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
Book Icon
Chapter 4, Problem 4.7.3P
To determine

(a)

If the compression member is adequate to support the loads using LRFD.

Expert Solution
Check Mark

Answer to Problem 4.7.3P

W12×79 is adequate

Explanation of Solution

Given information:

Dead load,DL=180kips

Live load,LL=320kips

Length of the column,L=28ft

W12×79Kx=1.0Ky=1.0bf=12.1intf=0.735inh/tw=20.7

Steel type=A572

Yield stress=60ksi

Concept used:

Slenderness ratio=KLr

Fe=π2E(KL/r)2

ϕcPn=ϕc×Fcr×Ag

Pu=1.2(DL)+1.6(LL)

KEffective length factorrRadius of gyrationFeBuckling stressFcrCritical stress

EModulus of elasticityPnNominal compressive strengthAgGross areaϕcResistance factor=0.90ΩcSafety factor=1.67

Calculation:

From AISC manual, the radius of gyration for the given material is 5.34 in. and 3.05 in.

KxLrx=1.0×28ft×12in1ft5.34=62.92

KyLry=1.0×16ft×12in1ft3.05=62.95

Taking the higher value of slenderness ratio is 62.95.

Fe=π2E(KL/r)2=π2×29,000(62.95)2=72.23ksi

Check:

4.71EFy=4.7129,00060=103.55

KyLry<4.71EFy62.95<103.55

Use AISC equation 3-2.

Fcr=0.658(Fy/Fe)Fy=0.658(60/72.23)×60=42.38ksi

From AISC manual, the gross area for the given material is 23.2in2.

ϕcPn=ϕc×Fcr×Ag=0.9×42.38×23.2=884.89kips

Check for slender:

λ=bf2tf=12.12×0.735=8.23

λr=0.56EFy=0.5629,00060=12.31

λ<λr8.23<12.31

The flange is non-slender.

λ=htw=20.7

λr=1.49EFy=1.4929,00060=32.76

λ<λr20.7<32.76

The web is non-slender.

Pu=1.2(DL)+1.6(LL)=1.2(180)+1.6(320)=728kips

ϕcPn>Pu884.89kips>728kips

Hence, the section is safe.

Conclusion:

The section W12×79 is adequate.

To determine

(b)

The compression member is adequate to support the loads using ASD.

Expert Solution
Check Mark

Answer to Problem 4.7.3P

W12×79 is adequate

Explanation of Solution

Given information:

Dead load,DL=180kips

Live load,LL=320kips

Length of the column,L=28ft

W12×79Kx=1.0Ky=1.0bf=12.1intf=0.735inh/tw=20.7

Steel type=A572

Yield stress=60ksi

Concept used:

Slenderness ratio=KLr

Fe=π2E(KL/r)2

PnΩc=Fcr×AgΩc

Pa=DL+LL

KEffective length factorrRadius of gyrationFeBuckling stressFcrCritical stress

EModulus of elasticityPnNominal compressive strengthAgGross areaϕcResistance factor=0.90ΩcSafety factor=1.67

Calculation:

From AISC manual, the radius of gyration for the given material is 5.34 in. and 3.05 in.

KxLrx=1.0×28ft×12in1ft5.34=62.92

KyLry=1.0×16ft×12in1ft3.05=62.95

Taking the higher value of slenderness ratio is 62.95.

Fe=π2E(KL/r)2=π2×29,000(62.95)2=72.23ksi

Check:

4.71EFy=4.7129,00060=103.55

KyLry<4.71EFy62.95<103.55

Use AISC equation 3-2.

Fcr=0.658(Fy/Fe)Fy=0.658(60/72.23)×60=42.38ksi

From AISC manual, the gross area for the given material is 23.2in2.

PnΩc=Fcr×AgΩc=42.38×23.21.67=588.75kips

Check for slender

λ=bf2tf=12.12×0.735=8.23

λr=0.56EFy=0.5629,00060=12.31

λ<λr8.23<12.31

The flange is non-slender.

λ=htw=20.7

λr=1.49EFy=1.4929,00060=32.76

λ<λr20.7<32.76

The web is non-slender.

Pa=DL+LL=180+320=500kips

ϕcPn>Pu500kips>588.75kips

Hence, the section is safe.

Conclusion:

The section W12×79 is adequate.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Determine the maximum axial compressive service load that can be supported if the live load is twice as large as the dead load. Use AISC Equation E3-2 or E3-3. a. Use LRFD. b. Use ASD
Question 4: A W12 x 65 of A572 Grade 60 steel is used as a compression member. It is 26 feet long, pinned at each end, and has additional support in the weak direction at a point 12 feet from the top. Can this member resist a service dead load of 180 kips and a service live load of 320 kips? a. Use LRFD. b. Use ASD.
Select all zero-force members in the truss shown below. Check the box for zero- force members 3 m 3 m 12 m, 8 @ 1.5 m DE O LK ЕР O HF O BC BM EF OM CD BN LO O DK FI O co
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning