First of all fores will be calculated
on resolving forces P₁ + P₃ = P₂ + P₄
So P₂ = 45 + 450 ? 130 I.e. P₂ = 365kN
So forces on three sections will be 45kN, 320kN and 130kN respectively.
After that increase in length = PL/AE for all three sections will be calculated.

Compatibility equation insists that change in length of bar must be compatible with boundary conditions. Here W₁ + W₂ = W it is also correct but it is equilibrium equation.

As change in length = PL/AE
P = 10x 1000N, L = 1m, A = ?d²/4 = 1.963 x 10^-5 m², E = 200 x 10⁹ N/m².

In a body loaded under plane stress conditions, number of independent stress components is 3 I.e. two normal components and one shear component.

Coefficient of rmal expansion = 3 x coefficient of volume expansion.

For a system to be in equilibrium force polygon and funicular polygon must close. If force polygon does not close n forces will reduce to a resultant force. If funicular polygon does not close, n re is resultant moment on system.

The stress is load per unit area. After addition of weight in bar due to its loading on lower end force will increase in upper cross-section resulting in maximum stress at built-in upper cross-section.

Aluminium has large value of rmal coefficient among m of value 24 x 10_-6. whereas brass and copper has 19×10^-6 and 17×10.

In a bar of different sections, resultant strain is algebraic sum of individual stresses.

When a beam is in bending layer in direction of bending will be In compression and or will be in tension. One side of neutral axis will be shortened and or will be elongated.