Question 4: The beam in Figure P5.5-2 has continuous lateral support. If the live load is twice the dead load, what is the maximum total service load, in kips/ft, that can be supported? A992 steel is used. a. Use LRFD. b. Use ASD. W33 × 141 25'-

Question 4: The beam in Figure P5.5-2 has continuous lateral support. If the live load is twice the dead load, what is the maximum total service load, in kips/ft, that can be supported? A992 steel is used. a. Use LRFD. b. Use ASD. W33 × 141 25'-

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter10: Plate Girders

Section: Chapter Questions

Problem 10.4.5P

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Concept explainers

Types and properties of steel

Steel is an alloy of iron. It is formed by heating iron and combining carbon (0.1-1.7%) with it and minor amounts of other elements such sulfur, chromium, manganese, nickel, etc. Thousands of steel grades have been published, registered, or regulated around the world, each with its chemical composition, and special numbering systems have been devised in various countries to classify the vast number of alloys.

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Question 3: The beam shown in Figure P5.5-1 is a W14 ×61 of A992 steel and has continuous lat- eral support. The load P is a service live load. What is the maximum permissible value of P? a. Use LRFD. b. Use ASD. 15'- 15'- FIGURE P5.5-1
2. The beam shown in Figure P5.8-3 is a W16 × 31 of A992 steel and has continuous lateral support. The two concentrated loads are service live loads. Neglect the weight of the beam and determine whether the beam is adequate for flexure and shear. a. Use LRFD. b. Use ASD.
The given girder has beams framing into it at the ends and at every L/3 point. The beam carries a service live load of 20 kips as shown and superimposed uniformly distributed service dead load of 10 kip/ft. Select the lightest A992 W-section that can carry the load. Do not check for deflection. P, = 20 kips LL PLL W. = 10 kip/ft DL = 20 ft WDL L/3 BEAMS FRAMING INTO GIRDER L
The cantilever beam shown in Figure P5.8-4 is a W10 × 77 of A992 steel. There isno lateral support other than at the fixed end. Use an unbraced length equal to the spanlength and determine whether the beam is adequate. The uniform load is a servicedead load that includes the beam weight, and the concentrated load is a service liveload.a. Use LRFD.b. Use ASD.
5.5-9 The beam shown in Figure P5.5-9 is a W36 x 182. It is laterally supported at A and B. The 300 kip load is a service live load. Using the unfactored service loads, a. Compute Cb. Do not include the beam weight in the loading b. Compute Cb, Include the beam weight in the loading
.
H.W-4: For the beam of sections and detail shown in Fig. find the maximum live load (PL) that the beam can be carry it. Neglect self-weight? Ans. PL=195.74 kN Use f 35 = - MPa fy 420 -2.0m- 70mm 2PL 2020mm 5025mm 350 mm sct.a-a 600 mm- a a -2.0m- -LHHH] .... .... 8025mm Soemy 3025mm 350 mm sct.b-b b b -2.0 m- 600 mm PL
The beam cross-section shown is design to resist a nominal moment M₁ of 320 kN m, if f = 35 MPa and fy = 420 MPa, determine the required area of steel reinforcement Ag. 75,75,75 As 225- 75 565 60 700mm 0.85f I N2 0.85f d 1-17/22 T₁ T = T₁+T₂ d C₂ a-75 I a-75 2 T₂
1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E345 MPa and Fu= 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. IR₁ 0.90m 70,KN * W14 x 68 1.20m 70 KN * 0.90m R₂ W14x68 70 KN
Problem 1. A W30 x 116 beam hasthe top flange restrained againstlateral displacement and rotationand has an unstiffened web. A pointload is applied to the top flange at2m. From the left support. The beamspans 12 m. Use A36 steel with yieldstrength of Fy = 248 Mpa. a. Compute the maximum value of point loadthat may be applied without web yieldingoccurring. b. Compute the maximum value of point loadthat may be applied without web cripplingoccurring. Properties of W30 x 116d = 762 mmK = 41.28 mmbf = 266.7 mmtf = 21.59 mmtw = 14.33 mm
Solve by LRFD (AISC ) masnual
A compression member is built up from a W14x90 and a W10x49, both of A992 steel.a. Compute rx and ry for the built-up shape.b. Neglect flexural torsional buckling and compute the available strength for KxL = KyL = 30 feetusing LRFD.