Crane Girder Design

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DESIGN OF CRANE GANTRY GIRDER IN TG BAY

FORM T9-P REV-A

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Description TABLE OF CONTENTS Sl No. Description

Page No.

1.0

GENERAL

4

2.0

REFERENCES

4

3.0

DESIGN CRITERIA

4

4.0

DESIGN METHODOLOGY

4

5.0

LOADS

4

6.0

DETAILED DESIGN

5

7.0

TEMPERATURE ANALYSIS

5

ANNEXURES 1.0

ANNEXURE-A

6

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Description 1.0 GENERAL 1.1 Introduction

1.2 Units of Measurement All the units used in the analysis and designs are in SI system unless noted otherwise 2.0 Design standards and Codes Design is prepared in accordance with Indian standard of codes of practices Various codes of practices being referred to are listed below 1 IS:800 - 1984 Code of practice for General construction of Steel 2 IS:875 - part 2 Code of practice for design loads (other than earthquake) Imposed loads 3 IS:816 - 1969 Code of practice for use of metal arc welding for general (Reaffirmed 1998) construction in mild steel 3.0 Design criteria Design criteria for Power House Building have been used in general for all analysis and designs 4.0 Analysis and design methodology Analysis and design has been carried out based on inputs provided in detailed specification. The crane girder is a plate girder of welded construction and designed as a simply supported beam of single span The maximum bending moment under any load occurs when the load and the resultant of all the loads are located equidistant from the centre of the span The maximum shear force at a given section occurs when one of the load is at the section itself 5.0 Loads considered for design Crane capacity as mentioned in dwg no XXX-YYY-ZZZ-000-M000 is 85 T The crane wheel loads considered for the design are based on the data furnished in dwg no XXX-YYY-ZZZ-000-M000 Wheel loads and spacing considered is as fallows

P 0 mm

P 6950 mm

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Description 1. Lifting capacity of crane 2. Weight of crab 3. Self weight of crane bridge 4. Number of wheels on each side 5. Wheel spacing (end to end) 6. Nearest point from crane girder to which crab can come 7. Width of walkway 8. DL of walk way 9. LL on walkway 10. Crane rail size 11. Span of crane bridge (Assuming side clearance as 0.75 m) 12. Max wheel load from given data 13. Max wheel load from given data(with impact)

Cc Cw Sw n a (Ds)

Lc

= = = = = = = = = = =

850 240 700 2 6950 1450 1200 0.75 5.00 CR80 23400

kN kN kN

= =

686.62 kN 857.75 kN

mm mm mm kN/Sqm kN/Sqm mm

6.0 Detail design calculations for the worst combination is reported in Annexure-A 7.0 Temperature Analysis acting on the Structure In the following pages Design Temperature Load is calculated in accordance with the Design Memorandum. Average Annual Minimum Temperature = -3.30 Deg Average Annual Maximum Temperature = 47.70 Deg Design Temperature Differential = 17.00 Deg { (47.7 - -3.3) X (0.5 X (2/3))} length from Grid 1 to grid 8 (TG bay) = 72.00 m Gap of expansion joint = 1.50 m length from Grid 8a to 12 (Electrical bay) = 42.00 m As per 3.3.4 α = The co-efficient of expansion for steel shall be taken ‘as 0.000 012 per degree centigrade per unit length. of IS 800-1984 In TG bay Change in length due to temrature δl = α X L X δt = 0.000012 X 72750 X 17.00 Deg = 14.84mm In Electrical bay bay Change in length due to temrature δl = α X L X δt = 0.000012 X 42750 X 17.00 Deg = 8.72mm Total gap of expansion joint required

Let us provide expansion gap of

= = 10 mm

14.84mm 19.20mm

+ (

8.72mm )/2

Both sides

between girders in grid 8- 8a & 12-12a

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ANNEXURE - A

FORM T9-P REV-A

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Description DESIGN OF CRANE GIRDER FOR DATA

10.0 M SPAN (TG BAY)

1. Lifting capacity of crane 2. Weight of crab 3. Self weight of crane bridge 4. Number of wheels on each side 5. Wheel spacing (end to end) Wheel spacing when more than 2 wheels

Cc Cw Sw n

(B'n 1&2) (B'n 2&3) (B'n 3&4) 6. Nearest point from crane girder to which crab can come a 7. Width of walkway (Ds) 8. DL of walk way 9. LL on walkway 10. Span of runway girder l 11. Crane rail size 12. Span of crane bridge Lc (Side clearance as 0.8 m) 13. Max wheel load from given data 14. Max wheel load from given data(with impact) DESIGN

= = = = = = = = = = = = = = =

850 240 700 2 6950 0 6950 0 1450 1200 0.75 5.00 10000 CR80 23400

kN kN kN

= =

640 800

kN kN

Refer Dwg no

mm mm mm mm mm mm kN/Sqm kN/Sqm mm

Assumed

mm

STEP 1 Determination of maximum wheel load Maximum static wheel load P = 1/n [Sw/2 + {(Cc+Cw)(Lc-a)/Lc}] P = 686.23 kN Maximum static wheel load

P =

Considering an impact factor of I

= 25 % P = 857.79 kN

Pmax =

857.79

686.23 kN

(max of calculated and given data) As per IS : 875 ( Part 2 ) - 1987 (6.1,a)

kN

STEP 2 Determination of maximum bending moment (Single crane operation) The maximum bending moment under moving loads Taking moment about A RB = 857.79 x( 4.00 +

4.00m 1.70m 1.8 0.9 0.9 1.70m 857.79 5.7

+

###

+

=

=

2564.78

x

3480.04

C

kN 4.30

A 857.79 Ax(

###

+

###

C

)

kNm

Hence max bending moment at C= Mv =

-0.95m

KN

)

5.0 M

10.00

= 2564.78 BM at C

###

C.G of girder

5.0 M

B

C.G of loads

B

10.00m

2144.46

kNm FORM T9-P

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Description To account for uniformly distributed dead load of runway girder, the design moment is calculated as M = Km Mv M =

Km = 1.05

2251.69

Assumed as 5 %

kNm

STEP 3 Determination of horizontal loads due to lateral surge Horizontal load per wheel is given by Hl = Ks (Cc+Cw) / 100 n = Hl =

10

x( 857.79 + 54.89

Ks =(impact factor) 240 )/( 100x

10

%

As per IS : 875 ( Part 2 ) - 1987 (6.1,c)

2 )

kN

Horizontal bending moment Mh =

54.88929 x 2144.46 = 137.22 857.786 Axial force at top flange of surge girder(Fs) = Mh = 114.353 kN Ds where Ds=depth of surge girder assumed Ds=width of walk way Considering horizontal span lh = Mh = Hl * lh / 4=

1200

16.467

mm

kNm

STEP 4 Adapting proper I section An I section is considered as the girder,

mm

500 mm

Assume a trail section as shown below 500 25 16 1150

mm mm mm mm

The sectional properties are as follows Ax = 43400 Ixx = 10658041666.667 Iyy = 521225866.66667 rx = 495.557259 ry = 109.589291 Iyyt (of top flange plate) =

16 mm

###

Flange width = Flange thickness = Web thickness =(tw) Web depth = (dw)

kNm

25

mm

mm2 mm4 Zxx = 17763402.7778 mm3 mm4 Zyy= 2084903.46667 mm3 mm Af (Area of top flange) = 12500 mm2 mm 260416667 mm4 Zyyt = 1041666.66667 mm3 FORM T9-P ORIGINATOR/DEPT.

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STEP 5 Design check Check for bending stresses ly =

lh / ry =

1200

/

109.5893 =

10.950

where λ=slenderness ratio From IS 800 permissible bending stress is Vertical : sbcv = 158.20 Mpa Horizontal sbch = 0.66fy = 158.40 Mpa Axial sac = 0.6 fy = 144.00 Mpa

Calculated as per bending stress calculation. For details refer sheet 1 sheet no 18

Actual bending stress in compression due to Vertical load fbcv = M/Zxx =

where M = design bending moment where Zxx = moduli of section

2251687700.3205 = 126.760 17763402.777778

Horizontal load fbch = Mh/Zyyt =

where Mh = horizontal bending moment whereZyyt = moduli of section of flange plate

137223223.82479 = 1041666.6666667

Axial stress fac = Fs/Af =

Mpa

15.808

Mpa

where Fs = Axial force at top flange of surge girder where Af = Area of top flange

114352.68652066 = 12500

9.148

Mpa

Stress ratio is fbcv SR = sbcv

fbch + sbch

126.760 + 158.1994

SR 0.959

< 1.1

15.808 158.4

fac + sac +

As per clause 7.1.1 of I S 800-1984 9.148 144

SAFE

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Description Check for tensile stresses due to bending fbt = M/Zxx where M = design bending moment where Zxx = moduli of section 2251687700.3205 17763402.777778 =

126.76

< (0.66 fy)

158.4

Mpa

SAFE CG of loads

1.53m

1.70m

3.55m

1.70m

1.53m

Check for deflection

10.00m

A. Self weight of the girder = Self weight of the rail = Self weight of the walkway = LL on walkway =

CG of girder

3.41 0.89 0.4500 3.000

kN/m kN/m kN/m kN/m

7.747

kN/m

Total = w =

Deflection at center due to above loads d1 = d1 =

5/384 * wl4 / EI 0.45 mm

Deflection of girder due to wheel load is maximum considering moment criteria Deflection at centre due to a concentrated load placed at a distance < span/2

d2 =

S

PL3 48EIxx

3(b/L) - 4(b/L)3

P (Wheel load without impact) = 1 x 48 = =

686228.6

x

210000

x

6.38749964 x

10000

^3

X

10658041666.7 1

reference steel designers manual

686.229 kN

+

3

5.00

1

10

-

4

5.00 10

^3

+

3

3.23 10

-

4

3.23

^3

10

0.83333194

6.39mm

Total deflection is

6.84

Allowable deflection is l/1000 =

mm 10

mm

Hence

SAFE FORM T9-P

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References / Remarks

Description Check for Shear

6.95m

Maximum shear will occur when one wheel is on the support

10.00m

Shear force is maximum when 2&3 wheels are in the span and one of the wheel is on the suport taking moment about B RB = 857.79 x RB = 596.16 RA = 1119.41 Vmax = 1119.41 Shear due to self weight = = 7746.900 x 10000 Total shear =

( 6.95m kN kN kN (wL/2)

)/

=

/2

38.73

10.00m

kN

1158.1 kN

Actual stress due to shear = V/twdw where V = Total shear tw = Web thickness dw = Web depth 1158144.9567308 16 X 1150 =

62.94

fy of web

Mpa 250

Mpa

Permissible shear = 0.4 fy =

100 Mpa >

62.94

SAFE

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Description Intermediate Stiffener where dw=

Web depth

d w/85

=

13.53

dw (tva, cal)1/2 / 816

=

11.18

mm

dw (fy)1/2 / 1344

=

13.26

mm

mm

As per 6.7.3.1 (a) of IS 800 -1984

Hence to avoid intermediate stiffener the web thickness shall be min Provided web thickness

=

16

13.53

mm

mm

Since provided thickness is greater than required thickness intermediate stiffeners are not required. Vertical stiffeners are provided to take up vertical point load at a spacing given by 0.33 d < spacing < 1.5 d 379.5

where d=depth of web

< Spacing <

1725

Consider panel dimension as (s) =

1250

mm

As per Cl 6.7.3.1 (b) of IS 800 - 1984 , Min required thickness of web stiffener is greater of = s/180 dw / 180 dw (fy)1/2 / 3200

= = =

6.94 6.39 5.57

mm mm mm

dw / 200

=

5.75

mm

Hence Min t

=

6.94

mm

Min I required for web of stiffener is

= 1.5 x d3 x t3 / c2

As per 6.7.4.2 of IS 800 -1894

where t = the minimum required thickness of the web c = the maximum permitted clear distance between vertical stiffener for thickness t. now t= c= I min =

6.94 1725

mm mm

256755.11545 mm4

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Consider vertical stiffeners of size b = d = I

=

10 mm 150 mm 2812500

mm4 > I min

SAFE

End Stiffener Max Shear s

=

1158.14 kN

Permissible bearing stress = 0.75 fy =

180 Mpa

As per 6.3 of IS 800 -1984

Min Area of stiffener required 1158144.9567308 180

=

6434.14

Assume the size of stiffener as b = 500 mm t = 20 mm A Stiff = 10000 sqmm

sqmm

> req SAFE

a. Check for Compressive stress for end stiffener Total effective width of web acting for the bearing is 20 times thickness of web As per 6.7.5.3 of IS 800-1984 the effective length of web of the section for compression flange =20 X tw 20 X 16 = 320 mm hence effective area of web = 5120 sqmm 320 X 16 = total effective area = effective area of web + area of stiffener (Astiff) 5120 + 10000 Area = 15120 sqmm Total I of the section I of end plate+I of effective web tb^3/12

+ 20*tw*tw^3/12)

2.1E+08 + I

=

109226.66667

208442560

mm4

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References / Remarks

Description

Radius of gyration

1

I/area

208442560 15120 r min =

117.41

l =0.7d / r min s ac

=

mm =

143.93

Actual stress =

6.86 Mpa

Max Shear (s ) Area

Actual stress =

76.60

=

Mpa <

ref compressive stress calculation sheet 2 sheet no 19

1158144.95673 15120 143.93 Mpa

SAFE

b. Check for bearing A Stiff =

15120

Bearing stress = Bearing stress =

sqmm

(Consider 75% area of contact)

Max Shear (s ) 0.75 X Area

=

102.13 Mpa <

(Permissible bearing stress = 0.75 fy =

As per 6.3 of IS 800-1984

1158144.957 11340 180 180

SAFE )

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Description Connections a. Web to Top flange Provide Full penetration Butt weld b. Web to bottom flange H = Hor Shear Force on bottom flange = V Af y' / 2 IXX H=

1119410 x 12500 x 587.5 21316083333.333

H= 385.66 N / mm Allowable stress as per 7.5.1.2 of IS 816 -1969 = 110 Mpa Thickness of weld required = H / 0.707 * 110 = 385.66 / /0.707*110 = 4.96 mm Provide 8 mm thick weld c. Vertical stiffener Shear Per mm of weld is SL = Pmax / 4(d-dn), where dn = 20mm = 857786 4520

=

189.78 N/mm

Thickness of weld required = SL / 0.707 * 110 = Allowable stress as per 7.5.1.2 of IS 816 -1969 = 110 Mpa = 2.44 mm Provide 6 mm thick weld

189.78 /0.707*110

d. End Stiffener : Shear on end member is SL = Vmax / 2 ( d - dn) = 857785.7906 = 495.31 N/mm 2260 Thickness of weld required = SL / 0.707 * 110 495.31 / 0.707*110 Thickness of weld required = 6.37 mm Provide

8

mm thick weld

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References / Remarks

Description DESIGN OF SURGE GIRDER

=

54.88929 kN

=

1.20m

54.88929 kN

857.79 KN

=

5.0 M

5.0 M

54.88929 kN

C

=

54.88929 kN

=

54.88929 kN

=

54.88929 kN

=

54.88929 kN

=

54.88929 kN

=

54.88929 kN

=

A

54.88929 kN

Horizontal bending moment (one span) Mhs =

B

10.00m

16.467

kNm

(Note : As only one wheel can be accommodated in the span for horizontal surge moment is calculated accordingly) Over all bending moment =

54.88929 kN X 2144.464 kNm 857.79 kN

Axial force in top flange of surge girder =

Maximum force in the diagonal bracing =

137.22 1.2

=

=

137.22

kNm

114.35

54.88929 x 1119.41 = 101.3162 857.79 x sin 45

kN FORM T9-P

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References / Remarks

Description Assume a trial section of ISA 65X65X6(Double angle back to back) for end diagonal The sectional properties are as follows Area Ixx Zxx rxx ryy rmin L=length

= = = = = = =

1488 582000 1240 19.8 30.4 19.8 1733

sqmm mm^4 sqmm mm mm mm mm

effective length (l) = 0.85 X L = 1473.05 mm slenderness ratio = l/rmin = 74.40 now from table 5.1 IS:800-1984 s ac = 104.00 N/sqmm (Permissible direct compressive stress) Section capacity = area X stress= 154.75 kN Maximum force in the diagonal bracing = 101.32 kN < Section capacity provided is safe Assume a trial section of ISA 65X65X6 for intermediate diagonals A B

C

54.88929

kN

Area rxx ryy rmin L=length

= = = = =

744 19.6 19.6 19.6 1733

sqmm mm mm mm mm

D Applying

S

Fx=0

FAB = 101.32 kN

FBC = 101.3162 x cos 45 cos 45

54.88929

FBC = 23.69098 kN effective length (l) = 0.85 X L = 1473.05 mm slenderness ratio = l/rmin = 75.16 now from table 5.1 IS:800-1984 s ac = 106.00 N/sqmm (Permissible direct compressive stress) Section capacity = area X stress= 78.86 kN Maximum force in the diagonal bracing = 23.69 kN < Section capacity provided is safe Same section is provided through out the surge girder FORM T9-P ORIGINATOR/DEPT.

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References / Remarks

SHEET 1 permissible bending stress calculation for 10 m span The calculation is as per IS 800 cluase 6.2.4 Flange width = 500 mm Flange thickness = 25 mm Web thickness = 16 mm Web depth = 1150 mm ly = ly = slenderness ratio from "main calculation" 10.95 where Y = 22101.39 Mpa where Y= (26.5 X 10E5)/(λy^2) under root 1.002602 under root =

where

1

1 + 20

lT 2 ry D

1= D=

effective length of compression flange overall depth of beam; ry = radius of gyration of the section about its axis of minimum strength (y-y axis ); T = mean thickness of the compression flange, is equal to the area of horizontal portion of flange divided by width; X= Y X under root X= 22130.13 Mpa elastic critical stress fcb = k1(X+k2Y)(c1/c2) where k1= a coefficient to allow for reduction in thickness or breadth of flanges between points of effective lateral restraint and depends k1 = 1 on #, the ratio of the total area of both flanges at the point of least bending moment to the corresponding area at the point of greatest bending moment between such points of restraint. Values of kl for different values of # are given in Table 6.3. k2 = a coefficient to allow for the inequality of flanges, and depends on W, the ratio of the moment of inertia of the compression k2 = 0 flange alone to that of the sum of the moments of inertia of the flanges, each calculated about its own axis parallel to the y-y axis of the girder, at the point of maximum bending moment. Values of k, for different values of o are given in Table 6.4. C1 C2 =1,1 C1 , C2 = respectively the lesser and greater distances from the section neutral axis to the extreme fibers. fcb = 22130.13 Mpa Now as per 6.2.3 of IS 800 fcb X fy sbc = 1 ((fcb)^n + (fy)^n)^(1/n)) where

fy =

sbc =

f cb = elastic critical stress in bending, fy = yield stress of the steel in MPa; n= a factor assumed as 1.4 240.00 Mpa 1210535.32 2149.29 22158.19 158.20 Mpa FORM T9-P REV-A

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References / Remarks

SHEET 2 compressive stress calculation for 10 m span The calculation of compressive stress is calculated according to 5.1 of IS 800-1984 fy = 240 Mpa foe= ehtic critical stress in compression,

π^2 E λ^2 E = modulus of elasticity of steek 2 x 10^6 MPa; λ ( = l/r) = slenderness ratio of the member

λ= 6.86 (slenderness ratio) hence fce = 44127.71 n-a&factor assumed as 1.4. n = 1.4 fy ^n = 2149 fce^ n = 3181237 σac =

0.6

(fce *fy)/((fcc^n+fyy^n)^(1/n))

3183386.05147 44149.0065062 stress =(σac)

=

143.93

Mpa

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References / Remarks

Description DESIGN OF SUPPORTING MEMBER FOR WALK WAY FOR Width of walk way 1.2 Dead load of walk way 0.75 live load of walk way 5.00 Maximum distance between supports = 1.25 Total load per sqm 5.75 KN/Sqm Load intensity per m run = 7.1875 KN/m Bending moment (M)= wl^2/8 1.29 KNm compression flange is literally unsupported Let us adapt ISMC 150 with fallowing properties Designation Weight per m Sectional area Moduli of section ryy d1 = ( 200 -

ISMC 200 22.1 2821 18190 22.3 2x 11.40 )

kg/m sqmm sqmm mm = 177.2

10.0 m SPAN

m KN/Sqm KN/Sqm m

Thickness of web(t) = Thickness of flange(T) = Depth of section (D)=

6.10 mm 11.40 mm 200 mm

mm

D/T = ( 200 / 11.40 ) = 17.54 T/t = ( 11.40 / 6.10 ) = 1.869 < 2 d1/t = ( 177.2 / 6.10 ) = 29 < 85 l/ryy = ( 1020 / 22.3 ) = 45.74 As per tabel no 6.1 B of IS 800-1984 permissible bending stresses σbc = = 157 N/sqmm Stress due to bending = ( 1293750 / 18190 ) = 71.12424 Mpa Stress due to direct compression compressive fource = 54.89 kN effective length (l) = 0.85 X L = 1020 mm slenderness ratio = l/rmin = 45.73991031 ( 1020 / 22.3 ) now from table 5.1 IS:800-1984 s ac = 136.00 N/sqmm (Permissible direct compressive stress) actuval compressive stress = load/area pro s ac cal = ( 54889.28953 / 2821 ) = 19.45739 s ac cal 19.45739 = = 0.143 < 0.15 s ac 136.00 Hence Combined Axial ,Compression and Bending s ac cal s bc cal + < 1.00 s ac s bc 19.45739 71.12424 + = 0.596 < 1.00 136.00 157.00 Hence provided is

stress

safe

FORM T9-P REV-A

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Span of girder l = No of wheels on 1 side n= Max wheel load Pmax = Wheel spacing (B'n 1&2) (B'n 2&3) (B'n 3&4)

10000 2 857.78579 0 6950 0

mm KN mm mm mm

Case 1. CG of loads for 1 wheel occurs at center M max = wl/4 = 2144.4644764957 KNm Maximum bending moment occurs when 1 wheel is at center of span Case 2. CG of loads for 2 wheels

(B'n 1&2) (B'n 2&3) (B'n 3&4)

0 mm 3475 mm 0 mm

distance between cg of span and first load Rb= 857.7858 Ra= 857.7858 M2= M1=

0

1308.123 KNm 1308.123 KNm

Mmax= 1308.123 KNm Maximum bending moment occurs when first two wheels are in the span distance between cg of span and cg of wheels Rb= 1155.866 Ra= 559.7052 M2= M1=

1737.5

-245.622 KNm 1826.038 KNm

Mmax= 1826.038 KNm Maximum bending moment occurs when second and third wheels are in the span Max of 2= 1826.038 Case 3 When 3 wheels are acting the cg of loads from first wheel occurs at Ra= Rb=

988.5981 KN 1584.759 KN

M1= M2= M3=

3797.864 KNm 3797.864 KNm -1254.6 KNm

Mmax=

3797.864 KNm

988.59812

Maximum bending moment occurs when first 3 wheels are in the span

2316.667 mm Pmax l a b c x l/2 x-a ('x-a)/2 b-(x-a)/2+c

857.7858 10000 0 6950 0 2316.667 5000 2316.667 1158.333 5791.667

Case 4.

3475

CG of loads for 4 wheels from first wheel Ra= Rb=

2311.733 KN 1119.41 KN

M1= M2= M3= M4=

-491.243 -491.243 3652.077 3652.077

Mmax=

3652.077 KNm

3475

KNm KNm KNm KNm

Maximum bending moment occurs when all four wheel are in span & moment occurs near second wheel Ra= Rb=

1119.41 KN 2311.733 KN

M1= M2= M3= M4=

3652.077 3652.077 -491.243 -491.243

Mmax=

3652.077 KNm

KNm KNm KNm KNm

Maximum bending moment occurs when all three / four wheel are in span & moment occurs near first wheel

2144.464 1308.123 1826.038 3797.864 3652.077 3652.077

Maximum bending moment occurs when 1 wheel is at center of span Maximum bending moment occurs when first two wheels are in the span Maximum bending moment occurs when second and third wheels are in the span Maximum bending moment occurs when first 3 wheels are in the span Maximum bending moment occurs when all four wheel are in span & moment occurs near second wheel Maximum bending moment occurs when all four wheel are in span & moment occurs near first wheel

Maximum bending moment occurs when 1 wheel is at center of span Some error Some error Some error Some error Some error

Pmax l a b c x l/2 a+b-x a+b-x/2 x-a x/2

857.7858 10000 0 6950 0 3475 5000 3475 1737.5 3475 1737.5

l= Wheel span B'n 1&2 B'n 2&3 B'n 3&4 Pmax

10000 mm 0 6950 0 857.7858

mm mm mm KN

case 1. 2 wheels are in 1 span Shear force when 2 wheels are in 1 span ( 1&2) V= 857.7858 KN Shear force when 2 wheels are in 1 span ( 2&3) V= 1119.41 KN Shear force when 2 wheels are in 1 span (3&4) V= 857.7858 KN Vmax=

1119.41 KN

case 2. 3 wheels are in 1 span Shear force is max when first 3 wheels in one span V= 1977.196 KN Shear force is max when last 3 wheels are in one span V= 1381.035 KN Vmax=

1977.196 KN

case 2. 4 wheels are in 1 span Shear force is max when all wheels are in 1 span V= 2238.821 KN Vmax= 2238.821 KN max shear=

857.7857906 1119.410457 857.7857906 1977.196247 1381.035123 2238.820913

1119.41

Shear force is maximum when first 2 wheels are in the span and one of the wheel is on the suport Shear force is maximum when 2&3 wheels are in the span and one of the wheel is on the suport Shear force is maximum when 3&4 wheels are in the span and one of the wheel is on the suport Shear force is maximum when first 3 wheels are in the span and one of the wheel is on the suport Shear force is maximum when Last 3 wheels are in the span and one of the wheel is on the suport Shear force is max when all wheels are in 1 span and that wheel is on the suport

Shear force is maximum when 2&3 wheels are in the span and one of the wheel is on the suport Shear force is maximum when 2&3 wheels are in the span and one of the wheel is on the suport Some error Some error Some error Some error

100