Project 2

Project -2

Ans of question-1

From staad model of I girder we got the reaction for DL-

From staad model of I girder we got the reaction for SIDL-

From staad model of I girder we got the maximum reaction for Generator load-

Ans of question-2

Design of bearings

1. Mean pressure- serviceability state

Sigma m= P/A’≤11 MPA

where, P= Pc + Ps

Pc= 585544 N.

Ps= 0 N.

A’=169320.25 MM^2

We know in serviceability state, Factor for DL=1 & Factor for SIDL = 1.20.

So, Pc= [(Pdl * 1) + (Psidl * 1.20)]

= [(440782*1) + (144762* 1.20)]

=614496.4 N.

P= Pc + Ps

=614496.4 + 0

=614496.4 N.

Sigma m= 614496.4/169320.25=3.63 N/mm^2 ≤11 N/mm^2 [ condition verified]

1. B1. No slip serviceability state-

Sigma m min= Pmin/A≥2 MPA

Pm=614496.4 N.

Sigma m min=614496.4/176400=3.48 MPA ≥ 2MPA [ condition verified]

2. B2.

H < f*P

Where f= friction.

H= horizontal force.

P= vertical force.

F= 0.12 + (0.12 Pc + 0.2A)

For slow loads,

T > G*Uc*A/(0.12Pc+ 0.2A)

G*Uc*A/(0.12Pc+ 0.2A)=(0.9*16.86*176400)/(0.12*614496.4 + 0.2*176400)=24.55 mm.

T= (12*5) + (6*2) =72 mm.>24.55 mm [ condition verified]

For rapid loads,

T > G.A(Uc+2Us)/ (0.12P+ 0.2A)

G.A(Uc+2Us)/ (0.12P+ 0.2A)

=0.9*176400(16.86+2*0)/ (0.12*585544 + 0.2*176400) =25.36 mm <T. [ condition verified]

1. Limitation of distortion- serviceability state-

tn/G ≤ 0.7 or,

T> (Uc+ Us) /0.7

(Uc+ Us) /0.7=24.08 mm<T [ condition verified]

1. No uplift at the edge with the least load- fracture state-

D1. Tanαc < 6 summation eic/a

For internal layer

Si= a.b/2ti (a + b)

= (420*420)/2*12(420+420)

= 8.75

Sigma mc= Pc/A’

For fracture state, DL=1.20, SIDL= 1.80

Pc=P(DL)+ P(SIDL) = (440782*1.20 + 144762*1.80) =789510 N

Sigma mc= Pc/A’=789510/169320.25=4.66 MPA.

Eic1= ti. Sigma mc/4* G * Si^2 +3*sigma mc

=12*4.66/4*0.9*8.75^2 + 3*4.66

=0.193.

For top & bottom layer

Si= a.b/2te (a + b)

= (420*420)/2*6(420+ 420)

=17.50

Eic1= te. Sigma mc/4* G * Si^2 +3 *sigma mc

=6*4.66/4*0.9 * 17.50^2+ 3* 4.66

=0.025

Total eic= summation eic1 + summation eic2

= 0.193+ 0.025

= 0.218

6 summation eic/a= (6* 0.218/420) =0.003114

Tanαc= (αc+αs) = 0.01+ 0.00= 0.01.

D2. Tanαc+1.5tanαs<6 summation ei/a

Sigma m= (Pc+Ps)/ A’

=4.66 MPA.

For internal layers,

Ei1= ti* sigma m/ (4*G*Si^2 + 3 sigma m)

=12*4.66/4*0.9*8.75^2+ 3* 4.66

=0.193

For top & bottom layers

ei2= te* sigma m/ (4*G*Si^2 + 3 sigma m)

=0.025

6 summation eia=6*0.218/420=0.003114

Tanαc+ 1.5 tanαs= (0.01+0.00*1.5)

=0.01

1. shear and strength of plates- no buckling

a/10 ≤ T ≤ a/5

T=72 mm.

a/10=42 mm.

a/5=84 mm.

42mm ≤ 72mm ≤ 84mm [condition verified]

1. shear serviceability state.

TP + TH + TS < 5G

TP=1.5 (Pc+ 1.5 Ps)/Si*a*b

=1.5*614496.4/8.75*420*420

=0.60 MPA.

TH= (G*Uc+ Gs*Us)/T

= (0.9*16.86+1.8*0)/72

= 0.21 MPA.

TS=G*a^2(tanαc+1.5tanαs)/2*T*ti

=0.9*(420) ^2[0.01+1.5*0]/2*72*12

=0.91 MPA

TP + TH + TS=1.72 MPA

5G=5*0.9=4.5 MPA > 1.72 MPA [ condition verified]

1. strength of plates- fracture state-

Ts> a/Si * sigma m/sigma s*gs

a/Si=420/8.75=48 mm

sigma m/sigma s=4.66/240=0.019

gs=1.20

a/Si * sigma m/sigma s*gs=48* 0.019* 1.20=1.0944mm<4mm [ condition verified]

Ans of question-3

Design of open foundation-

From staad model, we calculate the total load which comes on pier bottom-

For Dead load, P(DL)= R1+R2+R3+R4+R5+R6

= 485.548+580.430+485.549+485.548+580.430+485.549

= 3103.054 KN

= 316.315 TON

For Dead load, P(SIDL)= R1+R2+R3+R4+R5+R6

= 326.351+207.937+326.351+326.351+207.937+326.351

= 1721.278 KN

= 175.46 TON

For maximum reaction of load generator= R1+R2+R3+R4+R5+R6

= 237.904+52.356+(-26.431) +189.755+34.326+(-16.910)

= 471 KN

=48 TON

Calculating selfweight of pier cap-

Sectional view

Volume of truncated portion= 1/3(A1+A2+√A1*A2) *h

=1/3(6.9+32+√6.9*32) * 1.2

= 21.5 m^3

Volume of rectangular portion= (10*3.2*0.8) = 25.6 m^3

Total volume of pier cap= (21.5+25.6) =47.1 m^3

Density= 2.55 t/m^3

So, self weight I,e DL of pier cap= (2.55*47.1)= 120.105 ton

Total load that comes on pier(P)= [316.315+175.46+48+120.105]

= 659.88 TON

Calculation for longitudinal moments that comes on pier bottom-

ML(DL)= (158.16*1) – (158.16*1) = 0 t-m

ML(SIDL)= (87.73*1) – (87.73*1) = 0 t-m

ML(DL)= (158.16*1) – (158.16*1) = 0 t-m

ML(LL)= (26.89*1) – (21.11*1) = 5.78 t-m

ML for selfweight of pier cap= (5.78+102.089) = 107.87 t-m

Calculation for transverse moments that comes on pier bottom-

MT= (219.96 * 4.29) – (219.96 * 4.29) = 0 t-m

Now we get-

P (t)	ML (t-m)	MT (t-m)
659.88	107.87	0

Foundation details-

Foundation parameters-

A1	2.89 m^2
A2	44.89 m^2
Total wt of foundation	171.7 T
Area of foundation base	44.89 m^2
ZL or ZT	50.127 m^3

Calculations-

A1= (1.7*1.7) = 2.89 m^2

A2= (6.7*6.7) = 44.89 m^2

Total wt of foundation= (44.89*2.25*1.7) = 171.7 T

Area of foundation base= (6.7*6.7) = 44.89 m^2

ZL or ZT = BD^3/12/D/2= BD^2/6= 6.7*6.7^2/6= 50.127 m^3

Total vertical compressive load= P/A= 659.88/44.89= 14.7 t/m^2

Longitudinal moment (ML)= 107.87 t-m

ML/ZL= 107.87/50.127= 2.152 t/m^2

Longitudinal analysis-

Pressure value comes from previous data I,e 16.852 & 12.548 t/m^2, but after considering the selfweight of foundation and also add the factors the above value comes to be 19.052 & 15.848 t/m^2

Calculation-

(6.7-2.5) / (2.5-0) = (19.052-X)/(X-15.848)

X= 17.04 t/m^2

Calculation-

(6.7-4.2) / (4.2-0) = (19.052-X)/(X-15.848)

X= 17.85 t/m^2

Calculation-

(6.7-2.25) / (2.25-0) = (19.052-X)/(X-17.85)

X= 18.25 t/m^2

BM at the face of pier-

= [17.85*(2.5)^2+(19.052-17.85)/2*2.5]*2/3*2.5*8

=763.57 t-m (Ans)

SF at distance d from the face of the pier-

=18.25*(2.5-2.25) +(19.052-18.25)/2*(2.5-2.25)

=4.66 t (Ans)

Ans of question-4

Calculate the loads at the base of pier-

1) Wind load-

Calculation of wind load is based on IS-875(part-3)-1987.

The design wind speed at height Z is considered as follows-

Vz= Vb* K1* K2* K3

Where Vb= regional basic wind speed, taken 47 m/sec for Varanasi.

Total structure height above ground, H= 13.6 m

Probability factor (K1) = 1.07

Terrain height and structure size factor= 0.856 (assuming category 3 terrain)

Topography factor, (K3) = 1.0 (class c structure)

The design wind speed at height Z is considered as follows-

Vz= Vb* K1* K2* K3= 47* 1.07*0.856*1 = 43.1 m/sec

Wind pressure Pz= (0.6*Vz^2) = 1112.41 N/m^2 = 0.113 T/m^2

Diameter of pier = b= 1.7 m

Height of pier above ground,h= 10 m

Wind in longitudinal and transverse direction

Ratio of heighrt/ bredth= 5.88

Force co-efficient, Cf= 0.80

Projected area of structure, At= (h*b) = 17 m^2

Wind force, F= (0.80* 17* 0.113) = 1.54 ton

Acting @ above pier bottom= 5.28 m

2). Seismic acceleration-

Ah= (Z/2 * I/R * Sa/g)

Where,

Ah= horizontal seismic co-efficient to be considered in design.

Z= zone factor=0.16(for Varanasi reason)

I= importance factor= 1.50

R= response reduction factor=4 for pier

Sa/g=1+15T FOR 0.0<T<0.10

Sa/g=2.5 FOR 0.10<T<0.55

Sa/g=1.36/T FOR 0.55<T<4

3). LWR force-

Longitudinal lwr forces=0.8 t/m/track for two tracks.

Transverse force on top of pier due to LWR =[ΔT*a*Arail*Erail]*[no of rails]*[span length]/R

Where,

ΔT= Temperature difference= 34 degree celcius

α= Temperature co-efficient= 1.17* 10^-5 /degree celcius

Arail= area of one rail= 7670 mm^2/rail.

Erail= elastic modulus of rail= 200000 mpa.

Number of rails=4

Span length-1= 27 m

Span length-2= 27 m

Radious of curvature=1000m

Transverse force acting @ rail level= 0.07 ton

LWR =[ΔT*α*Arail*Erail]*[no of rails]*[span length]/R

=[34*1.17*10^-5*7670*200000]*[4]*[27]/1000000

= 65.90MPA.

4). Breaking load-

It takes as 18% of the unfactored vertical load, I,e 18% of unfactored vertical load comes due to breaking effect.

5). Centrifugal force-

C.F= V^2/127R

Where,

V= design speed of the train in km/hour= 75 km/hour

R= radius of cord span in meter= 1000 meter.

C.F= V^2/127R

= 75^2/127*1000

=0.044

Ans of question-5

Component of well foundations and function of each component with relevant drawings-

Well foundation is a type of deep foundation, it is provided for structures with foundation beneath rivers or other water bodies. It receives the load of whole structure and transfer it to the ground.

Various components of well foundation are-

1. i) Cutting edge.
2. ii) Well curb

iii) Bottom plug

1. iv) Steining
2. v) Top plug
3. vi) Well cap

Well curb- This transfer loads to ground through cutting edge while sinking.

Steining- it is the walls of well foundation, brick/ stone masonry

PCC or RCC materials are used. The design relies on the unit weight of well and skin friction.

Top plug- Top plug is provided after sand filling, it transfer the load of pier and superstructure to steining. It is constructed generally Using PCC(1:2:4) or (1:3:6).

Well cap- it is constructed as a slab resting on the well. It transfer

Load of the pier to well.