Question

Problem 3: Determine the maximum load P that frame can support without buckling member AB made of steel. (Est = 2003 MPa, Sy = 360 MPa). 4m A P x 40 mm -X 40 mm 5 m 40 mm у 7 m B

Answer 1

um 3 MPa P Est= 200x10 Sy= 360 Mpu E 80 mm 5m 7m I UD u B 20 FBD P T P=FACX Loso FAC = 1.6P (4) FAC EFY=0 o= tant (514) O » 51.34 FAC sin 51.34 + FAB = 0 FAB = 1.25P (1) yoxsoliz Ho 6666.66 mmu Ix= 42 6666.66 mm 4 Iya soxuoli? 7m B 2. Critical buckling for both and Pin P= n²E/? 3 68151.53 N. r²x x 200 x 10 x 170 6666.66 2 Pcx cabout x clocis): (1000) about y ancis - rix (200x103) (426666.66) - 17187.88N (1000)? Per is least of two .. Posa 17187.88 N

Now FAB = 1.75P 17187. 88 = 1.25 p P= 13.750 KN nou cules formula valid only if 6cr2 by Per A 17187.88 CuO X 80) 5.37 12 MPa 2360MPa - COK)
the maximum load P that frame can support without buckling of AB is 13.750 KN.

I hope this will help you.????