   # A bracket cut from a WT-shape is connected to a column flange with 10 Group A slip-critical bolts, as shown in Figure P8.3-8. A992 steel is used. The loads are service loads, consisting of 3 0 % dead load and 7 0 % live load. What size bolt is required? a . Use LRFD. b . Use ASD. ### 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 8.3.8P
Textbook Problem
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## A bracket cut from a WT-shape is connected to a column flange with 10 Group A slip-critical bolts, as shown in Figure P8.3-8. A992 steel is used. The loads are service loads, consisting of 3 0 % dead load and 7 0 % live load. What size bolt is required?a. Use LRFD.b. Use ASD. (a)

To determine

The size of bolt on the basis of load resistance and factor design.

### Explanation of Solution

Given:

The number of group A bolt used is 10.

The factored load along x-direction is 47kips.

The factored load along y-direction is 60kips.

Calculation:

Write the equation to obtain the load factor along x-direction.

Pux=1.2PD+1.6PL ..... (I)

PD=30100×47kips=14.1kips

PL=70100×47kips=32.9kips

Substitute 32.9kips for PL and 14.1kips for PD in Equation (I).

Pux=1.2(14.1kips)+1.6(32.9kips)=16.92kips+52.64kips=69.56kips

Write the equation to obtain the load factor along y-direction.

Puy=1.2PD+1.6PL ..... (II)

PD=30100×60kips=18kips

PL=70100×60kips=42kips

Substitute 42kips for PL and 18kips for PD in Equation (II).

Puy=1.2(18kips)+1.6(42kips)=21.6kips+67.2kips=88.8kips

Write the expression for the direct tensile strength per bolt.

td=Fuxn ..... (III)

Here, the direct tensile strength per bolt is td the factored load in y-direction is Fuy and the total number of bolts is n.

Substitute 69.56kips for Fux and 10 for n in Equation (III).

td=69.56kips10=6.956kips

Write the expression moment caused by eccentricity.

Mu=Puyex+Puxey ..... (IV)

Here, the moment caused by eccentricity is Mu, eccentricity in x-direction is ex and the eccentricity in y-direction is ey.

Calculate the eccentricity in y-direction.

ey=(3.5+3.5+2)in=9in

Substitute, 69.56kips for Pux, 8in for ex, 88.8kips for Puy and 9in for ey in Equation (IV).

Mu=(88.8kips×8in)+(69.56kips×9in)=1336in-kips

Write the expression to obtain the bolt force in tension.

M=n'rtdm ..... (V)

Here, resisting moment is M, the number of bolts above the neutral axis is n', the bolt force in tension is rt and the distance between the centroid of compression and tension bolts is dm.

Calculate the distance between the centroid of compression and tension bolts.

dm=[2×(3.5+3.52)]in=10.5in

Substitute, 1336in-kips for M, 4 for n' and 10.5in for dm in equation (V).

1336in-kips=4×rt×10.5inrt=1336in-kips4×10.5inrt=31.81kips

Write the expression for the total tension load per bolt.

t=td+rt ..... (VI)

Here, the total tension load per bolt is t.

Substitute 6.956kips for td and 31.81kips for rt in equation (VI).

t=6.956kips+31.81kips=38.77kips 

The total tension load per bolt is equal to the nominal shear strength per bolt.

Write the expression to obtain the nominal shear strength per bolt.

ϕRnv=ϕFnvAb ..... (VII)

Here, resistance factor is ϕ, area of bolt is Ab and nominal shear strength of the bolt is Fnv.

Substitute, 38.77kips for ϕRnv, 0

(b)

To determine

The size of bolt on the basis of Allowed Stress Design.

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