Let's solve the problem step by step: **Given:** - Two point loads \( P = 3.5 \) kN each at points \( C \) and \( D \) - \( AC = 1.5 \) m, \( CD = 4.5 \) m, \( DB = 1.5 \) m - Upward reaction (support) is **uniformly distributed**. - Find the **maximum absolute value of the bending moment** (in kN·m). --- ### **Step 1: Find the Uniformly Distributed Reaction Load** **Total Length of Beam:** \[ L = 1.5 + 4.5 + 1.5 = 7.5 \text{ m} \] **Total Downward Force:** \[ \text{Total load} = P + P = 3.5 + 3.5 = 7.0 \text{ kN} \] **Uniformly Distributed Reaction (\( w \)):** \[ w = \frac{7.0 \text{ kN}}{7.5 \text{ m}} = 0.933 \text{ kN/m} \] **Explanation:** The uniformly distributed upward reaction must balance the total downward force of the two point loads. --- ### **Step 2: Draw Shear and Bending Moment Diagram** We know: - Loads at \( C \) (1.5 m from A) and \( D \) (1.5 m from B) - Uniform upward load everywhere. Let’s consider a section between \( C \) and \( D \) (i.e., between 1.5 m and 6.0 m from \( A \)), as that's where the maximum moment is likely to occur (between the loads). **At a section between \( C \) and \( D \), say at distance \( x \) from \( A \) where \( 1.5 < x < 6.0 \):** \[ \text{Shear just right of } C: V = \text{Upward reaction to left} - \text{point load at } C \] \[ \text{Upward reaction to left of } C = w \cdot 1.5 = 0.933 \times 1.5 = 1.4 \text{ kN} \] \[ \text{Shear just right of } C = 1.4 - 3.5 = -2.1 \text{ kN} \] **Between \( C \) and \( D \), shear increases linearly by \( w \) per meter:** \[ \text{Shear at } x = -2.1 + w(x-1.5) \] **Set shear = 0 to find location of maximum moment:** \[ -2.1 + 0.933(x-1.5) = 0 \] \[ 0.933(x-1.5) = 2.1 \] \[ x-1.5 = \frac{2.1}{0.933} = 2.25 \] \[ x = 1.5 + 2.25 = 3.75 \text{ m from } A \] **Explanation:** Maximum bending moment occurs where shear crosses zero, i.e., between \( C \) and \( D \), 3.75 m from \( A \). --- ### **Step 3: Calculate the Maximum Bending Moment** **Sum moments to the left of \( x = 3.75 \) m:** - Upward reaction: \( w \times 3.75 \) - Downward load at \( C \): 3.5 kN at \( x=1.5 \) \[ M = (\text{Upward force}) \times (\text{distance to } x/2) - (\text{Point load}) \times (\text{distance from C to x}) \] \[ \text{Upward:} \quad w \cdot 3.75 \cdot \frac{3.75}{2} = 0.933 \cdot 3.75 \cdot 1.875 = 6.57 \text{ kN}\cdot\text{m} \] \[ \text{Downward:} \quad 3.5 \cdot (3.75 - 1.5) = 3.5 \cdot 2.25 = 7.88 \text{ kN}\cdot\text{m} \] \[ M = 6.57 - 7.88 = -1.31 \text{ kN}\cdot\text{m} \] **Check at supports and under the loads (at C or D):** - At \( C \) (1.5 m from \( A \)): \[ M_{C} = w \cdot 1.5 \cdot \frac{1.5}{2} = 0.933 \cdot 1.5 \cdot 0.75 = 1.05 \text{ kN}\cdot\text{m} \] - At \( D \) (6.0 m from \( A \)): \[ M_{D} = w \cdot 6.0 \cdot \frac{6.0}{2} - 3.5 \cdot (6.0 - 1.5) \] \[ = 0.933 \cdot 6.0 \cdot 3.0 - 3.5 \cdot 4.5 = 16.8 - 15.75 = 1.05 \text{ kN}\cdot\text{m} \] So, the **maximum absolute value** is at \( x = 3.75 \) m, and is **1.31 kN·m**. --- ## **Final Answer** **The maximum absolute value of the bending moment is \(\boxed{1.31 \text{ kN}\cdot\text{m}}\).**