1. Definition of Truss 2. Planar Trusses A structure composed of slender members jointed together at their end points by bolting, welding or pinning.

1. Definition of Truss

2. Planar Trusses

A structure composed of slender members jointed together at their end points by bolting, welding or pinning.

2F

Bolting

1F3F

planar trusses lie in a single and one often need to support roofs and bridges.

6.1 Simple Trusses

Roof truss Bridge truss

3. Assumption for Truss

(2) members are jointed together by smooth pin (No friction forces)

(1) all loading are applied at the jointsa. members’ weights are neglected (W << external force)b. members’ weights are included

T

T T T

C

C Compression force

Tensile force

Because of the two assumptions, the truss member is a two-force member

Note: Compression member must be made thicken than tension member because of the buckling or column effect in compression member.

A simple truss is constructed by starting with a basic triangular element and connecting two members to form an additional element.

unstable truss

PC

DA

E

D B

CA

Simple form of rigid or stable truss

4. Simple Truss

1. Objective Evaluate the force acting on each member of a truss structure to analyze or design a truss

2. Concept A truss is in equilibrium, so each of its joint also is in equilibrium Force system acting at pin or joint is coplanar and concurrent

6.2 the Method of Joints

Therefore, we have(A)member equilibrium is automatically satisfied at each joint(B)Only force equilibrium ΣFx=0 and ΣFy=0 is necessary

3. Analysis Procedure

Support A is pin constraint.Support C is roller constraint.

(1) Draw the free-body diagram of a joint having at least one known force and at most two unknown forces.

For the given truss, only joint B is satisfied.

F.B.D of joint B

(2) Establish the sense of the unknown forces (a) assume the unknown member forces to be in tension.

(b) Determine correct sense of unknown member forces by inspection

(3) Apply the force equilibrium equations

ΣFx=0 500+FBC sin45°=0

FBC=- 500/sin45°=-500√2N (in compression)FAB=-FBCcos45°=500N (in tension)

ΣFy=0 -FAB-FBCcos45°=0

(4) Solve the unknown forces and check their correct sense

(5) Continue to analyze then joints by repeating steps (1) to (4)

choose joint C first, then joint A

equilibrium equation at joint C

ΣFx=0 500√2 cos45°-FAC=0

ΣFy=0 Cy-500√2 sin45°=0

FAC=500N (in tension); Cy=500N

F.B.D joint A

F.B.D of joint C

Ax=Ay=500N

6-3 zero-force members1. zero-force member

A member supports no loading, which is used to increase the stability of the truss and to provide support if the applied loading is changed.

Ex.

2. Rules for determining zero-force members

(1) If only two members form a truss joint and no external load or support reaction is applied to the joint.

F E

C

BA

D

Joint A and joint D are formed by two members AF &AB, ED&CD respectively and

no loading is applied. Hence, members AF, A, ED and CD are zero-force members.

F.B.D of joint A

FAF

FAB

F.BD of joint D

FED

FDCx y

•ΣFx=0, FED+FDC cosθ=0

•ΣFy=0, FDC sinθ=0

So, FDC=0,FED=0

•ΣFx=0 , so FAB=0

• ΣFy=0 , so FAF=0A

Equations of equilibrium

D

Equations of equilibrium

The equivalent system is given as shown.

FE

C

P

b

B

P

D

C

B

E

A

(2) If three members form a truss joint for which two of the members are collinear , the third member is zero-force members provided no external force or support reaction is applied to the joint.

EX:

Joint D :3 members AD, ED and CD (ED &DC collinear)

Joint C :3 members AD, DC and CB (DC& CB collinear)

No external force or support reaction is applied to the joint D & C.

∴AD and AC zero-force members

6-4 The method of section1. Purpose

Determine the loadings acting within a body .

2. Principle

If a body is equilibrium , then any part of the body is also in equilibrium.

T

T

In equilibrium T

FF

0F In equilibrium

0F In equilibrium

F=T

F=T

Internal tension force

Apply the equations of equilibrium to the sectioned part to determine the loading at the section.

Imaginary section

T

3. Analysis procedure

(1) Determined the external reactions at the constraints F.B.D of entire truss.

CA

B

2m

2m45

500N

Equations of equilibrium500N

C

CyAy

x

y

AxA

2m

B

0xF 500-Ax=0+

F.B.D. of entire truss

+ 0yF Cy-Ax=0

+ 0MA 500x2-Cyx2=0

(2) Cut or section the truss through not more than three members in which the forces are unknown. (Because three independent equilibrium equations for three unknowns.)

three possible ways of section for given truss

C

500N

500N

500N

A

500N

(1)(2)

(3)

(3) Draw the free-body diagram of the part of sectioned truss which has the least number of forces acting on it.

500N500N

500N

500NB

A C

Section(1)

(4) Establish the sense of the unknown member force

y

x

F FBC

500N45

AB

(5) Apply the equations of equilibrium and check the correct sense of solve forces.

+ 0Fx 500N-FBCsin45=0

+ 0Fy FBCcos45-FAB=0

FBC=500 N2 (in compression)

FAB=500N (in tension)

(a) Moments should be summed about a point lying at the intersection of the lines of actions of two unknown forces. The third unknown is determined directly from the moment equation.

(b) Forces may be summed perpendicular to the direction of the two unknown forces which are parallel.

(6) Continue to analyze other new sections by repeating steps(1)to(5).

CA

B

500N

N2500

x

y

FAC

+ 0Fy 045cos2500 FAC

+ 0Fy 050045sin2500 So, FAC=500N

Section (2)

F.B.D. of right part of section (2)

Equations of equilibrium

6.6 Frames and Machines

1. Definition (1)Frames A stationary structure composed of pin-connected multiforce members is used to support loads.

2F 1F

Ex:

bicycle frame hoist

E

Engine jig.油壓缸

(2) Machine A structure composed of pin-connected multiforce members with moving parts is designed to transmit and alter the effect of forces.

CAN

FMoving part Ex:crusher

2. Assumptions

(1) The structure is properly supported.

(2) The structure contains no more supports or members than necessary to prevent its collapse.

(3) The joint reactions of the structure can be determined from the equilibrium equations.

3. Analysis procedure

2m 2m

200N

C

B

A

3m

60

(1) Draw the free body Diagram of entire structure, a part of structure, or each of its members. (a) Isolate each part by drawing its outline shape. Show all forces and/or couple moments act on the part. (b) Identify all the two-force member in the structure. (c) Forces common to any two contacting members act with equal magnitudes but opposite sense on respective members.

By inspection, member AB is a two-force member.

60 0

A

ABF

ABF

2m 2m

200NCx

60Cy

Free body diagrams of members AB & BC.

(2) Apply the equations of equilibrium (a) Count the total number of unknowns to make sure that an equivalent number of equilibrium equations can be written for solutions.

ABF , Cx , Cy (3)Unkoown:

(b) Moments should be summed about a point that lies at the intersection of the lines of action of as many unknown forces as possible.

Equations of equilibrium for member BC.

0

0

0

c

y

x

M

F

F

022004)60sin(

020060sin

060cos

oAB

oABy

xo

AB

F

FC

CF

NC

NC

NF

y

x

AB

1000

577

7.1154

(2) Check the correct sense of the unknown forces.

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