This article is explanation of videos, stored at here.

When students start to learn strength of materials he is getting

information like :

This is very simplified approach, and maybe enough to know for the overall development, but not enough to solve practical problems.

Basically there are many different types of limit states and the above formula is not a criteria of failure, it is just general formula for start of yielding (when stress at some point comes to yield surface.

For example, stress can be lower than yield strength to get fatigue failure after many cycles and it can be easily much more than yield strength to resist sufficient amount of cycles for your problem.

In practical applications people use code requirements. The history of technical codes has more than 100 years with huge experience of service under the different loads and conditions.

Let's take ASME BTH-1–2005 "Design of Below-the-hook lifting devices"

It uses allowable stress method, but it says:

Actually this code (as many other national codes) uses common formulas of beam theory, like P/A+M/Sx

what means that it operates with

Let's solve simple example to show the difference between fiber stress and peak stress ( https://yadi.sk/d/paoQtcq133zorf )

I-beam (b=h=6'', tw=tf=0.5''), fixed at one end and loaded with concentrated moment M=250000 lb*inch at other end.

Moment of inertia: Ix=2*6*0.5^3/12+2*6*0.5*2.75^2+0.5*5^3/12=50.7 inch^4

Section modulus: Sx=50.7/(6/2)=16.9 inch^3

Fiber bending stress =M/Sx=250000/16.9=14800 psi

Fig.1 - I-beam with hole

Fig.2 - Section properties

Fig 3 - Distribution of fiber stress (von Mises)

Fig 4- Peak stress at stress concentration point

Let's explain all values:

It is what ASME BTH-1–2005 "Design of Below-the-hook lifting devices" says about that

When students start to learn strength of materials he is getting

information like :

**Max. equivalent stress <= [Limit stress]**This is very simplified approach, and maybe enough to know for the overall development, but not enough to solve practical problems.

Basically there are many different types of limit states and the above formula is not a criteria of failure, it is just general formula for start of yielding (when stress at some point comes to yield surface.

For example, stress can be lower than yield strength to get fatigue failure after many cycles and it can be easily much more than yield strength to resist sufficient amount of cycles for your problem.

In practical applications people use code requirements. The history of technical codes has more than 100 years with huge experience of service under the different loads and conditions.

Let's take ASME BTH-1–2005 "Design of Below-the-hook lifting devices"

It uses allowable stress method, but it says:

*The allowable stresses and stress*

ranges defined in paras. <...>, and <...> are to be com-

pared to average or nominal calculated stresses due to

the loads defined in para. <...>. It is not intended that

highly localized peak stresses that may be determined

by computer-aided methods of analysis, and which may

be blunted by confined yielding, must be less than the

specified allowable stresses.ranges defined in paras. <...>, and <...> are to be com-

pared to average or nominal calculated stresses due to

the loads defined in para. <...>. It is not intended that

highly localized peak stresses that may be determined

by computer-aided methods of analysis, and which may

be blunted by confined yielding, must be less than the

specified allowable stresses.

Actually this code (as many other national codes) uses common formulas of beam theory, like P/A+M/Sx

what means that it operates with

**averaged fiber stress**, not with peak values.Let's solve simple example to show the difference between fiber stress and peak stress ( https://yadi.sk/d/paoQtcq133zorf )

I-beam (b=h=6'', tw=tf=0.5''), fixed at one end and loaded with concentrated moment M=250000 lb*inch at other end.

Moment of inertia: Ix=2*6*0.5^3/12+2*6*0.5*2.75^2+0.5*5^3/12=50.7 inch^4

Section modulus: Sx=50.7/(6/2)=16.9 inch^3

Fiber bending stress =M/Sx=250000/16.9=14800 psi

Fig.1 - I-beam with hole

Fig.2 - Section properties

Fig 3 - Distribution of fiber stress (von Mises)

Fig 4- Peak stress at stress concentration point

Let's explain all values:

*1) Fiber stress is subject of technical codes, based on allowable stress design**It should be less than yield strength in most of application (the exception is LRFD - load resistance factor design, where M/Sx>yield stress**LRFD is not applicable to structures with cyclic load**Allowable stress can be less, than yield stress (for example for elevators it is equal to 14 ksi for normal load and 27 ksi for emergency stop)**In CalculiX you can investigate fiber stress using 'max' card in CGX**2) Peak stress at stress concentration point can be much more than yield stress in many applications. It is subject for fatigue analysis, based on total deformations (elastic+plastic)**3) Situation when peak stress = yield stress don't make sense practically. It doesn't provide protection against fatigue because yield stress > fatigue threshold. Peak stress can't be used for allowable stress design, because usually it gives huge unnecessarily safety margin for most of practical applications and it doesn't provide complete safety in general. It is mostly some limit that make sense in theory (start of yielding at some point)**Peak stress is infinite in many FEA models due to singularity (solution with plastic card is not helpful in this case, because stress and strain stay infinite).**It is subject of Fracture mechanics and it uses other values, not stress or strains but sort of averaged stress)**In this case hydraulic analogy can be useful for understanding. it doesn't matter what is peak maximum of velocity, but flow rate is result of integration over the stream).*It is what ASME BTH-1–2005 "Design of Below-the-hook lifting devices" says about that

*"While the use of such methods is not prohibited, modeling of the device and interpretation of the results demands suitable expertise to assure the*

requirements of this standard are met without creating unnecessarily conservative limits for static strength and fatigue life."requirements of this standard are met without creating unnecessarily conservative limits for static strength and fatigue life."

*Many commercial programs, like ANSYS, Solidworks Simulation, and etc, use this approach when calculate "Safety Factor" but it is not perfect and doesn't provide accuracy (as people think when they buy expensive software)**The great benefit of using FEA, is getting accurate numerical solution of stress distribution.**Elasticity Theory proves the**theorem*of Existence and*Uniqueness*of*Solution, that means all (properly tested) FEA programs should give the same results (which depend on boundary conditions, applied by user, under the user's responsibility). But knowing correct stress distribution don't give answer for things like safety and strength.**Technical codes is right source for these issues.*

## No comments:

## Post a Comment