Let's solve for the **deflection and slope** at points \( C \) and \( D \) for the given beam using the **method of superposition**, given: - \( w = 7.50 \) kip/ft (uniform load over \( AB \)) - \( P = 6.50 \) kip (concentrated load at \( C \)) - \( M = 5.00 \) kip·ft (moment at \( C \)) - \( a = 6.00 \) ft (length \( AB \) and \( BC \)) - \( b = 2.00 \) ft (length \( CD \)) - \( EI = 51500 \) kip·ft\(^2\) ## **Step 1: System Description** Beam \( ABCD \) is cantilevered at \( A \) with: - Uniform load \( w \) over \( AB \) - Point load \( P \) and moment \( M \) at \( C \) - Lengths as marked We need deflection (\( \delta \)) and slope (\( \theta \)) at points \( C \) and \( D \). --- ## **Step 2: Superposition Principle** The total slope and deflection at any point is **the sum of the effects** of each load acting independently. \[ \theta_{\text{total}} = \theta_w + \theta_P + \theta_M \] \[ \delta_{\text{total}} = \delta_w + \delta_P + \delta_M \] --- ## **Step 3: Formulas for Cantilever Beams** ### **A. Uniform Load (\( w \)) Over Length \( L \)** - Slope at free end: \[ \theta_{w, \text{end}} = -\frac{w L^3}{6EI} \] - Deflection at free end: \[ \delta_{w, \text{end}} = -\frac{w L^4}{8EI} \] **For a point at distance \( x \) from fixed end:** - Slope: \[ \theta_w(x) = -\frac{w}{6EI} x^2 (3L-x) \] - Deflection: \[ \delta_w(x) = -\frac{w}{24EI} x^2 (6L^2 - 4Lx + x^2) \] ### **B. Point Load (\( P \)) at Distance \( a \)** - Slope at load: \[ \theta_{P,a} = -\frac{P a^2}{2EI} \] - Deflection at load: \[ \delta_{P,a} = -\frac{P a^3}{3EI} \] - For a point a distance \( x \) from fixed end, for \( x \geq a \) (beyond the point load): \[ \theta_{P}(x) = -\frac{P a^2}{2EI} \] \[ \delta_P(x) = -\frac{P a^2}{6EI}(3x - a) \] ### **C. Moment (\( M \)) at Distance \( a \)** - Slope at moment: \[ \theta_{M,a} = -\frac{M a}{EI} \] - Deflection at moment: \[ \delta_{M,a} = -\frac{M a^2}{2EI} \] - For \( x \geq a \): \[ \theta_{M}(x) = -\frac{M a}{EI} \] \[ \delta_M(x) = -\frac{M a}{2EI}(2x - a) \] --- ## **Step 4: Apply to Given Beam** ### **At Point \( C \) (\( x = 2a = 12\,\text{ft} \))** #### **A. Due to \( w \), over \( \leq x \leq a \), at \( C \), x = 12 ft** But uniform load only over \( \leq x \leq a = 6\,\text{ft} \): - **Slope at \( x=12\,\text{ft} \) due to \( w \) (use value at \( x=a \)):** \[ \theta_{w,C} = -\frac{w a^2}{2EI} \] - **Deflection at \( x=12\,\text{ft} \) due to \( w \) (use value at \( x=a \)):** \[ \delta_{w,C} = -\frac{w a^3}{3EI} \] #### **B. Due to \( P \) at \( x=12\,\text{ft} \)** - **Slope at \( C \) due to \( P \):** \[ \theta_{P,C} = -\frac{P a^2}{2EI} \] - **Deflection at \( C \) due to \( P \):** \[ \delta_{P,C} = -\frac{P a^3}{3EI} \] #### **C. Due to \( M \) at \( x=12\,\text{ft} \)** - **Slope at \( C \) due to \( M \):** \[ \theta_{M,C} = -\frac{M a}{EI} \] - **Deflection at \( C \) due to \( M \):** \[ \delta_{M,C} = -\frac{M a^2}{2EI} \] --- ### **At Point \( D \) (\( x = 2a + b = 14\,\text{ft} \))** #### **A. Due to \( w \), at \( x=14\,\text{ft} \)** For cantilever, the effect past the loaded length is same as at the loaded end (\( x=a \)): - **Slope at \( D \) due to \( w \):** \[ \theta_{w,D} = -\frac{w a^2}{2EI} \] - **Deflection at \( D \) due to \( w \):** \[ \delta_{w,D} = -\frac{w a^2}{6EI} (3x - a),\ \text{for}\ x \geq a \] Here, \( x = 14\,\text{ft} \), \( a = 6\,\text{ft} \): \[ \delta_{w,D} = -\frac{w a^2}{6EI} (3 \times 14 - 6) \] #### **B. Due to \( P \) at \( x=12\,\text{ft} \)** For \( x > a \): - **Slope at \( D \) due to \( P \):** \[ \theta_{P,D} = -\frac{P a^2}{2EI} \] - **Deflection at \( D \) due to \( P \):** \[ \delta_{P,D} = -\frac{P a^2}{6EI}(3x - a) \] \( a = 12 \), \( x = 14 \) #### **C. Due to \( M \) at \( x=12\,\text{ft} \)** For \( x > a \): - **Slope at \( D \) due to \( M \):** \[ \theta_{M,D} = -\frac{M a}{EI} \] - **Deflection at \( D \) due to \( M \):** \[ \delta_{M,D} = -\frac{M a}{2EI}(2x - a) \] \( a = 12 \), \( x = 14 \) --- ## **Step 5: Substitute Values** ### **At Point C: \( x = 12 \) ft, \( a = 6 \) ft** #### **A. Due to \( w \):** \[ \theta_{w,C} = -\frac{7.5 \times 6^2}{2 \times 51500} = -\frac{7.5 \times 36}{103000} = -\frac{270}{103000} \approx -.00262\ \text{rad} \] \[ \delta_{w,C} = -\frac{7.5 \times 6^3}{3 \times 51500} = -\frac{7.5 \times 216}{154500} = -\frac{162}{154500} \approx -.0105\ \text{ft} \] #### **B. Due to \( P \):** \[ \theta_{P,C} = -\frac{6.5 \times 6^2}{2 \times 51500} = -\frac{6.5 \times 36}{103000} = -\frac{234}{103000} \approx -.00227\ \text{rad} \] \[ \delta_{P,C} = -\frac{6.5 \times 6^3}{3 \times 51500} = -\frac{6.5 \times 216}{154500} = -\frac{1404}{154500} \approx -.00909\ \text{ft} \] #### **C. Due to \( M \):** \[ \theta_{M,C} = -\frac{5. \times 6}{51500} = -\frac{30}{51500} \approx -.000583\ \text{rad} \] \[ \delta_{M,C} = -\frac{5. \times 6^2}{2 \times 51500} = -\frac{5. \times 36}{103000} = -\frac{180}{103000} \approx -.00175\ \text{ft} \] #### **Total at C:** \[ \boxed{\theta_C = -.00262 - .00227 - .000583 = -.00547\ \text{rad}} \] \[ \boxed{\delta_C = -.0105 - .00909 - .00175 = -.0213\ \text{ft}} \] --- ### **At Point D: \( x = 14 \) ft, \( a = 12 \) ft, \( b = 2 \) ft** #### **A. Due to \( w \):** Slope (same as at C): \[ \theta_{w,D} = -\frac{7.5 \times 6^2}{2 \times 51500} \approx -.00262\ \text{rad} \] Deflection: \[ \delta_{w,D} = -\frac{7.5 \times 6^2}{6 \times 51500}(3 \times 14 - 6) = -\frac{7.5 \times 36}{309000}(42-6) = -\frac{270}{309000}\times 36 = -.000874 \times 36 \approx -.0315\ \text{ft} \] #### **B. Due to \( P \) at \( x=12 \) ft:** Slope: \[ \theta_{P,D} = -\frac{6.5 \times 12^2}{2 \times 51500} = -\frac{6.5 \times 144}{103000} = -\frac{936}{103000} \approx -.00909\ \text{rad} \] Deflection: \[ \delta_{P,D} = -\frac{6.5 \times 12^2}{6 \times 51500}(3 \times 14 - 12) = -\frac{6.5 \times 144}{309000} \times 30 = -\frac{936}{309000} \times 30 = -.00303 \times 30 \approx -.0909\ \text{ft} \] #### **C. Due to \( M \) at \( x=12 \) ft:** Slope: \[ \theta_{M,D} = -\frac{5. \times 12}{51500} = -\frac{60}{51500} \approx -.00117\ \text{rad} \] Deflection: \[ \delta_{M,D} = -\frac{5. \times 12}{2 \times 51500}(2 \times 14 - 12) = -\frac{60}{103000}(16) = -.000583 \times 16 = -.00933\ \text{ft} \] #### **Total at D:** \[ \boxed{\theta_D = -.00262 - .00909 - .00117 = -.0129\ \text{rad}} \] \[ \boxed{\delta_D = -.0315 - .0909 - .00933 = -.1317\ \text{ft}} \] --- ## **Final Answers** ### **At Point C:** \[ \boxed{ \begin{aligned} \theta_C &= -.00547\ \text{rad} \\ \delta_C &= -.

# Beam Deflection and Slope Calculation ## Problem Overview We are tasked with determining the deflection and slope at two positions along a cantilever beam (points \( C \) and \( D \)) subjected to a uniform load, a point load, and a moment using the method of superposition. ### Given Data - Uniform load, \( w = 7.50 \) kip/ft - Point load, \( P = 6.50 \) kip - Moment, \( M = 5.00 \) kip-ft - Distances: \( a = 6.00 \) ft, \( b = 2.00 \) ft - Flexural rigidity, \( EI = 51500 \) kip-ft\(^2\) ## Step 1: System Description - Beam \( ABCD \) is cantilevered at \( A \). - Loads are applied at points \( B \) and \( C \). ## Step 2: Superposition Principle The total deflection (\( \delta \)) and slope (\( \theta \)) at any point is the sum of the contributions from each individual load: \[ \theta_{\text{total}} = \theta_w + \theta_P + \theta_M \] \[ \delta_{\text{total}} = \delta_w + \delta_P + \delta_M \] ## Step 3: Formulas for Cantilever Beams ### A. Uniform Load \( w \) - Slope at the free end: \[ \theta_{w,\text{end}} = -\frac{w L^3}{6EI} \] - Deflection at the free end: \[ \delta_{w,\text{end}} = -\frac{w L^4}{8EI} \] - At distance \( x \) from the fixed end: \[ \theta_w(x) = -\frac{w}{6EI} x^2 (3L - x) \] \[ \delta_w(x) = -\frac{w}{24EI} x^2 (6L^2 - 4Lx + x^2) \] ### B. Point Load \( P \) - Slope at the load: \[ \theta_{P,a} = -\frac{P a^2}{2EI} \] - Deflection at the load: \[ \delta_{P,a} = -\frac{P a^3}{3EI} \] - For \( x \geq a \): \[ \theta_P(x) = -\frac{P a^2}{2EI} \] \[ \delta_P(x) = -\frac{P a^2}{6EI}(3x - a) \] ### C. Moment \( M \) - Slope at the moment: \[ \theta_{M,a} = -\frac{M a}{EI} \] - Deflection at the moment: \[ \delta_{M,a} = -\frac{M a^2}{2EI} \] - For \( x \geq a \): \[ \theta_M(x) = -\frac{M a}{EI} \] \[ \delta_M(x) = -\frac{M a}{2EI}(2x - a) \] ## Step 4: Calculate at Points C and D ### At Point C (\( x = 12 \) ft) - For \( w \): \[ \theta_{w,C} = -\frac{7.5 \cdot 6^2}{2 \cdot 51500} = -0.00262\ \text{rad} \] \[ \delta_{w,C} = -\frac{7.5 \cdot 6^3}{3 \cdot 51500} = -0.0105\ \text{ft} \] - For \( P \): \[ \theta_{P,C} = -\frac{6.5 \cdot 6^2}{2 \cdot 51500} = -0.00227\ \text{rad} \] \[ \delta_{P,C} = -\frac{6.5 \cdot 6^3}{3 \cdot 51500} = -0.00909\ \text{ft} \] - For \( M \): \[ \theta_{M,C} = -\frac{5 \cdot 6}{51500} = -0.000583\ \text{rad} \] \[ \delta_{M,C} = -\frac{5 \cdot 6^2}{2 \cdot 51500} = -0.00175\ \text{ft} \] - **Total at C:** \[ \theta_C = -0.00262 - 0.00227 - 0.000583 = -0.00547\ \text{rad} \] \[ \delta_C = -0.0105 - 0.00909 - 0.00175 = -0.0213\ \text{ft} \] ### At Point D (\( x = 14 \) ft) - For \( w \): \[ \theta_{w,D} = -0.00262\ \text{rad} \] \[ \delta_{w,D} = -\frac{7.5 \cdot 6^2}{6 \cdot 51500}(3 \cdot 14 - 6) \approx -0.0315\ \text{ft} \] - For \( P \): \[ \theta_{P,D} = -\frac{6.5 \cdot 12^2}{2 \cdot 51500} = -0.00909\ \text{rad} \] \[ \delta_{P,D} = -\frac{6.5 \cdot 12^2}{6 \cdot 51500}(3 \cdot 14 - 12) \approx -0.0909\ \text{ft} \] - For \( M \): \[ \theta_{M,D} = -\frac{5 \cdot 12}{51500} = -0.00117\ \text{rad} \] \[ \delta_{M,D} = -\frac{5 \cdot 12}{2 \cdot 51500}(2 \cdot 14 - 12) \approx -0.00933\ \text{ft} \] - **Total at D:** \[ \theta_D = -0.00262 - 0.00909 - 0.00117 = -0.0129\ \text{rad} \] \[ \delta_D = -0.0315 - 0.0909 - 0.00933 = -0.1317\ \text{ft} \] ## Final Answers ### At Point C: - Slope: \( \theta_C = \boxed{-0.00547\ \text{rad}} \) - Deflection: \( \delta_C = \boxed{-0.0213\ \text{ft}} \) ### At Point D: - Slope: \( \theta_D = \boxed{-0.0129\ \text{rad}} \) - Deflection: \( \delta_D = \boxed{-0.1317\ \text{ft}} \)