Method of Joints –Example

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Method of Joints –Example . Using the method of joints, determine the force in each member of the truss. Method of Joints –Example . Draw the free body diagram of the truss and solve for the equations. Method of Joints –Example . Solve the moment about C. Method of Joints –Example . - PowerPoint PPT Presentation

Transcript of Method of Joints –Example

Page 1: Method of Joints –Example
Page 2: Method of Joints –Example

Method of Joints –Example

Using the method of joints, determine the force in each member of the truss.

Page 3: Method of Joints –Example

Method of Joints –Example

Draw the free body diagram of the truss and solve for the equations

x x

x

y y

y

0

0 lb

0 2000 lb 1000 lb

3000 lb

F C

C

F E C

E C

Page 4: Method of Joints –Example

Method of Joints –Example

Solve the moment about C

C

y

0 2000 lb 24 ft 1000 lb 12 ft 6 ft

10000 lbC 3000 lb 10000 lb 7000 lb

M E

E

Page 5: Method of Joints –Example

Method of Joints –Example

Look at joint A

y AD

AD AD

x AD AB AB

AB AB

40 2000 lb5

2500 lb 2500 lb C

3 30 2500 lb5 5

1500 lb 1500 lb T

F F

F F

F F F F

F F

Page 6: Method of Joints –Example

Method of Joints –Example

Look at joint D

y AD DB DB

DB DB

x AD DB DE

DE

DE DE

4 4 4 40 2500 lb5 5 5 5

2500 lb 2500 lb T3 305 5

3 32500 lb 2500 lb5 53000 lb 3000 lb C

F F F F

F F

F F F F

F

F F

Page 7: Method of Joints –Example

Method of Joints –Example

Look at joint B

y BD BE

DE

DE DE

x BD BA BE BC

BC

BC DE

4 40 1000 lb5 5

4 42500 lb 1000 lb5 53750 lb 3750 lb C

3 305 5

3 32500 lb 1500 lb 3750 lb5 5

5250 lb 5250 lb T

F F F

F

F F

F F F F F

F

F F

Page 8: Method of Joints –Example

Method of Joints –Example

Look at joint E

y EB EC

DE

EC EC

x EB ED EC

EC

EC EC

4 40 10000 lb5 5

4 43750 lb 10000 lb5 58750 lb 8750 lb C

3 305 5

3 33750 lb 3000 lb5 58750 lb 8750 lb C

F F F

F

F F

F F F F

F

F F

Page 9: Method of Joints –Example

Method of Joints –Example

Look at joint C to check the solution

y CE

x CE CB x

40 7000 lb5

4 8750 lb 7000 lb 0 OK!5

305

3 8750 lb 5250 lb 0 05

F F

F F F C

Page 10: Method of Joints –Example

Method of Joints –Class Problem

Determine the forces BC, DF and GE. Using the method of Joints.

Page 11: Method of Joints –Example

Method of Sections -Truss

The method of joints is most effective when the forces in all the members of a truss are to be determined. If however, the force is only one or a few members are needed, then the method of sections is more efficient.

Page 12: Method of Joints –Example

Few simple guidelines of section truss analysis:• Pass a section through a maximum of 3 members of

the truss, 1 of which is the desired member where it is dividing the truss into 2 completely separate parts,

• At 1 part of the truss, take moments about the point (at a joint) where the 2 members intersect and solve for the member force, using ∑ M = 0,

• Solve the other 2 unknowns by using the equilibrium equation for forces, using ∑ Fx = 0 and ∑ Fy = 0.

Page 13: Method of Joints –Example

Method of Sections -Truss

If we were interested in the force of member CE. We can use a cutting line or section to breakup the truss and solve by taking the moment about B.

Page 14: Method of Joints –Example

Method of Sections – Example

Determine the forces in members FH, GH and GI of the roof truss.

Page 15: Method of Joints –Example

Method of Sections – Example

Draw a free body diagram and solve for the reactions.

RAx

RAy

L

x Ax

Ax

y

Ay

0

0 kN

0

20 kN

F R

R

F

L R

Page 16: Method of Joints –Example

Method of Sections – Example

Solve for the moment at A.

RAx

RAy

L

A

Ay

6 kN 5 m 6 kN 10 m 6 kN 15 m

1 kN 20 m 1 kN 25 m 30 m

7.5 kN12.5 kN

M

L

LR

Page 17: Method of Joints –Example

Method of Sections – Example

Solve for the member GI. Take a cut between the third and fourth section and draw the free-body diagram.

HIHI

HI

1 o

8 m 10 m 8 m15 m 10 m 15 m

5.333 m8 mtan 28.1

15 m

l l

l

Page 18: Method of Joints –Example

Method of Sections – Example

The free-body diagram of the cut on the right side.

H GI

GI GI

1 kN 5 m 7.5 kN 10 m 5.333 m

13.13 kN 13.13 kN T

M F

F F

Page 19: Method of Joints –Example

Method of Sections – Example

Use the line of action of the forces and take the moment about G it will remove the FGI and FGH and shift FFH to the perpendicular of G.

Page 20: Method of Joints –Example

Method of Sections – Example

Take the moment at G

G

oFH

FH FH

1 kN 5 m 1 kN 10 m 7.5 kN 15 m

cos 28.1 8 m

13.82 kN 13.82 kN C

M

F

F F

Page 21: Method of Joints –Example

Method of Sections – Example

Use the line of action of the forces and take the moment about L it will remove the FGI and FFH and shift FGH to point G.

1 o5 mtan 133.25.333 m

Page 22: Method of Joints –Example

Method of Sections – Example

Take the moment at L

oL GH

GH GH

1 kN 5 m 1 kN 10 m cos 43.2 15 m

1.372 kN 1.372 kN C

M F

F F

Page 23: Method of Joints –Example

Method of Sections – Class Problem

Determine the forces in members CD and CE using method of sections.

Page 24: Method of Joints –Example

Homework (Due 2/24/03)

Problems:

6-34, 6-37, 6-38, 6-40, 6-45, 6-63

Page 25: Method of Joints –Example

Truss –Bonus Problem

Determine whether the members are unstable, determinate or indeterminate.

Page 26: Method of Joints –Example

Truss –Bonus Problem

Determine the loads in each of the members.

Page 27: Method of Joints –Example

Truss –Bonus Problem

Determine the loads in each of the members.

Page 28: Method of Joints –Example

Truss –Bonus Problem

Determine the loads in each of the members.