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 . 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 × 0.525 d maximum = 1. 3125 in . Comparing the diameter of the stud and maximum diameter, we have: Maximum diameter is d maximum = 1. 3125 in and stud diameter is d = 0.75 in 1. 3125 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 Calculate the spacing of studs using the following equation. S = L N Where, S is the spacing. Substitute the values, we have S = L N S = 40 × 12 34 S = 14.12 i n . Implying that one stud at each location is satisfactory. 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 × 0.4417 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 ' = N 2 Where, S is the spacing. Substitute the values, we have: N ' = N 2 N ' = 34 2 N ' = 17. 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 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 = 11.8 i n c h e s 2 . C S = 11.8 i n c h e s 2 × 50 K s i C S = 590 K i p s . We have the effective flange width of the following section as follows: b = min of [ 12L AB t inches , 12L p inches ] b = min of [ 12×40 4 inches, 12×10 inches ] b = min of [ 120 inches, 120 inches ] b = 120 inches . Calculate the compressive force in concrete below: 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 × 120 i n × ( 4.5 − 1.5 ) i n C C = 0.85 × 4 k s i × 120 i n × ( 3 ) i n C C = 1224 k i p s Compute the shear strength of the studs using the equation. ∑ Q n = Q n × N ' Where, ∑ Q n is the combined shear strength of the studs. ∑ Q n = Q n × N ' ∑ Q n = 17.23 × 17 kips . ∑ Q n = 292.91 kips . Therefore, the compressive force is as follows: C = min ( C C , C S , Q n ) C = min ( 1224 k i p s , 590 k i p s , 292.91 k i p s ) C = 292.91 k i p s . The plastic neutral axis is determined by compressive strength of concrete as it lies in the steel beam. The distance of the plastic neutral axis from the top of the steel beam by equating the compressive force and tensile force as given below: C + C S − T = 0 Rearranging the equation as follows: C + F y b f t ' − F y ( A s t u d − b f t ' ) = 0 Substituting the required values, we get: C + F y b f t ' − F y ( A s t u d − b f t ' ) = 0 292.91 + 50 K s i × 6.02 i n × t ' − 50 K s i ( 11.8 i n 2 − 6.02 i n × t ' ) = 0 292.91 + 301 t ' − 590 K i p s + 301 t ' = 0 602 t ' − 297.09 K i p s = 0 503 t ' = 297.09 K i p s t ' = 297.09 503 t ' = 0.5906 t ' = 0.591 i n . Now, the compressive strength Imparted by the steel beam using the equation as follows: C S = F y b f t ' Where, C S is the compressive strength imparted by the steel beam. Substituting the values, we get: C S = F y b f t ' C S = 50 K s i × 6 To determine (b) The adequate strength of the composite section by ASD method.

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…

Question

Chapter 9, Problem 9.5.2P

To determine

(a)

The adequate strength of the composite section by LFRD method.

To determine

(b)

The adequate strength of the composite section by ASD method.

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

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(a) The adequate strength of the composite section by LFRD method.

Solution Summary: The author compares the stud diameter and the maximum diameter of the composite section by LFRD method.