144623938 Menghitung Momen Menggunakan Metode Takabeya

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


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


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


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


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


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


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


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


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


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


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

Documents

144623938 Menghitung Momen Menggunakan Metode Takabeya