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Design a three-plate moment connection of the type shown in Problem 8.6-2 for the connection of a W 18 × 35 beam to a W 14 × 99 column for the following conditions: The service dead-load moment is 42 ft-kips, the service live-load moment is 104 ft-kips, the service dead-load beam reaction is 8 kips, and the service live-load beam reaction is 21 kips. Use Group A Bearing-type bolts and E70 electrodes. The beam and column are of A992 steel, and the plate material is A36. a . Use LRFD. b . Use ASD.

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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 8, Problem 8.6.4P
Textbook Problem
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Design a three-plate moment connection of the type shown in Problem 8.6-2 for the connection of a W 18 × 35 beam to a W 14 × 99 column for the following conditions: The service dead-load moment is 42 ft-kips, the service live-load moment is 104 ft-kips, the service dead-load beam reaction is 8 kips, and the service live-load beam reaction is 21 kips. Use Group A Bearing-type bolts and E70 electrodes. The beam and column are of A992 steel, and the plate material is A36.

a. Use LRFD.

b. Use ASD.

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.

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.

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

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

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Chapter 8 Solutions

Steel Design (Activate Learning with these NEW titles from Engineering!)
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