Kumpulan Soal Mekrek

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 1

KUMPULAN SOAL DAN PENYELESAIAN

MEKANIKA REKAYASA

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G E R B E R

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 3 16,33kN JAWABAN :  Ms Kiri A _B S C D q = 4 kN/m 8 m 2 m 6 m 4 m P = 8 kN P = 5 kN BMD A _B _S C D 33,32 kNm 2,64 kNm -7,99 kN -33,32 kN -0,02 kNm

+

-0,33 kN - 8 kN -13,327 kN 2,677 kN 5,33 kN SFD

S O A L 1

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 4 Rav.10+Rbv.2-4(8)(6)-4(2)(1) = 0 Rav.10+Rbv.2-192-8 = 0 Rav.10+Rbv.2 = 200……….(1)  Ms Kanan -8(10)+Rcv.6 = 0 80= Rcv.6 Rcv = 13.333 kN  ∑H = 0 Rah-5 =0 Rah = 5  ∑V = 0 Rav+Rbv+Rcv-4(10)-8 = 0 Rav+Rbv+13,33-40-8 =0 Rav+Rbv = 34,67 kN………..(2) Eliminasi (1) dan (2) 10.Rav +2.Rbv = 200 ..………..(1) Rav + Rbv = 34,67 ………..……..(2) 10.Rav +2.Rbv = 200 ..………..(1) 2.Rav +2.Rbv = 69.34 ………..……..(2) 8.Rav = 130.66 Rav = 16.333 kN Menentukan Rbv Rav + Rbv = 34,67 ………..……..(2) 16.333 + Rbv = 34,67 Rbv = 18.34 kN

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GAYA-GAYA DALAM

INTERVAL MOMEN LINTANG

A – B 0 - 8 Mx = Rav.x – qx2 = 16.333 x - (4)x2 = 16.333 x - 2x2 x = 0 M0 = 0 x = 4 M4 = 33,332 kNm x = 8 M8 = 2,64 kNm Dx = = 16,333 – 4x x = 0 D0 = 16.333 kN x = 4 D4 = 0,33 kN x = 8 D8 = -15,677 kN B – S 8 - 10 Mx = Rav.x – qx2 + Rbv.(x-8) – q(x-8)2 =16.333 x - (4)x2 + 18.34(x-8) - (4) (x-8)2 = 16.333x - 2x2 + 18.34(x-8) – 2(x-8)2 = 16.333x- 2x2 + 18.34x – 146,72 – 2(X2 - 16x +64) = 16.333x - 2x2 + 18.34x – 146,72 – 2X2 + 32x – 128 = - 4x2 + 66.673 x -274.72 x = 8 M8 = 2,64 kNm x = 10 M10 = -7,99 kNm = Dx = = -8x + 66.673 x = 8 D8 = 2,673 kNm x = 10 D10 = -13.327 kNm D – C 0 - 4 Mx = -8x x = 0 M0 = 0 kNm x = 4 M4 = -32 kNm Dx = -8 x = 0 M0 = -8 kNm x = 4 M4 = -8 kNm

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 6 C – S 4 - 10 Mx = -8x + Rcv (x-4) = -8x + Rcv.x -4Rcv = -8x + 13,333x – 4(13,333) = 5,333x – 53,332 x = 4 M4 = -32 kNm x = 10 M10 = -0.02 kNm Dx = Dx = 5,333 x = 4 M4 = 5,333 kNm x = 10 M10 = 5,333 kNm

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P O R T A L

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 8 P=5 KN 3 m 3 m 1 m A B C D Q=2 KN/m RAV RBV RAH

Contoh 2: Hitung dan gambarkan diagram gaya-gaya

dalam yang terjadi pada Struktur Portal berikut ini.

PENYELESAIAN :

a. Menghitung Reaksi Perletakan:

 



H 0 PRAH 0  RAH 5KN

 

  



MA 0

R

BV

.

3 

R

BH

.

1 

2 .

3 .

1 ,

5 

5 .

3 

0 kN

R

_BV

3

29 

  MB 0

R

AV

.

3 

5 .

4 

2 .

3 .

1 ,

5 

0 ₍

₎

3

11 _



kN

R

_AV  



V 0

R

_AV



R

_BV



 

Q

.

3

KN R_AV 6 3 29 3 11_ _  

S O A L 1

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 9 Q2 = 2 29/3 P=5 29/3 11/3 11/3 0 11/3 29/3 5 15

FBD

5 20 5 20 ₅

b. Free Body Diagram dan Perhitungan GGD:

Tabel Gaya-Gaya Dalam:

Bagian/ Interval Momen (M) Geser (V) Normal (N) A - C 0 - 3 (dari kiri)

0 

X

M

0 0   MA X

0

3 



M

_C

X

0  dx dMX 3 11   X N C - D 0 - 3 (dari 2 5 , 0 3 / 11 X Q X  

0

0 



M

_C

X

20 3   MD X QX dx dMX ₁₁_/₃ 0 0   VC X Nx = -5

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 10 kiri) X3VD29/3 Tdk Ada Mmak B - D 0 – 4

4 .

BH X

R

M





0

0 



M

_B

X

20 4   M_D X 5   X V NX 29/3

c. Diagram Gaya-Gaya Dalam:

BMD

15 A B C D 15 2

NFD

8 A B C D 5

SFD

A B 2 C D 8

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 11 PENYELESAIAN ∑ H = 0 RBH – 2.3 = 0 RBH = 6 t (→) ∑ MB = 0 RAV.6 – 5.2.5 -10.2 – 2.3.1,5 = 0 RAV.6 – 50 – 20 – 9 = 0 RAV = 79/6 = 13,11 t ∑ MA = 0 RBV.6 - RBH.1 + 2.3.2,5 + 10.4 – 5.2.1 = 0 RBV.6 – 6.1 – 15 – 40 – 10 = 0 RBV =41/6 = 6,8 t Q1 = 5 t/m P = 10 t Q2 = 2t/m E A C D F B 3 m 1 m 2 m 2 m 2 m RBV RBH RAV

S O A L 2

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 12 DIAGRAM BENDA BEBAS

E A C D F B 3 m 1 m 6,8 6 13,11 Q1 = 5 t/m P = 10 t 13,11 13,11 6,8 6 0 9 6,8 0

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 13 Tabel Gaya-Gaya Dalam:

Bagian/ Interval Momen (M) Geser (V) Normal (N) A - C 0 - 4

0 

X

M

0 0   MA X

0

3 



M

_C

X

0  dx dMX NA-C = 13,11 B - F 0 - 3 Mx =6x – ½ Qx2 kNm M X M X F B 9 3 0 0       ) 6 ( QX dx dMX  ) 2 6 ( X dx dMX__ _ 6 0   VB X X3VF0 NB-F = -6,8 C - D 0 – 2 2 2 1 11 . 13 x qx M_X   0 0   MC X 2 ) 2 )( 5 ( 2 1 ) 2 ( 11 . 13 2    MD X x VX (5)(2) 2 1 11 . 13   11 . 13 0   VC X 11 . 3 2   VD X 0  D C N D – E 0 - 2 Mx = 13.11(x+2)-10(x+1) 34 . 16 ) 2 ( 10 2 ) 11 . 13 ( 0     MD X 68 . 22 ) 3 ( 10 ) 4 )( 11 . 13 ( 2     ME X 10 11 . 13   X V VC-D = 3.17 N D-E = 0 F – E 0 - 2 Mx = 6.83 x + 9 9 0   MF X 68 . 22 2   ME X 83 . 6   X V V F – E = -6.83 N F-E = -0

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 14 DIAGRAM BIDANG MOMEN (BMD)

DIAGRAM BIDANG GESER (SFD)

E A C D F B E A C D F B

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 15 DIAGRA BIDANG NORMAL (NFD)

E

A

C D F

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 18 Gambar 5.4 Diagram gaya-gaya dalam portal sendi-rol akibat beban merata dan beban lateral

terpusat

Struktur portal tiga sendi dengan ukuran dan pembebanan seperti Gambar di bawah, hitunglah : 1. Reaksi Perletakan

2. Persamaan Gaya-Gaya Dalam dan Diagram Benda Bebas (FBD) 3. Diagram Gaya-Gaya Dalam (Momen, Geser, dan Normal)

P = 5kN Q1 = 5 kN/m Q2 = 3 kN/m A C S D B E 4 m 2 m 2 m 2 m 4 m

S O A L 1

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SOLUSI

DIAGRAM BENDA LEPAS (DBL)

MENENTUKAN BESARNYA REAKSI PERLETAKAN

∑MA = 0 -RBV.6 + RBH.2 – 12.4 – 20.4 + 10.1 - 5.2 = 0 -RBV.6 + RBH.2 = -32 ………..(1) ∑MB = 0 -RAV.6 + RAH.2 –5.8 – 10.5 – 20.2 – 12.2 = 0 -RAV.6 - RAH.2 = 154 ………..(2) S D C A B R1 = 10 kN R3 = 12 kN P = 5 kN 2 m 2 m R2 = 20 kN 2 m 1 m 1 m 2 m 2 m 2 m Rah Rav Rbh Rbv

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 20 ∑Ms kanan = 0 -RBV.4 + RBH.4 + 12. 2 +20.2 = 0 -RBV.4 - RBH.4 = -64………..(3) ∑Ms kiri = 0 RAV.2 - RAH.6 – 5.4 – 10.1 = 0 RAV.2 - RAH.6 = 30………..(4)

ELIMINASI PERSAMAAN (1) DAN (3)

-RBV.6 + RBH.2 = -32 ×2 -RBV.12 + RBH.4 = -64 -RBV.4 - RBH.4 = -64 ×1 -RBV.4 - RBH.4 = -64 +

-16 RBV = -128 RBV = 8 kN Substitusi nilai RBV ke persamaan 1

-RBV.6 + RBH.2 = -32 → -6(8) + RBH.2 = -32 → RBH = 8 kN Eliminasi Pers. 2 dan Pers 4

-RAV.6 - RAH.2 = -154 ×1 -RAV.6 - RAH.2 = -154 RAV.2 - RAH.6 = 30 ×3 RAV.6 - RAH.18 = 90 -

16 RAH = 154 RAH = 4 kN Substitusi nilai RAH ke Persamaan 2

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DIAGRAM BENDA BEBAS

S D C A B Rah = 4 Rav = 27 Rbh = 8 Rbv = 8 q2 = 3 kN/m q1 = 5 kN/m P = 5 kN E 5 10 C 4 27 24 _C 34 4 22 R1 = 30 kN

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Tabel Gaya-Gaya Dalam:

Daerah Interval

Momen (M

x

)

Geser (V

x

)

Normal

(N

x

)

A – C 0 – 6 Mx = –R

AH

. x

= –4x

x = 0, M

A

= 0

x = 3, M

C

= – 24 kN m

V

x

= – R

AH

V

A-C

= –4 kN

–27 kN

E – C

0 – 2 Mx = –P . x

= –5x

x = 0, M

E

= 0

x = 2, M

C

= – 10 kN m

Vx = – P

V

E-C

= – 5 kN

0 C – D 0 – 6

Mx = 22x – 0,5qx

2

– 34

= 22x – 2,5x

2

– 34

x = 0, M

A

= – 34 kNm

x = 2, M

s

= 0

x = 6, M

D

= 8 kN m

Mmax = dMx/dx = 0

= 22 – 5x = 0 x = 4,4 m

Mmax = 22x – 2,5x

2

– 34

V

x

= 22 – 5x

x = 0, V

B

= 22 kN

x = 5, V

D

= – 8

kN

Jika Lintang sama dengan nol maka Momen Mak.

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= 14,4 kN

B – D 0 – 4 Mx = R

BH

. x – 0,5qx

2

= 8x – 1,5x

2

x = 0, M

B

= 0

x = 4, M

D

= 8 kN m

Mmax = dMx/dx = 0

= –8 + 3x = 0 x = 2,67 m

Mmax = 8x – 1,5x

2

= 10,67 kN

V

x

= – 8 + 3x

x = 0, V

B

= – 8

kN

x = 4, V

D

= 4 kN

Jika Lintang sama dengan nol maka Momen Mak.

–8 kN

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A

B

E

C

S

D

A

B

E

C

S

D

A

B

E

C

S

D

+

–

+

–

+

5

4

8

4

8

4

27

8

34

10

24 14,

4 10,67

+

4,4

2,6

7 4,4

8

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 26 Pertanyaan : Hitung GGD dititik C dan D

SOLUSI : Y = ( )( ) 15 = ( )( )( ) 15L2 = 8000(L-100) 3L2 - 1600L + 160.000 =0 L1,2 = ( ) √( ) ( )( ) L1 = 400 m (tidak memenuhi) L2 = 133,3 m Reaksi Perletakan 5 m 15 m Rbh Rbv D S C 20 m 10 m 10 m P=20 kN P = 10 kN Rah Rav 100 m

S O A L 1

25 m 10 m 25 m

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 27 ∑MA = 0 -Rbv(100)+Rbh(15)+20(56,665)+10(36,665) = 0 -100 Rbv+15Rbh =-1499,95 …...1) ∑MA = 0 Rav(100)+Rah(15) -10(63,335)-20(43,335) = 0 100Rav + 15 Rah = 1500,05 ……….2) ∑Ms kiri = 0 Rav(66,665)+Rah (20)-10(30)-20(10) = 0 66,665Rav +20Rah = 500 ………3) Pers(2) dan (3) 100 Rav+15Rah = 1500,05 ×4 400Rav+60Rah = 6000,2 66,665Rav+20Rah = 500 ×3 199,995Rav+60 Rah = 1500 200,005Rav = 4500,2 Rav = 22,5 kN (↑) 2250+15Rah =1500,05 15 Rah = -749,95 Rah =-49,997 kN (←) Rah = 49,997 kN(→) ∑V =0 → Rav+Rbv – 10-20 = 0 22,5+Rbv – 30 =0 Rbv = 7,5 kN ∑H =0 → Rah –Rbh = 0 49, 997+Rbh = 0

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 28 Rbh = 49,997 kN (←)

Gaya Gaya Dalam

Y = ( ) ( )_{( )} Y = ( ) Y = ( ) Y = 0.5998x – 0.0045x2 X = 35 > Y = 15.48 X = 35 → tan θ = 0,4423 → θ = 23.86 0 sin θ = 0,404 cos θ = 0,91 Vx = Rav – 10 = 22,5 – 10 = 12,5 kN (↓) Hx = Rah = 49,997 kN = 49,997 kN (←)  SFc = V cos θ – H sin θ = (12,5)( 0,91) – (49,997)( 0,404) = 11.375 – 20.198 = - 8.823 kN  NFc = V sin θ + H cos θ = (12,5)( 0,404)+( 49,997)( 0,91)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 29 = 50.547 kN

 Mc = Rav (35) – Rah(15.48) – 10(10) = 22.5(35) – 49.997 (15.48) - 100 = - 86.45 kNm

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 30 HITUNG GGD DI TITIK C Q1 = 3 kN/m Rbv Rbh S C A B Rav Rah 5 m Y =10 m 30 m 20 m Q2 = 1,5 kN/m C Q1 = 3 kN/m Q2 = 1,5 kN/m V = 33,15 kN H = 107,625 kN 93,15 kN 86,125 kN 20 m

S O A L 2

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SOLUSI

∑MA = 0 → (3)(50)(25) - (1,5)(5)(2,5) + (1,5)(10)(5) + Rbh(10) + Rbv(50) = 0 3750 – 18,75 + 75 + 10 Rbh -50 Rbv = 0 10 Rbh -50 Rbv = 3806,25 ……….(1) ∑MB = 0 → (3)(50)(25) – (1,5)(15)(15/2 ) + Rah(10) + Rav(50) = 0 -3750 – 168,75 + 10 Rah + 50 Rav = 0 10 Rah + 50 Rav = 3198,75 ……..………(2)

PERSAMAAN LENGKUNG PARABOLA, Y = 10, X = 50, h = 15 Y = ( )( ) 10 = ( )( )( ) 10L2 = 3000 L – 150.000 L2 = 300L + 15000 = 0 L1,2 = ( ) √( ) ( )( ) ( )( ) L1 = 63,40 m L2 = 236,61 m (tidak memenuhi)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 32 ∑Ms kanan = 0 (3) (31,7)(15,85) + (1,5)(15)(15/2) – Rbv(31,7) + Rbh(15) = 0 1507,335 + 168,75 – 31,7 Rbv + 15 Rbh = 0 31,7 Rbv – 15 Rbh = 1676,085 ……….(3) PERSAMAAN (1) DAN (3) 50Rbv – 10 Rbh = 3806,25 ×3 150 Rbv – 30 Rbh = 11418,75 31,7 Rbv – 15 Rbh = 1676,085 ×2 63,4 Rbv – 30 Rbh = 3352,17 86,6 Rbv = 8066,58 Rbv = 93,15 kN (↑) 50 Rbv – 10 Rbh = 3806,25 50(93,15) – 10 Rbh = 3806,25 4657,5 – 3806,25 = 10 Rbh Rbh = 85,125 kN Untuk x = 20, h = 15 m, L = 63,4 m Y = ( )( ) Y = ( )( )( ) Y = 0,95x – 0,01 x2 X = 20 → tan θ = 0,55 → θ = 28,81 0

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 33 sin θ = 0,48 cos θ = 0,88 Vx = Rbv – 3(x) = 93,15 – 3(20) = 33,15 kN (↓) Hx = Rbh + (1,5)(15) = 85,125 + 22,5 = 107, 625 kN (→)  SFc = V cos θ – H sin θ = (33,15)(0,88) – (107,625)(0,48) = 29,172 – 51,66 = -22,488 kN  NFc = V sin θ + H cos θ = (33,15)(0,48)+(107,625)(0,88) = 110,622 kN  Mc = Rbv (x) - Rbh(y) – 0,5(Q2)y2 – 0,5(Q1)x2 = 93,15(20) – 85,125(15) – 0,5 (1,5) (15)2 – 0,5 (3) (20)2 = 182,625 kNm

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 34 Pertanyaan : Hitung GGD titik C (8 meter dari A) dan titik D (15 meter dari A)

SOLUSI : h = 10 m x = 40 m Y = 6 m Y = ( )( ) 6 = ( )( )( ) 6 L2 – 1600L + 64000 = 0 L1,2 = ( ) √( ) ( )( ) ( ) L1,2 = √ ) P = 5 kN P = 10 kN S D C 4 m 6 m Rbh Rbv 40 m Rah Rav A B 5 m 5 m 8 m 15 m

S O A L 3

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 35 L1 = 217,6607 m (tidak mungkin) L2 = 49,0059 m Reaksi Perletakan  ∑MB = 0 Rav(40) – Rah(6) – 5(35) – 10(30) = 0 40 Rav – 6Rah – 475 = 0 ………(1)  ∑MS kiri = 0 Rav (L/2) – Rah.10 – 5(L/2-5) – 10(L/2 – 10) = 0 Rav (24,5030) – Rah.10 – 97,515 – 145,03 = 0 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0 ………(2)

ELIMINASI PERSAMAAN (1) DAN (2) :

40 Rav – 6Rah – 475 = 0 ×10 400Rav – 60Rah = 4750 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0 ×6 147,018 Rav – 60 Rah = 1455,27

252,982 Rav = 3294,73 Rav = 13,0236 kN(↑) Substitusi Rav ke persamaan (1)

40 Rav – 6Rah – 475 = 0 40(13,0236) – 6(Rah) – 475 = 0 Rah = 7,6573 kN (→)  ∑H = 0 Rah + Rbh = 0 Rah = - Rbh = -7,6573 kN (→)  ∑MA = 0 -Rbv.40 + Rbh.6 + 10.10 + 5.5 = 0 -Rbv.40 +(-7,6573)6 + 125 = 0 Rbv = 1,9764 kN (↑)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 36  ∑V = 0 Rav + Rbv -10 - 5 = 0 13,0236 + 1,9764 – 15 = 0 …..ok!! GAYA-GAYA DALAM Untuk h = 10 m , L = 49,0059 m Persamaan Parabola : Y = ( )( ) Y = ( ) ( ) Y = ( ) Untuk x = 8 m, maka y = Y = ( ( ) ( ) ) Y = 5.4693 m

Titik c, untuk x = 8 m dari A, maka :

= ( ) X = 8 → = 0,54790 tan θ = 28,7976 sin θ = 0,4817 cos θ = 0,8763  Titik C (8; 5,4693) Vx = Rav – 5 = 13, 0236 – 5 = 8,0236 kN (↓) Hx = Rah = 7,6573 kN (←) Gaya Lintang (SFx) SFx = Vcos θ – H sin θ = (8,0236)( 0,8763) – (7,6573)( 0,4817) = 3,3426 kN ≈ 3 kN Gaya Normal (NFx)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 37 NFx = -(Vcos θ + H sin θ) = - ((8,0236)( 0,8763) + (7,6573)( 0,4817)) = - 8,4762 kN Momen di titik C Mc = Rav 8 - Rah. 5,4639 – 5.3 = 13,0236. 8 – (7,6573)(5,4639) – 15 = 47,3501 kNm  Titik D Untuk x =15 m, maka Y = ( ( ) ( ) ) = = ( ) X=15 m → = ( ( )) m tan θ = 0,3166 → θ = 17,56780 sin θ =0,3018, cos θ = 0,9534  Titik D (15; 8,4959) Vx = Rav – 5 – 10 = 13,0236 – 15 = -1,9764 kN (↓) Hx = Rah = 7,6573 kN (←) o Gaya Lintang (SFx) SFx = Vcos θ – H sin θ = (-1,9764)(0,9534) – (7,6573)( 0,3018) = -4,1953 kN ≈ -4 kN o Gaya Normal (NFx) NFx = -( Vcos θ + H sin θ) = -((-1,9764)( 0,9534)+(7,6573)(0,3018)) = -1,8843 +2,3110 = -0,4267 kN o Momen di Titik D MD = Rav 15 - Rah.(8,4959) – 5(10) – 10(5) MD = (13,0236)15 – (7,6573) .(8,4959) – 50 – 50 MD = 30.298 kNm

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Pertanyaan : Hitung GGD pada titik C (5 meter dari B) dan titik D (2 meter dari A) Penyelesaian : Untuk x = 20 m, y = 4 m, h = 6 m Y = ( )( ) 4 = ( )( )( ) 4L2_{– 480 L + 9600 = 0} L1,2 = ( ) √( ) ( )( ) ( ) L1,2 = √ ) L1 = 94,6410 m (tidak mungkin) L2 = 25,3590 m

S O A L 4

q = 2 t/m’ q = 1 t/m’ D S C B A Rah Rav 2 m 4 m Rbv Rbh 2 m _{5 m} 20 m S

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 39 Reaksi Perletakan  ∑MA = 0 (2)(20)(10) + (1) (4)(2) – (1)(2)(1) - Rbv.20 + Rbh.4 = 0 400 + 8 – 2 - Rbv.20 + Rbh. = 0 - Rbv.20 + Rbh.4 + 406 = 0 Rbv.20 - Rbh.4 – 406 = 0 ……….………(1)  ∑MB = 0 -(2)(20)(10) – (1)(6)(3) + Rav.20 + Rah.4 = 0 -400 – 18 + Rav.20 + Rah.4 = 0 Rav.20 + Rah.4 - 418 = 0………..…(2)  ∑Ms kanan = 0 -Rbv. 12,6795 + Rbh.6 +(2)(12,6795)(6,3398) + (1)(6)(3) = 0 -12,6795 Rbv +6 Rbh + 178,7710 = 0………..……(3)  ∑Ms kiri = 0 -Rav.7,3025 – Rah.2 –(2)(7,3205)(3,6603) = 0 7,3205 Rav – 2Rah – 53,5905 = 0………..(4) ELIMINASI PERSAMAAN (1) DAN (3)

20 Rbv – 4Rbh = 406 ×6 120 Rbv - 24Rbh = 243,6 -12,6795 Rbv + 6 Rbh = -178,7710 ×4 -50,718 Rbv + 24Rbh = -715,084 + 69,282 Rbv = 1720,916 Rbv = 24,8393 t (↑) Substitusi Rbv = 24,8393 t ke persamaan (1) 20 Rbv – 4 Rbh = 406 20(24,8393) – 4Rbh = 406 Rbh = 22,6965 t (←)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 40 Eliminasi (2) dan (4) : Rav.20 + Rah.4 - 418 = 0 ×1 Rav.20 + Rah.4 = 418 7,3205 Rav – 2Rah – 53,5905 = 0 ×2 14,641 Rav – 4 Rah = 107,181 + 34,641 Rav = 525,81 Rav = 15,1607 t (↑) Substitusi Rav = 15,1607 t ke (2) 2Rav + 4 Rah = 418 20(15,1607) + 4 Rah = 418 Rah = 28,6965 t (→)  ∑ V = 0 Rav + Rbv – 40 = 0 15,1607 + 24,8393 – 40 = 0……….ok!  ∑ H = 0 Rah – Rbh – 6 = 0 28,6965 – 22,6965 – 6 = 0……….ok!

MENENTUKAN GAYA-GAYA DALAM

Untuk h = 6 m, L = 25,3590 m → Y = ( )( ) ( )( )( )

( )

Untuk titik c, x = 5 m dari B → y ( ( ) ( ) ) { } ( ) X = 5 m → { } ( ( )) = 0,5372 tan θ = 0,53720 → θ = 29,8213o sin θ = 0,4937 cos θ = 0,8676

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 41  Titik C (5; 3,7990) dari B Vx = Rbv – qx =24,8393 – (2)(5) = 14, 8393 t (↓) Hx = Rbh +q.y = 22,6965 +(1)(3,7990) = 26,4955 t (→) Gaya Lintang (SFx) SFx = V cos θ – H sinθ = (14,8393)(0,4973) + (26,4955)(0,8676) = 30,3671 t Mc = Rbv. 5 – Rbh. 3,7990 +(2)(5)(2,5) = 24,8393 (5) – 22,6965(3,7990) +25 = 62,9725 tn

Untuk titik D, x = 2m dari A, h = 2m , L = 25,3590 m, x = 2 m Y = ( )( ) ( )( )( ) = ( ) Untuk x = 2m , y = ( ( ) ( ) ) = 0,5812 m { } ( ) { } ( ( )) tan θ = 0,26570 → θ = 14,8797o sin θ = 0,2567 cos θ = 0,9665  TITIK D (2 ; 0,5812) Vx = Rav – q.x = 15,1607 – 2(2) = 11,1607 t (↓) Hx = Rah = 28,6965 t (←)  GAYA LINTANG (SFx) SFx = Vcos θ – H sin θ = (11,1607(0,9665))-(28,6965(0,2567)) = 3,4204 t ≈ 3t  Gaya Normal (NFx) NFx= -(Vsin θ + Hcos θ) = -((11,1607(0,2567)) +((28,6965)(0,9665)) = -30,6001 t MD = RAv.2 – Rah. 0,5812 – (2)(2)(1) = 15,1607(2) – 28,6965(0,5812) – 4 = 19,1128 tm

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 42 X = 0 → y = 0 { } →( ) Tan θ = 0,3155 → θ = 17,5105 sin θ = 0,3009 cos θ = 0,9537  TITIK A (0,0) Vx = Rav = 15,1607 t (↓) Hx = Rah = 28,6965 t (←)  Gaya Lintang (SFx) SFx = Vcos θ - Hsin θ (15,1607)(0,9537) – (28,6965)(0,3009) = 5,8240 t ≈ 6 t  Gaya Normal (NFx) NFx = -(Vsin θ + H cos θ) = -((15,1607) (0,3009) + (28,6965)(0,9537) = -31,9297 t M dititik A → Ma = 0

 Untuk titik S (12,6795;6) dari B {

} =

( ( )( ))

Tan θ = 0 → sin θ = 0, cos θ = 1 S = (12,6795;6) Vx = Rbv – q.x = 24,8393 –(2)(12,6795) = -0,5197 t (↓) Hx = Rbh + q.x = 22,q.y = 22.,6965 +(1)(6) = 28,6965 t (→) Gaya Lintang (SFx) SFx = Vcos θ – Hsin θ = (-0,5197.1) – (28,6965.0) =-0,5197 t Gaya Normal (NFx) NFx = Vsin θ + Hcos θ = (-0,5197.0) – (28,6965.1) = 28,6965 t  ntuk titik B (0,0) dari B

{ } = ( ( )( )) Tan θ = 0,9464 → θ = 43,4226 Sin θ = 0,6874 Cos θ = 0,7263  Titik B (0,0) Vx = Rbv = 24,8393 t (↓) HX = Rbh = 22,6965 t (→) Gaya Lintang (SFx) SFx = Vcos θ – Hsin θ = (24,8393)(0,7263) – (22,6965)(0,6874) =2,4392 t Gaya Normal (NFx)

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 43 NFx = Vsin θ + Hcos θ = (-0,5197)(0,6874) – (28,6965)(0,7263) = 28,6965 t

Pertanyaan : Tentukan Reaksi perletakan soal berikut ini :

Persamaan dasar parabola yang digunakan adalah : ( )( )

Dimana :

Y = tinggi titik yang ditinjau dari tumpuan H = tinggi puncak parabola dari tumpuan X = jarak mendatar dari tumpuan terdekat L = jarak mendatar dua tumpuan

SOLUSI : Y : ( )( ) = ( )( )( ) = ( )  ∑MA = 0

S O A L 5

S q = 2 kN/m C A B 10 m P1 = 5kN P2 = 4 kN 20 m 10 m 10 m

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 44 (-Rbv.40)-(4.7,5) + (5.30) + (q.20.10) = 0 -40 Rbv – 30 +150 + 400 = 0 -40 Rbv = -520 Rbv = 13 kNm  ∑MB = 0 (-Rav.40) – (q.20.30) – (P1.10) – (P2.7,5) = 0 (-Rav.40) – 1200 – 50 – 30 = 0 40 Rav = 1280 Rav = 32 kN  ∑Ms kanan = 0 (-Rbv.20) + (Rbh.10) + (4.2,5) + (5.10) = 0 (-20.13) + (10 Rbh) + 10 + 50 = 0 -260 + 10 Rbh + 60 = 0 Rbh = 20 kN  ∑Ms kiri = 0 (-Rav.20) - (Rah.10) - (q.20.10) = 0 (-20.32) - (10 Rah) – 400 = 0 -10 Rah = 0 Rah = 24 kN  ∑V = 0 Rav + Rbv = 40 +5 32 + 13 = 45 ………..ok!  ∑H = 0 Rah - Rbh = 4 24 – 20 = 4 …………..ok!

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 46 Pertanyaan : Hitung gaya gaya batang yang terjadi pada struktur rangka batang tegrambar di bawah ini menggunakan metode kesetimbanan titik :

m = 2j – r 9 = 2(6) – 3 9 = 9….(statis tentu) 1. REAKSI PERLETAKAN  ∑MA = 0 (-Rbv.3) + (5.3) + (-1.8) +(-1.4) = 0 1 kN 5 kN 5 kN 1 kN 1 kN 4 m 4 m 1 2 3 4 5 6 F2 F3 F1 F4 3 m Rav Rbv Rah _R_bh F5 F8 F6 F7 F9 M= jumlah batang J = jumlah titik buhul R = jumlah reaksi perletakan

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 47 -3Rbv – 15 – 8 – 4 = 0 Rbv = 1  ∑MB = 0 (-Rav.3) + (-5.3) + (-1.8) +(-1.4) = 0 Rav = 9  ∑V = 0 Rav + Rbv = 0 5 + 5 = 0………...ok!  ∑H = 0 Rah = 1+1+1 Rah = 3………...ok!

II. GAYA DI TITIK BUHUL BUHUL 1 BUHUL 2  ∑ V = 0 α Rah Rav F1 F2 4 m 3 m ∑V = 0 Rav + F2 = 0 9 + F2 = 0 F2 = -9 kN ∑H = 0 Rah + F1 = 0 3 + F1 = 0 F1 = -3 kN F4 F3y F3 F3x F1 F3x F3y 3 m 4 m 5 m α F3x = F3 sin α = F3 F3y = F3 cos α = F3

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 48 Rbv + F4 + F3y = 0 1 + F4 + 3,33 = 0 F4 + 3,66 = 0 → F4 = - 3,66  ∑ H = 0 F1 + F3x + 1 = 0 -3 + F3x + 1 = 0 F3 = 3,333  BUHUL 3  ∑ V = 0 F2 + F3y – F6 = 0 -9 + 3,33 = 0 → F6 = - 6,33  ∑ H = 0 F3x + F5 = 0 → F5 = - 1,99  BUHUL 5 o ∑ V = 0 F6 F2 F3y F3 F3x F5 F3x = F3 cos α = F3 F3y = F3 sin α = F3 F9 F7x F7 F7y F6 5 kN F7x = F7 cos α = F7 F7y = F7 sin α = F7

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Kumpulan Soal Mekanika Rekayasa 1 dan 2 Page 49 F6 + F7y + 5 = 0 -6,33 + F7 . + 5= 0 → F7 = 1,66 o ∑ H = 0 F7x + F9 = 0 → F9 = - 0,99  BUHUL 6 o ∑ V = 0 F8 + 5 = 0 F8 = -5 BUHUL 4 (kontrol) 5 kN F9 F8 1 kN F8 F7y F7 F7x F5 F4 F7x = F7 cos α = 1,66 =0,99 F7y = F7 sin α = F7 = 1,33 o ∑ V = 0 F6 + F7y - F4 = 0 -5 +1,33 – (-3,36) = 0 → ok o ∑ H = 0 F5 + 1 + F7x = 0 → ok

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