anstruk no 1.xls

Tentukan gaya-gaya dan displacement yang terjadi pada struktur berikut!!

Analogi struktur

Analogi struktur

1. Struktur Aksial

500 cm

400 cm

400 cm

400 cm

d1

d2

d3

d4

u1,f1

u2,f2 u2,f2

u3,f3 u3,f3

u4,f4 u4,f4

u5,f5

batang 1

batang 2

batang 3

batang 4

þ Kekakuan ( lokal )

= 2100000 2100000.00 -2100000.00-2100000.00 2100000.00

= 2100000 2100000.00 -2100000.00-2100000.00 2100000.00

= 2100000 1837500.00 -1837500.00-1837500.00 1837500.00

= 2100000 1575000.00 -1575000.00-1575000.00 1575000.00

þ Kekakuan Global

2100000.00 -2100000.00-2100000.00 4200000.00

K = 0 -2100000.000 00 0

² Penataan Ulang

2100000.00 -2100000.00-2100000.00 4200000.00

K = 0 -2100000.000 00 0

baris 1 & kolom 1 tidak dipakai(di ikutkan dalam perhitungan displacement) karena joint 1 merupakan tumpuan jepit(tidak terjadi displacemet) hanya digunakan dalam penghitungan reaksi.

U1 U2

k1

U2 U3

k2

U3 U4

k3

U4 U5

k4

U1 U2

U1 U2

[k1 ]= AEL [ 1 −1

−1 1 ]=E[ AL

−AL

− AL

AL

]

4200000.00 -2100000.00-2100000.00 3937500.00

K = E 0 -1837500.000 0

0.00000047619048 0.000000476190480.00000047619048 0.00000095238095

= E 0.00000047619048 0.000000952380950.00000047619048 0.00000095238095

þ Displacement

0.00000047619048 0.00000047619048D = 0.00000047619048 0.00000095238095

0.00000047619048 0.000000952380950.00000047619048 0.00000095238095

0.00107142857D = 0.00166666667

0.001938775510.00209750567

² Reaksi

AR = -1500 -2100000.00 0

= -3750 kg

þ Besar Gaya-gaya Batang Akhir

= 2100000.00 -2100000.00-2100000.00 2100000.00

U2 U3

U2 U3

K -1

D2

D3

D4

D5

U1 U2

F1 U1

U2

{ D }=[ K ]−1 {F }

{ F }= [K ] { D }

= 2100000.00 -2100000.00-2100000.00 2100000.00

= 1837500.00 -1837500.00-1837500.00 1837500.00

= 1575000.00 -1575000.00-1575000.00 1575000.00

þ Gambar Diagram Normal

U2 U3

F2 U2

U3

U3 U4

F3 U3

U4

U4 U5

F4 U4

U5

500 cm

400 cm

400 cm

400 cm

d1

d2

d3

d4

Gambar Diagram Normal

500 kg

750 kg

1000 kg

1500 kg3750 kg

2250 kg

1250 kg

500 kg

250 kg

Tentukan gaya-gaya dan displacement yang terjadi pada struktur berikut!!

Penampang berbentuk Lingkaran Panjang

500.000

400.000

350.000

300.000

E = 2100000

Penomoran sb. Lokal

Penomoran sb. Lokal

A1 = cm2 l1 =

A2 = cm2 l2 =

A3 = cm2 l3 =

A4 = cm2 l4 =

kg/cm3

u1,f1

u2,f2 u2,f2

u3,f3 u3,f3

u4,f4 u4,f4

u5,f5

batang 1

batang 2

batang 3

batang 4

0 0 0-2100000.00 0 03937500.00 -1837500.00 0.00-1837500.00 3412500.00 -1575000.00

0 -1575000.00 1575000.00

0 0 0-2100000.00 0 03937500.00 -1837500.00 0-1837500.00 3412500.00 -1575000.00

0 -1575000.00 1575000.00


U1

U2

U2

U3

U3

U4

U4

U5

U3 U4 U5

U1

U2

U3

U4

U5

U3 U4 U5

U1

U2

U3

U4

U5

[k1 ]= AEL [ 1 −1

−1 1 ]=E[ AL

−AL

− AL

AL

]

0 0-1837500.00 03412500.00 -1575000.00-1575000.00 1575000.00

0.00000047619048 0.000000476190480.00000095238095 0.00000095238095

0.00000149659864 0.000001496598640.00000149659864 0.00000213151927

0.00000047619048 0.00000047619048 10000.00000095238095 0.00000095238095 7500.00000149659864 0.00000149659864 2500.00000149659864 0.00000213151927 250

0.00107142860 0 0.0016666667

0.00193877550.0020975057

0+

-1500 = -3750 kg0.0010714286 0 2250 kg

U4 U5

U2

U3

U4

U5

U4 U5

U2

U3

U4

U5

0.0010714286 = -1250 kg0.0016666667 1250 kg

0.0016666667 = -500 kg0.0019387755 500 kg

0.0019387755 = -250 kg0.0020975057 250 kg

500 cm

400 cm

400 cm

400 cm

d1

d2

d3

d4

Gambar Diagram Normal

500 kg

750 kg

1000 kg

1500 kg3750 kg

2250 kg

1250 kg

500 kg

250 kg

500 cm

400 cm

400 cm

400 cm

analis beban

1500 0 1500 1

1000 0 1000 2

0 250 250 4

0 250 250 5

1000 2

AFC = 750 3ARC =

1500 1

250 4

250 5

-22502250

KEKAKUAN LOKAL

u1 u2K1 = E 1 -1 u1

-1 1 u2

u2 u3K2 = E 1 -1 u2

-1 1 u3

u3 u4K3 = E 0.875 -0.875 u3

-0.875 0.875 u4

u4 u5K4 = E 0.75 -0.75 u4

-0.75 0.75 u5

u5 u6K5 = E 0.75 -0.75 u5

-0.75 0.75 u6

KEKAKUAN GLOBALu1 u2 u3 u41 -1 0 0-1 2 -1 0

K = E 0 -1 1.875 -0.8750 0 -0.875 1.6250 0 0 -0.750 0 0 0

[k1 ]= AEL [ 1 −1

−1 1 ]=E[ AL

−AL

− AL

AL

]

u1 u2 u3 u41 -1 0 0-1 2 -1 0

K = E 0 -1 1.875 -0.8750 0 -0.875 1.6250 0 0 -0.750 0 0 0


u2 u3 u4 u52 -1 0 0-1 1.875 -0.875 0

K = E 0 -0.875 1.625 -0.750 0 -0.75 1.50 0 0 -0.75

= 1/E #VALUE! #VALUE! #VALUE! #VALUE!#VALUE! #VALUE! #VALUE! #VALUE!#VALUE! #VALUE! #VALUE! #VALUE!#VALUE! #VALUE! #VALUE! #VALUE!#VALUE! #VALUE! #VALUE! #VALUE!

DISPLACEMENT

1 1 1 1 11 2 2 2 2

D = 1 2 3 3 31 2 3 4 41 2 3 4 5

D = 1/E -2250 -0.009375-3500 -0.014583-4000 -0.016667-4500 -0.01875-5000 -0.020833

REAKSI

AR = 1500 -1 0 0 0

K -1

D2

D3

D4

D5

D6

{ D }=[ K ]−1 {F }

{ F }= [K ] { D }

= 3750 kg

BESAR GAYA-GAYA BATANG AKHIR

u1 u2F1 = 1 -1 u1 0 =

-1 1 u2 -2250 =

u2 u3F2 = 1 -1 u2 -2250 =

-1 1 u3 -3500 =

u3 u4F3 = 0.875 -0.875 u3 -3500 =

-0.875 0.875 u4 -4000 =

u4 u5F4 = 0.75 -0.75 u4 -4000 =

-0.75 0.75 u5 -4500 =

u5 u6F5 = 0.75 -0.75 u5 -4500 =

-0.75 0.75 u6 -5000 =

GAMBAR DIAGRAM NORMAL

500 kg

1000 kg

500 kg

500 kg

2000 kg

1000 kg

500 kg

1000 kg

u5 u60 0 u10 0 u20 0 u3

-0.75 0 u41.5 -0.75 u5

-0.75 0.75 u6

u5 u60 0 u10 0 u20 0 u3

-0.75 0 u41.5 -0.75 u5

-0.75 0.75 u6


u60 u20 u30 u4

-0.75 u50.75 u6

#VALUE!#VALUE!#VALUE!#VALUE!#VALUE!

-1000-750

1/E 00

-500

cmcmcmcmcm

0 -2250

-3500-4000-4500-5000

2250-2250

1250-1250

437.5-437.5

375-375

375-375

2000 kg

1000 kg

1000 kg

anstruk no 1.xls

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