Design of Steel Structures (2180610) IMP materials FOOT ...

Faculty of Degree Engineering - 083 Department of Civil Engineering - 06

Design of Steel Structures (2180610) IMP materials

FOOT OVER BRIDGE DESIGN

INTRODUCTION

A footbridge or person on foot connect is a scaffold intended for walkers and now

and again cyclists, creature movement and stallion riders, as opposed to vehicular

activity. Footbridges supplement the scene and can be utilized beautifully to

outwardly connect two particular zones or to flag an exchange. In numerous created

nations, footbridges are both useful and can be lovely masterpieces and figure. For

poor provincial groups in the creating scene, a footbridge might be a group's just

access to medicinal centers, schools and markets, which would some way or another

be inaccessible when waterways are too high to cross. Straightforward suspension

connect outlines have been created to be reasonable and effectively constructible in

such rustic zones utilizing just neighborhood materials and work.

An encased footbridge between two structures is in some cases known as a skyway.

Scaffolds accommodating the two people on foot and cyclists are regularly alluded to

as green extensions and frame a vital piece of economical transport development

towards more reasonable urban areas. Footbridges are frequently arranged to

enable people on foot to cross water or railroads in territories where there are no

adjacent streets to require a street connect. They are likewise situated crosswise

over streets to give walkers a chance to cross securely without backing off the

activity. The last is a kind of passerby detachment structure, cases of which are

especially found close schools, to help counteract kids running before moving autos.


Design data of foot over bridge

Clear span = 21m Width of carriageway = 25m Cross girder spaced at centers = 2.1m Type of truss configuration = N type

Assumptions:

Dead load and live load loads IS 875 part I &II Wind loading IS 875 part III Pedestrian loading = 4000 N/m2 Flooring to be made timber planks

Design of planking:

Span = 2.1 m

Assume the thickness of planking = 60 mm

Dead load of planking:

= (60×1000) ×8000 [dead load of wood = 8000 N/m3]

= 480 N/m2

Live load = 4000 N/m2

Total load = 4480 say 4500 N/m2

For 1 m width of timber planking

Total load = 4500 N/m2

Maximum bending moment = wl2/8

= [4500×(2.1)2]/8 =2480.625 N/m

Equating this moment of resistance, we have

(1/6)fbd2 = 2480.62×1000

(1/6)×f×1000×(60)2 = 2480.625×1000

f = 2480.625/600 = 4.135 N/m2

Where permissible value of bending stress

f = 10 N/mm2

Maximum shear force = wl2/2


= (4500×2.1)/2 = 4725 N

Mean shear stress = w/bd

= 4500/(60×1000) = 0.075 N/mm2

Maximum shear stress = 1.5× mean shear stress

= 1.5× 0.075 = 0.1125 N/mm2

Safe shear stress = 0.8 N/mm2

M.I of section of strip = bd3/12

=1000×(60)3/12 = 18×106 mm4

Maximum deflection seen δ =

=

= 6.33 mm

Allowable/permissible deflection = span/325

= 2100/325 =6.46mm

Design of cross beams

Clear width of footway = 2.5m

Assume the center to center distance between two girders

= 2.5+0.3 = 2.8m

Load transmitted to one girder from the flooring

= 2×2.5×4500 = 23625N

Self weight of the beam (assumed)

=300 N

Total load = 23925 N

Bending moment = wl/8

= (23925×2.8)/8 = 8373.75 Nm

Permissible bending stress = .66×fy

σbc = 0.66×250 = 165 N/m


As we know that section modulus (Z) is given by

Z = M/ σbc

= (8373.75×1000)/165 = 50750 mm3

Using steel table we found that the permissible channel section that can be used for Z =

50.750 cm3 is

ISLC 125 with value of Zxx = 57.1 cm3

Seismic coefficient method using IS 1893

αh = βIαo

Zone IV αo = 0.05

β = 1.0 (for type II medium soil) I

= 1.5 (important bridges)

αh = 0.05×1×1.5 = .075

Fh = Horizontal seismic force

Fh = αh × wm

= 6200 × 0.075 = 465 N/m

Dead load transmitted from cross beam

= 696.42 N/m or say 700 N/m

Weight of one truss assumed

= 400 N/m

Total dead load = 700 + 400 = 1100 N/m

Live load on one truss = (4000 × 2.5)/2 = 5000 N/m

2.1

2.1

2.1

2.1

2.1

2.1

2.1

2.1

2.1

2.1


= (8.4 × 12.6)/(21 × 1.8)

= 2.80

= (8.4 × 12.6)/(21 × 2)

= 2.5

Maximum tension in the member L3 L4

= Area of Influence line diagram × load intensity

= (1/2)×21×2.80×6100 = (1/2)×21×2.5×6100

= 179340/150 = 1195 m2 = 161406/150 = 1076 m2

Allowing a stress of 150 N/m , Net area required for this member

= 162750/150 = 1085 m2

Weight of two truss including bracing

= 1085 × 0.785 = 851 N/m

Weight of one truss including bracing

= 851/2 = 425.5 N/m

Hence safe


U0 U1 U2 U3 U4 U5 U6 U7 U8 U10 U10

L0 L1 L2 L3 L4 L5 L6 L7 L8 L9 L10

(9/10)*(4.2/2) = 0.945

U0U1 & L0L2 = 0.945

(8/10)*(6.3/2) = 1.68

U1U2 & L1L2 = 1.68

(7/10)*(6.3/2) = 2.205

U2U3 & L2L3 = 2.205

(6/10)*(8.4/2) = 2.5

U3U4 & L3L4 = 2.5

(5/10)*(8.4/2) = 2.625

U4U5 & L4L5 = 2.625

Figure 1: Influence line Diagram for Horizontal Member


U0 U1 U2 U3 U4 U5 U6 U7 U8 U10 U10

L0 L1 L2 L3 L4 L5 L6 L7 L8 L9 L10

x= 0

y = 2.1

y

2.1

x

4.2

y

6.3

x

y

x

y

x 8.4

16.8

14.7

y

12.6

10.5

x= 0.23

y = 1.87

x= 0.47

y = 1.63

x= 0.70

y = 1.40

x= 0.93

y = 1.17

Figure 2: Influence line Diagram for Vertical Member


U0 U1 U2 U3 U4 U5 U6 U7 U8 U10 U10

L0 L1 L2 L3 L4 L5 L6 L7 L8 L9 L10

h1 = 1.32

h1= 1.178

y h2 = 0.145

x

h2

y

x

h1

y

h1= 1.03

h2 = 0.29

h1= 0.885

h3 = 0.435

x

h1= 0.745

y h2 = 0.585

x

Figure 3: Influence line Diagram for Diagonal Member


Forces in various members of truss

1. For loading condition 1.

Dead load + Live Load that had been already worked out i.e. = 6200 N/m

Top chord members

All members are compression members, maximum force will be there when whole span is loaded Member

U0U1

= 6200×(1/2)×21×0.94 = 61519.5 N

Member U1U2

= 6200×(1/2)×21×1.68 = 109368 N

Member U2U3

= 6200×(1/2)×21×2.205 = 143545.5 N

Member U3 U4

= 6200×(1/2)×21×2.5 = 162750 N

Member U4 U5

= 6200×(1/2)×21×2.625 = 170887.5 N

So maximum compressive force in top chord will be 170887.5 N

Bottom chord members

These are the tension members

Member L0 L1

There will be no force in this member

Member L1 L2

The tensile force in this member will be numerically equal to the force in compressive member U0 U1 i.e.

= 61519.5 N (Tensile) Member

L2 L3


The tensile force in this member will be numerically equal to the force in

compressive member U1 U2 i.e.

= 109368 N (Tensile) Member

L3 L4


= 143545.5 N (Tensile)

Member L4 L5


= 162750 N (Tensile)

So maximum compressive force in bottom chord will be 162750 N

Vertical Members

Member U0 L0

This is the compression member. Maximum compression in this member will occur when the whole span is loaded

Maximum compression in U0 L0

= 6200×(1/2)×21×(9/10) = 58590 N (Compressive)

Member U1 L1

Area of +ve zone of ILD

= (1/2)×(16.8+1.87)×(8/10) = 7.468

Area of -ve zone of ILD

= (1/2)×(2.1+0.23)×(1/10) = 0.1165

Dead load in this member

= 1200(7.468 – 0.1165) = 8521.8 N (Compressive)

Live load compression

= 5000×7468 =37240 N (Compressive)

Live load tension

= 5000×1.165 = 582.5 N (Tension)


Extreme forces in this member

8821.8 + 37340 = 46161.8 (compressive)

8821.8 – 582.5 = 8239.3 (compressive)

Member U2 L2

Area of +ve zone

= (1/2)×(14.7 + 1.63)× (7/10) = 5.71units2

Area of –ve zone

= (1/2)×(4.2+0.47)×(2/10) = 0.467 units2

Dead load

= 1200×(5.71-0.467) = 6291.6 N (compressive)


= 5.71×5000 =28550 N

Live load tensile

= 0.467 × 5000 = 2335 N

Load on extreme values

6291.6 + 28550 = 34841.6 N (compressive)

6291.6 – 2335 = 3956.6 N (compressive)

Member U3 L3

Area of +ve zone

= (1/2)×(14)× (6/10) = 4.2

Area of –ve zone

=(1/2)×(7)×(3/10) = 1.05

Dead load

= 1200×(4.2-1.05) = 3780 N (compressive)


= 5000×4.2 = 21000 N (compressive)

Live load tensile

5000×1.05 = 5250 N (Tension)



3780 + 21000 = 24780 (compressive)

3780 – 5250 = 1470 N (tensile)

Member U4 L4

Area of +ve zone

=(1/2)×(11.67)×(5/10) = 2.91

Area of –ve zone

= (1/2)×(9.33)× (4/10) = 1.866

Dead load

1200×(2.91+1.86) = 1269 N


= 5000×2.91 = 14550 N

Live load tensile

5000×1.86 = 9300 N


1260 + 14550 = 15810 (compressive)

1260 – 9330 = 8040 N (tensile)

Diagonal Members

Force in U0 L1

= Force in U0 L0 cosecθ

= 58590×1.45 = 84955.5 N (Tensile)

Force in U1 L2


= 46161.8 × 1.45 = 66934.610 N (Tensile)

Force in U2 L3


= 34841.6 × 1.45 = 50520.320 N (Tensile)

Force in U3 L4



Tensile force = 247870 ×1.45 = 39846 N

Compressive force = 1470 × 1.45 = 2131.5 N

Force in U4 L5


Tensile force = 15810×1.45 = 22924.5 N

Compressive force = 8040×1.45 =11658 N

Wind load calculations

Basic wind speed for delhi zone is vb = 47 m/s

Design wind speed at any height z is given by

Vz = Vb K1 K2 K3 K4

K1 = Probablity factor (risk factior) K2 =

Terrain roughness and HF

K3 = Topography factor K4

= importance factor

Using IS 875 we came to know that

From table 1: Risk factor for different classes of structure in different wind speed zones.

K1 = 1.07

From table 2:

K2 = 0.8

Topography factor (K3) = 1.00 for as slope < 30

Importance factor for cyclonic region (K4)

As our structure is a post cyclonic importance

Pz = 0.6 Vz2

Vz = 47×1.07×0.5×1.0 =40.232 m/s

Pz = 0.6×40.232 = 971.168 N/m2

But design wind pressure (Pd) is calculated as follows Pd


= Kd Ka Kc Pz

Kd = wind directionality factor

Ka = area average ring factor Kc

= combination factor

Pd = 0.9×0.9×1.0×971.168 = 786.65 N/m2

Table 1: Dimensions of chords

S.No. Details of exposed area

Width (mm) Depth (mm) Face (mm) Total length (mm)

1. Top chord 60 120 60 21

2. Bottom chord 60 120 60 21

3. End post 60 120 60 2×2.0

4. Verticals 60 60 60 9×2.0

5. Diagonals 60 60 60 10×2.9

Gusset for top chord @ 0.25 m2

11×0.005 = 0.055m2

Same for bottom chord @ 0.25 m2

11×0.005 = 0.055m2

Wind load on windward and leeward girder both is as follows Wind

Load

On top girder chord

= 786.65 × 21 × .06 = 1982 N

On bottom chord

= 786.65 × 21 × .06 = 1982 N

On end posts

On diagonal

= 4×2.0×0.06×786.65 =755.185 N

= 10×2.0×0.06×786.65 = 2737.5 N


On verticals in between

= 9×2×0.06×786.65 = 1699.16 N

Gusset for top chord

= 11×.005×786.65×2 = 86.53 N

For bottom chord

= 11×.005×786.65×2 = 86.53 N

Basic design wind pressure for unloaded conditions

Pd = 2354.4 N/m2

Wind load on top chord

= (1982×2354.4)/786.65 = 5932 N (296 N)

Wind load on bottom chord

= 5932 N (296 N)

Wind load on vertical end posts = 2260 N

On diagonal member = 8185 N

On verticals in between = 5080 N


Table 2: Design forces in Newton

Member Combination 1

Dead load + Live load

Comp. Tension

Combination 2

Dead load + wind load

Comp. Tension

Combination 3 Dead + Live + wind Comp. Tension

Design loads

Comp. Tension

Top chord

170987.50

U0 U1 61519.5 12234.78 61619.5

U1 U2 109368 21517.32 109468.00

U2 U3 143545.5 28291.00 143645.5

U3 U4 162750 323036.25 16289.50

U4 U5 170887.5 33623.00 170987.50

Bottom chord

L0 L1 5932.00 100.00

162850.00 L1 L2 61519.5 17866.70 61619.50

L2 L3 109368 27144.32 109468.00

L3 L4 143545.5 33923.00 143645.50

L4 L5 162750 39255.00 162850.00

Verticals

U0 L0 58590.00 11625.00 58685.00

58685.00

8040.00 U1 L1 46161.80 9106.00 46256.80

U2 L2 34841.60 6756.60 34936.60

U3 L3 24780.00 1470.00 4065.00 24875.00 1470.00

U4 L4 15810.00 8040.00 1545.00 15905.00 8040.00

U5 L5 285 95

Diagonals

U0 L1 84955.50 16976.32 850.50

11658.00

85050.00 U1 L2 66934.61 13201.01 67029.61

U2 L3 50520.34 9532.82 50615.32

U3 L4 2131.50 39846.00 5891.00 2131.50 39041.00

U4 L5 11658.00 22924.50 2237.00 11658.00 23019.5


The final Design of Girder of Foot Over Bridge

Top chord of the Girder

Design force = 170987.5 N (Compressive)

Length of Top chord = 2.1 m or 2100mm

Try Double L‟s 60×60×10

Area = 2200 mm2

r = 1.78 cm or 17.8 mm

Slenderness ratio = l/r

= (0.85×2100)/17.8 = 100.2

or say l/r = 100

For value of l/r = 100, safe compressive stress = 80 N/mm2 Safe

load

= 80×2200 = 176000 N

But the maximum load that chord has to bear is 170987.5 N

Hence safe

Other top chord member will be designed in the same way

Bottom chord of the Girder

Design force = 162850.00 N

Safe tensile force = 150 N/mm2

Net area required

A(req) = 162850.00/150 = 1085.66mm2

Try Double L‟s 60×60×10 Gross

area

A(g) = 2200 mm2

Diameter of rivets = 20mm

Making allowance for 4 rivet hole


4 ×d×t = 4×21.5×10 = 860mm2

So, Net area available

Anet = 2200 – 860 = 1340 mm2

A(req) = 1085 mm2

Hence safe

Member U0 L0

Vertical members

Compressive force = 58685.00 N

Length of member = 2.0 m

Since the end vertical members it is desirable to provide a double section Try

Double L‟s 60×60×10

Area = 2200 mm2

r = 1.78 cm or 17.8 mm


= (0.85×2100)/17.8 = 95.50

for value of l/r = 95.50, safe compressive stress = 84.5 N/mm2

Safe load

= 84.5×2200 = 185900 N

Hence safe Member

U1 L1

Compressive force = 46256.8 N

Length of member = 2.0 m

Try single L‟s 60×60×10

Area = 1100 mm2

r = 1.15cm or 11.5 mm


= (0.85×2000)/11.5 = 147.5

Permissible compressive stress 50.22 N/mm2


so the total allowable load is

50.22 × 1100 = 55242 N

Hence safe

Member U2L2 , U3L3 , U4 L4 and U5 L5 can be designed in the same way with one LS 60×60×10 for compressive force

For Tension force in U4 L4

Maximum tension = 1470 N

Effective area for single angle Tension

= A1 + A2K

A1 = Net section area of connected leg

= (60-5)×10-21×10 = 340 mm2

A2 = Net section area of outstanding leg

= (60-5)×10 = 550mm2

K = 3A1/(3A1+A2)

= 1020/(1020+550) = 0.649

Net area provided

= 0.649×550+340 = 697.324 mm2

Permissible tensile stress = 150 N/mm2

Safe tensile load

= 649.324×150 = 104598.72 N

But the maximum tension in member is just 1470 N

The section is safe

Same for tension force U5 L5

Member U0 L1

Diagonal member

Maximum tension = 85050 N

Permissible tensile stress = 150 N/mm2

Try single L‟s 60×60×10


Safe tension for this section = 104598.72 Hence the section is safe

Member U1 L2

Maximum tension in this member = 67029.61 N So,

again let us provide single LS 60×60×10

hence safe

Member U2 L3

Maximum tension in this member = 50615.32 N

So, again let us provide single LS 60×60×10

which is safe

Member U3 L4

Design compressive force = 2131.5 N

Design tensile force = 39941.0 N

So, both for compression and tension single LS 60×60×10 will be safe to be provided Member U4 L5

Design compressive force = 23019.5 N

Design tensile force = 11658 N

So, both for compression and tension single LS 60×60×10 will be enough

Design of joints

Assume the thickness of gusset plate = 8mm

Diameter of rivets = 20 mm

Finished diameter of rivets

= 20 + 1.5 = 21.5mm

Rivet value in single shear

= τvf × (π/4)d2 [τvf = 100 N/mm2 ]

= 100 × (π/4) × 21.52 = 36305 N

Rivet value in double shear

= 2 × 36305 = 72610 N


Rivet value in bearing on 8mm plate

= σcb × d × t

= 300 × 21.5 × 8 = 51600 N

Joint U0

Riveting in U0 U1 = 61619.5/51600 = 2 rivets

Riveting in U0 L0 = 58685/51600 = 2 rivets


Joint U1

U1 U2 as this is continuous chord

Riveting in U1 U2 = 2 × No. of members meeting the continuous member -1

= 2×2-1 = 3

No. of rivets in U1 L1 = 46256.80/36305 = 2 rivets

No. of rivets in U1 L2 = 67029.6/36305 = 2 rivets

Joint U2

Joint U3

Again U2 U3 is continuous member so 3 rivets will be enough

Riveting in U2 L2 = 34936.6/36305 = 2 rivets


Again U3 U4 = 3 rivets same as in U1 U2



Joint U4

Again U2 U3 = 170987.5/51600 = 3 rivets


Riveting in U4 L5 = 23019.5/36305 = 2 rivets


Joint U5

No. of joints in top continuous chord

= 2 ×2 – 1= 3

No. of rivets in U5 L5 = 2 rivets

Bottom chord

Joint L0

Joint L1

Joint L2

Joint L3

Joint L4

Joint L5

Riveting in L0 U0 = 2 rivets

Provide 3 rivets in bottom chord


Provide 3 rivets in bottom chord


Riveting in L3 L2 = 109468/51600 = 3 rivets


Riveting in L3 L4 = 143645.5/51600 = 3 rivets

No. of rivets in L4 L3 = 3 rivets

No. of rivets in L4 L5 = 162850/51600 = 4 rivets

No. of rivets in L4 U5 = 2 rivets

Bottom chord

= 2×3-1 = 5 rivets




2.8

θ

2.0 m

0.5 m

Rakers Assume

the rakers be provided at an angle θ with the vertical For the

arrangement shown in fig.

tan θ = 0.5/2.0 = 1/4

θ = 14002‟

H in rakers is assumed to be 2.5 % of maximum compressive force in top girder T =

thrust in rakers

Tsinθ = (2.5/100) × 170987.50 = 4274.68 N

T = 4274.68/sin14002

‟ = 17628.50

Length of rakers

= 2.0 sec14002‟ = 2.062m

The rakers can be designed as compression member of 2.062 long for 17628.50 N So,

max. force in rakers = 17628.5 (compressive).

Length of raker

= 2.062m = 2062mm

Effective length „l‟

0.85 ×L = 0.85 × 2062 = 1752.7 mm

Try single L‟s for this;


So, Slenderness ratio = l/r

= (0.85×2062)/16.9 = 103.7

or say l/r = 100

So, for this value, the safe compressive strength of the section

= 80 N/mm2 for [fy = 225 MPa]

Therefore total compressive load taken up by the section

=

Which is for above from total max. Compressive of 1728.5N.

Hence section is safe.

BEARING PLATE:

Maximum end reaction

=

[Dead load + live load = 6200 / and wind load

=

So, total load = 5.199N/m]

Total end reaction = 65595.58

Area of bearing plate

=

Try a bearing plate of

So, cantilever projection

Upward pressure on bearing plate =

B.M per mm width of plate =

Allowable bending stress of bearing plate = 185


Equating moment of resistance to maximum bending moment

=

Therefore;

So, provide a bearing plate of 10 thickness.

Drawings of joints


2. Numerical Problem Design a Gantry Girder to be used in an industrial building carrying a Manually Operated Overhead Travelling Crane, for the following

data in Table 1:


Gantry Girders

Table 1: Data for Numerical Problem

Sr. No. Crane Property Magnitude

1 Crane Capacity 300 kN

2 Self-Weight of Crane Girder excluding Trolley 200 kN

3 Self-Weight of Trolley, Electric Motor, Hook, etc. 40 kN

4 Approximate Minimum Approach of Crane Hook to the

Gantry Girder

1.20 m

5 Wheel Base 3.5 m

6 Centre-to-Centre Distance between Gantry Rails 18 m

7 Centre-to-Centre Distance between Columns (Span of

Gantry Girder)

10 m

8 Self-Weight of Rail Section 300 N/m

9 Diameter of Crane Wheels 150 mm


3. Solution For Fe 410 grade of steel: fu = 410 MPa,

fy = fyw = fyf = 250 MPa

For hand-operated OT crane

Lateral loads = 5% of maximum static wheel load

Longitudinal loads = 5% of weight of crab and weight

lifted Maximum permissible deflection = L/500

3.1. Partial safety factors

γm0 = 1.10

γmw = 1.50 (for site welds)

Load factor γm1 = 1.50

3.2. Design forces

Maximum wheel load:

Maximum concentrated load on crane=300 + 40 = 340 kN

Maximum factored load on crane = 1.5 × 340 = 510 kN


The crane will carry the self-weight as a uniformly distributed load = 11.11 kN/m

Factored uniform load = 1.5 × 11.11 = 16.67 kN/m

For maximum reaction on the gantry girder the loads are placed on the crane girder

as shown in Fig. 2.

Taking moment about B,

RA × 18 = 510 × (18-1.2) + or, RA = 626 kN

Similarly, RB = 184 kN

The reaction from the crane girder is distributed equally on the two wheels at the

end of the crane girder.

Therefore, maximum wheel load on each wheel of the crane = = 313 kN

3.3 Maximum bending moment

Assume self weight of gantry girder as 2.2 kN/m.

For maximum bending moment, the wheel loads shall be placed as shown in Fig. 4.

Total dead load = 2200 + 300 = 2500 N/m = 2.5 kN/m

Factored dead load = 1.5 × 2.5 = 3.75 kN/m

The position of one wheel load from the midpoint of span

= wheel base4 = 3.54 = 0.875 m

Bending moment due to live load only:

Taking moment about D,

RC × 10 = 313 × (10-2.375) + 313 × 4.125

RC = 367.78 kN

Taking moment about C,

RD × 10 = 313 × 2.375 + 313 × 5.875

RC = 258.22 kN


Maximum bending moment due to live load

= 258.22 × 4.125

= 1065.16 kNm

Bending moment due to impact = 0.1 × 1065.16

= 106.52 kNm

Total bending moment due to live and impact loads

= 1065.16 + 106.52 = 1171.68 kNm

3.4 Maximum shear force

For maximum shear force, wheels are placed as shown in Fig. 5.

Taking moment about D,

RC × 10 = 313 × 10 + 313 × 6.5

RC = 516.45 kN

Hence maximum shear force due to wheel loads = 516.45 kNm

3.5 Lateral forces

Lateral force transverse to the rails = 5% of the weight of crab and weight lifted =

0.05 × 340 = 17 kN

Factored lateral force = 1.5 × 17 = 25.5 kN

Lateral force on each wheel = 12.75 kN

Design of Steel Structures (2180610) IMP materials FOOT ...

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