Curved Beams

Moment of resistance is a term in structural engineering. It is found from the moment of inertia and the distance from the outside of the object concerned to its major axis.

A piston ring is a split ring that fits into a groove on the outer diameter of a piston in a reciprocating engine such as an internal combustion engine or steam engine.

Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.

Castigliano'\b{s method} is a method for determining the displacements of a linear-elastic system based on the partial derivatives of the strain energy.

An analysis of stresses and strains in curved beams.

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

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

Then strain,

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

(1)

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

Substituting in equation (#1)

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

(2)

Total normal stress = 0, i.e.

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

(3)

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

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

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

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

(4)

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 og 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)

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

(5)

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

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

(6)

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 $= \displaystyle\frac{24\,R^2}{t_0^3}$ 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 (#5), 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 it's 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

$f = E\;X\;S = E= \frac{QQ^'}{PQ}= \frac{E\,y\,\delta \theta }{(R_0 + y)\theta }$

where $S$ is the strain, $y$ is the distance from the neutral axis as before and $R_0$ is the initial $d$ radius of the neutral surface.

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

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

(7)

Moment of resistance,

$M = \int f\;y\;dA= \frac{E\;\delta \theta }{\theta }\int \frac{y^2\;dA}{R_0 + y} = 0$

(8)

But $\displaystyle\int \displaystyle\frac{y^2\;dA}{R_0 + y} = \displaystyle\int\displaystyle\frac{[y(y + R_0) - R_0\;y]}{R_0 + y}\;dA$ $= \int y\;dA - R_0\;y\int \frac{y\;dA}{r_0 + y}= A\;e - 0$

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

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

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

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

Example 1 [imperial]

Problem

A curved bar, initially unstressed, of square cross section, has $3\;in.$ sides and a mean radius of curvature of $4.5\;in.$

If a bending moment of $30\;tons-in.$ is applied to the bar tending to straighten it, find the stresses at the outer and inner faces.

$R_m = 4.5\;in.$ and $d = 3\;in.$

Workings

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

$\therefore\;\;\;e = 4.5 - \frac{3}{Ln\;2} = 0.172\;in.$

But $R_0 = R_m - e = 4.328\;in.$

$M\;= - 30\;tons.in.$ and $f = \displaystyle\frac{My}{Ae(R_0\;+\;y)}$

At the inside face, $y\;= - (\displaystyle\frac{d}{2}\;-\;e) = -4.328\; in.$

$\therefore\;\;\;f\;=\frac{-\,30\;\times \;(-1.328)}{9\;\times \;0.172(4.328\,-1.328)}$

Thus, $f = 8.6\;tons/sq.in.$ Tension

At the outside face, $y = \displaystyle\frac{d}{2} + e = 1.672\;in.$

$\therefore\;\;\;f = (-30\,\times \;1.672)\;\div\;[9\;\times \;0.172(4.328\;+\;1.672)]$

$= 5.45\;tons/sq.in.$ compression

The actual stress distribution is shown in the diagram.

Solution

$f = 8.6\;tons/sq.in.$ Tension
$f= 5.45\;tons/sq.in.$ Compression

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 (#6) 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, $e = R_m - \displaystyle\frac{A}{\displaystyle\int \displaystyle\frac{dA}{(R_m\,+\,z)}}$

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

(9)

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

From which,

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

(10)

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

$e$ can be evaluated from equations (#8) and (#9).

Example 1 [imperial]

Problem

A crane hook whose horizontal cross-section is trapezoidal, $2\;in.$ wide on the inside and $1\;in.$ wide on the outside by $2\;in.$ thick, carries a vertical load of one ton whose line of action is $2.5\;in.$ from the inside edge of this section. The centre of curvature is $2\;in.$ from the inside edge.

Calculate the maximum tensile and compressive forces set up.

Workings

Referring to the last figure.

$d_1 = \frac{2}{3}\left(\frac{2\,+\,2\,\times \,1}{2\,+\,1} \right)} = \frac{8}{9}\,in.$ $d_2 = \frac{2}{3}\left(\frac{2\,\times \,2\,+ \,1}{2\,+\,1} \right)} = \frac{10}{9}\,in.$ $R_m = 2 + d_1$

From equation (9)

$\int \frac{dA}{R_m\,+\,z} = \left(1\,+\,\left[\frac{2\,-\,1}{2} \right]\left[ \frac{26}{9}\,+\,\frac{10}{9} \right]\right)Ln\frac{26\,+\,10}{26\,-\,8}\;-(2\,-\,1)$ $= 3\,(\ln\,2 )\,-\,1\;=1.0793$ $A = \left(\frac{2\,+\,1}{2} \right)\times 2 = 3in^2$ $\therefore\;\;\;e = \frac{26}{9} - \frac{3}{1.0793} = 0.11 in.$

Direct stress = load / area = $\displaystyle\frac{1}{3}\;ton/sq.$ in. tensile

Bending stress = $\displaystyle\frac{M\;y}{A\;e\;(R_m\;+\;y)}= \displaystyle\frac{M(z\;+\;e)}{A\;e\;(R_m\;+\;z)}$

At the inside edge, $z = -d_1 = 0.89\,in.$

$M\;= - (2.5\,+\,d_1)\;= - 3.39\,ton.in.$ (tending to decrease the curvature)

Bending stress = $\left(\displaystyle\frac{- 3.39(- 0.89 + 0.11)}{3\;\times \;0.11(2.89 - 0.89\0}\right)=4.02\;tons/sq.in.$ in tension

The combined stress = $4.02 + 0.33 = 4.35\;tons/sq.in.$ tensile.

At the outside edge, $z = d_2 = 1.11\;in.$

Bending stress = $\displaystyle\frac{-\;3.39(1.11\;+\;0.11)}{3\;\times 0.11(2.89\;+\,1.11)}\;= - 3.14\,Tons/sq.in.$

Combined stress = $3.14 - 0.33 = 2.81\;tons/sq.in.$ in compression

Circular Cross section

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

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

And, $\displaystyle\int \displaystyle\frac{dA}{R_m + z} = 2\displaystyle\int_{-r}^{r}\;\displaystyle\frac{\sqrt{(r^2 - z^2)}}{R_m + z}\;dy= 2\pi \left(R_m - \sqrt{(R_m^2 - r^2)} \right)$

$\therefore\;\;\;e = R_m - r^2\div 2 \left(R_m - \sqrt{(R_m^2 - r^2)} \right)$ To evaluate the above expand: $\sqrt{(R_m^2 - r^2)} = R_m\left[1\,-\,\frac{1}{2}\frac{r^2}{R_m^2}\,-\,\frac{1}{8}\frac{r^4}{R_m^4}\,-\,.... \right]$ $= \frac{R_m[(\displaystyle\frac{r^2}{R_m^2})\,+\,\frac{1}{2}(\displaystyle\frac{r^4}{R_m^4})\,+\,\frac{1}{4}(\displaystyle\frac{r^6}{16R_m^6})\,+...]}{4\,+\,(\displaystyle\frac{r^2}{R_m^2})\,+\,\frac{1}{2}(\displaystyle\frac{r^4}{R_m^4})\,+\,....}$ And $f = \displaystyle\frac{M\;y}{A\;e\;(R_0 + y)}$

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 $B$ .

Due to the action of $M$ on $\delta s$ at $C$ only, the length $CB$ is rotated through an angle $\delta \phi = \displaystyle\frac{M\;\delta_s}{E\;I}$ . $B$ moves to $B$ ', where $BB' = CB\;\delta \phi$

The vertical deflection of $B = BB'\;cos\theta= CB\;cos\theta\;\delta \phi = x\;\delta \phi$

The horizontal deflection of $B = BB'\;sin\;\theta= y\;\delta \phi$

Due to the bending of all the elements along $AB$ ,

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

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

Example 1 [imperial]

Problem

A steel tube having an outside diameter of $2\;in.$ and a bore of $1.5\;in.$ is bent into a quadrant of $6\;ft.$ radius. One end is rigidly attached to a horizontal base plate to which the tangent at that end is perpendicular.

If the free end supports a load of $100\;lb.$ , determine the vertical and horizontal deflection of the free end.

$E = 30\times 10^6\;lb\,in^{-2}$

Workings

$I = \left(\frac{\pi }{64} \right)\left(2^4 - 1.5^4 \right) = 0.537\,in.^4$

$x = 72\;sin\theta$ and $y = 72(1 - cos\;\theta )$

$M = 100x = 100\,\times \;72\.sin\theta$ $\delta s = 72\,\delta \theta$

Vertical deflection = $\;\displaystyle\int \displaystyle\frac{M\;x}{E\;I}\;ds$

$= \int_{0}^{\frac{\pi }{2}}\frac{100\times 72^3\,sin^2\theta }{30\times 10^6\times 0.537}\;d\theta = 2.32\;\int_{0}^{\frac{\pi }{2}}\frac{1 - cos\,2\theta }{2}\;d\theta = 1.82 in.$

Horizontal deflection = $\displaystyle\int \displaystyle\frac{M\;y}{E\;I}ds$

$= \frac{100\times 72^3}{30\times 10^6\times 0.537}\;\int_{0}^{\frac{\pi }{2}}\frac{1 - cos\,\theta }{2}d\theta$ $= 2.32\left[-\,cos\,\theta + cos\,2\theta \right]{_0^\frac{\pi }{2}} = 1.16\;in.$

Solution

Vertical deflection is $1.82 in.$
Horizontal deflection is $1.16\;in.$

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, $W_1\;W_2\;...$ at $O_1\;O_2\;...$ in the direction $\;O_1X_1\;O_2X_2$ 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 $\displaystyle\frac{\delta U}{\delta W_1}$ and $\displaystyle\frac{\delta U}{\delta W_2}$ and the angular rotations of the couples are $\displaystyle\frac{\delta\;U}{\delta\;M_1}$ , $\displaystyle\;\frac{\delta\;U}{\delta\;M-2}$ 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,

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

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

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

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

(11)

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

Where,

$\delta x_1\;\delta x _2\;\delta x_3$

(12)

are increases in $x_1\;x_2\;x_3$

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

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

Subtracting equation (#10) and neglecting the products of small quantities,

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

Subtracting equation (#11), $\delta U = \delta W_1x_1$ or $\displaystyle\frac{\delta U}{\delta W_1} = x_1$

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

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.

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