32 CHAPTER 1 Basic Elasticity
An examination of Eq. (1.54) shows thatν≤0.5, since a body cannot increase in volume under
pressure.Also,thelateraldimensionsofabodysubjectedtouniaxialtensioncannotincreasesothat
ν>0.Therefore,foranisotropicmaterial0≤ν≤0.5andformostisotropicmaterials,νisintherange
0.25to0.33belowtheelasticlimit.Abovethelimitofproportionality,νincreasesandapproaches0.5.
Example 1.4
Arectangularelementinalinearlyelasticisotropicmaterialissubjectedtotensilestressesof83and
65N/mm^2 onmutuallyperpendicularplanes.Determinethestraininthedirectionofeachstressandin
thedirectionperpendiculartobothstresses.Findalsotheprincipalstrains,themaximumshearstress,
themaximumshearstrain,andtheirdirectionsatthepoint.TakeE=200000N/mm^2 andv=0.3.
Ifweassumethatσx=83N/mm^2 andσy=65N/mm^2 ,thenfromEqs.(1.52)
εx=
1
200000
( 83 −0.3× 65 )=3.175× 10 −^4
εy=
1
200000
( 65 −0.3× 83 )=2.005× 10 −^4
εz=
−0.3
200000
( 83 + 65 )=−2.220× 10 −^4
Inthiscase,sincetherearenoshearstressesonthegivenplanes,σxandσyareprincipalstressesso
thatεxandεyaretheprincipalstrainsandareinthedirectionsofσxandσy.ItfollowsfromEq.(1.15)
thatthemaximumshearstress(intheplaneofthestresses)is
τmax=
83 − 65
2
=9N/mm^2
actingonplanesat45◦totheprincipalplanes.
Further,usingEq.(1.50),themaximumshearstrainis
γmax=
2 ×( 1 +0.3)× 9
200000
sothatγmax=1.17× 10 −^4 ontheplanesofmaximumshearstress.
Example 1.5
At a particular point in a structural member, a two-dimensional stress system exists where
σx=60N/mm^2 ,σy=−40N/mm^2 ,andτxy=50N/mm^2. If Young’s modulusE=200000N/mm^2 and
Poisson’s ratioν=0.3, calculate the direct strain in thexandydirections and the shear strain at the
point.Alsocalculatetheprincipalstrainsatthepointandtheirinclinationtotheplaneonwhichσxacts;
verifytheseanswersusingagraphicalmethod.