The document analyzes the forces, moments, and reactions in a beam-column structure. It calculates the internal forces including shear forces and bending moments at various points along the beams. Diagrams are drawn showing the shear force and bending moment diagrams. The maximum bending moment of 51.68 tm occurs at a point 3.46 m from end C of the beam.

1. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
BALOK GERBER
S1
S2
S1
RVS2
RVS1
M = 10 tm
P = 7 t
C
MC
Q = 10 t/m’
RVC
BA
P = 6 t
RVA RVB
S2
M = 10 tm
Q = 10 t/m’
P = 7 t
P = 6 t
C
6 44222 2 2 22
BA

2. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
MENGHITUNG REAKSI PERLETAKAN
𝑀𝑆2 = 0
RVS1 (6) – P (4) + M = 0
RVS1 (6) – 7 (4) + 10 = 0
RVS1 (6) – 18 = 0
RVS1 (6) = 18
RVS1 = 3 t
2 4
S2
S1
RVS2
RVS1
M = 10 t/m’
P = 7 t
𝑀𝑆1 = 0
-RVS2 (6) + P (2) + M = 0
-RVS2 (6) + 7 (2) + 10 = 0
-RVS2 (6) + 24 = 0
RVS2 (6) = 24
RVS2 = 4 t
𝐾𝑜𝑛𝑡𝑟𝑜𝑙 ∶ V = 0
RVS1 + RVS2 – P = 0
3 + 4 – 7 = 0
44
P’
𝑀 = 0
MC – P’ (6) – RVS2 (8) = 0
MC – 40 (6) – 4 (8) = 0
MC – 272 = 0
MC = 272 tm
𝑉 = 0
RVC – RVS2 – P’ = 0
RVC – 4 – 40 = 0
RVC – 44 = 0
RVC = 44 t
22
RVS2 = 4 t
C
MC
Q = 10 t/m’
RVC
P’ = q . l
P’ = 10 (4)
P’ = 40 t

3. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
𝑀𝐵 = 0
RVA (10) – P’ (4) +P (2) + RVS1 (2) = 0
RVA (10) – 36 (4) + 6 (2) + 3 (2) = 0
RVA (10) – 120 = 0
RVA (10) =120
RVA = 12 t
𝑀𝐴 = 0
RVS1 (14) + P (12) – RVB (10) + P’ (6) = 0
3 (14) + 6 (12) – RVB (10) + 36 (6) = 0
330 – RVB (10) = 0
RVB (10) = 330
RVB = 33 t
P’ = ½ q.l
P’ = ½ 12 (6)
P’ = 36 t
𝐾𝑜𝑛𝑡𝑟𝑜𝑙 ∶ 𝑉 = 0
RVA + RVB – P’ – P – RVS1 = 0
12 + 33 – 36 – 6 – 3 = 0
24
P’
BA
RVS1 = 3 t
P = 6 t
RVA RVB
2 6 22 2

4. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
MENGGAMBAR DIAGRAM GAYA LINTANG, GAYA NORMAL, & MOMEN
R
X1 (0° ≤ α𝑥1 ≤ 48,18°) (S1 – R)
NX1 = - RVS1 cos α𝑥1 = - 3 cos α𝑥1
α𝑥1 = 0° NS1 = - 3 cos 0°
NS1 = - 3 t
α𝑥1 = 48,18° NR = - 3 cos 48,18°
NR = - 2 t
X2 (0° ≤ α𝑥2 ≤ 41,81°) (S2 – R)
NX2 = RVS2 sin α𝑥2 = 4 sin α𝑥2
α𝑥2 = 0° NS2 = 4 sin 0°
NS2 = 0 t
α𝑥2 = 41,81° NR = 4 sin 41,81°
NR = 2,66 t
PNPQ
X2
X1
R
α
cos α =
4
6
= 0,66 sin α = 0,74
α = 48,18° 90°- α = 41,81°
α
X1 (0° ≤ α𝑥1 ≤ 48,18°) (S1 – R)
QX1 = RVS1 sin α𝑥1 = 3 sin α𝑥1
α𝑥1 = 0° QS1 = 3 sin 0°
QS1 = 0 t
α𝑥1 = 48,18° QR = 3 sin 48,18°
QR = 2,23 t
2 4
M = 10 t/m’
RVS2 = 4 t
S2
S1
RVS1 = 3 t
P = 7 t
X2 (0° ≤ α𝑥2 ≤ 41,81°) (S2 – R)
QX2 = - RVS2 cos α𝑥2 = - 4 cos α𝑥2
α𝑥2 = 0° QS2 = - 4 cos 0°
QS2 = - 4 t
α𝑥2 = 41,81° QR = - 4 cos 41,81°
QR = -2,98 t
R S2
2,23
S1
-2,98
-4
-3
-2
S1 S2
2,66
R
GAYA NORMAL (N)
GAYA LINTANG (Q)

5. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
T
0 ≤ 𝑥1 ≤ 2 (S1 – R)
MX1 = RVS1 𝑥1 = 3 𝑥1
𝑥1 = 0 MS1 = 3 (0)
MS1 = 0 tm
𝑥1 = 2 MR = 3 (2)
MR = 6 tm
0≤ 𝑥2 ≤ 2 (S2 – T)
MX2 = RVS2 𝑥2 = 4 𝑥2
𝑥2 = 0 MS2 = 4 (0)
MS2 = 0 tm
𝑥2 = 2 MT = 4 (2)
MT = 8 tm
X3
PNPQ
X2
X1
R
α
α
2 4
M = 10 t/m’
RVS2 = 4 t
S2
S1
RVS1 = 3 t
P = 7 t
0≤ 𝑥3 ≤ 2 (T – R)
MX2 = RVS2 3 – M = 4 3+2) – 10
𝑥3 = 0 MT = 4 (0+2) – 10
MT = 8 – 10
MT = - 2 tm
𝑥3 = 2 MR = 4 (2+2) – 10
MR = 16 – 10
MR = 6 tm
8
-2
6
R
T
S2S1
MOMEN (M)

6. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
B
BC = 42 + 62
BC = 52
BC = 7,21
α
A
X1
6
44
P’= 40 t
0 ≤ 𝑥1≤4 (A – B)
QX1 = - RVS2 – P’x = - 4 – 10x
𝑥1 = 0 QA = - 4 – 10 (0)
QA = - 4 t
𝑥1 = 4 QB = - 4 – 10 (4)
QB = - 44 t
0 ≤ 𝑥2≤7,21 (C – B)
QX2 = - RVC cos α = - 44
4
7,21
= - 24,41 t
𝑥2 = 0 QC = - 24,41 t
𝑥2 = 7,21 QB = - 24,41 t
22
RVS2 = 4 t
C
MC = 272 tm
Q = 10 t/m’
RVC = 44 t
P’x = q.x
P’x = 10x
P’x
cos α =
4
7,21
sin α =
6
7,21
-24,41
-44
B CA
-4
0 ≤ 𝑥1≤4 (A – B)
NX1 = 0 t
𝑥1 = 0 NA = 0
𝑥1 = 4 NB = 0
(tidak ada gaya yang sejajar sumbu batang)
0 ≤ 𝑥2≤7,21 (C – B)
NX2 = - RVC sin α = - 44
6
7,21
= - 36,52 t
𝑥2 = 0 NC = - 36,52 t
𝑥2 = 7,21 NB = - 36,52 t
-36,52
B CA
GAYA NORMAL (N)
GAYA LINTANG (Q)

7. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
B
X2
A
X1
44
P’= 40 t
0 ≤ 𝑥1≤4 (A – B)
MX1 = - RVS2 (𝑥1) – P’x (½ 𝑥1) = - 4(𝑥1) – 10x (½ 𝑥1)
𝑥1 = 0 MA = - 4 (0) – 10 (0)(0)
MA = 0 tm
𝑥1 = 4 MB = - 4 (4) – 10 (4)(2)
MB = - 96 tm
0 ≤ 𝑥2≤ 4 (C – B)
MX1 = RVC (𝑥2) – MC = 44(𝑥2) - 272
𝑥2 = 0 MC = 44 (0) – 272
MC = - 272 tm
𝑥2 = 4 MB = 44 (4) – 272
MB = - 96 tm
22
RVS2 = 4 t
C
MC = 272 tm
Q = 10 t/m’
RVC = 44 t
P’x = q.x
P’x = 10x
P’x
CBA
-272
-96
MOMEN (M)

8. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
X5 X4
X3X2X1
P’x
DC E F
0≤𝑥2≤6 (C – D)
QX2 = RVA – P’x = 12 – 𝑥2
𝑥2 = 0 QC = 12 – 02
QC = 12 t
𝑥2 = 6 QD = 12 – 62
QD = - 24 t
P’ = ½ q.l
P’ = ½ 12 (6)
P’ = 36 t
0≤𝑥3≤2 (D – B)
QX3 = RVA – P’ = 12 –36
= - 24
𝑥3 = 0 QD = - 24 t
𝑥3 = 2 QB = - 24 t24
P’ = 36 t
BA
RVS1 = 3 t
P = 6 t
RVA = 12 t RVB = 33 t
2 6 22 2
0≤𝑥1≤2 (A – C)
QX1 = RVA = 12
𝑥1 = 0 QA = 12 t
𝑥1 = 2 QC = 12 t
P’x = ½ qx(
𝑥
𝑙
)
P’x = ½ 12 x(
𝑥
6
)
P’x = 𝑥2
0≤𝑥4≤2 (F – E)
QX1 = RVS1 = 3
𝑥4 = 0 QF = 3 t
𝑥4 = 2 QE = 3 t
0≤𝑥5≤2 (E – B)
QX2 = RVS1 + P = 3 + 6 = 9
𝑥5 = 0 QE = 9 t
𝑥5 = 2 QB = 9 t
3
9
-24
12
y
𝑥2
= 12
Y = 2 + 𝑥2
QX2 = 0
QX2 = 12 – 𝑥2
12 – 𝑥2
= 0
x = 3,46
GAYA LINTANG (Q)

9. NAMA : ANDRE CHANDRA ANALISIS STRUKTUR - MEKANIKA TEKNIK
NIM : 1422201013 DOSEN : IR. VIRGO TRISEP HARIS M.T
GAYA NORMAL (N) Tidak ada Gaya Normal yang terjadi karena tidak ada
gaya yang sejajar sumbu batang.
MOMEN (M)
0≤𝑥1≤2 (A – C)
MX1 = RVA(𝑥1) = 12(𝑥1)
𝑥1 = 0 MA = 12 (0) = 0 tm
𝑥1 = 2 MC = 12 (2) = 24 tm
0≤𝑥2≤6 (C – D)
MX2 = RVA(2 + 𝑥2) – P’x(
1
3
𝑥2 )
MX2 = 12(2 + 𝑥2) – 𝑥2
(
1
3
𝑥2 )
𝑥2 = 0 MC = 12(2+0) – 02
(
1
3
(0))
MC = 24 tm
𝑥2 = 6 MD = 12(2+6) – 62
(
1
3
(6))
MD = 24 tm
𝑥2 = 3,46 𝑀 𝑚𝑎𝑥 = 12(2+3,46) – 3,462
(
1
3
(3,46))
𝑀 𝑚𝑎𝑥 = 51,68 tm
0≤𝑥4≤2 (F – E)
MX4 = - RVS1(𝑥4) = - 3(𝑥4)
𝑥4 = 0 MF = - 3 (0) = 0 tm
𝑥4 = 2 ME = - 3 (2) = - 6 tm
0≤𝑥3≤6 (D – B)
MX3 = RVA(8 + 𝑥2) – P’(2 + 𝑥2)
MX3 = 12(8 + 𝑥2) – 36 (2 + 𝑥2)
𝑥2 = 0 MD = 12(8+0) – 36 (2+0)
MD = 24 tm
𝑥2 = 6 MB = 12(8+2) – 36 (2+2)
MB = -24 tm 0≤𝑥5≤2 (E – B)
MX3 = - RVS1(2 + 𝑥5) – P(𝑥5)
MX3 = - 3(2 + 𝑥5) – 6(𝑥5)
𝑥2 = 0 ME = -3(2+0) – 6(0)
ME = -6 tm
𝑥2 = 2 MB = 12(2+2) – 36 (2)
MB = -24 tm
-24
-6
FB E
DCA
24
51,68
24