Tag Archives: Mechanics of Materials

Torsion Featured Image: Chapter 3 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 3: Torsion

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Simple Strain Featured Image: Chapter 2 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 2: Simple Strain

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Simple Stress Featured Image: Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer

Chapter 1: Simple Stress

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Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual

Chapter 1: Stress
Chapter 2: Strain
Chapter 3: Mechanical Properties of Materials
Chapter 4: Design Concepts
Chapter 5: Axial Deformation
Chapter 6: Torsion
Chapter 7: Equilibrium of Beams
Chapter 8: Bending
Chapter 9: Shear Stress in Beams
Chapter 10: Beam Deflections
Chapter 11: Statically Indeterminate Beams
Chapter 12: Stress Transformations
Chapter 13: Strain Transformations
Chapter 14: Thin-Walled Pressure Vessels
Chapter 15: Combined Loads
Chapter 16: Columns
Chapter 17: Energy Methods
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Mechanics of Materials 3rd Edition by Timothy A. Philpot, P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

\displaystyle \sigma =\frac{P}{A}

\displaystyle A_{min}=\frac{P}{\sigma }

\displaystyle A_{min}=\frac{27\:kips}{18\:ksi}

\displaystyle A_{min}=1.500\:in.^2

The cross-sectional area of the aluminum tube is given by

\displaystyle A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter d 

\displaystyle \frac{\pi }{4}\left[\left(2.50\:in\right)^2-d^2\right]=1.500\:in^2

\displaystyle \left(2.50\:in\right)^2-d^2=\frac{4}{\pi }\left(1.500\:in^2\right)

\displaystyle \left(2.50\:in\right)^2-\frac{4}{\pi }\left(1.500\:in^2\right)=d^2

\displaystyle d_{max}=2.08330\:in

The outside diameter D, the inside diameter d, and the wall thickness t are related by 

D=d+2t

Therefore, the minimum wall thickness required for the aluminum tube is 

\displaystyle t_{min}=\frac{D-d}{2}

\displaystyle t_{min}=\frac{2.50\:in-2.08330\:in}{2}

\displaystyle t_{min}=0.20835\:in

\displaystyle t_{min}=0.208\:in

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Maximum Load of a Member | Stress | Mechanics of Materials | 3rd Edition | Timothy Philpot | Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

SOLUTION:

The cross-sectional area of the stainless steel tube is

\displaystyle A=\frac{\pi }{4}\left(D^2-d^2\right)

\displaystyle A=\frac{\pi }{4}\left[\left(60\:mm\right)^2-\left(50\:mm\right)^2\right]

\displaystyle A=863.938\:mm^2

The normal stress in the tube can be expressed as 

\displaystyle \sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

\displaystyle P_{max}=\sigma _{max}A

\displaystyle P_{max}=\left(200\:MPa\right)\left(863.938\:mm^2\right)

\displaystyle P_{max}=\left(200\:\frac{N}{mm^2}\right)\left(863.938\:mm^2\right)

\displaystyle P_{max}=172\:788\:N

\displaystyle P_{max}=172.8\:kN

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