Tag Archives: Normal Stress

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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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

Problem 202

Problem 203

Problem 204

Problem 205

Problem 206

Problem 207

Problem 208

Problem 209

Problem 210

Problem 211

Problem 212

Problem 213

Problem 214

Problem 215

Problem 216

Problem 217

Problem 218

Problem 219

Problem 220

Problem 221

Problem 222

Problem 223

Problem 224

Problem 225

Problem 226

Problem 227

Problem 228

Problem 229

Problem 230

Problem 231

Problem 232

Problem 233

Problem 234

Problem 235

Problem 236

Problem 237

Problem 238

Problem 239

Problem 240

Problem 241

Problem 242

Problem 243

Problem 244

Problem 245

Problem 246

Problem 247

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Problem 249

Problem 250

Problem 251

Problem 252

Problem 253

Problem 254

Problem 255

Problem 256

Problem 257

Problem 258

Problem 259

Problem 260

Problem 261

Problem 262

Problem 263

Problem 264

Problem 265

Problem 266

Problem 267

Problem 268

Problem 269

Problem 270

Problem 271

Problem 272

Problem 273

Problem 274

Problem 275

Problem 276

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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

Problem 103

Problem 104

Problem 105

Problem 106

Problem 107

Problem 108

Problem 109

Problem 110

Problem 111

Problem 112

Problem 113

Problem 114

Problem 115

Problem 116

Problem 117

Problem 118

Problem 119

Problem 120

Problem 121

Problem 122

Problem 123

Problem 124

Problem 125

Problem 126

Problem 127

Problem 128

Problem 129

Problem 130

Problem 131

Problem 132

Problem 133

Problem 134

Problem 135

Problem 136

Problem 137

Problem 138

Problem 139

Problem 140

Problem 141

Problem 142

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