A light, horizontal beam, ABCDE, with A on the left-hand side is 26 m long. It is supported at two points, A and D. A point load of 20 kN is at B, 12 m from A. A point load of 30 is at C, 6 m from B. A uniform distributed load of 8 kN/m is between A and B. D is 4 m from C.

A light, horizontal beam ABCDE, 26 m long, with A on the left-hand side.

• Support at A (0 m) with reaction $R_A$.
• Uniformly Distributed Load (UDL) of $8 \text{ kN/m}$ from A to B (0 m to 12 m).
• Point load of $20 \text{ kN}$ downwards at B (12 m from A).
• Point load of $30 \text{ kN}$ downwards at C (6 m from B, so $12+6=18 \text{ m}$ from A).
• Support at D (4 m from C, so $18+4=22 \text{ m}$ from A) with reaction $R_D$.
• End of beam at E (26 m from A).

      <------------------ 26 m ------------------>
      A       UDL=8kN/m       B       C       D       E
      |-----------------------|-------|-------|-------|
      ^                       |       |       ^
      RA                      20kN    30kN    RD
      0m                      12m     18m     22m     26m

• At A (x=0 m): Shear force $V_A = R_A = 84.36 \text{ kN}$.
• From A to B (0 m to 12 m): Shear force decreases linearly due to UDL.
  • At B (just left, x=12 m): $V_{B,left} = R_A - (8 \text{ kN/m} \times 12 \text{ m}) = 84.36 - 96 = -11.64 \text{ kN}$.
• At B (x=12 m): Shear force drops due to $20 \text{ kN}$ point load.
  • At B (just right): $V_{B,right} = -11.64 - 20 = -31.64 \text{ kN}$.
• From B to C (12 m to 18 m): Shear force is constant.
  • At C (just left, x=18 m): $V_{C,left} = -31.64 \text{ kN}$.
• At C (x=18 m): Shear force drops due to $30 \text{ kN}$ point load.
  • At C (just right): $V_{C,right} = -31.64 - 30 = -61.64 \text{ kN}$.
• From C to D (18 m to 22 m): Shear force is constant.
  • At D (just left, x=22 m): $V_{D,left} = -61.64 \text{ kN}$.
• At D (x=22 m): Shear force jumps up due to $R_D$.
  • At D (just right): $V_{D,right} = -61.64 + R_D = -61.64 + 61.64 = 0 \text{ kN}$.
• From D to E (22 m to 26 m): Shear force is constant.
  • At E (x=26 m): $V_E = 0 \text{ kN}$.

The shear force diagram starts at $84.36 \text{ kN}$ at A, decreases linearly to $-11.64 \text{ kN}$ at B, drops to $-31.64 \text{ kN}$, remains constant to C, drops to $-61.64 \text{ kN}$, remains constant to D, jumps to $0 \text{ kN}$, and remains $0 \text{ kN}$ to E.
The point of zero shear force occurs between A and B where $R_A - 8x = 0 \implies 84.36 - 8x = 0 \implies x = 10.55 \text{ m}$ from A.

• Bending Moment at A ($M_A$): $M_A = 0 \text{ kN} \cdot \text{m}$ (support at end of beam).

• Bending Moment at B ($M_B$, at x=12 m): Consider forces to the left of B: $R_A$ and the UDL.

  $$M_B = (R_A \times 12 \text{ m}) - (8 \text{ kN/m} \times 12 \text{ m} \times \frac{12 \text{ m}}{2})$$ $$M_B = (84.36 \text{ kN} \times 12 \text{ m}) - (96 \text{ kN} \times 6 \text{ m})$$ $$M_B = 1012.32 \text{ kN} \cdot \text{m} - 576 \text{ kN} \cdot \text{m}$$ $$M_B = \boxed{\text{436.32 kN} \cdot \text{m}}$$

• Bending Moment at C ($M_C$, at x=18 m): Consider forces to the left of C: $R_A$, UDL, and $20 \text{ kN}$ point load at B.

  $$M_C = (R_A \times 18 \text{ m}) - (96 \text{ kN} \times (18-6) \text{ m}) - (20 \text{ kN} \times (18-12) \text{ m})$$ $$M_C = (84.36 \text{ kN} \times 18 \text{ m}) - (96 \text{ kN} \times 12 \text{ m}) - (20 \text{ kN} \times 6 \text{ m})$$ $$M_C = 1518.48 \text{ kN} \cdot \text{m} - 1152 \text{ kN} \cdot \text{m} - 120 \text{ kN} \cdot \text{m}$$ $$M_C = \boxed{\text{246.48 kN} \cdot \text{m}}$$

• Bending Moment at A ($M_A$): $0 \text{ kN} \cdot \text{m}$.
• Maximum Bending Moment (between A and B): Occurs at $x = 10.55 \text{ m}$ (where shear force is zero). $$M_{max} = (R_A \times 10.55) - (8 \times 10.55 \times \frac{10.55}{2})$$ $$M_{max} = (84.36 \times 10.55) - (4 \times 10.55^2)$$ $$M_{max} = 889.70 - 445.21 = 444.49 \text{ kN} \cdot \text{m}$$
• Bending Moment at B ($M_B$): $436.32 \text{ kN} \cdot \text{m}$.
• Bending Moment at C ($M_C$): $246.48 \text{ kN} \cdot \text{m}$.
• Bending Moment at D ($M_D$, at x=22 m): $$M_D = (R_A \times 22) - (96 \times (22-6)) - (20 \times (22-12)) - (30 \times (22-18))$$ $$M_D = (84.36 \times 22) - (96 \times 16) - (20 \times 10) - (30 \times 4)$$ $$M_D = 1856.00 - 1536 - 200 - 120 = 0 \text{ kN} \cdot \text{m}$$
• Bending Moment at E ($M_E$): $0 \text{ kN} \cdot \text{m}$ (free end).

The bending moment diagram starts at $0 \text{ kN} \cdot \text{m}$ at A, curves parabolically upwards to a maximum of $444.49 \text{ kN} \cdot \text{m}$ at $x=10.55 \text{ m}$, then decreases to $436.32 \text{ kN} \cdot \text{m}$ at B. From B to C, it decreases linearly to $246.48 \text{ kN} \cdot \text{m}$. From C to D, it decreases linearly to $0 \text{ kN} \cdot \text{m}$. From D to E, it remains $0 \text{ kN} \cdot \text{m}$.