Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
Question
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Chapter 3, Problem 3.2.1P
To determine

(a)

The design strength using Load and Resistance Factor Design(LRFD).

Expert Solution
Check Mark

Answer to Problem 3.2.1P

85.05kips.

Explanation of Solution

Given:

A36 steel PL 38×7 tension member with three 1inch diameter bolt.

The effective area, Ae=An.

Concept Used:

Write the expression for the factored strength in yielding.

P=Φy×Pny ...... (I)

Write the expression for the factored strength in rupture.

P=Φf×Pnf ...... (II)

Herethe strength of the material is P, the resistance factor for yielding is Φy, the resistance factor for rupture is Φf, the nominal strength in yielding is Pny, and the nominal strength in rupture is Pnr

The design strength of LRFD is the minimum of Equation (I) and (II).

Write the expression for the nominal strength in yielding.

Pny=Fy×Ag ...... (III)

Here, the yield strength in yielding is Fy, the gross area of the member is Ag.

Write the expression for the gross area of the member.

Ag=L×t ...... (IV)

Here, the length of the member is L, the thickness of the member is t.

Write the expression for the nominal strength in rupture.

Pnf=Fu×Ae ...... (V)

Here, the yield strength in rupture is Fu, the effective area is Ae.

Given that, Ae=An, hence,

Ae=An=AgAholes ...... (VI)

Here, the area of the holes is Aholes, the net area is An.

Write the expression for the area of the holes.

Aholes=t×dholes ...... (VII)

Here, the diameter of the holes is dholes.

Write the expression for the diameter of the holes.

dholes=d+18in ...... (VIII)

Here, the diameter of the bolts is d and 18in is the allowance for side clearance for bolts less than 1in.

For the A36 steel resistance factors and the yield strength values are,

Φy=0.90Φf=0.75Fy=36ksiFu=58ksi

Calculation:

Calculate the gross area.

Substitute 7in for L and 38in for t in Equation (IV).

Ag=7in×38in=2.625in2

Calculate the diameter of holes.

Substitute 1in for d in Equation (VIII).

dholes=1in+18in=98in

Calculate the area of holes.

Substitute 38in for t and 98in for dholes in Equation (VII).

Aholes=38in×98in=0.422in2

Calculate the effective area.

Substitute 2.625in2 for Ag and 0.422in2 for Aholes in Equation (VI).

Ae=2.625in20.422in2=2.203in2

Calculate the yielding strength.

Substitute 36ksi for Fy, 2.625in2 for Ag in the Equation (III).

Pny=36×2.625=94.5kips

Calculate the strength at rupture.

Substitute 58ksi for Fu, 2.203in2 for Ae in the Equation (IV).

Pnf=58×2.203=127.77kips

Calculate the design strength in yielding.

Substitute 94.5kips for Pny and 0.9 for Φy in the Equation (I).

P=0.9×94.5kips=85.05kips

Calculate the design strength in rupture.

Substitute 127.77kips for Pnf and 0.75 for Φf in the Equation (II).

P=0.75×127.77kips=95.83kips

Conclusion:

Compare the design strengthin yielding and rupture. The design strength using LRFD is the smaller of the two.

Thus, the design strength using LRFD is 85.05kips.

To determine

(b)

The allowable strength usingAllowable Strength Design (ASD).

Expert Solution
Check Mark

Answer to Problem 3.2.1P

The allowable strength using ASD is 56.59kips.

Explanation of Solution

Write the expression for the allowable strength inyielding.

P=PnyΩy ...... (IX)

Here, the safety factor in yielding is Ωy.

Write the expression for the allowable strength for rupture.

P=PnfΩf ...... (X)

Here, the safety factor in rupture is Ωf.

The allowable strength is the minimum of Equation (IX) and (X).

Calculation:

Calculate the allowable strength in yielding.

Substitute 94.5kips for Pny and 1.67

Ωy for in Equation(IX).

P=94.51.67kips=56.59kips

Calculate the allowable strength in rupture.

Substitute 127.77kips for Pnf and 2 for Ωf in Equation (X).

P=127.772kips=63.89kips

Conclusion:

Compare the allowable strength in yielding and rupture. The smaller values are the allowable strength using ASD.

Thus, the allowable strength using ASD is 56.59kips.

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Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning