Certainly! Here is the ** format** of handwritten notes using LaTeX-style rendering and structured for clarity: # Structural Analysis --- ## Given Beam: - Uniform distributed load (UDL): \( 3 \text{ k/ft} \) on AB, \( 2 \text{ k/ft} \) on BC - Lengths: \( AB = 10' \), \( BC = 10' \) --- ## Degree of Indeterminacy \[ D_{10} = r - 3 \] \[ r = (3 + 1 + 1) = 5 \] \[ D_{10} = 5 - 3 = 2 \] --- ## Moment Distribution Diagram (Mo-diagram) ### For AB (A to B): \[ M_{B} = \frac{3x^2}{2} \] ### For BC (B to C): \[ M_{C} = \frac{2x^2}{2} \] \[ M_ = -2 \times 10(x-5) - 3(x-10)^2/2 \] \[ = -20(x-5) - \frac{3}{2}(x-10)^2 \] --- ## Moment Diagram for 1 kN Load \[ M_{11} = -x \] \[ M_{12} = -1(x-10) \] --- ## Tabular Summary | Member | Origin | Limit | \( M_ \) | \( M_1 \) | \( M_2 \) | |--------|--------|---------|---------------------------|-------------|---------------| | BC | C | - 10 | \(-x^2\) | \(-x\) | | | AB | C | 10 - 20 | \(-20(x-5) - \frac{3}{2}(x-10)^2\) | \(-x\) | \(10 + x\) | --- # Calculation of Deflections (\( \Delta \)) \[ \Delta_{10} = \frac{1}{EI} \int m_1 m_ \, dx \] \[ = \frac{1}{EI} \left[ \int_^{10} [(-x^2)(-x)] dx + \int_{10}^{20} [-20(x-5) - 1.5(x-10)^2](-x) dx \right] \] \[ = \frac{1}{EI} \left[ \int_^{10} x^3 dx + \int_{10}^{20} [-20x + 100 - 1.5(x^2 - 20x + 100)](-x) dx \right] \] \[ = \frac{1}{EI} \left[ \frac{x^4}{4} \Big|_^{10} + \int_{10}^{20} [1.5x^3 - 30x^2 + 130x + 100x ] dx \right] \] \[ = \frac{1}{EI} \left[ 250 + \left( \frac{1.5x^4}{4} - 30x^3 + 130x^2 + 100x \right) \Bigg|_{10}^{20} \right] \] \[ = \frac{1}{EI} \left[ 250 + (120000 - 24000 + 520 + 200) - (15000 - 300 + 130 + 100) \right] \] \[ \Delta_{10} = \frac{42916.67}{EI} \] --- ## Second Deflection Calculation \[ \Delta_{20} = \frac{1}{EI} \int m_2 m_ \, dx \] \[ = \frac{1}{EI} \int_{10}^{20} [-20(x-5) - 1.5(x-10)^2](10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} [-20x + 100 - 1.5(x^2 - 20x + 100)](10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} [-20x + 100 - 1.5x^2 + 30x - 150](10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} [-1.5x^2 + 10x - 50](10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} [-1.5x^3 - 5x^2 + 50x - 500] dx \] \[ \Delta_{20} = -\frac{65416.67}{EI} \] --- ## Stiffness Coefficients ### \( f_{11} \) \[ f_{11} = \frac{1}{EI} \int m_1 m_1 dx \] \[ = \frac{1}{EI} \left[ \int_^{10} (-x)(-x) dx + \int_{10}^{20} (-x)(-x) dx \right] \] \[ = \frac{1}{EI} \left[ \int_^{10} x^2 dx + \int_{10}^{20} x^2 dx \right] \] \[ = \frac{1}{EI} \left[ \frac{x^3}{3} \Big|_^{10} + \frac{x^3}{3} \Big|_{10}^{20} \right] \] \[ = \frac{1}{EI} \left[ \frac{100}{3} + \frac{800-100}{3} \right] \] \[ = \frac{2666.67}{EI} \] ### \( f_{22} \) \[ f_{22} = \frac{1}{EI} \int_{10}^{20} (10 + x)(10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} (100 + 20x + x^2) dx \] \[ = \frac{1}{EI} \left[ 100x + 10x^2 + \frac{x^3}{3} \right]_{10}^{20} \] \[ = \frac{6333.33}{EI} \] ### \( f_{12} \) \[ f_{12} = \frac{1}{EI} \int_{10}^{20} -x(10 + x) dx \] \[ = \frac{1}{EI} \int_{10}^{20} (-10x - x^2) dx \] \[ = \frac{1}{EI} \left[ -5x^2 - \frac{x^3}{3} \right]_{10}^{20} \] \[ = -\frac{3833.33}{EI} \] --- ## Compatibility Equations \[ f_{11} X_1 + f_{12} X_2 + \Delta_{10} = \] \[ f_{12} X_1 + f_{22} X_2 + \Delta_{20} = \] Substituting: \[ 2666.67 X_1 - 3833.33 X_2 + 42916.67 = \] \[ -3833.33 X_1 + 6333.33 X_2 - 65416.67 = \] --- ## Solving for \( X_1 \) and \( X_2 \): \[ X_1 = 9.588 \text{ kN} \] \[ X_2 = 4.5254 \text{ kN} \] --- ## Support Reactions \[ \sum V_A = -4.5254 + 9.588 - 3 \times 10 - 2 \times 10 = 44.9374 \text{ kN} \] \[ \sum M_A = -4.5254 \times 10 + 9.588 \times 20 - 2 \times 10 \times 15 - 3 \times 10 \times \frac{10}{2} \] \[ M_A = 303.494 \text{ kN-m} \] --- ## Final Moment Diagram (Insert your moment diagram here as per your sketch.) --- > **Note:** All values are per calculations shown in the images. If you want the exact moment diagram, you can plot the calculated values using your preferred graphing tool. Let me know if you need the diagrams in digital format too!