Section Modulus calculation and optimization

Section modulus

section modulus is ageometric property for a given cross secton used in the design of beam or members, its also at the polar modulus or the torsional section modulus .in other case geometric are used to the design included area for the tension and shear ,radius of gyration foe compression and moment of inertia and polar moment of interia stiffness. there are two types of the modulus are given and that are elatic and plastic modulus.

s=I/Y

Here

s=section modulus

I=MOMENT OF INERTIA (UNIT mm4)

Y=distance from the neutral axis to the extreme end(unit mm)

the unit of "section modulus" is(mm 3)

section modulus is two types

1.elastic modulus

2.plastic modulus

1.elastic modulus

the elastic design modulus is used, applying up to the yeild point for most metal and other common material.

The elastic section modulus is defined as the ratio of the second moment of area (or moment of inertia) and the distance from the neutral axis to any given (or extreme) fiber. It can also be defined in terms of the first moment of area.

Generally, elastic section modulus is denoted by Zₑ.

So,

Zₑ = I/ymax

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,

ymax = d/2

So,

Zₑ = I/ymax

Zₑ = (bd³/12)/(d/2)

After calculating them,

Section modulus of rectangular section,

Zₑ = bd²/6

Hollow 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)

Zₑ = (BD³ - bd³)/6D

Solid Circular

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

Zₑ = (π/32) × d³

Hollow 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)

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

ymax = 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 to the where the elastic yielding is acceptable and plastic behavior is assummed to be an acceptable limit. the plastic section modulus is depends on the plastic neutral axis . The PNA is the axix spilts the cross section such that compression force from the area in compression equal the tension froce from the area in tension .

FRIST DESIGN STUDY

The intersection of the design was taken and section inertia Analysis was done

moment of inertia (max)= 4.8747*10^6*mm^4

moment of inertia(min)=1.40655*10^4*mm^4

Y=882.1/2=441.05mm

S=I/Y

S=1.40655*10^4/441.05=31.89mm^4

HOOD DESIGN OPTIMIZATION

By increasing the depth of the inner panel by 0.5 mm

Here