To solve this problem, we need to create the shear and moment diagrams for the beam shown. Here’s a step-by-step breakdown:

1. Identifying Forces and Loads

• Distributed Load: There is a uniform distributed load of $20 \, \text{kN/m}$ over a span of $3 \, \text{m}$ from $A$ to $B$.
  • Total load from this section: $20 \times 3 = 60 \, \text{kN}$ acting at the midpoint of the distributed load (i.e., at $1.5 \, \text{m}$ from $A$).
• Point Load at $C$: A concentrated load of $40 \, \text{kN}$ is applied at point $C$ (at a distance of $5 \, \text{m}$ from $A$).
• Support Reactions:
  • At $A$: Assume a vertical reaction $R_A$.
  • At $D$: Since point $D$ is a fixed support, it provides both a vertical reaction $R_D$ and a moment $M_D$.

2. Static Equilibrium Equations

To solve for the reactions at the supports, we use the conditions for static equilibrium:

• Sum of vertical forces $\sum F_y = 0$
• Sum of moments about point $A$, $\sum M_A = 0$

a) Sum of Moments about $A$:

Let’s take moments around point $A$, where counterclockwise moments are positive.

$M_A = (R_D \times 7) - (60 \times 1.5) - (40 \times 5) = 0$ $M_A = 7R_D - 90 - 200 = 0$ $7R_D = 290 \implies R_D = \frac{290}{7} = 41.43 \, \text{kN}$

b) Sum of Vertical Forces:

Now apply the vertical equilibrium equation:

$R_A + R_D - 60 - 40 = 0$ $R_A + 41.43 = 100$ $R_A = 100 - 41.43 = 58.57 \, \text{kN}$

3. Shear Force Diagram (SFD)

• From $A$ to $B$ (0 to 3 m):

  • Start with $V_A = +58.57 \, \text{kN}$.
  • Subtract the distributed load of $20 \, \text{kN/m}$, which gives a linear drop in shear.
  • Shear at $B$ (after 3 m): $58.57 - 20 \times 3 = -1.43 \, \text{kN}$.

• From $B$ to $C$ (3 to 5 m):

  • Constant shear because no load is applied in this section.
  • Shear remains at $-1.43 \, \text{kN}$.

• At $C$ (5 m):

  • Apply the point load of $40 \, \text{kN}$: $-1.43 - 40 = -41.43 \, \text{kN}$.

• From $C$ to $D$ (5 to 7 m):

  • Again, no load in this section, so shear remains constant at $-41.43 \, \text{kN}$.

4. Bending Moment Diagram (BMD)

• At $A$ (0 m): Moment is zero because it's a pin support.

• From $A$ to $B$ (0 to 3 m):

  • Moment is quadratic (due to the distributed load). The moment at $B$ is: $M_B = 58.57 \times 3 - \frac{1}{2} \times 20 \times 3^2 = 175.71 - 90 = 85.71 \, \text{kNm}$

• From $B$ to $C$ (3 to 5 m):

  • Moment at $C$: $M_C = 85.71 + (-1.43) \times 2 = 85.71 - 2.86 = 82.85 \, \text{kNm}$
  • At $C$, subtract the effect of the point load: $M_C = 82.85 - 40 \times 2 = 82.85 - 80 = 2.85 \, \text{kNm}$

• At $D$ (7 m):

  • Moment at $D$ is zero since it’s a fixed support.

With these calculations, you can now sketch the shear and moment diagrams.

Let me know if you'd like a detailed sketch of the diagrams or any further clarifications!

Related Questions:

1. How do distributed loads affect the shape of moment diagrams?
2. What is the significance of zero-shear points in shear force diagrams?
3. How do you determine the maximum bending moment in a beam?
4. What is the difference between a pinned and a fixed support?
5. How do point loads alter both shear and moment diagrams?

Tip:

When drawing moment diagrams, remember that the slope of the moment diagram at any point corresponds to the value of the shear force at that point.