A fully composite floor system consists of 40 -foot-long, W 21 × 57 steel beams spaced at 9 feet center-to-center and supporting a 6 -inch-thick reinforced concrete floor slab. The steel is A992 and the concrete strength is f c ' = 4 ksi. There is a construction load of 20 psf and a live load of 250 psf. a. Determine whether this beam is adequate for LRFD. b. Determine whether the beam is adequate for ASD. c. How many 3 4 − i n c h × 3 − i n c h stud anchors are required for full composite behavior?

(a)

Whether the beam W  21×57 is satisfactory by using LFRD method.

Given:

Thickness of slab, t = 6.0 inches, spacing = 9.0 ft, span length, L = 40 feet, yield stress = 50 Ksi

Construction load = 20 psf, and live load = 250 psf.

The value of fc'=4 ksi.

Calculation:

Using LRFD method, we have

To check whether or not W 21 X 57 shape is 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=6.012(150)×9.0WS=675.00  lbft.

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

WD=  W+WS

Where, WD is the dead load on the beam and W is the self-weight.

Neglect the beam weight and check for it later.

WD=  W+WSWD=675lbft+57lbftWD=732lbft

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

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=250 psf×9.0ftWD=2250.00 lbft.

Calculate the strength of the section as 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=16.7  inches2.

CS=16.7  inches2×50 KsiCS= 835.00 Kips.

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(12 LABt, 12 Lp) b=Min  (12×404,12×9.0) b= Min(120 in,108 in) b= 108 in.

Substitute the values in the above equation, we get

CC=0.85×4ksi× 108in×4 inCC=1468.8  kips.

Therefore, the compressive force is as follows:

C=min (CC, CS)C=min (1468.8 kips,835.00kips )C=  835.00  kips.

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

0.85 fC' a b=ASFy0.85 fC' a b=C

0.85 ×4Ksi× a× 108=835.00 kipsa =2.2739  ina =2.274  in

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+t−a2y=21.1  in2+6.0 in  −2.274  in2y=10.55+6.0−1.137y=16.55 in−1.137  in.y=15.413 in.

Substitute the values, we get

Mn=Cy

Mn=835.00 Kips ×  15.413 inMn=12869.855 in −KipsMn=1072.5ft −Kips

Calculate the factored uniformly distributed load before curing is completed by 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

(b)

Whether the beam W 21×57 is satisfactory by using ASD method.

(c)

Number of studs required.