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
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Question
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Chapter 3, Problem 3.6.1P
To determine

(a)

The design of a single-angle tension member using Load Resistance Factor Design (LRFD).

Expert Solution
Check Mark

Answer to Problem 3.6.1P

The single angle tension member using LRFD is L8×4×12.

Explanation of Solution

Given:

The member is 18ft long with 1in diameter bolts.

Dead Load is 28kips.

Live load is 84kips.

Concept Used:

Write the expression for factored load.

Pu=1.2DL+1.6LL …… (I)

Here, the factored load is Pu, dead load is DL and live load is LL.

Write the expression for the required gross area.

Ag=Pu0.9Fy …… (II)

Here, the gross area is Ag and the yield stress is Fy.

Write the expression for the required effective area.

Ae=Pu0.75Fu …… (III)

Here, the required effective area is Ae and the ultimate stress is Fu.

Write the expression for the effective area of the section.

Aes=AnsU …… (IV)

Here, the effective area of the section is Aes, the net area of the section is Ans, and the area reduction factor is U.

Write the expression for the net area of the section.

Ans=Agst(dbolt+18in) …… (V)

Here, the gross area of the section is Ags, thickness of the member is t, and the diameter of bolt is dbolt.

Calculation:

Calculate the factored load.

Substitute 28kips for DL and 84kips for LL in the Equation (I).

Pu=(1.2×28kips)+(1.6×84kips)=(33.6+134.4)kips=168kips

Calculate the required gross area of the tension member.

Substitute 168kips for Pu and 36ksi for Fy in the Equation (II).

Ag=168kips0.9(36ksi)=16832.4in2=5.185in2

Calculate the required net area of the tension member.

Substitute 168kips for Pu and 58ksi for Fu in the Equation (III).

Ae=168kips0.75(58ksi)=16843.5in2=3.862in2

Before choosing the sections calculate the minimum radius of gyration.

Write the expression for the radius of gyration.

rmin=L300 …… (VI)

Here, the minimum radius of gyration is rmin, the length of the member is denoted as L.

Substitute 18ft for L in Equation (VI).

rmin=18ft300×(12in1ft)=18×12in300=0.72in

Choose the section that has parameters slightly above the required values.

Trial 1:

Choose the section L8×4×12.

The properties of the section are,

Ags=5.80in2rmins=0.863in

Here, the gross area of the section is Ags and the minimum radius of gyration of the section is rmins.

Compare the gross area.

(Ags=5.80in2)>(Ag=5.185in2)

Hence, the section is feasible.

Compare the radii of gyration.

(rmins=0.863in)>(rmin=0.72in)

Hence, the section is feasible.

Calculate the net area of the section.

Substitute 5.80in2 for Ags and 1in for dbolt in the Equation (V).

Ans=5.80in212(1in+18in)=5.80in20.5×1.125in2=(5.800.563)in2=5.233in2

Calculate the effective area of the section.

Substitute 5.233in2 for Ans and 0.80 for U in the Equation (IV).

Aes=5.233×0.80=4.19in2

Compare the effective area required and that of the section.

(Aes=4.19in2)>(Ae=3.862in2)

Hence, the section is feasible for design.

Adopt the section L8×4×12.

Conclusion:

Thus, the single angle tension member using LRFD is L8×4×12.

To determine

(b)

The single angle tension member using Allowable Strength Design (ASD).

Expert Solution
Check Mark

Answer to Problem 3.6.1P

The single angle tension member using ASD is L8×4×12.

Explanation of Solution

Concept used:

Write the expression for factored load,

Pu=DL+LL …… (VII)

Here, the factored load is Pu, dead load is DL, and live load is LL.

Write the expression for the required gross area.

Ag=Pu0.6Fy …… (VIII)

Write the expression for the required effective area.

Ae=Pu0.5Fu …… (IX)

Write the expression for the effective area of the section.

Aes=AnsU …… (X)

Here, the effective area of the section is Aes, the net area of the section is Ans, and the area reduction factor is U.

Calculation:

Calculate the factored load.

Substitute 28kips for DL and 84kips for LL in the Equation (VII).

Pu=28kips+84kips=112kips

Calculate the required gross area of the tension member.

Substitute 168kips for Pu and 36ksi for Fy in the Equation (VIII).

Ag=112kips0.6(36ksi)=11221.6in2=5.185in2

Calculate the required net area of the tension member.

Substitute 112kips for Pu and 58ksi for Fu in the Equation (IX).

Ae=112kips0.5(58ksi)=11229in2=3.862in2

Before choosing the sections calculate the minimum radius of gyration.

Substitute 18ft for L in Equation (VI).

rmin=18ft300×(12in1ft)=18×12in300=0.72in

Choose the section that has parameters slightly above the required values.

Trial 1:

Choose the section L8×4×12.

The properties of the section are,

Ags=5.80in2rmins=0.863in

Here, the gross area of the section is Ags and the minimum radius of gyration of the section is rmins.

Compare the gross area.

(Ags=5.80in2)>(Ag=5.185in2)

Hence, the section is feasible.

Compare the radii of gyration.

(rmins=0.863in)>(rmin=0.72in)

Hence, the section is feasible.

Calculate the net area of the section.

Substitute 5.80in2 for Ags and 1in for dbolt in the Equation (V).

Ans=5.80in212(1in+18in)=5.80in20.5×1.125in2=(5.800.563)in2=5.233in2

Calculate the effective area of the section.

Substitute 5.233in2 for Ans and 0.80 for U in the Equation (IV).

Aes=5.233×0.80=4.19in2

Compare the effective area required and that of the section.

(Aes=4.19in2)>(Ae=3.862in2)

Hence, the section is feasible for design.

Adopt the section L8×4×12.

Conclusion:

Thus, the single angle tension member using ASD is L8×4×12.

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