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2/6/2013 1 CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS
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CHAPTER 5 Statically Determinate
Plane Trusses
TYPES OF ROOF TRUSS
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TYPES OF ROOF TRUSS
ROOF TRUSS SETUP
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ROOF TRUSS SETUP
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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• A truss is a structure of slender members joined at their end points by bolting or welding to gusset plate.
• Members commonly used: wooden struts, metal bars, angles or channels
INTRODUCTION
• Transmitting loading from roof to columns – by means 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
ROOF TRUSSES
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BRIDGE TRUSSES
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 and concurrent 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
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 two members will increase one number of joint
STABILITY & DETERMINACY
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
STABILITY & DETERMINACY
Compound Truss
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STABILITY & DETERMINACY
Complex Truss
• 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 (Fx = 0 and Fy =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 internally and externally
• The external determinacy is given by:
R = 3 (provided that the support reactions have no lines 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
• The internal determinacy is given by
m = 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
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
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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EXAMPLE 3
Determine the stability and determinacy of the truss shown in the figure below.
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
• Suitable to be used to determine all the member forces 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 Fx = 0 and Fy =0, we can start and solve the problems.
METHOD OF JOINTS
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Determine all the member forces for the given truss below.
EXAMPLE 4
150 kN 50 kN
B C
D A
F G
100 kN
E 6m
3m 3m 8m
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 ( )
+ Fy = 0 ; RA – 150 – 50 – 167.9 kN ;
RA = 32.1 kN ( )
+ Fx = 0 ; HA = 100 kN ( )
EXAMPLE 4 – Solution
3. Analyse Every Joints
At Joint A
+Fx = 0 ; 100 + FAD = 0 FAD = 100 kN (T)
+ Fy = 0 ; FAB + 32.1 = 0
FAB = 32.1 kN (C)
A
FAB
FAD
32.1
100
EXAMPLE 4 – Solution
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At Joint B
+Fy = 0 ; 32.1 – FBD (6/10) = 0 FBD = 53.5 kN (T)
+ Fx = 0 ; FBC + 100 + 53.5 (8/10) = 0
FBC = 142.8 kN (C)
B FBC
32.1
100
FBD
EXAMPLE 4 – Solution
At Joint C
+Fx = 0 ; 142.8 + FCE (3/√18) = 0 FCE = 201.9 kN (C)
+ Fy = 0 ; FCD – (201.9) (3/√18) = 0
FCD = 142.8 kN (T)
C 142.8
FCE FCD
EXAMPLE 4 – Solution
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At Joint D
+Fy = 0 ; 142.8 + 53.5 (6/10) – 150
+ FDE (3/√18) = 0
FDE = 35.2 kN (C)
+ Fx = 0; 100 – 53.5 (8/10) + (-35.2)(3/√18) + FDF = 0
FDF = + 166.7 kN (T)
D
142.8
FDF
150
100
53.5 FDE
EXAMPLE 4 – Solution
At Joint E
+Fx = 0 ; FEG (3/√18) + 201.9 (3/√18)
+ 35.2 (3/√18) = 0 FEG = 237.1 kN (C)
+ Fy = 0; FEF 201.9 (3/√18) + 35.2 (3/√18) – (237.1)(3/√18) = 0
FEF = 49.8 kN (T)
E 142.8
FEG
FEF
35.2
EXAMPLE 4 – Solution
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At Joint F
+Fx = 0 ; 167.7 + FFG = 0 FFG = 167.7 kN (T)
F
49.8
FFG
50
166.7
EXAMPLE 4 – Solution
At Joint G (Checking)
+Fy = 0 ;
167.9 – 237.1 (3/√18) = 0 OK !
+ Fx = 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 C
AD +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 C
DF +167.7 T
EF +49.8 T
EG 237.1 C
FG +167.7 T
Summary: Internal Member Forces
• When only some of the member forces need to be calculated, 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 the truss into two parts, provided that the unknown values are not more than three
• The unknown forces will be assumed to be either in tension or compression
• Three equilibriums (Fx = 0, Fy = 0, M = 0) will be used to solve the problems.
METHOD OF SECTIONS
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Determine the member forces for BD, DE and CE.
EXAMPLE 5
100 kN
80 kN
40 kN
A B
C
D
E F
4m 4m 4m
2 m 2
1
Stability and Determinacy
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 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) = 0
RA = 82.6 kN ( )
+ Fy = 0 ; RF + (2/√5)(82.6) – 100 – 40 = 0 ;
RF = 66.2 kN ( )
+ Fx = 0 ; HF + (1/√5)(82.6) – 80 = 0 ;
HF = 43.1 kN ( )
EXAMPLE 5 – Solution
Section 1:
MD = 0
FEC (2) – 43.1 (2) – 66.2 (4) = 0 ; FEC = 175.5 kN (T)
Fy = 0 ; 40 + 66.2 – FDC (2/√20) = 0 ; FDC = 58.6 kN (T)
Fx = 0 ; FDB – 58.6 (4/√20) – 175.5 + 43.1 = 0 ;
FDB = 184.8 kN (C)
EXAMPLE 5 – Solution
40 kN
FEC
D
FDC
FDB
66.2
43.1 F E
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Section 2:
MF = 0 ; FED (4) – 40 (4) = 0 ; FED = 40 kN (T)
40 kN
FEC
FFD FED
66.2
43.1
EXAMPLE 5 – Solution
F E
• This is an extended version from the method of joints
• Every single joint is carefully analysed by considering not more than two unknowns at each joints.
• In this method, we do not have to write all the equations and calculations. All member forces are solved directly on the diagram.
METHOD OF FORCE RESOLUTIONS
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Determine the member forces of the truss
EXAMPLE 6
100 kN
1.5 m
A
B
C
D
E
F
50 kN 50 kN
20 kN
G
H
2m 2m 2m 2m
• Truss analysis using method of joints can greatly be simplified if one can first determine those member that support no loading (zero force member)
• The zero-force members can be determine by inspection of the joints. Normally, there are two cases where zero-force member can be identified
ZERO FORCE MEMBERS
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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
• 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
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‘Actual Structure’
A
B
C E
F
G H
DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS
D
• 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’
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• Both structures will then combined, thus producing the concept of ‘work’. Therefore, the external work:
W = 1 .
• Say P is the member force due to external loads and u is the 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
• Through combination, the amount of internal work is
given by = 𝑢𝑃𝐿
𝐴𝐸
• According to Energy Work Method:
External load = Internal Load
1 ∙ ∆= 𝑢𝑃𝐿
𝐴𝐸
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: 𝒖𝑷𝑳
𝑨𝑬
SOLUTION PROCEDURE
• When a structure is loaded, its stressed elements deform. 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 a displacement 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 of External
Load
Work of Internal
Load
𝑷 ∙ ∆ = 𝒖 ∙ 𝜹
• When a bar is loaded axially, it will deform and store strain 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 behaves elastically (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
xP
F
VIRTUAL WORK METHOD
𝑑𝐿 =𝑃𝐿
𝐴𝐸 From Hooke’s Law:
L
dL
P
F
• We can use the method of virtual work to determine the 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, and resulted a displacement of Δ, the external work = 1Δ.
DISPLACEMENT OF TRUSSES
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• Due to the unit force, each truss member will carry an 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
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 loads removed 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 are negative.
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
• Apply the equation of virtual work, to determine the desired displacement. It is important to retain the algebraic sign for ach of the corresponding internal forces when substituting these terms into the equation.
Member Virtual Force, u
Real Force, P (kN)
L (m) u.PL
(kN.m)
AB 0.601 50 1.8 2
CB -0.507 -30 1.8 -3
DB 0.515 20 1.8 5
Total 4
STEPS FOR ANALYSIS
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• If the resultant sum of displacement is positive, the direction 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
Determine the vertical displacement of joint C of the steel 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
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
Calculate the final deformation
EXAMPLE 6 – Solution
1 ∙ ∆= 𝑢 ∙ 𝑑𝐿
1 𝑘𝑁 ∙ ∆𝑐𝑣= 𝑢 ∙ 𝑃𝐿
𝐴𝐸=369.6
𝐴𝐸
∆𝑐𝑣=369.6
300 × 10−6 200 × 103
∆𝑐𝑣 = 6.16 𝑚𝑚
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In some cases, truss members may change their length due to temperature. If α is the coefficient of thermal expansion for a member and ΔT is the change in its temperature, the change in length of a member is:
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 external virtual unit load
Δ = External joint displacement caused by the temperature change
α = Coefficient of thermal expansion of member
ΔT = Change in temperature of member
L = Length of member
DISPLACEMENT OF TRUSSES (Due to Temperature Changes & Fabrication Error)
𝟏 ∙ ∆= 𝒖 ∙ 𝜶 ∙ ∆𝑻 ∙ 𝑳
Occasionally, errors in fabricating the lengths of the members of a truss may occur. 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 external virtual 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)
𝟏 ∙ ∆= 𝒖 ∙ ∆𝑳
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Determine the vertical displacement of joint C of the steel truss as shown in the Figure. Due to radiant heating from the wall, member AD is subjected to an increase in temperature of ΔT = +60°C. Take α = 1.08 × 10-5/°C and E = 200 GPa. The cross-sectional area of each member is indicated in the figure.
EXAMPLE 7
1200 mm2 1200 mm2
1200 mm2
1200 mm2
900 mm2
2.4 m
1.8 m
A B
D C Wall
400 kN
300 kN
Virtual force, u
EXAMPLE 7 – Solution
Calculate the Member Forces due to “Virtual Force”
0 1
0.75
0
-1.25
1.0 1
0.75
0.75
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Real force, N
Calculate the Member Forces due to “Actual Force”
EXAMPLE 7 – Solution
400 kN 400 kN
600 kN
0
-500 kN
400 kN
400 kN
600 kN
300 kN
300 kN
Both loads and temperature affect the deformation, therefore:
EXAMPLE 7 – Solution
1 ∙ ∆= 𝑢 ∙ 𝑑𝐿 + 𝑢 ∙ 𝛼 ∙ ∆𝑇 ∙ 𝐿
1 𝑘𝑁 ∙ ∆𝑐𝑣=0.75 600 1.8
1200 × 10−6 200 × 106+
1 400 2.4
1200 × 10−6 200 × 106
+−1.25 −500 3
900 × 10−6 200 × 106+ 1 1.08 × 10−5 60 2.4
∆𝑐𝑣 = 0.0193 𝑚 = 19.3 𝑚𝑚
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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 (Fx, Fy) 2. Method of Sections (Fx, Fy, M) 3. Force Resolution (Fx, Fy)
Stability and Determinacy (External & Internal) If OK, calculate reactions (3 equilibriums)
A+
SUMMARY
