# Lap Fillet Weld Calculations and FEA Techniques

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2 Table of Contents Introduction... 4 Definition... 4 Longitudinal Shear Stress Method... 5 Tensile Stress Method... 5 Modified Equation Method... 6 Example... 7 Material Properties... 8 Hand Calculations... 9 Finite Element Method Analysis Solving the Problem by using Shell (plate) Element Model Solving the Problem by using 2 nd Order Tetrahedral Solid Elements Model Conclusion CAE Studies By: Ahmad A. Abbas Page 2

3 Table of Illustrations FIGURE 1 LAP FILLET WELDED JOINT... 4 FIGURE 2 LAP FILLET WELDED JOINT CROSS SECTION VIEW... 5 FIGURE 3 LAP WELD JOINT MODEL... 7 FIGURE 4 MATERIAL ASSIGNMENTS OF THE MODEL... 8 FIGURE 5 THE CONSTRAINTS ON THE SHELL MODEL FIGURE 6 APPLIED LOAD ON THE SHELL MODEL FIGURE 7 VON MISES STRESSES IN THE SHELL MODEL FIGURE 8 VON MISES STRESSES IN THE SHELL MODEL FIGURE 9 THE CONSTRAINTS ON THE SOLID MODEL FIGURE 10 APPLIED LOAD ON THE SOLID MODEL FIGURE 11 VON MISES STRESSES IN THE SOLID MODEL FIGURE 12 VON MISES STRESSES IN THE SOLID MODEL Index of Tables TABLE 1 STEEL 1020 MECHANICAL PROPERTIES... 8 TABLE 2 WELD FILL MECHANICAL PROPERTIES... 8 CAE Studies By: Ahmad A. Abbas Page 3

4 Introduction The objective of the study is to evaluate the strength of a lap fillet welded joint under an axial load situation, and determine the welded joint capability. Usually in different structural analysis text books the Lap Joint Weld is analyzed as a joint that is solely subjected to longitudinal shear only. However the demonstration below will show a better method to calculate Lap welded Joint capability. Definition Lap welded Joint is a type of weld joint between two overlapping metal parts in parallel planes. Figure 1 Lap Fillet welded joint CAE Studies By: Ahmad A. Abbas Page 4

5 Longitudinal Shear Stress Method The simple calculation for the shear stress can be given by: Simple Where: F 2a * W F is the force. a is the weld's throat thickness ( a= Leg Length *.707) W is the length of the weld Figure 2 Lap Fillet welded joint cross section view Tensile Stress Method The simple calculation for the tensile stress can be given by: Simple 2* F a * LegLength CAE Studies By: Ahmad A. Abbas Page 5

6 Modified Equation Method The modified equation will take into account some excremental modifications and both tensile and shear stress in the joint, and could be written as: Modified Total 2 * Simple 3*( Simple ) 2 Where: is a constant between.7 and 1 based on the strength of material. * Note: It was assumed that the weld joint will fail before the plates. CAE Studies By: Ahmad A. Abbas Page 6

7 Example The below welded joint model will be analyzed using the following techniques: a) Hand Calculations. b) Finite element analysis using shell elements and no fill material. c) Finite Element method using 2 nd order tetrahedral solid elements, with modeling of the fill material. Figure 3 Lap weld joint model Dimensions: Large plate dimensions: 12x40X70 mm Small plate dimensions: 12X70X70 mm The weld Dimensions: 50X6 mm CAE Studies By: Ahmad A. Abbas Page 7

8 Material Properties For simplification all solids were modeled using the two main isotropic materials, the two main materials are steel 1020 and Weld fill material with tensile strength of 346 Mpa. In Figure 4 the material assignment of the assembly is illustrated Figure 4 Material assignments of the model Steel Density: 7200 kg/m 3 Elastic Modulus: 207 GPa Poisson's Ratio: 0.34 Tensile Strength: 455 MPa Yield Strength: 350 MPa Percent Elongation: 50% Hardness: 45 (HB) Table 1 Steel 1020 mechanical properties Weld Fill 2 Density: 7139 kg/m 3 Elastic Modulus: 201 GPa Poisson's Ratio: 0.32 Tensile Strength: 397 MPa Yield Strength: 346 MPa Table 2 Weld fill mechanical properties CAE Studies By: Ahmad A. Abbas Page 8

9 Hand Calculations If we only consider the shear stress in our calculations; from the material properties we can write: A= Throat area =.707*6*50*2 =424 mm 2 S y =350 MPa. Shear stress using distortion energy theory could be obtained by: S sy =.58 S y =203.0 MPa. F= S sy *A= N Hand calculations using previously introduced equations will give us the following results: Inputs Weld length 50 mm Number of weld Sides 2 # Weld leg Length 6 mm Force N 0.85 # Outputs Shear stress Mpa. Tensile stress Mpa. Equivalent Stress Mpa. Modified Equivalent Stress Mpa. Since the modified equivalent stresses exceed the yield stress of the material, evaluating the shear stress only is not sufficient. CAE Studies By: Ahmad A. Abbas Page 9

10 Let s consider that we would like to stay below 250 MPa to satisfy the Factor of Safety of 1.4; then the above calculations could be repeated and we will have the following results: Inputs Weld length 50 mm Number of weld Sides 2 # Weld leg Length 6 mm Force N 0.85 # Outputs Shear stress Mpa. Tensile stress Mpa. Equivalent Stress Mpa. Modifed Equivalent Stress Mpa. CAE Studies By: Ahmad A. Abbas Page 10

11 Finite Element Method Analysis The model could be analyzed using linear static simulations. This problem will be solved by using two types of modeling, first will be shell model, where plane stress theory will be applied, second we will model the problem using 2 nd order tetrahedral elements. Solving the Problem by using Shell (plate) Element Model Boundary Condition The boundary condition is fixed, that would mean there are zero degrees of freedom at the shown locations (Curve). Fixed Figure 5 The constraints on the shell model This will apply that the nodes at the curve will have: (Translation along x-axis) (Translation along y-axis) (Translation along z-axis) (Rotation about x-axis) (Rotation about y-axis) (Rotation about z-axis) CAE Studies By: Ahmad A. Abbas Page 11

12 Load The load will be a distributed force along the curve as shown below: Applied Force Figure 6 Applied load on the shell model CAE Studies By: Ahmad A. Abbas Page 12

13 Stress Results When applied force is equal to N, we will have the following stress results: Figure 7 Von Mises stresses in the shell model When applied force is equal to N, we will have the following stress results: Figure 8 Von Mises stresses in the shell model CAE Studies By: Ahmad A. Abbas Page 13

14 Solving the Problem by using 2 nd Order Tetrahedral Solid Elements Model Boundary Condition The boundary condition is fixed, that would mean there are zero degrees of freedom at the shown locations (Surface). Fixed Figure 9 The constraints on the solid model This will apply that the nodes on the surface will have: (Translation along x-axis) (Translation along y-axis) (Translation along z-axis) (Rotation about x-axis) (Rotation about y-axis) (Rotation about z-axis) CAE Studies By: Ahmad A. Abbas Page 14

15 Load The load will be a distributed force on the surface as shown below: Applied Force Figure 10 Applied load on the solid model CAE Studies By: Ahmad A. Abbas Page 15

16 Stress Results When applied force is equal to N, we will have the following stress results: Figure 11 Von Mises stresses in the solid model When applied force is equal to N, we will have the following stress results: Figure 12 Von Mises stresses in the solid model CAE Studies By: Ahmad A. Abbas Page 16

17 Conclusion Usually, engineers are looking to the shear stress state of the welded joint and most of the effort of the analysis is to determine the shear stress in the weld. Most hand books suggest 33% to 27% max shear stress vs. yield stress as a limiting factor in welded joints. The reasoning of such approximations could be explained with looking to more advanced failure theories such as Von Mises criterion. As it could be seen in the equation above the shear stress has a larger contribution to the overall Von Mises stress but is not the only contributor. Furthermore, looking to the shear stress only will be misleading if axial stresses are conspicuous in the welded joint. CAE Studies By: Ahmad A. Abbas Page 17

18 *The geometry was taken from a standard part library and modified for this study; also all data are assumptions for proof of concept only. MS.ME Ahmad A. Abbas Sr. Analysis Engineer CAE Studies By: Ahmad A. Abbas Page 18

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