Symmetrical Structures The following structures are symmetrical, with symmetrical loading, and as a result of the symmetry one would like to analyze only one half of the structure.

The same structures could have anti-symmetrical loading:

In both cases the symmetry of the structure can be used and the analysis can be restricted to a half structure. Sway structures will still stay sway structures. The sway may occur in the symmetrical loading case or in the anti-symmetrical loading case. Slope deflection equations or moment-distribution may be used to obtain the bending moments. To decide how the analysis can be restricted to one half only, it is necessary to investigate the effect of symmetry and anti-symmetry on a member AB.

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Note that in the symmetrical case θB is equal in size to θA but has the opposite sign, whereas in the anti-symmetrical case the signs are the same. The moments on either side of the axis will have the opposite sign in the case of symmetry and the same sign in the case of anti-symmetry. This can be seen from the slope-deflection equations: Symmetry:

( ) IABABBA

AB

ABAB M

LEIM +−+= ψθθ 322

( ) IABAA

AB

ABAB M

LEIM +−−= 022

θθ

( ) IABA

AB

ABAB M

LEIM += θ

2

Note that with symmetry the fixed end moment at B is the negative of the fixed end moment at A.

( ) IBAABBA

AB

ABBA M

LEIM +−+= ψθθ 322

( ) IABAA

AB

ABBA M

LEIM −−= 022

θθ

( ) IABA

AB

ABBA M

LEIM −−= θ

2

Note that in this case the moments have the opposite sign. Anti-symmetry: With anti-symmetry the rotation at B is equal in sign and size to the rotation at A.

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( ) IABABBA

AB

ABAB M

LEIM +−+= ψθθ 32

2

( ) IABAA

AB

ABAB M

LEIM +−+= 02

2θθ

( ) IABA

AB

ABAB M

LEIM += θ

6

The fixed end moment at B is equal in size and sign in the anti-symmetrical loading case.

( ) IBAABBA

AB

ABBA M

LEIM +−+= ψθθ 322

( ) IABAA

AB

ABBA M

LEIM +−+= 022

θθ

( ) IABA

AB

ABBA M

LEIM += θ

6

Note that in this case the moments have the same sign. Example 1: Determine the bending moment diagramme of the following symmetrical structure. Use the slope-deflection equations.

Unknowns are θB and θC. Use the symmetry of the structure and the loading, i.e., θC = -θB. Sum of the moments are B are equal to 0. MBA + MBC = 0

( ) IBAABBA

AB

ABBA M

LEIM +−+= ψθθ 322

( )12

8*60208

2 2

−−+= BBAEIM θ

325,0 −= BBA EIM θ

( ) 025

2+−= BBBC

EIM θθ

BBC EIM θ4,0=

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MBA + MBC = 0 0,5EIθB – 32 + 0,4EIθB = 0 θB = 35,5556/EI

( ) IABABBA

AB

ABAB M

LEIM +−+= ψθθ 32

2

( )12

8*60/5556,3508

2 2

+−+= EIEIM AB

MAB = 40,889 kN.m MDC = - 40,889 kN.m

32/5556,35*5,0 −= EIEIM BA MBA = -14,222 kN.m MCB = 14,222 kN.m

Structure with Anti-symmetrical loading

In this case the angle θC has the same sign as θB. Sum of the moments are B are equal to 0. MBA + MBC = 0

( ) IBAABBA

AB

ABBA M

LEIM +−+= ψθθ 322

( )12

8*60208

2 2

−−+= BBAEIM θ

325,0 −= BBA EIM θ

( ) 025

2++= BBBC

EIM θθ

BBC EIM θ2,1= MBA + MBC = 0

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0,5EIθB – 32 + 1,2EIθB = 0 θB = 18,8235/EI

( ) IABABBA

AB

ABAB M

LEIM +−+= ψθθ 322

( )12

8*60/8235,1808

2 2

+−+= EIEIM AB

MAB = 36,706 kN.m MDC = 36,706 kN.m

32/8235,18*5,0 −= EIEIM BA MBA = -22,588 kN.m MCB = - 22,588 kN.m

Moment distribution applied to symmetrical structures. The stiffness of the member is defined as the moment that is required for unit rotation at the node.

For symmetrical case:

( ) IABABBA

AB

ABAB M

LEIk +−+= ψθθ 322

1,,0,0 =−=== AABABIABM θθθψ

AB

ABAB L

EIk 2=

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For the anti-symmetrical loading case:

1,,0,0 ==== AABABIABM θθθψ

AB

ABAB L

EIk 6=

Use example above. For symmetrical loading case:

mkNwLMM IABAB .32

12

20 ===

mkNwLMM IBABA .32

12

20 −=−==

5,08

44===

EILEIk

AB

ABBA 5556,0

9,05,0===

∑kkD BA

BA

4,05

22===

EILEIk

BC

BCBC 4444,0

9,04,0===

∑kkD BC

BC

MAB MBA MBC 0,5556 0,4444 +32,00 -32,00 +8,890 +17,779 +14,221 +40,89 -14,221 +14,221 Use example above. For anti-symmetrical loading case:

mkNwLMM IABAB .32

12

20 ===

mkNwLMM IBABA .32

12

20 −=−==

5,08

44===

EILEIk

AB

ABBA 2941,0

7,15,0===

∑kkD BA

BA

2,15

66===

EILEIk

BC

BCBC 7059,0

7,12,1===

∑kkD BC

BC

MAB MBA MBC 0,2941 0,7059 +32,00 -32,00 +4,706 +9,411 +22,589 +36,706 -22,589 +22,589

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Loading Most loading cases can be converted into a symmetrical loading case and an anti-symmetrical case. The following are a few examples of this type of conversion.

Example 2: Determine the bending moment diagram of the following sway-structure.

Determine the symmetrical and anti-symmetrical loading cases. Symmetrical

Anti-symmetrical

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The deformed shape of the structure for the two loading cases is as follows:

The equivalent half structures for the two loading cases are as follows: Symmetrical loading:

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There is no sway with this structure. Anti-symmetrical loading:

In this case there will be sway. Use moment-distribution to solve for the bending moments. Distribution at B

EIEILEIk

AB

ABBA 125,1

83*33

=== 3856,0918,2125,1

===∑kkD BA

BA

EIEILEIk

BC

BCBC 60,0

103*22

=== 2056,0918,260,0

===∑kkD BC

BC

EIEILEIk

BE

BEBE 193,1

708,62*44

=== 4088,0918,2193,1

===∑kkD BE

BE

Initial moment

mkNLwMM IBCBC .667,41

1210*5

12* 22

0 ====

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MEB MBE MBA MBC 0,4088 0,3856 0,2056 +41,667 -8,517 -17,033 -16,067 -8,567 -8,517 -17,033 -16.067 33,100 Anti-symmetrical loading: Distribution at B

EIEILEIk

AB

ABBA 125,1

83*33

=== 2732,0118,4125,1

===∑kkD BA

BA

EIEILEIk

BC

BCBC 80,1

103*66

=== 4371,0118,480,1

===∑kkD BC

BC

EIEILEIk

BE

BEBE 193,1

708,62*44

=== 2897,0118,4193,1

===∑kkD BE

BE

Initial moment

mkNLwLwMMM ICB

IBCBC .625,15

245*5

125*5

24*

12*

21 2222

0 =+=+=−=

MEB MBE MBA MBC 0,2897 0,2732 0,4371 +15,625 -2,263 -4,526 -4,269 -6,830 -2,263 -4,526 -4,269 +8,795

Force, P, that prevents sway: Take moments about E. VAB * 5 + P * 6 –MEB + 5 * 5*5,5 –VCB * 8 = 0

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kNMV BAAB 5336,0

8269,4

8−=

−==

kNLwM

V BCCB 741,10

52

5*5795,85

2* 22

=+−

=+−

=

-0,5336 * 5 + P * 6 +2,263 + 137,5 –10,741 * 8 = 0 P = -8,5278 kN Arbitrary sway of the anti-symmetrical case.

Initial moments as a result of arbitrary sway MBA:MBC:MBE:MEB

BEBE

BEBE

BE

BEBC

BC

BCAB

AB

AB

LEI

LEI

LEI

LEI ψψψψ 6:6:3:3

−−−−

ψψψψ −−−−−−− *708,62*6:*

708,62*6:6,0*

53*3:375,0*

83*3 EIEIEIEI

0,4219 : -1,080 : 1,789 : 1,789 8,438 : -21,60 : 35,78 : 35,78

Symmetrical Structures Page 11 2/14/02

MEB MBE MBA MBC 0,2897 0,2732 0,4371 35,78 35,78 8,438 -21,60 -3,276 -6,552 -6,179 -9,886 32,504 29,228 2,259 -31,486 Force, Q, that induces sway: Take moments about E. VAB * 5 + Q * 6 –MEB–VCB * 8 = 0

kNMV BAAB 2824,0

8259,2

8===

kNMV BCCB 297,6

5486,31

5==

−=

0,2824 * 5 +Q * 6 – 32,504 - 6,297 * 8 = 0 Q = 13,578 kN P + xQ = 0

62806,0578,13

5278,8==−=

QPx

Final moments are the sum of Msymmetrical + Msway prevented + x * Marbitrary sway MEB MBE MBA MBC MCB MCD MCF MFC -8,517 -17,033 -16.067 +33,100 -33,100 +16,067 +17,033 +8,517 -2,263 -4,526 -4,269 +8,795 +8,795 -4,269 -4,526 -2,263 20,414 +18,357 +1,419 -19,775 -19,775 +1,419 +18,357 +20,414 +9,634 -3,202 -18,917 +22,120 -44,080 +13,217 +30,864 +26,668

Symmetrical Structures Page 12 2/14/02

Example 3: Use the symmetry of the structure and determine the bending moment diagram.

Change all the nodes to hinges. s = 4, r = 4 with s + r =8 n = 5 therefore 2 x n = 10. There are thus two independent sway mechanisms. For the symmetrical loading case the half structure will be represented by the following: s = 2, r = 3 with s + r =5 n = 3 therefore 2 x n = 6. There is thus one independent sway mechanisms.

Use the slope deflection method to solve the unknowns, θB and ψ.

( ) EIEIM BBAB )2,14,0(35

2 ψθψθ −=−=

( ) EIEIM BBBA )2,18,0(325

2 ψθψθ −=−=

( ) 333,53)677,18944,0(12

8*1025,1*329443,82*2 2

++=+−−= EIEIM BBBC ψθψθ

( ) 333,53)677,14472,0(12

8*1025,1*39443,82*2 2

−+=−−−= EIEIM BBCB ψθψθ

Use the equilibrium of the node B

Symmetrical Structures Page 13 2/14/02

Σ MΒ = 0 MBA + MBC = 0 (1,6944 θB + 0,477 ψ)EI = -53,333 (1) Take moments about the momentary centre of rotation. - MAB – MCB + VAB*9 + 10*8*4 = 0 (2)

5)4,22,1(

5EIMM

V BBAABAB

ψθ −=

+=

Equation (2) then becomes (-0,4θB + 1,2ψ)EI + (-0,4472θB – 1,677ψ)EI + 53,333 + 9/5*(1,2θB – 2,4ψ)EI + 320 = 0 + 1,3128θB –4,797ψ = -373,333/EI (3) Use equations (1) and (3) and solve the unknown rotations. θB = -49,5666/EI ψ = 64,2614/EI

mkNEIM BAB .940,96)2,14,0( −=−= ψθ mkNEIM BBA .767,116)2,18,0( −=−= ψθ

mkNEIM BBC .767,116333,53)677,18944,0( =++= ψθ mkNEIM BCB .267,32333,53)677,14472,0( =−+= ψθ

Anti-symmetrical loading case. The structure can be simplified as follows: Change nodes to hinges, then S = 2, r = 3 s + r = 5 N = 3 2 n = 6 we thus have one independent sway mechanism.

Unknowns are θB and ψ.

Symmetrical Structures Page 14 2/14/02

( ) EIEIM BBAB )2,14,0(*35

2 ψθψθ +=−−=

( ) EIEIM BBBA )2,18,0(*325

2 ψθψθ +=−−=

( ) 000,80)67082,0(12

8*102

112

8*100*39443,82*3 22

+=

−−+−= EIEIM BBBC θθ

Σ MB = 0 MBA + MBC = 0 (1,47082 θB + 1,2 ψ)EI = - 80,00 (4) Σ forces in X – direction = 0 VAB = 0

05

)4,22,1(5

=+

=+

=EIMM

V BBAABAB

ψθ (5)

Solve for θB and ψ θB = -91,8674/EI ψ = 45,93372/EI

mkNEIM BAB .374,18)2,14,0( =+= ψθ 374,18)2,18,0( −=+= EIM BBA ψθ

374,18000,80)67082,0( =+= EIM BBC θ The two effects must now be added together: Type MAB MBA MBC MCB MCD MDC MDE MED Symm -96,94 -116,77 +116,77 +32,27 -32,27 -116,77 +116,77 +96,94 Anti +18,37 -18,37 +18,37 0 0 +18,37 -18,37 +18,37 Final -78,57 -135,14 135,14 32,27 -32,27 -98,40 +98,40 +115,31

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