Section Modulus calculation and optimization
SECTION MODULUS CALCULATION AND OPTIMIZATION
Objective: To determine the section modulus calculation of hood by using NX12
Section modulus:
Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. The section modulus of the cross-sectional shape is of significant importance in designing beams. It is the direct measure of strength in the beam, the larger the section modulus greater the load-bearing capacity, and the higher the resistance to bending. From the above condition, the section modulus is calculated by obtaining the Moment of Inertia in NX using Section Inertia Analysis
section modulus s=I\Y
S = Section Modulus
I = Moment of Inertia
y = Distance between the neutral axis and the extreme end of the object
case 1
Import the hood panel creates a datum plane on the entire assembly of the hood and makes intersection curve (select both inner and outer panel of the hood ) create the sketch on datum plane trim the inside sketch lines. Draw the lines where gaps in a curve as shown in fig. Go to analyze and select the section inertia calculate the sections modulus
From the above analysis following minimums' and maximum moment of inertia are obtained
maximum moment of inertia Imix=4.8832e+06(mm^4)
minimum moment of inertia Imin=1.5126e+04(mm^4)
section modulus s=I\y
Y=885.2\2=442.6
smax=4.8832e+06\442.6
smax=11.0328*10^3mm^3
smin=34.17mm^3
case 2:improved section
In the second case increase the total hood panel thickness to 1mm for examination increase the strength comparing case1. Again creates a datum plane on the entire assembly of the hood and makes an intersection curve (select both inner and outer panel of the hood ) create the sketch on the datum plane trim the inside sketch lines. Draw the lines where gaps in a curve as shown in fig. Go to analyze and select the section inertia calculate the sections modulus
0
From the above analysis following minimums' and maximum moment of inertia are obtained
maximum moment of inertia Imix=4.the ihood 30322+06(mm^4)
minimum moment of inertia Imin=1.6679e+04(mm^4)
section modulus s=I\y
Y=885.2\2=442.6
smax=4.930322+06\442.6
smax=11.288*10^3mm^3
smin=37.70mm^3
conclusions:The strengths of the material is directly proportional to the moment of inertia.as we increase sectional area,the moment of inertia & section modulus increased