chapter 5: equilibrium of beams - …me.eng.usc.ac.ir/files/1509793732356.pdf · chapter 5:...
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CHAPTER5:EQUILIBRIUMOFBEAMS
Ourobjectiveinthischapteristodeterminethediagramof internalmoments(M)andshearforces(V) inbeamssupporting transverse loading. The bending couple Mcreates normal stresses in the cross section, while theshearforceVcreatesshearingstressesinthatsection.Inmost cases the dominant criterion in the design of abeam for strength is themaximumvalueof thenormalstress in the beam. The determination of the normalstresses( / )inabeamwillbethesubjectofthischapter,whileshearingstresseswillbediscussedinthenextchapter.
ShearandBending‐MomentDiagrams
There are three methods to determine the diagram ofinternal shear forces and moments in a beam undertransverseloading:
1‐ EquilibriumMethod2‐ Load‐Shear‐MomentRelationshipMethod3‐ SingularityFunctionMethod
EquilibriumMethod
The shear and bending‐moment diagrams will beobtained by determining the values of V and M atselectedpointsof thebeam.Thesevalueswillbefoundin the usualway, i.e., by passing a section through thepointwheretheyaretobedeterminedandconsideringthe equilibrium of the portion of beam located oneither sideof thesection.TheshearVand thebendingmoment M at a given point of a beam are said to bepositivewhenthe internal forcesandcouplesactingoneachportionofthebeamaredirectedasshown.
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Example1:Drawtheshearandbending‐moment diagrams forthebeamandloadingshownanddeterminethemaximumnormalstressduetobending.
AyBy
Ax
Findingsupportreactionforces:
→ 0 → 0.
↑ 0 → 24 2 0.
↷ 0 → 64 4 24 2 5 0.
76 , 28
EquilibriumofpartAC:
↑ 0 → 28
↷ 0 → 28 0.→ 28
EquilibriumofpartCB:
↑ 0 → 28
↷ 0 → 28 64 0.
→ 28 64
EquilibriumofpartBD:
↑ 0 → 24 4 0
28 76 24 96 → 24 144
↷ 0 → 28 64 76 4
24 4 42
0.
→ 12 144 432
Maximumnormalstress:
56000000 254/261.2 10
116.2
x x
x
V(kN)
‐28
48
M(kNm)
‐56
8
‐48
x
x
48
‐56 ‐56
B
D
B
D
B
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Load‐Shear‐MomentRelationshipMethod
The construction of the shear diagram and,especially, of the bending‐moment diagram willbe greatly facilitated if certain relations existingamong load, shear, and bending moment aretakenintoconsideration.
↑ ∑ 0 → ∆ 0 →
∆ ∆ → →
→
VD‐VC=‐(areaunderloadcurvebetweenCandD)
↶ 0 → ∆ ∆∆2
∆ 0
∆ ∆∆2
∆ →
→
MD‐MC=(areaundershearcurvebetweenCandD)
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Example 2: Draw the shear andbending‐moment diagrams for thebeam and loading shown anddetermine the maximum normalstressduetobending.
↑ 0 → 2 3 4
14
↶ 0 →2 1 4 12 2 0
→ 5.5 , 8.5
3 4 → 12 6.55.5
2 → 2 0 2
7.04 → 7.04 2 5.04
: 0
5.04 10 80
4 80 70
30.28
V(kN)
x
‐2
6.5
‐2
6.5×2.16×0.5=7.04
2.16
‐5.5
5.5×1.83×0.5=‐5.04
M(kNm)
x
‐2
O
5.04
2.16
0.34
.
→ .
x
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Example3:Drawtheshearandbending‐moment diagrams forthebeamandloadingshownanddetermine themaximumnormalstressduetobending.
ThesystemisstaticallyindeterminateandweneedtoconsiderequilibriumofpartsABandBEseparatelytodeterminereactionforces:
: ↷ 0 → 2.4 80 2.4 1.2 0. → 96
: ↶ 0 → 3.6 0.6 80 0.6 0.3 160 2.1
0. → 3.6 0.6 350.4
: ↑ 0 → 160 80 3 400
→ 304 → 56 , 248
PartAC:
96 80
96 40
PartCD:
96 248 80 3 104
96 248 3 80 3 1.5104 384
PartDE:
96 248 80 3 160 56
96 248 3 80 3 1.5160 4.5 56 336
RA RC RE
V(kN)
x
96
‐144
104
‐56
x
‐72
8457.6
M(kNm)
96×1.2/2=57.6
1.2
‐144×1.8/2=‐129.6
‐96
104×1.5=156
‐ 56×1.5=‐84
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TBR 1: Draw the shear and bending‐moment diagrams for the beam andloadingshownusingLoad‐Shear‐MomentRelationshipMethod.
FromStatics:
↑ 0 →2
↶ 0 →2
23 0
6,
3
Load‐ShearRelationship
2→
6 2
3
AtpointA: 0
63 → 0 /√3
AreaunderAO:
63
9√3
/√
AreaunderOB:√
Shear‐momentRelationship
9√3→
9√3
6
/√
√
√
√
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SingularityFunctionMethod
⟨ ⟩ 0
0
⟨ ⟩ 0
1 0
⟨ ⟩11⟨ ⟩ 0
⟨ ⟩ ⟨ ⟩ 1
⟨ ⟩ 0
⟨ ⟩ ⟨ ⟩
⟨ ⟩ 0 1
⟨ ⟩ 0
(‐1isconsideredsothatunitforwbecomesN/m)
⟨ ⟩ 0 1
⟨ ⟩ 0
(‐2isconsideredsothatunitforwbecomesN/m)
⟨ ⟩ ⟨ ⟩ 0
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Example 4: For the beam andloading shown and usingsingularity functions, express theshear and bending moment asfunctionsofthedistancexfromthesupportatA.
2.6⟨ 0⟩ 1.2⟨ 0.6⟩ 1.5⟨ 0.6⟩ 1.5⟨ 1.8⟩ ⟨ 2.6⟩
2.6⟨ 0⟩ 1.2⟨ 0.6⟩1.51⟨ 0.6⟩
1.51⟨ 1.8⟩ ⟨ 2.6⟩
2.61⟨ 0⟩
1.21⟨ 0.6⟩
1.52⟨ 0.6⟩
1.52⟨ 1.8⟩ ⟨ 2.6⟩
0 2.6⟨ 0⟩ 1.2⟨ 0.6⟩1.51⟨ 0.6⟩
1.51⟨ 1.8⟩ ⟨ 2.6⟩ 2.6 0 0 0 0
2.6
02.61⟨ 0⟩
1.21⟨ 0.6⟩
1.52⟨ 0.6⟩
1.52⟨ 1.8⟩ ⟨ 2.6⟩ 2.6 0
VEandMEshouldbecalculatedinasimilarway.
FindingshearandmomentatpointDforexample:
1.8 2.6⟨1.8 0⟩ 1.2⟨1.8 0.6⟩1.51⟨1.8 0.6⟩
1.51⟨1.8 1.8⟩ ⟨1.8 2.6⟩
2.6 1.2 1.5 1.8 0.6 0 0 0.4
1.82.61⟨1.8 0⟩
1.21⟨1.8 0.6⟩
1.52⟨1.8 0.6⟩
1.52⟨1.8 1.8⟩ ⟨1.8 2.6⟩
2.6 1.8 1.2 1.8 0.61.52
1.8 0.6 0 0 2.16
↶ 0 → 1.44 1.5 1.2 2.4
1.2 3 3.6 0→ 2.6
⟨ ⟩ ⟨ ⟩ ⟨ ⟩
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Example 5: For the beam andloading shown and usingsingularity functions, express theshear and bending moment asfunctionsofthedistancexfromthesupportatA.
FromStatics:RA=2.75kN
2.75⟨ 0⟩ 1.5⟨ 3⟩ 3⟨ 3⟩ 1⟨ 3⟩
2.75⟨ ⟩ 1.5⟨ 3⟩31⟨ 3⟩
12⟨ 3⟩
2.751
⟨ ⟩ 1.5⟨ 3⟩32⟨ 3⟩
16⟨ 3⟩
Example 6: For the beam andloading shown and usingsingularity functions, express theshear and bending moment asfunctionsofthedistancexfromthesupportatA.
⟨ 0⟩ ⟨ ⟩ ⟨ ⟩
2⟨ ⟩
1⟨ ⟩
2⟨ ⟩
6⟨ ⟩
2⟨ ⟩
6⟨ ⟩
3m 3m
3kN/m
6kN/m
AB1.5kN.m
TBR2: For the beam and loadingshown and using singularityfunctions, express the shear andbending moment as functions ofthe distance xfrom the support atA.
⟨ 0⟩ ⟨ ⟩ ⟨ ⟩
⟨ ⟩2
⟨ ⟩2
⟨ ⟩
2⟨ ⟩
6⟨ ⟩
6⟨ ⟩