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Chapter 4: Problem 45

The bolt has a diameter of \(20 \mathrm{mm}\) and passes through a tube that has an inner diameter of \(50 \mathrm{mm}\) and an outer diameter of \(60 \mathrm{mm}\). If the bolt and tube are made of A-36 steel, determine the normal stress in the tube and bolt when a force of \(40 \mathrm{kN}\) is applied to the bolt. Assume the end caps are rigid.

Short Answer

Expert verified
The normal stress in the bolt is approximately \(127.32 \mathrm{MPa}\) and the normal stress in the tube is approximately \(56.60 \mathrm{MPa}\).

Step by step solution

01

Calculate the cross-sectional area of the bolt

The cross-sectional area \(A_b\) of a circle (the bolt) can be calculated using the formula \(A=\pi r^2\), where \(r\) represents radius. But diameter = 2*radius, so we have radius = \(20 \mathrm{mm}\) / 2 = \(10 \mathrm{mm}\). Therefore, \(A_b=\pi (10)^2 = 314.16 \mathrm{mm}^2\).
02

Calculate the cross-sectional area of the tube wall

The cross-sectional area \(A_t\) of a cylinder (the tube wall) is determined by subtracting the area of the inner circle from the area of the outer circle. Since the inner diameter is \(50 \mathrm{mm}\), the inner radius is \(50 \mathrm{mm}\) / 2 = \(25 \mathrm{mm}\). Similarly, the outer radius is \(60 \mathrm{mm}\) / 2 = \(30 \mathrm{mm}\). Therefore, \(A_t= \pi (30)^2 - \pi (25)^2 = 706.86 \mathrm{mm}^2\).
03

Calculate the normal stress in the bolt

The normal stress \(\sigma_b\) is the force \(F_b\) divided by the cross-sectional area \(A_b\) of the bolt. Since the bolt directly receives the applied force, \(F_b = 40 \mathrm{kN}\). Therefore, converting the force to Newtons for consistency of units, the stress in the bolt is \(\sigma_b = F_b/A_b = 40000 N / 314.16 \mathrm{mm}^2 = 127.32 \mathrm{MPa}\). This is the stress in the bolt.
04

Calculate the normal stress in the tube

The normal stress \(\sigma_t\) is the force \(F_t\) divided by the cross-sectional area \(A_t\) of the tube. Since the tube indirectly receives the force exerted by the bolt, \(F_t = 40 \mathrm{kN}\). Thus, converting the force to Newtons again, the stress in the tube wall is \(\sigma_t = F_t/A_t = 40000 N / 706.86 \mathrm{mm}^2 = 56.60 \mathrm{MPa}\). This is the stress in the tube.

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Key Concepts

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

Mechanics of Materials
Mechanics of materials, also known as strength of materials, focuses on the relationship between the external loads applied to a body and the resulting internal forces and material deformations. This field ensures that structures and components can withstand various forces without failure. For example, when a force is applied to a bolt and tube, we need to understand how these components accommodate the force. This understanding helps in predicting whether the materials will fail or remain functional under different circumstances.

Key considerations in mechanics of materials include:
  • Stress: A measure of the intensity of internal forces acting within a deformable body.
  • Strain: The deformation of the material due to applied stress.
  • Elastic Modulus: A material property that measures its tendency to deform under stress.
  • Boundary conditions: Determines how the components interact with other structures they connect to or over which they are applied (like the rigid end caps here).
Cross-Sectional Area Calculation
The cross-sectional area is fundamental in calculating how stress is distributed in components. This area refers to the surface or area that is exposed by making an orthogonal cut through any form of material. Understanding the cross-sectional area helps in determining how much force a material can handle per unit area, a pivotal concept in mechanics of materials.

To find the cross-sectional area:
  • For a circle (like a bolt), use the formula: \(A = \pi r^2\), where the radius (\(r\)) is half the diameter of the bolt.
  • For a tube, subtract the inner circle's area from the outer circle's area using: \(A = \pi R_o^2 - \pi R_i^2\), where \(R_o\) and \(R_i\) are the outer and inner radii, respectively.
These calculations are crucial for understanding and devising solutions that ensure the safety and functionality of various engineering applications.
Steel Properties
Steel, an alloy of iron and carbon, is prized for its strength, durability, and versatility. These properties make it a cornerstone in construction and manufacturing fields. When we analyze the properties of steel, especially in mechanics and stress scenarios, several factors are important:

  • Density: Influences weight and structural support.
  • Tensile Strength: The maximum stress steel can withstand while being stretched or pulled before necking, or cross-sectional area begins to significantly contract.
  • Yield Strength: The point of stress at which a material begins to deform plastically.
  • Elasticity: Ability to return to its original shape after deformation when stress is removed.
These attributes inform decisions on whether steel is suitable for specific load-bearing applications, including bolts and tubes subjected to large forces.
A-36 Steel
A-36 steel is a standard grade of mild carbon steel that is widely used for structural applications. Its popularity stems from its balanced properties in terms of strength, ductility, and ease of fabrication. Here are a few characteristics of A-36 steel to consider:

  • Tensile Strength: Around 400 to 550 MPa, making it suitable for general structural purposes.
  • Yield Strength: Approximately 250 MPa, giving it adequate flexibility for forming and welding.
  • Good machinability and weldability, which facilitates its use in various types of civil engineering and construction projects.
  • Cost-effective: Offers a balance of performance and cost that makes it accessible for many industries.
Understanding these properties helps engineers and builders decide whether A-36 steel is appropriate for the specific demands of their projects, particularly when considering factors like strength and budget.

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Most popular questions from this chapter

The wires \(A B\) and \(A C\) are made of steel, and wire \(A D\) is made of copper. Before the 150 -lb force is applied, \(A B\) and \(A C\) are each 60 in. long and \(A D\) is 40 in. long. If the temperature is increased by \(80^{\circ} \mathrm{F}\), determine the force in each wire needed to support the load. Take \(E_{\mathrm{w}}=29\left(10^{3}\right) \mathrm{ksi}, E_{\mathrm{cm}}=17\left(10^{3}\right) \mathrm{ksi}, \alpha_{\mathrm{st}}=8\left(10^{-6}\right) /^{\circ} \mathrm{F}\) \(\alpha_{\mathrm{cu}}=9.60\left(10^{-6}\right) /^{\circ} \mathrm{F}\). Each wire has a cross-sectional area of 0.0123 in \(^{2}\).

The load of 2800 lb is to be supported by the two essentially vertical \(A-36\) steel wires. If originally wire \(A B\) is 60 in. long and wire \(A C\) is 40 in. long, determine the crosssectional area of \(\mathcal{A B}\) if the load is to be shared equally between both wires Wire \(A C\) has a cross-sectional area of \(0.02 \mathrm{in}^{2}\).

The bar has a cross-sectional area \(A\), length \(L\) modulus of elasticity \(E,\) and coefficient of thermal expansion \(\alpha .\) The temperature of the bar changes uniformly along its length from \(T_{A}\) at \(A\) to \(T_{B}\) at \(B\) so that at any point \(x\) along the bar \(T=T_{A}+x\left(T_{B}-T_{A}\right) / L .\) Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temperature of \(T_{A}\).

The center rod \(C D\) of the assembly is heated from \(T_{1}=30^{\circ} \mathrm{C}\) to \(T_{2}=180^{\circ} \mathrm{C}\) using electrical resistance heating. Also, the two end rods \(A B\) and \(E F\) are heated from \(T_{1}=30^{\circ} \mathrm{C}\) to \(T_{2}=50^{\circ} \mathrm{C}\). At the lower temperature \(T_{1}\) the gap between \(C\) and the rigid bar is \(0.7 \mathrm{mm}\). Determine the force in rods \(A B\) and \(E F\) caused by the increase in temperature. Rods \(A B\) and EF are made of steel, and each has a cross-sectional area of \(125 \mathrm{mm}^{2} . \mathrm{CD}\) is made of aluminum and has a cross-sectional area of \(375 \mathrm{mm}^{2} . E_{\mathrm{st}}=200 \mathrm{GPa}, E_{\mathrm{al}}=70 \mathrm{GPa}, \alpha_{\mathrm{st}}=12\left(10^{-6}\right) /^{\circ} \mathrm{C}\) and \(\alpha_{a d}=23\left(10^{-6}\right)^{\circ} \mathrm{C}\).

The rod has a slight taper and length \(L\). It is suspended from the ceiling and supports a load \(\mathbf{P}\) at its end. Show that the displacement of its end due to this load is \(\delta=P L /\left(\pi E r_{2} r_{1}\right) .\) Neglect the weight of the material. The modulus of elasticity is \(E\).
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