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THE STRESSSTRAIN DIAGRAM

The strength of a material is not the only criterion that must be considered in designing structures The stiffness of a

material is frequently of equal importance To a lesser degree mechanical properties such as hardness toughness and ductility determine the selection of the material

THE STRESSSTRAIN DIAGRAM STRAIN

To obtain the unit deformation or strain we divide theelongation by the length L in which it was measuredthereby obtaining

L

=

The strain so computed however measures only theaverage value of strain The correct expression for strain atany position is

dL

d =

where d is the differential elongation of the differentiallength dL

STRAIN

Under certain conditions the strain may be assumedconstant and these conditions are as follows

The specimen must be of constant cross section! The material must be homogeneous" The load must be axial that is produce uniform stress

Since strain represents a change in length divided by theoriginal length strain is a dimensionless quantity

PROPORTIONAL LIMIT

From the origin O to a point called the proportional limit thefigure shows the stress strain diagram to be a straight line

From this we deduce the well known relation first postulatedby Robert Hooke in )*+ that stress is proportional to strainNotice carefully that this proportionality does not extendthroughout the diagram- it ends at the proportional limit

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THE STRESSSTRAIN DIAGRAM PROPORTIONAL LIMIT

Other concepts developed from the stress strain diagramcurve are the following

The elastic limit is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set

! The yield point is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load

" The yield strength is closely associated with the yield point

PROPORTIONAL LIMIT

Other concepts developed from the stress strain diagramcurve are the following

. The ultimate stress or ultimate strength as it is more commonly called is the highest ordinate on the stress strain curve

/ The rupture strength is the stress at failure

THE STRESSSTRAIN DIAGRAM

WORKING STRESS AND FACTOR OF SAFETY

The working stress also called the allowable stress is themaximum safe stress a material can carry In design the

working stress w should be limited to values not exceedingthe proportional limit so as not to invalidate the stress strainrelation of Hookes law on which all subsequent theory isbased However since the proportional limit is difficult to

determine accurately it is customary to base the workingstress on either the yield point or the ultimate strengthdivided by a suitable number N called the factor of safety

or

ypN

ypw

=

ultNult

w

=

HOOKES LAW( AXIAL AND SHEARING DEFORMATIONS

The slope of the straight line portion of the stress straindiagram is the ratio of stress to strain It is called the

modulus of elasticity and is denoted by E

=ESlope of stress strain curve 3

which is usually written in the form

E=In this form it is known as Hookes law Originally Hookes law specified merely that stress was proportional to strain but Thomas Young in +5* introduced a constant of proportionality that came to be known as Youngs modulus

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HOOKES LAW( AXIAL AND SHEARING DEFORMATIONS

The slope of the straight line portion of the stress straindiagram is the ratio of stress to strain It is called the

modulus of elasticity and is denoted by E

=ESlope of stress strain curve 3

which is usually written in the form

E=Eventually this name was superseded by the phrase modulus of elasticity

HOOKES LAW( AXIAL AND SHEARING DEFORMATIONS

The units for modulus of elasticity E are identical to the units

for stress 6 recall that strain is a dimensionless quantityAs an illustration the modulus of elasticity for steel in SI isapproximately !55x58 N9m! :!55x58 Pa

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Compute the total elongation caused by an axial load of 55 kN applied to a flat bar !5mm thick tapering from a width of !5mm to .5mm in a length of 5 m as shown below Assume E 3 !55 GPa

A steel wire "5 ft long hanging vertically supports a load of /55lb Neglecting the weight of the wire determine the required diameter if the stress is not to exceed !5 ksi and the total elongation is not to exceed 5!5 in Assume that

modulus of elasticity E 3 !8x5) psi

An aluminium bar having a cross sectional area of 5/ in!

carries the axial loads applied at the position shown Compute the total change in length of the bar if E 3 5x5)

psi Assume the bar is suitably braced to prevent lateral

buckling

The rigid bar ABC shown is hinged at A and supported by a steel rod at B Determine the largest load P that can be applied at C if the stress in the steel rod is limited to "5 ksi and the vertical movement of end C must not exceed 55

in

The rigid bar AB attached to two vertical rods is horizontal before the load P is applied Determine the vertical movement of P if its magnitude is /5 kN

STATICALLY INDETERMINATE MEMBERS

There are certain combinations of axially loaded members inwhich equations of static equilibrium are not sufficient for a

solution This condition exists in structures where thereactive forces or the internal resisting forces over a crosssection exceed the number of independent equations ofequilibrium Such cases are called statically indeterminate

and require the use of additional relations that depend on theelastic deformations in the member

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STATICALLY INDETERMINATE MEMBERS

The cases are so varied that they can best be described bysample problems illustrating the following general principles

To a free body diagram of the structure or a part of itapply the equations of static equilibrium

! If there are more unknowns than independent equationsof equilibrium obtain additional equations from the

geometric relations between the elastic deformationsproduced by the loads To define these relations clearlyyou will find it helpful to draw a sketch that exaggerates

the magnitudes of the elastic deformations

The short concrete post in the figure below is reinforced axially with six symmetrically placed steel bars each )55 mm! in area If the applied load P is 555 kN compute the stress developed in each material Use the following moduli

of elasticity for steel Es 3 !55 GPa- for concrete Ec 3 . GPa

In the preceding problem assume the allowable stresses to

be s 3 !5 MPa and c 3 ) MPa Compute the maximum safe axial load P that may be applied

A copper rod is inserted into a hollow aluminum cylinder The copper rod projects 555/ in as shown What maximum load P may be applied to the bearing plateF Use the data in the following table

A horizontal bar of negligible mass hinged at A and assumed rigid is supported by a bronze rod !5m long and a steel rod 5m long Using the data in the accompanying table compute the stress in each rod

A steel bar /5 mm in diameter and ! m long is surrounded by a shell of cast iron / mm thick Compute the load that will compress the combined bar a total of 5+ mm in the length of !m For steel E 3 !55 GPa and for cast iron E 3 55

GPa

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A rigid block of mass M is supported by three symmetrically spaced rods as shown below Each copper rod has an area of 855 mm!- E 3 !5 GPa- and the allowable stress is *5 MPa The steel rod has an area of !55 mm!- E 3 !55

GPa- and the allowable stress is .5 MPa Determine the largest mass M which can be supported

THERMAL STRESSES

It is well known that changes in temperature cause bodies to

expand or contract the amount of linear deformation Tbeing expressed by the relation

T 3 L:T