DRC D 1000LF, Operation Manual

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Example 2
Butt X-weld, thickness T = 60 mm
Groove Angle on the top = 30°
Groove Angle on the bottom = 60°
Groove Bevel = 15°
Weld Gap = 3 mm (5 % of T, i.e. 2.5 % on both sides of the weld axis)
Depth of Groove = 50 mm, i.e. 83 %
15°
60 (T)
+X
Y
-X
0
1
0.247
1 -1
0.025 -0.025
X1; Y1
X2; Y2
X’1; Y’1
X3; Y3
30°
X’3; Y’3
X’2; Y’2
0.83
3
50
0.123 -0.123
-0.247
This weld geometry will be described by 4 segments:
1
st
segment – X1 Y1 X2 Y2:
X1 = (Weld Gap)/2 + T*(Depth of Groove)*tg15° = 0.0 25 + 1*0.83*0.268 = 0.247;
Y1 = 0 [weld top];
X2 = (Weld Gap)/2 = T*2,5% = 1*0.025 = 0.025;
Y2 = T*(Depth of Groove) = 1*83% = 0.83.
2
nd
segment – X2 Y2 X3 Y3:
X2 = (Weld Gap)/2 = T*2,5% = 1*0.025 = 0.025;
Y2 = T*(Depth of Groove) = 1*83% = 0.83;
X3 = (Weld Gap)/2 + [T – T*(Depth of Groove)]*tg30° = 0.025 + (1 – 0.83)*0.577 = 0.123;
Y3 = T = 1.
3
rd
segment – X’1 Y’1 X’2 Y’2:
X’1 = -(Weld Gap)/2 - T*(Depth of Groove)*tg15° = - 0.025 - 1*0.83*0.268 = -0.247;
Y’1 = 0 [weld top];
X’2 = -(Weld Gap)/2 = -T*2,5% = -1*0.025 = -0.025;
Y’2 = T*(Depth of Groove) = 1*83% = 0.83.
2
nd
segment – X2 Y2 X3 Y3:
X’2 = -(Weld Gap)/2 = -T*2,5% = -1*0.025 = -0.025;
Y’2 = T*(Depth of Groove) = 1*83% = 0.83.
X’3 = -(Weld Gap)/2 - [T – T*(Depth of Groove)]*tg30° = -0.025 - (1 – 0.83)*0.577 = -0.123;
Y’3 = T = 1.
The resulting section configuration will be as follows:
[Weld 2 Name]
0.247
0 0.025 0,83
0.025 0,83 0.195 1
-0.247 0 -0.025 0.83
-0.025 0.83 -0.195 1