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
snip
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%--0.2
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_forwardPlease Answer fast as u know... im needed in 15-30 minutes thank u... So pleasearrow_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_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_forwardSolve itarrow_forwardWhere is/are the location(s) of the maximum compressive flexural stress? A simple I-beam is loaded as shown. 20 mm P KN PKN P KN B 20 mm- B с D L/4 m L/4 m L/4 m Pin support at the NA Midspan at point B Mid span at point D Midspan at the top fiber Roller support at top fiber Section D at the top fiber Section C at +170 mm from the NA L/4 m 20 mm C 250 mm 150 mm 150 mm D Aarrow_forward
- Columnarrow_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_forwardSHOW CLEAR AND COMPLETE SOLUTION. SOLVE FOR 1 HRarrow_forward
- DESIGN FOR TENSION ONLY (ACI METHOD fe=21mPa fy=276 MPa trength Design 2015 NSCP use ba 24 TL I SPAN 1: DL= 8 KN LL=12 KW/m L = 4.50m 2: DL= 10x/m LLIG W/m L=450m Ⓒ: DL=10 b/m LL 15/m L=4.70 m (4): DL = 8 KN/ 1 =12xx/m L = 4.60m TL ITarrow_forwardplease use only set 1 data table thank you this is non gradedarrow_forwardUSE NSCP 2010 A rectangular beam reinforced for both tension and compression barshas an area of 1450 mm² for compression bars and 4350 mm² fortension bars. The tension bars and compression bars are placed at a distance of 600 mm and 62.5 mm respectively from the top of thebeam. The beam width 300 mm, fc’ = 21 MPa, fy = 415 MPa andtension steel covering is 60 mm.If it is 8 m-simply-supported beam that carries three concentratedservice live loads P applied at three quarter points of the beam (exceptat the supports), neglecting the self weight of the beam, determine themaximum value of service load P in kiloNewtons.arrow_forward
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