Momen inertia dan Stress

  • #1

Homework Statement

Density : 7833,000 kg/m^3
Coef. of Thermal Exp. : 0,0000 /C
Thermal Conductivity : 0,032 kW/m-C
Specific Heat : 481,000 J/kg-C
Modulus of Elasticity : 199947953,000 kPa
Poisson's Ratio : 0,290
Yield Stress : 262000,766 kPa
Ultimate Stress : 358527,364 kPa
Elongation % : 0,000
Mass :1879,920 kg
Volume :240000000,000 mm^3
Area : 3040000,00 mm^2
Center of Mass = Center of Volume = 1000 mm (Y direction)
Mass Moments of Inertia :
Ixx = 2531,626 kg-m^2
Iyy = 39,165 kg-m^2
Izz = 2520,659 kg-m^2

Principal Moments of Inertia :
I1 = 651,706 kg-m^2
I2 = 640,739 kg-m^2
I3 = 39,165 kg-m^2

Radii of Giration :
K1 = 588,78 mm
K2 = 583,81 mm
K3 = 144,34 mm

Tekhnical Drawing : Rectangular Beam (0,3 m x 0,4 m x 2 m)


Homework Equations

Stress = MC/I

The Attempt at a Solution

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  • #2
If I assume the objective of the question is to find the maximum stress acting on the cantilever beam shown in the figure, the proposed relevant equation is valid.
Would you try to prune the irrelevant mass of data supplied and substitute appropriate values for M, C and I shown in your formula?
  • #3
The question is:
How much load must be put, so the beam is not broken??

For Tim-Tim :
The question is all about the load. Sorry that I have forgeted put it. :biggrin:

For mathmate:
Can you explained more specific about your suggestion Mathmate?? Because I still not understand.

Btw guys,
I have tried it guys. You can see in here :
M = F*L
C = 400/2 = 200 mm
I = [tex]\int[/tex]y^2 dA
= [tex]\frac{bh^3}{12}[/tex]

Some data Input :
b = 300 mm
h = 400 mm
L = 2000 mm
A = 120000 mm
Su = 358527,364 kPa

And put the data to the equation, I get the result.

The load : F = 1.4341.10^6 N

But I still not statisfied, because I found an artickel from Mechanical Engineering Handbook that explained the same equations about Mass moment of inertia and Area moment of Inertia. See attachment.

Sorry guys to makes you busy about my question because now I have 2 question :
1. Mass moment of Inertia
2. Maximum load

Any other suggestion guys??


  • Moment of Inertia.pdf
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  • #4
You have correctly interpreted the geometry of the cross section and calculated some parameters, such as I=bh^3/12, c=200 mm. You only need to identify correctly the maximum bending moment to be used to substitute into the equation of

stress = Mc/I

A. Loading

The permissible stress at a critical section is obtained by the formula you posted, Mc/I,
M=bending moment
c=distance of the extreme beam fibre from the neutral axis
I=second moment of area as you have obtained, bH^3/12.

The critical section for a beam of uniform cross section is where the bending moment is the greatest. You will need to find the bending moment with respect to the value of the force F and the distance from the applied force F, from which you will find the critical section.
It is available in most strength of materials books or web pages.

The neutral axis for a homogeneous rectangular section is at mid-depth.

B. Definitions of 'maximum' load
The 'maximum load' you can put on the beam is either the maximum value of F that one can put on the beam without the beam breaking, or the maximum load that will be permitted to be put on the beam.
There are three cases possible:
1. Ultimate load
When the beam actually breaks, this is called the ultimate load, which correspond to a certain section of the beam subject to the ultimate stress. This load is usually for experimental purposes, because at this point the beam will be subject to excessive deflection and the structure will not be serviceable at all. The stress corresponds to the ultimate stress.
2. Maximum elastic load
This load will cause the critical section of the beam to reach or stay just below the yield stress, in which case when the load is removed, it will 'bounce' back to the original state. Any load above this value will cause permanent plastic deformation to occur, even if the load is removed. The stress corresponds to the yield stress.
3. Working load.
This is frequently obtained by applying a safety factor to the maximum elastic load to reduce the maximum stress to a lower value than the yield point. The factor of safety depends on the country and the applicable building code. The stress corresponds to the yield stress divided by the factor of safety.

You will have to tell us which is the 'maximum' load for which you are looking. Once this is determined, it will be relatively easy to apply the equations and find the required load F.
  • #5
ricky_fusion: Your answer is correct. You computed a conservative estimate of the cantilever ultimate load (F = 1434 kN). You correctly computed and used the area moment of inertia for this calculation. Nice work.
  • #6
Thanxs guys, for your information.
But I still not statisfied about Mass moment of Inertia.

I will tried it first maybe there is some explanation about it.
  • #7
For Mathmate :
Thanxs for your lesson.

This is about the maximum elastic load.
Because, I want to make the max capacity load that can handle by the beam.

From your explanation about "load", is there a connection to mass moment of inertia??
  • #8
ricky_fusion: To compute the maximum elastic load, perform the same calculation you computed previously, except change Stu = 358.53 MPa to Sty = 262 MPa. Mass moment of inertia is not used in this problem; only area moment of inertia is used. Mass moment of inertia is applicable only in dynamic problems, when a component is accelerating.
  • #9
Mass moment of inertia is the mathematical quantity
[tex]\int[/tex]x2 dm
and is use as a measure of the resistance of a solid body to rotation.
For more information, google the term or try:

Second moment of area is a property of a cross section of a beam (I-beam, concrete beam, etc.) that is required for the calculation of rotations and deflections of beams, such as the value of I in the formula
stress([tex]\sigma[/tex]) = My/I,
or deflection = [tex]\int[/tex][tex]\int[/tex]M/EI dx dx
In civil engineering, the term moment of inertia is often (erroneously) used to mean second moment of area, hence the confusion.
It is obtained by the formulae
[tex]\int[/tex](y2 dA)
Google the term or see:

Back to the calculations, as you have correctly worked out
Let Mu=M(ultimate)
Sigma(Ultimate)=Mu . c /I
gives F(ultimate) = 1434

It would therefore be relatively easy for you to re-calculate F(yield) if that is what you need.
  • #10

I see...So the mass moment of inertia will effect in dynamic analysis. So, my conclusion from your statements that it's more to rotation effect and have relationship with acceleration.

Btw, I have other question :
To calculated elastic load, I should changed the Stu -> Sty. So the formula will be :

Stu = 0.5 Sty

The question are :
1. Why ?
2. Any reference book??
3. Is it acceptable to appy this formula to other kind of material?? Such as, Hyperelastic, Orthotrophic or Fluid?? I only have info that this formula acceptable to Calculated Stu from Sty. But for detail info, nope.
  • #11
No, Sty = 262 MPa and Stu = 358,53 MPa are given in post 1. There is no general formula relating them.
  • #12

I see.. Thnxs a lot nvn.. I thinks my problem have solved now. Thanxs a lot to you all guys..

Best Regards,



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