Project 1
PROJECT-1
Ans of question-1
Metro rail is needed in countries with developing economy because of overcome the problems like congestion and pollution. So that people reached their destination timely.
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Conventional rail system |
Metro rail system |
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Don’t have any special track |
Special track available for metro rail |
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Travelling distance covers the city and suburbs |
Travelling distance covers within the city |
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Large no of seats available as compare to metro |
Less no of seats available as compare to conventional rail system |
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Proper drainage system not available |
Proper drainage system available |
Ans of question-2
Longitudinal analysis of 3*25 m prestressed I girder.
Set up the units from search bar.
Create the grillage model of span length of 25 m.
Now we need to create the supports
Supports-1
Go to support-create-fixed but-release moment Mx, My, Mz -add
For assign the supports we select the mid supports at left side -assign
Supports-2
Go to support-create-fixed but-release moment Mx, My, Mz , Fx- add
For assign the supports we select the mid supports at right side -assign
Supports-3
Go to support-create-fixed but-release moment Mx, My, Mz , Fz- add
For assign the supports we select the side supports at left side -assign
Supports-4
Go to support-create-fixed but-release moment Mx, My, Mz , Fz, Fx- add
For assign the supports we select the side supports at right side -assign
Now we need to create section properties
End properties:-
Go to properties- go define-select general-then put the bellow value
Ax=1.75808 m2
Ix=0.1818 m4
Iy=0.6802 m4
Iz=0.285 m4
Material select as concrete-ADD
Mid properties: -
Go to properties- go define-select general-then put the bellow value
Ax=1.41702 m2
Ix=0.07 m4
Iy=0.625 m4
Iz=0.260 m4
Material select as concrete-ADD
Then we need to create a composite properties as dummy
Go to define-rectangular-ZD=1, YD=1-ADD
Then we need to assign all properties one by one.
Then we need to apply load-
Go to load-vehicle definations-vehicle type ref=1, width=1.51 m
Then we put the value of load comes on one wheel and the distance.
Click on add- close
Calculation-
Load comes on axle=16 T
Load comes on one wheel= 16/2=8 T=(8*9.81)=78.5 KN
Then we need to create DL-
Go to load case details-go to add-select title as DL & loading type dead-add
Click on dead load – add- global direction Y and factor be (-1) – add- close
Click on self wt – assign to view- assign
Then we need to create SIDL
Go to load case details-go to add-select title as SIDL & loading type dead-add
- for plinth load
Click on SIDL click on add-go to member load- go to uniform force
W1=-8.34 KN/M and select direction (GY)
Click on add
- parapet load
Click on SIDL click on add-go to member load- go to uniform force
W1=-23.5 KN/M and select direction (GY)
Click on add
- Moment of parapet at one side
Click on SIDL click on add-go to member load- go to uniform force
W1=-11.75 KN/M and select direction (GX)
Click on add
- Moment of parapet at another side
Click on SIDL click on add-go to member load- go to uniform force
W1=-17.75 KN/M and select direction (GX)
Click on add
Then we need to assign all the loades
First go to view- top view
- assign load case 1
Select 3 I girder-select load case 1, I,e plinth load under SIDL- assign selected beams-assign-yes
- assign load case 2
Select both end side 2 I girder -click on parapet loading case-assign to selected beams-assign-yes
- assign load case 3
Select the downward longitudinal I girder-click on moment case (11.75 kn/m)-assign to selected beams-assign-yes
- assign load case 4
Select the top longitudinal I girder-click on moment case (-11.75 kn/m)-assign to selected beams-assign-yes
Then from loading -click over load generator- no of loads to be generated=250 & predefined load to be added=None- add – close
Calculation-
Total length of train with two bogies=2.3+12.4+2.3+4.4+2.3+12.4+2.3=38.5 m
Span of girder=23.9 m
So total distance= (38.5+23.9) =62.4 m
If we applied the load at every 0.25 m then total no of load applied=250
Then from load case details select 250 generate load-add-and put the bellow value
Type=1, range=none
Initial position of load
X coordinate=-38.65
Y coordinate=0
Z coordinate=-4.6
Load increment
X coordinate=0.25- add
Then another load case would be added under 250 generate load
Type=1, range=none
Initial position of load
X coordinate=-38.65
Y coordinate=0
Z coordinate=0
Load increment
X coordinate=0.25- add
Go to analysis- run analysis- postprocessing mode
Bending moment diagram under dead load-
BM diagram under SIDL-
250 generate load as an increment of 0.25 mm
SF diagram under DL
Ans of question-3
Transverse STAAD model for U- girder.
At first set up the units, length as meter and load as kilonewton from serch bar.
Create the nodes-
Slab width= 4000mm
Height -1.8 m
Mid of the web=0.9 m
After creating all nodes – select the node cursor-select the nodes which are created- geometry- select translational repeat- then put the bellow value
No of steps=1
Default step speacing=0.7
Global direction= Y
Link steps – ok
We know the clear span of U- girder = 30 meter.
Each support= 0.7 meter.
Effective span of U- girder= 28.6 meter.
We need to add plate over the span of 28.6 meter.
So we devide the total span in 16 parts, so each span = 1.7875 meter, I,e each plate = 1.7875 meter.
Then go to view option- select the top view-select the right side node – go to geometry – translation repeat- put the bellow value
No of steps=1
Default step speacing= 1.7875
Link steps- ok
Go to view option – select isometric view.
Then go to top view- use of plate cursor we need to create one plate
Go to view option select isometric view.
Then go to top view - select plate cursor- select the first plate- go to geometry- translation repeat – put the bellow value-
No of steps= 15
Default step speacing= 1.7875
Global direction= Y
Link steps- ok.
Then we need to create right side support portion
So go to top view- go geometry- transational repeat- put the bellow value
No of steps=1
Default step spacing= 29.3
Link steps- ok
Go to view see the isometric view
But we see that the longitudinal beam has to be created so we need to create the transverse beam
Go to geometry- add beam- manually create the last left side transverse beam- go to view – select top view-from geometry – go to transitional repeat – put the bellow value
No of steps= 18
Default step spacing= 1.7875 m
Link steps- ok
Then we need to delete the duplicate beam
Select node cursor- select all the nodes- geometry- dead beams- shows that 245 new beams created- then select the structure- utilities- beam tools- duplicate beams- shows all duplicate beams- remove all duplicate beam.
Then we need to create the supports-
Go to support-create- pinned – add
For assign the supports- click over created support – use cursor to assign-assign at particular point.
Then we need to create the properties-
Go to properties- thickness- create slab thickness-materials as concrete-add
Go to properties- thickness- create web thickness-materials as concrete-add
Then we need to assign the properties
Select first plate I,e 11.81- go to select- select plate cursor-select the slab portion- assign to selected plates-assign- yes.
Then click over the second plate I,e 9.45- select the plate cursor- select both side top web plate-assign to selected pltes-assign-yes.
Then click over the third plate I,e 9.45 variable- select the plate cursor- select left side slunting web plate-assign to selected pltes-assign-yes.
Then click over the fourth plate I,e 11.81 variable- select the plate cursor- select right side slunting web plate-assign to selected pltes-assign-yes.
Go to isometric view- see the bellow picture.
Then click over 3D rendering & see how the U girder look like
Then from properties- define- rectangular- YD=1, ZD=1- material choose as concrete-add- close
For assigning select the whole structure-assign to selected beams- assign -close.
Then we need to apply the loads-
Go to load case details- add- primary
Number=1
Loading type=dead
Title=dead load
Add
Number=2
Loading type=dead
Title=SIDL
add
Number=3
Loading type=live
Title=live load
Add
Click over dead load- self weight-direction-Y and factor=-1 – add
For assigning the dead load
Click over self weight -1 – assign to view- assign – yes.
Then click over SIDL – add- plate loads-partial plate pressure load
say the load of SIDL is 20 Kn/m, which is transmitted over two rails, so for 1 rail it is 10 Kn/m
w1=-10 Kn/m2
X1=-0.8M
X2=0.8M
Y1=-0.5M
Y2=0.5M
Select global direction=Y- ADD
Then click over the created load under SIDL – go to select- click on plate cursor-select all mid plates-click on created loads- assign to selected plates-assign- yes.
Then we need to add the live load-
The live load on a axle of a train= 16 tones
So for one wheel=16/2=8 tones= 8*9.81=78.48 KN.
Wheels are placed at 1.51 meter.
Go to plate load- concentrated load
Force=-78.48 KN
X=0.48125M (Calculation- x= 1.7875/2-0.4125= 0.48125)
Y= 0.245 M (calculation-Y= 1-1.51/2=0.245 m)
ADD
Force=-78.48 KN
X=-0.48125M
Y= 0.245 M
ADD
Force=-78.48 KN
X=0.48125M
Y=- 0.245 M
ADD
Force=-78.48 KN
X=-0.48125M
Y=- 0.245 M
ADD
Then we need to assign the loads.
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Plate 1 |
Plate 2 |
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Plate 3 |
Plate 4 |
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Yellow marking- centre line of the U-girder.
Red marking- position of plate for assign the load.
For load case-1
Select plate cursor- select the plate 1 shown in above – assign to selected plates- assign
For load case-2
Select plate cursor- select the plate 1 shown in above – assign to selected plates- assign
For load case-3
Select plate cursor- select the plate 1 shown in above – assign to selected plates- assign
For load case-4
Select plate cursor- select the plate 1 shown in above – assign to selected plates- assign
Then go to run analysis- go postprocessing
BM under SIDL
Reaction under dead load
Beam results
Plate results
Ans of question-4
We know, A=2.1735mm2
I= 0.74035 m4
No of stand(n)= 72
Prestressed force applied(P)=0.765fpu*As*n(1-Loss)/1000 kn
Where fpu= nominal ultimate stress=1860 MPA
So, max jacking stress= 0.765*fpu=0.765*1860=1422.9 MPA
Nominal area of stand AS= 140 mm2
P=1422.9*140*72(82.5%)/1000
=11832.8 Kn. (Ans)
Section modulus(Z)=I/Ybot
Where, I=Moment of inertia=0.74035m4
Ybot= stress at bottom=0.594 m
So, Z= 0.74035/0.594=1.2464 m3 (Ans)
Consider C.G of PT cables= 0.140 m from soffit level
S0, eccentricity of prestress tendons from C.G (e) = 0.594-0.140= 0.454 m (Ans)
Stress at bottom=P/A+P*(e/z)-M/Z
= (11832.8/2.1735) + 11832.8*(0.454/1.2464)-10628.58/1.2464
=1.607 N/MM2(Ans)