Section Modulus calculation and optimization

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

 

Mathematically,

 

It is defined as the ratio of the moment of inertia of a section about its centroidal axis and to the distance of the extreme layer from the neutral axis.

 

It is denoted by Z.

 

 

Mathematically By bending stress equation,

 

M/I = σmax/ymax

 

M = σmax × (I/ymax)

 

 

Where,

 

I = Moment of Inertia(m⁴)

 

y = distance from NA to extreme layer(m)

 

The unit of section modulus is m³.

 

There are two types of section modulus:

• Elastic Section Modulus
• Plastic 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,

 

 

Where,

 

I = Moment of inertia 

 

ymax = Distance from NA to extreme layer

 

There are following different elastic section modulus:

 

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

I = bd³/12

 

And the distance from the neutral axis,

 

ymax = d/2

 

So,

 

Zₑ = I/ymax

 

Zₑ = (bd³/12)/(d/2)

 

After calculating them,

 

Section modulus of rectangular section,

 

 

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

 

I = (BD³/12) - (bd³/12)

 

ymax = D/2

 

After putting these values,

 

Zₑ = I/ymax

 

(BD³/12) - (bd³/12)/(D/2)

 

 

For solid Circular moment of inertia,

 

I = (π/64) × d⁴

 

ymax = d/2

 

Then,

 

Zₑ = {(π/64) × d⁴}/(d/2)

 

After calculating then find values of section modulus of solid circular

 

 

The moment of inertia of hollow Circular,

 

I = (π/64) × (D⁴ - d⁴)

 

And,

 

ymax = D/2

 

After putting these values,

 

Zₑ = (π/64) × (D⁴ - d⁴)/(D/2)

 

 

Moment of inertia for the I & C section,

 

I = (BD³/12) - {(B - b)d³}/12

 

ymax = D/2

 

So,

 

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

 

 

 

Elastic section modulus for diamond is,

 

 

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.

 

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,

 

 

And,

 

 

 

There are following section modulus for different 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

 

 

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

 

 

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

 

Circular hollow section,

 

 

First Design Study

The Intersection of the design was taken and then section Inertia Analysis was done-

 

Moment of inertia(max) = 8.7659394×10^7×mm^4

Moment of inertia(min) = 2.2257727×10^5×mm^4

Y=864.9076/2=432.4538 mm

S=I/Y

S=2.2257727×10^5/432.4538= 514.68 mm^3

Hood Design Optimization

By increasing the depth of the inner panel by 0.5 mm

 

Here,

Moment of inertia(max) = 1.460070983×10^8×mm^4

Moment of inertia(min) = 3.800278008×10^5×mm^4

Y=864.9076/2=432.4538 mm

S=I/Y

S= 3.800278008×10^5/432.4538=878.77 mm^3

Conclusion

So, here the difference can be seen in the moment of inertia and the change in the section modulus can be seen by the change in the design(depth).