Slope‐Deflection Method, Lecture notes of Theory of Structures

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Slope‐Deflection Method

Theory of Structures‐II^ M Shahid Mehmood Department of Civil Engineering Swedish College^ of Engineering & Technology,

Wah Cantt

Slope‐Deflection Method •^ Previously we have discussed

Force/Flexibility methods

of analysis of statically indeterminate structures. • In force method, the unknown redundant forces aredetermined first by solving the structure’s compatibilityti^ th^ th^

h^ t^ i ti^ f th

equations; then other response characteristics of thestructure^ are^ evaluated^

by^ equilibrium^ equations

or

superpositionsuperposition. • An alternative approach can be used for analyzing is• An alternative approach can be used for analyzing istermed the displacement or stiffness method.

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Slope‐Deflection Method •^ This^ method^ takes^ into

account^ only^ the^ bending

deformations. • This^ method^ gives^ an^ understanding

of^ the^ Matrix‐

Stiffness^ Method,^ which

forms^ the^ basis^ of^ mostt ft tl^ d f^ t^ t^ l^ l^

i

computer software currently used for structural analysis.

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Slope‐Deflection Equations •^ When a continuous beam or a frame is subjected to external loads,internal moments generally develop at the ends of its individualinternal moments generally develop at the ends of its individualmembers.“ The slope‐deflection equations relate the moments at the ends ofthe member to the rotations and displacements of its end and theexternal loads applied to the member”external loads applied to the member. •^ Let us consider an arbitrary member AB of the continuous beam.y

5 A^ BL

P w A L B w M AB

M^ BA

-^ Double‐subscript notation is used for member end moments, withp^

the first subscript identifying the member end at which themoment acts and the second subscript indicating the other end ofthe memberthe member. • Mdenotes the moment at end A of the member AB.AB^ AB • MBAdenotes the moment at end B of the member AB.^

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Pw (^) M AB M^ BA L A^

B AB Un‐deformed position^

Tangent at ABA Elastic curveElastic curveB’ θA θB A’ Tangent at B

-^ θ&^ θdenote,^ respectively,A^ B^

the^ rotations^ of^ end^ A^ and^ B

with θ&^ θdenote,^ respectively,^ A^ B^

the^ rotations^ of^ end^ A^ and^ B^

with respect to the un‐deformed (horizontal) position of the member.

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Pw (^) M AB M^ BA L A^

B AB Un‐deformed position^

Tangent at ABA Elastic curveElastic curveB’Ψ θA ΔθB^ ΨA’ Cord Tangent at B

-^ Ψ^ denotes^ the^ rotation^ of

the^ member’s chord (straight lineΨ denotes the rotation of the member s chord (straight lineconnecting the deformed positions of the member ends) due tothe relative translation Δ.

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Pw (^) M AB M^ BA L A^

B AB Un‐deformed position^

Tangent at ABA Elastic curveElastic curveB’Ψ θA ΔθB^ ΨA’ Cord Tangent at B

-^ Since^ the^ deformations^ are

assumed^ to^ be^ small,^ the^

chord Since^ the^ deformations^ are^

assumed^ to^ be^ small,^ the^ chord rotation can be expressed as

∆^ (1) =^11^ L

ψ

Pw (^) M AB M^ BA L A^

B AB Un‐deformed position^

Tangent at A^ ΔBBAA Elastic curveElastic curveB’ ΨθA ΔθB ΨA’ Tangent at B Cord ΔAB • From figure we can see that

∆∆∆∆ ABBA^ (2) B LL^13

A

∆+∆∆+∆ ==θ θ

-^ By substituting^ Δ/L=Ψ^ into the preceding equation we have,

AB^ ( )

∆∆ BA θ θ

(3) L

∆∆ ABBA =−=−ψθψθ BAL • Δis tangential deviation of end B from the tangent to the elasticBA (^) curve at end A and Δis the tangential deviation of end A fromAB (^) the tangent to the elastic curve at end Bthe tangent to the elastic curve at end B.A di t th d t th th^ i

-^ According to the second‐moment area theorem, the expressionsfor the tangential deviations

Δand^ Δcan be obtained byAB^ BA^ summing the moments about the ends A and B, respectively, of thearea under M/EI diagram between the two ends.

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-^ Assuming that the member is prismatic (EI is constant along thelength of the member) we sum the moments of the area under theM/EI diagram about the ends B and A, respectively, to determineM/EI diagram about the ends B and A, respectively, to determinethe tangential deviations.

^  BAAB^ ∆ 

LLMLLg

M^^21

 ^

BAABB BA^

g

EI^

323222 gLML (4a)

M^ BBAAB ∆

(4a)

gBBAAB −−=∆ BA 63 EIEIEI  LLMLLM^21  LLMLLM^21   BAAB ++−=∆ g  AAB  EI^3232  22 LMLM

(4b)^16

22 gLMLM ABAAB ++−=∆ AB 36^ EIEIEI

-^ In which gand gare the moments about the ends B and A,B^ A^ respectively, of the area under the simple

‐beam bending moment diagram due to external loading (M

diagram).L^ diagram due to external loading (M

diagram).L^

-^ The^ three^ terms^ in^ equations

(^4 a^ &^4 b)^ represent^ the^ tangentialThe three terms in equations (^4 .a & 4.b) represent the tangentialdeviations due to MAB, M , and the external loading, actingBA^ separately on the member, with a negative term indicating that thedi i l d i^ i^ i^ i^ h^ di^ i^

i corresponding tangential deviation is in the direction opposite tothat shown on the elastic curve of the member.^2 LM^ AB^ EI^6 M^ AB

(^2) LM AB EI 3 17 M^ AB A^ Tangential deviation due to M

AB

-^ By substituting the expressions for

Δand^ Δinto Eq. 3, we haveBA^ AB^ (^

) 5

a

gLMLM BBAAB =−ψθ A

(^ ) (5b)

63^ gLMLM^ EILEIEI

a

ψθ AEILEIEI ABAAB^ ++−=−ψθ B • To express^ the^ member^ end^ moments^ in^ terms^ of

the^ end B^36 EILEIEI rotations, the chord rotation, and the external loading, we solveEq. 5 simultaneously for M^ and MBA. Rewriting Eq. 5a asAB^ gLMLM^^22 BABBA^ (^ )ψθ^ −−−=^2 AEILEIEI^33

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EILEIEI 33

-^ By substituting this equation into Eq. 5b and solving the resultingequation for M^ , we haveAB^ EI^

(^22) ( ) (^ )^ (

) agg

EI LL

M^

ABBA AB^

22 32 −+−+=ψθθ^2

and by substituting Eq. 6a into either Eq. 5a or 5b, we have^ EI^

(^22) ( ) (^ )^ (

) bgg

EI LL

M^

ABBA BA^

22 32 −+−+=ψθθ^2

-^ It indicates that the moments develop at the ends of a memberdepend on the rotations and translations of member’s ends as wellas on the external loading applied between the endsas on the external loading applied between the ends.

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