06-Strength of Double Angle Welded Tension Members (Steel Structural Design & Prof. Shehab Mourad)

1. 1 Prepared by Prof. 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 by Prof. 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 by Prof. 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