Ch4 Bridge Floors (Steel Bridges تصميم الكباري المعدنية & Prof. Dr. Metwally Abu-Hamd)

Chapter 4: Bridge Floors
CHAPTER 4
BRIDGE FLOORS

Steel Bridges
CHAPTER 4
BRIDGE FLOORS
4.1 INTRODUCTION
The principal function of a bridge deck is to provide support to local
vertical loads (from highway traffic, railway or pedestrians) and transmit
these loads to the primary superstructure of the bridge, Figure 4.1. In addition
to this, the overall structural actions may include:
2. Contributing to the top flange of the longitudinal girders
3. Contributing to the top flange of cross girders at supports and, where
present in twin girder and cross girder structures, throughout the span,
4. Stabilizing stringers and cross girders in the transversal direction,
5. Acting as a diaphragm to transmit horizontal loads to supports,
6. Providing a means of distribution of vertical load between longitudinal
girders.
It may be necessary to take account of these combined actions when
verifying the design of the deck. This is most likely to be the case when there
are significant stresses from the overall structural actions in the same
direction as the maximum bending moments from local deck actions, e.g. in
structures with cross girders where the direction of maximum moment is
along the bridge.

Fig 4.1 Structural Actions of a Roadway Bridge Deck
4.2 STRUCTURAL SYSTEMS OF BRIDGE FLOORS
Structural systems used in bridge floors vary according to the bridge usage
as follows:
4.2.1 ROADWAY BRIDGE FLOORS
Three main types of transverse structural systems may be used in roadway
bridge floors:
a) Slab
b) Beam-Slab (slab with floor beams)
c) Orthotropic plate floor

Steel Bridges
a) In the Slab cross-sections, Fig. 4.2a, a reinforced concrete deck slab
about 20 to 30 cm thick is supported directly on the bridge main girders. This
system is economical for small spans, generally below 25m, where multiple
girders are used for the longitudinal structural system at spacing of 2.5 – 4 m.
b) In the Beam-Slab cross-sections, Fig 4.2b, the deck slab is supported on
longitudinal floor beams (called stringers) and /or transversal floor beams
(called cross-girders). This system is generally adopted for medium spans
below 80 m where the spacing of main girders exceeds about 4 m.
In both cases, the slab may act independently of the supporting beams (a
very uneconomic solution for medium and large spans) or it may work
together with the supporting beams (composite bridge deck). The composite
action requires the shear flow between the slab and the girders to be taken by
shear connectors as shown in Fig. 4.2a.
c) In the Orthotropic Plate Deck, Fig. 4.2c, a stiffened steel plate covered
with a light wearing surface is welded on top of the main girder webs to
provide a deck surface. The deck plate, acting as the top flange of the main
girders, gives a very efficient section in bending. The steel plate is
longitudinally stiffened by ribs, which may be of open or closed section.
Transversally, the ribs are connected through the transverse floor beams
(cross girders) yielding a complex grillage system where the main girders, the
steel plate, the ribs and the floor beams act together.
Fig. 4.2 Roadway Bridge Floors: a) Slab Type Floor

Fig. 4.2 Roadway Bridge Floors: b) Beam Slab Type Floor
Fig. 4.2 Roadway Bridge Floors: c) Orthotropic Plate Floors
When compared to concrete slab decks, the biggest disadvantage of
orthotropic steel plate decks is their high initial cost and the maintenance
required. Concrete decks are therefore usually more economic than
orthotropic steel plates. The latter are only adopted when deck weight is an
important component of loading, i.e. for long span and moveable bridges.

Steel Bridges
3.2.2 RAILWAY BRIDGE FLOORS
Tracks of railway bridges are normally carried on timber sleepers which
are 260 cm long and spaced at not more than 50 cm between centers. The
sleepers are then supported on the bridge floor system, which may be of the
open timber floor type, Fig. 4.3a, or of the ballasted floor type, Fig. 4.3b:
a) The Open Floor type consists of longitudinal beams, called stringers,
spaced at 1.5 to 1.8 meters, and transversal beams, called cross
girders, spaced at 4.0 to 6.0 meters.
Fig. 4.3 Railway Bridge Floors: a) Open Timber Floor

b)The Ballasted floor type consists of a 20 cm layer of ballast carried on
an R.C. slab which is supported on steel floor beams, e.g.; stringers
and/or cross girders as shown in Fig. 4.3b:
Fig. 4.3 Railway Bridge Floors: b) Ballasted Floor

Steel Bridges
4.3 DESIGN CONSIDERATIONS
4.3.1) ALLOWABLE STRESSES FOR STEEL St 52 (ECP 2001)
4.3.1.1) Allowable Stress in Bending Fb
1- Tension and compression due to bending on extreme fibers of “compact”
sections symmetric about the plane of their minor axis:
Fbx = 0.64 Fy = 2.304 t/cmP
2
In order to qualify under this section:
i- The member must meet the compact section requirements of Table 2.1
of ECP.
Note that most rolled sections satisfy these requirements.
ii-The laterally unsupported length (Lu) of the compression flange is
limited by
y
f
u1
F
20b
L ≤
bC
Fd
L
y
f
u2
1380A
≤
2- Compression on extreme fibers of flexural members meeting the “non-
compact” section requirements of Table 2.1of ECP:
Fbx = Fltb < 0.58 Fy = 2.1 t/cmP
2
Usually Fltb is governed by: yb
fu
1ltb
F58.0C
A/d.L
800
F ≤=
3- Tension and compression due to bending on extreme fibers of doubly
symmetrical I-shape members meeting the “compact” section
requirements of Table 2.1(c) of ECP, and bent about their minor axis:
Fby = 0.72 Fy = 2.592 t/cmP
2
4.3.1.2) Allowable Stress In Shear:
qall = 0.35 Fy = 1.26 t/cmP
2

The effective area in resisting shear of rolled shapes shall be taken as the
full height of the section times the web thickness while for fabricated shapes
it shall be taken as the web height times the web thickness.
4.3.2) DESIGN OF STRINGER CROSS SECTION
Stringers are usually designed as beams simply supported on the cross
girders. The maximum straining actions for design are computed from the
load positions and load combinations producing the maximum effect on the
member considered. The maximum bending stress in the flanges or the
maximum shear stress in the web usually governs the cross section size. The
stringers are usually connected at their ends to the cross girder by two
framing angles which are designed to transmit the maximum end reaction of
the stringer to the cross girder, Fig. 4.4a.
Stringers may also be designed as continuous beams. In this case the
connection between the stringer and the cross girder is designed to carry also
the negative moment at the stringer supports, Fig. 4.4b.
a) Simple Stringer b) Continuous Stringer
Fig. 4.4 Connection between Stringer and Cross Girder
In addition to the effect of vertical loads, stringers in open railway bridge
floors should be designed to carry the effect of the horizontal loads caused by
the lateral shock of the running wheels, see section 2.2 (f). This lateral load is
transmitted from the rails to the sleepers and then to the upper flange of the
stringer. This effect causes double bending of the stringer cross section.
Alternatively, a system of horizontal bracing, called lateral shock bracing,
can be arranged between the stringers upper flanges to reduce the effect of
lateral loads, Fig. 4.5.

Steel Bridges
Fig. 4.5 Stringer (Lateral Shock) Bracings
4.3.3) DESIGN OF CROSS GIRDER CROSS SECTION
Similarly, cross girders are usually designed as beams simply supported on
the main girders. The maximum bending stress in the flanges or the
maximum shear stress in the web usually governs the cross section size. The
cross girders are usually connected at their ends to the main girder by two
framing angles which are designed to transmit the maximum end reaction of
the cross girder to the main girder.
Cross girders of open railway bridge floors are designed to carry the effect
of the horizontal loads caused by the longitudinal braking forces. This lateral
load is transmitted from the rails to the sleepers and then to the upper flange
of the stringers and the cross girders. This effect causes double bending of
the cross girder section. Alternatively, a system of horizontal bracing, called
braking force bracing, can be arranged between the stringers and the cross
girders to eliminate the effect of longitudinal loads, see Fig. 4.6
Fig. 4.6 Cross Girder (Braking Force) Bracings

4.4 DESIGN EXAMPLES
4.4.1) EXAMPLE 1: ROADWAY BRIDGE FLOOR

Steel Bridges
4.4.1.1) STRINGER
Structural System: Beam supported on cross girders,
Span = 4.50 m, Spacing = 1.75 m.
1) Straining Actions:
1.1) Dead Load:
• 22 cm Deck Slab = 0.22 × 2.5 = 0.55 t/mP
2
• 5 cm Asphalt = 0.05 × 2.0 = 0.10 t/mP
2
Total D.L = 0.65 t/mP
2
Own wt of stringer (assumed) = 0.10 t/mP
/
P
uniform load on stringer = 0.65 × 1.75 + 0.1 = 1.238 t/mP
/
Dead Load Actions: QDL = 1.238 × (4.5) / 2 = 2.784 t
MDL = 1.238 × (4.5)P
2
P / 8 = 3.132 mt
1.2) Live Load & Impact:
• Impact factor I = 0.4 - 0.008 * L = 0.4-0.008*4.5= 0.364
(L = Loaded Length of main traffic lane = 4.5m)
Impact is applied to Main Lane Loads only
• Maximum LL Reaction on Intermediate Stringer:
Place 10P
t
P on stringer and add effect of 5 P
t
P @ 1 m:
P = 10 * (1 + 0.364) + 5 × (1.75-1)/1.75 = 15.783 t

Loads position for Max Moment:
Note:
For Lst < 2.6 m: Mmax occurs at the middle section with one LL reaction load
acting in the middle.
For Lst > 3.4 m: Mmax occurs at the middle section with all three loads acting
as shown;
For 2.6 < Lst < 3.4 m: Mmax occurs with two LL reaction loads placed such
that the stringer centerline bisects the distance between
the resultant and one load.
MLL & I = 23.674 × 2.25 – 15.783 × 1.5 = 29.592 mt
• Loads position for Max Shear:
Notes:
For L > 3.0 m: Qmax occurs at support with two loads acting on span
For L > 3.0 m: Qmax occurs at support with all three loads acting as shown:
QLL&I = 15.783 + 15.783 × (3/4.5) + 15.783 × (1.5/4.5) = 31.566 t
Bending
Moment
Shear Force

Steel Bridges
1.3) Design Straining Actions:
The total design moment on an intermediate stringer is:
at middle section: Mdesign = 3.132 + 29.592 = 32.724 mt
The total design shear on an intermediate stringer is:
At support: Qdesign = 2.784 + 31.566 = 34.35 t
2) Design of Cross Section:
2.1) Case of Simple Stringer:
Straining Actions: Mx = 32.724 m.t. (Maximum near middle)
Qy = 34.504 t (Maximum at support)
Design for Bending then check shear.
Section is compact w.r. to both local buckling requirements (being a
rolled section), and lateral torsional buckling requirements (compression
flange supported by deck slab); i.e.
Fbx = 0.64 Fy = 2.304 t/cm2
Req. Zx = Mx / Fbx 32.72 x 100 / 2.304 = 1420 cm3
--------- use IPE 450
Bending Stress: fbx = 32.72 x 100 / 1500 = 2.18 t/cm2
< 2.304 t/cm2
OK
Check of Fatigue: Actual Stress Range = fsr =(0.5 x 29.592)x100/1500
= 0.9984 t/cm2
< Allowable Stress Range = Fsr = 1.26 t/cm2
(Assuming Class B detail under 2x106
cycles (Case 1.2 of Group 1 ECP)
Check Shear: qy = Q / Aw net = 34.504 / (0.85 x 45 x 0.94) = 0.95 t/cm2
< 1.26 t/cm2
OK
2.2) Case of Continuous Stringer:
i) Section at mid span :
Mx = 0.80 x32.72 = 26.20 m.t.,

Section is compact (see above): Req. Zx = Mx / Fbx = 26.20 x 100 / 2.304
= 1137 cm3
Use IPE 400
Check is similar to case 2.1 above.
ii) Section at support :
Mx = 0.75 x 32.72 = 24.54 m.t.
Compression flange (being at the bottom) is laterally unsupported,
therefore the section is assumed non-compact for simplicity. (Usually Lu >
Lu1, Lu2) i.e., Fbx = 0.583 Fy = 2.10 t/cm2
.
Use IPE 450:
fbx = Mx / Zx net = 24.54 x 100 / (0.85x1500) = 1.925 t/cm2
< 2.10 t/cm2
OK
(Note: Net section properties were used to account for the moment bolted
connection)
Check Shear:
q = Q / Aw net = 34.504 / (0.85 x 45 x 0.94) = 0.95 t/cm2
< 1.26 t/cm2
OK
Equivalent Stresses due to combined shear and bending:
all
22
e F1.1q3ff ≤+= = 2.436 t/cm2
> 1.1 x 2.1 = 2.31 N.G., Use IPE 500.
4.4.1.2) CROSS GIRDER
Structural System: Beam supported on main girders
Span = 7 m, Spacing = 4.5 m
1.1) Dead Load Effect:
Concentrated reaction from stringers = 2 × 2.784 = 5.568 t
Own weight of Cross Girder (assumed) = 0.3 t/m/

Steel Bridges
QDL = 3 × 5.568/2 + 0.3 × 7 / 2 = 9.402 t
MDL = 11.70 × 3.5 – 5.568 ×1.75 - 0.3 × (3.5)2
/2 = 21.326 mt
1.2) Live Load & Impact Effect:
Impact I = 0.4 - 0.008 L = 0.4 × 0.008 (2×4.5) = 0.328
{L = larger of 2 Lst (directly loaded members)
or Lxg (indirectly loaded member)}
Impact is applied to Main Lane only
• Max LL Reactions on Cross Girder:
a) From Main Truck:
P60 = 10 x (1+0.328) + 2 x 13.28 x 3/4.5
= 30.987 t
b) From Main Lane Uniform Load:
w60 = 2 [ 0.5x1.328 x 0.75/4.5] = 0.331 t/m/
c) From Secondary Truck:
P30 = 5+ 2 x 5x 3/4.5 = 11.667 T
d) From Secondary Lane Uniform Load:
w30 = 2 [ 0.3x 1.5 x 0.75/4.5] = 0.15 t/m/
N.B.: Uniform load on Lane Fractions on both sides of trucks is to be
neglected.

• Loads position for Max Moment:
Mmax occurs at the middle with loads placed as shown:
MLL & I = 96.88 mt
Qmax occurs at support with loads placed as shown:
QLL&I = 64.24 t
The total design moment on an intermediate cross girder is:
At the middle section: Mdesign = 21.326 + 96.88 = 118.206 mt
And the total design shear on an intermediate cross girder is:
At the support: Qdesign = 9.402 + 64.24 = 73.74 t

Steel Bridges
2) Design of Cross Section:
Straining Actions:
Mx = 118.206 m.t. (Maximum near middle)
Qy = 73.74 t (Maximum at support)
rolled section), and lateral torsional buckling requirements (comp flange
supported by deck slab); i.e.
Fbx = 0.64 Fy = 2.304 t/cm2
Req. Zx = Mx / Fbx = 118.206 x 100 / 2.304 = 5121 cm3
--------- use HEA 650
fbx = 118.206 x 100 / 5470 = 2.157 t/cm2
< 2.304 t/cm2
OK
Check of Fatigue: Actual Stress Range = fsr =(0.5 x 96.88)x100/5470
= 0.886 t/cm2
< Allowable Stress Range = Fsr = 1.26 t/cm2
Check Shear: q = Q / Aw net = 73.74 / (0.85 x 64 x 1.35) = 1.00 t/cm2
< 1.26 t/cm2

4.4.2) EXAMPLE 2: RAILWAY BRIDGE FLOOR

Steel Bridges
4.4.2.1) STRINGER
Structural System:
Beam supported on cross girders,
Span = 4.50 m, Spacing = 1.80 m.
1.1) Dead Load:
Track (rails, sleepers, conn. ) = 0.6 t/m/
of track
= 0.3 t/m/
of stringer
Own wt of stringer (assumed) = 0.15 t/m/
uniform load on stringer =0.3 + 0.15 = 0.45 t/m/
Dead Load Actions: QDL = 0.45 × (4.5) / 2 = 1.013 t
MDL = 0.45 × (4.5)2
/ 8 = 1.139 mt
1.2) Live Load & Impact:
• Impact factor I = 24/(24+ L) = 24/(24+4.5)= 0.842 (max 0.75)
(L = Loaded Length of track = 4.5 m)
• Loads position for Max Moment:
For L < 3.4 m: Mmax occurs with single wheel on stringer
For 3.4 < L < 4.4 m: Mmax occurs with two wheels on stringer
For L > 4.4 m: Mmax occurs with three wheels on stringer
MLL & I = ((3x12.5/2)x2.25-12.5x2) x(1+I) = 30.08 mt

For L > 3.0 m: Qmax occurs at support with three loads acting as shown:
QLL&I = (12.5+12.5x2.5/4.5+12.5x.5/4.5) x (1+I) = 36.46 t
1.3) Lateral Shock Effect:
a) If No Stringer Bracing is used:
My = 6 x 4 /4 = 6 m.t. ( at middle)
(Corresponding Mx = 25.51 m.t. )
b) If Stringer Bracing is used:
My = 6 x 2 / 4 =3 m.t. ( at quarter point)
Corresponding Mx = 0.9 + 21.875 = 22.775 m.t. )

Steel Bridges
The total design moment on an intermediate stringer is:
a) If No Stringer Bracing is Used: Critical Section at Middle
Mx = 1.139 + 30.08 = 31.219 mt
My = 6.75 mt
b) If Stringer Bracing is Used: Critical Section at Quarter Point
Mx = 25.38 mt
My = 3.375 mt
And the total design shear on an intermediate stringer is:
At support: Qy = 1.013 + 36.46 = 37.473 t
1.5) Design of Cross Section:
1.5.1) Case of Simple Stringer without Lateral Shock (stringer) Bracing:
Straining Actions: Mx = 31.94 m.t., My = 6.75 m.t.
(carried by top flange only)
Qy = 37.47 t
Section is compact w.r. to local buckling requirements (being a rolled
section), but not compact w.r. to lateral torsional buckling requirements
(comp flange unsupported for Lun= 4.5m); i.e., Fbx = 0.583 Fy = 2.10 t/cm2
Fby = 0.72 Fy = 2.592 t/cm2
(Minor axis bending)
Section HEB 400
fbx = 31.94 x 100 / 2880 = 1.1 t/cm2
< 2.10 t/cm2
OK
fby = 6.75x 100 / (721/2) = 1.87 t/cm2
< 2.592 t/cm2
OK
Combined Bending: fbx / Fbx + fby / Fby = 1.1/2.10 + 1.87/2.592= 1.24
< 1 x 1.2
(Factor 1.2 accounts for additional stress of Case II loads)
Unsafe  then use HEB 450
Check of Fatigue:
From Mx : Actual Stress Range = fsr =( 30.08)x100/3551 = 0.847 t/cm2
From My : Actual Stress Range = fsr =( 6.75)x100/(781/2) = 1.728 t/cm2
Total Stress Range = 0.847 + 1.728 = 2.575 t/cm2
> Allowable Stress Range = Fsr = 1.2x 1.26 = 1.512 t/cm2
Fatigue Check is UNSAFE: increase cross section

< 1.26 t/cm2
1.5.2) Case of Simple Stringer with Lateral Shock (stringer) Bracing:
Straining Actions: Mx = 28.25 m.t., My = 3.375 m.t. (at quarter point)
Qy = 37.47t at support & Qy = 22.325t at quarter point
rolled section), and w.r. to lateral torsional buckling requirements (comp.
flange supported at Lun= 2.25 m by stringer bracing);
i.e., Fbx = 0.64 Fy = 2.304 t/cm2
Fby = 0.72 Fy = 2.592 t/cm2
Section HEA 360 fbx = 28.25 x 100 / 1890 = 1.495 t/cm2
< 2.304 t/cm2
OK
fby = 3.375 x 100 / (526/2) = 1.283 t/cm2
< 2.592 t/cm2
OK
Combined Bending: fbx / Fbx + fby / Fby = 1.495/2.304 + 1.283/2.592
= 1.144
< 1 x 1.2
Check of Fatigue is similar to case above.
Check Shear at support: q = Q / Aw net = 37.47 / (0.85 x 35 x 1.0)
= 1.259 t/cm2
< 1.26 t/cm2
OK
1.5.3) Case of Continuous Stringer without Lateral Shock Bracing:
i) Section near mid span: Mx = 0.8 x 31.94= 25.552 m.t., My = 6.75 m.t.
(not affected by continuity)
section), but not compact w r to lateral torsional buckling requirements
(comp flange unsupported for Lun= 4.5 m); i.e., Fbx = 0.583 Fy = 2.10 t/cm2
Fby = 0.72 Fy = 2.592 t/cm2
Section HEB 400
fbx = 25.552 x 100 / 2880 = 0.887 t/cm2
< 2.10 t/cm2
OK

Steel Bridges
fby = 6.75 x 100 / (721/2) = 1.872 t/cm2
< 2.592 t/cm2
OK
Combined Bending: fbx / Fbx + fby / Fby = 0.887/2.10 + 1.872/2.592= 1.145
<1x 1.2
ii) Section at support : Mx = 0.75 x 31.94 = 23.955 m.t., My = 0, Qy = 37.47 t
Compression flange (being at the bottom) is laterally unsupported,
therefore the section is non-compact;
Use HEB360: for
y
f
u
F
20b
L ≤
= 316 cm,
therefore: Fbx = 0.583 Fy = 2.10 t/cm2
fbx = Mx / Zx net =23.955 x 100 / (0.85 x 2400) = 1.174 t/cm2
< 2.10 t/cm2
OK
q = Q / Aw net = 37.47 / (0.85 x 36 x 1.25) = 0.980 t/cm2
< 1.26 t/cm2
OK
Equivalent Stresses: all
22
e
F1.1q3ff ≤+= = 1.993 t/cm2
< 1.1 x 2.1
= 2.31 t/cm2
1.5.4) Continuous Stringer with Lateral Shock (stringer) Bracing:
i) Section at mid span: Mx = 0.8 x 28.25= 22.60 m.t.
My = 3.375 m.t (at quarter point)
Section is compact w.r. to both local buckling requirements (being a rolled
section), and lateral torsional buckling requirements (comp. flange
supported at Lun=2 m by stringer bracing); i.e., Fbx = 0.64 Fy = 2.304 t/cm2
Fby = 0.72 Fy = 2.592 t/cm2
Section HEB 320
fbx = 22.60 x 100 / 1930 = 1.171 t/cm2
< 2.304 t/cm2
OK
fby = 3.375 x 100 / (616/2) = 1.096 t/cm2
< 2.592 t/cm2
OK
Combined Bending: fbx / Fbx + fby / Fby = 1.171/2.304 + 1.096/2.592
= 0.931 < 1.2

< 1.26 t/cm2
ii) Section at support : See ((1.3) ii) above.
4.4.2.2) CROSS GIRDER
Structural System:
Beam supported on main girders
Span = 5.30 m, Spacing = 4.50 m
2.1) Dead Load Effect:
Concentrated reaction from stringers = 2 × 1.013 = 2.026 t
Own weight of X.G. (assumed) = 0.3 t/m/
QDL = 2.026 + 0.3 × 5.3 / 2 = 2.821 t
MDL = 2.821 x 5.3/2 – 2.026x0.9-0.3 × (2.65)2
/2= 4.6 mt
2.2) Live Load & Impact Effect:
• Impact factor I = 24/(24+ L) = 24/(24+ 9)=0.727
(L = Loaded Length of tracks= 2 x 4.5 = 9 m)

Steel Bridges
• Reactions on Cross Girder:
P = {12.5 + 2 x 12.5 x 2.5 / 4.5 + 6.25x0.75/4.5}x (1+I)
= {12.5 + 2 x 12.5 x 2.5 / 4.5 + 6.25x0.75/4.5}x (1.727)
= 47.37t
Mmax occurs at the stringer location
MLL & I = 47.37 x 1.75 = 82.90 mt
Qmax occurs at support:
QLL&I = 47.37 t
So the total design moment on an intermediate XG is:
Mdesign = 4.6 + 82.90 = 87.5 mt
And the total design shear on an intermediate XG is:
Qdesign = 2.821 + 47.33 = 50.191 t
2.3) Braking Force Effect:
a)If Braking Force Bracing is used:
Braking force is carried by the braking force bracing without any bending
in the Cross Girders. (i.e. My=0)
b) If No Braking Force Bracing is used:
Total Braking Force on the bridge:
B = Sum of train loads on bridge / 7 = 295 / 7 = 42.1 t
Braking force is equally divided between cross girders:
Braking force/ XG = 42.1 /no of XGs = 42.1/7= 6.02 t

My = ( 6.02 / 2) x 1.75 = 5.26 mt
2.4) Design of Cross Section:
2.4.1) Without Braking Force Bracing:
Straining Actions: Mx = 87.50 m.t., My = 5.26 m.t. ( at stringer location)
Qy = 50.191 t (at support)
section), but not compact w.r. to lateral torsional buckling requirements
(comp. flange unsupported for Lun=5.3 m); i.e., Fbx = 0.583 Fy = 2.10 t/cm2
Fby = 0.72 Fy = 2.592 t/cm2
Section HEB 600 :
fbx = 87.50 x 100 / 5700 = 1.535 t/cm2
< 2.10 t/cm2
OK
fby = 5.26 x 100 / (902/2) = 1.166 t/cm2
< 2.592 t/cm2
OK
Combined Bending: fbx / Fbx + fby / Fby = 1.535/2.10 + 1.166/2.592 = 1.074
<1.2 OK
Check Fatigue as before.
< 1.26 t/cm2
OK
2.4.2) With Braking Force Bracing:
My is carried by axial forces in the braking force bracing with the XG
subjected to Mx only:
Req. Zx = Mx / Fbx = 7.50 x 100 / 2.304 = 3798 cm3
--------- use HEB 550
< 1.26 t/cm2

Steel Bridges
4.4.3) CONNECTIONS OF BRIDGE FLOOR BEAMS
4.4.3.1 Calculations of Bolt Resistance:
High strength bolts of Grade 10.9 or 8.8 are normally used in bridge
constructions. Connections may be designed as Bearing Type (easier in
execution) or Friction Type (when slip is not allowed).
For Bearing Type Connections:
Bolt Resistance = Smaller of Rshear and Rbearing;
Rshear = n x (Bolt area x Allowable bolt shear stress) = n x (π d2
/4) x 2
Rbearing = Bolt diameter x Allowable bearing stress x tmin = d x (0.8 Fult) x tmin
(n = no. of shear planes, d= bolt diameter, Edge distance≥ 2 d, F ult = 5.2
t/cm2
)
For Friction Type Connections: Bolt Resistance = n x Ps
Bolt
Diameter
Bearing Type Connections Friction Type
Connections
RS.Shear Rbearing RD. Shear Bolt 8.8 Bolt 10.9
M20 6.28 8.32 tmin 12.56 3.37 4.82
M22 7.60 9.15 tmin 15.20 4.17 5.96
M24 9.04 9.98 tmin 18.08 4.85 6.94
4.4.3.2 Design of Connection between Stringer and Cross Girder:
Railway Bridge Floor Data:
Stringer: Shear Force = 37.47 t
, -ve Moment = 23.955 m.t.,
section is HEB 360.
Cross Girder: HEB 600
2.1) Case of Simple Stringer:
Connection is designed for the max shear of stringer using framing angles as
shown:.
Using M20 HSB Grade 10.9 (Bearing Type):

a)Bolts between stringer web (tw = 1.25 cm) and angle legs:
Double shear bolts Rb = 8.32 x 1.25 = 10.4 t
= Rleast
Number of bolts = Q / Rleast = 37.47 / 10.4 = 3.3 bolts
Fatigue Considerations: Case 27.1 of Group 3 of ECP: for Class C;
Allowable stress range = 0.91 t/cm2
Rsh = 2 x 2.45 x 0.91 = 4.46 ton = Rsr
Number of bolts = Qsr / Rsr = 36.46 / 4.46 = 8.17 too many
Either use bolts with larger diameter or use Friction Type Joint:
i) Use M24 bolts: Rsh = 2x3.53x0.91 = 6.43 ton = Rsr
Number of bolts = Qsr / Rleast = 36.46 / 6.43 = 5.8
Use 6 bolts M24 (Bearing Type)
ii) Friction Type Joint: Rsh = 2 x 4.82 = 9.64 ton = Rleast
Number of bolts = Qt / Rleast = 37.47 / 9.64 = 3.88
Use 4 bolts M24 (Friction Type)
b) Bolts between angle legs and cross girder web:
Single shear bolts on two sides: Fatigue governs the design
Number of bolts = Qsr / Rsr = 36.46 / (3.53x0.91) = 11.35
Use 12 bolts M24 (6 bolts each side)
(Alternative Using Friction Type Bolts: No. of Bolts = 37.47/4.82= 7.77,
i.e.; 4 bolts M20 (Friction Type) Each Side)
2.2) Case of Continuous Stringer:
Continuity is achieved by using top and bottom plates designed to transmit
the flange force:
C = T = M-ve / hstr = (0.75x30.08) / 0.36 = 62.667 t
Number of bolts = 62.667 / 6.43 = 9.746
use 10 bolts M24 (Bearing Type) each side.
(Alternative Using Friction Type Bolts: No. of Bolts = 66.54/4.82=13.8,
i.e.; 14 bolts M20 (Friction Type) Each Side)
Compute Plate Thickness from:
(30 – 4 x 2.6) x t x 2.1 = 66.54 which gives tpl = 1.617 cm
Use tpl = 1.80 cm

Ch4 Bridge Floors (Steel Bridges تصميم الكباري المعدنية & Prof. Dr. Metwally Abu-Hamd)

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