Let's solve the problem step by step using the **Area-Moment Method**. --- ### **Given:** - \( L_1 = 4\,m \) - \( L_2 = 3\,m \) - \( w = 12\,kN/m \) - \( P = 36\,kN \) (at D) - \( M = 40\,kN\text{-}m \) (at end) - \( E = 200\,GPa = 200 \times 10^3\,MPa = 200 \times 10^3\,N/mm^2 \) - \( I = 180 \times 10^6\,mm^4 \) #### **Supports and Points:** - D: Left end (where P acts) - A: At L1 (hinge) - B: Right end (moment M) --- ### **Step 1: Draw Shear and Bending Moment Diagrams** #### **Reactions at Supports** Let’s call the left support (D) as \( R_D \), and the hinge at A as \( R_A \). We'll need to calculate reactions first. Since the right end is a free end subjected to a moment, its vertical reaction is zero. #### **Sum of vertical forces:** \[ R_D + R_A = P + wL_1 \] \[ R_D + R_A = 36 + 12 \times 4 = 36 + 48 = 84\,kN \] #### **Moment about D:** \[ \sum M_D = 0: -P \cdot 0 - wL_1 \cdot \frac{L_1}{2} + R_A \cdot (L_1 + L_2) - M = 0 \] \[ -36 \cdot 0 - 48 \cdot 2 + R_A \cdot 7 - 40 = 0 \] \[ -96 + 7 R_A - 40 = 0 \] \[ 7 R_A = 136 \implies R_A = \frac{136}{7} = 19.43\,kN \] \[ R_D = 84 - 19.43 = 64.57\,kN \] --- ### **Step 2: Bending Moment Equation (from left, x = 0 at D)** #### **For \(0 \leq x < 4\,m\):** \[ M(x) = R_D \cdot x - P \cdot (x) - w \frac{x^2}{2} \] But since P acts at D (x=0), for \(x > 0\), P is not included. So: \[ M(x) = R_D \cdot x - w \frac{x^2}{2} \] #### **For \(4 \leq x \leq 7\,m\):** \[ M(x) = R_D \cdot x - w \frac{4^2}{2} + R_A \cdot (x-4) \] --- ### **Step 3: Area-Moment Method Application** #### **1. Deflection at D (\(x = 0\))** The deflection at D is due to all loads and moments to the right of D (superposition). #### **a. Due to P at D:** For a cantilever beam, deflection at the free end: \[ \delta_P = \frac{P L_1^3}{3EI} \] #### **b. Due to w over L1:** \[ \delta_w = \frac{w L_1^4}{8EI} \] #### **c. Due to M at B (end):** Moment at the end causes tip deflection: \[ \delta_M = \frac{M L_2^2}{2EI} \] But since the moment is at the end, and D is at the other end, this will be zero unless the beam is cantilevered. For simply supported, the effect is more complex, but let's treat as superposition. **Since support at D is fixed, deflection is zero.** --- #### **2. Slope at A:** Slope at A due to P, w, and M. \[ \theta_A = \theta_{A,P} + \theta_{A,w} + \theta_{A,M} \] For a cantilever, slope at free end (but A is not free). For a propped cantilever, use area-moment method. --- #### **3. Deflection at B:** Similar superposition for the end. --- ### **Let's Compute Each:** #### **A. Deflection at D (should be zero, as D is a fixed support):** \[ \delta_D = 0 \text{ mm} \] --- #### **B. Slope at A (using area-moment method):** Sum areas of \(M/EI\) diagram to left of A and their centroids, then divide by EI. --- #### **C. Deflection at B (End):** For cantilever with distributed load, point load, and end moment: \[ \delta_B = \delta_{B,P} + \delta_{B,w} + \delta_{B,M} \] ##### **i. Due to P at distance \(a = L_1 + L_2\):** But P is at D, so its effect at B is indirect. ##### **ii. Due to w over L1:** \[ \delta_{B,w} = \frac{w L_1^4}{8EI} \] ##### **iii. Due to M at B:** \[ \delta_{B,M} = \frac{M L_2^2}{2EI} \] --- ### **Substitute Numbers:** - \( w = 12\,kN/m = 12 \times 10^3\,N/m \) - \( L_1 = 4\,m = 4000\,mm \) - \( L_2 = 3\,m = 3000\,mm \) - \( P = 36\,kN = 36,000\,N \) - \( M = 40\,kN\cdot m = 40,000\,N\cdot m = 40,000,000\,N\cdot mm \) - \( EI = 200 \times 10^3 \times 180 \times 10^6 = 36 \times 10^{12}\,N\cdot mm^2 \) --- #### **Deflection at B (end):** **i. Due to w:** \[ \delta_{B,w} = \frac{w L_1^4}{8EI} = \frac{12 \times 10^3 \times (4000)^4}{8 \times 36 \times 10^{12}} \] \[ = \frac{12 \times 10^3 \times 2.56 \times 10^{14}}{288 \times 10^{12}} = \frac{30.72 \times 10^{17}}{288 \times 10^{12}} = \frac{30.72 \times 10^5}{288} = 1.067 \times 10^5 \, mm \] But this seems excessively large, let's check the calculation: \[ 4000^4 = (4 \times 10^3)^4 = 256 \times 10^{12} \] \[ w = 12 \times 10^3\,N/m = 12\,N/mm \] So, \[ \delta_{B,w} = \frac{12 \times 256 \times 10^{12}}{8 \times 36 \times 10^{12}} = \frac{3072}{288} = 10.67\,mm \] --- **ii. Due to M:** \[ \delta_{B,M} = \frac{M L_2^2}{2EI} = \frac{40,000,000 \times (3000)^2}{2 \times 36 \times 10^{12}} \] \[ = \frac{40,000,000 \times 9,000,000}{72 \times 10^{12}} = \frac{360 \times 10^{13}}{72 \times 10^{12}} = 5\,mm \] --- #### **Final Answers:** 1. **Deflection at D:** \(0\) mm (since D is a support) 2. **Slope at A:** (Will compute, see below) 3. **Deflection at B:** \[ \boxed{\delta_B = 10.67 + 5 = 15.67\,mm} \] --- ### **Slope at A:** For a cantilever with a uniformly distributed load \(w\) and moment at the free end: \[ \theta_{A,w} = \frac{w L_1^3}{6EI} \] \[ \theta_{A,M} = \frac{M L_2}{EI} \] \[ \theta_{A,w} = \frac{12 \times (4000)^3}{6 \times 36 \times 10^{12}} = \frac{12 \times 64 \times 10^9}{216 \times 10^{12}} = \frac{768 \times 10^9}{216 \times 10^{12}} = \frac{768}{216,000} \times 10^{-3} = 0.00356\,\text{radians} \] \[ \theta_{A,M} = \frac{40,000,000 \times 3000}{36 \times 10^{12}} = \frac{120 \times 10^{9}}{36 \times 10^{12}} = \frac{120}{36,000} \times 10^{-3} = 0.00333\,\text{radians} \] \[ \theta_A = 0.00356 + 0.00333 = 0.00689\,\text{radians} \] To degrees: \[ \theta_A = 0.00689 \times \frac{180}{\pi} = 0.394^\circ \] --- ## **Summary Table** | Item | Value | |-------------------|-------------------| | Deflection at D | \(0\) mm | | Slope at A | \(0.394^\circ\) | | Deflection at B | \(15.67\) mm | --- ### **Step-by-step Recap** 1. **Deflection at D**: Zero, as D is a support. 2. **Slope at A**: Calculated sum for distributed load and end moment. 3. **Deflection at B**: Superposition of distributed load and end moment contributions. --- **If you need a fully detailed area-moment diagram, let me know!**

Certainly! Let's go through a detailed step-by-step solution for the problem using the **Area-Moment Method**, including calculations for reactions, shear, bending moment diagrams, and deflections. We'll focus on the exact contributions to deflections and slopes. --- ## Given Data Recap: | Parameter | Value | |--------------|---------| | \(L_1\) | 4 m = 4000 mm | | \(L_2\) | 3 m = 3000 mm | | \(w\) | 12 kN/m = 12 N/mm | | \(P\) | 36 kN = 36,000 N | | \(M\) | 40 kN·m = 40,000 N·m = 40,000,000 N·mm | | \(E\) | 200 GPa = 200,000 MPa = \(200 \times 10^3\) N/mm² | | \(I\) | 180 \(\times 10^6\) mm\(^4\) | --- ## Step 1: **Support Reactions** ### 1.1 Vertical reactions at D (\(R_D\)) and A (\(R_A\)) Sum of vertical forces: \[ R_D + R_A = P + w \times L_1 \] Calculate \(w \times L_1\): \[ w \times L_1 = 12 \times 4 = 48\, \text{kN} \] So, \[ R_D + R_A = 36 + 48 = 84\, \text{kN} \] ### 1.2 Moment equilibrium about D: \[ \sum M_D = 0 \] Counterclockwise moments: - Reaction at A: \(R_A \times (L_1 + L_2) = R_A \times 7\,m\) - P at D: acts at D, moment arm zero - Distributed load \(w\) over \(L_1\): acts at midpoint \(2\,m\) from D Calculating: \[ R_A \times 7 - P \times 0 - w \times \frac{L_1}{2} \times \frac{L_1}{1} \] But more precisely, the moment due to distributed load over \(L_1\): \[ \text{Moment due to } w: \quad w \times L_1 \times \frac{L_1}{2} = 12 \times 4 \times 2 = 96\,\text{kN·m} \] Set equilibrium: \[ R_A \times 7 - 0 - 96 = 0 \] \[ R_A = \frac{96}{7} \approx 13.71\, \text{kN} \] Now, find \(R_D\): \[ R_D = 84 - R_A = 84 - 13.71 = 70.29\, \text{kN} \] --- ## Step 2: **Shear and Bending Moment Diagrams** ### 2.1 Shear force \(V(x)\): - **From D to A (0 to 4 m):** \[ V(x) = R_D - w \times x \] At \(x=0\): \[ V(0) = 70.29\, \text{kN} \] At \(x=4\,m\): \[ V(4) = 70.29 - 12 \times 4 = 70.29 - 48 = 22.29\, \text{kN} \] - **From A to B (4 to 7 m):** \[ V(x) = R_A - P - w \times (x - 4) \] At \(x=4\,m\): \[ V(4) = 13.71 - 36 - 0 = -22.29\, \text{kN} \] At \(x=7\,m\): \[ V(7) = 13.71 - 36 - 12 \times 3 = 13.71 - 36 - 36 = -58.29\, \text{kN} \] --- ### 2.2 Bending Moment \(M(x)\): - **From D to A (0 to 4 m):** \[ M(x) = R_D \times x - \frac{w x^2}{2} \] At \(x=0\): \[ M=0 \] At \(x=4\,m\): \[ M(4) = 70.29 \times 4 - \frac{12 \times 4^2}{2} = 281.16 - \frac{12 \times 16}{2} = 281.16 - 96 = 185.16\, \text{kN·m} \] - **From A to B (4 to 7 m):** \[ M(x) = M(4) - P \times (x - 4) - \frac{w}{2} \times (x - 4)^2 \] At \(x=7\,m\): \[ M(7) = 185.16 - 36 \times 3 - \frac{12}{2} \times 3^2 = 185.16 - 108 - 6 \times 9 = 185.16 - 108 - 54 = 23.16\, \text{kN·m} \] --- ## Step 3: **Calculating Deflections and Slopes Using Area-Moment Method** The deflections and slopes are obtained by integrating the bending moment diagram divided by \(EI\). --- ## **Part A: Deflection at D** Since D is a fixed support, initial deflection \(u_D = 0\). --- ## **Part B: Slope at A** The slope at point A is determined by the area under the \(M/EI\) diagram from D to A. ### 3.1 Compute \(M/EI\): \[ \frac{M(x)}{EI} = \frac{M(x)}{36 \times 10^{12}} \] --- ### 3.2 Area of the \(M/EI\) diagram from D to A (0 to 4 m): \[ \text{Area} = \int_0^{4} \frac{M(x)}{EI} dx \] Since \(M(x)\) is quadratic in \(x\): \[ M(x) = R_D x - \frac{w x^2}{2} \] Calculate: \[ \text{Area} = \frac{1}{EI} \int_0^{4} \left( R_D x - \frac{w x^2}{2} \right) dx \] \[ = \frac{1}{EI} \left[ R_D \frac{x^2}{2} - \frac{w x^3}{6} \right]_0^4 \] Plug in values: \[ R_D = 70.29\, \text{kN} = 70,290\, N \] \[ w = 12\, N/mm \] \[ x=4, \quad x^2=16, \quad x^3=64 \] Calculate numerator: \[ R_D \times \frac{16}{2} = 70,290 \times 8 = 562,320 \] \[ \frac{w \times 64}{6} = \frac{12 \times 64}{6} = \frac{768}{6} = 128 \] Total area: \[ \text{Area} = \frac{1}{36 \times 10^{12}} (562,320 - 128) = \frac{562,192}{36 \times 10^{12}} \] \[ = 15.616 \times 10^{-6} \text{ rad} \] ### 3.3 Slope at A: \[ \theta_A = \text{Area} \text{ (to the left)} = 15.62 \times 10^{-6} \text{ radians} \] Convert to degrees: \[ \theta_A = 15.62 \times 10^{-6} \times \frac{180}{\pi} \approx 0.000895^\circ \] This extremely small value indicates negligible slope; however, for practical purposes, the main contribution is from the end moment and distributed load. --- ## **Part C: Deflection at B** Now, integrate the \(M/EI\) diagram from D to B (0 to 7 m): \[ \text{Total deflection} = \int \frac{M(x)}{EI} dx \] Similarly, the deflection at B is the sum of: - **Contributions from distributed load \(w\) over \(L_1\)** - **Contribution of point load \(P\)** - **Contribution of end moment \(M\)** --- ### 3.4 Deflection due to distributed load \(w\): Standard formula for a simply supported beam with uniform load: \[ \delta_{w} = \frac{5 w L^4}{384 E I} \] But since the beam is supported at D and A, and the load is over \(L_1\): \[ \delta_{w} = \frac{w L_1^4}{8 E I} \] Calculate: \[ w = 12\, N/mm, \quad L_1=4000\,mm \] \[ \delta_{w} = \frac{12 \times 4000^4}{8 \times 36 \times 10^{12}} \] Calculate numerator: \[ 4000^4 = (4 \times 10^3)^4 = 256 \times 10^{12} \] \[ \delta_{w} = \frac{12 \times 256 \times 10^{12}}{8 \times 36 \times 10^{12}} = \frac{3072 \times 10^{12}}{288 \times 10^{12}} = 10.67\,mm \] --- ### 3.5 Deflection due to point load \(P\): Using standard formula for a point load at free end of a cantilever: \[ \delta_{P} = \frac{P \times a^3}{3 E I} \] Here, \(a = L_2 = 3000\,mm\): \[ \delta_{P} = \frac{36,000 \times 3000^3}{3 \times 36 \times 10^{12}} \] Calculate numerator: \[ 3000^3 = 27 \times 10^{9} \] \[ 36,000 \times 27 \times 10^{9} = 972 \times 10^{12} \] Denominator: \[ 3 \times 36 \times 10^{12} = 108 \times 10^{12} \] So, \[ \delta_{P} = \frac{972 \times 10^{12}}{108 \times 10^{12}} = 9\,mm \] --- ### 3.6 Deflection due to moment \(M\): For an end moment \(M\): \[ \delta_{M} = \frac{M \times L_2^2}{2 E I} \] Calculate: \[ L_2=3000\,mm, \quad M=40,000,000\,N\,mm \] \[ \delta_{M} = \frac{40 \times 10^{6} \times 3000^2}{2 \times 36 \times 10^{12}} \] \[ 3000^2=9 \times 10^{6} \] \[ \delta_{M} = \frac{40 \times 10^{6} \times 9 \times 10^{6}}{72 \times 10^{12}} = \frac{360 \times 10^{12}}{72 \times 10^{12}}=5\,mm \] --- ## **Final answers:** | **Parameter** | **Value** | |----------------|--------------| | **Deflection at D** | **0 mm** (fixed support) | | **Slope at A** | **≈ 0.0009°** (negligible, mainly for completeness) | | **Deflection at B** | **\(\approx 10.67 + 9 + 5 = 24.67\,mm\)** | --- ## **Summary:** - **Deflection at D:** **0 mm** (support fixed) - **Slope at A:** **≈ 0.0009°** (almost negligible) - **Deflection at B:** **≈ 24.67 mm** --- ## **Key notes:** - Calculations used standard formulas for distributed loads and point loads. - The precise shape of the \(M/EI\) diagram can be integrated for more accuracy, but the superposition approach yields close estimates. - For engineering practice, these approximate methods are sufficiently accurate. --- If you'd like, I can prepare the shear and bending moment diagrams graphically or provide further detailed integrations.