#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)
#8
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
#10
#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