The image contains a structural engineering problem about a beam subjected to various loads and moments. Here's the transcription suitable for an educational website:---**Beam Analysis Problem****Beam Setup:**- A beam is supported at points A and B.- The beam is 25 meters long, divided into segments of 5 meters each.- There are specific loads and a moment applied to the beam: - A couple moment of 750 kN·m is applied at point C. - A point load of 300 kN is located between points D and E. - A uniformly distributed load of 15 kN/m extends from point E to point F.**Objective:**1. **Determine the Shear Function V(x):** - Analyze the shear force along the beam over the following domains: - AC (0 ≤ x ≤ 5 m) - CD (5 m ≤ x ≤ 10 m) - DE (10 m ≤ x ≤ 15 m) - EB (15 m ≤ x ≤ 25 m) - Draw the shear force diagram for the entire beam. *Note: Please use the reference axis x as indicated in the figure.*2. **Determine the Moment Function M(x):** - Analyze the bending moment along the entire beam for each x domain. - Draw the bending moment diagram for the entire beam.---This description specifies the problem requirements clearly and provides a structured approach for analyzing the shear and moment functions.

Consider the free body diagram of the beam. Consider the summation of forces in the horizontal direction. HB=0 Consider the moment equilibrium about point B. RA25+750=300×15+15×10×10225RA=4500RA=450025=180 kN Consider the summation of forces in the vertical direction. RA+RB=300+15×10RB=450-180=270 kNConsider a section between A and B at a distance x from end A. The shear force at the section is given by, Vx=RA The shear force at point A is, VA=RA=180 kN The shear force at point C is, VC=RA=180 kNConsider a section between C and D at a distance x from end A. The shear force at the section is given by, Vx2=RA=180 kN The shear force at point C is, VC=180 kN The shear force just to the left of point D is, VDL=180 kNConsider a section between D and E at a distance x from end A. The shear force at the section is given by, Vx3=RA-300=180-300=-120 kN The shear for just right of point D is, VDR=-120 kN The shear force just to the left of point E is, VEL=-120 kNConsider a section between E and B at a distance x from end A. The shear force at the section is given by, Vx4=RA-300-15x-15=180-300-15x-15=-120-15x-15 The shear force at point E is, VE=-120-1515-15=-120 kN The shear force at point B is, VB=-120-1525-15=-270 kN The shear force diagram is shown below.