Let's proceed step by step for the beam problem: --- ## (a) **Determine the Reaction Forces at the Supports** ### **Step 1: Identify Loads and Supports** - **Support A:** Pin (can take vertical and horizontal, but only vertical acts here) - **Support B:** Roller (vertical reaction only) - **At C:** Overhang (free end, no reaction) - **Loads:** - Uniformly distributed load (UDL): \( w = 300\, \text{lbf/in} \) over \( AB = 30\, \text{in} \) - Point load: \( P = 180\, \text{lbf} \) at \( x = 40\, \text{in} \) (point C) Let: - \( R_A \) = reaction at A (vertical) - \( R_B \) = reaction at B (vertical) ### **Step 2: Calculate Total Load from UDL** \[ w \times L = 300\, \text{lbf/in} \times 30\, \text{in} = 900\, \text{lbf} \] ### **Step 3: Equilibrium Equations** #### **Sum of Vertical Forces** \[ R_A + R_B = 900 + 180 = 10800\, \text{lbf} \] #### **Sum of Moments about A** Take moments about point A (counterclockwise positive): - UDL acts at center of \( AB \): \( 15\, \text{in} \) from A - Point load at C: \( 40\, \text{in} \) from A - \( R_B \) acts at \( 30\, \text{in} \) from A \[ \sum M_A = : \\ -R_B \times 30 + 900 \times 15 + 180 \times 40 = \] \[ -R_B \times 30 + 135000 + 72000 = \] \[ -R_B \times 30 + 207000 = \] \[ R_B = \frac{207000}{30} = 690\, \text{lbf} \] Now use \( R_A + R_B = 10800 \): \[ R_A = 10800 - 690 = 390\, \text{lbf} \] --- ### **Final Answers (a):** \[ \boxed{ R_A = 390\, \text{lbf} \\ R_B = 690\, \text{lbf} } \] --- ## (b) **Draw the Shear Force and Bending Moment Diagram** ### **Step 1: Shear Force Calculation** - At \( x = \) (just left of A): \( V = \) - Just right of A: \( V = +390 \) lbf (upward reaction) - From A to B (\( < x < 30\)): - Slope of shear diagram decreases by \( w = 300 \) lbf/in - At \( x \) from A: \[ V(x) = 390 - 300x \] - At \( x = 30 \) (just left of B): \[ V(30^-) = 390 - 300 \times 30 = 390 - 900 = -510\, \text{lbf} \] - At B: Add \( R_B \) upwards: - \( V(30^+) = -510 + 690 = 180\, \text{lbf} \) - From B to C (\(30 < x < 40\)): - No distributed load, so shear is constant: \[ V = 180\, \text{lbf} \] - At C: Subtract point load: \[ V(40^+) = 180 - 180 = \] ### **Step 2: Bending Moment Calculation** - At \( x = \): \( M = \) (pinned support) - For \( < x < 30 \): \[ M(x) = 390x - 300\frac{x^2}{2} \] - At \( x = 30 \): \[ M(30) = 390 \times 30 - 300 \times \frac{30^2}{2} = 117000 - 135000 = -18000\, \text{lbf} \cdot \text{in} \] - For \( 30 < x < 40 \): - Shear is constant, so moment is linear: \[ M(x) = M(30) + 180(x - 30) \] - At \( x = 40 \): \[ M(40) = -18000 + 180 \times (10) = -18000 + 18000 = \] --- ### **Shear and Moment Diagram Key Points**: - Shear starts at \( +390 \) lbf at A, decreases linearly to \( -510 \) lbf at B, jumps to \( +180 \) lbf at B, constant to C, then drops to . - Moment: Parabolic from A to B (due to UDL), then linear from B to C. --- ## (c) **Determine the Neutral Axis for the Cross Section** ### **Step 1: Locate Centroid (Neutral Axis)** For the inverted T-section: - Web: \(1 \text{ in} \) wide, \(3 \text{ in} \) tall (vertical) - Flange: \(3 \text{ in} \) wide, \(1 \text{ in} \) tall (horizontal at base) Assume base of flange as \( y = \). #### **Area of Each Part** - Web: \( A_w = 1 \times 3 = 3 \text{ in}^2 \) - Flange: \( A_f = 3 \times 1 = 3 \text{ in}^2 \) #### **Centroids from base:** - Flange: \( y_f = .5 \text{ in} \) - Web: \( y_w = 1 + 1.5 = 2.5 \text{ in} \) #### **Neutral Axis (ȳ):** \[ \bar{y} = \frac{A_f y_f + A_w y_w}{A_f + A_w} = \frac{3 \times .5 + 3 \times 2.5}{6} = \frac{1.5 + 7.5}{6} = \frac{9}{6} = 1.5 \text{ in} \] --- ### **Final Answer (c):** \[ \boxed{\text{The neutral axis is 1.5 in above the bottom of the flange.}} \] --- ## **To Continue...** Let me know if you want to proceed with parts (d)-(g), including bending stresses, deflection, and plotting the diagrams!