Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem From the Figure shown, diameter of holes is 20mm. Steel plates is made up of 10 mm x 250 mm with staggered sections. USE LRFD. Fy = 248.8 MPa; Fu = 449.3 MPa Resistance factor for the fracture limit state is 0.75 Questions: a) Compute the net area An for the plate. b) Compute the design strength of the connection. c) Compute the maximum value of service load T when A36 steel is used, the live load is four times the Dead load.
Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem From the Figure shown, diameter of holes is 20mm. Steel plates is made up of 10 mm x 250 mm with staggered sections. USE LRFD. Fy = 248.8 MPa; Fu = 449.3 MPa Resistance factor for the fracture limit state is 0.75 Questions: a) Compute the net area An for the plate. b) Compute the design strength of the connection. c) Compute the maximum value of service load T when A36 steel is used, the live load is four times the Dead load.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
*Use latest NSCP/NSCP 2015 formula to solve this problem
*Please use hand written to solve this problem
From the Figure shown, diameter of holes is 20mm. Steel plates is made up of 10 mm x 250 mm with staggered sections. USE LRFD.
Fy = 248.8 MPa; Fu = 449.3 MPa
Resistance factor for the fracture limit state is 0.75
Questions:
a) Compute the net area An for the plate.
b) Compute the design strength of the connection.
c) Compute the maximum value of service load T when A36 steel is used, the live load is four times the Dead load.
Transcribed Image Text:T
50
50
50
50
50
B
A
E
50
D
60
с
75
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