Where: g
Where: gBBand gand gAA are the moments of the bending moment diagrammes of the statically determinate beamare the moments of the bending moment diagrammes of the statically determinate beam
about B and A respectively. about B and A respectively.
Example: Determine the fixed end moments of a beam with a point load. 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. Simply supported beam with bending moment diagramme. Centroid in accordance with standard tables.
( (
))
2 2 2 2 2 2 A ABB BB AA F FEEMM gg gg L L=
=
⋅ ⋅ −−
2 2 2 2 2 2 2 2 33 22 33 AB AB L W L Waabb bb LL L WL Waabb aa LL FEM FEM L L LL L L
+
+
++
=
=
⋅
⋅ ⋅
⋅ ⋅
⋅
−
− ⋅
⋅ ⋅⋅
2 2 2 2 22 22 3 3 AB AB W Waabb bb LL aa LL FEM FEM L L⋅
⋅
+
+ −
− −−
==
2 2 2 2 22 3 3 AB AB W Waabb bb aa bb aa FEM FEM L L⋅
⋅
+
+ +
+ −−
==
2 2 2 2 2 2 AB AB Wab Wab FEM FEM L L⋅⋅
==
In a similar way FEM
In a similar way FEMBABAmay be calculated.may be calculated.
S
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Use of slope-deflection equations:
Use of slope-deflection equations:
Example 1: Determine the bending moment diagramme of the following statically indeterminate beam. Example 1: Determine the bending moment diagramme of the following statically indeterminate beam.
The unknowns are as follows:
The unknowns are as follows:
θθ
AA,,θθ
BB,,θθ
CC= 0,= 0,ψ
ψ
ABAB= 0,= 0,ψ
ψ
BCBC = 0= 0We require two equations to solve the
We require two equations to solve the two unknown rotations:two unknown rotations: 0 0 00 A A AABB M M
=
= ∴
∴
M M==
∑
∑
( (
))
2 2 2 2 33 A ABB AA BB AABB AABB EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
))
2 2 1100 44 2 2 4 4 88 A ABB AA BB EI EI M M=
=
⋅ ⋅
⋅
⋅ +
θ θ+ ++
θ θ⋅⋅
0 0,,55 55 A ABB AA BB M M= ⋅ +
= ⋅
EEII θ θ+ ⋅
⋅ ⋅
EEII⋅ ++
θ θ 0 0 A A M M==
∑
∑
EIEI A A + 0,5 EI + 0,5 EI BB++ 55== 00 ((11))
0 0 00 B B BBAA BBCC M M
=
= ∴
∴ +
MM+ ==
M M∑
∑
( (
))
2 2 2 2 33 B BAA AA BB AABB BBAA EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
+
+ ⋅
⋅ −
− ⋅
⋅ ++
( (
))
2 2 1100 44 2 2 4 4 88 B BAA AA BB EI EI M M=
=
⋅ ⋅
θ θ+
+ ⋅
⋅ −−
θ θ⋅⋅
0 0,,55 11 55 B BAA AA BB M M= ⋅ ⋅
=
⋅ ⋅ +
EEII θ θ+ ⋅ ⋅
⋅ ⋅ −−
EEII θ θ( (
))
2 2 2 2 33 B BCC BB CC BBCC BBCC EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
))
2 2 2 2 55 66 2 2 6 6 1212 BC BC BB EI EI M M=
=
⋅ ⋅
⋅ ⋅
θ θ++
⋅⋅
0 0,,6666666677 1155 BC BC BB M M=
=
⋅ ⋅ ++
θ θ M M BABA+ M + M BC BC = 0 = 0 0,5 EI0,5 EI A A+ 1,66667 EI + 1,66667 EI BB++1100==00 ((22))
Solve the unknowns: Solve the unknowns:
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M M BC BC = = 00,,6666667 7 x x – – 55,,229944112 2 + + 1155 = = + + 1111,,44771 1 kkNN..mm
( (
))
2 2 2 2 33 C CBB BB CC BBCC CCBB EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
+
+ ⋅
⋅ −
− ⋅
⋅ ++
( ( ))
2 2 2 2 55 66 6 6 1212 CB CB BB EI EI M M=
=
⋅ ⋅
θ θ++
−
− ⋅⋅
M M CBCB= = 00,,3333333 3 x x ––55,,229944112 2 – – 1155 = = --1166,,66775 5 kkNN..mmDraw the bending moment diagramme. Draw the bending moment diagramme.
The Modified Slope-Deflection Equation with a Hinge at A:
The Modified Slope-Deflection Equation with a Hinge at A:
We would like to
We would like to eliminateeliminate
θθ
AA from the equation as we know that Mfrom the equation as we know that MABAB= 0.= 0.( (
))
2 2 2 2 33 A ABB AA BB AABB AABB EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
Solve for Solve forθθ
AA.. 3 3 2 2 22 22 22 A ABB BB AABB A A FEM FEM LL EI EI θ θ ψ ψ θ θ=
=
−
−
⋅
⋅
−
− ++
⋅⋅
⋅⋅
( (
))
2 2 2 2 33 B BAA AA BB AABB BBAA EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
+
+ ⋅
⋅ −
− ⋅
⋅ ++
ReplaceReplace
θθ
AA in this equation.in this equation.3 3 2 2 11 2 2 33 2 2 22 22 B B ABAB B BAA BB AABB BBAA AABB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ θ θ
⋅⋅
ψ ψ⋅⋅
=
=
⋅
⋅ −
− +
+
−
− ⋅
⋅ +
+
−
− ⋅⋅
( (
))
3 3 11 2 2 B BAA BB AABB BBAA AABB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
−
− +
+
−
− ⋅⋅
This equation may be used to reduce the number of unknown rotations. This equation may be used to reduce the number of unknown rotations.
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The unknowns are as follows:
The unknowns are as follows:
θθ
AAuse the modified slope-deflection equation,use the modified slope-deflection equation,θθ
BB,,θθ
CC = 0,= 0,ψ
ψ
ABAB= 0,= 0,ψ
ψ
BCBC = 0= 0The total number of unknowns is reduced to The total number of unknowns is reduced to
θθ
BB0 0 00 B B BBAA BBCC M M
=
= ∴
∴ +
MM+ ==
M M∑
∑
( (
))
3 3 11 2 2 B BAA BB AABB BBAA AABB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
−
− +
+
−
− ⋅⋅
( ( ))
3 3 1100 44 11 1100 44 4 4 88 22 88 BA BA BB EI EI M M=
=
⋅ ⋅
θ θ−
−
⋅ ⋅
−−
⋅⋅
0 0,,7755 77,,55 BA BA AA M M=
=
⋅
⋅ ⋅
EI EI θ⋅ −−
θ( (
))
2 2 2 2 33 B BCC BB CC BBCC BBCC EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
))
2 2 2 2 55 66 2 2 6 6 1212 BC BC BB EI EI M M=
=
⋅ ⋅
⋅ ⋅
θ θ++
⋅⋅
0 0,,6666666677 1155 BC BC BB M M=
=
⋅ ⋅ ++
θ θ M M BABA+ M + M BC BC = 0 = 0 1,41667 EI 1,41667 EI BB++77,,55==00 ((11)) B B= - 5,29412/EI = - 5,29412/EIBending moments are as calculated previously. Bending moments are as calculated previously.
Example 2: Determine the bending moment diagramme of the following structure. Example 2: Determine the bending moment diagramme of the following structure.
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( (
))
3 3 11 2 2 B BAA BB AABB BBAA AABB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
−
− +
+
−
− ⋅⋅
No Loads No FEM: No Loads No FEM:( ( ))
3 3 22 1 1,5,5 4 4 B BAA BB BB EI EI M M=
=
⋅⋅
θ θ=
= ⋅
⋅ ⋅⋅
EI EI θ θ( (
))
2 2 2 2 33 B BCC BB CC BBCC BBCC EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
))
2 2 33 2200 44 2 2 4 4 88 B BCC BB CC EI EI M M=
=
⋅ ⋅
⋅
⋅ +
θ θ+ ++
θ θ⋅⋅
3 3,,00 11,,55 1100 B BCC BB CC M M= ⋅
=
⋅ ⋅
EEII⋅ +
θ θ+ ⋅
⋅ ⋅ +
EIEI⋅ +
θ θ( (
))
3 3 11 2 2 B BEE BB BBEE BBEE EEBB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
−
− +
+
−
− ⋅⋅
( ( ))
3 3 1 1,0,0 3 3 B BEE BB BB EI EI M M=
=
⋅⋅
θ θ=
= ⋅
⋅ ⋅⋅
EI EI θ θ 0 0 11,,55 33,,00 1,,515 1100 11,,00 00 B B BB BB CC BB MM
= ∴
=
∴ ⋅
⋅ ⋅
EEII⋅ +
θθ+ ⋅
⋅ ⋅ + ⋅
EEII⋅ + ⋅ ⋅ + + ⋅ ⋅ =
θθ EIIE⋅ + + ⋅ ⋅ =
θθ EIEI θθ∑
∑
5 5,,55⋅ ⋅ +
⋅
EEII⋅ + ⋅
θ θ B B 11,,55⋅ ⋅ +
EIEI⋅ + ==
θ θ C C 1010 00 (1)(1) 0 0 00 C C CCBB CCDD M M=
= ∴
∴
MM+
+ ==
M M∑
∑
( (
))
2 2 2 2 33 C CBB BB CC BBCC CCBB EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
+
+ ⋅
⋅ −
− ⋅
⋅ ++
( (
))
2 2 33 2200 44 2 2 4 4 88 C CBB BB CC EI EI M M=
=
⋅ ⋅
θ θ+
+ ⋅
⋅ −−
θ θ⋅⋅
1 1,,55 33,,00 1100 C CBB BB CC M M= ⋅
=
⋅ ⋅
EEII⋅ +
θ θ+ ⋅
⋅ ⋅ −
EIEI⋅ −
θ θ( (
))
2 2 2 2,,00 33 C CDD CC DD CCDD CCDD EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
))
2 2 22 2 2,,00 0 30 3 00 00 4 4 CD CD C C EI EI M M=
=
⋅⋅
⋅
⋅ +
θ θ+ −
− ⋅
⋅ ++
2 2,0,0 CD CD C C M M=
=
⋅ ⋅ ⋅⋅
EI EI θ θ 0 0 11,,55 33,,00 1100 22 00 C C BB CC CC M M= ∴
=
∴ ⋅ ⋅
⋅ ⋅ +
EIIE θ θ+ ⋅ ⋅ − +
⋅ ⋅ − + ⋅ ⋅ =
EEII θ θ⋅ ⋅ =
EIEI θ θ∑
∑
1 1,,55⋅ ⋅ +
⋅ ⋅
EEII θ θ B B+ ⋅ ⋅
55⋅ ⋅ −
EIEI θ θ C C− ==
1010 00 (2)(2) Solve forSolve for
θθ
AAandandθθ
BB..2,57426 2,57426 2,77228 2,77228 B B C C EI EI EI EI θ θ θ θ
−−
==
++
==
2,57426 2,57426−−
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2 2,,00 22,,7777222288 55,,554455 .. CD CD M M
=
=
⋅ ⋅
==
kkNN mm( (
))
2 2 2 2,,00 33 22,,777722 .. D DCC CC DD CCDD DDCC CC EI EI M M FFEEMM EEII kkNN mm L L θθ θθ ψψ θθ⋅⋅
=
=
+
+ ⋅
⋅ −
− ⋅
⋅ +
+
=
= ⋅
⋅ ==
Sway Structures
Sway Structures
One of the ways in which can calculate whether a structure can sway and the number of independent sway 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 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.
instability. The degree of instability will also be the number of independent sway mechanisms. Example:
Example:
Structure with
Structure with bar-hinged elements:bar-hinged elements: s = 5 s = 5 r r== 77 ss++rr = = 1122 n n==66 22nn ==1122 2 2n n – – ((s s + + rr)) = = 00 NNo o iinnddeeppeennddeennt t sswwaay y mmeecchhaanniissmm
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Example: Example:
Structure with
Structure with bar-hinged elements:bar-hinged elements: s = 5 s = 5 r r== 66 ss++rr = = 1111 n n==66 22nn ==1122 2 2n n – – ((s s + + rr)) = = 11 OOnne e iinnddeeppeennddeennt t sswwaay y mmeecchhaanniissmm
Example: Determine the bending moment diagramme of the following structure: Example: Determine the bending moment diagramme of the following structure:
s = 3 s = 3 r r== 44 ss++rr ==77 n n==44 22nn ==88 2 2 (( )) 11 OO ii dd dd t t hh ii
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But BB’ = 6
But BB’ = 6
ψ
ψ
ABABthereforethereforeψ
ψ
CDCD= CC’/4 = 6= CC’/4 = 6ψ
ψ
ABAB/4 = 1,5/4 = 1,5ψ
ψ
ABAB..Call
Call
ψ
ψ
ABAB, -, -ψ
ψ
..Unknown in this case:
Unknown in this case:
θθ
AA= 0= 0θθ
BB??θθ
CC??θθ
DDuse the modified slope deflection equation.use the modified slope deflection equation.ψ
ψ
??We require three equations to solve the unknowns. We require three equations to solve the unknowns.
0 0 B B M M
==
∑
∑
0 0 C C M M==
∑
∑
For the third equation, one must investigate all the
For the third equation, one must investigate all the external forces that are applied to the external forces that are applied to the structure.structure.
The axial forces Y
The axial forces YABABand Yand YDCDCare usually difficult to determine, whereas the shear forces are usually difficult to determine, whereas the shear forces VVABABand Vand VDCDCcan becan be
calculated by taking moments about B of the member AB and C of the member CD respectively. calculated by taking moments about B of the member AB and C of the member CD respectively. The third equation is obtained by:
The third equation is obtained by:
∑
∑
Y Y==
00 0 0 00 B B BBAA BBCC M M=
= ∴
∴ +
MM+ ==
M M∑
∑
( (
))
2 2 2 2 33 B BAA AA BB AABB BBAA EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
+
+ ⋅
⋅ −
−
++
( (
))
2 2 2200 66 2 2 33 (( )) 6 6 88 BA BA BB EI EI M M=
=
⋅ ⋅
⋅
⋅ −
θ θ− ⋅
⋅ − −
− −
ψ ψ⋅⋅
( (
00,,66666677 33))
1155 BA BA BB M M=
= ⋅
EI EI⋅
⋅
⋅ +
θ θ+ −−
ψ ψThe world’s largest digital library
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( (
))
2 2 1100 1100 2 2 10 10 88 C CBB BB CC EI EI M M=
=
⋅ ⋅
θ θ+
+ ⋅
⋅ −−
θ θ⋅⋅
( (
00,,22 00,,44))
1122,,55 C CBB BB CC M M=
= ⋅
EI EI⋅ ⋅
⋅ +
θ θ+ ⋅
⋅ −−
θ θ( (
))
3 3 11 2 2 C CDD CC CCDD CCDD DDCC EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
−
− +
+
−
− ⋅⋅
( (
))
3 3 (( 11,,55 )) 4 4 CD CD C C EI EI M M=
=
⋅⋅
θ θ−
− −
− ⋅⋅
ψ ψ( (
00,,7755 11,,112255))
CD CD C C M M=
= ⋅
EI EI⋅
⋅
⋅ +
θ θ+
⋅⋅
ψ ψ( (
))
0 0 00,,22 11,,1155 1,,1112255 1122,,55 00 C C B B C C M M=
= ∴
∴ ⋅
EI EI⋅ ⋅
⋅ +
θ θ+ ⋅
⋅ +
θ θ+
⋅
⋅ −
ψ ψ−
==
∑
∑
(2)(2)Take moments about B. Take moments about B.
20 20 33 6 6 A AB B BBAA AB AB M M M M V V
==
+
+
+
+ ⋅⋅
( (
))
2 2 2200 66 3 3 ((00,,33333333 )) 1155 6 6 88 B BAA BB BB EI EI M M=
=
⋅ ⋅
θθ+ ⋅
+
⋅ +
ψψ+
⋅⋅
= ⋅
=
EI EI⋅
⋅
⋅ +
θθ+ ++
ψψ ((00,,33333333 )) 1155 ((00,,66666677 )) 1155 6600 6 6 B B BB AB AB E EII EEII V V==
⋅
⋅
⋅ +
⋅
θθ+ +
ψψ+ +
+ ⋅
⋅
⋅
⋅ +
θθ+ −
ψψ− ++
2 2 10 10 6 6 66 B B AB AB V V= ⋅
=
EI EI⋅ +
θ θ+
⋅⋅
ψ ψ
++
In a similar fThe world’s largest digital library
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Cancel Anytime. M MBCBC = 3,144 kN.m= 3,144 kN.m M MCBCB = - 20,063 kN.m= - 20,063 kN.m M MCDCD = 20,063 kN.m= 20,063 kN.m
Momentary Centre of Rotation
Momentary Centre of Rotation
When two points on a rigid body undergo a small displacement, the body rotates about a momentary centre 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:
of rotation and the following angles are equal:
Example:
Example: Determine the sway angles of the following structure in Determine the sway angles of the following structure in terms of the sway terms of the sway angleangle
ψ
ψ
DBDB of theof thefollowing structure. following structure.
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D is a f
D is a fixed point so that the point B may only move vertical to the member BD. B moves from B to ixed point so that the point B may only move vertical to the member BD. B moves from B to B’. In aB’. In a similar way E is a f
similar way E is a f ixed point and C can only move vertically to the member to C’. A may move horizontally.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
If one looks at the member AB both ends may move so we will find a momentary centre of rotation O22verticalvertical
to the direction of
to the direction of movemenmovement. Both ends of t. Both ends of member BC can move so we member BC can move so we will find a will find a momentary centre of momentary centre of rotation, O
rotation, O11vertical to the direction of movement of B and C.vertical to the direction of movement of B and C.
Because movements are small relative to the length of the member, the tan
Because movements are small relative to the length of the member, the tan
ψ
ψ
= the angle= the angleψ
ψ
.. BB’ = 5 x BB’ = 5 xψ
ψ
BDBD 1 1 11 1 1 5 5 '' 0,6 0,6 8,3333 8,3333 BD BD B BCC OOBB OO CC BBDD O O BB BB BB L L ψ ψ ψ ψ=
= =
ψψ= =
ψψ= =
=
⋅⋅
=
=
−
− ⋅⋅
ψψ CC’ = 6,667 x CC’ = 6,667 xψ
ψ
BCBC = 4 x= 4 xψ
ψ
BDBD 4 4 '' 4 4 BD BD C CEE BBDD CE CE CC CC L L ψ ψ ψ ψ=
=
=
=
⋅⋅
==
ψ ψThe world’s largest digital library
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Structure with sway
Structure with sway mechanismmechanism..
Determine the number of independent sway mechanisms: Determine the number of independent sway mechanisms: s s = = 44 r r = = 55 s s + + r r = = 99 n n = = 55 22n n = = 1100 2n –(s + r) = 1 2n –(s + r) = 1
1 independent sway mechanism! 1 independent sway mechanism! Unknowns
Unknowns
θθ
AA use modified slope-deflection equationuse modified slope-deflection equationθθ
BB ??θθ
CC use modified slope-deflection equationuse modified slope-deflection equationθθ
BB 00θθ
EE 00ψ
ψ
??2 unknown – we require 2 equations 2 unknown – we require 2 equations Set
Set
ψ
ψ
DBDB = -= -ψ
ψ
BB’ = 4 BB’ = 4
ψ
ψ
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( (
))
( (
))
2 2 2 2 33 4 4 BD BD BB EI EI M M=
=
⋅⋅
⋅
⋅ −
θ θ− ⋅
⋅ −−
ψ ψ 1 1,,00 11,,55 BD BD BB M M=
= ⋅
⋅ ⋅ + ⋅ ⋅
EEII⋅ + ⋅ ⋅
θ θ EEII ψ ψ 0 0 22 11,,2255 3333,,7755 00 B B BB M M=
= ∴
∴ ⋅
⋅ ⋅
EIIE⋅ +
θ θ+ ⋅
⋅ ⋅
EEII⋅ −
ψ ψ−
==
∑
∑
(1)(1)To determine the second equation one must view all the external forces on the structure: To determine the second equation one must view all the external forces on the structure:
As it is difficult to
As it is difficult to determine YDB and YEC we will take moments about a point where their momedetermine YDB and YEC we will take moments about a point where their moment is knownnt is known to be 0.
to be 0. The momentary centre of rotation, OThe momentary centre of rotation, O11, is such a point., is such a point.
Determine the unknown forces in terms of the unknown rotations and translational angles. Determine the unknown forces in terms of the unknown rotations and translational angles.
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( (
))
3 3 (( )) 00,,66 5 5 EC EC EI EI M M=
=
⋅⋅
−
− − =
− = ⋅
ψ ψ⋅ ⋅⋅
EI EI ψ ψ 0,6 0,6 5 5 55 EC EC EC EC M M EI EI V V=
=
==
⋅ ⋅ ⋅⋅
ψ ψTake moments about the momentary centre of rotation: Take moments about the momentary centre of rotation: V
VABABx 6 + Vx 6 + VDBDBx 12 + Vx 12 + VECECx 15 – 30 x 3 – Mx 15 – 30 x 3 – MDBDB – M– MECEC = 0= 0
0
0,,55
⋅ ⋅ +
⋅
EEII⋅ +
θθB B 5566,,2255+
+ ⋅
4,,545⋅ ⋅ +
EIIE⋅ + ⋅
θθB B 99⋅ ⋅
EEII⋅ +
ψψ+ ⋅
1,,818⋅ ⋅
EIIE⋅ −
ψψ− −
9090 0− ⋅
0,,55⋅ ⋅ − ⋅ ⋅
EIIE⋅ − ⋅ ⋅ −
θθBB 11,,55 EEII ψψ− ⋅
0,,606⋅ ⋅
EIEI⋅ ==
ψψ 001 1 00 44,,55 88,,77 3333,,7755 00 O O BB M M
=
= ∴
∴ ⋅
⋅ ⋅
EIIE⋅ +
θ θ+ ⋅
⋅ ⋅
EEII⋅ −
ψ ψ−
==
∑
∑
(2)(2)Solve the unknowns: Solve the unknowns:
21,3535 21,3535 7,16561 7,16561 B B EI EI EI EI θ θ ψ ψ
==
−−
==
M MBABA = -23,073 kN.m= -23,073 kN.m M MBCBC = +12,468 kN.m= +12,468 kN.m M MBDBD = +10,605 kN.m= +10,605 kN.m M MDBDB = -0,072 kN.m= -0,072 kN.m M MECEC = -4,299 kN.m= -4,299 kN.mThe world’s largest digital library
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We have three unknowns, namely
We have three unknowns, namely
θθ
BB,,θθ
DD andandψ
ψ
. We require three equations to solve . We require three equations to solve these unknowns.these unknowns.=
= ∴
∴ + +
+ + ==
∑
∑
MMBB 00 MMBBAA MMBBCC MMBFBF 00( (
))
3 3 11 2 2 B BAA BB BBAA BBAA AABB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
⋅
⋅ −
− +
+
−
− ⋅⋅
( (
))
2 2 22 3 3 22 66 55 11 66 55 0 0 5 5 1122 22 1122 BA BA BB EI EI M M=
=
⋅ ⋅
⋅ −
⋅ − −
θ θ−
⋅ ⋅
−
− ⋅
⋅ ++
⋅⋅
M M BABA= 1,2 EI = 1,2 EI BB– 18,75 – 18,75( (
))
3 3 11 2 2 B Bcc BB BBCC BBcc CCBB EI EI M M FFEEMM FFEEMM L L θ θ ψ ψ⋅⋅
=
=
⋅
⋅ −
− +
+
−
− ⋅⋅
( (
))
3 3 22 (( )) 00 00 3 3 Bc Bc BB EI EI M M=
=
⋅⋅
⋅
⋅ −
θ θ− − +
− + −−
ψ ψ MThe world’s largest digital library
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Cancel Anytime. M M DG DG = = EI EI DD 0 0 33 22 2200 00 D D DD M M
= ∴
= ∴ ⋅
⋅ ⋅
EIIE⋅ −
θ θ− ⋅
⋅ ⋅
EEII⋅ +
ψ ψ+ ==
∑
∑
(2)(2)Third equation may be obtained from the vertical equilibrium of node C Third equation may be obtained from the vertical equilibrium of node C
-V -VCBCB- V- VCDCD– 20 = 0– 20 = 0 2 2 22 3 3 33 BC BC BB CB CB M M EEII EEII V V
=
=
−−
==
−
− ⋅
⋅ ⋅
⋅ −
θ θ− ⋅
⋅ ⋅⋅
ψ ψ 2 2 22 3 3 33 DC DC DD CD CD M M EEII EEII V V=
=
++
==
+
+ ⋅
⋅ ⋅
⋅ −
θ θ− ⋅
⋅ ⋅⋅
ψ ψ -V -VCBCB- V- VCDCD– 20 = 0– 20 = 0 2 2 22 44 20 20 00 3 3 B B DD E EII θ θ EEII θ θ EEII ψ ψ+
+ ⋅
⋅ ⋅
⋅ −
− ⋅
⋅ ⋅
⋅ −
− ⋅
⋅ ⋅⋅
−
−
==
+ 2 EI+ 2 EI BB- 2 EI - 2 EI DD- 4 EI - 4 EI = 60 = 60 (3)(3)
Solve the three simultaneous equations: Solve the three simultaneous equations:
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The support D of the structure undergoes the following displacement, 10 mm vertically down and 20 mm 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
horizontally to the left. E = 200 GPa and I = 50 x 10-6-6mm44..
If one determines the number of independent sway mechanisms we see that there is one. The unknowns are If one determines the number of independent sway mechanisms we see that there is one. The unknowns are thus
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The total sway as
The total sway as a result of a result of the displacementhe displacements is ts is the sum of the the sum of the individual sway angles. Therefore:individual sway angles. Therefore:
3 3 3 3 33 2 2,,55 1100 33,,7755 1100 66,,2255 1100 AB AB xx xx xx ψ ψ
=
=
−
−
− −−
−
− −=
=
−−
−− 3 3 3 3 33 1 1,,11111111 1100 11,,66666677 1100 22,,77777788 1100 BC BC xx xx xx ψ ψ=
=
− −+
+
− −==
−− ψ ψ BDBD= 5,0 x 10= 5,0 x 10-3-3To determine the relative sway angles: To determine the relative sway angles:
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Cancel Anytime. M M BABA= 6 666,667 = 6 666,667 θ θ BB– 2 222,22 ψ – 2 222,22 ψ + 82,7313+ 82,7313
( (
))
2 2 2 2 33 B BDD BB DD BBDD BBDD EI EI M M FFEEMM L L θ θ θ θ ψ ψ⋅⋅
=
=
⋅
⋅ +
+ −
− ⋅
⋅ ++
( (
33))
2 2 220000 5500 2 2 33 (( 55 1100 )) 5 5 BD BD BB M M=
=
⋅ ⋅
⋅⋅
⋅ −
⋅
θ θ− ⋅
⋅ ++
ψ ψ x x −− M M BDBD= 8 000 = 8 000 θ θ BB– 12 000 – 12 000 ψ ψ - 60,00 - 60,00 0 0 229 69 6666666,,6677 22 997722,,2222 111166,,44881133 00 B B BB M M=
=
∴
∴
⋅
⋅
θ θ−
−
⋅
⋅
ψ ψ+
+
==
∑
∑
(1)(1)For the second equation one must look at all the external forces that are applied to the structure. For the second equation one must look at all the external forces that are applied to the structure.
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Cancel Anytime. M MBCBC = + 58,418 kN.m= + 58,418 kN.m M MBDBD = - 77,990 kN.m= - 77,990 kN.m M MDBDB = - 61,804 kN.m= - 61,804 kN.m
