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Chapter 6: Problem 56

Upon what three criteria are factors of safety based?

Short Answer

Expert verified
Answer: The three main criteria for factors of safety in engineering systems are material strength, load magnitude and variability, and consequences of failure. Material strength is important for understanding a material's ability to withstand stress or load without failure, while load magnitude and variability account for uncertainties in the loads that a system will experience during its service life. Consequences of failure help determine the necessary safety margins for different systems, prioritizing critical applications where failure could result in loss of life or significant damage. These criteria are crucial for minimizing the risk of system failure and ensuring the safety and reliability of engineering systems.

Step by step solution

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1. Material Strength

The first criterion is material strength. Material strength refers to the ability of a material to withstand an applied stress or load without failure. The factor of safety for material strength considers any uncertainties in determining the actual strength of the material and accounts for possible imperfections, such as defects in manufacturing or other sources of variability in the material's performance. It is essential to know and understand the material's strength so that it can be designed and used safely within its capabilities.
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2. Load Magnitude and Variability

The second criterion is the magnitude and variability of the loads that the engineering system will be subjected to during its service life. These loads can be static (constant over time) or dynamic (changing over time), and their magnitude can be uncertain due to factors such as manufacturing tolerances or variations in the environment. The factor of safety for load magnitude and variability accounts for these uncertainties and ensures that the system is designed to handle the range of possible loads it may experience.
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3. Consequences of Failure

The third criterion is the consequences of failure. The consequences of failure vary depending on the specific application and can range from minor inconveniences to catastrophic events, resulting in loss of life or significant property damage. The factor of safety accounts for the potential consequences of failure by requiring higher safety margins for systems where the consequences of failure are more severe. This helps ensure that the risk of failure is minimized, particularly in critical applications where lives may depend on the system's performance.

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Most popular questions from this chapter

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(20.000\) and \(20.025 \mathrm{~mm}\), respectively, and its final length is \(74.96 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(105 \mathrm{GPa}\) and \(39.7 \mathrm{GPa}\), respectively.

A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{~mm}(0.505\) in.) and gauge length of \(50.80 \mathrm{~mm}(2.000 \mathrm{in} .)\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(6.60 \mathrm{~mm}(0.260 \mathrm{in} .)\), and the fractured gauge length is \(72.14 \mathrm{~mm}\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

A cylindrical specimen of aluminum having a diameter of \(19 \mathrm{~mm}\) (0.75 in.) and length of 200 \(\mathrm{mm}(8.0 \mathrm{in}\).) is deformed elastically in tension with a force of \(48,800 \mathrm{~N}\left(11,000 \mathrm{lb}_{\mathrm{f}}\right)\). Using the data in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen, Will the diameter increase or decrease?

A brass alloy is known to have a yield strength of \(275 \mathrm{MPa}(40,000 \mathrm{psi})\), a tensile strength of \(380 \mathrm{MPa}(55,000 \mathrm{psi})\), and an elastic modulus of \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\). A cylindrical specimen of this alloy \(12.7 \mathrm{~mm}(0.50\) in.) in diameter and \(250 \mathrm{~mm}\) (10.0 in.) long is stressed in tension and found to elongate \(7.6 \mathrm{~mm}(0.30 \mathrm{in}\).). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.

A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure \(6.21\) is subjected to a tensile load; the specimen is 300 mm (12 in.) long and has a square cross section \(4.5 \mathrm{~mm}(0.175 \mathrm{in}\).) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of \(0.45 \mathrm{~mm}\) \((0.018\) in.). (b) What will be the deformation after the load has been released?
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