Section Modulus calculation and optimization

OBJECTIVE: SECTION MODULUS CALCULATION AND OPTIMIZATION

SECTION MODULUS:

Section modulus is geometric property used to calculate the strength of beam or given cross sectional area and it has significant importance in designing beams.

It is defined as the moment of inertia (I) of cross sectional area of the beam about neutral axis to the distance (y_max) of the extreme fibre from the neutral axis.

Section modulus is denoted by the letter “Z”.

So, Z = I / y_max.

Beams having larger section modulus than another will be stronger and capable of supporting greater loads.

Moment of Inertia (I):

Moment of inertia is a measure of how resistant an object to changes in its rotational motion.

Moment of inertia of the cross section of a body is its resistance to changes in its rotation.

Calculations of Section Modulus (Z):

Formula: Z = I / y_max

For symmetrical sections the value of the Z is the same above or below the centroid.

For asymmetrical sections the value of the Z is different that is Z_max & Z_min.

Section Modulus for Rectangular shape:

Z_xx = I_xx / y where I_xx = bd³/ 12 mm³

And y = d/2, where d is depth.

So, Z = bd³/ 6 mm³

Centre of Area (Centroid):

The centroid of a section in structural design is like centre of the gravity of the shape. Structural shape is made up of more than one structural section. In these cases a calculation is needed to calculate sentroid.

Below picture shows formula to calculate section modulus of different shapes.

Here, we are going to calculate section modulus of Hood for 0Y plane in NX Software.

Case I : Before Optimization

MOI_max = 1.15489 15762e+07(mm⁴)

MOI_min = 1.150668948e+05(mm⁴)

SO, Section Modulus S,

S = I / Y

I = MOI_min value is considered. = 1.150668948e+05(mm⁴)

Y = (Total Distance) / 2; Y = 487.4051/2 = 243.70255 mm

So,

S = 1.150668948e+05 / 243.70255

S = 472.16161 mm³

Case II : After Optimization

In this case we have offset the section curve by 0.5 mm outwards and section analysis is done for closed curve.

MOI_max = 1.563006281e+07(mm⁴)

MOI_min = 1.230723419e+05(mm⁴)

SO, Section Modulus S,

S = I / Y

I = MOI_min value is considered. = 1.230723419e+05(mm⁴)

Y = (Total Distance) / 2; Y = 488.4025/2 = 244.20125 mm

So,

S = 1.230723419+05 / 244.20125

S = 503.97916 mm³

Now, we are going to calculate section modulus of Hood from offset plane from 0X plane at a distance of 950mm in NX Software.

Case I : Before Optimization

MOI_max = 9.371867034e+07(mm⁴)

MOI_min = 2.792730687e+05(mm⁴)

SO, Section Modulus S,

S = I / Y

I = MOI_min value is considered. = 2.792730687e+05(mm⁴)

Y = (Total Distance) / 2; Y = 886.7439/2 = 443.37195 mm

So,

S = 2.792730687e+05 / 443.37195

S = 629.88439 mm³

Case II : After Optimization

In this case we have offset the section curve by 0.1 mm outwards and section analysis is done for closed curve.

MOI_max = 9.380876454e+07(mm⁴)

MOI_min = 2.810683724e+05(mm⁴)

SO, Section Modulus S,

S = I / Y

I = MOI_min value is considered. = 2.810683724e +05(mm⁴)

Y = (Total Distance) / 2; Y = 886.9420 / 2 = 443.471 mm

So,

S = 2.810683724e +05 / 443.471

S = 633.7920 mm³

Conclusion:

From above both cases of 0Y and offset of 0X section we have concluded that,

Section modulus value of case II (after optimization) is higher than Case I.

Hence, as we increase the cross sectional area, moment of inertia and section modulus is increases, means strength of the component increases.

Thus, section modulus for Hood is calculated and optimization also done to increase the strength of the Hood.