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A fully composite floor system consists of 36 -foot-long steel beams spaced at 6 feet 6 inches center-to center supporting a 4 1 2 -inch-thick reinforced concrete floor slab. The steel yield stress is 50 ksi and the concrete strength is f c ' = 4 ksi. There is a construction load of 20 psf and a live load 175 psf. Select a W 16 shape. a. Use LRFD. b. Use ASD. c. Select stud anchors and show the layout on a sketch similar to Figure 9.13.

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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 9, Problem 9.5.1P
Textbook Problem
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A fully composite floor system consists of 36 -foot-long steel beams spaced at 6 feet 6 inches center-to center supporting a 4 1 2 -inch-thick reinforced concrete floor slab. The steel yield stress is 50 ksi and the concrete strength is f c ' = 4 ksi. There is a construction load of 20 psf and a live load 175 psf. Select a W 16 shape.

a. Use LRFD.

b. Use ASD.

c. Select stud anchors and show the layout on a sketch similar to Figure 9.13.

To determine

(a)

Use LFRD method to select the W16X satisfactory shape.

Explanation of Solution

Given:

Thickness of slab, t = 4.5 inches, spacing = 6.5 ft, span length, L = 36 feet, yield stress = 50 Ksi

Construction load = 20 psf, and live load = 175 psf.

The value of fc'=4ksi.

Concept Used:

Calculation:

Using LRFD method, we have

To select the suitable W 16 shape as follows:

Calculate the loads on the beam as follows:

After curing we have,

WS=t12(150)Lp

Where, t is the thickness of the slab, WS the self-weight of slab on one beam, Lp is the distance

between two adjacent beams.

WS=t12(150)Lp

WS=4.512(150)×6.5WS=365.625lbft.

The dead load on the beam after the concrete has cured is:

WD=WS

Where, WD is the dead load on the beam.

Neglecting the beam weight and check for it later.

WD=WS=365.625lbft

Calculate the live load on the beam using the following equation:

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=175psf×6.5ftWD=1137.5lbftWD=1138lbft

Calculate the factored uniformly distributed load after curing has completed by the following formula:

Wu=((1.2×WD)+(1.6×WL))Kipsft

Where, WD is the uniformly distributed dead load on the beam and WL is the applied live load on the beam.

Substitute the values, we get

Wu=((1.2×WD)+(1.6×WL))KipsftWu=((1.2×0.3656)+(1.6×1.138))KipsftWu=2.2595KipsftWu=2.26Kipsft

Calculate the bending moment on the beam;

Mu=wuLAB28

Where, Mu is the bending moment and LAB is the length of the beam.

Mu=wuLAB28Mu=(2.26×362)8Kipsftft2Mu=366.1Kips-ft.

Let’s try a 16-inch deep beam.

Select a shape with the limiting self-weight given by the following formula:

W=3.4MuϕFY[d2+ta2]

Where, t is the thickness of the concrete slab, d is the depth of the steel beam, a is the distance of the neutral axis from the top, FY is the yield strength of steel and ϕ is the reduction factor.

Estimating weight per unit foot of the steel beam as:

W=3.4MuϕFY[d2+ta2]W=3.4×366.1Kips-ft0.9×50Ksi[16inch2+4.5inch1inch2]W=27.66lbft.

Let’s try for W 16 X 31 and note the properties and dimensions from the Manual.

Calculate the strength of the section as following below:

C=min(CC,CS)

Where, the compressive force is C, the compressive force of concrete is CC, and the compressive force of steel beam is CS.

Calculate the compressive force in steel as follows:

CS=ASFY

Where, the area of steel section is AS and the yield strength of steel is FY

Get the value of AS from properties and dimensions table from part 1 of the manual.

AS=9.13inches2.

CS=9.13inches2×50KsiCS=456.5Kips.

Calculate the compressive force in concrete as following:

CC=0.85fC'bt

Where, b is the width of the concrete slab, t is the thickness of the concrete beam and fC' is the compressive strength of the concrete.

The effective flange width is as follows:

b=Min(12LABt,12Lp)b=Min(12×364,12×6.5)b=Min(108in,78in)b=78in.

Substitute the values in the above equation, we get

CC=0.85×4ksi×78in×4.5inCC=1193.4kips.

Therefore, the compressive force is as follows:

C=min(CC,CS)C=min(1193.4kips,456.5kips)C=456.5kips.

Therefore, the plastic neutral axis lies in the slab.

Calculate the tensile force (T) and compressive force (C) and the position of plastic neutral axis from the top of concrete slab by following formula:

C=T

0.85fC'ab=ASFy0.85fC'ab=C

0.85×4Ksi×a×78=456.5kipsa=1.721in

Compute the flexural strength as:

Mn=Cy

Where, Mn is the flexural strength and y is the distance of action of tensile force from plastic

neutral axis.

Compute the value of Y as following below:

y=d2+ta2y=15.9in2+4.5in1.721in2y=7.95+4.500.8605y=11.59in.

Substitute the values, we get

Mn=Cy

Mn=456.5Kips×11.59inMn=5290.835inKipsMn=440.90ftKips

Calculate the design strength of the section as follows:

Md=(ϕb×Mn)

Where, Md is the design strength of the beam

To determine

(b)

Use ASD method to select the W16X satisfactory shape.

To determine

(c)

Selecting the stud anchors for the given section.

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Chapter 9 Solutions

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