A simplified approach to calculating truss forces. This presentation is built upon a wonderful presentation which comes from Fazirah Abdul Ghafar. This presentation uses Power Point Animations and is best viewed using Power Point Slide Show to present.

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