Please show all work! ### Problem Statement2) Use superposition and the deflections given in Appendix E to determine the deflection of the beam illustrated.### Diagram ExplanationThe diagram shows a beam fixed at one end and subject to various loads and moments. Here's a detailed breakdown:- **Beam Configuration:** - The beam is horizontally aligned and fixed on the left side. - The fixed end is depicted with a hatched vertical line.- **Forces and Moments:** - **Uniformly Distributed Load (UDL):** A series of downward-pointing arrows represent a uniformly distributed load across the entire length of the beam. - **Concentrated Moment (\( M_0 \)):** At the free end (right side), there is a clockwise moment \( M_0 \).- **Notations:** - \( E \): Modulus of Elasticity of the beam. - \( I \): Moment of Inertia of the beam's cross-section. - \( L \): Length of the beam. - \( w_0 \): Intensity of the uniformly distributed load. - The axis notation indicates horizontal and vertical axes with \( x \) (horizontal) and \( y^+ \) (vertical upwards).This setup is typical for analyzing beam deflection and deformation under load, requiring the principles of superposition to consider combined effects of distributed load and moments.

Given:E, I, L, Mo, woRequired:y ( deflection of the beam )Solution:Deflection of the beam due to load wo alone,ywo = ( downward )Deflection of the beam due to moment Mo alone,yMo = ( upward )Using superposition method,y = ywo - yMoy = - y = - )Deflection of the beam is