(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). 

 

 

SymmetricPrismaticMemberinPureBending(Equilibrium)

Assumptions:(1)Sectionhasatleastoneplaneofsymmetry,(2)Thebendingmomentisappliedinplaneofsymmetryand(3)thebeamisprismatic

Conclusions: (1) Only normal stress (uniaxial stress) exists in bending (from theory andexperiment),(2)Thereexistsaneutralaxis,and(3)Thedeflectioncurveofthebentbeamformsacirculararc(fromtheoryandexperiment),seebelowfordetails.

Consideringequilibrium:Wehavetotalof3equilibriumequationsasfollows:

0 → 0 , 0 → 0 , 0 → 0  

Asthevariationof onthesection(A)isunknowntheseequilibriumequationscannotberesolvedandthereforethesystemisstatisticallyindeterminateandweneedcompatibility.

(75)   

   

DeformationofSymmetricMemberinPureBending

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).

   

0 → 0 → 0 →  Neutralaxispassesthroughthecentroidofthesection

→ → → → →

:

→

112/2

16

16

 

→ → →  

,

→ → | | →  

→  

  

(76)   

   

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.

 

→ 24 40

 

112

100 80425 3959871  

→ 24 40

3959871→ 2375922  

→ 2.38  

 

(77)   

   

 

M(x)

x

55.5kNm

Example2:Twoverticalforcesareappliedtoabeamofthecross section shown. Determine themaximum tensile andcompressivestressesinportionBC.

Position of neutral axis (considering origin of thecoordinatesystematthebase):

∑∑

12.5 100 25 100 150 25 187.5 200 25100 25 150 25 200 25

→ 119.44

Momentofinertiawithrespecttotheneutralaxis:

. .

112

100 25

100 25 119.44 12.5112

25 150

25 150 119.44 100112

200 25

200 25 200 119.44 12.560.6 10

Maximumtensilestressatthelowermostcorner:

55.5 10 119.4460.6 10

109.4

Maximumcompressivestressattheuppermostcorner:

55.5 10 200 119.44 60.6 10

73.8

1→1 55500000

200000 60.6 106 4

→ 218378.4 218.4

 

(78)   

   

Example 3: Knowing that for the extruded beamshowntheallowablestressis120MPaintensionand150 MPa in compression, determine the largestcoupleMthatcanbeapplied.

 

 

Positionofneutral axis (consideringoriginofthecoordinatesystematthebase):

∑∑

125 150 250 125 50 50150 250 50

→ 138.24

Moment of inertia with respect to theneutralaxis:

. . →

. .112

150 250

150 250 125 138.24

450 50

138.24 75165567042

Maximum tensile stress at the uppermostcorner:

→

120250 138.24

165567042

→ 177774204 177.78

Maximum compressive stress at theuppermostcorner:

→

150138.24

165567042

→ 179651739 179.65

(Controls)

(79)   

   

Example 4: Determine the

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

 

(80)   

   

EccentricAxialLoadinginaPlaneofSymmetryWe now analyze the distribution of stresses when the line of action of the loads does notpassthroughthecentroidofthecrosssection,i.e.,whentheloadingiseccentric.

 

(81)   

   

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.

CalculatingPandMata‐asection:

0 → 20000

0 → 240

20000 2404800000

Calculatingsectionproperties:

60 80 40 60 2400

. .112

60 80112

40 60

1840000

StressatpointA:

200002400

4800000 401840000

8.33 104.35 112.7

StressatpointB:

200002400

4800000 401840000

8.33 104.35 96

M

112.7MPa

‐96MPa

43.2mm 36.8mmN.A.

 

(82)   

   

TBR4:DeterminethemagnitudesandlocationsofthemaximumtensionandcompressionnormalstresseswithintheverticalportionBCofthepost(P=25kN).

 

MB

Answer:

Maximaltensilestress:82.2MPaatsectionB

Maximalcompressivestress:‐79.3MPaatsectionC

MCZ

101 , 10 761 666.67  

 

(83)   

   

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  

(84)   

   

GeneralCaseofEccentricAxialLoading

 

 

 

Example 11: The tube shown has a uniform wallthicknessof12mm.Fortheloadinggiven,determine(a) thestressatpointsAandB, (b) thepointwheretheneutralaxisintersectslineABD.

125 75 125 24 75 24 4224

14 1252

2 28 1252

28 752

14 28 752

2625 , 525 112

75 125112

75 24 125 24

7828252 112

125 75112

125 24 75 24

3278052

700004224

26250001252

7828252525000

752

327805231.52

700004224

26250001252

7828252525000

752

327805210.39

700004224

2 625 0001252

7828252

525 000752

32780521.62

AlternativelytofindN.A.:

0 →

16.57 0.335 0.16 0 . .

x

y

⊚ 

A

BD

N.A.

94mm

64.8mm

(85)   

   

TBR6:Thebasketballplayerappliestheforcesshowntothebasketring.Theposthasacircularcrosssectionwithinternalandexternalradiusof150and200mm.Findstressatpointsa,b,andcontheoutersurfaceofthepostatsectionAB(1393).

1500 ① 

1500 200 50 1000 250000 ③ 

1500 1000 1500000 ③ 

1500200 150

250000 200

4 200 150⑤ 

0.085 85 ① 

1500200 150

1500000 200

4 200 150⑤ 

0.38 380 ① 

200 cos 45° 200 sin 45°

1500200 150

250000 141.4

4 200 150

1500000 141.4

4 200 150⑤ 

0.26 260 ① 

 

(86)   

   

BendingofMembersMadeofSeveralMaterials(E2>E1)

 Todeterminepositionoftheneutralaxisweconvertonematerialtoanothersothat:

1

2‐StraindistributionremainsunchangedsotohavethesameN.A.forthetransformedsection

formaterial2onlyassuming that material 1 does not exist

Soformaterial2alonetheN. A. passesthrough its centroid. As material 1 is weakerweexpect that:

Thetransformedsectionbecomesbiggerinordertohavethesamebendingresistance

→

→ →

Forrectangularcrosssection:

112112

→

AspositionofNAisdeterminedwecancalculatestress:

. .,

. .

Variationofstrainislinearregardlessofthematerialproperties

(87)   

   

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

(88)   

   

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.

 

105210

12 

70210

13 

NewwidthforBrass:1/2×40=20mm

NewwidthforAluminium:1/3×40=13.33mm

 

Neutralaxispassesthroughthecentroidofthesectionwhichislocatedatitsmiddleduetothesymmetry.

. .112

40 20 2

112

20 10 20 10 15 2

112

13.33 10 13.33 10

25 288888.9

13.33mm

20mm

40mm

NeutralAxis

10mm

10mm

20mm

1800 10 10

288888.962.3  

1800 10 20

288888.962.3  

1800 10 30

288888.962.3  

1 1800000210000 288888.9 4 2.967 10

1→ 33703 33.7  

(89)   

   

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  

 

(90)   

   

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

(91)   

   

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

 

 

 

(92)   

   

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.

ThepositionoftheneutralaxisisobtainedbydeterminingthedistancexfromtheupperfaceofthebeamtothecentroidCofthetransformedsection.Denotingbybthewidthofthebeam,andbydthedistancefromtheupperfacetothecenterlineofthesteelrods,wewritethatthefirstmomentofthetransformedsectionwithrespecttotheneutralaxismustbezero.Sincethefirstmomentofeachofthetwoportionsofthetransformedsectionisobtainedbymultiplyingitsareabythedistanceofitsowncentroidfromtheneutralaxis,wehave:

2 0 →

12

0

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.

. .,

. .

(93)   

   

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

(94)   

   

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

 

(95)   

   

STRESSCONCENTRATIONS

 

 

 

 

Review cross sectionalpropertiesfromStatics:moment of inertia (Ix,Iy) and product ofinertia for an area(Ixy)! Mohr’s circle todetermine principalaxesofanarea!

(96)   

   

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.

(97)   

   

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:

Tofindthepositionofneutralaxis:

→

Thus,theangleφthattheneutralaxisformswiththezaxisisdefinedbytherelation:

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.

(98)   

   

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

(99)   

   

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:

29.386.8

→29.3 86.8

86.8

116.186.8

80→ 59.8

59.8 20

4541.5°

A

E

29.3

86.8

O

CΦ=41.5°

Findingpositionoftheneutralaxis(Method2):

tan tan16640000 4

5040000 4 tan 15° → 41.5°

 

Method3:

BestMethodtodetermineNA:

24.15 10 16640000

6.47 10 5040000

0

Y=0.8845Z(NAequation)

OR

24.15 10 16640000

6.47 10 5040000

0

Y=0.8845Z(NAequation)

(100)   

   

 

Example13:ThecoupleMisappliedtoabeamofthecrosssectionshown.FindstressatpointA.Calculatingsectionproperties:

112

10 90 2112

40 10 40 10 40

1894167 112

90 10 2112

10 40 40 10 25

614166.7

0 0 10 90 25 40 40 10

25 40 40 10 800000 Findingprincipalaxesofthesection(Mohr’scircle):

2 2 229666.8

2 2 2278667

tan 2

2

8000001894167 614166.7

2

0.625 → 2

51.34° → 25.67°Decomposingthemomentontheprincipalaxes:

1.2 cos 25.67° 1.0816  

1.2 sin 25.67° 0.5198  

FindingcoordinateofpointAinuvsystem:

cos sinsin cos

cos 25.67° sin 25.67°sin 25.67° cos 25.67°

4545

60.0521.07

CalculatingstressatpointA:

1.0816 10 21.07229666.8

0.5198 10 60.052278667

112.97 CalculatingstressatpointB:

cos sinsin cos

cos 25.67° sin 25.67°sin 25.67° cos 25.67°

450

40.5619.5

1.0816 10 19.5229666.8

0.5198 10 40.562278667

80.3

,

 

 

2

B

112.9780.3

→ 

112.97 80.380.3

45 

18.7  

45 25.67° 21.63  

tan tan 25.67°

229666.8227 8667

tan 25.67°  

→ 2.77°

D

112.97

80.3

O

A

B

D

N.A. φ 

vN.A.

(101)   

   

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.