(74) CHAPTER 4: BENDING OF BEAMS This chapter will be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' acting in the same longitudinal plane. Such members are said to be in pure bending. An example of pure bending is provided by the bar of a typical barbell as it is held overhead by a weight lifter as shown. The results obtained for pure bending will be used in the analysis of other types of loadings as well, such as eccentric axial loadings and transverse loadings (see examples below).Symmetric Prismatic Member in Pure Bending (Equilibrium) Assumptions: (1) Section has at least one plane of symmetry, (2) The bending moment is applied in plane of symmetry and (3) the beam is prismatic Conclusions: (1) Only normal stress (uniaxial stress) exists in bending (from theory and experiment), (2) There exists a neutral axis, and (3) The deflection curve of the bent beam forms a circular arc (from theory and experiment), see below for details. Considering equilibrium: We have total of 3 equilibrium equations as follows: ܨ௫ ൌ 0→න ߪ௫ ܣൌ 0 ሺሻ, ܯ௬ ൌ 0→න ߪݖ௫ ܣൌ0 ሺሻ , ܯ௭ ൌ 0 → නሺ ߪݕ௫ ܣሻ ܯൌ 0 ሺሻAs the variation of ߪ௫ on the section (A) is unknown these equilibrium equations cannot be resolved and therefore the system is statistically indeterminate and we need compatibility.
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(74)
CHAPTER4:BENDINGOFBEAMS
This chapterwill be devoted to the analysis of prismaticmemberssubjectedtoequalandoppositecouplesMandM'actinginthesamelongitudinalplane.Suchmembersaresaidtobeinpurebending. Anexampleofpurebendingisprovidedbythebarofatypicalbarbellasitisheldoverheadbyaweightlifterasshown.Theresultsobtainedfor pure bending will be used in the analysis of other types ofloadings as well, such as eccentricaxial loadingsand transverseloadings(seeexamplesbelow).
Conclusions: (1) Only normal stress (uniaxial stress) exists in bending (from theory andexperiment),(2)Thereexistsaneutralaxis,and(3)Thedeflectioncurveofthebentbeamformsacirculararc(fromtheoryandexperiment),seebelowfordetails.
Sinceall the faces represented in the twoprojectionsareat90°toeachother,weconcludethatγxy=γzx=0and,thus,thatτxy= τxz=0. Also, σy, σz, and τyz,wenote that theymust bezero on the surface of themember. Thus, at any point of aslendermemberinpurebending,wehaveastateofuniaxialstress. Recalling that, for M > 0, lines AB and A'B' areobserved,respectively,todecreaseandincreaseinlength,wenote that the strain ɛxand the stress σxare negative in theupper portion of themember (compression) and positive inthe lower portion (tension). Therefore there must exist asurfaceparalleltotheupperandlowerfacesofthemember,where ɛxand σxare zero. This surface is called the neutralsurface. For two reasons it is important to determineposition of the neutral axis: 1) to compute maximalstress and 2) tomake holds (if needed during design)alongtheneutralaxis(toavoidstressconcentration).
POINT1:Sinceinpure‐bendingMisconstantalongtheentirelengthofthebeam,deformationof a beam is uniform along the length of the segment undergoing pure bending; sowhateverhappensatatypicalcross‐sectionalsohappensatanyothersection.Forexample,thecurvatureofthedeflectioncurveatanysectionisthesameasthecurvatureatanyothersection.Therefore,thedeflectioncurveformsacirculararc,withcenterofcurvatureatC.POINT2: 0 → 0 → 0 → 0 →
→ DeformationsinaTransverseCrossSectionAsmentioned the transverse cross section of amember inpure bending remains plane but we will have somedeformationswithintheplaneofthesection.
, →
The relations we have obtained show that the elementslocatedabovetheneutralsurface(y>0)willexpandinboththeyandzdirections,whiletheelementslocatedbelowtheneutralsurface(y<0)willcontract.Inthecaseofamemberof rectangular cross section, the expansionand contractionof the various elements in the vertical direction willcompensate,andnochangeintheverticaldimensionofthecrosssectionwillbeobserved.Asfarasthedeformationsinthe horizontal transverse z direction are concerned,however, the expansion of the elements located above theneutral surface and the corresponding contraction of theelementslocatedbelowthatsurfacewillresultinthevarioushorizontallinesinthesectionbeingbentintoarcsofcircle.
→1′
Example1:A nylon spacing bar has the crosssection shown. Knowing that the allowablestress for the grade of nylon used is 24 MPa,determine the largest couple Mz that can beappliedtothebar.
Example 3: Knowing that for the extruded beamshowntheallowablestressis120MPaintensionand150 MPa in compression, determine the largestcoupleMthatcanbeapplied.
maximum tensile andcompressive stresses in thebeamdue to the uniform load (crosssectionofthebeamisshown).
2 40 80 12 74 12 2762 12 80 12 276
61.52
. . 2112
12 80 12 80
21.52112
276 12
276 12 12.482.469 10
FromStatics:
2.025 , 3.6
Bendingstressesdueto :
2.025 10 61.522.469 10
50.5
2.025 10 18.482.469 10
15.2
Bendingstressesdueto :
3.6 10 18.482.469 10
26.9
3.6 10 61.522.469 10
89.7
→ 50.5
→ 89.7
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EccentricAxialLoadinginaPlaneofSymmetryWe now analyze the distribution of stresses when the line of action of the loads does notpassthroughthecentroidofthecrosssection,i.e.,whentheloadingiseccentric.
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Example 10: The vertical portionof the press shown consists of arectangulartubeofwallthicknesst= 10 mm. Knowing that the presshas been tightened on woodenplanksbeinggluedtogetheruntilP=20kN,determinethestressat(a)pointA,(b)pointB.
TBR5:Maximaltensileandcompressivestressesata‐a are equal to 47 and 67MPa, respectively. Findmaximal allowable value of P. Based on thecalculatedPfindthepositionofneutralaxis(1392).
10 4 20 18 24 20 2
11.086
112
4 20 4 20 11.086 1042 2 8.91 2 3373.6
32 8.91 40.91
_ 4 20 2240.91 8.91
3373.6 4 47 → 395.5
_ 4 20 2240.91 11.086
3373.6 4 67 → 541.9
→ 395.5
NeutralAxis:
395.5
4 20 2240.91 395.5 8.91
3373.6 4 47
395.5
4 20 2240.91 395.5 11.086
3373.6 4 48.9
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GeneralCaseofEccentricAxialLoading
Example 11: The tube shown has a uniform wallthicknessof12mm.Fortheloadinggiven,determine(a) thestressatpointsAandB, (b) thepointwheretheneutralaxisintersectslineABD.
Example5: A steel bar and an aluminiumbar are bonded together to form thecomposite beam shown. The modulus ofelasticity for aluminum is 70 GPa and forsteel is200GPa.Knowing that thebeam isbentaboutahorizontal axisbya coupleofmoment M=1500 Nm, determine themaximum stress in (a) the aluminums, (b)thesteel.
200
702.857
Newwidth=30×2.857=85.71mm
20 40 85.71 50 20 3040 85.71 20 30
24.47
. .112
85.71 40 85.71 40
24.47 20112
30 20
30 20 50 24.47936717.3
40mm
85.71mm
30mm
20mm
NeutralAxis
24.47mm
1500 10 60 24.47
936717.356.9
. 1500 10 24.47
936717.3111.9
1 150000070000 936717.3
2.287 101
→ 43713.5
43.7
Attention:
. 1500 10 40 24.47
936717.371.1
1500 10 40 24.47
936717.324.9
B
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Example 6: Five metal strips, each 40 mmwide, are bonded together to form thecomposite beam shown. The modulus ofelasticityis210GPaforthesteel,105GPaforthe brass, and 70 GPa for the aluminium.Knowing that the beam is bent about ahorizontal axis by a couple of moment 1800Nm,determine(a)themaximumstressineachofthethreemetals,(b)theradiusofcurvatureofthecompositebeam.
Example 7: A steel pipe and analuminiumpipearesecurelybondedtogethertoformthecompositebeamshown. The modulus of elasticity is210GPaforthesteeland70GPaforthe aluminium. Knowing that thecomposite beam is bent by a coupleof moment 500 Nm, determine themaximum stress (a) in thealuminium,(b)inthesteel.
→ 34 19 16
→ 152.64 10
416 10 196.26 10
500000 16196.26 10
40.8
3 500000 19196.26 10
145.2
1 500000
70000 196.26 103 43.639 10
1→ 27476
27.5
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TBR 7: A wood beam reinforced by analuminium channel section is shown in thefigure. The beam has a cross section ofdimensions 150 mm by 250 mm, and thechannelhasauniformthicknessof6mm.Iftheallowablestressesinthewoodandaluminiumare 8.0MPa and 38MPa, respectively, and iftheirmodulusofelasticityareintheratio1to6, what is the maximum allowable bendingmomentforthebeam?
Answer: 108.92
297.35 10
Mallow=16.2kNm
. .
.
131 25 250 2 20 40 6 3 150 625 250 2 40 6 150 6
108.92
. .112
25 250 25 250 131 108.92 2
112
6 40 6 40 108.92 20
112
150 6 150 6 108.92 3
49558213
108.92
4955821338 → 17.3
16
256 108.9249558213
8 → 16.2
150mm/6=25mm
N.A.
108.9mm
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TBR8:Thelowstrengthconcretefloor slab (σY = 10MPa, E = 22.1GPa) is integrated with a wide‐flangeA‐36 steel beam (σY = 165MPa, E = 200 GPa) using shearstuds (not shown) to form thecomposite beam. If the allowablebending stress for the concrete isand allowable bending stress forsteel is determine the maximumallowableinternalmomentMthatcan be applied to the beam. Alsofind the curvature based on thecalculated maximal moment(1390).
Answer:Mallow=330kNm
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ReinforcedConcreteBeams
An important example of structural members made of two different materials is furnished byreinforcedconcretebeams. Thesebeams,whensubjectedtopositivebendingmoments,arereinforcedby steel rods placed a short distance above their lower face. Fortunately, there is a natural bondbetweenconcreteandsteel, sothatnoslippingoccursbetweenthemduringbending. Sinceconcreteisveryweakintension,itwillcrackbelowtheneutralsurfaceandthesteelrodswillcarrytheentiretensile load,whiletheupperpartof theconcretebeamwillcarrythecompressive load.Tobemosteffective,theserodsarelocatedfarthestfromthebeam’sneutralaxissothattheyresistthegreatestpossible tensile moment. The diameters of the rods are small compared to the depth of the crosssection.
Solvingthisquadraticequationforx,weobtainboththepositionoftheneutralaxisinthebeam,andtheportionofthecrosssectionoftheconcretebeamthatiseffectivelyused.Thedeterminationofthestresses in the transformed section is carried out as explained before. The distribution of thecompressive stresses in the concrete and the resultant Fs of the tensile forces in the steel rods areshown.
. .,
. .
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Example9:The reinforced concrete beamshown is subjected to a positive bendingmoment of 175 kNm. Knowing that themodulus of elasticity is 25 GPa for theconcrete and 200 GPa for the steel,determine(a)thestressinthesteel,(b)themaximumstressintheconcrete.
300mm
=15707.96
480mm
x
480‐x
N.A.
20025
8
8 4425
15707.96
Findingtheneutralaxis:
3002
15707.96 480 0.
104.71 50265.472 0.
. 282.6
Calculatingthemomentofinertia:
. .13300 177.87 15707.96
480 177.871.996 10
Stressinsteelmembers:
Transformedsection(concrete)
175000 000 480 177.87
1.996 109 4211.9
Stressinconcretememberiscompressiveandisequalto:
175000000 177.87
1.996 109 415.59
C
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TBR9:Abeamhasthecrosssectionshowninfigure,and is subject to a positive bending moment thatcausesatensilestressinthesteelof20ksi(20000psi=20000lb/in2).Ifn=12(elasticmodulusofsteelis12 timesgreater than thatof concrete) calculate thebendingmomentappliedtothebeam(1391).
28″
6″
3″3″ 12″
TotalAs=3.0in2
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STRESSCONCENTRATIONS
Review cross sectionalpropertiesfromStatics:moment of inertia (Ix,Iy) and product ofinertia for an area(Ixy)! Mohr’s circle todetermine principalaxesofanarea!
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UNSYMMETRICBENDING
Ouranalysisofpurebendinghasbeenlimitedsofartomemberspossessingatleastoneplaneofsymmetryandsubjectedtocouplesactinginthatplane.We found that the neutral axis of the cross section in symmetric bendingpasses through centroid of the section and coincides with the axis of thecouple.Now consider situationswhere the bending couples donotact in aplane of symmetry of the member, either because they act in a differentplane, or because thememberdoes not possess anyplane of symmetry. Insuchsituations,wecannotassumethatthememberwillbendintheplaneofthe couples. As shown, the couple exerted on the section has again beenassumedtoactinaverticalplaneandhasbeenrepresentedbyahorizontalcouple vector M. However, since the vertical plane is not a plane ofsymmetry,wecannotexpectthemembertobendinthatplane,ortheneutralaxisofthesectiontocoincidewiththeaxisofthecouple.
We assume that N.A. is directed toward an arbitrary z‐axis. Anarbitrary directed moment has a component toward z and acomponenttowardy.WeinitiallyonlyconsiderMtowardz.
0 → 0 → . .
0 → →
0 → 0 → 0 →
0 → 0 → yandzmustbeprincipalaxesofthecrosssection
The first equation indicates that the N.A. passes through thecentroid of the section and the third equation determine thedirectionof theN.A. (directed towardprincipal axiswhereM isapplied). The samemethod is used to determine theN.A.whenonlythecomponentofMtowardyisconsidered.
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The principle of superposition can be used to determine stresses in the most general case ofunsymmetricbending.Considerfirstamemberwithaverticalplaneofsymmetry,whichissubjectedtobendingcouplesMandM’actinginaplaneforminganangleθwiththeverticalplane.
Since the y and z axes are the principal centroidal axes of the crosssection,wecanuse theequation / todetermine thestressesresultingfromtheapplicationofeitherofthecouplesrepresentedbyMzandMy:
whereθistheanglethatthecouplevectorMformswiththesameaxis.Since Iz and Iy are both positive, φ and θ have the same sign.Furthermore,we note thatφ > θwhen Iz > Iy, andφ< θwhen Iz < Iy.Thus,theneutralaxisisalwayslocatedbetweenthecouplevectorMandtheprincipalaxiscorrespondingtotheminimummomentofinertia.
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Example12:ThecoupleM isapplied toabeamofthecrosssectionshowninaplaneforming an angle b with the vertical.Determine the stress at (a) point A, (b)pointB,(c)pointD.
Decomposingthemomentontheprincipalaxes:
25 cos 15° 24.15
25 sin 15° 6.47
Calculatingmomentofinertiaabouttheprincipalaxes:
∑∑
100
112
30 80 30 80 60112
90 80 90 80 20
16640000
112
80 30112
80 90 5040000
CalculatingstressatpointA:
24.15 10 6016640000
6.47 10 455040000
29.3
CalculatingstressatpointB:
24.15 10 6016640000
6.47 10 455040000
144.8
CalculatingstressatpointD:
24.15 10 10016640000
6.47 10 155040000
125.7
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Example12(continued):
Findingpositionoftheneutralaxis(Method1):
We know that the neutral axis passesthroughpointCwherestress iszero. Ifwecanfindasecondpointwherestressis zero we can find the position ofneutralaxis.
24.15 10 2016640000
6.47 10 455040000
86.8
As thestressatpointA isnegativeandat point E positive there should be apoint in between where the stress iszero.As thevariationof stress is linearthis point of zero stress can be foundeasily:
TBR 10: The moment acting on the cross section of the unequal-leg angle has a magnitude of 14 kNm and is oriented as shown. Determine: (a) the bending stress at point H, (b) the bending stress at point K, (c) the maximum tension and the maximum compression bending stresses in the cross section, (d) the orientation of the neutral axis relative to the +z axis. Show its location on a sketch of the cross section.
Answer:Centroidlocation:64.18mm(frombottomofshapetocentroid)and39.18mmfromrightedge of shape to centroid.Moment of inertia about the zaxis (Iz): 25,059,086.23mm4.Moment ofinertiaabout theyaxis (Iy):12,133,386.23mm4.Productof inertiaabout thecentroidalaxes (Iyz):10,207,907.81mm4. BendingstressatK:‐82.6MPacompression.Maximumtensionandcompressionbendingstresses:101MPaand‐82.6MPa.Orientationofneutralaxisisshown.