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Tutorial

Analisis Struktur Statis Tertentu ITA. 2011/2012

Dosen Mata Kuliah:Prof. Ir. Hrc. Priyosulistyo, M.Sc., Ph.D. Dr.-Ing. Ir. Andreas TriwiyonoIr. Suprapto Siswosukarto, Ph.D. Dr.-Ing. Ir. Djoko Sulistyo

Ali Awaludin, S.T., M.Eng., Ph.D.

Asisten Tutorial: N M Setyo Nugroho, S.T.

Korps Asisten Teknik SipilJurusan Teknik Sipil dan Lingkungan

Fakultas Teknik Universitas Gadjah Mada Yogyakarta

tutorial asst1-nmsn-2011

E

10 T

2 T/m

satuan dalam Ton dan meter

C balok gerber (1)struktur

ontoh

A

S1

5 T

B C D

S2

5 5 2 26 4 6

F

5

3

4


E



ontoh

A

5 T

B C D

2 T/m

S1 S2

5 5 2 26 4 6

F

5

3

42 T/m2 T/m

5 T

3 T

4 T

3 T

4 T

10 T




ontoh 2 T/m

S1 S2

6( ) ( ) ( )

( ) ( ) ( )

TRS

RS

RS

RSRS

MS

6

366

06360

063620

0

2

2

2

21

1

=

=

=−+

=×−××+×

=∑ RS2RS1

( ) ( ) ( )( ) ( ) ( )

TRS

RS

RS

RSRS

MS

6

366

00366

003626

0

1

1

1

21

2

=

=

=+−

=×+××−×

=∑( ) ( ) ( )

( ) ( ) ( )( )!00

06126

062

0

21

OK

RSRS

FvCEK

=

=−−

=+×−

=∑


E



ontoh

A B C D

2 T/m

S1 S2

5 5 2 26 4 6

F

2 T/m2 T/m

3 T

4 T

66

6610 T


E



ontoh

A B

5 5 2

2 T/m

3 T

4 T6

RBvRAv

( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )TRBv

RBv

RBv

RBvRAhRAv

MA

6,13

010136

072441002000

01261102210035400

0

=

=−

=++−−++

=×++××+×−×−×+×+×

=∑

RAh


E



ontoh

A B

5 5 2

2 T/m

3 T

4 T6

RBvRAv

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )TRBv

RAv

RAv

RBvRAhRAv

MB

4,0

0410

01240020010

02612200354010

0

=

=−

=+++−−+

=×+××+×+×−×−×+×

=∑

RAh


E



ontoh

A B

5 5 2

2 T/m

3 T

4 T6

RBvRAv

( ) ( ) ( ) ( ) ( )( ) ( ) ( )

)!(00

06,13144,0

06224

0

OK

RBvRAv

Fv

=

=+−

=+−×−−

=∑

RAh


E



ontoh

A B

5 5 2

2 T/m

3 T

4 T6

RBvRAv

( ) ( )( ) TRAh

RAh

Fh

3

03

0

=

=−

=∑

RAh




ontoh

C D

2 4 6

F

2 T/m

6 10 T

RDvRCv

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )TRBv

RBv

RDv

RDvRCv

MC

4,2

01024

010400412

010410012226

0

=

=−

=−++−−

=×−×+×+××−×−

=∑




ontoh

C D

2 4 6

F

2 T/m

6 10 T

RDvRCv

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )TRCv

RCv

RCv

RDvRCv

MD

6,17

010176

0060104472

00610101122126

0

=

=+−

=+−+−−

=×+×−×+××−×−

=∑




ontoh

C D

2 4 6

F

2 T/m

6 10 T

RDvRCv

( ) ( ) ( ) ( ) ( )( ) ( ) ( )

)!(00

04,2206,17

010226

0

OK

RBvRCv

Fv

=

=+−

=+−×−−

=∑




ontoh

E

10 T

2 T/m

A

S1

B C D

S2 F

2,4T17,6T13,6T0,4T

Reaksi Tumpuan

3 T

4 T

3T




ontoh

E

10 T

2 T/m

A

S1

B C D

S2 F

2,4T17,6T13,6T0,4T

3 T

4 T

3T

NFD

3T

‐




ontoh

E

10 T

2 T/m

A

S1

B C D

S2 F

2,4T17,6T13,6T0,4T

3 T

4 T

3T

SFD

0,4T+

‐3,6T

10T

10T

+

‐

+

7,6T

‐ 2,4T




ontoh

E

10 T

2 T/m

A

S1

B C D

S2 F

2,4T17,6T13,6T0,4T

3 T

4 T

3T

SFD

2Tm+

16Tm

10T

14,4Tm

‐

+

‐

‐

+

16Tm

‐

+


dimensi dalam meter

Q balok gerber (2)struktur

uiz

10 kN

2 kN/m

A

S1

B C D

4 kN/m 5

3

453

4

5

15 kN

10 10 5 510 17 35

S2


dimensi dalam meter


uiz

2 kN/m

A

S1

B C D

4 kN/m

5

9kN

10 10 5 510 17 35

S2

5kN 10kN 10kN 15kN 20kN 20kN 14,8kN 3,2kN

6kN12 kN

8kN

6 kN

5 5 5 5 5 2


dimensi dalam meter


uiz

A

S1

B C D

10 10 5 510 17 35

S2

5kN 10kN 10kN

15kN 20kN 20kN 14,8kN 3,2kN

6kN

9kN

12 kN

10kN 15kN

5

5 5 5 5



uiz

A B

10 10 55

9kN

12 kN

10kN

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( )kNRBv

RBv

RBv

RBvRAhRAv

MA

17

34020

02025090000

020251010901200

0

=

=

=−++++

=×−×+×+×+×+×

=∑

RAh

RAv RBv



uiz

A B

10 10 55

9kN

12 kN

10kN

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( )kNRAv

RAv

RAv

RBvRAhRAv

MB

2

4020

0050900020

00510109012020

0

=

=

=−+−++

=×−×+×−×+×+×

=∑

RAh

RAv RBv



uiz

C D

5 17 3

15kN 20kN 20kN 14,8kN 3,2kN

6kN

15kN

RCv RDv

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )kNRDv

RDv

RDv

RDvRCv

MC

55,25

51120

0206402222001000075

020202,306158,1410205200150515

0

=

=

=−+++++++−

=×−×+×+×+×+×+×+×+×−

=∑



uiz

C D

5 17 3

15kN 20kN 20kN 14,8kN 3,2kN

6kN

15kN

RCv RDv

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )kNRBv

RCv

RCv

RDvRCv

MD

45,62

249.120

00007420030030020375

0002,30658,14102015202015202515

0

=

=

=+++−−−−+−

=×+×+×+×−×−×−×−×+×−

=∑


dimensi dalam meter


uiz

2 kN/m

A

S1

B C D

4 kN/m

5

10 10 5 510 17 35

S2

9kN

12 kN

8kN

6 kN

RAh

( ) ( ) ( )kNRAh

RAh

Fh

6

0612

0

=

=−+−

=∑



uiz

2 kN/m

A

S1

B C D

4 kN/m

15

10 10 5 510 17 35

S2

62,45kN 25,55 kN2 kN 17kN

9kN

12 kN

8kN

6 kN

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) )!(0107107

055,25815445,621521792

0

OK

Fv

CEK

=−

=+−×−+×−+−

=∑

15



uiz

10 10 5 510 15 55

6

6

A S1B C DS2


6kN

9kN

12 kN

62,45kN 25,55 kN2 kN 17kN

00

6

NFD (kN)



uiz

10 10 5 510 15 55

2

7

105

515

A S1B C DS2


6kN

9kN

12 kN

62,45kN 25,55 kN2 kN 17kN32,45

12,45

7,55

22,35

00

6

SFD (kN)



uiz

10 10 5 510 15 55

2

25

87,25

A S1B C DS2


6kN

9kN

12 kN

62,45kN 25,55 kN2 kN 17kN

75

149,5

00

6

BMD (kNm)

20

50

111,75

TERIMA KASIH

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