Curved Beams

Introduction

If a beam is originally curved, before a bending moment is applied to it, it is termed as a Curved Beams. A curved beam differs from a straight beam in that the neutral axis, in the unloaded condition, does not pass through the centroid of the cross-section. Rather, it is shifted toward the center of curvature of the beam. As a result, the distribution of the bending stress is non- linear or hyperbolic.

In this section we analyze the stresses on curved beams of different cross sections, including rectangular, trapezoidal, and circular cross sections.

Applications of curved beams include 'C'- Clampers, crane hooks, frames of presses, as well as piston rings, chains, links, and rings etc.

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

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 $R(\theta +\delta \theta ) = R_0\theta$ that $\delta \theta = \left(\displaystyle\frac{R_0-R}{R} \right)\theta$

Then strain,

Neglecting lateral stress, the normal stress, $f = E\times$ strain

Substituting in equation (1)

Total normal stress = 0, i.e.

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

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

$= E\left(\displaystyle\frac{1}{R} - \displaystyle\frac{1}{R_0} \right)\displaystyle\int y^2\;dA$ from equation (3)

Combining equations (2) and (4), $\frac{f}{y} = \frac{M}{I} = E\,\left(\frac{1}{R} - \frac{1}{R_0} \right)$ the strain energy of a short length $\delta ,s$ (measured along the neutral surface) under the action of bending moment $M$ is:

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

From equation (3)

Application to the design of a piston ring

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:

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

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

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

i.e. $\;\;\;\;\;\;\displaystyle\frac{p\,R^2\;d(1 + cos\theta )}{d\,\displaystyle\frac{t^3}{12}}$ = constant $= \displaystyle\frac{24\,p\,R^2}{t_0^3}$ when $\theta = 0$ and $t=t_0$

$\therefore\;\;\;\;\frac{t}{t_0} = \sqrt[3]{\left(\frac{(1 + cos\,\theta )}{2} \right)}$ Which is the required variation of thickness. Using equation (6). The maximum bending stress at any section $= \left(\frac{M}{I} \right)\left(\frac{t}{2} \right) = \left(\frac{6\,p\,R^2}{t^2} \right)\left(1 + cos\theta \right)= \frac{12\,p\,R^2\,t}{t_0^3}$ which has its greatest value when $\theta = 0$ i.e. $\hat{f} = \displaystyle\frac{12\,p\,R^2}{t_0^2}$

From which, $\;\;\;\displaystyle\frac{1}{R} - \displaystyle\frac{1}{R_0} = \displaystyle\frac{f}{E\,y} = \displaystyle\frac{24\,p\,R^2}{E\,t_0^3}$ $\therefore\;\;\;\frac{1}{R_0} = \left(\frac{1}{R} \right)\left(1 - \frac{2\,\hat{f}\, R}{E\,t_0} \right)$ which determines the initial radius when values for $t_0$ and $\hat{f}$ 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

$f = E\;X\;S = E\times \frac{QQ^'}{PQ}= \frac{E\,y\,\delta \theta }{(R_0 + y)\theta }$ where $S$ is the strain,$y$ is the distance from the neutral axis as before and $R_0$ is the initial radius of the neutral surface.

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

Moment of resistance,

But $\displaystyle\int \displaystyle\frac{y^2\;dA}{R_0 + y} = \displaystyle\int\displaystyle\frac{[y(y + R_0) - R_0\;y]}{R_0 + y}\;dA$ $= \int y\;dA - R_0\int \frac{y\;dA}{R_0 + y}= A\;e - 0$

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 (8) $M = \left(\frac{E\;\delta \theta }{\theta } \right)A\;e= \left[\int \frac{R_0 + y}{y} \right]\;A\;e$

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

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 (7), $\displaystyle\int \displaystyle\frac{y\;dA}{R_0 + y} = 0$

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

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

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

Hence, $\;\;\;d - (R_m - e)Ln\displaystyle\frac{R_m\;+\displaystyle\frac{d}{2}}{R_m - \displaystyle\frac{d}{2}} = 0$

Giving, $\;\;\;\;e = R_m - d\;\div Ln\displaystyle\frac{R_m\;+\displaystyle\frac{d}{2}}{R_m\;-\displaystyle\frac{d}{2}}$

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

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

Trapezoidal Cross-section.

By Moments, $d_1 = \frac{(B_1\,+\,2B_2)}{(B_1\,+\,B_2)}\left(\frac{D}{3} \right)$ $d_2 = \frac{(2B_1\,+\,B_2)}{(B_1\,+\,B_2)}\left(\frac{D}{3} \right)$

By putting $z = y - e$ and $R_m = R_0 + e$ equation (7) becomes $\int \frac{z + e}{R_m + z}\;dA = 0$

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

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

From which,

And since, $A = \left(\displaystyle\frac{B_1\,+\,B_2}{2} \right)\times D$

$e$ can be evaluated from equations (9) and (10).

Circular Cross section

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

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 (4) that:

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

But $\displaystyle\frac{\delta s}{R} - \displaystyle\frac{\delta s}{R_0}$ 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. $\delta \phi = \displaystyle\frac{M\;\delta s}{E\;I}$

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 $B$.

Deflection from Strain Energy ( Castigliano's Theorem)

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. $\displaystyle\int \displaystyle\frac{M^2}{2\,E\,I}ds$)

• 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 (i.e. $x\;=\left[ \displaystyle\frac{\delta U}{\delta W} \right]_{W=0}$)

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.