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 8, Problem 8.6.4P
To determine

(a)

The design of a three-plate moment connection of a W18×35 beam to a W14×99 column for the given conditions by using LRFD.

Expert Solution
Check Mark

Answer to Problem 8.6.4P

PL34×7×2'412''with 20 bolts.

Use a PL14×312×1012with four 5/8-in. diameter Group A bolts for the web plate.

1/8-in. fillet weld on both sides of the plate.

Use a PL34×7×2'412''with twenty ½-in. diameter Group A bolts for each flange plate.

Explanation of Solution

Given:

Service dead-load moment = 42 ft-kips

Service live-load moment = 104 ft-kips

Service dead-load beam reaction = 8 kips

Service live-load beam reaction = 21 kips

Group A bearing type bolts

E70 electrodes

A992 steel- Beam and column

A36 steel- Plate material

Calculation:

Reaction:

Ru=1.2D+1.6L=1.2(8)+1.6(21)=43.2kips

Moment:

Mu=1.2MD+1.6ML=1.2(42)+1.6(104)=216.8ftkips

Web plate:

Neglect eccentricity.

Try 5/8-in. diameter bolts.

Assume that threads are in the plane of shear.

Ab=πd24=π(5/8)24=0.3068in.2

Shear capacity of one bolt is

ϕRn=ϕFnvAb=0.75(54)(0.3068)=12.43kips/bolt

Number of bolts required is

43.212.43=3.48

Try 4 bolts.

Determine plate thickness required for bearing. Assume that

ϕRn=ϕ(2.4dtFu)=0.75(2.4)(5/8)t(58)=65.25t

Load resisted by each bolt = 43.24=10.8kips

Let ϕRn=10.8=65.25t

t=0.166in.

Try t=316in.

Determine whether plate or beam web controls bearing. For the plate,

tFu=316(58)=10.9kips/in.

For the beam web, twFu=0.3(65)=19.5kips/in.>10.9kips/in.

Therefore, plate controls.

Check bearing strength assumption:

h=58+116=1116in.

For the hole nearest the edge, minimum le=78in.

Try 112in.

lc=leh2=1.511/162=1.156in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.156)(3/16)58=11.31kipsϕ(2.4dtFu)=0.75(2.4)(5/8)(3/16)58=12.23kips>11.31kips

Therefore, use ϕRn=11.31kips/bolt

For other bolts, minimum s=2.667(5/8)=1.667in.

Use s=212in.

lc=sh=2.51116=1.813in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.813)(3/16)(58)=17.74kipsϕ(2.4dtFu)=12.23kips<17.74kips

Therefore, use ϕRn=12.23kips/bolt

Bearing controls over shear at each bolt location. Total strength is

11.31+3(12.23)=48kips>43.2kips (OK)

Use four 5/8-in. diameter Group A bolts.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  1

Determine plate thickness required for shear:

Shear yielding strength is

ϕRn=1.0(0.6FyAgv)

Let 43.2=1.0(0.6)(36)(10.5t)

t=0.191in.

Try t=14in.

Check shear rupture strength:

Use hole diameter = 58+18=34in.

Anv=(10.54×34)14=1.875in.2ϕRn=ϕ(0.6FuAnv)=0.75(0.6)(58)(1.875)

=48.9kips>43.2kips (OK)

Check block shear:

Shear areas:

Agv=1.5+3×2.54=2.25in.2Anv=93.5(3/4)4=1.594in.2

Tension area:

Ant=1.50.5(3/4)4=0.2813in.2Rn=0.6FuAnv+UbsFuAnt=0.6(58)(1.594)+1.0(58)(0.2813)=71.79kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(2.25)+1.0(58)(0.2813)=64.92kips

The nominal block shear strength is therefore 64.92 kips. The design block shear strength is

ϕRn=0.75(64.92)=48.7kips>43.2kips (OK)

Use a PL14×312×1012 as shown above.

Connection of shear plate to column flange:

Use E70 electrodes

Minimum weld size, based on the plate thickness, is 1/8-in. try w=18in.

Weld strength = 1.392×2=2.784kips/in.

Base metal (plate) shear strength:

Yielding:

ϕRn=1.0(0.6Fyt)=1.0(0.6)(36)(1/4)=5.4kips/in.>2.784kips/in.

Rupture:

ϕRn=0.75(0.6Fut)=0.45Fut=0.45(58)(1/4)=6.525kips/in.>2.784kips/in.

Total length required = 43.22.784=15.5in.

Use a continuous 1/8-in. fillet weld, both sides of plate.

Flange plate:

From M=Hd,

H=Md=216.8(12)17.7=147kips

Try 7/8-in. diameter bolts.

Assume that threads are in the plane of shear.

Ab=π(7/8)24=0.6013in.2ϕRn=0.75(54)(0.6013)=24.35kips/bolt

Number of bolts required for shear is

14724.35=6.04

Try 8 bolts (4 pair)

Determine plate thickness required for bearing:

h=78+116=1516in.

Minimum le=112in.

Use 112in.

Minimum s=2.667(78)=2.33in.

Use 212in.

For the hole nearest the edge,

lc=leh2=1.515/162=1.031in.

ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.031)t(58)=53.82tkipsϕ(2.4dtFu)=0.75(2.4)(78)t(58)=91.35tkips>53.82tkips

Therefore, use ϕRn=53.82tkips/bolt

For other bolts,

lc=sh=2.51516=1.563in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.563)t(58)=81.59tkipsϕ(2.4dtFu)=91.35tkips>81.59tkips

Therefore, use ϕRn=81.59tkips/bolt

Total connection strength = 2(24.35t)+6(24.35t)=170.5t (Shear controls)

Let 543.4t=170.5

t=0.3138in.

Design top flange plate as a tension connection element

Tension on gross area:

Required Ag=H0.90Fy=1470.9(36)=4.537in.2

Tension on net area:

Required Ae=H0.75Fu=1470.75(58)=3.379in.2

Try a plate width of wg=7in.

For gross area requirement,

t=Agwg=4.5377=0.648in.

For net area requirement,

Hole diameter = 78+18=1in.

t=Anwn=Anwgdhole=3.37972(78+18)=0.676in. (Controls)

Try a plate 34×7

An=[72(1.0)](34)=3.75in.2

Check compression in the bottom plate:

Assume that the plate acts as a fixed-end compression member between the end fastener and the weld. Use KL=0.65(1.5+0.5)=1.3in.

r=IA=7(5/8)3/127(5/8)=0.1804in.KLr=1.30.1804=7.21

For compression elements with KLr<25, the nominal strength is

Pn=FyAg

Therefore, ϕPn=ϕFyAg=0.9(36)(7×34)=170kips>147kips (OK)

Check block shear on the plate using the dimensions and bolt layout shown.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  2

Shear areas:

Agv=34(9)×2=13.5in.2

Anv=34[93.5(1.0)]×2=8.25in.2 (3.5-hole diameters each side)

Tension area:

Ant=34[3.51(1.0)]=1.875in.2 (using 0.5+0.5 hole diameters)

Rn=0.6FuAnv+UbsFuAnt=0.6(58)(8.25)+1.0(58)(1.875)=395.9kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(13.5)+1.0(58)(1.875)=400.4kips

The nominal block shear strength is therefore 395.9 kips. The design block shear strength is

ϕRn=0.75(395.9)=297kips>147kips (OK)

Check block shear strength of beam flange:

Transverse spacing = gauge distance = 3.5 in.

Transverse edge distance = 63.52=1.25in.

Longitudinal spacing and edge distance same as for plate.

Shear areas:

Agv=0.425(9)×2=7.65in.2Anv=0.425[93.5(1.0)]×2=4.675in.2

Tension area:

Ant=0.425[2.51(1.0)]=0.6375in.2Rn=0.6FuAnv+UbsFuAnt=0.6(58)(4.675)+1.0(58)(0.6375)=199.7kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(7.65)+1.0(58)(0.6375)=202.2kips

The nominal block shear strength is therefore 199.7 kips.

The design block shear strength is

ϕRn=0.75(199.7)=149.8kips>147kips (OK)

Check beam for the effect of bolt holes in the tension flange:

The gross area of one flange is

Afg=tfbf=0.425(6.0)=2.55in.2

The effective hole diameter is

dh=78+18=1in.Afn=Afgtfdh=2.550.425(2×1)=1.7in.2FuAfn=65(1.7)=110.5kipsYtFyAfg=1.1(50)(2.55)=140.3kips

Since FuAfn<YtFyAfg, the holes must be accounted for.

Mn=FuAfnAfgSx=110.52.55(57.6)=2496in.kips=208ftkips

ϕbMn=0.90(208)=187ftkips<216.8ftkips (N.G.)

Try a smaller diameter bolt. Try ½-in. diameter bolts.

Ab=πd24=π(12)24=0.1963in.2

Normal shearing strength = ϕRn=0.75(54)(0.1963)=7.95kips/bolt

Number of bolts required for shear is

1477.95=18.49

Try 20 bolts (10 pair)

Bearing and block shear will be satisfactory.

Check reduction in beam flange area:

Use a hole diameter = 12+18=58in.

Afn=Afgtfdh=2.550.425(2×58)=2.019in.2FuAfn=65(2.019)=131.2kipsYtFyAfg=1.1(50)(2.55)=140.3kips

Since FuAfn<YtFyAfg, the holes must be accounted for.

Mn=FuAfnAfgSx=131.22.55(57.6)=2964in.kips=247ftkips

ϕbMn=0.90(247)=222.3ftkips>216.8ftkips (OK)

Plate length = 10(2.5)+1.5+1.5+0.5=28.5in.

Conclusion:

Use a PL34×7×2'412'' with 20 bolts. For the web plate, use a PL14×312×1012 with four 5/8-in. diameter Group A bolts as shown. Attach to column with a continuous 1/8-in. fillet weld on both sides of the plate. For each flange plate, use a PL34×7×2'412'' with twenty ½-in. diameter Group A bolts as shown.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  3

To determine

(b)

The design of a three-plate moment connection of a W18×35 beam to a W14×99 column for the given conditions by using ASD

Expert Solution
Check Mark

Answer to Problem 8.6.4P

PL34×7×2'412''with 20 bolts.

Use a PL14×312×1012with four 5/8-in. diameter Group A bolts for the web plate.

1/8-in. fillet weld on both sides of the plate.

Use a PL34×7×2'412''with twenty ½-in. diameter Group A bolts for each flange plate.

Explanation of Solution

Given:

Service dead-load moment = 42 ft-kips

Service live-load moment = 104 ft-kips

Service dead-load beam reaction = 8 kips

Service live-load beam reaction = 21 kips

Group A bearing type bolts

E70 electrodes

A992 steel- Beam and column

A36 steel- Plate material

Calculation:

Reaction:

Ra=D+L=8+21=29kips

Moment:

Ma=MD+ML=42+104=146ftkips

Web plate:

Neglect eccentricity.

Try 5/8-in. diameter bolts.

Assume that threads are in the plane of shear.

Ab=πd24=π(5/8)24=0.3068in.2

Shear capacity of one bolt is

RnΩ=FnvAbΩ=54(0.3068)2.0=8.28kips/bolt

Number of bolts required is

298.28=3.5

Try 4 bolts.

Determine plate thickness required for bearing. Assume that

RnΩ=2.4dtFuΩ=2.4(5/8)t(58)2.0=43.5t

Load resisted by each bolt = 294=7.25kips

Let RnΩ=7.25=43.5t

t=0.167in.

Try t=316in.

Determine whether plate or beam web controls bearing. For the plate,

tFu=316(58)=10.9kips/in.

For the beam web, twFu=0.3(65)=19.5kips/in.>10.9kips/in.

Therefore, plate controls.

Check bearing strength assumption:

h=58+116=1116in.

For the hole nearest the edge, minimum le=78in.

Try 112in.

lc=leh2=1.511/162=1.156in.RnΩ=1.2lctFuΩ=1.2(1.156)(3/16)582.0=7.54kips2.4dtFuΩ=2.4(5/8)(3/16)582.0=8.156kips>7.54kips

Therefore, use RnΩ=7.54kips/bolt

For other bolts, minimum s=2.667(5/8)=1.667in.

Use s=212in.

lc=sh=2.51116=1.813in.RnΩ=1.2lctFuΩ=1.2(1.813)(3/16)(58)2.0=11.83kips2.4dtFuΩ=8.156kips<11.83kips

Therefore, use RnΩ=8.156kips/bolt

Bearing controls over shear at each bolt location.

Total strength is

7.54+3(8.156)=32.008kips>29kips (OK)

Use four 5/8-in. diameter Group A bolts.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  4

Determine plate thickness required for shear:

Shear yielding strength is

RnΩ=0.6FyAgvΩ

Let 29=0.6(36)(10.5t)1.5

t=0.192in.

Try t=14in.

Check shear rupture strength:

Use hole diameter = 58+18=34in.

Anv=(10.54×34)14=1.875in.2RnΩ=0.6FuAnvΩ=0.6(58)(1.875)2.0

=32.625kips>29kips (OK)

Check block shear:

Shear areas:

Agv=1.5+3×2.54=2.25in.2Anv=93.5(3/4)4=1.594in.2

Tension area:

Ant=1.50.5(3/4)4=0.2813in.2Rn=0.6FuAnv+UbsFuAnt=0.6(58)(1.594)+1.0(58)(0.2813)=71.79kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(2.25)+1.0(58)(0.2813)=64.92kips

The nominal block shear strength is therefore 64.92 kips. The design block shear strength is

RnΩ=64.922.0=32.46kips>29kips (OK)

Use a PL14×312×1012 as shown above.

Connection of shear plate to column flange:

Use E70 electrodes.

Minimum weld size, based on the plate thickness, is 1/8-in. try w=18in.

Weld strength = 1.392×2=2.784kips/in.

Base metal (plate) shear strength:

The allowable shear yield strength per unit length is

RnΩ=0.6Fyt1.5=0.4Fyt=0.4(36)(1/4)=3.6kips/in.>2.784kips/in.

The base metal allowable shear rupture strength per unit length is

RnΩ=0.6Fut2.0=0.3Fut=0.3(58)(1/4)=4.35kips/in.>2.784kips/in.

Total length required = 292.784=10.42in.

Use a continuous 1/8-in. fillet weld, both sides of plate.

Flange plate:

From M=Hd,

H=Md=146(12)17.7=98.983kips

Try 7/8-in. diameter bolts.

Assume that threads are in the plane of shear.

Ab=π(7/8)24=0.6013in.2RnΩ=54(0.6013)2.0=16.235kips/bolt

Number of bolts required for shear is

98.98316.235=6.097

Try 8 bolts (4 pair)

Determine plate thickness required for bearing:

h=78+116=1516in.

Minimum le=112in.

Use 112in.

Minimum s=2.667(78)=2.33in.

Use 212in.

For the hole nearest the edge,

lc=leh2=1.515/162=1.031in.

RnΩ=1.2lctFu2.0=1.2(1.031)t(58)2.0=35.88tkips2.4dtFuΩ=12.0(2.4)(78)t(58)=60.9tkips>35.88tkips

Therefore, use RnΩ=35.88tkips/bolt

For other bolts,

lc=sh=2.51516=1.563in.RnΩ=1.2lctFu2.0=1.2(1.563)t(58)2.0=54.4tkips2.4dtFuΩ=60.9tkips>54.4tkips

Therefore, use RnΩ=54.4tkips/bolt

Total connection strength = 2(16.235t)+6(16.235t)=98.983 (Shear controls)

Let 129.88t=98.983

t=0.762in.

Design top flange plate as a tension connection element

Tension on gross area:

Required Ag=1.67HFy=1.67(98.983)36=4.592in.2

Tension on net area:

Required Ae=2.0(H)Fu=2.0(98.983)58=3.413in.2

Try a plate width of wg=7in.

For gross area requirement,

t=Agwg=4.5927=0.656in.

For net area requirement,

Hole diameter = 78+18=1in.

t=Anwn=Anwgdhole=3.41372(78+18)=0.6826in. (Controls)

Try a plate 34×7

An=[72(1.0)](34)=3.75in.2

Check compression in the bottom plate:

Assume that the plate acts as a fixed-end compression member between the end fastener and the weld. Use KL=0.65(1.5+0.5)=1.3in.

r=IA=7(5/8)3/127(5/8)=0.1804in.KLr=1.30.1804=7.21

For compression elements with KLr<25, the nominal strength is

Pn=FyAg

Therefore, PnΩ=FyAg1.67=11.67(36)(7×34)=113.17kips>98.983kips (OK)

Check block shear on the plate using the dimensions and bolt layout shown.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  5

Shear areas:

Agv=34(9)×2=13.5in.2

Anv=34[93.5(1.0)]×2=8.25in.2 (3.5-hole diameters each side)

Tension area:

Ant=34[3.51(1.0)]=1.875in.2 (using 0.5+0.5 hole diameters)

Rn=0.6FuAnv+UbsFuAnt=0.6(58)(8.25)+1.0(58)(1.875)=395.9kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(13.5)+1.0(58)(1.875)=400.4kips

The nominal block shear strength is therefore 395.9 kips. The design block shear strength is

RnΩ=395.92.0=197.95kips>98.983kips (OK)

Check block shear strength of beam flange:

Transverse spacing = gauge distance = 3.5 in.

Transverse edge distance = 63.52=1.25in.

Longitudinal spacing and edge distance same as for plate.

Shear areas:

Agv=0.425(9)×2=7.65in.2Anv=0.425[93.5(1.0)]×2=4.675in.2

Tension area:

Ant=0.425[2.51(1.0)]=0.6375in.2Rn=0.6FuAnv+UbsFuAnt=0.6(58)(4.675)+1.0(58)(0.6375)=199.7kips

with an upper limit of

0.6FyAgv+UbsFuAnt=0.6(36)(7.65)+1.0(58)(0.6375)=202.2kips

The nominal block shear strength is therefore 199.7 kips.

The design block shear strength is

RnΩ=199.72.0=99.85kips>98.983kips (OK)

Check beam for the effect of bolt holes in the tension flange:

The gross area of one flange is

Afg=tfbf=0.425(6.0)=2.55in.2

The effective hole diameter is

dh=78+18=1in.Afn=Afgtfdh=2.550.425(2×1)=1.7in.2FuAfn=65(1.7)=110.5kipsYtFyAfg=1.1(50)(2.55)=140.3kips

Since FuAfn<YtFyAfg, the holes must be accounted for.

Mn=FuAfnAfgSx=110.52.55(57.6)=2496in.kips=208ftkips

MnΩb=2081.67=124.55ftkips<146ftkips (N.G.)

Try a smaller diameter bolt. Try ½-in. diameter bolts.

Ab=πd24=π(12)24=0.1963in.2

Normal shearing strength = RnΩ=54(0.1963)2.0=5.3kips/bolt

Number of bolts required for shear is

98.9835.3=18.7

Try 20 bolts (10 pair)

Bearing and block shear will be satisfactory.

Check reduction in beam flange area:

Use a hole diameter = 12+18=58in.

Afn=Afgtfdh=2.550.425(2×58)=2.019in.2FuAfn=65(2.019)=131.2kipsYtFyAfg=1.1(50)(2.55)=140.3kips

Since FuAfn<YtFyAfg, the holes must be accounted for.

Mn=FuAfnAfgSx=131.22.55(57.6)=2964in.kips=247ftkips

MnΩb=2471.67=147.9ftkips>146ftkips (OK)

Plate length = 10(2.5)+1.5+1.5+0.5=28.5in.

Conclusion:

Use a PL34×7×2'412'' with 20 bolts. For the web plate, use a PL14×312×1012 with four 5/8-in. diameter Group A bolts as shown. Attach to column with a continuous 1/8-in. fillet weld on both sides of the plate. For each flange plate, use a PL34×7×2'412'' with twenty ½-in. diameter Group A bolts as shown.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 8, Problem 8.6.4P , additional homework tip  6

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Chapter 8 Solutions

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