CALCS | PROC067 Fillet Weld Strength 170527 |

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

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.

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

a | thickness thru assumed failure plane | |

b c | geometric constructs used to find 'a' | |

alpha | angle of assumed failure plane | |

fd | direct stress across assumed failure plane | |

fs | vector sum shear stress across assumed failure plane | |

fsxy | shear stress (in x-y plane) across assumed failure plane | |

fvM | von Mises stress across assumed failure plane | |

F_EXX | weld classificaton stress (AISC notation) | |

leg | weld leg length | |

plt | plots to be output: 1 yes 0 no | |

px py pz | line forces (P/L) applied to weld | |

Rn | nominal strength (AISC notation) | |

steps | for calculating assumed failure plane stresses | |

theta | angle between shear resultant and (weld longitudinal) z-axis |

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).

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

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 |

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 |

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 |

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 |