The document discusses Castigliano's Theorem, a method for determining deflection in structures such as trusses, beams, or frames, introduced by Alberto Castigliano in 1879. It outlines how to compute displacements and slopes using strain energy principles, applicable to structures with certain conditions. The document includes detailed procedures for analyzing forces in trusses along with example problems to illustrate the application of the theorem.

1. CED 426
Structural Theory II
Lecture 2
Castigliano’s Theorem
Castigliano’s Theorem for Trusses
Mary Joanne C. Aniñon
Instructor

2. Castigliano’s Theorem
• Alberto Castigliano published a book in 1879 in which he outlined a
method for determining the deflection or slope at a point in a
structure, be it a truss, beam, or frame.
• This method is also referred to as Castigliano’s second theorem, or
the method of least work
• It only applies to structures that have constant temperature,
unyielding supports, and linear elastic material response.

3. Castigliano’s Theorem
• The theorem states that:
• The displacement at a point in a structure is equal to the first partial
derivative of the strain energy in the structure with respect to a force acting
at the point and in the direction of displacement.
• The slope at a point in a structure is equal to the first partial derivative of the
strain energy in the structure with respect to a couple moment acting at the
point and in the direction of rotation.

4. Castigliano’s Theorem for Trusses
• To compute the vertical displacement Δ at a point in a structure:
∆ = 𝑁(
𝜕𝑁
𝜕𝑃
)
𝐿
𝐴𝐸
Δ = external joint displacement of the truss
𝑃 = external force applied to the truss joint in the direction of Δ
𝑁 = internal normal force in a member caused by both the force 𝑃 and the
loads on the truss.
𝐿 = length of the member
𝐴 = cross-sectional area of a member
𝐸 = modulus of elasticity of a member

5. Castigliano’s Theorem for Trusses
• Note:
In order to determine the partial derivative in this equation, it will
be necessary to treat 𝑃 as a variable (not a specific numerical
quantity), and furthermore, each member force 𝑁 must be
expressed as a function of 𝑃.

6. Procedure for Analysis
External Force, 𝑷
• Place a force 𝑷 on the truss at the joint where the desired
displacement is to be determined. This force is assumed to have a
variable magnitude in order to obtain the change ∂N/∂P.
• Be sure 𝑷 is directed along the line of action of the displacement.

7. Procedure for Analysis
Internal Forces, 𝑵
• Determine the force 𝑁 in each member caused by both the real
(numerical) loads and the variable force 𝑃.
• Assume tensile forces are positive and compression forces are
negative.
• Calculate the partial derivative ∂N/∂P for each member of the truss.
• After 𝑁 and ∂N/∂P have been determined, assign 𝑃 its numerical
value if it has replaced a real force on the truss. Otherwise, set 𝑃
equal to zero.

8. Procedure for Analysis
Castigliano’s Theorem
• Apply Castigliano’s theorem to determine the desired displacement
∆. It is important to retain the algebraic signs for corresponding values
of 𝑁 and ∂N/∂P when substituting these terms into the equation.
• If the resultant sum ∑ 𝑁(∂N/∂P) L/AE is positive, ∆ is in the same
direction as 𝑃. If a negative value results, ∆ is opposite to 𝑃.

9. Example 1
Problem:
Determine the vertical displacement of joint C of the truss shown
below. The cross-sectional area of each member is A = 400mm2 and
E=200 GPa.

10. EXAMPLE 1
Solution:
• External Force, 𝑷
A vertical force 𝑷 is applied to the truss at joint
C, since this is where the vertical displacement is to be
determined. 4 kN
P
A B
C

11. EXAMPLE 1
Solution:
• Internal Force, 𝑵
𝑀𝐴 = 0 :
P(4)+4(3)-RB(8)=0
RB= 0.5P + 1.5 kN
𝐹𝑌 = 0:
RAY + RB -P=0
RAY= 0.5P - 1.5 kN
𝐹𝑥 = 0:
4 – RAX = 0
RAX = 4 kN
A B
C
RB
RAY
RAX
4 kN
P

12. EXAMPLE 1
Solution:
• Internal Force, 𝑵
𝐹𝑌 = 0 @ pt. B:
RBC (3/5) + RB = 0
RBC= -0.833P – 2.5 kN
𝐹𝑥 = 0@ pt. B:
- RBC (4/5)– RBA = 0
RBA = 0.667P + 2 kN
RBC
B
RB
RBA

13. EXAMPLE 1
Solution:
• Internal Force, 𝑵
𝐹𝑌 = 0 @ pt. A:
RAC (3/5) + RAY = 0
RAC= -0.833P + 2.5 kN
RAC
A
RAY
RAX RAB

14. EXAMPLE 1
Solution:
• Castigliano’s Theorem
Arrange the data in tabular form:
Members N ∂N/∂P N (P=0) L N(∂N/∂P)L
AB 0.667P + 2 0.667 2 8 10.672
AC -0.833P + 2.5 -0.833 2.5 5 -10.413
BC -0.833P - 2.5 -0.833 -2.5 5 10.413
∑ = 10.672

15. EXAMPLE 1
Solution:
• Castigliano’s Theorem
Applying the formula:
∆ = 𝑁(
𝜕𝑁
𝜕𝑃
)
𝐿
𝐴𝐸
1𝑘𝑁 ∙ Δ𝐻𝑣 =
5657.28 𝑘𝑁2 ∙ 𝑚
𝐴𝐸
Δ𝐻𝑣 =
(5657.28 𝑘𝑁 ∙ 𝑚)(103 𝑁
𝑘𝑁
)
(3𝑥10−3 𝑚2)(200𝑥109 𝑁
𝑚2)
Δ𝐻𝑣 = 0.0094288 m or 9.4288 mm

16. Example 2
Problem:
Determine the vertical displacement of joint C of the truss shown
below. Assume that A = 300mm2 and E = 200 GPa.

17. Example 2
Solution:

18. Example 2
Solution:

19. Example 2
Solution:

20. Try to solve this…
P
Where P = 0
Problem:
Determine the vertical
displacement of joint C of the truss
shown below. Assume that A =
300mm2 and E = 200 GPa.