Chapter 3: Chapter 3:

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Chapter 3:

Analysis of Statically Determinate Trusses

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Common Types of Trusses

• A truss is a structure composed of slender A truss is a structure composed of slender members joined together at their end points members joined together at their end points

• The joint connections are usually formed by The joint connections are usually formed by

bolting or welding the ends of the members to a bolting or welding the ends of the members to a

common plate called gusset common plate called gusset

• Planar trusses lie in a single plane & is often used Planar trusses lie in a single plane & is often used to support roof or bridges

to support roof or bridges

Chapter 3: Analysis of Statically Determinate Trusses

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Common Types of Trusses

• Roof Trusses Roof Trusses

• They are often used as part of an industrial building They are often used as part of an industrial building frame

frame

• Roof load is transmitted to Roof load is transmitted to the truss at the joints by the truss at the joints by

means of a series of purlins means of a series of purlins

• To keep the frame rigid & thereby capable of To keep the frame rigid & thereby capable of

resisting horizontal wind forces, knee braces are

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Common Types of Trusses

• Roof Trusses Roof Trusses

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Common Types of Trusses

• Bridge Trusses Bridge Trusses

• The load on the deck is first transmitted to the The load on the deck is first transmitted to the

stringers -> floor beams -> joints of supporting side stringers -> floor beams -> joints of supporting side

truss truss

• The top & bottom cords of these side trusses are The top & bottom cords of these side trusses are connected by top & bottom lateral bracing resisting connected by top & bottom lateral bracing resisting

lateral forces

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Common Types of Trusses

• Bridge Trusses Bridge Trusses

• Additional stability is provided by the portal & sway Additional stability is provided by the portal & sway bracing

bracing

• In the case of a long span truss, a roller is provided In the case of a long span truss, a roller is provided at one end for thermal expansion

at one end for thermal expansion

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Common Types of Trusses

• Assumptions for Design Assumptions for Design

• The members are joined together by smooth pins The members are joined together by smooth pins

• All loadings are applied at the joints All loadings are applied at the joints

• Due to the 2 assumptions, each truss member acts Due to the 2 assumptions, each truss member acts as an axial force member

as an axial force member

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Simple , Compound or Complex Truss Simple , Compound or Complex Truss

• Simple Truss Simple Truss

• To prevent collapse, the framework of a truss must To prevent collapse, the framework of a truss must be rigid

be rigid

• The simplest framework that is rigid or stable is a The simplest framework that is rigid or stable is a triangle

triangle

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Simple Truss Simple Truss

• A simple truss is the basic “stable” triangle element A simple truss is the basic “stable” triangle element is ABC

is ABC

• The remainder of the joints D, E & F are established The remainder of the joints D, E & F are established in alphabetical sequence

in alphabetical sequence

• Simple trusses do not have to consist entirely of Simple trusses do not have to consist entirely of triangles

triangles

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Compound Truss Compound Truss

• It is formed by connecting 2 or more simple truss It is formed by connecting 2 or more simple truss together

together

• Often, this type of truss is used to support loads Often, this type of truss is used to support loads acting over a larger span as it is cheaper to

acting over a larger span as it is cheaper to

construct a lighter compound truss than a heavier construct a lighter compound truss than a heavier

simple truss simple truss

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Compound Truss Compound Truss

• Type 1 Type 1

• The trusses may be connected by a common joint & The trusses may be connected by a common joint &

bar bar

• Type 2 Type 2

• The trusses may be joined by 3 bars The trusses may be joined by 3 bars

• Type 3 Type 3

• The trusses may be joined where bars of a large The trusses may be joined where bars of a large

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Compound Truss Compound Truss

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Complex Truss Complex Truss

• A complex truss is one that cannot be classified as A complex truss is one that cannot be classified as being either simple or compound

being either simple or compound

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Classification of Coplanar Trusses Classification of Coplanar Trusses

• Determinacy Determinacy

• Total unknowns = forces in b no. of bars of the Total unknowns = forces in b no. of bars of the truss + total no. of external support reactions truss + total no. of external support reactions

• Force system at each joint is coplanar & concurrent Force system at each joint is coplanar & concurrent

• Rotational or moment equilibrium is automatically Rotational or moment equilibrium is automatically satisfied

satisfied

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• Determinacy Determinacy

• Therefore only Therefore only

• By comparing the total unknowns with the total no. By comparing the total unknowns with the total no.

of available equilibrium eqn, we have:

Classification of Coplanar Trusses Classification of Coplanar Trusses

ate indetermin statically

2 e determinat statically

2 j r

b

j r

b









 

 F

_x

 0 and F

_y

0

(16)

• Stability Stability

• If b + r < 2j => collapse If b + r < 2j => collapse

• A truss can be unstable if it is statically determinate A truss can be unstable if it is statically determinate or statically indeterminate

or statically indeterminate

• Stability will have to be determined either through Stability will have to be determined either through inspection or by force analysis

inspection or by force analysis

Classification of Coplanar Trusses Classification of Coplanar Trusses

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• Stability Stability

• External Stability External Stability

• A structure is externally unstable if all of its reactions A structure is externally unstable if all of its reactions are concurrent or parallel

are concurrent or parallel

• The trusses are externally unstable since the support The trusses are externally unstable since the support reactions have lines of action that are either

reactions have lines of action that are either concurrent or parallel

concurrent or parallel

Classification of Coplanar Trusses

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• Internal Stability Internal Stability

• The internal stability can be checked by careful The internal stability can be checked by careful inspection of the arrangement of its members inspection of the arrangement of its members

• If it can be determined that each joint is held fixed If it can be determined that each joint is held fixed so that it cannot move in a “rigid body” sense wrt so that it cannot move in a “rigid body” sense wrt

the other joints, then the truss will be stable the other joints, then the truss will be stable

• A simple truss will always be internally stable A simple truss will always be internally stable

• If a truss is constructed so that it does not hold its If a truss is constructed so that it does not hold its joints in a fixed position, it will be unstable or have a joints in a fixed position, it will be unstable or have a

“critical form”

Classification of Coplanar Trusses Classification of Coplanar Trusses

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• Internal Stability Internal Stability

• To determine the internal stability of a compound To determine the internal stability of a compound

truss, it is necessary to identify the way in which the truss, it is necessary to identify the way in which the

simple truss are connected together simple truss are connected together

• The truss shown is unstable since the inner simple The truss shown is unstable since the inner simple

truss ABC is connected to DEF using 3 bars which are truss ABC is connected to DEF using 3 bars which are

concurrent at point O concurrent at point O

Classification of Coplanar Trusses

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• Internal Stability Internal Stability

• Thus an external load can be applied at A, B or C & Thus an external load can be applied at A, B or C &

cause the truss to rotate slightly cause the truss to rotate slightly

• For complex truss, it may not be possible to tell by For complex truss, it may not be possible to tell by inspection if it is stable

inspection if it is stable

• The instability of any form of truss may also be The instability of any form of truss may also be noticed by using a computer to solve the 2j

noticed by using a computer to solve the 2j simultaneous eqns for the joints of the truss simultaneous eqns for the joints of the truss

• If inconsistent results are obtained, the truss is If inconsistent results are obtained, the truss is unstable or have a critical form

unstable or have a critical form

Classification of Coplanar Trusses Classification of Coplanar Trusses

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Classify each of the trusses as stable, unstable, statically

determinate or statically indeterminate. The trusses are subjected to arbitrary external loadings that are assumed to be known &

can act anywhere on the trusses.

Example 3.1

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For (a),

• Externally stable

• Reactions are not concurrent or parallel

• b = 19, r = 3, j = 11

• b + r =2j = 22

• Truss is statically determinate

• By inspection, the truss is internally stable

Solution Solution

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For (b),

• Externally stable

• b = 15, r = 4, j = 9

• b + r = 19 >2j

• Truss is statically indeterminate

• By inspection, the truss is internally stable

Solution

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For (c),

• Externally stable

• b = 9, r = 3, j = 6

• b + r = 12 = 2j

• Truss is statically determinate

• By inspection, the truss is internally stable

Solution Solution

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For (d),

• Externally stable

• b = 12, r = 3, j = 8

• b + r = 15 < 2j

• The truss is internally unstable

Solution

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• Satisfying the equilibrium eqns for the forces Satisfying the equilibrium eqns for the forces exerted on the pin at each joint of the truss exerted on the pin at each joint of the truss

• Applications of eqns yields 2 algebraic eqns that Applications of eqns yields 2 algebraic eqns that can be solved for the 2 unknowns

can be solved for the 2 unknowns

The Method of Joints The Method of Joints

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• Always assume the unknown member forces Always assume the unknown member forces acting on the joint’s free body diagram to be in acting on the joint’s free body diagram to be in

tension tension

• Numerical solution of the equilibrium eqns will Numerical solution of the equilibrium eqns will yield positive scalars for members in tension &

yield positive scalars for members in tension &

negative for those in compression negative for those in compression

• The correct sense of direction of an unknown The correct sense of direction of an unknown

member force can in many cases be determined member force can in many cases be determined

The Method of Joints

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• A +ve answer indicates that the sense is correct, A +ve answer indicates that the sense is correct, whereas a –ve answer indicates that the sense whereas a –ve answer indicates that the sense

shown on the free-body diagram must be reversed shown on the free-body diagram must be reversed

The Method of Joints The Method of Joints

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Determine the force in each member of the roof truss as shown.

State whether the members are in tension or compression. The State whether the members are in tension or compression. The reactions at the supports are given as shown.

reactions at the supports are given as shown.

Example 3.2

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Only the forces in half the members have to be determined as the truss is symmetric wrt both loading & geometry,

Solution Solution

) ( 93

. 6

0 30

cos 8

; 0

) ( 8

0 30

sin 4

; 0

A, Joint

0 0

T kN F

F F

C kN F

F F

AB

AB x

AG

AG y



  





  





(31)

Solution Solution

) ( 50

. 6

0 30

sin 3 8

; 0

) ( 60

. 2

0 30

cos 3

; 0 G,

Joint

0

C kN F

F F

C kN F

F F

GF

GF x

GB

GB y





  





  





(32)

Solution Solution

) ( 33

. 4

0 93

. 6 60

cos 60

. 2 60

cos 60

. 2

; 0

) ( 60

. 2

0 60

sin 60

. 2 60

sin

; 0 B,

Joint

0 0

T kN F

F F

T kN F

F F

BC

BC x

BF

BF y







  





  





(33)

• Truss analysis using method of joints is greatly Truss analysis using method of joints is greatly simplified if one is able to first determine those simplified if one is able to first determine those

members that support no loading members that support no loading

• These zero-force members may be necessary for These zero-force members may be necessary for the stability of the truss during construction & to the stability of the truss during construction & to provide support if the applied loading is changed provide support if the applied loading is changed

• The zero-force members of a truss can generally The zero-force members of a truss can generally be determined by inspection of the joints & they be determined by inspection of the joints & they

Zero-Force Members

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• Case 1 Case 1

• The 2 members at joint C are connected together at The 2 members at joint C are connected together at a right angle & there is no external load on the joint a right angle & there is no external load on the joint

• The free-body diagram of joint C indicates that the The free-body diagram of joint C indicates that the force in each member must be zero in order to

force in each member must be zero in order to maintain equilibrium

maintain equilibrium

Zero-Force Members Zero-Force Members

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• Case 2 Case 2

• Zero-force members also occur at joints having a Zero-force members also occur at joints having a geometry as joint D

geometry as joint D

Zero-Force Members

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• Case 2 Case 2

• No external load acts on the joint, so a force No external load acts on the joint, so a force

summation in the y-direction which is perpendicular summation in the y-direction which is perpendicular

to the 2 collinear members requires that F

_DF_DF

= 0 = 0

• Using this result, FC is also a zero-force member, as Using this result, FC is also a zero-force member, as indicated by the force analysis of joint F

indicated by the force analysis of joint F

Zero-Force Members Zero-Force Members

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Using the method of joints, indicate all the members of the truss that have zero force.

Example 3.4

(38)

We have,

Solution Solution

0 0 0

; 0 0

0 sin

; 0

D, Joint



  





  





DE

DE x

DC

DC y

F

F F

F

F 

(39)

Solution Solution

G, Joint

0 ; 0 H,

Joint

0 ; 0 E, Joint

  





  



HB y

EF x

F F

F

(40)

• If the forces in only a few members of a truss are If the forces in only a few members of a truss are to be found, the method of sections generally

to be found, the method of sections generally

provide the most direct means of obtaining these provide the most direct means of obtaining these

forces forces

• This method consists of passing an imaginary This method consists of passing an imaginary section through the truss, thus cutting it into 2 section through the truss, thus cutting it into 2

parts parts

• Provided the entire truss is in equilibrium, each of Provided the entire truss is in equilibrium, each of the 2 parts must also be in equilibrium

the 2 parts must also be in equilibrium

The Method of Sections The Method of Sections

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• The 3 eqns of equilibrium may be applied to either The 3 eqns of equilibrium may be applied to either one of these 2 parts to determine the member

one of these 2 parts to determine the member forces at the “cut section”

forces at the “cut section”

• A decision must be made as to how to “cut” the A decision must be made as to how to “cut” the truss

truss

• In general, the section should pass through not In general, the section should pass through not more than 3 members in which the forces are more than 3 members in which the forces are

unknown unknown

The Method of Sections

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• If the force in GC is to be determined, section aa If the force in GC is to be determined, section aa will be appropriate

will be appropriate

• Also, the member forces acting on one part of the Also, the member forces acting on one part of the truss are equal but opposite

truss are equal but opposite

• The 3 unknown member forces, F The 3 unknown member forces, F _BC _BC , F , F _GC _GC & F & F _GF _GF can can be obtained by applying the 3 equilibrium eqns

be obtained by applying the 3 equilibrium eqns

The Method of Sections The Method of Sections

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• When applying the equilibrium eqns, consider When applying the equilibrium eqns, consider

ways of writing the eqns to yield a direct solution ways of writing the eqns to yield a direct solution

for each of the unknown, rather than to solve for each of the unknown, rather than to solve

simultaneous eqns simultaneous eqns

The Method of Sections

(44)

Determine the force in members CF and GC of the roof truss.

State whether the members are in tension or compression. The reactions at the supports have been calculated.

Example 3.5 Example 3.5

(45)

The free-body diagram of member CF can be obtained by considering the section aa,

Solution Solution

) ( 73

. 1

0 )

31 . 2 ( 50 . 1 ) 4 ( 30 sin

0 ve,

as moments clockwise

- anti With

. simplicity for

C point to

slide is

ibility, transmiss

of Principal Applying

0 applying

by obtained

be can F

for solution

direct

A

_CF

C kN F

F

M F

M

CF

o CF

E CF

E







 



 

(46)

The free-body diagram of member GC can be obtained by considering the section bb,

Solution Solution

) ( 73

. 1

0 )

4 ( 30 sin 73 . 1 ) 4 ( )

31 . 2 ( 50 . 1

0 ve,

as moments clockwise

- anti With

. simplicity for

C point to

slide is

: have we

C, point to

Sliding .

and unknowns

the eliminate

order to in

A point about

summed be

will Moments

T kN F

F

M F

F F

F

GC

o GC

A CF

CF CD

HG









 



(47)

Determine the force in member GF and GD of the truss. State whether the members are in tension or compression. The

reactions at the supports have been calculated.

Example 3.6

(48)

The distance EO can be determined by proportional triangles or realizing that member GF drops vertically 4.5 – 3 = 1.5m in 3m.

Hence, to drop 4.5m from G the distance from C to O must be 9m

Solution Solution

(49)

The angles F

_GD

and F

_GF

make with the horizontal are tan

^-1

(4.5/3) = 56.3

^o

tan

^-1

(4.5/9) = 26.6

^o

Solution Solution

0 )

3 ( 7 ) 6 ( 6 . 26 sin

0 ve,

as moments clockwise

- anti With

O. point to

slide is

0 applying by

directly determined

be can GF

in force The

F

M F

M

o GF

D GF

D







 



 

(50)

Solution Solution

) ( 80

. 1

0 )

6 ( 3 . 56 sin )

6 ( 2 )

3 ( 7

0 ve,

as moments clockwise

- anti With

D. point to

slide is

0 applying by

directly determined

be can GD

in force The

C kN F

F

M F

M

GD

o GD

O GD

O







 



 