Principal stresses at web-flange junction

The point is at a distance $y = 100$ mm from the neutral axis (the junction of web and flange).

Bending stress

$$\sigma = \frac{My}{I} = \frac{60\times10^6\times100}{86.13\times10^6} = 69.7\ \text{MPa (compression at the top, for a sagging moment)}$$

Shear stress

$Q$ of the flange about the NA: $150\times20\times110 = 330\,000\ \text{mm}^3$. Width at this level (web) $b = 20$ mm.

$$\tau = \frac{VQ}{Ib} = \frac{100\times10^3\times330\,000}{86.13\times10^6\times20} = 19.2\ \text{MPa}$$

Principal stresses

Take $\sigma_x = -69.7$ MPa (axial direction of the beam), $\sigma_y = 0$, $\tau_{xy} = 19.2$ MPa.

$$\frac{\sigma_x}{2} = -34.8\ \text{MPa}, \qquad R = \sqrt{34.8^2 + 19.2^2} = 39.75\ \text{MPa}$$ $$\sigma_1 = -34.8 + 39.75 = +4.9\ \text{MPa (tension)}$$ $$\sigma_2 = -34.8 - 39.75 = -74.6\ \text{MPa (compression)}$$

Maximum shear stress $\tau_{max} = R = 39.75\ \text{MPa}$.

Direction: $\tan2\theta = \dfrac{2\tau}{|\sigma_x|} = 0.551$, so $2\theta = 28.8^\circ$ and the compressive principal stress acts at $\theta = 14.4^\circ$ to the beam axis.

Comparison with the extreme fibre

At the extreme fibre ($y = 120$ mm) there is no shear stress and

$$\sigma_{max} = \frac{60\times10^6\times120}{86.13\times10^6} = 83.6\ \text{MPa}$$

The principal stress at the junction ($74.6$ MPa) is smaller than the extreme-fibre stress ($83.6$ MPa), so for this loading the extreme fibre governs. For deep plate girders, or where a large shear force and large moment act together, the junction can govern, so it must be checked.

Answer: $\sigma = 69.7$ MPa, $\tau = 19.2$ MPa; $\sigma_1 = 4.9$ MPa, $\sigma_2 = -74.6$ MPa; $\tau_{max} = 39.8$ MPa; $\sigma_{max,fibre} = 83.6$ MPa.