lecture 7 plane truss full page

8/18/2019 Lecture 7 Plane Truss Full Page

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TYPES OF ROOF TRUSS


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TYPES OF ROOF TRUSS


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ROOF TRUSS SETUP


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ROOF TRUSS SETUP


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OBJECTIVES

• To determine the STABILITY and

DETERMINACY of plane trusses

• To analyse and calculate the FORCES in truss

members

• To calculate the DEFORMATION at any joints


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• Transmitting loading from roof to columns – bymeans of series of purlins.

• Trusses used to support roofs are selected on the

basis of the span, slope and roof materials.

ROOF TRUSSES

Metal Gusset

Top Chord

Truss Web

Bottom Chord


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ROOF TRUSSES


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ROOF TRUSSES


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BRIDGE TRUSSES


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BRIDGE TRUSSES


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• The members are joined together by smooth pins

(NO friction)

• All loadings and reactions are applied at the joints

• The centroid for each members are straight andconcurrent at a joint

Each truss member acts as an axial force member.

If the force tends to elongate → tensile (T)

If the force tends to shorten → compressive (C)

ASSUMPTIONS IN DESIGN


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Tensile (positive)

Compressive (negative)

SIGN CONVENTION


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• In terms of stability, the most simple truss can be

constructed in triangle using three members.

• This shape will provide stability in both x and y

direction. Each additional element of twomembers will increase one number of joint

STABILITY & DETERMINACY


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There are 3 types of stable trusses:

1. Simple Truss

2. Compound Truss – combination of two or more

simple trusses together

3. Complex Truss – one that cannot be classified as

being either simple or compound



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Simple Truss


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Compound Truss


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Complex Truss


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• Let consider a simple truss.

• There will be 6 unknown values: 3 internal member

forces and 3 reactions

m + R → m + 3• And for every joint, 2 equilibriums can be written (F x

= 0 and F y =0) – no rotation or moment at joint 2

j

• By comparing the total unknowns with total number

of available equation, we can check the determinacy.

DETERMINACY


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• The determinacy of truss should be checked internallyand externally

• The external determinacy is given by:

R = 3 (provided that the support reactions have nolines of action that are either concurrent or

parallel)

• If R > 3 Statically indeterminate (external)

R = 3 Statically determinate (external)

R < 3 Unstable truss system

DETERMINACY


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• The internal determinacy is given bym = 2j – 3 (provided that the components of the

truss do not form a collapsible mechanism)

• If m > 2j – 3 Statically indeterminate (internal)

m = 2j – 3 Statically determinate (internal)

m < 2j – 3 Unstable truss system

DETERMINACY


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Determine the stability and determinacy of the

truss shown in the figure below.

EXAMPLE 1


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Externally:R = 3 ; R – 3 = 3 – 3 = 0 OK

Internally:m = 9, j = 6,

9 = 2(6) – 3 = 9 OK

Therefore, the truss is determinate

(externally and internally)

EXAMPLE 1 – Solution


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Determine the stability and determinacy of the

truss shown in the figure below.

EXAMPLE 2


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Externally:R = 3 ; R – 3 = 3 – 3 = 0 OK


9 = 2(6) – 3 = 9 OK

Therefore, the truss is determinate

(externally and internally)



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Externally:

R = 4 ; R – 3 = 4 – 3 = 1 1 degree redundant


10 > 2(6) – 3 : 10 > 9 1 degree redundant

Therefore, the truss is internally and externally

indeterminate (1 degree redundant)



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There are several methods of calculating the member

forces for the truss

i. Method of Joints

ii. Method of Sections

iii. Method of Force Resolution

MEMBER FORCES


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• Suitable to be used to determine all the memberforces in the truss

• In this method, every joint will be analysed by drawing

the Free Body Diagram, limiting the unknown values

to TWO only.

• The selected joints must only consisted concurrent

and coplanar forces

• Using the equilibrium of F x = 0 and F y =0, we canstart and solve the problems.

METHOD OF JOINTS


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Determine all the member forces for the given trussbelow.

EXAMPLE 4

150 kN 50 kN

B C

D

A

F

G

100 kN

E6m

3m3m8m


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1. Check the Stability and Determinacy

Externally:

R = 3 ; R – 3 = 3 – 3 = 0 OK

Internally:

m = 11, j = 7,

11 = 2(7) – 3 = 11 OK

Therefore, truss is determinate (externally and

internally)



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2. Calculate the Reactions at the Support

+ MA = 0

100 (6) + 150 (8) + 50 (11) – RG (14) = 0

RG = 167.9 kN ( )

+ F y = 0 ; RA – 150 – 50 – 167.9 kN ;

RA = 32.1 kN ( )

+ F x = 0 ; H A = 100 kN ( )



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3. Analyse Every Joints

At Joint A

+F x = 0 ; 100 + F AD = 0F AD = 100 kN (T)

+ F y = 0 ; F AB + 32.1 = 0

F AB = 32.1 kN (C)

A

F AB

F AD

32.1

100



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At Joint B

+F y = 0 ; 32.1 – F BD (6/10) = 0

F BD = 53.5 kN (T)

+ F x = 0 ; F BC + 100 + 53.5 (8/10) = 0

F BC = 142.8 kN (C)

B F BC

32.1

100

F BD



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At Joint C

+F x = 0 ; 142.8 + F CE (3/√18) = 0

F CE =

201.9 kN (C)

+ F y = 0 ; F CD – (201.9) (3/√18) = 0

F CD = 142.8 kN (T)

C142.8

F CEF CD



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At Joint D

+F y = 0 ; 142.8 + 53.5 (6/10) – 150

+ F DE (3/√18) = 0F DE = 35.2 kN (C)

+ F x = 0; 100 – 53.5 (8/10) + (-35.2)(3/√18) + F DF = 0

F DF = + 166.7 kN (T)

D

142.8

F DF

150

100

53.5 F DE



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At Joint E

+F x = 0 ; F EG (3/√18) + 201.9 (3/√18)

+ 35.2 (3/√18) = 0

F EG = 237.1 kN (C)

+ F y = 0; F EF 201.9 (3/√18) + 35.2 (3/√18) –

(237.1)(3/√18) = 0

F EF = 49.8 kN (T)

E 142.8

F EG F EF

35.2



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At Joint F

+F x = 0 ; 167.7 + F FG = 0

F FG = 167.7 kN (T)F

49.8

F FG

50

166.7



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At Joint G (Checking)

+F y = 0 ;

167.9 – 237.1 (3/√18) = 0 OK !

+ F x = 0; 167.7 + 237.1 (3/√18) = 0 OK !


G

237.1

167.9

166.7


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Member Force

AB

32.1 CAD +100 T

BC 142.8 C

BD +53.5 T

CD +142.8 T

CE 201.9 C


Member Force

DE

35.2 CDF +167.7 T

EF +49.8 T

EG 237.1 C

FG +167.7 T

Summary: Internal Member Forces


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• When only some of the member forces need to becalculated, it is suitable to use this method. However, it

can also used to determine all the member forces in

truss.

• The method of sections consists of cutting through thetruss into two parts, provided that the unknown values

are not more than three

•The unknown forces will be assumed to be either intension or compression

• Three equilibriums (F x = 0, F y = 0, M = 0) will be

used to solve the problems.

METHOD OF SECTIONS


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Stability and DeterminacyExternally:

R = 3 ; R – 3 = 3 – 3 = 0 OK

Internally:

m = 9, j = 6,

9 = 2(6) – 3 = 9 OK

Therefore, the truss is determinate (externally and

internally)



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Calculate the Reactions at the Support

+ MF = 0

(2/√5)RA (12) + (2/√5)RA (2) –

100 (8) – 40 (4) = 0RA = 82.6 kN ( )

+ F y = 0 ; RF + (2/√5)(82.6) – 100 – 40 = 0 ;

RF

= 66.2 kN ( )

+ F x = 0 ; HF + (1/√5)(82.6) – 80 = 0 ;

H F = 43.1 kN ( )



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Section 1:

MD = 0

F EC (2) – 43.1 (2) – 66.2 (4) = 0 ; F EC = 175.5 kN (T)

F y = 0 ; 40 + 66.2 – F DC (2/√20) = 0 ; F DC = 58.6 kN (T)

F x = 0 ; F DB – 58.6 (4/√20) – 175.5 + 43.1 = 0 ;

F DB = 184.8 kN (C)


40 kN

F EC

D

F DC

F DB

66.2

43.1

F E


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Section 2:

MF = 0 ; F ED (4) – 40 (4) = 0 ; F ED = 40 kN (T)

40 kN

F EC

F FD F ED

66.2

43.1


F E


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Case 1:

If only two members form a truss joint, and no external

load or support is applied, the members must be zero-

force members

Case 2:

If three members form a truss joint for which two of the

members are collinear, the third member will be a zero-

force member (provided no external load or support

reaction acting at the joint).

ZERO FORCE MEMBERS

DEFORMATION OF STATICALLY


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• The deformation of statically determinate plane truss

can be determined using Virtual Work Method

•Consider to determine vertical deformation at joint C

• Due to external loads, point C will deform – producing


DETERMINATE PLANE TRUSS



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‘ Actual Structure’

A

B

C E

F

G H



D



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• Now, eliminate all external loads and assign 1 unit load(vertical) at joint C. This structure is known as ‘Virtual

Structure’



C

‘Virtual Structure’



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• Both structures will then combined, thus producing theconcept of ‘work’. Therefore, the external work:

W = 1 .

• Say P is the member force due to external loads and u isthe member force due to unit load

• If we consider one of the truss members, having force

of P , this member will produce certain deformation

which can be calculated, given as:

= PL/ AE





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• Through combination, the amount of internal work is

given by = • According to Energy Work Method:

External load = Internal Load

1 ∙ ∆ =




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1. Calculate the member forces for “ Actual Structure”.2. Eliminate all external loads and assign ONE (1) unit load

in the same direction of deformation. Then, calculate

the member forces of “Virtual Structures”.

3. The deformation at point C can be then calculated

using:

SOLUTION PROCEDURE


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•When a structure is loaded, its stressed elementsdeform. As these deformations occur, the structure

changes shape and points on the structure displace.

• Work – the product of a force times a displacement

in the direction of the force

VIRTUAL WORK METHOD


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•External Work – when a force F undergoes adisplacement dx in the same direction as the force.

• Internal Work – when internal displacements δ occur

at each point of internal load u.

VIRTUAL WORK METHOD

Work ofExternal

Load

Work ofInternal

Load

∙ ∆ = ∙


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•When a bar is loaded axially, it will deform and storestrain energy u.

• A bar (as shown in the figure) subjected to the

externally applied load P induces an axial force F of

equal magnitude (F = P). If the bar behaveselastically (Hooke’s Law), the magnitude of the strain

energy u stored in a bar by a force that increases

linearly from zero to a final value F as the bar

undergoes a change in length dL.

VIRTUAL WORK METHOD


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F

x

Δ

P

x P

F

VIRTUAL WORK METHOD

= From Hooke’s Law:

L

dL

P

F


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• We can use the method of virtual work to determinethe displacement of a truss joint when the truss is

subjected to an external loading, temperature

change, or fabrication errors.

• When a unit force acting on a truss joint, andresulted a displacement of Δ, the external work =

1Δ.

DISPLACEMENT OF TRUSSES


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•Due to the unit force, each truss member will carryan internal forces of u, which cause the deformation

of the member in length dL. Therefore, the

displacement of a truss joint can be calculated by

using the equation of:

dLu..1

Virtual

loadings

Real

displacements



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1. Place the unit load on the truss at the joint where

the desired displacement is to be determined. The

load should be in the same direction as the specified

displacement; e.g. horizontal or vertical.

1

50 kN

20 kN

STEPS FOR ANALYSIS

A

B

C

D


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2. With the unit load so placed, and all the real loadsremoved from the truss, use the method of joints or

the method of sections and calculate the internal

force in each truss member. Assume that tensile

forces are positive and compressive forces arenegative.

3. Use the method of joints or the method of sections

to determine the internal forces in each member.

These forces are caused only by the real loads

acting on the truss. Again, assume tensile forces are

positive and compressive forces are negative.

STEPS FOR ANALYSIS


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• If the resultant sum of displacement is positive, thedirection is same as the unit load or vice-versa.

• When applying any formula, attention should be paid

to the units of each numerical quantity. In particular,

the virtual unit load can be assigned any arbitrary

unit (N, kN, etc.).

STEPS FOR ANALYSIS


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Determine the vertical displacement of joint C of thesteel truss shown in Figure. The cross-sectional area of

each member is A = 300 mm2 and E = 200 GPa.

20 kN 20 kN

3 m

3 m 3 m 3 m

A B C D

F E

EXAMPLE 6


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Calculate the Member Forces due to “Virtual Force”

0.667

1.0

-0.943

0.333

0.333

0.667

-0.471

0.333 1

0.667

-0.333

-0.471

Virtual Structure: Virtual force, u



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Calculate the Member Forces due to “ Actual Forces”

Actual Structure: Real force, N

20 kN

20 kN

-28.3 kN

20 kN

20 kN

20 kN

-28.3 kN

20 kN 20 kN

20 kN

-20 kN

0

20 kN



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Calculate the total deformation

Member Virtual force, u Real force, P (kN) L (m) u.PL (kN.m)

AB 0.333 20 3 20

BC 0.667 20 3 40

CD 0.667 20 3 40

DE -0.943 -28.3 4.24 113

FE -0.333 -20 3 20

EB -0.471 0 4.24 0

BF 0.333 20 3 20

AF -0.471 -28.3 4.24 56.6

CE 1 20 3 60

Σ 369.6



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Calculate the final deformation


1 ∙ ∆ = ∙

1 ∙ ∆= ∙ =369.6

∆= 369.6300×10− 200×10

∆ = 6.16


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Occasionally, errors in fabricating the lengths of the members of a truss mayoccur. Also, in some cases truss member must be made slightly longer or

shorter in order to give the truss a camber. If a truss member is shorter or

longer than intended, the displacement of a truss joint from its expected

position can be determined from direct application:

1 = External virtual unit load acting on the truss joint in the stated direction

of Δ

u = Internal virtual normal force in a truss member caused by the externalvirtual unit load

Δ = External joint displacement caused by the fabrication errors

ΔL = Difference in length of the member from its intended size as caused by

a fabrication error

DISPLACEMENT OF TRUSSES(Due to Temperature Changes & Fabrication Error)

∙ ∆ = ∙ ∆


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Virtual force, u


Calculate the Member Forces due to “Virtual Force”

01

0.75

0

-1.25

1.01

0.75

0.75


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Both loads and temperature affect the deformation,therefore:


1 ∙ ∆ = ∙ + ∙ ∙ ∆ ∙

1 ∙ ∆=0.75 600 1.8

1200×10− 200×10 +1 400 2.4

1200×10− 200×10

+ 1.25 500 3900×10− 200×10 + 1 1.08 × 10− 60 2.4

∆ = 0.0193 = 19.3


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To determine

vertical/horizontal

displacement

Virtual Work Method• Actual structure (P )

• Virtual structure (u)

Forces in Member

1. Method of Joints (F x, F y)2. Method of Sections (F x, F y, M )

3. Force Resolution (F x, F y)

Stability and Determinacy(External & Internal)

If OK, calculate reactions (3

equilibriums)A+

SUMMARY

lecture 7 plane truss full page

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