Answer according to n4 level

The beam is $8 \text{ m}$ long.
Support $R_A$ is at $1 \text{ m}$ from the left end ($x=1 \text{ m}$).
Support $R_B$ is at $2 \text{ m}$ from the right end, which is $8-2 = 6 \text{ m}$ from the left end ($x=6 \text{ m}$).

Loads (all acting downwards):

• Point load $P_1 = 450 \text{ kN}$ at $x=1 \text{ m}$.
• Concentrated load $P_2 = 700 \text{ kN}$ at $x=6 \text{ m}$.
• Uniformly Distributed Load (UDL) $w = 12 \text{ kN/m}$ over the first $5 \text{ m}$ from the right end. This means from $x=8-5=3 \text{ m}$ to $x=8 \text{ m}$.
  • The total force from the UDL is $F_{UDL} = 12 \text{ kN/m} \times 5 \text{ m} = 60 \text{ kN}$.
  • This equivalent force acts at the centroid of the UDL, which is at $x = 3 + \frac{5}{2} = 5.5 \text{ m}$ from the left end.
• Concentrated load $P_3 = 40 \text{ kN}$ at $x=8 \text{ m}$.

Sum of vertical forces $\sum F_y = 0$ (taking upward forces as positive):

Sum of moments about $R_A$ (at $x=1 \text{ m}$) $\sum M_A = 0$ (taking clockwise moments as positive):

• $P_1$ is at $R_A$, so its moment arm is 0.
• $R_B$ creates a counter-clockwise moment: $-R_B \times (6-1) = -5 R_B$.
• $P_2$ creates a clockwise moment: $+P_2 \times (6-1) = +700 \times 5 = 3500 \text{ kNm}$.
• $F_{UDL}$ creates a clockwise moment: $+F_{UDL} \times (5.5-1) = +60 \times 4.5 = 270 \text{ kNm}$.
• $P_3$ creates a clockwise moment: $+P_3 \times (8-1) = +40 \times 7 = 280 \text{ kNm}$.

The reactions at the supports are:

• $R_A = \boxed{\text{440 kN}}$
• $R_B = \boxed{\text{810 kN}}$

7.1.2. Draw a fully labelled shear force diagram.

We will calculate the shear force ($V$) at key points along the beam, starting from the left end ($x=0$). Upward forces are positive.

• At $x=0 \text{ m}$: $$V_0 = 0 \text{ kN}$$
• Just before $x=1 \text{ m}$: $$V_{1^-} = 0 \text{ kN}$$
• At $x=1 \text{ m}$ (after $P_1$ and $R_A$): $$V_{1, \text{after } P_1} = V_{1^-} - P_1 = 0 - 450 = -450 \text{ kN}$$ $$V_{1, \text{after } R_A} = V_{1, \text{after } P_1} + R_A = -450 + 440 = -10 \text{ kN}$$
• From $x=1 \text{ m}$ to $x=3 \text{ m}$: (No loads, shear is constant) $$V_x = -10 \text{ kN}$$ $$V_{3^-} = -10 \text{ kN}$$
• From $x=3 \text{ m}$ to $x=6 \text{ m}$: (UDL of $12 \text{ kN/m}$ starts at $x=3 \text{ m}$) $$V_x = V_{3^-} - w(x-3) = -10 - 12(x-3)$$ $$V_{3^+} = -10 \text{ kN}$$ $$V_{6^-} = -10 - 12(6-3) = -10 - 36 = -46 \text{ kN}$$
• At $x=6 \text{ m}$ (after $P_2$ and $R_B$): $$V_{6, \text{after } P_2} = V_{6^-} - P_2 = -46 - 700 = -746 \text{ kN}$$ $$V_{6, \text{after } R_B} = V_{6, \text{after } P_2} + R_B = -746 + 810 = 64 \text{ kN}$$
• From $x=6 \text{ m}$ to $x=8 \text{ m}$: (UDL continues) $$V_x = V_{6, \text{after } R_B} - w(x-6) = 64 - 12(x-6)$$ $$V_{8^-} = 64 - 12(8-6) = 64 - 24 = 40 \text{ kN}$$
• At $x=8 \text{ m}$ (after $P_3$): $$V_{8, \text{after } P_3} = V_{8^-} - P_3 = 40 - 40 = 0 \text{ kN}$$

Shear Force Diagram (SFD) values:

• $V_{0} = 0 \text{ kN}$
• $V_{1^-} = 0 \text{ kN}$
• $V_{1, \text{after } P_1} = -450 \text{ kN}$
• $V_{1, \text{after } R_A} = -10 \text{ kN}$
• $V_{3} = -10 \text{ kN}$
• $V_{6^-} = -46 \text{ kN}$
• $V_{6, \text{after } P_2} = -746 \text{ kN}$
• $V_{6, \text{after } R_B} = 64 \text{ kN}$
• $V_{8^-} = 40 \text{ kN}$
• $V_{8, \text{after } P_3} = 0 \text{ kN}$

Shear Force Diagram (SFD)

   0 kN
   |
   |
   |
   +-------------------------------------------------------------------- (x=0)
   |
   |                                                                    (x=1, before P1)
   |
   | -450 kN (at x=1, after P1)
   |--------------------------------------------------------------------
   |
   | -10 kN (at x=1, after RA)
   +-------------------------------------------------------------------- (x=3)
   |
   |                                                                    (x=6, before P2)
   | -46 kN
   |--------------------------------------------------------------------
   |
   | -746 kN (at x=6, after P2)
   |--------------------------------------------------------------------
   |
   | 64 kN (at x=6, after RB)
   +-------------------------------------------------------------------- (x=8, before P3)
   |
   | 40 kN
   |--------------------------------------------------------------------
   |
   | 0 kN (at x=8, after P3)
   +--------------------------------------------------------------------

(Note: A graphical representation is required for a "fully labelled shear force diagram". Due to text-based limitations, the key values and their positions are listed above. The diagram would show a horizontal line from 0 to 1m, a drop to -450kN, a jump to -10kN, a horizontal line to 3m, a linear decrease to -46kN at 6m, a drop to -746kN, a jump to 64kN, a linear decrease to 40kN at 8m, and a final drop to 0kN.)

7.1.3. Draw a fully labelled bending moment diagram.

We will calculate the bending moment ($M$) at key points along the beam. The change in bending moment between two points is the area under the shear force diagram between those points. Sagging moments are positive.

• At $x=0 \text{ m}$ (Left end): $$M_0 = 0 \text{ kNm}$$
• At $x=1 \text{ m}$ (at $P_1$ and $R_A$): $$M_1 = M_0 + \text{Area of SFD from 0 to 1} = 0 + (0 \times 1) = 0 \text{ kNm}$$
• At $x=3 \text{ m}$ (Start of UDL): $$M_3 = M_1 + \text{Area of SFD from 1 to 3} = 0 + (-10 \text{ kN} \times (3-1) \text{ m})$$ $$M_3 = 0 - 20 = -20 \text{ kNm}$$
• At $x=6 \text{ m}$ (at $P_2$ and $R_B$): $$M_6 = M_3 + \text{Area of SFD from 3 to 6}$$ The SFD from 3 to 6 is a trapezoid with values $-10 \text{ kN}$ and $-46 \text{ kN}$ over a length of $3 \text{ m}$. $$\text{Area} = \frac{-10 + (-46)}{2} \times 3 = \frac{-56}{2} \times 3 = -28 \times 3 = -84 \text{ kNm}$$ $$M_6 = -20 - 84 = -104 \text{ kNm}$$
• At $x=8 \text{ m}$ (Right end, at $P_3$): $$M_8 = M_6 + \text{Area of SFD from 6 to 8}$$ The SFD from 6 to 8 is a trapezoid with values $64 \text{ kN}$ and $40 \text{ kN}$ over a length of $2 \text{ m}$. $$\text{Area} = \frac{64 + 40}{2} \times 2 = \frac{104}{2} \times 2 = 52 \times 2 = 104 \text{ kNm}$$ $$M_8 = -104 + 104 = 0 \text{ kNm}$$

Bending Moment Diagram (BMD) values:

• $M_0 = 0 \text{ kNm}$
• $M_1 = 0 \text{ kNm}$
• $M_3 = -20 \text{ kNm}$
• $M_6 = -104 \text{ kNm}$
• $M_8 = 0 \text{ kNm}$

Bending Moment Diagram (BMD)

   0 kNm (at x=0)
   +--------------------------------------------------------------------
   |
   | 0 kNm (at x=1)
   +--------------------------------------------------------------------
   |
   | -20 kNm (at x=3)
   |--------------------------------------------------------------------
   |
   | -104 kNm (at x=6)
   |--------------------------------------------------------------------
   |
   | 0 kNm (at x=8)
   +--------------------------------------------------------------------

(Note: A graphical representation is required for a "fully labelled bending moment diagram". Due to text-based limitations, the key values and their positions are listed above. The diagram would show a horizontal line from 0 to 1m, then a linear decrease to -20kNm at 3m, then a parabolic curve decreasing to -104kNm at 6m, and finally a parabolic curve increasing to 0kNm at 8m.)