   # a . A W 18 × 35 beam is to be connected to a W 14 × 53 column. F y = 50 ksi for both the beam and the column. The connection is at a distance from the end of the column that is more than the depth of the column. Use A36 steel for the web plate and use LRFD to design a connection similar to the one shown in Figure 8.37a for a factored moment of 220 ft-kips and a factored reaction of 45 kips. If column stiffeners are needed, use A36 steel and specify the required dimensions. b . If the factored column shear adjacent to the connection is V u = 0 , and P u / P y = 0.6 , determine whether panel zone reinforcement is required. If it is, provide two alternatives: (1) a doubler plate of A36 steel and (2) diagonal stiffeners of A36 steel. ### Steel Design (Activate Learning wi...

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

#### Solutions

Chapter
Section ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 8, Problem 8.7.3P
Textbook Problem
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## a. A W 18 × 35 beam is to be connected to a W 14 × 53 column. F y = 50 ksi for both the beam and the column. The connection is at a distance from the end of the column that is more than the depth of the column. Use A36 steel for the web plate and use LRFD to design a connection similar to the one shown in Figure 8.37a for a factored moment of 220 ft-kips and a factored reaction of 45 kips. If column stiffeners are needed, use A36 steel and specify the required dimensions.b. If the factored column shear adjacent to the connection is V u = 0 , and P u / P y = 0.6 , determine whether panel zone reinforcement is required. If it is, provide two alternatives: (1) a doubler plate of A36 steel and (2) diagonal stiffeners of A36 steel.

To determine

(a)

The design of a connection similar to the given connection by using LRFD.

### Explanation of Solution

Given:

W18×35 beam

W14×53 column

Fy = 50 ksi

A36 steel

Factored moment = 220 ft-kips

Factored reaction = 45 kips

And the given connection is

Formula used:

Rn=6.25Fyft2f

Rn is nominal strength.

Fyf is yield stress of the column flange.

tf is thickness of the column flange.

Rn=Fywtw(5k+lb)

k is distance from the outer flange surface of the column to the toe of the fillet in the column web.

lb is length of applied load 5 thickness of beam flange or flange plate.

Fyw is yield stress of the column web.

tw thickness of the column web.

ϕRn=ϕ{0.80tw2[1+3(lbd)(twtf)1.5]EFywtftw}

d is total column depth

Ast=PbfϕRnminϕstFystϕst= 0.90 (since this is a yielding limit state)

ϕst= 0.90 (since this is a yielding limit state)

Pbf is applied factored load from the beam flange or flange plate

ϕRnmin is smallest of the strengths corresponding to the three limit states

Calculation:

Web plate:

Neglect eccentricity

Try ¾ in. diameter Group A bearing type bolts. Assume that threads are in the plane of shear.

Ab=πd24=π(3/4)24=0.4418in.2

Shear capacity of one bolt is

ϕRn=ϕFnvAb=0.75(54)(0.4418)=17.89kips/bolt

Number of bolts required is

4517.89=2.52

Try 4 bolts

Determine plate thickness required for bearing. Assume that

ϕRn=ϕ(2.4dtFu)=0.75(2.4)(3/4)t(58)=78.3t

Load resisted by each bolt = 454=11.25kips

Let ϕRn=11.25=78.3t

t=0.144in.

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=0.30(65)=19.5kips/in.>10.9kips/in.

Therefore, plate controls.

Check bearing strength assumption:

h=34+116=1316in.

For the hole nearest the edge, minimum le=1in.

Try 112in..

lc=leh2=1.513/162=1.094in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.094)(3/16)(58)=10.71kipsϕ(2.4dtFu)=0.75(2.4)(3/4)(3/16)(58)=14.68kips>10.71kips

Therefore, use ϕRn=10.71kips/bolt

For other bolts, minimum s = 2.667(3/4)=2.0in.

Try s=212in.

lc=sh=2.51316=1.688in.ϕRn=ϕ(1.2lctFu)=0.75(1.2)(1.688)(3/16)(58)=16.52kipsϕ(2.4dtFu)=14.68kips<16.52kips

Therefore, use ϕRn=14.68kips/bolt

Total shear/ bearing strength (bearing controls at each bolt location):

ϕRn=10.71+3(14.68)=54.8kips>45kips (OK)

Total length of connection = 2(1.5)+3(2.5)=10.5in.

Use four ¾ in. diameter Group A bolts.

Determine plate thickness required for shear:

Shear yielding strength is

ϕRn=1.0(0.6FyAg)

Let 45=1.0(0.6)(36)(10.5t)

t=0.198in.

Try t=14in.

Check shear rupture strength:

Use hole diameter = 34+18=78in.

Anv=(10.54×78)(14)=1.75in.2

ϕRn=ϕ(0

To determine

(b)

Whether panel zone reinforcement is required.

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