Question

The steel pipe is filled with concrete and subjected to a compressive force of 80 kn. determine the average normal stress in the concrete and the steel due to this loading. the pipe has an outer diameter of 100 mm and an inner diameter of 50 mm. est = 200 gpa, ec = 30 gpa.

Answer 1

The first thing you should do in this case is to realize that you have two different materials.
 First we make a free body diagram.
 Then, we use the definitions of mechanics to find the forces.
 Finally we substitute the values to find the normal stress due to the load.
 Solution attached.

Answer 2

The stress on the steel is 3.88mPa and the stress on the concrete is 1.94mPa

Data;

• Compressive force = 80kn
• outer diameter = 100m = 0.1m
• inner diameter = 50mm = 0.05m

Summation of Forces

$\sum f_y = 0 = Pst + Pc - \delta 0 = 0...equation(i)\\\delta st = \delta c$

We can solve for Pst and Pc

$\frac{Pst * L}{\pi/4 * (0.1^2 - 0.05^2) * 200*10^9} = \frac{Pc * L}{\pi/4 * (0.05)^2 * 30*10^9}\\ 0.075Pst = 1.5Pc\\Pst = 20Pc...equation(ii)$

From equation(i) and (ii)

$Pst + Pc + 0 = 80\\Pst = 20Pc\\20Pc + Pc = 80\\21Pc = 80\\Pc = \frac{80}{21} \\Pc = 3.81kN$

Let's solve for Pst

$Pst = 20Pc\\Pst = 3.81 * 20\\Pst = 76.2kN$

Normal Stress

The normal stress between the concrete and steel can be calculated as

The stress on the steel

$\sigma st = \frac{Pst}{A} \\\sigma st = \frac{76.2*10^3}{\frac{\pi }{4} * (0.1^2 * 0.05^2}\\\sigma st = 3.88mPa$

The stress on the concrete

$\sigma c=\frac{3.81*10^3}{\frac{\pi }{4}* (0.05)^2 } \\\sigma c = 1.94mPa$

The stress on the steel is 3.88mPa and the stress on the concrete is 1.94mPa

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