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Link \(B C\) is \(6 \mathrm{mm}\) thick and is made of a steel with a \(450-\mathrm{MPa}\) ultimate strength in tension. What should be its width \(w\) if the structure shown is being designed to support a 20 -kN load \(\mathbf{P}\) with a factor of safety of \(3 ?\)

Short Answer

Expert verified
The width \(w\) should be at least \(22.22\,\text{mm}\).

Step by step solution

01

Understand the Problem

We need to find the width \(w\) of link \(BC\) in a steel structure that supports a load \(P = 20\,\text{kN}\). The link should withstand an ultimate tensile strength of \(450\,\text{MPa}\) with a factor of safety of \(3\). The thickness \(t\) of the link is given as \(6\,\text{mm}\).
02

Calculate Allowable Stress

To ensure safety, the allowable stress \(\sigma_{\text{allow}}\) is calculated by dividing the ultimate tensile strength by the factor of safety: \[\sigma_{\text{allow}} = \frac{450\,\text{MPa}}{3} = 150\,\text{MPa}.\]
03

Relate Load to Stress and Cross-Sectional Area

The relationship between the applied load, allowable stress, and cross-sectional area \(A\) is \(A = \frac{P}{\sigma_{\text{allow}}}\). Given \(P = 20\,000\,\text{N}\): \[A = \frac{20,000\,\text{N}}{150\,\text{MPa}} = \frac{20,000\,\text{N}}{150 \times 10^3\,\text{N/m}^2} = \frac{20,000}{150,000}\,\text{m}^2 = 0.1333\times 10^{-3}\,\text{m}^2 = 133.3\,\text{mm}^2.\]
04

Calculate Required Width

Using the cross-sectional area \(A = w \times t\), where \(t = 6\,\text{mm}\), solve for width \(w\): \[w = \frac{A}{t} = \frac{133.3\,\text{mm}^2}{6\,\text{mm}} \approx 22.22\,\text{mm}.\] Thus, the width should be at least \(22.22\,\text{mm}\) to satisfy the design requirements.

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Key Concepts

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

Factor of Safety
The factor of safety (FoS) is a crucial concept in engineering design. It acts as a buffer or safeguard to ensure that materials and structures can handle unexpected conditions or variances without failing. But how exactly does it work?
Simply put:
  • It is a ratio that compares the ultimate strength to the permissible stress of a material.
  • By dividing the ultimate tensile strength by the factor of safety, we determine the allowable stress a material can safely withstand.
For example, in the context of the original exercise, with a factor of safety of 3, the ultimate tensile strength of 450 MPa is reduced to a safe, allowable stress of 150 MPa.
This ensures that even if unaccounted factors arise, the structure will remain intact and operational. Using a factor of safety provides peace of mind, preventing overloading and extending the lifespan of components.
Tensile Strength
Tensile strength is an inherent property of materials that defines their ability to withstand tension - that is, the force that attempts to pull them apart. Imagine trying to stretch a steel rod: tensile strength is the amount of pull or tension that the rod can take before it breaks.
For materials like steel, which are used in many construction applications, tensile strength is crucial because it indicates how much load they can safely support in service. In the exercise, the steel link must support tensile forces, and its tensile strength is given as 450 MPa.
When designing and analyzing structures, engineers need to ensure that the material's tensile strength is not exceeded under the working load, factoring in an appropriately chosen factor of safety. The role of tensile strength, therefore, is to inform these safety calculations and ensure structures maintain integrity under loads.
Cross-Sectional Area
The cross-sectional area is a pivotal factor in load-bearing calculations. It represents the area through which a load or force is distributed, affecting how stress acts within a material. A larger cross-sectional area can reduce stress concentration, allowing more force to be safely handled.
In the exercise, the aim is to calculate the required width of a link, utilizing the relationship between cross-sectional area, stress, and internal loads. Using the formula \[ A = \frac{P}{\sigma_{\text{allow}}} \]we find the needed area to support the given load at a safe stress level. Then, the cross-sectional area, derived as 133.3 mm², tells us the distribution of load is key in sizing the material dimensions, in this case, the width of the steel link.
Achieving the correct cross-sectional area ensures the structural component will perform as required, without yielding or breaking under normal service conditions.
Load Calculation
Load calculation is an essential part of stress analysis, involving determining the forces exerted on a structure and ensuring that they are within safe limits. It begins with identifying the types of loads involved, such as tension, compression, or shear, and their magnitudes.
In the exercise, we specifically dealt with calculating the load bearing requirements to determine the necessary width of steel used in the link to withstand a 20 kN force. The procedure followed involves:
  • Understanding the total applied load, here denoted as 20,000 N (since 1 kN = 1000 N).
  • Relating the load to the allowable stress, which is calculated from tensile strength divided by the factor of safety.
  • Employing formulas to find out dimensions needed to support the load, utilizing concepts like cross-sectional area.
Each load component contributes to the safety and functionality of the overall structure. Proper load calculation ensures the design is neither overbuilt (wasting materials) nor underbuilt (compromising safety).

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Most popular questions from this chapter

The hydraulic cylinder \(C F,\) which partially controls the position of rod \(D E,\) has been locked in the position shown. Member \(B D\) is \(15 \mathrm{mm}\) thick and is connected at \(C\) to the vertical rod by a 9-mm-diameter bolt. Knowing that \(P=2 \mathrm{kN}\) and \(\theta=75^{\circ},\) determine \((a)\) the average shearing stress in the bolt, \((b)\) the bearing stress at \(C\) in member \(B D\).

A steel loop \(A B C D\) of length 5 ft and of \(\frac{3}{8}\) -in. diameter is placed as shown around a 1 -in.-diameter aluminum rod \(A C\). Cables \(B E\) and \(D F\), each of \(\frac{1}{2}\) -in. diameter, are used to apply the load \(Q\). Knowing that the ultimate strength of the steel used for the loop and the cables is \(70 \mathrm{ksi}\), and that the ultimate strength of the aluminum used for the rod is 38 ksi, determine the largest load \(\mathbf{Q}\) that can be applied if an overall factor of safety of 3 is desired.

Link \(A C\) is made of a steel with a 65 -ksi ultimate normal stress and has a \(\frac{1}{4} \times \frac{1}{2}\) -in. uniform rectangular cross section. It is connected to a support at \(A\) and to member \(B C D\) at \(C\) by \(\frac{3}{4}\) -in. -diameter pins, while member \(B C D\) is connected to its support at \(B\) by a \(\frac{5}{16}\) -in.- diameter pin. All of the pins are made of a steel with a 25 -ksi ultimate shearing stress and are in single shear. Knowing that a factor of safety of 3.25 is desired, determine the largest load \(\mathbf{P}\) that can be applied at \(D\). Note that link \(A C\) is not reinforced around the pin holes.

An aircraft tow bar is positioned by means of a single hydraulic cylinder connected by a 25 -mm-diameter steel rod to two identical arm-and-wheel units \(D E F\). The mass of the entire tow bar is \(200 \mathrm{kg},\) and its center of gravity is located at \(G\). For the position shown, determine the normal stress in the rod.

Link \(A B,\) of width \(b=2\) in. and thickness \(t=\frac{1}{4}\) in., is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is -20 ksi and that the average shearing stress in each of the two pins is 12 ksi determine ( \(a\) ) the diameter \(d\) of the pins, (b) the average bearing stress in the link.

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