lecture 7 plane truss full page
TRANSCRIPT
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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
STABILITY & DETERMINACY
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STABILITY & DETERMINACY
Simple Truss
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STABILITY & DETERMINACY
Compound Truss
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STABILITY & DETERMINACY
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
Internally:m = 9, j = 6,
9 = 2(6) – 3 = 9 OK
Therefore, the truss is determinate
(externally and internally)
EXAMPLE 2 – Solution
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Externally:
R = 4 ; R – 3 = 4 – 3 = 1 1 degree redundant
Internally:m = 10, j = 6,
10 > 2(6) – 3 : 10 > 9 1 degree redundant
Therefore, the truss is internally and externally
indeterminate (1 degree redundant)
EXAMPLE 3 – Solution
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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)
EXAMPLE 4 – Solution
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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 ( )
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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
EXAMPLE 4 – Solution
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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 !
EXAMPLE 4 – Solution
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
EXAMPLE 4 – Solution
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)
EXAMPLE 5 – Solution
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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 ( )
EXAMPLE 5 – Solution
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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)
EXAMPLE 5 – Solution
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
EXAMPLE 5 – Solution
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
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
DEFORMATION OF STATICALLY
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‘ Actual Structure’
A
B
C E
F
G H
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
D
DEFORMATION OF STATICALLY
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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’
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
C
‘Virtual Structure’
DEFORMATION OF STATICALLY
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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
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
DEFORMATION OF STATICALLY
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• Through combination, the amount of internal work is
given by = • According to Energy Work Method:
External load = Internal Load
1 ∙ ∆ =
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
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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
DISPLACEMENT OF TRUSSES
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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
EXAMPLE 6 – Solution
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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
EXAMPLE 6 – Solution
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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
EXAMPLE 6 – Solution
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Calculate the final deformation
EXAMPLE 6 – Solution
1 ∙ ∆ = ∙
1 ∙ ∆= ∙ =369.6
∆= 369.6300×10− 200×10
∆ = 6.16
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DISPLACEMENT OF TRUSSES
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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
EXAMPLE 7 – Solution
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:
EXAMPLE 7 – Solution
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