Steel Design (Activate Learning with these NEW titles from Engineering!)
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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Question
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Chapter 9, Problem 9.5.3P
To determine

(a)

Use LFRD method to select the W?X? satisfactory shape.

Expert Solution
Check Mark

Answer to Problem 9.5.3P

Therefore, W16X31 is satisfactory to use.

Explanation of Solution

Given:

Thickness of slab, t = 5.0 inches, spacing = 7.0 ft, span length, L = 30 feet, yield stress = 50 Ksi

Construction load = 20 psf, and live load = 800 psf.

The value of fc'=4ksi.

Concept Used:

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 9, Problem 9.5.3P

Calculation:

Using LRFD method, we have

To select the suitable W 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=5.012(150)×7.0WS=437.5lbft.

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=437.5lbft

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

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=800psf×7.0ftWD=5600lbft

Calculate the factored uniformly distributed load after curing has 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=((1.2×WD)+(1.6×WL))KipsftWu=((1.2×0.4375)+(1.6×5.600))KipsftWu=0.525Kipsft+8.96KipsftWu=9.485Kipsft

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=(9.485×302)8Kipsftft2Mu=1067.06Kips-ft.Mu=1067.00Kips-ft.

Let’s try 18-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=3.4×1067Kips-ft×120.9×50Ksi[18inch2+5.0inch1inch2]W=71.663lbft.

Let’s try for W 18 X 86 and note the properties and dimensions from the Manual.

DesignationImperial (in x lb/ft) Depthh (in) Widthw (in) Web Thicknesstw (in) Flange Thicknesstf (in) Sectional Area (in2) Weight (lbf/ft) Static Parameters
Moment of Inertia Elastic Section Modulus
Ix (in4) Iy (in4) Sx (in3) Sy (in3)
W 18 x 86 18.4 11.09 0.480 0.770 25.3 86 1530 175 166 31.6

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=25.3inches2.

CS=25.3inches2×50KsiCS=1265.00Kips.

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(12×304,12×7.0)b=Min(90in,84in)b=84in.

Substitute the values in the above equation, we get

CC=0.85×4ksi×84in×7inCC=1428kips.

Therefore, the compressive force is as follows:

C=min(CC,CS)C=min(1265.00kips,1428.00kips)C=1265.00kips.

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×84=1265.00kipsa=4.43in

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=18.4in2+5.00in4.43in2y=9.20+5.002.215y=11.985in.

Substitute the values, we get

Mn=Cy

Mn=1265.00Kips×11.985inMn=15161.48inKipsMn=1263.457ftKipsMn=1263.50ftKips

Calculate the design strength of the section as follows:

Md=(ϕb×Mn)

Where, Md is the design strength of the beam.

Substitute the values, we get

Md=(ϕb×Mn)Md=(0.90×1264)Kips.ftMd=1137.60Kips.ft

The flexural strength of the beam after curing is given as:

Md=(ϕb×Mn)=1138Kips.ft

Where, (ϕb×Mn) is the nominal flexural strength and Md is the design strength.

Comparing the values of (ϕbMn) and Mu, We have

1067Kips-ft<1138Kips.ftMu<(ϕbMn)

Thus, the beam is satisfactory in bending after the curing of concrete is complete.

Check for beam weight.

Where, WS is the self-weight of the beam, 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(WS+WD)+(1.6×WL))KipsftWu=((1.2(0.4375psf+0.086psf))+(1.6×(5.6psf)))KipsftWu=0.6282Kipsft+8.96KipsftWu=9.588Kipsft

Calculate the maximum bending moment on the beam;

Mul=wuL28

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

Mul=wuL28Mul=(9.588×302)8Kipsftft2Mul=1078.67Kips-ft.

Check the flexural strength of the beam.

Compare the maximum bending moment and the nominal flexural strength

Comparing the values of (ϕbMn) and Mul, We have

1078.67Kips-ft<1138Kips-ftMul<(ϕbMn)

Thus, the section is safe in flexure including its self-weight.

Check for the shear:

Checking the value of nominal value of shear strength of W18×86 from the AISC Manual.

ϕvVn=265kips

Where, ϕvVn is the nominal shear strength.

The maximum shear force is as following for the above conditions:

Vn=wuL2

Substitute the values, we have

Vn=wuL2Vn=9.588kips ft-1×30ft2Vn=143.82kipsVn=144.00kips

Now comparing the two we have

Vn=144kips<ϕvVn=265kips

Therefore, the beam is safe in shear and we can use W18×86.

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

Wu=((1.2×WD)+(1.6×WL))Kipsft

Load before curing:

Calculate the self-weight of the slab to be allowed by a single beam as

WS=t12(150)Lp

Where, t is the thickness of the slab, Lp is the distance between two adjacent beams.

WS=5.012(150)×7.0ftWS=437.50lbft.

The dead load on the beam before the concrete has cured

Wd=(w+ws)

Where, w is the self-weight of the beam

Substitute the values, we have

Wd=(437.50+86.00)lbftWd=523.50lbft

Now, calculate the live load on the beam as follows:

WL=(wsl)Lp

Where the construction load on the slab is wsl

WL=20psf×7.0ftWL=140.00lbft

The uniformly distributed factored load can be found as

Wu=1.2Wd+1.6Wl

Where, Wd is the uniformly distributed dead load on the beam and applied live load on the beam is Wl.

Substitute the values in the above equation, we have

Wu=[1.2×0.5235]+[1.6×0.140]Wu=0.6282+0.224Wu=0.8522lbft

Calculate the maximum bending moment on the beam;

Mul=wuLAB28

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

Mul=wuLAB28Mul=(0.8522×302)8Kipsftft2Mul=95.8725Kips-ft.Mul=95.90Kips-ft.

Check the value of the nominal flexural strength of W- sections from the Manual:

Flexural strength of the beam before curing is given as follows:

ϕb×Mn=ϕb×MP=698Kips.ft

Where, (ϕb×Mn) is the nominal flexural strength and ϕb×MP is the plastic moment.

Comparing the values of (ϕbMn) and Mu, We have

95.90Kips-ft<698.00Kips.ftMu<(ϕbMn)

Thus, the beam is satisfactory before the curing of concrete is complete.

Therefore, W 18 X 86 is satisfactory to use.

Conclusion:

Therefore, W18X86 is satisfactory to use.

To determine

(b)

Use ASD method to select the W?X? satisfactory shape.

Expert Solution
Check Mark

Answer to Problem 9.5.3P

Therefore, W16X31 is satisfactory to use.

Explanation of Solution

Calculation:

Applying Allowable stress design:

Select appropriate W shape

Calculate the loads on the beam as follows:

Before curing, we have

Calculate the self-weight of the slab to be allowed by a single beam as follows:

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=5.012(150)×7.00WS=437.50lbft.

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=437.50lbft

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

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=800psf×7ftWD=5600.00lbft

Calculate the allowable uniformly distributed load as follows:

Wa=WD+WL

Substitute the values as follows

Wa=0.4375kips-ft+5.600kips-ftWa=6.0375kips-ftWa=6.038kips-ft

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=(6.038×302)8Kipsftft2Mu=679.30Kips-ft.

Let’s try a 18-inch deep beam.

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

W=3.4ΩbMuFY[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 Ωb is the reduction factor.

Estimating weight per unit foot of the steel beam as :

W=3.4ΩbMuFY[d2+ta2]W=3.4×1.67×366.1Kips-ft×1250Ksi[18inch2+5.0inch1inch2]W=68.57lbft.

Let’s try for W 18 X 86 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=25.3inches2.

CS=25.3inches2×50KsiCS=1265.00Kips.

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(12×304,12×7.0)b=Min(90in,84in)b=84in.

Substitute the values in the above equation, we get

CC=0.85×4ksi×84in×5inCC=1428.00kips.

Therefore, the compressive force is as follows:

C=min(CC,CS)C=min(1265kips,1428kips)C=1265.00kips.

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×84=1265kipsa=4.43in

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=18.40in2+5.0in4.43in2y=9.20+5.002.215y=11.985in.y=12.00in.

Substitute the values, we get

Mn=Cy

Mn=1265.00Kips×12.00inMn=15180.00in-KipsMn=1265.00ft-Kips.

Calculate the design strength of the section as follows:

Md=MnΩb

Where, Md is the design strength of the beam

Substitute the values, we get

Md=MnΩbMd=1265.001.67Kips.ftMd=757.485Kips.ftMd=757.50Kips.ft

Check the nominal value of the flexural strength of W- sections from the manual.

Md=MnΩbMd=757.50Kips.ft

Where, MnΩb is the allowable flexural strength.

Comparing the values, we get

679.30Kips.ft<757.50Kips.ftMa<MnΩb.

Thus, the beam is satisfactory after curing of concrete has completed.

Check for beam weight

Calculating the max bending mo0ment by busing the following formula:

Mal=Ma+WaL28

Substitute the values as follows:

Mal=Ma+WaL28Mal=679.30Kipsft+0.086Kipsft[30ft]28Mal=679.30Kipsft+9.675Kipsft.Mal=688.975Kipsft.

Now, checking the flexural strength with the beam taken into consideration:

Compare the maximum bending moment and the nominal flexural strength

Comparing the values of MnΩb and Mu, We have

689.00Kips-ft<757.50Kips-ftMu<MnΩb

Thus, the section is safe in flexure including its self-weight.

Check for the shear:

Checking the value of nominal value of shear strength of W18×86 from the AISC Manual.

VnΩv=177kips

Where, VnΩv is the nominal shear strength.

The maximum shear force is as following for the above conditions:

Va=(wa)L2

Substitute the values, we have

Va=(wa)L2Vn=(6.124)kips ft-1×30ft2Vn=91.86kips

Now comparing the two we have

Va=91.86kips<VnΩv=177kips

Therefore, the beam is safe in shear.

Loading before curing:

Calculate the self-weight of the slab to be allowed by a single beam as follows:

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=5.012(150)×7.0WS=437.50lbft.

The dead load on the beam before the concrete has cured

Wd=(w+ws)

Where, w is the self-weight of the beam

Substitute the values, we have

Wd=(437.50+86.00)lbftWd=523.50lbft

Now, calculate the live load on the beam as follows

WL=(wsl)Lp

Where the construction load on the slab is wsl

WL=20psf×7.0ftWL=140.00lbft

The uniformly distributed factored load can be found as

Wu=1.2Wd+1.6Wl

Where, Wd is the uniformly distributed dead load on the beam and applied live load on the beam is Wl

Substitute the values in the above equation, we have

Wu=[0.5235]+[0.140]Wu=0.6635Kipsft

Calculate the maximum bending moment on the beam;

Mul=wuLAB28

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

Mul=wuLAB28Mul=(0.6635×302)8Kipsftft2Mul=74.60Kips-ft.

Check the value of the nominal flexural strength of W- sections from the Manual:

Flexural strength of the beam before curing is given as follows:

MnΩb=MPΩb=464.00Kips.ft

Where, MnΩb is the allowable flexural strength

Comparing the values of MnΩb and Ma, We have

74.60Kips-ft<464.00Kips.ftMa<MnΩb

Thus, the beam is satisfactory before the curing of concrete is complete.

Conclusion:

Therefore, W18X86 is satisfactory to use.

To determine

(c)

Selecting the stud anchors.

Expert Solution
Check Mark

Answer to Problem 9.5.3P

we will use 118 Studs with 34inch×3inch dimension spaced at 6.0inch.

Explanation of Solution

Calculation:

We have to check whether the studs satisfy the ASIC specifications.

From AISC specifications the maximum stud diameter will be equal to

dmaximum=2.5tf

Where, dmaximum is the maximum diameter and tf is the flange thickness.

Substitute the values in the equation, we get

dmaximum=2.5×0.770dmaximum=1.93in.

Try for, 34in×3in stud is a satisfactory stud element.

The area of stud using the following equation as follows:

Astud=π4d2

Where, the diameter of the stud is d.

Substitute 34in for d

Astud=π4[34]2in2Astud=0.4418in2

We have the modulus of elasticity of concrete as follows:

EC=(wC)32fC'ksi

Where, the modulus of elasticity of concrete is EC,

unit weight of concrete is wC and

the 28-day compressive strength of concrete is fC'.

Substitute 145lbft3 for wC and 4ksi for fC'.

EC=(145lbft3)324ksiEC=1746.03lbft3×2ksiEC=3492.06ksi.

Calculating the shear strength of stud with the following formula:

Qn=0.5AstudfC'EC

Substitute the values, we get

Qn=0.5×0.44184ksi×3492ksiQn=26.11kips

Now, the upper limit of the shear strength is as follows:

Qnn=RgRpAstudFuQnn=1.00×0.75×0.4418in2×65ksiQnn=21.54kips.

As, the calculated value is less than the upper limit given by AISC, take the shear strength as:

Qn=21.54kips.

The number of studs required for the half beam are as follows:

N'=VQn

Substitute the values, we have

N'=1265Kips21.54KipsN'=58.73N'=59

The number of studs required are as follows:

N=2N'

Substitute the values, we have

N=2×59N=118

Calculate the spacing of the studs as following :

The minimum longitudinal spacing for the studs will be as follows:

St,minimum=6d

Where, the minimum longitudinal spacing for the studs is St,minimum and d is the diameter of the stud.

St,minimum=6d

Substitute the values, we have

St,minimum=6dSt,minimum=6×0.75inchSt,minimum=4.5inch.

The minimum transversal spacing for the studs will be as follows:

St,minimum=4d

Where, the minimum transversal spacing for the studs is St,minimum and d is the diameter of the stud.

St,minimum=4d

Substitute the values, we have

St,minimum=4dSt,minimum=4×0.75inchSt,minimum=3inch.

The maximum longitudinal spacing for the studs will be as follows:

St,maximum=8t

Where, the maximum longitudinal spacing for the studs is St,maximum and t is the thickness of the concrete slab.

St,maximum=8t

Substitute the values, we have

St,maximum=8tSt,maximum=8×4.0inchSt,maximum=32inch.

Therefore, the upper limit of the spacing is 32 inches.

Calculating the spacing for a stud at each section by the following formula:

S1=LN

Where, S is the spacing required

S1=LNS1=30×12118S1=3.051inch

Calculating the spacing for two studs at each section by the following formula:

S=2S1

Where, S is the spacing required for two studs.

S=2S1S=2×3.051S=6.102inch

Conclusion:

Therefore, we will use 118 Studs with 34inch×3inch dimension spaced at 6.0inch.

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