1. **Problem Statement:** Draw the shear force and bending moment diagrams for the first beam: a simply supported beam of length 3m with a moment of 3 KN.m applied at 2m from the left support. 2. **Given Data:** - Beam length, $L = 3$ m - Applied moment at $x = 2$ m, $M_0 = 3$ KN.m - Simply supported beam (supports at $x=0$ and $x=3$ m) 3. **Key Concepts:** - Shear force ($V$) is the internal force perpendicular to the beam axis. - Bending moment ($M$) is the internal moment causing bending. - At a point moment, shear force is continuous but bending moment has a jump equal to the applied moment. - For simply supported beams, reactions at supports balance loads and moments. 4. **Step 1: Calculate support reactions** - Since only a moment is applied (no vertical loads), vertical reactions at supports are zero: $$ R_A = 0, \quad R_B = 0 $$ 5. **Step 2: Shear Force Diagram (V)** - No vertical loads, so shear force is zero everywhere: $$ V(x) = 0 \quad \text{for} \quad 0 \leq x \leq 3 $$ 6. **Step 3: Bending Moment Diagram (M)** - Bending moment at left support $A$ is zero: $$ M(0) = 0 $$ - From $0$ to $2$ m, no loads, so moment is constant zero: $$ M(x) = 0 \quad \text{for} \quad 0 \leq x < 2 $$ - At $x=2$ m, a moment of $3$ KN.m is applied, causing a jump in bending moment: $$ M(2^-) = 0, \quad M(2^+) = M(2^-) + 3 = 3 $$ - From $2$ m to $3$ m, no loads, so moment remains constant: $$ M(x) = 3 \quad \text{for} \quad 2 < x \leq 3 $$ - At right support $B$, moment must be zero: $$ M(3) = 0 $$ - This implies a reaction moment at support $B$ to balance the applied moment, but since it is simply supported, the moment at $B$ is zero and the beam is statically indeterminate if only moment is applied. For this problem, we consider the moment diagram as described. 7. **Summary:** - Shear force diagram is zero along the beam. - Bending moment diagram is zero from $0$ to $2$ m, jumps to $3$ KN.m at $2$ m, and remains $3$ KN.m up to $3$ m. **Final answer:** Shear force diagram: $$ V(x) = 0 $$ Bending moment diagram: $$ M(x) = \begin{cases} 0, & 0 \leq x < 2 \\ 3, & 2 \leq x \leq 3 \end{cases} $$