Let's solve the given problem step by step: --- ## **a. Determine the Reaction Forces at the Supports** Let: - \( A \): Pin support (vertical reaction \( A_y \)) - \( B \): Roller support (vertical reaction \( B_y \)) - The overhang at \( C \) is free. ### **Step 1: Calculate Loads** - **Distributed load:** \( w = 300 \) lbf/in over \( 30 \) in. - **Point load:** \( P = 180 \) lbf at \( C \) (at 40 in from A). ### **Step 2: Draw Free Body Diagram** - Span \( AB = 30 \) in, \( BC = 10 \) in. ### **Step 3: Equilibrium Equations** #### **Sum of Vertical Forces:** \[ A_y + B_y = (300)(30) + 180 = 900 + 180 = 10800 \text{ lbf} \] #### **Sum of Moments about \( A \):** Take moments counterclockwise as positive. - Distributed load acts at \( 15 \) in from \( A \). - Point load acts at \( 40 \) in from \( A \). \[ \sum M_A = = - (300)(30)(15) - 180(40) + B_y(30) \] \[ = -900 \times 15 - 180 \times 40 + B_y \times 30 \] \[ = -135000 - 72000 + 30 B_y \] \[ = -207000 + 30 B_y \] \[ 30 B_y = 207000 \] \[ B_y = \frac{207000}{30} = 690 \text{ lbf} \] #### **Now solve for \( A_y \):** \[ A_y = 10800 - 690 = 390 \text{ lbf} \] --- ### **Summary of Reactions** \[ \boxed{ A_y = 390 \text{ lbf}, \quad B_y = 690 \text{ lbf} } \] --- ## **b. Shear Force and Bending Moment Diagram** ### **Shear Force Calculation** - From \( A \) to \( B \) ( to 30 in), distributed load only. - At \( B \), point reaction. - From \( B \) to \( C \) (30 to 40 in), no distributed load, but point load at \( C \). #### **Shear Just to the Right of A:** \[ V_A^+ = A_y = 390 \text{ lbf} \] #### **Shear at \( x \) (where \( < x < 30 \) in):** \[ V(x) = 390 - 300x \] #### **Shear Just Left of \( B \) (\( x = 30^- \)):** \[ V(30^-) = 390 - 300 \times 30 = 390 - 900 = -510 \text{ lbf} \] #### **Shear Just Right of \( B \):** \[ V(30^+) = -510 + 690 = 180 \text{ lbf} \] #### **Shear at \( C \) (\( x = 40 \) in):** \[ V(40^-) = 180 \] \[ V(40^+) = 180 - 180 = \] --- ### **Bending Moment Calculation** #### **Moment at \( A \):** \[ M_A = \] #### **Moment at \( x \) ( < x < 30 in):** \[ M(x) = 390x - 300 \frac{x^2}{2} \] #### **Moment at \( B \) (\( x = 30 \) in):** \[ M_B = 390 \times 30 - 300 \times \frac{30^2}{2} \] \[ = 117000 - 300 \times 450 = 117000 - 135000 = -18000 \text{ lbf-in} \] #### **Moment between \( B \) and \( C \) (30 < x < 40 in):** \[ M(x) = M_B + 690(x - 30) \] #### **Moment at \( C \) (\( x = 40 \) in):** \[ M_C = -18000 + 690 \times 10 = -18000 + 69000 = 51000 \text{ lbf-in} \] --- ### **Summary Table** | Location | Shear (lbf) | Moment (lbf-in) | |----------|-------------|-----------------| | \(A\) | 390 | | | \(B\) | -510/+180 | -18000 | | \(C\) | | 51000 | --- ## **c. Neutral Axis for the Cross Section** The cross section is a T-shape. - Flange: \( 3 \) in wide, \( 1 \) in thick - Web: \( 1 \) in wide, \( 3 \) in tall (but part overlaps flange) #### **Step 1: Find Areas** - Flange: \( A_1 = 3 \times 1 = 3 \) in² - Web: \( A_2 = 1 \times 3 = 3 \) in² #### **Step 2: Find Centroids from Bottom** - Flange centroid: \( y_1 = 3 + .5 = 3.5 \) in from bottom - Web centroid: \( y_2 = 1.5 \) in from bottom #### **Step 3: Neutral Axis Location (\( \bar{y} \))** \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{3 \times 3.5 + 3 \times 1.5}{3 + 3} = \frac{10.5 + 4.5}{6} = \frac{15}{6} = 2.5 \text{ in} \] --- ## **d. Maximum Tensile and Compressive Bending Stresses and Locations** \[ \sigma = \frac{M y}{I} \] - **Neutral axis:** \( y_{NA} = 2.5 \) in from bottom - **Top fiber (compression):** \( y_{top} = (4 - 2.5) = 1.5 \) in above NA - **Bottom fiber (tension):** \( y_{bot} = (2.5) \) in below NA ### **Step 1: Moment of Inertia (I)** Calculate \( I \) about neutral axis. - Flange: \( I_1 = \frac{b h^3}{12} + A_1 d_1^2 \) - \( b = 3 \) in, \( h = 1 \) in - \( d_1 = 3.5 - 2.5 = 1 \) in above NA \[ I_1 = \frac{3 \times 1^3}{12} + 3 \times 1^2 = .25 + 3 = 3.25 \text{ in}^4 \] - Web: \( I_2 = \frac{b h^3}{12} + A_2 d_2^2 \) - \( b = 1 \) in, \( h = 3 \) in - \( d_2 = 2.5 - 1.5 = 1 \) in below NA \[ I_2 = \frac{1 \times 3^3}{12} + 3 \times 1^2 = \frac{27}{12} + 3 = 2.25 + 3 = 5.25 \text{ in}^4 \] - Total \( I = I_1 + I_2 = 3.25 + 5.25 = 8.5 \) in\(^4\) ### **Step 2: Maximum Moment** Maximum moment occurs at support \( B \) or at \( C \). - At \( C \): \( M_{max} = 51000 \) lbf-in ### **Step 3: Stresses** \[ \sigma_{top} = \frac{M_{max}(y_{top})}{I} = \frac{51000 \times 1.5}{8.5} = 900 \text{ psi (compression)} \] \[ \sigma_{bottom} = \frac{51000 \times 2.5}{8.5} = 15000 \text{ psi (tension)} \] - **Top fiber (compression):** \( \sigma = 900 \) psi - **Bottom fiber (tension):** \( \sigma = 15000 \) psi --- ## **e. Slope and Deflection Using Superposition** For beam with overhang, use standard beam formulas and superposition. Let \( E \) and \( I \) be modulus of elasticity and moment of inertia. - For distributed load: Use standard tables for simply supported beam with overhang. - For point load at end: Use standard tables for simply supported beam with overhang. \[ y(x) = y_{w}(x) + y_{P}(x) \] - \( y_{w}(x) \): deflection due to uniform load over \( < x < 30 \) - \( y_{P}(x) \): deflection due to point load at \( x = 40 \) Refer to beam tables, or if needed, equations can be written out for each section. --- ### **Final Answers Summary** #### **a. Reaction Forces** \[ \boxed{ A_y = 390 \text{ lbf}, \quad B_y = 690 \text{ lbf} } \] #### **b. Shear and Moment Diagrams** - Shear and moment equations as derived above. #### **c. Neutral Axis** \[ \boxed{ \bar{y} = 2.5 \text{ in (from bottom)} } \] #### **d. Maximum Stresses** \[ \boxed{ \text{Top (compression): } 900 \text{ psi} \\ \text{Bottom (tension): } 15000 \text{ psi} } \] #### **e. Slope and Deflection** - Use superposition, with standard beam solutions for overhang (refer to beam tables for explicit formulas). --- Let me know if you'd like the detailed equations for part (e) and a plot for part (f)!

# Step-by-Step Solution for Steel Beam Problem ## a. Determine the Reaction Forces at the Supports ### Step 1: Calculate Loads - **Distributed Load**: \( w = 300 \) lbf/in over \( 30 \) in - **Point Load**: \( P = 180 \) lbf at point \( C \) (40 in from \( A \)) ### Step 2: Free Body Diagram - Span \( AB = 30 \) in, \( BC = 10 \) in. ### Step 3: Equilibrium Equations #### Sum of Vertical Forces: \[ A_y + B_y = (300)(30) + 180 = 10800 \text{ lbf} \] #### Sum of Moments about \( A \): \[ \sum M_A = - (300)(30)(15) - 180(40) + B_y(30) \] \[ = -135000 - 72000 + 30 B_y \] \[ 30 B_y = 207000 \] \[ B_y = \frac{207000}{30} = 690 \text{ lbf} \] #### Solve for \( A_y \): \[ A_y = 10800 - 690 = 390 \text{ lbf} \] ### Summary of Reactions \[ \boxed{ A_y = 390 \text{ lbf}, \quad B_y = 690 \text{ lbf} } \] --- ## b. Shear Force and Bending Moment Diagram ### Shear Force Calculation 1. **From \( A \) to \( B \)**: - \( V_A^+ = 390 \text{ lbf} \) - At \( x < 30 \): \[ V(x) = 390 - 300x \] 2. **At \( B \)**: \[ V(30^-) = 390 - 900 = -510 \text{ lbf} \] - Just right of \( B \): \[ V(30^+) = -510 + 690 = 180 \text{ lbf} \] 3. **At \( C \)**: \[ V(40^-) = 180 \text{ lbf} \] \[ V(40^+) = 180 - 180 = 0 \] ### Bending Moment Calculation 1. **At \( A \)**: \[ M_A = 0 \] 2. **At \( x \) (where \( 0 < x < 30 \))**: \[ M(x) = 390x - 150x^2 \] 3. **At \( B \)**: \[ M_B = 390 \times 30 - 150 \times (30^2) = 11700 - 13500 = -1800 \text{ lbf-in} \] 4. **At \( C \)**: \[ M_C = -1800 + 690 \times 10 = -1800 + 6900 = 5100 \text{ lbf-in} \] ### Summary Table | Location | Shear (lbf) | Moment (lbf-in) | |----------|-------------|-----------------| | \(A\) | 390 | 0 | | \(B\) | -510 / 180 | -1800 | | \(C\) | 0 | 5100 | --- ## c. Neutral Axis for the Cross Section ### Step 1: Find Areas - **Flange**: \( A_1 = 3 \times 1 = 3 \) in² - **Web**: \( A_2 = 1 \times 3 = 3 \) in² ### Step 2: Find Centroids - **Flange centroid**: \( y_1 = 3 + 0.5 = 3.5 \) in from bottom - **Web centroid**: \( y_2 = 1.5 \) in from bottom ### Step 3: Neutral Axis Location (\( \bar{y} \)) \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{3 \times 3.5 + 3 \times 1.5}{3 + 3} = \frac{15}{6} = 2.5 \text{ in} \] ### Summary \[ \boxed{ \bar{y} = 2.5 \text{ in (from bottom)} } \] --- ## d. Maximum Tensile and Compressive Bending Stresses and Locations ### Step 1: Moment of Inertia (I) - **Flange**: \[ I_1 = \frac{3 \times 1^3}{12} + 3 \times (3.5 - 2.5)^2 = \frac{3}{12} + 3 \times 1 = 0.25 + 3 = 3.25 \text{ in}^4 \] - **Web**: \[ I_2 = \frac{1 \times 3^3}{12} + 3 \times (2.5 - 1.5)^2 = \frac{27}{12} + 3 \times 1 = 2.25 + 3 = 5.25 \text{ in}^4 \] - **Total Moment of Inertia**: \[ I = I_1 + I_2 = 3.25 + 5.25 = 8.5 \text{ in}^4 \] ### Step 2: Maximum Moment - Occurs at \( C \): \( M = 5100 \) lbf-in ### Step 3: Bending Stresses - **Top fiber (compression)**: \[ \sigma_{top} = \frac{M y_{top}}{I} = \frac{5100 \times 1.5}{8.5} \approx 900 \text{ psi} \] - **Bottom fiber (tension)**: \[ \sigma_{bottom} = \frac{M y_{bot}}{I} = \frac{5100 \times 2.5}{8.5} \approx 15000 \text{ psi} \] ### Summary \[ \boxed{ \text{Top (compression)}: 900 \text{ psi} \\ \text{Bottom (tension)}: 15000 \text{ psi} } \] --- ## e. Slope and Deflection Using Superposition ### General Approach 1. **For Uniform Load**: - Use standard beam deflection equations for a uniform load over a span. 2. **For Point Load**: - Use standard beam deflection equations for a point load at the end of the overhang. ### Combined Deflection \[ y(x) = y_{w}(x) + y_{P}(x) \] Where: - \( y_{w}(x) \): deflection due to the uniform load - \( y_{P}(x) \): deflection due to the point load ### Example Formulas - **For a uniform load**: \[ y_{w}(x) = \frac{w x^2 (3L - x)}{6EI} \] - **For a point load** at the end: \[ y_{P}(x) = \frac{P(L^3)}{3EI} \] --- ## f. Plot Slope and Deflection Use software like MATLAB or Python to plot \( y(x) \) based on the combined equations derived. --- ## g. Maximum Deflection and Location ### Maximum Deflection - Evaluate \( y(x) \) at key points (i.e., \( x = 0, 30, 40 \)). - Identify maximum from calculated deflections. ### Location - Measure from \( A \) to find where maximum deflection occurs. ### Summary - **Maximum Deflection**: Calculated value from \( y(x) \) - **Location**: \( x = \) position of maximum deflection along the beam. --- This concludes the step-by-step solution for the steel beam problem. If you need further calculations or clarifications, please let me know!