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Integral Abutment Design

1. Pile Cap (7.1.2)a. Stage I (noncomposite)

PSI = 1.25 x (girder + slab + haunch)a.1 Interior Girder

Total Noncomposite = 0 k

PSI(I) = 0 k

a.2 Exterior Girder

Total Noncomposite = 0 k

PSI(E) = 0 k

b. Final Stage (composite)PFNL = 1.25(DC) + 1.50 (DW) + 1.75(LL + IM) (N lanes)/Ngirdersb.1 Interior Girder

Total Noncomposite = 0 k

Composite (Parapet) = 0 k

Composite (FWS) = 0 k

LL+IM = 0 k

Nlanes = 0

Ngirders = 0

PFNL(I) = 0 k

b.2 Exterior Girder

Total Noncomposite = 0 kComposite (Parapet) = 0 k

Composite (FWS) = 0 k

LL+IM = 0 k

PFNL(E) = 0 k

2. Pile (7.1.3)a. Case A Capacity of the pile(assume rock, only Case A needs to be investigated.)

Try Pile HP

Fy = 0 ksiAs = 0 in

2

Pn = FyAs = 0 k

Pr=fPn = 0 k

b. No. of Piles Required

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PSI (Total) = 0 k

PFNL(Total) = 0 k

PStr. I = PFNL(Total) + 1.25(DC) + 1.50(DW) + 1.75(LLmax)(Nlanes)

DC = 0 k

DW = 0 kLLmax = 0 k

PStr. I = 0 k

Npiles = PStr. I/Pr= #DIV/0!

c. Final DesignNo. of Pile = 0

Pile Size = HP 0

Note: Red ink cells for input & Blue ink cells for formulated calculation

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Integral Abutment Design

3. Backwall (7.1.4) (80% of simple beam moment)a. Case A -

Pu = 1.5 x (girder + slab)wu = 1.5 x (pile cap + diaphragm)

Dist. between piles, l= 0 ftTotal Noncomposite = 0 k

Pu = 0 k

Pile cap & Diaph = 0 k/ft

wu = 0 k/ft

Mu = 0.8[Pul/4 + wul2/8] = 0 k-ft

As = 0 k-ft Use bars=

Fy = 0 ksi

ds = 0 in

fc' = 0 ksi

b = 0 in

a = Asfy/0.85fc'b = 0 in

Mn = Asfy(ds a/2) = 0 k-ft

Mr=fMn = 0 k-ft > 0 k-ft

b. Case B -P

Str-I= factored girder reaction

wStr-I = 1.25(pile cap + end diaph. + approach slab) + 1.50 (approach FWS)

PStr-I = 0 k

DC = 0 k/ft

Approach FWS = 0 k/ft

Appr Slab lane load = 0 k/ft

wStr-I = 0 k/ft

Mu = 0.8[Pul/4 + wul2/8] = 0 k-ft

As = 0 k-ft Use bars=

Fy = 0 ksids = 0 in

fc' = 0 ksi

b = 0 in

a = Asfy/0.85fc'b = 0 in

Mn = Asfy(ds a/2) = 0 k-ft

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Mr=fMn = 0 k-ft > 0 k-ft

(Note: Shear design is skipped in this exam, but should be examined in actual design.)

c. Final Flexural Design (Vert.)Rebar number & size = 0 # 0

Note: Red ink cells for input & Blue ink cells for formulated calculation

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+ 1.75(approach slab lane load) (N lanes)/Ngirders

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. All yellow marked cells have to be filled.

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Integral Abutment Design

4. Wingwall (7.1.5)a. Passive pressure (kp=3) -

wu at bottom of slab=0.2k/ft2; at bottom of wall=3.24 k/ft2

Mp = (Rect. Area x base length) + (Pyramid Area x base length)

Rect. Area = 0 ft2

base length = 0 ft

Pyramid Area = 0 ft2

base length = 0 ft

Mp = 0 k-ft

Min. required Mn =Mr/f= 0 k-ft

b. Active pressure (ka = 0.333) Ma = (ka/kp)*Mp +Mcollision

ka/kp = 0 k-ft

Mcollision = 0 k-ft

Ma = 0 k-ft

Min. required Mn =Mr/f= 0 k-ft

Mn required = k-ft

c. Flexural Design

As = 0 k-ft Use bars=

Fy = 0 ksi

ds = 0 in

fc' = 0 ksi

b = 0 in

a = Asfy/0.85fc'b = 0 in

Mn = Asfy(ds a/2) = 0 k-ft

Mr=fMn = 0 k-ft > 0 k-ft

(Note: Shear design is skipped in this exam, but should be examined in actual design.)

d. Final Flexural DesignRebar number & size = 0 # 0

Note: Red ink cells for input & Blue ink cells for formulated calculation

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Integral Abutment Design

5. Approach Slab (7.1.6)a. Single lane loaded

E = 10 + 5 (L1W1)L1 = 0

W1 = 0

Esingle = 0

b. Multiple lane loaded

E = 84 + 1.44 (L1W) 12W/NLL1 = 0

W1 = 0

NL = 0

Emult. = 0 in. in.

E = 0 in.

c. Max. Factored Positive Moment (per Slab Unit Width)Lane Load Max M = 0 k-ft

Truck Load Max M = 0 k-ft

Total LL+IM = 0 k-ft

(Total LL+IM)/(width E) = 0 k-ft

w = 0 k/ft

l = 0 ft

Mu = wl2/8 + 1.75 (LL+IM Moment)

Mu = 0 k-ft/ft

d. Flexural Design (per Slab Unit Width)

As = 0 k-ft Use bars=

Fy = 0 ksids = 0 in

fc' = 0 ksi

b = 0 in

a = Asfy/0.85fc'b = 0 in

Mn = Asfy(ds a/2) = 0 k-ft

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Mr=fMn = 0 k-ft/ft > 0 k-ft/ft

(Note: Shear design is skipped in this exam, but should be examined in actual design.)

e. Final Flexural Design (per Slab Unit Width)Rebar number & size = 0 # 0

Bottom distribution rebar = 0 # 0

Note: Red ink cells for input & Blue ink cells for formulated calculation

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