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
Question
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Chapter 5, Problem 5.15.6P
To determine

(a)

Whether a W6×12 is adequate or not by using LRFD

Expert Solution
Check Mark

Answer to Problem 5.15.6P

Adequate

Explanation of Solution

Given:

Total gravity load = 40 psf of roof surface

W6×12 of A992 steel

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  1

Formula used:

Lp=1.76ryEFy

Lpis unbraced length in an inelastic behavior

c=h02IyCw

Lr=1.95rtsE0.7FyJcSxh0+(JcSxh0)2+6.76(0.7FyE)2

Lris unbraced length in an elastic behavior

Mn=Cb[Mp(Mp0.7FySx)(LbLpLrLp)]Mp

Mn is nominal moment strength

Mpis plastic moment capacity

Calculation:

Determine the nominal flexural strength about x and y axes:

Neither the beam design charts nor the Z tables include shapes smaller than W8, so the flexural strength of the W6×12 must be computed.

From the dimensions and properties tables, the shape is compact.

The following properties of a W6×12 are below:

A=3.55in.2rts=1.08in.h0=5.75in.Sx=7.31in.3Zx=8.30in.3Iy=2.99in.4ry=0.918in.J=0.0903in.4Sy=1.50in.3

A is Cross-sectional area

Sxis Elastic section modulus about X -axis

Zxis Plastic section modulus about X -axis

Iyis Moment of inertia about Y -axis

ryis Radius of gyration about Y -axis

Syis Elastic section modulus about Y -axis

rts is IyCwSx

Cwis Warping constant

h0is Distance between centroid of flanges

J is Torsional moment of inertia

Lp=1.76ryEFy=1.76(0.918)2900050=38.91in.=3.243ftLr=1.95rtsE0.7FyJcSxh0+(JcSxh0)2+6.76(0.7FyE)2=1.95(1.08)290000.7(50)0.0903×17.31×5.75+(0.0903×17.31×5.75)2+6.76(0.7(50)29000)2=134.6in.=11.22ft.

For Lb=14ft, Lp<Lb<Lr, so

Mn=Cb[Mp(Mp0.7FySx)(LbLpLrLp)]Mp (Inelastic Lateral torsional buckling)

From the below given figure in the textbook, Cb=1.14

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  2

Mp=FyZx=50(8.30)=415inkips=34.58ftkipsMn=1.14[415(4150.7×50×7.31)(103.2311.223.243)]=319.4inkips=26.62ftkips<Mp

For the y axis, since the shape is compact, there is no flange local buckling

Mny=Mpy=FyZy=50(2.32)=116inkips=9.667ftkips

Check the upper limit:

ZySy=2.321.50=1.55<1.6Mny=Mpy=9.667ftkipsϕbMnx=0.90(26.62)=23.96ftkipsϕbMny=0.90(9.667)=8.7ftkips

Roof load: Combination 3 controls

wu=1.2D+1.6S=1.2×402+1.6×402=56psf

where,D is dead load and S is snow load

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  3

Tributary width = 174(6)=6.185ft

Purlin load = 56(6.185)=346.4lb/ft

Component normal to roof = wux=417(346.4)=336.1lb/ft

Component parallel to roof = wuy=117(346.4)=84.01lb/ft

Calculate factored bending moment about x axis and y axis

Mux=wuxL28=0.3361(10)28=4.201ftkipsMuy=wuyL28=0.08401(10)28=1.05ftkips

Use ½ of weak-axis bending strength in the interaction equation:

MuxϕbMnx+MuyϕbMny=4.20123.96+1.058.7/2=0.417<1.0 (OK)

Conclusion:

W6×12 is adequate.

To determine

(b)

Whether a W6×12 is adequate or not by using ASD

Expert Solution
Check Mark

Answer to Problem 5.15.6P

Adequate

Explanation of Solution

Given:

Total gravity load = 40 psf of roof surface

W6×12 of A992 steel

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  4

Formula used:

Lp=1.76ryEFy

Lpis unbraced length in an inelastic behavior

c=h02IyCw

Lr=1.95rtsE0.7FyJcSxh0+(JcSxh0)2+6.76(0.7FyE)2

Lris unbraced length in an elastic behavior

Mn=Cb[Mp(Mp0.7FySx)(LbLpLrLp)]Mp

Mn is nominal moment strength

Mpis plastic moment capacity

Calculation:

Determine the nominal flexural strength about x and y axes:

Neither the beam design charts nor the Z tables include shapes smaller than W8, so the flexural strength of the W6×12 must be computed.

From the dimensions and properties tables, the shape is compact.

The following properties of a W6×12 are below:

A=3.55in.2rts=1.08in.h0=5.75in.Sx=7.31in.3Zx=8.30in.3Iy=2.99in.4ry=0.918in.J=0.0903in.4Sy=1.50in.3

A is Cross-sectional area

Sxis Elastic section modulus about X -axis

Zxis Plastic section modulus about X -axis

Iyis Moment of inertia about Y -axis

ryis Radius of gyration about Y -axis

Syis Elastic section modulus about Y -axis

rts is IyCwSx

Cwis Warping constant

h0is Distance between centroid of flanges

J is Torsional moment of inertia

Lp=1.76ryEFy=1.76(0.918)2900050=38.91in.=3.243ftLr=1.95rtsE0.7FyJcSxh0+(JcSxh0)2+6.76(0.7FyE)2=1.95(1.08)290000.7(50)0.0903×17.31×5.75+(0.0903×17.31×5.75)2+6.76(0.7(50)29000)2=134.6in.=11.22ft.

For Lb=14ft, Lp<Lb<Lr, so

Mn=Cb[Mp(Mp0.7FySx)(LbLpLrLp)]Mp (Inelastic Lateral torsional buckling)

From the below given figure in the textbook, Cb=1.14

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  5

Mp=FyZx=50(8.30)=415inkips=34.58ftkipsMn=1.14[415(4150.7×50×7.31)(103.2311.223.243)]=319.4inkips=26.62ftkips<Mp

For the y axis, since the shape is compact, there is no flange local buckling

Mny=Mpy=FyZy=50(2.32)=116inkips=9.667ftkips

Check the upper limit:

ZySy=2.321.50=1.55<1.6Mny=Mpy=9.667ftkipsMnxΩb=26.621.67=15.94ftkips

Roof load: Combination 3 controls

wa=D+S=402+402=40psf

where, D is dead load and S is snow load

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 5, Problem 5.15.6P , additional homework tip  6

Tributary width = 174(6)=6.185ft

Purlin load = 40(6.185)=247.4lb/ft

Component normal to roof = wax=417(247.4)=240lb/ft

Component parallel to roof = way=117(247.4)=60lb/ft

Calculate factored bending moment about x axis and y axis

Max=waxL28=0.24(10)28=3ftkipsMay=wayL28=0.06(10)28=0.75ftkips

Use ½ of weak - axis bending strength in the interaction equation:

MaxMnx/Ωb+MayMny/Ωb=315.94+0.755.789/2=0.447<1.0 (OK)

Conclusion:

W6×12 is not adequate.

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Chapter 5 Solutions

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ISBN:9781337094740
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Publisher:Cengage Learning