Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
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Question
Chapter 10, Problem 10.4.5P
To determine
(a)
Whether flexural strength is adequate or not by using LRFD.
To determine
(b)
Whether flexural strength is adequate or not by using ASD.
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A plate girder cross section consists of two flanges, 11⁄2 inchesx 15 inches, and a 5⁄16-inch 3 66-inch web. A572 Grade 50 steel is used. The span length is 55 feet, the service live load is 2.0 kips/ft, and the dead load is 0.225 kips/ft, including the weight of the girder. Bearing stiffeners are placed at the ends, and intermediate stiffeners are placed at 69-20 and 12,-9,, from each end. Does this girder have enough shear strength?
a. Use LRFD.
b. Use ASD
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.
A 66-foot long plate girder must support a uniformly distributed load and concentrated loads at the one-third points. The uniform load consists of a 1.3-kipyft dead load and a 2.3 kip/ft live load. Each concentrated load consists of a 28-kip dead load and a 49-kip live load. There is lateral support at the ends and at the one-third points. Use A572 Grade 50 steel, a total depth of 80 inches, and LRFD.
a. Select the girder cross section and the required spacing of intermediate stiffeners.
b. Determine the size of intermediate and bearing stiffeners. c. Design all welds
Chapter 10 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
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- 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 built up section of A992 steel, F, = 345 mPa is made from plates fully welded together. The flanges consist of PL16X380 and PL12X500 for the web. Use the NSCP 2015 specifications. PL16x380 WL = 1.5wp %3D A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m PL12x500arrow_forward5. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2arrow_forward
- 4/47 Compute the force in member HN of the loaded truss. BE 2 m A 4/40 n L L D R Q L L F 6 m L P O N -8 panels at 3 m Problem 4/47 age to L BODJEN H M L I L L 2- J [2 m EM Karrow_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_forwardA built up section of A992 steel, F, = 345 mPa is %3D made from plates fully welded together. The flanges consist of PL16×380 and PL12×500 for the web. Use the NSCP 2015 specifications. PL16×380 W̟ = 1.5wp A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? O 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m B. PL12x500arrow_forward
- PROBLEM 2: The steel beam loaded below has the built-up cross section as shown. Determine the maximum permissible value of the load w so as not to exceed allowable bending stresses of 110 MPa. -125 mm- w (including beam weight) 20 mm 20 mm 100 mm .-- NA A B - 2 m - 6 m 20 mm FINAL ANSWER: Max. allowable w kN/marrow_forward5.5-3 A simplrted beam is subjected to a uniform service dead load of 1.0 kip cluding the weight of the beam), a uniform service live load of 0 kipft, and a concentrated service dead load of 40 kips. The beam feet long, and the concentrated load is located 15 feet from the left The beam has continuous lateral support, and A572 Grade 50 steel is used. Is a W30 x 108 adequate a. Use LRFD. b. Use ASD.arrow_forwardQ2 1- Check the following weather one way or two way 2- Calculate the design ultimate load Wu on the on the beams B3, B6 Assume slab load: Live load = 5 kN/m2 Dead load = 8 kN/m? Beam size 300x500 mm %3D B5 B6 B7 3.6 m |L 4.8 4 m 5m 5m BI B2 B4arrow_forward
- A W36x210 can be used for Transfer Girder ‘A’ as shown in the elevation next page.Assume the LRFD loads on the ‘Transfer Girder’ are as follows:Pu= 200 kips from the column above. Assume also that the loads from self-weight of thewu = 2.0 kip/ft (girder self weight and beams framing into this transfer girder are equivalentto a uniformly distributed load). These loads are shown in the last figure.a) Check whether the W36x210 ‘Transfer Girder’ is adequate for bending and shear.Assume the unbraced length Lb =20 ft.b) Suppose that service loads, w and P, are both 50% dead load and 50% live load. Does theW36x210 satisfy a service live load deflection limit of L/480?arrow_forwardA beam must be designed to the following specifications: Span length = 35 ft Beam spacing = 10 ft 2-in. deck with 3 in. of lightweight concrete fill (wc=115 pcf) for a total depth of t=5 in. Total weight of deck and slab = 51 psf Construction load = 20 psf Partition load = 20 psf Miscellaneous dead load = 10 psf Live load = 80 psf Fy=50 ksi, fc=4 ksi Assume continuous lateral support and use LRFD. a. Design a noncomposite beam. Compute the total deflection (there is no limit to be checked). b. Design a composite beam and specify the size and number of stud anchors required. Assume one stud at each beam location. Compute the maximum total deflection as follows: 1. Use the transformed section. 2. Use the lower-bound moment of inertia.arrow_forward1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E 345 MPa and Fy= 450 MPa). The only uniform service dead load is the weight of the beam. The 70 KN service loads are 30%DL and 70%LL. Use LRFD. 70 KN 70, KN 70 KN W 14 x 68 0.90m 1.20m 0.90m WI4 x68 IR, R2arrow_forward
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