Home Work ASMM Frame 2D

1

HOMEWORK ASMMPLANE FRAME

96 kN2 3

2I Tentukan gaya-gaya batang 1,2 dan 3 denganmetode Matriks Kekakuan Langsung

3 m Jika : E = 21000 48 kN I 5 m b = 15 cm

h = 20 cm

I Maka : A = b.h = 300

4 10000 4,5 m

EA untuk semua elemen sama = 6E+06 kN

3 m 3 m

PENYELESAIAN :

A. MATRIKS KEKAKUAN ELEMEN

ELEMEN 1 (dipilih nodal i = 2 dan j = 1 750 cm

Y Y

X X b b I

I

Koordinat lokal

Koordinat global

kN/cm2

cm2

I = (1/12).b.h3 = cm4

L1 =

V2 V1

U2 U1

b = -90o

cos b = 0sin b = -1

Matriks Kekakuan Dalam Koordinat Lokal : [kl(1)]

(3)

(1)

1

(1)

(2)

(1)2

1

u2

v2 y

1

y

x

x

u1

v1

X

Y

2

q1

q2 q2

q1

2

0 0 0 0 8400 0 0 -8400 0 0

=0 0

=0 5.973 2240 0 -5.973 2240

0 4EI/L 0 4EI/L 0 2240 1E+06 0 -2240 560000

0 0 0 0 -8400 0 0 8400 0 0

0 0 0 -5.9733 -2240 0 5.973 -2240

0 4EI/L 0 4EI/L 0 2240 560000 0 -2240 1E+06

0 0 0 0 0 -1 0 0 0 0

0 0 0 0 1 0 0 0 0 0

=0 0 1 0 0 0

= 0 0 1 0 0 0

0 0 0 0 0 0 0 0 -1 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 1 0 0 0 0 0 1

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

1 2 3 4

1 2 3 4 5 6

0 6 2240 0 -6 2240 0 -1 0 0 0 0 1

-8400 0 0 8400 0 0 1 0 0 0 0 0 2

=0 2240 ### 0 -2240 560000

x0 0 1 0 0 0 3

0 -6 -2240 0 6 -2240 0 0 0 0 -1 0 4

8400 0 0 -8400 0 0 0 0 0 1 0 0 5

0 2240 560000 0 -2240 1120000 0 0 0 0 0 1 6

6 0 2240 -6 0 2240 1

0 8400 0 0 -8400 0 2

=2240 0 ### -2240 0 560000 3

-6 0 -2240 6 0 -2240 4

0 -8400 0 0 8400 0 5

2240 0 560000 -2240 0 1120000 61 2 3 4 5 6

u21 v2

1 q21 u1

1 v11 q1

1 u21 v2

1 q21 u1

1 v11 q1

1

AE/L1 -AE/L1 u21 u2

1

[kl(1)]

12EI/L3 6EI/L2 -12EI/L3 6EI/L2 v21 v2

1

6EI/L2 -6EI/L2 q21 q2

1

-AE/L1 AE/L1 u11 u1

1

-12EI/L3 -6EI/L2 12EI/L3 -6EI/L2 v11 v1

1

6EI/L2 -6EI/L2 q11 q1

1

Matriks Transforamsi Elemen 1 : [T(1)] , dan Transposenya : [T(1)]T

u21 v2

1 q21 u1

1 v11 q1

1 u21 v2

1 q21 u1

1 v11 q1

1

cos a sin a u21 u2

1

-sin a cos a v21 v2

1

[T(1)]q2

1 q21

cos a sin a u11 u1

1

-sin a cos a v11 v1

1

q11 q1

1

u21 v2

1 q21 u1

1 v11 q1

1

u21

v21

[T(1)]T = q2

1

u11

v11

q11

Matriks Kekakuan Dalam Koordinat Global : [kg(2)]

[kg(1)] = [T(1)]T [kl

(1)] [T(1)]

[kg(1)]

[T(2)]T [kl(2)] [T(2)]

U21 V2

1 q21 U1

1 V11 q1

1

U21

V21

[kg(1)]

q21

U11

V11

q11

3


Y y = Y

y Koordinat lokal = koordinat global

x X x = X 2I 2I

0 0 0 0 10500 0 0 -10500 0 0

=0 0

=0 23.333 7000 0 -23.333 7000

0 4EI/L 0 2EI/L 0 7000 3E+06 0 -7000 1E+06

0 0 0 0 -10500 0 0 10500 0 0

0 0 0 -23.333 -7000 0 23.3333 -7000

0 2EI/L 0 4EI/L 0 7000 1E+06 0 -7000 3E+06


Y Y

X X b b I

I

Koordinat lokalKoordinat global

0 0 0 0 12600 0 0 -12600 0 0

=0 0

=0 20.16 5040 0 -20.16 5040

0 4EI/L 0 2EI/L 0 5040 2E+06 0 -5040 840000

0 0 0 0 -12600 0 0 12600 0 0

0 0 0 -20.16 -5040 0 20.16 -5040

L2 =

V2 V3

U2 U3

Matriks Kekakuan Dalam Koordinat Lokal = Global : [kl(2)] = [kg

(2)]

u22 v2

2 q22 u3

2 v32 q3

2 U22 V2

2 q22 U3

2 V32 q3

2

AE/L2 -AE/L2 u22 U2

2

[kg(2)]

12EI/L3 6EI/L2 -12EI/L3 6EI/L2 v22 V2

2

6EI/L2 -6EI/L2 q22 q2

2

-AE/L2 AE/L2 u32 U3

2

-12EI/L3 -6EI/L2 12EI/L3 -6EI/L2 v32 V3

2

6EI/L2 -6EI/L2 q32 q3

2

L3 =

V3 V4

U3 U4

b = -90o

cos b = 0sin b = -1

Matriks Kekakuan Dalam Koordinat Lokal : [kl(3)]

u33 v3

3 q33 u4

3 v43 q4

3 u33 v3

3 q33 u4

3 v43 q4

3

AE/L3 -AE/L3 u33 u3

3

[kl(3)]

12EI/L3 6EI/L2 -12EI/L3 6EI/L2 v33 v3

3

6EI/L2 -6EI/L2 q33 q3

3

-AE/L3 AE/L3 u43 u4

3

-12EI/L3 -6EI/L2 12EI/L3 -6EI/L2 v43 v4

3

(2) (2)2 3

u2

v2

3

v3

2u3

y = Y

x = X

(3)3

4

(3)

u3

v3 y

4

y

x

x

u4

v4

X

Y

3

q4

q2 q3 q2 q3

q3 q3

q4

4

0 2EI/L 0 4EI/L 0 5040 840000 0 -5040 2E+06

0 0 0 0 0 -1 0 0 0 0

0 0 0 0 1 0 0 0 0 0

=0 0 1 0 0 0

= 0 0 1 0 0 0

0 0 0 0 0 0 0 0 -1 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 1 0 0 0 0 0 1

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

1 2 3 4

1 2 3 4 5 6

0 20 5040 0 -20 5040 0 -1 0 0 0 0 1

-12600 0 0 12600 0 0 1 0 0 0 0 0 2

=0 5040 ### 0 -5040 840000

x0 0 1 0 0 0 3

0 -20 -5040 0 20 -5040 0 0 0 0 -1 0 4

12600 0 0 -12600 0 0 0 0 0 1 0 0 5

0 5040 840000 0 -5040 1680000 0 0 0 0 0 1 6

20 0 5040 -20 0 5040 1

0 12600 0 0 -12600 0 2

=5040 0 ### -5040 0 840000 3

-20 0 -5040 20 0 -5040 4

0 -12600 0 0 12600 0 5

5040 0 840000 -5040 0 1680000 6

1 2 3 4 5 6

6EI/L2 -6EI/L2 q43 q4

3

Matriks Transforamsi Elemen 1 : [T(1)] , dan Transposenya : [T(1)]T

u33 v3

3 q33 u4

3 v43 q4

3 u33 v3

3 q33 u4

3 v43 q4

3

cos b sin b u33 u3

3

-sin b cos b v33 v3

3

[T(3)]q3

3 q33

cos b sin b u43 u4

3

-sin b cos b v43 v4

3

q43 q4

3

u33 v3

3 q33 u4

3 v43 q4

3

u33

v33

[T(3)]T = q3

3

u43

v43

q43

Matriks Kekakuan Dalam Koordinat Global : [kg(3)]

[kg(3)] = [T(3)]T [kl

(3)] [T(3)]

[kg(3)]

[T(3)]T [kl(3)] [T(3)]

U33 V3

3 q33 U4

3 V43 q4

3

U33

V33

[kg(3)]

q33

U43

V43

q43

5

B. MATRIKS KEKAKUAN STRUKTUR

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

6 0 -2240 -6 0 -2240 0 0 0 0 0 0

0 8400 0 0 -8400 0 0 0 0 0 0 0

-2240 0 1120000 2240 0 560000 0 0 0 0 0 0

-6 0 2240 10506 0 2240 -10500 0 0 0 0 0

0 -8400 0 0 8423 7000 0 -23 7000 0 0 0

-2240 0 560000 2240 7000 3920000 0 -7000 1400000 0 0 0

0 0 0 -10500 0 0 10520 0 5040 -20 0 5040

0 0 0 0 -23 -7000 0 12623 -7000 0 -12600 0

0 0 0 0 7000 1400000 5040 -7000 4480000 -5040 0 840000

0 0 0 0 0 0 -20 0 -5040 20 0 -5040

0 0 0 0 0 0 0 -12600 0 0 12600 0

0 0 0 0 0 0 5040 0 840000 -5040 0 1680000

Overall Stiffness Matrix ==> Matriks Kekakuan Akhir : [K] = [kg(1)] + [kg

(2)] + [kg(3)]

U1 V1 q1 U2 V2 q2 U3 V3 q3 U4 V4 q4

U1U11 V1U1

1 q1U11 U2U1

1 V2U11 q2U1

1 U1

U1V11 V1V1

1 q1V11 U2V1

1 V2V11 q2V1

1 V1

U1q11 V1q1

1 q1q11 U2q1

1 V2q11 q2V1

1 q1

[K] =U1U2

1 V1U21 q1U2

1 U2U21 + U2U2

2 V2U21 + V2U2

2 q2U21 + q2U2

2 U3U22 V3U2

2 q3U22 U2

U1V21 V1V2

1 q1V21 U2V2

1 + U2V22 V2V2

1 + V2V22 q2V2

1 + q2V22 U3V2

2 V3V22 q3V2

2 V2

U1q21 V1q2

1 q1q21 U2q2

1 + U2q22 V2q2

1 + V2q22 q2q2

1 + q2q22 U3q2

2 V3q22 q3q2

2 q2

U2U32 V2U3

2 q2U32 U3U3

2 + U3U33 V3U3

2 + V3U33 q3U3

2 + q3U33 U4U3

3 V4U33q4U3

3 U3

U2V32 V2V3

2 q2V32 U3V3

2 + U3V33 V3V3

2 + V3V33 q3V3

2 + q3V33 U4U3

3 V4V31q4V3

3 V3

U2q32 V2q3

2 q2q32 U3q3

2 + U3q33 V3q3

2 + V3q33 q3q3

2 + q3q33 U4q3

3 V4q31q4q3

3 q3

U3U43 V3U4

3 q3U44 U4U4

3 V4U43q4U4

3 U4

U3V43 V3V4

3 q3V44 U4V4

3 V4V41q4V4

3 V4

U3q43 V3q4

3 q3q43 U4q4

3 V4q41q4q4

3 q4

U1 V1 q1 U2 V2 q2 U3 V3 q3 U4 V4 q4

U1

V1

q1

U2

V2

q2

[K] =U3

V3

q3

U4

V4

q4

6

{F} = [K] {U} :

6 0 -2240 -6 0 -2240 0 0 0 0 0 0

0 8400 0 0 -8400 0 0 0 0 0 0 0

-2240 0 ### 2240 0 560000 0 0 0 0 0 0

-6 0 2240 10506 0 2240 -10500 0 0 0 0 0

0 -8400 0 0 8423 7000 0 -23 7000 0 0 0

=-2240 0 560000 2240 7000 3920000 0 -7000 1400000 0 0 0

0 0 0 -10500 0 0 10520 0 5040 -20 0 5040

0 0 0 0 -23 -7000 0 12623 -7000 0 -12600 0

0 0 0 0 7000 1400000 5040 -7000 4480000 -5040 0 840000

0 0 0 0 0 0 -20 0 -5040 20 0 -5040

0 0 0 0 0 0 0 -12600 0 0 12600 0

0 0 0 0 0 0 5040 0 840000 -5040 0 ###

{F} [K] {U}

C. MATRIKS BEBAN {F}

elemen batang, maka harus ditransformasikan ke titik nodalPada contoh di atas : 48 kN 48 kN

96 kN2 3

2I 31.104 kN

3 m 48 kN I 5 m

ekivalen ke titik nodal, I menjadi :

F1 U1

G1 V1

M1 q1

F2 U2

G2 V2

M2 q2

F3 U3

G3 V3

M3 q3

F4 U4

G4 V4

M4 q4

Pada metode matriks digunaan konsep beban titik nodal ekivalen (equivalent joint load), dimana bila ada beban pada

(2)

(3)

(1)

(2)

(1)(3)

5184 kN.cm

7200 kN.cm 7200 kN.cm

7

4 4,5 m

16.896 kN3 m 3 m

dimana : {F}* --> adalah matriks beban akibat gaya dalam --> Beban nodal ekivalent{F} --> adalah matriks beban akibat gaya luar

16.9 -F1 6 0 -2240 -6 0 -2240 0 0 0 0 0 0

0 0 8400 0 0 -8400 0 0 0 0 0 0 0

0 0 -2240 0 1120000 2240 0 560000 0 0 0 0 0 0

31.1 0 -6 0 2240 10506 0 2240 -10500 0 0 0 0 0

-48 0 0 -8400 0 0 8423 7000 0 -23 7000 0 0 0

-2016 0=

-2240 0 560000 2240 7000 ### 0 -7000 ### 0 0 0

0 + 0 0 0 0 -10500 0 0 10520 0 5040 -20 0 5040

-48 0 0 0 0 0 -23 -7000 0 12623 -7000 0 -12600 0

7200 0 0 0 0 0 7000 ### 5040 -7000 ### -5040 0 840000

0 0 0 0 0 0 0 -20 0 -5040 20 0 -5040

0 0 0 0 0 0 0 0 -12600 0 0 12600 0

0 0 0 0 0 0 0 0 5040 0 840000 -5040 0 ###

{F}* {F} [K] {U}

6 0 -2240 -6 0 -2240 0 0 0 0 0 0 0

0 8400 0 0 -8400 0 0 0 0 0 0 0 0

0 -2240 0 1120000 2240 0 560000 0 0 0 0 0 0

31.104 -6 0 2240 10506 0 2240 -10500 0 0 0 0 0

-48 0 -8400 0 0 8423 7000 0 -23 7000 0 0 0

-2,016=

-2240 0 560000 2240 7000 ### 0 -7000 ### 0 0 0

0 0 0 0 -10500 0 0 10520 0 5040 -20 0 5040

-48 0 0 0 0 -23 -7000 0 12623 -7000 0 -12600 0

7200 0 0 0 0 7000 ### 5040 -7000 ### -5040 0 840000

0 0 0 0 0 0 -20 0 -5040 20 0 -5040 0

Kondisi batas gaya dan displacement, yaitu :

Titik nodal 1 (tumpuan sendi) : F1* = 16.896 kN 0 ; G1 ≠ 0 ; dan M1 = 0 ; U1 = 0 ; V1 = 0 ; dan q1 ≠ 0

Titik nodal 2 (bebas) : F2* = 31.104 kN ; G2 = -48 kN ; dan M2* =(-7200-5184)=-2016 kN.cm ; U2 ≠ 0; V2 ≠ 0 ; dan q2 ≠ 0

Titik nodal 3 (bebas) : F3 = 0 ; G3* = -48 kN ; dan M3* = 7200 kN.cm ; U2 ≠ 0 ; V2 ≠ 0 ; dan q2 ≠ 0

Titik nodal 4 (tumpuan sendi) : F4 ≠ 0 ; G4 ≠ 0 ; dan M4 = 0 ; U4 = 0 ; V4 = 0 ; dan q4 ≠ 0

U1

G1 V1

q1

U2

V2

q2

U3

V3

q3

F4 U4

G4 V4

q4

U1 V1 q1 U2 V2 q2 U3 V3 q3 U4 V4 q4

F1 - 16.896

G1

q1

U2

V2

q2

U3

V3

q3

F4

1

(1)

8

0 0 0 0 0 0 0 -12600 0 0 12600 0 0

0 0 0 0 0 0 0 5040 0 840000 -5040 0 ###{U}

0 ### 2240 0 560000 0 0 0 0 -2240 0 0 0

31.104 2240 10506 0 2240 -10500 0 0 0 -6 0 0 0

-48 0 0 8423 7000 0 -23 7000 0 0 -8400 0 0

-2,016 560000 2240 7000 3920000 0 -7000 1400000 0 -2240 0 0 0

0 0 -10500 0 0 10520 0 5040 5040 0 0 -20 0

-48=

0 0 -23 -7000 0 12623 -7000 0 0 0 0 -12600

7200 0 0 7000 1400000 5040 -7000 4480000 840000 0 0 -5040 0

0 0 0 0 0 5040 0 840000 ### 0 0 -5040 0

-2240 -6 0 -2240 0 0 0 0 6 0 0 0 0

0 0 -8400 0 0 0 0 0 0 8400 0 0 0

0 0 0 0 -20 0 -5040 -5040 0 0 20 0 0

0 0 0 0 0 -12600 0 0 0 0 0 12600 0

{U}

= x

0 ### 2240 0 560000 0 0 0 0

31.104 2240 10506 0 2240 -10500 0 0 0

-48=

0 0 8423 7000 0 -23 7000 0

-2016 560000 2240 7000 3920000 0 -7000 ### 0

0 0 -10500 0 0 10520 0 5040 5040

-48 0 0 -23 -7000 0 12623 -7000 0

7200 0 0 7000 1400000 5040 -7000 ### 840000

0 0 0 0 0 5040 0 840000 1680000

1 2 3 4 5 6 1

0.00 0.00 0.00 0.00 0.00 0.00 3E-07 1E-06 0 -0.010

0.00 0.20 0.00 0.00 0.20 0.00 0 -0.0005 31.104 5.458

=0.00 0.00 0.00 0.00 0.00 0.00 -2E-07 -2E-07 -48 = -0.003

0.00 0.00 0.00 0.00 0.00 0.00 -1E-07 1E-07 x -2016 -0.002

0.00 0.20 0.00 0.00 0.20 0.00 0 -0.0005 0 5.455

G4

q4

D. Matriks Kekakuan Struktur Disusun Kembali (Rearrangement)

q1 U2 V2 q2 U3 V3 q3 q4 U1 V1 U4 V4

q1

U2

V2

q2

U3

V3

q3

q4

F1 - 16.896

G1

F4

G4

Fe K11 K12 Uu

Fr K21 K22 Uk

E. HITUNG DISPLACEMENTS DAN REAKSI, {U}

q1 U2 V2 q2 U3 V3 q3 q4

q1

U2

V2

q2

U3

V3

q3

q4

{Fe} [K11] {Uu}

Untuk memperoleh displacement {Uu} dengan operasi sbb : {Uu} = [K]-1 {F}

q1

U2

V2

q2

U3

9

0.00 0.00 0.00 0.00 0.00 0.00 1E-07 2E-07 -48 -0.005

0.00 0.00 0.00 0.00 0.00 0.00 4E-07 2E-07 7200 -0.001

0.00 0.00 0.00 0.00 0.00 0.00 2E-07 2E-06 0 -0.016

F. Reaksi Perletakan Diperoleh :

1 1-2240 -6 0 -2240 0 0 0 0 -0.010 -6.064 kN

0 0 -8400 0 0 0 0 0 5.458 28.193 kN= 0 0 0 0 -20 0 -5040 -5040 x -0.003 = -25.040 kN

0 0 0 0 0 -12600 0 0 -0.002 67.807 kN5.455

### kN -0.005

= 28.193 kN -0.001

### kN -0.016

67.807 kN

Check perletakan :

Jadi : 0 kN……….. Ok !

0 kN……….. Ok !

G. MENCARI GAYA-GAYA BATANG TIAP ELEMEN

Vektor gaya-gaya dalam tiap elemen batang harus diintrepretasikan dalam koordinat lokal

Elemen 1 (nodal i = 2 dan j = 1)1 2 3 4 1

1 2 3 4 5 6 1

0 -8400 0 0 8400 0 1 5.458 28.193 kN

6 0 2240 -6 0 2240 2 -0.003 6.064 kN

=2240 0 1120000 -2240 0 560000 3 -0.002

=4548.02 kN.cm

0 8400 0 0 -8400 0 4 0 -28.193 kN

-6 0 -2240 6 0 -2240 5 0 -6.064 kN

2240 0 560000 -2240 0 1120000 6 -0.010 0 kN.cm

Elemen 2 (nodal i = 2 dan j = 3)1 2 3 4 1

10500 0 0 -10500 0 0 1 5.458 25.040 kN

0 23 7000 0 -23 7000 2 -0.003 -19.807 kN

V3

q3

q4

--> {Fr} = [K21] {Uk}

F1 - 16.896

G1

F4

G4

F1

G1

F4

G4

F1 + F4 + 48 kN =

G1 + G4 - 96 kN =

M1 = 0

M4 = 0

{f}* + {f} = [T(1)] [kg(1)] {U(1)] ==> {f} = [T(1)] [kg

(1)] {U(1)] - {f}*

f2

g2

m2[T(1)] [kg

(1)]{U(1)} = f1

g1

m1

f2

g2

10

=0 7000 2800000 0 -7000 1400000 3 -0.002

=### kN.cm

-10500 0 0 10500 0 0 4 5.455 -25.040 kN

0 -23 -7000 0 23 -7000 5 -0.005 19.807 kN

0 7000 1400000 0 -7000 2800000 6 -0.001 ### kN.cm

Elemen 3 (nodal i = 3 dan j = 4)1 2 3 4 5 6 1

0 -12600 0 0 12600 0 1 5.455 67.807 kN

20 0 5040 -20 0 5040 2 -0.005 25.040 kN

=5040 0 1680000 -5040 0 840000 3 -0.001

=### kN.cm

0 12600 0 0 -12600 0 4 0 -67.807 kN

-20 0 -5040 20 0 -5040 5 0 -25.040 kN

5040 0 840000 -5040 0 1680000 6 -0.016 0 kN.cm

HASIL ALHIR GAYA-GAYA DALAM TIAP ELEMEN Elemen 1 (nodal i = 1 dan j = 2) H. Free Body Diagram :

-28.193 16.896 -45.089 kN

-6.064 0 -6.064 kN

=0

-0

=0 kN.cm

28.193 31.104 -2.911 kN

6.064 0 6.064 kN 28.193 kN 67.807 kN

4548.02 5184 -635.98 kN.cm 2.911 kN 67.807 kN

Elemen 2 (nodal i = 2 dan j = 3)6.064 kN 25.04

25.040 31.104 -6.064 kN 500 cm

-19.807 -48.000 28.193 kN

=-6564.02

--7200

=635.98 kN.cm 750 cm

-25.040 0 -25.040 kN 67.807 kN

19.807 -48.000 67.807 kN

-5319.99 7200 ### kN.cm 45.089 kN 25.040 kN

Elemen 3 (nodal i = 3 dan j = 4) 6.064 kN

67.807 0 67.807 kN

25.040 0 25.040 kN 600 cm

m2[T(2)] [kg

(2)]{U(2)} = f3

g3

m3

f3

g3

m3[T(3)] [kg

(3)]{U(3)} = f4

g4

m4

f1

g1

m1

f2

g2

m2

f2

g2

m2

f3

g3

m3

f3

g3

I(3)

2 3

4

1

2 3

(2)

(1)

635.98 kN.cm

635.98 kN.cm 12519.99 kN.cm

12519.99 kN.cm

I

2I

11

=12519.99

-0

=12519.99 kN.cm

-67.807 0 -67.807 kN

-25.040 0 -25.040 kN

0 0 0 kN.cm

m3

f4

g4

m4

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Home Work ASMM Frame 2D