CodeCogs

An analysis of stresses and strains in curved beams

You're viewing an older version of this page (#3358). View the current version.

Stress in bars of small initial curvature.

Where the radius of curvature is large compared to the dimensions of the cross section, the analysis of stress is similar to that for pure bending.

Let $R_)$ be the initial ( unstrained) radius of curvature of the neutral surface and R the radius of curvature under the action of a pure bending moment M

Then the strain in a n element at a distance y from the neutral axis is given by:-

$Strain\;=\;\frac{PQ^{'}\;-\;PQ}{PQ}\;=\;\frac{(R\,+\,y)(\theta \,+\,\delta \theta)\,-\,(R_0\,+\,y)\theta }{(R_0\,+y)\theta }$

(1)

$=\;\frac{R(\theta \,+\,\delta \theta )\,-\,R_0\theta \,+\,y\delta \theta }{(R_)\,+\,y)\theta }$

(2)

$\frac{y\delta \theta }{(R_0\,+\,y)\theta }$

(3)

since $R(\theta \,+\,\delta \theta )\;=\;R_0\theta \;=\;length\;along\;the\;neutral\;axis$

If y is neglected in comparison with $R_0\;and\;noting\;from\;R(\theta \,+\,\delta \theta )\;=\;R_0\;that\; \delta \theta \;=\;\left(\frac{R_0\,-\,R}{R} \right)\theta$

$then\;\;\;\;strain\;=\;\left(\frac{y}{R_0} \right)\left(\frac{(R_0\,-\,R)}{R} \right)\;=\;y\left(\frac{1}{R}\,-\,\frac{1}{R_0} \right)$

(4)

Neglecting lateral stress the normal stress

$f\;=\;E\;\times \;strain$

(5)

substituting in equation (4)

$f\;=\;E\,y\left(\frac{1}{R}\;-\;\frac{1}{R_)} \right)$

(6)

Total normal stress = 0

$i.e.\;\;\;\;\;\int f\,dA\;=\;E\left(\frac{1}{R}\,-\,\frac{1}{R_0} \right)\int y\,dA\;=\;0$

(7)

which shows that the neutral axis passes through the centroid of the section.

$Moment\;of\;resistance\;M\;=\int f\,y\,dA$

(8)

$=\;E\,I\left(\frac{1}{R}\;-\;\frac{1}{R_0} \right)\int f\,y^2\,dA\;\;\;\;from\; equation (7)$

(9)

$=\;E\,I\left(\frac{1}{R}\;-\;\frac{1}{R_0} \right)$

(10)

Combining equations (6)and(10)

$\frac{f}{y}\;=\;\frac{M}{I}\;=\;E\,\left(\frac{1}{R}\;-\;\frac{1}{R_0} \right)$

(11)

the strain energy of a short length $\delta \,s$ (measured along the neutral surface) under the action og bending moment M is:-

$\delta U\;=\;\frac{1}{2}\,M\,\delta \theta$

(12)

$\therefore\;\;\;\;\;\delta \: U\;=\;\frac{1}{2}\,M\,\left(\frac{R_0\;-\;R}{R} \right)\theta$

(13)

From equation ( 7)

$\delta \: U\;=\;\frac{1}{2}\,M\,\delta\, s\;X\;\frac{M}{E\.I}\;=\;\left(\frac{M^2}{2\,E\,I} \right)$

(14)

Application to the design of a piston ring

Blank

Suppose it is required to design a split ring so that its outside surface will be circular in both the stressed and unstressed conditions and the radial pressure exerted will be uniform. If p is the uniform pressure on the outside then the bending moment at be is given by:-

$M\;=\;\int_{0}^{{\pi \;-\;\theta }}(p\,.\,d\:R\,.\,d\phi )R\,sin\phi \;\;\;approx$

(15)

where d is the depth of the ring in the axial direction integrating

$M\;=\;p\,R^2\,d(I\;+\;cot\,\theta )$

(16)

$But\;\;\;\;\;\frac{M}{I}\;=\;E\left(\frac{1}{R}\;+\;\frac{1}{R_0}} \right)\;=\;a\;constant\;for\;a\;given\;condition$

(17)

$i.e.\;\;\;\;\;\;\frac{p\,R^2\;d(1\;+\;cos\theta )}{d\,t^3/12}\;=\;constant$

(18)

$=\;\frac{24\,R^2}{t_0^3}\;\;\;\;\;\;when \;\theta \;=\;0\; and\;t\/=\/t_0$

(19)

$\therefore\;\;\;\;\frac{t}{t_0}\;=\;\sqrt[3]{\left(\frac{(1\;+\;cos\,\theta )}{2} \right)}$

(20)

Which is the required variation of thickness Using equation (16). The maximum bending stress at any section

$=\;\left(\frac{M}{I} \right)\left(\frac{T}{2} \right)\;=\;\left(\frac{6\,p\,R^2}{t^2} \right)\left(1\;+\;cos\theta \right)$

(21)

$=\;\frac{12\,p\,R^2\.t}{t_0^3}\;\;\;\;from\; equation( )$

(22)

which has it's greatest value when $\theta \;=\;0$

$i.e.\;\;\;\;\hat{f}\;=\;\frac{12\,p\,r^2}{t_0^2}$

(23)

$From\;which\;\;\;\frac{1}{R}\;-\;\frac{1}{R_0}\;=\;\frac{f}{E\,y}\;=\;\frac{24\,p\,R^2}{E\,t_0^3}$

(24)

$\therefore\;\;\;\frac{1}{R_0}\;=\;\left(\frac{1}{R} \right)\left(1\;-\;\frac{2\,\hat{f}\, R}{E\,t_0} \right)$

(25)

which determines the initial radius when values for $t_0\;amd\;\hat{f}$ are assumed.

Stresses in Bars of Large Initial Curvature.

When the radius of curvature is of the same order as the dimensions of the cross section, it is no longer possible to neglect y in comparison to R and it will be found that the neutral axis does not pass through the centroid. Further the stress is NOT proportional to the distance from the neutral axis

$f\;=\;E\;X\;strain\;=\;E.\=\;\frac{QQ^'}{PQ}$

(26)

$=\;\frac{E\,y\,\delta \theta }{(R_0\;+\;y)\theta }$

(27)

where y is the distance from the neutral axis as before and $R_)$ is the initial d radius of the neutral surface.

For pure bending the Total normal force on the cross section = 0

$\int f\,dA\;=\;\frac{E\,\delta \theta }{\theta }\int \frac{y\,dA}{R_0\,+\,y}\;=\;0$

(28)

$Moment\;of\;resistance\;M\;=\;\int f\,y\,dA$

(29)

$=\;\frac{E\,\delta \theta }{\theta }\int \frac{y^2\,dA}{R_0\,+\,y}\;=\;0\;\;\;from ( )$

(30)

$But\;\;\;\;\int \frac{y^2\,dA}{R_0\,+\,y}\;=\;\int \frac{[y(y\,+\,R_0)\;-\;R_0\,y]}{R_0\,+\,y}\;dA$

(31)

$=\;\int y\,dA\;-\;R_0\,y\int \frac{y\,dA}{r_0\,+\,y}$

(32)

$=\;A\,e\;-\;0\;\;\;\;from\; equation ( )$

(33)

Where e is the distance between the neutral axis and the principle axis which is through the centroid ( e is positive when the neutral axis is on the same side of the centroid as the centre of curvature)

Substituting in equation (30)

$M\;=\;\left(\frac{E\,\delta \theta }{\theta } \right)A\,e$

(34)

$=\;\left[\int \frac{R_0\,+\,y}{y} \right]\;A\,e\;\;\;\;\;from \;equation ( )$

(35)

Rearranging

$f\;=\;\frac{M\,y}{A\,e(R_0\,+\,y)}$

(36)

In this equation y is positive measured outwards, a positive bending moment being one that tends to increase the curvature.

Rectangular Cross-section

from equation (28)

$\int \frac{y\,dA}{R_0\,+\,y}\;=\;0$

(37)

Let z = y - e = the distance from the centroid. Also the mean radius of curvature $R_m\;=\;R_0\;+\;e \;\;\; and\;\;\; dA \;= \;bdz$

$Then\;\;\;\int \frac{z\,+\,e}{R_m\,+\,z}\;bdz\;=\;0$

(38)

$i.e.\;\;\;\int_{-\frac{d}{2}}^{\frac{d}{2}}\frac{R_m\,+\,z\,-(R_m\,-\,e)}{R_m\,+\,z}.\,dz\;=\;0$

(39)

$\therefore\;\;\;\int_{-\frac{d}{2}}^{\frac{d}{2}}dz\,-\,(R_m\,-\,e)\int_{-\frac{d}{2}}^{\frac{d}{2}}\frac{dz}{R_m\,+\,z}\;=\;0$

(40)

$hence\;\;\;d\,-\,(R_m\,-\,e)Ln\frac{R_M\,+\,d/2}{R_m\,-\,d/2}\;=\;0$

(41)

$giving\;\;\;\;e\;=\;R_m\;-\;d\;\div \Ln\frac{R_M\,+\,d/2}{R_m\,-\,d/2}\;=\;0$

(42)

As e is small compared to $R_m$ and d, itis difficult to calculate with sufficient accuracy from this equation and the expansion of the log term into a convenient series is of advantage.

then

$e\;=\;R_m\;-\;\frac{d}{2\left[d/2R_m\,+\,\frac{1}{3}(d /2R_m)^3\,+\,\frac{1}{5}(d/2R_m)^5\,+\,.... \right]}$

(43)

Example 1

More...

Trapezoidal Cross-section.

By Moments

$d_1\;=\;\frac{(B_1\,+\,2B_2)}{(B_1\,+\,B_2)}\left(\frac{D}{3} \right)$

(55)

$d_2\;=\;\frac{(2B_1\,+\,B_2)}{(B_1\,+\,B_2)}\left(\frac{D}{3} \right)$

(56)

By putting $z\;=\;y\;-\;e\;\;\;andR_m\;=\;R_0\;+\;e\;\;\; equation (28)\;becomes$

$\int \frac{z\;+\;e}{R_m\;+\;z}\;dA\;=\;0$

(57)

$i.e.\;\;\;A\;-\;(R_m\;-\;e)\int \frac{dA}{(R_m\;+\;z)}\;=\;0$

(58)

$or\;\;\;\;e\;=\;R_m\;-\;\frac{A}{\int \frac{dA}{(R_m\,+\,z)}}$

(59)

$dA\;=\;b\,.\,dz\;=\;\left(B_2\;+\;\left[\frac{(B_1\,-\,B_2)}{D} \right](d_2\,-\,z) \right)dz$

(60)

$=\;\int_{-d_1}^{d_2} \frac{B_2\,+\,\frac{B_1\,-\,B_2}{D}d_2\;+\;\frac{B_1\.-\.B_2}{D}\times R_m\,-\,\frac{B_1\.-\.B_2}{D}(R_m\,+\,z)}{R_M\;+\;z}\;dz$

(61)

from which

$\int \frac{dA}{R_m\,+\,z}\;=\;\left(B_2\,-\,\left[\frac{B_1\,+\,B_2}{D} \right](R_m\,+\,d_2) \right)Ln\;\frac{R_m\,+\,d_2}{R_m\,-\,d_1}{\;-(B_1\;-\;B_2)$

(62)

and since

$A\;=\;\left(\frac{B_1\,+\,B_2}{2} \right)\times D\;\;\;\;\;"e"\;\;can \;be\;evaluated\;from\;equations\;(59)\;and\;(62)$

(63)

Example 2

More...

Circular Cross section

The analysis follows the same method as was used in the previous section on Trapezoidal cross sections.

$Hence \;\;\;\;\;e\;=\;R_m\;-\;A\;\div \int \frac{dA}{R_m\;+\;z}$

(81)

$and\;\;\;\;\int \frac{dA}{R_m\;+\;z}\;=\;2\int_{-r}^{r}\; \frac{\sqrt{(r^2\,-\,z^2)}}{R_m\,+\,z}\;dy$

(82)

$=\;2\pi \left(R_m\;-\;\sqrt{(R_m^2\;-\;r^2)} \right)$

(83)

$\therefore\;\;\;e\;=\;R_m\;-\;r^2\div 2 \left(R_m\;-\;\sqrt{(R_m^2\;-\;r^2)} \right)$

(84)

To evaluate the above expand

$\sqrt{(R_m^2\;-\;r^2)}\;=\;R_m\left[1\,-\,\frac{1}{2}\frac{r^2}{R_m^2}\,-\,\frac{1}{8}\frac{r^4}{R_m^4}\,-\,.... \right]$

(85)

$=\;\frac{R_m[(r^2/R_m^2)\,+\,\frac{1}{2}(r^4/R_m^4)\,+\,\frac{1}{4}(r^6/16R_m^6)\,+...]}{4\,+\,(r^2/R_m^2)\,+\,\frac{1}{2}(r^4/R_m^4)\,+\,....}$

(86)

$and\;\;\;\;f\;=\;\frac{M\,y}{A\,e\,(R_0\,+\,y)}$

(87)

Deflection of Curved Beams (Direct Method)

If the length $\delta s$ of an initially curved beam is acted upon by a bending moment M it follows from equation (10) that;-

$\frac{M\,\delta s}{E\,I}\;=\;\delta s\left(\frac{1}{R}\;-\;\frac{1}{R_0} \right)$

(88)

But $\frac{\delta s}{R}\;-\;\frac{\delta s}{R_0}$ is the change of angle subtended by $\delta s$ at the centre of curvature and consequently is the angle through which the tangent at one end of the element rotates relative to the tangent at the other end.

i.e.

$\delta \phi \;=\;\frac{M\,\delta s}{E\,I}$

(89)

The diagram shows a loaded bar which is fixed in direction at A and it is required to find the deflection at the other end B

Due to the action of M on $\delta s$ at C only, the length CB is rotated through an angle

$\delta \phi \;=\;\frac{M\;\delta s}{E\,I.}.\;\;\;\;B\;moves\;to\;B',\;where\;BB{'}\;=\;CB\,\delta \phi$

(90)

The vertical deflection of B = BB' cos $\theta$

$=\;CB\; cos\theta \;.\;\delta \phi \;=\;x\,\delta \phi$

(91)

The horizontal deflection of B = BB' sin $\theta$

$=\;y\;\delta \phi$

(92)

Due to the bending of all the elements along AB

The vertical deflection at B $=\;\int x\,\delta \phi \;=\;\int \frac{M\,x}{E\,I}\;ds$

And the horizontal deflection =

$\;\int y\,\delta \phi \;=\;\int \frac{M\,y}{E\,I}\;ds$

(93)

Example 3

More...

Deflection from Strain Energy ( Castigliano's Theorem)

Theorem If U is the total strain energy of any structure due to the application of external loads

$\f$W_1\,W_2\,...at\;O_1\,O_2\,...in\; the \;direction\;O_1X_1\,O_2X_2\;$

(105)

and to the couples $M_1\;M_2$ ...then the deflections at $O_1\;O_2$ ...in the directions $O_1X_1\;O_2X_2$ ... are $\frac{\delta U}{\delta W_1\;}and\frac{\delta U}{\delta W_2}$ and the angular rotations of the couples are $\frac{\delta\,U}{\delta\;M_1},\;frac{\delta\;U}{\delta\;M-2}$ at their applied points.

Proof for concentrated loads

If the displacements ( in the directions of the loads) produced by gradually applied loads $W_1\;W_2\;W_3 ...are\;x_1\;x_2\;x_3\;\;\; then\;$

$U\;=\;\frac{1}{2}W_1x_1\;+\frac{1}{2}\;W_2x_2\;+\;\frac{1}{2}W_3x_3\;+\;...$

(106)

Let $W_1$ alone be increased by $\delta\,W_1$

then

$\delta U\,=\,increase\;in\;external\;work\;done$

(107)

$=\;\left(W_1\;+\;\frac{\delta \,W_1}{2} \right)\delta \,x_1\;+\;W_2\,\delta \,x_2\;+\;W_3\;$

(108)

$\delta \,u\;=\;W_1\delta x_1\;+\;W_2\delta x_2\;+\;W_3\delta x_3$

(109)

where

$\delta x_1\;\delta x _2\;\delta x_3\;are\;increases\;in\;x_1\;x_2\;x_3$

(110)

But if the loads $W_1\;+\;\delta W_1\;W_2\;W_3\f\;were\; applied\; gradually \;from\; zero,\; the\; total\; strain\; energy.\;$

$U\;+\;\delta U\;=\;\frac{1}{2}(W_1\;+\;\delta W_1)\;+\;\frac{1}{2}W_2(x_2\;+\;\delta x_2)\;+\;\frac{1}{2}W_3(x_3\;+\;\delta x_3)$

(111)

Subtracting equation ( 108 ) and neglecting the products of small quantities

$\delta U\;=\;\frac{1}{2}W_1\delta x_1\;+\;\frac{1}{2}\delta W_1\,x_1\;+\;\frac{1}{2}W_2\delta x_2\;+\;\frac{1}{2}W_3\delta x_3\;+...$

(112)

Subtracting equation ( 110 )

$\delta U\;=\;\delta W_1x_1$

(113)

$or\;\;\;\;\;\frac{\delta U}{\delta W_1}\;=\;x_1$

(114)

Similarly for $x_2\;and\;x_3$ and the proof can be extended to incorporate couples.

It is important to stress that U is the total strain energy, expressed in terms of loads and not including statically determinate reactions and the partial derivative with respect to each load in turn ( treating the others as constant) gives the deflection at the load points in the direction of the load.

The following principles should be observed in applying the theorem

1) In finding the deflection of curved beams and similar problems, only strain energy due to bending need normally be taken into account
$(i.e.\;\int \frac{M^2}{2\,E\,I}ds)$
(115)
2) Treat all loads as variables initially carry out the partial differentiation and integration and only putting in numerical values at the final stage.
3) If the deflection is to be found at a point where, or in a direction there is no load, a load may be put in where required and given a value of zero in the final reckoning $\left(i.e.\;x\;=\left[ \frac{\delta U}{\delta W}\ \right]_{W=0} \right)$

Generally it will be found that the strain energy method requires less thought in application than the direct method, it being only necessary to obtain an expression for the bending moment; also there is no difficulty over the question of sign as the strain energy is bound to be positive and deflection is positive in the direction of the load. The only disadvantage occurs when a case such as mentioned in note 3 above has to be dealt with in which case the direct method will probably be shorter.

Example 4

More...