{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1250_Mechanics Homework Mechanics of Materials Solution

1250_Mechanics Homework Mechanics of Materials Solution -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 10 6 6 10.66 A column of 21-11 efi'ective length is obtained by connecting two C10 X 20 ° steel channels with lacing bars as shown. Using allowable stress design, determine the allowable centric load for the column. Use 0i! m 36 ksi and E m 29 x 10" psi. SOLUTION Cloxzc: A=b‘.881'111 “0.4.01; w. L: 78.91%" 13: 2.8! 1.." at: 3.5—x = 2.2514». Po» +1.: “flow: A = (2)633): 11.71:. 111‘ T251 I,I - (13(78515 r 157.8 .‘»’_ I:l - 2E2.81+ (maximum! = 104.11 {M Le z 21431-252111. I 126.”) 3 “(0-67165) 1&(0. 671(5): L880? 3. Al"- i(%f£)‘]=,1%%;=;[l - {CO-671657]: H.821 1a.; - (Ramona) -: 179.3 11:95 10. 67 A compression member of 2 3—m effective length IS obtained by bolting PROBLEM 10.67 together two 127 x 76 x 12. 7-mm steel angles as shown. Using allowable stress design, determine the allowable centric load for the column. Use ay= 250 MPa and E = 200 GPa. .Y SOLUTION . . .. 232E _ /1n"(21noxlo‘5i _ Sine}. Cc— 5y _ mib’oxto‘ — 125.66 L I27>¢76x 12.7 MM «FE/laye- jive: AL: 2920 MM: Ix:- 3_Q3XIOE “MY 3: 44.4mm) = Locxto‘m”, x: new, 13.— Fiw‘ Column 2(I7que -' (2)0. 06 “’06): 2 £23406 MM I...,- - I,r 2.12%!0‘1‘4». = 2.12xio“m*‘ new m. = 4390x10“"m ‘ n‘ - ————lf;3"f;":d6 20.93 x10 ‘ w. = 107.91: < C,_ 35/} = 0.87455" LI 3?: + 5(0 37455i-J'(o 87455)3 -: 1.61110 ' 1.411ol"2(L/rl]= 2,—-—-—-_"’;,°,‘,'f‘[l— -(o.8741$s)‘]= so.7c1-xzo‘Pa. pm : 6:,1A = (80,7‘iaulo‘)(llsiloxlo‘) = 39111103 11) = 391 w ...
View Full Document

{[ snackBarMessage ]}