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Homework 3 in MEE 322 Structural Mechanics | Normal and Shear Stresses Under Combined Loading Part 2

Problem 1

Three forces act on the tip of a L-shaped rod with a cross-sectional radius of 0.5 in.

(a) Determine the normal and shear stress at points A and B and draw the stress cube at those points based on the given coordinate system.

(b) Determine the maximum normal stress on the cross-section and locate the point at which it occurs.

Problem 2

The simply supported solid shaft has a radius of 15 mm and is under static equilibrium. Pulley C has a diameter of 100 mm. The pulleys B and D have the same diameter as each other. The forces on pulley B are at an angle of 45 to the negative z-axis. The forces on pulley C and pulley D are in the z and -y direction. The shaft dimensions are in mm.

(a) Determine the maximum bending and torsional stresses in the shaft.

(b) Locate the point(s) on the cross-section where the bending stress is maximum.

Problem 3

The structural part of a setup to measure net belt tensions in pulleys is shown in the figure. The belt tensions at both sides of the pulley at B (radius 10 cm) are P and F=0.1*P along z and a reaction force is measured from the pulley at C (radius 2 cm), which is connected to a load cell at E with an axial member parallel to x. Pulleys are rigidly attached to rod AD, which is made with a ductile steel rod 60 cm long and 1.27 cm in diameter. Length AB=0.20 m, and length DC=0.15 m. There is a spherical hinge at A and a plane hinge at D. The latter constrains motion in the x-z plane only.

(a) Draw bending moment and torsion diagrams for this structure as functions of the unknown tension P and use them to draw a diagram of the critical section showing internal loads (bending and torsion moments) and the critical points.

(b) Use your results from part a to determine the maximum normal stress due to bending and the maximum shear stress due to torsion in terms of the unknown tension P. Calculate the maximum value that P can have if only bending stresses are considered (with σallow = 350 MPa) and then if only torsion stresses are considered (with τallow = 175 MPa).

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Homework #2 in MEE 322 Structural Mechanics

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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

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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

Problem 101

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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: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem 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:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D & = 2.50\ \text{in} \\
\text{Axial Load}, P & = 27\ \text{kips} \\
\text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\
\text{Inside Diameter}, d & = D-2t
\end{align*}

We are solving for the unknown wall thickness of the tube, t.

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

\begin{align*}
\sigma & =\frac{P}{A} \\ 
A_{\text{min}} & =\frac{P}{\sigma } \\ 
A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ 
A_{\text{min}} & = 1.500\:\text{in}^2 
\end{align*}

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

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

\begin{align*}
A & =\frac{\pi }{4}\left(D^2-d^2\right) \\
A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
\frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\
d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\
d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\
d & = 2.08330\ \text{in}
\end{align*}

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

\begin{align*}
t & =\frac{D-d}{2} \\
t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\
t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}