Diketahui Struktur portal bidang (plane frame) seperti gambar di bawah :
Data Batang :
L1 = 500 cm
720000 cm4
E = sama kg/cm2
Data Beban :
P = 2000 kg
Penyelesaian :
Penomoran Lokal Penomoran Global
Sistim Koordinat Global
Tabel Perhitungan :
( i ) ( j ) ( k ) L (cm) E (kg/cm2) A (cm2) I (cm4) α (O)
EA/L 12EI/L3 6EI/L2
4EI/L 2EI/L
1 3 1 500 1.5 1600 213333.3 90 4.800 0.031 0.015 2560.000 1280.000
2 1 2 300 1 2400 720000.0 0 8.000 0.320 0.160 9600.000 4800.000
Matriks Kekakuan Batang Lokal [SM] :
Batang 1 :
4.800 0.000 0.000 -4.800 0.000 0.000 7
0.000 0.031 0.015 0.000 -0.031 0.015 8
0.000 0.015 2560.000 0.000 -0.015 1280.000 9
[SM]1 = E -4.800 0.000 0.000 4.800 0.000 0.000 1
0.000 -0.031 -0.015 0.000 0.031 -0.015 2
0.000 0.015 1280.000 0.000 -0.015 2560.000 3
7 8 9 1 2 3
Batang 2 :
8.000 0.000 0.000 -8.000 0.000 0.000 1
0.000 0.320 0.160 0.000 -0.320 0.160 2
0.000 0.160 9600.000 0.000 -0.160 4800.000 3
Analisislah struktur di atas dengan menggunakan metode Kekakuan Langsung (Direct
Stiffness Method)
0 0 1 0 0 0
[RT]2 = 0 0 0 1 0 0
0 0 0 0 1 0
Matriks Rotasi Transformasi Transpose [RT]T :
Batang 1 (α = 90O) :
0 -1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
[RT]T1 = 0 0 0 0 -1 0
0 0 0 1 0 0
0 0 0 0 0 1
Batang 2 (α = 0O) :
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
[RT]T2 = 0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
Matriks Kekakuan Batang Global [SMS] = [RT] T
[SM] [RT]
Batang 1 :
0.031 0.000 -0.015 -0.031 0.000 -0.015 7
0.000 4.800 0.000 0.000 -4.800 0.000 8
-0.015 0.000 2560.000 0.015 0.000 1280.000 9
[SMS]1 = E -0.031 0.000 0.015 0.031 0.000 0.015 1
0.000 -4.800 0.000 0.000 4.800 0.000 2
-0.015 0.000 1280.000 0.015 0.000 2560.000 3
7 8 9 1 2 3
Batang 2 :
8.000 0.000 0.000 -8.000 0.000 0.000 1
0.000 0.320 0.160 0.000 -0.320 0.160 2
0.000 0.160 9600.000 0.000 -0.160 4800.000 3
[SMS]2 = E -8.000 0.000 0.000 8.000 0.000 0.000 4
0.000 -0.320 -0.160 0.000 0.320 -0.160 5
0.000 0.160 4800.000 0.000 -0.160 9600.000 6
1 2 3 4 5 6
Catatan :
Perakitan Matriks Kekakuan Struktur [SJ] :
1 2 3 4
[SJ] = E 5
6 7 8 9
1 2 3 4 5 6 7 8 9
Penataan Ulang Matriks Kekakuan Struktur [SJ] (Rearrangement) :
1 2 3 6
[SJ] = E 4
5 7 8 9
1 2 3 6 4 5 7 8 9
Analisis Beban Joint Ekivalen {AC} = {AJ} + {AE} :
Untuk batang 2 tidak perlu ditransformasikan karena Sumbu Lokal sama dengan Sumbu Global. Hal ini juga berlaku untuk semua batang yang mempunyai Sumbu Lokal sama dengan Sumbu Global.
S
FF
S
FR
S
RF
S
RR
[ ]
+
α
+
α
+
α
−
α
+
+
α
+
α
+
α
−
α
+
=
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
Cos
Sin
Sin
Cos
Cos
Sin
Sin
Cos
Matriks Perpindahan {DF} = [SFF]-1 {AFC}
0.125 0.000 0.000 0.000 2000 249.04 1
0.000 0.195 0.000 0.000 0 0.00 2
{DF} = 1/E 0.000 0.000 0.000 0.000 * 0 = 1/E 0.00 3 (rad - cm)
0.000 0.000 0.000 0.000 0 0.00 6
Matriks Reaksi Tumpuan {AR} = - [ARC} + [SRF] {DF}
0.00 -8.00 0.00 0.00 0.00 249.04
0.00 0.00 -0.32 -0.16 -0.16 0.00
{AR} = - 0.00 + -0.03 0.00 -0.02 0.00 * 0.00
0.00 0.00 -4.80 0.00 0.00 0.00
0.00 0.02 0.00 1280.00 0.00
0.00 -1992.35 -1992.35 4
0.00 0.00 0.00 5
= - 0.00 + -7.65 = -7.65 7 (kg-cm)
0.00 0.00 0.00 8
0.00 -3825.31 -3825.31 9
Kontrol Reaksi : Σ
ΣΣ ΣH = 0
H4 + H7 + P = 0
-1992.35 + -7.65 + 2000.00 = 0.00
-2000.00 + 2000.00 = 0.00
0 = 0 … Ok.
Σ ΣΣ
ΣMterhadap titik 3 = 0 P.500 - H4.500 + M9 = 0
1000000.00 + -996174.69 + -3825.31 = 0.00
1000000.00 + -1000000.00 = 0.00
0 = 0 … Ok.
Gaya Ujung Batang :
{AM}i = {AML}i + [SM]i {DM}i
{AM}i = {AML}i + [SM]i .[RT].{DJ}i
Batang 1 :
0 4.8 0 0 -4.8 0 0 7
0 0 0.03072 0.01536 0 -0.03072 0.01536 8
{AM}1 = 0 + 0 0.01536 2560 0 -0.01536 1280 9
0 -4.8 0 0 4.8 0 0 1
0 0 -0.03072 -0.01536 0 0.03072 -0.01536 2
0 0 0.01536 1280 0 -0.01536 2560 3
0 -1 0 0 0 0 0 7
1 0 0 0 0 0 0 8
x 0 0 1 0 0 0 x 0 9
0 0 0 0 -1 0 37036.19 1
0 0 0 1 0 0 -62332.24 2
0 0 0 0 0 1 -3207.69 3
0 4.8 0 0 -4.8 0 0 7 0
0 0 0.03072 0.01536 0 -0.03072 0.01536 8 0
= 0 + 0 0.01536 2560 0 -0.01536 1280 9 x 0
0 -4.8 0 0 4.8 0 0 1 62332.2427
0 0 -0.03072 -0.01536 0 0.03072 -0.01536 2 37036.1937
0 0 0.01536 1280 0 -0.01536 2560 3 -3207.6917
299194.76 7 -1088.48 8
= -4105276.47 9 -299194.76 1 1088.48 2 -8211121.82 3
7
8 9 1 2
3
1
Batang 2 :
0 6 0 0 -6 0 0 1
300000 0 0.135 0.0675 0 -0.135 0.0675 2
{AM}2 = 20000000 + 0 0.0675 7200 0 -0.0675 3600 3
0 -6 0 0 6 0 0 4
300000 0 -0.135 -0.0675 0 0.135 -0.0675 5
-20000000 0 0.0675 3600 0 -0.0675 7200 6
1 0 0 0 0 0 37036.1937 1
0 1 0 0 0 0 -62332.243 2
x 0 0 1 0 0 0 x -3207.6917 3
0 0 0 1 0 0 36884.27 4
0 0 0 0 1 0 -56400.98 5
0 0 0 0 0 1 3140.81 6
0 911.5182725 1
300000 -805.235024 2
= 20000000 + -11788878.2 3
0 -911.518273 4
300000 805.2350241 5
-20000000 11065714.14 6
911.52 1 299194.76 2
= 8211121.82 3 -911.52 4 300805.24 5 -8934285.86 6
Batang 3 :
0 5.333333333 0 0 -5.33333333 0 0 10
0 0 0.0237037 0.011851852 0 -0.023704 0.011851852 11
{AM}3 = 0 + 0 0.0118519 2844.444444 0 -0.011852 1422.222222 12
0 -5.33333333 0 0 5.333333333 0 0 4
0 0 -0.023704 -0.01185185 0 0.0237037 -0.01185185 5
0 0 0.0118519 1422.222222 0 -0.011852 2844.444444 6
0 -1 0 0 0 0 0 10
1 0 0 0 0 0 0 11
x 0 0 1 0 0 0 x 0 12
0 0 0 0 -1 0 36884.27 4
0 0 0 1 0 0 -56400.98 5
0 0 0 0 0 1 3140.81 6
0 5.333333333 0 0 -5.33333333 0 0 10 0
0 0 0.0237037 0.011851852 0 -0.023704 0.011851852 11 0
= 0 + 0 0.0118519 2844.444444 0 -0.011852 1422.222222 12 x 0
0 -5.33333333 0 0 5.333333333 0 0 4 56400.9816
0 0 -0.023704 -0.01185185 0 0.0237037 -0.01185185 5 36884.274
0 0 0.0118519 1422.222222 0 -0.011852 2844.444444 6 3140.80619
300805.24 10 -911.52 11
= 3336220.47 12 -300805.24 4
911.52 5 8934285.86 6
Catatan : Untuk Gambar Bidang M, D dan N dapat dicoba sendiri dengan cara Superposisi.
2
1
3
5
4 6
2
1 2
4 5
6
10
11
12 3
