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2 14 2

42,6

1,4

Cross-Section

1) Draw a general layout

(1)

Example:The shown figure illustrated

to :-

2) Calculate dead and live loads forfor each element

3) Design the stringer4) Design the cross-girder5) Design the main girder6) Calculate wind loads7) Ccalculate the internal forces in the

bracing system

Given steel used is St.52

8) Design the bearing

Solution1) General layout of Deck Bridge

a simple span deck bridge. it is required

Lectures Notes (Steel III- CT225) by Dr. Essam Abdelaty Mohamed Amoush

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14

Floor system

2

Sec.Elevation

6

6

6

6

222222222

1,4

4

36

(2)

6

6


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41

Upper bracing Lower Bracing

14

36

14

(3)


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Cross-Section

2

2) Loads

2-1) loads on stringer

Dead loads

WDL= o.w.t of stringer + o.w.t of R.c slab+ o.w.t

of asphalt

= 0.05+0.1x2.5x2+0.05x2.2x2=0.77 t/m

2

6

WdL=0.77t/m

(4)


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For New Loads ECP2008

60

40

20

15

15

15

15

10

10

10

10

5 5

5 5

0.5

2

1.2

0.9t/m2

0.25

t/m2

0.9t/m2

0.25t/m2

0.25t/m2

0.2

5t/m2

0.25t/m2

0.25t/m2

If sidewalk width

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Transverse direction2 14 2

3 3 3

15t 10t 5 t 5t15t 10t

0.25 t/m 0.5 t/m0.5 t/m

0.25 t/m0.5 t/m0.5 t/m 0.9 t/m

Path I-I

Path II-II

A

A

I= 0.4-0.008x6=0.352

15t10t

15t

A0.50

2.00

0.50 0.50

2.00

A0.50

2.00

1.0 1.0

2.00

0.9 t/m 0.2 5 t/m

15t10t

15t

0.25

1.0 0.75 0.25

15t10t

15t

0.25

1.0 0.75 0.25

Ra=15x0.5/2+15x1.5/2)x1.352

10x0.5/2=22.78 t

Path I-I

Path II-II

Wa=(0.9x2x12+0.9x1x1.5/2)x1.352

+0.25x1.0x0.5/2=2.192 t/m

(6)


0.9 t/m

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Longitudnal direction

22.78 t 22.78 t2.192t/m2.192 t/m

0.9 4.2

0.9

B.M.D

55.44

21.27

21.27

55.44

22.78 t 22.78 t2.192 t/m

6.00

ML+I=54.34 t.m

a b

M@b=0.0

Ra=(22.78x6+22.78x4.8+

2.192x3.3x1.65)/6=42.99 t

QL+I=42.99 t.m

For maximum moment

For maximum shear

(7)


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Design of StringerProcedure to design the stringer:-

1) Determine the straining actions MD+L+I,QD+L+I2) Assume section of stringer {I.P.E, S.I.B, HEA, HEB}

3) Check bending stress. Pages (16-21) code ECP20014) Check shear stress. Pages (13-15)

5) Check fatigue stress Pages (35-50)6) Check deflection. Page (132)

GivenMD=0.77x6^2/8=3.465 t.m ML+I=54.34 t.m

QD=0.77x6/2=2.31 t QD+L+I= 42.74 t

assume: the depth of section= span/10=600/10= 60 cmchoose HEA 600

h=590 mm bf=300 mm tf=25 mm d w=486 mm tw=13 mmr=27 mm Ix=141208 cm^4 Sx=4787 cm^3

i) classification of sectionbf/2tf= 300/(2x25)=6 < 16.9/v3.6=8.9 compact flange

dw/tw=486/13=37.38 < 127/v3.6=66.93 compact web

section is compact section

ii) check the laterally supported

Lc = smaller from {20bf/vFyor 1380 Af. cb/(dFy)}

= smallre from {316.23 cm or 487.3 cm)= 316.23 cm

Lu= larger from {1380 Af. cb/(dFy) or 84rTv(cb/Fy)}

= Larger from { 487.3 or 356.83}= 487.3 cm

Luns = 600 cm > LuFltb1=800Af. cb/(Luns.d)=

800x30x2.5x1/(600x59)=1.67 t/cm^2

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Fltb2= [0.64-(Luns/rT

)^2 Fy/117600. cb] Fy = [0.64-(74.44^2)x 3.6/117600 x 1)x 3.6=1.69 t/cm^2

= 2.08 t/cm^2

Fltb=v(Fltb1^2+Fltb2^2)=v(1.69^2+1.69^2)=2.39 t/cm^2

= 2.08 t/cm^2

Check bending

fact= MD+L+I/Sx=57.8x 100/4787=1.21 t/cm^2 = 2.08 t/cm^2

O.K

Check shearFor unstiffened web

dw/tw=486/13=37.38 < 105/v3.6=55.34

qb=0.35Fy=1.26 t/cm^2

q act= 45.05 /(48.6x1.3)=0.713 t/cm^2 < 1.26 t/cm^2 O.KCheck Fatigue

Detail Category is Detail A, N=2000,000

Fsr=1.68 t/cm^2Fmax= (M D+0.6ML+I)/Sx= 36.07 x 100/ 4787=0.753 t/cm^2

Fmin = MD/Sx= 3.465x100/4787=0.072 t/cm^2

Fmax-Fmin= 0.753-0.072=0.68 t/cm^2 < Fsr = 1.68 t/cm^2O.K

Check Deflection

dmax = dall= span/600=600/600= 1.0 cm

dmax=d1+d2

d1=(PL^3)[3a/4L-(a/L)^3]/6EI = 22.78x600^3x

[3x90/4x600-(90/600)^3] /(6x2100x141208)= 0.3 cm

d2= 2wa(3L^2-2a^2)/96EI

= (2x2.192/100)x90x(3x600^2-2x90^2)

/96x2100x141208=0.000147

dmax=0.3+0.000127=0.300147 cm

< span/600=1.0 cm

22.78 t 22.78 t

0.9 4.2 0.9

2.192 t/m2.192 t/m

0.9 4.2 0.9

(9)


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Cross Girder1- Dead Loads

M@mid-span

M+ve=17.07x7-2.31x9-0.1x9x4.5-4.62x7-4.62x5-4.62x3

-4.62x1=20.73 t.m

Max Shear

QD=17.07-2.31-0.1x2=14.56 t

(10)


A

4.624.624.624.624.624.62 2.312.31

0.1 t/m

17.07 t 17.07 t

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2-Live LoadsTransverse Direction

I=0.4-0.008X 2X 6=0.304RI=2X15X0.9X(1+0.304)=35.21 t

RII=2X10X0.9 = 18 tRIII=2X5X0.9 = 9 tWI= 2X0.9X(0.5X3.9X0.65)X (1+0.304)=2.98 t/m

WII=WIII= 2X0.25X (0.5 X3.9X0.65) = 0.634 t/m

WIV=2X0.25X(0.5X6X1)=1.5 t/mWV=2X0.5X(0.5X6X1)=3 t/m

(11)


14

2

6

6

6

60 40 20

15 15

15

10

10

10

10

5

5

5

5

0.5 2

1.2

0.9 t/m2 0.25 t/m2

0.9 t/m2

0.25 t/m2

0.25 t/m20.25 t/m2

0.25 t/m20.25 t/m2

Ifsidewalkwidth

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Longitudinal Direction

R=35.2 x 2+18x2+2x9=124.4 t

X=(35.2x2+18x3+18x5+9x6+9x8)/124.4=2.736 m

For max.+ve Moment

RA=1.5x[0.5x(1+0.705)x4.13]+2.98x[0.5x(0.705+0.491)x3]

+ 0.634x[0.5x(0.491+0.062)x6]+1.5x(0.5x0.87x0.062) +

+ 35.2x0.669+35.2x0.526 +18x0.455+18x0.312

+ 9x0.276+9x0.098=70.96 t

M@ zero shearM+ve=70.96x6.63-1.5x4.13x(4.13/2+2.5)-2.98x2.5x1.25 -35.2x2=362.47 t.m

For M-ve

M-ve = (3x2^2)/2 = 6 m.t

For max shear Qmax

Qmax=RA= 3x{0.5x(1.143+1)x2}+2.98x{0.5x(1+0.786)x3}

+0.634x {0.5x(0.786+0.357)x6}+1.5x(0.5x5x0.357)

+ 35.2x0.964+35.2x0.821+18x0.75+18x0.607 + 9x0.536 + 9x0.393=113.544 t

(12)


35.2 35.2 18 9 918

2,74

35.2 35.2 18 9 918

124.4 t

2,74

5,63

77

6,63

0,37

124.4 t

2 2

4,13

0,87

1.50 t/m 2.98 t/m 0.634 t/m 1.5 t/m

10.705

0.669

0.526

0.491

0.455

0.312

0.276

0.098

0.062

I.L.RA

A

35.2 35.2 18 9 918

5

3 t/m 2.98 t/m0.634 t/m 1.5 t/m

10.96

4

0.8

21

0.786

0

.750.607

0.5360.393

0.357

1.143

I.L.RA

A 5,5

8,5

10,5

1111,5

13,5

142

7,5

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1499 Bridge NReq