ef 202, module 3, lecture 1 trusses - 1 ef 202 - week 9
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EF 202, Module 3, Lecture 1
Trusses - 1EF 202 - Week 9
2
EF 202, Module 3, Lecture 1
Truss - Definition
•A truss is a framework of members joined at ends with frictionless pins to form a stable structure.
•Basic shape is a triangle.
3
EF 202, Module 3, Lecture 1
Truss Bridges
Knoxville Railroad Bridges
Highway Bridge
CSX
Norfolk Southern
4
EF 202, Module 3, Lecture 1
Joints
Pinned Joint Riveted Joint
Frictionless pin is good assumption if members concentric.
5
EF 202, Module 3, Lecture 1
Types of Truss
Named after inventor:
Howe Pratt
Named after railroad company:
Baltimore
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EF 202, Module 3, Lecture 1
Methods of Truss Analysis
•Method of joints (today)
•Method of sections (next week)
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EF 202, Module 3, Lecture 1
Method of Joints - Concepts•Draw FBD’s of pins.
•Show forces acting on pins (not forces acting on members).
•Concurrent, coplanar systems.
•Two independent equations (force summations) at each pin.
•No more than two unknowns at a pin.
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EF 202, Module 3, Lecture 1
The method of joints does not employ moment summations.
No-No
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EF 202, Module 3, Lecture 1
Method of Joints - Steps
1. Solve for structure’s overall support reactions.
Follow process for nonconcurrent, coplanar system.
2. Start at a joint with 1 or 2 unknowns.
Follow process for concurrent, coplanar system.
3. Repeat step 2 as necessary.
Example – Method of Joints
Given:
Required: Force in each member using method of joints.
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
DE
1. Solve for support reactions of whole structure.
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
DE
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
Ey
Ex
Cx
€
ΣME = 0⇒ 2000 16( ) +1000 10( ) −Cx 8( ) = 0
€
ΣFy = 0⇒ −1000 − 2000 + Ey = 0
x
y
+
€
ΣFx = 0⇒ Cx − Ex = 0⇒ Ex = 5,250 lb
€
Cx = 5,250 lb
€
Ey = 3,000 lb
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint A.
x-comp. y-comp.
Σ 0 0€
FBA
€
FDA
€
FBA
€
−0.8FDA
€
0.6FDA
€
ΣFy = 0
−0.8FDA − 2000 = 0
FDA = −2500 lb
€
ΣFx = 0
FBA + 0.6FDA = 0
FBA =1500 lb
x
y
+
2000 lb1000 lb
6 ft 10 ft
8 ft
A B C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint B.
x-comp. y-comp.
Σ 0 0
x
y
+
€
FBA
€
FBA
€
FBD
€
FBD
€
FBC
€
FBC €
ΣFy = 0
1000 + FBD = 0
FBD = −1000 lb
€
ΣFx = 0
FBA + FBC = 0
FBC =1500 lb
2000 lb1000 lb
6 ft 10 ft
8 ft
AB
C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint D.
x-comp. y-comp.
2500 1500 2000
1000 0 1000
Σ 0 0
x
y
+
€
FDE
€
FDE
€
441FDC
€
FDC
€
541FDC
2000 lb1000 lb
6 ft 10 ft
8 ft
AB
C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint E.
x-comp. y-comp.
5250 +5250 0
FEC 0 FEC
5250 -5250 0
3000 0 +3000
Σ 0 0
x
y
+
2000 lb1000 lb
6 ft 10 ft
8 ft
AB
C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint C.
x-comp. y-comp.
1500 -1500 0
4800 -3750 -3000
3000 0 3000
5250 5250 0
Σ 0 0
x
y
+
2000 lb
1000 lb
6 ft 10 ft
8 ft
AB C
DE
1500 lb T 1500 lb T
5250 lb C
3000
lb C
1000
lb C
4800 lb T2500 lb C
18
EF 202, Module 3, Lecture 1
Bonus!
•Some truss members carry no force and are called zero-force members. (See section 6-6 in the text.)
•Some structures are externally statically indeterminate. (See Online Homework 3--1-1.)
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