Alternatively, since bending always occurs about the N.A., the deflection equation can be written in the form


where /n.a. is the second moment of area about the N.A. and W' is the component of the load perpendicular to the N.A. The value of /n.a. may be found either graphically using Mohr's circle or the momental ellipse, or by calculation using

where au is the angle between the N.A. and the principal U axis.

^ E.J. Hearn, Mechanics of Materials /, Butterworth-Heinemann, 1997.


Example 1.1

A rectangular-section beam 80 mm x 50 mm is arranged as a cantilever 1.3 m long and loaded at its free end with a load of 5 kN inclined at an angle of 30° to the vertical as shown in Fig. 1.13. Determine the position and magnitude of the greatest tensile stress in the section. What will be the vertical deflection at the end? E = 210 GN/m2.

5 kN

5 kN


In the case of symmetrical sections such as this, subjected to skew loading, a solution is obtained by resolving the load into its components parallel to the two major axes and applying the bending theory simultaneously to both axes, i.e.

Ir h

Now the most highly stressed areas of the cantilever will be those at the built-in end where

Mxx = 5000 cos 30° x 1.3 = 5629 Nm Myy = 5000 sin 30° x 1.3 = 3250 Nm The stresses on the short edges AB and DC resulting from bending about XX are then

= 105.5 MN/m tensile on AB and compressive on DC.

The stresses on the long edges AD and BC resulting from bending about YY are

Myy _ 3250 x 25 x 10"3 x 12 777* ~ 80 x 503 x 10-12

97.5 MN/m tensile on BC and compressive on AD.

The maximum tensile stress will therefore occur at point B where the two tensile stresses add, i.e.

The deflection at the free end of the cantilever is then given by

Therefore deflection vertically (i.e. along the YY axis) is

_ (IV cos 30°)L3 _ 5000 x 0.866 x 1.33 x 12 " ~~ 3EIXX ~ 3 x 210 x 109 x 50 x 803 x 10-'2 = 0.0071 =7.1 mm

Example 12

A cantilever of length 1.2 m and of the cross section shown in Fig. 1.14 carries a vertical load of 10 kN at its outer end, the line of action being parallel with the longer leg and arranged to pass through the shear centre of the section (i.e. there is no twisting of the section, see §7.5^). Working from first principles, find the stress set up in the section at points A, B and C, given that the centroid is located as shown. Determine also the angle of inclination of the N.A.

Ixx = 4 x 10""6 m4, I yy = 1.08 x 10"6 m4

Centroid Structural Steel Angle
0 0

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