structural theory 2

STRUCTURAL TIMBER DESIGN

THEORY2 Structural Theory 2Chapter 3

Slope Deflection Method;

slope deflection is a method that takes into account the flexural deformations of beams and frames (that is, rotations, settlements, etc.), but that neglects shear and axial deformations. Although this classical method is generally considered to be obsolete, its study can be useful for several reasons, which include;

slope deflection is convenient for hand analysis of some small structures.

a knowledge of the method provides an excellent background for understanding moment distribution method.

it is a special case of the displacement or stiffness method of analysis and provides a very effective introduction for the matrix formulation of structures.

the slopes and deflections determined by slope deflection enable the analyst to sketch the deformed shape of a particular structure, and therefore, will have a better feel for the behaviour of the structure.

Slope Deflection Method for Beams;

the name slope deflection comes from the fact that the moments at the ends of the members in statically indeterminate structures are expressed in terms of the rotations (or slopes) and deflections of the joints. For developing the equations, members are assumed to be of constant cross section between each pair of supports. Although it is possible to derive expressions for members of varying section, the results are so complex as to be of little practical value. It is also assumed that the joints in a structure may rotate or deflect, but the angles between the members meeting at a joint remain unchanged.

Imagine a span AB of a continuous beam and consider it for the following discussion. If the span is completely fixed at each end, the slope of the elastic curve of the beam at the ends is zero. External loads produce fixed-end moments, and these moments cause the span to take its own shape. Joints A and B are actually not fixed and will rotate slightly under load to some position. In addition to the rotation of the joints, there may be possibly be some settlement of one or both of the supports, which will cause a chord rotation of the member, where support B is assumed to have settled by a certain amount, .

From the study, the values of the final end moments at A and B ( and ) are seen to be equal to the sum of the moments caused by the following;

fixed-end moments ( and ) which can be determined with the moment-area theorems.

moments caused by chord rotation if one or both of the joints settles or deflects.

moments caused by the rotations of joints A and B ( and )

Rotations of the joints in a structure cause changes in the slopes of the tangents to the elastic curves at those points. For a particular beam, the change in the slope equals the end shear of the beam when it is loaded with M/EI diagram. The beam is assumed to have the fixed-end moments, and . The end reactions or end slopes are as follows;

=

=

(2 - )

=

=

(2 - )

If one of the supports of the beam settled or deflected by an amount, , the angles and caused by a joint rotation would change by . Adding the chord rotation to the expression results in the following total values for the slopes of the tangents to the elastic curves at the ends of the beams.

=

(2 - ) +

=

(2 - ) +

Solving the equations simultaneously for and gives the values of the end moments due to slopes and deflections.

=2EK(2 + - 3)

=2EK( + 2 - 3)

where,

=

andK =

The final end moments are equal to the moments due to slopes and deflections plus the fixed-end moments. The slope-deflection equations are as follows;

=

+ 2EK(2 + - 3)

=

+ 2EK( + 2 - 3)

where:

-actual moment at end A

-actual moment at end B

-fixed-end moment at end A

-fixed-end moment at end B

K-stiffness factor of the member

-displacement factor

-rotation of end A

-rotation of end B

With these equations, it is possible to express the end moments in a structure in terms of joint rotations and settlements. Slope deflection appreciable reduces the amount of work involved in analyzing multi-redundant structures because the unknown moments are expressed in terms of only a few unknown joint rotations and settlements.

Sign Conventions; in applying the slope-deflection equations, it is simpler to use a convention in which signs are given to clockwise and counterclockwise moments at the member ends. Once these moments are determined, their signs can be easily converted to the usual convention for drawing moment diagrams (known as beam notation).

The following convention is used; should a member cause a moment that tends to rotate a joint clockwise, the joint moment is considered negative; if counterclockwise, it is positive. In other words, a clockwise resisting moment on the member is considered positive, and a counterclockwise resisting moment is considered negative.

EX. Determine all the support moments of the structure shown.

relative stiffness factors,

=

=

;

=

=1

=

=

;

=

=1

fixed-end moments,

=-

=-

=-37.5 kN-m

=+

=+

=+37.5 kN-m

=-

=-

=-26.67 kN-m

=+

=+

=+13.33 kN-m

slope deflection equations,

=-37.50 + 1(2 + )

=-37.50 + 2 +

=+37.50 + 1( + 2)

=+37.50 + + 2

=-26.67 + 1(2 + )

=-26.67 + 2 +

=+13.33 + 1( + 2)

=+13.33 + + 2

using joint conditions,

@ joint B;

+

=0

0

=+37.50 + + 2 - 26.67 + 2 +

0

=4 + 10.83;

=-2.71 rad

final end moments,

=-37.50 2.71

=-40.208 kN-m

=+37.50 - 2(2.71)

=+32.083 kN-m

=-26.67 - 2(2.71)

=-32.083 kN-m

=+13.33 2.71

=+10.625 kN-m

considering span AB,

(+) = 0;

(6) - 40.21 - 50(3) + 32.08 = 0;

=26.36 kN

considering span BC,

(+) = 0;

30(2) + 10.63 32.08 - (6) = 0;

=6.42 kN

considering the whole structure,

(+) = 0;

50(3) + 30(8) + 10.63 40.21 6.42(12) - (6) = 0;

=47.23 kN

Check;

((+) = 0;

26.36 + 47.23 + 6.42 50 30 = 0;(ok



=

=

;

=

=1

=

=

;

=

=1

fixed-end moments,

=-

=-

=-50.0 kN-m

=+

=+

=+50.0 kN-m

=-

=-

=-26.67 kN-m

=+

=+

=+26.67 kN-m


=-50.00 + 1(2 + )

=-50.00 + 2 +

=+50.00 + 1( + 2)

=+50.00 + + 2

=-26.67 + 1(2 + )

=-26.67 + 2 +

=+26.67 + 1( + 2)

=+26.67 + + 2


@ joint B;

+

=0

0

=+50 + + 2 - 26.67 + 2 +

0

=+23.33 + 4 +

-eqn. 1

@ joint C;

=0

0

=+26.67 + + 2

-eqn. 2

solving simultaneously,

=-2.86 rad

=-11.91 rad

final-end moments,

=-50 - 2.86

=-52.857 kN-m

=+50 - 2(2.86)

=+44.286 kN-m

=-26.67 - 2(2.86) - 11.91

=-44.286 kN-m

=+26.67 - 2.86 - 2(11.91)

=0


(+) = 0;

(4) 52.86 - 100(2) + 44.29 = 0;

=52.14 kN

considering span BC,

(+) = 0;

20(4)(2) 44.29 - (4) = 0;

=28.93 kN


(+) = 0;

100(2) + 20(4)(6) 52.86 28.93(8) - (4) = 0;

=98.93 kN

Check;

((+) = 0;

52.14 + 98.93 + 28.93 100 - 20(4) = 0;(ok



=

=

;

=

=1

=

=

;

=

=2

=

=

;

=

=2

fixed-end moments,

=-

=-

=-30.0 kN-m

=+

=+

=+30.0 kN-m

=-

=-

=-45.0 kN-m

=+

=+

=+45.0 kN-m

=-

=-

=-37.5 kN-m

=+

=+

=+37.5 kN-m


=-30.0 + 2(2 + )

=-30.0 + 4 + 2

=+30.0 + 2( + 2)

=+30.0 + 2 + 4

=-45.0 + 2(2 + )

=-45.0 + 4 + 2

=+45.0 + 2( + 2)

=+45.0 + 2 + 4

=-37.5 + 1(2 + )

=-37.5 + 2 +

=+37.5 + 1( + 2)

=+37.5 + + 2


@ joint A;

=0

0

=-30.0 + 4 + 2

-eqn. 1

@ joint B;

+

=0

0

=+30.0 + 2 + 4 - 45.0 + 4 + 2

0

=-15.0 + 2 + 8 + 2

-eqn. 2

@ joint C;

+

=0

0

=+45.0 + 2 + 4 - 37.50 + 2 +

0

=+7.5 + 2 + 6 +

-eqn. 3

@ joint D;

=0

0

=+37.50 + + 2

-eqn. 4


=+7.83 rad

=-0.65 rad

=+2.28 rad

=-19.89 rad

final end moments

=-30.0 + 4(7.83) - 2(0.65)

=0

=+30.0 + 2(7.83) - 4(0.65)

=43.04 kN-m

=-45.0 - 4(0.65) + 2(2.28)

=-43.04 kN-m

=+45.0 - 2(0.65) + 4(2.28)

=52.83 kN-m

=-37.50 + 2(2.28) - 19.89

=-52.83 kN-m

=+37.50 + 2.28 - 2(19.89)

=0


(+) = 0;

(6) + 43.04 10(6)(3) = 0;

=22.83 kN

considering span CD,

(+) = 0;

50(3) 52.83 - (6) = 0;

=16.20 kN


(+) = 0;

22.83(12) + 50(3) + (6) 10(6)(9) 20(6)(3) 16.2(6) = 0;

=95.54 kN

(+) = 0;

22.83(6) + 20(6)(3) + 50(9) 10(6)(3) 16.2(12) - (6) = 0;

=95.43 kN

Check;

((+) = 0;

22.83 + 95.54 + 95.43 + 16.20 - 10(6) - 20(6) 50 = 0;(ok



=

=

=

;

=

=9

=

=

=

;

=

=8

=

=

=

;

=

=6

fixed-end moments,

=-

=-

=-50.0 kN-m

=+

=+

=+50.0 kN-m

=- -

=- -

=-40.0 kN-m

=+ +

=+

+

=+40.0 kN-m

=-

=-

=-30.0 kN-m

=+

=+

=+30.0 kN-m


=-50 + 9(2 + )

=-50 + 18 + 9

=+50 + 9( + 2)

=+50 + 9 + 18

=-40 + 8(2 + )

=-40 + 16 + 8

=+40 + 8( + 2)

=+40 + 8 + 16

=-30 + 6(2 + )

=-30 + 12 + 6

=+30 + 6( + 2)

=+30 + 6 + 12


@ joint A;

=0

0

=-50.0 + 18 + 9

-eqn. 1

@ joint B;

+

=0

0

=+50.0 + 9 + 18 - 40.0 + 16 + 8

0

=+10.0 + 9 + 34 + 8

- eqn. 2

@ joint C;

+

=0

0

=+40.0 + 8 + 16 - 30.0 + 12 + 6

0

=+10.0 + 8 + 28 + 6

-eqn. 3

@ joint D;

=10(3)

=30.0 kN-m

+30.0

=+30.0 + 6 + 12

0

=6 + 12

-eqn. 4


=+3.37 rad

=-1.18 rad

=-0.02 rad

=+0.01 rad

final-end moments,

=-50.0 + 18(3.37) - 9(1.18)

=0

=+50.0 + 9(3.37) - 18(1.18)

=+59.06 kN-m

=-40.0 - 16(1.18) - 8(0.02)

=-59.06 kN-m

=+40.0 - 8(1.18) - 16(0.02)

=+30.20 kN-m

=-30.0 - 12(0.02) + 6(0.01)

=-30.20 kN-m

=+30.0 - 6(0.02) + 12(0.01)

=+30.0 kN-m


(+) = 0;

(8) + 59.06 50(4) = 0;

=17.62 kN

considering span CE,

(+) = 0;

10(6)(3) + 10(9) 30.20 - (6) = 0;

=39.97 kN


(+) = 0;

17.62(17) + 10(6)(3) + 10(9) + (9) 50(13) 20(6) 20(3)

39.97(6) = 0;

=55.59 kN

(+) = 0;

17.62(8) + 20(3) + 20(6) + 10(6)(12) + 10(18) 50(4) 39.97(15)

(9) = 0;

=46.82 kN

Check;

((+) = 0;

17.62 + 55.59 + 46.82 + 39.97 50 20 20 - 10(6) 10 = 0; ( ok

EX. Find the moment at support B in the beam if support B settles by 10 mm.

Given:E=200 GPa

I=2.0 x

absolute stiffness factors,

2EK

=

=20000 kN-m

fixed-end moments,

=-

=-

=-50.0 kN-m

=+

=+

=+50.0 kN-m

=-

=-

=-25.0 kN-m

=+

=+

=+25.0 kN-m

relative displacement factor,

=

=+

=+0.0025

=

=-

=-0.0025


=-50 + 10000(2 + - 0.0075)

=-125 + 20000 + 10000

=+50 + 10000( + 2 - 0.0075)

=-25 + 10000 + 20000

=-25 + 10000(2 + + 0.0075)

=+50 + 20000 + 10000

=+25 + 10000( + 2 + 0.0075)

=+100 + 10000 + 20000


@ joint A;

=0

0

=-125 + 20000 + 10000

-eqn. 1

@ joint B;

+

=0

0

=-25 + 10000 + 20000 + 50 + 20000 + 10000

0

=+25 + 10000 + 40000 + 10000

- eqn. 2

@ joint C;

=0

0

=+100 + 10000 + 20000

-eqn. 3


=+0.00688 rad

=-0.00125 rad

=-0.00438 rad

final-end moments,

=-125 + 20000(0.00688) + 10000(-0.00125)

=0

=-25 + 10000(0.00688) + 20000(-0.00125)

=+18.80 kN-m

=+50 + 20000(-0.00125) + 10000(-0.00438)

=-18.80 kN-m

=+100 + 10000(-0.00125) + 20000(-0.00438)

=0


(+) = 0;

(4) + 18.8 100(2) = 0;

=45.30 kN

considering span CE,

(+) = 0;

50(2) 18.8 - (4) = 0;

=20.30 kN


(+) = 0;

45.30(8) 100(6) 50(2) + (4) = 0;

=84.40 kN

Check;

((+) = 0;

45.30 + 84.40 + 20.30 100 50 = 0; ( ok

EX. Determine all moments for the beam shown, which is assumed to have the following support settlements: A = 40 mm, B = 60 mm C = 70 mm, D = 35 mm.

Given:E=200 GPa

I=4.0 x

absolute stiffness factors,

2EK

=

=20000 kN-m

fixed-end moments,

=-

=-

=-50.0 kN-m

=+

=+

=+50.0 kN-m

=-

=-

=-50.0 kN-m

=+

=+

=+50.0 kN-m

=-

=-

=-46.875 kN-m

=+

=+

=28.125 kN-m

relative displacement factor,

=

=+

=+0.00250

=

=+

=+0.00125

=

=-

=-0.00500


=-50 + 20000(2 + - 0.0075)

=-200 + 40000 + 20000

=+50 + 20000( + 2 - 0.0075)

=-100 + 20000 + 40000

=-50 + 20000(2 + - 0.00375)

=-125 + 40000 + 20000

=+50 + 20000( + 2 - 0.00375)

=-25 + 20000 + 40000

=-46.875 + 20000(2 + + 0.0150)

=-253.125 + 40000 + 20000

=+28.125 + 20000( + 2 + 0.0150)

=+328.125 + 20000 + 40000


@ joint A;

=0

0

=-200 + 40000 + 20000

-eqn. 1

@ joint B;

+

=0

0

=-100 + 20000 + 40000 - 125 + 40000 + 20000

0

=-225 + 20000 + 80000 + 20000

-eqn. 2

@ joint C;

+

=0

0

=-25 + 20000 + 40000 - 253.13 + 40000 + 20000

0

=-278.125 + 20000 + 80000 + 20000 -eqn. 3

@ joint D;

=0

0

=+328.125 + 20000 + 40000

-eqn. 4


=+0.00501 rad

=-0.00002 rad

=+0.00632 rad

=-0.01136 rad

final-end moments,

=-200 + 40000(0.00501) - 20000(0.00002);

=0

=-100 + 20000(0.00501) - 40000(0.00002);

=-0.625 kN-m

=-125 - 40000(0.00002) + 20000(0.00632);

=+0.625 kN-m

=-25 - 20000(0.00002) + 40000(0.00632);

=+227.5 kN-m

=-253.125 + 40000(0.00632) - 20000(0.01136);

=-227.5 kN-m

=+328.125 + 20000(0.00632) - 40000(0.01136);

=0


(+) = 0;

(8) 50(4) 0.625 = 0;

=25.08 kN

considering span CD,

(+) = 0;

40(3) 227.5 - (8) = 0;

=-13.44 kN


(+) = 0;

25.08(16) + 40(3) + 13.44(8) + (8) 50(12) 40(4) = 0;

=16.40 kN

(+) = 0;

25.08(8) + 40(4) + 40(11) + 13.44(16) - 50(4) - (8) = 0;

=101.96 kN

Check;

((+) = 0;

25.08 + 16.40 + 101.96 - 13.44 40 50 40 = 0; ( ok

23.59

26.36

23.64

30 kN

10.63

32.08

EMBED Equation.3

40.21

32.08

50 kN

EMBED Equation.3

100 kN

EMBED Equation.3

100 kN

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

50 kN

20 kN/m

30 kN

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

4 m

2 m

2 m

100 kN

20 kN/m

A

C

B

40.21

38.87

32.08

15.13

10.63

6.42

4 m

3 m

2 m

3 m

A

C

B

50 kN

30 kN

44.29

52.86

EMBED Equation.3

44.29

20 kN/m

28.93

51.07

47.86

52.14

20.91

44.29

51.42

52.86

EMBED Equation.3

50 kN

52.83

20 kN/m

52.83

43.04

10 kN/m

EMBED Equation.3

43.04

B

C

D

A

10 kN/m

50 kN

20 kN/m

2I

2I

6 m

3 m

3 m

6 m

I

10 kN/m

50 kN

20 kN/m

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

22.83

16.20

58.37

61.63

33.80

37.17

48.60

52.80

42.16

43.02

22.83

20 kN

20 kN

30.20

59.06

50 kN

EMBED Equation.3

59.06

EMBED Equation.3

10 kN/m

10 kN

30.20

B

C

D

A

E

50 kN

10 kN/m

20 kN

20 kN

10 kN

2I

I

2I

3 m

4 m

3 m

6 m

3 m

3 m

4 m

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

50 kN

10 kN/m

20 kN

20 kN

10 kN

EMBED Equation.3

10

29.97

30.03

16.79

3.21

23.21

32.38

17.62

59.04

70.48

10.59

20.22

30.20

14.94

30

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

100 kN

50 kN

100 kN

B

C

A

50 kN

2 m

2 m

2 m

2 m

100 kN

EMBED Equation.3

18.8

50 kN

18.8

EMBED Equation.3

45.3

54.7

29.7

20.3

90.6

18.8

40.6

40 kN

50 kN

40 kN

B

C

A

D

4 m

3 m

4 m

4 m

4 m

5 m

EMBED Equation.3

EMBED Equation.3

40 kN

50 kN

40 kN

EMBED Equation.3

EMBED Equation.3

227.5

40 kN

0.625

50 kN

EMBED Equation.3

0.625

227.5

EMBED Equation.3

40 kN

24.92

25.08

8.52

48.52

53.44

13.44

0.64

100.32

33.44

227.52

67.2

61Engr. I.R. Bonzon

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structural theory 2

Documents