The document presents a simplified approach to truss analysis, focusing on calculating forces using Cartesian coordinates to minimize confusion between tension and compression. It outlines methods for analyzing truss structures, including equilibrium checks and joint analysis, using trigonometric functions to resolve vector components. Practical examples demonstrate the application of these principles to determine member forces and structural stability.

1. Truss Analysis
A Simplified Approach to
Calculating Truss Forces

2. What’s different
• Minimize confusion caused by two reference systems
– Tension/Compression vs Cartesian Coordinates
• Arrow radiating outwards from joint are positive (tension), negative for
compression.
• Calculating member force use Cartesian coordinates (left/down is negative,
up/right is positive)
– Uses familiar mathematical conventions
• Arrow swizzling occurs after all member forces are known

3. Truss Rules
Structure Analysis
(whole structure)
• Static Determinacy
 2J = R + M
• Check Equilibrium
 Only external forces are
used when summing
moments.
 CCW Moment: Positive
 CW Moment: Negative
 Static Forces
 Right/Up Force: Positive
 Down/Left Force: Negative
Methods of Joints
(one joint at a time)
• All joints are positive
 One joint at a time
 Start at the joint with the
least unknowns
 All arrows radiate outwards
from their respective pin
 Arrow directions are
changed after all member
forces are known.
• Cartesian conventions
 Right/Up Force: Positive
 Down/Left Force: Negative

4. Use cosine and sine to determine x and y vector
components.
Assume all members are in tension. A positive answer will
mean the member is in tension, and a negative number will mean the
member is in compression.
B

Method of Joints
The diagonal is
the hypotenuse
𝑨
𝐴 𝑦 = 𝐻 ∙ sin 𝜃
𝐴 𝑥 = 𝐻 ∙ cos 𝜃
𝐻

5. Initial Analysis
D
A B
C
5000 𝑁
5m
6m
5m
45 𝑜
3.54m
3.54m 2.46m
5𝑚 ∙ sin 450 = 𝐷𝐸 5𝑚 ∙ 0.7071 = 3.54𝑚
45 𝑜
𝜃1
tan 𝜃1 = 2.46𝑚 3.54𝑚 = 2.46 3.54 = 0.695
tan−1 0.695 = 34. 8 𝑜
34. 8 𝑜
55. 2 𝑜
45 𝑜55. 2 𝑜
34. 8 𝑜
45 𝑜
3.54m2.46m
2J = M + R
8 = 5 + 3
E
Is the structure statically determinate?Now find the angles
Did anyone
say geometry
?

6. Structure Analysis: Reactive Forces
D
B
C
5m
6m
6m
5m
45 𝑜
45 𝑜
34. 8 𝑜
55. 2 𝑜
45 𝑜55. 2 𝑜
34. 8 𝑜
45 𝑜
3.54m
A
𝑅 𝐴𝑥
𝑅 𝐴𝑦
𝑀𝐴 = 0
0 = 6𝑚 ∙ 𝑅 𝐵𝑦 − (6𝑚 + 3.54𝑚) ∙ 5000𝑁 6𝑚 ∙ 𝑅 𝐵𝑦 = (9.54𝑚) ∙ 5000𝑁
𝑅 𝐵𝑦 = 7950 N
7950 N
𝑅 𝐵𝑦
𝑅 𝐴𝑦 + 𝑅 𝐵𝑦 − 5000 𝑁 = 0𝐹𝑦 = 0 𝑅 𝐴𝑦 + 7950 − 5000 𝑁 = 0 𝑅 𝐴𝑦 = −2950 N
-2950 N
𝐹𝑥 = 0 𝑅 𝐴𝑥 = 0
𝑅 𝐵𝑦 = 9.54𝑚 ∙ 5000𝑁 6𝑚
5000 𝑁
E

7. Joint Analysis
D
B
C
5m
6m
5m
45 𝑜
45 𝑜
34. 8 𝑜
55. 2 𝑜
45 𝑜55. 2 𝑜
34. 8 𝑜
45 𝑜
3.54m
A
𝑅 𝐴𝑦
5000 𝑁
7950 N
𝑅 𝐵𝑦
-2950 N
−5000𝑁 − 𝐵𝐶 𝑦 = 0
𝐵𝐶 ∙ sin 45 𝑜 = −5000 𝑁 𝐵𝐶 = −7071 N
−𝐶𝐷 − 𝐵𝐶 𝑥 = 0
𝑐 𝑥 = 0
−𝐶𝐷 = 𝐵𝐶 𝑥 = BC ∙ cos 45 𝑜
𝑐 𝑌 = 0
𝐵𝐶 𝑦 = 𝐵𝐶 ∙ sin 45 𝑜𝐵𝐶 𝑦 = −5000𝑁
−𝐶𝐷 = −7071 ∙ cos 45 𝑜
N −𝐶𝐷 = −5000 N 𝐶𝐷 = 5000 N
6m
E
𝐵𝐶 = −7071 N
𝐶𝐷 = 5000 N

8. Joint Analysis
D C
7950 N-2950 N
𝐵𝐶 = −7071 N
𝐶𝐷 = 5000 N
𝐵 𝑌 = 0
𝑅 𝐵𝑦 + BC ∙ cos 45 + 𝐷𝐵 ∙ cos 34.8 = 0
𝐵𝑥 = 0
0 = −𝐴𝐵 − DB ∙ sin 34.8 + 𝐵𝐶 ∙ sin 45
𝐴𝐵 = − (6089) ∙ sin 34.8 + (−7071) ∙ sin 45 𝐴𝐵 = −8475 N
7950 + (−7071) ∙ cos 45 + 𝐷𝐵 ∙ cos 34.8 = 0
7950 + (−5000) +𝐷𝐵 ∙ 0.821 = 0 2950 = −𝐷𝐵 ∙ 0.821 𝐷𝐵 = −3593N
B
45 𝑜
45 𝑜
34. 8 𝑜
55. 2 𝑜
45 𝑜55. 2 𝑜
34. 8 𝑜
45 𝑜
𝑅 𝐴𝑦
5000 𝑁
𝑅 𝐵𝑦
A
𝐴𝐵 = −8475 N
𝐷𝐵 = −3593 N

9. Joint Analysis
D C
𝐵𝐶 = −7071 N
B
45 𝑜
45 𝑜
34. 8 𝑜
55. 2 𝑜
45 𝑜55. 2 𝑜
34. 8 𝑜
45 𝑜
𝑅 𝐴𝑦
5000 𝑁
7950 N
𝑅 𝐵𝑦
-2950 N
𝐶𝐷 = 5000 N
𝐴𝐵 = −8475 N
𝐷𝐵 = −3593 N𝐴 𝑦 = 0
𝑅 𝐴𝑦 + 𝐴𝐷 ∙ sin 45 = 0 𝑅 𝐴𝑦 = −𝐴𝐷 ∙ sin 45
−2950 𝑁 = −𝐴𝐷 ∙ sin 45 𝐴𝐷 = 4172 𝑁
𝐷 𝑦 = 0
−AD ∙ cos 45 − DB ∙ cos 34.8 = 0
−AD ∙ cos 45 = −3593 ∙ cos 34.8 𝐴𝐷 = 4172 𝑁
𝐴𝐷 = 4172 N
Tension: CD AD
Compression: BC DB AB
A