Slope Deflection Examples

20
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: ( ) 2 2 2 AB B FEM g g L = A ( ) 2 2 2 BA B A FEM g g L = Where: g B and g A are the moments of the bending moment diagrammes of the statically determinate beam about B and A respectively. 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. ( ) 2 2 2 AB B FEM g g L = A 2 2 2 2 3 2 3 AB L Wab b L L Wab a L FEM L L L + + = 2 2 2 2 3 AB Wab b L a L FEM L + = 2 2 2 3 AB Wab b a b a FEM L + + = 2 2 2 AB Wab FEM L = In a similar way FEM BA may be calculated. Slope-deflection Page 1 of 20 7/23/2003

Transcript of Slope Deflection Examples

Page 1: 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:

( )2

2 2AB BFEM g gL

= ⋅ − A

( )2

2 2BA B AFEM g gL

= − ⋅

Where: gB and gA are the moments of the bending moment diagrammes of the statically determinate beam about B and A respectively. 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.

( )2

2 2AB BFEM g gL

= ⋅ − A

2

2 22 3 2 3ABL Wab b L L Wab a LFEM

L LL + + = ⋅ ⋅ ⋅ − ⋅ ⋅

2

2 2 23AB

Wab b L a LFEML

⋅ + − =

2

2 23AB

Wab b a b aFEML

⋅ + + =

2

2

2AB

WabFEML

⋅=

In a similar way FEMBA may be calculated.

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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: θA, θB, θC = 0, ψAB = 0, ψBC = 0 We require two equations to solve the two unknown rotations:

0 0A ABM M= ∴ =∑

( )2 2 3AB A B ABEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + ABEM

( )2 124 8AB A BEIM θ θ⋅ ⋅

= ⋅ + +0 4

0,5 5AB A BM EI EIθ θ= ⋅ + ⋅ ⋅ +

0AM =∑ EI θA + 0,5 EI θB + 5 = 0 (1)

0 0B BA BCM M M= ∴ + =∑

( )2 2 3BA A B AB BAEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

( )2 124 8BA A BEIM θ θ⋅ ⋅

= + ⋅ −0 4

0,5 1 5BA A BM EI EIθ θ= ⋅ ⋅ + ⋅ ⋅ −

( )2 2 3BC B C BC BCEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )22 52

6 1BC BEIM θ⋅ ⋅

= ⋅ +62

0,66667 15BC BM θ= ⋅ + MBA + MBC = 0 0,5 EI θA + 1,66667 EI θB + 10 = 0 (2) Solve the unknowns: θA = - 2,35294/EI θB = - 5,29412/EI Calculate the values of the moments: MBA = 0,5 x –2,35294 – 5,29412 – 5 = -11,471 kN.m

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MBC = 0,6667 x – 5,29412 + 15 = + 11,471 kN.m

( )2 2 3CB B C BC CBEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

( )22 5

6 1CB BEIM θ⋅ −

= +6

2⋅

MCB = 0,3333 x –5,29412 – 15 = -16,675 kN.m Draw the bending moment diagramme.

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

We would like to eliminate θA from the equation as we know that MAB = 0.

( )2 2 3AB A B ABEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + ABEM

Solve for θA.

32 2 2 2

AB BA

FEM LEI

θ ψθ

⋅= − ⋅ − +

⋅AB

( )2 2 3BA A B AB BAEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

Replace θA in this equation.

32 12 32 2 2B AB

BA B AB BA ABEIM F

Lθ ψ

θ ψ⋅⋅ = ⋅ − + − ⋅ + − ⋅

EM FEM

( )3 12BA B AB BA AB

EIM FEML

θ ψ⋅= − + − ⋅ FEM

This equation may be used to reduce the number of unknown rotations. Solve the previous problem using the modified slope-deflection equation.

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The unknowns are as follows: θA use the modified slope-deflection equation, θB, θC = 0, ψAB = 0, ψBC = 0 The total number of unknowns is reduced to θB

0 0B BA BCM M M= ∴ + =∑

( )3 12BA B AB BA AB

EIM FEML

θ ψ⋅= − + − ⋅ FEM

( )3 10 4 1 10 44 8 2BA BEIM θ⋅ ⋅ = − −

8⋅

0,75 7,5BA AM EI θ= ⋅ ⋅ −

( )2 2 3BC B C BC BCEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )22 52

6 1BC BEIM θ⋅ ⋅

= ⋅ +62

0,66667 15BC BM θ= ⋅ + MBA + MBC = 0 1,41667 EI θB + 7,5 = 0 (1) θB = - 5,29412/EI Bending moments are as calculated previously. Example 2: Determine the bending moment diagramme of the following structure.

Unknowns: θA – use modified SD equation, θB, θC, θE – use modified SD equation, ΨAB = ΨBC = ΨCD = 0

0 0B BA BC BEM M M M= ∴ + + =∑

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( )3 12BA B AB BA AB

EIM FEML

θ ψ⋅= − + − ⋅ FEM

No Loads No FEM:

( )3 2 1,54BA B B

EIM EIθ θ⋅= = ⋅ ⋅

( )2 2 3BC B C BC BCEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )2 3 20 424 8BC B C

EIM θ θ⋅ ⋅= ⋅ + +

3,0 1,5 10BC B CM EI EIθ θ= ⋅ ⋅ + ⋅ ⋅ +

( )3 12BE B BE BE EB

EIM FEML

θ ψ⋅= − + − ⋅ FEM

( )3 1,03BE B BEIM EIθ θ⋅

= = ⋅ ⋅

B

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

5,5 1,5 10 0B CEI EIθ θ⋅ ⋅ + ⋅ ⋅ + = (1)

0 0C CB CDM M M= ∴ + =∑

( )2 2 3CB B C BC CBEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

( )2 3 20 424 8CB B C

EIM θ θ⋅ ⋅= + ⋅ −

1,5 3,0 10CB B CM EI EIθ θ= ⋅ ⋅ + ⋅ ⋅ −

( )2 2,0 3CD C D CD CDEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )2 2 2,0 0 3 0 04CD C

EIM θ⋅= ⋅ + − ⋅ +

2,0CD CM EI θ= ⋅ ⋅

0 1,5 3,0 10 2C B CM EI EI EIθ θ θ= ∴ ⋅ ⋅ + ⋅ ⋅ − + ⋅ ⋅ =∑ 0C 1,5 5 10 0B CEI EIθ θ⋅ ⋅ + ⋅ ⋅ − = (2) Solve for θA and θB.

2,57426

2,77228

B

C

EI

EI

θ

θ

−=

+=

2,574261,5 1,5 3,861 .BA BM EI EI k

EIθ −

= ⋅ ⋅ = ⋅ ⋅ = − N m

N mN m

N m

3,0 2,57426 1,5 2,77228 10 6,4355 .BCM k= ⋅ − + ⋅ + = + 1,0 2,57426 2,574 .BEM k= ⋅ − = − 1,5 2,57426 3,0 2,77228 10 5,545 .CBM k= ⋅ − + ⋅ − = −

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2,0 2,77228 5,545 .CDM k= ⋅ = N m

( )2 2,0 3 2,772 .DC C D CD DC CEIM FEM EI kN m

Lθ θ ψ θ⋅

= + ⋅ − ⋅ + = ⋅ =

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 s + r = 7 n = 4 2n = 8 2n – (s + r) = 1 One independent sway mechanism

Sway of the structure.

One assumes that the member BC does not deform as AE is so large that 0P LA E

⋅⇒

If this is true, BB’ must = CC’.

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But BB’ = 6 ψAB therefore ψCD = CC’/4 = 6 ψAB/4 = 1,5 ψAB. Call ψAB, - ψ. Unknown in this case: θA = 0 θB ? θC ? θD use the modified slope deflection equation. ψ ? We require three equations to solve the unknowns.

0BM =∑ 0CM =∑

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: 0Y =∑

0 0B BA BCM M M= ∴ + =∑

( )2 2 3BA A B AB BAEIM F

Lθ θ ψ⋅

= + ⋅ − + EM

( )2 22 3 ( )6 8BA BEIM θ ψ⋅ ⋅

= ⋅ − ⋅ − −0 6

( )0,6667 3 15BA BM EI θ ψ= ⋅ ⋅ + −

( )2 2 3BC B C BC BCEIM F

Lθ θ ψ⋅

= ⋅ + − + EM

( )2 1210 8BC B C

EIM θ θ⋅ ⋅= ⋅ + +

0 10

( )0,4 0,2 12,5BC B CM EI θ θ= ⋅ ⋅ + ⋅ +

0 (1,0667 0,2 ) 2,5 0B B CM EI θ θ ψ= ∴ ⋅ ⋅ + ⋅ + − =∑ (1)

0 0C CB CDM M M= ∴ + =∑

( )2 2 3CB B C BC CBEIM F

Lθ θ ψ⋅

= + ⋅ − + EM

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( )2 1210 8CB B C

EIM θ θ⋅ ⋅= + ⋅ −

0 10

( )0,2 0,4 12,5CB B CM EI θ θ= ⋅ ⋅ + ⋅ −

( )3 12CD C CD CD DC

EIM FEML

θ ψ⋅= − + − ⋅ FEM

( )3 ( 1,5 )4CD CEIM θ ψ⋅

= − − ⋅

( )0,75 1,125CD CM EI θ ψ= ⋅ ⋅ + ⋅

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

Take moments about B.

20 36

AB BAAB

M MV + + ⋅=

( )2 20 63 (0,3333 ) 156 8BA B BEIM EIθ ψ θ ψ⋅ ⋅

= + ⋅ + = ⋅ ⋅ + +

(0,3333 ) 15 (0,6667 ) 15 606

B BAB

EI EIV θ ψ θ ψ⋅ ⋅ + + + ⋅ ⋅ + − +=

2 106 6B

ABV EI θ ψ⋅ = ⋅ + +

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

0,75 1,1254

CDCV EI θ ψ⋅ + ⋅ = ⋅

0 20 AB DCY V V= ∴ + − − =∑ 0 ( 2 2,25 7,375 ) 120 0B CEI θ θ ψ⋅ − ⋅ − ⋅ − ⋅ + = (3)

Solve the unknowns: θB = -18,582 / EI θC = - 9,61572 / EI ψ = 24,24397 / EI Substitute in the equations for the moments: MAB = 33,051 kN.m MBA = -3,144 kN.m

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MBC = 3,144 kN.m MCB = - 20,063 kN.m MCD = 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 ψDB of the following structure.

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D is a fixed 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 fixed 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 O2 vertical to the direction of 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 ψ = the angle ψ. BB’ = 5 x ψBD

1 1

1

5' 0,68,3333

BDBC O B O C BD

O B

BBL

ψψ ψ ψ ψ

⋅= = = = = − ⋅

CC’ = 6,667 x ψBC = 4 x ψBD 4'

4BD

CE BDCE

CCL

ψψ ψ

⋅= = =

2 1

2

5'' 0,510

BDAB O B O A B

O B

CCL

ψDψ ψ ψ ψ

⋅= = = = = + ⋅

The direction of the angle is important. If it is clock-wise it is negative. ψBD as shown is negative so ψBC will be positive. Example: Determine the bending moment diagramme of the following sway structure.

Structure with sway 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 use modified slope-deflection equation θB ? θC use modified slope-deflection equation θB 0 θE 0 ψ ? 2 unknown – we require 2 equations Set ψDB = -ψ BB’ = 4 ψ

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1 1

1

' 4 0,58BC BO CO

O B

BBL

ψψ ψ ψ ⋅= = = = = ⋅ψ

CC’ = 10 ψBC = 5 ψ ' 5

5CECE

CCL

ψψ ψ⋅= − = − = −

0 0B BA BC BDM M M M= ∴ + + =∑

( )3 12BA B AB BA AB

EIM FEML

θ ψ⋅= − + − FEM

( )3 30 6 1 30 66 8 2BA BEIM θ⋅ ⋅ = − −

8⋅

0,5 33,75BA BM EI θ= ⋅ ⋅ −

( )3 12BC B BC BC CB

EIM FEML

θ ψ⋅= − + − FEM

( )36BC BEIM θ ψ⋅

= −

0,5 0,25BC BM EI EIθ ψ= ⋅ ⋅ − ⋅ ⋅

( )2 2 3BD B D BD BDEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )( )2 2 34BD BEIM θ ψ⋅

= ⋅ − ⋅ −

1,0 1,5BD BM EI EIθ ψ= ⋅ ⋅ + ⋅ ⋅

0 2 1,25 33,75B BM EI EIθ ψ= ∴ ⋅ ⋅ + ⋅ ⋅ − =∑ 0 (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 is known to be 0. The momentary centre of rotation, O1, is such a point. Determine the unknown forces in terms of the unknown rotations and translational angles.

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Member AB

0 6 30 3B AB BAM V M= ∴ ⋅ − ⋅ − =∑ 0 0,5 56,25

6B

ABEIV θ⋅ ⋅ +

=

Member BD

0 4B DB DB BDM V M M= ∴ ⋅ − − =∑ 0

( )2 2 3DB B D BD DBEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

I

0,5 1,5DB BM EI Eθ ψ= ⋅ ⋅ + ⋅ ⋅ 1,5 3

4 4DB BD B

DBM M EI EIV θ ψ+ ⋅ ⋅ + ⋅

= =⋅

0

Member CE

0 5C EC ECM V M= ∴ ⋅ − =∑

( )3 12EC E EC EC CE

EIM FEML

θ ψ⋅= − + − FEM

( )3 ( ) 0,65ECEIM EIψ ψ⋅

= − − = ⋅ ⋅

0,65 5EC

ECM EIV ψ⋅ ⋅

= =

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

10 4,5 8,7 33,75O BM EI EIθ ψ= ∴ ⋅ ⋅ + ⋅ ⋅ − =∑ 0 (2)

Solve the unknowns:

21,3535

7,16561B EI

EI

θ

ψ

=

−=

MBA = -23,073 kN.m MBC = +12,468 kN.m MBD = +10,605 kN.m MDB = -0,072 kN.m MEC = -4,299 kN.m

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Bending moment diagramme Calculate the bending moments and draw the bending moment diagramme of the following structure.

Change the nodes to hinges and calculate the number of independent sway mechanisms. s = 6 r = 6 (s + r) = 12 n = 7 2 n = 14 therefore 2 n – (s+r) = 2 with two independent sway mechanisms

We have three unknowns, namely θB, θD and ψ. We require three equations to solve these unknowns.

= ∴ + + =∑ 0 0B BA BC BFM M M M

( )3 12BA B BA BA AB

EIM FEML

θ ψ⋅= ⋅ − + − ⋅ FEM

( )2 23 2 6 5 1 6 50

5 12 2BA BEIM θ

⋅ ⋅= ⋅ − − − ⋅ +

12⋅

MBA = 1,2 EI θB – 18,75

( )3 12Bc B BC Bc CB

EIM FEML

θ ψ⋅= ⋅ − + − ⋅ FEM

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( )3 2 ( ) 0 03Bc B

EIM θ ψ⋅= ⋅ − − + −

MBC = 2 EI θB + 2 EI ψ

( )2 2 3BF B F BF BFEIM F

Lθ θ ψ⋅

= ⋅ + − + EM

( )2 2 0 04BF BEIM θ⋅

= ⋅ + − + 0

0

MBF = 1 EI θB

0 4,2 2 18,75B BM EI EIθ ψ= ∴ ⋅ ⋅ + ⋅ ⋅ − =∑ (1) 0 0D DC DE DGM M M M= ∴ + + =∑

( )3 12DC D DC DC CD

EIM FEML

θ ψ⋅= ⋅ − + − ⋅ FEM

( )3 23DC D

EIM θ ψ⋅= ⋅ −

MDC = 2 EI θD - 2 EI ψ MDE = + 10 x 2 = + 20

( )2 2 3DG D G DG DGEIM F

Lθ θ ψ⋅

= ⋅ + − + EM

( )2 24DG DEIM θ⋅

= ⋅

MDG = EI θD

0 3 2 20D DM EI EIθ ψ= ∴ ⋅ ⋅ − ⋅ ⋅ + =∑ 0 (2) Third equation may be obtained from the vertical equilibrium of node C

-VCB - VCD – 20 = 0

2 23 3

BC BCB

M EI EIV θ ψ− − ⋅ ⋅ − ⋅ ⋅= =

2 2

3 3DC D

CDM EI EIV θ ψ+ + ⋅ ⋅ − ⋅ ⋅

= =

-VCB - VCD – 20 = 0

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2 2 4 20 03

B DEI EI EIθ θ ψ+ ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅− =

+ 2 EI θB - 2 EI θD - 4 EI ψ = 60 (3) Solve the three simultaneous equations:

θB -6.0185/EI θD 8.0093/EI ψ 22.0139/EI

MBA = - 25,972 kN.m MBC = + 31,991 kN.m MBF = - 6,0185 kN.m MFB = - 3,009 kN.m MDC = - 28,009 kN.m MDG = + 8,009 kN.m MGD = + 4,005 kN.m

Example 2: 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 θB and ψ. The sway angles will consist of a known angle as a result of the displacement of D and the unknown, ψ. Determine the known angles for the 10 mm and 20 mm displacement individually and add them together. In order to do this the unknown sway must be prevented.

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For the 10 mm displacement, B drops vertically by 10 mm. The sway angles are thus equal to BB’/Length of the member:

30,010 2,5 104AB xψ −= − = −

30,010 1,1111 109BC xψ −= =

For the 20 mm displacement, B may only move vertically so that both ends of member BD move and in this way we will find a momentary centre of rotation.

30,20 5 104

OB BD OD

OD x

ψ ψ ψ

ψ −

= =

= =

Therefore, ψBD = 5,0 x 10-3 BB’ = ψOB x 3 m = 0,015 m

30,015 3,75 104AB xψ −= − = −

30,015 1,66667 109BC xψ −= =

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

3 32,5 10 3,75 10 6,25 10AB x xψ − −= − − = − 3x −

3x −

3 31,1111 10 1,6667 10 2,7778 10BC x xψ − −= + =

ψBD = 5,0 x 10-3 To determine the relative sway angles:

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Set ψBD = ψ, then BB’ = 5 ψ

11

' 5 0,756,6667AB O B

O B

BBL

ψψ ψ ψ⋅= = − = − = − ⋅

22

' 5 0,333315BC O B

O B

BBL

ψψ ψ ψ⋅= = + = + = + ⋅

Equations required to solve the unknowns:

0 0B BA BC BDM M M M= ∴ + + =∑

( )3 12BA B BA BA AB

EIM FEML

θ ψ⋅= ⋅ − + − ⋅ FEM

( )6 6

33 2 (200 10 50 10 ) ( 0,75 6,25 104BA B

x xM xθ ψ −⋅ ⋅ ⋅= ⋅ − − ⋅ −

MBA = 15 000 θB + 11 250 ψ + 93,75

( )3 12BC B BC BC CB

EIM FEML

θ ψ⋅= ⋅ − + − ⋅ FEM

( )2 2

33 2 (200 50) 10 9 1 10 9(0,3333 2,7778 109 1BC BM xθ ψ − ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ − ⋅ + + − ⋅ − 2 2 12

MBA = 6 666,667 θB – 2 222,22 ψ + 82,7313

( )2 2 3BD B D BD BDEIM F

Lθ θ ψ⋅

= ⋅ + − ⋅ + EM

( )32 200 50 2 3 ( 5 105BD BM xθ ψ −⋅ ⋅

= ⋅ − ⋅ + )

MBD = 8 000 θB – 12 000 ψ - 60,00

0 29 6666,67 2 972,22 116,4813 0B BM θ ψ= ∴ ⋅ − ⋅ + =∑ (1)

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

As it is very difficult to determine YDB take moments about a point where the moment of YDB = 0, i.e., O2. Take moments about B of the member AB:

15000 11250 93,75 3750 2812,5 23,43754

BA BAB B

AB

MVL

θ ψθ ψ

⋅ + ⋅ += = = ⋅ + ⋅ +

To determine the force VDC, take moments about B of the member BD:

BD DBDB

BD

M MVL+

=

( )2 2 3DB B D BD DBEIM F

Lθ θ ψ⋅

= + ⋅ − ⋅ + EM

( )32 200 50 3 ( 5 10 )5BD BM xθ ψ −⋅ ⋅

= − ⋅ +

MDB = 4 000 θB – 12 000 ψ - 60,00

12 000 24 000 240 2 400 4 800 245

BD DB BDB B

BD

M MVL

θ ψθ ψ

+ ⋅ − ⋅ −= = = ⋅ − ⋅ −

2

210 90 13 10

2O AB DB DBM V V M ⋅= ∴ ⋅ − ⋅ − − =∑ 0

20 750 θB + 96 562,5 ψ + 199,6875 = 0 (2) Solve the two simultaneous equations: θB = - 0,0040464 ψ = - 0,00119844 MBA = + 19,572 kN.m MBC = + 58,418 kN.m MBD = - 77,990 kN.m MDB = - 61,804 kN.m

Slope-deflection Page 19 of 20 7/23/2003

Page 20: Slope Deflection Examples

Slope-deflection Page 20 of 20 7/23/2003