EF 102 Module 2 Lecture 4
1
Module 2, Lecture 4 - Trusses
• Trusses– Definition
– Various types and examples
– Two-force members
• Method of Joints
Trusses - Definition
• Framework of members joined at endswith frictionless pins to form a stablestructure
• Basic shape is a ______________
EF 102 Module 2 Lecture 4
2
Truss Bridges
Knoxville Railroad BridgesCSX
Norfolk Southern
Highway Bridge
Joints
_______ Joint ________ Joint
Frictionless pin is good assumption if members concentric
EF 102 Module 2 Lecture 4
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Types of TrussesNamed after inventor
Howe Pratt
Named after Railroad
Baltimore
Two-Force Members
Line of action of forces is along __________________________________________________________________
EF 102 Module 2 Lecture 4
4
Methods of Truss Analysis
• Method of __________
• Method of ___________
Method of Joints:
Draw FBD of pins. Show forces acting ______ pins, not members
∑∑ == 0 0 yx FF
∑ = 0zM Does not giveindependentequation
x
y
Method of Joints
1. Solve for support reactions of whole structure¸ Follow 8-step process; FBD of whole structure¸ Occasionally not necessary, but it will never hurt
2. Start at a joint with only 1 or 2 unknown forces¸ Follow 8-step process; FBD of joint, or pin¸ Member forces obtained using 3rd law partners¸ Suggest using Tension for all UNKNOWN members
3. “Step” through structure by repeating Step 2 atall joints.
EF 102 Module 2 Lecture 4
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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 supportreactions of whole structure
2000 lb1000 lb
6 ft 10 ft
8 ft
AB
C
DE
EF 102 Module 2 Lecture 4
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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
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint ??
00Σ
y-comp.x-comp.θF F
BAF
BDF
BCF
x
y
îĵ
k̂
+
∑ ⇒=→+
0xF
∑ ⇒=↑+ 0yF
EF 102 Module 2 Lecture 4
7
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint ??
-100090-1000
00Σ
0FDE0FDE
+.624FDE+.781FDE38.6FDC
-2000+1500126.9-2500
y-comp.x-comp.θF F
DAF
DBF
DCF
DEF
x
y
îĵ
k̂
+
∑ ⇒=→+
0xF
06.38cos9.36sin2500 =+°+°+ DEDC FFlb
∑ ⇒=↑+ 0yF06.38sin10009.36cos2500 =°+−°− DCFlblb
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint ??
FEC090FEC
00Σ
+3000090+3000
0-5250180+5250
0+5250180-5250
y-comp.x-comp.θF F
EDF
ECF
xE
yE
x
y
îĵ
k̂
+
∑ ⇒=→+
0xF
052505250?
=−+ lblb
∑ ⇒=↑+ 0yF03000 =++ lbFEC
EF 102 Module 2 Lecture 4
8
2000 lb1000 lb
6 ft 10 ft
8 ft
AB C
D
E
3000 lb
5250 lb
5250 lb
2. Draw FBD of Joint ??
-3000-3750218.64800
00Σ
0525005250
30000270-3000
0-15001801500
y-comp.x-comp.θF F
CBF
CDF
CEF
xC
x
y
îĵ
k̂
+
∑ ⇒=→+
0xF
05250lb8.64800lbcos3-1500lb-?
=+°
∑ ⇒=↑+ 0yF03000lb8.64800lbsin3-
?
=+°
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
