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06-Strength of Double Angle Welded Tension Members (Steel Structural Design & Prof. Shehab Mourad)
1. The document discusses the strength of double angle welded tension members. It describes three failure modes: yielding of the gross area, fracture of the effective area, and block shear rupture in the angle. 2. Formulas are provided to calculate the nominal resistance (Rn) for each failure mode based on the yield strength, ultimate tensile strength, and dimensions of the angles. 3. An example problem is worked through to determine the governing strength of a specific welded double angle configuration based on the provided dimensions and material properties. The yielding of the gross area controls with a nominal resistance of 459 kN.
06-Strength of Double Angle Welded Tension Members (Steel Structural Design & Prof. Shehab Mourad)
- 1. 1 Prepared byProf. Shehab Mourad – Department of Civil Eng. - KSU Strength of Double Angle Welded Tension Members Fig. Welded tension member 1- Yielding of gross area Ag 2- Fracture of effective area Ae a) If only transverse weld is used A = area connection element U = 1.0 (a) b) If only longitudinal weld are used A = Ag (of the member) if - Lc/w > 2.0 , U = 1.0 if - 2 > Lc/w > 1.5 , U = 0.87 if - 1.5 > Lc/w > 1.0 , U = 0.75 (b) c) Transverse & longitudinal weld used A = Ag (of the member) U = ( 1- x / Lc ) < 0.9 x = c.g along horizontal leg (c) Fig . Different welded layouts w Lc w b t h hg tg section (1-1) Pu Pu 1 1 L2 L1 Lg L3 Longitudinal weldsTransversal weld Ø Rn = 0.9 * Ag * Fy Ø Rn = 0.75 * Ae * Fu L w Ag = gross area of angles Ae = A * U
- 2. 2 Prepared byProf. Shehab Mourad – Department of Civil Eng. - KSU 3) Block shear rupture in angle Fig. Block shear failure of welded angle Lv = L1 , Lt = h - tangle Av = 2 * tangle * Lv , At = 2 * tangle * Lt At = gross area of tensile plane Av = gross area of shear plane Fu = Ultimate Tensile Strength of angles Fy = Yield Tensile Strength of angles Lt Lv If 0.6 * Fu * Av > Fu * At Ø Rn = 0.75 *(0.6 * Fu * Av + Fy * At) If 0.6 * Fu * Av < Fu * At Ø Rn = 0.75 *( Fu * At + 0.6 * Fy * Av)
- 3. 3 Prepared byProf. Shehab Mourad – Department of Civil Eng. - KSU Example Determine the factor tensile resistance of the given welded double angles Fig. Welded tension member for example Given : Fy = 250 MPa Fu = 400 MPa Solution 1- Yielding of Ag. Ag = 2 * 1020 = 2040 mm2 Ø Rn = 0.9 * Fy * (2*Ag) = 0.9 * 250 * 2040 / 1000 = 459 kN 2- Fracture on Ae. Ae = Ag*U U = ( 1 – x / Lc) < 0.9 Lc = L1 = 114 mm U = ( 1 – 19.8 / 114) = 0.826 < 0.90 Ø Rn = 0.75 * Ae * Fu = 0.75 * (2040*0.826) *400 / 1000 = 505.7 kN 3- Block shear rupture. Therefore the governing strength of welded angles is given by yielding of gross area = 459 kN t b h hg tg section (1-1) Lv Lt 2L 89 x 76 x 6.4 mm for single angle Ag = 1020 mm2 x' = 19.8 mm y' = 26.2 mm Pu Pu 1 1 L2 L1 Lg L3 Longitudinal weldsTransversal weld L1 = 114 mm L2 = 38 mm L3 = 89 mm Lv = L1 = 114 mm Av = Lv * t = 114 * 2 * 6.4 = 1459.2 mm2 Lt = leg – t = 89 – 6.4 = 82.6 mm At = 82.6 * 2 * 6.4 = 1057.28 mm2 Fu*At = 400 * 1057.28 = 422912 N 0.6* Fu *Av = 0.6 * 400 * 1459.2 = 350208 N < Fu*At Ø Rn = 0.75*(422912+ 0.6*250*1459.2)/1000 = 481.3 kN