16-9. Determine the equations of the elastic curve for thebeam using the x; and 1z coordinates. El is constant.2.

Answered: 16-9. Determine the equations of the…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

16-9. Determine the equations of the elastic curve for the
beam using the x; and 1z coordinates. El is constant.
2.

Expert Solution

Step 1

To determine:

Equation of Elastic curve.

Step 2

To determine equation of elastic curve, moment equation is needed to be determine. First we will determine support reactions as follows:

$\sum F_{x} = 0 \Rightarrow H = 0 \sum F_{y} = 0 \Rightarrow V - P = 0 \Rightarrow V = P \sum M = 0 \Rightarrow M - P \cdot \frac{L}{2} = 0 \Rightarrow M = \frac{P \cdot L}{2}$

Macaulay Method is improvement over double integration method which is suitable for prismatic beams having discontinuous loading.

Now using Macaulay Method, writing moment equation in terms of x₁ and x₂ as follows:

$M_{x} = V \times x_{1} + M \times {x_{1}}^{0} - P (x_{2} - \frac{L}{2}) \Rightarrow M_{x} = P \times x_{1} + \frac{P L}{2} \times {x_{1}}^{0} - P (x_{2} - \frac{L}{2})$

As per double integration method

$E I \frac{d^{2} y}{d x^{2}} = M_{x}$

By integrating the above equation twice, we will get the equation of the elastic curve:

$E I \frac{d^{2} y}{d x^{2}} = P \cdot x_{1} + \frac{P \cdot L \cdot {x_{1}}^{0}}{2} - P \cdot (x_{2} - \frac{L}{2}) i n t e g r a t i n g i t, w e g e t E I \frac{d y}{d x} = \frac{P \cdot {x_{1}}^{2}}{2} + C_{1} + \frac{P \cdot L \cdot x_{1}}{2} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2} a t x_{1} = 0, \frac{d y}{d x} = 0 \Rightarrow C_{1} = 0 \Rightarrow E I \frac{d y}{d x} = \frac{P \cdot {x_{1}}^{2}}{2} + \frac{P \cdot L \cdot x_{1}}{2} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2} i n t e g r a t i n g i t, w e g e t E I \cdot y = \frac{{P \cdot x_{1}}^{3}}{6} + C_{2} + \frac{P \cdot L \cdot {x_{1}}^{2}}{4} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2} a t x_{1} = 0, y = 0 \Rightarrow C_{2} = 0 \Rightarrow E I \cdot y = \frac{{P \cdot x_{1}}^{3}}{6} + \frac{P \cdot L \cdot {x_{1}}^{2}}{4} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2} \Rightarrow y = \frac{{P \cdot x_{1}}^{3}}{6 E I} + \frac{P \cdot L \cdot {x_{1}}^{2}}{4 E I} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2 E I}$

Equation of elastic curve is $y = \frac{{P \cdot x_{1}}^{3}}{6 E I} + \frac{P \cdot L \cdot {x_{1}}^{2}}{4 E I} - \frac{P \cdot {(x_{2} - \frac{L}{2})}^{2}}{2 E I}$ .