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
A 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
PL12x500
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_forwardPlease Answer fast as u know... im needed in 15-30 minutes thank u... So pleasearrow_forwardQ2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m Darrow_forward
- Please solve question carrow_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_forwardASAP PLSS I GIVE UPVOTEE ? ♥arrow_forward
- round off final answer to 4 decimal placesarrow_forwardPROBLEM 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_forward
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ISBN:9781337094740
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Publisher:Cengage Learning