Let's break down **Problem No. 1** step by step: ### **Given Beam:** - Supports at A (pin) and F (roller) - **Loads:** - Point Load: 7 kN at B (2 m from A) - Point Load: 2z kN at D (6 m from A, since D is 2 m from C, and C is 4 m from A) - Distributed Load: 4y kN/m from C to E (length = 4 m) - **Spans:** - AB = 2 m - BC = 2 m - CD = 2 m - DE = 2 m - EF = 2 m --- ## **Step 1: Calculate Reactions at Supports** Let’s denote: - Reaction at A: \( R_A \) - Reaction at F: \( R_F \) **Sum of vertical forces:** \[ R_A + R_F = 7 + 2z + (4y \times 4) \] \[ R_A + R_F = 7 + 2z + 16y \] **Take moments about A:** \[ \sum M_A = 0 \] \[ -7 \times 2 - (2z) \times 6 - (4y \times 4) \times (4 + 2) + R_F \times 12 = 0 \] - 7 kN at 2 m - 2z kN at 6 m - 4y kN/m over 4m, acting at 4 + 2 = 6 m from A (center of distributed load at 4 m from C, which starts at 4 m from A, so center is at 4 + 2 = 6 m from A) **Distributed load:** - Total load = 4y kN/m × 4 m = 16y kN - Acts at midpoint of 4 m span from C (so at 4 m + 2 m = 6 m from A) So, sum of moments about A: \[ -7 \times 2 - 2z \times 6 - 16y \times 6 + R_F \times 12 = 0 \] \[ -14 - 12z - 96y + 12R_F = 0 \] \[ 12R_F = 14 + 12z + 96y \] \[ R_F = \frac{14 + 12z + 96y}{12} \] **Now, sum of vertical forces:** \[ R_A + R_F = 7 + 2z + 16y \] \[ R_A = 7 + 2z + 16y - R_F \] \[ R_A = 7 + 2z + 16y - \frac{14 + 12z + 96y}{12} \] --- ## **Step 2: Shear Force Equations** Break the beam into sections: 1. **AB:** \( 0 < x < 2 \) 2. **BC:** \( 2 < x < 4 \) 3. **CD:** \( 4 < x < 6 \) 4. **DE:** \( 6 < x < 8 \) 5. **EF:** \( 8 < x < 10 \) **Write shear equations for each span (using x measured from A):** ### 1. **Section AB (0 < x < 2):** \[ V(x) = R_A \] ### 2. **Section BC (2 < x < 4):** \[ V(x) = R_A - 7 \] ### 3. **Section CD (4 < x < 6):** \[ V(x) = R_A - 7 - 4y(x-4) \] (Distributed load starts at C) ### 4. **Section DE (6 < x < 8):** \[ V(x) = R_A - 7 - 4y(x-4) - 2z \] (Point load at D, 2z kN at x = 6) ### 5. **Section EF (8 < x < 10):** \[ V(x) = R_A - 7 - 4y(x-4) - 2z \] (Distributed load ends at x = 8) --- ## **Step 3: Moment Equations** Integrate shear equations for moment: ### 1. **Section AB (0 < x < 2):** \[ M(x) = R_A x \] ### 2. **Section BC (2 < x < 4):** \[ M(x) = R_A x - 7(x-2) \] ### 3. **Section CD (4 < x < 6):** \[ M(x) = R_A x - 7(x-2) - 2z(x-6) - 2y(x-4)^2 \] (The moment due to distributed load is \(-2y(x-4)^2\)) ### 4. **Section DE (6 < x < 8):** \[ M(x) = R_A x - 7(x-2) - 2z(x-6) - 2y(x-4)^2 \] ### 5. **Section EF (8 < x < 10):** \[ M(x) = R_A x - 7(x-2) - 2z(x-6) - 2y(4)^2 \] (Distributed load no longer acts beyond E) --- ## **Step 4: Shear and Moment Diagrams** 1. **Plot the shear and moment equations along the length of the beam.** 2. **Indicate discontinuities at points of concentrated loads.** 3. **Show slope changes where distributed loads start and end.** --- ### **Summary Table** | Section | Shear \( V(x) \) | Moment \( M(x) \) | |------------|-----------------------------------------------------|--------------------------------------------------------| | AB (0-2) | \( R_A \) | \( R_A x \) | | BC (2-4) | \( R_A - 7 \) | \( R_A x - 7(x-2) \) | | CD (4-6) | \( R_A - 7 - 4y(x-4) \) | \( R_A x - 7(x-2) - 2y(x-4)^2 \) | | DE (6-8) | \( R_A - 7 - 4y(x-4) - 2z \) | \( R_A x - 7(x-2) - 2z(x-6) - 2y(x-4)^2 \) | | EF (8-10) | \( R_A - 7 - 4y(4) - 2z \) | \( R_A x - 7(x-2) - 2z(x-6) - 2y(4)^2 \) | --- ### **To Draw the Diagrams:** - Start from left support and use the equations to plot shear and moment. - Use calculated reaction values, and substitute x for each segment. - Indicate jumps and slopes as appropriate. --- If you provide values for \( y \) and \( z \), I can plug them in and give you the exact diagrams and equations! If you'd like a step-by-step drawing of the diagrams, let me know!

Given the values: \[ x = 10\, \text{m}, \quad y = 5\, \text{kN/m}, \quad z = 9\, \text{kN} \] Let's proceed step-by-step to compute reactions, shear force, and bending moment, then outline the shear and moment diagrams. --- ### Step 1: Calculate Reactions at Supports Recall: \[ R_A + R_F = 7 + 2z + 16 y \] \[ R_A + R_F = 7 + 2 \times 9 + 16 \times 5 = 7 + 18 + 80 = 105\, \text{kN} \] Moment about A: \[ 12 R_F = 14 + 12z + 96 y \] \[ 12 R_F = 14 + 12 \times 9 + 96 \times 5 = 14 + 108 + 480 = 602 \] \[ R_F = \frac{602}{12} \approx 50.17\, \text{kN} \] Now, reaction at A: \[ R_A = 105 - R_F = 105 - 50.17 \approx 54.83\, \text{kN} \] --- ### **Reactions:** \[ \boxed{ \begin{aligned} R_A & \approx 54.83\, \text{kN} \\ R_F & \approx 50.17\, \text{kN} \end{aligned} } \] --- ### Step 2: Shear Force Equations Using \(x\) from A: | Section | \(x\) Range | Shear \(V(x)\) | |---------|--------------|----------------| | AB | \(0 < x < 2\) | \(V(x) = R_A \approx 54.83\, \text{kN}\) | | BC | \(2 < x < 4\) | \(V(x) = R_A - 7 \approx 54.83 - 7 = 47.83\, \text{kN}\) | | CD | \(4 < x < 6\) | \(V(x) = R_A - 7 - 4 y (x - 4) = 54.83 - 7 - 4 \times 5 \times (x - 4)\) <br> Simplify: \(47.83 - 20 (x - 4)\) | | DE | \(6 < x < 8\) | \(V(x) = V_{CD} - 2z = [47.83 - 20(x - 4)] - 2 \times 9 = 47.83 - 20(x - 4) - 18\) <br> = \(29.83 - 20 (x - 4)\) | | EF | \(8 < x \leq 10\) | Same as DE, since no additional loads: \(V(x) = 29.83 - 20 (x - 4)\) | --- ### Step 3: Shear Force Values at Key Points - At \(x=0\): \(V = 54.83\, \text{kN}\) - At \(x=2\): \(V = 54.83\, \text{kN}\) - At \(x=4\): \(V = 47.83\, \text{kN}\) - At \(x=6\): \(V = 47.83 - 20(6-4) = 47.83 - 40 = 7.83\, \text{kN}\) - At \(x=8\): \(V = 29.83 - 20(8-4) = 29.83 - 80 = -50.17\, \text{kN}\) - At \(x=10\): \(V = -50.17 - 20(10-8) = -50.17 - 40 = -90.17\, \text{kN}\) --- ### Step 4: Moment Equations Using the integrated shear: | Section | \(M(x)\) expression | |---------|---------------------| | AB | \(M(x) = R_A x = 54.83 x\) | | BC | \(M(x) = R_A x - 7(x - 2)\) | | CD | \(M(x) = R_A x - 7(x - 2) - 2 y (x - 4)^2\) <br> \(= 54.83 x - 7(x-2) - 2 \times 5 (x-4)^2\) <br> \(= 54.83 x - 7x + 14 - 10 (x-4)^2\) <br> \(= (54.83 - 7) x + 14 - 10 (x-4)^2\) <br> \(= 47.83 x + 14 - 10 (x-4)^2\) | | DE | Same as CD, minus the effect of \(2z\) load, which contributes to moments at \(x \geq 6\): \[ M(x) = 47.83 x + 14 - 10 (x-4)^2 - 2z (x - 6) \] \[ = 47.83 x + 14 - 10 (x-4)^2 - 2 \times 9 (x - 6) = 47.83 x + 14 - 10 (x-4)^2 - 18 (x - 6) \] At \(x=8\): \[ M(8) = 47.83 \times 8 + 14 - 10 \times (8-4)^2 - 18 \times (8 - 6) \] Calculate: \[ 47.83 \times 8 = 382.64 \] \[ (8-4)^2 = 16 \] So, \[ M(8) = 382.64 + 14 - 10 \times 16 - 18 \times 2 = 396.64 - 160 - 36 = 200.64\, \text{kNm} \] At the free end \(x=10\): \[ M(10) = 47.83 \times 10 + 14 - 10 \times (10-4)^2 - 18 \times (10 - 6) \] \[ = 478.3 + 14 - 10 \times 36 - 18 \times 4 = 492.3 - 360 - 72 = 60.3\, \text{kNm} \] --- ### **Summary of Key Values:** | Point | Shear \(V\) (kN) | Moment \(M\) (kNm) | |--------|------------------|---------------------| | 0 m | 54.83 | 0 | | 2 m | 54.83 | \(54.83 \times 2 = 109.66\) | | 4 m | 47.83 | \(47.83 \times 4 + 14 - 10 \times 0 = 191.32 + 14 = 205.32\) | | 6 m | 7.83 | \(47.83 \times 6 + 14 - 10 \times 4^2 - 18 \times 0 = 287 - 160 = 127\) | | 8 m | -50.17 | \(\approx 200.64\) (from above) | | 10 m | -90.17 | \(\approx 60.3\) | --- ### **Final Notes:** - **Shear diagram:** starts high (~54.83 kN), decreases linearly, crosses zero somewhere between 6 m and 8 m, then becomes negative. - **Moment diagram:** parabolic shape due to distributed load, with maximum around mid-span (near 8 m). --- ### **Summary:** - Reactions: \( R_A \approx 54.83\, \text{kN} \), \( R_F \approx 50.17\, \text{kN} \) - Shear force varies from about +55 kN at the left to about -90 kN at the right end. - Bending moment peaks around 200–300 kNm near mid-span, with values depending on exact positions. --- If you need detailed diagrams, I recommend plotting these equations over the beam length, marking key points and discontinuities. Let me know if you'd like a drawn schematic or further calculations!