8.17. Given the 40-ft simply supported span carrying two symmetrically placed concentrated loads, 20% dead load and 80% live load, as shown in the accompa- nying figure. If the loads are eccentric to the web by 5 in., and the member is torsionally simply sup- ported, select the lightest W14 section suitable using the b modified flexure analogy method. 14k 14k 12'- 12'-0" -5" 40'-0" W14 section

Tutorial 7 Вeams 1. A simply supported beam must carry the factored ultimate loads (two 15kN point loads) shown in the figure. Do not consider the self-weight of the beam. The beam is laterally and torsional supported only at A, B, C and D. For a 203 x 133 x 25 I- section: (Grade 350 W steel) (a) Determine the class of the section. (b) Determine whether the section has sufficient flexural resistance. 15 kN 15 kN 4,0 m 2,0 m 4,0 m

PLATE #3 PROBLEM: A built up section of A992 steel is made from plates fully welded together. The flanges consist of PL16x380 and PL12x500 for the web. Check the beam for flexure. Total load Deflection is limited to L/240. Cite all applicable specifications you used to solve the problem from NSCP 2015. 1. For full lateral support 2. For Lateral bracing at supports and midspan of AB. WD 3. For lateral bracing at supports only, what maximum w and w₁ can the beam carry for WL A WD = 26 kN/m_Includes self wt. WL = 40 kN/m B 10m ee m 2m PL16x380 PL12x500 X = 1.5.

A two-story frame, is braced in y axis (Ky-1) and is connected as moment frame for x axis as shown below. Assuming elastic behavior, and all steel has Fy-50ksi, determine following for column BF. a) Kx for column. b) The governing buckling axis. c) Failure mode in column (Inelastic or elastic global buckling). d) Design strength of the column according to LRFD method (Pn). e) If the column is adequate for the service dead load of 180K and live load of 240K. Use the kx value obtained in part a and select lightest HSS rectangular to replace W12x72. Compare the weight of selected section f) E A W18 X 35 W18 X 35 F 20' B J W12 X 50 W12 X 72 W18 X 40 W18 X 50 30' G K 30' H 20¹ 15'

The simply supported W24x55 steel beam (E 50 ksi) is braced laterally at the ends and its third points (represented by X). The beam's flexural strength (kip-ft) is most nearly 30 ft (a) 310 (b) 385 (c) 455 (d) 200

Verify the bending resistance and global stability of the beam. Gravity load is 0.9kN/m, concentrated load 100 kN is applied in the midpoint of the span. Q235 steel is used. a. No brace in the middle of the span. Concentrated load is applied on the top flange. b. No brace in the middle of the span. Concentrated load is applied on the bottom flange. (optional) C. Braced in the middle of the span. Concentrated load is applied on the top flange. d. Same as (a), use Q345B steel. 0.9kN/m 200×12 5000 100KN 5000 I 776x8 200×12

6-1.2-1: Find lightest W14 for L = 20', P = 600k E = 29,000ksi, Fb = 30ksi and Fv = 20ksi. 30. CONTINUOUS BEAM - TWO EQUAL SPANS-CONCENTRATED LOAD AT CENTER OF ONE SPAN Total Equiv. Uniform Load 13 p 8 P 2 R,= V₁.. 32 R R₂ R₁ R₂= V₁ + V₂. 11 p 16 V₁ R= V₁ V₂ Mmax (at point of load). M. (at support R)..... .......... Amax (at 0.480 / from R.).... M Imax Shear Moment M₁ V₂ ********* ***************** 11 = 11 P 32 - 19 p 32 13 64 = 3. Pl PI 32 0.015 P³ El

Select the lightest W section of A36 steel (E250 MPa and F₂-400 MPa) to serve as a column 9.15 m long to carry an axial compressive service load of 220 KN dead load and 490 KN live load in a braced structure. The member is assumed pinned at top and bottom and in addition has weak direction support at mid-height. Consider also local stability. Use LRFD.

A W 16 x50 is used as a beans to carry a uniform we and dead load including own weight of 35.7 kN/m and an axial tension load of T acting through the centroid, of the member. It has a simple span of 5.9 m. The compression flange of the member is laterally supported against local buckling.Use A 36 steel, Fy = 248 MPa. Properties of W 16 x 50A= 9483.85 mm^2d= 412.75 mmbf= 179.65 mmtf= 15.95 mmtw= 9.65 mmSx= 1324 x 10^3 mm^3Sy = 172 × 10^3 mm^3 Questions:Determine the Safe axial load T that the beam could carry.Hint Answer: 399.65KN

A built up l-section has a total unsupported length of 6m. It is made up of 2 - 400mm * 12mm plates as flanges and 450mm x 12 mm plate as web. The plates are of Grade 40 steel (fy = 276MPa). Considering the local buckling of the stiffened and unstiffened slender elements, determine the ultimate axial load capacity of the built-up column for both LRFD and ASD. Use NSCP 2015 Specification and effective length factor k = 1.0. Properties of the built-up section: A = 15,000 mm² ly = 128.06 x 106 mm4 Ix = 6.035 x 108 mm² rx = 200.58 mm ry = 92.40 mm What is the Slenderness Ratio? KL/r Using LRFD, what is the ultimate load capacity of the column? Express your answer in KN Using ASD, what is the allowable load capacity of the column? Express your answer in KN Note: Width to Thickness ratio for flanges and web: 555 4 Uniform compression in flanges of built-up I-shaped sections and plates or angle legs projecting from built- up 1-shaped sections b/t ΝΑ 0.64.JK.E/F, lal kc ,but shall not be taken…