Theory of Structures II Castiglianos Theorem for Brams and Frames Reference: Structural Analysis 10th Edition in SI Units by R.C.

1. CED 426
Structural Theory II
Lecture 5
Castigliano’s Theorem for
Beams and Frames
Mary Joanne C. Aniñon
Instructor

2. CASTIGLIANO’S THEOREM FOR BEAMS & FRAMES

3. CASTIGLIANO’S THEOREM FOR BEAMS & FRAMES

4. PROCEDURE FOR ANALYSIS

5. PROCEDURE FOR ANALYSIS

6. PROCEDURE FOR ANALYSIS

7. Example 1
• Determine the displacement of point B of the beam shown below.
Take E = 200 GPa, I = 71.1(106) mm4.

8. Example 1
• EXTERNAL FORCE, P
• A vertical force P is placed on the beam at B as shown below.
• Since there is no discontinuities of loading between A and B, a single x
coordinate is assigned as shown below.

9. Example 1
• INTERNAL MOMENTS, M
• Solve the internal moments, M
• Solve ∂M/∂P
• Solve M when P=0
↻ +∑𝑀 = 0
𝑀 + 12𝑥
𝑥
2
+ 𝑃 𝑥 = 0
𝑀 = −6𝑥2 − 𝑃𝑥 = 0
𝜕𝑀
𝜕𝑃
= −𝑥
M= −6𝑥2

10. Example 1
• CASTIGLIANO’S THEOREM

11. Example 2
• Determine the slope at point B of the beam shown below. Take E =
200 GPa, I = 60(106) mm4.

12. Example 2
• EXTERNAL COUPLE MOMENT, M’
• Since the slope at point B is to be determined, an external couple M’ is placed
n the beam at point B.
• Two coordinates, x1 and x2, must be used to determine the internal moments
within the beam since there is a discontinuity, M′, at B.
• As shown below, x1 ranges from A to B and x2 ranges from B to C.

13. Example 2
• INTERNAL MOMENTS, M
• Solve the internal moments, M
• Solve ∂M/∂P
• Solve M when M’=0
↻ +∑𝑀 = 0
−𝑀1 − 3(𝑥1) = 0
𝑀1 = −3 𝑥1
𝜕𝑀1
𝜕𝑀′
= 0
𝑀1 = −3 𝑥1
−𝑀2 − 3 2 + 𝑥2 + 𝑀′ = 0
𝑀2 = −3 2 + 𝑥2 + 𝑀′
𝜕𝑀2
𝜕𝑀′
= 1
𝑀2 = −3 2 + 𝑥2

14. Example 2
• CASTIGLIANO’S THEOREM
The negative sign indicates that
𝜃𝐵 is opposite to the direction
of the couple moment M’.

15. Example 3
• Determine the slope at point C of the two-member frame shown
below. The support at A is fixed. Take E = 200 GPa, I = 145(106) mm4.

16. Example 3
• EXTERNAL COUPLE MOMENT, M’
• A variable moment M’ is applied to the frame at point C, since the slope at
this point is to be determined.
• Due to discontinuity of internal loading at B, two coordinates, x1 and x2, are
chosen.

17. Example 3
• INTERNAL MOMENTS, M
• Solve the internal moments, M
• Solve ∂M/∂P
• Solve M when M’=0
↻ +∑𝑀 = 0
𝑀1 + 30𝑥1
𝑥1
2
+ 𝑀′ = 0
𝑀1 = −(15𝑥1
2
+ 𝑀′)
𝜕𝑀1
𝜕𝑀′
= −1
𝑀1 = −15𝑥1
2
𝑀2 + (30)(2) 𝑥2 cos 60 + 1 + 𝑀′ = 0
𝑀2 = −60 𝑥2 cos 60 + 1 − 𝑀′
𝜕𝑀2
𝜕𝑀′
= −1
𝑀2 = −60 𝑥2 cos 60 + 1

18. Example 3
• CASTIGLIANO’S THEOREM