Q.1 Figure 1 shows a simply supported steel beam with design span of 24 ft. with its flange fully restrained against local buckling by a concrete slab. The beam supports uniformly distributed loads dead (D) = 1.2 kips/ft and live (L) = 3.0 kips/ft. Using W14 x 53 lb/ft. Grade A992 (Fy 50 ksi) W-steel check the suitability of the beam is safe for the following LRFD criteria by Calculation using the relevant equation below: = a) Strength Design i. Yielding (Flexure ii. Shear b) Serviceability (Deflection) M=FZ (% = 0.90) (Eqn. 1) Eqn. 2-3 Eqn. 2.4 V₁ = 0.6F, AwCv QVn>Vu Aw = dtw k 24 ft. FIG. 1 WD=1.2 kips/ft L = 3.0 kips/ft WL
Q.1 Figure 1 shows a simply supported steel beam with design span of 24 ft. with its flange fully restrained against local buckling by a concrete slab. The beam supports uniformly distributed loads dead (D) = 1.2 kips/ft and live (L) = 3.0 kips/ft. Using W14 x 53 lb/ft. Grade A992 (Fy 50 ksi) W-steel check the suitability of the beam is safe for the following LRFD criteria by Calculation using the relevant equation below: = a) Strength Design i. Yielding (Flexure ii. Shear b) Serviceability (Deflection) M=FZ (% = 0.90) (Eqn. 1) Eqn. 2-3 Eqn. 2.4 V₁ = 0.6F, AwCv QVn>Vu Aw = dtw k 24 ft. FIG. 1 WD=1.2 kips/ft L = 3.0 kips/ft WL
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter9: Composite Construction
Section: Chapter Questions
Problem 9.2.1P: A W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The...
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
Transcribed Image Text:Q.1 Figure 1 shows a simply supported steel beam with design span of 24 ft. with its flange fully
restrained against local buckling by a concrete slab. The beam supports uniformly distributed
loads dead (D) = 1.2 kips/ft and live (L) = 3.0 kips/ft. Using W14 x 53 lb/ft. Grade A992 (Fy
50 ksi) W-steel check the suitability of the beam is safe for the following LRFD criteria by
Calculation using the relevant equation below:
=
a) Strength Design
i. Yielding (Flexure
ii. Shear
b) Serviceability (Deflection)
M=FZ (% = 0.90)
(Eqn. 1)
Eqn. 2-3
Eqn. 2.4
V₁ = 0.6F, AwCv
QVn>Vu
Aw = dtw
k
24 ft.
FIG. 1
WD=1.2 kips/ft
L = 3.0 kips/ft
WL
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