(102)   

   

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.

 

 

(103)   

   

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

(104)   

   

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)

(105)   

   

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

(106)   

   

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

(107)   

   

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

 

/√  

 √ 

√

√ 

(108)   

   

SingularityFunctionMethod

⟨ ⟩ 0

0

⟨ ⟩ 0

1 0

⟨ ⟩11⟨ ⟩ 0

⟨ ⟩ ⟨ ⟩ 1

 

⟨ ⟩ 0

 

 

 

⟨ ⟩ ⟨ ⟩  

 

 

 

⟨ ⟩ 0 1

⟨ ⟩ 0

(‐1isconsideredsothatunitforwbecomesN/m)

⟨ ⟩ 0 1

⟨ ⟩ 0

(‐2isconsideredsothatunitforwbecomesN/m)

 

 

⟨ ⟩ ⟨ ⟩ 0

 

 

(109)   

   

 

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  

 

 

⟨ ⟩   ⟨ ⟩   ⟨ ⟩  

(110)   

   

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⟨ ⟩