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.

Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740

Chapter
Section

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 live-load moment = 104 ft-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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