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How to calculate beam with slope deflection. Discuss the application of slope-deflection method to beams and frames. Analyse statically indeterminate structures using slope-deflection method
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tabel momen primer metode crossFull description
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Analisa StrukturFull description
Analisa Struktur Statis Tak Tentu Metode Slope DeflectionDeskripsi lengkap
Full description
Slope‐Deflection Equations •
Fixed‐End Moment
Slope‐Deflection Equations •
ope‐ e ect on equat ons –
The resultant moment (adding all equations together) M A B M BA
= 2E ⎜ ⎟ ⎢ 2θ A + θ B − 3⎜ ⎟ ⎥ + (FEM ) A B ⎝ L ⎠ ⎣ ⎝ L ⎠⎦ ∆ ⎤ I ⎡ = 2E ⎛ ⎞ 2θ + θ − 3⎛ ⎞ + (FEM ) B
A
BA
•
Lets represent the member stiffness as k = I/L &
•
The s an ration due to dis laceme lacement nt as
•
Referring to one end of the of the span as near end (N) & the other end as the far end (F).
=∆ L
–
M N
= 2EK [ 2θ N + θ F − 3ψ ] + (FEM )N
M F
= 2EK [ 2θ F + θ N − 3ψ ] + (FEM )F
Slope‐Deflection Equations •
ope‐ e ect on equat ons or p n supporte En –
If the If the far end is a pin or a roller support
M N
= 2EK [ 2θ N + θ F − 3ψ ] + (FEM )N 0 = 2 EK [ 2θ F
–
pan
+ θ N − 3ψ ] + 0
Multiply the first equation by 2 and subtracting the second equation rom
M N
= 3EK [θ N −ψ ] + (FEM ) N
‐ •
• •
MN, MF = the internal moment in the near & far end of the of the span. Considered positive when acting in a clockwise direction ,
= mo u us o e as c y o ma er a
span s
ne s s
=
of the span at the supports in radians. θN, θF = near & far end slope of the Considered positive when acting in a clockwise direction
•
if the right end of ψ = span ration due to a linear displacement ∆ / L. if the member sinks with respect the the left end the sign is positive
•
Steps to analyzing beams using this method –
Fined the fixed end moments of each of each span (both ends left & right)
–
Apply the slope deflection equation on each span & identify the unknowns
– –
Solve the equilibrium equations to get the unknown rotation & deflections
–
Determine the end moments and then treat each span as simply supported beam the bending moment & shear force diagram
a
Example 1 •
Examp e 1 –
xe
Draw the bending moment & shear force diagram.
n
oment
FEM A B
=−
FEM BC
=−
4× 4 8
× 12
= −2t .m = −6t .m
t FEM BA
= 2t .m
FEM CB
= 6t .m
2t/m
4m
6m
Slope Deflection Equations M A B
M BC M CB
I = 2E ⎛⎜ ⎞⎟ [ 2θ A + θ B − 0] − 2 ⎝4⎠ I = ⎛ ⎞ − 4 ⎝ ⎠ 2 I = 2E ⎛⎜ ⎞⎟ [ 2θ B + θ C − 0] − 6 ⎝ 6 ⎠ 2 I = 2E ⎛⎜ ⎞⎟ [θ B + 2θ C − 0 ] + 6 ⎝ 6 ⎠
A=
0.
θA = 0. θC = 0. θC = 0.
M A B
= 0.5EI θ B − 2
M
= EI θ + 2
M BC
=
4
M CB
=
2
3 3
EI θ B
−6
EI θ B
+6
Joint E uilibrium E uations Joint B M BA
+ M BC = 0
EI θ B
+2+
3
EI θ B
− 6 = 0.
θ B
=
. EI
Substituting in slope deflection equations .
M A B
= 0.5EI
M BA
= EI
M BC
= −3.7t .m =
M
2 3
EI
1.7
EI
− 2 = −1.15t .m
+ 2 = 3.7t .m
1.7 EI
4t
+ 6 = 7.1t .m
3.7
1.15
Computing The Reactions A
=
1.15 + 4 × 2 − 3.7
4 R B 1 = 4 − 1.36 = 2.64t
=
.
A
B1
R B 2
=
× × +
.
−
.
6 R C = 2 × 6 − 5.43 = 6.57t
m
= 5.43t
RB2
5.43 1.36
+
+ -
-
2.64
. 7.1 3.7 1.15
-
-
- +
1.57
7.1
3.7
+
3.79
RC
Example 2 •
.
Fixed End Moment FEM AB A B
FEM BA FEM
=−
=
60 × 4 × 2 2
60 × 2 × 4 2
62 30 × 62 =− 8
2
= −26.67k N .m 60kN
= 53.33k N .m
30kN/m
4m
= −135k N .m
B
I
I
6m
6m
Slope Deflection Equations I = 2E ⎛⎜ ⎞⎟ [ 2θ A + θ B − 0] − 26.67 ⎝6⎠ I = ⎛ ⎞ − 6 ⎝ ⎠ ⎛ I ⎞ M BC = 3E ⎜ ⎟ [θ B − 0] − 135 ⎝6⎠
M A B
A=
0.
θA= 0.
M A B
=
1
M
=
2
3 3
M BC
EI θ B
− 26.67
EI θ
+ 53.33
=
1 2
EI θ B
− 135
Joint E uilibrium E uations Joint B M BA
+ M BC = 0
3
EI θ B
EI θ B
+ 53.33 +
2
EI θ B
− 135 = 0
= 70
Substituting in slope deflection equations
=
1
M BA
=
2
M BC
=
1
3 3 2
−
=−
(70) + 53.33 = 100k N .m (70) − 135 = −100k N .m
6
EI θ B
= 81.67
Example 3 •
Examp e : Determine the internal moments in the beam at the supports
Support A; downward movement of 0.3cm of 0.3cm & clockwise rotation of 0.001 of 0.001 rad. Support B; downward movement of 1.2cm. of 1.2cm. Support C downward movement of 0.6cm. of 0.6cm. EI = 5000 t.m2 Fixed End Moment –
⎛ I ⎞ ⎡ ⎛ 0. ⎞⎤ − − BC B C ⎝ 15 ⎠ ⎣ ⎝ 15 ⎠⎦ ⎛ I ⎞ M CB = 2E ⎜ ⎟ [ 2θC + θ B − 0] + 375 ⎝ 15 ⎠ =
M CD M DC
=
EI θ B 3
=
4
−
=
B
=
2 1 9
+
EI θ C
−
EI θ C
−
9
EI ∆
2
15 2 = EI θ B 15
∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θ C + 0 − 3 ⎛⎜ ⎞⎟⎥ + 0. ⎝ 18 ⎠ ⎣ ⎝ 18 ⎠⎦ ∆ I = 2E ⎜ ⎟ ⎢0 + θ C − 3 ⎜ ⎟ ⎥ + 0 ⎝ 18 ⎠ ⎣ ⎝ 18 ⎠ ⎦
24
15 4 15 1
54 1 54
C
EI θ C
EI ∆ EI ∆
E uilibrium E uations Joint B ΣMB = 0 M BA
+ M BC = 0
1 3
B
−
1 24
4
2 B
15 15 − EI ∆ + 14.4EI θ B
C
−
.
+ 3.2EI θ C = 9000
1
− + 375
=
M CB
+ M CD = 0
2
EI θ B 15
+
4 15
EI θC
+ 375 +
2
9 − EI ∆ + 7.2EI θ B
−
1
EI ∆ = 0. 54 + 26.4EI θ C = −20250
EI θ C
2
Three unknown & just two equations so we need another equilibrium equation. Let take ΣFx = 0.
40 − H A H A
− H D = 0. M + M A B = − BA
H D
=−
12
M CD
+ M DC
18
40 × 18 + 1.5 ( M BA
⎛1 720 + 1.5 ⎜ EI θ B ⎝3 ⎛2
+ M A B ) + ( M CD + M DC ) = 0. − −
1 24 1
EI ∆ +
1 6 1
⎞ 24 ⎠ 1 ⎞= − 54 ⎠ + 0.333EI θ C = −720
EI θ B
54 9 ⎝9 −0.162 EI ∆ + 0.75EI θ B
−
1
EI ∆ ⎟
3
⎡ −1 ⎢ −1 ⎢ − .
⎤ ⎧ EI ∆ ⎫ ⎧ 9000 ⎫ ⎪ ⎪ ⎪ ⎪ R1-R2 26.4 ⎥ ⎨ EI θ B ⎬ = ⎨ −20250 ⎬ ⎥ 0.162R1-R3 . − C
14.4
3.2
7.2 .
⎡ −1 ⎢0
14 14.4
⎢⎣ 0
1.58
7.2
⎡ −1
14 14.4
0.
7.2
⎢ ⎢⎣ 0.
0.
⎤ ⎧ EI ∆ ⎫ ⎧ 9000 ⎫ ⎪ ⎪ ⎪ ⎪ −23.2 ⎥ EI θ = 29250 ⎪ ⎪ ⎪ 0.185 ⎦⎥ ⎪ ⎩ EI θ C ⎭ ⎩ 2178 ⎭ 3.2
⎤ ⎧ EI ∆ ⎫ ⎧ 9000 ⎫ −23.2 ⎥ EI θ B = 29250 −5.29 ⎦⎥ ⎪⎩ EI θ C ⎪⎭ ⎪⎩ 4257 ⎪⎭ 3.2
u st tut ng n s ope e ect on equat ons M A B
=
M BA
=
1 6 3
0.22R2-R3
1467.2 −
1469.6 −
1 24 24
9587.2 = −155k N .m 9587.2 = 90 k N .m
C
=
4257
−5.29
= 1469.6 EI ∆ = 587.2 EI θ B
=−
.
M BC M CB
=
1469.6 + (−804.7) − 375 = −90 k N .m 15 15 2 4 = 1469.6 + ( −804.7) + 375 = 356k N .m 15 15
M CD
=
M DC
=
2 9 1 9
( −804.7) − ( −804.7) −
1 54 1 54
9587.2 = −356k N .m
356
9587.2 = −267 k N .m
-
90
-
90
-
155 267
Example 8
Example 9 – Draw the bending
moment diagram. EI constant
As there is no span loading in any of the member FEM for all the members is zero Slope Deflection Equations M A B
∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢0. + θ B − 3 ⎛⎜ 1 ⎞⎟ ⎥ + 0 =
M BA
2 EI θ B 5 I
−
6 25
EI ∆1
= 2E ⎜ ⎟ ⎢ 2θ B + 0 − 3 ⎜ ⎝ 5 ⎠⎣ ⎝ 4 6 = EI θ B − EI ∆1
∆1 5
⎟⎥ + 0 ⎠⎦
M BE M EB
M FE M EF
M BC M CB
= 2E ⎜ ⎟ ⎢ 2θ B + θ E − 3 ⎜ ⎟⎥ + 0 ⎝ 7 ⎠⎣ ⎝ 7 ⎠⎦ 0 ⎤ I ⎡ = 2E ⎛ ⎞ 2θ + θ − 3 ⎛ ⎞ + 0 E
B
∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢0. + θ E − 3 ⎛⎜ 1 ⎞⎟⎥ + 0 ⎝ 5 ⎠⎣ ⎝ 5 ⎠⎦ ∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θ E + 0 − 3 ⎛⎜ 1 ⎞⎟ ⎥ + 0 ⎝ 5 ⎠⎣ ⎝ 5 ⎠⎦ = 2E ⎜ ⎟ ⎢ 2θ B + θ C − 3 ⎜ 2 ⎟ ⎥ + 0 ⎝ 5 ⎠⎣ ⎝ 5 ⎠⎦ ∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θC + θ B − 3 ⎛⎜ 2 ⎞⎟⎥ + 0
0 ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θC + θ D − 3 ⎛⎜ ⎞⎟ ⎥ + 0 ⎝ 7 ⎠⎣ ⎝ 7 ⎠⎦ ⎛ I ⎞ ⎡ ⎛ 0 ⎞⎤ M DC = 2E ⎜ ⎟ ⎢ 2θ D + θ C − 3 ⎜ ⎟ ⎥ + 0 ⎝ 7 ⎠⎣ ⎝ 7 ⎠⎦
M CD
=
EI θ B
+
EI θ B
+
4
EI θ E
−
6
EI θ E
−
EI θ B
+
=
2 EI θ B
+
4
=
4
EI θC
+
2
EI θC
+
7
=
2
=
2
= =
=
5 4 5
5
7 2 7
EI θ E
7
EI θ E
25 6 25
5
7 4 7
EI ∆1 EI ∆1
EI θ C
−
EI θ C
−
EI θ D EI θ D
25 6
EI ∆ 2 EI ∆ 2
M DE M ED
=
= 2E ⎜ ⎟ ⎢ 2θ D + θ E − 3 ⎜ 2 ⎟ ⎥ + 0 ⎝ 5 ⎠⎣ ⎝ 5 ⎠⎦ ∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θ E + θ D − 3 ⎛⎜ 2 ⎞⎟ ⎥ + 0
=
EI θ D
+
2 EI θ D
+
5
5 4
EI θ E
−
EI θ E
−
25 6
EI ∆ 2 EI ∆ 2
Equilibrium Equations Joint B ΣM = 0 M BA
+ M BC + M BE = 0
4 EI θ B
−
380 EI θ B
6
EI ∆1 +
4
EI θ B
+
2
EI θC
−
6
EI ∆ 2
+
4
EI θ B
+
2
+ 70EI θC + 50EI θ E − 42EI ∆1 − 42EI ∆ 2 = 0
EI θ E
=0
1
E=
M EB
+ M ED + M EF = 0
2 EI θ B
+
4
50 EI θ B
+ 70EI θ D + 380EI θ E − 42EI ∆1 − 42EI ∆ 2 = 0
EI θ E
+
2
EI θ D
+
4
EI θ E
−
6
EI ∆ 2
+
4
EI θ E
6
−
EI ∆1
2
=0
on
C=
M CB + M CD = 0 2 4 EI θ B + EI θC
70 EI θ B
−
6
EI ∆ 2
+
4
EI θC
+
2
EI θ D
=0
+ 240EI θC + 50EI θ D − 42EI ∆ 2 = 0
3
= M DC + M DE = 0 2 4 EI θC + EI θD 7 7
50 EI θC
+
4 5
EI θD
+
2 5
EI θ E
6
−
25
EI ∆ 2
+ 240EI θ D + 70EI θ E − 42EI ∆ 2 = 0
=0 4
Top story s tory ΣFX = 0
−
B
−
H B
=−
H E
=−
= M CB + M BC
B
E
5 DE
ED
5
E
200 + EI θ B 5 4 + EI θ D 5 6 EI θ B
+ +
5 2 5
EI θC
−
EI θ E
−
EI ∆ 2
25 6 25
EI ∆ 2
+
EI θC − EI ∆ 2 5 25 4 6 + EI θ D + EI θ E − EI ∆ 2 5 5 25
5 2
EI θ B
+
+ 6EI θC + 6EI θD + 6EI θ E − 4.8EI ∆ 2 = −1000
ottom story
=0 5
X=
40 + 80 − H A
− H F = 0 M BA + M A B
H
=−
H F
=−
+ +
5 M EF
4 5 4 5
B
−
EI θ E
−
30 EI θ B
+ M FE 5 6
25 6 25
1 +
EI ∆1 +
2 5 2 5
B
−
EI θ E
−
6 25 6 25
+ 30EI θ E − 24EI ∆1 = −15000
1
EI ∆1
=0 6
HA
HF
un nown an
equat on
⎡380 ⎢ 50
70 70
0
50
0
70
38 380
⎢ 70 ⎢ 0 ⎢
240
50
0
50
240
70
⎢ ⎣ 30
0
0
30
−42 −42 ⎤ ⎧ EI θ B ⎫ ⎧ 0 ⎫ −42 −42 ⎥ ⎪ EI θ ⎪ ⎪ 0 ⎪ 0 −42 ⎥ ⎪ EI θ D ⎪ ⎪ 0 ⎪ ⎨ ⎬=⎨ ⎬ ⎥ 0 −42 ⎥ ⎪ EI θ E ⎪ ⎪ 0 ⎪ − . − 1 ⎪ ⎪ ⎪ ⎥⎪ −24 0 ⎦ ⎩ EI ∆ 2 ⎭ ⎩−15000 ⎭
= 171.79 EI θ C = 79.80 EI θ D = 79.80 EI θ B
EI θ E
= 171.79 .
1
EI ∆ 2
= 837.29
Substituting in slope deflection equations M A B
= −184.4k N .m
M BE
= 147.2k N .m
M FE
= −184.4k N .m
M BA
= −115.6k N .m
M EB
= 147.2k N .m
M EF
= −115.6k N .m
M BC
= −31.6k N .m
M CD
= 68.4k N .m
M DE
= −68.4k N .m
CB
−
.
.
DC
.
.
ED
−
.
.
-
68.4 + 68.4
68.4 68.4
-
147.2 31.6
-
-
115.6 +
+
31.6 -
.
184.4
+
184.4
115.6
Example 10 – Draw the bending
moment diagram. EI constant
Degree of freedom DOF = 3x4-6-3 = 3 That means we got three unknown & we need three equations Before we start let us discus the relative displacement (∆) of each span
∆CD
The relative displacement ( ∆) for span AB AB is equal ∆AB – 0. = ∆AB (clockwise)
B
C
equal 0. – ∆BC = – ∆BC (counterclockwise) The relative displacement (∆) for span CD is equa equall 0. 0. – ( – – ∆CD)= ∆CD (clockwise)
∆BC A
60o
∆AB
Let us build a relationship between ∆AB, ∆BC & ∆CD a e AB =
∆BC = ∆AB × sin30 = 0.5∆ ∆CD = ∆AB × cos30 = 0.866∆ So in the slope deflection equations we will use;
D B
AB. ∆ as the relative displacement of span AB. –0.5 ∆
as the relative displacement of span BC.
0.866 ∆
as the relative displacement of span CD.
60o
60o ∆
CD
θ =30o AB
∆BC
xe
n
omen
= FEM BA = 0. 2 × 42 − − . BC
FEM A B
12
FEM CB
= 2.667.m
FEM
= FEM = 0.
.
Slope Deflection Equations M
∆ ⎤ I ⎡ = 2E ⎛ ⎞ 0. + θ − 3 ⎛ ⎞ + 0
=
2 EI θ B
−
2
EI ∆
M BA
∆ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θ B + 0 − 3 ⎛⎜ ⎞⎟ ⎥ + 0 ⎝ 3 ⎠⎣ ⎝ 3 ⎠⎦
=
4
EI θ B
−
2
EI ∆
= EI θ B +
1
M BC CB
−0.5∆ ⎞ ⎤ I ⎡ = 2E ⎛⎜ ⎞⎟ ⎢ 2θ B + θ C − 3 ⎛⎜ − 2.667 ⎟ ⎥ ⎝ 4 ⎠⎣ ⎝ 4 ⎠⎦ ⎛ I ⎞ ⎡ ⎛ −0.5∆ ⎞ ⎤ . − B C ⎝ 4 ⎠⎣ ⎝ 4 ⎠⎦
3
2
3
EI θ C
1 2
+
3 16 3
B
C
16
EI ∆ − 2.667
.
M CD M DC
0.866∆ I = 2E ⎜ ⎟ ⎢ 2θ C + 0 − 3 ⎜ ⎟ +0 ⎝ 6 ⎠⎣ ⎝ 6 ⎠⎥⎦ 0.866∆ ⎞ ⎤ I ⎡ = 2E ⎛ ⎞ 0 + θ − 3 ⎛ +0
=
2
EI θ C
3 1 = EI θ C
C
− −
Equilibrium Equations Joint B ΣM = 0 M BA
+ M BC = 0
4 2 1 3 EI θ B − EI ∆ + EI θ B + EI θ C + EI ∆ − 2.667 = 0 3 3 2 16 112 EI θ B + 24EI θ C − 23EI ∆ = 128 1
Joint C ΣMC = 0 M CB
1
+ M CD = 0
3 2 EI θ B + EI θC + EI ∆ + 2.667 + EI θ C 2 16 3 24 EI θ B + 80EI θ C + 2.072EI ∆ = −128
−
0.866 6
EI ∆ = 0
2
0.866 6 0.866
EI ∆ EI ∆
r
qu
r um um qua ons
M A B + M BA + M DC − ⎛⎜ AB 3 ⎝ −8 2 = 0.0
M A B
e o
⎞ 11 − ⎛ M DC + M CD ⎟( ) ⎜ 6 ⎠ ⎝
⎞ 12.92 ) ⎟( ⎠
−21 M AB − 22M BA − 12.92M CD − 11.92M DC − 96 = 0 A B 34.5 EI θ B
− 3.1EI θ C − 38.375EI ∆ = −144
8.0 6.92m
.
.
.
3
6.0m 3.0
6.0
Third Equilibrium Equations (Method 2) M CD
X
M BA
− H A − H D = 0 From the free body diagram for column CD H D
=−
CD
m 6 . 2
DC
6
Free body diagram for column AB H A
HA
A
× 2.6 + ( M BA + M A B ) +V A ×1.5 = 0
1.5m
M A B
VA
Free body diagram for Beam BC
× 4 + ( M BC + M CB ) − 2 × 4 × 2 = 0 ( M BC + M CB ) +4 V A =V B = −
M DC
M BC
4
∴ H A =
( M BA + M A B ) 2.6
2.6
+
1.5
⎢− 2.6 ⎣
−
1.5 ⎡
− ⎢ 2.6 ⎣
( M BC + M CB )
( M BC + M CB ) 4
HD
D
V B
− ( M BA + M A B )
C
B
4
+ 4⎥ + ⎦
M CD
⎤ + 4⎥ ⎦ + M DC 6
2t/m
C
B M CB
VB
= 0.
BA
+
⎛4 24 ⎜ EI θ B ⎝3 3
AB
−
2 3
− EI ∆ +
BC
2 3
+
EI θ B
CB
−
2 3
+
.
⎞ ⎠
CD
⎛ ⎝
+
EI ∆ ⎟ − 9 ⎜ EI θ B
DC
+
1 2
=−
EI θC
1 3 EI ∆ − 2.667 + EI θ B + EI θ C + EI ∆ + 2.667 ⎟ 16 2 16 ⎠ 2 0.866 1 0.866 ⎞ +10.4 ⎛ EI θ − EI ∆ + EI θ − EI ∆ = −144 3 6 3 6
+
34.5 EI θ B
− 3.1EI θ C − 38.375EI ∆ = −144
3
Solving the three equation
⎡ 112 ⎢
24
⎢⎣34.5 −3.1
−23 ⎤ ⎧ EI θ B ⎫ ⎧ 128 ⎫ ⎪ ⎪ ⎪ ⎥⎪ = − . C ⎪ −38.375⎥⎦ ⎩ EI ∆ ⎪⎭ ⎪⎩−144 ⎪⎭
EI θ B C
= 3.11 =−
.
EI ∆ = 6.77
Solving the three equation
⎡ 112
24
⎢ ⎢⎣34.5 −3.1
−23 ⎤ ⎧ EI θ B ⎫ ⎧ 128 ⎫ ⎪ ⎪ ⎪ ⎪ = − . C ⎥ ⎪ −38.375⎥⎦ ⎩ EI ∆ ⎪⎭ ⎪⎩−144 ⎪⎭
EI θ B
=
EI θ C
= −2.71
=
.
.
Substituting in slope deflection equations M A B BA
= −2.44t .m
−
.
2.78
.
_
M BC
= 0.36t .m
B
M CB
= 2.78t .m
. 0.36
CD
M DC
=−
.
.
= −1.88t .m
2.44
C _
+
≈2.8 A
1.88
+
D
2.78
Example – Draw the bending
moment diagram. EI constant
Before we start let us discus the relative displacement (∆) of each span
The relative displacement ( ∆) for span AB AB is equal ∆AB – 0. = ∆AB (clockwise)
The relative displacement (∆) for span CD is equa equall 0. 0. – ( – – ∆CD)= ∆CD (clockwise)
∆AB
Let us build a relationship between ∆AB, ∆BC & ∆CD a e AB =
∆BC = 2(∆AB × cosθ) = 2∆ × 5/8.6 = 1.163∆ ∆CD = ∆AB = ∆ So in the slope deflection equations we will use; AB. ∆ as the relative displacement of span AB. –1.163 ∆
as the relative displacement of span BC.
∆ as the relative displacement of span CD.
∆CD θ
B
θ
∆AB
θ δ2 θ δ1
∆BC
δ1 = δ2 & ∆CD = ∆AB because of the symmetry in the geometry
4 2 6.978 −1.163∆ ⎞ ⎤ I ⎡ = + + θ θ EI EI EI ∆ − 16.33 = 2E ⎛⎜ ⎞⎟ ⎢ 2θB + θ C − 3 ⎛⎜ − 1 6 . 3 3 B C ⎟ ⎥ 7 7 49 ⎝ 7 ⎠⎣ ⎝ 7 ⎠⎦ 2 4 6.978 ⎛ I ⎞ ⎡ ⎛ −1.163∆ ⎞ ⎤ − . . C B B C 7 7 7 7 49 ⎝ ⎠⎣ ⎝ ⎠⎦
M CD M DC
∆ I = 2E ⎜ ⎟ ⎢ 2θ C + 0 − 3 ⎜ ⎟ ⎥ + 0 ⎝ 8.6 ⎠ ⎣ ⎝ 8.6 ⎠ ⎦ ∆ ⎤ I ⎡ = 2E ⎛ ⎞ 0 + θ − 3 ⎛ ⎞ + 0 C
.
= =
.
4 8.6 2 .
EI θ C
−
EI θ C
−
6 73.96 6 .
Equilibrium Equations Joint B ΣM = 0 M BA
+ M BC = 0
4
6 4 2 6.978 EI θ B − EI ∆ + EI θ B + EI θ C + EI ∆ − 16.33 = 0 8.6 73. 6 7 7 4 103.65 EI θ B + 28.57EI θ C + 6.13EI ∆ = 1633 1
Joint C ΣMC = 0 M CB
2
+ M CD = 0
EI θ B
+
4
7 7 28.57 EI θ B
EI θC
+
6.978
49 + 103.65EI θ C
EI ∆ + 16.33 +
4
EI θ C
8.6 + 6.13EI ∆ = −1633
−
6 73.96
EI ∆ = 0
2
EI ∆ EI ∆
X
6 − H A
− H D = 0 B
Free body diagram for column AB H A
× 7 + ( M BA + M A B ) +V A × 5 = 0
m 7
Free body diagram for Beam BC B
V A
BC
=V B = −
∴ H A = −
CB
−
BA
7 AB
7
A
.
( M BC + M CB ) +
HA M A B
+ 14
5
− ⎢− 7⎣
5m
M BC
VA BC
+ 7
CB
+ 14 ⎥ ⎦
4kN/m
B VB
VC
× 7 + ( M CD + M DC ) −V D × 5 = 0
VC
× 7 − ( M BC + M CB ) − 4 × 7 × 3.5 = 0
V D
=V C =
BC
CB
7
+ 14
M CB
M CD
From the free body diagram for column CD H D
C
m 7
M DC m
VD
HD
∴ H D = − CD − H D = 0 Q 6 − H A
DC
7
BA
7 ( M BA
⎛ 2 ⎝ 8.6
AB
+ ⎢ 7⎣
BC
+ 14⎥ ⎦
CB
7
−
BC
CB
CD
−
DC
BC
CB
+ M A B ) + 7 ( M CD + M DC ) − 10 ( M BC + M CB ) = −294 6 ⎛ 2 − 73.96 8.6 73.96 73.96 ⎝ 8.6 4 6 2 6.978 ⎞ ⎛ 4 + EI θC − EI ∆ ⎟ − 10 ⎜ EI θ B + EI θ C + EI ∆ − 16.33 8.6 73.96 7 49 ⎠ ⎝ 7
−
6
4
−
+
7
⎞ ⎠
6
EI θ B
+
−3.688 EI θ B − 3.688EI θ C − 5.12EI ∆ = − 294
7
EI θ C
.
+
49
EI ∆ + 16.33 ⎟ = −294
⎠
3
Solving the three equation
⎡103.65
28.57
. ⎢ . ⎢⎣ −3.688 −3.688
⎤ ⎧ EI θ B ⎫ ⎧ 1633 ⎫ ⎪ ⎪ ⎪ ⎪ . = − C ⎥ −5.12 ⎥⎦ ⎪⎩ EI ∆ ⎪⎭ ⎪⎩ −294 ⎭⎪ 6.13
EI θ B C
= 18.897 −
.
EI ∆ = 61.532
Substituting in slope deflection equations M AB A B BA
