I-section: bending stress 80 mm above base

Section: flanges 300 mm × 30 mm, web 20 mm × 140 mm, overall depth 200 mm. It is symmetrical, so the neutral axis is at mid-depth, 100 mm above the base.

Bending moment

$$M = \frac{wL^2}{8} = \frac{3\times 4^2}{8} = 6\ \text{kNm}$$

Moment of inertia (about neutral axis)

$$I = 2\left[\frac{300\times30^3}{12} + 300\times30\times(85)^2\right] + \frac{20\times140^3}{12}$$ $$I = 2\,(0.675\times10^6 + 65.025\times10^6) + 4.573\times10^6 = 135.97\times10^6\ \text{mm}^4$$

Stress 80 mm above the base

Distance from neutral axis $y = 100 - 80 = 20$ mm, below the NA, so the fibre is in tension (sagging beam).

$$\sigma = \frac{My}{I} = \frac{6\times10^6\times 20}{135.97\times10^6} = 0.88\ \text{N/mm}^2$$

Answer: bending stress at 80 mm above base = 0.88 N/mm² (tensile).