Let's solve the problem for **Fig. P7.35** as an example, following your instructions. --- # Step 1: **Shear Force and Bending Moment Diagrams** ### **Given Beam:** - Point loads: 10 N (left), 30 N (middle), 20 N (right) - Distances: .5 m between each load ### **A. Find Support Reactions** Let’s assume the beam is simply supported at A and E. Let’s call: - \( R_A \) = Reaction at A - \( R_E \) = Reaction at E **Sum of Vertical Forces:** \[ R_A + R_E = 10 + 30 + 20 = 60\,N \] **Sum of Moments about A:** \[ (30 \times 1.) + (20 \times 1.5) - (R_E \times 1.5) = \] \[ 30 + 30 - 1.5R_E = \] \[ 60 = 1.5R_E \implies R_E = 40\,N \] \[ R_A = 60 - 40 = 20\,N \] --- ### **B. Shear Force Diagram (SFD)** - **Just right of A:** \(+20\,N\) - **Just right of C:** \(20 - 10 = 10\,N\) - **Just right of D:** \(10 - 30 = -20\,N\) - **Just right of E:** \(-20 - 20 = -40\,N\) - **Just right of support E:** \(-40 + 40 = \,N\) (balances out) --- ### **C. Bending Moment Diagram (BMD)** - **At A:** \(M_A = \) - **At C (.5 m):** \(M_C = 20 \times .5 = 10\,N\cdot m\) - **At D (1. m):** \(M_D = 10 \times .5 + 10 = 15\,N\cdot m\) - **At E (1.5 m):** \(-20 \times .5 + 15 = 5\,N\cdot m\) - **At E (end):** \(M_E = \) --- ## **Shear and Moment Diagrams** ``` Shear Force (N) | 20 |--------| | | 10 | |--------| | | -20|-----------------| | |--------| -40|--------------------------| .5 1. 1.5 (m) Bending Moment (N·m) | |----------------| 10 | / | 15 | / | 5 | / | |----------------| .5 1. 1.5 (m) ``` *Note: These are rough sketches for understanding.* --- # Step 2: **Maximum Absolute Values** - **Maximum Shear:** \( |20|, |10|, |{-20}|, |{-40}| \) → **40 N** - **Maximum Moment:** \( |15| \) N·m ### **Final Answers:** - **Maximum Shear Force:** \(\boxed{40\,N}\) - **Maximum Bending Moment:** \(\boxed{15\,N\cdot m}\) --- ## **Summary Table** | Location | Shear (N) | Moment (N·m) | |----------|-----------|--------------| | A | 20 | | | C | 10 | 10 | | D | -20 | 15 | | E | -40 | 5 | --- If you need the solution for **Fig. P7.36**, let me know!