Tugas Mekanika Teknik

Tugas Mekanika Teknik

Tugas Besar


Buatlah perhitungan Analisis Struktur portal 3 sendi seperti yang tergambar t ergambar berikut. q2 = 5 t/m'

P2 = 3 t

q1 = 3 t/m'

q3 = 2 t/m'

m

Q1

0 0 . 3 = 2 h

D

S

C Q2

Q3

m 0 0 . 4 = 3 h

P1 = 6 t

B m 0 0 . 4 = 1 h

m 0 0 . 3 = 0 h

A

L1 = 2.00 m

L2 = 4.00 m

L3 = 3.00 m

L4 = 2.00 m

L = 11.00 m

Penyelesaian Penyelesaian : A. Beban – beban yang bekerja Q1 = 21 q1L1

= 21 ( 3)( 2)

= 3 ton

Q2 = 12 q2L2

= 21 ( 5)( 4)

= 10 ton

Q3 = 21 q3h3

= 21 ( 2)( 4)

= 4 ton


C. Menghitung Reaksi Perletakan q2 = 5 t/m'

P2 = 3 t

q1 = 3 t/m'

q3 = 2 t/m'

m

Q1

0 0 . 3 = 2 h

D

S

C Q2

Q3

m 0 0 . 4 = 3 h

P1 = 6 t

RBH

B

m 0 0 . 4 = 1 h

RBV

RAH

m 0 0 . 3 = 0 h

A

RAV L1 = 2.00 m

L2 = 4.00 m

L3 = 3.00 m

L4 = 2.00 m

L = 11.00 m

Beda tinggi perletakan (ho): h0 = h1 + h 2 − h3 = 4 + 3 − 4 = 3 m

ΣMB = 0 ⇒

R

(L )

R

(h ) + P ( h

h

)

Q (L

2

L) Q

(1L + L + L )

P (L

)

Q

(2 h )

0


ΣM A = 0 ⇒

−RBV ( L ) + R BH ( h0 ) + P1 ( h1 ) + Q1 ( 23 L1 ) + Q2 ( L1 + 23 L2 ) + P2 ( L − L4 ) − Q3 ( h0 + 23 h3 ) = 0

⇒

−RBV (11) + RBH ( 3 ) + 6 ( 4 ) + 3 ( 23 (2) ) + 10( 2 + 23 ( 4)) + 3( 11− 2 ) − 4 ( 3 + 23 ( 4)) = 0

⇒

−RBV (11) + RBH ( 3 ) + 24 + 4 + 46, 66667 + 27 − 22, 66667 = 0

⇒

−11RBV + 3R AH + 79 = 0

⇒

−11R

+ 3R

AV

= − 79

AH

.....pers( 3)

ΣMS ( kanan ) = 0 ⇒

−RBV ( L3 + L4 ) − RBH ( h3 ) + P2 ( L3 ) + Q3 ( 31 h3 ) = 0

⇒

−RBV ( 3 + 2 ) − RBH ( 4 ) + 3( 3) + 4 ( 31 (4 (4) ) = 0

⇒

−RBV ( 5 ) − RBH ( 4 ) + 9 + 5, 33333 = 0

⇒

−5RBV − 4RBH + 14, 33333 = 0

⇒

5RBV + 4RBH = 14, 33333

Eliminasi pers(3) dan (4) diperoleh : RBV = + 6,08475ton ( ↑ ) RBH = − 4.02260 ton ( ← )

Hasil perhitungan reaksi perletakan digambarkan sebagai berikut :

.....pers ( 4)


Kontrol keseimbangan Statis :

ΣV = 0 ⇒

R

+ R BV− Q1 − Q2 − P2 = 0

AV

9, 91525 + 6, 08475 − 3 − 10 − 3 = 0 0 .0 0 0 0 0 = 0

ΣH = 0 ⇒

R

.....OK .!!

− R BH+ P1 − Q3 = 0

AH

2, 02260 − 4, 02260 + 6 − 4 = 0 0 .0 0 0 0 0 = 0

.....OK .!!

Kontrol keseimbangan keseimbangan momen pada setiap setiap titik kumpul : Syarat : ΣM = 0 -

Titik kumpul C

ΣM C (kiri )

ΣM C (kanan)

=

RAV ( L1 ) − RAH ( h1 + h2 ) − P1( h2 ) − Q1 ( 31 L1 )

=

9, 91525 ( 2 ) − 2, 02260( 4 + 3) − 6(3) − 3( 31 ( 2))

=

−14,32770 tm

=

−RBV ( L − L1 ) + RBH ( h3 ) + Q2 ( 23 L2 ) + P2 ( L2 + L3 ) + Q3 ( 31 h3 )

=

−6, 08475 (11 1 1 − 2 ) + 4, 02260( 4) + 10( 23 ( 4)) + 3( 4 + 3) + 4( 31 ( 4))

= =

−54,76 ,76275 + 16,09 ,09040 + 26,66 ,66667 + 21 + 5, 33333 14,32765 tm


-

Titik kumpul S

ΣM S (kiri )

ΣM S (kanan)

=

1 1 RAV ( L1 + L2 ) − RAH AH ( h1 + h2 ) − P1 ( h2 ) − Q1 ( 3 L1 + L2 ) − Q2 ( 3 L2 )

=

9, 91525 ( 2 + 4 ) − 2, 02260( 4 + 3) − 6(3) − 3( 31 ( 2) + 4) − 10( 31 ( 4))

= =

59, 49150 − 14,15820 − 18 − 14 − 13, 33333

=

−RBV ( L3 + L4 ) + RBH ( h3 ) + Q3 ( 31 h3 ) + P2 ( L3 )

=

−6, 08475 ( 3 + 2) + 4, 02260( 4) + 4( 31 ( 4)) + 3( 3)

= =

−30, 42 42375 + 16, 09040 + 5, 33333 + 9

−0, 00003 tm ≈ 0

0, 00002 tm



0

... OK .!

....OK .!

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


Q1 sinα = 21 qax LAC

2Q1 sinα 2(3)sin74,05460 0 qax = = = 0, 79245 t / m L AC 7,28011

Arah Lateral : Q1 cosα = 21 qlt LAC

2Q1 cosα 2( 3) cos 74, 054600 q lt = = = 0, 22 22642 t / m L AC 7,28011

Hasil transformasi beban digambarkan sebagai berikut : x2 d1 = 4,16006 m d1 = 4,16006 m

d2 = 3,12005 m

P1 sin

x1

α

qlt

A Q1 cos LAC Rlt

α

C

Q1 sin

α

A

P1 cos α LAC Arah Aksial

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

= Rlt ( x ) −

qlt x 3

6L

qax

Rax

Arah Lateral

M x1

d2 = 3,12005 m

x1

........0 ≤ x ≤ 4,16006

C


Mx2

= Rlt ( x ) − ( P1 sinα ) ( x − d1 ) −

q lt x 3

........4,16006 ≤ x ≤ 7, 28011

6L AC

= 0, 77915 ( x ) − ( 6 sin 74, 05460 ) (x − 4,16006) − 0

= 24 − 4, 98999 ( x ) − 0, 005184 ( x ) Dx 2 = Nx2

dM x 2 dx

= −4,98 ,98999 − 0,01 ,015551( x )

= −Rax − P1 cosα +

0,22642( x )

6(7,28011)

3

2

qax x 2

2L AC

= −10,0 10,089 8940 40 − 6cos74 6cos74,0 ,054 5460 60 + 0

= − 11,7 11,737 3773 73 + 0,054 ,05443 43( x )

0,79245( x )

3

2

2(7,28011)

2

Tabel perhitungan MDN batang AC Interval

x

Mx (ton.m)

Dx(ton)

Nx(ton)

0 ≤ x ≤ 4,16006

0.000 1.000 2.000 3.000 4.000 4.16006

0.000 0.774 1.517 2.197 2.785 2.868

0.779 0.764 0.717 0.639 0.530 0.510

-10.089 -10.035 -9.872 -9.600 -9.219 -9.147


Nx

q2 = 5 t/m'

= − 2, 02260 − 6 = − 8, 02260 ton(tekan )

q1 = 3 t/m'

S

C m

x

Q1

0 0 . 3 = 2 h

= − R AH − P1

Check apakah Mmax terdapat

Q2

dalam interval P1 = 6 t

Syarat : Mmax  Dx = 0 2

6,91525 − 0, 625 ( x ) = 0

m 0 0 . 4 = 1 h

x= A

RAH = 2,02260 t

L1 = 2.00 m

L2 = 4.00 m

RAV =9,91525 t

Tabel perhitungan MDN batang CS Interval

x 0.000

Mx (ton.m) -14.328 -14.328

Dx(ton) 6.915

Nx(ton) -8.023

6,91525 = 3, 32632 m (memenuhi ) 0,625


3. Batang DS (Ditinjau sebelah kanan S) Interval 0 ≤ x ≤ 2

x2

P2 = 3 t

M x1

x1 q3 = 2 t/m'

S

= − {−6, 08475 ( x ) + 4, 02260 ( 4 ) + 4 ( 31 (4) )}

D

Q3

RBH = 4,02260 t

= − {−RBV ( x ) + RBH ( h3 ) + Q3 ( 31 h3 )}

m 0 0 . 4 = 3 h

B m 0 0 . 3 = 0 h

= − {−6, 08475 ( x ) + 21, 42373} Dx 1 = −

dMx

Nx1

= −RBH − Q4 = −4, 02260 − 4 = −8, 02260 ton(tekan)

dx

= −6,08475 ton

RBV = 6,08475 t

L3 = 3.00 m

L4 = 2.00 m

Interval 2 ≤ x ≤ 5 Mx2

1 = −{−RBV ( x ) + R BH BH ( h3 ) + Q3 ( 3 h3 ) + P2 ( x − L4 )}

= −{−6, 08475 ( x ) + 4, 02260 ( 4 ) + 4 ( 13 ( 4) ) + 3 ( x − 2 )} = −{−6, 08475 ( x ) + 16, 09040 + 5, 33333 + 3 ( x ) − 6}

Tugas Tugas Mekanika Teknik

4. Batang BD q3 = 2 t/m' D

Q3

Mx

m 0 0 . 4 = 3 h

{

B

Dx = −

RBV = 6,08475 t

Nx

Check apakah Mmax terdapat dalam interval Syarat : Mmax  Dx = 0 2

4, 02260 + 0, 25 ( x ) = 0 x= −

4,02260 0,25

.................(tdk memenuhi )

Tabel perhitungan MDN batang BD

} = {4, 02 02260 + 0, 25 25 ( x ) } 3

= − 4, 02260 ( x ) + 0, 08333 ( x )

x

RBH = 4,02260 t

⎧ q3 x 3 ⎫ = − ⎨RBH ( x ) + ⎬ 6LBD ⎭ ⎩ ⎫ 2x 3 ⎪ ⎪⎧ = − ⎨4,02260 ( x ) + ⎬ 6( 4) ⎭ ⎪ ⎩⎪

dMx dx

2

= −RBV = −6,08475 ton(tekan )


E. Gambar Bidang Momen -21,424 tm

-14,328 tm -9,254 tm -14,328 tm -21,424 tm C

+0,998 tm

D

S

+2,868 tm

BIDANG MOMEN SKALA JARAK 0

0

A

1m

6 t.m

2m

12 t.m

SKALA GAYA

3m

18 t.m

B


F. Gambar Bidang Lintang

+6,915 tm

S

C

D

+8,023 t

-5,814 t

-3,085 t

-6,085 t +0,510 t -5,259 t B

DIAGRAM GAYA GESER SKALA JARAK 0

+0,779 t

A

0

1m

2m

3t

6t

SKALA GAYA

3m

9t

+4,023 t


G. Gambar Bidang Normal

S

C

D

-8,023 t

-9,147 t -10,796 t -6,875 t

DIAGRAM GAYA NORMAL SKALA JARAK 0

0

A

1m

2m

4t

8t

SKALA GAYA

-10,089 t

3m

12 t

B


VERIFIKASI HASIL PERHITUNGAN MENGGUNAKAN PROGRAM SAP 2000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:07:56 S T A T I C

L O A D

Ton-m Units

C A S E S

STATIC CASE

CASE TYPE

SELF WT FACTOR

LOAD1

DEAD

1.0000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:07:56 J O I N T

PAGE 1

Ton-m Units

PAGE 2

D A T A

JOINT

GLOBAL-X

GLOBAL-Y

GLOBAL-Z

1 2 3 4 5

-6.00000 -4.00000 0.00000 5.00000 5.00000

0.00000 0.00000 0.00000 0.00000 0.00000

0.00000 7.00000 7.00000 7.00000 3.00000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:07:56 F R A M E

E L E M E N T

Ton-m Units

RESTRAINTS 1 0 0 0 1

1 0 0 0 1

1 0 0 0 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

ANGLE-A

ANGLE-B

ANGLE-C

0.000 0.000 0.000 0.000 0.000

0.000 0.000 0.000 0.000 0.000

0.000 0.000 0.000 0.000 0.000

PAGE 3

D A T A

FRAME

JNT-1

JNT-2

SECTION

ANGLE

RELEASES

SEGMENTS

R1

R2

FACTOR

LENGTH

1 2 3 4

1 2 3 4

2 3 4 5

SECTION SECTION SECTION SECTION

0.000 0.000 0.000 0.000

000000 000222 000111 000000

2 4 4 2

0.000 0.000 0.000 0.000

0.000 0.000 0.000 0.000

1.000 1.000 1.000 1.000

7.280 4.000 5.000 4.000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:08:25 J O I N T JOINT

Ton-m Units

PAGE 1

D I S P L A C E M E N T S LOAD

U1

U2

U3

R1

R2

R3

1

LOAD1

0.0000

0.0000

0.0000

0.0000

14.2615

0.0000

2

LOAD1

84.8286

0.0000

-24.2368

0.0000

25.3342

0.0000

3

LOAD1

84.8286

0.0000

-177.0988

0.0000

0.0000

0.0000

4

LOAD1

84.8286

0.0000 0.0000

0.0000 0.0000

0.0000

-4.5134

0.0000 0.0000

5

LOAD1

0.0000

0.0000

0.0000

0.0000

33.0008

0.0000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:08:25 J O I N T JOINT

Ton-m Units

PAGE 2

R E A C T I O N S LOAD

F1

F2

F3

M1

M2

M3

1

LOAD1

2.0226

0.0000

9.9152

0.0000

0.0000

0.0000

5

LOAD1

-4.0226

0.0000

6.0848

0.0000

0.0000

0.0000

SAP2000 v7.40 File: SAP_GAFAR 1/26/10 13:08:25 F R A M E FRAME 1

E L E M E N T LOAD LOAD1

LOC

Ton-m Units

PAGE 3

F O R C E S P

V2

V3

T

M2

M3

SAP2000

SAP2000 v7.40 - File:SAP_Gafar File:SAP_Gafa r - Moment 3-3 Diagram

1/26/10 13:12:14

(LOAD1) - Ton-m Units

SAP2000

SAP2000 v7.40 - File:SAP_Gafar File:SAP_Gafa r - Shear Force 2-2 Diagram

1/27/10 21:58:34


SAP2000

SAP2000 v7.40 - File:SAP_Gafar File:SAP_Gafa r - Axial Force Diagram

1/26/10 13:09:23


Tugas Mekanika Teknik

Recommend Documents