A W12 X 65 is used as a simply supported, uniformly loaded beam with aspan length of 50 feet and continuous lateral support. The yield stress, F,is 50 ksi. If the ratio of live load to dead load is 3, compute the availablestrength and determine the maximum total service load, in kips/ft, that canbe supported.a. Use LRFD.b. Use ASD.

It is required to determine the upper limit for compact. It is required to determine the upper…

Answered: A W12 X 65 is used as a simply…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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The rigid frame shown is unbraced. The members are oriented so that bending is about the strong axis. Support conditions in the direction perpendicular to the plane of the frame are such that Ky = 1.0. The beams are W410 x 85, and the columns are W250 × 89. A992 steel is used, Fy=345 mPa. The axial compressive dead load is 750 kN and the axial compressive live load is 900 kN. Use the AISC alignment chart or Magdy I. Salama Formula for Kx. B 4.5 m 5.5 m 6.0 m Sx mm^2 mm^4 mm^3 1100 L A Ix Sy Zx Cw ly mm mm^4 mm^3 112 48.4 Zy mm mm^3 mm^3 mm^3 mm19 1040 SECTION ry 65.2 1230 574 713 W250x89 4500 11400 W410x85 5500 10800 315 143 378 1510 171 18 199 40.8 1730 310 926 716 Which of the following best gives the allowable axial strength of column AB? Use the NSCP 2015 LRFD specifications. O 1,647 kN 2,471 kN O 900 kN
For the column shown below, create an interaction diagram on Excel for bending about the strong axis and email it to the instructor. Show the nominal strength as well as the ACI strength including the strength reduction factor, . Utilize the full capacity of the ACI 318. For the graph, the y-axis shall be the axial load and the x-axis shall be the moment. Plot the factored loads, as points, for the various combinations and indicate them in a key. Additionally, plot the same load on the appropriate interaction diagram from the textbook and the provided a copy of that interaction diagram along with the spreadsheet. All work must be your own. Do not collaborate with classmates. Given: f= 4,000 psi fy = 60,000 psi Comb. 120 Design Loads Pu K 780 525 404 570 160 3 4 5 Muy FT-K 62 180 275 145 320 18" 2.0"-> ..... V 20" 10-#7 Total 2.0" X
AW14 x 43 column in a braced frame has a yield stress of 50 ksi and a height of 10 ft. The column is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column considering the critical stress for flexural buckling. From AISC Manual Table 1-1, the properties of a W14 x 43 are: A = 12.6 in^2 tw = 0.305 in bf/2tf = 7.54 h/tw = 37.4 rx = 5.82 in ry = 1.89 in Ix = 428 in4 ly = 45.2 in4 G= 11,200 ksi J = 1.05 in^4 Cw=1950 in^6 What is the nominal axial strength considering flexural buckling stress of the column section (in kips). O 310 O 468 O 447 O 375 O 420 O 436 O 517 O 280
Topic: BEAM BEARING PLATES AND COLUMN BASE PLATES - STEEL DESIGN
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A lightly welded steel beam shown in Figure 5 is continuously supported along its length. A downward concentrated load of 4P acts at Point C, and two downward concentrated loads P act at Points A and E. The beam with full lateral restraint is bent about its major principal x-axis. Determine the followings: a. Section type based on section slenderness; b. Effective section modulus; c. The maximum design value of P. A k B fu 6 m d tw bf 6 m tf 4P d bf tw tf = - = fyf= fyw 6 m D um * 550 mm 206 mm 6 mm 10 mm 320 MPa = 340 MPa Zx=1362985 mm³ Sx 1533750 mm³ Figure 5 Steel beam under design loads. 6 m P E
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SOLVE please Determine the design bending strength of ISLB 350 @ 486 N/m, considering the beam to be laterally supported. Assume factored design shear force will not exceed 0.6 times the design shear strength of the cross-section. If it is used as simply support beam of unsupported length 5 m then determine the factored distributed load the beam can carry. Use Fe410 grade steel For ISLB 350 @ 486 N/m, section properties are as follows: h = 350 mm; b= 165 mm; t₁ = 11.4 mm; t = 7.4 mm; R₁ = 16 mm; Izz = 13158.3 × 104 mm²; I = 631.9 × 10¹ mm²; Zpz = 851.11 × 10³ mm³; Zez = 751.9 x 10³ mm³
Figure 1 shows the cross-section of a heavily welded I-section steel column under axial compression. The yield stress of the steel section is 300 MPa. (a) Check the slenderness of the flanges and webs. (b) Determine the effective and gross areas of the cross-section. (c) Determine the design axial capacity of the cross-section. 380 E 10 425 -8 Figure 1. Cross-section of the rectangular hollow section steel column (Units: mm)
Questions: Civil Engineering - Steel Find the maximum axial load that the W200 x 46 column can safely carry if its effective length is 6m. Fy = 248 MPa and E = 200 GPa. Use ASD. Section Properties A = 5860 mm2 I= 15.3 x 106 mm4 r = 51.1 mm Final Answerri Pn - 703.87 こ 22 1-67 = Design 421.481N Pn= For Ag 120.11X580O (n = 703.87kW :. Please show solutions on how to get the answers 421.48 kW
AW10x30, 3.1 m long A36 steel is used as a column with fixed ends to support a service dead load of 267 kN. Calculate the maximum service live load the column can sustan uning LRFD Use theoretical value of k. Write your answer rounded to the nearest whole number in KN Use properties Ag = 5.703 mm2, rx = 111.25 mm, ry = 34.8 mm. Use stored value. Asnwer in 5 decimal places.
Question 1: (Use NSCP 505.3 provisions and ASEP Steel Handbook for effective length factors) Use AISC Equation E3-2 or E3-3 and determine the nominal axial compressive strength for the following cases: a. L= 15 ft b. L=20 ft W10 × 33 A992 steel FIGURE P4.3-1

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