Roof challenge

OBJECTIVE:

• The objective of this project is to design car roof with essentail flanges and reinforement like ditch area ,front roof rail,bow roof ,central roof rail and rear roof rail for the given styling surface
• perform the curvature study on the roof and peform the calculation for the heat distortion and snow load criteria to determine wheather the bow roofs are at the correct positions
• Finding movement of intertia and section modulus for bow roofs ,central roof rail,front roof rail and rear roof rail.
• perform the draft analysis for each reinforcements

CAR ROOF:

An automobile roof is the portion of automoble that sits above the passenger compartment ,protecting the vehicle occupants from sun,wind ,rain and other external things .Roof is the stucture above the wind shield glass having the reinforcement to provide the strength to all the overall structure to overall structure and to with stand the over head force like snow,reinforcements are required because ,if the roof is flat it is very weak and cannot able to with stand the forces like snow,so in order to increase the strength of then roof these reinforcement like front roof rail,central roof rail,rear roof rail and the bow roofs are introduced and joined to the flat surface of the roof with the different joining process

Different types of reinforcement that the car has are:

• front roof rail
• bow roofs
• central roof rail
• rear roof rail

FRONT ROOF RAIL:
Front roof rail is the one which joins the wind shield glass ,body side outer and the inner panel also

REAR ROOF RAIL:

Rear roof rail is the one which joins the back door and the body side outer

CENTRAL ROOF RAIL:

Central roof rail helps in providing efectively support to the flat area of the roof as it is more susceptible to failure under the action of the load . Generally central roof rail is placed at the centre of the roof which is connected to B-piller support structure .This central roof rail helps in adding strengh to the roof doring the rool-over test .The thickness of the central roof rail is 1.25mm

BOW ROOFS:

The bow roofs are give to improve the torsional stifness and load bearing capacity of the roof stucture . The no of bow roof present is depends on the overall size of the roof

JOINIJNG PROCESS:

Mastic are introduced between the roof outer panel the reinforcement for the joining .where ever the mastic are introduced there is remasrkable improve of the strength and spot welding is also intoduced to jointhe reinforcements and the outer panel

DESIGN CONSIDERATIONS FOR A ROOF:

• visibility criteria
• head clearance
• curvature study
• heat distortion and snow load criteria
• draft analysis

ROOF CRASH TEST:

Roof crush test is the design given by the IIHC .After analysing the many accidents IIHS has then fixed some regulations ,that how to perform the crush test

IIHS gives the rating based on this test.For this they consider two points while rating

• How fast the car can reforms to its original shape after the roll over
• How deep is the protrusion after the roll over

GOAL OF ROOF CRUSH TEST:

To minimize the protursion into the occupant compartment ,so that the seat belt and airbag can do their job in protecting the occupant

HEAT DISTRORTION STUDY ON THE ROOF:

The heat distortion study plays a major role in steet metal usage .The heat distortion temperature is a temperature limit above wwhich the material cannot be used for the structutral applications.This study is used to predict the distortion temperature at where the material start to soften when exposed to a fixed load at elevated temperature .In order to avoid bending or damages on the roof ,based on the heat distortion temperature , this study will predict the bow roof position which helps to strengthen the roof

W = [1.73 x 10^(-3) x L] + [1.85 x 10^(-8) x (R^2)/t] + [ 1.10x10^(-3) x l] - 2.68

Judgement Condition : OK< 2.7> 3.1

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]

HEAT DISTORTION CALCULATIONS:

ROOF RAIL	Rx(mm)	Ry(mm)	Rx*Ry(mm)	Rx+Ry(mm)	R(mm)	L(mm)	t(mm)	l(mm)	W(mm)
FR BOW	5601	2001	11207601	7602	2948.5	2037.33	.75	493	1.5
BOW-1	4769	5901	28141869	10670	5274.9	2037.33	.75	404	1.9
CR BOW 2	3834	9413	36089442	13247	5448.6	2037.33	.75	369	2.2
BOW 2	3144	13523	42516312	16667	5101	2037.33	.75	486	2.0

From the above table it can be concluded that all values of W<2.7 and also not w>3.1 so thus infers that current positionig of bow row are good in state as per design and found ok

SECTION MODULUS:

Section modulus is a geometrical property for a given cross section used in the design of beams or flexuL members .Other geometric properties used in the design include area for tension and shear ,radius of gyration for compression and moment of inertia and polar momemt of inertia for stiffness .Any relationship between these properties is highly dependent on the shape in question .Equation for the section moduli of common shapes are given below

Section modulus has appeared as fallows:

S = I/Y

here Y - is the detachment from theneautral axis to the top or lower part of the fiber .There are 2 sorts of section modulus,the elastic section modulus and the plastic section modulus.The section modulus is of diferent profiles can moreover be found as mathematical characteristics for ordinary profiles in the tables posting properties of such .section modulus is the numerical property of a given cross section used in the design of flexual people . section modulus depends just upon the cross section condition of the bar or development .For symmetric section,the section modulus is like both aobve and under the centriod of the cross section .An Upside -down section has two characteristics for the top and lower parts of the centroid ,the maximum section modulus and minimum section modulus

FRONT ROOF 1:

Moment of inertia maximum I max= 15.73

Moment of inertia of minimum I min = 0.92

y = 34.95

so

S max = 0.45

S min =0.026

BOW ROOF RAIL:

Moment of inertia maximum I max= 0.31

Moment of inertia of minimum I min = .O05

y = 34.95

so

S max = 0.0088

S min =0.0014

Moment of inertia maximum I max= 12.08

Moment of inertia of minimum I min = 0.99

y = 34.95

so

S max =0.34

S min =0.028