bridge

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1. PROBLEM STATEMENT AND ASSUMPTIONS:

A single span noncomposite I-girder bridge has span length of 10.668m and a 14.184 deck

width. The steel girders have Fy= 350 Mpa and all concrete has a 28-day compressive

strength of f’c= 28 The concrete slab is 203 mm thick. A typical 2” haunch was used in the

section properties. γc = 24 KN/m3 , γwearing =22.5 KN/m

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Concrete barrier 380 mm. wide and weighs 9.34 KN/m and Allow for a future wearing surface of 75mm. thick bituminous overlay.

Consider the outline of AASHTO (2005) LRFD Bridge Specifications, Section 6.

Solution

A. Develop Typical Section

1. I-Girder

a. Composite or Noncomposite Section [A6.10.1.1] This bridge

is noncomposite, does not have shear connectors, and the shear

strength should follow [A.6.10.10].

b. Nonhybrid [A6.10.1.3] This cross section is a rolled beam and

the same material properties are used throughout the cross

section. The section is nonhybrid.

c. Variable Web Depth [A6.10.1.4] The section depth is prismatic

and variable-depth provisions are not applicable.

d. Section Properties:

Girder spacing, S = (2438 mm)

Span length, L = (10,668 mm)

Deck thickness, ts = (203 mm)

Deck modulus of elasticity, Ec = (24,827.6 MPa)

Girder modulus of elasticity, Es = (200,000 MPa)

Modular ratio, n = Es /Ec =200,000 /24,827.6 = 8.05; use 8

Girder area, Ag = (20452 mm2)

Girder moment of inertia, Ig = 4470 in.4 (1860.55×10

6 mm

4)

2

Girder eccentricity, eg= 0 (for noncomposite section) .

Stiffness parameter, Kg = n(Ig + eg2 Ag ) = 8×(1860.55×10

6) = 14884.4×10

6 mm

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B. Design Conventionally Reinforced Concrete Deck : The deck will be designed later on

C. Select Resistance Factor

. Strength Limit State φ [A6.5.4.2]

Flexure 1.00

Shear 1.00

D. Select Load Modifiers η. Strength Service Fatigue

1. Ductility, ηD [A1.3.3] 1.0 1.0 1.0

2. Redundancy, ηR [A1.3.4] 1.0 1.0 1.0

3. Importance, ηI [A1.3.5] 1.0 N/A N/A

η = ηDηRηI [A1.3.2.1] 1.0 1.0 1.0

E. Select Load Combination and Load Factors

1. Strength I Limit State: U = η[1.25DC + 1.50DW + 1.75(LL + IM) + 1.0FR + γTGTG]

2. Service I Limit State: U = η[1.0(DC + DW ) + 1.0(LL + IM) + 0.3(WS + WL) + 1.0FR]

3. Service II Limit State: U = η[1.0(DC + DW ) + 1.3(LL + IM)]

4. Fatigue and Fracture Limit State: U = η[0.75(LL + IM)]

F. Calculate Distribution Factors: .

1. Distribution Factor for Moment [A4.6.2.2.2]

a. Interior Beams [A4.6.2.2.2b] (Table 6.5)

One design lane loaded:

𝑚𝑔𝑀𝑆𝐼 = 0.06 +

𝑆

4300

0.4

𝑆

𝐿

0.3

𝑘𝑔

𝐿𝑡𝑠3

0.1

𝑚𝑔𝑀𝑆𝐼 = 0.06 +

2438

4300

0.4

2438

10668

0.3

14884.4 × 106

10668 × 203 3

0.1

= 0.4878

Two or more design lanes loaded:

𝑚𝑔𝑀𝑀𝐼 = 0.075 +

𝑆

2900

0.6

𝑆

𝐿

0.2

𝑘𝑔

𝐿𝑡𝑠3

0.1

𝑚𝑔𝑀𝑀𝐼 = 0.075 +

2438

2900

0.6

2438

10668

0.2

14884.4 × 106

10668 × 203 3

0.1

= 0.6357

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b. Exterior Beams [A4.6.2.2.2d] (Table 6.5) [Table A4.62.2.2d-1]

One design lane loaded—lever rule:

𝑅 =𝑃

2

638+2438

2438 = 0.631𝑃

𝑚𝑔𝑀𝑆𝐸 = 1.2 × 0.631 = .757

Two or more design lanes loaded:

𝑚𝑔𝑀𝑀𝐸 = 𝑒 × 𝑚𝑔𝑀

𝑀𝐼

𝑒 = 0.77 +𝑑𝑒

2800 ≥ 1

e=0.77+610/2800=0.987 ≤ 1 use 1

𝑚𝑔𝑀𝑀𝐸 = 0.6357

2. Distribution Factor for Shear [A4.6.2.2.3]:

a. Interior Beams [A4.6.2.2.2a]

One design lane loaded

𝑚𝑔𝑉𝑆𝐼 = 0.36 +

𝑆

7800= 0.36 +

2438

7800= 0.68

Two or more design lanes loaded:

𝑚𝑔𝑉𝑀𝐼 = 0.2 +

𝑆

3600−

𝑆

10700

0.2

= 0.36 +2438

7800−

2438

10700

0.2

= 0.83

b. Exterior Beams [A4.6.2.2.2b]

One design lane loaded—lever rule

𝑚𝑔𝑉𝑆𝐸 = 1.2 × 0.631 = .757

Two or more design lanes loaded

𝑚𝑔𝑉𝑀𝐸 = 𝑒 × 𝑚𝑔𝑉

𝑀𝐼

𝑒 = 0.6 +𝑑𝑒

3000= 0.8

𝑚𝑔𝑉𝑀𝐸 = 0.8 × 0.83 = 0.667

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G. Calculate Live-Load Force Effects:

1. Select Live Loads [A3.6.1] and Number of Lanes [A3.6.1.1.1] Select

Number of Lanes [A3.6.1.1.1]:

NL = INT(w/3.6) = INT(13.424/3.6) =3.73 ≈ 3 lanes

2. Multiple Presence [A3.6.1.1.2] (Table 4.6)

No. of Loaded Lanes M

1 1.20

2 1.00

3 0.85

3. Dynamic Load Allowance [A3.6.2] (Table 4.7)

Component IM (%)

Deck joints 75

Fatigue 15

All other 33

Not applied to the design lane load

4. Max Moment :

The following design vehicular live load cases described in AASHTO-LRFD are

considered:

MLL+IM =mg ((MTruck or MTandem)(1+IM)+ MLane )

MTruck = 145(2.667)+(145 + 35)(0.5334)=482.73 KN.m

MTandem = 110(2.667+2.05) = 520.74 KN.m (governs)

MFatigue = 145(2.667)+ 35(0.5334) = 405.4 KN.m (used later)

MLane =(9.37*10.6682)/8 =133.29 KN.m

1. The effect of a design tandem combined with the effect of the lane loading. The

design tandem consists of two 110 KN axles spaced 1.2 m apart. The lane loading

consists of a 9.37 KN/m uniform load on all spans of the bridge. (HL-93M)

2. The effect of one design truck with variable axle spacing combined with the effect of

the 9.37 KN/m lane loading. (HL-93K)

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The absolute moment due to the tandem actually occurs under the wheel closest to the resultant

when the cg of the wheels on the span and the critical wheel are equidistant from the centreline of

the span. For this span, the absolute maximum moment is 526.15 KN.m. However, the value of

520.74 KN.m is used because the moments due to other loads are maximum at the centreline and

thus can be added to the tandem load moment:

MLL+IM =0.757*(1.33*520.74+133.29) =625 KN.m

Mfatigue+IM = (0.757/1.2)[ 405.4 (1.15)]= 294.1 KN.m

Truck, tandem, and lane load placement for maximum moment at

location 105.

Fatigue truck placement for maximum moment.

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5. Max Shear :

VLL+IM = mg((VTruck or VTandem)*(1 + IM/100)+ VLane

VTruck = 238.335 KN governs

VTandem = 207.62 KN

VLane = 50 KN

VFatigue = 165.9 KN (used later)

VLL+IM = 0.83[238.335 (1.33) + 50] = 367 KN

VFatigue+IM = (0.757/1.2)[ 165.9 (1.15)] = 120.35 KN (used later)

H. Calculate Force Effects from Dead Loads: Analysis for a uniformly distributed load w

Mmax = M105 = wL2/8 =14.225w KN.m

Vmax = V100 = wL/2=5.334w KN

Truck, tandem, and lane load placement for maximum shear at

location 100

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component W (KN/m/Girder) Mmax

(KN.m/Girder)

Vmax

(KN/Girder)

Steel Girder(DC1) 1.567*1.15 25.79 9.67

Deck slab (DC1) . 203 × 24 × 14.184

6= 11.517 163.838 61.43

Barriers(DC2) 2 × 9.34

6= 3.11 44.307 16.61

Wearing

surface(Dw)

. 075 × 22.5 × 13.424

6= 3.775

53.702 20.135

LL+IM N/A 625 304.6

Fatigue + IM N/A 294.1 120.35

I. Factored Loads:

1. Strength I Limit State: U = η[1.25DC + 1.50DW + 1.75(LL + IM) ]

Mu=1[1.25(25.79+163.838+44.307)+1.5(53.702)+1.75(625)]=1466.72KN.m governs

Vu=1[1.25(9.67+61.43+16.61)+1.5(20.135)+1.75(304.6)] = 672.9 KN governs

2. Service II Limit State: U = η[1.0(DC + DW ) + 1.3(LL + IM)]

Mu =1[1.0(25.79+163.838+44.307+53.702) + 1.3(625)] =1131.387 KN.m

Vu=1[1.0(9.67+61.43+16.61+20.135)+ 1.3(304.6)] =503.825 KN

3. Fatigue and Fracture Limit State: U = η[0.75(LL + IM)]

Mu =1[0.75*(294.1)= 220.5 KN.m

Vu =1[0.75*(120.35) = 90.3 KN

Unfactored Moment and Shear

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Check design section W30x108 : from AISC manual:

A= 31.7 in2 I=4470 in

4 bf =10.475 in

,tw =0,545 in tf = 0.76 in D = 29.83 in

Z = 346 in3 Mp = 1440 ksi S =299 in

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Compactness :

All rolled sections in AISC have compact webs for Fy ≤ 50ksi

Based on this the section is compact

Flexural :

φf Mn ≥ Mu , φf =1.0 Mn = Mp = ZFy

Mp = 1953 KN.m > Mu = 1466.72KN.m OK. For Flexural.

Cross-Section Proportion Limits [A6.10.2] :

𝑑

tw ≤ 150 ,

29.83 − 2 × 0.76

0,545= 51.9 𝑂𝐾

𝑏𝑓

2𝑡𝑓≤ 12 ,

10.475

2 × 0.76= 6.9 𝑂𝐾

𝑏𝑓 ≥𝐷

6 , 10.475 ≥

29.83

6= 4.9 𝑂𝐾

Shear :

Vu ≤ φVcr = φCVp = φC(0.58)FyDtw

Vp =0.58*50*29.83*0.545 =471.5 kips =2100 KN

Vu =672.9 KN < 1*1*2100 OK

Fatigue [A6.10.5.1] [A6.6.1] :

Allowable fatigue stress range depends on load cycles and connection details.

(a) Stress Cycles

Assuming a rural interstate highway with 20,000 vehicles per lane per day, Fraction of

trucks in traffic = 0.20 (Table 4.4) [Table C3.6.1.4.2-1]

ADTT = 0.20 × ADT = 0.20(20 000)(2 lanes) = 8000 trucks/day

p = 0.85 (Table 4.3) [Table A3.6.1.4.2-1]

ADTTSL = p × ADTT = 0.85(8000) = 6800 trucks/day

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From Table 8.4 [Table A6.6.1.2.5-2], cycles per truck passage, for a simple-span girder of

span 35 ft, is equal to n = 2.0

N = (365 days/year)(75 years)(2.0 cycles/pass) (6800 trucks/day)

= 372×106 cycles

(b) Allowable Fatigue Stress Range—Category A

∆𝑓𝑛 = 𝐴

𝑁

13

= 250 × 108

372 × 106

13

= 4.1 𝑘𝑠𝑖

0.5(∆f)Th =0.5*24 = 12 ksi < 4.1ksi

There for (∆f)n =12 ksi

(c) The Maximum Stress Range [C6.6.1.2.5]

The maximum stress range is assumed to be twice the live-load stress range due to the

passage of the fatigue load. However, the stress range need not be multiplied by 2

because the fatigue resistance is divided by 2.

For fatigue, U = 0.75(LL + IM).

Dynamic load allowance for fatigue is IM = 15%.

MLL+IM is maximum in the exterior girder, no multiple presence (live-load range only):

Mfatigue = 0.75 (294.1)= 220.5 KN.m = 168.8 K.ft

𝑓 =𝑀

𝑆=

168.8 × 12

229= 6.77 𝑘𝑠𝑖 < 12 𝑘𝑠𝑖 𝑂𝐾