SECTION MODULUS CALCULATION AND OPTIMIZATION FOR CAR HOOD

SECTION MODULUS

• SECTION MODULUS
• TYPES OF SECTION MODULUS
• HOOD FOR SECTION MODULUS CALCULATION 
• SECTION MODULUS CALCULATION OF FRONT SECTION OF HOOD
• SECTION MODULUS CALCULATION OF SIDE SECTION OF HOOD

SECTION MODULUS

Section Modulus is a geometric property used to calculate the bending stresses in a structural member.

Sectional modulus: The ratio of moment of inertia of beam section about neutral axis and the distance of extreme fiber from neutral axis.

Mathematically, It is denoted by Z.

Z = I/y

Mathematically By bending stress equation,

M/I = σ/y

M = σ × (I/y)

M = σ × Z

Where,

I = Moment of Inertia(m⁴ or mm⁴)

Y = distance from NA to extreme layer(m or mm)

The unit of section modulus is m³ or mm³ .

TYPES OF SECTION MODULUS

There are two types of section modulus:

• Elastic Section Modulus
• Plastic Section Modulus

ELASTIC SECTION MODULUS

Elastic section modulus is calculated when we are to use the material behavior till the yield point. 

For general design purposes, we use elastic section modulus.

Generally, elastic section modulus is denoted by Zₑ.

So,

Zₑ = I/y

Where,

I = Moment of inertia 

ymax = Distance from NA to extreme layer

ELASTIC SECTION MODULUS OF DIFFERENT SECTION

There are following different elastic section modulus:

RECTANGULAR SECTION

As we know the moment of inertia of the rectangular section, 

I = bd³/12

And the distance from the neutral axis,

y= d/2

So,

Zₑ = I/y

Zₑ = (bd³/12)/(d/2)

After calculating them,

Section modulus of rectangular section,

Zₑ = bd²/6

HOLLOW RECTANGUAR SECTION

As we know the moment of inertia of the hollow rectangular section,

I = (BD³/12) - (bd³/12)

y= D/2

After putting these values,

Zₑ = I/y

(BD³/12) - (bd³/12)/(D/2)

Zₑ = (BD³ - bd³)/6D

SOLID CIRCULAR

For solid Circular moment of inertia,

I = (π/64) × d⁴

y= d/2

Then,

Zₑ = {(π/64) × d⁴}/(d/2)

After calculating then find values of section modulus of solid circular

Zₑ = (π/32) × d³

HOLLOW CIRCULAR

The moment of inertia of hollow Circular,

I = (π/64) × (D⁴ - d⁴)

And,

y= D/2

After putting these values,

Zₑ = (π/64) × (D⁴ - d⁴)/(D/2)

Zₑ = π/32 × {(D⁴ - d⁴)/D}

SYMMETRIC I&C SECTION

Moment of inertia for the I & C section,

I = (BD³/12) - {(B - b)d³}/12

y= D/2

So,

Zₑ = [(BD³/12) - {(B - b)d³}/12]/(D/2)

Zₑ =  {BD³ - (B - b)d³}/6D

DIAMOND

Elastic section modulus for diamond is,

Zₑ = bd²/24

 

PLASTIC SECTION MODULUS

The plastic section modulus is used for materials where the elastic yield is acceptable and the plastic behavior is considered to be an acceptable range.

It depends on the location of the plastic neutral axis.  The PNA is the axis that bisects the cross-section such that the compressive force from the region in compression is equal to the tension force from the region under tension.  So, for sections with a constant yield stress, the area above and below the PNA will be equal, but for mixed sections, this is not necessarily the case.

Mathematically,

The plastic section modulus is the sum of the areas of cross-sections on each side of the plastic neutral axis that may or may not be equal and multiplied by the distance from the local centroid of the two regions to the plastic neutral axis.

Zₚ = (Ac × yc) + (Aₜ + yₜ)

Where,

Ac = Area of cross-section under compression.

Aₜ = Area of cross-section under tension.

yc = Distance from the plastic neutral axis to the center of gravity of the areas under compression.

yₜ = Distance from the plastic neutral axis to the center of gravity of areas under tension.

Since on the neutral axis, the area under compression is equal to the area under tension.

So,

Ac = Aₜ

And,

yc = yₜ

PLASTIC SECTION MODULUS OF DIFFERENT SECTION

There are following section modulus for different section:

RECTANGULAR SECTION

As we know,

On the neutral axis, the area under compression is equal to the area under tension.

So,

Ac = Aₜ

And,

yc = yₜ

Ac = Aₜ = bd/2

yc = yₜ = d/4

So,

Zₚ = (Ac × yc) + (Aₜ + yₜ)

Zₚ = (bd/2 × d/4) + (bd/2 × d/4))

After calculating

Zₚ = bd²/4

Similarly, we can calculate the section modulus of the rectangular hollow section,

Zₚ = (bd²/4) - (b- 2t) × {(b/2) - t}²

 

CIRCULAR SECTION

As we know, on the neutral axis, the area under compression is equal to the area under tension.

So,

Ac = Aₜ = πd²/8

yc = yₜ = 2d/3π

Zₚ = (πd²/8 × 2d/3π) + (πd²/8 × 2d/3π) 

After calculating,

Zₚ = d³/6

Similarly, we can calculate the section modulus of the circular hollow section,

Circular hollow section,

Zₚ = (D³ - d³)/6

HOOD FOR SECTION MODULUS CALCULATION 

SECTION MODULUS CALCULATION OF FRONT SECTION OF HOOD

INTERSECTION OF FRONT SECTION

SECTION INERTIA ANALYSIS (Hallow Thickness - 0.75mm)

 

SECTION MODULUS OF FRONT SECTION

Z = I/y

Ex:1.0e+07 means 1.0 x 10⁷ or 10,000,000

I(max)= 9.047771404e+07  mm⁴

I(min)= 2.885258248e+05  mm⁴

y = d/2

   = 876.8112/2

   = 438.4056 mm

Z(max) = 90477714.04/438.4506 

             = 209220.92382344 mm³

Z(min)  = 288525.8248/438.4506     

             = 658.0577716167 mm³

 OPTIMIZED FRONT SECTION (0.5mm - offsetted the lower curve into Outward Direction)

SECTION INERTIA ANALYSIS (Hallow Thickness - 0.75mm)

SECTION MODULUS OF OPTIMIZED FRONT SECTION

Z = I/y

I(max)= 9.030419635e+07  mm⁴

I(min)= 2.920473222e+05  mm⁴

y  = 438.4056 mm

Z(max) = 90304196.35/438.4506 

             =  208819.68102253 mm³

Z(min)  = 292047.3222/438.4506     

             = 675.3310602413 mm³

For Comparison I took a Z(min), Modulus of Front Section

Z(min) = 658.0577716167 mm³

Modulus of Optimized Front Section

Z(min) = 675.3310602413 mm³

Therefore,

Section Modulus is Increased a value of,

Z(min) = 17.2732886246  mm³

• So, Optimiation is helped to increas a section modulus, section modulus is deirectly proportional to the strength.
• The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.
• Bending stress is inversely proportional to the Section Modulus.
• Area is directly proportional to section modulus.
• Thickness is also directly proportional to the section modulus.

SECTION MODULUS CALCULATION OF SIDE SECTION OF HOOD

INTERSECTION OF SIDE SECTION

SECTION INERTIA ANALYSIS (Hallow Thickness - 0.75mm)

SECTION MODULUS OF FRONT SECTION

Z = I/y

Ex:1.0e+07 means 1.0 x 10⁷ or 10,000,000

I(max)= 1.543872215e+07  mm⁴

I(min)= 9.871585940e+04  mm⁴

y = d/2

   =484.2694/2

   =242.1347 mm

Z(max) = 15438722.15/242.1347

             = 63760.882475746 mm³

Z(min)  = 98715.85940/242.1347   

             = 405.4596858691 mm³

OPTIMIZED SIDE SECTION (0.5mm - offsetted the lower curve into Outward Direction)

SECTION INERTIA ANALYSIS (Hallow Thickness - 0.75mm)

SECTION MODULUS OF OPTIMIZED FRONT SECTION

Z = I/y

I(max)= 1.539186659e+07  mm⁴

I(min)= 1.020265051e+05  mm⁴

y  = 242.1347 mm

Z(max) = 15391866.59/242.1347

             = 63567.372169293 mm³

Z(min)  = 102026.5051/242.1347    

             =  421.3625932177 mm³

For Comparison I took a Z(min), Modulus of Front Section

Z(min) = 405.4596858691 mm³

Modulus of Optimized Front Section

Z(min) = 421.3625932177 mm³

Therefore,

Section Modulus is Increased a value of,

Z(min) = 15.9029073486 mm³

• So, Optimiation is helped to increas a section modulus, section modulus is deirectly proportional to the strength.
• The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.
• Bending stress is inversely proportional to the Section Modulus.
• Area is directly proportional to section modulus.
• Thickness is also directly proportional to the section modulus.