Adji Sutama

θ2 = 0

θ1 = 0

½ h1½ h1

θ10

21

P1

P1

2

1

M°21

M°12

h1

θ20

¼ P1 h1

RV12 RV21

X1 X2

A/EI = (½ b h)/EIA/EI = (½ h1 ¼ P1 h1)/EIA/EI = (1/8 P1 h1

2)/EI

θ10 θ20

½ h1½ h1

Pembuktian Momen Primer(Conjugate Beam Method/Metode Balok Sepadan)

Batang 1-2

Akibat Beban Luar

Bidang Momen

Reaksi Perletakan

RV12 ΣM2 = 0

RV12 . h1 – P1 . ½ h1 + RV21 . 0 = 0

RV12 =.

= ½ P1 ( )

RV21 ΣM1 = 0

- RV21 . h1 + P1 . ½ h1 + RV12 . 0 = 0

RV21 =.

= ½ P1 ( )

Adji Sutama

θ11 21

RV12 RV21

X1

M°12

h1

θ21 θ11 θ21

A/EI = (½ b h)/EIA/EI = (- ½ h1 Mº12)/EI

- Mº12

1/3 h1 2/3 h1

Bidang Momen

Daerah 1 – P1 ( 0 ≤ X1 ≤ ½ h1 )

MX1 = RV12 . X1 = ½ P1 . X1X1 = 0 M1 = 0

X1 = ½ h1 MP1 = ¼ P1 h1

Daerah 2 – P1 ( 0 ≤ X2 ≤ ½ h1 )

MX2 = RV21 . X2 = ½ P1 . X2X2 = 0 M2 = 0

X2 = ½ h1 MP1 = ¼ P1 h1

θ10 ΣM2 = 0

θ10 . h1 – . ½ h1 + θ20 . 0 = 0

θ10 =.

= = ( )

θ20 ΣM1 = 0

- θ20 . h1 + . ½ h1 + θ10 . 0 = 0

θ20 =.

= = ( )

Akibat Momen Mº12

Bidang Momen

Adji Sutama

Reaksi Perletakan

RV12 ΣM2 = 0

RV12 . h1 – Mº12 + RV21 . 0 = 0

RV12 =°

( )

RV21 ΣM1 = 0

RV21 . h1 – Mº12 + RV12 . 0 = 0

RV21 =°

( )

Bidang Momen

Daerah 1 – 2 ( 0 ≤ X1 ≤ h1 )

MX1 = RV12 . X1 - Mº12 =°

. X1 - Mº12

X1 = 0 M1 = - Mº12

X1 = h1 M2 = 0

θ11 ΣM2 = 0

- θ11 . h1 – (°

. 2/3 h1) + θ21 . 0 = 0

θ11 =°

=°

=°

=°

( )

θ21 ΣM1 = 0

θ21 . h1 + (°

. 1/3 h1) + θ11 . 0 = 0

θ21 =°

=°

( )

Adji Sutama

M°21

θ12 21

RV12 RV21

X1

h1

θ22 θ12 θ22

- Mº21

1/3 h12/3 h1

A/EI = (½ b h)/EIA/EI = (- ½ h1 Mº21)/EI

Akibat Momen Mº21

Bidang Momen

Reaksi Perletakan

RV12 ΣM2 = 0

- RV12 . h1 + Mº21 + RV21 . 0 = 0

RV12 =°

( )

RV21 ΣM1 = 0

- RV21 . h1 + Mº21 + RV12 . 0 = 0

RV21 =°

( )

Bidang Momen

Daerah 1 – 2 ( 0 ≤ X1 ≤ h1 )

MX1 = - RV12 . X1 = -°

. X1

X1 = 0 M1 = 0

X1 = h1 M2 = - Mº21

θ12 ΣM2 = 0

- θ12 . h1 – (°

. 1/3 h1) + θ22 . 0 = 0

Adji Sutama

θ12 =°

=°

( )

θ22 ΣM1 = 0

θ22 . h1 + (°

. 2/3 h1) + θ12 . 0 = 0

θ22 =°

=°

( )

θ1 = 0

θ10 - θ11 - θ12 = 0

-°

-°

= 0°+

°= ..... Pers (I)

θ2 = 0

θ20 – θ21 – θ22 = 0

-°

-°

= 0°+

°= ..... Pers (II)

°+

°= x 1

°+

°=°

+°

= x ½°

+°

=° °=°=

96 Mº12 EI h1 = 12 P1 h12 EI

Adji Sutama

Mº12 =

Mº12 =

°+

°= x ½

°+

°=°

+°

= x 1°

+°

=° – °=

–°

=

- 96 Mº21 EI h1 = - 12 P1 h12 EI

Mº21 =

Mº21 =

Adji Sutama

q2

L

1 4

M°14 M°41θ1 = 0 θ4 = 0

L

1 4

θ10RV14 RV41

X1

q2

θ40

½ L

(q2 L2)/8

A/EI = (2/3 b h)/EIA/EI = (2/3 L (q2 L2/8))/EIA/EI = (1/12 q2 L3)/EI

θ10 θ40

Batang 1-4

Akibat Beban Luar

Bidang Momen

Reaksi Perletakan

RV14 ΣM4 = 0

RV14 . L – (q2 . L) ½ L + RV41 . 0 = 0

RV14 = = ½ q2 L ( )

RV41 ΣM1 = 0

- RV41 . L + (q2 . L) ½ L + RV14 . 0 = 0

RV41 = = ½ q2 L ( )

Adji Sutama

Bidang Momen

Daerah 1 – 4 ( 0 ≤ X1 ≤ L )

MX1 = RV14 . X1 – (q2 . X1) ½ X1 = ½ q2 L X1 – ½ q2 X12

X1 = 0 M1 = 0

X1 = L M4 = ½ q2 L2 - ½ q2 L2 = 0

Letak Mmax pada Qx = 0

= 0

½ q2 L - q2 X1 = 0

X1 = = ½ L

X1 = ½ L Mmax = ¼ q2 L2 – 1/8 q2 L2 =

θ10 ΣM4 = 0

θ10 . L – . ½ L + θ40 . 0 = 0

θ10 = = ( )

θ40 ΣM1 = 0

- θ40 . L + . ½ L + θ10 . 0 = 0

θ40 = = ( )

Adji Sutama

θ11 41

RV14 RV41

X1

M°14

L

θ41 θ11 θ41

- Mº14

1/3 L 2/3 L

A/EI = (½ b h)/EIA/EI = (- ½ L Mº14)/EI

Akibat Momen Mº14

Bidang Momen

Reaksi Perletakan

RV14 ΣM4 = 0

RV14 . L – Mº14 + RV41 . 0 = 0

RV14 =°

( )

RV41 ΣM1 = 0

RV41 . L – Mº14 + RV14 . 0 = 0

RV41 =°

( )

Bidang Momen

Daerah 1 – 4 ( 0 ≤ X1 ≤ L )

MX1 = RV14 . X1 - Mº14 =°

. X1 - Mº14

X1 = 0 M1 = - Mº14

X1 = L M4 = 0

θ11 ΣM4 = 0

- θ11 . L – (°

. 2/3 L) + θ41 . 0 = 0

θ11 =°

=°

( )

Adji Sutama

M°41

θ12 41

RV14 RV41

X1

L

θ42θ12 θ42

- Mº41

1/3 L2/3 L

A/EI = (½ b h)/EIA/EI = (- ½ L Mº41)/EI

θ41 ΣM1 = 0

θ41 . L + (°

. 1/3 L) + θ11 . 0 = 0

θ41 =°

=°

( )

Akibat Momen Mº41

Bidang Momen

Reaksi Perletakan

RV14 ΣM4 = 0

- RV14 . L + Mº41 + RV41 . 0 = 0

RV14 =°

( )

RV41 ΣM1 = 0

- RV41 . L + Mº41 + RV14 . 0 = 0

RV41 =°

( )

Bidang Momen

Daerah 1 – 4 ( 0 ≤ X1 ≤ L )

MX1 = - RV14 . X1 = -°

. X1

X1 = 0 M1 = 0

Adji Sutama

X1 = L M4 = - Mº41

θ12 ΣM4 = 0

- θ12 . L – (°

. 1/3 L) + θ42 . 0 = 0

θ12 =°

=°

( )

θ42 ΣM1 = 0

θ42 . L + (°

. 2/3 L) + θ12 . 0 = 0

θ42 =°

=°

( )

θ1 = 0

θ10 - θ11 - θ12 = 0

-°

-°

= 0°+

°= ..... Pers (I)

θ4 = 0

θ40 – θ41 – θ42 = 0

-°

-°

= 0°+

°= ..... Pers (II)

Adji Sutama

°+

°= x 1

°+

°=°

+°

= x ½°

+°

=° °=°=

144 Mº14 EI L = 12 q2 L3 EI

Mº14 =

Mº14 =

°+

°= x ½

°+

°=°

+°

= x 1°

+°

=° – °=

–°

=

- 144 Mº41 EI L = - 12 q2 L3 EI

Mº41 =

Mº41 =

Adji Sutama

1/16 L

q1

2 3M°23

M°32

½ L ½ L

L

θ2 = 0 θ3 = 0

q12 3

½ L ½ L

L

θ20

θ30

X1X2

RV23 RV32

C

3/8 L1/8 L

1/16 q1 L2

A/EI = (1/2 b h)/EIA/EI = (1/2 ½ L 1/16 q1 L2)/EIA/EI = (1/64 q1 L3)/EI

θ20 θ30

1/16 L

9/128 q1 L2

A/EI = (2/3 b h)/EIA/EI = (2/3 3/8 L 9/128 q1 L2)/EIA/EI = (9/512 q1 L3)/EIA/EI = (b h)/EI

A/EI = (1/8 L 1/16 q1 L2)/EIA/EI = (1/128 q1 L3)/EI

A/EI = (2/3 b h)/EIA/EI = (2/3 1/8 L 1/128 q1 L2)/EIA/EI = (1/1536 q1 L3)/EI

3/8 L = 3/8 (3/8 L) = 9/64 L

5/8 L = 5/8 (3/8 L) = 15/64 L

3/8 L = 3/8 (1/8 L) = 3/64 L

5/8 L = 5/8 (1/8 L) = 5/64 L

1/3 L = 1/3 (1/2 L) = 1/6 L

2/3 L = 2/3 (1/2 L) = 2/6 L

Batang 2-3

Akibat Beban Luar

Bidang Momen

Adji Sutama

Reaksi Perletakan

RV23 ΣM3 = 0

RV23 . L – (q1 . ½ L) . ¼ L + RV32 . 0 = 0

RV23 = = 1/8 q1 L ( )

RV32 ΣM2 = 0

- RV32 . L + (q1 . ½ L) . ¾ L + RV23 . 0 = 0

RV32 = = 3/8 q1 L ( )

Bidang Momen

Daerah 2 – C ( 0 ≤ X1 ≤ ½ L )

MX1 = RV23 . X1 = 1/8 q1 L . X1X1 = 0 M2 = 0

X1 = ½ L MC = 1/16 q1 L2

Daerah 3 – C ( 0 ≤ X2 ≤ ½ L )

MX2 = RV32 . X2 – (q1 . X2) ½ X2 = 3/8 q1 L . X2 – ½ q1 X22

X2 = 0 M3 = 0

X2 = ½ L MC = 3/16 q1 L2 – 1/8 q1 L2 = 1/16 q1 L2

Letak Mmax pada Qx = 0

= 0

3/8 q1 L - q1 X2 = 0

X2 = = 3/8 L

X2 = 3/8 L Mmax = 9/64 q1 L2 – 9/128 q1 L2 = 9/128 q1 L2

Adji Sutama

θ21 32

RV23 RV32

X1

M°23

L

θ31 θ21 θ31

A/EI = (½ b h)/EIA/EI = (- ½ L Mº23)/EI

- Mº23

1/3 L 2/3 L

θ20 ΣM3 = 0

θ20 . L – . 4/6 L - . 7/16 L - .

27/64 L - . 15/64 L + θ30 . 0 = 0

θ20 =

θ20 =

θ20 = = ( )

θ30 ΣM2 = 0

- θ30 . L + . 2/6 L + . 9/16 L + .

37/64 L + . 49/64 L + θ20 . 0 = 0

θ30 =

θ30 =

θ30 = = ( )

Akibat Momen Mº23

Bidang Momen

Adji Sutama

Reaksi Perletakan

RV23 ΣM3 = 0

RV23 . L – Mº23 + RV32 . 0 = 0

RV23 =°

( )

RV32 ΣM2 = 0

RV32 . L – Mº23 + RV23 . 0 = 0

RV32 =°

( )

Bidang Momen

Daerah 2 – 3 ( 0 ≤ X1 ≤ L )

MX1 = RV23 . X1 - Mº23 =°

. X1 - Mº23

X1 = 0 M2 = - Mº23

X1 = L M3 = 0

θ21 ΣM3 = 0

- θ21 . L – (°

. 2/3 L) + θ31 . 0 = 0

θ21 =°

=°

( )

θ31 ΣM2 = 0

θ31 . L + (°

. 1/3 L) + θ21 . 0 = 0

θ31 =°

=°

( )

Adji Sutama

M°32

θ22 32

RV23 RV32

X1

L

θ32θ22 θ32

- Mº32

1/3 L2/3 L

A/EI = (½ b h)/EIA/EI = (- ½ L Mº32)/EI

Akibat Momen Mº32

Bidang Momen

Reaksi Perletakan

RV23 ΣM3 = 0

- RV23 . L + Mº32 + RV32 . 0 = 0

RV23 =°

( )

RV32 ΣM2 = 0

- RV32 . L + Mº32 + RV23 . 0 = 0

RV32 =°

( )

Bidang Momen

Daerah 2 – 3 ( 0 ≤ X1 ≤ L )

MX1 = - RV23 . X1 = -°

. X1

X1 = 0 M2 = 0

X1 = L M3 = - Mº32

θ22 ΣM3 = 0

- θ22 . L – (°

. 1/3 L) + θ32 . 0 = 0

θ22 =°

=°

( )

Adji Sutama

θ32 ΣM2 = 0

θ32 . L + (°

. 2/3 L) + θ22 . 0 = 0

θ32 =°

=°

( )

θ2 = 0

θ20 – θ21 – θ22 = 0

-°

-°

= 0°+

°= ..... Pers (I)

θ3 = 0

θ30 – θ31 – θ32 = 0

-°

-°

= 0°+

°= ..... Pers (II)

°+

°= x 1

°+

°=°

+°

= x ½°

+°

=° °=°=

2304 Mº23 EI L = 60 q1 L3 EI

Mº23 =

Mº23 =

Adji Sutama

°+

°= x ½

°+

°=°

+°

= x 1°

+°

=° – °=

–°

=

- 2304 Mº32 EI L = - 132 q1 L3 EI

Mº32 =

Mº32 =

Adji Sutama

P1

q2

q1

2EI

2EI

2EI 2EI

3EI 3EI

L

h2

h1

½ L

½ h1

A

1

B

4

2 3H°2

H°1

TUGAS ANALISA STRUKTUR I

Hitung bidang M, Q, dan N pada struktur gedung

bertingkat berikut dengan cara Cross?

Diketahui :

q1 = 1,1 t/mq2 = 1,2 t/m

P1 = 3,5 t h1 = 3,75 mh2 = 3,75 m

L = 3,75 m

Adji Sutama

Penyelesaian :

a) Akibat beban luar (Pendel mendatar H°1 dipasang dititik

1 dan H°2 dipasang dititik 2).

Menghitung koefisien distribusi (µ)

- Titik kumpul 1

µ1A : µ14 : µ12 = : :

µ1A : µ14 : µ12 =( ), :

( ), :( ),

µ1A : µ14 : µ12 = 1,600EI : 2,130EI : 3,200EI

µtotal = 6,930EI

µ1A =,, = 0,231

µ14 =,, = 0,307

µ12 =,, = 0,462

- Titik kumpul 2

µ21 : µ23 = :

µ21 : µ23 =( ), :

( ),µ21 : µ23 = 3,200EI : 2,130EI

µtotal = 5,330EI

µ21 =,, = 0,600

µ23 =,, = 0,400

Cek :µ1A + µ14 + µ12 = 10,231 + 0,307 + 0,462 = 1

Cek :µ21 + µ23 = 10,600 + 0,400 = 1

Adji Sutama

- Titik kumpul 3

µ32 : µ34 = :

µ32 : µ34 =( ), :

( ),µ32 : µ34 = 2,130EI : 3,200EI

µtotal = 5,330EI

µ32 =,, = 0,400

µ34 =,, = 0,600

- Titik kumpul 4

µ43 : µ41 : µ4B = : :

µ43 : µ41 : µ4B =( ), :

( ), :( ),

µ43 : µ41 : µ4B = 3,200EI : 2,130EI : 1,600EI

µtotal = 6,930EI

µ43 =,, = 0,462

µ41 =,, = 0,307

µ4B =,, = 0,231

Cek :µ32 + µ34 = 10,400 + 0,600 = 1

Cek :µ43 + µ41 + µ4B = 10,462 + 0,307 + 0,231 = 1

Adji Sutama

P1 = 3,5 t

2

1

M°21

M°12

h1 = 3,75 m

q1 = 1,1 t/m

2 3M°23 M°32

1,875 m 1,875 m

L = 3,75 m

Menghitung momen primer (M°)

- Batang 1-2

M°12 = + = +, ,

= + 1,641 tm

M°21 = - = -, ,

= - 1,641 tm

- Batang 2-3

M°23 = + = +, ( , )

= + 0,403 tm

M°32 = - = -, ( , )

= - 0,886 tm

Momen Batang

Adji Sutama

q2 = 1,2 t/m

L = 3,75 m

1 4

M°14M°41

- Batang 1-4

M°14 = + = +, ( , )

= + 1,406 tm

M°41 = - = -, ( , )

= - 1,406 tm

Jumlah momen primer di titik kumpul :

M°1 = M°1A + M°14 + M°12 = 0 + 1,406 + 1,641 = 3,047 tm

M°2 = M°21 + M°23 = -1,641 + 0,403 = -1,238 tm

M°3 = M°32 + M°34 = -0,886 + 0 = -0,886 tm

M°4 = M°43 + M°41 + M°4B = 0 + (-1,406) + 0 = -1,406 tm

Adji Sutama

P1

q2

q1

3,75 m

3,75 m

3,75 m

A

1

B

4

2H°2

H°1

3

M°1A = 0,925 tm

M°14 = 0,486 tm

M°12 = 0,439 tm

M°21 = 1,267 tm

M°23 = 1,267 tm M°32 = 0,459 tm

M°41 = 1,399 tm

M°34 = 0,459 tm

M°43 = 0,932 tm

M°4B = 0,467 tm

Didapatkan hasil momen ujung :

“Arah momen cross dibalik (diambil nilai mutlaknya).”

M°1A = 0,925 tm ( ) M°41 = 1,399 tm ( )

M°14 = 0,486 tm ( ) M°4B = 0,467 tm ( )

M°12 = 0,439 tm ( )

M°21 = 1,267 tm ( )

M°23 = 1,267 tm ( )

M°32 = 0,459 tm ( )

M°34 = 0,459 tm ( )

M°43 = 0,932 tm ( )

Adji Sutama

h2 = 3,75 m

A

1

M°1A = 0,925 tm

HA1

H1A

Menghitung reaksi pendel : H°1 dan H°2- Batang A-1

HA1 ∑M1A = 0 (Misal HA1 ke kanan)

-HA1.h2 + M°1A + H1A.0 = 0

HA1 =°

=, , = 0,247 ton ( )

H1A ∑MA1 = 0 (Misal H1A ke kiri)

-H1A.h2 + M°1A + HA1.0 = 0

H1A =°

=, , = 0,247 ton ( )

Cek :∑H = 0HA1 – H1A = 00,247 – 0,247 = 0

Adji Sutama

P1 = 3,5 t

2

1

h1 = 3,75 m

M°12 = 0,439 tm

M°21 = 1,267 tm

H12

H21

- Batang 1-2

H12 ∑M21 = 0 (Misal H12 ke kiri)

H12.h1 – P1.1,875 - M°12 + M°21 + H21.0 = 0

H12 =, . , , ,, =

, , = 1,529 ton ( )

H21 ∑M12 = 0 (Misal H21 ke kiri)

-H21.h1 + P1.1,875 - M°12 + M°21 + H12.0 = 0

H21 =, . , , ,, =

, , = 1,971 ton ( )

Cek :∑H = 0- H12 – H21 + P1 = 0- 1,529 – 1,971 + 3,5 = 0

Adji Sutama

h2 = 3,75 m

B

4

M°4B = 0,467 tm

HB4

H4B

- Batang 4-B

H4B ∑MB4 = 0 (Misal H4B ke kanan)

H4B.h2 – M°4B + HB4.0 = 0

H4B =°

=, , = 0,125 ton ( )

HB4 ∑M4B = 0 (Misal HB4 ke kiri)

HB4.h2 – M°4B + H4B.0 = 0

HB4 =°

=, , = 0,125 ton ( )

Cek :∑H = 0- HB4 + H4B = 0- 0,125 + 0,125 = 0

Adji Sutama

h1 = 3,75 m

4

3

H43

H34

M°34 = 0,459 tm

M°43 = 0,932 tm

- Batang 3-4

H34 ∑M43 = 0 (Misal H34 ke kanan)

H34.h1 – M°34 – M°43 + H43.0 = 0

H34 =, ,, =

, , = 0,371 ton ( )

H43 ∑M34 = 0 (Misal H43 ke kiri)

H43.h1 – M°34 – M°43 + H34.0 = 0

H43 =, ,, =

, , = 0,371 ton ( )

Cek :∑H = 0H34 – H43 = 00,371 – 0,371 = 0

Adji Sutama

AB

4

2

H°2

H°1

3

2 3

4

41

1

1 0,371 t

0,371 t

0,371 t

1,971 t

1,529 t

0,125 t

0,125 t

0,247 t

0,247 t

1,971 t

0,247 t

1,529 t

0,125 t

0,371 t

q1

M°23 = 1,267 tm M°32 = 0,459 tm

P1

M°12 = 0,439 tm

M°21 = 1,267 tmM°34 = 0,459 tm

M°43 = 0,932 tm

q2

M°14 = 0,486 tm M°41 = 1,399 tm

M°1A = 0,925 tm M°4B = 0,467 tm

Pendel H°1 (Misal H°1 ke kiri)

∑H = 0

1,529 + 0,247 + 0,371 – 0,125 - H°1 = 0

H°1 = 2,022 ton ( )

Pendel H°2 (Misal H°2 ke kiri)

∑H = 0

1,971 - 0,371 - H°2 = 0

H°2 = 1,600 ton ( )

Adji Sutama

L = 3,75 m

A

1

B

2 3

HX1

δ δ

MX12

MX21

MX34

4

MX43

b) Akibat goyangan I (karena pendel 1 dipasang dan pendel

2 dilepas).

Menghitung koefisien distribusi (µ)

Koefisien distribusi sama dengan keadaan tanpa

goyangan.

µ1A = 0,231 ; µ14 = 0,307 ; µ12 = 0,462

µ21 = 0,600 ; µ23 = 0,400

µ32 = 0,400 ; µ34 = 0,600

µ43 = 0,462 ; µ41 = 0,307 ; µ4B = 0,231

Adji Sutama

Harga x ini yang akan

dicari.

Menghitung momen primer akibat goyangan I

MX12 = MX21 = +δ

= +( )δ, = +

δ,MX34 = MX43 = +

δ= +

( )δ, = +δ,

Nilai EIδ diambil = 0,78125 (untuk mempermudah

perhitungan).

MX12 = MX21 = + 1,000

MX34 = MX43 = + 1,000

Misalkan : MX12 = 1,000 x tm

MX21 = 1,000 x tm

MX34 = 1,000 x tm

MX43 = 1,000 x tm

Jumlah momen primer dititik kumpul :

MX1 = MX1A + MX14 + MX12 = 0 + 0 + 1,000 = 1,000 x

MX2 = MX21 + MX23 = 1,000 + 0 = 1,000 x

MX3 = MX32 + MX34 = 0 + 1,000 = 1,000 x

MX4 = MX43 + MX41 + MX4B = 1,000 + 0 + 0 = 1,000 x

Adji Sutama

3,75 m

3,75 m

3,75 m

A

1

B

4

2

HX1

3

MX1A = 0,158 x

MX14 = 0,315 x

MX12 = 0,473 x

MX21 = 0,421 x

MX23 = 0,421 x MX

32 = 0,421 x

MX41 = 0,315 x

MX34 = 0,421 x

MX43 = 0,473 x

MX4B = 0,158 x

Didapatkan hasil momen ujung :

MX1A = 0,158 x tm ( ) MX41 = 0,315 x tm ( )

MX14 = 0,315 x tm ( ) MX4B = 0,158 x tm ( )

MX12 = 0,473 x tm ( )

MX21 = 0,421 x tm ( )

MX23 = 0,421 x tm ( )

MX32 = 0,421 x tm ( )

MX34 = 0,421 x tm ( )

MX43 = 0,473 x tm ( )

Adji Sutama

MX1A = 0,158 x

h2 = 3,75 m

A

1

HX

A1

HX

1A

Menghitung reaksi pendel : HX1 dan HX2- Batang A-1

HXA1 ∑M1A = 0 (Misal HXA1 ke kanan)

-HXA1.h2 + MX1A + HX1A.0 = 0

HXA1 = =, , = 0,042 x ton ( )

HX1A ∑MA1 = 0 (Misal HX1A ke kiri)

-HX1A.h2 + MX1A + HXA1.0 = 0

HX1A = =, , = 0,042 x ton ( )

Cek :∑H = 0HXA1 – HX1A = 00,042 x – 0,042 x = 0

Adji Sutama

2

1

h1 = 3,75 m

HX12

HX21

MX12 = 0,473 x

MX21 = 0,421 x

- Batang 1-2

HX12 ∑M21 = 0 (Misal HX12 ke kiri)

HX12.h1 - MX12 - MX21 + HX21.0 = 0

HX12 = =, ,, = 0,238 x ton ( )

HX21 ∑M12 = 0 (Misal HX21 ke kanan)

HX21.h1 - MX12 - MX21 + HX12.0 = 0

HX21 = =, ,, = 0,238 x ton ( )

Cek :∑H = 0- HX12 + HX21 = 0- 0,238 x + 0,238 x = 0

Adji Sutama

h2 = 3,75 m

B

4

HXB4

HX4B

MX4B = 0,158 x

- Batang 4-B

HX4B ∑MB4 = 0 (Misal HX4B ke kiri)

-HX4B.h2 + MX4B + HXB4.0 = 0

HX4B = =, , = 0,042 x ton ( )

HXB4 ∑M4B = 0 (Misal HXB4 ke kanan)

-HXB4.h2 + MX4B + HX4B.0 = 0

HXB4 = =, , = 0,042 x ton ( )

Cek :∑H = 0- HX4B + HXB4 = 0- 0,042 x + 0,042 x = 0

Adji Sutama

h1 = 3,75 m

4

3

HX43

HX34

MX34 = 0,421 x

MX43 = 0,473 x

- Batang 3-4

HX34 ∑M43 = 0 (Misal HX34 ke kanan)

HX34.h1 - MX34 - MX43 + HX43.0 = 0

HX34 = =, ,, = 0,238 x ton ( )

HX43 ∑M34 = 0 (Misal HX43 ke kiri)

HX43.h1 - MX34 - MX43 + HX34.0 = 0

HX43 = =, ,, = 0,238 x ton ( )

Cek :∑H = 0HX34 - HX43 = 00,238 x – 0,238 x = 0

Adji Sutama

AB

HX2

HX1

3

2 3

4

41

10,238 x

t

0,238 x

0,238 x

0,238 x

0,238 x

0,042 x

0,042 x

0,042 x

0,042 x

0,238 x

0,042 x

0,238 x

0,042 x

0,238 x

2

1 4

MX23 = 0,421 x MX

32 = 0,421 x

MX12 = 0,474 x

MX21 = 0,421 x

MX34 = 0,421 x

MX43 = 0,473 x

MX1A = 0,158 x

MX14 = 0,315 x MX

41 = 0,315 x

MX4B = 0,158 x

Pendel HX1 (Misal HX1 ke kiri)

∑H = 0

0,238 x + 0,042 x + 0,238 x + 0,042 x -

HX1 = 0

HX1 = 0,560 x ton ( )

Pendel HX2 (Misal HX2 ke kanan)

∑H = 0

- 0,238 x - 0,238 x + HX2 = 0

HX2 = 0,476 x ton ( )

Adji Sutama

L = 3,75 m

A

1

B

2 3

HY2

δ

MY12

4

δ

MY21

MY1A

MY34

MY43

MY4B

c) Akibat goyangan II (karena pendel 1 dilepas dan pendel

2 dipasang).

Menghitung koefisien distribusi (µ)

Koefisien distribusi sama dengan keadaan tanpa

goyangan.

µ1A = 0,231 ; µ14 = 0,307 ; µ12 = 0,462

µ21 = 0,600 ; µ23 = 0,400

µ32 = 0,400 ; µ34 = 0,600

µ43 = 0,462 ; µ41 = 0,307 ; µ4B = 0,231

Adji Sutama

Harga y ini yang akan

dicari.

Menghitung momen primer akibat goyangan II

MY1A = +δ

= +( )δ, = +

δ,MY4B = +

δ= +

( )δ, = +6EIδ

14,0625MY12 = MY21 = -

6EIδh12 = -

6(3EI)δ3,752 = -

18EIδ14,0625

MY34 = MY43 = -6EIδh12 = -

6(3EI)δ3,752 = -

18EIδ14,0625

Nilai EIδ diambil = 0,78125 (untuk mempermudah

perhitungan).

MY1A = + 0,333

MY4B = + 0,333

MY12 = MY21 = - 1,000

MY34 = MY43 = - 1,000

Misalkan : MY1A = + 0,333 y tm

MY4B = + 0,333 y tm

MY12 = - 1,000 y tm

MY21 = - 1,000 y tm

MY34 = - 1,000 y tm

MY43 = - 1,000 y tm

Jumlah momen primer dititik kumpul :

MY1 = MY1A + MY14 + MY12 = 0,333 + 0 + (-1,000) = - 0,667 y

MY2 = MY21 + MY23 = (-1,000) + 0 = - 1,000 y

MY3 = MY32 + MY34 = 0 + (-1,000) = - 1,000 y

MY4 = MY43 + MY41 + MY4B = (-1,000) + 0 + 0,333 = - 0,667 y

Adji Sutama

3,75 m

3,75 m

3,75 m

A

1

B

4

2HY

23

MY1A = 0,421 y

MY14 = 0,175 y

MY12 = 0,596 y

MY21 = 0,456 y

MY23 = 0,456 y MY

32 = 0,456 y

MY41 = 0,175 y

MY34 = 0,456 y

MY43 = 0,596 y

MY4B = 0,421 y

Didapatkan hasil momen ujung :

MY1A = 0,421 y tm ( ) MY41 = 0,175 y tm ( )

MY14 = 0,175 y tm ( ) MY4B = 0,421 y tm ( )

MY12 = 0,596 y tm ( )

MY21 = 0,456 y tm ( )

MY23 = 0,456 y tm ( )

MY32 = 0,456 y tm ( )

MY34 = 0,456 y tm ( )

MY43 = 0,596 y tm ( )

Adji Sutama

h2 = 3,75 m

A

1

HY

A1

HY

1A

MY1A = 0,421 y

Menghitung reaksi pendel : HY1 dan HY2- Batang A-1

HYA1 ∑M1A = 0 (Misal HYA1 ke kiri)

HYA1.h2 - MY1A + HY1A.0 = 0

HXA1 = =, , = 0,112 y ton ( )

HY1A ∑MA1 = 0 (Misal HY1A ke kanan)

HY1A.h2 - MY1A + HYA1.0 = 0

HY1A = =0,421 y3,75 = 0,112 y ton ( )

Cek :∑H = 0- HYA1 + HY1A = 0- 0,112 y + 0,112 y = 0

Adji Sutama

2

1

h1 = 3,75 m

HY12

HY21

MY12 = 0,596 y

MY21 = 0,456 y

- Batang 1-2

HY12 ∑M21 = 0 (Misal HY12 ke kanan)

-HY12.h1 + MY12 + MY21 + HY21.0 = 0

HY12 =MY12 MY21

h1=

0,596 y 0,456 y3,75 = 0,281 y ton ( )

HY21 ∑M12 = 0 (Misal HY21 ke kiri)

-HY21.h1 + MY12 + MY21 + HY12.0 = 0

HY21 =MY12 MY21

h1=

0,596 y 0,456 y3,75 = 0,281 y ton ( )

Cek :∑H = 0HY12 - HY21 = 00,281 y - 0,281 y = 0

Adji Sutama

h2 = 3,75 m

B

4

HYB4

HY4B

MY4B = 0,421 y

- Batang 4-B

HY4B ∑MB4 = 0 (Misal HY4B ke kanan)

HY4B.h2 - MY4B + HYB4.0 = 0

HY4B =MY4Bh2

=0,421 y3,75 = 0,112 y ton ( )

HYB4 ∑M4B = 0 (Misal HYB4 ke kiri)

HYB4.h2 – MY4B + HY4B.0 = 0

HYB4 =MY4Bh2

=0,421 y3,75 = 0,112 y ton ( )

Cek :∑H = 0HY4B - HYB4 = 00,112 y - 0,112 y = 0

Adji Sutama

h1 = 3,75 m

4

3

HY43

HY34

MY34 = 0,456 y

MY43 = 0,596 y

- Batang 3-4

HY34 ∑M43 = 0 (Misal HY34 ke kiri)

-HY34.h1 + MY34 + MY43 + HY43.0 = 0

HY34 =MY34 MY43

h1=

0,456 y 0,596 y3,75 = 0,281 y ton ( )

HY43 ∑M34 = 0 (Misal HY43 ke kanan)

-HY43.h1 + MY34 + MY43 + HY34.0 = 0

HY43 =MY34 MY43

h1=

0,456 y 0,596 y3,75 = 0,281 y ton ( )

Cek :∑H = 0- HY34 + HY43 = 0- 0,281 y + 0,281 y = 0

Adji Sutama

AB

HY2

HY1

3

2 3

4

41

10,281 y

0,281 y

0,281 y

0,281 y

0,281 y

0,112 y

0,112 y

0,112 y

0,112 y

0,281 y

0,112 y

0,281 y

0,112 y

0,281 y

2

1 4

MY1A = 0,421 y

MY14 = 0,175 y

MY12 = 0,596 y

MY21 = 0,456 y

MY23 = 0,456 y MY

32 = 0,456 y

MY41 = 0,175 y

MY34 = 0,456 y

MY43 = 0,596 y

MY4B = 0,421 y

Pendel HY1 (Misal HY1 ke kanan)

∑H = 0

- 0,281 y - 0,112 y - 0,281 y - 0,112 y

+ HY1 = 0

HY1 = 0,786 y ton ( )

Pendel HY2 (Misal HY2 ke kiri)

∑H = 0

0,281 y + 0,281 y – HY2 = 0

HY2 = 0,562 y ton ( )

Adji Sutama

Untuk x Untuk y

Hasil reaksi pendel yang didapatkan :

H°1 = 2,022 ton ( )

H°2 = 1,600 ton ( )

HX1 = 0,560 x ton ( )

HX2 = 0,476 x ton ( )

HY1 = 0,786 y ton ( )

HY2 = 0,562 y ton ( )

Jadi :

- H°1 - HX1 + HY1 = 0

- 2,022 – 0,560 x + 0,786 y = 0

- 0,560 x + 0,786 y = 2,022 …… (Pers. I)

- H°2 + HX2 - HY2 = 0

- 1,600 + 0,476x – 0,562 y = 0

0,476 x – 0,562 y = 1,600 …… (Pers. II)

Eliminasi pers. I dan II :

- 0,560 x + 0,786 y = 2,022 x - 0,476 x 0,562

0,476 x – 0,562 y = 1,600 x 0,560 x - 0,786

0,26656 x – 0,374136 y = - 0,962472

0,26656 x – 0,31472 y = 0,896

– 0,059416 y = - 1,858472

y = 31,279

- 0,31472 x + 0,441732 y = 1,136364

- 0,374136 x + 0,441732 y = - 1,2576

0,059416 x = 2,393964

x = 40,292

Adji Sutama

Jadi, momen ujung akhir sebenarnya ialah :

M1A = 5,877 tm ( )

M14 = 6,732 tm ( )

M12 = 0,855 tm ( )

M21 = 1,433 tm ( )

M23 = 1,433 tm ( )

M32 = 3,159 tm ( )

M34 = 3,159 tm ( )

M43 = 1,348 tm ( )

M41 = 8,617 tm ( )

M4B = 7,269 tm ( )

Adji Sutama

q1

P1

q2

L

h2

h1

A

1

B

4

2 3M23 = 1,433 tm M32 = 3,159 tm

M12 = 0,855 tm

M21 = 1,433 tmM34 = 3,159 tm

M43 = 1,348 tm

M14 = 6,732 tm M41 = 8,617 tm

M1A = 5,877 tm M4B = 7,269 tm

Gambar momen ujung akhir sebenarnya :

q1 = 1,1 t/mq2 = 1,2 t/m

P1 = 3,5 t h1 = 3,75 mh2 = 3,75 m

L = 3,75 m

Adji Sutama

Q

M

1,742 m

M23 = 1,433 tm C

1,875 m ( q1 . X2 )( q1 . 1,875 )

2 3

1,2019 t1,2019 t

M32 = 3,159 tm

RV23 RV32

q1 = 1,1 t/m

1,875 m

N

Freebody 2–3

00

00

00

- 1,2019 t- 1,2019 t

- 2,7714 t

- 0,7089 t- 0,7089 t

- 3,159 tm

0

0,104 tm

1,433 tm

X1X2

Adji Sutama

Reaksi Perletakan

RV23 ΣM3 = 0

- RV23 . 3,75 – (q1 . 1,875) 0,9375 + M23 + M32 + RV32 . 0 = 0

RV23 =( . , ) , , = 0,7089 ton ( )

RV32 ΣM2 = 0

- RV32 . 3,75 + (q1 . 1,875) 2,8125 + M23 + M32 + RV23 . 0 = 0

RV32 =( . , ) , , = 2,7714 ton ( )

ΣV = 0

- RV23 + RV32 – (q1 . 1,875) = 0

- 0,7089 + 2,7714 – (1,1 . 1,875) = 0

0 = 0

Bidang Momen

Daerah 2 – C (0 ≤ X1 ≤ 1,875)

MX1 = - RV23 . X1 + M23 = - 0,7089 X1 + 1,433

X1 = 0 M2 = 1,433 tm

X1 = 1,875 Mc = 0,104 tm

Daerah 3 – C (0 ≤ X2 ≤ 1,875)

MX2 = RV32 . X2 – (q1 . X2) ½ X2 - M32MX2 = 2,7714 X2 – 0,55 X22 - 3,159

X2 = 0 M3 = - 3,159 tm

X2 = 1,875 Mc = 0,104 tm

Letak Mmax pada QX = 0

= 0

2,7714 – 1,1 X2 = 0

Adji Sutama

X2 =, , = 2,520 m (Tidak Memenuhi)

Letak Momen MX = 0

2,7714 X2 – 0,55 X22 - 3,159 = 0

– 0,55 X22 + 2,7714 X2 – 3,159 = 0

X1,2 =± –

X1,2 =, ± ( , ) – ( , ) ( , )( , )

X1,2 =, ± ,– ,

X1 =, ,– , = 1,742 m

X2 =, ,– , = 3,297 m

Bidang Lintang

Daerah 2 – C (0 ≤ X1 ≤ 1,875)

QX1 = - RV23 = - 0,7089

X1 = 0 Q2 = - 0,7089 ton

X1 = 1,875 Qc = - 0,7089 ton

Daerah 3 – C (0 ≤ X2 ≤ 1,875)

QX2 = - RV32 + (q1 . X2) = - 2,7714 + (1,1 X2)

X2 = 0 Q3 = - 2,7714 ton

X2 = 1,875 Qc = - 0,7089 ton

Bidang Normal

Daerah 2 – C (0 ≤ X1 ≤ 1,875)

NX1 = - 1,2019

Adji Sutama

X1 = 0 N2 = - 1,2019 ton

X1 = 1,875 Nc = - 1,2019 ton

Daerah 3 – C (0 ≤ X2 ≤ 1,875)

NX2 = - 1,2019

X2 = 0 N3 = - 1,2019 ton

X2 = 1,875 Nc = - 1,2019 ton

Adji Sutama

N

2,149 m

( q2 . X1 ) ( q2 . 3,75 )

41

0,7929 t 0,7929 t

q2 = 1,2 t/m

M14 = 6,732 tm M41 = 8,617 tmRV14 RV41

3,75 m

M

Q

Freebody 1–4

- 0,7929 t- 0,7929 t

00

- 6,3431 t

- 1,8431 t

00

00

- 8,617 tm

0

6,732 tm

X1

Adji Sutama

Reaksi Perletakan

RV14 ΣM4 = 0

- RV14 . 3,75 – (q2 . 3,75) 1,875 + M14 + M41 + RV41 . 0 = 0

RV14 =( . , ) , , = 1,8431 ton ( )

RV41 ΣM1 = 0

- RV41 . 3,75 + (q2 . 3,75) 1,875 + M14 + M41 + RV14 . 0 = 0

RV41 =( . , ) , , = 6,3431 ton ( )

ΣV = 0

- RV14 + RV41 – (q2 . 3,75) = 0

- 1,8431 + 6,3431 – (1,2 . 3,75) = 0

0 = 0

Bidang Momen

Daerah 1 – 4 (0 ≤ X1 ≤ 3,75)

MX1 = - RV14 . X1 – (q2 . X1) ½ X1 + M14MX1 = - 1,8431 X1 – 0,6 X12 + 6,732

X1 = 0 M1 = 6,732 tm

X1 = 3,75 M4 = - 8,617 tm

Letak Mmax pada QX = 0

= 0

- 1,8431 – 1,2 X1 = 0

X1 =, , = - 1,536 m (Tidak Memenuhi)

Letak Momen MX = 0

- 1,8431 X1 – 0,6 X12 + 6,732 = 0

Adji Sutama

– 0,6 X12 – 1,8431 X1 + 6,732 = 0

X1,2 =± –

X1,2 =( , ) ± ( , ) – ( , ) ( , )( , )

X1,2 =, ± ,– ,

X1 =, ,– , = - 5,221 m

X2 =, ,– , = 2,149 m

Bidang Lintang

Daerah 1 – 4 (0 ≤ X1 ≤ 3,75)

QX1 = - RV14 – (q2 . X1) = - 1,8431 – (1,2 X1)

X1 = 0 Q1 = - 1,8431 ton

X1 = 3,75 Q4 = - 6,3431 ton

Bidang Normal

Daerah 1 – 4 (0 ≤ X1 ≤ 3,75)

NX1 = - 0,7929

X1 = 0 N1 = - 0,7929 ton

X1 = 3,75 N4 = - 0,7929 ton

Adji Sutama

N

M

Q

2

1

1,875 m

RH12

RH21

P1 = 3,5 t

M12 = 0,855 tm

M21 = 1,433 tm

0,7089 t

0,7089 t

1,875 m

Freebody 1–2

0,7089 t

0,7089 t

2,3601 t

2,3601 t- 1,1399 t

- 1,1399 t

- 0,855 tm

3,570 tm

0,362 m

1,433 tm

0

Y 1Y 2

Adji Sutama

Reaksi Perletakan

RH12 ΣM2 = 0

RH12 . 3,75 – P1 . 1,875 - M12 - M21 + RH21 . 0 = 0

RH12 =. , , = 2,3601 ton ( )

RH21 ΣM1 = 0

- RH21 . 3,75 + P1 . 1,875 - M12 - M21 + RH12 . 0 = 0

RH21 =. , , = 1,1399 ton ( )

ΣH = 0

- RH12 - RH21 + P1 = 0

- 2,3601 – 1,1399 + 3,5 = 0

0 = 0

Bidang Momen

Daerah 1 – P1 (0 ≤ Y1 ≤ 1,875)

MY1 = RH12 . Y1 – M12 = 2,3601 Y1 – 0,855

Y1 = 0 M1 = - 0,855 tm

Y1 = 1,875 MP1 = 3,570 tm

Letak Momen MY = 0

2,3601 Y1 – 0,855 = 0

Y1 =,, = 0,362 m

Daerah 2 – P1 (0 ≤ Y2 ≤ 1,875)

MY2 = RH21 . Y2 + M21 = 1,1399 Y2 + 1,433

Y2 = 0 M2 = 1,433 tm

Y2 = 1,875 MP1 = 3,570 tm

Adji Sutama

Bidang Lintang

Daerah 1 – P1 (0 ≤ Y1 ≤ 1,875)

QY1 = RH12 = 2,3601

Y1 = 0 Q1 = 2,3601 ton

Y1 = 1,875 QP1 = 2,3601 ton

Daerah 2 – P1 (0 ≤ Y2 ≤ 1,875)

QY2 = - RH21 = - 1,1399

Y2 = 0 Q2 = - 1,1399 ton

Y2 = 1,875 QP1 = - 1,1399 ton

Bidang Normal

Daerah 1 – P1 (0 ≤ Y1 ≤ 1,875)

NY1 = 0,7089

Y1 = 0 N1 = 0,7089 ton

Y1 = 1,875 NP1 = 0,7089 ton

Daerah 2 – P1 (0 ≤ Y2 ≤ 1,875)

NY2 = 0,7089

Y2 = 0 N2 = 0,7089 ton

Y2 = 1,875 NP1 = 0,7089 ton

Adji Sutama

h1 = 3,75 m

4

3

RH43

RH34

M34 = 3,159 tm

M43 = 1,348 tm

2,7714 t

2,7714 t

N

M

Q

Freebody 3-4

- 2,7714 t

- 2,7714 t

1,2019 t

1,2019 t

2,628 m

1,348 tm

0

Y1

- 3,159 tm

Adji Sutama

Reaksi Perletakan

RH34 ΣM4 = 0

RH34 . 3,75 – M34 – M43 + RH43 . 0 = 0

RH34 = , = 1,2019 ton ( )

RH43 ΣM3 = 0

RH43 . 3,75 – M34 – M43 + RH34 . 0 = 0

RH43 = , = 1,2019 ton ( )

ΣH = 0

RH34 – RH43 = 0

1,2019 – 1,2019 = 0

0 = 0

Bidang Momen

Daerah 3 – 4 (0 ≤ Y1 ≤ 3,75)

MY1 = RH34 . Y1 – M34 = 1,2019 Y1 – 3,159

Y1 = 0 M3 = - 3,159 tm

Y1 = 3,75 M4 = 1,348 tm

Letak Momen MY = 0

1,2019 Y1 – 3,159 = 0

Y1 =,, = 2,628 m

Bidang Lintang

Daerah 3 – 4 (0 ≤ Y1 ≤ 3,75)

QY1 = RH34 = 1,2019

Y1 = 0 Q3 = 1,2019 ton

Y1 = 3,75 Q4 = 1,2019 ton

Adji Sutama

Bidang Normal

Daerah 3 – 4 (0 ≤ Y1 ≤ 3,75)

NY1 = - 2,7714

Y1 = 0 N3 = - 2,7714 ton

Y1 = 3,75 N4 = - 2,7714 ton

Adji Sutama

h2 = 3,75 m

A

1

RHA1

RH1A

M1A = 5,877 tm

2,5520 t

2,5520 t

N

M

Q

Freebody A-1

2,5520 t

2,5520 t

1,5672 t

1,5672 t

5,877 tm

0

Y 1

Adji Sutama

Reaksi Perletakan

RHA1 ΣM1 = 0

RHA1 . 3,75 – M1A + RH1A . 0 = 0

RHA1 = , = 1,5672 ton ( )

RH1A ΣMA = 0

RH1A . 3,75 – M1A + RHA1 . 0 = 0

RH1A = , = 1,5672 ton ( )

ΣH = 0

- RHA1 + RH1A = 0

- 1,5672 + 1,5672 = 0

0 = 0

Bidang Momen

Daerah A – 1 (0 ≤ Y1 ≤ 3,75)

MY1 = RHA1 . Y1 = 1,5672 Y1Y1 = 0 MA = 0 tm

Y1 = 3,75 M1 = 5,877 tm

Bidang Lintang

Daerah A – 1 (0 ≤ Y1 ≤ 3,75)

QY1 = RHA1 = 1,5672

Y1 = 0 QA = 1,5672 ton

Y1 = 3,75 Q1 = 1,5672 ton

Bidang Normal

Daerah A – 1 (0 ≤ Y1 ≤ 3,75)

NY1 = 2,5520

Adji Sutama

Y1 = 0 NA = 2,5520 ton

Y1 = 3,75 N1 = 2,5520 ton

Adji Sutama

h2 = 3,75 m

B

4

RHB4

RH4B

M4B = 7,269 tm

9,1145 t

9,1145 t

N

M

Q

Freebody 4-B

1,9384 t

1,9384 t

- 7,269 tm

- 9,1145 t

- 9,1145 t

0

Y1

Adji Sutama

Reaksi Perletakan

RH4B ΣMB = 0

RH4B . 3,75 – M4B + RHB4 . 0 = 0

RH4B = , = 1,9384 ton ( )

RHB4 ΣM4 = 0

RHB4 . 3,75 – M4B + RH4B . 0 = 0

RHB4 = , = 1,9384 ton ( )

ΣH = 0

RH4B – RHB4 = 0

1,9384 - 1,9384 = 0

0 = 0

Bidang Momen

Daerah 4 – B (0 ≤ Y1 ≤ 3,75)

MY1 = RH4B . Y1 – M4B = 1,9384 Y1 – 7,269

Y1 = 0 M4 = - 7,269 tm

Y1 = 3,75 MB = 0 tm

Bidang Lintang

Daerah 4 – B (0 ≤ Y1 ≤ 3,75)

QY1 = RH4B = 1,9384

Y1 = 0 Q4 = 1,9384 ton

Y1 = 3,75 QB = 1,9384 ton

Bidang Normal

Daerah 4 – B (0 ≤ Y1 ≤ 3,75)

NY1 = - 9,1145

Adji Sutama

Y1 = 0 N4 = - 9,1145 ton

Y1 = 3,75 NB = - 9,1145 ton

Adji Sutama

(q2 . 3,75)

(q1 . 1,875)

q1 = 1,1 t/m

A B

4

2 3

2 3

4

41

1

1

M23 = 1,433 tm M32 = 3,159 tm

q2 = 1,2 t/m

M14 = 6,732 tm M41 = 8,617 tm

0,7089 t

0,7089 t

2,7714 t

2,7714 t

0,7089 t 2,7714 t

1,8431 t 6,3431 t

2,5520 t 9,1145 t

2,5520 t 9,1145 t

1,875 m 1,875 m

q2 = 1,2 t/m

Superposisi Vertikal

Adji Sutama

AB

4

2 3

2 3

4

41

1

1 1,2019 t

1,2019 t

1,2019 t

2,3601 t

1,9384 t

1,9384 t

1,5672 t

1,5672 t

1,2019 t

0,7929 t 0,7929 t

P1 = 3,5 t

M12 = 0,855 tm

M21 = 1,433 tmM34 = 3,159 tm

M43 = 1,348 tm

M1A = 5,877 tm M4B = 7,269 tm

3,75 m

1,875 m

1,875 m

1,1399 t

Superposisi Horizontal

Adji Sutama

Tinjau seluruh konstruksi :

ΣV = 0

- (q1 x 1,875) – (q2 x 3,75) – 2,5520 + 9,1145 = 0

- (1,1 x 1,875) – (1,2 x 3,75) – 2,5520 + 9,1145 = 0

- 2,0625 – 4,5 – 2,5520 + 9,1145 = 0

0 = 0

ΣH = 0

3,5 – 1,5672 – 1,9384 = 0

0 = 0

Adji Sutama

2,149 m

1,742 m

Gambar Keseluruhan Bidang Momen

- 7,269 tm

0

5,877 tm

0

0

1,348 tm

2,628 m

- 3,159 tm

- 0,855 tm

3,570 tm0

1,433 tm

- 8,617 tm

0

6,732 tm

- 3,159 tm

0

0,104 tm1,433 tm

0,362 m

Adji Sutama

Gambar Keseluruhan Bidang Lintang

1,2019 t

1,9384 t

1,9384 t

2,3601 t

1,5672 t

1,5672 t

1,2019 t- 0,7089 t

2,3601 t - 1,1399 t

- 1,1399 t

- 6,3431 t

- 1,8431 t

- 2,7714 t

- 0,7089 t

Adji Sutama

Gambar Keseluruhan Bidang Normal

2,5520 t

0,7089 t

- 9,1145 t

- 9,1145 t

- 1,2019 t

- 2,7714 t

- 2,7714 t0,7089 t

- 0,7929 t- 0,7929 t

- 1,2019 t

2,5520 t