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7/27/2019 Copy of Ce717abutment_calc
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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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. All yellow marked cells have to be filled.
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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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. All yellow marked cells have to be filled.
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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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. All yellow marked cells have to be filled.
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