Project 2

Qu 1)Calculate the loads to be applied on the abutment as per IRC 6

Ans -

1) Calculate the loads to be applied on the abutment as per IRC 6

1) Calculation of load due to DL of Superstructure

i)SW of slab = 15*12*1*2.5=450 t

Total Reaction of deck slab on abutment = 450/2=225 t

Longitudinal eccentricity from A of pile cap=4.3-0.2-0.02-0.39=3.69 m

ML =450*3.69=1660.5 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

ii) Back Wall

assume, back wall thickness=0.3m

back wall height = 1.0m

back wall length = 9.0m

SW of back wall = 9*1*0.2*2.5=4.5 t

Total Reaction of deck slab on abutment = 4.5 t

Longitudinal eccentricity from A of pile cap=4.3-0.1=4.20 m

ML =4.50*4.2=18.90 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

2) Calculation of Load due to SIDL (CB,footpath)

i) CB width = 0.5m each side

intensity of SIDL (CB)=1.5 t/m

CB reaction on abutment =1/2*(15*1.5*2)=22.5 t

ii) Footpath width = 1.5m each side

intensity of SIDL (footpath)=1 t/m

footpath reaction on abutment =1/2*(15*1*2)=15 t

total reaction on abutment = 22.5+15 =37.5 t

Longitudinal eccentricity from A of pile cap=3.69 m

ML =37.5*3.69=138.38 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

3) Calculation of Load due to SIDL (WC)

thickness of WC= 0.065m

WC reaction on abutment =1/2*(15*8*0.065*2.2)=8.58 t

Longitudinal eccentricity from A of pile cap=3.69 m

ML =8.58*3.69=31.66 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

4) Calculation of load due to DL of Substructure

SW of Abutment = 9*1*5*2.5=112.5 t

Reaction =112.5 t

Longitudinal eccentricity from A of pile cap=4.3/2=2.15 m

ML =112.5*2.15=241.88 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

5) Calculation of load due to returnwall of Substructure

Assume,

length of return wall = 1.65m

height of return wall = 6m

thickness of return wall = 0.3m

SW of return wall = 2*1.65*6*0.3*2.5=14.85 t

Longitudinal eccentricity from A of pile cap=1.65/2=0.825 m

ML =14.85*0.825=12.25 t-m

Transeverse eccentricity from C/L of Pile Cap =4.35 m

since both MT acts opposite to each other ,hence nullify the effects

MT = 0 t-m

6) Calculation of load due to Backfill of Substructure

assume backfill soil density = 20kN/m3= 2t/m3

SW of Backfill soil = 8.4*1.65*6*2=166.32 t

Longitudinal eccentricity from A of pile cap=1.65/2=0.825 m

ML =166.32*0.825=137.21 t-m

Transeverse eccentricity from C/L of Pile Cap =0 m

MT = 0 t-m

7) Calculation of Earth Pressure

k=0.33

lateral Earth pressure = Pa=0.5*Ka*Y*H^2=0.5*0.33*20*6^2

=118.8 kN/m = 11.88 t/m

Total Earth force = 11.88*9=106.92 t

Acting at H/3 =2m

H = 106.92 t

HL= 106.92*2 =213.84 t-m

HT= 0 t-m

8)LL Surcharge load

live load surcharge pressure = k*Y*1.2=0.33*2*1.2=0.79 t/m2

lateral force = 0.79*9*6=42.66 t

point of action = 6/2=3m

ML = 42.66*3=127.98 t-m

MT = 0 t-m

9) Live Load Analysis and effective width calculation for 70R and Cl A to be done to calculate the reactions on the abutment

Ans -

CW = 8m

abutment width = 1 m

effective span = 15 m

Number of lanes = 2

vehicle combination - 1. 1 lane of class 70 R (T)

2. 2 Lane of class A

9.a) 1 lane of class 70 R(T)

impact factor = 10% for span of 15m

Total Load = 1.1*70 = 77 t

Max Moment Case

dispersion of load along the span ----

leff= 4.57+2(1+0.065) =6.57m

effective width of load----------

bef= α*a(1-a/lo)+b1

b/lo =12/15 = 0.8

α = 2.24

a = 7.5 m

b1= 0.84+2(0.065)=0.97 m

bef= 2.24*7.5(1-7.5/15)+0.97 =9.37 m = 2*4.685m ...................for one track

hence bef of 70R T vehicle =0.5+1.5+1.2+2.9-0.84/2+4.685=10.365m

intensity of load = 77 /(6.57*10.365)=1.13 t/m2

vertical reaction on abutment = 1.13*6.57 /2= 3.72 t

ML = (3.72*7.5)- (1.13*3.285*3.285/2) =218.73 kNm

P = 3.72 t

Transeverse eccentricity from C/L of Pile Cap =4.5-0.5-1.5-1.2-2.9/2=0.15 m

MT= 3.72*0.15=0.56 t-m

Max Reaction case

Vehicle is at support to get the max vertical reaction

P = 77 t

Longitudinal eccentricity from A of pile cap=3.69 m

ML =77*3.69=284.13 t-m

Transeverse eccentricity from C/L of Pile Cap =4.5-0.5-1.5-1.2-2.9/2=0.15 m

MT = 77*0.15=11.55 t-m

9.b) 2 lane of class A

impact factor = 21.4% for span of 15m

Total Load = 1.214*2*11.4 = 27.68 t

Max Moment Case

dispersion of load 11.4 t one wheel along the span ----

leff= 0.25+2(1+0.065) =2.38m=2*1.19m

dispersion length of load of 2 wheels=1.19+1.2+1.19=3.58m

effective width of one vehicle load----------

bef= α*a(1-a/lo)+b1

b/lo =12/15 = 0.8

α = 2.24

a = 7.5-1.2/2=6.9 m

b1= 0.5+2(0.065)=0.63 m

bef= 2.24*6.9(1-6.9/15)+0.63 =8.976 m = 2*4.49m ...................for one track

hence bef of Cl A 2 vehicle =0.5+1.5+0.15+0.25+1.8+0.25+1.2+0.25+1.8+4.49=12.19m

Hence bef = 12m

intensity of load = 27.68 /(3.58*12)=0.64 t/m2

vertical reaction on abutment = 0.64*3.58 /2= 1.15 t

ML = (1.15*7.5)- (0.64*1.79*1.79/2) =7.6 t-m

P = 1.15 t

Transeverse eccentricity from C/L of Pile Cap =0.5+1.5+0.15+(2.3+1.2+2.3)/2=5.05 m

MT = 1.15*5.05=5.81 t-m

Max Reaction case

Vehicle is at support to get the max vertical reaction

1.2 4.3 3.0 3.0 3.0 =14.50m

11.4 11.4 6.8 6.8 6.8 6.8 =50 t

CG of vehicle = 5.7 m

P = 2*50=100 t

impact factor =1.214

P = 1.214*100=121.4 t

Reaction on abutment = Rb =121.4*5.7/15=46.13 t

Ra= 121.4-46.13 =75.27 t

Longitudinal eccentricity from A of pile cap=2.15-0.39=1.76 m

ML =75.27*1.76=132.48 t-m

Transeverse eccentricity from C/L of Pile Cap =0.5+1.5+0.15+(2.3+1.2+2.3)/2=5.05 m

MT = 75.27*5.05=380.11 t-m

Load Item	Pmax	Pmin	ML	MT	HL	HT
Live Load	t	t	t-m	t-m	t-m	t-m
Cl 70R (T)
Max Moment case	3.72	3.72	218.73	0.56	0	0
Max Reaction case	77	77	284.13	11.55	0	0
Cl A -2 Lane
Max Moment case	1.15	1.15	7.6	5.81
Max Reaction case	75.27	75.27	132.48	380.11

10) Braking force

braking force =0.20*70=14 t

point of application = 6 m

ML=14*6=84 t-m

MT=0

11) Footway Live Load

Intensity of footway loading = 400 kg/m2

for 15m effective span

P=P'-(40*15-300)/9 =366.67 kg/m2 =0.37 t/m2

footway load = 2*0.37*15*1.5=16.65 t

vertical reaction on abutment = 8.33 t

Longitudinal eccentricity from A of pile cap=2.15-0.39=1.76 m

ML = 8.33*1.76 =14.66 t-m

Transeverse eccentricity from C/L of Pile Cap =5.125 m

since both MT acts opposite to each other ,hence nullify the effects

MT = 0 t-m

12) Buoyancy

volume of abutment= 9*4*1=36 m3

Buoyancy force = 36*1t/m3 =-36 t

ML = -36*2.15= 77.4 t-m

Summary of Load

Pile Cap bottom level

Load Item	Pmax	Pmin	ML	MT	H	HL	HT
	t	t	t-m	t-m	t	t-m	t-m
DL Superstructure	225+4.5=229.5	229.5	1660.5+18.90=1679.4	0		0	0
SIDL (CB,Footpath)	37.5	37.5	138.38	0		0	0
SIDL (WC)	8.58	8.58	31.66	0		0	0
DL Substructure	112.5+14.85=127.35		241.88+12.25=254.13
Backfill	166.32		137.21
Earth Pressure					106.92	213.84
LL surcharge					42.66	127.98
LL	77		284.13	380.11
Braking Force					14	84
Footway LL	8.33		14.66
Buoyancy	-36		77.4

For Load Combination

Load Item	for Axial force	for Moment ML	for Moment MT
DL(Superstructure+(CB+Footpath)+Substructure)	229.5+37.5+127.35=394.35	1679.4+138.38+254.13=2071.91
SIDL(WC)	8.58	31.66
Backfill	166.32	137.21
EP		213.84
LL Surcharge		127.98
LL(LL+Braking+Footway LL)	77+8.33=85.33	284.13+84+14.66=382.79	380.11
Buoyancy	36	77.40

1) ULS = 1.35 DL+1.75 SIDL-WC+1.5 Backfill+1.0 EP(Relieving)+1.2 LL surcharge+1.5 LL+0.15 Buoyancy
for axial force = 930.26 t

for ML = 4011.51 t-m

for MT = 570.17 t-m

2) SLS rare =1.0 DL+1.2 SIDL-WC+1.0 EP+1.0 LL+0.80 LL surcharge+0.15 Buoyancy

for axial force = 495.38 t

for ML = 2820.53 t-m

for MT = 380.11 t-m

3) SLS Quasi =1.0 DL+1.2 SIDL-WC+1.0 EP+0.15 Buoyancy

for axial force = 410.05 t

for ML = 2335.35 t-m

for MT = 0 t-m