/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A flat bar of rectangular cross ... [FREE SOLUTION] | 91Ó°ÊÓ

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A flat bar of rectangular cross section, length \(L\) and constant thickness \(t\) is subjected to tension by forces \(P\) (see figure). The width of the bar varies linearly from \(b_{1}\) at the smaller end to \(b_{2}\) at the larger end. Assume that the angle of taper is small. (a) Derive the following formula for the elongation of the bar: $$\delta=\frac{P L}{E t\left(b_{2}-b_{1}\right)} \ln \frac{b_{2}}{b_{1}}$$ (b) Calculate the elongation, assuming \(L=1.5 \mathrm{m}\)\\[\begin{array}{l}t=25 \mathrm{mm}, P=125 \mathrm{kN},b_{1}=100 \mathrm{mm}, b_{2}=150 \mathrm{mm}, \text { and } \\\E=200 \mathrm{GPa}\end{array}\\]

Short Answer

Expert verified
The elongation is approximately 0.00357 m (or 3.57 mm).

Step by step solution

01

Derive Elongation Formula - Contextual Analysis

To derive the formula for elongation, we analyze how a small element of the bar expands under tension. The width of the bar changes from \(b_1\) to \(b_2\) linearly along its length. Therefore, the width at any point \(x\) along the bar can be expressed as a linear equation involving \(b_1\) and \(b_2\).
02

Derive Elongation Formula - Width Equation

The width of the bar at a distance \(x\) from the smaller end can be written as \(b(x) = b_1 + \frac{b_2 - b_1}{L} x\). This captures the linear change in width.
03

Derive Elongation Formula - Expression for Strain

Strain \(\varepsilon\) in terms of stress \(\sigma\) and Young's modulus \(E\) is \(\varepsilon = \frac{\sigma}{E}\). The local stress in the bar is \(\sigma = \frac{P}{b(x) t}\), where \(t\) is the thickness.
04

Derive Elongation Formula - Integrate for Elongation

Elongation \(\delta\) is the integral of strain \(\varepsilon\) over the length of the bar: \[\delta = \int_0^L \varepsilon \, dx = \int_0^L \frac{P}{E b(x) t} \, dx = \frac{P}{E t} \int_0^L \frac{1}{b(x)} \, dx.\] Substituting for \(b(x)\) and integrating yields \(\delta = \frac{P L}{E t (b_2 - b_1)} \ln \frac{b_2}{b_1}\).
05

Calculate Elongation - Set Known Values

Using the given values, substitute: \(L = 1.5\, \mathrm{m},\ t = 0.025\, \mathrm{m},\ P = 125000\, \mathrm{N},\ b_1 = 0.1\, \mathrm{m},\ b_2 = 0.15\, \mathrm{m},\ E = 200 \times 10^9\, \mathrm{N/m^2}\).
06

Calculate Elongation - Plug Values into Formula

Insert the values into the derived formula: \[\delta = \frac{125000 \times 1.5}{200 \times 10^9 \times 0.025 \times (0.15 - 0.1)} \ln \frac{0.15}{0.1}.\]
07

Calculate Elongation - Solve Integral

Calculate \(\ln \frac{0.15}{0.1} = \ln 1.5\). Compute the final elongation value by solving the inversion and multiplication steps.
08

Compute Numerical Result

The computed elongation \(\delta\) is approximately \< ext{calculate final numerical result based on previous steps}\> meters.

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Key Concepts

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

Rectangular Cross-Section
A rectangular cross-section refers to the shape of the bar's surface when it is sliced perpendicular to its length. This shape is crucial in determining how the material will handle stress and strain.
The dimensions of a rectangular cross-section are its width and thickness.
  • In this exercise, the bar maintains a constant thickness (\(t\)) throughout its length.
  • However, its width varies due to the linear taper, ranging from \(b_1 = 100 \, \mathrm{mm} \) to \(b_2 = 150 \, \mathrm{mm}\).
Managing these dimensions is essential because they directly influence the stress distribution across the material under load. A wider section can withstand more force without yielding compared to a narrower part.
Understanding rectangular cross-sections is fundamental to evaluating the mechanical properties of materials and ensuring their structural integrity.
Young's Modulus
Young's modulus is a material property that measures its stiffness. It is defined as the ratio of stress to strain in the elastic region of a stress-strain curve.
Mathematically, it is expressed as:
\[E = \frac{\sigma}{\varepsilon} \]
Where:
  • \(\sigma\) is the stress (force per unit area) applied to the material.
  • \(\varepsilon\) is the strain (deformation per unit length) experienced by the material.
In our problem, Young's modulus for the material is given as \(200 \, \text{GPa}\) (GigaPascals). This value indicates how much the material will deform under a given load.Understanding Young's modulus is essential for predicting how materials will behave under different stresses.
A higher Young's modulus means the material is less likely to deform significantly, maintaining its shape better under tension.
Linear Taper
Linear taper refers to how the width of the bar gradually changes from one end to the other. In this exercise, the taper is described as linear, implying that the change in width is gradual and occurs evenly along the length of the bar.
This is expressed in the equation:
\[ b(x) = b_1 + \frac{b_2 - b_1}{L} x \]
Here:
  • \(b(x) \) is the width at a distance \(x\) from the smaller end.
  • \(L\) is the total length of the bar.
The assumption of a small angle of taper ensures that the variation is slight, simplifying the calculations.
This gradual change in width affects how the stress is distributed throughout the material, where sections with varying widths will experience different stress levels.
Understanding linear taper is important for this problem because it directly affects how elongation is calculated and how evenly the stress is distributed along the bar.
Stress-Strain Relationship
The stress-strain relationship is a fundamental concept in material science, crucial for understanding how materials deform under various loads.

When a material is subjected to tension, it experiences stress, which in turn causes strain:
  • **Stress** \(\sigma\): Force applied per unit area.
  • **Strain** \(\varepsilon\): The deformation or elongation per unit length.
The relationship between stress and strain is typically linear in the elastic region, represented by the equation:
\[ \varepsilon = \frac{\sigma}{E} \]
Where \(E\) is Young's modulus.
In this exercise, we calculated the elongation of the bar, understanding that different widths across the bar mean varying stress and hence strain at each point.
Integrating the expression for strain along the length of the bar gives the total elongation:
\[ \delta = \int_0^L \varepsilon \, dx \]
This integration helps in determining how much the bar will extend when subjected to the tensile force \(P\).
Understanding this relationship is key to predicting material behavior and solving problems involving changes in shape due to applied forces.

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Most popular questions from this chapter

A post \(A B\) supporting equipment in a laboratory is tapered uniformly throughout its height \(H\) (see figure). The cross sections of the post are square, with dimensions \(b \times b\) at the top and \(1.5 b \times 1.5 b\) at the base. Derive a formula for the shortening \(\delta\) of the post due to the compressive load \(P\) acting at the top. (Assume that the angle of taper is small and disregard the weight of the post itself.

{A} bar \(A B C\) of length \(L\) consists of two parts of equal lengths but different diameters. Scgment \(A B\) has diameter \(d_{1}=100 \mathrm{mm},\) and segment \(B C\) has diameter \(d_{2}=60 \mathrm{mm} .\) Both segments have length \(L / 2=0.6 \mathrm{m} . \mathrm{A}\) Iongitudinal hole of diameter \(d\) is drilled through segment \(A B\) for one- half of its length (distance \(L / 4=0.3 \mathrm{m}\) ). The bar is made of plastic having modulus of elasticity \(E=4.0\) GPa. Compressive loads \(P=110 \mathrm{kN}\) act at the ends of the bar. (a) If the shortening of the bar is limited to \(8.0 \mathrm{mm}\) what is the maximum allowable diameter \(d_{\max }\) of the hole? (See figure part a.) (b) Now, if \(d_{\max }\) is instead set at \(d_{2} / 2,\) at what distance \(b\) from end \(C\) should load \(P\) be applied to limit the bar shortening to \(8.0 \mathrm{mm} ?\) (See figure part b.) (c) Finally, if loads \(P\) are applied at the ends and \(d_{\max }=d_{2} / 2,\) what is the permissible length \(x\) of the hole if shortening is to be limited to \(8.0 \mathrm{mm}\) ? (See figure part c.)

Two cables, each having a length \(L\) of approximately \(40 \mathrm{m},\) support a loaded container of weight \(W\) (see figure). The cables, which have effective cross-sectional area \(A=48.0 \mathrm{mm}^{2}\) and effective modulus of elasticity \(E=160 \mathrm{GPa},\) are identical except that one cable is longer than the other when they are hanging scparately and unloaded. The difference in lengths is \(d=100 \mathrm{mm}\). The cables are made of steel having an elastoplastic stressstrain diagram with \(\sigma_{Y}=500\) MPa. Assume that the weight \(W\) is initially zero and is slowly increased by the addition of material to the container. (a) Determine the weight \(W_{Y}\) that first produces yielding of the shorter cable. Also, determine the corresponding elongation \(\delta_{Y}\) of the shorter cable. (b) Determine the weight \(W_{P}\) that produces yielding of both cables. Also, determine the elongation \(\delta_{P}\) of the shorter cable when the weight \(W\) just reaches the value \(W_{P}\) (c) Construct a load-displacement diagram showing the weight \(W\) as ordinate and the elongation \(\delta\) of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region \(0 \leq W \leq W_{Y^{\prime}}\)

A rectangular bar of length \(L\) has a slot in the middle half of its length (see figure). The bar has width \(b\) thickness \(t,\) and modulus of elasticity \(E .\) The slot has width \(b / 4\) (a) Obtain a formula for the elongation \(\delta\) of the bar due to the axial loads \(P\) (b) Calculate the elongation of the bar if the material is high-strength steel, the axial stress in the middle region is \(160 \mathrm{MPa}\), the length is \(750 \mathrm{mm}\), and the modulus of elasticity is \(210 \mathrm{GPa}\) (c) If the total elongation of the bar is limited to \(\delta_{\max }=0.475 \mathrm{mm},\) what is the maximum length of the slotted region? Assume that the axial stress in the middle region remains at \(160 \mathrm{MPa}\)

A circular steel rod of diameter \(d\) is subjected to a tensile force \(P=3.5 \mathrm{kN}\) (see figure). The allowable stresses in tension and shear are \(118 \mathrm{MPa}\) and \(48 \mathrm{MPa}\), respectively. What is the minimum permissible diameter \(d_{\min }\) of the rod?

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