Stress, strength and safety


Stress concentration

We shall use a tightened nut and bolt, in which the detailed stress variation is extremely complicated, to illustrate some of the simplifying assumptions which are usually made in routine analysis and design.
The leftmost sketch below shows an assembly comprising a nut and bolt which fasten two components together. The nut and bolt pair can be regarded as a sub-assembly or component in its own right. Before any identification of the stresses can be attempted, all external effects on this sub-assembly must be known. A free body of the nut and bolt demonstrates that the external load on them is the force P due to contact over annular areas with the two fastened components.

The flow analogy is useful when visualising 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 - fluid enters and leaves the channel at locations which correspond to the areas where the external loads are applied to the component. [/right]


A complicated streamline pattern infers a complex stress situation.
Stresses are low where the streamlines are widely spaced.
Stresses are high where the streamlines are bunched together due to geometric shape variations - the more sudden these variations, the higher the local stresses. This last is known as stress concentration. Geometric irregularities give rise to non-uniform stresses.
A free body of portion of the bolt which includes the bolt head shows that the external load P is equilibrated by stress σ in the shank. If the free body boundary cuts the shank in the middle then the stress is uniform across the shank cross-section of area A - that is σ = P/A.
But if the boundary lies close to the head or to the thread and nut, then stress concentration occurs and the maximum stress is greater than P/A.

If it is required to calculate the elastic extension, δ, of a bolt whose length L is relativey large, then a first approximation neglects stress concentration to obtain δ = Lε = L(σ/E) = LP/AE = P/k where the stiffness , k, of the bolt is defined as k = AE/L.
If the accent is on bolt safety rather than on deflection, then the core area Ac should be used for stress calculations - this minimum area in the force transmission path is formed by a transverse plane cutting the bolt thread between the plain shank and the nut’s contact face.
The rightmost sketch above employs similar reasoning to illustrate details of the complex transfer of force across the thread into the bolt, together with the high stress concentration in way of the thread root, which may lead to localised yielding.

Static indeterminacy

The external effects on each member of an assembly - the loads and deformations - must be resolved before stresses and strains internal to the member can be examined. If the assembly is statically indeterminate then resolution is carried out by a three-pronged attack :-


Equilibrium of all the members must be assured, so free bodies of the individual members are drawn, complete with known and unknown loads consistent with the physical contact between the various members. The equilibrium equations are written down.
Compatibility considers all the members’ deformations, and the constraints between these deformations which are necessary to preserve the geometric integrity of the assemblage. [/li]This step is carried out by sketching the assembly in the free and in the loaded conditions with member deformations exaggerated for clarity; the compatibility equations which state the integrity constraints are then written down by inspection.
A very common error is not to pay sufficient attention to compatibility.

Definition of the constitutive law for each of the members which undergoes significant deformation enables resolution of the indeterminacy. The constitutive law for each member inter-relates the member’s external load and deformation. If the member is elastic, the constitutive law is based upon :-
COLOR=blue εx = [ σx - ν ( σy + σz ) ] / E + αt ; γxy = τxy / G etc. [/color]
which includes a thermal expansion term due to the temperature rise t.
If a component is loaded uniaxially and uniformly then, like the bolt above, equations (2) degenerate to :-

COLOR=blue δ = P / k + L α t [/color]
where δ is the extension of the component whose length is L and cross-sectional area is A, and on which the tensile load is P; the axial stiffness of the component is k = AE/L.
If the load is compressive or if the deformation is a shortening then the signs of the various terms of ( 2a) will have to be changed by inspection or by the positive tensile convention.
If a component is bent then its stiffness (load/deflection) follows from beam theory


An eyebolt 1 and tubular sleeve 2 are supported by the plate 3 as in (a), (b) below.
Initially the nut is just finger tight ( ‘snubbed’ - there are no gaps in, and no appreciable loads on the assembly).
Subsequently the nut is tightened by screwing it a known distance Δ along the thread, then the temperature of the sleeve only is increased by t, and finally the load P is suspended from the eye.
Given the axial stiffnesses k1 and k2, resolve the indeterminacy in the final loaded state.

The washer and support plate 3 are thin - ie. very short axially compared to the bolt and sleeve - and therefore relatively rigid since k is proportional to 1/L. It follows that the only deformations of significance are those of the bolt 1 and sleeve 2.
The sequence of load application (heat, tightening, external load) is irrelevant in this elastic analysis - what matters is the final loaded state after all three have been applied. The significant deformations (δ1, δ2) in this final loaded state are assumed to be as shown in (b) in which deformations are grossly exaggerated in order to clarify compatibility constraints. Other deformation assumptions may be equally valid - the result should not be affected by the assumption provided any subsequent analysis is consistent with the assumption. More of this anon.
Emphasising this point once more -


In figure (a) all components are completely free, ie. unloaded
In figure (b) loading is complete - initial tightening, heating, and application of external load, P, have all been applied.
By the assumptions of sketch (b) :
δ1 represents the unknown extension of the (great majority of the) eyebolt 1
δ2 represents the unknown extension of the cylindrical sleeve 2, and
Δ is the known nut movement along the thread.
For the components to remain in contact - ie. for compatibility - the geometry of (a) and (b) requires that :-

COLOR=blue δ1 - δ2 = Δ [/color]
Before the components are examined individually, all externals on the assembly must be known. The given external load P applied to the eye is supported by the plate 3, hence a known upward load P appears on the free body (b) of the assembly. The support must also exert a bending moment on the plate to equilibrate the P-couple, but this is neglected here as having no bearing on resolution of the indeterminacy - eg. if the cantilevered plate 3 droops then the other components undergo rigid body motion which does not affect the relative motion between 1 and 2.
Free bodies of the separate components are now prepared - ©, (d), (e) - in which the unknown action/reaction at each contact is identified conveniently by the indices of the two components in contact. No components are stuck together so the action/reactions must be ‘pushes’ on the contacting components - hence F12 is shown in free body © pushing on component 1, and on free body (d) pushing on component 2. This force is the resultant of (presumably uniform) pressure over the annular contact area - the washer’s purpose is to distribute this pressure evenly.
There are three contacts 1-2, 2-3, 3-1 between the three components, hence the three contact resultants F12, F23, F31 appear on the three free bodies. Equilibrium of all three free bodies requires that :-

COLOR=blue F12 - F31 = P
( iii) F12 = F23 [/color]

It is noted particularly from the free bodies ©, (d) that :


F12 (or P + F31 from ( ii)) represents the unknown tension in the (great majority of the) eyebolt 1
F12 (or F23 from ( iii)) represents the unknown compression of the tubular sleeve 2
If it is not clear from free body © that the eyebolt is tensioned by F12 and not by F31, then the flow analogy should be applied or free bodies drawn cutting the shank of © and embracing either end of the bolt.

Constitutive Laws
Equation ( 2a) is applied to each deformable component, ensuring that the signs in the constitutive equations tally with the senses of loads and deformations used in the previous two steps.

Eyebolt 1 From above, the eyebolt is subjected to a tensile load F12 and suffers a tensile deformation δ1, so the constitutive law ( 2a) consistent with compatibility, equilibrium and constant temperature is :-
COLOR=blue δ1 = F12 / k1 ; k1 = ( A E / L )1 [/color]
Sleeve 2 From above, the sleeve is subjected to a compressive load F12 and a temperature rise t, and suffers a tensile deformation δ2, so by inspection or by adhering to a positive tensile convention, the form of ( 2a) consistent with compatibility, equilibrium and temperature rise of this component is :-

COLOR=blue δ2 = - F12 / k2 + L2α2t ; k2 = ( A E / L )2 [/color]
Resolving the indeterminacy by solving the five equations for the three unknown forces and two unknown deformations gives :-
COLOR=blue F12 = Fo where the known initial load is defined as Fo = ke ( Δ + L2α2t )
in which the equivalent series stiffness ke is given by 1/ke = 1/k1 + 1/k2
Fo is called the initial load because it is due to nut tightening and heating before the external load P is applied. [/color]

This result seems ridiculous since it has F12 (the only failure agent in both 1 and 2) completely independent of the applied load P - ie. the load may be raised out-of-hand with no adverse effects on bolt or sleeve !!!

??? What’s wrong with the analysis ???

The channel for the flow analogy is shown at (f) here; the restrictive annular areas are arrowed. Two distinct force paths are suggested in (g) - that of P from the eye directly into the support and byepassing the bolt shank and sleeve, and that of the self-equilibrating initial load Fo in the bolt shank and sleeve. These make physical sense as no external support is necessary to cater for a temperature rise of the sleeve or for a tightening of the nut - these effects, embodied in Fo, are resisted internally in the assemblage. Note that the force paths have been greatly simplified here to avoid confusion in the sketch.
The two forces F12 and F31 are graphed against increasing external load P in (h).
F12 is constant at Fo from ( vi) while from ( ii) F31 drops off from this value in proportion to P, until it becomes zero when P reaches a value of Fo. But F31 is the force between eyebolt and support; it cannot be negative (there is no adhesive to enable the eyebolt to pull down on the support). The physical interpretation of F31 reaching zero is that contact between eyebolt and support is lost as indicated by ( j).
This then is the reason for the apparent anomaly in the foregoing analysis. The analysis is perfectly correct - but only within the limitations of its assumption, implicit in the free bodies © and (e), that contact between 1 and 3 is maintained and F31 > 0.
This assumption should not be confused with the unrelated deformation assumption of (b).

If it is desired to increase the load P at which contact is lost, then (h) indicates that Fo must be raised, probably by further nut tightening, Δ. After contact is lost (P > Fo) the arrangement is statically determinate and component loads can be found immediately from equilibrium only. Setting F31 = 0 and F12 = P in the free bodies ©, (d) and (e) demonstrates that all components transmit the external load P. This is confirmed physically by the force path ( j).

Having found the force F12 transmitted through the bolt eg. the bolt’s safety can be assessed from the tensile design equation ( 1b) applied to the component 1 :
n1 = (AS/P)1 = Ac1S1/F 12 ; F12 = maximum( P, Fo) from (h

Stress resolution

An element’s stress state is essentially three-dimensional, generally with both normal and shear components in each of the three dimensions. The components are usually deduced by superposition of load building blocks as in the stress example of the previous section. The two- and one-dimensional cases illustrated here are particularisations of what is essentially three-dimensional.

The element may be rotated about all three axes into a unique principal orientation in which all shear stresses vanish. The corresponding normal stresses in this principal orientation are termed the principal stresses. A stress state is characterised most succintly by the principal stresses, say ( σ1, σ2, σ3 ), and failure theories - the next step in failure assessment - are expressed in terms of principal stresses. It is therefore necessary to examine how principal stresses are derived from Cartesian components.

Stress is a tensor entity, so complex tensor arithmetic must be applied in the three- dimensional case to evaluate stress components as the element rotates. We consider here only the simpler two dimensional case in which one direction is known to be principal, and resolution consists of element rotation only about that principal axis. Thus in the 2-dimensional sketch above, the z-axis is principal and x-y stress components vary as the element rotates about that axis. Fortunately the great majority of practical cases are two -dimensional since the relatively simple loading in conjunction with the natural choice of axes leads to one of the axes being principal automatically. For example there can be no stress on the free surface of a body, so the surface normal is an obvious choice for one of the three Cartesian axes - as there can be no shear on this surface, the axis is automatically a principal axis.
Although we consider only “two-dimensional stresses”, it is important to remember always that stress states are essentially three dimensional.

In this course the positive convention adopted for the orientation of a plane (characterised by its normal) is shown at (a) below, and for normal and shear stress and strain at (b) - positive shear is counter-clockwise. The two sketches © illustrate the consistency of the positive senses for shear stresses and strains; the total shear strain (distortion) is γ = 2( γ/2).

Using these conventions, the plane stress state at ‘A’ in the example of the previous section is :
σx = -93, σy = 0, τxy = +34, ( τyx = -34) MPa

We now examine the variation of normal and shear stress components as the inclination of the face on which they occur changes.
Consider the elemental unit cube ( size 1x1x1 ) under the known positive 2-dimensional stress components shown in ( i) below - the third dimension (z) is principal.

Rotational equilibrium requires complementary shear, that is τyx must equal -τxy. This necessity has been incorporated into ( ii), from which it is apparent that the three components ( σx, σy, τxy ) together with the third principal are necessary to define the stress state. The three components are called the Cartesian triad.
We wish to evaluate - in terms of the Cartesian triad - the stress components ( σ, τ) in the general direction θ, so we consider force equilibrium of the wedge element ( iii), one of whose faces is inclined at θ. The height of the wedge remains 1 unit, however the dimension in the x-sense becomes 1.tanθ, and the length of the hypotenuse is 1.secθ.
The force components on each face of the wedge are the stress components multiplied by the face area - these are shown in ( iv). For force equilibrium of the wedge :
in the σ-sense :- σ.secθ - ( σx + τxy.tanθ ) cosθ - ( σy.tanθ + τxy ) sinθ = 0
in the τ-sense :- τ.secθ + ( σx + τxy.tanθ ) sinθ - ( σy.tanθ + τxy ) cosθ = 0
which on simplification give the required resolution equations :-
σ = 1/2 ( σx + σy ) + 1/2 ( σx - σy ) cos 2θ + τxy sin 2θ
τ = - 1/2 ( σx - σy ) sin 2θ + τxy cos 2θ

Similar equations but with different signs are encountered in the literature - sign differences arise from positive conventions other than the above.

The simultaneous occurence of sine and cosine terms in these equations makes it difficult to visualise how the resolved components ( σ, τ) vary as the direction θ changes. More easily interpreted equations result if the stress state is defined by the basic triad ( σm, σa, θp ) rather than by the Cartesian triad. θp is the inclination of the plane of the maximum principal stress with respect to the x-reference.
The basic triad is related formally to the Cartesian as follows :-
( 3a) σm = 1/2 ( σx + σy )
σa cos 2θp = 1/2 ( σx - σy ) that is σa = √[ {1/2 ( σx - σy ) }2 + τxy2 ]
σa sin 2θp = τxy θp = 1/2 arctan [ 2 τxy /( σx - σy ) ]
. . . . though these equations seldom need to be implemented.

Making these substitutions leads to resolution equations in the more meaningful form :-
( 4a) σ = σm + σa cos 2 ( θp - θ ) {
nomenclature explanation }
τ = σa sin 2 ( θp - θ )

It is apparent that normal and shear stress components vary sinusoidally with direction θ (not unlike vector components) however the variation is second harmonic - that is stress components are the same along axes which lie at 180o to one another. The two sinusoids are of the same amplitude σa and out of phase by 45o.
σm is the constant component of the normal stress.

The principal and maximum shear stresses follow immediately as :-
( 5) σmax = σm + σa ( at θp )
σmin = σm - σa ( at θp - π/2 )
τmax = σa ( at θp - π/4 )

These relations are often expressed graphically via Mohr’s stress circle, in which σm and σa represent the circle’s centre location and radius respectively. The conventions require that angles on the circle, reckoned from the X-radius, are double the corresponding angles on the element (which are measured from the x-reference), and in the opposite sense. The reader should confirm that this construction satisfies equation ( 4a

EXAMPLE An element is stressed as in Fig A below. All stresses are in MPa.
Sketch the element oriented principally.

Employing the conventions outlined above, one technique for constructing Mohr’s circle is as follows :-
Fig B By the convention, the stress state defined by Cartesian components is σx = -420, σy = -280, τxy = -240 MPa; so the points X ( σx, τxy) and Y ( σy, τyx= -τxy = 240 MPa) are plotted in τ-σ space. Fig C Since the line X-Y is the diameter of the circle, the trigonometry of the circle requires :
circle centre : σm = ( -280 -420)/2 = -350 MPa
circle radius : σa = √( 702 + 2402) = 250 MPa
inclination of X-Y diameter to vertical = arctan( 70/240 ) = 16.2o
Fig D The circle is completed, noting that the two principals from ( 5) are σmax = -100, σmin = -600 MPa. We choose here to define the principal orientation by reckoning σmax at 73.8o clockwise from Y on the circle, so . . . Fig E . . . σmax on the element lies at 73.8/2 = 36.9o anticlockwise from y, which enables completion of the sketch of the principally oriented element. Alternatively the minimum might be reckoned from y, or from x, etc.
The variations of the normal and shear stresses with inclination θ, calculated from ( 4a), are plotted below together with the Mohr’s circles corresponding to the inclinations of principals, of maximum shear etc. The reader should appreciate the trends indicated.
The foregoing should not be construed as being the only way for resolving a two-dimensional stress state, but it lends itself to rapid interpretation.

You are walking in the countryside when you come across a
creek which is too wide to jump. Problem: How can you cross to the far side without a soaking? Applying creativity, you may come up with the following possible solutions, depending upon the circumstances :-


roll up your jeans and wade across
dam the creek
adapt a couple of tent-poles as stilts to walk dry-shod through the creek
bend a sapling, secure it with a rope, climb the sapling, cut the rope and catapult across
construct a bridge from a convenient tree branch;
use a broken branch to pole-vault across
position piles of boulders to use as stepping stones
attach a rope to a tree, lassoo another tree on the far side, and swing across hand-over-hand
build a raft and float across . . . . and so on.
[/li][/ul]These and other solutions, apart from the first, require implements - ropes, trees and the like - whose failure may be through either


fracture, ie. they break, or
excessive deformation eg. they bend or stretch too much.
[/li][/ul]We may examine an implement’s safety (absence of failure) by one of three approaches :-


Try it with care, an approach which is not possible if failure could be hazardous or cause significant financial loss, or
Analyse it via a mathematical model of the implement, from which the stresses and deflections may be calculated for the known loading and the existing implement’s size (dimensions) and material (strength, modulus), thus indicating whether the implement is safe or not; or
Synthesise it, ie. analyse in reverse, where the material is chosen and the minimum dimensions necessary to avoid failure are calculated before the implement is later made to suit.
Analysis and synthesis thus involve four major aspects of the implement :-

An implement in mechanical engineering is more complex than those above, and requires careful design to ensure that everyone who is associated with it is satisfied with it. A well designed artefact is cheap to manufacture, and is easy and safe to use and to maintain, among other things.

Although safety is just one aspect of design, it is a fundamental necessity for all designs
Implement safety can of course be assured by building a physical model and testing it, but this is usually uneconomical and so one of the major aspects of this course is to demonstrate the formation of mathematical models of various mechanical components - bits and pieces such as shafts, brakes, welds, bearings and the like, which are assembled into machines for transforming mechanical power outside the human performance envelope. These models may be analysed to predict the prototypes’ behaviour and safety before they are built, and in conjunction with sketches enable component design to be carried out.
It must be appreciated that the techniques of (mathematical) model building which are introduced in the context of one particular component are usually applicable to many other components which cannot be considered in the course, and therefore the course emphasis is as much on how we arrive at a result as on what that result is.

In this chapter we trace the steps necessary for setting up the mathematical model (the design equation) of any component and derive the equation for one of the most common components in mechanical engineering - the round shaft. Static failure only is considered at this stage, but the same approach is used later in the course for other failure mechanisms such as fatigue and fracture. Safety factor

A component subjected to a solitary load will be considered in the first instance. This load is interpreted in the context of the component’s nature and duty - thus load usually implies a transverse force in the case of a beam component, or a longitudinal compressive force in a column, or a torque in the case of a shaft, or a pressure in a fluid containment vessel, and so on.

There are two completely different manifestations of the load, which have important consequences for the component :


the extrinsic actual load is the load exerted on the component by its surrounds, and
the intrinsic maximum load is the largest load that the component can withstand without failure; the maximum load is a property of the component, a function of its dimensions and material.
Each of these two loads is expressed usually by its nominal value. ( “Nominal” means “in name only”; a nominal value of something is merely a convenient label. For example, 500mm pipes are made with an outside diameter of 508mm and various wall thicknesses - and hence come in various internal diameters - but they are all referred to by the convenient rounded “500” nominal diameter as a descriptor of size. )
There are many reasons why the actual load on a particular component at a particular time will differ from the nominal value, including -


the inherent variability of the load (eg. in practice the mass of a “ten tonne truck” will depend on the load it’s carrying),
static indeterminacy (when components share the load in proportion to their elastic responses - more of this anon),
dynamic (or shock ) effects - for example if a weight W is dropped from a height h onto an elastic component of stiffness k, then the peak force in the component is [/li]Fdyn = dynamic magnification factor (dmf ) * W
Elementary energy methods give dmf = [ 1+ ( 1 +2hk/W)1/2 ] for this type of loading, demonstrating that releasing a load W suddenly even from zero height induces a peak force of at least twice that arising from gradual release - the effective actual load is at least twice its nominal value,
. . . . and so on.

Similarly the maximum load which a particular component can withstand at a particular moment in time will differ from the nominal maximum load due to many factors including -


dimensions differing from their nominal or expected values - eg. we might look at our creek-bridging branch, calling it 50 mm and figuring the nominal maximum load on that basis; but the diameter may in fact be 47 mm, or 54 mm, or more likely vary along the branch;
material strength differing from its nominal value due in turn to variations in material composition or heat treatment, to unsuspected flaws (is there an undetected split or other anomaly in our bridging branch?), [/li]. . . . and so on.

Deviations may be highly significant or they may be insignificant - it’s important that we have an idea of their magnitudes. Sometimes we haven’t much of a clue and therefore have to fear the worst and make suitable allowance for our ignorance.

Clearly a component is safe only if the actual load applied to the component does not exceed the component’s inherent maximum sustainable load. The degree of safety is usually expressed by the safety factor, n :-
COLOR=blue n = maximum load / actual load = Fmax / F [/color]
. . . . and it follows that :

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.
The safety factor is usually expressed as a ratio of nominal loads. A higher value of the safety factor seems to indicate a safer component - however this is not necessarily the case as the inevitable variations must be kept in mind.
In Fig.A, n = 1.25 based on nominal values, but because of the relatively large variations in both actual and maximum loads, there is a significant probability of the actual load exceeding the maximum, and hence of failure.
For a negligible failure probability with these levels of variations, the nominal safety factor must be increased by eg. reducing the actual load applied to the component, as indicated in Fig.B - alternatively the maximum load could be increased, by increasing the material’s strength or the component’s dimensions.
Fig.C illustrates how the probability of failure may be decreased also by reducing the variations of the actual and/or maximum loads. This can be accomplished by a better understanding of what’s going on - we’ll amplify this later.

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 its weakest link’. If a component or an assemblage is subjected to a number of different simultaneous loads, then the concept of a single safety factor may be inappropriate - but nevertheless all potential failure mechanisms must be investigated when deciding whether an implement is safe to use or not.
IF - and only IF - the stress in a component is proportional to the actual load on the component, then the safety factor may be interpreted also as a stress ratio :-

COLOR=blue n = S / σ [/color]where S is the strength of the component’s material, and
σ is the stress in the component due to the actual load, F.

Such an interpretation is evidently inapplicable to assemblages. Stresses are proportional to load for the majority of practical elements, so we usually assume that the stress ratio interpretation ( 1a) applies. But this is NOT the case with some not uncommon failure mechanisms (examined later) such as buckling and Hertzian contact - in these cases we must fall back on the fundamental load interpretation ( 1).
The strength of a material is the maximum stress it can withstand without failure; it is obtained from a tensile test on a specimen of the material. Common metals follow stress-strain relations of the forms illustrated :

The stress in a brittle material, Fig.D, cannot exceed the ultimate strength, Su.
Ductiles also display such an upper limit, Fig.E, but the yield strength, Sy, is often more relevant as it forms a bound above which plasticity and excessive deformation may occur. Most modern ductiles do not possess a distinct yield so in this case an artificial value is usually defined - the offset yield, Fig.F - based upon some maximum acceptable permanent deformation (eg 0.2%) remaining after load release. Yields and ultimates are material properties; representative values for many materials are cited in the literature.

Let us derive a design equation for the simplest of all components - the tensile bar.
The bar’s cross-sectional area is A, it is subjected to a tensile force P and the strength of its material is S - which may be the ultimate if fracture is important, or the yield if the material is ductile and excessive deformation is relevant.
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 is Pmax = A.S - this expression when combined with ( 1) leads to the design equation for direct normal stress, namely :-

COLOR=blue n P = A S [/color]
The design equation embodies conveniently and directly the four aspects mentioned previously - safety, load, dimensions and material (strength). The equation may be used either :


for analysis to determine the degree of safety n = A S /P for given S, A, P, or
for synthesis (design) to ascertain the dimensions required A ≥ n P/S to withstand a given P with material of strength S and specified degree of safety n.
Codes often cite a material’s allowable or design stress, σd = S/n as a characteristic in its own right, rather than stipulating both the strength and the necessary safety factor. The required area would then be A ≥ P/σd.

Equation ( 1b) applies also to compressive loading, however components in compression are usually more likely to fail by buckling than by any strength limitation being exceeded. Buckling (ie. geometric instability) is dictated as much by the component’s overall geometry as by the inherent strength of the component’s material. A cardboard tube can withstand a larger compressive load than a steel wire. The design equation for buckling is derived in a later chapter.
Suitable nominal factors of safety for use in elementary design work are as follows :

[RIGHT]SUGGESTED SAFETY (DESIGN) FACTORS FOR ELEMENTARY WORK based on yield strength - according to Juvinall & Marshek op cit. 1 1.25 - 1.5 for exceptionally reliable materials used under controllable conditions and subjected to loads and stresses that can be determined with certainty - used almost invariably where low weight is a particularly important consideration 2 1.5 - 2 for well-known materials under reasonably constant environmental conditions, subjected to loads and stresses that can be determined readily. 3 2 - 2.5 for average materials operated in ordinary environments and subjected to loads and stresses that can be determined. 4 2.5 - 3 for less tried materials or for brittle materials under average conditions of environment, load and stress. 5 3 - 4 for untried materials used under average conditions of environment, load and stress. 6 3 - 4 should also be used with better-known materials that are to be used in uncertain environments or subject to uncertain stresses. 7 Repeated loads : 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. 8 Impact forces : the factors given in items 3 to 6 are acceptable, but an impact factor (the above dynamic magnification factor ) should be included. 9 Brittle materials : where the ultimate strength is used as the theoretical maximum, the factors presented in items 1 to 6 should be approximately doubled. 10 Where higher factors might appear desirable, a more thorough analysis of the problem should be undertaken before deciding on their use.

The table illustrates clearly the need to increase safety factors when designing with loads either actual or maximum which are not known with certainty. In order to avoid waste of material to cater for ignorance, we try to forecast loads as accurately as possible. For example, rather than use a large design factor to allow for unknown shock effects, we may obtain a realistic dynamic magnification factor from the literature citing other folks’ experience with similar components, and then increase the nominal actual load by this dmf. Statically determinate assemblies are generally preferred to indeterminate because component loads are predictable from simple statics and do not rely on a complex interaction between the components.
For the same reason we employ mathematical models which describe real behaviour as accurately as possible - rather than use simplistic models which are known to be poor predictors. Don’t confuse a model’s descriptive accuracy with its numerical accuracy - a mathematical model might give an answer to 53 decimal places, but it may describe a component’s actual behaviour so simplistically that only one decimal place is significant.

Design factors are increased also when the consequences of failure - economic, social, environmental or political - are serious. For example, a local manufacturer of processing machinery destined for the headwaters of the Yangtze River, doubled the size of every motor predicted by the design process because of the delay in providing replacements if failure should occur due to overload. This is another example of the designer having to foresee the whole future life of an artefact.

The tensile component is the simplest of all machine members. Investigation into the safety of a more complex component involves searching over all elements of the component (ie. over all locations in the component) to find the weakest link - the element with the smallest safety factor. In practice for a given component only a few elements are potentially critical, so an intelligent search is not necessarily protracted. At each element, safety is assessed by means of the following general approach :-


Find the loads on the component, given the loads on the machine or structure (assemblage). If the assemblage is statically determinate then free bodies will give the components’ loads immediately, but if it is statically indeterminate then a three-pronged attack based on equilibrium, compatibility and the constitutive laws of the assemblage’s individual members must be carried out.
Ascertain the stress components (often Cartesian) at the element in question again by the use of free bodies. This is essentially a superposition of load building blocks such as tension (or compression), shear, bending and torsion - though others will be encountered. If the member is itself indeterminate then the three-pronged attack mentioned above must be applied to elements within the member - it will be recalled that the elementary equations for bending and torsional stresses were deduced in this manner based on an assumed deformation geometry.
Resolution of the stress components into principal stresses, using either analytic techniques or Mohr’s circles. We shall not consider tensor resolution of the general 3-dimensional state - it is somewhat complex and seldom encountered in routine design.
Implementation of an appropriate failure theory whereby the biaxial or triaxial stresses at the element may be correlated with material strengths, which are derived from uniaxial tests, to evaluate the safety factor.
[/li][/ol]These four steps will be examined individually in the following sections, and will be examplified in setting up the design equation for a very common component - a shaft of circular cross-section subjected to simultaneous bending and torsion.

But before embarking on this and writing down stress equations such as σ = P/A, we’ve got to appreciate the limitations of such mathematical models - we must be able to visualise qualitatively how stresses vary throughout a body and to appreciate the concept of stress concentration.
For example consider equation ( 1b) applied to the assembly comprising two short coaxial bars, 1 and 2, of cross-sectional areas 100 & 200 and strengths 600 & 400 MPa respectively ( recall that 1 MPa is equivalent to 1 N/
The assembly is compressed by 50 kN distributed uniformly over each end.