Explanation Given data: W 10 × 77 sec t i o n Y i e l d s t r e s s = 50 k s i L e n g t h o f b e a m − c o l u m n = 20 f t K y = 1.0 K x = 1.25 P n t = 100 k P l t = 150 k M n t x = 60 f t − k M l t x = 90 f t − k C b = 1.0 P s t o r e y = 4800 k P m f P s t o r e y = 1.0 H = 120 k Δ H L = 0.0025 From the W shape properties tables in the AISC manual, The sectional properties of the given member W 10 × 77 are as follows: Plastic length of the section, L p = 9.18 f t Unbraced length corresponding to a lateral torsional buckling moment, L r = 45.3 f t Based on LRFD, the maximum available strength of section, ϕ b M p x = 366 k i p − f t Based on LRFD, the flexural strength of unbraced length, ϕ b B F = 3.90 k i p s Resistance factor for bending, ϕ b = 0.9 Safety factor bending, Ω b = 0.9 Area of the section, A g = 22.6 i n . 2 Moment of inertia about x-x axis, I x x = 455 i n . 4 Radius of gyration about x-axis, r x = 4.49 i n . Radius of gyration about y-axis, r y = 2.60 i n . Design strength, ϕ P n = P c = 546 k i p s The final axial strength is P r = P n t + P l t w h e r e , P r i s f i n a l a x i a l s t r e n g t h P n t i s a x i a l s t r e n g t h i n b e a m c o l u m n d u e t o g r a v i t y l o a d P l t i s a x i a l s t r e n g t h i n b e a m c o l u m n d u e t o l a t e r a l l o a d Substituting the values, P r = 100 + 150 = 250 k i p s The cross-section compressive strength is P n s = F y × A g w h e r e , P n s i s c o m p r e s s i v e s t r e n g t h F y i s y i e l d s t r e s s A g i s g r o s s sec t i o n a r e a Substituting the values, P n s = 50 × 22.6 = 1130 k i p s The effective length in the x-direction is ( L e ) x = K x × L w h e r e , ( L e ) x i s e f f e c t i v e l e n g t h i n x − d i r e c t i o n K x i s e f f e c t i v e l e n g t h f a c t o r i n x − d i r e c t i o n L i s l e n g t h o f b e a m − c o l u m n Substituting the values, ( L e ) x = 1.25 × 20 = 25 f t The effective length in the y-direction is ( L e ) y = K y × L = 1.0 × 20 = 20 f t The equivalent effective length is [ ( L e ) y ] e q = ( L e ) x r x r y w h e r e , [ ( L e ) y ] e q i s e q u i v a l e n t e f f e c t i v e l e n g t h ( L e ) x i s e f f e c t i v e l e n g t h i n x − d i r e c t i o n r x i s r a d i u s o f g y r a t i o n a b o u t x − a x i s r y i s r a d i u s o f g y r a t i o n a b o u t y − a x i s Substituting the values, [ ( L e ) y ] e q = 25 4.49 2.60 = 14.48 f t Hence, the effective length is control by the value of 20 ft. The elastic buckling load is given by P e l x = π 2 × E × I x x ( L e ) 2 x w h e r e , P e l x i s e l a s t i c b u c k l i n g l o a d E i s mod u l u s o f e l a s t i c i t y I x x i s m o m e n t o f i n e r t i a a b o u t x − x a x i s ( L e ) x i s e f f e c t i v e l e n g t h Substituting the values, P e l x = π 2 × 29000 × 455 ( 20 × 12 1 ) 2 = 2260.93 k i p s The modification factor for moment is C m = 0.6 − 0.4 ( − M 1 M 2 ) w h e r e , C m i s m o m e n t mod i f i c a t i o n f a c t o r M 1 i s s m a l l e r m o m e n t M 2 i s l arg e r m o m e n t Substituting the values, C m = 0.6 − 0.4 ( − 3 4 ) = 0.9 The magnification factor is B 1 x = C m 1 − α × P r P e l x w h e r e , B 1 x i s m a g n i f i c a t i o n f a c t o r C m i s mod i f i c a t i o n f a c t o r α i s c o n s tan t w i t h v a l u e o f 1

6th Edition

ISBN: 9780134589657

Author: Jack C. McCormac, Stephen F. Csernak

Publisher: PEARSON

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Chapter 11, Problem 11.15PFS

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Structural Steel Design (6th Edition)

Ch. 11 - Prob. 11.1PFS Ch. 11 - Prob. 11.11PFS Ch. 11 - Prob. 11.12PFS Ch. 11 - Prob. 11.15PFS Ch. 11 - Design problem in an unbraced frame—using loads...

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The adequacy of member using LRFD method.

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