Section 8 Div 1 Appendix 13 question

[CadLady]

I am trying to work through the example in Sec Vii Appendix 13 , section 13-17, ligament efficiency of multidiameter holes in plates.

I can not figure out how the k value of 1.82 was arrived at.

Any help would be appreciated.

 

[dsi]

K = (I2/I1) * Alpha
I1 = t1^3 / 12 = 0.625^3 / 12 = 0.0203
I2 = t2^3 / 12 = 1.000^3 / 12 = 0.0833
Alpha = H/h = 6 / 13/5 = 0.444
K = 1.82
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[CadLady]

Thank you for your help.