A steel bar of length L=2 m, with modulus E=200 GPa and constant cross sectional area A=200 mm2 is constrained between two walls at its two ends A and B. A distributed axial load, fx(x)=−p0(1−3x2L), with p0=200 kN/m, is applied to the bar only in the 0≤x≤23L section (no load is applied on right third of the bar).

Find a symbolic expression for the axial force in the bar, N(x), in terms of p0,L, and x for 0≤x≤23L & for 2L3≤x≤L:

For for 2L/3 ≤ x ≤ L:

N(x) = p_0*2*L/27
u(x) = p_0*2*L/(27*E*A)*(x-L)

To find the displacement of the bar under the applied load, we can use the principles of mechanics and solve the differential equation governing the deflection of the bar.

The equation for the deflection of an axially loaded bar can be represented as follows:

d²u/dx² = -fx(x) / (EA)

Where:
- u(x) represents the displacement at position x along the length of the bar.
- fx(x) is the distributed axial load applied on the bar.
- E is the modulus of elasticity of the material.
- A is the cross-sectional area of the bar.

In this case, we have the expression for the distributed axial load:

fx(x) = -p0(1 - 3x²/L)

Substituting this into the differential equation, we have:

d²u/dx² = p0(1 - 3x²/L) / (EA)

To solve this differential equation, we need to integrate it twice. The first integration will give the first derivative of the displacement, and the second integration will give the displacement itself.

Once the displacement equation is obtained, we can substitute the given values for L, E, A, and p0, and calculate the displacement at any point along the length of the bar.