Direct Stress
Tension is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object. It is the opposite of compression. As tension is the magnitude of a force, it is measured in newtons (or sometimes pounds-\b{force}) and is always measured parallel to the string on which it applies.
Compression is the result of the subjection of a material to compressive stress, which results in reduction of volume as compared to an uncompressed but otherwise identical state.
Velocity is the measurement of the rate and direction of change in the position of an object. It is a vector physical quantity; both magnitude and direction are required to define it.
The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Weight is the force that gravitation exerts upon a body, equal to the mass of the body times the local acceleration of gravity: commonly taken, in a region of constant gravitational acceleration, as a measure of mass.
Temperature is a measurement of the average kinetic energy of the molecules in an object or system and can be measured with a thermometer or a calorimeter. It is a means of determining the internal energy contained within the system.
Resilience and Direct Stress in bars, both composite and of varying cross section.
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Introduction
A structural member is said to be under Direct Stress when the loads acting on it are either compressive or tensile. Any other forms of loading, including bending and shear, cannot be said to cause Direct Stress.
This section highlights the Direct Stress and the resilience in bars of varying cross section as well as in compound bars; made with more than one material, each having its coefficient of thermal expansion.
Strain Energy, Resilience
When either a Tensile or Compressive Load is applied to a bar, there will be a change in length $x$, which for a material that obeys Hooke's law, is proportional to the load.
The Strain Energy $U$ of the bar is defined as the work done by the load in altering the length (i.e. Producing Strain).
For a gradually applied or "Static" load the work done is represented by the shaded area on the diagram.
The Strain Energy can be expressed in terms of the Stress and dimensions. For a bar of uniform cross section and length $l$.
$P=f\;A$
$x=\frac{f\;l}{E}$
$\therefore\;\;\;\;\;\;U=\frac{1}{2}\;f\;A\;\frac{f\;l}{E}=\left(\frac{f^2}{2\;E} \right)\times A\;l$
But $A\;l$ is the volume of the bar and can be expressed in words as:
The Strain Energy per unit volume (Usually called Resilience) in either simple Tension or Compression is :
- Proof Resilience is the value at the elastic limit or for non-ferrous materials, the Proof Stress.
- Strain Energy is always a positive quantity and is expressed in units of work. (In the imperial system in.lb. ; ft.lbs or in.tons)
Example 1 [imperial]
Calculate the Strain Energy of the following Bolt which is under a Tensile Load of $1\;ton$.
Show that the Strain Energy is increased when the bolt is under maximum Stress, if the shank of the Bolt is turned down to the root diameter of the thread.
$E = 30\times 10^6\;lb.in.^{-2}$
It is normal practice to assume that the load is distributed evenly over the core of the screwed portion ( i.e. In this case the core diameter is $0.622\;in.$ and the Area of the core is $0.304\;sq.\;in$. The Cross sectional Area of the shank is $0.442\;sq.in.$
The Stress in the screwed portion is given by: $\displaystyle\frac{2240}{0.304} = 7380\;lb.in^{-2}$
(Note $\; 1\;ton = 2240\;lb.$)
The Stress in the Shank is given by: $\displaystyle\frac{2240}{0.442} = 5070\;lb.in^{-2}$
Thus, the Total Strain Energy is given by: $U_T = \frac{1}{2}\left(\frac{7380^2\times 0.304\times1 + 5070^2\times 0.442\times 2}{30\times 10^6} \right)$ $= \frac{16.6 + 22.7}{60} = 0.655\;in.lb.$
If the shank is now reduced in diameter to $0.662\;in.$ the Stress in the bolt will be constant and have a value of $7380\;lb./sq.in.$ and the Strain Energy will be given by: $U = \frac{1}{2}\times 7380^2\times0.304 \times \frac{3}{30\times 10^6} = 0.827\;in.lb.$
- The Strain Energy is $0.827\;in.lb.$
Impact Loads.
If a weight $W$ falls through a height $h$ onto a collar attached at one end of a uniform bar (See Diagram),
the bar will extend. The extent of this extension will be greater than if the load had been gradually applied.
This must be so since the weight will have Kinetic Energy at the point of impact and this energy is absorbed by an increase in the strain energy of the bar. Assuming that the bar does not fail the weight will oscillate about and finally come to rest at the Normal equilibrium position.
In the diagram the maximum extension of the bar is shown as $x$. $P$ is the equivalent static or gradually applied load which would produce an extension of $x$.
The Strain Energy in the bar at the point of maximum extension is $\displaystyle \frac{1}{2}\;P\;x$
If the loss of energy at Impact is neglected then the following equation can be written:
Loss of Potential Energy by the Weight = Increase in Strain Energy in the Bar i.e. $W(h+x)=\displaystyle\frac{1}{2}\;P\;x$ (see equations (1) and (2))
Hence, $W\;\left(h+\frac{P\;l}{A\;E} \right)=\frac{1}{2}\;\left(\frac{P^2\;l}{A\;E} \right)$
Rearranging and multiplying through by $\displaystyle \frac{A\;E}{l}$ a quadratic in $P$ is produced. i.e. $\frac{P^2}{2}-W\;P-\frac{W\;h\;A\;E}{l}=0$
Solving and discarding the negative root:
Using the resulting value for $P$ it is now possible to calculate both the value of the maximum extension $x$ and the resulting Direct Stress.
The particular case where $h = 0$ (i.e. the Load is suddenly applied) gives the value $P = 2\;W$ i.e. The Stress produced by a suddenly applied load is twice the Static Stress.
The above simple analysis assumes that the whole of the rod attains the same value of maximum Stress at the same moment. This is however not strictly true. A wave of stress is set up by the Impact which is propagated along the rod.
The actual maximum Stress set up will then depend upon the dimensions of the rod, its density and the velocity of the load at impact. Usually this approximate analysis gives results that err on the "Safe side" but this is not always the case.
Example 2 [imperial]
If in the previous problem the bar was turned down to $\displaystyle\frac{1}{2}\;in.$ diameter over half its length what would be the maximum Stress and extension caused by the $200\;lb.$ load falling 2 inches?
Let $P$ be the equivalent gradually applied load to cause the same maximum Stress. The corresponding extension is made up of two parts. $x = \frac{P\;l_1}{A_1\;E} + \frac{P\;l_2}{A_2\;E}$
Substituting given values, $x = \displaystyle\frac{P\;(5\times 12)}{\displaystyle\frac{\pi}{16}\;E} + \displaystyle\frac{P\;(5\times 12)}{\displaystyle\frac{\pi}{4}\;E}$
(Note. The lengths of each part of the bar have been converted to inches)
$x = \frac{P\times 20\times 60}{\pi \times 30\times 10^6} = \frac{4\;P}{\pi \times10^5}$
Applying the Energy Equation, $W\;(h+x)=\displaystyle\frac{1}{2}P\;x$
Using the values for $x$ from above: $200\;\left(2+\frac{4\;P}{\pi \times 10^5} \right)=\frac{1}{2}P\;\left(\frac{4\;P}{\pi \times10^5} \right)$
Multiplying through by $\displaystyle \frac{\pi \times10^5}{4}$ and rearranging: $\frac{P^2}{2} - 200\;P - \frac{400\times \pi \times10^5}{4}=0$
Solving this quadratic and ignoring the negative root: $P = 200 + \sqrt{200^2\;+\frac{400\;\pi \times 10^5}{2}} = 200\times 40.7\; lb.$
The Maximum Stress will occur in the smaller section giving: $f = \frac{P}{A} = \frac{200\times 40.7}{\displaystyle\frac{\pi}{16}} = 41.500\;lb.in.^{-2}$
The Maximum extension is given by: $x = \displaystyle\frac{4\;P}{\pi \times 10^5}$ $\therefore\;\;\;\;\;\;x = \frac{4\times 200\times 40.7}{\pi \times 10^5}=0.1035\;in.$
If the bar is already stressed before Impact e.g. If the collar in the previous examples had been given a weight, it would be correct to allow for the loss of Potential Energy of the collar after the impact and equate the total loss of Potential Energy to the difference between the initial and final Strain Energies.
If $W$' is the weight of the collar and $LM$ represents the further extension after impact, then the area $ALMB$ represents the increase in Strain Energy. But the area $ALMC$ being $W$' times the added extension represents the loss of Potential Energy of the collar after impact. This leaves $ABC$ to be equated to the loss of energy of the weight alone.
Consequently the Stress due to impact may be calculated without any consideration of the initial Stress and the final Total Stress is then found by adding the initial Stress.
i.e. $P=P'+W'$
Where $P$' is calculated on the assumption of no initial Stress.
- The maximum Stress is $41.500\;lb.in.^{-2}$
- The maximum extension is $0.1035\;in.$
Example 3 [imperial]
The loads carried by a lift may be dropped $4\;in.$ onto the floor. The cage itself weighs $2\;Cwt.$ and is supported by $80\;ft.$ of wire rope weighing $0.6\;lb./ft.$ and consisting of $49$ wires each of $\displaystyle\frac{1}{16}\;in.$ diameter. The Maximum Stress in the wire is limited to $13000\;lb/sq.in.$ and $E$ for the rope is $\displaystyle 10.5\times .10^6\;lb.in.^{-2}$.
Find the maximum Load which can be carried.
Note: $1\;Cwt. = 112\;lbs.$
The maximum Stress will occur at the top of the rope and the Initial Stress will be found from the weight of the cage and wire rope.
The Initial Stress $=\displaystyle\frac{2\times 112 + 80\times0,6}{49\times \left( \displaystyle\frac{\pi }{4} \right)\left(\displaystyle\frac{1}{16} \right)^2}=1810\;lb.in^{-2}$
Subtracting this from the Permissible Stress of $13000\;lb/sq.in.$, the increased Stress due to impact is $11190\;lb./sq.in.$ This would be caused by an equivalent Static load of:
$11190\times 49\times \left( \frac{\pi }{4} \right)\left(\frac{1}{16} \right)^2 = 1680\;lb.$
This would produce an extension of: $\displaystyle\frac{11190\times 80\times 12}{10.5\times 10^6}=1.015\.in.$
If $W$ is the load dropped , then applying the Energy Equation gives: $W(h+x)=\displaystyle\frac{1}{2}\;P\;x$
Substituting in Values: $W(4+1.025)=\displaystyle\frac{1}{2}\times1680 \times1.025$
$\therefore\;\;\;\;\;\;\;W = \frac{1680\times1.025}{2\times 5.025} = 171\;lb.$
- The maximum Load is $W= 171\;lb.$
Varying Cross-section and Load
It is usual to assume that the load is uniformly distributed over the cross-section and that therefore the Stress is inversely proportional to the area.
The load may also vary as in the case of a column where its own weight needs to be taken into account and of course the case of inertia loadings on members in motion.
Example 1 [imperial]
A rod of length $l$ tapers uniformly from a diameter $D$ at one end to a diameter $d$ at the other.
Find the extension caused by an axial load $P$.
At a distance $x$ from the small end the diameter is given by: $d_x=d+(D-d)\times \frac{x}{l}$
At $x$ the load $P$ will extend a short length $dx$ by: $\frac{4\;P\;dx}{\pi [d + (D - d)\displaystyle\frac{x}{l}]^2\;E}$
And for the whole rod the extension is given by: $=\int_{0}^{l}{\frac{4\;P\;dx}{\pi [d+(D-d)\displaystyle\frac{x}{l}]^2\;E}}=-\frac{l}{D-d}\times \frac{4\;P}{\pi \;E}\left[ \frac{1}{d+(D-d)\;\displaystyle\frac{x}{l}} \right]_0^l$ $=\frac{4\;P\;l}{\pi \;E\;(D-d)}\times \left(\frac{1}{d}-\frac{1}{D} \right)=\frac{4\;P\;l}{\pi \;D\;d\;E}$
- The extension is $\displaystyle\frac{4\;P\;l}{\pi \;D\;d\;E}.$
Example 2 [imperial]
What is the condition that a column will have uniform strength (i.e. a constant maximum stress) when under the action of its own weight and a longitudinal Stress $f$ which is applied to the top?
Let the cross-section at the top be $a$ and at a distance $x$ from the top be $A$. The diagram shows the forces acting on a slice of thickness $dx$ where $w$ is the density and $f$ the uniform Stress.
Equating Forces: $f\;(A+dA)-f\;A=w\;A\;dx$
Separating the variables: $\frac{dA}{A}=\frac{w}{f}dx$
Integrating: $\ln A=\frac{w}{f}\;x+C$
When $x = 0\;A = a$, $\therefore\;\;\;\;\;\;\;C = \ln a$ Thus, $\ln \frac{A}{a}=\frac{w\;x}{f}$ Or $A=a\;e^{\displaystyle\frac{wx}{f}}$
- $C = \ln\;a$
Example 3 [imperial]
A steel rod of uniform section $30\;inches$ long is rotated about a vertical axis through one end at right angles to its length, at $1000\;r.p.m.$
If the density of the material is $0.28\;lb./cu.in.$, find the maximum Stress.
Let the Stress at a distance $x$ from the axis be $f$ and a distance $x + dx$ ,$f + df$. If the area is $A$, the density of the rod is $w$ and the angular velocity $\displaystyle \alpha$
Then the forces acting on the element are as shown. The Centrifugal Force is $\displaystyle \left(\frac{w\;A\;dx}{g} \right)\;x\;\omega ^2$.
For equilibrium, $(f + df)\;A - f\;A + \left(\frac{w\;A\;dx}{g} \right)\;x\;\omega ^2 = 0$
$\therefore\;\;\;\;\;\;\;\;df\;= - \left(\frac{w\;A\;dx}{g} \right)\;x\;\omega ^2$
Integrating, $f\;= - \left(\displaystyle\frac{w\;x^2\;\omega ^2}{2g} \right)\:+\;C$
When $x = l\;f = 0$
$\therefore\;\;\;\;\;\;\;\;C = \left(\frac{w\;l^2\;\omega ^2}{2g} \right)$
Thus, $\;\;\;\;\;\;\;\;f = \left(\displaystyle\frac{w\;\omega ^2}{2g} \right)\left(l^2 - x^2 \right)$
The maximum Stress will occur at the axis where $x = 0$
$\therefore\;\;\;\;\;\;\;\;f = \left(\frac{w\;l^2\;\omega ^2}{2g} \right) = \frac{0.28\times 1000^2\times (2\pi )^2\times30^2}{2\times 32.2\times 12\times60^2}= 3580\;lb.in^{-2}$
Notes.
- $\displaystyle \omega$ is measured in radians per second and $\displaystyle 1000\;r.p.m.\equiv 1000\times2\pi \div 60$
- $g$ is usually taken as $32.2\;ft/sec.sq.$ but in this example must be multiplied by 12 to bring it inches/sec. sq.
- The maximum Stress is $f=3580\;lb.in^{-2}.$
Compound Bars
Any Tensile or Compressive member which consists of two or more bars or tubes in parallel is called a Compound Bar.
The bars are usually of different materials. The method of analysis is shown in the following examples.
Temperature Stresses
If a Compound bar is made of several different materials with different coefficients of thermal expansion, then if it is subjected to a change in temperature, then the different parts will tend to expand by different amounts.
If the parts are constrained to remain together, then the actual change in length must be the same for each. This change will be the result ( taking into account positive and negative Strains) of the effects due to both temperature and Stress conditions.