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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 3, Problem 3.4.3P
Textbook Problem
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The tension member shown in Figure P3.4-3 is a PL 3 8 × 8 . The bolts are 1 2 inch in diameter and A36 steel is used.

a. Compute the design strength.

b. Compute the allowable strength.

Chapter 3, Problem 3.4.3P, The tension member shown in Figure P3.4-3 is a PL3 8 8. The bolts are 1 2 inch in diameter and A36

To determine

(a)

The design strength.

Explanation of Solution

Given:

Tension member is PL38×8.

Steel used is A36.

Diameter of bolt is 12in.

Calculation:

The properties of A36 steel are as follows:

The yield strength (Fy) is 36ksi.

The ultimate strength (Fu) is 58ksi.

Calculate the gross area of the plates.

Ag=38in×8in=3in2

Here, the gross section is Ag.

Write the expression to calculate the nominal strength in yielding.

Pn=FyAg      ...... (I)

Here, the yield strength is Fy and the nominal strength is Pn.

Substitute 3in2 for Ag and 36ksi for Fy in Equation (I).

Pn=36ksi×3in2=108kips

Write the expression to calculate the diameter of the hole.

dh=db+sc      ...... (II)

Here, the diameter of the hole is dh the side clearance is sc and the diameter of the bolt is db.

Substitute 18in for sc and 12in for db in Equation (II).

dh=12in+18in=58in=0.625in

Write the expression to calculate the area of the hole.

Ah=t×dh×2      ...... (III)

Here, the thickness of the leg is t and the area of the hole is Ah.

Substitute 0.625in for dh and 38in for t in Equation (III).

Ah=0.625in×38in×2=0.46875in2

Write the expression to calculate the net area of the plate.

An=AgAh      ...... (IV)

Here, the net area of the plate is An

To determine

(b)

The allowable strength.

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Chapter 3 Solutions

Steel Design (Activate Learning with these NEW titles from Engineering!)
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