| 2W [(a² − y²) {2a² (1 + σ) + b²} − z²a² (1 − 2σ)] | . |
| πa³b (1 + σ) (3a² + b²) |
The resultant of these stresses is W, but the amount at the centroid, which is the maximum amount, exceeds the average amount, W/πab, in the ratio
{4a² (1 + σ) + 2b²} / (3a² + b²) (1 + σ).
If σ = ¼, this ratio is 7⁄5 for a circle, nearly 4⁄3 for a flat elliptic bar with the longest diameter vertical, nearly 8⁄5 for a flat elliptic bar with the longest diameter horizontal.
In the same problem the horizontal shearing stress T or Zx at any point on any cross-section is of amount
| − | 4Wyz {a² (1 + σ) + b²σ} | . |
| πa³b (1 + σ) (3a² + b²) |
The resultant of these stresses vanishes; but, taking as before σ = ¼, and putting for the three cases above a = b, a = 10b, b = 10a, we find that the ratio of the maximum of this stress to the average vertical shearing stress has the values 3⁄5, nearly 1⁄15, and nearly 4. Thus the stress T is of considerable importance when the beam is a plank.
As another example we may consider a circular tube of external radius r0 and internal radius r1. Writing P, U, T for Xx, Xy, Zx, we find
| P = − | 4W | (l − x)y, |
| π (r04 − r14) |
| U = | W | [ (3 + 2σ) { r0² + r1² − y² − | r0² r1² | (y² − z²) } − (1 − 2σ) z² ] |
| 2(1 + σ) π (r04 − r14) | (y² + z²)² |