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 ???
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.