Menghitung Momen Menggunakan Metode Takabeya

7/16/2019 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



2

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



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 = 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



10

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’



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

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



12

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


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



14

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


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’



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

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



16

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



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

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



18

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


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’



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

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



20

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


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



22

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



23

=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



25

2) Gambar bidang gaya lintang



3) Gambar bidang gaya normal

Menghitung Momen Menggunakan Metode Takabeya

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