   # Same as Problem 9.8-5, but use ASD. ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740

#### Solutions

Chapter
Section ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 9, Problem 9.8.6P
Textbook Problem
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## Same as Problem 9.8-5, but use ASD.

To determine

(a)

Selection of W16× shape from ASD method.

### Explanation of Solution

Given:

Thickness of slab, t = 4 inches, spacing = 8.0 ft, span length, L = 36 feet, height of the deck rib, H =2 Inches.

The value of fc'=4ksi.

Calculation:

WS=t12×(150)(LP)

Where, the thickness of the slab is t=4.0in and the unit weight of concrete is γC=150pcf.

WS=412×(150)(8in)

WS=400.00lbft

WD=WS+WPLp+WCLP

Where, WD is the dead load on the beam, WC is the ceiling load, WP is the partition load and Lp is the distance between the two adjacent beams.

Substitute the values in the above equation, we get:

WDWS + WPLp+WCLPWD=400lbft + (20psf × 8ft)+(8psf × 8ft)WD=400lbft + 160lbft + 64lbftWD=624lbft.

Compute the live load on the beam using following equation.

WL=(wsl)Lp

Where, wsl is the service load on the slab and wsl is the partition load.

Substitute the values, we get:

WL= 100 × 8.0lbftWL= 800 lbft.

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

Wa=[WD+WL]lbft

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=((WD)+(WL))lbftWu=((624)+(800))lbftWu=1424lbft

Calculate the bending moment acting at the center.

Ma=wuL28

Where, Ma is the bending moment and L is the length of the beam.

Ma=wuL28Ma=(1.424×362)8Kipsftft2Ma=231Kips-ft.

Let’s try for W16×31

From the table 3-19 for the plastic neutral axis at 4 inches from top:

Qn=335ft-kips

We have the effective flange width as follows:

b=min[12Ltinches,12Lpinches]

Where, L is the length of the beam and Lp is the distance between the two beams.

b=min[12Ltinches,12Lpinches]b=min[12×364inches,12×8inches]b=min[108inches,96inches]b=96inches.

Now for selecting the shape of the composite section as follows:

Calculate the distance from the top of steel shape to compressive force in concrete :

Y2=ta2

Location of the plastic neutral axis from the top of the slab is as follows;

Substitute the values, we have:

Y2=ta2

Y2=412Y2=40.5Y2=3.5inches.

Compute the distance of the plastic neutral axis from the top of the concrete slab as follows;

C=0.85fC'ab

As the neutral axis lies in concrete slab, the full area of steel is in tension and the tensile force of

concrete below the neutral axis is neglected.

Substitute the values in the above equation, we get:

0.85×4ksi×a×96in=335kipsa=335kips0.85×4ksi×96ina=1.0264in.

Compute the value of Y2 (distance between top of the steel to the centroid of the concrete)

below:

y2=ta2y2=4in0

To determine

(b)

Number of shear studs required.

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