ME 201Engineering Mechanics: Statics

Unit 7.1

Simple Trusses

Method of Joints

Zero Force Members

Simple Truss

A structure composed of slender members

joined together at their end points

“Planar Truss”

Simple Truss

To Design members and connections of a

truss, necessary to determine forces

developed in each member

Assumptions:

All loadings are applied at the joints

Members are joined together by smooth pins

Assumptions satisfactory for bolted or welded joints

Simple Truss

Each truss member acts as a 2-force member

Force directed along axis of member

Tension – pull, elongation

Compression – push, shorten

Several analytical approaches to solving

Free Body Diagrams

Which of the following is NOT a “ simple truss”?

A B

C D

1. A

2. B

3. C

4. D

5. More

than 1

Method of Joints

If truss is in equilibrium, each joint must also be in equilibrium

Approach:

Sketch Free Body Diagram of Entire Truss

Solve for Reactions

Sketch Free Body Diagrams of Each Joint

2 unknown limit per FBD

∑ Fx =0 ∑ Fy =0 ∑ Mo =0

Example Problem Solution

0 yF

0 CM

0 xF

600 N

4 m

B

Cy

Ay

3 m

400 N

3 m D

Cx

Given:

F1 = 400 N

F2 = 600 N

Find:

Force in each member

Solution:

1-FBD

2-Support Reactions

4 m

BC

A3 m

F1

F23 m D

0600 xC NorCx 600600

0460063400 yA

NAy 600

0600400 yC

NorCy 200200

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Example Problem Solution

0 yF

0 xF

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

Joint A

600 N

AD

AB

3

4

Joint D

600 N

BD

3

4

AD

450 N

CD

0 yF

0 xF

05

4600 AB

CNAB 750750

07505

3AD

TNAD 450

05

3450600 BD

TNBD 250

02505

4CD

CNCD 200200

Example Problem Solution

0 xF

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

Joint C

CD

200 N

BC 600 N

200 N

B

200 C

450 TD

600 C C

A

Force Diagram

0600 BCCNBC 600

Zero Force Members

Zero Force Members support NO loading

Used for:

Stability

Changing Applied Loads

Zero Force Members

If only 2 members form a truss joint AND no

external load or support reaction is applied to

the joint, then the members must be Zero

Force Members

X

X

XX

Zero Force Members

If 3 members form a truss joint for which 2

members are collinear, the 3rd member is a Zero

Force Member, provided no external force or

support reaction is applied to the joint.

X

X X

Simple Truss

A structure composed of slender members

joined together at their end points

“Planar Truss”

Simple Truss

To Design members and connections of a

truss, necessary to determine forces

developed in each member

Assumptions:

All loadings are applied at the joints

Members are joined together by smooth pins

Assumptions satisfactory for bolted or welded joints

Simple Truss

Each truss member acts as a 2-force member

Force directed along axis of member

Tension – pull, elongation

Compression – push, shorten

Several analytical approaches to solving

Free Body Diagrams

Which of the following is NOT a “ simple truss”?

A B

C D

1. A

2. B

3. C

4. D

5. More

than 1

Method of Joints

If truss is in equilibrium, each joint must also be in equilibrium

Approach:

Sketch Free Body Diagram of Entire Truss

Solve for Reactions

Sketch Free Body Diagrams of Each Joint

2 unknown limit per FBD

∑ Fx =0 ∑ Fy =0 ∑ Mo =0

Example Problem

Given:

F1 = 400 N

F2 = 600 N

Find:

Force in each

member

4 m

BC

A

3 m

F1

F23 m D

Example Problem SolutionGiven:

F1 = 400 N

F2 = 600 N

Find:

Force in each member

4 m

BC

A3 m

F1

F23 m D

Example Problem Solution

0 yF

0 CM

0 xF

600 N

4 m

B

Cy

Ay

3 m

400 N

3 m D

Cx

Given:

F1 = 400 N

F2 = 600 N

Find:

Force in each member

Solution:

1-FBD

2-Support Reactions

4 m

BC

A3 m

F1

F23 m D

0600 xC NorCx 600600

0460063400 yA

NAy 600

0600400 yC

NorCy 200200

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Example Problem Solution

0 yF

0 xF

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

Joint A

600 N

AD

AB

3

4

Joint D

600 N

BD

3

4

AD

450 N

CD

0 yF

0 xF

05

4600 AB

CNAB 750

07505

3AD

TNAD 450

05

3450600 BD

TNBD 250

02505

4CD

CNCD 200

Example Problem Solution

0 xF

600 N

4 m

B

200 N

600 N

3 m

400 N

3 m D

600 NC

A

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

Joint C

CD

200 N

BC 600 N

200 N

B

200 C

450 TD

600 C C

A

Force Diagram

0600 BCCNBC 600

Zero Force Members

Zero Force Members support NO loading

Used for:

Stability

Changing Applied Loads

Zero Force Members

If only 2 members form a truss joint AND no

external load or support reaction is applied to

the joint, then the members must be Zero

Force Members

X

X

XX

Zero Force Members

If 3 members form a truss joint for which 2

members are collinear, the 3rd member is a Zero

Force Member, provided no external force or

support reaction is applied to the joint.

X

X

In Class Exercise

Given:

F = 3 kN

θ = 30º

Find:

Force in each member

2 m

B

CA

2 m

F

D

2 m

θ

B

D

A C4.1

0 T

Solution

Given:

F = 3 kN

θ = 30º

Find:

Force in each member

Solution:

1-FBD

2-Support Reactions

0 yF

0 AM

0 xF

2 m

B

Ax

2 m

3 kN

D

2 m

30º

Ay Cy

2 m

B

CA

2 m

F

D

2 m

θ

03 xA

kNorAx 33

0234 yC

kNCy 5.1

05.1 yA

kNorAy 5.15.1

Solution

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

0 yF

TkNAD

CkNAB

10.4

777.

0 xF

2 m

B

3 kN

2 m

3 kN

D

2 m

30º

1.5 kN 1.5 kN

A C

Joint A

1.5 kN

AD

AB

30º

45º3 kN

Solving simultaneous eqn:

BCAB

.777 kN

45º

3 kN

BD

0 yF

0 xF

Joint B

330cos45cos ADAB

5.130sin45sin ADAB

0345sin45sin777. BCCkNBC 02.5

045cos02.545cos777. BDTkNBD 10.4

Solution

Solution:

1-FBD

2-Support Reactions

3-Joint FBDs

0 xF

Joint D

AD

4.10 kN

BD

4.10 kN

30ºDC

B

D

A C

4.1

0 T

Force Diagram2 m

B

3 kN

2 m

3 kN

D

2 m

30º

1.5 kN 1.5 kN

A C

030cos10.430cos DCTkNDC 10.4

In Class Exercise

Solution

In Class Exercise

Solution