Welded Connections.
Example Nr 1. (Checking problem).
In the beam to column connection shown in the figure, the steel plate is supporting a support factored reaction of
525 kN from the beam. If the size of the weld is 8 mm and steel Grade Fe 360, check if the connection is adequate.
Geometry, materials and loading.
Plate t = 16 mm. Fe 360, fy = 235 N/mm2 and fu = 360 N/mm2.
Size of the welds s = 8 mm.
Throat thickness: a = 0.707 s = 0.707 x 8 = 5.66 mm.
Length of the weld l = 2 x 295 + 260 – 4 x 8 = 818 mm.
1). Check the dimension of the chords:
Minimum weld length l = 40 mm or 6 x a = 6 x 5.66 = 34 mm < 260 mm. OK.
Maximum weld length l = 150 a = 150 x 5.66 = 849 mm > 295 OK.
2). The design strength per unit length is Fw, Rd = f vw, d a
0.63 f ye 0.65 fu
where f vw, d = ≤
λMw γ Mw
In most practical cases, the quality of the electrodes are greater than that of the base metal and then the ultimate
tensile strength of the weaker part joined, that is, the base metal govern the design. Therefore the second statement
of the equation is checked.
0.65 x360
Fw, Rd = x5.66 = 1059.55 N / mm
1.25
−3
And the total resistance force F = 1059.55 N / mmx818 mmx10 = 866 kN > 525 kN OK .
Example Nr 2. (Design problem).
Special case (Eccentrically loaded joint. Unsymmetrical section).
When securing an unsymmetrical section, for example two angles to a plate, attention is paid to uneven distribution
of the load between the welds transmitting the force field from the angles to the plate.
The Force F is discomposed into F1 and F2
Taken moment with respect to point o.
2 2 F
F b = F1b ⇒ F1 = F ; Thus F2 =
3 3 3
therefore; F1 is taken by two chord (filled weld) length L1each and F2 is taken by the lower two fillet weld length L2
each.
Example: Compute the welds required for connecting two angles 75x75x8 mm to a gusset plate with a thickness 10
mm. The factored tensile force in the angle is 450 kN. The material is steel Grade 430.
Geometry, Materials.
Plate and angles Grade 430 steel, fy = 275 N/mm2 ; fu = 430 N/mm2
Size of the welds ≤ 8 mm; take s = 6 mm.
Throat thickness a = 0.707 x 6 = 4.24 mm.
1). Acting Force F1 and F2.
2 2
F1 = F1 = x 450 kN = 300 kN
3 3
450
F2 = = 150 kN
3
2). Design strength per unit length. (Suppose the weaker part is the base steel).
0.65 x 430
Fw, Rd = x 4.24 = 948 N / mm.
1.25
3). Length of fillets:
F1 300 x10 3 N
L1 = = = 158.2 mm
2 ( fillet welds ) Fw, Rd 2 x948 N / mm
Actual length if no round a corner L1 = 158.2 mm + 2 s = 158.2 + 2 x 6 = 170 mm each side.
And L2 = 158.2 + 2 x6 = 91.1 mm (take 92 mm)
4). Check the dimension of the fillets.
6 x 4.24 < 92 and 170 < 150 x 4.24 OK
- Joints in beams under the action of bending moment and shear force.
Suppose a beam such that:
Principle: The flanges take the acting bending moment and the web takes shear force.
M is discomposed into a couple of forces F.
F = M/h acting on the flange levels.
284 kN − m
Then F= = 532 kN .
0.533 m
Design of cover plates.
Af y
The design plastic resistance of the gross section N pl , Rd = = 532 kN
γ Mo
γ Mo x532 x103 1.1x532 x103
A = 180tc , p = = = 2128 mm 2
fy 275
2128 mm 2
tc , p = = 11.82 mm take 12 mm.
180 mm
- suppose size of the weld 8 mm < 12 mm.
- throat thickness a = 0.707 x8 mm = 5.66 mm
0.65 x 430
- strength per unit length Fw, Rd = x5.66 = 1266 N / mm
1.25
F 532 x103 N
The length of the filled weld L1 = = = 210 mm each sides + (round 2 x6 mm)
2 Fw, Rd 2 x1266 N / mm
Design of the central plate for shear.
0.65 x 430
Strength per unit length Fw, Rd = x 4.24 = 948 N / mm
1.25
−3
The total resistance force = 948 N / mm x 350 mm x 10 = 331.8 kN > 142 kN OK .
