This worksheet is for checking weld sizes References: Pressure Vessel Handbook, 8th ed, Megyesy Mechanical Engineering Design, 7th ed, Shigley By: WBL Date: 8/4/2014 Job: 10 Description of weld: Vaporizer skid cross beam to main welds
WELD OUTLINE CASE
1
2
DEFINITION OF SYMBOLS Aw = Length of weld, in f = Allowable unit force on weld, kip per lin in of weld M = Bending moment, kip-in l = Bending moment arm length, in P = Tension or Compression Load Sw = Section Modulus of weld lines subjected to bending moment, in^2
3
V = Vertical load for bending and/or shear, kip w = Fillet weld leg dimension, in Wn = Tensive or compressive force on weld, kip per lineal inch of weld Ws = Average vertical shear on fillet weld, kip per lineal inch of weld
4
Wb = Bending force on weld, kip per lineal inch of weld Wr = Resultant load on fillet weld, kip per lineal inch of weld FORMULAS FOR FORCES ON WELD TENSION OR COMPRESSION VERTICAL SHEAR
BENDING 5
Wn =
P Aw
V Aw
Ws =
Wb =
M Sw
RESULTANT FORCE: Wr = √(W + Ws + Wb ) 2 n
2
2
Weld is adequate & stress ratio = 0.51 INPUT Weld Outline Case: 5 Strength Level of Weld Metal:
6
* To 1/4 incl.
70*
Loading condition for bending, ie…Cantilever Mid-Span or Mid-Span of Simply Supported Beam w= 3/8 in P= 12.50 kip If no tension or compression is present, P = 0 V= 12.50 kip If only tension or compression is present, V = 0 b= 4 in Weld dimension; refer to figure on right d= 10 in Weld dimension; refer to figure on right l= 48 in If bending is not to be considered, l = 0 Aw = 18.00 in Sw =
56.67 in^2 M= 150 lb-in f= 5.57 kip/lin in Calculating Tensive or Compressive Force Wn = 0.69 kip/lin in
Over 1 1/2 Over 2 1/4 7
8
Calculating Shear Force Ws = 0.69 kip/lin in
=P/Aw
=V/Aw
=12.5kip/18lin in
=12.5kip/18lin in
Calculating Bending Force Wb = 2.65 kip/lin in =M/Sw =150kip-in/56.67in^2
Calculating Resultant Force Wr = 2.82 kip/lin in =SQRT(Wn^2 + Ws^2 + Wb^2) =SQRT(0.69^2 + 0.69^2 + 2.65^2)
WELDED BEAM SPLICE DESIGN 1. DESIGN CRITERIA: Section Properties: UC152X152X30 d = 6.20 in.
= 157.6 mm
tf = 0.37 in.
= 9.4 mm
tw = 0.26 in.
= 6.5 mm
bf = 6.02 in.
= 152.9 mm
T = 4.87 in.
= 123.6 mm
Design Standard: ASTM A 992 ASTM A36 Fy = 40 ksi
= 275 N/mm²
Welding Strength: E70XX
FIGURE - 1 2. CALCULATION: a. Flange Force: Pu = 53.28 kips
= 237 kN
Pu = f b M p * d
fbMp = 103 ft kips Tension on gross plate area:
fPn = f Ag Fy f = 0.90 Ag = 1.48 in2 Plate Cross Sectional Area: Top Flange:
Bf= 11.81 in.
= 300 mm
Lf = 6.30 in. tf = 0.39 in.
Bw = 3.94 in.
= 100 mm
= 160 mm
Lw = 6.30 in.
= 160 mm
= 10 mm
tw = 0.39 in.
= 10 mm
Af = 2.48 in²
Aw = 2.48 in²
n= 2 b = 6.02 in.
joint gap, g = 0.39 in. L > b, safe!
TABLE 10.2 Maximum and Minimum fillet weld sizes Maximum Fillet Weld Size Connected part thickness, t1
Maximum Weld Size, w
t < 1/4
w=t
t > 1/4
w = t - 1/16 Minimum Fillet Weld Size2
1
Web Plate:
t < 1/4
1/8
1/4 < t < 1/2
3/16
1/2 < t < 3/4
4/16
t > 3/4
5/16
The term is the thickness of the thicker connected part.
= 10 mm
2
Single-pass welds must be used. The maximum weld size that can be made in a single pass is 5/16 in.
From Table 10.2: Minimum Weld size = 1/4 in. Dmin = t < 4/1
= tf - w
Maximum Weld size = 0.308 in. Threfore, Dmax = 5
fRn = fRwl + fRwt
From Equation 10.7:
Since: fRn = Puf f = 0.85 From Eq. 10.5: f Rn = 1.392DL t 53.28 = 1.392DL t + L = 5 in
1.392DL t
min. Required Weld length
Lt = 2 in. =
b
127 mm
b. Shear Strength:
FIGURE - 2 fVn = 13.76 kips Since, T = 4.87 in.
= 102 kN
60% shear capacity considered.
maximum depth of shear plate.
fvVn = fv0 .6 Fy Aw f v = 1.0 Aw = T* tp
tp = plate thickness
tp = 1/7 in For practical reasons, a plate has a thickness equal to or greater than the web thickness should be used.
use, tp = 3/8 in. From Table 10.2: Minimum Weld size = 3/16 in. Dmin = 3 Maximum Weld size = t - 1/16 in.
since: t = tw
Threfore, Dmax = 0.193 in.
Interpolation, k
therefore, Dmax = 3 b = 2.95 in.
refer from Figure-2.
0.70
2.16
1.85
0.75
k1
k2
0.80
2.43
2.08
d = 3.94 in. Interpolation, C
CG = b²/(2b + d)
Refer from Table 10-3
CG = 0.89 in.
1
eccentricity, e = 4.63 in. l = 3.94 in.
2
a
k
1.00
2.3
1.177
C
1.20
1.97
From AISC Manual, Table 8-8: a = 1.177 in.
a=e/l
f = 0.75
k = 0.75
k=b/l
C = 2.26 C1 = 1.0
Dmin = 2.062
<
3
Ok!
by interpolation from Table 8-8 AISC electrode strength coefficient from Table 8-3 AISC
Note the the final design detail includes a 3/8 in. gap between the ends of the beam added for construction tolerance.
