1. Analisis Struktur I
Reaksi Perletakan
∑Ms2 Kanan = 0 - VE∙30 - VD∙10 + Q2∙30∙15 = 0
- 30∙VE - 10∙VD + 450 = 0
- 30∙VE - 10∙VD = -450 Pers 1
∑Ms1 Kanan = 0 - VE∙50 - VD∙30 + Q2∙30∙15 + P2∙12 = 0
- 50∙VE - 30∙VD + 1050 + 600 = 0
- 50∙VE - 30∙VD = -1650 Pers 2
Substitusi Pers 1 & 2
- 30∙VE - 10∙VD = -450 3 - 90 VE - 30VD = -1350
- 50∙VE - 30∙VD = -1650 1 - 50 VE - 30VD = -1650 _
- 40VE = 300
VE = -7.7 Ton ( )
- 30∙VE - 10∙VD = -450
- 30∙(-7.7) - 10∙VD = -450
225 - 10∙VD = -450
- 10∙VD = -675
VD = 67.5 Ton ( )
∑MA = 0 VE∙78 - VD∙58 - VB∙20 + Q1∙20∙10 + P1∙24 + P2∙40 + Q2∙30∙63 = 0
(7.5∙78) – (67.5∙58) – 20VB + 200 + 240 + 2000 + 1890 = 0
- 20VB = -1000
VB = 50 Ton ( )
∑V = 0 VA + VB + VD – VE - Q1∙20 - P1 - P2 - Q2∙30 = 0
VA + 50 + 67.5 – 7.5 – 20 – 10 – 50 – 30 = 0
VA = 0
∑H = 0 HA = 0
Kontrol ke titk C
∑Mc = 0 VE∙30 - VD∙10 + VB∙28 + VA∙48 - Q1∙20∙38 - P1∙24 - P2∙8 + Q2∙30∙15 = 0
7.5∙30 – 67.5∙10 + 50∙28 + 0∙48 – 760 – 240 – 400 + 450 = 0
0 = 0
OKEE

2. Analisis Struktur I
Gaya Dalam
Balok A-B interval 0 ≤ x ≤ 20
Mx = VA∙x - q∙x∙1
/2x
= - q∙x∙1
/2x
= - x2
/2
X = 0 Mx= 0 X = 10 Mx= -50
X = 2 Mx= -2 X = 12 Mx= -72
X = 4 Mx= -8 X = 14 Mx= -92
X = 6 Mx= -18 X = 16 Mx= -128
X = 8 Mx= -32 X = 18 Mx= -162
X = 20 Mx= -200
Lx = VA - q∙x
= - q∙x X = 0 Lx= 0
= - x X = 20 Lx= -20
Balok B-S1 Interval 0 ≤ x ≤ 4
Mx = VA∙(20+x) + VB∙x – q∙20∙(10+x)
= VB∙x - 20∙(10+x)
= 50∙x – 200+20∙x
X = 0 Mx= -200
X = 4 Mx= -80
Lx = VB - q∙20
= 50 – 20
= 30
Balok B-S1 (P-S1) Interval 0≤ x ≤4
Mx = VA∙(24+x) + VB∙(4+x) – q∙20∙(14+x) - P∙x
= VB∙(4+x) - 20∙(14+x) - P∙x
= 50∙(4+x) – 280+20∙x - 10∙x
= 200+50∙x - 280+20∙x - 10∙x
X = 0 Mx= -80
X = 4 Mx= 0
Lx = VB - q∙20 – P
= 50 – 20 – 10
= 20
Balok S1-S2 (S1-P) Interval 0≤ x ≤12
Mx = VA∙(28+x) + VB∙(8+x) - q∙20∙(18+x) –
P1∙(4+x)
= 50∙(8+x) - 20∙(18+x) - 10∙(4+x)
= 400+50∙x -360+20∙x – 40+10∙x
X = 0 Mx= 0
X = 12 Mx= 240

3. Analisis Struktur I
Lx = VB - q∙20 – P1
= 50 – 20 – 10
= 20
Balok S1-S2 (P-S2) Interval 0≤ x ≤8
Mx = VA∙(40+x) + VB∙(20+x) - q∙20∙(30+x) – P1∙(16+x) – P2∙x
= 50∙(20+x) - 20∙(30+x) - 10∙(16+x) - 50∙x
= 1000+50∙x – 600+20∙x – 160+10∙x - 50∙x
X = 0 Mx= 240
X = 8 Mx= 0
Lx = VB - q∙20 – P1 – P2
= 50 – 20 – 10 - 50
= -30
Balok E-D Interval 0≤ x ≤20
Mx = - VE∙x - q∙x∙1
/2x
= - 7.5∙x – x2
/ 2
X = 0 Mx= 0 X = 10 Mx= -125
X = 2 Mx= -17 X = 12 Mx= -162
X = 4 Mx= -38 X = 14 Mx= -203
X = 6 Mx= -63 X = 16 Mx= -248
X = 8 Mx= -92 X = 18 Mx= -297
X = 20 Mx= -350
Lx = - VE - q∙x
X = 0 Lx= -7.5
X = 20 Lx= -27.5
Lx = 0
-7.5= q∙x
-7.5= x x = 7.5
Mmax = - 7.5∙x – x2
/ 2
= -84.375

4. Analisis Struktur I
Balok D-C Interval 0≤ x ≤10
Mx = - VE∙(20+x) + VD∙x - q∙(20+x) ∙1
/2(20+x)
= -7.5∙(20+x) + 67.5∙x - 1∙(20+x) ∙1
/2(20+x)
= - 150+7.5∙x + 67.5∙x – 20+x∙10+x
/2
= -350 + 40x – x2
/2
X = 0 Mx = -350 X = 6 Mx = -128
X = 2 Mx = -272 X = 8 Mx = -62
X = 4 Mx = -198 X = 10 Mx = 0
Lx = -VE + VD - q∙(20+x)
= -7.5 + 67.5 - 1∙(20+x)
= -7.5 + 67.5 – 20+x
X = 0 Lx = 40
X = 10 Lx = 30
Lx = 0
-20+x = 7.5 - 67.5
-20x = - 60
X = 3
Mmax = -234.5

5. Analisis Struktur I
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