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The tension member is an L 6 × 3 1 2 × 5 16 . It is connected to a 5 16 -inch-thick gusset plate with 3 4 -inch-diameter, Group A bolts. Both the tension member and the gusset plate are of A36 steel. What is the total service load that can be supported, based on boll shear and hearing, if the ratio of live load to dead load is 2.0? The belt threads are in the plane of shear. 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
BuyFindarrow_forward

Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 7, Problem 7.4.5P
Textbook Problem
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The tension member is an L 6 × 3 1 2 × 5 16 . It is connected to a 5 16 -inch-thick gusset plate with 3 4 -inch-diameter, Group A bolts. Both the tension member and the gusset plate are of A36 steel. What is the total service load that can be supported, based on boll shear and hearing, if the ratio of live load to dead load is 2.0? The belt threads are in the plane of shear.

a. Use LRFD.

b. Use ASD.

Chapter 7, Problem 7.4.5P, The tension member is an L631 2 5 16 . It is connected to a 5 16 -inch-thick gusset plate with 3 4

To determine

(a)

The service load using Load and Resistance Factor Design (LRFD).

Explanation of Solution

Given:

The tension member is L6×312×516.

The thickness of the gusset plate is 516inch.

The diameter of the bolt is 34inch.

The live load and dead load ratio is 2.

Concept Used:

Write the expression for the factored load in LRFD.

Pu=1.2D+1.6L     ...... (I)

Here, the factored load is Pu, the dead load is D, and the live load is L.

Rewrite the equation in terms of the design strength.

ϕRn=1.2D+1.6L     ...... (II)

Here, the load reduction factor is ϕ and the design strength is Rn.

Write the expression for the live load.

LD=2     ...... (III)

Write the expression for the service load.

S=D+L     ...... (IV)

Here, the service load is S.

Write the expression for the strength of the bolt in shear.

Rns=FmvAbns     ...... (V)

Here, the shear capacity is Fmv, the area of the bolts is Ab, and the number of shear failure is ns.

Write the expression for the area of the bolts.

Ab=π4d2     ...... (VI)

Here, the diameter of the bolt is d.

Write the expression for the strength in bearing.

Rnb=2.4dtFu     ...... (VII)

Here, the thickness of the member is t and the ultimate stress of the member is Fu.

Write the expression for the nominal strength of the bolts.

Rn=neRne+niRnv     ...... (VIII)

Here the bearing strength of the inner bolt is Rni, the number of inner bolts is ni, the bearing strength of the edge bolt is Rne, and the number of edge bolts is ne.

Write the expression for the bearing strength of the bolts.

Rne=Rni=1.2lctFu     ...... (IX)

Here, the clear distance between the edge bolts is lc, the thickness of the member is t, and the ultimate shear stress is Fu.

Write the expression for the upper limit of the nominal strength.

Ru=2.4tdFu     ...... (X)

Write the expression for the clear distance of the edge bolts.

lc=leh2     ...... (XI)

Here the diameter of the hole is h.

Write the expression for the whole diameter.

h=d+116in     ...... (XII)

Write the expression for the clear distance for other bolts.

lc=sh     ...... (XIII)

Calculation:

Calculate the area of the bolts.

Substitute 34in for d in the Equation (VI).

Ab=π4×(34in)2=0.442in2

Calculate the strength of the bolt in shear.

Substitute 0.442in2 for Ab, 54ksi for Fmv

1 for ns in the Equation (V).

Rns=54ksi×0.442in2×1=23.868kips

Calculate the hole diameter of the bolts.

Substitute 34in for d in the Equation (XII).

h=34in+116in=1316in

Calculate the clear distance of the edge bolts.

Substitute 112in for le, 1316in for h in the Equation (XI).

lc=(112in)(1316in)2=32in1332in=3532in

Calculate the clear distance of the inner bolts

To determine

(b)

The service load using Allowable Strength Design (ASD).

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