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1 3102 ‫دورة اب‬ Using the slope-deflection method, determine the reaction and draw the bending moment diagram for the structure shown below. 3m B C EI 20kN/m' 2EI 3m 3EI 4m A D Solution: 1- S.S=1   B C  AB  DC A D 2- FEM : AB ‫العنصر احململ هو‬ AB : S-D Equation : 2 AB: [ ] BC: CD: 3- Equil...

Description

1

3102 ‫دورة اب‬

Using the slope-deflection method, determine the reaction and draw the bending moment diagram for the structure shown below. 3m B

C

2EI

3EI

4m

20kN/m'

3m

EI

A

D

Solution: 1-

S.S=1

B



C

 AB



 DC

A

D

2- FEM : AB ‫العنصر احململ هو‬ AB : S-D Equation :

2

AB: [

]

BC:

CD:

3- Equilibrium Equation : Node B:

∑ ( )

Node C:

∑ ( )

30kN

M AB  M BA L AB

B

C

M CD  M DC LCD

∑ ( ) :)3(‫)و‬2(‫)و‬1( ‫حبل‬

4- End Equation :

13

B

-

12.19 +

12.19

C 13

-

+

22,5

B.M.D

5- B.M.D

-A 49.5 + D 24.3

3

6 ‫ المسألة‬6102/8/62 ‫دورة‬ Use the moment distribution method to determine the end moments of the members of the frame given below and draw the bending moment diagram. 10kN / m ' 20kN

C

B

6m

EI  const

6m

A

Solution: 1) FEM:

q

?

C

B

q qL 2 12

M

MF BC

M

C

B

F BC

qL 2 12

1 F qL2 1 qL2 qL2 10*62  M CB    ( )   45kN .m 2 12 2 12 8 8

2) K ij : joint B:

K BC

3) DF : joint B:

4EI 2  EI 6 3 3EI 1   EI 6 2

K BA 

DFBA 

K BA 4  K BA  K BC 7

DFBC 

K BC 3  K BA  K BC 7

4

0.5 Joint member DF FEM balance C.O.M M0

A AB

B BA 0.571

12.857 12.857

6.428kN

20kN 25.714

25.714  12.857

25.714

BC 0.429 -45.000 19.286

25.714

-25.714

C CB

0

R0 25.714  12.857  6.428 6

25.714  12.857

25.714  12.857 6

12.857

 X  0  6.428  20  R 0  0  R 0  26.428kN  B

 C

B'

:‫نعطي المنشأ انزياح‬ C'

A

M AB  M BA   let

6EI  62

6EI   100  M AB  M BA  100kN .m 62

5

0.5 Joint member DF FEM balance C.O.M M1

A AB

B

-100 28.571 -71.429

BA 0.571 -100 57.143

BC 0.429 42.857

-42.857

42.857

C CB

0

R1

19.048 42.857

42.857  71.429  19.048 6

42.857  71.429

42.857  71.429

42.857  71.429 6

71.429

 X  0  19.048  R1  0  R1  19.048kN

R 0  R1 *  0    1.387 M fin  M 0  M 1 * M0 M1 M_fin

12.857 -71.429 -86.248

25.714 -42.857 -33.749

-25.714 42.857 33.749

0 0 0

33.749 33.749





qL2  45 8

B .M .D 86.248

6

‫ المسألة الثالثة‬٢٠١٣-٨-٢٦ ‫الدورة‬ Draw the influence lines for reactions at supports
 ,
 
 ,the shear just to the right of supports C,(  ) ,the bending moment at point B,  ,  ℎ  ℎ. Solution:

A

C 2

E

4

4

2

+

I.L.RA

-

I.L.RC

+

1

1

+

1

1

1

+

I.L.RE

I.L.(Vc)R

+

I.L.MB 2

1

1

2

D

7

-

3102 ‫دورة تموز‬ Q1 : The cantilevered beam AB is additionally supported using a tie rod BD .By using virtual work method determine the force in the rod and draw the bending moment diagram .Neglect the effect of axial compression and shear in the beam. For the beam ,

(

and for the tie rod, ,take

.

3m

D

4kN/m' B

A

80kN

4m Solution: 1- DS ‫ → حندد درجة عدم التقرير‬DS=5-1-3=1 (FS redundant

)

DB ‫لتقرير املنشأ نقطع الشداد‬

D

X1 X1 4kN/m' B

A

F.s+redundent X1 2-

( FS Loads)

8

80kN

)

352

A 352

B

M0:F.s+loads

A

B 8

+

‫للمنحين‬ (

3-

)

X1 =1

1*(3/5)

1*(4/5) B

A

M1:F.s+X1=1 A

B

+

2.4

∑

∫

4∫

[

(

[ ∫

( )(

)

) → (

) ( )]

]

∑ [

9

(

)(

)

]

[( )( )( )]

[

] ( (

)

5)(

(

)

D

+6

2.2

6

202.58

A

93.29 B

B.M.D

10

) (

)

‫ المسألة الثانية‬3102/7/4 ‫دورة‬ Use the moment distribution method to determine the end moments of the members of the frame given below and draw the bending moment diagram. 18kN/m'

100kN.m B

A

D

5m

EI=const

C

5m

5m

Solution:

1) FEM:

2) Stiffness factor (K):

3) DF: Joint B:

∑

0.5 Joint member DF FEM Balance C.O.M M0 M() Balance C.O.M M1 M_Fin

BA 0.3 56.25 13.125

B BD 0.3 13.125

M 0 -100 0

69.375

13.125

-100

30

30

0

30 75

30 18.75

11

0 -100

BC 0.4

C CB 0

17.5 17.5 -100 40 -60 6.25

8.75 8.75 -100 20 -80 -6.25

‫مالحظة‪ :‬ال داعي لوضع النقطة ‪ A‬والنقطة ‪ D‬ضمن الجدول ألن العزم فيهم معدوم ‪.‬‬ ‫نعطي المنشأ انزياحا ‪:‬‬

‫‪‬‬ ‫'‪D‬‬

‫‪‬‬

‫‪‬‬ ‫'‪B‬‬

‫‪D‬‬

‫‪B‬‬

‫‪A‬‬

‫'‪A‬‬

‫‪C‬‬

‫حساب ‪ R0‬و ‪ R1‬و‪:α‬‬

‫'‪18kN/m‬‬

‫‪100kN.m‬‬ ‫‪R0‬‬

‫‪B‬‬

‫‪D‬‬

‫‪A‬‬ ‫‪M BC  17.5‬‬

‫‪EI=const‬‬

‫‪M BC  M CB‬‬ ‫‪5‬‬

‫‪Xc ‬‬

‫‪C‬‬

‫‪M CB  8.75‬‬

‫∑‬ ‫‪R1‬‬

‫‪D‬‬

‫‪A‬‬

‫‪M BC  60‬‬

‫‪M BC  M CB‬‬ ‫‪5‬‬

‫‪Xc ‬‬

‫‪M CB  80 C‬‬

‫‪12‬‬

∑

75

q *l 2 8

A

-

18.75

D

+

+

-

B .M .D

6.25

C

13

For the frame shown use the slope-deflection

method to

determine the end moments of the members and draw the bending moment diagram.

is the same throughout. 40kN/m'

2

3m

2m

3

60kN

2m

4

1

4m

Solution: −



2

3



 43  12 4

1

1-

FEM : 12 ‫ و‬23 ‫العناصر احململة‬ 12: 23: S-D Equation : 12: [

]

14

23: 34: 1 2- Equilibrium Equation : Node 2 : ∑ ( ) Node 3 : ∑ ( )

M 12  M 21 L12 2

3

M 34  M 43 L34

∑ ( ) :‫)جند‬3(‫)و‬2(‫)و‬1( ‫حبل كل من‬

3- Ends Moments :

4- B.M.D

50.5 36.7

2

36.7

-

-

80

30kN

-

-

+

60

B.M.D

+

+ 40.6

-

1

35.3

15

50.5

3

4

3102 ‫دورة شباط‬ Q1: Use the virtual work principle to determine the reaction and draw the bending moment diagrams for the structure shown. EI is constant for all120kN.m members.

B C

6m

D 2kN/m'

A

4m

4m

Solution: 1- DS=5-3=2 D‫و‬C ‫لتقرير هذا املنشأ نزيل املساند عند‬ (FS + redundant

B

)

X2

X1

C

D

F.s+redundent X1,X2

A

2-

:(

)

16

( )

120 + B 120

B 120

D

C

+

+

180

M0:F.s+loads

-

+

A

600

A

600

3-

(

)

X 1 =1

8

-

8 B

-

D

C

M1:F.s+X1=1

A

(

2-

)

X 2 =1

4

-

4

B

-

D

C

M2:F.s+X2=1

A

4-

( ) …(2)

17

(

(

)(

(

)

)(

[(

∫ )(

)

)

(

)

(

(

)(

)

(

)

(

)(

)

(

)(

)

(

)(

)

))]

∫ (

[( (

(

)(

)

(

∫

) (

) )

)(

(

)(

)

[(

∫

)

(

)(

) )

)]

[( ∫

(

∫

)

(((

)( )) (

)

((

) (

[(

⇒

))] ))] (

)

)(

( ) ( )

⇒

( ) ( )

( )

⇒

( ) 580.92

-

203.72 B

C +

+

D 120

B.M.D 203.72

A 516.28

18

)]

50.23

B

D

C

121.39

A

240

516.28 71.16

Q2: Use the slop deflection method to determine the end moments of the members of the frame given below and draw the bending moment diagram.

3m

20kN/m'

C

3EI

2EI

4m

3m

B

2EI

A

D Solution: 1- Number of degree of freedom (rotations displacement) S.S=1 ⇒

(‫)وثاقة‬

(‫)مفصل‬





C

B

 AB

 DC

A

19

D

2- FEM: BC ‫ و‬DC )‫(عناصر غري حمملة‬ AB : )‫⇒ (فقط احململ‬

Slope-deflection equation: AB: (

‫معادلة عنصر موثوق الطرفني‬ [

)

[

] ⇒

]

[

⇒

]

BC: (

)

[

]

‫معادلة عنصر موثوق من طرف ومتمفصل من طرف‬ [

] [

⇒ ]

CD: [

]

⇒

3- Equilibrium equation : Node B : ∑ ⇒ ⇒

( ) 30

B

30

M AB  M BA 3 B M BA M AB  M BA B 3

M DC 4 C

M DC 4

C

60

A

30

A

M AB  M BA 3

M DC 4

M AB

D

M DC ∑

20

⇒

‫بعد التعويض بقيم العزوم اليت اوجدناها سابقا حنصل على املعادلة‪:‬‬ ‫) (‬ ‫حبل ) ( و) ( جند‬

‫⇒‬

‫و ‪:‬‬

‫‪4- End Moments‬‬

‫‪5- B.M.D:‬‬

‫‪16.56‬‬ ‫‪B‬‬ ‫‪+‬‬

‫‪C‬‬

‫‪16.56‬‬ ‫‪+‬‬ ‫‪q*(l^2)/8‬‬

‫‪B.M.D‬‬ ‫‪A‬‬

‫‬‫‪53.92‬‬

‫‪+‬‬ ‫‪D‬‬ ‫‪26.03‬‬

‫‪21‬‬

‫ المسألة الثالثة‬٢٠١٣-٢-٣ ‫دورة‬ Construct the Influence lines for -

The reaction at A,C and E, the shear and moment at D.

-

Shear before and after support C, the moment at point C.

Solution:

B

A

C

I .LV . CR

I .LV . CL

-

0,67 0,33

+

+

0,33

+

0,33 +

-

2

I .L .M C

+

-

22

+

1

1,33

1

+

0,33

1,33

I .L .M D

+

1

I .LV . D

4

1

I .L .R E

2

E

+

1,33

I .L .RC

2

1

I .L .R A

1

2

D...