1
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
1 q l
2 = -
12
1(3.679)(5
2) = -7.664 TM
b. MFFE = MF
KJ = 12
1 q l
2 =
12
1(3.679)(5
2) = 7.664 TM
c. MFFG = MF
KL = -12
1 q l
2 = -
12
1(3.616)(4.85
2) = -7.088 TM
d. MFFG = MF
KL = 12
1 q l
2 =
12
1(3.616)(4.85
2) = 7.088 TM
2
e. MFOP = -12
1 q l
2 = -
12
1(1.787)( 5
2) = -3.723 TM
f. MFPO = 12
1 q l
2 =
12
1(1.787)( 5
2) = 3.723 TM
g. MFPQ = -12
1 q l
2 = -
12
1(1.757)(4.85
2) = -3.443 TM
h. MFQP = 12
1 q l
2 =
12
1(1.757)( 4.85
2) = 3.443 TM
2. Menentukan jumlah momen primer di titik kumpul
a. E = J = MF
EF = MF
JK = -7.664 TM
b. F = K = MF
FE + MF
FG = MF
KJ + MF
KL = 7.664 + (-7.088) = 0.576 TM
c. G = L = MF
GF = MF
LK = 7.088 TM
d. O = MF
OP = -3.723 TM
e. P = MF
PO + MF
PQ = 3.723 + (-3.443) = 0.280 TM
f. Q = MF
QP = 3.443 TM
3. Menentukan kekakuan balok dan kolom
IB =
12
1(b)(h
3) =
12
1(30)(50
3) = 312500 cm
4
IC =
12
1(b)(h
3) =
12
1(40)(40
3) = 213333 cm
4
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
3
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
FKk
= (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
4
4) Titik J
JE = J
JEk
= (0.533/9.689) = 0.055
JK = J
JKk
= (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
LKk
= (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
PKk
= (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
5
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
6
5. Pemberesan momen-momen parsil m(0)
7
6. Perhitungan Momen Akhir
a. ME
1) MEA = KEA (2 ME + MA) + M
FEA = 0.533 (2(0.758)+0)+0 = 0.808 tm
2) MEF = KEF (2 ME + MF) + MF
EF = 0.625 (2(0.758)-0.109)-7.664 = -6.784 tm
3) MEJ = KEJ (2 ME + MJ) + MF
EJ = 0.533 (2(0.758)+0.809)+0 = 1.240 tm
4) MED = KED (2 ME + MD) + MF
ED = 3.125 (2(0.758)+0)+0 = 4.736 tm
Jumlah = 0 Ok
b. MF
1) MFB = KFB (2 MF + MB) + M
FFB = 0.533 (2(-0.109)+0)+0 = -0.116 tm
2) MFG = KFG (2 MF + MG) + MF
FG = 0.644 (2(-0.109)-0.749)-7.088 = -7.710 tm
3) MFE = KFE (2 MF + ME) + MF
FE = 0.625 (2(-0.109)+0.758)+7.664 = 8.002 tm
4) MFK = KFK (2 MF + MK) + MF
FK = 0.533 (2(-0.109)-0.073)+0 = -0.175 tm
Jumlah = 0 Ok
c. MG
1) MGC = KGC (2 MG + MC) + M
FGC = 0.533 (2(-0.749)+0)+0 = -0.799 tm
2) MGF = KGF (2 MG + MF) + MF
GF = 0.644 (2(-0.749)-0.109)+7.088 = 6.053 tm
3) MGH = KGH (2 MG + MH) + MF
GH = 2.717 (2(-0.749)+0)+0 = -4.071tm
4) MGL = KGL (2 MG + ML) + MF
GL = 0.533 (2(-0.749)-0.718)+0 = -1.182 tm
Jumlah = 0 Ok
d. MJ
1) MJE = KJE (2 MJ + ME) + M
FJE = 0.533 (2(0.809)+0.758)+0 = 1.267 tm
2) MJK = KJK (2 MJ + MK) + MF
JK = 0.625 (2(0.809)-0.073)-7.664 = 5.054 tm
3) MJO = KJO (2 MJ + MO) + MF
JO = 0.561 (2(0.809)+0.383)+0 = -6.723 tm
4) MJI = KJI (2 MJ + MI) + MF
JI = 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
8
e. MK
1) MKF = KKF (2 MK + MF) + M
FKF = 0.533 (2(-0.112)-0.109)+0 = -0.177 tm
2) MKL = KKL (2 MK + ML) + MF
KL = 0.644 (2(-0.112)-0.718)-7.088 = 8.029 tm
3) MKJ = KKJ (2 MK + MJ) + MF
KJ = 0.625 (2(-0.112)-0.134)+7.664 = -7.695 tm
4) MKP = KKP (2 MK + MP) + MF
KP = 0.561 (2(-0.112)-0.057)+0 = -0.158 tm
Jumlah = 0 Ok
f. ML
1) MLG = KLG (2 ML + MG) + M
FLG = 0.533 (2(-0.718)-0.749)+0 = -1.166 tm
2) MLK = KLK (2 ML + MK) + MF
LK = 0.644 (2(-0.718)-0.112)+7.088 = 6.090 tm
3) MLM = KLM (2 ML + MM) + MF
LM = 2.717 (2(-0.718)+0)+0 = -3.903tm
4) MLQ = KLQ (2 ML + MQ) + MF
LQ = 0.561 (2(-0.718)-0.383)+0 = -1.021 tm
Jumlah = 0 Ok
g. MO
1) MOJ = KOJ (2 MO + MJ) + M
FOJ = 0.561 (2(0.383)+0.809)+0 = 0.884 tm
2) MOP = KOP (2 MO + MP) + MF
OP = 0.625 (2(0.383)-0.057)-3.723 = 2.395 tm
3) MON = KON (2 MO + MN) + MF
ON = 3.125 (2(0.383)+0)+0 = -3.280 tm
Jumlah = 0 Ok
h. MP
1) MPK = KPK (2 MP + MK) + M
FPK = 0.561 (2(-0.057)-0.112)+0 = -0.127 tm
2) MPQ = KPQ (2 MP + MQ) + MF
PQ = 0.644 (2(-0.057)-0.383)-3.443 = 3.891 tm
3) MPO = KPO (2 MP + MO) + MF
PO = 0.625 (2(-0.057)+0.383)+3.723 = -3.764 tm
Jumlah = 0 Ok
i. MQ
1) MQL = KQL (2 MQ + ML) + M
FQL = 0.561 (2(-0.383)-0.718)+0 = -0.833 tm
2) MQP = KQP (2 MQ + MP) + MF
QP = 0.644 (2(-0.383)-0.057)+3.443 = 2.913 tm
3) MQR = KQR (2 MQ + MR) + MF
QR = 2.717 (2(-0.383)+0)+0 = -2.080tm
Jumlah = 0 Ok
j. MAE = KAE (2 MA + ME) + M
FAE = 0.533 (2(0)+0.758)+0 = 0.404 tm
k. MBF = KBF (2 MB + MF) + M
FBF = 0.533 (2(0)-0.109)+0 = -0.058 tm
l. MCG = KCG (2 MC + MG) + M
FCG = 0.533 (2(0)-0.749)+0 = -0.400 tm
9
Gambar distribusi momen pada portal
10
7. Menentukan perletakan momen maksimum, menghitung momen maksimum, dan
menentukan perletakan momen minimum (M=0)
a. Batang EF
RE => MF = 0
RE (5) (3.679)(52) 6.784 + 8.002 = 0
5RE 45.9875 6.784 +8.002 = 0
RE = (44.7695 / 5) = 8.954 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) qx12 - MEF
= 8.954x1 1.8395x12 6.784
dx
dM max =0
8.954 -3.679x1 = 0 ==> x1 = (8.954/3.679) = 2.434 m
Dari titik F
Mmax = RF (x2) qx22 MFE
= 9.441x2 1.8395x22 8.002
dx
dM max =0
9.441 -3.679x2 = 0 ==> x2 = (9.441/3.679) = 2.566 m
RF => ME = 0
-RF (5) + (3.679)(52) 6.784 + 8.002 = 0
-5RF + 45.9875 6.784 +8.002 = 0
RF = (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
11
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
ac4b 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
ac4b 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
12
b. Batang FG
RF => MG = 0
RF (4.85) (3.616)(4.852) 7.710 + 6.053 = 0
4.85RF 42.529 7.710 + 6.053 = 0
RF = (44.186 / 4.85) = 9.111 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) qx22 MFG
= 9.111x2 1.808x22 7.710
dx
dM max =0
9.111 -3.616x2 = 0 ==> x2 = (9.111/3.616) = 2.52 m
Dari titik G
Mmax = RG (x1) qx12 MGF
= 8.427x1 1.808x12 6.053
dx
dM max =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
RG => MF = 0
-RG (4.85) + (3.616)(4.852) 7.710 + 6.053 = 0
-4.85RG +42.529 7.710 +6.053 = 0
RG = (40.872 / 4.85) = 8.427 T
13
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
ac4b 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
ac4b 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
14
c. Batang JK
RJ => MK = 0
RJ (5) (3.679)(52) 6.816 + 8.029 = 0
5RJ 45.9875 6.816 +8.029 = 0
RJ = (44.7745 / 5) = 8.955 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) qx12 MJK
= 8.955x1 1.8395x12 6.816
dx
dM max =0
8.955 -3.679x1 = 0 ==> x1 = (8.955/3.679) = 2.434 m
Dari titik K
Mmax = RK (x2) qx22 MKJ
= 9.440x2 1.8395x22 8.029
dx
dM max =0
9.440-3.679x2 = 0 ==> x2 = (9.440/3.679) = 2.566 m
RK => MJ = 0
-RK (5) + (3.679)(52) 6.816 + 8.029 = 0
-5RK + 45.9875 6.816 +8.029 = 0
RK = (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
15
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
ac4b 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
ac4b 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
16
d. Batang KL
RK => ML = 0
RK (4.85) (3.616)(4.852) 7.695 + 6.090 = 0
4.85RK 42.529 7.695 + 6.090 = 0
RK = (44.134 / 4.85) = 9.100 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) qx22 MKL
= 9.100x2 1.808x22 7.695
dx
dM max =0
9.100 -3.616x2 = 0 ==> x2 = (9.100/3.616) = 2.52 m
Dari titik L
Mmax = RL (x1) qx12 MLK
= 8.438x1 1.808x12 6.090
dx
dM max =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
RL => MK = 0
-RL (4.85) + (3.616)(4.852) 7.710 + 6.053 = 0
-4.85RL +42.529 7.695 +6.090 = 0
RL = (40.924 / 4.85) = 8.438 T
17
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
ac4b 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
ac4b 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
18
e. Batang OP
RO => MP = 0
RO (5) (1.787)(52) 3.280 + 3.891 = 0
5RO 22.3375 3.280 +3.891 = 0
RO = (21.7265 / 5) = 4.345 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) qx12 MOP
= 4.345x1 0.8935x12 3.280
dx
dM max =0
4.345 1.787x1 = 0 ==> x1 = (4.345/1.787) = 2.431 m
Dari titik P
Mmax = RP (x2) qx22 MPO
= 4.59x2 0.8935x22 3.891
dx
dM max =0
4.59 - 1.787x2 = 0 ==> x2 = (4.59/1.787) = 2.569 m
RP => MO = 0
-RP (5) + (1.787)(52) 3.280 + 3.891 = 0
-5RP + 22.3375 3.280 +3.891 = 0
RP = (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
19
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
ac4b 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
ac4b 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
20
f. Batang PQ
RP => MQ = 0
RP (4.85) (1.757)(4.852) 3.764 + 2.913 = 0
4.85RP 20.6645 3.764 + 2.913 = 0
RP = (21.5155 / 4.85) = 4.436 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) qx22 MPQ
= 4.436x2 0.8785x22 3.764
dx
dM max =0
4.436 -1.757x2 = 0 ==> x2 = (4.436/1.757) = 2.525 m
Dari titik Q
Mmax = RQ (x1) qx12 MQP
= 4.085x1 0.8785x12 2.913
dx
dM max =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
RQ => MP = 0
-RQ (4.85) + (1.757)(4.852) 3.764 + 2.913 = 0
-4.85RQ +20.6645 3.764 +2.913 = 0
RQ = (19.8135 / 4.85) = 4.085 T
21
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
ac4b 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
ac4b 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
22
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 = 4
)MMMM()MMM(M KFFKBFFBJEEJAEEA
= 4
0.177-0.175-0.058 116.01.1871.2400.4040.808
= 0.78 ton (tarik)
2) NFG = 4
)MMMM()MMM(M- LGGLCGGCKFFKBFFB
= 4
166.1182.1400.0799.0177.0175.0058.00.116-
= 1.02 ton (tarik)
3) NJK = 4
)MM()M(M FKKFEJJE +8.3
)MM()M(M PKKPOJJO
23
= 4
0.175 177.0240.1187.1
8.3
0.127 158.0844.0040.1
= 0.94 ton (tarik)
4) NKL = 4
)MM()M(M- GLLGFKKF
8.3
)MM()M(M- QLLQPKKP
= 4
182.1166.1175.00.177-
8.3
833.0021.1127.00.158-
= 1.24 ton (tarik)
5) NOP = 8.3
)MM()M(M KPPKJOOJ
= 8.3
0.158 127.0040.1844.0 = 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 = 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 = 6.221 T (tarik)
4) NJE = RJI + RJK P4 = 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)
24
10. Gambar bidang momen, gaya lintang, dan gaya normal
1) Gambar bidang momen
25
2) Gambar bidang gaya lintang
26
3) Gambar bidang gaya normal
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