Hi Dr, in the problem solution no.2, in finding the Net area for path 2, shouldnt 0.448 be substituted for t? It seems like there was a mis input there, since it shows 2 x 2²/10 x 7/8, while it should be 2 x 2²/10 x 0.448, if I am right. Through my calculation, the net area for path 2 is 5.0715 in², thus leaving this as my final effective area. Thanks
In exercise problem 2, the notation 4@2” means the spacing between two consecutive vertical lines (shown on the drawing) is 2 inches, and we have 4 of them, which makes the overall length of the connection 8 inches. That is why we use 8 in the denominator when determining U, since the length in the denominator for U is the overall length of the connection.
Nice teaching method
Appreciated
thanks for excellent materials
Thanks for these series of videos. Please continue all subjects about Steel design.
Always useful for building concepts
Plz keep it up
Excellent explaination.
Hi Dr, in the problem solution no.2, in finding the Net area for path 2, shouldnt 0.448 be substituted for t? It seems like there was a mis input there, since it shows 2 x 2²/10 x 7/8, while it should be 2 x 2²/10 x 0.448, if I am right. Through my calculation, the net area for path 2 is 5.0715 in², thus leaving this as my final effective area. Thanks
You are correct. We have updated the solution file. Thank you for bringing this error to our attention.
Nice lecture , which application used ? For tutorial?
can u explain the " 4@2'' " like what is the units for " 4 " and why is ur L value when calculating U 8 in the denominator why not 2''?
In exercise problem 2, the notation 4@2” means the spacing between two consecutive vertical lines (shown on the drawing) is 2 inches, and we have 4 of them, which makes the overall length of the connection 8 inches.
That is why we use 8 in the denominator when determining U, since the length in the denominator for U is the overall length of the connection.
Plz make video on dynamic analysis of structure