mekrek-1

Tugas Besar

Mekanika Teknik

Dikerjakan oleh :

Lionel ‘oneil’ Zakhy F 111 10 001

Jurusan Teknik Sipil - Fakultas Teknik Universitas Tadulako

2009

Buatlah perhitungan Analisis Struktur portal 3 sendi seperti yang tergambar berikut.

2.0 m 4.0 m 6.0 m

6.0 m

q1 = 3 t/m q2 = 4 t/m q3 = 3 t/m

q4 = 3 t/mA

C S D

B

h1 =

h2 =

h3 =

L1 = L2 = L3 =

Q1Q2 Q3

Q4

P = 6 t

α

3.0 m

2.0 m

Penyelesaian :

A. Beban – beban yang bekerja 1 1

1 1 1 3 3 32 21 1 1 1

2 2 2 4 4 32 2 2 2

(3)(2) 6 (3)(6) 18(4)(4) 8 (3)(6) 9

Q q L ton Q q L tonQ q L ton Q q h ton

= = = = = =

= = = = = =

B. Parameter-parameter pada batang miring (AC) 1. Panjang batang AC :

( ) ( ) ( )2 2 221 1 2 2 3 2 29 5,38517ACL L h h m= + + = + + =

1 2 1 2

1

1

5 5sin ; tan 2.5229

2cos29

AC

AC

h h h hL LLL

α α

α

+ += = = = =

= =

2. Panjang a1 dan a2 serta d1 dan d2 :

( )

( )

11

22

11 5

29

22 5

29

3 1.2tan 2.5

2 0.8tan 2.5

3 3.23110sin

2 2.15407sin

ha m

ha m

hd m

hd m

α

α

α

α

= = =

= = =

= = =

= = =

3.0 m

2.0 m

q1 = 3 t/m

A

C

α

α

2.0 m

a1 a2

d1

d2

P = 6 t

C. Menghitung Reaksi Perletakan

2.0 m 4.0 m 6.0 m

6.0 m

q1 = 3 t/m q2 = 4 t/m q3 = 3 t/m

q4 = 3 t/mA

C S D

B

h1 =

h2 =

h3 =

L1 = L2 = L3 =

Q1Q2 Q3

Q4

P = 6 t

α

3.0 m

2.0 m

RAH

RAV

RBV

RBH1.0 mh0 =

Beda tinggi perletakan (ho):

0 3 1 2 6 3 2 1h h h h m= − − = − − =

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

( ) ( ) ( ) ( )

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

( ) ( )

0 3 2 1

2 1 1 11 1 2 2 3 3 3 4 33 3 2 3

2 1 1 13 3 2 3

1763

sin cos0

5 212 1 6 6 2 6 12 1.229 29

3 12 2 8 4 6 18 6 9 6 0

120 129.612 1 32 54 18 029 29

12

AV AH

AV AH

AV AH

AV AH

R L R h P h h P L a

Q L L Q L L Q L Q h

R R

R R

R R

α α⇒ + + − − −

− − − + − − =

⎛ ⎞ ⎛ ⎞⇒ + + − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− − − + − − =

⎛ ⎞ ⎛ ⎞⇒ + + − − − − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ + 164.44934 012 164.44934 ..... (1)AV AHR R pers

− =⇒ + =

0BMΣ =

( )Σ = 0SM kiri

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

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

1 2 1 2 2 2 2

1 11 1 2 2 23 3

1 13 3

323

sin cos0

5 22 4 3 2 6 2 6 0.8 429 29

3 (2) 4 8 (4) 0

60 57.66 5 14 029 29

6 5 46.50444 06 5 46.50444 .

AV AH

AV AH

AV AH

AV AH

AV AH

R L L R h h P h P a L

Q L L Q L

R R

R R

R RR R

α α⇒ + − + − − +

− + − =

⎛ ⎞ ⎛ ⎞⇒ + − + − − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− + − =

⎛ ⎞ ⎛ ⎞⇒ − − − − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ − − =⇒ − = .... (2)pers

Eliminasi pers(1) dan (2) diperoleh :

( )( )

13.16290

6.49459AV

AH

R ton

R ton

= + ↑

= + →

Σ = 0AM

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

( ) ( ) ( ) ( )

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

( ) ( )

0 1 1

2 2 1 11 1 2 2 1 3 3 4 3 03 3 2 3

2 2 1 13 3 2 3

1123

sin cos0

5 212 1 6 3 6 1.229 29

3 2 8 4 2 18 12 6 9 6 1 0

90 14.412 1 4 162 9 029 29

12 213

BV BH

BV BH

BV BH

BV AH

R L R h P h P a

Q L Q L L Q L L Q h h

R R

R R

R R

α α⇒ − − + +

+ + + + − − − =

⎛ ⎞ ⎛ ⎞⇒ − − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ + + + − − − =

⎛ ⎞ ⎛ ⎞⇒ − − + + + + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ − − + .71993 012 213.71993 ..... (3)AV AHR R pers

=⇒ + =

( )Σ = 0SM kanan

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

1 23 3 3 3 4 32 3

1 22 3

0

6 6 18 6 9 6 0

6 6 54 36 06 6 90 0

6 6 90 ..... (4)

BV BH

BV BH

BV BH

BV BH

BV BH

R L R h Q L Q h

R R

R RR R

R R pers

⇒ − − + + =

⇒ − − + + =

⇒ − − + + =

⇒ − − + =⇒ + =

Eliminasi pers(3) dan (4) diperoleh :

( )( )

18.06545

3.06545BV

BH

R ton

R ton

= + ↑

= − ←

Hasil perhitungan reaksi perletakan digambarkan sebagai berikut :

2.0 m 4.0 m 6.0 m

6.0 m

q1 = 3 t/m q2 = 4 t/m q3 = 3 t/m

q4 = 3 t/mA

C S D

B

h1 =

h2 =

h3 =

L1 = L2 = L3 =

Q1Q2 Q3

Q4

P = 6 t

α

3.0 m

2.0 m

RAH = 6.49459 ton

RAV = 13.16290 tonRBV = 18.06545 ton

RBH = 3.06545 ton

1.0 mh0 =

Kontrol keseimbangan Statis :

1 2 3

4

0 cos 0213.16290 18.06545 6 3 8 18 029

0.00001 0 ..... .!!0 sin 0

56.49459 3.06545 6 9 029

0.00000 0 ..... .!!

AV BV

AH BH

V R R P Q Q Q

OKH R R P Q

OK

α

α

Σ = ⇒ + − − − − =

⎛ ⎞+ − − − − =⎜ ⎟⎝ ⎠

Σ = ⇒ − + − =

⎛ ⎞− + − =⎜ ⎟⎝ ⎠

D. Menghitung Gaya – gaya Batang

1. Batang AC Karena merupakan batang miring maka terlebih dahulu dilakukan transformasi gaya berdasarkan arah aksial dan arah lateral batang. - Transformasi reaksi perletakan

Arah Aksial : cos sin

2 56.49459 13.1629029 29

14.63348

ax AH AVR R R

ton

α α= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

Arah Lateral : cos sin

2 513.16290 6.4945929 29

1.14150

lt AV AHR R R

ton

α α= −

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

- Transformasi beban

Arah Aksial :

α =

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

11 2

12

sin53 2929

2 15 1.03448 /29 29

ax AC

ax

ax

Q q L

q

q ton m

Arah Lateral :

α =

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

11 2

12

cos23 2929

2 6 0.41379 /29 29

lt AC

lt

lt

Q q L

q

q ton m

q1 = 3 t/m

A

C

α

LAC

α

Q1

Q1 cos α Q1 si

n α

RAH

RAV

αα

RAH sin α

RAV cos α

RAH c

os α

RAV s

in α

Q1 cos αLAC

qlt

P = 6 t

Rlt

d2 = 2.15407 md1 = 3.23110 m

x1

x2

AC

Q1 sin α

LAC

qax

x1

A C

Rax

Arah LateralArah Aksial

Bidang Momen (Mx), Bidang Geser (Dx) dan Bidang Normal (Nx)

( )

( ) ( )

( ) ( )

( )

3

1 1

3

3

211

........06

0.413791.14150

6 291.14150 0.012806

1.14150 0.038419

ltx lt

AC

xx

q xM R x x dL

xx

x xdMD xdx

= − − ≤ ≤

= − −

= − −

= = − −

( )

( ) ( )

( ) ( )

( )

3

2 1 1

3

3

222

( ) ........6

0.413791.14150 6( 3.23110)

6 2919.38660 7.14150 0.012806

7.14150 0.038419

ltx lt AC

AC

xx

q xM R x P x d d x LL

xx x

x xdMD xdx

= − − − − ≤ ≤

= − − − −

= − −

= = − −

( )

( )

2

2

2

........02

1.0344814.63348

2 2914.63348 0.09605

axx ax AC

AC

q xN R x LL

x

x

= − + ≤ ≤

= − +

= − +

Tabel perhitungan MDN batang AC

Interval x Mx (ton.m) Dx(ton) Nx(ton) 0.000 0.000 -1.142 -14.633 0.500 -0.572 -1.151 -14.609 1.000 -1.154 -1.180 -14.537 1.500 -1.755 -1.228 -14.417 2.000 -2.385 -1.295 -14.249 2.500 -3.054 -1.382 -14.033 3.000 -3.770 -1.487 -13.769

10 x d≤ ≤

3.231 -4.120 -1.543 -13.631 3.231 -4.120 -7.543 -13.631 3.500 -6.158 -7.612 -13.457 4.000 -9.999 -7.756 -13.097 4.500 -13.917 -7.919 -12.688 5.000 -17.922 -8.102 -12.232

10 x d≤ ≤

5.385 -21.072 -8.256 -11.848

2. Batang CS (0 ≤ x ≤ 4)

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

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

( )

321

1 3

323

3

2

2 5 sin 2 cos 0.8 26

5 2 413.16290 2 6.49459 5 6 2 6 0.8 36 429 29

21.07155 7.93456( ) 0.16667( )

7.93456 0.5

x AV AHCS

xx

q xM R x R P P x Q xL

xx x x

x xdMD xdx

α α= + − − − + − + −

⎛ ⎞ ⎛ ⎞= + − − − + − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − + −

= = −

sin

56.49459 629

12.06545 ( )

x AHN R P

ton tekan

α= − −

⎛ ⎞= − − ⎜ ⎟⎝ ⎠

= −

Check apakah Mmax terdapat dalam interval Syarat : Mmax Dx = 0

( )2

2

7.93456 0.5 07.93456

0.53.98361

x

x

x m

− =

=

=

2.0 m 4.0 m

q1 = 3 t/m q2 = 4 t/m

A

C S

RAH

RAV

L1 = L2 =

x

α

P = 6 t

Tabel perhitungan MDN batang CS

Interval x Mx (ton.m) Dx(ton) Nx(ton) 0.000 -21.072 7.935 -12.065 1.000 -13.304 7.435 -12.065 2.000 -6.536 5.935 -12.065 3.000 -1.768 3.435 -12.065

3.98361 0.001 0.000 -12.065

(0 ≤ x ≤ 4)

4.000 0.000 -0.065 -12.065

3. Batang DS (Ditinjau sebelah kanan S)

6.0 m

6.0 m

q3 = 3 t/m

q4 = 3 t/m

S D

B

RBV

RBH

h3 =

x

( ) ( ) ( )( )

( ) ( ) ( ) ( )

( ) ( ){ }( ){ }

232

4 3

2

2

6 62

318.06545 3.06545 6 9 4

2

54.39270 18.06545 1.5

18.06545 3

x BV BH

x

q xM R x R Q

xx

x x

dMxD xdx

⎧ ⎫= − − + + +⎨ ⎬

⎩ ⎭⎧ ⎫⎪ ⎪= − − + + +⎨ ⎬⎪ ⎪⎩ ⎭

= − − +

= − = − +

4

3.06545 912.06545 ( )

x BHN R Q

ton tekan

= − −= − −= −


( )18.06545 3 06.02182 ( )

xx m tdk memenuhi− + =

=

Tabel perhitungan MDN batang DS

Interval x Mx (ton.m) Dx(ton) Nx(ton) 0.000 -54.393 -18.065 -12.065 1.000 -37.827 -15.065 -12.065 2.000 -24.262 -12.065 -12.065 3.000 -13.696 -9.065 -12.065 4.000 -6.131 -6.065 -12.065 5.000 -1.565 -3.065 -12.065

(0 ≤ x ≤ 6)

6.000 0.000 -0.065 -12.065

4. Batang BD

6.0 m

q4 = 3 t/m

D

B

RBV

RBH

h3 =

x

( )

( ) ( )

( ) ( ) ( ){ }( ) ( ){ }

2 34 4

2 3

2 3

2

2 6

3 33.065452 6 6

3.06545 1.5 0.08333

3.06545 3 0.25

x BHBD

x

q x q xM R xL

x xx

x x x

dMxD x xdx

⎧ ⎫= − + −⎨ ⎬

⎩ ⎭⎧ ⎫⎪ ⎪= − + −⎨ ⎬⎪ ⎪⎩ ⎭

= − + −

= − = + −

18.06545 ( )x BVN R ton tekan= − = −


( ) ( )( )( )( )

2

2

1

2

3.06545 3 0.25 0

3 3 4 0.25 3.065452 0.25

3 3.473540.5

3 3.47354 0.94708 ( )0.5

3 3.47354 12.94708 ( )0.5

x x

x

x

x m tdk memenuhi

x m tdk memenuhi

+ − =

− ± − −=

−

− ±=

−− +

= = −−

− −= =

−

Tabel perhitungan MDN batang BD Interval x Mx (ton.m) Dx(ton) Nx(ton)

0.000 0.000 3.065 -18.065 1.000 -4.482 5.815 -18.065 2.000 -11.464 8.065 -18.065 3.000 -20.446 9.815 -18.065 4.000 -30.928 11.065 -18.065 5.000 -42.411 11.815 -18.065

(0 ≤ x ≤ 6)

6.000 -54.393 12.065 -18.065

E. Gambar Bidang Momen

A

C S D

B

SKALA JARAK

0 1 m 2 m 3 m

0 8 t.m 16 t.m 24 t.m

SKALA MOMEN

BIDANG MOMEN

-21.072 tm

-54.393 tm

-54.393 tm

-21.072 tm

-4.120 tm

Mmax = +0.001 tm

F. Gambar Bidang Lintang

A

CS

D

B

BIDANG LINTANG

+7.935 t

-0.065 t

-8.256 t

-7.543 t

-1.543 t

-1.142 t

-18.065 t

+3.065 t

+12.065 t

SKALA JARAK

0 1 m 2 m 3 m

0 5 t 10 t 15 t

SKALA GAYA

G. Gambar Bidang Normal

A

C S D

B

BIDANG NORMAL

-12.065 t

-18.065 t

-14.633 t

-13.631 t

-11.848 t

-18.065 t

SKALA JARAK

0 1 m 2 m 3 m

0 5 ton 10 ton 15 ton

SKALA GAYA

VERIFIKASI HASIL PERHITUNGAN MENGGUNAKAN PROGRAM SAP 2000

SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 1 12/11/09 13:36:17 S T A T I C L O A D C A S E S STATIC CASE SELF WT CASE TYPE FACTOR MATI DEAD 0.0000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 2 12/11/09 13:36:17 J O I N T D A T A JOINT GLOBAL-X GLOBAL-Y GLOBAL-Z RESTRAINTS ANGLE-A ANGLE-B ANGLE-C 1 0.00000 0.00000 1.00000 1 1 1 0 0 0 0.000 0.000 0.000 2 2.00000 0.00000 6.00000 0 0 0 0 0 0 0.000 0.000 0.000 3 6.00000 0.00000 6.00000 0 0 0 0 0 0 0.000 0.000 0.000 4 12.00000 0.00000 6.00000 0 0 0 0 0 0 0.000 0.000 0.000 5 12.00000 0.00000 0.00000 1 1 1 0 0 0 0.000 0.000 0.000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 3 12/11/09 13:36:17 F R A M E E L E M E N T D A T A FRAME JNT-1 JNT-2 SECTION ANGLE RELEASES SEGMENTS R1 R2 FACTOR LENGTH 1 1 2 EI 0.000 000000 2 0.000 0.000 1.000 5.385 2 2 3 EI 0.000 000002 4 0.000 0.000 1.000 4.000 3 3 4 EI 0.000 000001 4 0.000 0.000 1.000 6.000 4 4 5 EI 0.000 000000 2 0.000 0.000 1.000 6.000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 4 12/11/09 13:36:17 F R A M E S P A N D I S T R I B U T E D L O A D S Load Case MATI FRAME TYPE DIRECTION DISTANCE-A VALUE-A DISTANCE-B VALUE-B 1 FORCE Z PROJ 0.0000 0.0000 1.0000 -3.0000 2 FORCE GLOBAL-Z 0.0000 0.0000 1.0000 -4.0000 3 FORCE GLOBAL-Z 0.0000 -3.0000 1.0000 -3.0000 4 FORCE GLOBAL-X 0.0000 0.0000 1.0000 -3.0000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 5 12/11/09 13:36:17 F R A M E S P A N P O I N T L O A D S Load Case MATI FRAME TYPE DIRECTION DISTANCE VALUE 1 FORCE LOCAL-2 0.4000 0.0000 1 FORCE LOCAL-2 0.6000 -6.0000

SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 1 12/11/09 13:36:46 J O I N T D I S P L A C E M E N T S JOINT LOAD U1 U2 U3 R1 R2 R3 1 MATI 0.0000 0.0000 0.0000 0.0000 71.5763 0.0000 2 MATI 397.4384 0.0000 -158.9753 0.0000 104.7394 0.0000 3 MATI 397.4384 0.0000 -670.4013 0.0000 0.0000 0.0000 4 MATI 397.4384 0.0000 0.0000 0.0000 -29.9469 0.0000 5 MATI 0.0000 0.0000 0.0000 0.0000 106.2330 0.0000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 2 12/11/09 13:36:46 J O I N T R E A C T I O N S JOINT LOAD F1 F2 F3 M1 M2 M3 1 MATI 6.4945 0.0000 13.1628 0.0000 0.0000 0.0000 5 MATI -3.0656 0.0000 18.0656 0.0000 0.0000 0.0000 SAP2000 v7.40 File: PORTAL GERBER Ton-m Units PAGE 3 12/11/09 13:36:46 F R A M E E L E M E N T F O R C E S FRAME LOAD LOC P V2 V3 T M2 M3 1 MATI 0.00 -14.63 1.14 0.00 0.00 0.00 0.00 2.69 -13.94 1.42 0.00 0.00 0.00 -3.32 5.39 -11.85 8.26 0.00 0.00 0.00 -21.07 2 MATI 0.00 -12.07 -7.93 0.00 0.00 0.00 -21.07 1.00 -12.07 -7.43 0.00 0.00 0.00 -13.30 2.00 -12.07 -5.93 0.00 0.00 0.00 -6.54 3.00 -12.07 -3.43 0.00 0.00 0.00 -1.77 4.00 -12.07 6.555E-02 0.00 0.00 0.00 0.00 3 MATI 0.00 -12.07 6.555E-02 0.00 0.00 0.00 0.00 1.50 -12.07 4.57 0.00 0.00 0.00 -3.47 3.00 -12.07 9.07 0.00 0.00 0.00 -13.70 4.50 -12.07 13.57 0.00 0.00 0.00 -30.67 6.00 -12.07 18.07 0.00 0.00 0.00 -54.39 4 MATI 0.00 -18.07 -12.07 0.00 0.00 0.00 -54.39 3.00 -18.07 -9.82 0.00 0.00 0.00 -20.45 6.00 -18.07 -3.07 0.00 0.00 0.00 0.00

SAP2000

SAP2000 v7.40 - File:portal gerber - Moment 3-3 Diagram (MATI) - Ton-m Units

12/11/09 13:38:24

SAP2000

SAP2000 v7.40 - File:portal gerber - Shear Force 2-2 Diagram (MATI) - Ton-m Units

12/11/09 13:40:15

SAP2000

SAP2000 v7.40 - File:portal gerber - Axial Force Diagram (MATI) - Ton-m Units

12/11/09 13:40:36

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