Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)
Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)
6th Edition
ISBN: 9780134441184
Author: Robert L. Mott, Edward M. Vavrek, Jyhwen Wang
Publisher: PEARSON
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Chapter 3, Problem 1P

A tensile member in a machine structure is subjected to a steady load of 4.50 kN. It has a length of 750 mm and is made from a steel tube having an outside diameter of 18 mm and an inside diameter of 12 mm. Compute the tensile stress in the tube and the axial deformation.

Expert Solution & Answer
Check Mark
To determine

The value of tensile stress in the tube.

The axial deformation in the tube.

Answer to Problem 1P

The value of stress in a tube is, σ=31.83MPa .

The axial deformation in the tube is, δ=0.12mm .

Explanation of Solution

Given:

Length of the tube, L=750mm

Inside diameter of the tube, di=18mm

Outside diameter of the tube, do=12mm

Steady load in the tensile member, F=4.50KN

Concept used:

Stress- It is the applied force per unit area.

  σ=Force(F)Area(A)

Unit of the stress is N/m2.

The axial deformation in the tube is, δ=σ×LE

Where,

  δ is the axial deformation in the tube.

  L is the length of the tube..

  E is the young’s modulus,

  E=200GPa=200Nmm2 (For steel).

Calculation:

The cross sectional area of the tube is calculated as, A=π4(D2d2) .

The cross sectional area of the tube is,

  A=π4(D2d2)=π4[( 18)2( 12)2]mm2=3.1424×[324144]mm2=141.39mm2

Now the value of tensile stress is calculated as:

  Stress(σ)=Force(F)Area(A)=4.5× 103N141.39mm2=31.83Nmm2

As, 1Nmm2 is converted as 1MPa .

Thus, 31.83Nmm2 is converted as:

  σ=431.83Nmm2=31.83MPa

Hence, the value of stress in a tube is, σ=31.83MPa .

Now calculate the axial deformation in the tube. δ=σ×LE

The axial deformation in the tube is calculated as,

  δ=σ×LE=31.83N m m 2 ×750mm200× 103N m m 2 =0.12mm

Hence, the axial deformation in the tube is, δ=0.12mm .

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Chapter 3 Solutions

Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)

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