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CHAPTER 2CHAPTER 2 (C01-P03)(C01-P03)
CASTIGLIANO’S 2ND THEOREMCASTIGLIANO’S 2ND THEOREM
Learning OutcomeLearning Outcome
At the end of this chapter, students should:At the end of this chapter, students should:
1)1) Able to determine deformation for statically determinate beams by usingAble to determine deformation for statically determinate beams by using Castigliano’s 2Castigliano’s 2ndnd^ theorem.theorem.
2)2) Able to determine deformation for statically determinate frames by using Castigliano’s 2Able to determine deformation for statically determinate frames by using Castigliano’s 2 ndnd^ theorem.theorem.
3)3) Able to determine deformation for statically determinate trusses by using Castigliano’s 2Able to determine deformation for statically determinate trusses by using Castigliano’s 2 ndnd^ ttheoremheorem..
Castigliano 2nCastigliano 2nd Theoremd Theorem (C2T) for Beams a(C2T) for Beams and Framesnd Frames
In 1879 Alberto Castigliano, an Italian railroad engineer, published a book in which he outlined a methodIn 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.for determining the deflection or slope at a point in a structure.
The formula for theThe formula for the Castigliano 2nd TheoremCastigliano 2nd Theorem for beamfor beam and frame deflections isand frame deflections is as follows.as follows.
WhereWhere
∆∆ == externalexternal displacementdisplacement ofof thethe pointpoint causedcaused byby thethe realreal loadsloads actingacting onon thethe beambeam oror
frame.frame.
PP == external force applied to the beam of frame in the direction of ∆.external force applied to the beam of frame in the direction of ∆.
MM == internalinternal momentmoment inin thethe beambeam oror frame,frame, expressedexpressed asas aa functionfunction ofof xx andand causedcaused byby bothboth
the force P and the real loads on the beam or frame.the force P and the real loads on the beam or frame.
EE == modulusmodulus ofof elasticityelasticity ofof thethe material.material.
II == momentmoment ofof inertiainertia ofof cross-scross-sectionalectional area,area, computedcomputed aboutabout thethe neutralneutral axis.axis.
In a similar manner, if the slopeIn a similar manner, if the slope at a point is to be determine, we must find the partial derivative of theat 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.internal moment M with respect to an external couple moment M’ acting at the point i.e.
Procedure for analysisProcedure for analysis
The following step-by-step procedure can be used to determine the slopes and deflections of beams andThe following step-by-step procedure can be used to determine the slopes and deflections of beams and
frames byframes by Castigliano’s 2nd theoremCastigliano’s 2nd theorem..
1.1. Place a force P on the beam or frame at the point and in the direction of the desired displacement. IfPlace a force P on the beam or frame at the point and in the direction of the desired displacement. If
the sthe slope is to be determined, place a couple moment M’ at the point. It is assumed that both P and M’lope is to be determined, place a couple moment M’ at the point. It is assumed that both P and M’ have a variable magnitude.have a variable magnitude.
2.2. Calculate reactions at the supportCalculate reactions at the support in terms of P /M’in terms of P /M’ by usingby using EquilibriumEquilibrium EquationEquations.s.
3.3. For each segment of the beam/frame, determine the equation expressing the variation of the bendingFor 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 coordinatemoment along the length of the segment in terms of a position coordinate xx.. ItIt isis usuallyusually convenientconvenient to consider the bending moments as positive in accordance with the beam sign convention.to consider the bending moments as positive in accordance with the beam sign convention.
4.4. Compute the partial derivativeCompute the partial derivative ^ oror for each coordinatefor each coordinate xx..
5.5. After M andAfter M and oror have been determined, assign P or M’ its numerical value if it hashave 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 zreplaced a real force or couple moment. Otherwise, set P or M’ equal to z ero.ero.
6.6. Apply Equation to determine the desired displacementApply Equation to determine the desired displacement
or slopeor slope
. If the resultant sum of all the. If the resultant sum of all the
definite integral is positive, displacementdefinite integral is positive, displacement or slopeor slope is in the same direction of P or M’.is in the same direction of P or M’.
Solve the reactions for the real beam/system using Equilibrium Equation.Solve the reactions for the real beam/system using Equilibrium Equation.
∑M∑MAA = 0,= 0,
25(20)(20/2) + P(10) + 150(25)25(20)(20/2) + P(10) + 150(25) – – Cy(20)Cy(20) == 00
CCyy == (8750(8750 ++ 10P)10P) // 20 20
CCyy == 437.5437.5 ++ 0.5P0.5P
∑F∑Fyy = 0,= 0,
AAyy + C+ Cyy - - 25(20)25(20) -- PP -- 150 150 == 00
AAyy ++ (437.5(437.5 ++ 0.5P)0.5P) -- 25(20)25(20) -- PP -- 150 150 == 00
AAyy == 212.5212.5 ++ 0.5P0.5P
∑F∑Fxx = 0,= 0,
AAxx^ == 0 0 kNkN
Step 2: Moment equation for each member(M)Step 2: Moment equation for each member(M)
Member AB (origin A) 0 ≤Member AB (origin A) 0 ≤ xx ≤ 10 m≤ 10 m
Free-body diagram (FBD)Free-body diagram (FBD)
∑M = 0,∑M = 0,
- –MM +(212.5 + 0.5P)(x)+(212.5 + 0.5P)(x) – – 25(x)(x/2)25(x)(x/2) == 00
MM == 21 212.52.5xx ++ 00.5Px.5Px – – 12.5x12.5x^22
25 kN/m25 kN/m
212.5 + 0.5P212.5 + 0.5P
AA
MM
xx
Member BC (origin A) 10Member BC (origin A) 10 ≤≤ xx ≤≤ 20 m20 m
Free-body diagram (FBD)Free-body diagram (FBD)
∑M = 0,∑M = 0,
- –M +(212.5M +(212.5 + 0.5P)(x+ 0.5P)(x)) – – 25(x)(x/2)25(x)(x/2) – – P (xP (x––10)10) == 00
MM == 21 212.52.5xx ++ 00.5Px.5Px (^) – – 12.5x12.5x^22 – – PxPx + 1+ 10P0P
Member CD (origin DMember CD (origin D) 0 ≤) 0 ≤ xx ≤≤ 5 m5 m
Free-body diagram (FBD)Free-body diagram (FBD)
∑M = 0,∑M = 0,
M + 100(x)M + 100(x) == 00
MM == -10-100x0x
xx
DD
10 m10 m ((^ xx^ – –^ 10) m10) m
25 kN/m25 kN/m
PP
BB
212.5 + 0.5P212.5 + 0.5P
MM
xx
MM
150 kN150 kN
b)b) Using Castigliano’sUsing Castigliano’s 2nd theorem, determine rotation at joint D.2nd theorem, determine rotation at joint D.
Step 1: Reactions in terms of M’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,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’.determine reactions at each support using equilibrium equation in terms of M’.
M’M’
25 kN/m25 kN/m
150 kN150 kN
AA
100 kN100 kN
BB
CC
DD
10 m10 m (^) 10 m10 m 5 m5 m
Step 2: Moment equation for each member(M)Step 2: Moment equation for each member(M)
Self-Learning ExerciseSelf-Learning Exercise
Question 1Question 1
Below is a cantilever beam ABCD subjected uniformly distributed load of 25 kN/m along member ABCBelow is a cantilever beam ABCD subjected uniformly distributed load of 25 kN/m along member ABC whereas joint B and Dwhereas joint B and D are subjected to a point load of 100 kN andare subjected to a point load of 100 kN and 150 kN, respectively.150 kN, respectively. The total lengthThe total length of the beam is 25 m.of the beam is 25 m.
a)a) Using Castigliano’s 2nd theoremUsing Castigliano’s 2nd theorem, determine vertical displacement at joint D in terms of, determine vertical displacement at joint D in terms of EIEI.. b)b) If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities,If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities, EIEI..
Figure 2.2Figure 2.
Solution:Solution:
a)a) Using Castigliano’s 2nd theorem, determine vertical displacement at joint D in terms of EI.Using Castigliano’s 2nd theorem, determine vertical displacement at joint D in terms of EI. Step 1: Reactions in terms of PStep 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, determineSince 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.reactions at each support using equilibrium equation in terms of P.
25 kN/m25 kN/m
150 kN150 kN
AA
100 kN100 kN
BB CC DD
10 m10 m 10 m10 m 5 m5 m
25 kN/m25 kN/m
PP
AA
100 kN100 kN
BB CC DD
10 m10 m (^) 10 m10 m 5 m5 m
Step 2: Moment equation for each member (M)Step 2: Moment equation for each member (M)
Lastly, apply the formula as below.Lastly, apply the formula as below.
b)b) If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities, EI.If rotation at joint C is 0.05 radian clockwise, determine value of flexural rigidities, EI.
Question 3Question 3
Below is a simply supported beam ABCD subjected uniformly distributed load of 25 kN/m along memberBelow is a simply supported beam ABCD subjected uniformly distributed load of 25 kN/m along member ABC whereas joint B andABC whereas joint B and D are subjected to a point load ofD are subjected to a point load of 100 kN and 150 kN, respectively.100 kN and 150 kN, respectively. The totalThe total length of the beam is 25 m.length of the beam is 25 m.
a)a) Using Castigliano’s 2nd theoremUsing Castigliano’s 2nd theorem, determine rotational displacement at joint D in terms of, determine rotational displacement at joint D in terms of EIEI ifif EIEI isis
constant for all members.constant for all members. b)b) If the rotational displacement at joint B is given as 0.05 rad counterclockwise, determine the flexuralIf the rotational displacement at joint B is given as 0.05 rad counterclockwise, determine the flexural rigidities,rigidities, EIEI valuevalue by using Casby using Castiglianotigliano’s 2nd theo’s 2nd theoremrem..
Figure 2.4Figure 2.
25 kN/m25 kN/m
150 kN150 kN
AA
100 kN100 kN
BB
CC
DD
10 m10 m (^) 10 m10 m 5 m5 m
Example 2: Unit Load Method for FrameExample 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.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 ofThere is a point load of 200 kN subjected at B. Beam CD is subjected to a uniformly distributed load of magnitude 50 kN/m.magnitude 50 kN/m. By taking EBy taking E is 200 GPa andis 200 GPa and I is 500 x 10I is 500 x 10^66 mmmm^44 ..
a)a) Show that the frame is statically determinate frame.Show that the frame is statically determinate frame.
b)b) Using Castigliano 2nd theorem, determine horizontal displacement at joint C.Using Castigliano 2nd theorem, determine horizontal displacement at joint C.
Figure 2.5Figure 2.
Solution:Solution:
a)a) Show that the frame is statically determinate frame.Show that the frame is statically determinate frame.
b)b) Horizontal displacement at joint C can be determined using the equation as below.Horizontal displacement at joint C can be determined using the equation as below.
Analysis of fAnalysis of frame is erame is equal to analyqual to analysis of besis of beam since wam since we only coe only consider bendnsider bending effecting effect in the frain the frame.me.
EIEI
2EI2EI
2EI2EI
4 m4 m
4 m4 m
8 m8 m
50 kN/m50 kN/m
200 kN200 kN
AA
DD
BB
CC
Procedure for AnalysisProcedure for Analysis
The following procedure provides a method that may be used to determine the displacement of any jointThe following procedure provides a method that may be used to determine the displacement of any joint of a truss usingof a truss using Castigliano’Castigliano’s 2s 2nd theoremnd theorem..
Example 3Example 3
Determine the vertical displacement of joint C of the steel truss shown in Figure 6Determine the vertical displacement of joint C of the steel truss shown in Figure 6 usingusing Castigliano’sCastigliano’s
second theoremsecond theorem. The cross-sectional area of each member is. The cross-sectional area of each member is AA = 300 mm= 300 mm^22 andand EE = 200 GPa.= 200 GPa.
Figure 2.6Figure 2.
FF EE
DD
BB CC
AA
20 kN20 kN 20 kN20 kN
3 3 mm 3 3 mm 3 3 mm
3 m3 m
Solution:Solution:
Step 1: External Force (P)Step 1: External Force (P)
Place a forcePlace a force PP on the truss at the joint where the desired displacement is to be determined. This force ison 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.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 CFigure 2.7: External Force (P) at point C
0.667P+6.670.667P+6.
00
FF (^) EE
DD BB CC
AA
0.333P+13.330.333P+13.33 20 kN20 kN PP
