 Ix h
lb 8
0
y
stiff f
t

6- Calculate the permissible buckling stress Fpb
 7 Check that F = d F / A  F
A
Ix
i 
i
h
i
l w
b 8
.
0



 7- Check that Fact = d.F / A  Fpb
 8- Design the weld between the stiffeners and the web plate such that the weld
in upper and lower thirds can resist the whole design force

30.
•Intermediate Vertical Stiffeners
 Design Steps:
 1- Define the design force d F = act
y
d Q
F
Q 






 1
35
.
0
65
.
0
 1 Define the design force d.F =

 2- Choosing single stiffeners x  dw/30 + 10 cm
 3- Required area of Intermediate vertical stiffeners
act
b
d Q
q
Q 



q
 Areq = d.F/1 t/cm2  Astiff + 25tw2
 Astiff = (Areq - 25tw2)
 4- Choose stiff. sections to avoid local buckling such that
f
t
x 21

 5- Calculate the slenderness ratio of the proposed column as follows
 calculate moment of inertia and cross sectional area Ix & A &

y
stiff f
t
I h
l 8
0


6- Calculate the permissible buckling stress Fpb
A
Ix
i 
i
h
i
l w
b 8
.
0



p
 7- Check that Fact = d.F / A  Fpb
 8- Design the weld between the stiffeners and the web plate such that the
weld in upper and lower thirds can resist the whole design force

31.
•Horizontal Stiffeners
 The Egyptian code of practice requires that the
hori ontal stiffeners satisfies the follo ing
horizontal stiffeners satisfies the following
stiffness:
F hl Stiff At d/5 t f i ti >4d(t )3
 For hl. Stiff. At d/5, moment of inertia >4d(tw)3
 For hl. Stiff. At d/2, moment of inertia>d(tw)3
 Local buckling should be checked to satisfy the
relation of x 21

y
stiff f
t


32.
4-Curtailment of Flange Plates
 Why curtailment is needed?
 Bending moment varies along bridge span, therefore it may be
i t h th l t i d ti t t i
economic to change the plate girder section to get maximum
utilization of steel strength.
Ho co ld e specif the c rtailment location?
 How could we specify the curtailment location?
 based on developed formulas by Johnson et al [1] to find the
minimum volume of steel that will yield minimum steel weight
y g

33.
1. For Parabolic variation of bending moment
for simple or continuous girder
6
.
1
w
b
f
A
F
d
M
A 

f2
A
f1
A
6
.
max
2
w
b
f
A
F
d
M
A 

L
x
 
2
.
.
3
2
2
3
1
2
L
A
x
Lx
x
L
L
M
x
A
x
L
A
Vol f
f



 




 M Mmax
6
.
2
.
. 2
max L
A
L
x
Lx
x
L
L
F
d
M
Vol w
b








 

4
0
.)
( 2
2






 L
x
L
Lx
x
vol
5
A
and
3
3
3
0
f1







x
Lx
x
x
9
Af2


34.
2. For Parabolic variation of bending
g
moment for cantilever girder
3
2
2
max






 

 wL
A
x
x
L
L
M
vol
3
0
)
(
6
.
2
2
2









b
L
x
L
x
vol
L
F
d
vol
1
3
3
0
)
(
1






f
A
x
L
x
x
3
1
2
1

f
f
A

35.
5- Connection between Flange
and web plates
 Connection is either by using bolts or welds
m
1.5mm

36.
Procedure
 ECP recommends continuous fillet weld between flange
and web plates in case of dynamic loading
 The size of welding is given by
s
f
QY
q 2

 Where
 Q = maximum shearing force at support
s
f
I
Q
q weld 2
.


g pp
 Y = first moment of area of flange plate about centroidal axis of the
section
I M t f i ti f l t i d
 I = Moment of inertia of plate girder
 s = size of welding
 f ld = allowable stresses in fillet weld = 0 2Fu
 fweld allowable stresses in fillet weld 0.2Fu
 Fu = ultimate strength of the base metal.

37.
Minimum sizes of weld
tmax S mm
tmax
mm
S mm
10 4
10-20 5
20 30
20-30 6
30-50 8
50-100 10
Fatigue strength of the fillet weld under the effect of shear flow due to
live load plus impact should be checked, the connection is detailed as
class D (Group 3 Fasteners)

38.
Example
 Given: Qt=125 25t& Ql l=57 15t I=4 267 213cm4 &
 Given: Qt 125.25t& Ql.l 57.15t , I 4,267,213cm4 &
M.G.2Pl#600X30/1900X16mm, St. 37
 Design of the connection between the flange plate and the web plate.
 Maximum shear flow is at sec 3 where Q=125 25t
 Maximum shear flow is at sec. 3 where Q 125.25t.
 Y=first moment of area of flange plate about N.A.
 3
370
,
17
)
95
5
.
1
(
3
60 cm
x
x
Y 


x
Y
Q 370
17
25
125
 cm
t
x
I
Y
Q
q /
51
.
0
213
,
267
,
4
370
,
17
25
.
125
.



 Assuming direct contact between the web plate and the flange
cm
x
F
Sx
q 1
2
.
0
2

therefore,
mm
cm
x
x
S
cm
x
F
Sx
q u
6
35
.
0
6
.
3
2
.
0
2
51
.
0
1
2
.
0
2




based on the maximum thickness of
flange plate = 30mm.
flange plate 30mm.
 Checking the fatigue resistance where 60% of live load is only
considered i.e. 0.6x57.15=34.29t.
 The shear flow due to this live load cm
t
x
q /
14
0
370
,
17
29
.
34
 The shear flow due to this live load cm
t
q I
ll /
14
.
0
213
,
267
,
4


 The allowable stress range based on No. of stress cycles of 2E6
and Detail D Group 3, is 0.71t/cm2
.
p

safe
O.k
1
.
0
14
.
0
71
.
0
2
2
cm
S
Sx
q
SxF I
ll
sr





 

39.
6- Splices
 why a plate girder may be spliced?
y p g y p
 Un-sufficient plates lengths (Web plates ≈ 6m long,
Flange plates ≈12-18m).
Th d i d i li d j i id
 The designer may desire to use spliced joints to aid
in cambering.
 Change the girder cross section to fit the actual
 Change the girder cross section to fit the actual
bending moment.
 Transportation of full length plate girders plays an
p g p g p y
important role in locating spliced joints. Highway
road system conditions, maximum limit of legal
loads over existed bridges and maximum crane
loads over existed bridges and maximum crane
capacities limits the maximum length and weight of
plate girders to be transmitted in one time.

40.
Shop Splicing
 Generally, flange and web plates are spliced using single or
double V-joints.
 J and U joints require the least amount of weld of metal but
 J and U joints require the least amount of weld of metal but
require the plates to be prepared by planning or milling which is
impractical in most structural fabricating shops.
This limits the preparation to flame beveling giving a V joint
 This limits the preparation to flame beveling giving a V-joint.
Single V-joint may be acceptable if the plate thickness is up to
25mm.
 For thicker plates double V-joints are preferred since they require
less weld metal.
 It should be remembered that a single V-joint will produce more
g j p
angular distortion which is increased rapidly as the flange
thickness increases.

41.
Design forces
 Splices in webs of plate girders must be
designed to resist the moments and shearing
designed to resist the moments and shearing
forces at that section. The principle stresses
in welds are determined from:
in welds are determined from:
 The greatest bending moment at the splice
d th di h i f
and the corresponding shearing force.
 The greatest shearing force at the splice and
the corresponding bending moment.

42.
q
f
Fig. (17) Staggering compression and tension flanges butt weld
The stresses in butt welds:
F th b
weld
Good
for
35
.
0

 y
F
A
Q
q
 For the web
weld
Excellent
for
0.385Fy

y
w
A
 For the flange
weld
Good
for
0.4F
0.7x0.58F
i.e.
7
.
0 y
y 


 pt
F
I
My
f
 For the flange
weld
Excellent
for
0.58Fy

I

43.
Field Splicing
 Arrangement: Generally each girder of a bridge is
assembled of two or three parts at the bridge
location cite
location cite.
 Methods: The girder assembly is carried out either
by welding or by bolting.
 Design forces: The ECP [2] cl. 7.5 recommends
that, splices should be designed on the maximum
bending resistance of the girder section and the
bending resistance of the girder section and the
actual shearing force at the splice location.
 Execution: Generally welded field splices requires
y p q
testing either by ultrasonic, X-ray or any other
testing technique which increases the cost.

44.
Welded field splices
the German window splices as shown in Fig. (18) Which is assembled as
follows:
(2)
(4)
(2)
(3)
(1)
(3)
(1)
(2)
(3)
(1)
(3)
(1)
German Window
Fig. (18) The German window for field splice
 A short portion of web is omitted and the adjacent parts of the web are
shop welded to the flanges (see ends of weld 1).
 The flanges welded first where they are not restrained by the web.
Moreover, the roots of these welds are fully accessible.
Th i d i th i i i f th b i th i t d It i
 The window i.e. the missing piece of the web is then inserted. It is
slightly curved so that no shrinkage stresses will arise when the piece is
welded to the remainder of the web.
 The remainder welds between the web and the flanges are deposited
 The remainder welds between the web and the flanges are deposited
last.

45.
Bolted field splice
Bolted field splice
Q Q Q
Q
Q Q
S2
S2
Q
Q
S2 S2
Detail of Field Splice in bolted Main Girder
Flange Pl. Spl. : Direct Method
m m
m
m
m
m
in welded plate girder
Detail of Field Splice
g p
Flange < Spl. : Two Cuts in welded plate girder

46.
Q
Q tsp2
S2 S2
m m m m m
m
Flange Pl Spl : Direct Method
Detail of Field Splice in bolted Main Girder
m m m m m
m
Flange Pl. Spl. : Direct Method
Flange < Spl. : Two Cuts

47.
Remember
Ignorance with Code
leads to
leads to

48.
Thank you
Thank you