Finding Shear Forces and Bending Moments for Beam Segments

Jul 29, 2024

Lecture Notes: Finding Shear Forces and Bending Moments

1. Introduction

  • Objective: Calculate the shear forces and bending moments for different segments of a beam.
  • Sign Conventions:
    • Moments: Positive for clockwise, negative for anticlockwise.
    • Shears: Consider the direction for calculating reactions.

2. Segment AB

  • End Moments:
    • At A: 62.475 (clockwise, positive)
    • At B: 125.25 (clockwise, positive)
  • Reactions:
    • Assumed upward at segment ends: Ra, Rb-
    • Equations:
      • ΣM about A = 0: 62.475 + 125.25 - (Rb- * 12) = 0
      • Solving: Rb- = 15.64 kN
      • ΣFy = 0: Ra + 15.64 = 0 → Ra = -15.64 kN

3. Segment BC

  • End Moments:
    • At B: -125.25 (anticlockwise, positive)
    • At C: 281.485 (clockwise, positive)
  • Reactions:
    • Assumed upward at segment ends: Rb+, Rc-
    • Equations:
      • ΣM about B = 0: -125.25 + (20 * 12 * 6) + 281.485 - (Rc- * 12) = 0
      • Solving: Rc- = 133 kN
      • ΣFy = 0: Rb+ - (20 * 12) + 133 = 0 → Rb+ = 107 kN

4. Segment CD

  • End Moments:
    • At C: -281.485 (anticlockwise, positive)
    • At D: 234.25 (clockwise, positive)
  • Reactions:
    • Assumed upward at segment ends: Rc+, Rd
    • Equations:
      • ΣM about C = 0: -281.485 + (250 * 4) + 234.25 - (Rd * 8) = 0
      • Solving: Rd = 119.1 kN
      • ΣFy = 0: Rc+ - 250 + 119.1 = 0 → Rc+ = 130.9 kN

5. Shear Force Diagram (SFD)

  • Procedure: Move from leftmost point to the rightmost point of the beam.
  • Key Points:
    • Start: Move downward by 15.64 kN (negative).
    • Remains constant till the end of AB.
    • Jump upward by 15.64 kN at end of AB, reaching zero.
    • Jump upward by 107 kN at start of BC.
    • Decrease linearly due to UDL (20 kN/m) over 12 m: Drop to 133 kN.
    • Then upward jump by 133 kN at end of BC.
    • Upward jump by 130.9 kN at start of CD.
    • Constant until a 250 kN downward jump, reaching -119.1 kN.
    • Constant until an upward jump of 119.1 kN at the end, reaching zero.
    • The SFD is now complete.

6. Bending Moment Diagram (BMD)

  • Procedure: Use SFD to derive BMD.
  • Key Points:
    • Start: 64.475 kNm positive (sagging).
    • Linear decrease in BMD for segment AB due to constant shear force.
    • Parabolic increase during BC due to triangular SFD segment.
    • Continues parabolic decrease due to opposite triangular SFD segment.
    • Linear increase in CD due to constant negative shear force.
    • Final linear decrease to match the end moments.

7. Support Reactions

  • End Supports:
    • RA = 15.64 kN (downward)
    • MA = 64.475 kNm (clockwise)
    • RD = 119.1 kN (upward)
    • MD = 234.25 kNm (clockwise)
  • Intermediate Supports:
    • RB = 122.64 kN
    • RC = 263.9 kN
  • Note that bending moments are different for internal and external hinges.

Conclusion: Shear forces and bending moments diagrams are essential for understanding beam segment behaviors under loading conditions and determining support reactions accurately.