Menghitung Momen Menggunakan Metode Takabeya
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Transcript of Menghitung Momen Menggunakan Metode Takabeya
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Menghitung Momen, Gaya Lintang, dan Gaya Normal pada Portal dengan
Menggunakan Metode Takabeya
Ukuran Balok (30/50), Kolom (40/40)
Langkah Penyelesaian:
1. Menentukan momen primer
a. MFEF = MF
JK = -12
1q l2 = -
12
1(3.679)(52) = -7.664 TM
b. MFFE = MF
KJ =12
1q l2 =
12
1(3.679)(52) = 7.664 TM
c. MFFG = MF
KL = -12
1q l2 = -
12
1(3.616)(4.852) = -7.088 TM
d.
M
F
FG = M
F
KL = 12
1
q l
2
= 12
1
(3.616)(4.85
2
) = 7.088 TM
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e. MFOP = -
12
1q l2 = -
12
1(1.787)( 52) = -3.723 TM
f. MFPO =
12
1q l2 =
12
1(1.787)( 52) = 3.723 TM
g. MF
PQ = -12
1q l
2
= -12
1(1.757)(4.85
2
) = -3.443 TM
h. MFQP =
12
1q l2 =
12
1(1.757)( 4.852) = 3.443 TM
2. Menentukan jumlah momen primer di titik kumpul
a. E = J = MFEF = MF
JK = -7.664 TM
b. F = K = MFFE + MF
FG = MFKJ + MF
KL = 7.664 + (-7.088) = 0.576 TM
c. G = L = MFGF = MF
LK = 7.088 TM
d. O = MFOP = -3.723 TM
e. P = MFPO + MF
PQ = 3.723 + (-3.443) = 0.280 TM
f. Q = MFQP = 3.443 TM
3. Menentukan kekakuan balok dan kolom
IB =12
1(b)(h3) =
12
1(30)(503) = 312500 cm4
IC =12
1 (b)(h3) =12
1 (40)(403) = 213333 cm4
K = 1000 cm3
a. Kekakuan balok bentang 500 cm
K b’ = IB/500 = 625 cm3 = 0.625
b. Kekakuan balok bentang 485 cm
K b’’ = IB/485 = 644.33 cm3 = 0.644
c. Kekakuan balok bentang 100 cm
K b’’’ = IB/100 = 3125 cm3 = 3.125
d. Kekakuan balok bentang 115 cm
K b’’’’ = IB/115 = 2717.4 cm3 = 2.717
e. Kekakuan kolom tinggi 400 cm
K c’ = IC/400 = 533.33 cm3 = 0.533
f. Kekakuan kolom tinggi 380 cm
K c’’ = IC/380 = 561.4 cm3 = 0.561
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4. Menentukan nilai , , dan m(0)
a. (jumlah kekakuan pada masing-masing titik kumpul)
1) E = 2
x
(0.625 + 3.125 + 0.533 + 0.533) = 9.633 2) F = 2 x (0.644 + 0.625 + 0.533 + 0.533) = 4.672
3) G = 2 x (2.717 + 0.644 + 0.533 + 0.533) = 8.857
4) J = 2 x (0.625 + 3.125 + 0.561 + 0.533) = 9.689
5) K = 2 x (0.644 + 0.625 + 0.561 + 0.533) = 4.728
6) L = 2 x (2.717 + 0.644 + 0.561 + 0.533) = 8.913
7) O = 2 x (0.625 + 3.125 + 0.561) = 8.623
8) P = 2 x (0.644 + 0.625 + 0.561) = 3.661
9) Q = 2 x (2.717 + 0.644 + 0.561) = 7.846
b. Menentukan nilai
1) Titik E
EF =E
EFk
= (0.625/9.633) = 0.065
EJ =E
EJk
= (0.533/9.633) = 0.055
2) Titik F
FE =F
FEk
= (0.625/4.672) = 0.134
FG =F
FGk
= (0.644/4.672) = 0.138
FK =F
FK k
= (0.533/4.672) = 0.114
3) Titik G
GF =G
GFk
= (0.644/8.857) = 0.073
GL =G
GLk
= (0.533/8.857) = 0.060
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4) Titik J
JE =J
JEk
= (0.533/9.689) = 0.055
JK =J
JK k
= (0.625/9.689) = 0.065
JO =J
JOk
= (0.561/9.689) = 0.058
5) Titik K
KF =K
KFk
= (0.533/4.728) = 0.113
KJ =K
KJk
= (0.625/4.728) = 0.132
KL = K
KLk
= (0.644/4.728) = 0.136
KP =K
KPk
= (0.561/4.728) = 0.119
6) Titik L
LG =L
LGk
= (0.533/8.913) = 0.060
LK =L
LK k
= (0.644/8.913) = 0.072
LQ =L
LQk
= (0.561/8.913) = 0.063
7) Titik O
OJ =O
OJk
= (0.561/8.623) = 0.065
OP =O
OPk
= (0.625/8.623) = 0.072
8) Titik P
PK =
P
PK k
= (0.561/3.661) = 0.153
PO =P
POk
= (0.625/3.661) = 0.171
PQ =P
PQk
= (0.644/3.661) = 0.176
9) Titik Q
QL =Q
QLk
= (0.561/7.846) = 0.072
QP = Q
QPk
= (0.644/7.846) = 0.082
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c. Menentukan nilai m(0)
1)
mE
(0)
= E
E
= 633.9
664.7
= 0.796
2) mF(0) =
F
F
=
672.4
576.0 = -0.123
3) mG(0) =
G
G
=
857.8
088.7 = -0.800
4) mJ(0) =
J
J
=
689.9
664.7 = 0.865
5) mK (0) =
K
K
=
728.4
576.0 = -0.122
6) mL(0) =
L
L
=
913.8
088.7 = -0.795
7) mO(0) =
O
O
=
623.8
723.3 = 0.432
8) mP(0) =
P
P
=
661.3
280.0 = -0.076
9) mQ(0) =
Q
Q
=
846.7
443.3 = -0.439
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5. Pemberesan momen-momen parsil m(0)
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6. Perhitungan Momen Akhir
a. ME
1) MEA = K EA (2 ME + MA) + MF
EA = 0.533 (2(0.758)+0)+0 = 0.808 tm
2) MEF = K EF (2 ME + MF) + MFEF = 0.625 (2(0.758)-0.109)-7.664 = -6.784 tm
3) MEJ = K EJ (2 ME + MJ) + MFEJ = 0.533 (2(0.758)+0.809)+0 = 1.240 tm
4) MED = K ED (2 ME + MD) + MFED = 3.125 (2(0.758)+0)+0 = 4.736 tm
Jumlah = 0 Ok
b. MF
1) MFB = K FB (2 MF + MB) + MF
FB = 0.533 (2(-0.109)+0)+0 = -0.116 tm
2) MFG = K FG (2 MF + MG) + MFFG = 0.644 (2(-0.109)-0.749)-7.088 = -7.710 tm
3) MFE = K FE (2 MF + ME) + MFFE = 0.625 (2(-0.109)+0.758)+7.664 = 8.002 tm
4) MFK = K FK (2 MF + MK ) + MFFK = 0.533 (2(-0.109)-0.073)+0 = -0.175 tm
Jumlah = 0 Ok
c. MG
1) MGC = K GC (2 MG + MC) + MF
GC = 0.533 (2(-0.749)+0)+0 = -0.799 tm
2) MGF = K GF (2 MG + MF) + MFGF = 0.644 (2(-0.749)-0.109)+7.088 = 6.053 tm
3) MGH = K GH (2 MG + MH) + MFGH = 2.717 (2(-0.749)+0)+0 = -4.071tm
4) MGL = K GL (2 MG + ML) + MFGL = 0.533 (2(-0.749)-0.718)+0 = -1.182 tm
Jumlah = 0 Ok
d. MJ
1) MJE = K JE (2 MJ + ME) + MF
JE = 0.533 (2(0.809)+0.758)+0 = 1.267 tm
2) MJK = K JK (2 MJ + MK ) + MFJK = 0.625 (2(0.809)-0.073)-7.664 = 5.054 tm
3) MJO = K JO (2 MJ + MO) + MFJO = 0.561 (2(0.809)+0.383)+0 = -6.723 tm
4) MJI = K JI (2 MJ + MI) + MFJI = 3.125 (2(0.809)+0)+0 = 1.123 tm
Jumlah = 0.721 tm
MJE = 1.267 – (2JE x 0.721) = 1.267 – (2 (0.055) x 0.721) = 1.187 tm
MJK = 5.054 – (2JK x 0.721) = 5.054 – (2 (0.065) x 0.721) = 4.589 tm
MJO = -6.723 – (2JO x 0.721) = -6.723 – (2 (0.058) x 0.721) = -6.816 tm
MJI = 1.123 – (2JI x 0.721) = 1.123 – (2 (0.323) x 0.721) = 1.040 tm
Jumlah = 0 Ok
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e. MK
1) MKF = K KF (2 MK + MF) + MF
KF = 0.533 (2(-0.112)-0.109)+0 = -0.177 tm
2) MKL = K KL (2 MK + ML) + MFKL = 0.644 (2(-0.112)-0.718)-7.088 = 8.029 tm
3) MKJ = K KJ (2 MK + MJ) + MFKJ = 0.625 (2(-0.112)-0.134)+7.664 = -7.695 tm
4) MKP = K KP (2 MK + MP) + MFKP = 0.561 (2(-0.112)-0.057)+0 = -0.158 tm
Jumlah = 0 Ok
f. ML
1) MLG = K LG (2 ML + MG) + MF
LG = 0.533 (2(-0.718)-0.749)+0 = -1.166 tm
2) MLK = K LK (2 ML + MK ) + MFLK = 0.644 (2(-0.718)-0.112)+7.088 = 6.090 tm
3)
MLM = K LM (2 ML + MM) + M
F
LM = 2.717 (2(-0.718)+0)+0 = -3.903tm 4) MLQ = K LQ (2 ML + MQ) + MF
LQ = 0.561 (2(-0.718)-0.383)+0 = -1.021 tm
Jumlah = 0 Ok
g. MO
1) MOJ = K OJ (2 MO + MJ) + MF
OJ = 0.561 (2(0.383)+0.809)+0 = 0.884 tm
2) MOP = K OP (2 MO + MP) + MFOP = 0.625 (2(0.383)-0.057)-3.723 = 2.395 tm
3) MON = K ON (2 MO + M N) + MFON = 3.125 (2(0.383)+0)+0 = -3.280 tm
Jumlah = 0
Ok
h. MP
1) MPK = K PK (2 MP + MK ) + MF
PK = 0.561 (2(-0.057)-0.112)+0 = -0.127 tm
2) MPQ = K PQ (2 MP + MQ) + MFPQ = 0.644 (2(-0.057)-0.383)-3.443 = 3.891 tm
3) MPO = K PO (2 MP + MO) + MFPO = 0.625 (2(-0.057)+0.383)+3.723 = -3.764 tm
Jumlah = 0 Ok
i. MQ
1) MQL = K QL (2 MQ + ML) + MFQL = 0.561 (2(-0.383)-0.718)+0 = -0.833 tm
2) MQP = K QP (2 MQ + MP) + MFQP = 0.644 (2(-0.383)-0.057)+3.443 = 2.913 tm
3) MQR = K QR (2 MQ + MR ) + MFQR = 2.717 (2(-0.383)+0)+0 = -2.080tm
Jumlah = 0 Ok
j. MAE = K AE (2 MA + ME) + MF
AE = 0.533 (2(0)+0.758)+0 = 0.404 tm
k. MBF = K BF (2 MB + MF) + MF
BF = 0.533 (2(0)-0.109)+0 = -0.058 tm
l. MCG = K CG
(2 MC + MG) + MF
CG = 0.533 (2(0)-0.749)+0 = -0.400 tm
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Gambar distribusi momen pada portal
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7. Menentukan perletakan momen maksimum, menghitung momen maksimum, dan
menentukan perletakan momen minimum (M=0)
a. Batang EF
R E => MF = 0
R E (5) – ½ (3.679)(52) – 6.784 + 8.002 = 0
5R E – 45.9875 – 6.784 +8.002 = 0
R E = (44.7695 / 5) = 8.954 T
Kontrol
R = Q
R E + R F = q.l
8.954+ 9.441 = (3.679 x 5)
18.395 = 18.395Ok
1) Posisi momen maksimum
Dari titik E
Mmax = R E (x1) – ½ qx12 - MEF
= 8.954x1 – 1.8395x12 – 6.784
dxdMmax =0
8.954 -3.679x1 = 0 ==> x1 = (8.954/3.679) = 2.434 m
Dari titik F
Mmax = R F (x2) – ½ qx22 – MFE
= 9.441x2 – 1.8395x22 – 8.002
dx
dMmax =0
9.441 -3.679x2 = 0 ==> x2 = (9.441/3.679) = 2.566 m
R F => ME = 0
-R F (5) + ½ (3.679)(52) – 6.784 + 8.002 = 0
-5R F + 45.9875 – 6.784 +8.002 = 0
R F = (47.2055 / 5) = 9.441 T
F
P = 3.85 T P = 4.2 T
6.784 tm 8.002 tm
5.00 m
E
q = 3.679 t/m’
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2) Momen maksimum
Dari titik E
Mmax = 8.954x1 – 1.8395x12 – 6.784
= (8.954)(2.434) - (1.8395)(2.4342) – 6.784
= 21.794 – 10.898 – 6.784 = 4.112 tm
Dari titik F
Mmax = 9.441x2 – 1.8395x22 – 8.002
= (9.441)(2.566) - (1.8395)(2.5662) – 8.002
= 24.226 – 12.112 – 8.002 = 4.112 tm
3) Posisi momen minimum (M=0)
Dari titik E
M(0) => 8.954x1 – 1.8395x12 – 6.784 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)8395.1(2
)784.6)(8395.1(4954.8(8.954) 2
x(a,b) =679.3
501.5(8.954)
xa = 0.94 m
xb = 3.93 m
Dari titik F
M(0) => 9.441x2 – 1.8395x22 – 8.002
x(a,b) =
a2
ac4 b b 2
x(a,b) =)8395.1(2
)002.8)(8395.1(4441.9(9.441) 2
x(a,b) =679.3
500.5(9.441)
xa = 1.07 m
xb = 4.06 m
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b. Batang FG
R F => MG = 0
R F (4.85) – ½ (3.616)(4.852) – 7.710 + 6.053 = 0
4.85R F – 42.529 – 7.710 + 6.053 = 0
R F = (44.186 / 4.85) = 9.111 T
Kontrol
R = Q
R F + R G = q.l
9.111+ 8.427 = (3.616 x 4.85)
17.538= 17.538Ok
1) Posisi momen maksimumDari titik F
Mmax = R F (x2) – ½ qx22 – MFG
= 9.111x2 – 1.808x22 – 7.710
dx
dMmax =0
9.111 -3.616x2 = 0 ==> x2 = (9.111/3.616) = 2.52 m
Dari titik G
Mmax = R G (x1) – ½ qx12 – MGF
= 8.427x1 – 1.808x12 – 6.053
dx
dMmax =0
8.427-3.616x1 = 0 ==> x1 = (8.427/3.616) = 2.33 m
G
P = 4.2 T P = 4 T
7.710 tm 6.053 tm
4.85 m
F
q = 3.616 t/m’
R G => MF = 0
-R G (4.85) + ½ (3.616)(4.852) – 7.710 + 6.053 = 0
-4.85R G +42.529 – 7.710 +6.053 = 0
R G = (40.872 / 4.85) = 8.427 T
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2) Momen maksimum
Dari titik F
Mmax = 9.111x2 – 1.808x22 – 7.710
= (9.111)(2.52) - (1.808)(2.522) – 7.710
= 22.96 – 11.482 – 7.710 = 3.768 tm
Dari titik G
Mmax = 8.427x1 – 1.808x12 – 6.053
= (8.427)(2.33) - (1.808)(2.332) – 6.053
= 19.635 – 9.815 – 6.053 = 3.767 tm
3) Posisi momen minimum (M=0)
Dari titik F
M(0) => 9.111x2 – 1.808x22 – 7.710 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)808.1(2
710.7)(808.1(4111.9(9.111) 2
x(a,b) =616.3
22.5(9.111)
xa = 1.08 m
xb = 3.96 m
Dari titik G
M(0) => 8.427x1 – 1.808x12 – 6.053 = 0
x(a,b) =
a2
ac4 b b 2
x(a,b) =)808.1(2
)053.6)(808.1(4427.8(8.427) 2
x(a,b) =616.3
22.5(8.427)
xa = 3.77 m
xb = 0.89 m
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c. Batang JK
R J => MK = 0
R J (5) – ½ (3.679)(52) – 6.816 + 8.029 = 0
5R J – 45.9875 – 6.816 +8.029 = 0
R J = (44.7745 / 5) = 8.955 T
Kontrol
R = Q
R J + R K = q.l
8.955+ 9.440 = (3.679 x 5)
18.395= 18.395Ok
1) Posisi momen maksimum
Dari titik J
Mmax = R J (x1) – ½ qx12 – MJK
= 8.955x1 – 1.8395x12 – 6.816
dxdMmax =0
8.955 -3.679x1 = 0 ==> x1 = (8.955/3.679) = 2.434 m
Dari titik K
Mmax = R K (x2) – ½ qx22 – MKJ
= 9.440x2 – 1.8395x22 – 8.029
dx
dMmax =0
9.440-3.679x2 = 0 ==> x2 = (9.440/3.679) = 2.566 m
R K => MJ = 0
-R K (5) + ½ (3.679)(52) – 6.816 + 8.029 = 0
-5R K + 45.9875 – 6.816 +8.029 = 0
R K = (47.201 / 5) = 9.440 T
K
P = 3.85 T P = 4.2 T
6.816 tm 8.029 tm
5.00 m
J
q = 3.679 t/m’
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2) Momen maksimum
Dari titik J
Mmax = 8.955x1 – 1.8395x12 – 6.816
= (8.955)(2.434) - (1.8395)(2.4342) – 6.816
= 21.796 – 10.898 – 6.816 = 4.082 tm
Dari titik K
Mmax = 9.440x2 – 1.8395x22 – 8.029
= (9.440)(2.566) - (1.8395)(2.5662) – 8.029
= 24.223 – 12.112 – 8.029 = 4.082 tm
4) Posisi momen minimum (M=0)
Dari titik J
M(0) => 8.955x1 – 1.8395x12 – 6.816 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)8395.1(2
)816.6)(8395.1(4955.8(8.955) 2
x(a,b) =679.3
481.5(8.955)
xa = 0.94 m
xb = 3.92 m
Dari titik K
M(0) => 9.440x2 – 1.8395x22 – 8.029 = 0
x(a,b) =
a2
ac4 b b 2
x(a,b) =)8395.1(2
)029.8)(8395.1(4440.9(9.440) 2
x(a,b) =679.3
481.5(9.440)
xa = 1.08 m
xb = 4.06 m
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d. Batang KL
R K => ML = 0
R K (4.85) – ½ (3.616)(4.852) – 7.695 + 6.090 = 0
4.85R K – 42.529 – 7.695 + 6.090 = 0
R K = (44.134 / 4.85) = 9.100 T
Kontrol
R = Q
R K + R L = q.l
9.100+ 8.438 = (3.616 x 4.85)
17.538= 17.538Ok
1) Posisi momen maksimumDari titik K
Mmax = R K (x2) – ½ qx22 – MKL
= 9.100x2 – 1.808x22 – 7.695
dx
dMmax =0
9.100 -3.616x2 = 0 ==> x2 = (9.100/3.616) = 2.52 m
Dari titik L
Mmax = R L (x1) – ½ qx12 – MLK
= 8.438x1 – 1.808x12 – 6.090
dx
dMmax =0
8.438-3.616x1 = 0 ==> x1 = (8.438/3.616) = 2.33 m
L
P = 4.2 T P = 4 T
7.695 tm 6.090 tm
4.85 m
K
q = 3.616 t/m’
R L => MK = 0
-R L (4.85) + ½ (3.616)(4.852) – 7.710 + 6.053 = 0
-4.85R L +42.529 – 7.695 +6.090 = 0
R L = (40.924 / 4.85) = 8.438 T
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2) Momen maksimum
Dari titik K
Mmax = 9.100x2 – 1.808x22 – 7.695
= (9.100)(2.52) - (1.808)(2.522) – 7.695
= 22.932 – 11.482 – 7.695 = 3.755 tm
Dari titik L
Mmax = 8.438x1 – 1.808x12 – 6.090
= (8.438)(2.33) - (1.808)(2.332) – 6.090
= 19.661 – 9.815 – 6.090 = 3.756 tm
3) Posisi momen minimum (M=0)
Dari titik K
M(0) => 9.100x2 – 1.808x22 – 7.695 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)808.1(2
)695.7)(808.1(4100.9(9.100) 2
x(a,b) =616.3
21.5(9.100)
xa = 1.08 m
xb = 3.96 m
Dari titik L
M(0) => 8.438x1 – 1.808x12 – 6.090 = 0
x(a,b) =
a2
ac4 b b 2
x(a,b) =)808.1(2
)090.6)(808.1(4438.8(8.438) 2
x(a,b) =616.3
21.5(8.438)
xa = 3.77 m
xb = 0.89 m
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e. Batang OP
R O => MP = 0
R O (5) – ½ (1.787)(52) – 3.280 + 3.891 = 0
5R O – 22.3375 – 3.280 +3.891 = 0
R O = (21.7265 / 5) = 4.345 T
Kontrol
R = Q
R O + R P = q.l
4.345+ 4.59 = (1.787 x 5)
8.935= 8.935Ok
1) Posisi momen maksimum
Dari titik O
Mmax = R O (x1) – ½ qx12 – MOP
= 4.345x1 – 0.8935x12 – 3.280
dx
dMmax =0
4.345 – 1.787x1 = 0 ==> x1 = (4.345/1.787) = 2.431 m
Dari titik P
Mmax = R P (x2) – ½ qx22 – MPO
= 4.59x2 – 0.8935x22 – 3.891
dx
dMmax =0
4.59 - 1.787x2 = 0 ==> x2 = (4.59/1.787) = 2.569 m
R P => MO = 0
-R P (5) + ½ (1.787)(52) – 3.280 + 3.891 = 0
-5R P + 22.3375 – 3.280 +3.891 = 0
R P = (22.9485 / 5) = 4.59 T
P
P = 3.4 T P = 3.7 T
3.280 tm 3.891 tm
5.00 m
O
q = 1.787 t/m’
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2) Momen maksimum
Dari titik O
Mmax = 4.345x1 – 0.8935x12 – 3.280
= (4.345)(2.431) - (0.8935)(2.4312) – 3.280
= 10.563 – 5.280 – 3.280 = 2.003 tm
Dari titik P
Mmax = 4.59x2 – 0.8935x22 – 3.891
= (4.59)(2.569) - (0.8935)(2.5692) – 3.891
= 11.792 – 5.897 – 3.891 = 2.004 tm
5) Posisi momen minimum (M=0)
Dari titik O
M(0) => 4.345x1 – 0.8935x12 – 3.280 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)8935.0(2
)280.3)(8935.0(4345.4(4.345) 2
x(a,b) =787.1
675.2(4.345)
xa = 0.93 m
xb = 3.93 m
Dari titik P
M(0) => 4.59x2 – 0.8935x22 – 3.891 = 0
x(a,b) =
a2
ac4 b b 2
x(a,b) =)8935.0(2
)891.3)(8935.0(459.4(4.59) 2
x(a,b) =787.1
676.2(4.59)
xa = 1.07 m
xb = 4.07 m
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f. Batang PQ
R P => MQ = 0
R P (4.85) – ½ (1.757)(4.852) – 3.764 + 2.913 = 0
4.85R P – 20.6645 – 3.764 + 2.913 = 0
R P = (21.5155 / 4.85) = 4.436 T
Kontrol
R = Q
R P + R Q = q.l
4.435+ 4.085 = (1.757 x 4.85)
8.52= 8.52Ok
1) Posisi momen maksimumDari titik P
Mmax = R P (x2) – ½ qx22 – MPQ
= 4.436x2 – 0.8785x22 – 3.764
dx
dMmax =0
4.436 -1.757x2 = 0 ==> x2 = (4.436/1.757) = 2.525 m
Dari titik Q
Mmax = R Q (x1) – ½ qx12 – MQP
= 4.085x1 – 0.8785x12 – 2.913
dx
dMmax =0
4.085-1.757x1 = 0 ==> x1 = (4.085/1.757) = 2.325 m
Q
P = 3.7 T P = 3.5 T
3.764 tm 2.913 tm
4.85 m
P
q = 1.757 t/m’
R Q => MP = 0
-R Q (4.85) + ½ (1.757)(4.852) – 3.764 + 2.913 = 0
-4.85R Q +20.6645 – 3.764 +2.913 = 0
R Q = (19.8135 / 4.85) = 4.085 T
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2) Momen maksimum
Dari titik P
Mmax = 4.436x2 – 0.8785x22 – 3.764
= (4.436)(2.525) - (0.8785)(2.5252) – 3.764
= 11.201 – 5.601 – 3.764 = 1.836 tm
Dari titik Q
Mmax = 4.085x1 – 0.8785x12 – 2.913
= (4.085)(2.325) - (0.8785)(2.3252) – 2.913
= 9.4976 – 4.7488 – 2.913 = 1.836 tm
3) Posisi momen minimum (M=0)
Dari titik P
M(0) => 4.436x2 – 0.8785x22 – 3.764 = 0
x(a,b) =a2
ac4 b b 2
x(a,b) =)8785.0(2
)764.3)(8785.0(4436.4(4.436) 2
x(a,b) =757.1
54.2(4.436)
xa = 1.08 m
xb = 3.97 m
Dari titik Q
M(0) => 4.085x1 – 0.8785x12 – 2.913 = 0
x(a,b) =
a2
ac4 b b 2
x(a,b) =)8785.0(2
)913.2)(8785.0(4085.4(4.085) 2
x(a,b) =757.1
54.2(4.085)
xa = 3.77 m
xb = 0.88 m
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8. Menghitung gaya lintang
a. DED = q . l – P7 = (1.56)(1) – 3.85 = - 2.29 T
b. DEF = R EF – P7 = 8.954 – 3.85 = 5.104 T
c. DFE = R FE – P8 = 9.441 – 4.2 = 5.241 T
d. DFG = R FG – P8 = 9.111 – 4.2 = 4.911 T
e. DGF = R GF – P9 = 8.427 – 4 = 4.427 T
f. DGH = q.l – P9 = (1.794)(1.15) – 4 = - 1.94 T
g. DJI = q.l – P4 = (1.56)(1) – 3.85 = - 2.29 T
h. DJK = R JK – P4 = 8.955 – 3.85 = 5.105 T
i. DKJ = R KJ – P5 = 9.440 – 4.2 = 5.240 T
j. DKL = R KL – P5 = 9.100 – 4.2 = 4.900 T
k. DLK = R LK – P6 = 8.438 – 4 = 4.438 T
l. DLM = q.l – P6 = (1.794)(1.15) – 4 = - 1.94 T
m. DON = q.l – P1 = (0.758)(1) – 3.4 = - 2.64 T
n. DOP = R OP – P1 = 4.345 – 3.4 = 0.945 T
o. DPO = R PO – P2 = 4.59 – 3.7 = 0.890 T
p. DPQ = R PQ – P2 = 4.436 – 3.7 = 0.736 T
q. DQP = R QP – P3 = 4.085 – 3.5 = 0.585 T
r. DQR = q.l – P3=(0.871)(1.15) – 3.5 = - 2.50 T
9. Menghitung gaya normal
a. Balok
1) NEF =4
)MMMM()MMM(M KFFK BFFBJEEJAEEA
=
4
0.177-0.175-0.058116.01.1871.2400.4040.808
= 0.78 ton (tarik)
2) NFG =4
)MMMM()MMM(M- LGGLCGGCKFFK BFFB
=4
166.1182.1400.0799.0177.0175.0058.00.116-
= 1.02 ton (tarik)
3)
NJK = 4
)MM()M(M FK KFEJJE
+ 8.3
)MM()M(M PK KPOJJO
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=4
0.175177.0240.1187.1
8.3
0.127158.0844.0040.1
= 0.94 ton (tarik)
4) NKL = 4
)MM()M(M- GLLGFK KF
8.3
)MM()M(M- QLLQPK KP
=4
182.1166.1175.00.177-
8.3
833.0021.1127.00.158-
= 1.24 ton (tarik)
5) NOP =8.3
)MM()M(M KPPK JOOJ
=
8.3
0.158127.0040.1844.0 = 0.42 T (tarik)
6) NPQ =8.3
)MM()M(M- LQQLKPPK
=8.3
021.1833.0158.00.127- = 0.56 T (tarik)
b. Kolom
1) NEA = R EF + R ED – P7 = 8.954 + 1.560 – 3.85 = 6.664 T (tarik)
2) NFB = R FE + R FG – P8 = 9.441 + 9.111 – 4.2 = 14.352 T (tarik)
3) NGC = R GF + R GH – P9 = 8.427 + 1.794 – 4 = 6.221 T (tarik)
4) NJE = R JI + R JK – P4 = 1.560 + 8.955 – 3.85 = 6.665 T (tarik)
5) NKF = R KJ + R KL – P5 = 9.440 + 9.100 – 4.2 = 14.34 T (tarik)
6) NLG = R LK + R LM – P6 = 8.438 + 1.794 – 4 = 6.232 T (tarik)
7) NOJ = R OP + R ON – P1 = 4.345 + 0.758 – 3.4 = 1.703 T (tarik)
8) NPK = R PO + R PQ – P2 = 4.59 + 4.436 – 3.7 = 5.326 T (tarik)
9) NQL = R QP + R QR – P3 = 4.085 + 0.871 – 3.5 = 1.456 T (tarik)
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10. Gambar bidang momen, gaya lintang, dan gaya normal
1) Gambar bidang momen
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2) Gambar bidang gaya lintang
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3) Gambar bidang gaya normal