Barrier Lift Fixture; If The Support Surface Already Exists; If The Support Surface Does Not Exist; Installing The Rod And The Accessories Provided - Nice M3BAR Instructions And Warnings For Installation And Use

TABLE 4
M3BAR / M5BAR
C B A A B C
M7BAR / L9BAR
B A A B
3 2 1 1 2 3
1) - Add up the numbers in brackets in
the column, choosing only those corre-
sponding to the accessories installed .
2) - Use the sum of these numbers to
identify the holes to which to anchor the
spring .

3.4 - Barrier lift fixture

3.4.1 - If the support surface already exists

01. Open the cabinet of the barrier (fig. 13);
02. Place the barrier on the fixing surface and trace the points where the slots
are to be fixed (fig. 14);
03. Move the barrier and drill the traced surface points; then insert 4 expansion
bolts, not supplied (fig. 15);
04. Position the barrier correctly and secure by means of the relative nuts and
washers not supplied (fig. 16) .

3.4.2 - If the support surface does not exist

01. Dig the foundation pit (*) to house the foundation plate;
02. Prepare ducting for connection cables (fig. 17);
03. On the foundation plate, fix the 4 bolts, placing a nut on the upper side of
each and one on the lower side of the plate (fig. 17) . Caution – The lower
nut must be tightened down to the threaded section;
04. Now cast the concrete, and before it sets, embed the foundation plate,
which must be positioned flush with the surface, parallel to the pole and
perfectly level (fig. 17) . Wait for the concrete to set completely; in general,
at least 2 weeks;
05. Remove the 4 upper nuts of the bolts;
06. Open the cabinet of the barrier (fig. 18);
07. Position the barrier correctly and secure it by means of the relative nuts
and washers supplied with the foundation plate and removed in point 04
(fig. 19) .
(*) Note - The fixing surface must be perfectly smooth and flat . If the surface is
in concrete, it must be at least 0 .15 m thick, and must be adequately reinforced
with steel cages . The concrete volume must be greater than 0 .2 m
ness of 0 .25 m corresponds to 0 .8 m
of approx . 0 .9 m per side) . Anchoring to the concrete can be by means of 4
A
M3BAR
M3BAR
? 2,65 m
XBA15
(3,15m) – 0,50m
A
XBA13
1
XBA13
A
1
XBA4 /
XBA6 / XBA18
B
XBA11
3
B
M5BAR
M5BAR
?
3,50 m
XBA14
(4,15m) – 0,65m
XBA13
(0)
XBA13
(1)
XBA4 /
XBA6 / XBA18
WA13
(1)
(5)
WA12
0 ÷ 1 =
2 ÷ 7 =
(a thick-
3
; in other words equal to a square base
2
3,15 m 3,50 m
XBA15 XBA14
(3,15m) (4,15 m) – 0,65m
A B
3 2
A B
3 2
B C
3 1
4,15 m 5,15 m
XBA14 XBA5
(4,15m) (5,15m)
(0) (0)
(1) (1)
(1) (2)
(4) (4)
B B A
0 ÷ 1 = 0 ÷ 2 = 0 ÷ 2 =
2 3 2
B C A
2 ÷ 4 = 3 ÷ 5 = 3 ÷ 5 =
3 1 2
C A
5 ÷ 6 = 6 ÷ 7 =
2 3
expansion bolts, fitted with 12 MA screws, which resist to a traction load of at
least 400 kg . If the fixing surface is in another material, the consistency must
be checked and ensure that the 4 anchoring points can resist a load of at least
1000 kg . For fixture, use 12 MA screws .

3.5 - Installing the rod and the accessories provided

3.5.1 - Pole support assembly

01. Insert the two plugs in the relative seats on the output motor shaft (fig. 20);
02. Position the support on the output motor shaft, placing it in the "vertical
pole" position and tighten the relative screws and washers fully down to
secure (fig. 21);
03. Position the pole cover and partially secure by means of the 6 screws sup-
plied (fig. 22) .

3.5.2 - Assembling a pole formed of a single piece, whole or cut.

Possible lengths:
M3BAR: 2,65 m = XBA15 (3,15 m -0,50 m = 2,65 m)
3,15 m = XBA15 (3,15 m)
M5BAR: 3,50 m = XBA14 (4,15 m -0,65 m = 3,50 m)
4,15 m = XBA14 (4,15 m)
5,15 m = XBA5 (5,15 m)
M7BAR: 5,15 m = XBA5 (5,15 m)
01. Assemble the two pole insertions (fig. 23);
02. Insert, from the same end of the pole, the insertions just assembled . Use a
rubber mallet (fig. 24);
03. Lightly grease the aluminium guide on both sides (fig. 25);
04. Perform this operation on both ends of the pole: insert the first part of im-
pact protection rubber in the slot, through to the end of the pole; then in-
sert the joint for the impact protection rubber (fig. 26) and repeat with all
parts;
M5BAR
M5BAR M7BAR
M7BAR
4,15 m 5,15 m 7,33 m
XBA14 XBA5 XBA15
(4,15m) (5,15 m) + XBA14
B C B
3 2 2
B C B
3 2 2
C
3
M7BAR
M7BAR
5,00 m 6,33 m 7,33 m
XBA15+XBA15 XBA15 XBA15
(6,30 m) –1,30m + XBA15 + XBA14
(0) (0) (0)
(1) (1) (1)
(1) (1) (2)
(3) (3) (3)
B B
0 ÷ 2 = 0 ÷ 2 =
1 1
B B
3 ÷ 5 = 3 ÷ 4 =
2 2
5 ÷ 6 =
L9BAR
L9BAR
9,33 m
XBA14
+ XBA5
B
1
B
1
L9BAR
L9BAR
8,33 m
XBA14
+ XBA14
(0)
(1)
(2)
(3)
A A
0 ÷ 2 =
1 3
A B
3 ÷ 6 =
2 1
A
3
English – 5