Slope Deflection Examples

Slope Deflection Examples: Fixed End Moments For a member AB with a length L and any given load the fixed end moments are given by: FEM

=

AB

=

2 L2 2

( 2⋅g − g )   B

A

( gB − 2 ⋅ g A )   L2 Where: gB and gA are the moments of the bending moment diagrammes of the statically determinate beam about B and A respectively. FEMBA

Example: Determine the fixed end moments of a beam with a point load.

Simply supported beam with bending moment diagramme. Centroid in accordance with standard tables. FEM

=

AB

FEM  AB FEM  AB FEM  AB

2 L2 2

( 2⋅g − g )   B

A

 L Wa Wab  b + L  L Wa Wab  a + L      2⋅ ⋅ ⋅ − ⋅ ⋅        L  3  2 L  3      L  2 2 ⋅ Wab  2b + 2L − a − L   =    3 L    2 ⋅ Wab  2b + a + b − a   =    3 L    2 ⋅ Wab = =

2

2

2

2

FEM  AB

L2

In a similar way FEMBA may be calculated.

Slope-deflection

Page 1 of 20

3/22/2010

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Use of slope-deflection equations: Example 1: Determine the bending moment diagramme of the following statically indeterminate beam.

The unknowns are as follows:

θ

,

A

θ

,

B

θ

= 0, ψ  AB = 0,

C

ψ  BC = 0

We require two equations to solve the two unknown rotations:

∑M

= 0 ∴ M  = 0 2 ⋅ EI  = ( 2 ⋅ θ + θ − 3 ⋅ψ  ) + FEM A

M

AB

M 

AB

M

=

AB

A

L 2 ⋅ EI 

A

 A

=0

EI 

 A

+ 0,5 EI 

MBA

=

M BA

=

B

+5=0

BA

(1)

BC

 

2 ⋅ EI 

( θ + 2 ⋅ θ − 3 ⋅ψ  ) + FEM A

L 2 ⋅ EI 

B

( θ + 2 ⋅ θ  ) − A

MBA MBC

=

M BC

10 ⋅ 4

B

= 0 ∴ M + M  = 0

4 = 0, 5 ⋅ EI ⋅ θ A

M BC

 

AB

8 +A 0, 5 ⋅ EI ⋅ θ   +B 5

∑ M 

B

AB

( 2 ⋅ θ + θ  ) +

4 = EI ⋅ θ AB

∑M

B

AB

 

BA

10 ⋅ 4

B

8 + 1⋅ EI ⋅ θ B  − 5

2 ⋅ EI 

( 2 ⋅ θ + θ − 3 ⋅ψ  ) + FEM

L 2 ⋅ EI 

B

C

BC

BC

 

 

5 ⋅ 62 = ( 2 ⋅ θ B ) + 6 12 = 0, 66667 ⋅ θ B + 15

M BA BA + M BC  BC  = 0 

0,5 EI 

 A

+ 1,66667 EI 

Solve the unknowns:

B

+ 10 = 0

(2)

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M BC  BC  = 0,6667 x – 5,29412 + 15 2 ⋅ EI  MCB = ( θ B + 2 ⋅ θ C − 3 ⋅ψ BC ) L 2 ⋅ EI  −5 ⋅ 62 M CB = θ  + ( B) 6 12 M CB CB = 0,3333 x –5,29412 – 15

= + 11,471 kN.m

+ FEM

 

CB

= -16,675 kN.m

Draw the bending moment diagramme.

The Modified Slope-Deflection Equation with a Hinge at A:

We would like to eliminate θ

=

2 ⋅ EI 

( 2 ⋅θ L Solve for θ A. M

AB

θ  A

=−

MBA

=

FEM 

2

=

MBA

=

+ θ − 3 ⋅ψ  ) + FEM A

B

L

2 ⋅ EI 

−

θ

2

AB

+

B

3 ⋅ψ 

 

AB

AB

2

2 ⋅ EI 

( θ + 2 ⋅ θ − 3 ⋅ψ  ) + FEM A

L

Replace θ MBA

⋅

AB

from the equation as we know that M AB = 0.

A

B

 2 ⋅θ − 

θB

B

2

+

3 ⋅ψ AB

1 − 3 ⋅ψ     + FEM − 2 ⋅ FEM   AB

2

3 ⋅ EI  L

 

BA

in this equation.

A

2 ⋅ EI  L

AB

BA

1

( θ − ψ  ) + FEM − ⋅ FEM B

AB

BA

2

AB

AB

 

 

This equation may be used to reduce the number of unknown rotations.

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The unknowns are as follows: θ A use the modified slope-deflection equation, The total number of unknowns is reduced to θ B

∑M

B

MBA

=

M BA

=

B

,θ

C

= 0, ψ  AB = 0,

= 0 ∴ M + M  = 0 BA

BC

 

3 ⋅ EI 

1 2 1  10 ⋅ 4  

( θ − ψ  ) + FEM − ⋅ FEM B

L 3 ⋅ EI 

AB

( θ  ) −

MBA MBC

=

M BC

=

BA

10 ⋅ 4

B

4 = 0, 75 ⋅ EI ⋅ θ A

M BC

θ

8 − 7, 5

−  2

8

 

AB

   

2 ⋅ EI 

( 2 ⋅ θ + θ − 3 ⋅ψ  ) + FEM B

L 2 ⋅ EI 

C

( 2 ⋅ θ  ) + B

6 = 0, 66667 ⋅ θ B

BC

BC

 

 

5 ⋅ 62

12 + 15

M BA BA + M BC  BC  = 0 

1,41667 EI  B

B

+ 7,5 = 0

= - 5,29412/EI 

Bending moments are as calculated previously. Example 2: Determine the bending moment diagramme of the following structure.

(1)

ψ  BC = 0

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MBA

=

MBC

=

3 ⋅ EI 

1 2

( θ − ψ  ) + FEM − ⋅ FEM B

AB

BA

L No Loads No FEM: 3 ⋅ 2EI  MBA = ( θ B ) = 1,5 ⋅ EI ⋅ θ B 4

2 ⋅ EI 

MBC

( 2 ⋅ θ B + θ C − 3 ⋅ψ BC ) + FEM BC L 2 ⋅ 3EI  20 ⋅ 4 = ( 2 ⋅ θ B + θ C ) +  4 8 = 3, 0 ⋅ EI ⋅ θ B + 1, 5 ⋅ EI ⋅ θ C +  10 

MBE

=

MBE

=

M BC

∑M

 

AB

3 ⋅ EI 

( θ − ψ  ) + FEM

L 3 ⋅ EI 

B

BE

BE

1 2

− ⋅ FEM

 

 

EB

 

( θ ) = 1,0 ⋅ EI ⋅ θ  B

3

B

= 0 ∴ 1, 5 ⋅ EI ⋅ θ + 3, 0 ⋅ EI ⋅ θ +1, 5 ⋅ EI ⋅θ +10 + 1, 0 ⋅ EI ⋅θ = 0 5, 5 ⋅ EI ⋅ θ + 1, 5 ⋅ EI ⋅ θ    + 10 = 0 B

B

B

∑M

C

MCB

=

MCD

=

B

C

B

C 

=0 ∴

MCB

CD

2 ⋅ EI 

 

2 ⋅ EI 

+ FEMCD ( 2, 0 ⋅ θ C + θ D − 3 ⋅ψ CD CD ) L 2 ⋅ 2EI  = ( 2, 0 ⋅ θ C  + 0 − 3 ⋅ 0 ) + 0 4 = 2,0 ⋅ EI ⋅ θ C 

MCD

∑M

 

= 0 ∴ 1, 5 ⋅ EI ⋅θ + 3, 0 ⋅ EI ⋅θ −10 + 2 ⋅ EI ⋅θ  =  0 1, 5 ⋅ EI ⋅ θ + 5 ⋅ EI ⋅ θ   − 10 = 0 C

B

B

Solve for θ θ B

=

θ C 

=

C

C 

A

(1)

+ M  = 0

( θ B + 2 ⋅ θ C − 3 ⋅ψ BC ) + FEMCB L 2 ⋅ 3EI  20 ⋅ 4 M CB = ( θ B + 2 ⋅ θ C ) −  4 8 MCB = 1, 5 ⋅ EI ⋅ θ B + 3, 0 ⋅ EI ⋅ θ C −  10  

M CD

 

and θ B.

−2,57426 EI  +2,77228 EI 

−2,57426

C

 

(2)

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MCD MDC

= 2, 0 ⋅ 2, 77228 = 5, 545 kN.m 2 ⋅ EI  = ( θ + 2, 0 ⋅ θ − 3 ⋅ψ ) + FEM L

C

D

CD CD

DC

= EI ⋅ θ = 2,  772 kN.m C

 

Sway Structures One of the ways in which can calculate whether a structure can sway and the number of independent sway mechanisms, is to convert the structural elements to bar hinged elements and to determine the degree of  instability. The degree of instability will also be the number of independent sway mechanisms. Example:

Structure with bar-hinged elements: s=5 r=7 s+r = 12 n=6 2n = 12 2n – (s + r) =0 No independent sway mechanism

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Example:

Structure with bar-hinged elements: s=5 r=6 s+r = 11 n=6 2n = 12 2n – (s + r) =1 One independent sway mechanism Example: Determine the bending moment diagramme of the following structure:

s=3 r=4 n=4 2 (

)

1

s+r =7 2n =8 O i d d

t

h

i

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But BB’ = 6 ψ  AB therefore ψ  CD = CC’/4 = 6 ψ  AB/4 = 1,5 ψ  AB. Call ψ  AB, - ψ  .

θ A=0 θ B? θ C? θ D use the modified slope deflection equation. ψ  ?

Unknown in this case:

We require three equations to solve the unknowns.

∑ M  = 0 ∑ M  = 0 B

C 

For the third equation, one must investigate all the external forces that are applied to the structure.

The axial forces YAB and YDC are usually difficult to determine, whereas the shear forces VAB and VDC can be calculated by taking moments about B of the member AB and C of the member CD respectively. The third equation is obtained by:

∑M

B

MBA

=

M BA

=

MBA

∑Y  = 0

= 0 ∴ M + M  = 0 BA

BC

 

2 ⋅ EI 

( θ + 2 ⋅ θ − 3ψ  ) + FEM

L 2 ⋅ EI 

A

B

AB

( 2 ⋅ θ − 3 ⋅ ( −ψ ) ) − B

6 = EI ⋅ ( 0, 6667 ⋅ θ B

20 ⋅ 6

8 + 3ψ  ) − 15

BA

 

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MCB

2 ⋅ EI  10 ⋅ 10 ( θ B + 2 ⋅ θ C ) −  10 8 = EI ⋅ ( 0, 2 ⋅ θ B + 0, 4 ⋅θ C ) − 1  2, 5

MCD

=

M CD

=

=

M CB

3 ⋅ EI 

( θ − ψ  ) + FEM

L 3 ⋅ EI 

C

CD

CD

1 2

− ⋅ FEM

DC

 

 

( θC  − ( −1, 5 ⋅ψ ) ) 4 = EI ⋅ ( 0, 75 ⋅ θ C  + 1,125 ⋅ψ )

MCD

∑M

C

= 0 ∴ EI ⋅ ( 0, 2 ⋅ θ + 1,15 ⋅θ + 1,125 ⋅ψ  ) − 12, 5 = 0 B

C 

Take moments about B. M

+ M  + 20 ⋅ 3

V  AB

=

MBA

=

V  AB

=

V AB

θ  2 ⋅ψ    = EI ⋅  +  + 10 6   6

AB

BA

6 2 ⋅ EI 

( θ + 3 ⋅ψ ) + B

6 EI ⋅ (0, 3333 ⋅ θ B

20 ⋅ 6

= EI ⋅ (0, 3333 ⋅θ B + ψ ) + 15  8 + ψ ) + 15 + EI ⋅ (0, 6667  ⋅θ B + ψ ) − 1 5 + 60 6

B

In a similar fashion f ashion one may calculate VDC in terms of the unknowns:

(2)

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MBC MCB MCD

= 3,144 kN.m = - 20,063 kN.m = 20,063 kN.m

Momentary Centre of Rotation When two points on a rigid body undergo a small displacement, the body rotates about a momentary centre of rotation and the following angles are equal:

Example: Determine the sway angles of the following structure in terms of the sway angle ψ  DB of the following structure.

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D is a fixed f ixed point so that the point B may only move vertical to the member BD. B moves from B to B’. In a similar way E is a f ixed point and C can only move vertically to the member to C’. A may move horizontally. If one looks at the member AB both ends may move so we will find a momentary centre of rotation O 2 vertical to the direction of movement. movement. Both ends of member BC can move so we will find a momentary centre of  rotation, O1 vertical to the direction of movement of B and C. Because movements are small relative to the length of the member, the tan BB’ = 5 x ψ  BD ψ BC

=ψ

O1B

=ψ

O1C

=

BB '

=

LO B 1

CC’ = 6,667 x ψ  BC CC ' 4 ⋅ ψ BD ψ CE = = 4 LCE 

5 ⋅ψ BD 8,3333

= −0,6 ⋅ψ

= 4 x ψ  BD

= ψ 

BD

BD

 

ψ  = the angle ψ  .

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Structure with sway mechanism. mechanism. Determine the number of independent sway mechanisms: s=4 r=5 s+r=9 n = 5 2n = 10 2n –(s + r) = 1 1 independent sway mechanism! Unknowns

θA θB θC θB θE ψ 

use modified slope-deflection equation ? use modified slope-deflection equation 0 0 ? 2 unknown – we require 2 equations

Set ψ  DB = -ψ  BB’ = 4 ψ 

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M BD MBD

∑M

2 ⋅ EI  ( 2 ⋅ θ B − 3 ⋅ ( −ψ  ) ) 4 = 1, 0 ⋅ EI ⋅ θ B + 1, 5 ⋅ EI ⋅ψ   

=

B

= 0 ∴ 2 ⋅ EI ⋅ θ + 1, 25 ⋅ EI ⋅ψ  − 3  3, 75 = 0 B

(1)

To determine the second equation one must view all the external forces on the structure:

As it is difficult to determine YDB and YEC we will take moments about a point where their moment moment is known to be 0. The momentary centre of rotation, O 1, is such a point. Determine the unknown forces in terms of the unknown rotations and translational angles.

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=

MEC  V EC 

=

3 ⋅ EI  ( −( −ψ )) = 0, 6 ⋅ EI ⋅ψ  5 M EC  0,6 ⋅ EI ⋅ψ  5

=

5

Take moments about the momentary centre of rotation: VAB x 6 + VDB x 12 + VEC x 15 – 30 x 3 – M DB – MEC = 0 0, 5 ⋅ EI ⋅ θ B + 56, 25 + 4, 5 ⋅ EI ⋅ θ B + 9 ⋅ EI ⋅ψ + 1, 8 ⋅ EI ⋅ψ − 90 − 0, 5 ⋅ EI ⋅θ B −1, 5 ⋅ EI ⋅ψ

∑M

O1

= 0 ∴ 4, 5 ⋅ EI ⋅ θ + 8, 7 ⋅ EI ⋅ψ  − 3 3, 75 = 0 B

Solve the unknowns:

= 21,3535 EI  ψ  = −7,16561 θ B

EI 

MBA MBC MBD MDB MEC

= -23,073 kN.m = +12,468 kN.m = +10,605 kN.m = -0,072 kN.m = -4,299 kN.m

− 0, 6 ⋅ EI ⋅ψ =  0

 

(2)

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We have three unknowns, namely

∑M

B

MBA

=

M BA

=

= 0 ∴ M +M + M BA

3 ⋅ EI  L

=

B

3 ⋅ 2EI 

M Bc

=

θ

D

⋅ ( θ  − 0 ) −

 

BA

6 ⋅ 52

B

12

and ψ  . We require three equations to solve these unknowns.

= 0  1 2

AB

2

– 18,75  1

B

BC

Bc

⋅ ( θ − ( −ψ )) + 0 − 0 B

B

 

1  6 ⋅ 5   − ⋅+   2  12  

⋅ ( θ − ψ  ) + FEM − ⋅ FEM

L 3 ⋅ 2EI 

3 M BC  BC  = 2 EI 

BA

B

5

3 ⋅ EI 

BF BF

,

B

⋅ ( θ − ψ  ) + FEM − ⋅ FEM

M BA BA = 1,2 EI  MBc

BC

θ

+ 2 EI 

2

CB

 

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M DG  DG  = EI 

∑M

D

= 0 ∴ 3 ⋅ EI ⋅ θ − 2 ⋅ EI ⋅ψ  + 20 = 0

D

D

(2)

Third equation may be obtained from the vertical equilibrium of node C

-VCB - VCD – 20 = 0 V CB

=

V CD

=

−M 

BC 

3

+M 

DC

3

= =

−2 ⋅ EI ⋅ θ − 2 ⋅ EI ⋅ψ   B

3

+2 ⋅ EI ⋅ θ − 2 ⋅ EI ⋅ψ   D

3

-VCB - VCD – 20 = 0

+2 ⋅ EI ⋅ θ − 2 ⋅ EI ⋅ θ − 4 ⋅ EI ⋅ψ    − 20 = 0 B

+ 2 EI 

3 B - 2 EI 

D

D

- 4 EI 

= 60 

Solve the three simultaneous equations:

(3)

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The support D of the structure undergoes the following displacement, 10 mm vertically down and 20 mm horizontally to the left. E = 200 GPa and I = 50 x 10 -6 m4.

If one determines the number of independent sway mechanisms we see that there is one. The unknowns are thus θ B and ψ  .

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The total sway as a result of the displacements displacements is the sum of the individual sway angles. Therefore: ψ  AB ψ BC 

  = −2, 5 x 10 − − 3, 75 x 10 − = −6, 25 x 10 − − − − = 1,1111 x 10 + 1, 6667 x 10 = 2, 7778 x 10 3

3

3

3

3

-3

ψ  BD = 5,0 x 10

To determine the relative sway angles:

3

 

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M BA BA = 6 666,667 θ 

MBD

=

MBD

=

2 ⋅ EI 

( 2 ⋅θB L 2 ⋅ 200 ⋅ 50

5 M BD BD = 8 000 θ 

B

– 2 222,22 ψ  + 82,7313

+ θ − 3 ⋅ψ  ) + FEM D

BD

 

10 − ) ) ( 2 ⋅ θ − 3 ⋅ (ψ  + 5 x 10 3

B

B

– 12 000 ψ  - 60,00 

6666, 67 ⋅ θ ∑ M  = 0 ∴ 29 66 B

BD

B

−2 972, 22 ⋅ψ  + 116, 4813 = 0

For the second equation one must look at all the external forces that are applied to the structure.

(1)

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MBC MBD MDB

= + 58,418 kN.m = - 77,990 kN.m = - 61,804 kN.m