#1

WELDING

Objective: At the end of the course, student must be able to do force and stress analysis on the group of welding and must also be able to decide the size of the welding under static loading.

Scope of the analysis Type welding: fillet

Terminologies Leg size: h Throat size (without reinforcement): t

thro

at

leg s

ize h

Good to have reinforcement, however it may create stress

concentration; therefore, grinding will create smoother surface..

Size is determined by leg size, analysis is using the throat size (detail discussion in Section 9-2). t = 0.707h

#2

Standard

#3

#4

FORCE AND STRESS ANALYSIS

Tensile Load (F)

A

F'

Shear Load (V)

C

V

’

A

V'

#5

Bending (M)

C ’’ M

FRONT SIDE

I

Mc"

Torsion (T)

C C

T

A

rA

rB B

rD

D

D"

A"

B"

J

Tr"

#6

CENTROID Each welding strip can be regarded as rectangular shape with length x throat ( A = dt)

n

nn

AAA

xAxAxAX

21

2211

n

nn

AAA

yAyAyAY

21

2211

Example 2

)0( b

AA

AbAX

2

2)

2(

d

AA

dA

dA

Y

AREA

Area = total length of the strip x throat length A = 2dt

= 2d(0.707h) = 1.414d h

C c2 c1

X

Y

d

b

d

t

#7

MOMENT OF AREA

3

12

1baI for rectangular b (base) x a (height)

C c2 c1

X

Y

d

b

For each strip

Rectangular d x t (d: height and t : base) 3

12

1tdI

Strip 1 3

12

1tdI I

Strip 2 3

12

1tdI II

Total IT

)707.0(6

)707.0(6

1

6

1

12

1

12

1

3

3

3

33

hIId

ILet

dh

td

tdtd

III

Uu

IIIT

IU= unit moment of area (Table 9-2 pp 488)

*Be careful with the orientation of Moment Axis (in the table, it is horizontal)

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POLAR MOMENT OF AREA

)(12

1 22 baabJ for rectangular b (base) x a (height)

C c2 c1

X

Y

d

b

For each strip

Rectangular d x t (d: height and t : base) )(12

1 22 dttdJ

For strip I )(12

1 22 dttdJ I @ c1

For strip II )(12

1 22 dttdJ II @ c2

Total J @ C Parallel Axis Theorem

Strip 1

)3(12

1

)4

(12

1

0

)4

()(12

1

22

23

2

222

2

bdtd

bdttd

t

bdtdttd

AdJJ IIC

#9

Total JT

U

U

u

ICICT

hJ

tJ

bddJLet

bdtd

bdtd

JJJ

707.0

)3(6

1

)3(6

1

)3(12

12

22

22

22

JU= unit moment of area (Table 9-1 pp 466)

Example: Calculate the primary and secondary forces for the following examples by replacing the threaded joints with welding

2515

a

b

c

d

1

2

3

4

5

P

e

l1

l1

l3

l2

l4

l5

10

l1 = 70 mm

l2 = 180mm

l3 = 60mm

l4 = 80mm

l5 = 300mm

P = 5kN

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#11

Procedure to analyze welding

1. FBB transfer all the force to its centroid

2. Classify the force on the centroid Primary

Direct shear load (V): A

V'

Axial load (F): A

F'

Secondary

Bending (M): I

Mc"

Torsion (T): J

Tr"

3. Determine the most critical point on the weld Use engineering judgment to select the most critical point on the welding. Primary tensile and shear has no effect on this selection as they are equally distributed. Moment and torsion are the key factor on this selection.

4. Calculate the total tensile and shear stresses

(*summation of vector)

"' T

"' T

For total shear, the total shear can be calculated using Cosines Rules or summation of force in X and Y axis

Cosines Rule

cos"'2"' 22 T

Summation of force

22

"'

"'

YTXTT

YYYT

XXXT

#12

5. Combine both tensile and shear using DET

22

TTe

6. Compare the induced stresses with the allowable shear stress

e

sy

SF

S

.

Ssy: Shear Yield strength Ssy = 0.577 Sy (Sy from Table 9-3 pp 472) Ssy = 0.3 Sut ( AISC Code)

h is always round number.

h min is 3 mm