## Unsymmetrical Bending

Similarly, repeating the process with OQ\ perpendicular to OQ gives the result

Further,

= (/„ sin 6 — Iv sin 6) cos 9 - I (/„-/„) sin 26 = Ixy

Thus the construction shown in Fig. 1.8 can be used to determine the second moments of area and the product second moment of area about any set of perpendicular axis at a known orientation to the principal axes.

### 1.7. Momental ellipse

Consider again the general plane surface of Fig. 1.7 having radii of gyration ku and kv about the U and V axes respectively. An ellipse can be constructed on the principal axes with semi-major and semi-minor axes ku and kv, respectively, as shown.

Thus the perpendicular distance between the axis UU and a tangent to the ellipse which is parallel to UU is equal to the radius of gyration of the surface about UU. Similarly, the radius of gyration kv is the perpendicular distance between the tangent to the ellipse which is parallel to the VV axis and the axis itself. Thus if the radius of gyration of the surface is required about any other axis, e.g. the N.A., then it is given by the distance between the N.A. and the tangent AA which is parallel to the N.A. (see Fig. 1.11). Thus

The ellipse is then termed the momental ellipse and is extremely useful in the solution of unsymmetrical bending problems as described in §1.10.

### 1.8. Stress determination

Having determined both the values of the principal second moments of area /„ and /,, and the inclination of the principal axes U and V from the equations listed below,

the stress at any point is found by application of the simple bending theory simultaneously about the principal axes,

l V 'u where Mv and Mu are the moments of the applied loads about the V and U axes, e.g. if loads are applied to produce a bending moment Mx about the X axis (see Fig. 1.14), then

the maximum value of Mx, and hence Mu and M„, for cantilevers such as that shown in Fig. 1.10, being found at the root of the cantilever. The maximum stress due to bending will then occur at this position.

### 1.9. Alternative procedure for stress determination

Consider any unsymmetrical section, represented by Fig. 1.9. The assumption is made initially that the stress at any point on the unsymmetrical section is given by a = Px + Qy

where P and Q are constants; in other words it is assumed that bending takes place about the X and Y axes at the same time, stresses resulting from each effect being proportional to the distance from the respective axis of bending. Now let there be a tensile stress a on the element of area dA. Then force F on the element = o dA

the direction of the force being parallel to the Z axis. The moment of this force about the X axis is then odAy.

Now, by definition,

the latter being termed the product second moment of area (see §1.1):

Mx = PIxy + QIXX Similarly, considering moments about the Y axis,

The sign convention used above for bending moments is the corkscrew rule. A positive moment is the direction in which a corkscrew or screwdriver has to be turned in order to produce motion of a screw in the direction of positive X or as shown in Fig. 1.9. Thus with a knowledge of the applied moments and the second moments of area about any two perpendicular axes, P and Q can be found from eqns. (1.20) and (1.21) and hence the stress at any point (x, y) from eqn. (1.19).

Since stresses resulting from bending are zero on the N.A. the equation of the N.A. is

where aN A. is the inclination of the N.A. to the X axis.

If the unsymmetrical member is drawn to scale and the N.A. is inserted through the centroid of the section at the above angle, the points of maximum stress can be determined quickly by inspection as the points most distant from the N.A., e.g. for the angle section of Fig. 1.10, subjected to the load shown, the maximum tensile stress occurs at R while the maximum compressive stress will arise at either S or T depending on the value of a.

1.10. Alternative procedure using the momenta! ellipse

Consider the unsymmetrical section shown in Fig. 1.11 with principal axes UU and VV. Any moment applied to the section can be resolved into its components about the principal axes and the stress at any point found by application of eqn. (1.18).

For example, if vertical loads only are applied to the section to produce moments about the OX axis, then the components will be M cos 6 about UU and M sin 0 about VV. Then

1U I, the value of 6 having been obtained from eqn. (1.14).

Alternatively, however, the problem may be solved by realising that the N.A. and the plane of the external bending moment are conjugate diameters of an ellipse^ - the momental

^ Conjugate diameters of an ellipse: two diameters of an ellipse are conjugate when each bisects all chords parallel to the other diameter.

Two diameters v = m\x and y = m2X are conjugate diameters of the ellipse + ^ = 1 if mimi =--=-.

ellipse. The actual plane of resultant bending will then be perpendicular to the N.A., the inclination of which, relative to the U axis (au), is obtained by equating the above formula for stress at P to zero,

u Iv k2

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