<p> Appendix-E Kingpin Bracket Weld Analysis</p><p>Summary Section</p><p>The objective of this analysis is to determine an adequate weld size for the bracket that holds the Kingpin in place under load. The weld size is specified to be ¾ inch. The following figure shows the part of the assembly where the weld is located: The result of this analysis would be a factor of safety against static failure of the weld bead under full pull force. The analysis indicates a factor of safety of 1.8 against weld failure.</p><p>Evaluation</p><p>The model represents a worst-case loading situation. The weld analysis is a conservative analysis. The assumptions do not cause much variation in results – lack of rigidity may just introduce some non-uniform loading. Since the cost of welding is low and the cost consequences of weld failure is high, a factor of safety of 1.5 is considered to be acceptable. The result of the analysis yields a factor of safety of 1.8 and therefore a weld size of ¾” is an acceptable design.</p><p>Formulation Section</p><p>Schematics 9</p><p>9</p><p>F=300000</p><p>4</p><p>1.25 9</p><p>The weld is all-around the bracket. </p><p>Given: Weld leg : 0.75 inch Weld length: 9 inch by 9 inch continuous weld Weld electrode: Yield strength is 100 ksi Force: F=300000 lbs</p><p>Find: Factor of safety guarding against static failure of the weld</p><p>Assumptions: Single-point loading. Rigid members. Uniform load distribution and centrally applied load.</p><p>Solution [See 1 for analysis method]:</p><p>Stresses due to direct shear 1.414F 1.414(300000) 1   15711 psi wAw 0.75(36)</p><p>Stresses due to bending is normal stress but is treated as shear stress in weld analysis procedures:</p><p>1.414M 1.414(300000)(4 1.25)  2    27494 psi wSw .75(108)</p><p>2 2 2 Sw  bd  d / 3  (9)(9)  9 / 3 108 in</p><p>Since these stresses act in directions 90 degrees apart, they are added vectorially to find the resultant stress:</p><p>2 2 2 2   1  2  15711  27494  31666 psi</p><p>The factor of safety is:</p><p>0.58(100000) FS   1.83 31666</p><p>Answer: FS = 1.8</p><p>Conclusion: The factor of safety is above the minimum acceptable level and therefore this weld is adequately designed.</p><p>References:</p><p>[1]: Robert L. Mott, Machine Elements in Mechanical Design, 3rd Edition, Prentice Hall, 1992, pp. 707-715</p>