Roof Design
ROOF DESIGN THEORY
Aim: Design of the front roof rail, rear roof rail, and bow roof for the roof of the vehicles with the help of the given skin data of the roof according to snow load condition and hear distortion criteria.
Objective:
- Study different type of the roof available in the market.
- Study about the heat distortion criteria to adjust the position of bows and center roof rail.
- Study about snow load conditions to check the strength of the roof according to bows positions.
Theory:
Roof: An automobile roof or car top is the portion of an automobile that sits above the passenger compartment, protecting the vehicle occupants from sun, wind, rain, and other external elements.
2.Front roof rail:
a.The front roof rail is used as the reinforcement to the roof surface to increase the strength and also used to mount the brackets for some features such as room lamp or map lamp and rear mirror.
b.The depth of the front roof rail is decided based on the visibility criteria line decided for the driver visibility.
c.The width of the front roof rail is decided in such a way that there should be enough headroom clearance between the head of the driver and the end surface of the roof rail to avoid injuries to the person.
d.At the front side we are doing spot welding and on the rear side of the front roof rail, we are using mastic points to join it with the roof surface.
e. The minimum radius for the fillet at edges is 4mm and 5mm.
3.Rear roof rail:
a.The rear roof rail is used as the reinforcement to the roof surface at the rear side.
b.The joining method used for this rail is the same as that of the front roof rail.
c.The minimum radius for the fillet is 4mm and 5 mm.
4.Bows:
a.We are making two bows one is in between the front roof rail and central roof rail and another is in between the central roof rail and rear roof rail.
b.Both ends of the bow are joined to the roof outer flanges by the spot welding and the inner sides of the bow are joined to the inner surface of the roof with the help of mastic points.
c.The minimum radius of the fillet is 4 and 5 mm.
d.The cross-section of the bow is made in such a way that we should get the maximum value of the minimum moment of inertia at which the bow fails.
5.Heat distortion criteria:
In this criteria, we check the heat distortion capacity of the roof at higher temperatures with the fixed load on it. So when the surrounding temperature is high and the roof is loaded with some weight on it then if the roof is deformed it fails, so to calculate whether the roof deform or not, according to the position of the bow we do these calculations as follows .
Bow–Roof Prediction Formula
W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68
Where,
L = Roof Length in X-Direction[mm](Roof dimension in 0-Y)
R = Roof curvature
R = 2(Rx*Ry)/(Rx+Ry)
Rx = X curvature
Ry = Y curvature
t = Roof plate thickness [mm]
l = Bow Roof Span [mm] .{ I = Distance between the two consecutive reinforcement added(Bow roof span mm}
Steps to calculate the Heat distortions as follows:
1.First we offset the center y-axis line to 300mm distance apart from it represented as (-3Y to 3Y red line).
2.After that we offset this -3Y to 3Y to 100mm offset each side represented with the center line.
3.We draw the vertical red line at the center of the two consecutive bow rood span. Next, we offset this red line to 100 on each side.
4. Now we have three horizontal lines and three 12 vertical lines we take the intersection points of the vertical red line with three horizontal lines and a horizontal red line we take the intersection of the vertical lines with that line.
5.After getting all the points we project these points on the roof surface and draw the three points curve along the x-axis point to get the radius of curvature of Ry and along y-axis points to get the radius of curvature of Rx.
Calculations:
1. For the Front Roof rail and Bow Roof .
Rx=3315.6524 mm
Ry =4558.9719 mm
L= 2030 mm
l = 354.7074 mm
{R = 2 ( Rx *Ry)/ (Rx+Ry) }
R =2( 3315.6524 * 4558.9719) / ( 3315.6524 + 4558.9719 )
R= 3839.156 mm
W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68
W = [1.73 x 10^(-3) x 2030] + [1.85 x 10^(-8) x ( 3839.159)^2 / 0.75] + [ 1.10x10^(-3) x 354.7074] - 2.68
W = [3511.9 x 10^(-3) ] + [36356493.021 10^(-8) ] + [ 390.178x10^(-3) ] - 2.68
W= [ 3.5119+.3635+.390]- 2.68
W= 4.2654- 2.68
W= 1.58
2-: For the Bow Roof Rail-1 & Center Bow Roof Rail.
Rx= 2860.3952 mm
Ry= 3184.8464 mm
L = 2030 mm
l =453.0215mm
t= 0.75mm
{R = 2 ( Rx *Ry)/ (Rx+Ry) }
R = 3013.913mm
W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68
w= [ 1.73x10^(-3) x 2030]+ [1.85x 10^(-8) x 12111562.09] +[ 1.1o x10^(-3) x453.021] - 2.68
w = [3511.9 x 10^(-3)] + [ 22406398.87 x10^(-8)] + [498.32 x 10^(-3)] - 2.68
w =[3.5119] + [ .224] + [.498]- 2.68
w= 4.23 - 2.68
w= 1.55 ANSWER
3-: For The C.T.R & Bow Roof -2.
Rx= 3104.4538mm
Ry = 5797.9662 mm
L= 2030 mm
l= 380.4459mm
t 0.75 mm
R= 4043.79 mm {R = 2 ( Rx *Ry)/ (Rx+Ry) }
W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68
w= [1.73 x10^(-3) x 2030] +[ 1.85 x 10^(-8) x 21802983.41] + [1.10 x 10^(-3) x 380.4459]- 2.68
w= [3511.9 x10^(-3)] + [40335519.30x 10^(-8)] + [418.49 x 10^(-3)]- 2.68
w= [3.511]+ [.403]+ [.418] - 2.68
w = 4.332- 2.68
w= 1.652 answer
4-: For The B.R.F-2 & R.R.R:-
Rx= 4558.9719mm
Ry= 3315.6525mm
L=2030mm
l= 401.7109 mm
R= 3839.16 mm
W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68
w= [1.37x 10^(-3) x 2030] + [1.85 x 10 ^ (-8)x 19652199.34]+ [ 1.10 x 10^ 9-3) x 401.71]- 2.68
w= [ 3511 x 10^(-3) ] + [ 36356568.77 x10 ^(-8)] + [441 x10^(-3)]- 2.68
w = [3.511]+ [.365]+ [.44]- 2.6
w= [4.316- 2.68]
w = 1.636 ANSWER
The condition for the heat distortion criteria is W>2.7 so the given is less than 2.7 so the position of the bow 1 is correct according to the hear distortion criteria.
Like this, we do the calculation for the remaining Bows positions as shown in the following table.
6. Snow load condition:
This snow load condition calculation is done to check the strength of the roof when there is dead weight on the roof. As we know that in some regions we have snowfalls so when snow falls on the roof then this act as dead weight because in some the rate of snowfall is more so the roof should sustain this dead weight for that we do this calculation to check the position of the bow and its sectional area.
The formula for snow load condition:
Qr = [Iy x t2] / [α x s x [(Rx + Ry)/2]2 x 10-8]
Note: The Qr value given over here is correct. There was a typo error in the formula given in the video.
Where
α = My x Lx2 x 10-12 , My = Y(Ly-Y)
Judgement condition = Qr ≥ 3.1
t = Roof plate thickness [mm]
Ly = Distance between the front and rear roof Rails on the Vehicle along with 0Y[mm]
Length of Roof panel with the center point between Roof rail Front /Rear as the front and rear reference point.
Lx = Distance between the Left and Right end of the roof on the Roof BOW [mm]
Width of the roof panel exposed on the surface.
Y = Distance front Front Roof Rail to Roof BOW[mm]
s = Distance for which Roof BOW bears divided load [mm]
s = L1/2 + L2/2
Iy = Geometrical moment of inertia of Roof BOW (Y cross-section )[mm4]
Rx = Lateral direction curvature radius of roof panel Y cross-section on Roof BOW [mm]
Roof panel curvature Radius of the Length Lx in Front view
Ry = Longitudinal Direction curvature radius of the Roof panel X cross-section on Roof BOW [mm]
Roof panel X curvature radius of length s in Side view.
Section analysis:
1-: FOR THE BOW ROOF RAIL & CENTER BOW ROOF RAIL-:
Qr = [Iy x t2] / [α x s x [(Rx + Ry)/2]2 x 10-8] ,α = My x Lx2 x 10-12 , My = Y(Ly-Y)
Iy= 1.717x 10^3 mm^4
t= 0.75mm
α= .832mm
Rx= 4060.78mm
Ry= 7310.55 mm
L1= 435.57 mm
L2= 519.23 mm
Ly= 1148.60mm
Lx= 1148.60 mm
S= 5685.67mm
Y= 435.57 mm
Qr = [Iy x t2] / [α x s x [(Rx + Ry)/2]2 x 10-8]
Qr = [1.717x10^3 x ( 0.75)^2] / [ 0.832 x{(4060.78+7310.55)/2}^2 x 10^(-8)]
Qr = [ 1.717 x10^3 x .5625] / [134.82]
Qr = [0.9755994 x10^3] / [134.82]
Qr =0.00732 x 10^3
Qr = 7.32 answer
2-: CENTER ROOF RAIL&BOW ROOF RAIL-2
Iy = 1.578 x10^3 mm^4
t = 0.75 mm
α= 0.781 mm
Rx = 3262.00 mm
Ry= 6009.07 mm
L1= 459.55 mm
L2 = 559.55 mm
Ly= 1974.345 mm
Lx= 1059.6891 mm
S= 509.55 mm
My = 696126.34mm
Qr = [Iy x t2] / [α x s x [(Rx + Ry)/2]2 x 10-8] ,α = My x Lx2 x 10-12 , My = Y(Ly-Y)
Qr =[1.578 x10^3 x (0.75)^2] / [0.781 x 509 .76 x { (3262.00+ 6009.07)/2}^2 x10^(-8)]
Qr = [.8245 x 10^ 3] / [ 85.558]
Qr = 0.0096367 x 10^3
Qr = 9.6 answer
3-: BOW ROOR RAIL -2 & REAR ROOF RAIL-:
Iy=1.637 x 10^3 mm^4
t= 0.75 mm
Rx = 4060.7807 mm
Ry = 7310.5573 mm
L1= 459.5506 mm
L2= 559.9963 mm
LY= 1974.345 mm
S = 509.785mm
Y= 1974.345 mm
α= - 1.998 mm
MY= -1840370.609 mm
Qr = [Iy x t2] / [α x s x [(Rx + Ry)/2]2 x 10-8] ,α = My x Lx2 x 10-12 , My = Y(Ly-Y)
Qr= [ 1.637 x 10^3 x(0.75)^2] / 1.998 x 509.785 x {(4060.78+7310.55)/2}^2 x 10^(-8)]
Qr= [0.921265 x 10^3] / [329.25]
Qr= 0.00279 x 10^3
Qr= 2.78 answer
Now lets see what are all the parts are there in roof:-
- OUTER ROOF
- FRONT ROOF RAIL
- BOW ROOF 1
- CENTER ROOF
- BOW ROOF 2
OUTER ROOF-:
An automobile roof or car top is the portion of an automobile that sits above the passenger compartment, protecting the vehicle occupants from sun, wind, rain, and other external elements.
- FRONT ROOF RAIL:--
Front roof rail is the one which joins the wind shield glass , body side outer and the inner panel also. Here the front roof rail is designed based up on the master section given
Front Roof Rail Master section
- BOW ROOF 1 and 2:--
The bow roofs are give to improve the torsional stiffness and load bearing capacity of the roof structure . The number of bow roof present is depends on the overall size of the roof . Presently in this project two bow roofs are added.
- CENTER ROOF
Central roof rail helps in providing effectively support to the flat area of the roof as it is more susceptible to failure under the action of load . generally central roof rail is placed at the center of the roof which is connected to the B-pillar support structure.This central roof rail helps in adding the strength to the roof during the rool- over test.
REAR ROOF RAIL :
Rear roof rail is the one which joins the back door and the body side outer
Now Lets see what is the procedure to Design the Roof :--
Open the given model in NX that is the styling surface of the car's roof.
In this challenge, we will be creating the Roof's outer panel, Front roof rail, Rear roof rail, Centre reinforcement and Bow roofs.
Roof Outer Panel:
- First, we will be extracting the top half portion of the roof surface. Since the roof surface is symmetrical in shape, we can easily work only on half of the portion and reflect it at the end to create the whole roof.
- Untrim this surface in order to avoid the holes. Extract the outer boundaries of the original half surface and project it on the untrim surface to get the exact outer edges of the roof.
- The corners of the roof surface are sharp so use the bridge curve and trim the sheet and trim the sharp edge.
- Now we will be creating the flanges by using the law extension command .
- From the edge of the law extension create an another law extension
- Similarly, by following the same procedure we can create a flange on the other edges as well.
- Using the Through Curve Mesh Command to create a surface on the corners by giving guide curves and support of the surfaces.
- Sew these Surfaces together and give the required fillets.
- Now Follow The Master section given for Rear roof rail and create the Additional area
- For the Rear end, there should be an additional area that helps to hold the back door without hitting the roof.
Draft Analisys for each part:--
Outer Roof Panel:-
Front Roof Rail:-
RESULTS:--
The Roof of a car has been successfully developed with all the required components and a draft analysis have also been performed for the same.
Final Roof Model:--
