Explanation Given information: A W 14X 26 composite beam, thickness, t = 4.5 inches, effective slab width = 66 inches, span length, L = 30feet. The value of f c ' = 4 k s i . Diagram of composite section : Calculation: We have to check whether the studs satisfy the ASIC specifications. From AISC specifications the maximum stud diameter will be equal to d maximum = 2.5 t f Where, d maximum is the maximum diameter and t f is the flange thickness. Substitute the values in the equation, we get d maximum = 2.5 × 0.420 d maximum = 1.05 in . Comparing the diameter of the stud and maximum diameter, we have: Maximum diameter is d maximum = 1.05 in = 1.05 in and stud diameter is d = 0.75 in . 1.05 in > 0.75 in d maximum > d From the formed steel deck, the maximum diameter according to the AISC specifications, we have: d maximum formed = 0.75 i n Comparing the diameter of the stud and maximum diameter for formed steel deck, we have 0.75 i n = 0.75 i n d = d maximum formed Therefore, 3 4 i n × 3 i n stud is a satisfactory stud element. The area of stud using the following equation as follows: A s t u d = π 4 d 2 Where, the diameter of the stud is d. Substitute 3 4 i n for d A s t u d = π 4 [ 3 4 ] 2 i n 2 A s t u d = 0.4417 i n 2 We have the modulus of elasticity of concrete as follows: E C = ( w C ) 3 2 f C ' k s i Where, the modulus of elasticity of concrete is E C , unit weight of concrete is w C and the 28-day compressive strength of concrete is f C ' . Substitute 145 l b f t 3 for w C and 4 k s i for f C ' . E C = ( 145 l b f t 3 ) 3 2 4 k s i E C = 1746.03 l b f t 3 × 2 k s i E C = 3492.06 k s i . Calculating the shear strength of stud with the following formula: Q n = 0.5 A s t u d f C ' E C Substitute the values, we get: Q n = 0.5 × 0.4418 4 k s i × 3492 k s i Q n = 26.11 k i p s Now, the upper limit of the shear strength is as follows: Q n n = R g R p A s t u d F u Q n n = 1 × 0.6 × 0.4418 i n 2 × 65 k s i Q n n = 17.23 k i p s . As, the calculated value is less than the upper limit given by AISC, take the shear strength as: Q n n = 17.23 k i p s . The number of studs required for the half beam are as follows: N ' = L 2 S Where, S is the spacing Substitute the values, we have: N ' = L 2 S N ' = 30 × 12 2 × 3 × 6 N ' = 10. Now, the compressive force of the beam will be selected in such a way that the smallest of the compressive force in steel, compressive force of concrete slab and combined shear strength of studs. Calculate the compressive force as following below: C = min ( C C , C S , ∑ Q n ) Where, the compressive force is C, the compressive force of concrete is C C , and the compressive force of steel beam is C S . Calculate the compressive force in steel as follows: C S = A S F Y Where, the area of steel section is A S and the yield strength of steel is F Y . Get the value of A S from properties and dimensions table from part 1 of the manual. A S = 7.69 i n c h e s 2 . C S = 7.69 i n c h e s 2 × 50 K s i C S = 384.5 K i p s . Calculate the compressive force in concrete as follows: C C = 0.85 f C ' b ( t − c ) Where, b is the width of the concrete slab, t is the thickness of the concrete beam, c is the thickness of the steel deck and f C ' is the compressive strength of the concrete. C C = 0.85 f C ' b ( t − c ) C C = 0.85 × 4 k s i × 66 i n × ( 4

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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Composite Construction

Composite construction is the construction of structures in which two or more materials are used. Regarding structural engineering, composite construction is a construction where two materials are bound together in a technical manner such that both the m…

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Chapter 9, Problem 9.8.1P

To determine

The flexural strength of the composite section.

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Chapter 9 Solutions

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

Ch. 9 - Prob. 9.1.1P Ch. 9 - Prob. 9.1.2P Ch. 9 - Prob. 9.1.3P Ch. 9 - Prob. 9.1.4P Ch. 9 - Prob. 9.1.5P Ch. 9 - Prob. 9.1.6P Ch. 9 - A W1422 acts compositely with a 4-inch-thick floor...Ch. 9 - Prob. 9.2.2P Ch. 9 - Prob. 9.3.1P Ch. 9 - Prob. 9.3.2P

Ch. 9 - Prob. 9.4.1P Ch. 9 - Prob. 9.4.2P Ch. 9 - Prob. 9.4.3P Ch. 9 - Prob. 9.4.4P Ch. 9 - Prob. 9.4.5P Ch. 9 - Prob. 9.5.1P Ch. 9 - Prob. 9.5.2P Ch. 9 - Prob. 9.5.3P Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Note For Problems 9.6-1 through 9.6-5, use the...Ch. 9 - Prob. 9.7.1P Ch. 9 - Prob. 9.7.2P Ch. 9 - Prob. 9.7.3P Ch. 9 - Prob. 9.7.4P Ch. 9 - Prob. 9.8.1P Ch. 9 - Prob. 9.8.2P Ch. 9 - A beam must be designed to the following...Ch. 9 - Prob. 9.8.4P Ch. 9 - Prob. 9.8.5P Ch. 9 - Prob. 9.8.6P Ch. 9 - Prob. 9.8.7P Ch. 9 - Prob. 9.8.8P Ch. 9 - Use the composite beam tables and select a W-shape...Ch. 9 - Prob. 9.8.10P Ch. 9 - Prob. 9.10.1P Ch. 9 - Prob. 9.10.2P

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