Kumpulan Soal Mekrek

KUMPULAN SOAL DAN PENYELESAIAN MEKANIKA REKAYASA 1 DAN 2

Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 1

GERBER


Page 2

SOAL 1 P = 8 kN q = 4 kN/m

A

B 8m

D

C

S 2m

6m

P = 5 kN

4m

-33,32 kN

-7,99 kN

-0,02 kNm BMD

A

+

B

C

S

D

2,64 kNm 33,32 kNm

16,33kN

0,33 kN

2,677 kN 5,33 kN SFD

-13,327 kN

- 8 kN

JAWABAN :  Ms Kiri Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 3

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 2.Rav +2.Rbv

= 200 ..……………………..(1) = 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


Page 4

GAYA-GAYA DALAM

INTERVAL A–B 0-8

MOMEN Mx = Rav.x – qx

2

= 16.333 x - (4) x2 = 16.333 x - 2 x2

B–S 8 - 10

LINTANG Dx = = 16,333 – 4x x=0

D0 = 16.333 kN

x=0

M0 = 0

x=4

D4 = 0,33 kN

x=4

M4 = 33,332 kNm

x=8

D8 = -15,677 kN

x=8

M8 = 2,64 kNm

Mx = Rav.x – qx2 + Rbv.(x-8) – q(x-8)2

= Dx =

= 16.333 x - (4) x2 + 18.34(x-8) - (4) (x-8)2

= -8x + 66.673

= 16.333x - 2x2 + 18.34(x-8) – 2(x-8)2

x=8

D8 = 2,673 kNm

= 16.333x- 2x2 + 18.34x – 146,72 – 2(X2 - 16x +64)

x = 10

D10 = -13.327 kNm

= 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

D–C

Mx = -8x

Dx = -8

0-4

x=0

M0 = 0 kNm

x=0

M0 = -8 kNm

x=4

M4 = -32 kNm

x=4

M4 = -8 kNm


Page 5

C–S 4 - 10

Mx = -8x + Rcv (x-4)

Dx =

= -8x + Rcv.x -4Rcv

Dx = 5,333

= -8x + 13,333x – 4(13,333)

x=4

M4 = 5,333 kNm

x = 10

M10 = 5,333 kNm

= 5,333x – 53,332 x=4

M4 = -32 kNm

x = 10

M10 = -0.02 kNm


Page 6

PORTAL


Page 7

SOAL 1 Contoh 2: gaya

Hitung dan gambarkan diagram gaya-

dalam yang terjadi pada Struktur Portal berikut ini.

Q=2 KN/m P=5 KN C

D

A

3m RAH

B

RAV

1m 3m

RBV

PENYELESAIAN : a. Menghitung Reaksi Perletakan:

H  0  M

A

P  RAH  0

0

RAH  5 KN 



R .3  R .1  2.3.1,5  5.3  0 BV

BH

R  BV

29 kN 3

 M  0  RAV .3  5.4  2.3.1,5  0 B

R  AV

V  0 

R  R  Q.3 AV

R  AV


11 kN () 3

BV

11 29   6 KN 3 3

Page 8

b. Free Body Diagram dan Perhitungan GGD: Q2 = 2

15

P=5 5

5 29/3

20 11/3 11/3

29/3

0

20

5

FBD 5

11/3 29/3

Tabel Gaya-Gaya Dalam:

Bagian/

Momen

Geser

Interval A-C

(M)

(V)

M 0

dM 0 dx X

X

0-3

X  0  MA  0

(dari kiri) C-D

X 3M 0

Normal (N) N  X

11 3

C

 11 / 3 X  0,5 Q X

2

X 0M 0

dM  11 / 3  QX dx X

Nx = -5

C

0-3 (dari

X  3  M  20 D

X  0 V  0


C

Page 9

kiri)

X  3  V  29 / 3

B-D

M   R .4

Tdk Ada Mmak VX  5

0–4

X 0M 0

D

X

BH

N  29 / 3 X

B

X  4  M  20 D

D

2 C

c. Diagram Gaya-Gaya Dalam:

8

5 A

15

15

C

SFD

D

B

A

BMD

B

C

D 2

8

A

NFD


B

Page 10

SOAL 2

Q1 = 5 t/m

C

P = 10 t

E

D

F Q2 = 2t/m

3m

B

RBH 1m A 2m

2m

2m

RBV

RAV 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


Page 11

DIAGRAM BENDA BEBAS Q1 = 5 t/m

P = 10 t

9

0

0 C

D

E

F 6,8

13,11 13,11

6,8

6 3m 6

B

1m A

6,8

13,11


Page 12

Tabel Gaya-Gaya Dalam: Bagian/

Momen

Geser

Interval

(M)

(V)

A-C

M 0

dM 0 dx X

X

0-4

Normal (N)

NA-C = 13,11

X  0  MA  0

X 3M 0 C

B-F

Mx =6x – ½ Qx2

0-3

X 0MB 0

C-D

1 M X  13.11x  qx 2 2

X  3  M F  9kNm

dM X  (6  QX ) dx

NB-F = -6,8

dM X   (6  2 X ) dx

X  0  VB  6

X  3  VF  0

1 V X  13.11  (5)(2) x 2

NC  D  0

0–2 X  0  MC  0 1 X  2  M D  13.11(2)  (5)(2) 2 2

D–E

Mx = 13.11(x+2)-10(x+1)

0-2

X  0  M D  (13 .11)2  10 (2)  16 .34

X  0  VC  13 .11

X  2  V D  3.11 V X  13 .11  10

N D-E = 0

VC-D = 3.17

X  2  M E  (13 .11)( 4)  10 (3)  22 .68

F–E

Mx = 6.83 x + 9

V X  6.83

0-2

X  0  MF  9

V F – E = -6.83

N F-E = -0

X  2  M E  22 .68


Page 13

DIAGRAM BIDANG MOMEN (BMD) C

D

E

F

B

A DIAGRAM BIDANG GESER (SFD)

C

D

E

F

B

A


Page 14

DIAGRA BIDANG NORMAL (NFD) C

D

E

F

B

A


Page 15


Page 16

PORTAL TIGA SENDI


Page 17

SOAL 1

Gambar 5.4 Diagram gaya-gaya dalam portal sendi-rol akibat beban merata dan beban lateral terpusat

P = 5kN Q1 = 5 kN/m

E

C

S

D Q2 = 3 kN/m

4m

B 2m A 2m

2m

4m

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)


Page 18

SOLUSI DIAGRAM BENDA LEPAS (DBL) P = 5 kN

R1 = 10 kN

C

R2 = 20 kN

D

S

2m R3 = 12 kN 2m

B

Rbh

2m

Rbv Rah

A

Rav 2m

1m

1m

2m

2m

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)


Page 19

∑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 RAH

-

= 154 = 4 kN

Substitusi nilai RAH ke Persamaan 2 -RAV.6 - RAH.2 = -154 → -6(RAV) – 2(4) = 154 → RAV = 27 kN Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 20

DIAGRAM BENDA BEBAS

R1 = 30 kN

P = 5 kN C

q1 = 5 kN/m

10

27

4

E 4

5 24

34

C 22

S

D

C q2 = 3 kN/m

Rbh = 8 B Rbv = 8 Rah = 4

A

Rav = 27


Page 21

Tabel Gaya-Gaya Dalam:

Daerah Interval A–C 0 – 6

Momen (Mx) Mx = –RAH . x

Geser (Vx) Vx = – RAH

= –4x

Normal (Nx)

VA-C = –4 kN –27 kN

x = 0, MA = 0 x = 3, MC = – 24 kN m E–C

0 – 2

Mx = –P . x

Vx = – P

= –5x

VE-C = – 5 kN

x = 0, ME = 0

0

x = 2, MC = – 10 kN m C–D

0 – 6

Vx = 22 – 5x

Mx = 22x – 0,5qx2 – 34

x = 0, VB = 22 kN

= 22x – 2,5x2 – 34

x = 5, VD = – 8 kN

x = 0, MA = – 34 kNm x = 2, Ms = 0 x = 6, MD = 8 kN m

Jika Lintang sama dengan nol maka

–4 kN

Momen Mak.

Mmax = dMx/dx = 0 = 22 – 5x = 0

x = 4,4 m

Mmax = 22x – 2,5x2 – 34


Page 22

= 14,4 kN B–D

0 – 4

Mx = RBH . x – 0,5qx2

Vx = – 8 + 3x

= 8x – 1,5x2

x = 0, VB = – 8 kN

x = 0, MB = 0 x = 4, VD = 4 kN x = 4, MD = 8 kN m Jika Lintang sama dengan nol maka

Mmax = dMx/dx = 0 = –8 + 3x = 0 22

x = 2,67 m

–8 kN

Momen Mak.

Mmax = 8x – 1,5x2 = 10,67 kN


Page 23

+

D –+

E C

– 5

S

4 –

E 4

C

S

D

8 4,4

– 8

– 4

27

– B

8

A

– B

A

34 10 E

– 24

– C

8 S

+

D 8

– 4,4

14, 10,67 4 + 2,6 7 B

A


Page 24

PELENGKUNG TIGA SENDI


Page 25

SOAL 1

P=20 kN P = 10 kN 5m

S

C

D

Rbh 15 m

Rah

Rbv

10 m

25 m

Rav

10 m 10 m

20 m

25 m

100 m

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 Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 26

∑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 ∑H =0 → Rah –Rbh 49, 997+Rbh

= 7,5 kN =0 =0


Page 27

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)


Page 28

= 50.547 kN 

Mc

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


Page 29

SOAL 2 Q1 = 3 kN/m

S C 5m

Q2 = 1,5 kN/m Rah

A

Rav Y =10 m Rbh

30 m

20 m

B

Rbv

HITUNG GGD DI TITIK C Q1 = 3 kN/m V = 33,15 kN H = 107,625 kN

C Q2 = 1,5 kN/m

86,125 kN 20 m


93,15 kN

Page 30

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)


Page 31

∑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


Page 32

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


Page 33

SOAL 3 P = 10 kN P = 5 kN D

S

B C

4m

Rbh 6m

A

Rah Rbv

Rav 5m

5m

8m 15 m 40 m

Pertanyaan : Hitung GGD titik C (8 meter dari A) dan titik D (15 meter dari A) SOLUSI : h = 10 m x = 40 m Y=6m Y

( )(

=

6 =

(

)(

)

)(

)

6 L2 – 1600L + 64000 = 0 L1,2 = L1,2 =

(

) √(

)

( )(

)

( ) √

)


Page 34

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



∑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

……………………………(1)

……………………………(2)

ELIMINASI PERSAMAAN (1) DAN (2) : 40 Rav – 6Rah – 475 =0 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0

×10 ×6

400Rav – 60Rah 147,018 Rav – 60 Rah 252,982 Rav Rav

= 4750 = 1455,27 = 3294,73 = 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 (↑)


Page 35



∑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) Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 36

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


Page 37

SOAL 4 q = 2 t/m’

SS D

2m A

Rah

q = 1 t/m’

C

4m Rav

B

Rbh

Rbv 2m

5m 20 m

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 Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 38

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 -12,6795 Rbv + 6 Rbh = -178,7710

×6 ×4

120 Rbv - 24Rbh = 243,6 -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 (←)


Page 39

Eliminasi (2) dan (4) : Rav.20 + Rah.4 - 418 = 0 7,3205 Rav – 2Rah – 53,5905 = 0

×1 ×2

Rav.20 + Rah.4 14,641 Rav – 4 Rah

= 418 = 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=5m→{ }

(

( )

) ( ) )

) (

( )) = 0,5372

tan θ = 0,53720 → θ = 29,8213o sin θ = 0,4937 cos θ = 0,8676 Kumpulan Soal Mekanika Rekayasa 1 dan 2

Page 40



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 UNTUK TITIK A (0,0) DARI A


Page 41

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)


Page 42

NFx = Vsin θ + Hcos θ = (-0,5197)(0,6874) – (28,6965)(0,7263) = 28,6965 t

SOAL 5

Pertanyaan : Tentukan Reaksi perletakan soal berikut ini : q = 2 kN/m P1 = 5kN S C

10 m

P2 = 4 kN

B

A 20 m

10 m

Persamaan dasar parabola yang digunakan adalah :

( )(

10 m

)

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


Page 43







(-Rbv.40)-(4.7,5) + (5.30) + (q.20.10) -40 Rbv – 30 +150 + 400 -40 Rbv Rbv

=0 =0 = -520 = 13 kNm

∑MB = 0 (-Rav.40) – (q.20.30) – (P1.10) – (P2.7,5) (-Rav.40) – 1200 – 50 – 30 40 Rav Rav

=0 =0 = 1280 = 32 kN

∑Ms kanan = 0 (-Rbv.20) + (Rbh.10) + (4.2,5) + (5.10) (-20.13) + (10 Rbh) + 10 + 50 -260 + 10 Rbh + 60 Rbh

=0 =0 =0 = 20 kN

∑Ms kiri = 0 (-Rav.20) - (Rah.10) - (q.20.10) (-20.32) - (10 Rah) – 400 -10 Rah Rah

=0 =0 =0 = 24 kN



∑V = 0 Rav + Rbv = 40 +5 32 + 13 = 45 ………..ok!



∑H = 0 Rah - Rbh = 4 24 – 20 = 4 …………..ok!


Page 44

RANGKA BATANG


Page 45

SOAL 1 5 kN

5 kN F9

6

5 F7

4m

F8

F6 F5

3

1 kN

4 1 kN

F3

F2

F4

4m

2

1

1 kN

F1

Rah Rav

Rbh Rbv

3m

Pertanyaan : Hitung gaya gaya batang yang terjadi pada struktur rangka batang tegrambar di bawah ini menggunakan metode kesetimbanan titik :

m = 2j – r

M= jumlah batang

9 = 2(6) – 3

J = jumlah titik buhul

9 = 9….(statis tentu)

R = jumlah reaksi perletakan

1. REAKSI PERLETAKAN 

∑MA = 0 (-Rbv.3) + (5.3) + (-1.8) +(-1.4)

=0


Page 46

-3Rbv – 15 – 8 – 4 Rbv



∑MB = 0 (-Rav.3) + (-5.3) + (-1.8) +(-1.4) Rav



∑V = 0 Rav + Rbv = 0 5 + 5 = 0…………………………...ok!



∑H = 0 Rah = 1+1+1 Rah = 3…………………………...ok!

=0 =1

=0 =9

II. GAYA DI TITIK BUHUL BUHUL 1 F2 4m

α

Rah

F1

3m

∑V = 0

∑H = 0

Rav + F2 = 0

Rah + F1 = 0

9 + F2 = 0

3 + F1 = 0

F2 = -9 kN

F1 = -3 kN

Rav

BUHUL 2 F3

F3y

F4

F3x 3 m

F3x = F3 sin α = F3 F3y

5m

F3x

4m

F3y = F3 cos α = F3

α

F1



∑V=0


Page 47

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

F6 F3x = F3 cos α = F3

F5 F3x

F2 

F3y = F3 sin α = F3

F3

F3y ∑V=0

F2 + F3y – F6 = 0 -9 + 3,33 = 0 → F6 = - 6,33 

∑H=0 F3x + F5 = 0 → F5 = - 1,99



BUHUL 5 5 kN F9 F7x F6

F7y o

F7x = F7 cos α = F7 F7y = F7 sin α = F7

F7

∑V=0


Page 48

F6 + F7y + 5 = 0 -6,33 + F7 . + 5= 0 → F7 = 1,66 o



∑H=0 F7x + F9 = 0 → F9 = - 0,99

BUHUL 6 5 kN 1 kN

F9

F8 o

∑V=0 F8 + 5 = 0 F8 = -5

BUHUL 4 (kontrol)

F8 F7

F7y

F7x = F7 cos α = 1,66 F7y = F7 sin α = F7 o

F7x

=0,99

= 1,33

∑V=0

F6 + F7y - F4 = 0

F5

-5 +1,33 – (-3,36) = 0 → ok o ∑H=0 F4

F5 + 1 + F7x = 0 → ok


Page 49

SOAL 2


Page 50

Kumpulan Soal Mekrek

Recommend Documents