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.

(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.

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)=43.2 kips

Moment:

Mu=1.2MD+1.6ML=1.2(42)+1.6(104)=216.8 ft−kips

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=0.3068 in.2

Shear capacity of one bolt is

ϕRn=ϕFnvAb=0.75(54)(0.3068)=12.43 kips/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=10.8 kips

Let ϕRn=10.8=65.25t

t=0.166 in.

Try t=316 in.

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

tFu=316(58)=10.9 kips/in.

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

Therefore, plate controls.

Check bearing strength assumption:

h=58+116=1116 in.

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

Try 112 in.

lc=le−h2=1.5−11/162=1.156 in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.156)(3/16)58=11.31 kipsϕ(2.4dtFu)=0.75(2.4)(5/8)(3/16)58=12.23 kips>11.31 kips

Therefore, use ϕRn=11.31 kips/bolt

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

Use s=212 in.

lc=s−h=2.5−1116=1.813 in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.813)(3/16)(58)=17.74 kipsϕ(2.4dtFu)=12.23 kips<17.74 kips

Therefore, use ϕRn=12.23 kips/bolt

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

11.31+3(12.23)=48 kips>43.2 kips (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=0.191 in.

Try t=14 in.

Check shear rupture strength:

Use hole diameter = 58+18=34 in.

Anv=(10.5−4×34)14=1.875 in.2ϕRn=ϕ(0.6FuAnv)=0.75(0.6)(58)(1.875)

=48.9 kips>43.2 kips (OK)

Check block shear:

Shear areas:

Agv=1.5+3×2.54=2.25 in.2Anv=9−3.5(3/4)4=1.594 in.2

Tension area:

Ant=1.5−0.5(3/4)4=0.2813 in.2Rn=0.6FuAnv+UbsFuAnt       =0.6(58)(1.594)+1.0(58)(0.2813)=71.79 kips

with an upper limit of

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

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

ϕRn=0.75(64.92)=48.7 kips>43.2 kips (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=18 in.

Weld strength = 1.392×2=2.784 kips/in.

Base metal (plate) shear strength:

Yielding:

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

Rupture:

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

Total length required = 43.22.784=15.5 in.

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

Flange plate:

From M=Hd,

H=Md=216.8(12)17.7=147 kips

Try 7/8-in. diameter bolts.

Assume that threads are in the plane of shear.

Ab=π(7/8)24=0.6013 in.2ϕRn=0.75(54)(0.6013)=24.35  kips/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=1516 in.

Minimum le=112 in.

Use 112 in.

Minimum s=2.667(78)=2.33 in.

Use 212 in.

For the hole nearest the edge,

lc=le−h2=1.5−15/162=1.031 in.

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

Therefore, use ϕRn=53.82 t kips/bolt

For other bolts,

lc=s−h=2.5−1516=1.563 in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.563)t(58)=81

(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