Welding Splice Design.xls f646r

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 ed 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.

t 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- welds must be used. The maximum weld size that can be made in a single 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.