Certainly! Here is everything written **as it is** from the given images, keeping the structure, equations, and formatting intact: --- ## Img 1 ``` soln given F / \ 20kN / \ 40kN / \ D-------E / \ / \ / \ / \ A--B-----C | | Aj Cj | | sign convention + -> tensile - -> compression Step1 - support reaction calculation ∑Mc = 0 => Aj × 8 = 40 × 4 + 20 × 6 Aj = 35 kN (↑↑) ∑Fy = 0 => Aj + Cj = 60 => Cj = 25 kN (↑↑) Step2 - calculation of member forces using method of joints joint A θ = tan⁻¹(2/2) = 45° ∑Fy = 0 => FAD sin45° = 35 => FAD = 49.5 kN (−) ∑Fx = 0 => FAB = 49.5 cos45° => FAB = 35 kN (−) joint C ∑Fy = 0 => FCE sin45° = 25 => FCE = 35.35 kN (−) ∑Fx = 0 => FCB = 35.35 cos45° => FCB = 25 kN (+) joint B ∑Fx = 0 => FBE cos45° − FBD cos45° = 10 —i) ∑Fy = 0 => FBE sin45° = − FBD sin45° => FBE = −FBD from eqn i) FBD = 7.07 kN (−) FBE = 7.07 kN (+) ``` --- ## Img 2 ``` joint F ∑Fx = 0 => FFE cos45° = FFD cos45° => FFE = FFD ∑Fy = 0 => FFD sin45° + FFE sin45° = 40 FFE = 28.28 kN (−) , FFD = 28.28 kN (−) joint E ∑Fx = 0 FED + FFE cos45° + FCE sin45° + FBE cos45° FED + 28.28 cos45° + 35.35 cos45° + 7.07 cos45° FED = 10 kN (−) ``` --- Let me know if you need it in another format, or if you want the solution explained in detail!

Certainly! Here is the content written **as it is** from the given images without using any rendering tools: --- ## Img 1 ``` soln given F / \ 20kN / \ 40kN / \ D-------E / \ / \ / \ / \ A--B-----C | | Aj Cj | | sign convention + -> tensile - -> compression Step1 - support reaction calculation ∑Mc = 0 => Aj × 8 = 40 × 4 + 20 × 6 Aj = 35 kN (↑↑) ∑Fy = 0 => Aj + Cj = 60 => Cj = 25 kN (↑↑) Step2 - calculation of member forces using method of joints joint A θ = tan⁻¹(2/2) = 45° ∑Fy = 0 => FAD sin45° = 35 => FAD = 49.5 kN (−) ∑Fx = 0 => FAB = 49.5 cos45° => FAB = 35 kN (−) joint C ∑Fy = 0 => FCE sin45° = 25 => FCE = 35.35 kN (−) ∑Fx = 0 => FCB = 35.35 cos45° => FCB = 25 kN (+) joint B ∑Fx = 0 => FBE cos45° − FBD cos45° = 10 —i) ∑Fy = 0 => FBE sin45° = − FBD sin45° => FBE = −FBD from eqn i) FBD = 7.07 kN (−) FBE = 7.07 kN (+) ``` --- ## Img 2 ``` joint F ∑Fx = 0 => FFE cos45° = FFD cos45° => FFE = FFD ∑Fy = 0 => FFD sin45° + FFE sin45° = 40 FFE = 28.28 kN (−) , FFD = 28.28 kN (−) joint E ∑Fx = 0 FED + FFE cos45° + FCE sin45° + FBE cos45° FED + 28.28 cos45° + 35.35 cos45° + 7.07 cos45° FED = 10 kN (−) ``` --- This reflects the text from the images exactly as requested. If you have any more specific requests, feel free to ask!

Sure! Here’s the solution to the problem presented in the images, structured step-by-step. --- # Step-by-Step Solution ## Problem Setup ### Given Data - Forces: - \( F_1 = 20 \, \text{kN} \) - \( F_2 = 40 \, \text{kN} \) ### Geometry ``` F / \ 20kN / \ 40kN / \ D-------E / \ / \ / \ / \ A--B-----C | | Aj Cj | | ``` ### Sign Convention - \( + \): Tensile - \( - \): Compression ## Step 1: Support Reaction Calculation Using the equations of equilibrium: ### Moment about Point C \[ \sum M_C = 0 \implies A_j \cdot 8 = 40 \cdot 4 + 20 \cdot 6 \] \[ A_j = 35 \, \text{kN} \, (\text{upward}) \] ### Vertical Forces \[ \sum F_y = 0 \implies A_j + C_j = 60 \] \[ C_j = 25 \, \text{kN} \, (\text{upward}) \] ## Step 2: Calculation of Member Forces Using Method of Joints ### Joint A Using the geometry, angle \( \theta \): \[ \theta = \tan^{-1}\left(\frac{2}{2}\right) = 45^\circ \] #### Vertical Forces \[ \sum F_y = 0 \implies F_{AD} \cdot \sin(45^\circ) = 35 \] \[ F_{AD} = 49.5 \, \text{kN} \, (\text{compression}) \] #### Horizontal Forces \[ \sum F_x = 0 \implies F_{AB} = 49.5 \cdot \cos(45^\circ) \] \[ F_{AB} = 35 \, \text{kN} \, (\text{compression}) \] ### Joint C #### Vertical Forces \[ \sum F_y = 0 \implies F_{CE} \cdot \sin(45^\circ) = 25 \] \[ F_{CE} = 35.35 \, \text{kN} \, (\text{compression}) \] #### Horizontal Forces \[ \sum F_x = 0 \implies F_{CB} = 35.35 \cdot \cos(45^\circ) \] \[ F_{CB} = 25 \, \text{kN} \, (\text{tensile}) \] ### Joint B #### Horizontal Forces \[ \sum F_x = 0 \implies F_{BE} \cdot \cos(45^\circ) - F_{BD} \cdot \cos(45^\circ) = 10 \quad (i) \] #### Vertical Forces \[ \sum F_y = 0 \implies F_{BE} \cdot \sin(45^\circ) = -F_{BD} \cdot \sin(45^\circ) \implies F_{BE} = -F_{BD} \] From equation (i): \[ F_{BD} = 7.07 \, \text{kN} \, (\text{compression}) \] \[ F_{BE} = 7.07 \, \text{kN} \, (\text{tensile}) \] ## Step 3: Joint F ### Forces in Joint F #### Horizontal Forces \[ \sum F_x = 0 \implies F_{FE} \cdot \cos(45^\circ) = F_{FD} \cdot \cos(45^\circ) \implies F_{FE} = F_{FD} \] #### Vertical Forces \[ \sum F_y = 0 \implies F_{FD} \cdot \sin(45^\circ) + F_{FE} \cdot \sin(45^\circ) = 40 \] \[ F_{FE} = 28.28 \, \text{kN} \, (\text{compression}), \, F_{FD} = 28.28 \, \text{kN} \, (\text{compression}) \] ## Step 4: Joint E ### Forces in Joint E \[ \sum F_x = 0 \implies F_{ED} + F_{FE} \cdot \cos(45^\circ) + F_{CE} \cdot \sin(45^\circ) + F_{BE} \cdot \cos(45^\circ) \] \[ F_{ED} + 28.28 \cdot \cos(45^\circ) + 35.35 \cdot \cos(45^\circ) + 7.07 \cdot \cos(45^\circ) = 0 \] \[ F_{ED} = -10 \, \text{kN} \, (\text{compression}) \] --- ## Final Answers - \( A_j = 35 \, \text{kN} \) (upward) - \( C_j = 25 \, \text{kN} \) (upward) - \( F_{AD} = 49.5 \, \text{kN} \) (compression) - \( F_{AB} = 35 \, \text{kN} \) (compression) - \( F_{CE} = 35.35 \, \text{kN} \) (compression) - \( F_{CB} = 25 \, \text{kN} \) (tensile) - \( F_{BD} = 7.07 \, \text{kN} \) (compression) - \( F_{BE} = 7.07 \, \text{kN} \) (tensile) - \( F_{FE} = 28.28 \, \text{kN} \) (compression) - \( F_{FD} = 28.28 \, \text{kN} \) (compression) - \( F_{ED} = -10 \, \text{kN} \) (compression) --- This solution captures the calculations and methods used to find the reactions and forces in the truss structure. If you need further clarification on any step, feel free to ask!

Sure! Let's break down the problem step-by-step to find the dimensions of the rectangular tank and the detention time for the wastewater treatment plant. --- ## Given Data - **Maximum flow rate**: \( Q = 6000 \, \text{m}^3/\text{d} \) - **Particle characteristics**: - Diameter: \( D_1 = 200 \times 10^{-6} \, \text{m} \) - Settling velocity constant: \( k = 0.05 \) - Friction factor: \( f = 0.025 \) - Particle density: \( \rho_p = 1.25 \times 10^3 \, \text{kg/m}^3 \) - **Classifier dimensions**: - Length to Width ratio: \( L/W > 6 \) - **Overflow rate**: At least 4 times the settling velocity. - **Horizontal velocity**: At most one-third of the scour velocity. ## Step 1: Calculate Settling Velocity The settling velocity \( V_s \) for a particle can be calculated using the formula: \[ V_s = k \cdot D_1^2 \cdot (\rho_p - \rho_f) \cdot g \] Where: - \( \rho_f \) is the density of water, approximately \( 1000 \, \text{kg/m}^3 \) - \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \) ### Substituting the Values 1. Calculate \( \rho_p - \rho_f \): \[ \rho_p - \rho_f = 1.25 \times 10^3 \, \text{kg/m}^3 - 1000 \, \text{kg/m}^3 = 250 \, \text{kg/m}^3 \] 2. Calculate \( V_s \): \[ V_s = 0.05 \cdot (200 \times 10^{-6})^2 \cdot (250) \cdot (9.81) \] \[ = 0.05 \cdot 4 \times 10^{-8} \cdot 250 \cdot 9.81 \] \[ = 0.05 \cdot 4 \times 10^{-8} \cdot 2452.5 \] \[ = 0.05 \cdot 9.81 \times 10^{-5} = 4.905 \times 10^{-7} \, \text{m/s} \] ### Settling Velocity So, the settling velocity \( V_s \) is approximately: \[ V_s \approx 4.905 \times 10^{-7} \, \text{m/s} \] ## Step 2: Calculate Overflow Rate Given that the overflow rate is at least 4 times the settling velocity: \[ Q_{overflow} = 4 \cdot V_s \] ### Calculating Overflow Rate \[ Q_{overflow} = 4 \cdot 4.905 \times 10^{-7} \, \text{m/s} = 1.962 \times 10^{-6} \, \text{m/s} \] ## Step 3: Calculate Dimensions of Rectangular Tank ### Flow Rate in m³/s Convert the maximum flow rate from m³/d to m³/s: \[ Q = \frac{6000 \, \text{m}^3/\text{d}}{86400 \, \text{s/d}} \approx 0.0694 \, \text{m}^3/\text{s} \] ### Area Calculation The area \( A \) of the tank can be calculated from the flow rate and the overflow rate: \[ A = \frac{Q}{Q_{overflow}} \] Substituting the values: \[ A = \frac{0.0694}{1.962 \times 10^{-6}} \approx 35323.12 \, \text{m}^2 \] ### Dimensions Let \( W \) be the width. Then, the length \( L \) can be expressed as: \[ L = 6W \] Substituting in the area equation: \[ A = L \cdot W = 6W \cdot W = 6W^2 \] Setting the two equations equal: \[ 6W^2 = 35323.12 \] \[ W^2 = \frac{35323.12}{6} \approx 5887.19 \] \[ W \approx \sqrt{5887.19} \approx 76.7 \, \text{m} \] Then, \[ L = 6W \approx 6 \cdot 76.7 \approx 460.2 \, \text{m} \] ## Step 4: Determine Detention Time The detention time \( t \) can be calculated using the volume of the tank and the flow rate: \[ V = L \cdot W \cdot h \] Where \( h \) is the height of the tank. Assuming an average height \( h \): \[ V = 35323.12 \cdot h \] The detention time \( t \) can be calculated as: \[ t = \frac{V}{Q} \] Substituting gives: \[ t = \frac{35323.12 \cdot h}{0.0694} \] Assuming a typical height \( h \) (e.g., \( 3 \, \text{m} \)): \[ t = \frac{35323.12 \cdot 3}{0.0694} \approx 1525629.36 \, \text{s} \approx 17.65 \, \text{days} \] --- ## Final Answers 1. **Dimensions of the rectangular tank**: - Width \( W \approx 76.7 \, \text{m} \) - Length \( L \approx 460.2 \, \text{m} \) 2. **Detention time**: - \( t \approx 17.65 \, \text{days} \) If you need further clarification or adjustments, feel free to ask!