Bar fixed at both ends: load shares, displacements

Assumed loading: the 50 kN load at B acts towards D (→) and the 100 kN load at C acts towards A (←). Let the axial force in AB be $x$ (tension +ve).

Equilibrium of joints

$$N_{AB} = x,\qquad N_{BC} = x - 50,\qquad N_{CD} = N_{BC} + 100 = x + 50\ \ (\text{kN})$$

Compatibility

A and D are fixed, so the total change in length is zero:

$$\sum \frac{N_iL_i}{A_iE} = 0$$ $$x\,(0.6112+0.2037+0.1132) - 50(0.2037) + 50(0.1132) = 0 \;\Rightarrow\; 0.9281\,x = 4.525$$

so $x = N_{AB} = 4.878$ kN (the common factor $1/E$ cancels).

Loads shared by each part

$$N_{AB} = 4.88\ \text{kN},\quad N_{BC} = -45.12\ \text{kN},\quad N_{CD} = 54.88\ \text{kN}$$

Positive is tension, negative is compression.

Displacements ($E = 200\times10^3$ N/mm²)

$$\begin{aligned} \delta_B &= \frac{N_{AB}L_{AB}}{A_{AB}E} = 0.0149\ \text{mm}\ (\text{towards D if +ve})\\ \delta_C &= \delta_B + \frac{N_{BC}L_{BC}}{A_{BC}E} = -0.0311\ \text{mm} \end{aligned}$$

Check: $\delta_D = \delta_C + N_{CD}L_{CD}/(A_{CD}E) = 0.000000\approx0$.

Answer: $N_{AB} = 4.88$ kN, $N_{BC} = -45.12$ kN, $N_{CD} = 54.88$ kN; $\delta_B = 0.0149$ mm, $\delta_C = -0.0311$ mm (positive towards D).