Given :
Width of plate,            b = 120 mm
Thickness of plate,     t = 20 mm
∴ Moment of inertia, I=\frac{b t^3}{12}=\frac{120 \times 20^3}{12}=8 \times 10^4  mm^4
Radius of curvature,  R = 10 m = 10 × 10³ mm
Young’s modulus,       E = 2 × 10^5  N/mm^2
Let                                  σ_{\max} = Maximum stress induced, and
M = Bending moment.

Using equation (7.2),      \frac{\sigma}{y}=\frac{E}{R}

∴                                          \sigma=\frac{E}{R} \times y     …(i)

Equation (i) gives the stress at a distance y from N.A.
Stress will be maximum, when y is maximum. But y will be maximum at the top layer or bottom layer.

∴                        y_{\max }=\frac{t}{2}=\frac{20}{2}=10  mm.

Now equation (i) can be written as

\sigma_{\max }=\frac{E}{R} \times y_{\max } \\ \space \\ =\frac{2 \times 10^5}{10 \times 10^3} \times 10= \pmb{2 0 0  N / m m ^2 .}

From equation (7.4), we have
(7.4):          \frac{M}{I}=\frac{\sigma}{y}=\frac{E}{R}

\frac{M}{I}=\frac{E}{R}

∴                  M=\frac{E}{R} \times I=\frac{2 \times 10^5}{10 \times 10^3} \times 8 \times 10^4 \\ \space \\ \quad \quad \quad \quad =16 \times 10^5  N mm = \pmb{1 . 6  k N m.}