Equipment for Rational Securing of Cargo on Railway ... - Vinnova

MariTerm AB 2004-01-30 jvgRASLA
a t
a v
Force
Acceleration
nF
nF
α
mg
F f
R
Equilibrium <strong>of</strong> <strong>for</strong>ces:
↑
→
R − m ⋅ g − 2 ⋅ n ⋅ F ⋅sinα
= m ⋅ a
F
f
= m ⋅ a
Force <strong>of</strong> friction:
F
f
t
⇒
a
t
=
F
f
m
v
⇒
[ m ⋅ ( a + g)
+ 2 ⋅ n ⋅ ⋅ α ]
= µ ⋅ R = µ ⋅
F sin
v
R = m ⋅
( a + g)
v
+ 2 ⋅ n ⋅ F ⋅sinα
Transverse acceleration as function <strong>of</strong> the pretension in the lashing F:
a
t
=
F
f
m
= µ ⋅
( a + g)
v
µ ⋅ 2 ⋅ n ⋅ F ⋅ sinα
+
m
n = Number <strong>of</strong> lashings
Example:
Sliding <strong>of</strong> top part in transverse direction:
The mass <strong>of</strong> the top part <strong>of</strong> the cargo is assumed to be 30 ton (30000 kg) and it is secured by 2
lashings with an angle (α) <strong>of</strong> 45°. The coefficient <strong>of</strong> friction is 0.3 and the vertical
acceleration acting at the time is -3 m/s 2 . According to UIC section 5.5.4, the breaking
strength <strong>of</strong> the lashings is assumed to be 40 kN, and the pretension <strong>for</strong>ce F, thus 4 kN (4000
N). These precautions give the following equation:
a
µ ⋅ 2⋅
n ⋅ F ⋅sinα
m
0.3⋅
2⋅
2⋅
4000⋅sin 45
30000
o
t
= µ ⋅
v
=
2
( a + g) +
= 0.3⋅( − 3 + 9.81) +
2.2 m / s
87