2nd Edition Example 4.6 page 4-18
Could anyone explain how the net area has been claculated? I do not see it has been multiplied by number of holes? It looks like Nholes =1 in this example?
An=Ag-Nholes*(Dbolt+0.125)*T
David's book is the best for SE preparation for bridge section.
David, could you please shed light on Problem#17(Page-61): how you have calculated Avf=4*0.2=0.8 in^2/ft?
