Original Loading

$M = 50P ~ \text{N}\cdot\text{mm}$

Equivalent Loading

**Top fiber (σ = 80 MPa compression)**

$\sigma_{top} = -\sigma_a - \sigma_f$

$\sigma_{top} = -\dfrac{P}{A} - \dfrac{6M}{bd^2}$

$-80 = -\dfrac{P}{40(200)} - \dfrac{6(50P)}{40(200d^2)}$

$-80 = -\dfrac{P}{3,200}$

$P = 256,000 ~ \text{N} = 256 ~ \text{kN}$

**Check the bottom fiber**

$\sigma_{bottom} = -\sigma_a + \sigma_f$

$\sigma_{bottom} = -\dfrac{P}{A} + \dfrac{6M}{bd^2}$

$\sigma_{bottom} = -\dfrac{256(1000)}{40(200)} + \dfrac{6(50 \times 256)(1000)}{40(200^2)}$

$\sigma_{bottom} = 16 ~ \text{MPa tension} \lt 40 ~ \text{MPa}$ (*okay*)

Thus,

$P = 256 ~ \text{kN}$ *answer*