Section Modulus calculation and optimization

section modulus;

section modulus is also referred to as the polar modulus or the torsional sectional modulus.it is denoted by Zp.section modulus is the geometric property of the of the given cross-section of any shaft or beam structures.it is the direct measure of the strength or the beam section.higer the section modulus of a structure,the more resistive it becomes the bending.

• section is defined as the ratio of polar moment of interia to the raius of shaft or the distance from the netural axis to the outer fibers.

Zp = IP/R

polar moment of interia ;

polar moment of interia mesures the resistance to torsion.polar moment of interia is also called the second polar moment of area .it is usally denoted by Lz.however sometimes j or Jz is also used.polar moment of interia can be represented by mathematically with the given by,

I = r^2DA.

location of netural axis;

for axi-symmetric,istropic sections the netural axis is located at centroideal axis.for curved sections teh neutrial axis is located at the point where tensile stresses or equal to the compressive stresses.

setion modulus of hood;

section modulus to the refers to the strength of the hood against loading conditions.for hood we obtain moment of interia of the desgin in sofware and we calculate section modulus in order to optimize the strentgh of the hood.

formula '

S = I/Y

where i = area of moment of interia

y = distance from neutrial axis to extreme layer .

case 1

finally i got my moment of interia (max&min)valves of hood part

moi(max) = (1.7714e-006 * 10^05)[mm^4]

moi(min) = (6.345e-004 *10^03)[mm^4]

below picture refers to theditance between hood from one end to extreme end.

the total distance of the hood y = (881.926)

distance between netural axis to extreme rod = (881.926/2) = 440.963

for as symmentric has two values for the top and bottom of the centroid ie,, s (max)and s(min)for s(max)and s(min)

s(max) = i(max)/y = (1.7714e06*10^05)/(440.963)=(401.698)[mm^3]

case 2

in this case 2 we are offseting hood inner panel about 0.5 downwards as shown in this picture

finally i got my moment of interia (max*min) values of hood part

moi(max) = (1.86144e-06*10^05)[mm^4]

moi(min) = (6.636e-004*10^03)[mm^4]

the total distance of the hood y = (881.926)

distance between the netural axis to the extreme rod = (881.926/2) = (440.963)

for as symmentric has two valus for the top and bottom of the centroid ie,,s(max)ands(min) for s(max)and s(min)

section modulus is the ratio os second moment of area and distance from N.A(netural axis)to the extremeend given by s= i/y

s(max) = I(max)/y = (1.86144e-06*10^05)/(440.963) = (422.1170)[mm^3]

s(min) = I(min)/y = (6.636e-004*10^03)/(440.963) = (15.03980[mm^3]

we considering the section modulus minimum value only

after calculating we got the values os case 1 and case 2 are

case 1; s(min) = (14.37980[mm^3]

case 2; s(min) = (15.0398)[mm^3]

• when comparing this two hood values.as you can see there is increasing in the values in both s max and s min
• when we see the values of s(max)there is slightly increase from (401.698)[mm^3]to (422.1170)[mm^3]
• hence comparing this two cases,we can say that when there is increase in cross sectional area similarlly there will be increase in strength.
• increase in cross sectional area by 0.5mm increased the minimum load carrying capacity or strength of the hood

conclusion;

thus i caluculated the section modulus of hood desgin and also there is improved in section modulus of the previous section finally i concluded that there is inrease in section area it increses the ovrall mass of an object and also strength of an object.