91Ó°ÊÓ

Plastic bar \(A B\) of rectangular cross section \((b=19 \mathrm{mm} \text { and } h=38 \mathrm{mm})\) and length \(L=0.6 \mathrm{m}\) is fixed at \(A\) and has a spring support \((k=3150 \mathrm{kN} / \mathrm{m})\) at \(C(\) see figure). Initially, the bar and spring have no stress. When the temperature of the bar is raised by \(48^{\circ} \mathrm{C}\), the compressive stress on an inclined plane \(p q\) at \(L_{\theta}=0.46 \mathrm{m}\) becomes \(5.3 \mathrm{MPa}\). Assume the spring is massless and is unaffected by the temperature change. Let \(\alpha=95 \times 10^{-6 /^{\circ} \mathrm{C}}\) and \(E=2.8 \mathrm{GPa}\) (a) What is the shear stress \(\tau_{\theta}\) on plane \(p q ?\) What is angle \(\theta ?\) (b) Draw a stress element oriented to plane \(p q,\) and show the stresses acting on all faces of this element. (c) If the allowable normal stress is ±6.9 MPa and the allowable shear stress is \(\pm 3.9 \mathrm{MPa}\), what is the maximum permissible value of spring constant \(k\) if allowable stress values in the bar are not to be exceeded? (d) What is the maximum permissible length \(L\) of the bar if allowable stress values in the bar are not be exceeded? (Assume \(k=3150 \mathrm{kN} / \mathrm{m} .\) ) (e) What is the maximum permissible temperature increase \((\Delta T)\) in the bar if allowable stress values in the bar are not to be exceeded? (Assume \(L=0.6 \mathrm{m}\) and \(k=\) \(3150 \mathrm{kN} / \mathrm{m}\)

Step by step solution

01

Calculate Thermal Strain

Given that the temperature increase \(\Delta T = 48^{\circ} C\) and coefficient of thermal expansion \(\alpha = 95 \times 10^{-6 / \circ C}\), calculate the thermal strain \(\varepsilon_t\) using the formula:\[\varepsilon_t = \alpha \times \Delta T = 95 \times 10^{-6} \times 48 = 0.00456.\]

02

Calculate Axial Compressive Stress

Using the modulus of elasticity \(E = 2.8 \text{ GPa} = 2800 \text{ MPa}\), calculate the axial compressive stress \(\sigma\) arising from thermal expansion resisted by the spring:\[\sigma = E \times \varepsilon_t = 2800 \times 0.00456 = 12.768 \text{ MPa}.\]

03

Solve for the Angle \(\theta\)

The compressive stress on plane \(pq\) is given as \(5.3 \text{ MPa}\). Since stress \(\sigma = 12.768 \text{ MPa}\) is the principal stress, use the transformation of stress formulas to find the angle \(\theta\):\[\cos 2\theta = \frac{\sigma_{pq} - \sigma_2}{\sigma_1 - \sigma_2} = \frac{5.3}{12.768} \]Solving gives \(\theta = 48.6^{\circ}\).

04

Calculate Shear Stress \(\tau_\theta\)

Calculate the shear stress \(\tau_\theta\) using the stress transformation equations. We know that for inclined plane:\[\tau_\theta = \frac{\sigma_x - \sigma_y}{2} \sin(2\theta) = \frac{12.768}{2} \sin(2 \times 48.6) = 3.71 \text{ MPa}\]

05

Construct Stress Element on Plane

Draw a stress element inclined to the plane \(pq\), labeling normal stress \(\sigma_{\theta} = 5.3 \text{ MPa}\) and shear stress \(\tau_\theta = 3.71 \text{ MPa}\) on all faces.

06

Calculate Maximum Permissible Spring Constant \(k\)

Use allowable stresses \(\pm 6.9 \text{ MPa}\) for normal and \(\pm 3.9 \text{ MPa}\) for shear to ensure that the bar stresses remain within limits. Given \(\sigma = k x \times \text{length}\), solve for spring constant using:\[\sigma_{allowed} = \frac{k \times \Delta L}{A}\], where \(A = b \times h\) and adjust \(k\) so \(\sigmae\) allowable stresses.

07

Calculate Maximum Permissible Bar Length \(L\)

Given spring constant \(k = 3150 \text{ kN/m} = 3150 \text{\ N/mm}\), adjust length \(L\) to ensure compliance with \(\sigma_\text{max} = 6.9 \text{ MPa}\) and \(\tau_\text{max} = 3.9 \text{ MPa}\). Use:\[\Delta L = \frac{\sigma_{allowed} \times b \times h}{k} \text{ and solve for } L \].

08

Maximum Permissible Temperature Increase (\(\Delta T\))

Calculate maximum permissible \(\Delta T\) using the allowable stresses. Use\(\tau_{allowed} = E \times \alpha \times \Delta T_{max}\) to solve for \(\Delta T_{max}\), ensuring \(\sigma_\text{max}\) and \(\tau_\text{max}\) are within limits.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Strain

Thermal strain measures how much a material expands or contracts when its temperature changes. It's influenced by the coefficient of thermal expansion and the amount of temperature change. In our exercise, the plastic bar expands because it is heated by 48°C. This strain, denoted by \( \varepsilon_t \), can be calculated using the formula \( \varepsilon_t = \alpha \times \Delta T \), where \( \alpha \) is the coefficient of thermal expansion and \( \Delta T \) is the temperature increase. Here, the coefficient \( \alpha = 95 \times 10^{-6}/^{\circ}C \), and \( \Delta T = 48^{\circ}C \). Applying these values, \( \varepsilon_t = 0.00456 \), meaning there is a 0.456% expansion of the bar. Understanding thermal strain is critical because it helps predict how materials behave when subjected to temperature variations, essential in designing structures subject to temperature fluctuations.

Shear Stress Calculation

In mechanics of materials, shear stress represents the element's ability to resist an external force applied parallel to its surface. Shear stress \( \tau \) can be calculated using the transformation of stress relationships which incorporate factors like inclined planes and the angle of application. For our plastic bar, we calculate shear stress on an inclined plane as part of the solution. Given the principal stress \( \sigma = 12.768 \text{ MPa} \) and using the angle \( \theta \) identified from stress transformation problems, shear stress is computed as \( \tau_{\theta} = \frac{\sigma_x - \sigma_y}{2} \sin(2\theta) \). With \( \theta = 48.6^{\circ} \) and using this in the formula, we calculate a shear stress \( \tau_{\theta} = 3.71 \text{ MPa} \). Proper calculations of shear stress are important to ensure the safety and integrity of materials under parallel forces.

Stress Transformation

Stress transformation is a key concept that allows us to evaluate stress components in different orientations. Specifically, for inclined planes within a stressed material, it helps assess how stresses change direction and intensity. The transformation involves mathematical equations that determine normal and shear stresses on any plane at an angle \( \theta \) to the material's principal planes. In our exercise, the principal stress was \( 12.768 \text{ MPa} \) with known stress on plane \( pq \) as \( 5.3 \text{ MPa} \). Using the equation \( \cos 2\theta = \frac{\sigma_{pq} - \sigma_2}{\sigma_1 - \sigma_2} \), the angle \( \theta \) was calculated to be \( 48.6^{\circ} \). Understanding stress transformation is crucial for predicting failure modes and designing elements that resist various force directions.

Elastic Modulus

The elastic modulus, or Young's modulus \( E \), describes a material's stiffness or rigidity – how much it will deform under stress. For calculations in our problem, the modulus of elasticity is given as \( 2.8 \text{ GPa} \). It directly relates the stress to strain in the elastic deformation region of the material's stress-strain curve, following Hooke's Law: \( \sigma = E \times \varepsilon \). Using this modulus allows us to determine the axial compressive stress in response to the thermal expansion of the spring-supported bar. With a calculated thermal strain \( \varepsilon_t = 0.00456 \), we find the axial stress as \( 12.768 \text{ MPa} \). Understanding the elastic modulus is pivotal for predicting how materials will respond under loads, ensuring they perform as expected without failure.

Thermal Expansion

Thermal expansion describes the tendency of matter to change its shape, area, and volume in response to a change in temperature. It is quantified by the coefficient of thermal expansion \( \alpha \), which indicates how much a material expands per degree change in temperature. In the illustrated problem, the temperature change causes the plastic bar to expand, creating stress because the expansion is constrained by the spring. This effect is predictable and calculated as part of thermal strain using \( \alpha = 95 \times 10^{-6/^{\circ} \mathrm{C}} \). Thermal expansion knowledge is vital for anticipating how materials will behave in different thermal environments, allowing engineers to design components that maintain their integrity despite temperature shifts.