Section Modulus calculation and optimization

SECTION MODULUS CALCULATION & OPTIMIZATION OF THE HOOD

Objective: To find the section modulus of the hood and optimization to improve the overall section modulus.

BENDING EQUATIONS FOR BEAMS-

M/I = σ/y = E/R

Where,

M = Bending moment

I = Moment of Inertia of the section about the bending axis.

σ=Bending stress

y= Distance between the neutral axis and extreme end of the object

E = Youngs modulus

R = Radius of gyration

From the bending equation

M/I = σ/y

M = σI/y = σ Z

Where, Z is the section modulus

Section Modulus may be defined as the ration of total moment resisted by the section to the stress in the extreme fiber which is equal to yield stress. The section modulus is geometric property of the cross section used for designing beams and flexural members. It does not represent anything physically.

Neutral axis: When beam is subjected to pure bending the top most fibers of the beam are subjected to maximum compression, while the bottom most fibers of the beams are subjected to maximum tension. The line of intersection of the neutral layer with any normal cross section of a beam is known as Neutral axis of that section.

The Area moment of Inertia of a beam cross sectional area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.

CASE 1: Frontal Crash

1. For studying the section modulus we need the cross section of hood. For that I created a cross section using Intersection Curve and then section inertia analysis is done.

We have to make sure that the curve obtained is a single/closed curve.

MOI (max) = 4.742592405e+06(mm⁴)

MOI (min) = 1.421104738e+04(mm⁴)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is to considered

Y = Total distance /2

Total distance =874.7380mm

Y = 874.7380 / 2

Y = 437.369 mm

Therefore,

S = 1.421104738 × 10⁴ (mm⁴) / 437.369 mm

S = 32.49212308 mm³

2. After optimization the hood inner panel’s curve is offset by 2.5mm outwards and section inertia analysis is done same as case 1 for a closed curve.

MOI (max) = 4.801076257e+06(mm⁴)

MOI (min) = 1.591378689e+04(mm⁴)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is to considered

Y = Total distance /2

Total distance =874.7380mm

Y = 874.7380 / 2

Y = 437.369 mm

Therefore,

S = 1.591378689 × 10⁴ (mm⁴) / 437.369 mm

S = 36.38526482 mm³

CASE 2- Side Crash

1. For studying the section modulus we need the cross section of hood. For that I created a cross section using Edit Section View and then section inertia analysis is done.

We have to make sure that the curve obtained is a single/closed curve.

MOI (max) = 7.623695757e+05(mm⁴)

MOI (min) = 3.293942312e+03(mm⁴)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is to considered

Y = Total distance /2

Total distance =472.0228mm

Y = 472.0228 / 2

Y =236.0114 mm

Therefore,

S = 3.293942312 × 10³ (mm⁴) / 236.0114 mm

S =13.9567085 mm³

2. After optimization the hood inner panel’s curve is offset by 2.5mm outwards and section inertia analysis is done same as case 2 for a closed curve.

MOI (max) = 7.810561708e+05(mm⁴)

MOI (min) = 4.134678981e+03(mm⁴)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is to considered

Y = Total distance /2

Total distance =472.0228mm

Y = 472.0228 / 2

Y =236.0114 mm

Therefore,

S = 4.134678981 × 10³ (mm⁴) / 236.0114 mm

S =17.51897993 mm³

Conclusion: In this study, we take the minimum of inertia moment to analyze the minimum limit of hood deformation. Thus the section modulus of the hood was calculated and compared for both frontal & side crash.

Result - The section modulus value of case 1 is higher than the case 2 means better section modulus then better strength of the hood for frontal crash.

Also increasing the section area increases the moment of inertia & section modulus which means strength and load material also increases.