conjugate beam method

Structural Analysis (II) Chapter 5: Deflection

1

Method (2) Conjugate beam Method ....بالكمرات. نقطه أي عند Slope(θ) Deflection (y)حساب في المستخدمه الطرق وإحدي

الكمرة. طول علي Slope(θ)Deflection (y) تحقق بحيث ( المعطي ) األصلية الكمرة من مستنتجه تخيليه كمرة هي

و

Conjugate beam:

قيم

(Conjugate beam)ب محمله ل( Elastic load)تكون ) مساوي -MEI )حيث ( M )شكل هو

: يكون الحاله هذه في و األصلية الكمرة علي العزم

Rotation or Slope(θ)or (y')= Shear force of elastic load = QelasticDeflection (δ)or (y)= Moment of elastic load =Melastic

Real(Original) beam Conjugate beam:

Load = d²M(x)dx² = w

Shear = dM(x)dx =Q

Moment = M(x)

Elastic load = d²(y)dx² =- M

EI

Rotation = d(y)dx =Qelastic

Deflection = y = Melastic

wt/m'

ملكات

ضلتفا

ملكات

ضلتفا

- M/EI

ملكات

ضلتفا

ملكات

ضلتفا

Rotation)

Deflection)


2

θδ = 0 , θ ≠ 0 M = 0 , Q ≠ 0

Q

Free end Fixed supportδ ≠ 0 , θ ≠ 0 M ≠ 0 , Q ≠ 0

Fixed support Free end

δ = 0 , θ = 0

QM

δ = 0 , θ ≠ 0 M = 0 , Q ≠ 0

Interior support (Roller or Hinge) Internal hinge

θL

θR

(QL=QR)

Internal hinge

δ ≠ 0 , θL ≠ θR ≠ 0

Interior support (Roller or Hinge)θL

θR

M ≠ 0 , QL ≠ QR

من (Conjugate beam)(Original beam)التحويل إلي

M = 0 , Q = 0

End support (Roller or Hinge) End support (Roller or Hinge)

Conjugate SupportReal Support

Examples

Real Beam Conjugate Beam

Indetreminate Beam

(θR = θL)

(???)


3

Steps of Solution :

بإستخدام نقطه أي عند (θ) (y)حساب عند التاليه الخطوات Conjugate beam method.أوتتبع

اإلتزان. معادالت بإستخدام األفعال ردود Reactionsحساب

FOR Real beam

شكل Bending Moment Diagramرسم (B.M.D).المساحه حساب يسهل بحيثExamples

4 t/m` 2 t/m`

15 t 21 t 14 t

WL²/8=32m.tWL²/8=9m.t

8 m.t 12 m.t

C

3t/m`

B.M.DWL²8 = 24m.t

6t

12t.m

=WL²8 = 24m.t

12t.m

(1)

(2)

B.M.D

WL²/8=32m.t WL²/8=9m.t

8 m.t 12 m.t

B.M.D

6t


4

(Inertia)(Inertia): كالتالي العزم شكل عندها الموجود

C

12mt

8t8t

B.M.D

Modified B.M.D

قيمه تغيير عند العزم شكل تعديل معامليتم علي القسمه أو الضرب طريق عن

12mt6mt

12mt 12mt

12mt

24mt

Or6mt

12mt

6mt 3mt

6mt

FOR Conjugate beam

األصليه الكمره تحويل

علي (Conjugate Beam)التأثير

(Conjugate Beam).سبق كما

(Elastic loads)ب

(Given)إلي

ل (Modified B.M.D).مساوي

محصله علي(Elastic loads)حساب (Conjugate Beam).المؤثر

( المعدل ) العزم شكل مساحه = المحصلة قيمه

مكان حسب المحصلة إتجاه

= المحصلة تأثير (B.M.D ).مساحه(C.g)نقطة

.( B.M.D)ألسفل العزم

ألعلي العزمR

R

أشكال و(B.M.D )بعض لها.(Elastic loads )الهامه المقابله

12ML

Datum

ML

Datum

8t

I

2II

2I


5

23ML

Datum

Datum

M1M2

12M2L

12M1L

Datum

M1

M2

Datum

M1

M2

=

12M1L

12M2L

Datum

M1M2

=

23ML

Datum

Datum

M1M2

+12M2L

12M1L

DatumM1

M2

= +

23ML

Datum

DatumM1

M212M2L

12M1L


6

قيمه اإلشارات.(Rotation)و( Deflection)حساب مراعاه مع كاألتي نقطة أي عند

Rotation ( θ )= Deflection (y) =

Qelastic منمحصلة محسوبه القطاع عند

Melastic

QQ

QQ MM

MM

-ve

+ve

+ve

-ve

Positive )

Shear force

Negtive ) Positive )

Moment

Negtive )

( θ )= + ve ( θ )= - ve ( y )= + ve ( y )= - ve

Qelastic Melastic

+

-θθ

العزوم منمحصلة محسوبه القطاع عند

Solved Examples

Using the Conjugate beam method, determine the rotation at points (a,b,c and d)and deflection at points(c,d and e).

Ex(1)

3t10t 10t

EI = ConstantSolutionReactions and Bending moment diagram (Original beam)

3t10t 10t

16.2t18.8t

10t 10t

2t/m'

2t/m'

6.0t.m

46.8t.m47.4t.m

2*3²8 =2.25 t.m

2*3²8 =2.25 t.m

ac d

be

a c db e

7

Elastic loads( Area of bending moment)

6.0

46.847.42.25 t.m

a c db e

6.012

93.671.1 70.271.1

4.5 4.5Conjugate beam

ae

a b e

Elastic loads on Conjugate beam

a b e

6.012

93.6

71.1

70.2

71.1

4.5

4.5

a cd

Elastic Reactions

a

b e

6.0

12

93.6

71.1

70.2

71.1

4.5

4.5

a cd

b

138.7

132.7

164.3 269.4


b

Modified B.M.D( MEI)

( MEI)

Required rotation and deflection

Point (a)

Rotation ( θ )= Deflection (y) =Qelastic Melastic

θa = 1EI[164.3] = + 164.3

EI (Clockwise)aa

164.3

Point (b)

b138.7

θb =1EI[-138.7] = - 138.7

EI (Anti-Clockwise)

Point (c)θc = 1

EI[164.3-4.5-71.1] = + 88.7EI (Clockwise)

71.1

4.5

c

164.3yc = 1EI[164.3(3)-4.5(1.5)-71.1(1)] = + 415.05

EI (Down)


Point (d)θd =

1EI[ -138.7 -12 +93.6] = - 57.1

EI (Anti-Clockwise)

12

93.6

db

138.7

Point (e)

e

132.7

269.4θe = 1EI[ -132.7] = - 132.7

EI (Anti-Clockwise)

ye = 1EI[ -268.8] = - 269.4

EI (Upward)

8

Left Right

+ve Sign summary+ ve


9

Using the Conjugate beam method, determine the rotation at points (a,b,c and d )and deflection at points(c and d ).

Ex(2)

4t 4t/m'

I I 2Ib

ca

d

Solution

bcad

4*6²8 =18.0 t.m

12t.m

bc

a

d

12

6t.m

18t.m

9t.m

63

91818

4.5

18

36

Elastic loads

adb

9 1818 4.5

1836

Elastic Reactionsa

d

b

9 18

18

4.5

1836

a

10.12

7.88 14.6212.36

Required rotation and deflection

Point (a)

θa = 1EI[+7.88] = + 7.88

EI (Clockwise)

Point (b)

θb =1EI[-14.62] = - 14.62

EI (Anti-Clockwise)

Point (c)

yc = 1EI[14.62(3) + 4.5(1.0) -18(1.125)]

= + 28.11EI (Downward)

θc = 1EI[18 - 14.62 - 4.5]

= - 1.12EI (Anti-Clockwise)

c

b

4.5

18

14.62

c

Point (d)

yd =1EI[+ 12.36] = + 12.36

EI (Downward)

θd =1EI[-10.12] = - 10.12

EI (Anti-Clockwise)

B.M.D

Conj.beam

I I 2I

I I I


10

Ex(3) Using the Conjugate beam method, determine : * the rotation at points (a ,b,d and e ), * the relative (change in) rotation at point( f ), * the deflection at points(d ,e ,f and j).

2t/m' 6ta

d bf

ec

Solution EI = Constant

2t/m'

6t

ad b

fe

c

3t4t

a d b f e c

9t.m

10t.m

16t.m1t.m

5t.m

40

85.33

1.3327

10

ad

b fe

c

j

6.75t

16t

16.25t

j

ad

b

fe

c

85.33

1.33 27

10

40 1629.33

24.13 10.2

16

271.3385.33

1040


11

Required rotations and deflectionsPoint (a)

θa = 1EI[+29.33]

= + 29.33EI (Clockwise)

Point (b)

θb =1EI[-16] =-16

EI(Anti-Clockwise)

a

29.33

b

16

θf/L =1EI[-16+10-1.33] = - 7.33

EI (Anti-Clockwise)

yf =1EI[-16(2)+10(1.33)-1.33(1)] = - 20

EI (Upward)

Point (f)(Internal Hinge )

θf/R = 1EI[-16+10-1.33+24.13 ] = + 16.8

EI (Clockwise)

θf/rel = [θf/R - θf/L ]= + 16.8EI - -7.33

EI = + 24.13EI = Rf

EI

f

1.33

10

16

24.13

Point (d)

yd=1EI[+29.33(4)+10(1.33)-42.7(1.5)]

=+66.6EI (Downward)

a d

16t.m

5t.m

42.7

10

29.33

Point (e)

ye=1EI[+10.2(3)-13.5(1)] = + 17.1

EI (Downward)

e c

9t.m

13.5

10.2

Point (j)

M j= 6.75(3) - 6(1.5) = 11.25t.m

2t/m'

a

6.75t

j11.25t.m

29.33

Real beam Conj. beam

2.25 t.m

16.88

4.5

y j =1EI[+29.33(3)-16.88(1) -4.5(1.5)] = + 64.36

EI (Downward)

θd =1EI[29.33+10-42.7] = - 3.37

EI (Anti-Clockwise)

θe = 1EI[13.5-10.2 ] = + 3.3

EI (Clockwise)

6t


12

Using the Conjugate beam method, determine : * the rotation at points (c and e ), * the relative (change in) rotation at point( d ), * the deflection at points(c ,e and d ). [take EI = 6000 t/cm²]

Ex(4)

2I 2III

4ta

d eb

c

2t/m'

Solution

Reactions(Original beam)2I 2II

I

4t

ad

eb

c

2t/m'

7t

13t7t

ad

eb c

7t.m 8t.m4t.m

16t.m B.M.D(Original beam)

a de b c

7t.m 4t.m4t.m

16t.m

2t.m

3.5 8.0 4.0 8.04.0

42.6721.33

Conjugate beam

Elastic Reactions

16.29

a de

c8.0 4.0 8.0

4.0

21.33

3.5

42.67

24.2

12.29

27.25

16t

7.0t.m


13

Required rotations and deflections

Point (c) c

12.29

27.25

θc = 16000[-12.29] = - 0.002rad (Anti-Clockwise)

θd/L =1

6000[+ 3.5] = 5.833*10-4 rad (Clockwise)

yd =1

6000[3.5(0.67)] = + 0.039 cm (Downward)

yc = 16000[-27.25] = - 0.454 cm (Upward)

Point (d)(Internal Hinge ) a d3.5 24.2

θd/R = 16000[+ 3.5 + 24.2 ] = + 4.6167*10-3 (Clockwise)

θd/rel = │θd/R - θd/L│= RdEI =

24.26000 = 4.033*10-3 rad

Point (e)

θe = 16000[-16.29-8-4+21.33] = - 0.00116 rad (Anti-Clockwise)

ye = 16000[16.29*4 + 8*2.67+ 4*1.33-21.33*1.5] = + 0.997 cm (Downward)

16.29

e

4.0 8.0

21.33

conjugate beam method

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