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 10, Problem 10.7.8P
To determine

(a)

The design of a plate girder for the given conditions, the selection of girder cross section and the required spacing of intermediate stiffeners by using LRFD.

Expert Solution
Check Mark

Answer to Problem 10.7.8P

12×107web and 112×26flanges

Four panels spaced at 58.25in.

Explanation of Solution

Given:

Span length =100ft

Uniformly distributed live load =0.7 kips/ft

Superimposed dead load =0.3 kips/ft

Concentrated dead load = 50 kips

Concentrated live load =150 kips

A572 Grade 50 steel

Formula used:

Af=Mu/ϕbhFyAw6

Mu/ϕb is required nominal flexural strength

Aw is web area

Af is flange area

h is the depth of web

Calculation:

Assume a girder weight of 450 lb/ft.

Determine the factored loads:

wu=1.2wD+1.6wL

wD is uniformly distributed dead load

wL is uniformly distributed live load

wu=1.2wD+1.6wL=1.2(0.3)+1.6(0.7)=1.48kips/ft

Pu=1.2PD+1.6PL

PD is concentrated dead load

PL is concentrated live load

Pu=1.2PD+1.6PL=1.2(50)+1.6(150)=300kips

The factored moment and shear are

Mu=524(50)1.48(5022)300(25)=16850ftkipsVu=VL=524kips

Determine the overall depth:

Spanlength10=100(12)10=120in.Spanlength12=100(12)12=100in.

Use the maximum permissible depth of 110 in.

Try tf=1.5in.

h=1102(1.5)=107in.

To determine the web thickness, first examine the limiting values of htw.

For ah1.5,

(htw)max=12.0EFy=12.02900050=288.998

Minimum tw=107288.998=0.370in.

For ah>1.5,

(htw)max=0.40(EFy)=0.40(2900050)=232

Minimum tw=107232=0.46in.

Try a 12in.×107in. web.

Aw=0.5(107)=53.5in.2

Determine whether the web is slender:

htw=1070.5=214

5.70EFy=5.702900050=137.274<214

Therefore, the web is slender.

Estimate required flange size:

Af=Mu0.9hFyAw6=16850(12)0.9(107)(50)53.56=33.077in.2bf33.0771.5=22.05in.

Try a 112in.×26in. flange,

Af=1.5×26=39in.2

Girder weight = [53.5+2(39)](0.490144)=0.447kips<ft<0.450kips/ft (OK)

Ix=twh312+2Af(h+tf2)2=0.5(107)312+2(39)(107+1.52)2=280602.33in.4Sx=Ixc=Ix(h/2+tf)=280602.33(107/2+1.5)=5101.86in.3

Compression flange:

Check flange local buckling (FLB):

λ=bf2tf=262(1.5)=8.67λp=0.38EFy=0.382900050=9.15

Since λ<λp, Fcr=Fy=50ksi

Compute the plate girder strength reduction factor:

Rpg=1aw1200+300aw(hctw5.7EFy)1.0aw=hctwbfctfc=107(0.5)26(1.5)=1.37<10Rpg=11.371200+300(1.37)(1070.55.72900050)=0.935<1.0Mn=RpgFcrSxc=0.935(50)(5101.86)=238448.716in.kips

ϕbMn=0.90(238448.716)12=17883.654ftkips>16850ftkips (OK)

Try a 12×107 web and 112×26 flanges.

Shear: At left end (end panel),

Required ϕvVnAw=52453.5=9.8ksi,

htw=214

From Table 3-17a in the Manual, ah0.623 by interpolation

a=0.623h=0.65(107)=66.7in.

Use a=67in.

This spacing will apply for the remaining distance to the centerline of the girder. This distance is

25×1267=233in.

For a spacing a of 67 in., the number of panels is 23367=3.5

Use 4 panels at 2334=58.25in.

At 100/4=25ft from the left end (to the right of the concentrated load),

Vu=5241.48(25)300=187kips

Required ϕvVnAw=18753.5=3.5ksi,

For ah=25×12107=2.8,ϕvVnAw<3

Therefore, stiffeners are needed in middle 14.

Conclusion:

Therefore, Use a 12×107 web and 112×26 flanges and four panels spaced at 58.25in.

To determine

(b)

The size of intermediate and bearing stiffeners.

Expert Solution
Check Mark

Answer to Problem 10.7.8P

2 PL 12×6×8'8''for intermediate stiffeners

2 PL 58×8with 1-in. cutouts for bearing stiffeners

Explanation of Solution

Given:

Span length =100ft

Uniformly distributed live load =0.7 kips/ft

Superimposed dead load =0.3 kips/ft

Concentrated dead load = 50 kips

Concentrated live load =150 kips

A572 Grade 50 steel

Calculation:

Intermediate stiffener size:

(bt)st0.56EFy

Available width:

bftw2=260.52=12.75in.

Try b=6in.

tb0.56EFy=60.562900050=0.4449

To determine the required moment of inertia, use the conservative approximation from the User Note in AISC G2.3:

Ist=Ist1Ist=h4ρst1.340(FywE)1.5ρst=max{Fyw/Fyw1}=1Ist=(107)4(1)1.340(5029000)1.5=234.603in.4

Try two 12×6 plates:

Ist(3/8)(6+0.5+6)312=61in.4

Length: From Figure 10.9 in the textbook (Steel design),

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 10, Problem 10.7.8P

c4tw=4(0.5)=2in.c6tw=6(0.5)=3in.

Assume a flange-to-web weld size of w=316in. (minimum size) and c=3in.

Length = hwc=1073163=103.8in.

c=107104316=2.8125in. (OK)

Use two PL 12×6×8'8'' for intermediate stiffeners.

Design the bearing stiffeners at the supports for a load of VL=524kips

Maximum stiffener width = bftw2=260.52=12.75in.

Try b=8in.

tb0.56EFy=80.562900050=0.593in.

Try two plates, 58×8in., with 1-in. cutouts.

Bearing strength:

Apb=(81)(58)×2=8.75in.2Rn=1.8FyApb=1.8(50)(8.75)=787.5kips

ϕRn=0.75(787.5)=590.6kips>VL=524kips (OK)

Compressive strength: The maximum permissible length of web is

12tw=12(0.5)=6in.

Compute the radius of gyration about an axis along the middle of the web:

I=6(0.5)312+2[(5/8)8312+8(58)(4+14)2]=234in.4A=6(0.5)+2(8)(58)=13in.2r=IA=23413=4.243in.

Compute the compressive strength:

KLr=0.75(107)4.243=18.9<25

Therefore, Fcr=Fy=50ksi

ϕcPn=0.90FcrAg=0.90(50)(13)=585kips>524kips (OK)

Use 2 PL 58×8 with 1-in. cutouts for bearing stiffeners at the supports.

Because there is a large difference between the reactions and the interior concentrated loads, use 2 PL 58×8 with 1-in. cutouts for the interior bearing stiffeners.

Conclusion:

Use two PL 12×6×8'8'' for intermediate stiffeners,two PL 58×8 with 1-in. cutouts for bearing stiffeners at the supports and two PL 58×8 with 1-in. cutouts for the interiorbearing stiffeners.

To determine

(c)

The design of the all welds

Expert Solution
Check Mark

Answer to Problem 10.7.8P

316in. continuous fillet welds for the first20ft

316in.×112in.E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners

316in.×112in.E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.

Explanation of Solution

Given:

Span length =100ft

Uniformly distributed live load =0.7 kips/ft

Superimposed dead load =0.3 kips/ft

Concentrated dead load = 50 kips

Concentrated live load =150 kips

A572 Grade 50 steel

Calculation:

Design the flange-to-web welds.

The shear flow is

Q=Af(h2+tf2)=(1.5×26)(1072+1.52)=2115.75in.3

At the support,

VuQIx=524(2115.75)280602.33=3.95kips/in.

Minimum weld size = 3/16 in. (AISC Table J2.4)

Minimum length = 4(316)=0.75in.<1.5in.

Use 1.5 in.

Use E70 electrodes, ϕRn=1.392D kips/in.

where D is weld size in sixteenths.

Try an 316in.×112in. intermittent fillet welds.

For two welds,

Weld strength = 2×1.392(3)=8.352kips/in.

Base metal shear yield strength (web plate controls) is

0.4Fyt=0.4(50)(12)=10kips/in.

Shear rupture strength is

0.3Fut=0.3(65)(12)=9.75kips/in.

Weld strength controls.

For a 1.5-in. length,

ϕRn=1.5(8.352)=12.53kips

Required spacing:

ϕRns=VuQIx12.53s=3.95s=3.172in.

Since this is less than twice the length of the weld, use a continuous weld.

For s=2(1.5)=3in.

Vu=ϕRnIxsQ=12.53×280602.333(2115.75)=553.932kips

This occurs when 5241.48x=553.93x=20.22ft

Maximum clear spacing: From AISC E6,

d0.75EFytf=0.752900050(1.5)=27.1in. (or 12 in.; 12 in. controls.)

Maximum s=12+1.5=13.5in.

For s=13.5in.,

ϕRns=VuQIx12.5313.5=Vu(2115.75)280602.33Vu=123.096kips

Shear at first interior load, left of load, = 5241.48(25)=487kips

So maximum spacing will not be used in the first quarter of the span.

Spacing required at left side of first interior load is

ϕRnIxVuQ=12.53(280602.33)487(2115.75)=3.4123in.

Check middle fourth of span. Shear on right side of load is

487300=187kips

s=ϕRnIxVuQ=12.53(280602.33)187(2115.75)=8.8866in.<13.5in. (OK)

Welds for intermediate stiffeners (12×6) :

Minimum weld size = 3/16in. (AISC Table J2.4)

Minimum length = 4(316)=0.75in.<1.5in.

Use 1.5 in.

Use E70 electrodes, ϕRn=1.392Dkips/in.

where D is weld size in sixteenths.

Try 316in.×112in. intermittent fillet welds.

For four welds, the weld strength is

4×1.392(3)=16.7kips/in.

The base metal shear yield strength is

0.6Fyt=0.6(50)(12)×2=30kips/in.

Shear rupture strength is

0.45Fut=0.45(65)(12)×2=29.25kips/in.

Weld strength controls.

For a 1.5-in. length,

ϕRn=1.5(16.7)=25.05kips

The shear to be transferred is

f=0.045hFy3E=0.045(107)(50)329000=9.9966kips/in.25.05s=9.9966s=2.506in.

A center-to-center spacing of 3 in. is equal to twice the length of the weld segment, so

either a continuous weld or an intermittent weld can be used. Use intermittent welds.

Maximum clear spacing: From AISC E6,

d0.75EFytf=0.752900050(1.5)=27.1in. (or 12 in.; 12 in.; controls.)

Maximum s=12+1.5=13.5in.

Use 316in.×112in. E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners.

Welds for bearing stiffeners at the supports (58×8) :

Minimum weld size = 3/16in. (AISC Table J2.4)

Minimum length = 4(316)=0.75in.<1.5in.

Use 1.5 in.

Use E70 electrodes, ϕRn=1.392Dkips/in.

where D is weld size in sixteenths.

Try 316in.×112in. intermittent fillet welds.

For four welds, the weld strength is

4×1.392(3)=16.7kips/in.

The base metal shear yield strength is

0.6Fyt=0.6(50)(12)×2=30kips/in.

Shear rupture strength is

0.45Fut=0.45(65)(12)×2=29.25kips/in.

Weld strength controls.

For a 1.5-in. length,

ϕRn=1.5(16.7)=25.05kips

The shear to be transferred is

Reaction ÷ length available for weld = 5241072(1.0)=4.99kips/in.

25.05s=4.99s=5.02in.

Use 316in.×112in. E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.

Conclusion:

Use 3/16 in. continuous fillet welds for the first 20 ft, 316in.×112in. E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners and 316in.×112in. E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.

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