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Three different materials, designated \(A, B,\) and \(C\) are tested in tension using test specimens having diameters of \(12 \mathrm{mm}\) and gage lengths of \(50 \mathrm{mm}\) (see figure). At failure, the distances between the gage marks are found to be \(54.5 \mathrm{mm}, 63.2 \mathrm{mm},\) and \(69.4 \mathrm{mm},\) respectively. Also, at the failure cross sections the diameters are found to be 11.46 \(9.48,\) and \(6.06 \mathrm{mm},\) respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.

Short Answer

Expert verified
Material A is borderline brittle, Material B is ductile, Material C is very ductile.

Step by step solution

01

Calculate Initial and Final Areas

The initial cross-sectional area for each material can be calculated using the formula for the area of a circle:\[ A_0 = \frac{\pi}{4} \times d^2 \]Where \(d = 12 \text{ mm}\). Therefore, \(A_0 = \pi \times 36\text{ mm}^2 \approx 113.1 \text{ mm}^2\).The final areas at failure are:- Material A: \(d = 11.46 \text{ mm} \Rightarrow A_f = \pi \times (11.46/2)^2 = \pi \times 32.83 \text{ mm}^2 \approx 103.1 \text{ mm}^2\)- Material B: \(d = 9.48 \text{ mm} \Rightarrow A_f = \pi \times (9.48/2)^2 = \pi \times 22.49 \text{ mm}^2 \approx 70.6 \text{ mm}^2\)- Material C: \(d = 6.06 \text{ mm} \Rightarrow A_f = \pi \times (6.06/2)^2 = \pi \times 9.18 \text{ mm}^2 \approx 28.8 \text{ mm}^2\)
02

Calculate Percent Elongation

Percent elongation is calculated as follows:\[ \text{Percent elongation} = \frac{\Delta L}{L_0} \times 100\]Where \(\Delta L = L_f - L_0\).- Material A: \(L_f = 54.5 \text{ mm}, L_0 = 50 \text{ mm} \Rightarrow \Delta L = 4.5 \text{ mm} \)\[ \text{Percent elongation for A} = \frac{4.5}{50} \times 100 = 9\%\]- Material B: \(L_f = 63.2 \text{ mm}, L_0 = 50 \text{ mm} \Rightarrow \Delta L = 13.2 \text{ mm}\)\[ \text{Percent elongation for B} = \frac{13.2}{50} \times 100 = 26.4\%\]- Material C: \(L_f = 69.4 \text{ mm}, L_0 = 50 \text{ mm} \Rightarrow \Delta L = 19.4 \text{ mm}\)\[ \text{Percent elongation for C} = \frac{19.4}{50} \times 100 = 38.8\%\]
03

Calculate Percent Reduction in Area

Percent reduction in area is given by:\[ \text{Percent reduction in area} = \frac{A_0 - A_f}{A_0} \times 100\]- Material A: \( A_0 = 113.1 \text{ mm}^2, A_f = 103.1 \text{ mm}^2 \)\[ \text{Percent reduction for A} = \frac{113.1 - 103.1}{113.1} \times 100 \approx 8.8\% \]- Material B: \( A_0 = 113.1 \text{ mm}^2, A_f = 70.6 \text{ mm}^2 \)\[ \text{Percent reduction for B} = \frac{113.1 - 70.6}{113.1} \times 100 \approx 37.6\% \]- Material C: \( A_0 = 113.1 \text{ mm}^2, A_f = 28.8 \text{ mm}^2 \)\[ \text{Percent reduction for C} = \frac{113.1 - 28.8}{113.1} \times 100 \approx 74.5\% \]
04

Classify Materials as Brittle or Ductile

Materials are typically considered ductile if they exhibit high percent elongation or high percent reduction in area. The threshold for considering a material as ductile is subjective but often set around 10%. - Material A: 9% elongation and 8.8% reduction - borderline brittle. - Material B: 26.4% elongation and 37.6% reduction - ductile. - Material C: 38.8% elongation and 74.5% reduction - very ductile.

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Key Concepts

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

Tensile Testing
Tensile testing is a fundamental procedure in material testing where a material is subjected to tension forces till it fails. This helps us determine various mechanical properties such as strength and ductility. The process involves gradually applying force to a specimen until it fractures. By doing so, engineers can record the deformation of the material and characterize its performance under stress.
Tensile tests are performed using standardized test samples with a specific initial length and diameter, ensuring consistency in results. During the test, important data such as the elongation at fracture and the reduction in cross-sectional area are collected. These measurements can then be used to calculate the percent elongation and percent reduction in area, offering insights into the material's physical properties.
Understanding tensile testing is crucial for engineers and designers to choose the appropriate materials for various applications.
Ductility
Ductility refers to a material's ability to deform under tensile stress without breaking. It is one of the key factors that dictate a material's usability in applications that involve deformation or bending. A ductile material can absorb stress and deform to a large extent before fracturing.
This property is typically assessed through tensile testing by measuring the percent elongation and percent reduction in area. When a material undergoes tension, an increase in elongation and reduction in area suggests high ductility. Materials with greater ductility are preferred in applications where flexibility and formability are essential, such as in metal forming processes.
  • High ductility means a material is likely more able to withstand changes without cracking.
  • Low ductility indicates brittleness, leading to potential failure with little deformation.
Understanding ductility aids in selecting materials that will endure specific mechanical stresses, thus ensuring reliability and safety in engineering designs.
Material Properties
The properties of materials are diverse and influence their performance in various engineering applications. Mechanical properties, such as tensile strength, ductility, hardness, and elasticity, define how a material behaves under different forces.
Tensile strength, for example, reflects the material's resistance to breaking under tension. Elasticity determines how well a material can return to its original shape after deformation. While hardness measures the resistance of a material to indentation, ductility, as discussed, shows how much a material can stretch.
  • Keen analysis of these properties is crucial in material selection for manufacturing and engineering.
  • Engineers use these characteristics to evaluate and compare materials, ensuring they meet the specific requirements of their project.
A profound comprehension of these properties allows for the development and innovation of systems and structures that are both safe and efficient.
Mechanical Engineering
Mechanical Engineering is a vast field that involves designing, analyzing, and manufacturing mechanical systems. A key part of this discipline is understanding the material aspects of the engineering process. Engineers require detailed knowledge of material properties to ensure that the systems they build will be robust and effective.
In material testing, the insights gained from procedures like tensile testing allow engineers to assess the suitability of materials for specific uses. This information is pivotal in designing components that need to endure stress, strain, and thermal variation.
  • Designing gears, for instance, necessitates choosing materials with high strength and durability.
  • For load-bearing structures, one would prioritize materials with excellent ductility.
Mechanical engineering relies heavily on the intersection of theoretical knowledge and practical testing to optimize solutions and innovate new technologies. Understanding the fundamental roles of materials and their properties enhances the capability to craft advanced, reliable, and efficient engineering solutions.

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Most popular questions from this chapter

A high-strength steel bar used in a large crane has diameter \(d=50 \mathrm{mm}\) (see figure). The steel has modulus of elasticity \(E=200 \quad\) GPa and Poisson's ratio \(v=0.3 .\) Because of clearance requirements, the diameter of the bar is limited to \(50.025 \mathrm{mm}\) when it is compressed by axial forces. What is the largest compressive load \(P_{\max }\) that is permitted?

A wire of length \(L=2.5 \mathrm{m}\) and diameter \(d=1.6 \mathrm{mm}\) is stretched by tensile forces \(P=600 \mathrm{N}\). The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: $$\sigma=\frac{124,020 \varepsilon}{1+300 \varepsilon} \quad 0 \leq \varepsilon \leq 0.03 \quad(\sigma=\mathrm{MPa})$$ in which \(\varepsilon\) is nondimensional and \(\sigma\) has units of MPa. (a) Construct a stress-strain diagram for the material. (b) Determine the elongation of the wire due to the forces \(P\) (c) If the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?

A sign of weight \(W\) is supported at its base by four bolts anchored in a concrete footing. Wind pressure \(p\) acts normal to the surface of the sign; the resultant of the uniform wind pressure is force \(F\) at the center of pressure. The wind force is assumed to create equal shear forces \(\mathrm{F} 14\) in the \(y\) direction at each bolt (see figure parts a and \(c\) ). The over- turning effect of the wind force also causes an uplift force \(R\) at bolts \(A\) and \(C\) and a downward force \((-R)\) at bolts \(B\) and \(D\) (sec figure part b). The resulting effects of the wind, and the associated ultimate stresses for each stress condition, are: normal stress in each bolt \(\left(\sigma_{u}=410 \mathrm{MPa}\right) ;\) shear through the base plate \(\left(\tau_{i p}=115 \mathrm{MPa}\right)\); horizontal shear and bearing on each bolt \(\left(\tau_{k}=170 \mathrm{MPa} \text { and } \sigma_{b v}=520 \mathrm{MPa}\right) ;\) and bearing on the bottom washer at \(B \text { (or } D)\left(\sigma_{k}=340 \mathrm{MPa}\right)\) Find the maximum wind pressure \(p_{\max }(P a)\) that \(\operatorname{can}\) be carried by the bolted support system for the sign if a safety factor of 2.5 is desired with respect to the ultimate wind load that can be carried. Use the following numerical data: bolt \(d_{b}=19 \mathrm{mm}\) washer \(d_{w}=38 \mathrm{mm} ;\) base plate \(t_{b p}=25 \mathrm{mm} ;\) base plate \\[ \text { dimensions } h=350 \mathrm{mm} \text { and } b=300 \mathrm{mm} ; W=2.25 \mathrm{kN} ; H= \\] \(5.2 \mathrm{m} ; \operatorname{sign}\) dimensions \(\left(L_{v}=3 \mathrm{m} \times L_{\mathrm{h}}=3.7 \mathrm{m}\right)\); pipe column diameter \(d=150 \mathrm{mm},\) and pipe column thickness \(t=10 \mathrm{mm}\).

A circular aluminum tube with a length of \(L=420 \mathrm{mm}\) is loaded in compression by forces \(P\) (see figure). The hollow segment of length \(L / 3\) had outside and inside diameters of \(60 \mathrm{mm}\) and \(35 \mathrm{mm}\), respectively. The solid segment of length \(2 L / 3\) has a diameter of \(60 \mathrm{mm} .\) A strain gage is placed on the outside of the hollow segment of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain in the hollow segment is \(\varepsilon_{h}=470 \times 10^{-6},\) what is the strain \(\varepsilon_{s}\) in the solid part? (Hint: The strain in the solid segment is equal to that in the hollow segment multiplied by the ratio of the area of the hollow to that of the solid segment.) (b) What is the overall shortening \(\delta\) of the bar? (c) If the compressive stress in the bar cannot exceed \(48 \mathrm{MPa},\) what is the maximum permissible value of load \(P ?\)

A tensile test is peformed on a brass specimen \(10 \mathrm{mm}\) in diameter using a gage length of \(50 \mathrm{mm}\) (see figure). When the tensile load \(P\) reaches a value of \(20 \mathrm{kN}\) the distance between the gage marks has increased by \(0.122 \mathrm{mm}\) (a) What is the modulus of elasticity \(E\) of the brass? (b) If the diameter decreases by \(0.00830 \mathrm{mm}\), what is Poisson's ratio?

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