Energy Deflection

1! External Work and Strain Energy! Principle of Work and Energy! Principle of Virtual Work! Method of Virtual Work:

! Trusses! Beams and Frames

! Castigliano's Theorem! Trusses! Beams and Frames

DEFLECTIONS: ENERGY METHODS

2Ue

Eigen work

External Work and Strain Energy

Most energy methods are based on the conservation of energy principle, which statesthat the work done by all the external forces acting on a structure, Ue, is transformed into internal work or strain energy, Ui.

Ue = Ui

L

F

x

F

P

xPF

=

As the magnitude of F is gradually increasedfrom zero to some limiting value F = P, the finalelongation of the bar becomes .

External Work-Force.

Eigen work

FdxdUe =

=x

e FdxU0

=

0

)( dxxPUe

=

=

PxPUe 21)

2(

0

2

3P

L

F'

Displacement work

x

F

P Eigen

work

(Ue)Total = (Eigen Work)P + (Eigen Work)F + (Displacement work) P

'

L

F' + P

)'()')('(21))((

21)( ++= PFPU Totale

410 mm

L

20 kN

L

x (m)

F

0.01 m

20 kN

mNUe == 100)1020)(01.0(21 3

5Displacement work

5 kN

x (m)

F

L

2.5 mm

15 kN

0.0075

Eigen

work

L

15 kN

7.5 mm

L

15 kN

7.5 mm 0.01

20 kN

)1015)(0025.0()105)(0025.0(21)1015)(0075.0(

21 333 ++=W

mN =++= 10050.3725.625.56

6 External Work - Moment.

dM

MddUe =

Displacement work

M

M Eigen

work

'

M' + M

=

0

MdUe -----(8-12)

MUe 21

= -----(8-13)

Eigen work

'''21

21)( MMMU Totale ++=

)148()')('(21)( ++= MMU Totale

721

=oU

Strain Energy-Axial Force.

L

N

=V

dV ))(21(

=V

dVE

)(21 2

=V

i dVUU 0

=V

dVAN

E2)(

21

=L

AdxAN

E2)(

21

=L

dxEA

N )2

(2

=E

AN

=

8 Strain Energy-Bending

21

=oU

x dx

wP

L

=V

dV ))(21(

=V

dVI

MyE

2)(21

M M

dx

d

d=V

dVI

yME

)(21 2

2

2

=L

AdxdAy

IM

E))((

21 2

2

2

=L

dxEIM )

2(

2

=V

i dVUU 0

=V

dVE

)(21 2

IMy

=

I

9

=G

dx

cd

J

TT

Strain Energy-Torsion

21

=oU =L

i dxGJTU

2

2

JT =

=V

i dVUU 0

=V

dV)21(

=V

dVG

)(21 2

=V

dVJ

TG

2)(21

=L A

dxdAJT

G))((

21 2

2

2

10

VV

dx

dy

AK

Strain Energy-Shear

=G

21

=oU

=V

dV ))(21(

=V

dVG

)(21 2

=V

i dVUU 0

=L A

dxdAIt

QG

V )(2

22

=L

i dxGAVKU2

2

= dVItVQ

G2)(

21

11

Principle of Work and EnergyP

L

-PL

M diagram

+ Mx = 0: 0= PxM

PxM =

x

P

xV

M

ie UU =

=L

EIdxMP

0

2

221

=L

EIdxPxP

0

2

2)(

21

L

EIxPP

0

32

621

=

EIPL3

3

=

12

dLudVUP +=+ 011 1)21(

Then apply real load P1.

Au

u

L

Principle of Virtual Work

Apply virtual load P' first.

P1

A

P' = 1

1 = u dLReal displacements

Virtual loadings

1 = u dLReal displacements

Virtual loadings

In a similar manner,

u

u

L

dL

ie UU =

1

Real Work

13

B

Method of Virtual Work : Truss

External Loading.

N 2

N1

N3 N4

N5N

6

N7 N8 N9

1kN

n 2

n1

n3 n 4 n5

n6

n7 n8 n9

Where:1 = external virtual unit load acting on the truss joint in the stated direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the real load on the trussN = internal normal force in a truss member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity of a member

P1

P2

B

AEnNL

=1

14

Where: = external joint displacement caused by the temperature change = coefficient of thermal expansion of member

T = change in temperature of member

Where: = external joint displacement caused by the fabrication errors

L = difference in length of the member from its intended size as caused by a fabrication error

LTn )(1 =

Ln=1

Temperature

1 = u dL

Fabrication Errors and Camber

1 = u dL

dL

dL

15

Example 8-15

The cross-sectional area of each member of the truss shown in the figure isA = 400 mm2 and E = 200 GPa.(a) Determine the vertical displacement of joint C if a 4-kN force is applied to thetruss at C.(b) If no loads act on the truss, what would be the vertical displacement of joint Cif member AB were 5 mm too short?(c) If 4 kN force and fabrication error are both accounted, what would be thevertical displacement of joint C.

A B

C

4 m 4 m

4 kN

3 m

16

A B

C4 kN

N(kN)

A B

C

n (kN)

SOLUTION

Virtual Force n. Since the vertical displacement of joint C is to bedetermined, only a vertical 1 kN load is placed at joint C. The n force ineach member is calculated using the method of joint.

1 kN

0.667-0.

833-0.833

2+2

.5 -2.5

1.5 kN1.5 kN

4 kN

0.5 kN0.5 kN

0

Real Force N. The N force in each member is calculated using themethod of joint.

Part (a)

17CV = +0.133 mm,

0.667-0.

833-0.833

2+2

.5 -2.5

8

5 5

10.67-10

.41 10.41

A B

C

n (kN)

1 kN

A B

C4 kN

N (kN)

A B

C

L (m)=A B

C

nNL (kN2m)

= AEnNLkN CV ))(1(

)10200)(10400(

67.10)67.1041.1041.10(1

2626

mkNm

mkNAECV

=++=

18

Part (b): The member AB were 5 mm too short

5 mm

CV = -3.33 mm,

Part (c): The 4 kN force and fabrication error are both accounted.

CV = 0.133 - 3.33 = -3.20 mm

CV = -3.20 mm,

A B

C

n (kN)

1 kN

0.667-0.

833-0.833

)())(1( LnCV =

)005.0)(667.0( =CV

19

Example 8-16

Determine the vertical displacement of joint C of the steel truss shown. Thecross-section area of each member is A = 400 mm2 and E = 200 GPa.

4 m 4 m 4 m

AB C

D

EF

4 m

4 kN4 kN

20

4 m 4 m 4 m

AB C

D

EF

4 m

n (kN)

4 m 4 m 4 m

AB C

D

EF

4 m

4 kN4 kNN(kN)

SOLUTION

Virtual Force n. Since the vertical displacement of joint C is to bedetermined, only a vertical 1 kN load is placed at joint C. The n force ineach member is calculated using the method of joint.

Real Force N. The N force in each member is calculated using themethod of joint.

1 kN

0.667-0.

471

-0.47

1

-0.943

0.6670.333

0

.

3

3

3

1

-0.333

4-5.

66 0

-5.66

444 4

-4

0.667 kN0.333 kN

0

4 kN4 kN

0

21CV = 1.23 mm,

0.667-0.

471

-0.47

1

-0.943

0.6670.3330

.

3

3

3

1

-0.333

AB C

D

EF

n (kN) 1 kN

4-5.

66 0

-5.66

444 4

-4

AB C

D

EF

4 kN4 kN N(kN)

45.6

6 5.66

5.6644

4 4

4

AB C

D

EF

L(m)

AB C

D

EF

nNL(kN2m)

=

10.6715

.07 0

30.1810.675.33

5

.

3

3

16

5.33

= AEnNLkN CV ))(1(

)10200)(10400(

4.72)]18.3016)67.10(2)33.5(307.15[(1

2626

mkNm

mkNAECV

=++++=

22

Example 8-17

Determine the vertical displacement of joint C of the steel truss shown. Due toradiant heating from the wall, members are subjected to a temperature change:member AD is increase +60oC, member DC is increase +40oC and member AC isdecrease -20oC.Also member DC is fabricated 2 mm too short and member AC3 mm too long. Take = 12(10-6) , the cross-section area of each member is A =400 mm2 and E = 200 GPa.

2 mA B

CD

3 m

20 kN

10 kNwall

23

2 mAB

CD

3 m

n (kN)

SOLUTION

1 kN

0.667

0

-1.2

01

13.33 kN

23.33 kN

20 kN

23.33

0

-24.04

2020

0.667 kN

0.667 kN

1 kN

Due to loading forces.

CV= 2.44 mm,

2 mAB

CD

3 m

20 kN

10 kN

N (kN)

2

2

3.61

33

AB

CD

L (m)

31.13

0

104.1

2

060

AB

CD

nNL(kN2m)

= AEnNLkN CV ))(1(

)12.10413.3160()200)(400(

1++=CV

24

Due to temperature change.

Due to fabrication error.

Total displacement .

1 kN0.667

0

-1.2

01

A B

CD

n (kN)

+40

-20

+60

A B

D

T (oC)

C 2

2

3.61

33

AB

CD

L (m) Fabrication error (mm)

-2

+ 3

A B

D C

LTnkN CV )())(1( =

=++= ,84.3)]61.3)(20)(2.1()2)(40)(667.0()3)(60)(1)[(1012( 6 mmCV

)())(1( LnkN CV =

=+= ,93.4)003.0)(2.1()002.0)(667.0( mmCV

=+= ,35.193.484.344.2)( mmTotalCV

25

Method of Virtual Work : Bending

wC

A BC

RBRA

==L

C dxEIMmdm )())((1

Virtual loadings

Real displacements

M M

dx

d

d

dds =

dxEIMdsd =

1

26

Method of Virtual Work : Beams and Frames

wC

A BC

RBRA

==L

C dxEIMmdm )())((1

wC

A B

RBRA

C

==L

C dxEIMmdm )())((1

Virtual loadings

Real displacements

Virtual loadings

Real displacements

27

Virtual unit load

CA B

wC

A B

Method of Virtual Work : Beams and Frames

C

Real load

1RA RBx1 x2

RBRA

x1 x2

B

RB

x2

v2

m2x1

RA

v1

m1

Vertical Displacement

w

B

x2

RB

V2

M2

x1

RA

V1

M1

=L

C dxEIMm )(1

28

Virtual unit couple

CA B

wC

A B

Slope

Real load

x1 x2RBRA

w

B

x2

RB

V2

M2

x1

RA

V1

M1B

RB

x2

v2

m2

RA

v1

m1

RA

x1 x21

C RB

=L

C dxEIMm )(1

29

Example 8-18

The beam shown is subjected to a load P at its end. Determine the slope anddisplacement at C. EI is constant.

2a a

AB C

P

C

30

Real Moment M

AB

C2a a

P

AB

C2a a

SOLUTION

Virtual Moment m

Displacement at C

1 kN

x1 x2

-a

m

m2 = -x2

x1 x2

-Pa

M

M2 = -Px2

23

21

21

1PxM =

23P

2P

+== a a

LC dxPxxEI

dxPxxEI

dxEI

Mm 2

0 02221

11 ))((1)2

)(2

(11

21

1xm =

=+=+=EI

PaEI

PaEI

PaEI

PxEI

Pxaa

aC 3312

8)

3()

12(

333

0

32

231

31

AB

C2a a

P

AB

C2a a

Virtual Moment m Real Moment M

Slope at C

x1 x2

-1

m

x1 x2

-Pa

M

M2 = -Px2

1 kNm

a21

a21

21

1PxM =

23P

2P

M2 = -Px2axm2

11 = a

xm2

11 =

12 =m 12 =m 21

1PxM =

+==aaL

C dxPxEIdxPx

ax

EIdx

EIMmmkN

0221

12

0

1

0

))(1(1)2

()2

(1))(1(

),(67)

2)(1()

38)(

4)(1()

2)(1()

3)(

4)(1(

223

0

22

2

0

31

EIPaPa

EIa

aP

EIPx

EIx

aP

EI

aa

C =+=+=

32

C2a a

AB C

P

),(67 2

EIPa

C =

=EI

PaC 3

3

C

Conclusion

33

Example 8-19

Determine the slope and displacement of point B of the steel beam shown in thefigure below. Take E = 200 GPa, I = 250(106) mm4.

A5 m

B

3 kN/m

34

SOLUTION

Virtual Moment m

A5 m

B

1 kNx

1 kNx

v

m-1x =

xReal Moment M

A5 m

B

3 kN/m

x

3x

2x

V

M= 23 2x

Vertical Displacement at B

-1x = = 23 2x

EImkNx

EIx

EIdxxx

EIdx

EIMmkN

L

B

325

0

45

0

35

0

2

0

375.234)8

3(12

31)2

3)((1))(1( =====

==

=

,69.400469.0

)10250)(10200(

375.234466

3

mmmm

mkN

mkNB

35

SOLUTION

Virtual Moment m

A5 m

B

-1 =

xReal Moment M

A5 m

B

3 kN/m

=2

3 2x

Slope at B

x 1 kNm

x

v

m 1 kNm

x

3x

2x

V

M-1 = = 23 2x

EImkNx

EIx

EIdxx

EIdx

EIMmmkN

L

B

325

0

35

0

5

0

22

0

5.62)6

3(12

31)2

3)(1(1))(1( =====

,00125.0)10250)(10200(

5.62466

2

radm

mkN

mkNB =

=

36

Example 8-20


A C D

B

5 kN14 kNm

2 m 2 m 3 m

37

AC

DB

5 kN14 kNm

2 m 2 m 3 m

AC DB

2 m 2 m 3 m

1 kNVirtual Moment m Real Moment M

0.5 kN0.5 kN

x3x2

6 kN1 kN

111 5.0 xm =

x1

22 5.0 xm =m M

14

x3x2x1

M1 = 14 - x1M2 = 6x2

11 5.0 xm = 22 5.0 xm =M1 = 14 - x1

M2 = 6x2

Displacement at B

=L

B dxEIMmkN

0

))(1(

++=3

03

2

0222

2

0111 )0)(0(

1)6)(5.0(1)14)(5.0(1 dxEI

dxxxEI

dxxxEI

2

0

32

2

0

31

21

2

02

22

2

01

211 )3

3)(1(35.0

27)(1()3(1)5.07(1 x

EIxx

EIdxx

EIdxxx

EI+=+=

==== ,72.100172.0)60)(200(

667.20667.20 mmmEIB

38

AC

DB

5 kN14 kNm

2 m 2 m 3 m

AC DB

2 m 2 m 3 m


0.25 kN

x3x2

6 kN1 kN

x1

mM

14

x3x2x1

M1 = 14 - x1M2 = 6x2

1 kNm 0.25 kN

0.5

-0.5

m1 = 0.25x1

m2 = -0.25x2

M1 = 14 - x1M2 = 6x2

m1 = 0.25x1

m2 = -0.25x2

Slope at B

=L

B dxEIMmmkN

0

))(1( ++=3

03

2

0222

2

0111 )0)(0(

1)6)(25.0(1)14)(25.0(1 dxEI

dxxxEI

dxxxEI

+=2

02

22

2

01

211 )5.1(

1)25.05.3(1 dxxEI

dxxxEI

2

0

32

2

0

31

21 )

35.1(1)

325.0

25.3(1 x

EIxx

EI+=

,000194.0)60)(200(

333.2333.2 radEIB

===

39

Example 8-21

From the structure shown. Determine the slope and displacement at C. Take E =200 GPa, I = 200(106) mm4.

20 kN

Hinge

30 kNm

AB

C

4 m 3 m

2EI EI

40

20 kN

Hinge

30 kNm

A B

C

4 m 3 m

2EI EI

30 kNm

BC

30/3 = 10 kN10 kN20 kN

10 kN30 kN

30 kN30 kN

120 kNmA B

M (kNm) x (m)

-120

30

41

Real Moment MDisplacement at B

A BC

4 m 3 m

20 kN 30 kNm

2EI EI

M (kNm) 30

-120 M1 = -30x1

x1 x2

M2 = 10x2

10 kN30 kN

120 kNm

Virtual Moment m

A

BC

4 m 3 m

2EI EI

1 kN

0 kN1 kN

4 kNm

M (kNm) x1 x2

-4 m1 = -x1 m2 = 0

=L ii

iiB dxIE

Mm 02

)30)((4

0

111 += EI

dxxx

4

0)

330(

21 3xEI

=

=

== mEI

008.01040

32323

42

Real Moment MSlope at the left of B

A BC

4 m 3 m

20 kN 30 kNm

2EI EI

M (kNm) 30

-120 M1 = -30x1

x1 x2

M2 = 10x2

10 kN30 kN

120 kNm

Virtual Moment m

A

BC

4 m 3 m

2EI EI

0 kN0

1 kNm

M (kNm) x1 x2

m1 = -1 m2 = 0

iL iiBL dxIE

mM= 02)30)(1(4

0

11 += EI

dxx

4

0)

230(

21 2xEI

=

1 kNm

-1 -1

radEI

003.01040

1201203 =

==

43

Real Moment MSlope at the right of B

A BC

4 m 3 m

20 kN 30 kNm

2EI EI

M (kNm) 30

-120 M1 = -30x1

x1 x2

M2 = 10x2

10 kN30 kN

120 kNm

Virtual Moment m

A

BC

4 m 3 m

2EI EI

1/3 kN1/3 kN

4/3 kNm

M (kNm) x1 x2

iL iiBR dxIE

mM= ++=3

0

22

24

0

11

1 )10)(3

1(2

)30)(3

(EIdxxx

EIdxxx

3

0

4

0)

910

210(1)

310(

21 32

22

31 xx

EIx

EI++=

1 kNm

-4/3 m1 = -x1/3 m2 = -1 + x2/3-1

radEIEI

0023.0104067.91)3045(167.106 3 =

=++=

44

20 kN

Hinge

30 kNm

A B

C

4 m 3 m

2EI EI

B = 8 mm radBR 0023.0=Deflected Curve

radBL 003.0=

45

Example 8-22

(a) Determine the slope and the horizontal displacement of point C on the frame.(b) Draw the bending moment diagram and deflected curve. E = 200 GPa

I = 200(106) mm4

A

B C5 m

6 m2 kN/m

4 kN

1.5 EI

EI

46

A

B C5 m

6 m2 kN/m

4 kN

x1

x2

1x2

x1

M2= 12 x212 kN

16 kN

12 kN

m2= 1.2 x2

m1= x1

1.2 kN

1 kN

1.2 kN

Real Moment M

M1= 16 x1- x12

M2= 12 x2 m2= 1.2 x2

m1= x1M1= 16 x1- x12

+== 6

0

5

02221

2111 )12)(2.1(

1)16)((5.111 dxxx

EIdxxxx

EIdx

EIMm

LCH

+=5

02

22

6

01

31

21 )4.14(

1)16(5.11 dxx

EIdxxx

EI

+==+=+= ,8.28)200)(200(

1152600552)34.14(1)

4316(

5.11

5

0

32

6

0

41

31 mm

EIEIx

EIxx

EICH

1.5 EIEI

A

C

Virtual Moment m

1.5 EIEI

47

A

C

A

B C5 m

6 m2 kN/m

4 kN

x1

x2x2

x1

M2= 12 x212 kN

16 kN

12 kN

m2= 1-x2/5

m1= 0

1/5 kN

0

Real Moment M Virtual Moment m

M1= 16 x1- x12

1 kNm

1/5 kN

M2= 12 x2 m2= 1-x2/5

m1= 0M1= 16 x1- x12

+==6

0

5

022

21

211 )12)(5

1(1)16)(0(5.111 dxxx

EIdxxx

EIdx

EIMm

LC

+=5

02

22

2 )51212(10 dxxx

EI

,00125.0)200)(200(

5050)35

122

12(15

0

32

22 rad

EIxx

EIC+===

=

1.5 EIEI

1.5 EIEI

48

+

+

60

60

M , kNm

16

-12-

+V , kN

4

A

B C5 m

6 m2 kN/m

4 kN

12 kN

16 kN

12 kN

CH = 28.87 mm

C = 0.00125 rad ,

49

Example 8-23

Determine the slope and the vertical displacement of point C on the frame.Take E = 200 GPa, I = 15(106) mm4.

5 kN

3 m

60o

2 mA

B

C

50


3 m

B

C

30o

Displacement at C

3 m

B

C

30ox1

1 kN

x1

1 kN

C

30o

n1

v1

m1 = -0.5x11.5 m1.5 kNm

1 kN

m1 = -0.5x1

x1

5 kN

1.5 m M1 = -2.5x1

x1

5 kN

C

30o

N1

V1

7.5 kNm

M1 = -2.5x1

x22 mA

1.5 kNm

m2 = -1.5

x2

5 kN

2 mA

7.5 kNm

M2 = -7.5

x2

1.5 kNm1 kN

v2n2 m2 = -1.5

x2

7.5 kNm5 kN

N2

V2

M2 = -7.5

=L

CV dxEIMm1 +=

2

0211

3

01 )5.7)(5.1(

1)5.2()5.0(1 dxEI

dxxxEI

)15)(200(75.3375.33)25.11(1)

325.1(1

2

0

3

0

22

31 ==+=

EIx

EIx

EICV= 11.25 mm ,

51


3 m

B

C

30ox1

1.5 m

1 kNm

x22 mA

1 kNm

m1 = -1

m2 = -1

2 mA

3 m

B

C

30ox1

5 kN

1.5 m

7.5 kNm

x2

7.5 kNm5 kN

M1 = -2.5x1

M2 =- 7.5

x1

5 kN

C

30o

N1

V1

=L

C dxEIMm1 +=

2

0211

3

0

)5.7)(1(1)5.2()1(1 dxEI

dxxEI

)15)(200(25.2625.26)5.7(1)

25.2(1

2

0

3

0 2

21 ==+=

EIx

EIx

EIC = 0.00875 rad,

Slope at C

1 kNm

x1 C

30o

n1

v1

1 kNm

m1 = -1 M1 = -2.5x1

x2

1 kNm

n2

v2

m2 = -1

x2

7.5 kNm5 kN

N2

V2

M2 = -7.5

52

Virtual Strain Energy Caused by Axial Load, Shear, Torsion, and Temperature

Axial Load

Wheren = internal virtual axial load caused by the external virtual unit loadN = internal axial force in the member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity for the material

d

==L

i dxEANndnU )(

53

Bending

Wheren = internal virtual moment cased by the external virtual unit loadM = internal moment in the member caused by the real loadsL = length of a memberE = modulus of elasticity for the materialI = moment of inertia of cross-sectional area, computed about the the neutral axis

d

==L

i dxEIMmdmU )(

54

Torsion

Where t = internal virtual torque caused by the external virtual unit loadT = internal torque in the member caused by the real loadsG = shear modulus of elasticity for the material J = polar moment of inertia for the cross section, J = c4/2, where c is the

radius of the cross-sectional area

d

==L

i dxGJTtdtU )(

55

Shear

Wherev = internal virtual shear in the member, expressed as a function of x and caused by the external virtual unit loadV = internal shear in the member expressed as a function of x and caused by the real loadsK = form factor for the cross-sectional area:

K = 1.2 for rectangular cross sectionsK = 10/9 for circular cross sectionsK 1 for wide-flange and I-beams, where A is the area of the web

G = shear modulus of elasticity for the materialA = cross-sectional area of a member

d

==L

i dxGAKVvdvU )(

56

Axial =L

i dxTnU )(

Bending

=L

i dxcTmU )

2

Temperature Displacement :

Where = Differential temperatures:

- between the neutral axis and room temperature, for axial - between two extreme fibers, for bending = Coefficient of thermal expansion

d

d

57

Temperature

dx

T1

T2

T2 > T1

dxycTyd )

2()( =

dxcTd )

2()( =

= mdUtemp

=L

temp dxcTmU

0

)2

(

T2

cc

T1T = T2 - T1

cT

2

=

T1

T2

yy

cT

2

y

T2 > T1

T1

O

d

221 TTTm

+= d

MM

58

Example 8-24

From the beam below Determine :(a) If P = 60 kN is applied at the mid-span C, what would be the displacement atpoint C. Due to shear and bending moment.(b) If the temperature at the top surface of the beam is 55 oC , the temperature atthe bottom surface is 30 oC and the room temperature is 25 oC.What would be the vertical displacement of the beam at its midpoint C and thethe horizontal deflection of the beam at support B.(c) if (a) and (b) are both accounted, what would be the vertical displacement ofthe beam at its midpoint C.

Take = 12(10-6)/oC. E = 200 GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2. The cross-section area is rectangular.

A BC

2 m 2 m

59

A B

1 kN x x

A BC

2 m 2 m

PSOLUTION

=L

iibending dxEI

Mm=

2/

0

)2

)(2

(2L

EIdxPxx 2/

0)

34(2

3 LPxEI

=)200)(200(48

)4(6048

33

==EI

PL= 2 mm,

=L

iishear dxGA

VKv=

2/

0

)2

)(21(2

L

GAdxPK

)35000)(80(4)4)(60(2.1

42

2/

0===

GAKPL

GAKPx L

= 0.026 mm,

shearbendingC += = 2 + 0.026 = 2.03 mm,

P/2P/2

Mdiagram

PL/4

x x

xP2 x

P2

P/2

P/2

Vdiagram

Part (a) :

0.5 kN0.5 kN

m diagram 0.5x 0.5x

1

0.5

0.5

vdiagram

60

Part (b) : Vertical displacement at CSOLUTION

A B

1 kN x x

m diagram

0.5 kN0.5 kN

0.5x 0.5x1

Troom = 25 oC ,

T1=55oC

T2=30oC

260 mm

Temperature profile

5.422

3055=

+=mT

C = -2.31 mm ,

=L

C dxcTmkN

0 2)())(1(

- Bending

=2

0

)5.0(2

)(2 dxxcT 2

0)

25.0(

)10260()25)(1012(2

2

3

6 x

=

A BC

2 m 2 m

55 oC,

30 oC

260 m

61

A B1 kN

Part (b) : Horizontal displacement at B

x

00

Troom = 25 oC ,

T1=55oC

T2=30oC

260 mm

Temperature profile

5.422

3055=

+=mT

BH = 0.84 mm ,

=L

BH dxTnkN )())(1(

- Axial

=4

0

)1()( dxT

4

0))(255.42)(1012( 6 x=

1 kN

1 1n diagram

BH = 0.84 mmCv = 2.31 mm ,

A BC

2 m 2 m

55 oC

30 oC

260 m

Deflected curveA BC

62

P

AB

C 55 oC

30 oC

260 m

Part (c) :

C = -2.03 + 2.31 = 0.28 mm,

A B C

C = 2.03 mm

P

=

A B55 oC,

30 oC

C = 2.31 mm

+

BH = 0.84 mm

63

Example 8-25

Determine the horizontal displacement of point C on the frame.If thetemperature at top surface of member BC is 30 oC , the temperature at the bottomsurface is 55 oC and the room temperature is 25 oC.Take = 12(10-6)/oC, E = 200GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2 for both members.The cross-section area is rectangular. Include the internal strain energy due toaxial load and shear.

A

B C5 m

6 m2 kN/m

4 kN

1.5 EI,1.5AE, 1.5GA

EI,AE,GA260 mm

64

B C

Moment, m (kNm)

A

B C

Shear, v (kN)

A

5 m

6 m

A

CB

Virtual load

1

x2

x1

1.2 kN

1 kN

1.2 kN

1.2

11.2

1

1

1-1.2-1.2

6

6

1x1

1.2x2

B C

Axial, n (kN)

A

+

+

65

B C

Shear, V (kN)

A

B C

Axial, N (kN)

AA

B C5 m

6 m2 kN/m

4 kN

x1

x2

12 kN

16 kN

12 kN

Real load

12

412

4

16

4

16 - 2x1 -12 -12

60

16x1 - x12

12x2

60 B C

Moment, M (kNm)

A

66

Due to Axial

x1

x2

AE

1.5AE

5 m

6 m

B C

Virtual Axial, n (kN)

A1.2

11.2

1B C

Real Axial, N (kN)

A12

412

4

=ii

iiiCH EA

LNnkN ))(1(

AEAE)5)(4)(1(

5.1)6)(12)(2.1(

+=

AEmkN

=26.77

==

=

,0111.0)10(109.1

)10200)(1035000(

6.77 5

2626

mmm

mkNm

mkNCH

67

Due to Shear

x1

x2

1.5GA

5 m

6 m

GAB C

Virtual Shear, v (kN)

A1

1-1.2-1.2

B C

Real Shear, V (kN)

A16

4

16 - 2x1 -12 -12

=L

CH dxGAVKkN

0

)())(1(

2

5

01

6

01 )12)(2.1(2.1

5.1)216)(1(2.1 dx

GAdx

GAx

+

=

GAmkNx

GAxx

GA

=+=25

02

6

0

21

14.134)4.14)(2.1()

2216)(

5.12.1(

==

=

,048.0)10(8.4

)1035000)(1080(

4.134 526

26

mmmm

mkN

mkNCH

68

Due to Bending

x1

x2

1.5EI

5 m

6 m

EIB C

Virtual Moment, m (kNm)

A

6

6

1x1

1.2x2B C

Real Moment, M (kNm)

A

60

16x1 - x12

12x2

60

=L

CH dxEImMkN

0

))(1(

+=5

0222

6

01

2111 )12)(2.1(

1)16)((5.11 dxxx

EIdxxxx

EI

EImkNx

EIxx

EI

325

0

32

6

0

41

31 1152)

34.14(1)

4316(

5.11

=+=

+==

=

,8.280288.0

)10200)(10200(

115246

26

3

mmmm

mkN

mkNCH

69

T1=30oC

T2=55oC

260 mm

Temperature profile

Due to Temperature

CH = 0.0173 m = 17.3 mm ,

Tm= 42.5oC

- Bending

- Axial

CH = 0.00105 m = 1.05 mm ,

A

B C

5 m

x1

x2260 mm

30oC

55oC

Troom = 25oC

B C

m (kNm)

A

6

6

1x1

1.2x2B

C

n (kN)

A1.2

11.2

1+

+

25

03

62

0 )10260()3055)(1012)(2.1(

2)())(1( dxxdx

cTmkN

L

CH

=

=

2

5

0

6

0

)255.42)(1012)(1()())(1( dxdxTnkNL

CH ==

70

A

B C

2 kN/m

4 kN

Total Displacement

TempCHBendingCHShearCHAxialCHTotalCH )()()()()( +++=

= 0.01109 + 0.048 + 28.8 + (17.3 + 1.05) = 47.21 mm

CH= 47.21 mm

71

P1 P2 Pi + dPi

P

Castiglianos Theorem

Pi + dPi

U

U*iidP

PUdU

=

U = U*

Piiii

dPdPPU

= )(

dU = dU*

Ui = f (P1, P2,, Pn)

iPi P

U

=

P1 P2 Pi

Pi

dPi

(dPi)Pi = dU*

ii

dPPUdU

=

P

72

Axial Load

=Li

Pi dxAEN

P)

2(

2

=L i

dxAEN

PN )(

n

Bending )2(

2

=Li

Pi dxEIM

P

= dxEIM

PM

i

)(

m

Shear

=Li

Pi dxGAKV

P)

2(

2

= dxGAV

PVK

i

)(

v

Load Displacement :

Where = external displacement of the truss, beam or frameP = external force applied to the truss, beam or frame in the direction of N = internal axial force in the member caused by both the force P and the loads on the truss, beam or frameM = 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 beamV = internal moment in the beam or frame caused by both the force P and the real loads on the beam

73

Axial ))((

=Li

Pi dxTNP

= dxT

PN

i

))((

n

Bending

=Li

Pi dxcTM

P))

2((

= dxcT

PM

i

)2

)((

Temperature Displacement :

Where = Differential temperatures:

- between the neutral axis and room temperature, for axial - between two extreme fibers, for bending = Coefficient of thermal expansion

m

74

Bending )2(

2

=Li

Mi dxEIM

M

=

L i

dxEIM

MM )(

m

Slope :

Where = external slope of the beam or frameMi = external moment applied to the beam or 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

iMi M

U

=

75

Castiglianos Theorem : Truss

N 2

N1

N3 N4

N5

N6

N7 N8 N9

P

= iii L

AEN

PN )(

Where: = external joint displacement of the trussP = external force applied to the truss joint in the direction of N = internal force in a member cause by both the force P and the loads on the trussL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity of a member

P1

P2

B

76

Example 8-26

Determine the vertical displacement of joint C of the truss shown in the figurebelow. The cross-sectional area of each member of the truss shown in the figureis A = 400 mm2 and E = 200 GPa.

A B

C

4 m 4 m

4 kN

3 m

77

=A B

C

LPNN )(

A B

C

N: Virtual Load P

AB

C 4 kN5 m

3 m

4 m 4 m

N: Real Load

SOLUTION

2+2

.5 -2.5

P

0.667P-0.

833P -0.833P

0

=AEL

PNNCV )(

1.5 kN1.5 kN

4 kN

0.5P0.5P

10.656-10

.41 10.41

)10200)(10400(

67.10)67.1041.1041.10(1

2626

mkNm

mkNAECV

=++=

CV = 0.133 mm,

2+2

.5 -2.5

0.667P-0.

833P -0.833P+

78

Example 8-27


4 m 4 m 4 m

AB C

D

EF

4 m

4 kN4 kN

79

=A

B CD

EF

LPNN )(

AB C

D

EF

N: Virtual Load P

AB C

D

EF

4 kN4 kN

N: Real Load

4 m 4 m4 m

4 m

5.657

m

SOLUTION

P 0.667P0.333P

04-5.

657

0

-5.657

444 4

-4

4 kN4 kN

0 0.667P-0.

471P

-0.47

1P

-0.943P0.667P0.333P 0

.

3

3

3

P

1P

-0.333P

=AEL

PNNCV )(

)10200)(10400(

4.72)]18.3016)67.10(2)33.5(307.15[1

2626

mkNm

mkNAECV

=++++=

CV = 1.23 mm,

4-5.

657

0

-5.657

444 4

-4

0.667P-0.

471P

-0.47

1P

-0.943P0.667P0.333P 0

.

3

3

3

P

1P

-0.333P

10.6715

.07 0

30.1810.675.33

5

.

3

3

16

5.33

+

80

Example 8-28


2 mA B

CD

3 m

20 kN

10 kNwall

81

2 mAB

CD

3 m

N: Virtual Load P

2 mAB

CD

3 m

20 kN

10 kN

N: Real Load

3.61 m

13.333 kN

23.333 kN

20 kN

23.333

0

-24.03

6

2020

SOLUTION

P

0.667P

0

-1.2P

01P

0.667 P

0.667P

1P

)12.10413.3160(1 ++=AECV

CV= 2.44 mm,

=AEL

PNNCV )(

)10200)(10400(

25.195

2626

mkNm

mkN

=

31.126

0

104.1

24

060

23.333

0

-24.03

6

2020

0.667P

0

-1.2P

01P+

LPNN )(

AB

CD

82

wC

A B

Castiglianos Theorem : Beams and Frames

C

=L

dxEIM

PM )(

Px1 x2

RBRA

Displacement

w

B

x2

RB

V2

M2

x1

RA

V1

M1

Where: = external displacement of the point caused by the real loads acting on the beam or frameP = external force applied to the beam or frame in the direction of M = internal moment in beam or frame , expressed as a function of x and cause by

both the force P and the loads on the beam or frame

83

w

A B

=L

dxEIM

MM )

'(

Slope

x1 x2RBRA

M

w

B

x2

RB

V2

M2

x1

RA

V1

M1

Where: = external displacement of the point caused by the real loads acting on the beam or frameM = external moment applied to the beam or frame in the direction of M = internal moment in beam or frame , expressed as a function of x and cause by

both the force P and the loads on the beam or frame

84

Example 8-29

The beam shown is subjected to a load P at its end. Determine the slope anddisplacement at C. EI is constant.

2a a

AB C

P

C

85

AB

C2a a

SOLUTION Displacement at C

=L

C dxEIM

PM )(

+

=aa

dxMP

MEI

dxMP

MEI 0

222

2

011

1 ))((1))((1

+=aa

dxPxxEI

dxPxxEI 0

222

2

01

11 ))((1)2

)(2

(1

,)3

)((1)3

)(4

(133

23

10

2

0 EIPaxP

EIxP

EI

aa

C =+=

x1 x2

-PaM2 = -Px22

11

PxM =

23P

2P

P

Mdiagram

86

AB

C2a a

PSlope at C

x1 x2aMP2

5.1 +aMP2

5.0 +

M

A

x1aMP2

5.0 +V1

M1 )25.0( 11 a

MxPx +=

C

P

x2

M

V2

M2 MPx = 2

+

=aa

C dxMMM

EIdxM

MM

EI 022

22

011

1 ))((1))((1

+=aa

dxMPxEI

dxa

MxPxa

xEI 0

22

2

01

11

1 ))(1(1)2

5.0)(2

(1

,6

723

2)2

)((1)3

)(4

(13232

23

10

2

0 EIPa

EIPa

EIPaxP

EIxP

EI

aa

C =+=+=

00

87

Example 8-30


A5 m

B

3 kN/m

88

SOLUTION

x

A5 m

B

3 kN/m

Displacement at B

=L

B dxEIM

PM )()(

EImkN 32375.234

=

)10250)(10200(

375.234466

3

mmkN

mkN

=

B = 0.00469 m = 4.69mm,

P

=2

3 2xPx

x

3x

2x

V

MP

=5

0

2

)2

3)((1 dxxPxxEI

0

=5

0

3

231 x

EI

)8

3(15

0

4xEI

=

89

x

A5 m

B

3 kN/m

slope at B

=L

B dxEIM

MM )

'(

EImkN 325.62

=

)10250)(10200(

5.62466

3

mmkN

mkN

=

B = 0.00125 rad,

=2

3'2xM

=5

0

2

)2

3')(1(1 dxxMEI

0

=5

0

2

231 x

EI

)6

3(15

0

3xEI

=x

3x

2x

V

M M

M

A BDeflected curve

B = 0.00125 rad

B = 4.69mm,

90

Example 8-31


A C D

B

5 kN14 kNm

2 m 2 m 3 m

91

AC

DB

14 kNm

2 m 2 m 3 m

Px3x2x1

Mdiagram

142

0

2

0)

33)(1()

35.0

27)(1(

32

31

21 x

EIxx

EI+=

)60)(200(667.20667.20

==EI

SOLUTION Displacement at B

=L

B dxEIM

PM )()(

5

111

2

0

1 )22

714()2

(1 dxPxxxEI

+=

++2

02

222 )22

7)(2

(1 dxPxxxEI

5

+=2

02

221

21

2

01 )3(

1)5.07(1 dxxEI

dxxxEI

B = 0.00172 m = 1.72 mm,

227 P

+22

7 P

Vdiagram

)22

7( P+)

227( P

227 22

2PxxM +=

22714 111

PxxM +=

+3

03)0)(0( dx

92

AC

DB

14 kNm

2 m 2 m 3 m

5 kNx3x2x1

Mdiagram

SOLUTION Slope at B

=L

B dxEIM

MM

0

)'

(

11

2

0

1 )4

'14()4

(1 dxMxxEI

+=

+3

03)0)(0( dx

+2

02

22

2 )4'6)(

4(1 dxxMxx

EI

0

04

'6 M

Vdiagram

M

4'1 M

)4

'1( M)

4'6( M

1411 )4

'1(14 xMM =

22 )4'6( xMM =

12

1

2

01 )25.05.3(

1 dxxxEI

=

)60)(200(333.2333.2

==EI

B = 0.000194 rad,

2

0

2

0)

35.1(1)

325.0

25.3(1

32

31

21 x

EIxx

EI+=

+2

02

22 )5.1(

1 dxxEI

B = 1.72 mm

B = 0.000194 rad

A C DB

93

Example 8-32

Determine the displacement of point B of the steel beam shown in the figurebelow. Take E = 200 GPa, I = 200(106) mm4.

20 kNHinge10 kNm

AB C

4 m 3 m 3 mI 2I

94

20 kNP

x2

= (22.5 + P)x3 - (75 + 6P)

= -(2.5 + P)x1

= 10 - 2.5x1

SOLUTION

75 + 6P

22.5 + P2.5 kN

0

P

10 kNm

2.5 kN

0

2.5 kN

10 kNm

2.5 kNV1

M1

x1 P

2.5 kN V2

M2

x2

75 + 6P

22.5 + PV3

M3

x3

x1 x320 kN

10 kNmA

B C4 m 3 m 3 mI 2I

95

20 kN

10 kNmA

B C4 m 3 m 3 mI 2I

= (22.5 + P)x3 - (75 + 6P)

= -(2.5 + P)x2

= 10 - 2.5x110 kNm

2.5 kNV1

M1

x175 + 6P

22.5 + PV3

M3

P

2.5 kN V2

M2

x2x3

x2P x3x1

=L

B dxEIM

PM )(

333

3

03 )6755.22()6(2

1 dxPPxxxEI

++

222

3

0211

4

0

)5.2()(2

1)5.210()0(1 dxPxxxEI

dxxEI

+= 0

0 0

+++=3

033

23

3

02

22 )4502105.22(2

1)5.2(2

10 dxxxEI

dxxEI

)200)(200(31531575.30325.11

==+=EIEIEIB = 7.875 mm,

0

96

Example 8-33

Determine the displacement of hinge B and the slope to the right of hinge Bof the steel beam shown in the figure below. Take E = 200 GPa, I = 200(106) mm4.

3 m 4 m

20 kNHinge

30 kNm

AB

C2EIEI

5 kN/m

97

20 kNSOLUTION

3 m 4 m

30 kNm

AB

C2EIEI

5 kN/mP

M

30 kNm

A B

5 kN/m

15 kN

17.5 kN2.5 kN

B

C

2EI

P

M17.5 kN

P + 17.5

4(P + 17.5) + M

98

20 kN

= (P + 17.5)x2 - 4(P+17.5) - M1

21 5.2

2530 xx =

3 m 4 m

30 kNm

AB

C2EIEI

5 kN/m

x1 x2P

M2.5 kN P + 17.5

4(P + 17.5) + M

x1

30 kNm

A

5x1

2.5 kN V1

M1

C2EI4(P + 17.5) + M

P + 17.5x2V3

M3

=L

B dxPMM

EI)(1

++=4

02222 )4)('7045.17(2

10 dxxMPxPxEI

020 20

The displacement of hinge B

==== ,1001.0)200)(200(2

8002800 mmm

EI

99

20 kN

= (P + 17.5)x2 - 4(P+17.5) - M1

21 5.2

2530 xx =

3 m 4 m

30 kNm

AB

C2EIEI

5 kN/m

x1 x2P

M2.5 kN P + 17.5

4(P + 17.5) + M

x1

30 kNm

A

5x1

2.5 kN V1

M1

C2EI4(P + 17.5) + M

P + 17.5x2V3

M3

=L

B dxMMM

EI)'

(1

++=4

0222 )1)('7045.17(2

10 dxMPxPxEI

020 20

The slope to the right of hinge B

radEI

31075.3)200)(200(2

3002300 ===

100

Example 8-34

Determine the slope and the horizontal displacement of point C on the frame.Take E = 200 GPa, I = 200(106) mm4

A

B C5 m

6 m2 kN/m

4 kN

1.5 EI

EI

101

A

B C5 m

6 m

2

k

N

/

m

1.5 EI

EI

12 kN

Horizontal Displacement at C

P

SOLUTION2)5

65

36( xP+=

12 + P

56

536 P

+

56

536 P

+

A

2x112 + P

56

536 P

+

x1V1

M1

C P

56

536 P

+V2

M2

x22

11)12( xxP +=

x1

x2

=L

iiCH dxEI

MP

M )( +++=5

02

2221

2111

6

01 )5

65

36)(5

6(1)12()(5.11 dxPxxx

EIdxxxPxx

EI

+=5

02

22

6

01

31

21 )4.14(

1)16(5.11 dxx

EIdxxx

EI

)200)(200(1152600552)

34.14(1)

4316(

5.11 5

0

6

0

32

41

31 =+=+=

EIEIx

EIxx

EICH = + 28.8 mm ,

44

102

A

BC

5 m

6 m

2

k

N

/

m

1.5 EI

EI

12 kN

4 kN

Slope C

5'12 M

x1

x2M

5'12 M

16

2)5'12(' xMM +=

4 N

5'12 M

V2

M2

x2

M

21116 xx =

A

2x116

5'12 M

x1V1

M1

=L

iiC dxEI

MMM

0

)'

( ++=5

02

22

21

211

6

0

)5'12')(

51(1)16()0(

5.11 dxxMxMx

EIdxxx

EI

+=5

02

22

2 )51212(10 dxxx

EI

)200)(200(5050)

3512

212(1

5

0

32

22 ==

=

EIxx

EIC = + 0.00125 rad ,

000

Energy Deflection

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