The displacement of end D with respect to end A is [tex]-0.488 \times 10^{-3}\;m[/tex]
Given the following data:
To determine the displacement of end D with respect to end A:
First of all, we would do a sectional cut of the circular rods at point A and then determine the sum of forces acting on it:
[tex]\sum F=0\\\\ 10000+N_1=0\\\\N_1 = -10000\;N[/tex]
Mathematically, the displacement of a circular rod is given by this formula:
[tex]d = \frac{NL}{AE}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]d_1 = \frac{-10000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_1 = \frac{-4350}{0.0003142 \times 68.9 \times 10^9}\\\\d_1 =-0.201 \times 10^{-3}\;meter[/tex]
For section cut BC:
[tex]\sum F=0\\\\ 10000-20000+N_2=0\\\\N_2 = 10000\;N[/tex]
[tex]radius = outer\;radius^2 - inner\;radius^2\\\\radius = 0.04^2 -0.03^2=0.0007\;m[/tex]
[tex]d_2 = \frac{10000 \times 0.435}{\frac{3.142 \times 0.0007^2}{4} \times 68.9 \times 10^9} \\\\d_2 = \frac{4350}{0.00054985 \times 68.9 \times 10^9}\\\\d_2 =0.115 \times 10^{-3}\;meter[/tex]
For section cut D:
[tex]\sum F=0\\\\ -20000-N_3=0\\\\N_3 =- 20000\;N[/tex]
[tex]d_3 = \frac{-20000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_3 = \frac{-8700}{0.0003142 \times 68.9 \times 10^9}\\\\d_3 =-0.402 \times 10^{-3}\;meter[/tex]
For the total displacement:
The total displacement is equal to the displacement of end D with respect to end A.
[tex]d_T = d_1 +d_2+d_3\\\\d_T = -0.201 \times 10^{-3} + 0.115 \times 10^{-3}-0.402 \times 10^{-3}\\\\d_T = -0.488 \times 10^{-3}\;m[/tex]
Read more on displacement here: https://brainly.com/question/25690629
