BMD of a simple beam and a frame

(i) Simply supported beam (span 4 m + 6 m = 10 m, 10 kN·m couple at 4 m from the left support)

Assume the couple is clockwise (if it is anticlockwise, reverse all signs).

Reactions: $\Sigma F_y = 0 \Rightarrow R_A + R_B = 0$. Moments about A: $R_B(10) - 10 = 0$ (the clockwise couple is balanced by the anticlockwise moment of $R_B$), so

$$R_B = +1\ \text{kN}\ (\uparrow),\qquad R_A = -1\ \text{kN}\ (\text{i.e. }1\ \text{kN}\ \downarrow)$$

Bending moment (sagging +ve):

• Left of the couple ($0 \le x < 4$): $M = R_Ax = -x$, so $M_A = 0$ and $M = -4$ kN·m just left of the couple (hogging).
• The clockwise couple causes a sudden rise of 10 kN·m: just right of the couple $M = -4 + 10 = +6$ kN·m.
• Right of the couple: $M = R_B(10 - x)$, falling linearly to 0 at B ($M = 1\times6 = 6$ at $x = 4$ ✓).

 BMD (kN·m, sagging +ve above the line)
                 +6
                  |`-._
                  |    `-._
                  |        `-._
 A o--------------+------------`-o B
    `-._          |
        `-._      |
            `-._  |
               -4 |
           x = 4 m (couple)

The shear force is constant, $V = -1$ kN throughout (the couple does not affect it). The BMD is two straight lines with a jump of 10 kN·m at the couple.

(ii) Frame (column 5 m, horizontal member 4 m, 20 kN at the free end)

Assume the column is fixed at the base (a hinged base would make this frame unstable) and the 20 kN load acts downward at the free end of the horizontal member.

• Horizontal member: $M = -20x$ (hogging, tension on top), from 0 at the free end to $-20\times4 = -80$ kN·m at the corner. The shear force is constant, $V = 20$ kN, and $N = 0$.
• Column: the load acts at a horizontal distance of 4 m from the column axis, so it causes an axial force $N = 20$ kN (compression) and a constant moment $M = 20\times4 = 80$ kN·m along the full height, with tension on the outer face (the same side as the top of the beam). The shear force in the column is zero.
• Reactions at the base: $V = 20$ kN (↑), $H = 0$, $M = 80$ kN·m.

  free end ●───────┐  BMD: beam   0 → -80 (linear)
  20 kN            │       column -80 (constant)
                   │
                   │
                 ▒▒▒ fixed base

Answer: (i) $M = -4$ kN·m (left of the couple) and $+6$ kN·m (right of the couple), with $R_A = 1$ kN ↓ and $R_B = 1$ kN ↑; (ii) the beam moment varies linearly from 0 to 80 kN·m (hogging), and the column carries a constant 80 kN·m.