WELD COEFFICIENT IN RAM CONNECTION.

i AM TRYING TO DESIGN A SIMPLE SHEAR PLATE USING RAM CONNECTION, AND WEHN I WAS LOOKING AT THE REPORT I COUDNT FIGYRE IUT HOW DID THE SOFTWARE CALCULATE THE VALUE OF THE WELD COEFFICIENT, WHEN I DID THE CALCULATION MANUALLY IT GAVE ME A DIFFERENT VALUE: HERE IS HOW THE SOFTWARE CALCULATES THE WELD STRENGTH:

Verification Unit Capacity Demand Ctrl EQ Ratio References

Weld capacity [KN]   219.42    50.00   1V - DL 0.23   Tables 8-4 .. 8-11     fRn = f*C*C1*D*L = 0.75*0.650122[KN/mm]*1*3*150[mm] = 219.42[KN]   Tables 8-4 .8-11

#AND ATTACHED IS THE DIMENSION OF THE CONNECTION

how did the software calculate the 0.650122 factor ???

Parents
  • Based the design equation and reference to Tables 8-4 to 9-11, it appears that you are designing the connection with the AISC 360 LRFD design code. Shear plate connections of this type are designed using the procedure outlined on pages 10-101 through 10-102 of the AISC Steel Construction Manual. When the connection uses less than 9 bolts and standard holes, eccentricity can be ignored. The coefficient C is taken from Table 8-4 for k=0 and a =0 (no eccentricity). The tabulated coefficient for these values is 3.71 in units of kips and inches. Converting to units of kN and mm gives 0.65 kN/mm.



Reply
  • Based the design equation and reference to Tables 8-4 to 9-11, it appears that you are designing the connection with the AISC 360 LRFD design code. Shear plate connections of this type are designed using the procedure outlined on pages 10-101 through 10-102 of the AISC Steel Construction Manual. When the connection uses less than 9 bolts and standard holes, eccentricity can be ignored. The coefficient C is taken from Table 8-4 for k=0 and a =0 (no eccentricity). The tabulated coefficient for these values is 3.71 in units of kips and inches. Converting to units of kN and mm gives 0.65 kN/mm.



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