Section Modulus calculation and optimization

Section modulus is the property of the cross-section that is used while the design of beams.

The section modulus is classified into two types:-

1] Elastic section modulus:- The elastic section modulus is applicable up to the yield point of material.

It is used in most engineering applications.

2] Plastic section modulus:- The plastic section modulus is generally used for material in which plastic behavior is observed.

Elastic section modulus is the ratio of moment of inertia about a neutral axis

Where,

I = Moment of inertia (m4)

Y = Distance between neutral axes and outermost fiber (m)

Steps to calculate elastic section modulus:-

• Step 1:- Find location of centroid of given cross section.
• Step 2:- Find moment of inertia about neutral axis (INA)
• Step 3:- Find  Ymax from the neutral axis.
• Step 4:- Find elastic section modulus using the formula provided above.

The plastic section modulus is necessary to calculate plastic moment capacity of the section.

The plastic section modulus is the addition of moment of area of area under compression about PNA (plastic neutral axis) and the moment of area of area under tensile stress.

Plastic section modulus equation:–

For the above cross-section, the plastic section modulus is given by,

ZP=∑ACYC+∑ATYT

Where,

AC = Area of cross-section under compression.

AT = Area of cross-section under tension.

YC = Distance from PNA to CG of the areas under compression.

YT = Distance from PND to CG of areas under tension.

Steps to calculate plastic section modulus:–

Step 1] Locate the plastic neutral axis.

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

AC = AT

By using this relation, we can find the position of the plastic neutral axis.

Step 2] Find the summation of the product of area of section undergoing compression and the distance of its CG from PNA that is 

∑ACYC

.

Step 3] Find the summation of the product of the area of cross-section undergoing tension and distance between CG of that area from PNA that is ∑ATYT

.

Step 4] Find the plastic section modulus using the formula.

The above figure shows the section modulus of the Hood which is having 

MOI(Max) : 4.866585070e+06(mm^4)

MOI(Min) : 1.546685958e+04(mm^4)

Section Area : 73.432575325(mm^2)

the distance between NA to extreme fiber is  = 876/2  = 438mm

Section modulus =    Z=  MOI/ Y  

Zmax= (4.8665850*e^6)/ 438 =    11110.92mm^3     

Zmin = (1.54668595958*e^4)/ 438  =  35.312mm^3

Hood Optimization by offsetting inner panel to 0.5mm downwards

MOI(Max) : 4.858479640e+06(mm^4)

MOI(Min) : 1.565342238e+04(mm^4)

Section Area : 73.387096353(mm^2)

the distance between NA to extreme fiber is 

Section modulus =    Z=  MOI/ Y  

Zmax= (4.858479640*e^6)/  438 =  11092.419 mm^3     

Zmin = (1.565342238*e^4)/ 438  =  35.738 mm^3

Conclusion 

compariosn of both the section modulus clearly indicates that increasing the depth by 0.5mm can increase the section modulus. the section modulus is directly depnds on the area momemnt of inertia.