1

D

CB

A

20

kN

/m'

3m

3m

4m3EI

EI

2EI

D

CB

A

دورة اب 3102

Using the slope-deflection method, determine the reaction and draw

the bending moment diagram for the structure shown below.

Solution:

1- S.S=1

2- FEM : AB العنصر احململ هو

AB :

S-D Equation :

ABDC

2

CB30kN

D

CB

A

22,5

49.5

12.19

12.19

13

13

24.3

B.M.D

-

--

+

+

+

AB:

[ ]

BC:

CD:

3- Equilibrium Equation :

Node B: ∑

( )

Node C: ∑

( )

∑

( )

(:3(و)2(و)1حبل )

4- End Equation :

5- B.M.D

AB BA

AB

M M

L

CD DC

CD

M M

L

3

6m

6m

6المسألة 62/8/6102دورة

Use the moment distribution method to determine the end

moments of the members of the frame given below and draw

the bending moment diagram.

Solution:

1) FEM:

2 2 2 21 1 10*6( ) 45 .

2 12 2 12 8 8

MF F F

BC BC CB

qL qL qLM M M kN m

2) :ijK joint B:

3) :DF joint B:

20kN

10 / 'kN m

EI const

q

q

2

12

qL 2

12

qL

?

A

BC

B C

CB

4 2

6 3

3 1

6 2

BA

BC

EIK EI

EIK EI

4

7

3

7

BABA

BA BC

BCBC

BA BC

KDF

K K

KDF

K K

4

C B A Joint

CB BC BA AB member

0.429 0.571 DF

-45.000 FEM

19.286 25.714 balance

12.857 C.O.M

0 -25.714 25.714 12.857 M0

0 00 6.428 20 0 26.428X R R kN

نعطي المنشأ انزياح:

2

2

6

6

6100 100 .

6

AB BA

AB BA

EIM M

EIlet M M kN m

0R20kN6.428kN

12.857

25.714

25.714 12.857 25.714 12.857

25.714 12.857

6

25.714 12.8576.428

6

A

B 'B C 'C

0.5

5

C B A Joint

CB BC BA AB member

0.429 0.571 DF

-100 -100 FEM

42.857 57.143 balance

28.571 C.O.M

0 42.857 -42.857 -71.429 M1

1 10 19.048 0 19.048X R R kN

0 1 * 0 1.387R R

0 1 *finM M M

0 -25.714 25.714 12.857 M0

0 42.857 -42.857 -71.429 M1

0 33.749 -33.749 -86.248 M_fin

42.857

71.429

42.857 71.429 42.857 71.429

42.857 71.42919.048

6

42.857 71.429

6

19.048 1R

86.248

33.749

33.749

. .B M D

2

458

qL

0.5

6

المسألة الثالثة ٢٠١٣- ٨-٢٦الدورة

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:

1

12

1

1

11

1

2 -

-

+

+

+

+

+I.L.RA

I.L.RC

I.L.RE

I.L.(Vc)R

I.L.MB

EA C D

2 2 4 4

7

A

3m

4m

4kN/m'

80kN

B

D

AB

D

X1

X1

F.s+redundent X1

4kN/m'

80kN

دورة تموز 3102

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 .

Solution:

1- DS حندد درجة عدم التقرير → DS=5-1-3=1

(FS redundant )

DB لتقرير املنشأ نقطع الشداد

2- ( FS Loads)

8

AB

AB

352

352

8

-

-+

M0:F.s+loads

AB

X =11 1*(3/5)

1*(4/5)

AB

+

M1:F.s+X1=1

2.4

لمنحينل

3- ( )

4- ∫

∑

( ) →

∫

[( )(

)

(

) (

)]

[ ]

∫

∑

[( ) ( )

]

[( )( )( )]

9

AB

D

202.58

- 93.29

B.M.D

+62.26

[

]

( )

( )

( )

5-

(

) ( )

10

A B D

C

18kN/m'100kN.m

EI=const5

m

5m 5m

المسألة الثانية 4/7/3102دورة

Use the moment distribution method to determine the end moments of the members of the frame given below and

draw the bending moment diagram.

Solution:

1) FEM:

2) Stiffness factor (K):

3) DF:

Joint B:

∑

C B Joint

CB BC M BD BA member

0 0.4 0 0.3 0.3 DF

-100

56.25 FEM

17.5 0 13.125 13.125 Balance

8.75

C.O.M

8.75 17.5 -100 13.125 69.375 M0

-100 -100

M()

40 0 30 30 Balance

20

C.O.M

-80 -60 0 30 30 M1

-6.25 6.25 -100 18.75 75 M_Fin

0.5

11

A B D

C

A' B' D'

A B D

C

18kN/m'100kN.m

EI=const

R0

A D

C

R1

ضمن الجدول ألن العزم فيهم معدوم . Dوالنقطة Aال داعي لوضع النقطة مالحظة:

نعطي المنشأ انزياحا :

:αو R1و R0حساب

∑

5

BC CBM MXc

17.5BCM

8.75CBM

5

BC CBM MXc

60BCM

80CBM

12

A D

C

-

-

+ +

∑

75

6.25

2*

8

q l 18.75

. .B M D

13

1

2 3

4

2m

2m

3m

4m

40kN/m'

60kN

1

4

2 3

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.

Solution:

−

1- FEM : 12 العناصر احململة 23 و

12:

23:

S-D Equation :

12:

[ ]

12

43

14

2 330kN

1

2 3

4

60

80

36.7

36.7

35.3

40.6

50.5

50.5

+

+

+

-

-

- -

-

B.M.D

23:

34:

1

2- Equilibrium Equation :

Node 2 : ∑

( )

Node 3 : ∑

( )

∑

( )

(جند:3(و)2(و)1حبل كل من )

3- Ends Moments :

4- B.M.D

12 21

12

M M

L

34 43

34

M M

L

15

2kN/m'6m

4m 4m

120kN.m

A

B

CD

A

B C D

XX 2 1

F.s+redundent X1,X2

دورة شباط 3102Q1:

Use the virtual work principle to determine the reaction and draw the

bending moment diagrams for the structure shown. EI is constant for all

members.

Solution:

1- DS=5-3=2

DوCلتقرير هذا املنشأ نزيل املساند عند

(FS + redundant )

2- : ( )

16

A

B C D

600

120

120

+

+

-

A

B

600

120

-

+

+

180M0:F.s+loads

A

B C D

X =11

8

8

-

-

M1:F.s+X1=1

A

B C D

X =12

-

-4

4

M2:F.s+X2=1

( )

3- ( )

2- ( )

4- ( )

…(2)

17

A

B C D+

+

-

-

120

80.92

203.72

203.72

516.28

B.M.D

∫

[(( )( )

( )( )

) (

( )( )

( )( )

( )( )

( )( )

) ( ( ))]

∫

[( ( )

( )

)

(( )( )

( )( )

( )( )

( )( )

)

( )]

∫

[(( )( )

) ((( )( )) ( ))]

∫

[(

) (( ) ( ))]

∫

∫

[(

) ( )( )]

⇒

( )

( )

⇒ ( )

( ) ⇒ ( )

( )

5-

18

A

BC

20kN/m'3m

3m4m

3EI

2EI2EI

D

A

B C D

50.23

121.39

516.28

71.16

240

A

BC

D

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.

Solution:

1- Number of degree of freedom (rotations displacement)

S.S=1 ⇒ (وثاقة) (مفصل)

AB

DC

19

A

B

BC

C

D

A

B

60

30

30

30

2- FEM: BC و DC )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 : ∑ ⇒

⇒ ( )

∑ ⇒

ABM

BAM

3

AB BAM M

3

AB BAM M

3

AB BAM M

DCM

4

DCM

4

DCM

4

DCM

20

A

BC

D

-

+

+

+53.92

26.03

16.56

16.56

B.M.D

q*(l^2)/8

:بعد التعويض بقيم العزوم اليت اوجدناها سابقا حنصل على املعادلة

⇒

( )

حبل ( ) و( ) جند و :

4- End Moments

5- B.M.D:

21

المسألة الثالثة ٢٠١٣- ٢- ٣دورة

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:

1

1,33 1

0,33 1

1,33

1,33

21 1

10,33

0,330,33

+

+

+

+

+

+

++

+

-

-

-

2 2 2 4

0,67

A B C D E

. . AI L R

. . CI L R

. . EI L R

. . DI LV

. . DI L M

. . RCI LV

. . LCI LV

. . CI L M

22