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
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
B
+ 10 = 0
(2)
Solve the unknowns: = - 2,35294/EI B = - 5,29412/EI
A
Calculate the values of the moments: M BA 0,5 x –2, –2,35 3529 294 4 – 5,29 5,294 412 – 5 BA = 0,5
Slope-deflection
= -11 -11,4 ,471 71 kN.m kN.m
Page 2 of 20
3/22/2010
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. Solve the previous problem using the modified slope-deflection equation.
Slope-deflection
Page 3 of 20
3/22/2010
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,
ψ BC = 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
(1)
= - 5,29412/EI
Bending moments are as calculated previously. Example 2: Determine the bending moment diagramme of the following structure.
Unknowns: θ
∑M
B
=0 ∴
Slope-deflection
A
– use modified SD equation, equation,
M BA
+M +M BC
BE
θ
,
B
θ
,θ
C
E
– use modified SD equation,
Ψ AB = Ψ BC = Ψ CD = 0
= 0
Page 4 of 20
3/22/2010
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
=
MBA MBC MBE MCB
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
(2)
and θ B.
−2,57426 EI +2,77228 EI
= 1, 5 ⋅ EI ⋅ θ = 1, 5 ⋅ EI ⋅ B
−2,57426
= −3, 861 kN.m
EI = 3, 0 ⋅ −2, 57426 + 1, 5 ⋅ 2, 77228 + 10
= +6, 4355 kN. m
= 1, 0 ⋅ −2, 57426 = −2, 574 kN.m = 1, 5 ⋅ −2, 57426 + 3, 0 ⋅ 2, 77228 − 10 = −5, 545 kN.m
Slope-deflection
Page 5 of 20
3/22/2010
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
Slope-deflection
Page 6 of 20
3/22/2010
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 2n – (s + r)
=1
s+r =7 2n =8 One independent sway mechanism
Sway of the structure. One assumes that the member BC does not deform as AE is so large that
P ⋅L A ⋅ E
⇒0
If this is true, BB’ must = CC’.
Slope-deflection
Page 7 of 20
3/22/2010
But BB’ = 6 ψ AB therefore ψ CD = CC’/4 = 6 ψ AB/4 = 1,5 ψ AB. Call ψ AB, - ψ . Unknown in this case:
θ A=0 θ B? θ C? θ D use the modified slope deflection equation. ψ ?
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
=
∑Y = 0
= 0 ∴ M + M = 0 BA
BC
2 ⋅ EI
( θ + 2 ⋅ θ − 3ψ ) + FEM A
L 2 ⋅ EI
B
AB
MBA
20 ⋅ 6 6 8 = EI ⋅ ( 0, 6667 ⋅ θ B + 3ψ ) − 15
MBC
=
M BC
=
( 2 ⋅ θ − 3 ⋅ ( −ψ ) ) − B
2 ⋅ EI
( 2 ⋅ θ + θ − 3ψ ) + FEM B
L 2 ⋅ EI
C
BC
BC
10 ⋅ 10 10 8 = EI ⋅ ( 0, 4 ⋅ θ B + 0, 2 ⋅ θ C ) + 1 2, 5
MBC
∑M
( 2 ⋅ θ + θ ) + B
C
B
= 0 ∴ EI ⋅ (1, 0667 ⋅ θ + 0, 2 ⋅ θ + ψ ) − 2, 5 = 0
C
= 0 ∴ M + M = 0
∑M MCB
BA
=
B
CB
CD
2 ⋅ EI L
( θ + 2 ⋅ θ − 3ψ ) + FEM
Slope-deflection
B
(1)
C
C
BC
CB
Page 8 of 20
3/22/2010
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
(2)
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: 0, 75 ⋅ θ + 1,125 ⋅ψ = EI ⋅ 4 ∑Y = 0 ∴ +20 − V − V = 0 EI ⋅ ( −2 ⋅ θ − 2, 25 ⋅ θ − 7, 375 ⋅ψ ) + 120 = 0 C
VDC
AB
B
DC
(3)
C
Solve the unknowns: = -18,582 / EI θB = - 9,61572 / EI θC = 24,24397 / EI ψ Substitute in the equations for the moments: MAB MBA
= 33,051 kN.m = -3,144 kN.m
Slope-deflection
Page 9 of 20
3/22/2010
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.
Slope-deflection
Page 10 of 20
3/22/2010
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 ψ
=
AB
ψ
=
2O B
ψ
=
1O A
5 ⋅ψ BD 8,3333
= −0,6 ⋅ψ
BD
ψ = the angle ψ .
= 4 x ψ BD
= ψ
BD
CC '' LO B 2
=
5 ⋅ψ BD 10
= +
0,5 ⋅ψ
BD
The direction of the angle is important. If it is clock-wise it is negative. ψ BD as shown is negative so be positive.
ψ BC will
Example: Determine the bending moment diagramme of the following sway structure.
Slope-deflection
Page 11 of 20
3/22/2010
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!
θA θB θC θB θE ψ
Unknowns
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 ψ BC
=ψ
BO1
=ψ
=
CO1
BB ' LO B
4 ⋅ψ 8
=
1
CC’ = 10 ψ BC = 5 ψ 5 ⋅ψ CC ' ψ CE = − =− 5 LCE
∑M
B
MBA
=
M BA
=
= −ψ
= 0 ∴M + M + M BA
= 0,5 ⋅ψ
BC BC
BD BD
= 0
3 ⋅ EI
1
( θ − ψ ) + FEM − FEM B
L 3 ⋅ EI
AB
( θ ) − B
MBA
6 = 0, 5 ⋅ EI ⋅ θ B
MBC
=
M BC
=
BA
30 ⋅ 6
2 1 30 ⋅ 6
−
8 2 − 33, 75
8
3 ⋅ EI
( θ − ψ ) + FEM
L 3 ⋅ EI
B
BC
BC
−
MBC
( θ B − ψ ) 6 = 0, 5 ⋅ EI ⋅ θ B − 0, 25 ⋅ EI ⋅ψ
MBD
=
AB
1 FEMCB 2
2 ⋅ EI L
( 2 ⋅ θ + θ − 3 ⋅ψ ) + FEM
Slope-deflection
B
D
BD
BD
Page 12 of 20
3/22/2010
2 ⋅ EI ( 2 ⋅ θ B − 3 ⋅ ( −ψ ) ) 4 = 1, 0 ⋅ EI ⋅ θ B + 1, 5 ⋅ EI ⋅ψ
=
M BD MBD
∑M
= 0 ∴ 2 ⋅ EI ⋅ θ + 1, 25 ⋅ EI ⋅ψ − 3 3, 75 = 0
B
(1)
B
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.
Member AB MB = 0 ∴VAB ⋅ 6 − 30 ⋅ 3 − M BA AB
∑
V AB
=
0, 5 ⋅ EI ⋅ θ B
+ 56, 25
6
Member BD MB = 0 ∴VDB ⋅ 4 − M DB
∑
MDB MDB V DB
−M
( θ B + 2 ⋅ θ D − 3 ⋅ψ BD ) L = 0, 5 ⋅ EI ⋅ θ B + 1, 5 ⋅ EI ⋅ψ MDB
+M
BD
4
=
1, 5 ⋅ EI ⋅ θ B
∑
=
L
DB
=0
( θ − ψ ) + FEM
Slope-deflection
+ FEM
+ 3 ⋅ EI ⋅ψ
3 ⋅ EI
E
= 0
4
Member CE MC = 0 ∴ VEC ⋅ 5 − M EC MEC
BD
2 ⋅ EI
= =
=0
EC
EC
−
1 2
FEMCE
Page 13 of 20
3/22/2010
=
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, 6 ⋅ EI ⋅ψ = 0
= 0 ∴ 4, 5 ⋅ EI ⋅ θ + 8, 7 ⋅ EI ⋅ψ − 3 3, 75 = 0
(2)
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
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
Slope-deflection
Page 14 of 20
3/22/2010
We have three unknowns, namely
∑M MBA M BA
B
= =
= 0 ∴ M +M + M BA
3 ⋅ EI
B
L
=
3 ⋅ 2EI
M Bc
=
=
M BF
=
∑M ∑M
B
D
MDC
=
M DC
=
D
= 0
BA
⋅ ( θ − 0 ) −
6 ⋅ 52
B
12
2 ⋅ EI L 2 ⋅ EI
and ψ . We require three equations to solve these unknowns.
1 2
AB
6 ⋅ 5 − ⋅+ 2 12 2
1
– 18,75 1
L 3 ⋅ 2EI
4 M BF BF = 1 EI
θ
⋅ ( θ − ψ ) + FEM − ⋅ FEM B
BC
Bc
2
CB
⋅ ( θ − ( −ψ )) + 0 − 0 B
3 M BC BC = 2 EI
M BF
BA
B
5
3 ⋅ EI
BF BF
,
B
⋅ ( θ − ψ ) + FEM − ⋅ FEM
M BA BA = 1,2 EI MBc
BC
θ
B
+ 2 EI
⋅ ( 2θ + θ − 3ψ ) + FEM B
F
BF
BF
⋅ ( 2θ + 0 − 0 ) + 0 B
B
= 0 ∴ 4, 2 ⋅ EI ⋅ θ +2 ⋅ EI ⋅ψ − 1 8, 75 = 0 = 0 ∴ M +M + M = 0
(1)
B
DC
3 ⋅ EI
DE
DG DG
1 2
⋅ ( θ − ψ ) + FEM − ⋅ FEM
L 3 ⋅ 2EI
D
DC
DC
CD
⋅ ( θ − ψ ) D
3 M DC DC = 2 EI
D
- 2 EI
M DE DE = + 10 x 2 = + 20
MDG
=
M DG
=
2 ⋅ EI L 2 ⋅ EI
4
⋅ ( 2θ + θ − 3ψ ) + FEM D
G
DG
DG
⋅ ( 2θ )
Slope-deflection
D
Page 15 of 20
3/22/2010
M DG DG = EI
∑M
D
= 0 ∴ 3 ⋅ EI ⋅ θ − 2 ⋅ EI ⋅ψ + 20 = 0
D
(2)
D
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
D
3 B - 2 EI
D
- 4 EI
= 60
(3)
Solve the three simultaneous equations:
θB θD ψ MBA MBC MBF MFB MDC MDG MGD
-6.0185/EI 8.0093/EI 22.0139/EI = - 25,972 kN.m = + 31,991 kN.m = - 6,0185 kN.m = - 3,009 kN.m = - 28,009 kN.m = + 8,009 kN.m = + 4,005 kN.m
Example 2:
Slope-deflection
Page 16 of 20
3/22/2010
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 . 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.
For the 10 mm displacement, displacement, B drops vertically by 10 mm. The sway angles are thus equal to BB’/Length of the member: ψ AB
=−
ψ BC
=
0,010
4 0,010 9
= −2, 5 x 10−
3
= 1,1111 x 10 −
3
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. ψ OB = ψ BD = ψ OD ψ OD
=
0,20
= 5 x 10−
3
4 Therefore, ψ BD = 5,0 x 10-3 BB’ = ψ OB x 3 m = 0,015 m 0,015 = −3, 75 x 10 −3 4 0,015 66667 x 10 −3 = 1, 66 9
ψ AB
=−
ψ BC
=
Slope-deflection
Page 17 of 20
3/22/2010
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 10-3
To determine the relative sway angles:
Set ψ BD = ψ , then BB’ = 5 ψ ψ
AB
ψ BC
=ψ
O 1B
=ψ
O 2B
=− =+
BB ' LO1B BB ' LO 2B
=−
5 ⋅ψ 6,6667
=+
5 ⋅ψ 15
= −0,75 ⋅ψ
= +0,3333 ⋅ψ
Equations required to solve the unknowns:
∑M
B
MBA
=
MBA
=
MBC
=
MBC
=
= 0 ∴ M +M + M BA
3 ⋅ EI
BC
BD BD
= 0 1 2
⋅ ( θ − ψ ) + FEM − ⋅ FEM B
BA
BA
L 3 ⋅ 2 ⋅ (200 x 10 6 ⋅ 50 x 10 6 )
AB
⋅ ( θ − ( −0, 75 ⋅ψ − 6, 25 x 10 − ) 3
B
4 M BA BA = 15 000 θ B + 11 250 ψ + 93,75 3 ⋅ EI L
1
⋅ ( θ − ψ ) + FEM − ⋅ FEM B
BC
3 ⋅ 2 ⋅ (200 ⋅ 50) ⋅ (θB 9
Slope-deflection
BC
2
CB
− (0, 3333 ⋅ψ + 2, 7778 x 10 − ) + 3
10 ⋅ 9 2 12
Page 18 of 20
1 10 ⋅ 9 − ⋅− 2 12 2
3/22/2010
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
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
B
−2 972, 22 ⋅ψ + 116, 4813 = 0
(1)
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 Y DB = 0, i.e., O2. Take moments about B of the member AB: V
=
M BA L AB
=
15000 ⋅ θ B
+ 11250 ⋅ψ + 93, 75
AB
= 3750 ⋅ θ + 2812, 5 ⋅ψ + 23, 4375 B
4
To determine the force V DC, take moments about B of the member BD: M + M DB V DB = BD LBD
V DB
=
2 ⋅ EI
( θ B + 2 ⋅ θ D − 3 ⋅ψ BD ) + FEMDB L 2 ⋅ 200 ⋅ 50 MBD = θ B − 3 ⋅ (ψ + 5 x 10 −3 ) ) ( 5 M DB DB = 4 000 θ B – 12 000 ψ - 60,00 MDB
=
MBD
+ M
LBD
DB
=
12 000 ⋅ θ B
− 24 000 ⋅ψ − 240
10 ⋅ 9 2 MO 2 = 0 ∴VAB ⋅ 13 − VDB ⋅10 − M DB − 2 20 750 B + 96 562,5 + 199,6875 = 0
∑
= 2 400 ⋅ θ − 4 800 ⋅ψ − 24 B
5
=0 (2)
Solve the two simultaneous equations: θ B ψ
= - 0,0040464 = - 0,00119844
MBA
= + 19,572 kN.m
Slope-deflection
Page 19 of 20
3/22/2010
MBC MBD MDB
= + 58,418 kN.m = - 77,990 kN.m = - 61,804 kN.m
Slope-deflection
Page 20 of 20
3/22/2010