The cantilever beam 8m supports a parabolic load (y=kx^n) and uniformly distributed load (10 kN/m) with total 353 kN/m. The vertex of the parabola is
located 1 meter from the free end. 

what is the deflection of the beam at the free end if EI=350x10^12 N-mm2




To determine the deflection of the cantilever beam at the free end, we will first need to approach this problem systematically, step by step.

### Given:
- **Length of the beam (L)** = 8 m = 8000 mm
- **Uniformly distributed load (UDL)** = 10 kN/m = 10 × 10³ N/m = 10 N/mm
- **Total load** = 353 kN/m
- **Parabolic load function**: \( y = kx^n \), where \( k \) and \( n \) are constants.
- **EI** = \( 350 \times 10^{12} \, \text{N-mm}^2 \)
- **Location of the vertex of the parabolic load** = 1 meter from the free end.
- **Units**: Converting all units to mm and N for consistency.

### Steps to calculate the deflection:
1. **Determine the parabolic load equation**:
   The load is given in the form \( y = kx^n \), and since the vertex is located 1 meter (1000 mm) from the free end, we need to adjust the load equation accordingly.

2. **Superposition principle**:
   The total load consists of:
   - Uniformly distributed load (UDL).
   - Parabolic load (which varies as \( y = kx^n \)).
   For the deflection, we will treat the deflections caused by these two loads separately and then sum them using the principle of superposition.

3. **Deflection due to UDL**:
   For a cantilever beam subjected to a uniformly distributed load \( w \), the deflection at the free end is given by the formula:
   \[
   \delta_{\text{UDL}} = \frac{wL^4}{8EI}
   \]
   where \( w \) is the UDL per unit length, \( L \) is the length of the beam, \( E \) is the Young's modulus, and \( I \) is the moment of inertia.

4. **Deflection due to parabolic load**:
   The parabolic load has a more complex effect on the beam, but it follows the general principle for distributed loads, involving integration over the length of the beam. If we assume the parabolic load varies as \( y = kx^2 \), we will integrate this load distribution along the length of the beam.

Let's compute the contributions from both the UDL and the parabolic load and find the total deflection at the free end.The deflection at the free end of the beam due to the uniformly distributed load (UDL) is approximately \( 14.63 \, \text{mm} \).

Next, let's compute the deflection due to the parabolic load and then sum both contributions to get the total deflection.To calculate the deflection due to the parabolic load, we will proceed under the assumption that the parabolic load has the form \( y = kx^2 \), and the vertex is located 1 meter (1000 mm) from the free end. The deflection contribution from a parabolic load is different from a uniform load and follows a separate beam deflection formula.

We can use the fact that the total load is 353 kN/m to estimate the contribution from the parabolic load and compute the deflection using the principle of superposition.

### Parabolic Load:
The deflection due to a parabolic load \( y = kx^n \) on a cantilever beam, for \( n = 2 \) (which is typically used in this type of problem), can be calculated using the known load distribution and adjusting for the position of the vertex.

Would you like me to calculate this for a parabolic load specifically (with load distribution)? Let me know if this assumption about the parabolic load is accurate.

The cantilever beam 8m supports a parabolic load (y=kx^n) and uniformly distributed load (10 kN/m) with total 353 kN/m. The vertex of the parabola is located 1 meter from the free end. What is the deflection of the beam at the free end if EI=350x10^12 N-mm^2?

Calculate the deflection of a cantilever beam subjected to both a uniformly distributed load and a parabolic load. This example involves structural mechanics concepts like beam deflection, parabolic load distribution, and superposition. The given problem uses a parabolic load with vertex 1 meter from the free end, total load of 353 kN/m, and flexural rigidity EI=350x10^12 N-mm^2.

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