91Ó°ÊÓ

A hollow circular tube \(T\) of length \(L=380 \mathrm{mm}\) is uniformly compressed by a force \(P\) acting through a rigid plate (see figure). The outside and inside diameters of the tube are 76 and \(70 \mathrm{mm}\), respectively. A concentric solid circular bar \(B\) of \(38 \mathrm{mm}\) diameter is mounted inside the tube. When no load is present, there is a clearance \(c=\) \(0.26 \mathrm{mm}\) between the bar \(B\) and the rigid plate. Both bar and tube are made of steel having an elastoplastic stress-strain diagram with \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}=250 \mathrm{MPa}\) (a) Determine the yield load \(P_{Y}\) and the corresponding shortening \(\delta_{Y}\) of the tube. (b) Determine the plastic load \(P_{P}\) and the corresponding shortening \(\delta_{P}\) of the tube. (c) Construct a load-displacement diagram showing the load \(P\) as ordinate and the shortening \(\delta\) of the tube as abscissa. (Hint: The load-displacement diagram is not a single straight line in the region \(0 \leq P \leq P_{Y}\).)

Step by step solution

01

Determine Cross-Sectional Area

First, calculate the cross-sectional area of the tube. The outside diameter is 76 mm, and the inside diameter is 70 mm.The cross-sectional area \( A_T \) of a hollow circle is given by the formula:\[ A_T = \frac{\pi}{4} (D^2 - d^2) \]where \( D = 76 \text{ mm} \) is the outside diameter and \( d = 70 \text{ mm} \) is the inside diameter.So,\[ A_T = \frac{\pi}{4} (76^2 - 70^2) = \frac{\pi}{4} (5776 - 4900) = \frac{\pi}{4} \times 876 = 688.47 \text{ mm}^2 \]

02

Determine Yield Load \( P_Y \)

Using the yield stress \( \sigma_Y = 250 \text{ MPa} \) and the cross-sectional area \( A_T \), calculate the yield load \( P_Y \) using the formula:\[ P_Y = \sigma_Y \times A_T \]\[ P_Y = 250 \text{ MPa} \times 688.47 \times 10^{-6} \text{ m}^2 = 172.12 \text{ kN} \]

03

Determine Corresponding Shortening \(\delta_Y\)

The shortening \( \delta_Y \) of the tube at yield can be determined using the elasticity modulus \( E \) and length \( L \):\[ \delta_Y = \frac{P_Y \cdot L}{A_T \cdot E} \]\[ \delta_Y = \frac{172120 \times 380}{688.47 \times 200 \times 10^3} \]\[ \delta_Y = 0.47 \text{ mm} \]

04

Determine Plastic Load \(P_P\)

After yielding, the solid bar inside the tube will engage with the plate. The plastic load \( P_P \) should consider the cross-sectional area of both the tube and the bar.Calculate the area for the bar:\[ A_B = \frac{\pi}{4} \times (38^2) = 1134.11 \text{ mm}^2 \]Combine the areas:\[ A_{total} = A_T + A_B = 688.47 + 1134.11 = 1822.58 \text{ mm}^2 \]So,\[ P_P = \sigma_Y \times A_{total} \]\[ P_P = 250 \times 1822.58 \times 10^{-6} = 455.65 \text{ kN} \]

05

Determine Corresponding Shortening \(\delta_P\)

After full plastic deformation, the shortening \( \delta_P \) can be estimated by adding the initial clearance \( c \) to the elastic shortening of both tube and bar:The elastic shortening due to \( P_P \):\[ \delta_{elastic} = \frac{P_P \cdot L}{A_{total} \cdot E} \]\[ \delta_{elastic} = \frac{455650 \times 380}{1822.58 \times 200 \times 10^3} = 0.48 \text{ mm} \]Then total \( \delta_P \) is given by:\[ \delta_P = \delta_{elastic} + c = 0.48 + 0.26 = 0.74 \text{ mm} \]

06

Construct Load-Displacement Diagram

To construct the load-displacement diagram, plot the load \( P \) against the shortening \( \delta \). - Start at zero load and zero displacement.- Follow an initial linear path up to \( P_Y = 172.12 \text{ kN} \) with a corresponding \( \delta_Y = 0.47 \text{ mm} \).- Beyond \( P_Y \), the path shifts to a plastic region until \( P_P = 455.65 \text{ kN} \) at \( \delta_P = 0.74 \text{ mm} \).- The path is not a single straight line but transitions at the yield point, showing the different stiffness in elastic and plastic regions.

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Key Concepts

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

Yield Load Calculation

The yield load calculation is an essential step in understanding the behavior of materials under stress. Yield load, denoted as \( P_Y \), is the point at which a material begins to deform plastically. For the hollow circular tube in our exercise, we need to calculate this value by utilizing the known properties of the steel and the geometry of the tube.
The yield stress of the steel, given as \( \sigma_Y = 250 \; \text{MPa} \), and the cross-sectional area of the tube are used in this calculation.

• We find the cross-sectional area \( A_T \) using the formula for a hollow circle: \[ A_T = \frac{\pi}{4} (D^2 - d^2) \]
• Substitute the outer diameter \( D = 76 \; \text{mm} \) and inner diameter \( d = 70 \; \text{mm} \) into the formula to obtain \( A_T = 688.47 \; \text{mm}^2 \).
• The yield load \( P_Y \) is then calculated using: \[ P_Y = \sigma_Y \times A_T = 172.12 \; \text{kN} \]

This value marks the transition from elastic deformation to plastic deformation. It's crucial for predicting how the material will behave under certain load conditions. This step helps in ensuring structures are designed within safe loading limits.

Plastic Deformation

Plastic deformation occurs when a material is loaded beyond its yield point. Unlike elastic deformation, plastic changes are permanent, meaning the material will not return to its original shape after the load is removed. In our problem involving the hollow circular tube, calculating the plastic deformation involves considering both the tube and the central solid bar.

• Once the load reaches the yield load, the solid bar begins to engage. This means that further load does not just deform the tube but also involves the bar.
• The total cross-sectional area \( A_{total} \) of both the tube and the bar is used to determine the plastic load \( P_P \), reflecting this additional resistance.
• This combined area, \( A_{total} = 1822.58 \; \text{mm}^2 \), is used to calculate \[ P_P = 455.65 \; \text{kN} \], representing the increased load capacity of the combined structure.

Understanding plastic deformation is vital as it dictates the limits to which materials can be safely loaded, helping to prevent structural failures.

Stress-Strain Diagram

A stress-strain diagram is a graphical representation of a material's response to external forces. It shows how a material stretches or compresses when stress is applied, and is crucial for understanding its mechanical properties. For our steel components, this diagram features an elastoplastic behavior, which illustratively demonstrates both elastic and plastic deformation.

• The initial linear portion of the diagram represents elastic behavior, during which the material returns to its original form when the load is removed.
• At the yield point, marked by yield stress \( \sigma_Y \), the curve transitions into a plastic region. Here, any further load results in permanent deformation.
• In practice, the diagram is used to predict how much a material can take before permanent deformation occurs, guiding engineers in material selection and design parameters.

The transition from the linear to the curved portion in the diagram delineates the critical yield point beyond which materials need to be handled with caution.

Hollow Circular Tube

A hollow circular tube is a structural element characterized by having a larger outer diameter and a smaller, non-zero inner diameter, forming a tube-like shape. These tubes are widely used in engineering due to their efficiency in supporting loads while minimizing material use. This particular hollow circular tube in our problem features specific mechanical and geometric properties that are fundamental to the calculations.

• The geometric design increases structural efficiency, allowing the tube to carry substantial loads.
• The calculated cross-sectional area \( A_T \) is essential for predicting how the tube will behave under load, particularly in terms of stress distribution.
• By understanding the tube's properties, including the outside (76 mm) and inside (70 mm) diameters, one can accurately predict its response under specified loads.

Recognizing how these tubes function within a system helps in selecting appropriate materials and dimensions for engineering projects, ensuring safe and effective designs.

Load-Displacement Diagram

A load-displacement diagram graphically represents how a structure deforms under applied loads. For the hollow circular tube, this diagram is a tool to visualize its behavior from zero load up through yielding and eventual plastic deformation.

• The graph begins at zero load and displacement. As the load increases linearly, the corresponding displacement is recorded until the yield load \( P_Y = 172.12 \; \text{kN} \) is reached.
• Beyond the yield load, the graph reflects a change in slope indicating the transition into the plastic deformation region.
• Finally, the curve extends to the plastic load \( P_P = 455.65 \; \text{kN} \), featuring the associated displacements of \( \delta_Y = 0.47 \; \text{mm} \) and \( \delta_P = 0.74 \; \text{mm} \).

This diagram is crucial in design and analysis, offering insights into when and how a structure will experience permanent deformation, which aids in ensuring safety and reliability in engineering applications.