10
1. GLOBLAL STIFFNESS MATRIX Elemen Batang 1 L = 500 cm x = 400 cm y = 300 cm E = 2,100 A = 100 I = 5,000 420 0 0 -420 0 0 0 1 252 0 -1 252 0 252 84,000 0 -252 42,000 -420 0 0 420 0 0 0 -1 -252 0 1 -252 0 252 42,000 0 -252 84,000 0.8 0.6 0.0 0.0 0.0 0.0 -0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 0.6 0.0 0.0 0.0 0.0 -0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.8 -0.6 0.0 0.0 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 -0.6 0.0 0.0 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.8 -0.6 0.0 0.0 0.0 0.0 420 0 0 -420 0 0 0.8 0.6 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0 1 252 0 -1 252 -0.6 0.8 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0 252 84,000 0 -252 42,000 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.8 -0.6 0.0 -420 0 0 420 0 0 0.0 0.0 0.0 0.8 0.0 0.0 0.0 0.6 0.8 0.0 0 -1 -252 0 1 -252 0.0 0.0 0.0 -0.6 0.0 0.0 0.0 0.0 0.0 1.0 0 252 42,000 0 -252 84,000 0.0 0.0 0.0 0.0 336.00 -0.60 -151.20 -336.00 0.60 -151.20 0.80 0.60 0.00 0.00 0.00 0.00 252.00 0.81 201.60 -252.00 -0.81 201.60 -0.60 0.80 0.00 0.00 0.00 0.00 0.00 252.00 ### 0.00 -252.00 42,000.00 0.00 0.00 1.00 0.00 0.00 0.00 -336.00 0.60 151.20 336.00 -0.60 151.20 0.00 0.00 0.00 0.80 0.60 0.00 -252.00 -0.81 -201.60 252.00 0.81 -201.60 0.00 0.00 0.00 -0.60 0.80 0.00 0.00 252.00 ### 0.00 -252.00 84,000.00 0.00 0.00 0.00 0.00 0.00 1.00 269.16 201.12 -151.20 -269.16 -201.12 -151.20 201.12 151.85 201.60 -201.12 -151.85 201.60 -151.20 201.60 ### 151.20 -201.60 42,000.00 -269.16 -201.12 151.20 269.16 201.12 151.20 -201.12 -151.85 -201.60 201.12 151.85 -201.60 -151.20 201.60 ### 151.20 -201.60 84,000.00 Elemen Batang 2 L = 300 cm 270 E = 2,100 A = 40 I = 1,000 280.00 0.00 0.00 -280.00 0.00 0.00 0.00 0.93 140.00 0.00 -0.93 140.00 0.00 140.00 ### 0.00 -140.00 14,000.00 -280.00 0.00 0.00 280.00 0.00 0.00 0.00 -0.93 -140.00 0.00 0.93 -140.00 0.00 140.00 ### 0.00 -140.00 28,000.00 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 t/cm 2 cm 2 cm 4 [Kl] = [T] = [T] T = [Kg] 1 = [T] T [Kl] [T] [Kg] 1 = [Kg] 1 = u1 v1 q1 u2 v2 q2 u1 v1 [Kg] 1 = q1 u2 v2 q2 a = o t/cm 2 cm 2 cm 4 [Kl] = [T] = 1 2 3 5 Tm 4 T 5 m 3 m E = 2100 t/cm 2 A = 100 cm 2 I = 5000 cm 4 E = 2100 t/cm 2 A = 40 cm 2 I = 1000 cm 4 4 m a Cos a = 4/5 Sin a = 3/5

Contoh Soal FRAME 2D

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Page 1: Contoh Soal FRAME 2D

1. GLOBLAL STIFFNESS MATRIX

Elemen Batang 1 L = 500 cm x = 400 cmy = 300 cm

E = 2,100

A = 100

I = 5,000

420 0 0 -420 0 00 1 252 0 -1 252

0 252 84,000 0 -252 42,000-420 0 0 420 0 0

0 -1 -252 0 1 -2520 252 42,000 0 -252 84,000

0.8 0.6 0.0 0.0 0.0 0.0-0.6 0.8 0.0 0.0 0.0 0.00.0 0.0 1.0 0.0 0.0 0.00.0 0.0 0.0 0.8 0.6 0.00.0 0.0 0.0 -0.6 0.8 0.00.0 0.0 0.0 0.0 0.0 1.0

0.8 -0.6 0.0 0.0 0.0 0.00.6 0.8 0.0 0.0 0.0 0.0

0.0 0.0 1.0 0.0 0.0 0.00.0 0.0 0.0 0.8 -0.6 0.00.0 0.0 0.0 0.6 0.8 0.00.0 0.0 0.0 0.0 0.0 1.0

0.8 -0.6 0.0 0.0 0.0 0.0 420 0 0 -420 0 0 0.8 0.6 0.0 0.0 0.0 0.00.6 0.8 0.0 0.0 0.0 0.0 0 1 252 0 -1 252 -0.6 0.8 0.0 0.0 0.0 0.0

0.0 0.0 1.0 0.0 0.0 0.0 0 252 84,000 0 -252 42,000 0.0 0.0 1.0 0.0 0.0 0.00.0 0.0 0.0 0.8 -0.6 0.0 -420 0 0 420 0 0 0.0 0.0 0.0 0.8 0.6 0.00.0 0.0 0.0 0.6 0.8 0.0 0 -1 -252 0 1 -252 0.0 0.0 0.0 -0.6 0.8 0.00.0 0.0 0.0 0.0 0.0 1.0 0 252 42,000 0 -252 84,000 0.0 0.0 0.0 0.0 0.0 1.0

336.00 -0.60 -151.20 -336.00 0.60 -151.20 0.80 0.60 0.00 0.00 0.00 0.00

252.00 0.81 201.60 -252.00 -0.81 201.60 -0.60 0.80 0.00 0.00 0.00 0.000.00 252.00 84,000.00 0.00 -252.00 42,000.00 0.00 0.00 1.00 0.00 0.00 0.00

-336.00 0.60 151.20 336.00 -0.60 151.20 0.00 0.00 0.00 0.80 0.60 0.00-252.00 -0.81 -201.60 252.00 0.81 -201.60 0.00 0.00 0.00 -0.60 0.80 0.00

0.00 252.00 42,000.00 0.00 -252.00 84,000.00 0.00 0.00 0.00 0.00 0.00 1.00

269.16 201.12 -151.20 -269.16 -201.12 -151.20

201.12 151.85 201.60 -201.12 -151.85 201.60

-151.20 201.60 84,000.00 151.20 -201.60 42,000.00

-269.16 -201.12 151.20 269.16 201.12 151.20

-201.12 -151.85 -201.60 201.12 151.85 -201.60

-151.20 201.60 42,000.00 151.20 -201.60 84,000.00

Elemen Batang 2 L = 300 cm

270

E = 2,100

A = 40

I = 1,000

280.00 0.00 0.00 -280.00 0.00 0.000.00 0.93 140.00 0.00 -0.93 140.00

0.00 140.00 28,000.00 0.00 -140.00 14,000.00-280.00 0.00 0.00 280.00 0.00 0.00

0.00 -0.93 -140.00 0.00 0.93 -140.000.00 140.00 14,000.00 0.00 -140.00 28,000.00

0 -1 0 0 0 01 0 0 0 0 00 0 1 0 0 00 0 0 0 -1 00 0 0 1 0 00 0 0 0 0 1

0 1 0 0 0 0

t/cm2

cm2

cm4

[Kl] =

[T] =

[T]T =

[Kg]1 = [T]T [Kl] [T]

[Kg]1 =

[Kg]1 =

u1 v1 q1 u2 v2 q2

u1

v1

[Kg]1 = q1

u2

v2

q2

a = o

t/cm2

cm2

cm4

[Kl] =

[T] =

1

2

3

5 Tm4 T

5 m

3 m

E = 2100 t/cm2

A = 100 cm2

I = 5000 cm4

E = 2100 t/cm2

A = 40 cm2

I = 1000 cm4

4 m

a

Cos a = 4/5Sin a = 3/5

Page 2: Contoh Soal FRAME 2D

-1 0 0 0 0 0

0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1

0 1 0 0 0 0 280 0 0 -280 0 0 0 -1 0 0 0 0-1 0 0 0 0 0 0 1 140 0 -1 140 1 0 0 0 0 0

0 0 1 0 0 0 0 140 28,000 0 -140 14,000 0 0 1 0 0 00 0 0 0 1 0 -280 0 0 280 0 0 0 0 0 0 -1 00 0 0 -1 0 0 0 -1 -140 0 1 -140 0 0 0 1 0 00 0 0 0 0 1 0 140 14,000 0 -140 28,000 0 0 0 0 0 1

0.00 0.93 140.00 0.00 -0.93 140.00 0 -1 0 0 0 0

-280.00 0.00 0.00 280.00 0.00 0.00 1 0 0 0 0 00.00 140.00 28,000.00 0.00 -140.00 14,000.00 0 0 1 0 0 00.00 -0.93 -140.00 0.00 0.93 -140.00 0 0 0 0 -1 0

280.00 0.00 0.00 -280.00 0.00 0.00 0 0 0 1 0 00.00 140.00 14,000.00 0.00 -140.00 28,000.00 0 0 0 0 0 1

0.93 0.00 140.00 -0.93 0.00 140.00

0.00 280.00 0.00 0.00 -280.00 0.00

140.00 0.00 28,000.00 -140.00 0.00 14,000.00

-0.93 0.00 -140.00 0.93 0.00 -140.00

0.00 -280.00 0.00 0.00 280.00 0.00

140.00 0.00 14,000.00 -140.00 0.00 28,000.00

2. OVERALL STIFFNESS MATRIX

=

269.16 201.12 -151.20 -269.16 -201.12 -151.20 0.00 0.00 0.00 0

201.12 151.85 201.60 -201.12 -151.85 201.60 0.00 0.00 0.00 0

-151.20 201.60 84,000.00 151.20 -201.60 42,000.00 0.00 0.00 0.00 0

-269.16 -201.12 151.20 270.10 201.12 291.20 -0.93 0.00 140.00

-201.12 -151.85 -201.60 201.12 431.85 -201.60 0.00 -280.00 0.00

-151.20 201.60 42,000.00 291.20 -201.60 112,000.00 -140.00 0.00 14,000.00

0.00 0.00 0.00 -0.93 0.00 -140.00 0.93 0.00 -140.00 0

0.00 0.00 0.00 0.00 -280.00 0.00 0.00 280.00 0.00 0

0.00 0.00 0.00 140.00 0.00 14,000.00 -140.00 0.00 28,000.00 0

3. REARRAGEMENT

-4

=

270.10 201.12 291.20 -269.16 -201.12 151.20 -0.93 0.00 140.00

0 201.12 431.85 -201.60 -201.12 -151.85 -201.60 0.00 -280.00 0.00

500 291.20 -201.60 112,000.00 -151.20 201.60 42,000.00 -140.00 0.00 14,000.00

-269.16 -201.12 -151.20 269.16 -201.12 151.20 0.00 0.00 0.00 0

-201.12 -151.85 201.60 201.12 -151.85 -201.60 0.00 0.00 0.00 0

151.20 -201.60 42,000.00 -151.20 201.60 42,000.00 0.00 0.00 0.00 0

-0.93 0.00 -140.00 0.00 0.00 0.00 0.93 0.00 -140.00 0

0.00 -280.00 0.00 0.00 0.00 0.00 0.00 280.00 0.00 0

140.00 0.00 14,000.00 0.00 0.00 0.00 -140.00 0.00 28,000.00 0

4. UNKNOWN DISPLACEMENT AND REACTIONS

Displacement diselesaikan dg Metode Eliminasi Gauss

-4

=

270.10 201.12 291.20

0 201.12 431.85 -201.60

500 291.20 -201.60 112,000.00

270 201 291 -4201 432 -202 0 Disusun seperti ini291 -202 112,000 500

1 0.74 1.08 -0.01481 2.15 -1.00 0.0000 Kolom pertama diusahakan menjadi satu1 -0.69 384.62 1.7170 dapat dilakukan dengan cara membagi dengan kolom pertama

1 0.74 1.08 -0.0148 Usahakan kolom pertama menjadi nol0 1.40 -2.08 0.0148 baris ke 2 dikurangi baris ke 10 -1.44 383.54 1.7318 baris ke 3 dikurangi baris ke 1

1 0.74 1.08 -0.0148 Kolom kedua diusahakan menjadi satu0 1 -1.48 0.0106 dapat dilakukan dengan cara membagi dengan kolom kedua0 1 -266.92 -1.2052

1 0.74 1.08 -0.01480 1 -1.48 0.0106 Usahakan kolom kedua baris 3 menjadi nol0 0 265.43 1.2158 baris ke 3 dikurangi baris ke 2

0.00458

0.01056 0.00679 = 0.01735

-0.01481 -0.00494 -0.01292 = -0.03267

=

-0.03267

0.01735

0.00458

Reaksi

[T]T =

[Kg]2 = [T]T [Kl] [T]

[Kg]2 =

[Kg]2 =

u2 v2 q2 u3 v3 q3

u2

v2

[Kg]2 = q2

u3

v3

q3

M2 = 5 Tm = 500 Tcm F2 = -4 T

u1 v1 q1 u2 v2 q2 u3 v3 q3

F1 u1

G1 v1

M1 q1

F2 = -4 u2 u2

G2 = 0 v2 v2

M2 = 500 q2 q2

F3 u3

G3 v3

M3 q3

u2 v2 q2 u1 v1 q1 u3 v3 q3

u2 u2

v2 v2

q2 q2

F1 u1

G1 v1

M1 q1

F3 u3

G3 v3

M3 q3

u2

v2

q2

q2 =

v2 =

u2 =

u2

v2

q2

Page 3: Contoh Soal FRAME 2D

=

-269.16 -201.12 -151.20 -0.03267

=

4.611 ton

-201.12 -151.85 201.60 0.01735 4.859 ton

151.20 -201.60 42,000.00 0.00458 183.941 ton cm

-0.93 0.00 -140.00 -0.611 ton

0.00 -280.00 0.00 -4.859 ton

140.00 0.00 14,000.00 59.553 ton cm

5. MEMBER FORCES

Elemen Batang 1

[T] [Kg] {u}

0.8 0.6 0.0 0.0 0.0 0.0 269.16 201.12 -151.20 -269.16 -201.12 -151.20 0

-0.6 0.8 0.0 0.0 0.0 0.0 201.12 151.85 201.60 -201.12 -151.85 201.60 0

0.0 0.0 1.0 0.0 0.0 0.0 -151.20 201.60 84,000.00 151.20 -201.60 42,000.00 0

0.0 0.0 0.0 0.8 0.6 0.0 -269.16 -201.12 151.20 269.16 201.12 151.20 -0.03267

0.0 0.0 0.0 -0.6 0.8 0.0 -201.12 -151.85 -201.60 201.12 151.85 -201.60 0.01735

0.0 0.0 0.0 0.0 0.0 1.0 -151.20 201.60 42,000.00 151.20 -201.60 84,000.00 0.00458

336.00 252.00 0.00 -336.00 -252.00 0.00 0

-0.60 0.81 252.00 0.60 -0.81 252.00 0

-151.20 201.60 84,000.00 151.20 -201.60 42,000.00 0

-336.00 -252.00 0.00 336.00 252.00 0.00 -0.03267

0.60 -0.81 -252.00 -0.60 0.81 -252.00 0.01735

-151.20 201.60 42,000.00 151.20 -201.60 84,000.00 0.00458

6.60 ton

1.12 ton

183.94 ton cm

-6.60 ton

-1.12 ton

376.32 ton cm

Elemen Batang 2

[T] [Kg] {u}

0 -1 0 0 0 0 0.93 0.00 140.00 -0.93 0.00 140.00 -0.0327

1 0 0 0 0 0 0.00 280.00 0.00 0.00 -280.00 0.00 0.0174

0 0 1 0 0 0 140.00 0.00 28,000.00 -140.00 0.00 14,000.00 0.0046

0 0 0 0 -1 0 -0.93 0.00 -140.00 0.93 0.00 -140.00 0

0 0 0 1 0 0 0.00 -280.00 0.00 0.00 280.00 0.00 0

0 0 0 0 0 1 140.00 0.00 14,000.00 -140.00 0.00 28,000.00 0

0.00 -280.00 0.00 0.00 280.00 0.00 -0.03267

0.93 0.00 140.00 -0.93 0.00 140.00 0.01735

140.00 0.00 28,000.00 -140.00 0.00 14,000.00 0.00458

0.00 280.00 0.00 0.00 -280.00 0.00 0

-0.93 0.00 -140.00 0.93 0.00 -140.00 0

140.00 0.00 14,000.00 -140.00 0.00 28,000.00 0

-4.86 ton

0.61 ton

123.68 ton cm

4.86 ton

-0.61 ton

59.55 ton cm

F1

G1

M1

F2

G2

M2

f1

g1

m1

f2

g2

m2

f1

g1

m1

f2

g2

m2

f1

g1

m1

f2

g2

m2

f1

g1

m1

f2

g2

m2

f2

g2

m2

f3

g3

m3

f2

g2

m2

f3

g3

m3

f2

g2

m2

f3

g3

m3

f2

g2

m2

f3

g3

m3

123,68 tcm

59,55 tcm

183,94 tcm

376,32 tcm

BMD

Page 4: Contoh Soal FRAME 2D
Page 5: Contoh Soal FRAME 2D
Page 6: Contoh Soal FRAME 2D

4 3 1 131 2 3 14 Dapat disusun seperti ini3 2 5 22

1.00 0.75 0.25 3.251.00 2.00 3.00 14.00 Kolom pertama diusahakan menjadi satu1.00 0.67 1.67 7.33 dapat dilakukan dengan cara membagi dengan kolom pertama

1.00 0.75 0.25 3.25 Usahakan kolom pertama menjadi nol0.00 1.25 2.75 10.75 baris ke 2 dikurangi baris ke 10.00 -0.08 1.42 4.08 baris ke 3 dikurangi baris ke 1

1.00 0.75 0.25 3.25 Kolom kedua diusahakan menjadi satu0.00 1.00 2.20 8.60 dapat dilakukan dengan cara membagi dengan kolom kedua0.00 1.00 -17.00 -49.00

1.00 0.75 0.25 3.250.00 1.00 2.20 8.60 Usahakan kolom kedua baris 3 menjadi nol0.00 0.00 19.20 57.60 baris ke 3 dikurangi baris ke 2

Z = 3.00Y = 8.60 -6.60 = 2.00X = 3.25 -0.75 -1.50 = 1.00

4X + 3Y + Z = 13 X + 2Y + 3Z = 143X + 2Y + 5Z = 22

Page 7: Contoh Soal FRAME 2D

Kolom pertama diusahakan menjadi satudapat dilakukan dengan cara membagi dengan kolom pertama

Usahakan kolom pertama menjadi nol

Kolom kedua diusahakan menjadi satudapat dilakukan dengan cara membagi dengan kolom kedua

Usahakan kolom kedua baris 3 menjadi nol