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 10, Problem 10.7.7P
To determine

(a)

The girder cross section and the spacing of intermediate stiffeners that is required.

Expert Solution
Check Mark

Answer to Problem 10.7.7P

Web dimensions 12in×83in

Flange 112in×24in

Explanation of Solution

Given information:

Span length L=66ft

Uniform dead load wD=1.3kip/ft

Uniform live load wL=2.3kip/ft

Dead load PD=28kip

Live load PL=49kip

Depth d=80in

Fy=50

Calculation:

Determine girder cross section and girder spacing of intermediate stiffeners.

wu=1.2wD+1.6wLwu=1.2(1.3)+1.6(2.3)wu=5.24kips/ftPu=1.2PD+1.6PLPu=1.2×28+1.6×49Pu=112kips

Calculate design strength.

Vu=wuL2+PuVu=5.24×662+112Vu=284.92kipsMu=284.92×335.24×3322112666Mu=5317.18ft-kips

Try tf=1.5in

Determine the trial web size.

h=d2tfh=802(1.5)h=83in

Consider the condition for the web to slender.

htw>5.70EFy455.7029,00050tw83137.3tw0.6045in

Estimate the web thickness from the following limitations:

For ah1.5htw12.0(EFy)htw12(29,00050)htw=289

Estimate the web thickness from the following limitations:

For ah1.5htw0.4(EFy)83tw0.4(29,00050)tw83232tw0.3578in

From the estimated limit of tw try a web size of 12in×83in

Determine area of web.

Aw=h×twAw=83×0.5Aw=41.5n2

Determine area of flange.

Af=MnhFyAw6Af=5317.18×120.9×83×5041.56Af=10.167in2

Determine the breadth of flange.

bfAftfbf10.1671.5bf6.778inAf=1.5×24Af=36in2

Conclusion:

Therefore, girder size of 12in×83in web and 112in×24in flanges should be used.

To determine

(b)

The size of intermediate and bearing stiffeners.

Expert Solution
Check Mark

Answer to Problem 10.7.7P

Web size 12in×83in

Bearing stiffeners 112in×24in

Explanation of Solution

Given information:

Span length L=66ft

Uniform dead load wD=1.3kip/ft

Uniform live load wL=2.3kip/ft

Dead load PD=28kip

Live load PL=49kip

Depth d=80in

Fy=50

Calculation:

Calculate the elastic section modulus by using the formula.

Sx=Ixc

Moment of inertia is given by,

Ix=112twh3+2Af(h+tf2)2Ix=112(0.5)(83)3+2(36)(83+(1.5)2)2Ix=1.523×104in4

Calculate the maximum distance whichis given by,

c=h2+tfc=832+1.5c=43in

Elastic section modulus about the axis Sx is calculated using the values of Ix and c.

Sx=IxcSx=1523×10443Sx=3541.86in3

From AISC B4, Table B4.1 the relevant slenderness parameters for local buckling are,

λ=bf2tfλ=242(1.5)λ=8.0λp=0.38EFyλp=0.3829,00050λp=9.152

λ<λpFcr=Fy=50ksi

Radius of gyration is given by,

rt=IyA

Calculating the moment of inertia.

Iy=112(h6)(tw)3+112(tf)(bf)3Iy=112(13.83)(0.5)3+112(1.5)(24)3Iy=1782.14in4

Determine area of cross-section.

A=(h6)(tw)+(tf)(bf)A=(13.83)(0.5)+(1.5)(24)A=42.92in2

Radius of gyration is given by,

rt=IyArt=1728.1442.92rt=6.35in

Unbraced length Lb=7ft.

Determine the LP as shown below,

Lp=1.1r1EFyLp=1.1×6.3529,00050Lp=88.38×(ft12in)Lp=14.02ft

Determine the Lr as shown below:

Lr=πr1E0.7FyLp=π×6.3529,0000.7×50Lp=47.85ft

Since LpLb<Lr

Fcr=Cb[Fy0.3Fy(LbLpLrLp)]FyFcr=1.046[500.3×50(23.3314.0247.8514.02)]50Fcr=45.87ksi<50ksi

Calculate compression flange strength.

aw=hctwbftfaw=83×(0.5)24×1.5aw=1.15<10

Calculate bending strength reduction factor.

Rpg=1-aw1200+300aw(hctw-5.7EFy)1Rpg=1-1.151200+300(1.15)(83316-5.729,00050)Rpg=0.9786<1

Determine compression flange strength.

Mn=RpgFcrSxMn=0.9786×45.87×3541.86Mn=1.5899×105kips-in

For proper flexural strength,

ϕbMn=0.90×1.5899×10512ϕbMn=11924.128kips-ft>5317.18kips-ft

Conclusion:

Therefore, stiffeners are not needed in the middle and use girder size of 12in×83in web and 112in×24in flanges should be used.

To determine

(c)

Design of all welds.

Expert Solution
Check Mark

Answer to Problem 10.7.7P

Web dimensions 12in×83in

Flanges 112in×24in

Explanation of Solution

Given information:

Span length L=66ft

Uniform dead load wD=1.3kip/ft

Uniform live load wL=2.3kip/ft

Dead load PD=28kip

Live load PL=49kip

Depth d=80in

Fy=50

Calculation:

Calculate the elastic section modulus by using the formula,

Sx=Ixc

Moment of inertia is given by,

Ix=112twh3+2Af(h+tf2)2Ix=112(0.5)(83)3+2(36)(83+(1.5)2)2Ix=1.523×104in4

Calculate the maximum distance which is given by,

c=h2+tfc=832+1.5c=43in

Elastic section modulus about the axis Sx is calculated using the values of Ix and c.

Sx=IxcSx=1523×10443Sx=3541.86in3

From AISC B4, Table B4.1 the relevant slenderness parameters for local buckling are,

λ=bf2tfλ=242(1.5)λ=8.0λp=0.38EFyλp=0.3829,00050λp=9.152

λ<λpFcr=Fy=50ksi

Radius of gyration is given by,

rt=IyA

Calculating the moment of inertia,

Iy=112(h6)(tw)3+112(tf)(bf)3Iy=112(13.83)(0.5)3+112(1.5)(24)3Iy=1782.14in4

Determine area of cross-section.

A=(h6)(tw)+(tf)(bf)A=(13.83)(0.5)+(1.5)(24)A=42.92in2

Radius of gyration is given by,

rt=IyArt=1728.1442.92rt=6.35in

Unbraced length Lb=66ft

Determine the LP as shown below.

Lp=1.1r1EFyLp=1.1×6.3529,00050Lp=88.38×(ft12in)Lp=14.02ft

Determine the Lr as shown below,

Lr=πr1E0.7FyLr=π×6.3529,0000.7×50Lr=47.85ft

Since, LpLb<Lr

Fcr=Cb[Fy0.3Fy(LbLpLrLp)]FyFcr=1.046[500.3×50(23.3314.0247.8514.02)]50Fcr=45.87ksi<50ksi

To calculate the plate girder strength reduction factor values, aw and Rpg is needed.

aw=hctwbftfaw=83×(0.5)24×1.5aw=1.15<10

Rpg=1-aw1200+300aw(hctw-5.7EFy)1Rpg=1-1.151200+300(1.15)(83316-5.729,00050)Rpg=0.9786<1

Calculate compression flange strength.

Mn=RpgFcrSxMn=0.9786×45.87×3541.86Mn=1.5899×105kips-in

Calculate proper flexural strength.

ϕbMn=0.90×1.5899×10512ϕbMn=11924.128kips-ft>5317.18kips-ft

Calculate nominal shear strength.

Vn=VuϕVn=284.920.90Vn=316.57kips

Check for design strength.

Vn=0.6AwFyCvCv=Vn0.6AwFyCv=316.57kips0.6×8.438×50Cv=1.25kips

Compute the value of kv.

Cv=1.51kvE(htw)2Fykv=Cv(htw)2Fy1.51Eykv=1.2(83316)2501.5×29,000kv=25.1

Assuming the equation G2-5  controls,

1.37kvEFy=1.3725.1×29,000501.37kvEFy=165.29<htw=240

Hence, equation G2-5  controls.

Determine the value of a from AISC equation G2-6 as shown below:

ah=5kv5ah=525.15a=0.497×ha=0.497×83a=41.33in

Calculate the value of required shear strength 15 in from the left end as shown below:

Vu1=Vu2(15inbf)Vu1=1682(15in12)Vu1=165.5kips

Compute the value of required shear 15 in from the left end as shown below:

ϕvVuAw=Vu1AwϕvVuAw=165.58.438ϕvVuAw=19.61ksi

Using the curves in the table 3-17b with the value of,

htw=45316htw=240

Values obtained are shown below,

For,ϕvVuAw=18ksi,ah=1.05For,ϕvVuAw=21ksi,ah=0.75

By interpolation determine the value of a, ϕvVuAw=19.61ksi

ah=1.05(19.91182118)(1.050.75)a=0.859×ha=0.859×45a=38.66in

Use the value a=38in which will be added to the distance to the next stiffener.

Compute the value required to shear strength from the left end,

Vu2=Vu2(15in+38inbf)Vu2=284.922(5312)Vu2=276.0kips

Compute the value of required shear 15+38in in from the left end as shown below,

ϕvVuAw=Vu1AwϕvVuAw=159.28.438ϕvVuAw=18.87ksi

Values obtained are shown blow,

For,ϕvVuAw=18ksi,ah=1.05For,ϕvVuAw=21ksi,ah=0.75

By interpolation we get the value of a, ϕvVuAw=18.87ksi

ah=1.05(18.87182118)(1.050.75)a=0.963×ha=0.963×45a=43.44in

Use the value a=43in which we can add to the distance of the next stiffener.

Calculate to shear strength from the left end.

Vu3=Vu2(15in+38in+43bf)Vu3=1682(9612)Vu3=152kips

Calculate shear 96in in from the left end as shown below:

ϕvVuAw=Vu3AwϕvVuAw=1528.438ϕvVuAw=18ksi

Values obtained are shown blow,

For,ϕvVuAw=18ksi,ah=1.05

By interpolation calculate the value of a,

ϕvVuAw=18ksi

ah=1.05a=1.05×ha=1.05×83a=87.15in

Use the value a=87.15in which we can add to the distance of the next stiffener.

We can use next stiffener will be used at 96+87.15in=180.15.

Calculate distance remaining concentrated load.

d=16(bf)-180.15ind=16(12)-180.15ind=11.85in

Calculate shear strength 16in from the left end is given by,

Vu4=Vu2(66)PuVu4=284.922(66)112Vu4=40.9kips

Calculate the required shear 16in from left end.

ϕvVuAw=Vu4AwϕvVuAw=168.438ϕvVuAw=1.896ksi

ah=16×bf45inah=16×12in45inah=4.267>3

Conclusion:

Therefore, girder size of 12in×83in web and 112in×24in flanges should be used.

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
Where is/are the location(s) of the maximum compressive flexural stress? A simple I-beam is loaded as shown. 20 mm P KN PKN P KN B 20 mm- B с D L/4 m L/4 m L/4 m Pin support at the NA Midspan at point B Mid span at point D Midspan at the top fiber Roller support at top fiber Section D at the top fiber Section C at +170 mm from the NA L/4 m 20 mm C 250 mm 150 mm 150 mm D A
Please Answer fast as u know... im needed in 15-30 minutes thank u... So please
Figure 1.5 m E -2 m- 3 kN D B -2 m 1 of 1 Identify the zero-force members in the truss. Check all that apply. BE DE CD AB AE BD BC Submit Drouido Foodbook Request Answer
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