question archive Assignment 5B: This assignment is due by end of the class Use the stiffness method to evaluate the deformations and draw the NFD (internal normal forces in the members ) for the following structure
Assignment 5B: This assignment is due by end of the class Use the stiffness method to evaluate the deformations and draw the NFD (internal normal forces in the members ) for the following structure
Subject:Civil EngineeringPrice:9.82 Bought3
Assignment 5B: This assignment is due by end of the class Use the stiffness method to evaluate the deformations and draw the NFD (internal normal forces in the members ) for the following structure. (1) Note that the Stiffness matrix for a truss members is EA - EA L -EA EA L L D 150 KN 1.256A 1.25 EA 4.0 8 ZEA C 3.0 3.D
Purchase A New Answer
Custom new solution created by our subject matter experts
GET A QUOTE
Answer Preview
I answer based on my understanding.
APPLYING STIFFNESS MATRIX ; BOUNDARY CONDITION 42= V2= 43 = V3- 4 4 = V4=0 0 -34 ( 1. 25 ) / 5 + 0 6-48 ( 1:25 ) /5 +0 TKGT= 6 .34 (1 .25 ) / 5 0-48 ( 1 . 25 ) 15 AE 0 - 48 ( 1 .25 ) / 5 + 0 1 . 34 ( 1 :25 ) 15 + 2/ 4 - 044 ( 1:25 ) / 5 0 -64 ( 1 -25 ) /5 = 10-18 TU (AE) 0 .25 ( f ] = 1 150 KN ? the ( UIT [ D - 18 0 7 / 1 ) 6 150 7 - 0:25 LAE (833.33 ( (HE) y1 = 833. 33 NOPAL DISPLACEMENT AE FOR FORCES : @ M1 WEB = 06 1= 5 Sing = 0-2 El = ( F-0-4 - 0.8 0-4 0. 8 ( ) 100 833.39 AE = EE - E ( 100/ AE ) - 100/ 4 HE HA = 100 ( A) = 100 KN F1 = 100 KN @ M2 (60 - 0 1= 4 Sing = 1 EZ = ( 14 ) (0 - 1 0 1) 60 833.33 (AE) 12 = EE = 0 8 +20 $2 = OKN @ M3 Coso = - 0-6 1 =5 CInD = 0-8 63 = ( 45 ) (06 - 08 - 0.468 ) 10 833-33 ( ) = -100KN AE AE 03 EZ = 100 : A ) 13 = JBA = - 100 KN $3= - 100 KN
Please see the attached file
