Engineering › Materials › Strength › Axial ›

bar diameter

Computes the diameter of a bar based on various stress conditions.

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Contents

Interface
Bar Diameter
Bar Diameter
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Private project under development, to help contact the author:

Interface

C++

Overview

Consider the case of a bar made of a certain material on which various forces are exerted along its longitudinal axis, in an upward or downward direction. This module computes the minimum diameter of the bar such that it will withstand the sum of forces that act upon it, also taking into account a safety coefficient.

<b>Technical details</b>

This module computes the diameter of a bar in the case of tensial and/or compressive stress, where each section of the bar is found between the origins of two consecutive forces.

Let $\inline \overrightarrow{F_1}, \overrightarrow{F_2}, \ldots, \overrightarrow{F_n}$ be the stress forces that act along the axis of the bar. We define $\inline N_k$ for $\inline k = 1, 2, \ldots, n$ the norm of the resultant of the first $\inline k$ forces, thus:

$N_k = \left| \sum_{i=1}^k \overrightarrow{F_i} \right|.$

(2)

For the sake of clarity, the forces should be given in a consecutive order such that the diameters will correspond to consecutive sections of the bar.

In the following image you may notice the final shape of the bar (the red contour at the right side of the image) based on the various axial forces that act upon it. The diameter of sufficient size is in this case equal to $\inline d_2$ corresponding to the maximum stress $\inline N_2$ .

The force $\inline R$ is the reaction force and $\inline l_1, l_2, \ldots, l_n$ are the lengths that give the origin of the axial forces.

MISSING IMAGE!

1/suff_shape-378.jpg cannot be found in /users/1/suff_shape-378.jpg. Please contact the submission author.

References:

Gh. Buzdugan - "Strength of Materials"

Authors

Grigore Bentea, Lucian Bentea (October 2006)

Bar Diameter

doublebar_diameter(	int	`n`
	double*	`forces`
	double	`sigma`	)

Considering $\inline \sigma$ the value of the maximum allowable stress per unit area, we compute the diameter of the bar using the following formula:

$d = \max \left\{ 2 \sqrt{\frac{N_k}{\pi \sigma}} \,\mid\, k = \overline{1,n} \right\}$

(3)

where the value of $\inline \sigma$ is expressed as:

$\sigma = \frac{\sigma_f}{s_c}$

(4)

with $\inline \sigma_f$ the failure stress per unit area and $\inline s_c \geq 1$ the safety coefficient. The values of $\inline \sigma_f$ for various materials are given in the following table.

<center> <table border="1"> <tr> <td><center>Material</center></td> <td><center> $\inline \sigma_f$ (N/sq.m)</center></td> </tr> <tr> <td>Steel</td> <td>6.2E+8 - 6.6E+8</td> </tr> <tr> <td>Iron</td> <td>2.6E+8 - 4.0E+8</td> </tr> <tr> <td>Copper</td> <td>3.2E+8</td> </tr> <tr> <td>Wood</td> <td>0.6E+8 - 1.0E+8</td> </tr> </table> </center>

The example code below computes the diameter of a copper bar on which axial forces with various values are exerted.

Example 1

#include <codecogs/engineering/materials/strength/axial/bar_diameter.h>
#include <stdio.h>
 
int main()
{
  // input data
  double F[4] = {32000, 11000, -8000, 2500}, sigma = 3.2E+8;
 
  // display the input data
  printf("Forces:\n\n");
  for (int i = 0; i < 4; i++)
    printf("F[%d] = %8.2lf N\n", i+1, F[i]);
  printf("\nSigma = %.2lf N/sq.m\n\n", sigma);
 
  // compute the diameter of the bar
  double diameter = Engineering::Materials::Strength::Axial::bar_diameter
  (4, F, sigma);
 
  // display the result
  printf("Diameter = %.10lf m\n\n", diameter);
 
  return 0;
}

Output

Forces:
 
F[1] = 32000.00 N
F[2] = 11000.00 N
F[3] = -8000.00 N
F[4] =  2500.00 N
 
Sigma = 320000000.00 N/sq.m
 
Diameter = 0.0130801974 m

Parameters

n	the number of axial forces
forces	an array containing the signed value of each force (<i>Newtons</i>)
sigma	the value of the maximum allowable stress per unit area (<i>Newtons per sq. meters</i>)

Returns

The diameter of the bar (<i>meters</i>).

Source Code

This module is private, for owner's use only.

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Bar Diameter

doublebar_diameter(	int	`n`
	double*	`forces`
	double	`E`
	double	`delta`	)

Considering $\inline \delta$ the value of the maximum allowable elongation/shortening and E the elastic modulus, we compute the diameter of the bar using the following formula:

$d = \max \left\{ 2 \sqrt{\frac{N_k}{\pi E \delta}} \,\mid\, k = \overline{1,n} \right\}.$

(5)

The values of the elastic modulus E for various materials are given in the following table.

<center> <table border="1"> <tr> <td><center>Material</center></td> <td><center>E (N/sq.m)</center></td> </tr> <tr> <td>Steel</td> <td>0.2E+12 - 0.21E+12</td> </tr> <tr> <td>Iron</td> <td>0.115E+12 - 0.16E+12</td> </tr> <tr> <td>Copper</td> <td>0.11E+12 - 0.13E+12</td> </tr> <tr> <td>Wood</td> <td>0.9E+10 - 1.2E+10</td> </tr> </table> </center>