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Chapter 3: Problem 13

Find the deflection of a uniformly loaded beam pinned at the ends, \(x=0, L\). Where is the maximum bending moment? What is the maximum bending stress?

Short Answer

Expert verified
The maximum bending moment is \(\frac{wL^2}{8}\), maximum deflection is \(\frac{5wL^4}{384EI}\), and the maximum bending stress is \(\frac{wL^2c}{8I}\).

Step by step solution

01

Understanding the Problem

We are given a beam of length \(L\) pinned at both ends, supporting a uniform load \(w\). The beam is simply supported, so the moments at the ends are zero. Our task is to find the deflection, maximum bending moment, and maximum bending stress.
02

Calculating Maximum Bending Moment

For a uniformly loaded beam with supports at both ends, the maximum bending moment occurs at the center of the beam, \(x = \frac{L}{2}\). The formula for the bending moment is \(M(x) = w \left( \frac{L}{2} - x \right) \cdot x\). At \(x = \frac{L}{2}\), the maximum moment is \(M_{max} = \frac{wL^2}{8}\).
03

Finding Maximum Bending Stress

The maximum bending stress can be calculated using the formula \(\sigma_{max} = \frac{M_{max} \cdot c}{I}\), where \(c\) is the distance from the neutral axis to the outermost fiber and \(I\) is the moment of inertia. Substituting \(M_{max} = \frac{wL^2}{8}\) gives \(\sigma_{max} = \frac{wL^2c}{8I}\).
04

Determine Deflection Equation

The deflection \(y(x)\) of a beam under uniform load can be found using the differential equation \(EI \frac{d^2y}{dx^2} = M(x)\). Solving these equations and applying boundary conditions leads to the deflection curve for a simply supported beam: \[ y(x) = \frac{w}{24EI} \left( -x^4 + 2Lx^3 - L^3x \right) \].
05

Find Maximum Deflection

The maximum deflection occurs at the center of the beam, \(x = \frac{L}{2}\). Substituting this into the deflection equation gives the maximum deflection as \[ \delta_{max} = \frac{5wL^4}{384EI} \].

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Key Concepts

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

Bending Moment
The bending moment is an important concept in structural engineering. It tells us how much a beam is likely to bend under a load. In our problem, the maximum bending moment occurs in the center of the simply supported beam. This is because the beam is experiencing a uniform load across its length.
The bending moment (M(x) = w \left( \frac{L}{2} - x \right) \cdot x) describes how the load affects the beam at a specific point, with the peak at the point where the forces creating the moment are perfectly balanced.
For a simply supported beam with a uniform load, the formula for the maximum bending moment at the center is M_{max} = \frac{wL^2}{8}, where \(w\) is the load per unit length and \(L\) is the length of the beam. This formula shows that the moment grows with both the load intensity and the square of the beam's length. Understanding this helps engineers design beams that won't fail under expected loads.
Bending Stress
Bending stress plays a critical role in determining if a beam can safely carry a given load. It occurs due to the bending moment and depends on the material's properties. The stress \sigma_{max} = \frac{M_{max} \cdot c}{I}, where \(c\) is the distance from the neutral axis to the outermost fiber and \(I\) is the moment of inertia of the cross-section.
This equation shows that the maximum bending stress is influenced by the bending moment, the shape of the cross-section, and the material properties. If the bending stress is too high, it can cause the beam to crack or fail.
By knowing the maximum bending moment and using the formula, engineers can ensure that the stress levels within the material remain safe. This helps in choosing appropriate materials and cross-sectional shapes for beams in buildings, bridges, and other structures.
Simply Supported Beam
A simply supported beam is a simple model ideal for situations where a beam is supported at both ends, allowing it to freely rotate but not translate. This type of support does not resist moment, which means moments at the support points are always zero. Such characteristics make it easier to analyze mathematically.
Simply supported beams are widely used in practical applications due to their simplicity and efficiency. They experience different forces and stresses along their length, reaching a peak at locations other than the support points.
  • Commonly used in bridges and structural supports.
  • They provide straightforward analysis and calculations.
  • Ideal for theoretical explorations and evaluations.
These beams serve as a basis for understanding more complex support structures and their responses to loads.
Uniform Load
A uniform load is a load that is distributed evenly across the length of a beam. It is usually measured as force per unit length, denoted by \(w\). This type of load creates a steady pressure on the beam, resulting in predictable deflection and stress patterns.
The uniform load simplifies calculations in beam design, as it creates symmetrical and repeatable force applications. As a result, it allows for more general solutions like those used in solving for the maximum deflection or bending moment in a simply supported beam.
Understanding the effects of a uniform load is essential, as many real-world scenarios approximate these conditions due to evenly distributed weights (like floor slabs or snow on a roof). In our problem, the uniform load directly influences both the deflection curve and the maximum bending moment.

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Most popular questions from this chapter

Plane Strain In the case of plane strain, \(\varepsilon_{3}=0,\) for example, and \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are nonzero. Figure \(3-7\) illustrates a plane strain situation. A long bar is rigidly confined between supports so that it cannot expand or contract parallel to its length. In addition, the stresses \(\sigma_{1}\) and \(\sigma_{2}\) are applied uniformly along the length of the bar. Equations \((3-1)\) to \((3-3)\) reduce to $$ \begin{aligned} \sigma_{1} &=(\lambda+2 G) \varepsilon_{1}+\lambda \varepsilon_{2} \\ \sigma_{2} &=\lambda \varepsilon_{1}+(\lambda+2 G) \varepsilon_{2} \\ \sigma_{3} &=\lambda\left(\varepsilon_{1}+\varepsilon_{2}\right) \end{aligned} $$. From Equation \((3-6)\) it is obvious that $$ \sigma_{3}=v\left(\sigma_{1}+\sigma_{2}\right) $$ This can be used together with Equations \((3-4)\) and (3-5) to find $$ \begin{array}{l} \varepsilon_{1}=\frac{(1+v)}{E}\left\\{\sigma_{1}(1-v)-v \sigma_{2}\right\\} \\\ \varepsilon_{2}=\frac{(1+v)}{E}\left\\{\sigma_{2}(1-v)-v \sigma_{1}\right\\} \end{array} $$

The Amazon River basin in Brazil has a width of \(400 \mathrm{~km}\). Assuming that the basin is caused by a line load at its center and that the elastic lithosphere is not broken, determine the corresponding thickness of the elastic lithosphere. Assume \(E=70\) \(\mathrm{GPa}, v=0.25,\) and \(\rho_{m}-\rho_{s}=700 \mathrm{~kg} \mathrm{~m}^{-3}\)

If all the principal stresses are equal \(\sigma_{1}=\sigma_{2}=\sigma_{3} \equiv p\), then the state of stress is isotropic, and the principal stresses are equal to the pressure. The principal strains in a solid subjected to isotropic stresses are also equal \(\varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=\frac{1}{3} \Delta ;\) each component of strain is equal to one-third of the dilatation. By adding Equations \((3-1)\) to \((3-3),\) we find $$ p=\left(\frac{3 \lambda+2 G}{3}\right) \Delta \equiv K \Delta \equiv \frac{1}{\beta} \Delta . $$ The quantity \(K\) is the bulk modulus, and its reciprocal is \(\beta,\) the compressibility. The ratio of \(p\) to the bulk modulus gives the fractional volume change that occurs under isotropic compression. Because the mass of a solid element with volume \(V\) and density \(\rho\) must be conserved, any change in volume \(\delta V\) of the element must be accompanied by a change in its density \(\delta \rho .\) The fractional change in density can be related to the fractional change in volume, the dilatation, by rearranging the equation of mass conservation $$ \delta(\rho V)=0 $$ which gives $$ \rho \delta V+V \delta \rho=0 $$ or $$ \frac{-\delta V}{V}=\Delta=\frac{\delta \rho}{\rho} $$ Equation \((3-53)\) of course assumes \(\Delta\) to be small. The combination of Equations \((3-50)\) and \((3-53)\) gives $$ \delta \rho=\rho \beta p $$ This relationship can be used to determine the increase in density with depth in the earth. Using Equations \((3-11)\) to \((3-13),\) we can rewrite the formula for \(K\) given in Equation \((3-50)\) as $$ K=\frac{1}{\beta}=\frac{E}{3(1-2 v)} $$ Thus as \(v\) tends toward \(1 / 2,\) that is, as a material becomes more and more incompressible, its bulk modulus tends to infinity.
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