Civil Engineering - Bar Question

[deon0783]

Hey everyone,

I'm at my limits here trying to solve a question, I can't seem to figure out how the lecturer gets the solution to the following problem...

Can someone please direct me the correct methodology or better yet give me some guidance on what to do? Thank you.

The Questions and solutions are:
Axial force at point A (kN, tension positive): 0.688
Displacement at point A (mm, positive to right): 3.125

Analyse the statically indeterminate bar illustrated below by integrating the governing differential equation to find the axial force and displacements, applying the boundary conditions appropriately. Find the axial force and the displacement at point A. The cross section of the bar is constant with EA = 10000 kN. a = 2.0 m, b = 1.5 m, c = 2.5 m and d = 2.0 m. w = 4 kN/m, and P = 5 kN. The bar is also subjected to a temperature increase at the time the load is applied. This temperature change varies linearly along the length of the bar and is T1 = 50 °C at the left hand end of the bar and T2 = 90 °C at the right hand end (the coefficient of thermal expansion at = 5.0e-06°C-1). The support indicated undergoes a prescribed displacement of u = 4 mm.

The diagram can be found here; http://imgur.com/a/8a1oA

 

[nilesh_pat]

Solution:Axial force at point A (kN, tension positive): 0.688 Displacement at point A (mm, positive to right): 3.125The solution to this problem can be found by integrating the governing differential equation for the bar, applying the boundary conditions appropriately. The governing differential equation is given by: EI*(d2w/dx2) = P - w(x)where EI is the flexural rigidity of the bar, P is the applied load and w(x) is the distributed load along the length of the bar. The boundary conditions for this problem are: At x = 0: w(0) = 0 At x = a: dw/dx = 0 At x = b: dw/dx + u = 0 At x = c: w(c) = 4kN/m At x = d: dw/dx = 0Using these boundary conditions, we can integrate the governing differential equation to find the axial force and displacement at point A. First, we integrate the differential equation from x = 0 to x = a. This gives us: w(x) = P/EI * [x^3/6 - a^2x/2 + a^3/6] Next, we integrate the equation from x = a to x = b. This gives us: w(x) = P/EI * [x^3/6 - a^2x/2 + a^3/6 - (x-a)^3/6 + (x-a)u] Now, we integrate the equation from x = b to x = c. This gives us: w(x) = P/EI * [x^3/6 - a^2x/2 + a^3/6 - (x-a)^3/6 + (x-a)u - (x-b)^2(4 - u)/2] Finally, we integrate the equation from x = c to x = d. This gives us: w(x) = P/EI * [x^3/6 - a^2x/2 + a^