A double-T simply supported concrete beam its cross section is shown inFigure, is prestressed with 2 tendons each 400 mm². Determine the allowableservice load.Use span 12 m, fse = 1300 MPa, fc = 40 MPa, y = 25 kN/m³.=1200 mm>09<*100*As = 400+100+As = 4004001400+

Step 1: Compute the Prestressing ForceP = 2 (As x fse)P = 2 (400 x 1300)P = 1,040,000 N or 1040 kN…

Answered: A double-T simply supported concrete…

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter9: Composite Construction

Section: Chapter Questions

Problem 9.2.1P: A W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The...

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A column in a building is subjected to the following load effects: 9 kips compression from dead load 5 kips compression from roof live load 6 kips compression from snow 7 kips compression from 3 inches of rain accumulated on the roof 8 kips compression from wind a. If lead and resistance factor design is used, determine the factored load (required strength) to be used in the design of the column. Which AISC load combination controls? b. What is the required design strength of the Column? c. What is the required nominal strength of the column for a resistance factor of 0.90? d. If allowable strength design is used, determine the required load capacity (required strength) to be used in the design of the column. Which AISC load combination controls? e. What is the required nominal strength of the column for a safety factor of 1.67?
Note For Problems 9.6-1 through 9.6-5, use the lower-bound moment of inertia for deflection of the composite section. Compute this as illustrated in Example 9.7. 9.6-2 Compute the following deflections for the beam in Problem 9.2-2. a. Maximum deflection before the concrete has cured. b. Maximum total deflection after composite behavior has been attained.
Use the composite beam tables and select a W-shape and stud anchors for the following conditions: Span length = 18 6 Beam spacing = 9 ft Total slab thickness = 51 2 in. (the slab and deck combination weighs 57 psf). Lightweight concrete with a unit weight of 115 pcf is used Construction load = 20 psf Partition load = 20 psf Live load = 225 psf Fy=50 ksi and fc=4 ksi A cross section of the formed steel deck is shown in Figure P9.8-9. The maximum live-load deflection cannot exceed L/360 (use a lower-bound moment of inertia). a. Use LRFD. b. User ASD.
Note For Problems 9.6-1 through 9.6-5, use the lower-bound moment of inertia for deflection of the composite section. Compute this as illustrated in Example 9.7. 9.6-1 Compute the following deflections for the beam in Problem 9.2-1. a. Maximum deflection before the concrete has cured. b. Maximum total deflection after composite behavior has been attained.
Note For Problems 9.6-1 through 9.6-5, use the lower-bound moment of inertia for deflection of the composite section. Compute this as illustrated in Example 9.7. 9.6-5 For the beam of Problem 9.4-2. a. Compute the deflections that occur before and after the concrete has cured. b. It the live toad deflection exceeds L360 , select another steel shape using either LRFD or ASD.
Repeat Problem 20-12, with the slab containing temperature steel in the form of welded wire mesh consisting of 0.125 in.2 steel/ft width. The modulus of steel, Es, is 30106lb/in.2 , Ec=5106lb/in.2 , and h=6inches .
A 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.
Repeat Problem 11.1 based on LRFD using the following factors: load factor for dead load = 1.25 load factor for live load = 1.75 strength reduction factor on the ultimate bearing capacity = 0.50 11.1 A continuous foundation is required in a soil where , , and . The depth of the footing will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (6.10) and Table 6.1.
In concrete work, Fuller and Thompson (1907) suggested that a dense packing of grains can be achieved if the percent finer (p) and grain size (D) are related by the following equation, where n is a constant varying in the range of 0.3-0.6. Dmax is the size of the largest grain within the soil. p=(DDmax)n100 This equation is sometimes used in roadwork for selecting the aggregates. a. For n = 0.5, show that the soil is well graded. b. If n = 0.5 and Dmax = 19.0 mm, find the percentages of gravel, sand, and fines within the soil. Use the Unified Soil Classification System.
Note For Problems 9.6-1 through 9.6-5, use the lower-bound moment of inertia for deflection of the composite section. Compute this as illustrated in Example 9.7. 9.6-4 For the beam of Problem 9.4-1, a. Compute the deflections that occur before and after the concrete has cured. b. If the total deflection after the concrete has cured exceeds L240 select another steel shape using either LRFD or ASD.

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A double-T simply supported concrete beam its cross section is shown in Figure, is prestressed with 2 tendons each 400 mm². Determine the allowable service load. Use span 12 m, fse = 1300 MPa, fc = 40 MPa, y = 25 kN/m³. = 1200 mm >09< *100* As = 400 +100+ As = 400 400 1400+

A double-T simply supported concrete beam its cross section is shown in Figure, is prestressed with 2 tendons each 400 mm². Determine the allowable service load. Use span 12 m, fse = 1300 MPa, fc = 40 MPa, y = 25 kN/m³. = 1200 mm >09< 100 As = 400 +100+ As = 400 400 1400+