Example:3 An aluminum bar 1.8 meters long has a25 mm square c/s over 0.6 meters of its length and25 mm circular c/s over other 1.2 meters . Howmuch will the bar elongate under a tensile loadP=17500 N, if E = 75000 MPa.17500 N25 mm sq.sect 25 mm cir..sect1.2 m0.6 mSolution :- 8 = EPL/AE=17500*600 / (252*75000) +17500*1200/(0.785*252*75000) =0.794 mm

Answered: Image /qna-images/answer/3228a85f-9206-4bd8-9e20-29470f5cb42a.jpg

Answered: An aluminum bar 1.8 meters

Engineering

Electrical Engineering

An aluminum bar 1.8 meters

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Antenna Characteristics

The antenna is an important concept in the domain of electronics and communication engineering. An antenna forms an integral part of many commonly used devices such as wireless modems, routers, game controllers, mobile phones, bluetooth devices, and so on. In simple terms, any wireless communication makes use of antennas to serve the purpose. An antenna, also known by the term 'arial', is an electrical device that converts electrical signals into electromagnetic waves or electromagnetic waves into electrical signals. An antenna contains a transmission line usually made up of co-axial cable, it then receives the signals and converts them into electromagnetic waves, which are then transmitted via space into their required destination. A communication system essentially makes use of two antennas, a transmission antenna, which converts electrical signals into electromagnetic waves, and a receiving antenna, which receives the electromagnetic waves and converts them back into electrical signals.

Question

Example:3 An aluminum bar 1.8 meters long has a
25 mm square c/s over 0.6 meters of its length and
25 mm circular c/s over other 1.2 meters . How
much will the bar elongate under a tensile load
P=17500 N, if E = 75000 MPa.
17500 N
25 mm sq.sect 25 mm cir..sect
1.2 m
0.6 m
Solution :- 8 = EPL/AE
=17500*600 / (252*75000) +
17500*1200/(0.785*252*75000) =0.794 mm

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