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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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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 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
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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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_forwardsniparrow_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_forward
- 5. 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_forward4/47 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_forwardPlease solve question carrow_forward
- Use 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_forwardASAP PLSS I GIVE UPVOTEE ? ♥arrow_forwardSolve it.arrow_forward
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