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A copper wire having a diameter of \(d=4 \mathrm{mm}\) is bent into a circle and held with the ends just touching (see figure) (a) If the maximum permissible strain in the copper is \(\varepsilon_{\max }=0.0024,\) what is the shortest length \(L\) of wire that can be used? (b) If \(L=5.5 \mathrm{m},\) what is the maximum acceptable diameter of the wire if the maximum normal strain must remain below yield? Assume \(E=120 \mathrm{GPa}\) and \\[ \sigma_{Y}=300 \mathrm{MPa} \\]

Short Answer

Expert verified
(a) Shortest wire length is approximately 5.24 m; (b) Maximum wire diameter is 4.375 mm.

Step by step solution

01

Understanding Bending and Strain

To find the shortest length of the wire that can be bent into a circle without exceeding the maximum permissible strain, we need to consider the relationship between curvature, strain, and the geometry of bending. Assuming small strains, the curvature of the wire when bent into a circle is inversely proportional to the radius: \( \varepsilon = \frac{c}{2R} \), where \( c \) is the wire diameter and \( R \) is the radius of the circle.
02

Calculate Minimum Radius and Length in Part (a)

Given that the maximum permissible strain \( \varepsilon_{max} = 0.0024 \) and the wire diameter \( d = 4 \mathrm{mm} = 0.004 \mathrm{m} \), the minimum radius of curvature \( R \) can be calculated using \( \varepsilon = \frac{c}{2R} \). Solving for \( R \), we get: \( R = \frac{c}{2\varepsilon_{max}} = \frac{0.004}{2 \times 0.0024} = 0.833 \, \mathrm{m} \). The circumference of the circle, which is the length of the wire, is \( L = 2\pi R \). Substituting \( R \), \( L = 2\pi \times 0.833 \approx 5.24 \mathrm{m} \).
03

Determine Maximum Diameter in Part (b)

For a wire length \( L = 5.5 \mathrm{m} \), we need to calculate the maximum allowable diameter \( d \) such that the strain doesn’t exceed the yield threshold. The strain \( \varepsilon \) as a function of \( d \) and \( R \) is again given by \( \varepsilon = \frac{d}{2R} \). First, solve for \( R \) using the perimeter equation \( 2\pi R = 5.5 \Rightarrow R = \frac{5.5}{2\pi} = 0.875 \mathrm{m} \). Rearrange the strain formula to find: \( d = 2 \varepsilon R \). Since the maximum strain should be \( \frac{\sigma_{Y}}{E} = \frac{300\times 10^6}{120\times 10^9} = 0.0025 \), substitute \( \varepsilon \) and \( R \) to get \( d = 2 \times 0.0025 \times 0.875 = 0.004375 \mathrm{m} = 4.375 \mathrm{mm} \).

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Key Concepts

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

Bending and Strain Relationships
When a material such as copper wire is bent into a shape, like a circle, the outer fibers elongate while the inner fibers compress, creating what is known as bending strain. Strain in this context measures how much deformation occurs relative to the material's original dimensions. The bending strain is defined by the formula \( \varepsilon = \frac{c}{2R} \), where \( c \) is the diameter of the wire and \( R \) is the radius of the circle. This relationship considers that the strain is evenly distributed throughout the material when curved.
  • Curvature: Directly affects strain; less radius means higher strain.
  • Material Diameter: Larger diameters increase potential strain for a given curvature.

Knowing this relationship is key when designing to avoid exceeding material strain limits, as it allows for predicting how different sizes and curves will impact the material's behavior.
Curvature and Geometry in Materials
Curvature is foundational when analyzing how materials behave under certain forms of loading or deformation. It describes how sharply a material bends or curves. For materials bent into circles, the curvature \( K \) is the reciprocal of the radius \( R \) (i.e., \( K = \frac{1}{R} \)). This is a helpful measure when determining how changes in geometry influence material properties.
Most significant is how this curvature relates to the bending strain. Note that as the curvature increases (which occurs when \( R \) decreases), the strain experienced by the material rises. Alternatively, a larger radius implies gentler curves and less stress.
  • Calculating Curvature: Direct math, \( K = \frac{1}{R} \), explains and predicts stress impact.
  • Design Implications: Changes in curvature should consider strain constraints.
Maximum Permissible Strain
The maximum permissible strain in a material like copper signifies the maximal extent of deformation it can sustain before failure or permanent deformation. Within engineering, staying below this limit prevents the material from breaking or deforming in a way that could jeopardize structural integrity.
In the computation of permissible strain, the wire’s diameter and its target geometry (like a circle in this problem) are pivotal. The exercise demonstrates calculating a safe radius and hence length, ensuring the copper wire remains below its strain threshold. This safeguard translates into detailed design and planning processes aiming to exploit without exceeding a material's properties.
  • Calculation Example: With \( \varepsilon_{max} = 0.0024 \), safe dimensions ensure strain control.
  • Utilization: Max permissible strain aligns design within material limits, presaging safety.
Yield Strain Analysis
The concept of yield strain relates to the point where a material begins to deform plastically and won’t return to its original shape after the stress is removed. Engineering materials often have characterizable yield points that help ensure durable designs.
In yield strain analysis, particularly for the copper wire here, the equation \( \varepsilon = \frac{d}{2R} \) helps in determining if a proposed diameter or configuration would exceed the yield strain. For example, given the material's yield point, computations in part (b) must ensure that \( \frac{300\times 10^6}{120\times 10^9} = 0.0025 \) isn't surpassed by the wire's strain.
  • Practical Application: Preventing strain levels from exceeding yield strain averts material damage.
  • Critical Calculations: Ensure design decisions maintain strains beneath yield thresholds for durability.

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Most popular questions from this chapter

A steel wire with a diameter of \(d=1.6 \mathrm{mm}\) is bent around a cylindrical drum with a radius of \(R=0.9 \mathrm{m}\) (see figure). (a) Determine the maximum normal strain \(\varepsilon_{\max }\) (b) What is the minimum acceptable radius of the drum if the maximum normal strain must remain below yield? Assume \(E=210 \mathrm{GPa}\) and \(\sigma_{Y}=690 \mathrm{MPa}\) (c) If \(R=0.9 \mathrm{m},\) what is the maximum acceptable diameter of the wire if the maximum normal strain must remain below yield?

A vertical pole of aluminum is fixed at the base and pulled at the top by a cable having a tensile force \(T\) (see figure \() .\) The cable is attached at the outer edge of a stiffened cover plate on top of the pole and makes an angle \(\alpha=20^{\circ}\) at the point of attachment. The pole has length \(L=2.5 \mathrm{m}\) and a hollow circular cross section with outer diameter \(d_{2}=280 \mathrm{mm}\) and inner diameter \(d_{1}=220 \mathrm{mm}\). The circular cover plate has diameter \(1.5 d_{2}\). Determine the allowable tensile force \(T_{\text {allow }}\) in the cable if the allowable compressive stress in the aluminum pole is \(90 \mathrm{MPa}\).

A so-called "trapeze bar" in a hospital room provides a means for paticnts to cxercise while in bed (sec figure). The bar is \(2.1 \mathrm{m}\) long and has a cross section in the shape of a regular octagon. The design load is \(1.2 \mathrm{kN}\) applied at the midpoint of the bar, and the allowable bending stress is \(200 \mathrm{MPa}\) Determine the minimum height \(h\) of the bar. (Assume that the ends of the bar are simply supported and that the weight of the bar is negligible.)

Each girder of the lift bridge (see figure) is \(50 \mathrm{m}\) long and simply supported at the cnds. The design load for each girder is a uniform load of intensity \(18 \mathrm{kN} / \mathrm{m}\). The girders are fabricated by welding three steel plates so as to form an I-shaped cross section (see figure) having section modulus \(S=46,000 \mathrm{cm}^{3}\) What is the maximum bending stress \(\sigma_{\max }\) in a girder due to the uniform load?

A square wood platform, \(2.4 \mathrm{m} \times 2.4 \mathrm{m}\) in area rests on masonry walls (see figure). The deck of the platform is constructed of \(50 \mathrm{mm}\) nominal thickness tongue-andgroove planks (actual thickness \(47 \mathrm{mm}\); see Appendix \(\mathrm{F}\) ) supported on two \(2.4-\mathrm{m}\) long beams. The beams have \(100 \mathrm{mm} \times 150 \mathrm{mm}\) nominal dimensions (actual dimensions \(97 \mathrm{mm} \times 147 \mathrm{mm}\) ). The planks are designed to support a uniformly distributed load \(w\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) acting over the entire top surface of the platform. The allowable bending stress for the planks is \(17 \mathrm{MPa}\) and the allowable shear stress is 0.7 MPa. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams. \(\begin{array}{lllll}\text { (a) } & \text { Determine } & \text { the } & \text { allowable } & \text { platform } & \text { load }\end{array}\) \(w_{1}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the bending stress in the planks. (b) Determine the allowable platform load \(w_{2}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load \(w_{\text {allow }}\) on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.)

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