Tapered bar analysis using two elements

Mesh and element areas

Area varies as $A(x) = 600 - x$ (mm$^2$, $x$ in mm from the fixed end).

 |#### 600 -> 400 -> 200
 |###################>---> 15 kN
 1        2          3

Element stiffness

$$\begin{aligned} k_1 &= \frac{A_1E}{L} = \frac{500(200\times10^3)}{200} = 5\times10^5\ \text{N/mm} \\ k_2 &= \frac{300(200\times10^3)}{200} = 3\times10^5\ \text{N/mm} \end{aligned}$$

Global equations with $u_1 = 0$

$$\begin{bmatrix} 8\times10^5 & -3\times10^5 \\ -3\times10^5 & 3\times10^5 \end{bmatrix}\begin{Bmatrix} u_2 \\ u_3 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 15\,000 \end{Bmatrix}$$

Both elements carry the same force, 15 kN, so:

$$u_2 = \frac{15\,000}{5\times10^5} = 0.03\ \text{mm},\qquad u_3 = u_2 + \frac{15\,000}{3\times10^5} = 0.03 + 0.05 = 0.08\ \text{mm}$$

Stresses

$$\sigma_1 = \frac{15\,000}{500} = 30\ \text{MPa},\qquad \sigma_2 = \frac{15\,000}{300} = 50\ \text{MPa}$$

Exact solution

$$u_{tip} = \int_0^{400}\frac{P\,dx}{E(600 - x)} = \frac{15\,000}{200\times10^3}\ln\frac{600}{200} = 0.075(1.0986) = 0.08240\ \text{mm}$$

The FE tip value is 0.0800 mm, which is 2.9 % below the exact value. The stress at the fixed end, $15\,000/600 = 25$ MPa, and at the tip, $75$ MPa, differ from the element values (30 and 50 MPa); the error reduces with more elements.

Answer: $u_2 = 0.030$ mm, $u_3 = 0.080$ mm; $\sigma_1 = 30$ MPa, $\sigma_2 = 50$ MPa; exact tip displacement 0.0824 mm.