Slope Deflection Method for the Analysis of Indeterminate Structures

ByProf. Dr. Wail Nourildean Al-Rifaie

All structures must satisfy:

Load-displacement relationship Equilibrium of forces Compatibility of displacements

Using the principle of superposition by considering separately the moments developed at each support of a typical prismatic beam (AB) shown in Fig. 1(a) of a continuous beam, due to each of the displacements , , and the applied loads. Assume clockwise moments are +ive.

1. Assume ends A and B are fixed, i., e., the rotations . This means that we have to apply counterclockwise moment at end A and clockwise moment at end B

due to the applied loads to cause zero rotation at each of ends A and B. Table (1) gives for different loading conditions.

Table (1)

2. Release end A against rotation at end A (rotates to its final position ) by applying clockwise moment while far end node B is held fixed as shown in Fig. 1.

3. Now, the clockwise moment - rotation relationship is:

4. The carry over moment at end B is:

5. In a similar manner, if end B of the beam rotates to its final position , while end A is held fixed. The clockwise moment – rotation relationship is:

6. The carry over moment at end A is:

7. If node B is displaced relative to as shown in Fig. (1), so that the cord of the member rotates clockwise i., e., positive displacement and yet both ends do not rotate, then equal but anticlockwise moments are developed in the member as shown in the figure.

Slope-Deflection Equation

Load-displacement relationship

If the end moments due to each displacement and the loading are added together, the resultant moments at the ends may then be written as:

For prismatic beam element, equation (1) may be written as:

The slope deflection equations (1 or 2) relate the unknown moments applied to the nodes

to the displacements of the nodes for any span of the structure.

To summarize application of the slope-deflection equations, consider the

continuous beam shown in Fig. (2) which has four degrees of freedom.

Now equation (2) can be applied to each of the three spans.

Fig. (2)

From Fig.(2):

Equilibrium conditions Compatibility conditions

These equations would involve the four unknown rotations .

Solving for these four unknown rotations. It may be noted that there is no relative deflection between the supports, so that The values of the obtained rotations may then be substituted in to the slope deflection equations to determine the internal moments at the ends of each member.

If any of the results are negative, they indicate counterclockwise rotation.

DCBA ,,,

Example (1)

Solution

Draw the shear and moment diagrams for the beam shown in Fig.(3). EI is constant.

1. Using the formulas for the tabulated in Table (1) for the given loadings:

Fig. (3)

2. There are two slopes at B and C, i., e., are unknowns. Since end A is fixed, Also,

since the supports do not settle, nor are they

displaced up or down

Now, by applying the equilibrium conditions:

Substituting the computed values in to moment equations (a), (b), (c), and (d):

By considering the values of support moments and the applied loads, the support reactions may then be determined:

RA = 8.3625 kN

RB = 10.2042 kN RC = 1.8333 kN

Shearing force and bending moment digrams are shown in Fig. (4).

Fig. (4) Shearing Force & Bending Moment Digrams

Example (2)

Determine the internal moments at the supports of the beam shown in Fig. (5). The support at B is displaced (settles) 12 mm.

Solution

1. Two spans must be considered. FEMs are determined using Table (1).

4

4

101.012

012.00

10667.618

0012.0

x

x

BC

AB

2. Using equation 2:

)(0)101.032()12

2(

)()101.032()12

2(

)()10667.632()18

2(

)()10667.63()18

2(

4

4

4

4

lxxEI

M

kxxEI

M

jxxEI

M

ixxEI

M

BCBCCB

CBBCBC

BABBA

BABAB

3. Equilibrium condition:

00 CB MandM

)(0)101.032()12

2(

)(0)101.032()12

2()10667.632()

18

2(

)()10667.63()18

2(

4

44

4

pxxEI

nxxEI

xxEI

mxxEI

M

BCBC

CBBCBAB

BABAB

In order to obtained the rotations equations (n) & (p) may then be solved simultaneously, it may be noted that since A is fixed support. Thus,

CB and

0A

.1047647.2.1065294.4 44 radxandradx CB

Substituting these values into equations (i to l) yields

Example (3)

If end A in example (1) is simply supported, and by applying the compatibility condition, their will be three unknown rotations,

Now,

),,( CBA

)(62.1)2()6.3

2(

)(62.1)2()6.3

2(

)(5.4)2()4.2

2(

)(5.4)2()4.2

2(

dEI

M

cEI

M

bEI

M

aEI

M

BCBCCB

CBBCBC

BABBA

BAABAB

Applying the equilibrium conditions:

)3(062.1)2()6.3

2(

)2(062.1)2()6.3

2(5.4)2()

4.2

2(

)1(05.4)2()4.2

2(

BCBCCB

CBBCBAB

BAABAB

EIM

EIEI

EIM

By solving equations (1, 2 & 3) for and substitute the values into equations (a, b, c, d):

CBA ,,

0

.158.4

.158.4

0

CB

BC

BA

AB

M

iseanticlockwmkNM

clockwisemkNM

M

Shearing force and bending moment diagrams are shown in the following figure.

Example (4)Determine the moments at each joint of the frame shown in Fig.(7). EI is constant.

Fig. (7)

MkNFEM

MkNFEM

CB

BC

.8096

)8)(24(5)(

.8096

)8)(24(5)(

2

2

Because ends A and D are fixed supports.

and since no sidesway will occur.

0 CDBCAB

CCCDDC

CCCDCD

BCBCBCCB

CBCBBCBC

BBABBA

BBABAB

EIl

EIM

EIEI

M

EIEIEI

M

EIEIEI

M

EIEI

M

EIEI

M

1667.0)()2(

3334.0)2()12

2(

8025.05.080)2()8

2(

8025.05.080)2()8

2(

3334.0)2()12

2(

1667.0)()12

2(

Equilibrium conditions:

)2(08025.08334.0

0803334.025.05.0

0

)1(08025.08334.0

08025.05.03334.0

0

BC

CBC

CDCB

CB

CBB

BCBA

EI

Or

EIEIEI

MM

EIEI

Or

EIEIEI

MM

Solving simultaneously yields

EIand

EI CB1.1371.137

Therefore,

seanticlokwimkNM

iseanticlockwmkNM

clockwisemkNM

iseanticlockwmkNM

clockwisemkNM

clockwisemkNM

DC

CD

CB

BC

BA

AB

.9.22

.7.45

.7.45

.7.45

.7.45

.9.22

Fig.(8)

The bending moment diagram is shown in Fig.(8).

Example (5)Determine the internal moments at each of the frame shown in Fig.(9). Solution

Fig.(9)

1. Fixed end moments:

12)(,

12)(:

12

)5.1()(,

12

)5.1()(:

22

22

wLFEM

wLFEMBCSpan

LwFEM

LwFEMABSpan

CBBC

BAAB

8)(,

8)(:

PlFEM

PlFEMBDSpan BDDB

2. Joint moments:

)(8

)2()2(

)(8

)2()2(

)(12

)2()2(

)(12

)2()2(

)(12

)5.1()2()

5.1

2(

)(12

)5.1()2()

5.1

2(

2

2

2

2

nPL

L

EIM

mPL

L

EIM

lwL

L

EIM

kwL

L

EIM

jLw

L

EIM

iLw

L

EIM

BDBDDB

DBBDBD

BCBCCB

CBBCBC

ABABBA

BAABAB

3. Equilibrium conditions:

Solving equations (1,2,3) simultaneously yields

)4(8

)()2(

)3(012

)2()2(

)2(08

)2()2(

12)2()

2(

12

)5.1()2()

5.1

2(

0

)1(012

)5.1()2()

5.1

2(

2

22

2

PL

L

EIM

wL

L

EIM

PL

L

EI

wL

L

EILw

L

EI

M

Lw

L

EIM

BBDDC

BCBCCB

DBBD

CBBCABAB

B

BAABAB

)4(8

)()2(

)3(012

)2()2(

)2(08

)2()2(

12)2()

2(

12

)5.1()2()

5.1

2(

0

)1(012

)5.1()2()

5.1

2(

2

22

2

PL

L

EIM

wL

L

EIM

PL

L

EI

wL

L

EILw

L

EI

M

Lw

L

EIM

BBDDC

BCBCCB

DBBD

CBBCABAB

B

BAABAB

CBA ,,Solving equations (1,2,3) simultaneously yieldsSubstituting the rotation values into equations (i to n) to

determine the joint moments.

Example (6)Determine the joint internal moments of the frame shown in

Fig.(10), both ends A and D are fixed. Assume

5.1)(1)()( CDBCAB L

EIand

L

EI

L

EI

Fig.(10)

Solution1. Fixed end moments:

mkNFEM

mkNFEMBCSpan

mkNFEM

mkNFEMABSpan

CB

BC

BA

AB

.96.1212

)2.7(3)(

.96.1212

)2.7(3)(:

.0.8)4.5(

)6.3)(8.1(10)(

.0.4)4.5(

)8.1)(6.3(10)(:

2

2

2

2

2

2

It is assumed that the axial deformation is neglected so that OO CCBB as shown in the following figure.

It may be noted that

2. Joint moments:

05.1 DABCABCD and

)(75.65.1

)5.4(5.1

)(75.63

)5.42(

)(96.12)2(5.1

)(96.12)2(

)(8)2()5.1

2(

)(4)3(

n

M

m

M

lM

kM

jL

EIM

iM

ABC

ABCDC

ABC

ABCCD

BCCB

CBBC

ABBABBA

ABBAB

3. Equilibrium conditions:

)2(96.1275.65

0:int

)1(96.434

0:int

ABCB

CDCB

ABCB

BCBA

Or

MMCJo

Or

MMBJo

Since a horizontal displacement entire frame in the x-direction. This yields

occurs, the summing forces on the

)3(667.1075.425.2

06.34.53

1010

6.3

4.53

10

:

010:0

ABCB

DCCDBAAB

DCCDD

BAABA

DAX

Or

MMMM

MMH

and

MMH

whichIn

HHF

Solving equations (1, 2, 3) yields

9194.1,565.0,8208.2 ABCB By substituting these values into moment equations (i to n):

mkNM

mkNM

mNkM

mkNM

mkNM

mkNM

DC

CD

CB

BC

BA

AB

.8035.13

.6509.14

..6509.14

.8833.7

.8833.7

.9374.6

Bending moment diagram is plotted in the following figure.