Example 1 - Method of Joints PDF

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MethodofJoints‐Workedexample 

Solvethetrussusingthemethodsofjoints. 

 



1 

SOLUTION: 1. Determinethesupportreactions. Disconnectthetrussfromthesupportsandapplyreactions.Numberthejoints.Drawthefreebody diagramofthetruss.

20N 17.3m

4

H

5

6

1

3 2

VL 10m

VR 10m 

Applythethreeequilibriumequations: SumofforcesinXdirection    0   SumofmomentsaroundJoint1    0   ∙ 20  20 ∙ 10 →    10 SumofforcesinYdirection    0      20 →    20    20  10  10 Drawnowthefreebodydiagramwithalltheforcesappliedonthetruss.

2 

20N 5

6

17.3m

4

1

3 2

10N

10N

10m

10m  

  tan

17.3  tan 1.73 ≅ tan √3  60° 10

2. Drawthefreebodydiagramforeachjoint. Assumethatallthemembersareintension.Disconnectthejointsfromthemembers.

20N F4-5

4

F5-4

F5-6

F6-5

6

5

F4-1

F4-2

F6-2

F5-2 F2-5

F1-4

F2-6

F2-4

F3-6

1

3

F1-2

F2-1

2

F2-3

10N 

F3-2

10N

 3



F6-3

Notationofforcesbynumbering

Fa-b=Fb-a Here‘a’isthejointatwhichtheforceisappliedand‘b’thejointatwhichtheforceisdirected.

Signconvention Wherethevalueofaninternalforceispositivethecorrespondingmemberisintension,whereitis negativethememberisincompression. Startsolutionwiththejoint1(or3)sinceithasonlytwounknownforces.

Joint1 Writethetwoequilibriumequationsforthejoint    0    →      0

F1-4 

1

   0    10 →      10

F1-2

10N

  

Joint3 Writethetwoequilibriumequationsforthejoint    0    →      0     0    10 →      10   

Joint4 Writethetwoequilibriumequationsforthejoint    0     sin     

10    11.55 sin  sin 60°

   0     cos       cos   11.55 cos 60°  5.77 

 4



Joint5 Writethetwoequilibriumequationsforthejoint    0      →        5.77     0    20 →      20   

Joint6 Writethetwoequilibriumequationsforthejoint    0     cos      

 5.77  11.55  cos 60° cos 

Check(F6‐3isknownalreadyfromthesolutionofthejoint3):    0     sin       sin   11.55 sin 60°  10 ThevaluesofF6‐3andF3‐6confirmthatthesolutionismadewithoutmistakesbynow.

Joint2 Alltheforcesareknownnow.Thesolutionofthisjointisneededtoconfirmthattherearenomistakes anywhereontheway. Writethetwoequilibriumequationsforthejoint Fromthesolutionofthejoints1and3isfoundthatF2‐1=0andF2‐3=0    0   cos    cos      11.55 Check(F2‐5isknownalreadyfromthesolutionofthejoint5):    0     sin    sin      󰇛   󰇜 sin   󰇛11.55  11.55󰇜 sin 60°  20 ThevaluesofF2‐5andF5‐2confirmthatthesolutionismadewithoutmistakes.

5 ...