Category Archives: Strength of Materials

Strength of Materials Problem 101 – Stress in each section of a composite bar

A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1-8a. Axial loads are applied at the positions indicated. Determine the stress in each section.

Strength of Materials by Andrew Pytel and Ferdinand Singer Problem 101
Figure 1.8a

Solution:

We must first determine the axial load in each section to calculate the stresses. The free-body diagrams have been drawn by isolating the portion of the bar lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered.

Solve for the internal axial load of the bronze

Free body diagram for the internal axial load of the bronze section for Problem 101 of Strength of Materials by Ferdinand Singer and Andrew Pytel
The free-body diagram of the bronze section
\begin{align*}
\sum_{}^{}F_x & = 0  \to  \\
-4000\ \text{lb}+P_{br} & = 0 \\
P_{br} & = 4000 \ \text{lb} \ \text{(tension)}
\end{align*}

Solve for the internal axial load of the aluminum

Free-body diagram of the aluminum section for problem 101 of Strength of materials by Andrew Pytel and Ferdinand Singer
The free-body diagram of the aluminum section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000 \ \text{lb} + 9000 \ \text{lb} - P_{al} & = 0 \\
P_{al} & = 5000 \ \text{lb} \ \text{(Compression)}
\end{align*}

Solve for the internal axial load of the aluminum

The free-body diagram of the steel section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000\ \text{lb} + 9000 \ \text{lb} + 2000\ \text{lb} - P_{st} & =0 \\
P_{st} & = 7000 \ \text{lb} \ \text{(Compression)}
\end{align*}

We can now solve the stresses in each section.

For the bronze

\begin{align*}
\sigma_{br} & = \frac{P_{br}}{A_{br}} \\
& = \frac{4000\ \text{lb}}{1.2 \ \text{in}^2} \\
& = 3330 \ \text{psi}\ \text{(Tension)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the aluminum

\begin{align*}
\sigma_{al} & = \frac{P_{br}}{A_{al}} \\
& = \frac{5000\ \text{lb}}{1.8 \ \text{in}^2} \\
& = 2780 \ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the steel

\begin{align*}
\sigma_{st} & = \frac{P_{st}}{A_{st}} \\
& = \frac{7000\ \text{lb}}{1.6 \ \text{in}^2} \\
& = 4380\ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Torsion Featured Image: Chapter 3 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 3: Torsion

Problem 301

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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

Problem 201

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Problem 211

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

Chapter 1: Simple Stress

Problem 101

Problem 102

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Problem 111

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Strength of Materials Fourth Edition by Andrew Pytel and Ferdinand Singer Featured Image

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer

Chapter 1: Simple Stress
Chapter 2: Simple Strain
Chapter 3:
Torsion
Chapter 4: Shear and Moment in Beams
Chapter 5: Stresses in Beams
Chapter 6: Beam Deflections
Chapter 7: Restrained Beams
Chapter 8: Continuous Beams
Chapter 9: Combined Stresses
Chapter 10: Reinforced Beams
Chapter 11: Columns
Chapter 12: Riveted, Bolted, and Welded Connections
Chapter 13: Special Topics
Chapter 14: Inelastic Action
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Mechanics of Deformable Bodies

Mechanics of Materials: An Integrated Learning System 4th Edition Cover by Timothy Philpot

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

Strength of Materials Fourth Edition by Andrew Pytel and Ferdinand Singer

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer

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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Engineering Sciences

Chemistry
Physics
Geology
Statics of Rigid Bodies
Dynamics of Rigid Bodies
Mechanics of Deformable Bodies
Fluid Mechanics
Thermodynamics
Engineering Economy
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Problem 1-2|Stress | Mechanics of Materials| Ninth Edition| R.C. Hibbeler|

Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.

Problem 1-2

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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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