Classic version

The sum of the forces in the x-axis doesn't give us any information, so we have two effective equilibrium equations. The unknowns are the forces in three rods, so the problem is statically indeterminate once, and to solve it we need to use an additional geometric condition (from the displacement plan). Equilibrium equations:
∑𝑀𝐶=0𝑁1⋅4−75⋅2−𝑁3⋅4=0𝑁1=150+4⋅𝑁34𝑁1=37.5+𝑁3∑𝑌=0𝑁1−75+𝑁2+𝑁3=037.5+𝑁3−75+𝑁3=−𝑁2𝑁2=37.5−2⁢𝑁3
Geometric condition Δ⁢𝑙3−Δ⁢𝑙18=Δ⁢𝑙2−Δ⁢𝑙14 Transform and solve the geometric condition Δ⁢𝑙3−Δ⁢𝑙1=2⁢(Δ⁢𝑙2−Δ⁢𝑙1)Δ⁢𝑙3+Δ⁢𝑙1−2⋅Δ⁢𝑙2=0Δ⁢𝑙=𝑁⋅𝑙𝐸⋅𝐴𝑁3⋅4𝐸3⋅𝐴3+𝑁1⋅3𝐸1⋅𝐴1−2⋅𝑁2⋅2𝐸2⋅𝐴2=0 From the problem statement: 𝐴1=𝐴2=2⁢𝐴3𝐸2𝐸1=105 𝐺⁡𝑃⁢𝑎210 𝐺⁡𝑃⁢𝑎=12⇒2⁢𝐸2=𝐸1𝐸1=𝐸3 Substitute these relationships into the previous equation 𝑁3⋅42⁢𝐸2⋅𝐴3+(37.5+𝑁3)⋅32⁢𝐸2⋅2⁢𝐴3−2⋅(37.5−2⁢𝑁3)⋅2𝐸2⋅2⁢𝐴3=0|⋅𝐸2⁢𝐴32⁢𝑁3+34⁢(37.5+𝑁3)−2⁢(37.5−2⁢𝑀3)=02⁢𝑁3+28.125+0.75⁢𝑁3−75+2⁢𝑁34.75⁢𝑁3=46.875𝑁3=9.868 𝑘⁢𝑁 Go back to the relationships from the static equilibrium equations and calculate the forces in the remaining rods 𝑁1=37.5+𝑁3=47.368 𝑘⁢𝑁𝑁2=37.5−2⁢𝑁3=17.764 𝑘⁢𝑁 Solve the strength condition for all rods 𝜎=𝑁𝐴𝜎1=47.368⋅103𝐴1≤120⋅106⇒𝐴1≥3.95⋅10−4 𝑚2𝜎2=17.764⋅103𝐴2≤30⋅106⇒𝐴2≥5.92⋅10−4 𝑚2𝜎3=9.868⋅103𝐴3≤120⋅106⇒𝐴3≥8.22⋅10−5 𝑚2 Finally, assume the cross-sectional area for the rods, keeping in mind the relationship from the problem statement 𝐴1=𝐴2=2⁢𝐴3𝐴1=𝐴2=6⋅10−4𝐴3=12⋅6⋅10−4=3⋅10−4 Calculate the stresses in the rods for the assumed cross-sectional area 𝜎1=47.368⋅1036⋅10−4=78.95 𝑀⁢𝑃⁢𝑎𝜎2=17.764⋅1036⋅10−4=29.60 𝑀⁢𝑃⁢𝑎𝜎3=9.868⋅1033⋅10−4=32.89 𝑀⁢𝑃⁢𝑎