Bidang Gaya Dalam Konstruksi Portal / Pelengkung Tiga Sendi
Hitung
dan gambar bidang Gaya Dalam (M, L, N)
Asumsi
arah reaksi VA dan VB ke atas, HA ke kanan,
dan HB ke kiri
tan
α = 4/3 sin α = 0.8 cos α = 0.6
Σ
MB = 0 → VA (9) -
HA (1) - 1200 (4) - (800 x 6) 3 - 600 (3) = 0
9
VA - HA = 21000
.........................................................................................................(a)
Σ
Ms kiri = 0 →
-(800 x 2) + VA (5) - HA (4) = 0
5VA
- 4 HA = 1600
............................................................................................................(b)
Dari
hasil eliminasi atau substitusi pers. (a) dan (b) diperoleh :
VA
= 2658 kg (↑) dan HA = 2922.5 kg (→)
Σ
MA = 0 → -VB (9) -
HB (1) + (800 x 6)(3+3) + 1200(5) - 600(4) = 0
9
VB + HB = 32400
.........................................................................................................(c)
Σ
Ms kanan = 0 → -VB
(4) + HB (3) + (800 x 4)2 = 0
4VB
- 3 HB = 6400 ............................................................................................................(d)
Dari
hasil eliminasi atau substitusi pers. (c) dan (d) diperoleh :
VB
= 3342 kg (↑) dan HB = 2322.6 kg (←)
Check
:
Σ
V = 0? → VA + VB - 1200 - (800 x 6) = 0 (OK)
Σ
H = 0? → HA - HB - 600 = 0 (OK)
Interval
untuk persamaan gaya dalam ditentukan sebagai berikut
Interval 1 (0 < x1
< 3 m)
Mx1
= VA x1 - HA x1 tan α = 2658x1
- 2922.5 (4/3 x1) = -1238.67 x1
x1
= 0 → Mx1 = 0
x1
= 3 m → Mx1 = -3716.01 kg m ≈ 3716 kg m (∩)
Lx1
= VA cos α - HA sin α = 2658 (3/5) - 2922.5 (4/5) =
-743.2 kg (⇃↾)
Nx1
= -VA sin α - HA cos α = -2650 (4/5) - 2922.5 (3/5) =
-3879.9 kg (tekan)
Interval 2 (0 < x2
< 2 m)
Mx2
= VA (3+x2) -
HA (4) - q x2 (0.5 x2) = 2658(3+x2)
- 2922.5 (4) - 400 x22
Mx2
= -3716 + 2658 x2 - 400
x22
x2
= 0 → Mx2 = -3716 kg m (∩)
x2
= 1 m → Mx2 = -1450 kg m (∩)
x2
= 2 m → Mx2 = 0 (momen lentur pada titik sendi dalam harus nol)
Lx2
= VA -q x2 = 2658-800x2
x2
= 0 → Lx2 = 2658 kg (↿⇂)
x2
= 2 m → Lx2 = 1058 kg (↿⇂)
Nx2
= -HA = 2922.5 kg (tekan)
Interval 3 (0 < x3
< 3 m)
Mx3
= -HB (x3) = -2322.6 x3
x3
= 0 → Mx3 = 0
x3
= 3 m → Mx3 = -6967.8 kg m
(∩)
Lx3
= +HB = 2322.6 kg (↿⇂)
Nx3
= -VB = -3342 kg (tekan)
Interval 4 (0 < x4
< 4 m)
Mx4
= VB x4 - HB
(3) - q x4 (0.5 x4)
Mx4
= -400 x42 +
3342 x4 - 6967.8
x4
= 0 → Mx4 = -6967.8 kg m (∩)
x4
= 2 m → Mx4 = -1883.8 kg m (∩)
x4
= 4 m → Mx4 = 0 (momen lentur pada titik sendi dalam harus nol)
Lx4
= -VB + q x4 =
-3342 + 800x4
x4
= 0 → Lx4 = -3342 kg (⇃↾)
x4
= 4 m → Lx4 = -142 kg (⇃↾)
Nx4
= -HB - 600 = -2922.6 kg (tekan)
Bidang
Gaya Dalam (M, L, N) *tidak berskala
Bidang Momen Lentur
Bidang Gaya Lintang
Bidang Gaya Normal