CALCS PROC067
Fillet Weld Strength
170527

CONTENTS

Guidance on Application
Theoretical Basis
Notation
Input Data Format
Project
Project Input
Project Output
PROC067 Theory

GUIDANCE ON APPLICATION

The procedure is applicable to strength verification of fillet welds in accordance with the requirements of recognised national design standards

Line forces for input the procedure are to be obtained from separate calculation.

The only geometric parameter used is weld leg length

The procedure is limited to joined plates being relatively orientated at 90deg.

The weld face is orientated at 45deg to each plate.

See also: BAS008

INPUT TO PROCEDURE

LET Ks=:KvM=:REM Design acceptance criteria
Design acceptance criteria are shear strength and von Mises equivalent stress of the weld filler material. The shear factor Ks=1/2, used in the project example, is taken from AISC ASD (omega=2) for fillet weld strength checks. The von Mises acceptance factor (KvM=1) is consistent with BS5400 & IIW-Commission XV. AISC does not consider von Mises stress for welds.

LET F_EXX=:Fy=:REM Weld characteristic strengths used in the procedure
AISC advises the weld classificaton stress may be taken as the UTS of the joined material. BS5400 & BS5950 advise the weld yield stress may be taken to be that of the joined material.

PROC067:REM This statement calls the procedure
The row containing data used to calculate the weld stresses shows non-zero default line forces. This is to avoid division by zero and they should be set to actual working loads (less than or greater than zero) for the case under consideration. Set plt=0 where output plots are not required.

THEORETICAL BASIS

The response of welds to applied loads is complex for reasons including the processes used in formation, material characteristics and stress distribution. This procedure is a practical means to determine stresses in fillet welds. Results from the procedure should not be taken as a comprehensve evaluation of the multivarious parameters that may need to be considered in design.

This is the main part of the explanation for the derivation of line forces being obtained by separate calculation. The scope of insight needs to be suitable to the class of structure of which particular welds are part. The responsible engineer needs to make this judgement.

Simplifying assumptions are used for the stress distribution along and through the thickness of welds. Design standards present stress concentration factors applicable for geometric aspect ratios of welds and these may be used as factors on the applied line forces. The theory is consistent with recognised national standards and design texts. These should be read to establish the design parameters which are relevant for particular structures joined by welds.

Eurocode 3: Design of Steel Structures. Part 1-8: Design of Joints (450)
AISC 360-10
Structural Steel Design - Steel Construction Institute
Shingley's Mechanical Engineering Design. 9th Ed.
Failure Criteria, Ryder (173)
IIW-Commission XV

NOTATION

athickness thru assumed failure plane
b cgeometric constructs used to find 'a'
alphaangle of assumed failure plane
fddirect stress across assumed failure plane
fsvector sum shear stress across assumed failure plane
fsxyshear stress (in x-y plane) across assumed failure plane
fvMvon Mises stress across assumed failure plane
F_EXXweld classificaton stress (AISC notation)
legweld leg length
pltplots to be output: 1 yes 0 no
px py pzline forces (P/L) applied to weld
Rnnominal strength (AISC notation)
stepsfor calculating assumed failure plane stresses
thetaangle between shear resultant and (weld longitudinal) z-axis

PROJECT

The project example consists of four load cases applied to the same weld size. The first three are for each of line forces px (transverse), py (transverse) and pz (longitudinal). The object of presenting these cases is to demonstrate the affect on stress distributions for the range of assumed failure planes from alpha=0 to alpha=90deg. The forth case is with the three forces being applied in combination.

The case of pz acting alone is interesting in view of the shear stress variation with assumed failure plane angle. This effect is due only to the variation in width 'a' across the weld.

The direction of direct stress is obviously important. According to the sign convention used in this procedure,, +ve direct stress implies tension, -ve compression. Design acceptance is the then different according to which load is acting. The direction of stress is generally considered irrelevant for shear. However, the plots output by the procedure show shear stress consistent with the sign convention used for algebraic addition. The importance of this is that for positive direction line forces (px & py), the respective shear stresses (fs) across the weld are in opposing directions. In consequence, it is evident that care must be teken in assigning the direction of forces where both px & py act concurrently because of the affect on resulting shear stresses.

It is known that weld strength across a failure plane is principally associated with shear stress. Inspection of the plots substantiates this argument as von Mises stress (fvM) can be seen to be driven by shear (fs) and not direct stress (fd).

PROJECT INPUT

LET Ks=dac(dac_i,Ks%):KvM=dac(dac_i,KvM%)
LET Ks=0.5:KvM=1
LET F_EXX=mgr(S355,Ftu%):Fy=mgr(S355,Fy%)
LET F_EXX=4.9:Fy=3.5
LET leg=.6:px=.5:py=.01:pz=.01:steps=10:plt=1
LET leg=0.6:px=0.5:py=0.01:pz=0.01:steps=10:plt=1
PROC067

LET px=.01:py=.5:pz=.01:steps=10
PROC067

LET px=.5:py=.5:pz=.7:steps=10
PROC067

LET px=.5:py=.5:pz=.7:steps=10
PROC067

PROJECT OUTPUT

PROC067 Fillet Weld Strength - px only

PROC067 Fillet Weld Strength - px only

Loading State Across Failure Plane
alpha= (step_fp/steps)*PI/2= 0.3491
alpha= (2/9)*3.142/2= 0.3491
b= leg/(TAN(alpha)+1)= 0.4399
b= 0.6/(TAN(0.3491)+1)= 0.4399
c= leg-b= 0.1601
c= 0.6-0.4399= 0.1601
a= SQR(b^2+c^2)= 0.4681
a= SQR(0.4399^2+0.1601^2)= 0.4681
fd= px*SIN(alpha)/a+py*COS(alpha)/a= 0.3854
fd= 0.5*SIN(0.3491)/0.4681+0.01*COS(0.3491)/0.4681= 0.3854
fsxy= -px*COS(alpha)/a+py*SIN(alpha)/a= -0.9964
fsxy= -0.5*COS(0.3491)/0.4681+0.01*SIN(0.3491)/0.4681= -0.9964
fsz= pz/a= 0.02136
fsz= 0.01/0.4681= 0.02136
fs= SQR(fsxy^2+fsz^2)= 0.9966
fs= SQR(-0.9964^2+0.02136^2)= 0.9966
fvM= SQR(fd^2+3*fs^2)= 1.769
fvM= SQR(0.3854^2+3*0.9966^2)= 1.769
theta= ASN(ABS(fsxy)/SQR(fsxy^2+fsz^2))= 1.549
theta= ASN(ABS(-0.9964)/SQR(-0.9964^2+0.02136^2))= 1.549
Rn= .6*F_EXX*(1+.5*SIN(theta)^1.5)= 4.409
Rn= .6*4.9*(1+.5*SIN(1.549)^1.5)= 4.409
UF_001= fs/(Rn*Ks)= 0.452
UF_001= 0.9966/(4.409*0.5)= 0.4521
UF_002= fvM/(Fy*KvM)= 0.5053
UF_002= 1.769/(3.5*1)= 0.5054
END_PROC067

PROC067 Fillet Weld Strength - py only




PROC067 Fillet Weld Strength - py only


Loading State Across Failure Plane
alpha= (7/9)*3.142/2= 1.222
b= 0.6/(TAN(1.222)+1)= 0.16
c= 0.6-0.1601= 0.4399
a= SQR(0.1601^2+0.4399^2)= 0.4681
fd= 0.01*SIN(1.222)/0.4681+0.5*COS(1.222)/0.4681= 0.3851
fsxy= -0.01*COS(1.222)/0.4681+0.5*SIN(1.222)/0.4681= 0.9965
fsz= 0.01/0.4681= 0.02136
fs= SQR(0.9964^2+0.02136^2)= 0.9966
fvM= SQR(0.3854^2+3*0.9966^2)= 1.769
theta= ASN(ABS(0.9964)/SQR(0.9964^2+0.02136^2))= 1.549
Rn= .6*4.9*(1+.5*SIN(1.549)^1.5)= 4.409
UF_003= 0.9966/(4.409*0.5)= 0.4521
UF_004= 1.769/(3.5*1)= 0.5054
END_PROC067

PROC067 Fillet Weld Strength 170523 - pz only


PROC067 Fillet Weld Strength- pz only

Loading State Across Failure Plane
alpha= (4/9)*3.142/2= 0.6982
b= 0.6/(TAN(0.6981)+1)= 0.3263
c= 0.6-0.3262= 0.2738
a= SQR(0.3262^2+0.2738^2)= 0.4259
fd= 0.01*SIN(0.6981)/0.4259+0.01*COS(0.6981)/0.4259= 0.03308
fsxy= -0.01*COS(0.6981)/0.4259+0.01*SIN(0.6981)/0.4259= -2.895E-3
fsz= 0.7/0.4259= 1.644
fs= SQR(-2.894E-3^2+1.644^2)= 1.644
fvM= SQR(0.03308^2+3*1.644^2)= 2.848
theta= ASN(ABS(-2.894E-3)/SQR(-2.894E-3^2+1.644^2))= 1.76E-3
Rn= .6*4.9*(1+.5*SIN(1.761E-3)^1.5)= 2.94
UF_005= 1.644/(2.94*0.5)= 1.118*
UF_006= 2.847/(3.5*1)= 0.8134
END_PROC067

PROC067 Fillet Weld Strength - px py & pz


PROC067 Fillet Weld Strength - px py & pz

Loading State Across Failure Plane
alpha= (4/9)*3.142/2= 0.6982
b= 0.6/(TAN(0.6981)+1)= 0.3263
c= 0.6-0.3262= 0.2738
a= SQR(0.3262^2+0.2738^2)= 0.4259
fd= 0.5*SIN(0.6981)/0.4259+0.5*COS(0.6981)/0.4259= 1.654
fsxy= -0.5*COS(0.6981)/0.4259+0.5*SIN(0.6981)/0.4259= -0.1448
fsz= 0.7/0.4259= 1.644
fs= SQR(-0.1447^2+1.644^2)= 1.65
fvM= SQR(1.654^2+3*1.65^2)= 3.302
theta= ASN(ABS(-0.1447)/SQR(-0.1447^2+1.644^2))= 0.08779
Rn= .6*4.9*(1+.5*SIN(0.08781)^1.5)= 2.978
UF_007= 1.65/(2.978*0.5)= 1.108*
UF_008= 3.302/(3.5*1)= 0.9434
END_PROC067

PROJECT END