**Problem Statement:**A metal bar of variable cross section is subjected to a force of 1000 kN. The bar consists of two sections:1. A 1 m long section with a cross-sectional area of 10,000 sq. mm.2. A 500 mm long section with a cross-sectional area of 5,000 sq. mm.Given that the elastic modulus is 200 GPa, calculate the total elongation of the bar.**Explanation:**To solve this problem, you will need to use the formula for elongation (\(\Delta L\)) of a material under stress, which is given by:\[\Delta L = \frac{F \cdot L}{A \cdot E}\]Where:- \(F\) is the force applied (1000 kN = 1,000,000 N),- \(L\) is the length of the section,- \(A\) is the cross-sectional area,- \(E\) is the elastic modulus (200 GPa = 200,000 N/mm²).The total elongation is the sum of the elongations of each section.

A metal bar of variable cross section as shown and length is subjected to 1000 kN of force. The sections are of 1 m long section is 10000 sq.mm. and that of 500 mm long section is 5000 sq. mm.. Assuming the elastic modulus of 200 GPa, the total elongation of the bar will be?

A metal bar of variable cross section as shown and length is subjected to 1000 kN of force. The sections are of 1 m long section is 10000 sq.mm. and that of 500 mm long section is 5000 sq. mm.. Assuming the elastic modulus of 200 GPa, the total elongation of the bar will be?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

**Problem Statement:**

A metal bar of variable cross section is subjected to a force of 1000 kN. The bar consists of two sections:

1. A 1 m long section with a cross-sectional area of 10,000 sq. mm.
2. A 500 mm long section with a cross-sectional area of 5,000 sq. mm.

Given that the elastic modulus is 200 GPa, calculate the total elongation of the bar.

**Explanation:**

To solve this problem, you will need to use the formula for elongation (\(\Delta L\)) of a material under stress, which is given by:

\[
\Delta L = \frac{F \cdot L}{A \cdot E}
\]

Where:
- \(F\) is the force applied (1000 kN = 1,000,000 N),
- \(L\) is the length of the section,
- \(A\) is the cross-sectional area,
- \(E\) is the elastic modulus (200 GPa = 200,000 N/mm²).

The total elongation is the sum of the elongations of each section.

Expert Solution

Step 1: Given :

Force = 1000 kN

Section :

Length 1 meter long

area = 10000 sq. mm

Section 2 :

Length = 500 mm

Area = 5000 sq. mm

Elastic modulus = 200 GPa

To Find :

Total elongation of bar ?

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