Welded Connection

Welded Connection Welding – process of joining metal by heating the metal to state of fusion permitting it to flow together into a solid joint. Gas Welding – type of welding which utilizes the heat of the flame which is produced by the combustion of a gas. The most commonly used are acetylene, hydrogen, and natural gas in combination with oxygen. Acetylene welding is widely used in welding thin plates and in welding gas, steam and hydraulic pipelines. Electric Arc Welding – type of welding in which heat is supplied by a continuous arc drawn between two electrodes, the work forms one electrode and the welding rod forms the other. Shielded arc welding uses coated welding rods to prevent oxidation of the metal. Thermit Welding – type of welding in which the weld metal is essentially cast steel fused in the parts welded. This process is principally used in repairing heavy machine parts and in building up defective casting. Atomic-Hydrogen and Helium Arc Welding – type of welding in which a jet of hydrogen of helium is forced through the arc drawn between two tungsten electrodes to prevent oxidation of the metal. Electric Resistance Welding – type of welding, requiring both heat and pressure, in which the parts to be welded are brought into contact and a heavy current at low voltage is passed through the junction which causes the metal to fuse. Example of electric resistance welding are spot welding, butt and flash welding, seam, projection and upset welding.

Types of Welds and Welded Joints:

Common welding joint types – (1) Square butt joint, (2) Single-V preparation joint, (3) Lap joint, (4) T-join

Butt Welds

Fillet Welds

Strength of Butt Welds: F = St t L

where: F = load St = tensile stress t = plate thickness L = length of weld

Strength of Fillet Welds: F = 2 Ss L (b cos 45O) where: F = load Ss = shearing stress L = length of weld b = leg dimension of weld which is the size of the fillet weld b cos 45O = throat thickness

Other Methods of Joining Metals: Soldering – method of joining metal by using an alloy of lead and tin (called the solder) applied between the two pieces in a molten state. Brazing – method of joining metal using a non ferrous filler (copper alloy) which is melted and applied to the pieces being joined.

Sample Problems in Welded Connection: Problem no. 1 A 16mm plate is lapped over and secured by means of fillet weld on the inside and outside to form a penstock 1.6 m in diameter. Determine the safe internal pressure assuming an allowable stress of 138 Mpa on the allowable stress of 90 Mpa on the throat side of the 14 mm weld.

Given: tp = 16mm D = 1.6 m St = 138 Mpa (on the plate) St = 90 Mpa (throat side) b = 14mm Solution: ∑Fv = 0] 2P = F eq. 1 Plate is subjected to tensile stress St = P/A

where: A = tpL thus; P = St * tpL eq. 2 but: F = ρ Ap = ρDL eq.3 Combining 1, 2 & 3 2 St* tpL = ρDL ρ = 2St * tp / D eq. 4 Where: St = 138 Mpa (considering plate side) tp = 16mm D = 1.6 m Substituting to equation 4 ρ = 138 N/m2 (16mm) / [1.6 m(1000mm/1 m)] ρ = 276*104 Pa

Considering Welded Side: Note: two ends are to be welded as shown in the figure. St = P/2A eq. 5 where: A = 2t * L thus; P = St (2t * L) eq. 6 since: t = 0.707 b eq.7 Combining 1, 5, 6 & 7 2 (St)(2*0.707b*L) = ρDL ρ = 4St (0.707b) / D where: St = 90 Mpa (welded side) b = 14mm D = 1.6 m thus: ρ = 4(90*106N/m2)(0.707*14mm) / 1.6 m ρ = 222.7*104 N/m2 Therefore: safe internal pressure is the smaller, so ρ = 222.7*104 N/m2 or Pa

Problem no. 2 For the welded connection shown a.) recommend the plate thickness to be used if maximum bending stress is not to exceed 90 Mpa. b.) Determine the maximum stress on the 10mm fillet weld. Given: Sf = 90 Mpa Throat (b) = 10mm Solution: Sf = Mmax*C / I eq. 1 where: Mmax = PL = 35,000(200) = 7*106 N-mm C = 200/2 = 100mm Sf = 90 Mpa I = tph3/12 = tp(200)3/12 Substituting to Eq. 1

90 N/mm2 = 7*103 N-mm(100mm) / tp(200)3 mm3/ 12 tp = 11.66 ; say use 12mm Weld Side: Primary Shear Stress S1 = P/As = P/tL = P/0.707 bL eq.2 = 35,000 / [(0.707)(10)(200)]*2 = 12.37 Mpa Secondary Torsional Shear Stress S2 = Tρ / J eq.3 where: ρ = [(75)2 + (100)2] ½ = 125mm T = P*d = 35,000 (275) = 9.625*106 N-m J1 = tL3/12 + t l d2 = 0.707(10)(200)3/12 + 0.707(10)(75)2(200) = 1.2667083*107 mm4 J1 = J2 = 1.2667083*107 mm4

Then from Equation 3: S2 = 35,000(275)(125) / 2(1.2667083*107) = 47.5 Mpa thus; Maximum Stress will be: Ŧ2 = S12 + S22 + 2 S1S2 cos θ = (12.37)2 + (47.5)2 + 2 (12.37)(47.5)(75/125) = 55.79 Mpa