To solve this problem, we need to determine the correct nodal forces for nodes 1 and 2 in a 2-node bar element subjected to a distributed load $T = cx^2$ over a length $L = 1 \, \text{m}$.

Steps to Solution

1. Shape Functions: The shape functions $N_1$ and $N_2$ for a linear 2-node element are given by: $N_1 = \frac{x_2 - x}{x_2 - x_1}, \quad N_2 = \frac{x - x_1}{x_2 - x_1}$ Since $x_1 = 0$ and $x_2 = L = 1$, these become: $N_1 = 1 - x, \quad N_2 = x$

2. Expression for the Nodal Force Vector: The nodal force vector is defined by the integral: $\{ f^{(e)} \} = \int_0^L \begin{bmatrix} N_1 \\ N_2 \end{bmatrix} T \, dx$ Substituting $T = cx^2$: $\{ f^{(e)} \} = \int_0^1 \begin{bmatrix} 1 - x \\ x \end{bmatrix} cx^2 \, dx = c \int_0^1 \begin{bmatrix} (1 - x)x^2 \\ x^3 \end{bmatrix} dx$

3. Evaluating Each Component:

   • For Node 1: $f_1 = c \int_0^1 (1 - x)x^2 \, dx$ Expanding $(1 - x)x^2$: $f_1 = c \int_0^1 (x^2 - x^3) \, dx = c \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = c \left( \frac{1}{3} - \frac{1}{4} \right) = c \cdot \frac{1}{12} = \frac{c}{12}$

   • For Node 2: $f_2 = c \int_0^1 x^3 \, dx = c \left[ \frac{x^4}{4} \right]_0^1 = c \cdot \frac{1}{4} = \frac{c}{4}$

4. Result: The nodal forces for nodes 1 and 2, respectively, are: $f_1 = \frac{c}{12} \quad \text{and} \quad f_2 = \frac{c}{4}$ Both forces act towards the right.

Answer

The correct answer is:

• Option 1: $\frac{c}{12}, \frac{c}{4}$ both acting towards right.

Would you like further details on any part of the solution?

Related Questions

1. How would the solution change if the distributed load were linear instead of quadratic?
2. What would happen to the nodal forces if the length $L$ were increased?
3. How would you apply this method to a 3-node element?
4. Can you derive the expression for $N_1$ and $N_2$ for a general element length $L$?
5. What if $c$ were negative? How would the direction of forces change?

Tip

For distributed loads, breaking down the load into components and applying shape functions systematically is key to finding nodal forces.