castigliano's 2nd theorem

Chapter 2 Castigliano’s 2nd Theorem 2012

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CHAPTER 2 (C01-P03)

CASTIGLIANO’S 2ND THEOREM

Learning Outcome

At the end of this chapter, students should:

1) Able to determine deformation for statically determinate beams by using Castigliano’s 2nd

theorem.

2) Able to determine deformation for statically determinate frames by using Castigliano’s 2nd

theorem.

3) Able to determine deformation for statically determinate trusses by using Castigliano’s 2nd

theorem.

Castigliano 2nd Theorem (C2T) for Beams and Frames

In 1879 Alberto Castigliano, an Italian railroad engineer, published a book in which he outlined a method

for determining the deflection or slope at a point in a structure.

The formula for the Castigliano 2nd Theorem for beam and frame deflections is as follows.

Where

∆ = external displacement of the point caused by the real loads acting on the beam or

frame.

P = external force applied to the beam of frame in the direction of ∆.

M = internal moment in the beam or frame, expressed as a function of x and caused by both

the force P and the real loads on the beam or frame.

E = modulus of elasticity of the material.

I = moment of inertia of cross-sectional area, computed about the neutral axis.

In a similar manner, if the slope at a point is to be determine, we must find the partial derivative of the

internal moment M with respect to an external couple moment M’ acting at the point i.e.


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Procedure for analysis

The following step-by-step procedure can be used to determine the slopes and deflections of beams and

frames by Castigliano’s 2nd theorem.

1. Place a force P on the beam or frame at the point and in the direction of the desired displacement. If

the slope is to be determined, place a couple moment M’ at the point. It is assumed that both P and M’

have a variable magnitude.

2. Calculate reactions at the support in terms of P /M’ by using Equilibrium Equations.

3. For each segment of the beam/frame, determine the equation expressing the variation of the bending

moment along the length of the segment in terms of a position coordinate x. It is usually convenient

to consider the bending moments as positive in accordance with the beam sign convention.

4. Compute the partial derivative or for each coordinate x.

5. After M and or have been determined, assign P or M’ its numerical value if it has

replaced a real force or couple moment. Otherwise, set P or M’ equal to zero.

6. Apply Equation to determine the desired displacement or slope . If the resultant sum of all the

definite integral is positive, displacement or slope is in the same direction of P or M’.


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Example 1: Castigliano’s 2nd theorem for Beam

Below is a simply supported beam ABCD subjected uniformly distributed load of 25 kN/m along member

ABC whereas joint B and D are subjected to a point load of 100 kN and 150 kN, respectively. The total

length of the beam is 25 m. By taking E is 200 GPa and I is 500 x 106 mm

4.

a) Using Castigliano’s 2nd theorem, determine vertical displacement at joint B.

b) Using Castigliano’s 2nd theorem, determine rotation at joint D.

Figure 2.1

Solution:

a) Using Castigliano’s 2nd theorem, determine vertical displacement at joint B.

Step 1: Reactions in terms of P

Since the vertical displacement is to be determined at B, place a vertical force P on the beam at B as

shown. Then, determine reactions at each support using equilibrium equation in terms of P.

25 kN/m

150 kN

Ay

P

B

Cy

D

Ay

10 m 10 m 5 m

25 kN/m

150 kN

A

100 kN

B

C

D 10 m 10 m 5 m


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Solve the reactions for the real beam/system using Equilibrium Equation.

∑MA = 0,

25(20)(20/2) + P(10) + 150(25) – Cy(20) = 0

Cy = (8750 + 10P) / 20

Cy = 437.5 + 0.5P

∑Fy = 0,

Ay + Cy - 25(20) - P - 150 = 0

Ay + (437.5 + 0.5P) - 25(20) - P - 150 = 0

Ay = 212.5 + 0.5P

∑Fx = 0,

Ax = 0 kN

Step 2: Moment equation for each member(M)

Member AB (origin A) 0 ≤ x ≤ 10 m

Free-body diagram (FBD)

∑M = 0,

–M +(212.5 + 0.5P)(x) – 25(x)(x/2) = 0

M = 212.5x + 0.5Px – 12.5x2

25 kN/m

212.5 + 0.5P

A M

x


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Member BC (origin A) 10 ≤ x ≤ 20 m


∑M = 0,

–M +(212.5 + 0.5P)(x) – 25(x)(x/2) – P (x–10) = 0

M = 212.5x + 0.5Px – 12.5x2 – Px + 10P

Member CD (origin D) 0 ≤ x ≤ 5 m


∑M = 0,

M + 100(x) = 0

M = -100x

x D

( x – 10) m 10 m

25 kN/m

P

B

212.5 + 0.5P

M

x

M

150 kN


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Step 3: Identify M (P = 100 kN) as well as and summarized in table form.

Member Origin Limit M M (P = 100 kN)

AB A 0 - 10 212.5x + 0.5Px – 12.5x2 262.5x – 12.5x

2 0.5x

BC A 10 – 20 212.5x + 0.5Px – 12.5x2 – Px

+ 10P 262.5x – 12.5x

2 –

100x + 1000 - 0.5x

CD D 0 - 5 -100x -100x 0

Lastly, apply the formula as below.

∆B =

+

=

=

1.∆D =

∆D =

∆D =

∆D =


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b) Using Castigliano’s 2nd theorem, determine rotation at joint D.

Step 1: Reactions in terms of M’

Since the rotation is to be determined at D, place a moment M’ on the beam at D as shown. Then,

determine reactions at each support using equilibrium equation in terms of M’.

M’

25 kN/m

150 kN

A

100 kN

B

C

D 10 m 10 m 5 m


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Step 2: Moment equation for each member(M)


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Step 3: Identify M (M’ = 0 kN) as well as and summarized in table form.

Member Origin Limit M M (M’ = 0 kN)

AB A 0 - 10

BC A 10 – 20

CD D 0 - 5



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Self-Learning Exercise

Question 1

Below is a cantilever beam ABCD subjected uniformly distributed load of 25 kN/m along member ABC

whereas joint B and D are subjected to a point load of 100 kN and 150 kN, respectively. The total length

of the beam is 25 m.

a) Using Castigliano’s 2nd theorem, determine vertical displacement at joint D in terms of EI.

b) If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities, EI.

Figure 2.2

Solution:

a) Using Castigliano’s 2nd theorem, determine vertical displacement at joint D in terms of EI.

Step 1: Reactions in terms of P

Since the deflection is to be determined at D, place P on the beam at D as shown. Then, determine

reactions at each support using equilibrium equation in terms of P.

25 kN/m

150 kN

A

100 kN

B C D 10 m 10 m 5 m

25 kN/m

P

A

100 kN

B C D 10 m 10 m 5 m


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Step 2: Moment equation for each member (M)


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Step 3: Identify M (P = 150 kN) as well as and summarized in table form.

Member Origin Limit M M (P = 150 kN)

AB

BC

CD


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b) If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities, EI.


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Question 2

Below is a cantilever beam ABCD subjected uniformly distributed load of 25 kN/m along member ABC

whereas joint B and D are subjected to a point load of 100 kN and 150 kN, respectively. The total length

of the beam is 25 m. Using Castigliano’s 2nd theorem determine vertical displacement at joint D in terms

of EI.

Figure 2.3

25 kN/m

150 kN

A

100 kN

B C D 2EI

10 m 5 m 10 m

2EI EI


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Question 3

Below is a simply supported beam ABCD subjected uniformly distributed load of 25 kN/m along member

ABC whereas joint B and D are subjected to a point load of 100 kN and 150 kN, respectively. The total

length of the beam is 25 m.

a) Using Castigliano’s 2nd theorem, determine rotational displacement at joint D in terms of EI if EI is

constant for all members.

b) If the rotational displacement at joint B is given as 0.05 rad counterclockwise, determine the flexural

rigidities, EI value by using Castigliano’s 2nd theorem.

Figure 2.4

25 kN/m

150 kN

A

100 kN

B

C

D 10 m 10 m 5 m


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Example 2: Unit Load Method for Frame

Below is a rigid jointed plane frame ABCD with pinned supported at A and supported on roller at D.

There is a point load of 200 kN subjected at B. Beam CD is subjected to a uniformly distributed load of

magnitude 50 kN/m. By taking E is 200 GPa and I is 500 x 106 mm

4.

a) Show that the frame is statically determinate frame.

b) Using Castigliano 2nd theorem, determine horizontal displacement at joint C.

Figure 2.5

Solution:

a) Show that the frame is statically determinate frame.

b) Horizontal displacement at joint C can be determined using the equation as below.

Analysis of frame is equal to analysis of beam since we only consider bending effect in the frame.

EI

2EI

2EI

4 m

4 m

8 m

50 kN/m

200 kN

A

D

B

C


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Castigliano 2nd theorem for trusses

In the general case where L, A and E are constant for a given member, we may write the equation for

truss as

AE

L

P

NN

Where

Δ = external joint displacement of the truss

P = external force applied to the truss join in the direction of Δ.

N = internal force in a member caused by both the force P and the loads on the truss.

L = length of a member

A = cross-sectional area of a member

E = modulus of elasticity of a member

This equation is similar to that used for the method of virtual work, except n is replaced byP

N

. Notice

that in order to determine this partial derivative, 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.

As a result, computing P

N

generally requires slightly more calculation than required to compute each n

force directly.

These terms will of course be the same since n or P

N

is simply the change of the internal member force

with respect to the load P, or the change in member force per unit load.


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Procedure for Analysis

The following procedure provides a method that may be used to determine the displacement of any joint

of a truss using Castigliano’s 2nd theorem.

Example 3

Determine the vertical displacement of joint C of the steel truss shown in Figure 6 using Castigliano’s

second theorem. The cross-sectional area of each member is A = 300 mm2 and E = 200 GPa.

Figure 2.6

F E

D

C B A

20 kN 20 kN

3 m 3 m 3 m

3 m


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Solution:

Step 1: 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 and should be directed along the line of action of the displacement.

Figure 2.7: External Force (P) at point C

0.667P+6.67

0

F E

D

C B

A

P 20 kN 0.333P+13.33


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Step 2: Internal Forces (N)

Determine the force N in each member caused by both the real loads and the variable force P. Assume

tensile forces are positive and compressive forces are negative.

Compute the respective partial derivative P

N

for each member.

After N and P

N

have been determined, assign P its numerical value if it has replaced a real force on the

truss. Otherwise, set P equal to zero.

Arranging the data in tabular form, we have

Member L (m) N

P

N

N (P = 20 kN) L

P

NN

AB

BC

CD

FE

BF

CE

AF

BE

DE

Step 3: Castigliano’s Theorem


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Apply Castigliano’s 2nd theorem to determine the desired displacement Δ. It is important to retain the

algebraic signs for corresponding of N and P

N

when substituting these terms into the equation.

AE

L

P

NNCv


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Self Learning Exercise

Question 4

A pin-jointed truss is shown in Figure 2.8. The truss has pinned support at A and D as shown. The truss

is subjected to a vertical load of 150 kN at joint C and 50 kN at joint B. By assuming that EA is constant

for all members, Take E = 200 GPa and A = 500 mm2 for all members.

(a) Show that the truss in Figure 2.8 is a determinate truss.

(b) Determine the vertical displacement at joint E by using Castigliano’s 2nd Theorem.

(c) If area of the member AB and BD varies between 500 mm2 and 2500 mm

2, sketch a graph of axial

rigidity, EA versus vertical deflection, ∆E. Then, made a conclusion of its relationship.

Figure 2.8

4 m 6 m

C A

E D

B

150 kN 50 kN

3 m


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Question 5

A pin-jointed truss is shown in Figure 2.9. The truss has pinned support at A and supported on roller at E

as shown. The truss is subjected to a vertical load of 150 kN at joint D and unknown horizontal load, Q at

joint C. By assuming that EA is constant for all members, Take E = 200 GPa and A = 500 mm2 for all

members.

(a) Show that the truss in Figure 2.9 is a determinate truss.

(b) Using Trigonometry, calculate length of ‘a’.

(c) If horizontal displacement at joint C is 0.025 m to the right direction (0.025 m), determine the value

of P by using Castigliano 2nd theorem.

Figure 2.9

a

4 m

6 m

C

A E

D B

150 kN

P

30o


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castigliano's 2nd theorem

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