# slope deflection bergoyang

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BAN NO.3 CONTOH SOAL  Kontr uksi Bergoya ng I I 1). Hitung Nilai K EI 15 2 EI 12  EI 10 2). Menentukan Momen Frimer (MF) 12 X 9 X 6  2 15  2 -12  X 9  2 X 6 15  2 24 X 12 8 = -  = 0 3). Tentukan nilai R ( ) 15 10 30 30 30 = 17.28 = -25.92 MF DC = = 3 R R BC R CD = MF CD X = 0 = MF AB = 2 R 3 K BC = x 30 R AB  = x 30 = K CB  = SOLUSI 5 K BD =  K DB  = X = 36 Kip ft Kip ft K AB =  K BA  = X = 2 Kip ft MF BC = - MF CB  = = MF BA SLOFE DEFLECTION 2I  10 M 24 k  A B C D 6 M 6 M 12 k 6 M 9 M

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8/13/2019 Slope Deflection Bergoyang

BAN NO.3

CONTOH SOAL   Kontr uksi Bergoyang

I

I

1). Hitung Nilai K

EI

15

2 EI

12

EI

10

2). Menentukan Momen Frimer (MF)

12 X 9 X 6   2

15   2

-12   X 9   2 X 6

15   2

24 X 12

8

= -   = 0

3). Tentukan nilai R ( ∆ )

15

10

30

30

30

= 17.28

= -25.92

MFDC

=

=

3 R

R BC

R CD

=

MFCD

X =

0

=

MFAB =

2 R

3

K BC =

x 30

R AB   = x 30

=

K CB   =

SOLUSI

5

K BD =   K DB   = X =

36 Kip ft

Kip ft

K AB =   K BA   = X = 2

Kip ft

MFBC = - MFCB   = =

MFBA

SLOFE DEFLECTION

2I

10 M

24 k

A

B C

D

6 M 6 M

12 k

6 M

9 M

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

Masukkan kembali nilai yang didapat kedalam rumus Slope Deflection

= + K ( -2 ΦA   - ΦB   + 2 R )

= + 2 { ( -2 x ) - ( + (   2 x   ) }

= + 2 ( + )

= + 2 ( )

= +

=

= + K ( -2 ΦB   - ΦA   + 2 R )

= + 2 { ( -2 x ) - ( ) +   2 x   ) }

= + 2 ( + )

= + 2 ( )

= +

=

= + K ( -2 ΦB   - ΦC   + 0 R )

= + 5 { ( -2 x ) - ( ) +   0 x   ) }

= + 5 ( + )

= + 5 ( )

= +

=

= + K ( -2 ΦC   - ΦB   + 0 R )

= + 5 { ( -2 x ) - ( ) +   0 x   ) }

= + 5 ( + )

= + 5 ( )

= +

=

= + K ( -2 ΦC   - ΦD   + 3 R )

= + 3 { ( -2 x ) - ( ) +   3 x   ) }

= + 3 ( + )= + 3 ( )

= +

=

= + K ( -2 ΦD   - ΦC   + 3 R )

= + 3 { ( -2 x ) - ( ) +   3 x   ) }

= + 3 ( + )

= + 3 ( )

= +

=

-MFDC

57.311

27.28

19.104

0.0 0

0.00

0.0 -54.55

0.0 0.00

57.311

27.275 2.76

-2.76

-1.99

19.104

38.207

19.104

38.207

1.99

73.545

-36 -5.521

2.7604

19.104

0

19.104

0

19.104

1.99

-27.28

-25.92 -3.976

-25.92

-7.51

-36 2.7604

-36

MDC

17.28 -8.64

17.28 -17.28

0.00

-25.92

0.0 -5.521

0.0

-25.92 23.603

-2.32

36

17.28 22.43

1.9881

17.28 -44.86 -1.99

2.76

36 -3.976 -2.76

MCD   MFCD

MBC   MFBC

1.9881 22.43

-22.43

MCB   -MFCB

36 -6.74

36

24.51

73.54

0.0

MAB   MFAB

2.32

MBA   -MFBA

-33.68

11.80

-36 -37.54

-73.54

0.0

0.0

8/13/2019 Slope Deflection Bergoyang

8). Lakukaan pemotongan pada masing-masing titik ( Free body )

B 2I C

I

I

a). potongan A & B

Kf

HB = Kip

HA = Kip

1. Reaksi perletakan

= 0

-   x 15 + P x   9 + = 0

- 15 HB   + 12 x 9 +   = 0- 15 HB   + +   = 0

- 15 HB   + = 0

=   /   -15

=   Kip ( )

= 0

HA   x 15   - P x 6 + = 0

15   HA   - 12 x 6 + = 0

15   HA   - +

15   HA   = 0

= / 15

=   Kip ( )

2). Momen dan Lintang

= x X = x X   - P ( X - 9 )

= X = X -   X +

= X +

= =

73.542.32

4.65

2.32

7.35

69.7

4.65

2.32

0

ΣHA

HB

0

12

2.32

73.54

-7.35   108

-7.35

108 2.32

110.3

2.32

2.32

72 2.32

-110.3

4.65

-69.7

ΣHB

Mx   4.65

Interval 0 ≤ x ≤ 9

Mx

108

Interval 9 ≤ x ≤ 15

DxDx   4.65

2.32

0

4.65

7.35

4.65

RHA

RHA

RHB

RHB

D

A

24 k

6 6

12 k

6 ft

9 ft

B

C

10 ft

12 k

6 ft

9 ft

A

B

8/13/2019 Slope Deflection Bergoyang

Tabel Momen dan Lintang pada potongan A & B

b. Potongan B & C

24 K

B C

= K = K

1. Reaksi perletakan

=   0-   x 12 + P x   6 + - = 0

- 12 RC   + 24 x 6 +   - = 0

- 12 RC   + +   - = 0

- 12 RC   +   = 0

=   /   -12

= K  ( )

=   0

R B   x 12 - P x   6 - + =

12   R B   - 24 x 6 -   + = 0

12   R B   - -   + = 0

12   R B   = 0

=   /

= K  ( )

2. Momen dan Lintang

= x X - =

= X -

73.54

73.54

6

2.32

2.32

2.32

Nilai Momen

Kip ft

41.81

Kip ft

23.23

6.06

6

Nilai LintangX Rumus Momen Rumus Lintang

0   0.00

9

Kip

Kip ft

Kip

2   9.29 Kip ft 4.65   Kip

1

Kip ft 4.65   Kip

4.65

13.94

Kip ft

Kip ft 4.65

37.16 Kip ft 4.65

7   32.52

4.65

4.65

4   18.58

Kip

Kip

Kip

4.65

4.65

4.65

Kip

6   27.87 Kip ft

13   12.39 Kip ft

9   41.81 Kip ft

8

10

12

Kip ft

Kip ft

27.10

4.65

Kip

Kip ft

Kip

Kip ft   Kip

Kip ft

Kip

-7.35

Kip

-7.35

-2.32

Kip

Kip ft   Kip

-7.35

11

14

3

5

-7.35   Kip

-7.35

-7.35

Kip

-7.35

-72.8

Interval 0 ≤ x ≤ 6

RC 73.54

2.32

15

6.06

Dx2.32

17.94

2.32

72.77 12

ΣMC

6.06 2.32

Mx   RBV

2.32   73.54

17.94R BV   6.06

R CV

R BV

73.54

2.32 73.54

144

R CV

73.54

2I

34.46

19.75

5.04

ΣMB

R CV   -215.23

215.23

144

R BV

7,35 X + 108 7,35

4,65 X 4,65

8/13/2019 Slope Deflection Bergoyang

= x X   - P ( X - 6 )   -

= X -   X +   -

= X +

=

Tabel Momen dan Lintang pada potongan A & B

C). Potongan C & D

= K

= K

1. Reaksi perletakan

= 0

x 10 -   = 0

=   /   10

=   kip ( )

= 0

-   HD   x 10   - = 0

HD   = / -10

HD =   kip ( )

2). Momen dan Lintang

= x X

= X

=

-7.35

Dx   7.35

7.35

Interval 0 ≤ x ≤ 10

Mx   7.35

73.5

Hc

Hc 7.35

73.5

ΣHD

ΣHC

Hc

73.5

73.5

0

73.5

Interval 6 ≤ x ≤ 12

Mx   RBV

6.06 24 144

0

1

Dx   -17.9

X Rumus Momen

Kip ft 6.06

3.75 Kip ft 6.06

-17.9   141.68

2   9.81 Kip ft 6.06

Kip

Nilai Momen Nilai Lintang

Kip-2.32

4   21.94 Kip ft 6.06

Kip

3   15.88 Kip ft 6.06   Kip

6   34.07 Kip ft 6.06

Kip

5   28.00 Kip ft 6.06   Kip

7   16.13

8   -1.80

-17.94   Kip

Kip

Kip

Kip ft -17.94

11   -55.61 Kip ft

Kip

12   -73.54

10

Kip ft -17.94

6

Kip ft -17.94

9   -19.74

Kip ft -17.94

-37.67

Kip ft

Kip

Kip ft -17.94   Kip

Kip-17.94

34.07

Kip

2.32

2.32

Rumus Lintang

7.35

Hc 7.35

Hc

6,06 X - 2,32 6,06

-17,9X + 141,68 -17,9

C

D

I 10 ft

8/13/2019 Slope Deflection Bergoyang

Tabel Momen dan Lintang pada potongan A & B

GAMBAR MOMEN, LINTANG DAN NORMAL

Kip ft

Kip ft

Kip ft

Kip ft

0

0

Gambar Momen

-2.32

10

9

1

X

Kip ft 7.35

Nilai Momen Nilai Lintang

Kip

Kip73.54

Kip

66.19 Kip ft 7.35   Kip

8   58.84 Kip ft 7.35

Kip

Kip7   51.48 Kip ft 7.35

6   44.13 Kip ft 7.35

5   36.77 Kip ft

3   22.06 Kip ft

14.71 Kip ft 7.35

7.35

2

Rumus Momen Rumus Lintang

0

7.35   Kip

4

Kip ft 7.35   Kip

Kip ft 7.35

Kip

Kip

Kip

0.00

73.54

34.07

73.54

7.35

29.42 Kip ft 7.35

41.81

-7,35 X -7,35

A

B

C

D

- 2,32 Kip ft

(+

(+

(-)

(-)

(+

(-)

8/13/2019 Slope Deflection Bergoyang

kip

7.35   kip

4.65 kip   -7.35 Kip

0   7.35   kip

kip

0

Gambar Lintang

Kip

kip

Gambar Normal

6.06

17.94

6.06

-17.94

A

B

C

D

(+

(+

(-)(-) (-)

A

B C

D

(-)

-

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

Detail Asumsi Pembebanan

ASUMSI PEMBEBANAN

q1 -   Beban Mati

Balok Ukuran 30 x 30 cm = x x =

q1   = x = 0.353   T/1.2 0.294

SLOPE DEFLECTION

SO US

0.35 0.35 2.4

4 M

q1

A B

C

D

E

4.5 M 4.5 M

EI  EI

0.01 M Keramik

0.02 M Spesi K

0.12 M Plat Lan

0.35 x 0.35 M B

2.25 M

2.25 M

EI EI

q2

2.25 M

4.5 M

8/13/2019 Slope Deflection Bergoyang

q2 -   Beban Hidup + Beban Mati

-   Beban Hidup

Untuk Ruang Pertemuan =

Wu1   = x =

-   Beban Mati

~ Plat Lantai = x = T/m2

~ Spesi Keramik = x = T/m2

~ Keramik = x = /m +T/m

Wu2   = x =

Wu =   Wu1 +   Wu2   =

PENYELESAIAN

Dari penyelesaian diatas dapat saya gambarkan sebagai berikut :

q2   = T/m' q2   = Wu x

=

=

q1   = T/m'

1). Menentukan Momen Frimer (MF)

0.353

2.756

1.225

1.2 0.354 0.425

2.756

2 0.021 0.042

1 0.024 0.024

0.5 0.8   T/

T/

1.225   T/m

0.354

500 Kg/m

0.12 2.4 0.288

1.6

4 M

A B

C

D

E

4.5 M 4.5 M

EI  EI

EI EI

2.25 M

8/13/2019 Slope Deflection Bergoyang

1 5

12 96

x   2 x

= +

= TM

1 5

12 96

x   2 x

= +

= TM

2). Rumus Slope Deflection

M AB   = 1 EI x θB   -   MBD   = x (

= EI θB   - = EI θB

MBA   = 1 EI x 2 θB   +   MDB   = x (

4.5

= EI θB   + = EI θD

MBC   = 1 EI x 2 θB   MDE   = 1 EI x 2 θD

4 4

= EI θB   = EI θD

MCB   = 1 EI x θB   MED   = 1 EI x θD

= EI θB   = EI θD

3). Joint Condition

MBA   +   MBC   +   MBD =   0

0.5 0.5

4 4

0.25 0.25

3.502 1 EI

4.5

0.44 3.502 0.44

3.502 1 EI

0.22 3.502 0.44

2.

9

0.5954 2.91

4.5 4.5

+

5

3.502

+   qL2

=

0.353 4.5

12

3.502

-   MFDB   =   MFBD   =   qL2

qL2

+

5 2.

9=

0.353 4.5

12

0.5954 2.91

=   qL2-   MF AB   =   MFBA    +

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

= TM

MBD   = x + x -

= - -

= TM

MDB   = x + x +

= + +

= TM

MDE   = x

= TM

MED   = x

= TM

6). Free Body

= T/m1

TM TM TM

= /m1

M

TM

4.5 4.

0.308

-3.85

-0.963

3.365 3.776 4.084

0.3528

2.756

-1.712 0.14 3.50

5.351

0.5 -3.85

-1.926

0.25

0.62 3.502

0.22   -3.85 3.502

-4.084

0.44 -3.85   0.22

0.274 0.86 3.50

0.154

0.44 0.62

q1

q2

A B q1

B

B

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

R  AV   = - = T

-

ΣV   = 0R  AV   + R BV   = R 1   + R 2

+ = +=

b). Momen dan Lintang

2. Titik potongan A dan B

= T/m

TM TM

= T/m1

T T

M M

M

~ Menghitung momen dari titik A ke titik BR 1   =

TM = T/m =

=

= T/m

R  AV   = T R 2   =

M

= X =

==

~ Bidang MomenMx = x X - - R  1   x - R 2   (

Interval 0 ≤ X ≥ 2,25

R  AV   3.365 X

2

0.3528

X

2.25q2'   2.76

2.25

3.365 2.756

x3.803   q2'

=2.25   2.76

3.803   R BV=   3.985

4.5

3.365 3.776

0.3528

17.112

2.25 2.25

7.788 7.788

2.756

R  AV=

3.803

4.5

3.803 3.985 1.59 6.201

q1

q2

A

q1

q2

A B

8/13/2019 Slope Deflection Bergoyang

= X - - X2

- X2

(

= X - X2

- X3

-

~ Bidang Lintang

Dx = - X - X2

~ Menghitung momen dari titik B ke titik A

TM

R BV   = T

M

~ Bidang Momen

Mx = - x -X - + R  1   x + R 2

= X - - X + X (

= X - X - X -

~ Bidang Lintang Dx = - f' (Mx) / Dx

Dx = + X + X2

3. Titik potongan B dan C

2.25

X Rumus Momen Rumus Lintang

0

-X

3.776

-3.985 0.3528 0.612

2

3.985 0.1764 0.204

x 3.985

2

3.985 3.776 0.353 0.612

Interval 0 ≤ X ≥ 2,25

R BV   3.776

2.25

2.25

3.776

0

3.803 0.3528 0.612

X Rumus Momen Rumus Lintang

0.1764 0.204

3.803 3.365 0.353 0.612

3.365

2

3.803

3.803 X - 0.1764 X2 - 0.204 X3- 3.365 3.803 - 0.3528 X - 0.612 X2

q1

q2

A

3.985 X - 0.1764 X2 - 0.204 X3- 3.776 - 3.985 + 0.3528 X + 0.612 X2

8/13/2019 Slope Deflection Bergoyang

T

TM

4 M

TM

T

a). Reaksi Perletakan

= 0- x 4 + + = 0

- 4 + = 0

- =

-

= 0

- x 4 + + = 0

- 4 + = 0

= T

b). Momen dan Lintang

T

TM

T

4 M

T

TM

T

7.598

0.308

0.116

0.116

0.154

7.598

0.116

-4

0.462 0.116

4

0.462

=  -0.462

ΣMC

0.308 0.154

ΣMB0.154 0.308

0.462

=

7.598

0.308

0.154

7.598

B

C

RCH

RCH

RCH

B

C

RCH

RBH

RBH

RBH

RCH

RBH

RBH

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

- R DV   = -

R DV   = - = T

-

ΣMD   = 0

R BV   x - R 1   x - R 2   x -R BV   - x - x -

R BV   - - - + = 0

R BV   - = 0

- R BV   = -

R BV   = - = T

-

ΣV   = 0R

BV

+ R DV

= R 1

+ R 2

+ = +

=

b). Momen dan Lintang

= T/m

TM TM

= T/m

T T

M M

M

R 1TM = T/m

1

= T/m1

R BV   = T R 23.612   q2'

4.084 2.756

X

0.3528

x

3.612   R DV=   4.176

4.5

16.256 3.612

4.5

2.25 2.25

7.788 7.788

2.756

4.084

R BV =

3.612 4.176 1.59 6.201

5.351

0.3528

5.351

4.5 16.256

4.5 16.256

4.5 3.572 13.95 4.084

4.0844.5 1.59 2.25 6.201 2.25 4.

18.791 4.176

4.5

4.5 2.25 2.25

4.5 18.791

q1

q2

B

q1

q2

B D

q2'

8/13/2019 Slope Deflection Bergoyang

M= X

~ Bidang Momen

Mx = x X - - R  1   x - R 2

= X - - X2

- X2

= X - X2   - X3 -

~ Bidang Lintang

Dx = - X - X2

~ Menghitung momen dari titik B ke titik D

TM

R DV   = T

M

~ Bidang MomenMx = - x -X - + R  1   x + R 2

= X - - X2

+ X2

(

2

4.176 5.351 0.353 0.612

Interval 0 ≤ X ≥ 2,25

R DV   5.351 -X

2.25

2.25

5.351

x 4.176

0

X Rumus Momen Rumus Lintang

0.353 0.612

2

3.612 0.1764 0.204 4.084

Interval 0 ≤ X ≥ 2,25

3.612 0.3528 0.612

2

3.612 4.084

R BV   4.084 X

2.25q2'   2.8

2.25

2.8 2.25

3.612 X - 0.1764 X2 - 0.204 X3- 4.084 3.612 - 0.3528 X - 0.612 X2

q1

q2

D

8/13/2019 Slope Deflection Bergoyang

= X - X - X -

~ Bidang Lintang Dx = - f' (Mx) / Dx

Dx = + X + X2

5. Titik potongan B dan C

T

TM

4 M

TM

T

a). Reaksi Perletakan

= 0

x 4 - - = 04 Rc - = 0

= T

= 0

x 4 - - = 0

4 - = 0

2.890 0.722

4

2.890

ΣME

1.926 0.963

ΣMD

0.963 1.9262.890

=

4.176

1.926

0.963

4.176

2.25

X Rumus Momen Rumus Lintang

0

5.351

-4.176 0.3528 0.612

2

4.176 0.1764 0.204

4.176 X - 0.3528 X2 - 0.612 X3- 5.351 - 4.176 - 0.3528 X - 0.612 X2

D

E

REH

RDH

REH

RDH

REH

RDH

8/13/2019 Slope Deflection Bergoyang

= T

b). Momen dan Lintang

TM

TMT

4 M

T

TMTM

~ Bidang Momen

Mx = x X -

= X -

~ Bidang Lintang

Dx =

Rumus Lintang

TM

TM

TM

9) Gambar

4   0.963 0.722

2.25   -0.301 0.722

1.926

0.722 1.926

Nilai Momen Nilai Lintan

0   -1.926 0.722

0.722

X Rumus Momen

4.176

1.9260.722

0.722

0.9634.176

Interval 0 ≤ X ≥ 4

0.722

4=

2.890

0.722 X - 1.926

D

E

0.722

RDH

RDH

REH

RDH

8/13/2019 Slope Deflection Bergoyang

TM TM

TM

TM

TM

TM

TM

Gambar Momen

T

T

T T

T

T

T

Gambar Lintang

-0.282

-0.116

-3.985

0.826

1.973

-0.154 0.

3.612

-0.091 -0.116

3.803

-3.365 -4.084

-3.776

0.308(+) (+)

(-) (-)(-)

(-)

(+)

(-)

(-)

(+)

A

B

C

A B

C

(+)

8/13/2019 Slope Deflection Bergoyang

T

Gambar Normal

7.598

(-)

A B

C

8/13/2019 Slope Deflection Bergoyang

/m1

0.294

Garis leleh

rmik

tai

alok

q1

q2

4.5 M

4.5 M

8/13/2019 Slope Deflection Bergoyang

t

x

T/m'

2.25

0

2

8/13/2019 Slope Deflection Bergoyang

x   2

x   2

θD

2 θB   + θD   ) -

+ EI θD   -

2 θD   + θB   ) +

+ EI θB   +

3.502

3.502

0.22 3.502

0.22 3.502

56 4.5

6

56 4.5

6

8/13/2019 Slope Deflection Bergoyang

EI θD   - ) = 0

……….. Pers ( 1 )

……….. Pers ( 2 )

3.50 2

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

4 M

TM

L

x = T

- = 0

- = 0

+ = 0

+ = 03.77665

3.776

76

4.5 6.201

3.365

3.365

5

0.963

C

8/13/2019 Slope Deflection Bergoyang

x

x

X

1 x al x t

20.5 x X x X

X x XX

X - 2 X )

3

2.25

0.6120.5 1.225

0.3528

2.76

0.3528 X

q1   X

8/13/2019 Slope Deflection Bergoyang

1 X )

3

( -X + 2 X )3

-1 X )

3

1.973 TM -0.091

-3.985   TTM

T

Nilai Momen Nilai Lintang

-3.776

1.973 TM -0.091 T

Nilai Momen Nilai Lintang

-3.365 TM 3.803   T

8/13/2019 Slope Deflection Bergoyang

8/13/2019 Slope Deflection Bergoyang

x L

x = T

- = 0

- = 04.08451

4.5

4.084

4.5

6 6.201

L

g

8/13/2019 Slope Deflection Bergoyang

+ = 0+ = 0

= x

= x

= X

= 1 x al x t

0.3528

0.3528 X

q1   X

5.35184 5.351

8/13/2019 Slope Deflection Bergoyang

2

= 0.5 x X x X

= X x X

=   X

( X + 2 X )

3

( 1 X )

3

( -X + 2 X )

3

-1 X )

T

T0.826 TM -0.282

3.612

Nilai Lintang

-4.084 TM

Nilai Momen

0.5 1.225

0.612

2.8

2.25

8/13/2019 Slope Deflection Bergoyang

3

( )

0.826 TM -0.282

-4.176   TTM

T

Nilai Momen Nilai Lintang

-5.351

8/13/2019 Slope Deflection Bergoyang

( )

TM-5.351

g

8/13/2019 Slope Deflection Bergoyang

TM

TM

T

T

TM

0.722

0.722

-4.176

63

-1.926

(+)

(-)

(-)

(+)

(-)

D

E

D

E

8/13/2019 Slope Deflection Bergoyang

T

11.774

3.803

15.576

7.788

7.788

15.576

4.176

(-)

D

E