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 1 1 q l2 = - (3.679)(52) = -7.664 TM 12 12 1 1 b. MFFE = MFKJ = q l2 = (3.679)(52) = 7.664 TM 12 12 1 1 c. MFFG = MFKL = q l2 = - (3.616)(4.852) = -7.088 TM 12 12 1 1 d. MFFG = MFKL = q l2 = (3.616)(4.852) = 7.088 TM 12 12
a. MFEF = MFJK = -
1
1 1 q l2 = - (1.787)( 52) = -3.723 TM 12 12 1 1 f. MFPO = q l2 = (1.787)( 52) = 3.723 TM 12 12 1 1 g. MFPQ = q l2 = - (1.757)(4.852) = -3.443 TM 12 12 1 1 h. MFQP = q l2 = (1.757)( 4.852) = 3.443 TM 12 12
e. MFOP = -
2. Menentukan jumlah momen primer di titik kumpul a. E = J = MFEF = MFJK = -7.664 TM b. F = K = MFFE + MFFG = MFKJ + MFKL = 7.664 + (-7.088) = 0.576 TM c. G = L = MFGF = MFLK = 7.088 TM d. O = MFOP = -3.723 TM e. P = MFPO + MFPQ = 3.723 + (-3.443) = 0.280 TM f. Q = MFQP = 3.443 TM
3. Menentukan kekakuan balok dan kolom IB =
1 1 (b)(h3) = (30)(503) = 312500 cm4 12 12
IC =
1 1 (b)(h3) = (40)(403) = 213333 cm4 12 12
K = 1000 cm3 a. Kekakuan balok bentang 500 cm Kb’ = IB/500 = 625 cm3
= 0.625
b. Kekakuan balok bentang 485 cm Kb’’ = IB/485 = 644.33 cm3
= 0.644
c. Kekakuan balok bentang 100 cm Kb’’’ = IB/100 = 3125 cm3
= 3.125
d. Kekakuan balok bentang 115 cm Kb’’’’ = IB/115 = 2717.4 cm3 = 2.717 e. Kekakuan kolom tinggi 400 cm Kc’ = IC/400 = 533.33 cm3
= 0.533
f. Kekakuan kolom tinggi 380 cm Kc’’ = IC/380 = 561.4 cm3
= 0.561
2
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 =
k EF
EJ =
k EJ
= (0.625/9.633)
E E
= (0.533/9.633)
= 0.065 = 0.055
2) Titik F FE = FG FK
k FE
= (0.625/4.672)
F k = FG = (0.644/4.672) F k = FK = (0.533/4.672) F
= 0.134 = 0.138 = 0.114
3) Titik G GF = GL
k GF
= (0.644/8.857)
G k = GL = (0.533/8.857) G
= 0.073 = 0.060
3
4) Titik J JE =
k JE
= (0.533/9.689)
J k JK = JK = (0.625/9.689) J k JO = JO = (0.561/9.689) J
= 0.055 = 0.065 = 0.058
5) Titik K KF =
k KF
KJ =
k KJ
K
= (0.533/4.728)
= 0.113
= (0.625/4.728)
= 0.132
K k KL = KL = (0.644/4.728) K k KP = KP = (0.561/4.728) K
= 0.136 = 0.119
6) Titik L LG =
k LG
= (0.533/8.913)
L k LK = LK = (0.644/8.913) L LQ =
k LQ
L
= 0.060 = 0.072
= (0.561/8.913)
= 0.063
= (0.561/8.623)
= 0.065
7) Titik O OJ =
k OJ
O k OP = OP = (0.625/8.623) O
= 0.072
8) Titik P PK = PO
k PK
= (0.561/3.661)
P k = PO = (0.625/3.661) P
PQ =
k PQ
P
= 0.153 = 0.171
= (0.644/3.661)
= 0.176
= (0.561/7.846)
= 0.072
= (0.644/7.846)
= 0.082
9) Titik Q QL =
k QL
QP =
k QP
Q Q
4
c. Menentukan nilai m(0) 1) mE(0) =
E 7.664 = = 0.796 E 9.633
2) mF(0) =
F 0.576 = F 4.672
= -0.123
3) mG(0) =
G 7.088 = G 8.857
= -0.800
4) mJ(0) =
J 7.664 = = 0.865 J 9.689
5) mK(0) =
K 0.576 = K 4.728
= -0.122
6) mL(0) =
L 7.088 = L 8.913
= -0.795
7) mO(0) =
O 3.723 = = 0.432 O 8.623
8) mP(0) =
P 0.280 = P 3.661
= -0.076
9) mQ(0) =
Q 3.443 = Q 7.846
= -0.439
5
5. Pemberesan momen-momen parsil m(0)
6
6. Perhitungan Momen Akhir a. ME 1) MEA = KEA (2 ME + MA) + MFEA = 0.533 (2(0.758)+0)+0
= 0.808 tm
2) MEF = KEF (2 ME + MF) + MFEF = 0.625 (2(0.758)-0.109)-7.664 = -6.784 tm 3) MEJ = KEJ (2 ME + MJ) + MFEJ = 0.533 (2(0.758)+0.809)+0
= 1.240 tm
4) MED = KED (2 ME + MD) + MFED = 3.125 (2(0.758)+0)+0
= 4.736 tm
Jumlah
= 0 Ok
b. MF 1) MFB = KFB (2 MF + MB) + MFFB = 0.533 (2(-0.109)+0)+0 2) MFG = KFG (2 MF + MG) +
MFFG =
= -0.116 tm
0.644 (2(-0.109)-0.749)-7.088 = -7.710 tm
3) MFE = KFE (2 MF + ME) + MFFE = 0.625 (2(-0.109)+0.758)+7.664 = 8.002 tm 4) MFK = KFK (2 MF + MK) + MFFK = 0.533 (2(-0.109)-0.073)+0 Jumlah
= -0.175 tm = 0 Ok
c. MG 1) MGC = KGC (2 MG + MC) + MFGC = 0.533 (2(-0.749)+0)+0 2) MGF = KGF (2 MG + MF) +
MFGF =
= -0.799 tm
0.644 (2(-0.749)-0.109)+7.088 = 6.053 tm
3) MGH = KGH (2 MG + MH) + MFGH = 2.717 (2(-0.749)+0)+0
= -4.071tm
4) MGL = KGL (2 MG + ML) + MFGL = 0.533 (2(-0.749)-0.718)+0
= -1.182 tm
Jumlah
= 0 Ok
d. MJ 1) MJE = KJE (2 MJ + ME) + MFJE = 0.533 (2(0.809)+0.758)+0 2) MJK = KJK (2 MJ + MK) +
MFJK =
= 1.267 tm
0.625 (2(0.809)-0.073)-7.664 = 5.054 tm
3) MJO = KJO (2 MJ + MO) + MFJO = 0.561 (2(0.809)+0.383)+0
= -6.723 tm
4) MJI = KJI (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
7
e. MK 1) MKF = KKF (2 MK + MF) + MFKF = 0.533 (2(-0.112)-0.109)+0
= -0.177 tm
2) MKL = KKL (2 MK + ML) + MFKL = 0.644 (2(-0.112)-0.718)-7.088 = 8.029 tm 3) MKJ = KKJ (2 MK + MJ) + MFKJ = 0.625 (2(-0.112)-0.134)+7.664 = -7.695 tm 4) MKP = KKP (2 MK + MP) + MFKP = 0.561 (2(-0.112)-0.057)+0 Jumlah
= -0.158 tm = 0 Ok
f. ML 1) MLG = KLG (2 ML + MG) + MFLG = 0.533 (2(-0.718)-0.749)+0
= -1.166 tm
2) MLK = KLK (2 ML + MK) + MFLK = 0.644 (2(-0.718)-0.112)+7.088 = 6.090 tm 3) MLM = KLM (2 ML + MM) + MFLM = 2.717 (2(-0.718)+0)+0
= -3.903tm
4) MLQ = KLQ (2 ML + MQ) + MFLQ = 0.561 (2(-0.718)-0.383)+0
= -1.021 tm = 0 Ok
Jumlah g. MO 1) MOJ = KOJ (2 MO + MJ) + MFOJ = 0.561 (2(0.383)+0.809)+0
= 0.884 tm
2) MOP = KOP (2 MO + MP) + MFOP = 0.625 (2(0.383)-0.057)-3.723 = 2.395 tm 3) MON = KON (2 MO + MN) + MFON = 3.125 (2(0.383)+0)+0
= -3.280 tm = 0 Ok
Jumlah h. MP 1) MPK = KPK (2 MP + MK) + MFPK = 0.561 (2(-0.057)-0.112)+0
= -0.127 tm
2) MPQ = KPQ (2 MP + MQ) + MFPQ = 0.644 (2(-0.057)-0.383)-3.443 = 3.891 tm 3) MPO = KPO (2 MP + MO) + MFPO = 0.625 (2(-0.057)+0.383)+3.723 = -3.764 tm Jumlah
= 0 Ok
i. MQ 1) MQL = KQL (2 MQ + ML) + MFQL = 0.561 (2(-0.383)-0.718)+0
= -0.833 tm
2) MQP = KQP (2 MQ + MP) + MFQP = 0.644 (2(-0.383)-0.057)+3.443 = 2.913 tm 3) MQR = KQR (2 MQ + MR) + MFQR = 2.717 (2(-0.383)+0)+0 Jumlah
= -2.080tm = 0 Ok
j. MAE = KAE (2 MA + ME) + MFAE = 0.533 (2(0)+0.758)+0
= 0.404 tm
k. MBF = KBF (2 MB + MF) + MFBF = 0.533 (2(0)-0.109)+0
= -0.058 tm
l. MCG = KCG (2 MC + MG) + MFCG = 0.533 (2(0)-0.749)+0
= -0.400 tm
8
Gambar distribusi momen pada portal
9
7. Menentukan perletakan momen maksimum, menghitung momen maksimum, dan menentukan perletakan momen minimum (M=0) a. Batang EF P = 3.85 T
P = 4.2 T
6.784 tm
q = 3.679 t/m’
8.002 tm
E
F 5.00 m
RE => MF = 0
RF => ME = 0
RE (5) – ½ (3.679)(52) – 6.784 + 8.002 = 0
-RF (5) + ½ (3.679)(52) – 6.784 + 8.002 = 0
5RE – 45.9875 – 6.784 +8.002 = 0
-5RF + 45.9875 – 6.784 +8.002 = 0
RE = (44.7695 / 5) = 8.954 T
RF = (47.2055 / 5) = 9.441 T
Kontrol R = Q RE + RF = q.l 8.954+ 9.441 = (3.679 x 5) 18.395 = 18.395 Ok
1) Posisi momen maksimum Dari titik E Mmax = RE (x1) – ½ qx12 - MEF = 8.954x1 – 1.8395x12 – 6.784
dM max =0 dx 8.954 -3.679x1 = 0
==> x1 = (8.954/3.679) = 2.434 m
Dari titik F Mmax = RF (x2) – ½ qx22 – MFE = 9.441x2 – 1.8395x22 – 8.002
dM max =0 dx 9.441 -3.679x2 = 0
==> x2 = (9.441/3.679) = 2.566 m 10
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
b b 2 4ac x(a,b) = 2a (8.954) 8.954 2 4(1.8395)(6.784) x(a,b) = 2(1.8395) x(a,b) =
(8.954) 5.501 3.679
xa
= 0.94 m
xb
= 3.93 m
Dari titik F M(0) => 9.441x2 – 1.8395x22 – 8.002
b b 2 4ac x(a,b) = 2a
(9.441) 9.4412 4(1.8395)(8.002) x(a,b) = 2(1.8395) x(a,b) =
(9.441) 5.500 3.679
xa
= 1.07 m
xb
= 4.06 m
11
b. Batang FG P = 4.2 T
P=4T
7.710 tm
6.053 tm
q = 3.616 t/m’
F
G 4.85 m
RF
=> MG = 0
RG
=> MF = 0
RF (4.85) – ½ (3.616)(4.852) –7.710 + 6.053 = 0
-RG (4.85) + ½ (3.616)(4.852) –7.710 + 6.053 = 0
4.85RF – 42.529 – 7.710 + 6.053 = 0
-4.85RG +42.529 – 7.710 +6.053 = 0
RF = (44.186 / 4.85) = 9.111 T
RG = (40.872 / 4.85) = 8.427 T
Kontrol R = Q RF + RG = q.l 9.111+ 8.427 = (3.616 x 4.85) 17.538= 17.538 Ok
1) Posisi momen maksimum Dari titik F Mmax = RF (x2) – ½ qx22 – MFG = 9.111x2 – 1.808x22 – 7.710
dM max =0 dx 9.111 -3.616x2 = 0
==> x2 = (9.111/3.616) = 2.52 m
Dari titik G Mmax = RG (x1) – ½ qx12 – MGF = 8.427x1 – 1.808x12 – 6.053
dM max =0 dx 8.427-3.616x1 = 0
==> x1 = (8.427/3.616) = 2.33 m
12
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
b b 2 4ac x(a,b) = 2a (9.111) 9.1112 4(1.808)(7.710 x(a,b) = 2(1.808) x(a,b) =
(9.111) 5.22 3.616
xa
= 1.08 m
xb
= 3.96 m
Dari titik G M(0) => 8.427x1 – 1.808x12 – 6.053 = 0
b b 2 4ac x(a,b) = 2a
(8.427) 8.427 2 4(1.808)(6.053) x(a,b) = 2(1.808) x(a,b) =
(8.427) 5.22 3.616
xa
= 3.77 m
xb
= 0.89 m
13
c. Batang JK
P = 3.85 T
P = 4.2 T
6.816 tm
q = 3.679 t/m’
8.029 tm
J
K 5.00 m
RJ
RK => MJ = 0
=> MK = 0
RJ (5) – ½ (3.679)(52) – 6.816 + 8.029 = 0
-RK (5) + ½ (3.679)(52) – 6.816 + 8.029 = 0
5RJ – 45.9875 – 6.816 +8.029 = 0
-5RK + 45.9875 – 6.816 +8.029 = 0
RJ = (44.7745 / 5) = 8.955 T
RK = (47.201 / 5) = 9.440 T
Kontrol R = Q RJ + RK = q.l 8.955+ 9.440 = (3.679 x 5) 18.395= 18.395 Ok
1) Posisi momen maksimum Dari titik J Mmax = RJ (x1) – ½ qx12 – MJK = 8.955x1 – 1.8395x12 – 6.816
dM max =0 dx 8.955 -3.679x1 = 0
==> x1 = (8.955/3.679) = 2.434 m
Dari titik K Mmax = RK (x2) – ½ qx22 – MKJ = 9.440x2 – 1.8395x22 – 8.029
dM max =0 dx 9.440-3.679x2 = 0
==> x2 = (9.440/3.679) = 2.566 m 14
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
b b 2 4ac x(a,b) = 2a (8.955) 8.955 2 4(1.8395)(6.816) x(a,b) = 2(1.8395) x(a,b) =
(8.955) 5.481 3.679
xa
= 0.94 m
xb
= 3.92 m
Dari titik K M(0) => 9.440x2 – 1.8395x22 – 8.029 = 0
b b 2 4ac x(a,b) = 2a
(9.440) 9.440 2 4(1.8395)(8.029) x(a,b) = 2(1.8395) x(a,b) =
(9.440) 5.481 3.679
xa
= 1.08 m
xb
= 4.06 m
15
d. Batang KL P = 4.2 T
P=4T
7.695 tm
6.090 tm
q = 3.616 t/m’
K
L 4.85 m
RK
=> ML = 0
RL
=> MK = 0
RK (4.85) – ½ (3.616)(4.852) –7.695 + 6.090 = 0
-RL (4.85) + ½ (3.616)(4.852) –7.710 + 6.053 = 0
4.85RK – 42.529 – 7.695 + 6.090 = 0
-4.85RL +42.529 – 7.695 +6.090 = 0
RK = (44.134 / 4.85) = 9.100 T
RL = (40.924 / 4.85) = 8.438 T
Kontrol R = Q RK + RL = q.l 9.100+ 8.438 = (3.616 x 4.85) 17.538= 17.538 Ok
1) Posisi momen maksimum Dari titik K Mmax = RK (x2) – ½ qx22 – MKL = 9.100x2 – 1.808x22 – 7.695
dM max =0 dx 9.100 -3.616x2 = 0
==> x2 = (9.100/3.616) = 2.52 m
Dari titik L Mmax = RL (x1) – ½ qx12 – MLK = 8.438x1 – 1.808x12 – 6.090
dM max =0 dx 8.438-3.616x1 = 0
==> x1 = (8.438/3.616) = 2.33 m
16
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
b b 2 4ac x(a,b) = 2a (9.100) 9.100 2 4(1.808)(7.695 ) x(a,b) = 2(1.808) x(a,b) =
(9.100) 5.21 3.616
xa
= 1.08 m
xb
= 3.96 m
Dari titik L M(0) => 8.438x1 – 1.808x12 – 6.090 = 0
b b 2 4ac x(a,b) = 2a
(8.438) 8.438 2 4(1.808)(6.090) x(a,b) = 2(1.808) x(a,b) =
(8.438) 5.21 3.616
xa
= 3.77 m
xb
= 0.89 m
17
e. Batang OP P = 3.4 T
P = 3.7 T
3.280 tm
q = 1.787 t/m’
3.891 tm
O
P 5.00 m
RO
RP => MO = 0
=> MP = 0
RO (5) – ½ (1.787)(52) – 3.280 + 3.891 = 0
-RP (5) + ½ (1.787)(52) – 3.280 + 3.891 = 0
5RO – 22.3375 – 3.280 +3.891 = 0
-5RP + 22.3375 – 3.280 +3.891 = 0
RO = (21.7265 / 5) = 4.345 T
RP = (22.9485 / 5) = 4.59 T
Kontrol R = Q RO + RP = q.l 4.345+ 4.59 = (1.787 x 5) 8.935= 8.935 Ok
1) Posisi momen maksimum Dari titik O Mmax = RO (x1) – ½ qx12 – MOP = 4.345x1 – 0.8935x12 – 3.280
dM max =0 dx 4.345 – 1.787x1 = 0
==> x1 = (4.345/1.787) = 2.431 m
Dari titik P Mmax = RP (x2) – ½ qx22 – MPO = 4.59x2 – 0.8935x22 – 3.891
dM max =0 dx 4.59 - 1.787x2 = 0
==> x2 = (4.59/1.787) = 2.569 m
18
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
b b 2 4ac x(a,b) = 2a (4.345) 4.345 2 4(0.8935)(3.280) x(a,b) = 2(0.8935) x(a,b) =
(4.345) 2.675 1.787
xa
= 0.93 m
xb
= 3.93 m
Dari titik P M(0) => 4.59x2 – 0.8935x22 – 3.891 = 0
b b 2 4ac x(a,b) = 2a
(4.59) 4.59 2 4(0.8935)(3.891) x(a,b) = 2(0.8935) x(a,b) =
(4.59) 2.676 1.787
xa
= 1.07 m
xb
= 4.07 m
19
f. Batang PQ P = 3.7 T
P = 3.5 T
3.764 tm
2.913 tm
q = 1.757 t/m’
P
Q 4.85 m
RP
=> MQ = 0
RQ
=> MP = 0
RP (4.85) – ½ (1.757)(4.852) –3.764 + 2.913 = 0
-RQ (4.85) + ½ (1.757)(4.852) –3.764 + 2.913 = 0
4.85RP – 20.6645 – 3.764 + 2.913 = 0
-4.85RQ +20.6645 – 3.764 +2.913 = 0
RP = (21.5155 / 4.85) = 4.436 T
RQ = (19.8135 / 4.85) = 4.085 T
Kontrol R = Q RP + RQ = q.l 4.435+ 4.085 = (1.757 x 4.85) 8.52= 8.52 Ok
1) Posisi momen maksimum Dari titik P Mmax = RP (x2) – ½ qx22 – MPQ = 4.436x2 – 0.8785x22 – 3.764
dM max =0 dx 4.436 -1.757x2 = 0
==> x2 = (4.436/1.757) = 2.525 m
Dari titik Q Mmax = RQ (x1) – ½ qx12 – MQP = 4.085x1 – 0.8785x12 – 2.913
dM max =0 dx 4.085-1.757x1 = 0
==> x1 = (4.085/1.757) = 2.325 m
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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
b b 2 4ac x(a,b) = 2a (4.436) 4.436 2 4(0.8785)(3.764 ) x(a,b) = 2(0.8785) x(a,b) =
(4.436) 2.54 1.757
xa
= 1.08 m
xb
= 3.97 m
Dari titik Q M(0) => 4.085x1 – 0.8785x12 – 2.913 = 0
b b 2 4ac x(a,b) = 2a
(4.085) 4.085 2 4(0.8785)(2.913) x(a,b) = 2(0.8785) x(a,b) =
(4.085) 2.54 1.757
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 = REF – P7 = 8.954 – 3.85
= 5.104 T
c. DFE = RFE – P8 = 9.441 – 4.2
= 5.241 T
d. DFG = RFG – P8 = 9.111 – 4.2
= 4.911 T
e. DGF = RGF – 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 = RJK – P4 = 8.955 – 3.85
= 5.105 T
i. DKJ = RKJ – P5 = 9.440 – 4.2
= 5.240 T
j. DKL = RKL – P5 = 9.100 – 4.2
= 4.900 T
k. DLK = RLK – 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 = ROP – P1 = 4.345 – 3.4
= 0.945 T
o. DPO = RPO – P2 = 4.59 – 3.7
= 0.890 T
p. DPQ = RPQ – P2 = 4.436 – 3.7
= 0.736 T
q. DQP = RQP – 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 = =
(M EA M AE M EJ M JE ) (M FB M BF M FK M KF ) 4 0.808 0.404 1.240 1.187 0.116 0.058 - 0.175 - 0.177 4
= 0.78 ton (tarik) 2)
NFG = =
- (M FB M BF M FK M KF ) (M GC M CG M GL M LG ) 4 - 0.116 0.058 0.175 0.177 0.799 0.400 1.182 1.166 4
= 1.02 ton (tarik) 3)
NJK =
(M JE M EJ ) (M KF M FK ) (M JO M OJ ) (M KP M PK ) + 4 3.8 22
=
1.187 1.240 0.177 0.175 1.040 0.844 0.158 0.127 4 3.8
= 0.94 ton (tarik) 4)
NKL = =
- (M KF M FK ) (M LG M GL ) - (M KP M PK ) (M LQ M QL ) 4 3.8 - 0.177 0.175 1.166 1.182 - 0.158 0.127 1.021 0.833 4 3.8
= 1.24 ton (tarik) 5)
NOP = =
6)
NPQ = =
(M OJ M JO ) (M PK M KP ) 3.8 0.844 1.040 0.127 0.158 = 0.42 T (tarik) 3.8
- (M PK M KP ) (M QL M LQ ) 3.8 - 0.127 0.158 0.833 1.021 = 0.56 T (tarik) 3.8
b. Kolom 1) NEA = REF + RED – P7 = 8.954 + 1.560 – 3.85 = 6.664 T (tarik) 2) NFB = RFE + RFG – P8 = 9.441 + 9.111 – 4.2 = 14.352 T (tarik) 3) NGC = RGF + RGH – P9 = 8.427 + 1.794 – 4 4) NJE = RJI + RJK – P4
= 6.221 T (tarik)
= 1.560 + 8.955 – 3.85 = 6.665 T (tarik)
5) NKF = RKJ + RKL – P5 = 9.440 + 9.100 – 4.2 = 14.34 T (tarik) 6) NLG = RLK + RLM – P6 = 8.438 + 1.794 – 4
= 6.232 T (tarik)
7) NOJ = ROP + RON – P1 = 4.345 + 0.758 – 3.4 = 1.703 T (tarik) 8) NPK = RPO + RPQ – P2 = 4.59 + 4.436 – 3.7
= 5.326 T (tarik)
9) NQL = RQP + RQR – 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
24
2) Gambar bidang gaya lintang
25
3) Gambar bidang gaya normal
26
