part 02 safety factor and stress concentration manf240 students

MANF240 Saf. Fac. & Str. Con. 26 February, 2011

Prof. Samy J. Ebeid ١

SJE Part 02 Saf. Fac. & Str. Con. ١

Safety Factor

and

Stress ConcentrationPart 02 – MANF240

SJE Part 02 Saf. Fac. & Str. Con. ٢

The term load can include:

-a transverse force in the case of a beam.

-a longitudinal compressive force in a column.

-a torque in the case of a shaft.

-a pressure in a fluid containment vessel.

-…etc.

Definition of load


Prof. Samy J. Ebeid ٢

SJE Part 02 Saf. Fac. & Str. Con. ٣

There are two completely different natures of loading:= the actual load is the load exerted on the component= the maximum load is the largest load that the component canwithstand without failure.

Actual and Maximum Loading Conditions

The maximum load is a property of the component, afunction of its dimensions and material.

There are many reasons why the actual load on a particularcomponent at a particular time can differ from the nominal actualvalue, due to:=the variability of loading (eg. in practice the mass of a "ten tontruck" will depend on the load it is carrying),=dynamic or shock effects eg. if a weight W is dropped from aheight h onto an elastic component.

SJE Part 02 Saf. Fac. & Str. Con. ٤

A component is safe only if the actual load applied to thecomponent does not exceed the component's maximumsustainable load.

Degree of Safety

The degree of safety is usually expressed by the safetyfactor, n :-n = maximum load / actual load = Fmax /F

if n = 1 then the component is on the point of failure.if n < 1 then the component is in a failed state.if n > 1 then the component is safe.

A higher value of the safety factor seems to

indicate a safer component.


Prof. Samy J. Ebeid ٣

SJE Part 02 Saf. Fac. & Str. Con. ٥

When an assemblage of components is

subjected to a single load, the assembly's

safety factor is the smallest of the

component safety factors.

'a chain is only as strong as itsweakest link'.

SJE Part 02 Saf. Fac. & Str. Con. ٦

The bar's cross-sectional area is A, it is subjected to a

tensile force P and the strength of its material is σ - whichmay be the ultimate if fracture is important, or the yield ifthe material is ductile and excessive deformation isrelevant.

Assuming uniform stress across the cross-section,the maximum load that the component can sustain occurs

when the stress σ reaches the material strength, and isPmax = A . σσσσ

this expression leads to the design equation for directnormal stress:

n P = A . σσσσ

Design equation for a tensile bar


Prof. Samy J. Ebeid ٤

SJE Part 02 Saf. Fac. & Str. Con. ٧

n P = A . σσσσ

The equation may be used either:

1. For analysis to determine the degree of safety

n = A .σσσσ / P for given σ, A, P, or

2. For synthesis (design) to estimate the

dimensions required A => n P/ σσσσ to withstand a

given P with material of strength σ and specified

degree of safety n.

SJE Part 02 Saf. Fac. & Str. Con. ٨

SUGGESTED SAFETY (DESIGN) FACTORS

FOR ELEMENTARY WORK

based on yield strength

CaseFactor

For reliable materials used under controllable conditions and

subjected to loads and stresses that can be determined with

certainty.

1.25 - 1.5 1

For well-known materials under reasonably constant environmental

conditions, subjected to loads and stresses that can be determined.

1.5 - 22

For average materials operated in ordinary environments and

subjected to loads and stresses that can be determined.

2 - 2.5 3

For less tried materials or for brittle materials under average

conditions of environment, load and stress.

2.5 - 3 4

For untried materials used under average conditions of

environment, load and stress.

3 - 4 5


Prof. Samy J. Ebeid ٥

SJE Part 02 Saf. Fac. & Str. Con. ٩

CaseFactor

Used with better-known materials that are to be used in uncertain

environments or subject to uncertain stresses.

3 - 4 6

The factors established in items 1 to 6 are acceptable but must be

applied to the endurance limit (ie. a fatigue

strength) rather than to the yield strength of the material.

Repeated

loads

7

The factors given in items 3 to 6 are acceptable, but an impact

factor (dynamic magnification factor) should be included.

Impact loads8

The ultimate strength is used as the theoretical maximum, the

factors presented in items 1 to 6 should be approximately doubled.

Brittle

materials

9

Where higher factors might appear desirable, a more thorough

analysis of the problem should be undertaken before deciding on

their use.

Higher factors 10

SUGGESTED SAFETY (DESIGN) FACTORS FOR ELEMENTARY WORKbased on yield strength

SJE Part 02 Saf. Fac. & Str. Con. ١٠

Analysis for Estimating the Factor of Safety

1. Find the loads on the component.

2. Find the stress components at the element in

question by the use of free bodies.

(tension, compression, shear, bending and torsion)

3. Resolution of the stress components into principal stresses, either analytically or by Mohr's circle.

4. Implementation of an appropriate failure theory.


Prof. Samy J. Ebeid ٦

SJE Part 02 Saf. Fac. & Str. Con. ١١

Selection of the Factor of Safety

The degree of uncertainty of loading, fl1

The degree of reliability of the material.2

The extent of the accuracy of the force analysis.3

The effect of any stress raisers (stress concentration), fc4

The surrounding environment.5

Risk of danger to human life in case of failure.6

Type of material whether ductile (yield stress) or brittle (fracture stress), fp7

The life of the material (wear conditions), ff8

Conditions of loading: steady, shocks, …., fv9

In case of speeds, inertia and dynamic effects will increase the stresses, fe10

Design stress = yield stress / (fl * fc * fp * ff * fv * fe * …)

SJE Part 02 Saf. Fac. & Str. Con. ١٢


Prof. Samy J. Ebeid ٧

SJE Part 02 Saf. Fac. & Str. Con. ١٣

Stress Concentration

SJE Part 02 Saf. Fac. & Str. Con. ١٤

Case of a Tightened Nut and Bolt

A free body of the nut and bolt demonstrates that theexternal load on them is the force P due to contact overannular areas with the two fastened components.


Prof. Samy J. Ebeid ٨

SJE Part 02 Saf. Fac. & Str. Con. ١٥

The flow analogy is useful when visualizing how

stress is transmitted through a loaded component.

In the analogy, lines of force (or force paths) in the

component are likened to streamlines in a fluid

channel whose shape is similar to that of the

component.

SJE Part 02 Saf. Fac. & Str. Con. ١٦

-Stresses are low where the streamlines are widely spaced.-Stresses are high where the streamlines are gathered together due to geometric shape variations - the more sudden these variations, the higher the local stresses.

This last is known as stress concentration.


Prof. Samy J. Ebeid ٩

SJE Part 02 Saf. Fac. & Str. Con. ١٧

Normal Stresses due to Normal Forces

(Variable Stress Case)

SJE Part 02 Saf. Fac. & Str. Con. ١٨

Stress Concentration

Stress distribution nearthe circular hole in a flatbar under Normal Axialloading.

Stress distribution nearthe fillet in a flat barunder Normal Axialloading.


Prof. Samy J. Ebeid ١٠

SJE Part 02 Saf. Fac. & Str. Con. ١٩

Stress Concentration Factor K

The average stress is

calculated at the critical section i.e. where the small hole is

found. Then the maximum stress is multiplied by

the factor K.

r/D

SJE Part 02 Saf. Fac. & Str. Con. ٢٠

Stress Concentration Factor K

The average stress is

calculated at the critical section

i.e. at d and not at D. Then the

maximum stress is multiplied by

the factor K.


Prof. Samy J. Ebeid ١١

SJE Part 02 Saf. Fac. & Str. Con. ٢١

Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, 40 mm and 60 mm wide, connected by fillets

of radius r = 8 mm.Assume an allowable normal stress of 165 MPa.

Exercise 02-1

SJE Part 02 Saf. Fac. & Str. Con. ٢٢

Exercise 02-1


Prof. Samy J. Ebeid ١٢

SJE Part 02 Saf. Fac. & Str. Con. ٢٣

Exercise 02-2

Knowing that the allowable normal stress = 120 MPa,determine the maximum allowable value of the centric axialload P.

SJE Part 02 Saf. Fac. & Str. Con. ٢٤

Exercise 02-2


Prof. Samy J. Ebeid ١٣

SJE Part 02 Saf. Fac. & Str. Con. ٢٥

Exercise 02-3

Two holes are drilledthrough a steel barthat is subjected to acentric axial load asshown. For P = 32 kN,determine themaximum value of thestress at A and at B.

SJE Part 02 Saf. Fac. & Str. Con. ٢٦

Exercise 02-3


Prof. Samy J. Ebeid ١٤

SJE Part 02 Saf. Fac. & Str. Con. ٢٧

Exercise 02-4

Knowing that P = 40 kN, determine the maximum stress when r = 12 mm

and r = 15 mm.

SJE Part 02 Saf. Fac. & Str. Con. ٢٨

Exercise 02-4


Prof. Samy J. Ebeid ١٥

SJE Part 02 Saf. Fac. & Str. Con. ٢٩

Exercise 02-5

Knowing that, for the plate shown, the

allowable stress is 110 MPa, determine

the maximum allowable value of P when r = 10 mm and

r = 18 mm.

SJE Part 02 Saf. Fac. & Str. Con. ٣٠

Exercise 02-5


Prof. Samy J. Ebeid ١٦

SJE Part 02 Saf. Fac. & Str. Con. ٣١

Exercise 02-6

For P = 35 kN, determine the minimum plate

thickness t required, if the

allowable stress is 125 MPa.

part 02 safety factor and stress concentration manf240 students

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