Mechanics of Materials - Formulas and Problems: Engineering Mechanics 2

219 Pages·2017·3.19 MB·English

by  Dietmar Gross

Dietmar Gross · Wolfgang Ehlers
Peter Wriggers · Jörg Schröder
Ralf Müller
Mechanics of
Materials –
Formulas and
Problems
Engineering Mechanics 2
123
–
Mechanics of Materials Formulas
and Problems
Dietmar Gross Wolfgang Ehlers
(cid:129)
ö
Peter Wriggers Jörg Schr der
(cid:129)
ü
Ralf M ller
Mechanics of
–
Materials Formulas
and Problems
Engineering Mechanics 2
123
Dietmar Gross Jörg Schröder
Division of Solid Mechanics Institute of Mechanics
TU Darmstadt Universität Duisburg-Essen
Darmstadt Essen
Germany Germany
WolfgangEhlers Ralf Müller
Institute of Applied Mechanics Institute of Applied Mechanics
Universität Stuttgart TU Kaiserslautern
Stuttgart Kaiserslautern
Germany Germany
PeterWriggers
Institute of Continuum Mechanics
LeibnizUniversität Hannover
Hannover
Germany
ISBN978-3-662-53879-1 ISBN978-3-662-53880-7 (eBook)
DOI 10.1007/978-3-662-53880-7
LibraryofCongressControlNumber:2016956827
©Springer-VerlagGmbHGermany2017
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Preface
Thiscollectionofproblemsresultsfromthedemandofstudentsforsup-
plementaryproblemsandsupportinthepreparationforexaminations.
With the present collection ’Engineering Mechanics 2 - Formulas and
Problems,MechanicsofMaterials’ weprovidemoreadditionalexercise
material.
The subject ’Mechanics of Materials’ is commonly taught in these-
cond course of Engineering Mechanics classes at universities. The pro-
blemsanalyzedwithinthesecoursesuseequilibriumconditionsandki-
nematicrelationsinconjunctionwithconstitutiverelations.Aswewant
concentrate more on basic concepts and solution procedures the focus
liesonlinearelasticmaterialbehaviorandthesmallstrainregime.Ho-
wever,thiscoversawiderangeofelasto-staticproblemswithrelevancy
inengineeringapplications.Specialattentionisgiventostructuralele-
ments like bars, beams and shafts as well as plane stress and strain
situations.
Following the warning in the first collection, we would like to make
thereaderaware that purereading and tryingtocomprehend thepre-
sentedsolutionswill notprovideadeeperunderstandingofmechanics.
Neither does it improve the problem solving skills. Using this collec-
tion wisely, one has to try to solve the problems independently. The
proposed solution should only be considered when experiencing major
problems in solving an exercise.
Obviously this collection cannot substitute a full-scale textbook. If
notfamiliarwiththeformulae,explanations,ortechnicaltermstherea-
der has to consider his or her course material or additional textbooks
on mechanics of materials. An incomplete list is provided on page IX.
Darmstadt, Hannover,Stuttgart, Essen and D.Gross
Kaiserslautern, Summer2016 P.Wriggers
W.Ehlers
J.Schr¨oder
R.M¨uller
Table of Contents
Literature,Notation.................................................. IX
1 Stress,Stain,Hooke’sLaw.......................................... 1
2 TensionandCompressioninBars................................. 29
3 BendingofBeams..................................................... 57
4 Torsion................................................................... 111
5 EnergyMethods....................................................... 143
6 BucklingofBars....................................................... 181
7 Hydrostatics............................................................ 195
Literature
Textbooks
Gross, D., Hauger, W., Schr¨oder, J., Wall, W., Bonet, J., Engineering
Mechanics 2: Mechanics of Materials, 1st edition, Springer 2011
Gross, D., Hauger, W., Wriggers, P., Technische Mechanik, vol 4:
Hydromechanics, Elements of Advanced Mechanics, Numerical
Methods (in German), 9th edition, Springer2014
Beer,F.P.,Johnston,E.R.,DeWolf,J.T.,Mazurek,D.F.,Mechanicsof
Materials, 7th edition, McGraw-Hill Education 2012
Hibbeler, R.C., Mechanics of Materials, 10th edition, Pearson 2016
Geer,J.M.,Goodno,B.J.,MechanicsofMaterials,8thedition,Cengage
Learning 2013
Ghavami, P., Mechanics of Materials, 1st edition, Springer2015
CollectionofProblems
Schaum’s Outlines Strength of Materials, 6th edition, McGraw-Hill
Education 2013
Beer,F.P.,Johnston,E.R.,DeWolf,J.T.,Mazurek,D.F.,Mechanicsof
Materials, 7th edition, McGraw-Hill Education 2012
Hibbeler, R.C., Mechanics of Materials, 10th edition, Pearson 2016
Notation
The following symbols are used in the solutions tothe problems:
↑: short notation for sum of all forces in the direction of the
arrow equals zero.
(cid:2)
A: short notation for sum of all moments with reference to
point A equals zero.
(cid:2) short notation for it follows.
1
Chapter 1
Stress, Strain, Hooke’s Law
2 Stress
1.1 1.1 Stress,Equilibriumconditions
Stressisrelatedtoforcesdistributedoverthe
area of a cross section. The stressvector t is
defined as
dF
dF dA
t= ,
dA n
where dF is the force acting on the area ele-
ment dA(unit: 1 Pa = 1 N/m2).
Note:Thestressvectoranditscomponentsdependontheorientation
of thearea element (with its normal n).
Components of the stress vector:
t
τ
σ–normalstress(perpendiculartotheplane)
σ n
τ – shear stress (in plane)
Sign convention:Positivestressesatapositive(negative)facepoint
in positive (negative) coordinate directions.
Spatial stress state: is uniquely defined
by the components of the stress vectors
in three mutually perpendicular sections.
z
Thestresscomponentsarethecomponents
of thestress tensor
τzxσz τzy
⎛ ⎞
⎜σx τxy τxz⎟ τxz τσyzy
σ=⎝τyx σy τyz⎠ σx τxy τyx
τzx τzy σz x y
Equilibrium of moments yieldsthe following relations
τxy =τyx, τxz =τzx, τyz =τzy .
Hencethestresstensorisasymmetrictensorofsecond order:τij =τji.
Planestressstate 3
Plane stress state: is uniquely defined σy
by the stress components of two mutual-
τyx
ly perpendicular sections. The stress com-
ponents in the third direction (here z- σx τxy
direction) vanish (σz =τyz =τxz =0)
(cid:8) (cid:9) τxy σx
σ= σx τxy . y τyx
τxy σy σy
x
Coordinate transformation τxy=τyx
y
σξ=σx+2 σy + σx−2 σy cos2ϕ+τxysin2ϕ, σx τξη σξ ξ
ση=σx+σy − σx−σy cos2ϕ−τxysin2ϕ, τxy ϕ
2 2
τξη=−σx−σy sin2ϕ+τxycos2ϕ. η ϕ
2 τyx x
σy
Principal stresses
(cid:10)(cid:11) (cid:12) σ2
σ1,2 = σx+2 σy ± σx−2 σy 2+τx2y σ1
∗ 2τxy
tan2ϕ = σx−σy σ1
y ϕ∗1 ϕ∗
Note:• The shear stresses vanish σ2 2
in these directions! x
• The principal directions are
perpendicular to each other: τmax
ϕ∗2 =ϕ∗1±π/2. σ0 σ0
Maximum shear stresses
(cid:10)
(cid:11) (cid:12)
τmax = σx−σy 2+τx2y, ϕ∗∗ =ϕ∗±π . y σ0 σ0
2 4 τmax
ϕ∗∗
In these sections the normal stresses x
reach thevalue σ0=(σx+σy)/2.
Invariants
Iσ=σx+σy =σξ+ση =σ1+σ2 ,
IIσ=σxσy−τx2y =σξση−τξ2η =σ1σ2 .