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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