   # A PL 3 8 × 6 tension member is welded to a gusset plate as shown in Figure P3.2-2. The steel has a yield stress F y = 50 ksi and an ultimate tensile stress F u = 65 ksi. Assume that A e = A g and compute a . the design strength for LRFD b . the allowable strength for 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 3, Problem 3.2.2P
Textbook Problem
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## A PL 3 8 × 6 tension member is welded to a gusset plate as shown in Figure P3.2-2. The steel has a yield stress F y = 50 ksi and an ultimate tensile stress F u = 65 ksi. Assume that A e = A g and computea. the design strength for LRFDb. the allowable strength for ASD To determine

(a)

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

### Explanation of Solution

Given:

A steel PL 38×6 welded tension member.

The effective area Ae=Ag.

The yield stress is 50ksi.

The tensile stress is 65ksi.

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)

Here the 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=Ag, hence the expression for the nominal strength in rupture is,

Pnf=Fu×Ag ...... (VI)

For the A36 steel resistance factors are,

Φy=0.90Φf=0.75

Calculation:

Calculate the gross area

To determine

(b)

The allowable strength for Allowable Strength Design (ASD).

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