Section Modulus calculation and optimization

SECTION MODULUS CALCULATION AND OPTIMIZATION

AIM:

• To increase the section modulus of a given hood model by necessary changes and then optimizing the hood panel for better section modulus.

TOOL USED:

• NX CAD.

INTRODUCTION:

• Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. Other geometric properties used in the design include an area for tension and shear, the radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness. Any relationship between these properties is highly dependent on the shape in question.
• Equation for the section moduli of common shapes is given below.

Z=I/Y

                        Z = section modulus

                        I = area moment of inertia

                        Y = distance from neutral axis to the outer most fibre of a material

 CASE 1 – BEFORE OPTIMIZATION:

            

CALCULATION:

Max MOI = 8.236793185e + 05[mm⁴]

Min MOI = 6.183850002e + 03[mm⁴]

            H = 440.2, y = H/2 – 220.1

Section modulus S = I / Y

For first analysis of max MOI S = I max/Y

                                                            =8236. 793185 / 220.1

                                                            =37.422 958570 mm³

Mini MOI   S = I mini / Y

                                                            = 6183.850002 / 220.1

                                                            = 28.0956383552 mm³

^{CASE 2 – AFTER OPTIMIZATION:}

 

 

 

Now, one more trail is that, here the inner panel curve offsetting outside by

  0.5 mm i.e., increasing in cross sectional area, the results shown below.                                     

CALCULATION:

            Max MOI = 8.214056116e + 05 [ mm⁴]

            Min I MOI = 6.398172995e + 03 {mm⁴]

H = 440.2, Y = H/2 – 220.1

Section modulus S = I / Y

For first analysis of max MOI S = I max/Y

                                                            =8214.056116 / 220.1

                                                            =37.3196mm³

 

Mini MOS   S = I mini / y

                                                            = 6398.172995 /220.1

                                                            = 29.66939 mm³

conclusion:

• Hence when comparing case 1 & case 2,
• The section modulus value of case 2 is higher than the case 1
• The strength of the material is directly proportional to the moment of inertia, hence increase in sectional area there is have increase in overall mass of an object, hence have increase in strength.