MEKANIKA BAHAN

DISUSUN OLEH :

AHMAD MUKHROJI WIRATAMA(11 2013 152.P)

DOSEN:Erwandi Permana , ST

FAKULTAS TEKNIK SIPILUNIVERSITAS MUHAMMADIYAH PALEMBANG

2015

No.1B

1A A

3

1

4 2

B1 kardus = 50 gr(40cmx50cm) = (0,4m x 0,5m) = 0,20 m

1 kardus ada 50 alas50 x 2 gr = 100 gr

1 kardus berisi 50 isi cair50 x 50 gr = 2500 gr

Jadi, berat kardus beserta isi = 50 gr + 100 gr + 2500 gr = 2650 gr

Jumlah kardus pada lantai adalah= ( )( , , ) = 210 buahKotak kardus berisi 3 tumpukan maka:210 buah x 3 = 630 buah

Berat seluruh bebanBanyak kardus x berat kardus total= 630 x 2,65 kg=1669,5 kg=1,669 tonJadi, q= 1,669 ton

Potongan A-A

4m 2m = , = 0,556

Jadi,beban yang dipotong oleh slope pada potongan A-A adalah sebesar 0,556 ton perslopePotongan B-B

3 m 1 m 1 m

= , = 0,41725

Jadi,beban yang dipotong oleh slope pada potongan B-B adalah sebesar 0,41725 ton perslope

No.2

Given: The car is towed at constant speed by the 600 N force and the angle is 25.Find: The forces in the ropes AB and AC.Plan:1. Draw a FBD for point A.2. Apply the EofE to solve for the forces in ropes AB and AC.

Answer:

Applying the scalar EofE at A, we get;+Fx = FAC cos 30 FAB cos 25 = 0+Fy = -FAC sin 30 FAB sin 25 + 600 = 0Solving the above equations, we get;FAB = 634 NFAC = 664 N

No.2

Given: The car is towed at constant speed by the 600 N force and the angle is 25.Find: The forces in the ropes AB and AC.Plan:1. Draw a FBD for point A.2. Apply the EofE to solve for the forces in ropes AB and AC.

Answer:

Applying the scalar EofE at A, we get;+Fx = FAC cos 30 FAB cos 25 = 0+Fy = -FAC sin 30 FAB sin 25 + 600 = 0Solving the above equations, we get;FAB = 634 NFAC = 664 N

No.2

Given: The car is towed at constant speed by the 600 N force and the angle is 25.Find: The forces in the ropes AB and AC.Plan:1. Draw a FBD for point A.2. Apply the EofE to solve for the forces in ropes AB and AC.

Answer:

Applying the scalar EofE at A, we get;+Fx = FAC cos 30 FAB cos 25 = 0+Fy = -FAC sin 30 FAB sin 25 + 600 = 0Solving the above equations, we get;FAB = 634 NFAC = 664 N

No 3

Given: Weight of the boom = 125 lb, the center of mass is at G, and the load = 600 lb.Find: Support reactions at A and B.Plan:1. Put the x and y axes in the horizontal and vertical directions, respectively.2. Determine if there are any two-force members.3. Draw a complete FBD of the boom.4. Apply the EofE to solve for the unknowns.

Answer:

Note: Upon recognizing CB as a two-force member, the number of unknowns at B arereduced from two to one. Now, using Eof E, we get:+ MA = 125 4 + 600 9 FB sin 40 1 FB cos 40 1 = 0FB = 4188 lb or 4190 lb

+ FX = AX + 4188 cos 40 = 0; AX = 3210 lb + FY = AY + 4188 sin 40 125 600 = 0; AY = 1970 lb

No 3

Given: Weight of the boom = 125 lb, the center of mass is at G, and the load = 600 lb.Find: Support reactions at A and B.Plan:1. Put the x and y axes in the horizontal and vertical directions, respectively.2. Determine if there are any two-force members.3. Draw a complete FBD of the boom.4. Apply the EofE to solve for the unknowns.

Answer:

Note: Upon recognizing CB as a two-force member, the number of unknowns at B arereduced from two to one. Now, using Eof E, we get:+ MA = 125 4 + 600 9 FB sin 40 1 FB cos 40 1 = 0FB = 4188 lb or 4190 lb

+ FX = AX + 4188 cos 40 = 0; AX = 3210 lb + FY = AY + 4188 sin 40 125 600 = 0; AY = 1970 lb

No 3

Given: Weight of the boom = 125 lb, the center of mass is at G, and the load = 600 lb.Find: Support reactions at A and B.Plan:1. Put the x and y axes in the horizontal and vertical directions, respectively.2. Determine if there are any two-force members.3. Draw a complete FBD of the boom.4. Apply the EofE to solve for the unknowns.

Answer:

Note: Upon recognizing CB as a two-force member, the number of unknowns at B arereduced from two to one. Now, using Eof E, we get:+ MA = 125 4 + 600 9 FB sin 40 1 FB cos 40 1 = 0FB = 4188 lb or 4190 lb

+ FX = AX + 4188 cos 40 = 0; AX = 3210 lb + FY = AY + 4188 sin 40 125 600 = 0; AY = 1970 lb

No.4

Jawab:

Q1= 1,5 kN/m . 3 m = 4,5 kNQ2= 1 kN/m . 3 m = 3 kNQ3= 2,5 kN . 1,5 m = 3,75 kN

MB=0-Q1 .6 Q2 . 3 Q3 . 0,75 + RVA . 7,5 = 0= (4,5 . 6) + (3 .3) + (3,75 .0,75)7,5

= 27 + 9 + 2,81257,5RVA = 5,175 kN

MA=0-RVB .7,5 Q2 . 1,5 Q2 . 4,5 + Q3 . 6,75 = 0

= (4,5 . 1,5) + (3 .4,5) + (3,75 .6,75)7,5= 6,75 + 13,5 + 25,31257,5

RVB = 6,075 kN

Pembuktian !!!!V=0RVA + RVB - Q1 - Q2 - Q3 = 05,175 + 6,075 4,5 3 3,75 =00 = 0 (oke!)Ruas kanan dan kiri = 0

No. 5

Pembuktian !!!!V=0RVA + RVB - Q1 - Q2 - Q3 = 05,175 + 6,075 4,5 3 3,75 =00 = 0 (oke!)Ruas kanan dan kiri = 0

No. 5

Pembuktian !!!!V=0RVA + RVB - Q1 - Q2 - Q3 = 05,175 + 6,075 4,5 3 3,75 =00 = 0 (oke!)Ruas kanan dan kiri = 0

No. 5

Penyelesaian:

MB = 0- Ay(20) + 0.6(10) 4 = 020Ay = 6 - 4Ay = 0.1 kip

Fy = 0Ay + By 600 = 00.1 + By 0.6 = 0By = 0.5 kip

Potongan 1-1

M1-1 = 0M 0.1(x) = 0M = 0.1(x)x = 0 ft MA = 0.1(0) = 0 kip-ftx = 10 ftMC = 0.1(10) = 1 kip-ft

Penyelesaian:

MB = 0- Ay(20) + 0.6(10) 4 = 020Ay = 6 - 4Ay = 0.1 kip

Fy = 0Ay + By 600 = 00.1 + By 0.6 = 0By = 0.5 kip

Potongan 1-1

M1-1 = 0M 0.1(x) = 0M = 0.1(x)x = 0 ft MA = 0.1(0) = 0 kip-ftx = 10 ftMC = 0.1(10) = 1 kip-ft

Penyelesaian:

MB = 0- Ay(20) + 0.6(10) 4 = 020Ay = 6 - 4Ay = 0.1 kip

Fy = 0Ay + By 600 = 00.1 + By 0.6 = 0By = 0.5 kip

Potongan 1-1

M1-1 = 0M 0.1(x) = 0M = 0.1(x)x = 0 ft MA = 0.1(0) = 0 kip-ftx = 10 ftMC = 0.1(10) = 1 kip-ft

F1-1 = 00.1 V = 0V = 0.1 kipx = 0 ft VA = 0.1 kipx = 10 ft VC = 0.1 kip

Potongan 2-2

M2-2 = 0M 0.1(x) + 0.6(x 10) = 0M 0.1(x) + 0.6(x) 6 = 0M = - 0.5(x) + 6x = 10 ft MC = - 0.5(10) + 6 = 1 kip-ftx = 15 ft MD = - 0.5(15) + 6 = - 1.5 kip-ft

F2-2 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 10 ft VC = - 0.5 kipx = 15 ft VD = - 0.5 kip

F1-1 = 00.1 V = 0V = 0.1 kipx = 0 ft VA = 0.1 kipx = 10 ft VC = 0.1 kip

Potongan 2-2

M2-2 = 0M 0.1(x) + 0.6(x 10) = 0M 0.1(x) + 0.6(x) 6 = 0M = - 0.5(x) + 6x = 10 ft MC = - 0.5(10) + 6 = 1 kip-ftx = 15 ft MD = - 0.5(15) + 6 = - 1.5 kip-ft

F2-2 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 10 ft VC = - 0.5 kipx = 15 ft VD = - 0.5 kip

F1-1 = 00.1 V = 0V = 0.1 kipx = 0 ft VA = 0.1 kipx = 10 ft VC = 0.1 kip

Potongan 2-2

M2-2 = 0M 0.1(x) + 0.6(x 10) = 0M 0.1(x) + 0.6(x) 6 = 0M = - 0.5(x) + 6x = 10 ft MC = - 0.5(10) + 6 = 1 kip-ftx = 15 ft MD = - 0.5(15) + 6 = - 1.5 kip-ft

F2-2 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 10 ft VC = - 0.5 kipx = 15 ft VD = - 0.5 kip

Potongan 3-3

M3-3 = 0M 0.1(x) + 0.6(x 10) 4 = 0M 0.1(x) + 0.6(x) 6 4 = 0M = - 0.5(x) + 10x = 15 ft MD = - 0.5(15) + 10 = 2.5 kip-ftx = 20 ft MB = - 0.5(20) + 10 = 0 kip-ft

F3-3 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 15 ft VD = - 0.5 kipx = 20 ft VD = - 0.5 kip

Potongan 3-3

M3-3 = 0M 0.1(x) + 0.6(x 10) 4 = 0M 0.1(x) + 0.6(x) 6 4 = 0M = - 0.5(x) + 10x = 15 ft MD = - 0.5(15) + 10 = 2.5 kip-ftx = 20 ft MB = - 0.5(20) + 10 = 0 kip-ft

F3-3 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 15 ft VD = - 0.5 kipx = 20 ft VD = - 0.5 kip

Potongan 3-3

M3-3 = 0M 0.1(x) + 0.6(x 10) 4 = 0M 0.1(x) + 0.6(x) 6 4 = 0M = - 0.5(x) + 10x = 15 ft MD = - 0.5(15) + 10 = 2.5 kip-ftx = 20 ft MB = - 0.5(20) + 10 = 0 kip-ft

F3-3 = 00.1 0.6 V = 0V = 0.1 0.6 kipV = - 0.5 kipx = 15 ft VD = - 0.5 kipx = 20 ft VD = - 0.5 kip

Diagram Geser dan MomenDiagram Geser dan MomenDiagram Geser dan Momen