Portalbergoyangmerekiiia-170516141313

TUGAS BESAR

MEKANIKA REKAYASA III q = 2 t/m' 1,5 t E

5

4

1,5 EI

D

1m

1t EI

EI 3m

3t

1t

3t

1t

2t F

1

2

EI

EI

3

G 1,5 m

3t EI

EI

EI

3m

A

B

1,5 m

3m

KA1 = K1A =

=

= 0,888

K12 = K21 =

=

= 0,888

KB2 = K2B =

=

= 0,888

KC3 = K3C =

=

= 0,888

K23 = K32 =

=

=1

K34 = K43 =

=

=1

K15 = K51 =

=

=1

K54 = K45 =

=

1m

METODE CROSS :  Hitung K :

= 0,705

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

C 3m

1m

1m

TUGAS BESAR

MEKANIKA REKAYASA III  Hitung Koefisien Distribusi : µA1 = 0 µ1A

=

= 0,320

µ12

=

= 0,320

µ15

=

= 0,360

µB2 = 0 µ2B

=

= 0,320

µ21

=

= 0,320

µ23

=

= 0,360

µC3 = 0 µ3C

=

= 0,308

µ32

=

= 0,346

µ34

=

= 0,346

µ43

=

= 0,587

µ45

=

= 0,413

µ54

=

= 0,413

µ51

=

= 0,587

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Hitung Momen Primer (Akibat Beban Luar) : q = 2 t/m' M5E

1,5 t

M45

M54

E

5

M4D

4

1,5 EI

4

D

M51

5

4

EI

EI

1t

1 3

M15

1t

3t

M12

3t

M23

1t

M3G

2t F 1

1

2

EI

M1F

M21

EI

3

3 M32

M2B

1

2

3 3t EI

EI

EI

A

B MB2

 M0A1 = M01A = M0C3 = M03C = M034 = M043 = 0  M0B2

=

 M02B

=-

= -2 t.m

 M012

=-

= -2 t.m

 M021

=

 M023

=

= - 0,5625 t.m

 M032

=

= 1,6875 t.m

 M01F

= 1 t.m

= 1 t.m

= 1 t.m

 M03G = - 1 t.m  M015

=

= - 0,1875 t.m

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

C

G

TUGAS BESAR

MEKANIKA REKAYASA III  M051

=

 M045

=+

= + 12,042 t.m

 M054

=-

= - 12,042 t.m

 M05E

= 1 t.m

= 0,5625 t.m

 M04D = - 1 t.m  Hitung Momen Primer (Akibat Pergoyangan) : Pendel Tingkat I : ?

?

M51

M43

4

5

1m

3m

M15

M34

1

2

3 1,5 m

3m

A 1m

1,5 m

B 3m

C 3m

 M015 = M051

=-

=

= - 0,375 t.m

 M034 = M043

=-

=

= - 0,375 t.m

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

1m

1m

TUGAS BESAR

MEKANIKA REKAYASA III  Hitung Momen Primer (Akibat Pergoyangan) : Pendel Tingkat II : M51

M43

4

5

1m

3m

M15

1

M34

M2B

2

M1A

3

M3C

1,5 m

3m

1m

A

B

C

MA1

MB2

MC3

1,5 m

3m

3m

 M0A1 = M01A

=-

=

= - 0,296 t.m

 M0B2 = M02B

=-

=

= - 0,296 t.m

 M0C3 = M03C

=-

=

= - 0,296 t.m

 M015 = M051

=

=

= + 0,375 t.m

 M034 = M043

=

=

= + 0,375 t.m

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

1m

1m

TUGAS BESAR

MEKANIKA REKAYASA III  Hitung H : Akibat Beban Luar : M43

M51

5

4

EI

EI

1t

H5

3

1 M15

H3 M34

M1A

M2B

1

M3C

2

3 3t

A

EI

EI

EI

HA

HB

B

MA1

MB2

∑M1 = 0 → HA . 4,5 – MA1 – M1A = 0 HA =

= 0,338 Ton

∑M2 = 0 → -HB . 4,5 + MB2 + 3 x 1,5 – M2B = 0 HB =

= 0,937 Ton

∑M3 = 0 → -HC . 4,5 + MC3 + M3C = 0 HC =

= 0,339 Ton

∑M4 = 0 → H3 . 4 - M34 - M43 = 0 H3 =

= 2,767 Ton

∑M5 = 0 → -H1 . 4 + M15 – 1 x 1 + M51 = 0 H1 =

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

= 2,507 Ton

HC

C MC3

TUGAS BESAR

MEKANIKA REKAYASA III Pendel I : M43

M51

5

4

H3'

H1'

3

1 M15

M34

M1A

M2B

1

2

HA'

A

3

B

MA1

HB' MB2

∑M1 = 0 → HA' . 4,5 – MA1 – M1A = 0 HA' =

= 0,136 Ton

∑M2 = 0 → HB' . 4,5 - MB2 – M2B = 0 HB' =

= 0,098 Ton

∑M3 = 0 → HCʹ . 4,5 - MC3 - M3C = 0 HCʹ =

= 0,136 Ton

∑M4 = 0 → -H3ʹ . 4 + M34 + M43 = 0 H3ʹ =

= 0,115 Ton

∑M5 = 0 → -H1ʹ . 4 + M15 + M51 = 0 H1ʹ =

M3C

= 0,115 Ton

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

C

HC' MC3

TUGAS BESAR

MEKANIKA REKAYASA III Pendel II : M43

M51

5

4

1

H1"

M15

M34

M1A

M2B

1

HA"

H3"

3

M3C

2

3

B

A

MA1

HB" MB2

∑M1 = 0 → -HAʺ . 4,5 + MA1 + M1A = 0 HAʺ =

= 0,034 Ton

∑M2 = 0 → -HBʺ . 4,5 - MB2 – M2B = 0 H Bʺ =

= 0,011 Ton

∑M3 = 0 → -HCʺ . 4,5 + MC3 + M3C = 0 HCʺ =

= 0,032 Ton

∑M4 = 0 → H3ʺ . 4 - M34 - M43 = 0 H3ʺ =

= 0,088 Ton

∑M5 = 0 → H1ʺ . 4 - M15 - M51 = 0 H1ʺ =

= 0,087 Ton

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

HC"

C MC3

TUGAS BESAR

MEKANIKA REKAYASA III  Hitung HPx : Lantai 2 : ∑MH

=0

1,5 + 1 + H1 – H3 + (H1’ + H3’) . X1 + (-H1” - H3”) . X2 = 0 1,5 + 1 + 2,507 – 2,767 + (0,115 + 0,115) . X1 + (-0,087 – 0,088) . X2 = 0 2,24 + 0,23 X1 – 0,175 X2 = 0 0,23 X1 – 0,175 X2 = - 2,24 . . . . ( I ) Lantai 1 : ∑MH

=0

1,5 + 1 + 2 – 3 – HA + HB + HC – (HA’ + HB’ + HC’) . X1 + (HA” - HB” + HC”) . X2 = 0 1,5 – 0,338 + 0,937 + 0,339 – (0,136 + 0,098 + 0,136) . X1 + (0,034 – 0,011 + 0,032) . X2 = 0 2,438 – 0,37 . X1 + 0,055 . X2 = 0 – 0,37 . X1 + 0,055 . X2 = - 2,438 . . . . (II) ELIMINASI PERSAMAAN I DAN II : 0,23 X1 – 0,175 X2

= - 2,24

⇨x1

– 0,37 . X1 + 0,055 . X2

= - 2,438

⇨ x -0,622

0,23 X1 – 0,175 X2

= - 2,24

0,23 X1 – 0,034 X2

= 1,516

-

- 0,141 X2 = - 3,756 X2 = 26,64 SUBTITUSI KE PERSAMAAN I : 0,23 X1 – 0,175 X2

= - 2,24

0,23 X1 – 0,175 x 26,64 = - 2,24 0,23 X1 – 4,662

= - 2,24

0,23 X1 X1

= - 2,24 + 4,662 = 10,53

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III METODE TAKABEYA :  Hitung K : KA1 = K1A =

=

= 0,222

K12 = K21 =

=

= 0,222

KB2 = K2B =

=

= 0,222

KC3 = K3C =

=

= 0,222

K23 = K32 =

=

= 0,25

K34 = K43 =

=

= 0,25

K15 = K51 =

=

= 0,25

K54 = K45 =

=

= 0,176

 Hitung ρ : ρ1 = 2 x (K1A + K12 + K15) = 2 x (0,222 + 0,222 + 0,25) = 1,388 ρ2 = 2 x (K21 + K2B + K23) = 2 x (0,222 + 0,222 + 0,25) = 1,388 ρ3 = 2 x (K32 + K3C + K34) = 2 x (0,25 + 0,222 + 0,25) = 1,444 ρ4 = 2 x (K43 + K45) = 2 x (0,25 + 0,176) = 0,852 ρ5 = 2 x (K54 + K51) = 2 x (0,25 + 0,176) = 0,852  Hitung Ɣ : Join 1 ⟹ Ɣ1A =

=

= 0,160

Ɣ12 =

=

= 0,160

Ɣ15 =

=

= 0,180

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III Join 2 ⟹ Ɣ21 =

=

= 0,160

Ɣ2B =

=

= 0,160

Ɣ23 =

=

= 0,180

Join 3 ⟹ Ɣ32 =

=

= 0,173

Ɣ3C =

=

= 0,154

Ɣ34 =

=

= 0,173

Join 4 ⟹ Ɣ43 =

=

= 0,293

Ɣ45 =

=

= 0,207

Join 5 ⟹ Ɣ51 =

=

= 0,293

Ɣ54 =

=

= 0,207

 Hitung Momon Primer :  M0A1 = M01A = M0C3 = M03C = M034 = M043 = 0  M0B2

=

 M02B

=-

= -2 t.m

 M012

=-

= -2 t.m

 M021

=

 M023

=

= 1 t.m

= 1 t.m = - 0,5625 t.m

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  M032

=

 M01F

= 1 t.m

 M03G

= - 1 t.m

 M015

=

= - 0,1875 t.m

 M051

=

= 0,5625 t.m

 M045

=+

= + 12,042 t.m

 M054

=-

= - 12,042 t.m

 M05E

= 1 t.m

 M04D

= - 1 t.m

= 1,6875 t.m

 Hitung τ : τ1 = M01A + M012 + M015 + M01F = O – 2 – 0,1875 + 1 = - 1,1875 τ2 = M021 + M02B + M023 = 1 – 2 – 0,5625 = - 1,5625 τ3 = M032 + M03C + M03G + M034 = 1,6875 + 0 – 1 + 0 = 0,6875 τ4 = M043 + M04D + M045 = 0 – 1 + 12,042 = 11,042 τ5 = M054 + M05F + M051 = - 12,042 + 1 + 0,5625 = -10,4795  Hitung Momen Rotasi Langkah Awal : M10 = -

=-

= 0,856

M20 = -

=-

= 1,126

M30 = -

=-

= - 0,476

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III M40 = -

=-

M50 = -

=-

= - 12,960 = 12,299

 Momen Displesmen : 

Tingkat Atas TI = 2 x (K15 + K34) = 2 x (0,25 + 0,25) = 1 t15 =

=

= 0,75

t34 =

=

= 0,75

mI0 = 

=-9

Tingkat Bawah TII = 2 x (K1A + K2B + K3C) = 2 x (0,222 + 0,222 + 0,222) = 1,333 t1A =

=

= 0,5

=

=

= 0,5

t3C =

=

= 0,5

t2B

mII0 =

= - 7,5938

 Pemberesan Momen Parsil ( Langkah 1) : M11 = M10 - Ɣ12 . M20 – Ɣ1A(MA0 + mII0) – Ɣ15(M50 + mI0) = 0,856 – 0,16 x 1,126 – 0,16 x (-7,5938) – 0,18 x (12,30 – 9) = 1,298 M21 = M20 – Ɣ21 . M11 – Ɣ23 . M30 – Ɣ2B(MB0 + mII0) = 1,126 – 0,16 x 1,298 – 0,18 x (-0,476) – 0,16 x - 7,5938 = 2,219 M31 = M30 – Ɣ32 . M21 – Ɣ34(M40 + mI0) - Ɣ3C(MC0 + mII0) = - 0,476 – 0,173 x 2,218 – 0,173 x (-12,960 - 9) – 0,154 x (-7,5938) = 4,108 FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III M41 = M40 – Ɣ45 . M50 – Ɣ43(M31 + mI0) = -12,960 – 0,207 x 12,30 – 0,293 x (4,108 – 9) = -14,073 M51 = M50 – Ɣ54. M41 – Ɣ51(M11 + mI0) = 12,299 – 0,207 x (-14,073) – 0,253 (1,298 – 9) = 17,162  Momen Displesmen Langkah 1 : mI1 = mI0 – t15 (M11 + M51) – t34 (M31 + M41) = -9 – 0,75 x (1,298 + 17,162) – 0,75 x (4,108 – 14,073) = - 15,371 mII1 = mII0 – t1A (M11 + MA1) – t2B (M21 + MB1) – t3C (M31 + MC1) = -7,5938 – 0,5 x 1,298 – 0,5 x 2,219 – 0,5 x 4,108 = - 11,406  Pemberesan Momen Parsil ( Langkah 2) : M12 = M10 - Ɣ12 . M21 – Ɣ1A(MA0 + mII1) – Ɣ15(M51 + mI1) = 0,856 – 0,16 x 2,219– 0,16 x (-11,406) – 0,18 x (17,162 – 15,371) = 2,003 M22 = M20 – Ɣ21 . M12 – Ɣ23 . M31 – Ɣ2B(MB0 + mII1) = 1,126 – 0,16 x 2,003 – 0,18 x 4,108 – 0,16 x - 11,406 = 1,891 M32 = M30 – Ɣ32 . M22 – Ɣ34(M41 + mI1) - Ɣ3C(MC0 + mII1) = - 0,476 – 0,173 x 1,891– 0,173 x (-14,073 - 15,371) – 0,154 x (-11,406) = 6,047 M42 = M40 – Ɣ45 . M51 – Ɣ43(M32 + mI1) = -12,960 – 0,207 x 17,162 – 0,293 x (6,047 – 15,371) = -13,780 M52 = M50 – Ɣ54. M44 – Ɣ51(M12 + mI1) = 12,299 – 0,207 x (-13,780) – 0,253 (2,003 – 15,371) = 18,534

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Momen Displesmen Langkah 2 : mI2 = mI0 – t15 (M12 + M52) – t34 (M32 + M42) = -9 – 0,75 x (2,003 + 18,534) – 0,75 x (6,047 – 13,780) = - 18,603 mII2 = mII0 – t1A (M12 + MA2) – t2B (M22 + MB2) – t3C (M32 + MC2) = -7,5938 – 0,5 x 2,003 – 0,5 x 1,891 – 0,5 x 6,047 = - 12,564  Pemberesan Momen Parsil ( Langkah 3) : M13 = M10 - Ɣ12 . M22 – Ɣ1A(MA0 + mII2) – Ɣ15(M52 + mI2) = 0,856 – 0,16 x 1,891 – 0,16 x (-12,564) – 0,18 x (18,534– 18,603) = 2,576 M23 = M20 – Ɣ21 . M13 – Ɣ23 . M32 – Ɣ2B(MB0 + mII2) = 1,126 – 0,16 x 2,576 – 0,18 x 6,047 – 0,16 x (- 12,564) = 1,635 M33 = M30 – Ɣ32 . M23 – Ɣ34(M42 + mI2) - Ɣ3C(MC0 + mII2) = - 0,476 – 0,173 x 1,635 – 0,173 x (-13,780- 18,603) – 0,154 x (-12,564) = 6,778 M43 = M40 – Ɣ45 . M52 – Ɣ43(M33 + mI2) = -12,960 – 0,207 x 18,534 – 0,293 x (6,778 – 18,603) = -13,332 M53 = M50 – Ɣ54. M43 – Ɣ51(M13 + mI2) = 12,299 – 0,207 x (-13,332) – 0,253 (2,576 – 18,603) = 19,114  Momen Displesmen Langkah 3 : mI3 = mI0 – t15 (M13 + M53) – t34 (M33 + M43) = -9 – 0,75 x (2,576 + 19,114) – 0,75 x (6,778 – 13,332) = - 20,487 mII3 = mII0 – t1A (M13 + MA3) – t2B (M23 + MB3) – t3C (M33 + MC3) = -7,5938 – 0,5 x 2,576 – 0,5 x 1,635 – 0,5 x 6,778 = - 13,088

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Pemberesan Momen Parsil ( Langkah Akhir) : M136

= 3,270

M236

= 1,433

M336

= 7,601

M436

= - 12,660

M536

= 20,754

 Momen Displesmen Langkah Akhir : mI36

= - 23,223

mII36 = - 13,746  Momen Akhir : MA1

= KA1 . (2 x MA + M136 + mII36) + M0A1 = 0,222 x (2 x 0 + 3,270 – 13,746) + 0 = - 2,326

M1A

= K1A . (2 x M136 + MA + mII36) + M01A = 0,222 x (2 x 3,270 + 0 – 13,746) + 0 = - 1,600

M12

= K12 . (2 x M136 + M236) + M012 = 0,222 x (2 x 3,270 + 1,433) – 2 = - 0,230

M15

= K15 . (2 x M136 + M536 + mI36) + M015 = 0,25 x (2 x 3,270 + 20,754 - 23,223) – 0,1875 = 0,830

M1F

= K1F . (2 x M136 + MF) + M01F = 0 x (2 x 3,270 + 0 ) + 1 =1

MB2

= KB2 . (2 x MB + M236 + mII36) + M0B2 = 0,222 x (2 x 0 + 1,433 – 13,746) + 1

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III = - 1,733 M21

= K21 . (2 x M236 + M136) + M021 = 0,222 x (2 x 1,433 + 3,270) + 1 = 2,362

M2B

= K2B . (2 x M236 + MB + mII36) + M02B = 0,222 x (2 x 1,433 + 0 – 13,746) -2 = - 4,415

M23

= K23 . (2 x M236 + M336) + M023 = 0,25 x (2 x 1,433 + 7,601) – 0,5625 = 2,054

MC3

= KC3 . (2 x MC + M336 + mII36) + M0C3 =0,222 x (2 x 0 + 7,601 -13,746) + 0 = - 1,364

M3C

= K3C . (2 x M336 + MC + mII36) + M03C = 0,222 x (2 x 7,601 + 0 -13,746) + 0 = 0,323

M32

= K32 . (2 x M336 + M236 ) + M032 = 0,25 x (2 x 7,601 + 1,433) + 1,6875 = 5,846

M34

= K34 . (2 x M336 + M436 + mI36 ) + M034 = 0,25 x (2 x 7,601 – 12,660 – 23,223) + 0 = - 5,170

M3G

= K3G . (2 x M336 + MG36) + M03G = 0 x (2 x 7,601 + 0) – 1 =-1

M43

= K43 . (2 x M436 + M336 + mI36 ) + M043 = 0,25 x {2 x (– 12,660) + 7,601 – 23,223} + 0

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III = - 10,236 M45

= K45 . (2 x M436 + M536) + M045 = 0,176 . {2 x (-12,660) + 20,754} + 12,042 = 11,238

M4D

= K4D . (2 x M436 + MD36) + M04D = 0 x ({2 x (-12,660) + 0} – 1 =-1

M51

= K51 . (2 x M536 + M136 + mI36 ) + M051 = 0,25 x (2 x 20,754 + 3,270 – 23,223) + 0,5625 = 5,951

M54

= K54 . (2 x M536 + M436) + M054 = 0,176 . (2 x 20,754 – 12,660) – 12,042 = - 6,965

M5E

= K5E . (2 x M536 + ME36) + M05E = 0 x (2 x 20,754 + 0) + 1 =1

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III A. DARI METODE CROSS MA1 = - 2,350 t.m M1A = - 1,584 t.m M1F = 1 t.m M15 = 0,736 t.m M12 = - 0,152 t.m MB2 = - 1,802 t.m M21 = 2,397 t.m M2B = - 4,487 t.m M23 = 2,090 t.m MC3 = - 1,386 t.m M32 = 5,930 t.m M34 = - 5,274 t.m M3G = - 1 t.m M3C = 0,344 t.m M43 = - 10,338 t.m M45 = 11,338 t.m M4D = - 1 t.m M54 = - 6,864 t.m M51 = 5,864 t.m M5E = 1 t.m B. MENGHITUNG REAKSI PERLETAKAN  Free Body A – 1 : ∑M1 = 0 ⇒ HA . 4,5 – MA1 – M1A = 0

M1A

H1

HA x 4,5 – 2,350 – 1,584 = 0

1

HA = HA = 0,874 Ton (←)

4,5 m

∑MA = 0 ⇒ H1 . 4,5 – MA1 – M1A = 0 H1 x 4,5 – 2,350 – 1,584 = 0

A

HA

MA1

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

H1 = H1 = 0,874 Ton (→)

TUGAS BESAR

MEKANIKA REKAYASA III  Free Body 1 – F :

1t

∑V = 0 ⇒ V1 – 1 = 0 M1F

2t F

V1 = 1 Ton (↑)

1

1m V1

 Free Body 1 – 2 : ∑M1 = 0 ⇒ -V2 . 4,5 + 3 . 1,5 – M12 + M21 = 0

3t

M12

-V2 . 4,5 + 4,5 – 0,152 + 2,397 = 0

M21

V2 =

1

2

1,5 m

V2 = 1,499 Ton (↑)

3m

V1

V2

∑M2 = 0 ⇒ V1 . 4,5 - 3 . 3 – M12 + M21 = 0 V1 . 4,5 - 9 – 0,152 + 2,397 = 0 V1 = V1 = 1,501 Ton (↑)

 Free Body 1 – 5 : M51

H5 5

∑M1 = 0 ⇒ -H5 . 4 + 1 . 3 + M51 + M15 = 0 -H5 . 4 + 3 + 5,864 + 0,736 = 0

1m

1t

H5 = H5 = 2,4 Ton (←) 3m

∑M5 = 0 ⇒ H1 . 4 - 1 . 1 + M51 + M15 = 0 H1 . 4 - 1 + 5,864 + 0,736 = 0

1

H1

M15

H1 = H1 = 1,4 Ton . . . Asumsi Terbalik (→)

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Free Body B – 2 : ∑M2 = 0 ⇒ HB . 4,5 + 3 . 1,5 –MB2 – M2B = 0

M2B

H2

HB . 4,5 + 4,5 – 1,802 – 4,487 = 0

2 1,5 m

HB =

3t HB = 0,398 Ton (←) ∑MB = 0 ⇒ H2 . 4,5 - 3 . 3 –MB2 – M2B = 0 3m

H2 . 4,5 - 9 – 1,802 – 4,487 = 0 H2 =

HB

B MB2

H2 = 3,398 Ton (→)

 Free Body 2 – 3 : ∑M2 = 0 ⇒ -V3 . 4 + 3 . 3 + M23 + M32 = 0

V2

3t 2

-V3 . 4 + 9 + 2,090 + 5,930 = 0

M32

V3 =

3

M23

3m

V3 = 4,255 Ton (↑)

1m V3

∑M3 = 0 ⇒ -V2 . 4 - 3 . 1 + M23 + M32 = 0 -V2 . 4 - 3 + 2,090 + 5,930 = 0 V2 = V2 = 1,255 Ton (↓)  Free Body C – 3 : M3C

H3

∑M3 = 0 ⇒ HC . 4,5 – MC3 + M3C = 0

3

HC . 4,5 – 1,386 + 0,344 = 0 HC = 4,5 m

C

HC MC3

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

HC = 0,232 Ton (←)

TUGAS BESAR

MEKANIKA REKAYASA III ∑MC = 0 ⇒ H3 . 4,5 – MC3 + M3C = 0 H3 . 4,5 – 1,386 + 0,344 = 0 H3 = H3 = 0,232 Ton (→)  Free Body 3 – G :

1t

M3G

∑V = 0 ⇒ V3 – 1 = 0 V3 = 1 Ton (↑)

G

3 1m V3

 Free Body 4 – D :

q = 2 t/m'

∑V = 0 ⇒ V4 – 2 x 1 = 0

M3G

V4 = 2 Ton (↑)

D

4 1m V4

 Free Body 3 – 4 : ∑M3 = 0 ⇒ H4 . 4 – M34 – M43 = 0

M43

H4

H4 . 4 – 5,274 – 10,338 = 0

4

H4 = 4m

H4 = 3,903 Ton (→) ∑M4 = 0 ⇒ H3 . 4 – M34 – M43 = 0 H3 . 4 – 5,274 – 10,338 = 0

H3

3

H3 =

M34

H3 = 3,903 Ton (←)

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Free Body 5 – E : ∑V = 0 ⇒ V5 – 2 x 1 = 0

q = 2 t/m' 2t

V5 = 2 Ton (↑)

M1F

5

E 1m

V5

 Free Body 4 – 5 : ∑M4 = 0 ⇒ V5 . 8,5 – ½ qL2 – M54 + M45 = 0

q = 2 t/m' M54

M45

V5 . 8,5 – 72,25 – 6,864 + 11,338 = 0 5

4

V5 =

8,5 m V5

V5 = 7,974 Ton (↑)

V4

∑M5 = 0 ⇒ -V4 . 8,5 + ½ qL2 – M54 + M45 = 0 -V4 . 8,5 + 72,25 – 6,864 + 11,338 = 0 V4 = V4 = 9,026 Ton (↑) C. MENGHITUNG GAYA DALAM  Free Body A – 1 : Momen : M1A

H1

MA = MA1 = - 2,350 t.m

1

M1 = HA x 4,5 – MA1 = 0,874 x 4,5 – 2,350 = 1,584 t.m Lintang :

4,5 m

A

DA

= HA = 0,874 Ton

D1

= HA – H1 = 0,874 – 0,874 = 0

HA

MA1

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

TUGAS BESAR

MEKANIKA REKAYASA III  Free Body 3 – 4 : Momen :

M43

H4 4

M3

= M34 = - 5,274 t.m

M4

= H3 x 4 – M34 = 3,903 x 4 – 5,274 = 10,338 t.m

Lintang : 4m

D3

= H3 = 3,903 Ton

D4

= H3 – H4 = 3,903 – 3,903 = 0

H3

3 M34

 Free Body C – 3 : Momen :

M3C

H3 3

MC

= MC3 = - 1,386 t.m

M3

= HC x 4,5 – MC3 = 0,232 x 4,5 – 1,386 = -0,342 t.m

Lintang : 4,5 m

C

DC

= HC = 0,232 Ton

D3

= HC – H3 = 0,232 0,232 = 0

HC MC3

 Free Body 1 – 2 : M1

3t

M12

M21

1

MH = V1 x 1,5 – M12 = 1,501 x 1,5 – 0,152 = 2,099 t.m

2

1,5 m

M2

3m

V1

= V1 x 4,5 – 3 x 3 – M12 = 1,501 x 4,5 – 9 – 0,152

V2 D1

= V1 = 1,501 Ton

D1-H

= 1,501 Ton

D1-2

= V1 – 3 = 1,501 – 3 = - 1,499 Ton

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG

= M12 = - 0,152 t.m

= - 2,397 t.m

TUGAS BESAR

MEKANIKA REKAYASA III

FERIYAL SUMARNO / 311 11 001 2 A GEDUNG