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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Chapter 9, Problem 9.8.7P
To determine

(a)

To select the W?X? satisfactory shape using LFRD method.

Expert Solution
Check Mark

Answer to Problem 9.8.7P

The beam W24×76 is safe.

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.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 9, Problem 9.8.7P

Calculation:

Using LRFD method, we select a suitable shape that will satisfy the given conditions:

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=437.50lbft.

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

WD=WSWD=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=800.00psf×7.0ftWD=5600.00lbft.

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+8.960KipsftWu=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.

Try for W24×76, we have from the manual of ASIC

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 24 x 76 23.9 9 0.440 0.680 22.4 76 2100 82.5 176 18.4

Qn=394Kips.ft

Calculate the distance of the plastic neutral axis from the top of the slab as follows:

CC=0.85fC'bt

Where, b is the width of the concrete slab, t is the thickness of the concrete beam, fC' is the compressive strength of the concrete and a is the distance of neutral axis from the top of the slab.

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, we have

0.85fC'ab=C

0.85×4Ksi×a×84=394.00kipsa=394.00kips0.85×4Ksi×84a=1.3795ina=1.380in

Compute the value of Y as shown below:

y=ta2y=5.0in1.380in2y=5.0in0.690in.y=4.310in.

From the manual the value of nominal flexural strength of the beam

ϕb×Mn=1102Kips.ft

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

1067.00Kips-ft<1102.00Kips.ftMu<(ϕbMn)

Thus, the beam is satisfactory excluding its self-weight.

Now for including the weight of the beam, we have

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

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+0.076psf))+(1.6(5.60psf)))KipsftWu=0.6162Kipsft+8.960KipsftWu=9.5762Kipsft

Calculate the maximum bending moment on the beam;

Mu=wuL28

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

Mu=wuL28Mu=(9.5762×302)8Kipsftft2Mu=1077.323Kips-ft.

Check the flexural strength of the beam including its weight.

1077.00Kips-ft<1102.00Kips.ftMul<(ϕbMn)

Thus, the beam is satisfactory including its self-weight.

Check for the shear:

Checking the value of nominal value of shear strength of W24×76 from the AISC Manual.

ϕvVn=315.00kips

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=9.576kips ft-1×30ft2Vn=143.00kips.

Now comparing the two we have

Vn=143.00kips<ϕvVn=315.00kips

Therefore, the beam is safe in shear and we can use W24×76.

Calculate the factored uniformly distributed load before 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.

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 is:

WD=WS+WWD=437.50lbft+76.00lbftWD=513.50lbft

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

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=20.00psf×7.0ftWD=140.00lbft.

Substitute the values, we get

Wu=((1.2×WD)+(1.6×WL))KipsftWu=((1.2×0.5135)+(1.6×0.140))KipsftWu=0.6162Kipsft+0.224KipsftWu=0.8402Kipsft

Calculate the maximum bending moment on the beam;

Mu=wuL28

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

Mu=wuL28Mu=(0.8402×302)8Kipsftft2Mu=94.523Kips-ft.

Check for the value of nominal flexural strength, the flexural strength of the beam before curing is

ϕb×Mn=ϕb×MP=315.00Kips.ft

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

94.00Kips-ft<315.00Kips.ftMu<(ϕbMn)

Therefore, the beam is satisfactory before the curing has completed.

Now calculating the maximum allowable live load deflection from the given beam using the formula as:

ΔMax=L360

Substitute the values, we have

ΔMax=L360ΔMax=30×12360ΔMax=360360inΔMax=1.00in

We have the value of lower bound moment of inertia for the given condition as follows:

Itb=3642in4

Calculating the total load on the beam using the following :

Δ=5wLL4384EItb

Where, Δ is the maximum deflection, L is the length of the beam, W is the uniformly distributed load on beam, E is the modulus of elasticity and I is the moment of inertia.

Δ=5wLL4384EItbΔ=5×5.60Kips.ft×(30×12)4384×12×29000Ksi×3642in4Δ=0.9663in.

Now by comparing the values, we have

ΔMax=1.00in>Δ=0.9663in.

Conclusion:

Hence, the beam W24×76 is safe.

To determine

(b)

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

Expert Solution
Check Mark

Answer to Problem 9.8.7P

The beam W24×76 is safe.

Explanation of Solution

Calculation:

Now, we will use allowable stress design

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=437.50lbft.

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

WD=WSWD=WS=437.50lbft

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

WL=WSLLP

Where, WSL is the construction load on the beam

WL=WSLLPWL=800.00psf×7.0ftWL=5600.00lbft.

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

Wa=(WD+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

Wa=(WD+WL)KipsftWu=0.4375Kipsft+5.60KipsftWu=6.0375Kipsft

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=(6.0375×302)8Kipsftft2Mu=679.22Kips-ft.

Compute the value of Y, which is the distance from the top of steel shape to compressive force

in concrete and is shown below:

y=ta2y=5.0in1.00in2y=5.0in0.50in.y=4.50in.

Try for W24×76, we have from the manual of ASIC

Qn=335.00Kips.ft

Calculate the distance of the plastic neutral axis from the top of the slab as follows:

CC=0.85fC'bt

Where, b is the width of the concrete slab, t is the thickness of the concrete beam, fC' is the compressive strength of the concrete and a is the distance of neutral axis from the top of the slab.

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, we have

0.85fC'ab=C

0.85×4Ksi×a×84=335.00kipsa=335.00kips0.85×4Ksi×84a=1.173in

Compute the value of Y as shown below:

y=ta2y=5.0in1.173in2y=5.0in0.5865in.y=4.414in.

From the table 3-19 of the ASIC manual:

Trying for W21×48

Check whether the section is safe in flexure if the self -weight is excluded

MnΩb=732.00ft-kips

Where, MnΩb is the allowable flexural strength.

Now comparing the values of Ma and MnΩb

Substitute the values, we have

MnΩb=732.00ft-kips  >  Mu=679.00ft-kips

Therefore, the section is safe in flexure if the self -weight is excluded.

Let’s check for the beam weight :

Calculation of the maximum bending moment as follows:

Mal=Ma+wsL28

Substitute the values, we get

Mal=Ma+wsL28Mal=679.00ft-kips+0.076kips ft-1×(30ft)28Mal=679.00ft-kips+8.55ft-kipsMal=687.55ft-kips

Now check for the flexural strength including the beam weight:

Comparing the values of maximum bending moment and the nominal flexural strength as follows:

Mal=687.55ft-kips<MnΩb=732.00ft-kips

Therefore, the section is safe in flexure including the self-weight of the beam.

Check for the shear:

Checking the value of nominal value of shear strength of W24×76 from the AISC Manual.

VnΩb=210.00kips

Where, VnΩb is the nominal shear strength.

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

Va=waL2

Substitute the values, we have

Va=waL2Va=6.114kips ft-1×30ft2Va=91.71kips

Now comparing the two we have

Va=91.71kips<VnΩb=210.00kips

Therefore, the beam is safe in shear and we can use W24×76.

Calculate the loads on the beam as follows:

Before 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=437.50lbft.

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

WD=WS+WWD=437.50lbft+76.00lbftWD=513.50lbft

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

WD=WSLLP

Where, WSL is the construction load on the beam.

WD=WSLLPWD=20.00psf×7.0ftWD=140.00lbft.

Substitute the values, we get

Wu=(WD+WL)KipsftWu=0.5135Kipsft+0.140KipsftWu=0.6535Kipsft

Calculate the maximum bending moment on the beam;

Mu=wuL28

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

Mu=wuL28Mu=(0.6535×302)8Kipsftft2Mu=73.52Kips-ft.

Checking the value of nominal flexural strength of W- section from the ZX table of the ASIC manual:

Check whether the section is safe in flexure if the self -weight is excluded

MnΩb=MPΩb=499.00ft-kips

Where, MnΩb is the allowable flexural strength and MPΩb is the plastic moment.

Now comparing the values of Ma and MnΩb

Substitute the values, we have

MnΩb=499.00ft-kips  >  Ma=73.52ft-kips

Therefore, the section is safe in before the curing of concrete.

Now calculating the maximum allowable live load deflection from the given beam using the formula as:

ΔMax=L360

Substitute the values, we have

ΔMax=L360ΔMax=30×12360ΔMax=360360inΔMax=1.00in

We have the value of lower bound moment of inertia for the given condition as follows:

Itb=3642in4

Calculating the total load on the beam using the following :

Δ=5wLL4384EItb

Where, Δ is the maximum deflection, L is the length of the beam, W is the uniformly distributed load on beam, E is the modulus of elasticity and I is the moment of inertia.

Δ=5wLL4384EItbΔ=5×5.60Kips.ft×(30×12)4384×12×29000Ksi×3642in4Δ=0.9663in.

Now by comparing the values, we have

ΔMax=1.00in>Δ=0.9663in.

Conclusion:

Hence, the beam W24×76 is safe.

To determine

(c)

Selecting the stud anchors.

Expert Solution
Check Mark

Answer to Problem 9.8.7P

We will use 54 Studs with 58inch×212inch dimension, with the spacing of 6.50in.

Explanation of Solution

Calculation:

From AISC specifications, compute the maximum stud diameter using the equation:

dmax=2.5tf

Where, tf id the flange thickness and dmax is the maximum diameter.

Substitute the values, we get

dmax=2.5tfdmax=2.5×0.680indmax=1.70in

Try for the studs of size 58in×212in studs.

From table 3-21 for lightweight concrete take one stud at each of the beam position.

Qn=15.00Kips

The number of studs for half beam can be found as follows:

N1=QnQn.

Substitute the values, we have

N1=QnQn.Qn=VN1=394.00Kips15.00Kips.N1=26.27N127

Compute the number of studs as follows:

N=2N1.

Substitute the value of N1, we have

N=2N1.N=2×27N=54

Calculate the spacing of the studs as follows:

Compute the minimum longitudinal spacing for studs using the equation

St,min=6d

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

Substitute the values

St,min=6dSt,min=6×58inSt,min=3.75in

Compute the minimum transverse spacing for studs using the equation

St,min=4d

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

Substitute the values

St,min=4dSt,min=4×58inSt,min=2.50in

Compute the maximum longitudinal spacing for studs using the equation

St,max=8t

Where, St,max is the minimum transverse spacing for studs and d is the thickness of the concrete slab.

Substitute the values

St,max=8tSt,max=8×5inSt,max=40in

But the upper limit of the spacing is 36 inches.

Calculate the require spacing for one stud at each of the section:

S=LN

Substitute the values, we have

S=30×1254S=6.670in.

Conclusion:

Therefore, we will use 54 Studs with 58inch×212inch dimension, with the spacing of 6.50in.

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