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Chapter 3: Problem 6

(a) \(\quad y_{\max }=y(L)=w_{0} L^{4} / 8 E I\) (b) Replacing both \(L\) and \(x\) by \(L / 2\) in \(y(x)\) we obtain \(w_{0} L^{4} / 128 E I,\) which is \(1 / 16\) of the maximum deflection when the length of the beam is \(L\). (c) \(y_{\max }=y(L / 2)=5 w_{0} L^{4} / 384 E I\) (d) The maximum deflection in Example 1 is \(y(L / 2)=\left(w_{0} / 24 E I\right) L^{4} / 16=w_{0} L^{4} / 384 E I,\) which is \(1 / 5\) of the maximum displacement of the beam in part c.

Short Answer

Expert verified
Deflection at \( L/2 \) is \( 1/16 \) of \( y(L) \). Maximum at \( L/2 \) is \( 1/5 \) of \( c \).

Step by step solution

01

Analyzing Maximum Deflection at End

According to the given expression, the maximum deflection at the end of the beam is given by \( y_{\max} = y(L) = \frac{w_0 L^4}{8 E I} \). This indicates the maximum deflection when the full length \( L \) of the beam contributes to the deflection.
02

Calculating Deflection at Half-Length

We are asked to replace \( L \) with \( L/2 \) in the expression for deflection \( y(x) \). Doing this, we get \( y\left(\frac{L}{2}\right) = \frac{w_0 (L/2)^4}{8 E I} = \frac{w_0 L^4}{128 E I} \). This shows that the deflection at half the length \( L/2 \) is \( \frac{1}{16} \) of the maximum deflection.
03

Maximum Deflection at Half-Length

The maximum deflection for the beam if it is evaluated at its mid-point \( L/2 \) is expressed as \( y_{\max} = y(L/2) = \frac{5 w_0 L^4}{384 E I} \). This indicates a different maximum condition when evaluating at the mid-span.
04

Relating Example's Deflection with Part (c)

The maximum deflection given in the example for \( y(L/2) \) is \( \frac{w_0 L^4}{384 E I} \). By comparing with part (c), we notice it equals \( \frac{1}{5} \) of the previously calculated maximum deflection \( \frac{5 w_0 L^4}{384 E I} \). This shows a relationship of the maximum displacement being proportionally reduced by a factor of \( \frac{1}{5} \).

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Key Concepts

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

Maximum Deflection
Maximum deflection in beams occurs when a beam bends under a load to its greatest extent. Understanding maximum deflection is crucial in structural engineering since it helps ensure that beams can safely support loaded forces without excessive bending.
For a beam with a uniform load, the formula for maximum deflection at the end of the beam is:
  • \( y_{\max} = \frac{w_0 L^4}{8 E I} \), where
    • \( w_0 \) is the load per unit length,
    • \( L \) is the length of the beam,
    • \( E \) is the modulus of elasticity, and
    • \( I \) is the moment of inertia.
This expression helps predict how much a beam will bend when subjected to a given load. By determining the maximum deflection, engineers can decide whether the deflection is within acceptable limits or if the beam needs reinforcement.
Also, when analyzing deflection at a beam's mid-point \( L/2 \), we see different behavior. It is essential to compare this with maximum deflection at the end, to understand the deflection profile across the beam.
Beam Theory
Beam theory provides a fundamental set of equations and concepts that describe how beams of different shapes and materials deflect under various loads. The foundation of beam theory is rooted in material science and physics, and it helps predict a beam's response to loads.
Key points about beam theory include:
  • It utilizes equations such as \( y(x) = \frac{w_0 x^4}{8 E I} \) to calculate deflection at any point \( x \) along the beam length.
  • The simplicity of beam theory allows it to provide accurate predictions for linear elastic materials.
  • The theory makes simplifying assumptions, such as small deflections and isotropic material properties. Real-world deviations may require more complex models.
  • Different boundary conditions, like fixed or free ends, affect how deflection develops across the beam.
Understanding beam theory is crucial for engineers to design structures that are both strong and efficient, ensuring that safety standards are met.
Structural Analysis
Structural analysis is the process of determining the effects of loads and forces on physical structures. It involves using mathematical and physical principles to ensure that structures can withstand the various forces they encounter while being functional and safe.
In structural analysis, engineers use the deflection formulas derived from beam theory to predict how structures like beams, floors, and bridges will behave under specific loads. Factors considered in structural analysis include:
  • Load types, such as point loads or distributed loads,
  • Material properties, including strength and elasticity,
  • The geometry and size of the structure,
  • Support conditions like fixed, roller, or pinned supports.
By employing structural analysis, engineers can evaluate potential points of failure, ensuring the safety and reliability of the structures they design. Maximum deflection calculations are a part of this holistic analysis, providing insights into the durability and performance of beams and other structural components.
Overall, structural analysis not only helps to optimize designs for cost-effectiveness but also to adhere to safety regulations and codes.

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Most popular questions from this chapter

For \(\lambda=\alpha^{4}, \alpha>0,\) the general solution of the boundary-value problem \\[ y^{(4)}-\lambda y=0, \quad y(0)=0, y^{\prime \prime}(0)=0, y(1)=0, y^{\prime \prime}(1)=0 \\] is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} \cosh \alpha x+c_{4} \sinh \alpha x\). The boundary conditions \(y(0)=0, y^{\prime \prime}(0)=0\) give \(c_{1}+c_{3}=0\) and \(-c_{1} \alpha^{2}+c_{3} \alpha^{2}=0,\) from which we conclude \(c_{1}=c_{3}=0 .\) Thus, \(y=c_{2} \sin \alpha x+c_{4} \sinh \alpha x .\) The boundary conditions \(y(1)=0, y^{\prime \prime}(1)=0\) then give \\[ \begin{aligned} c_{2} \sin \alpha+c_{4} \sinh \alpha &=0 \\ -c_{2} \alpha^{2} \sin \alpha+c_{4} \alpha^{2} \sinh \alpha &=0 \end{aligned} \\]. In order to have nonzero solutions of this system, we must have the determinant of the coefficients equal zero, that is, \\[ \left|\begin{array}{cc} \sin \alpha & \sinh \alpha \\ -\alpha^{2} \sin \alpha & \alpha^{2} \sinh \alpha \end{array}\right|=0 \quad \text { or } \quad 2 \alpha^{2} \sinh \alpha \sin \alpha=0 \\] But since \(\alpha>0,\) the only way that this is satisfied is to have \(\sin \alpha=0\) or \(\alpha=n \pi .\) The system is then satisfied by choosing \(c_{2} \neq 0, c_{4}=0,\) and \(\alpha=n \pi .\) The eigenvalues and corresponding eigenfunctions are then \\[ \lambda_{n}=\alpha^{4}=(n \pi)^{4}, n=1,2,3, \ldots \quad \text { and } \quad y=\sin n \pi x \\].

(a) From \(100 x^{\prime \prime}+1600 x=1600 \sin 8 t, x(0)=0,\) and \(x^{\prime}(0)=0\) we obtain \(x_{c}=c_{1} \cos 4 t+c_{2} \sin 4 t\) and \(x_{p}=-\frac{1}{3} \sin 8 t\) so that by a trig identity \\[x=\frac{2}{3} \sin 4 t-\frac{1}{3} \sin 8 t=\frac{2}{3} \sin 4 t-\frac{2}{3} \sin 4 t \cos 4 t\\] (b) If \(x=\frac{1}{3} \sin 4 t(2-2 \cos 4 t)=0\) then \(t=n \pi / 4\) for \(n=0,1,2, \ldots\) (c) If \(x^{\prime}=\frac{8}{3} \cos 4 t-\frac{8}{3} \cos 8 t=\frac{8}{3}(1-\cos 4 t)(1+2 \cos 4 t)=0\) then \(t=\pi / 3+n \pi / 2\) and \(t=\pi / 6+n \pi / 2\) for \(n=0,1,2, \ldots\) at the extreme values. Note: There are many other values of \(t\) for which \(x^{\prime}=0\) (d) \(x(\pi / 6+n \pi / 2)=\sqrt{3} / 2 \mathrm{cm}\) and \(x(\pi / 3+n \pi / 2)=-\sqrt{3} / 2 \mathrm{cm}\).

The auxiliary equation is \(m(m-1)(m-2)(m-3)-6 m(m-1)(m-2)+33 m(m-1)-105 m+169=0,\) so that \(m_{1}=m_{2}=3+2 i\) and \(m_{3}=m_{4}=3-2 i .\) The general solution of the differential equation is $$y=x^{3}\left[c_{1} \cos (2 \ln x)+c_{2} \sin (2 \ln x)\right]+x^{3} \ln x\left[c_{3} \cos (2 \ln x)+c_{4} \sin (2 \ln x)\right].$$

The auxiliary equation is \(m^{2}=0\) so that \(u(r)=c_{1}+c_{2} \ln r .\) The boundary conditions \(u(a)=u_{0}\) and \(u(b)=u_{1}\) yield the system \(c_{1}+c_{2} \ln a=u_{0}, c_{1}+c_{2} \ln b=u_{1} .\) Solving gives \\[ c_{1}=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)} \quad \text { and } \quad c_{2}=\frac{u_{0}-u_{1}}{\ln (a / b)} \\]. Thus \\[ u(r)=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)}+\frac{u_{0}-u_{1}}{\ln (a / b)} \ln r=\frac{u_{0} \ln (r / b)-u_{1} \ln (r / a)}{\ln (a / b)} \\].

The auxiliary equation is \(m^{2}+3 m+2=(m+1)(m+2)=0,\) so \(y_{c}=c_{1} e^{-x}+c_{2} e^{-2 x}\) and $$W=\left|\begin{array}{cc}e^{-x} & e^{-2 x} \\\\-e^{-x} & -2 e^{-2 x}\end{array}\right|=-e^{-3 x}$$ Identifying \(f(x)=\sin e^{x}\) we obtain $$\begin{aligned} &u_{1}^{\prime}=\frac{e^{-2 x} \sin e^{x}}{e^{-3 x}}=e^{x} \sin e^{x}\\\ &u_{2}^{\prime}=\frac{e^{-x} \sin e^{x}}{-e^{-3 x}}=-e^{2 x} \sin e^{x}\end{aligned}$$ Then \(u_{1}=-\cos e^{x}, u_{2}=e^{x} \cos x-\sin e^{x},\) and $$\begin{aligned}y &=c_{1} e^{-x}+c_{2} e^{-2 x}-e^{-x} \cos e^{x}+e^{-x} \cos e^{x}-e^{-2 x} \sin e^{x} \\\&=c_{1} e^{-x}+c_{2} e^{-2 x}-e^{-2 x} \sin e^{x}.\end{aligned}$$
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