**For cantilever beam BD**
From Case No. 1 of

beam loading cases, the maximum deflection at the end of cantilever beam due to concentrated force at the free end is given by

$\delta = \dfrac{PL^3}{3EI}$

Thus,

$\delta_B = \dfrac{1400(8^3)}{3EI} - \dfrac{R_B(8^3)}{3EI}$

$\delta_B = \dfrac{716\,800}{3EI} - \dfrac{512R_B}{3EI}$

**For the simple beam AC**

The deflection at distance x from Case No. 7 of different beam loadings is

$EI \, y = \dfrac{Pbx}{6L}(L^2 – x^2 – b^2)$ for $0 \lt x \lt a$

For $x = a$, the deflection equation will become

$EI \, y = \dfrac{Pab}{6L}(L^2 – a^2 – b^2)$

For beam AC; P = R_{B}, a = 8 ft, b = 4 ft, and L = 12 ft.

$\delta_B = \dfrac{R_B(8)(4)}{6(12)EI}(12^2 – 8^2 – 4^2)$

$\delta_B = \dfrac{4R_B}{9EI}(64)$

$\delta_B = \dfrac{256R_B}{9EI}$

**Solving for the contact force, R**_{B}

$\delta_B = \delta_B$

$\dfrac{716\,800}{3EI} - \dfrac{512R_B}{3EI} = \dfrac{256R_B}{9EI}$

$\dfrac{716\,800}{3EI} = \dfrac{1972R_B}{9EI}$

$R_B = 1200 \, \text{ lb}$

**Determining the maximum moment**

The maximum moment on cantilever beam will occur at D

$M_D = 1200(8) – 1400(8)$

$M_D = –1600 \, \text{ lb}\cdot\text{ft}$

The maximum moment on simple beam will occur at point B.

$M_B = \dfrac{Pab}{L} = \dfrac{1200(8)(4)}{12}$

$M_B = 3200 \, \text{ lb}\cdot\text{ft}$

Maximum moment is at point B

$M_{max} = 3200 \, \text{ lb}\cdot\text{ft}$

**Solving for maximum flexural stress**

The bending stress of rectangular beam is given by

$f_b = \dfrac{6M}{bd^2}$

Thus,

$( \, f_b \, )_{max} = \dfrac{6(3200)(12)}{3(8^2)}$

$( \, f_b \, )_{max} = 1200 \, \text{ psi} \,$ *answer*