Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 9, Problem 9.11PFS
To determine
The W section by using LRFD and ASD methods
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Students have asked these similar questions
2.
Use LRFD method for the following welded shape: The flanges are 1 in × 10 in, the web
is 5/16 in x 50 in, and the member is simply supported, uniformly loaded. Lateral support
of the compression flange is provided at the ends and at the midspan.
(a)
Determine the design flexural strength if A572 - Grade 50 steel is used.
(b)
If the span is 34 ft and supports the loads wp = 4.5 k/ft (not including the beam
weight) and w₁ = 4.5 k/ft, verify the moment capacity and select transverse stiffen-
ers as needed.
An 80-foot long plate girder (see below) is fabricated from a ½-inch x 78-inch web and two 3inch x 22-inch flanges. Continuous lateral support is provided. The steel is A992. The loading consists of a uniform service dead load of 1.0 kip/ft (including the self-weight), a uniform service live load of 2.0 kips/ft, and a concentrated service live load of 500 kips at midspan. Stiffeners are placed at each end and at 4 feet, 16 feet, and 28 feet from each end. Once stiffener is placed at midspan. Determine whether the flexural strength is adequate using LRFD.
For the plate girder shown below, check the flexural strength, then shear strength on end and
second panels. Lateral bracing is provided at each applied load (P). The girder is ASTM A1085
Gr. A (Fy=50 ksi). Using AISC-15th ed LRFD provisions. Pp=60 Kips, PL=30 Kips.
20"
0.375"
-105"
-105"
-285"
105"
105-
Chapter 9 Solutions
Structural Steel Design (6th Edition)
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Similar questions
- A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.arrow_forwardA girder supports concentrated dead and live loads. Determine the following required strengths for the girder based on the controlling load combinations.Use statics or Table 3-22a from the Steel Manual. The maximum bending moment, MD (kip-ft), due to dead load. The maximum bending moment, ML (kip-ft), due to live load. ASD required flexural strength, Ma (kip-ft) LRFD required flexural strength, Mu (kip-ft)arrow_forwardIf the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.arrow_forward
- Q4. [TYPE C] A loaded K-truss is supported by a roller at A and a pin at G as shown. Determine the force in members LM, MK, KC, and BC of the K-truss, and state whether they are in tension (T) or compression (C), or whether it is a zero-force member (ZFM). 500 lb 2 ft 2 ft A © 2023 Shanbe Huang 3 ft B K 3 ft M N C 3 ft D 3 ft P 3 ft H E F 3 ft Garrow_forwardThe truss shown is composed of steel sections. If the 2L150 x 90 x 8 A36 steel section was used in member CG, is it satisfactory? Use 2 lines of 4-20mm fasteners spaced 75 mm on center for all joints. The loads applied in the truss are. service loads. Use ASD only. You can use ASEP Manual. Multiply the loads by 10. 50 kN 30 kN- 3 m G 3 m 3 m 30 kN- D 30 kN- A m E H m Carrow_forwardUse NSCP 2015. The beam shown has continuous lateral support of both flanges. The uniform load consisting of 50% dead load and 50% live load. The dead load includes the weight of the beam. Used grade 50 steel. 6 k/ft |--0-0²- -18'-0"- fo 4-6- -6'-0" 1. Considering LRFD. Which of the following most nearly gives the design load Wu? 2. Considering LRFD. Which of the following most nearly gives the maximum negative design moment, Mu in ft-kips? 3. Considering W12x35 beam section, The section is 4. Considering ASD the W12x35 beam section, Which of the following most nearly gives allowable strength in ft-kipsarrow_forward
- Determine the adequacy of the tension members of the determinate bolted truss shown below based on the service loads shown. Use A50 steel. Assume there are three line of bolts in connection and neglect weight of members. Hint: There are three tension members in the truss.arrow_forwardFor the Integrated Project (office building), design the Floor Girders in the gravity-only system, assuming that the top flange is fully restrained with the floor diaphragm. 1. Use Table 3-2 from the Steel Manual to select the most economical compact W-shape section. Demands: Stories 1 and 2 Edge: 114.4 k.ft From Table 3-2, use W12x50 D/C = should be less than 0.95: D/C = Stories 1 and 2 Interior: 226.4 k.ft Use W12x50: D/C = Roof Edge: 33.3 k.ft Use W12x40: D/C = Roof Interior: 58.4 k.ft Use W12x40: D/C =arrow_forwardUse 2015 NSCP A simply supported beam is subjected to a uniform service dead load of 1.0 kip/ft(including the weight of the beam) a uniform service live load of 2.0 kips/ft and a concentrated service dead load of 40 kips. The beam is 40ft long and a concentrated load is located 15 ft from the left end. The beam has continuous lateral support and A36 steel is used. Is W30x108 adequate? a) Use LRFD b) use ASD P₁=40* -15- 25'- WD = 1.0km W₁ = 2.0³m W30 x 108 40'-arrow_forward
- Material properties are: - Concrete weight: wt = 150 pcf - Concrete compressive strength: f'c = 4000 psi - Steel tensile strength: fy = 60 ksi - Stirrup consists of #4 bar - Framing members are not exposed to weather or in contact with soil. A simplified line diagram of Girder G2 is shown below: Determine the tributary area for each concentrated load for G-2. 30-0¹ B4 B4 69 Colum Dea 30-0 - Mark 89 PD & PL 30 square feet 400 square feet 700 square feet 1200 square feet 90-0 BI -Ser PD & PL a=arrow_forwardPLATE #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.arrow_forwardA W14 x 61 must support a concentrated service live load of 665 KN applied to the top flange. Assume that the load is at a distance of at least half the beam depth from the support and design a bearing plate. Use E= 345 MPa for the beam, E= 250 MPa for the plate and fe'= 21 MPa. Use LRFDarrow_forward
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