Question: Types of Loading: 1) Point Load - A single load is acting on a point. Case 1: Free Body Diagram Beam W = mg mg N mg Ra Rb Case 2: Free Body Diagram mg mg mg Ra Rb Graph/diagram shapes and math, with position hints: top-left: title and case labels for loading diagrams; text reads 7Types of Loading:8 and 1) Point Load - A single load is acting on a point.8 and Case 1:8 plus Free Body Diagram8. center: a simply supported beam with a point load example; a car sits on the beam, with a free-body diagram showing downward $mg$ and upward $N$, and another beam diagram below with reactions $R_a$ and $R_b$. bottom-left: Case 2:8 and a second simply supported beam with three point loads; the free-body diagram shows three downward $mg$ forces and support reactions $R_a$ and $R_b$. top-right: handwritten note Free Body Diagram8 and FBD8 with small sketch of a block on a surface. right-center: small block diagram with downward arrow and label $W = mg$8. far-right: three separate free-body sketches showing a block on a surface, a block in a box with $N$ upward and $mg$ downward, and another block-on-surface sketch with $N$ upward and $mg$ downward. 2) Uniformly Distributed Load(UDL) The loadings are even distributed across the certain length. Case 1: Free Body Diagram $800$ N/m $5$ m $20$ m Equivalent Load $800 \times 5 = 4000$ N $2.5$ m $22.5$ m

1. **Problem Statement:** We are analyzing two types of loading on beams: Point Load and Uniformly Distributed Load (UDL). 2. **Point Load Case 1:** - A single point load $W = mg$ acts downward at a point on a simply supported beam. - The beam has two supports with reaction forces $R_a$ and $R_b$. 3. **Free Body Diagram (FBD) for Point Load Case 1:** - Downward force: $mg$ - Upward normal force: $N$ - Support reactions: $R_a$ and $R_b$ 4. **Point Load Case 2:** - Three point loads, each $mg$, act downward at different points on the beam. - Supports provide reactions $R_a$ and $R_b$. 5. **Uniformly Distributed Load (UDL) Case 1:** - Load intensity: $800$ N/m distributed over $5$ m length of the beam. - Total equivalent load: $$800 \times 5 = 4000$$ N acting at the centroid of the load distribution, which is at $2.5$ m from the left support. - Beam length: $20$ m 6. **Important Rules:** - For point loads, reactions are found by applying equilibrium equations: $$\sum F_y = 0$$ and $$\sum M = 0$$. - For UDL, the total load is the intensity multiplied by the length, acting at the midpoint of the distributed load. 7. **Intermediate Work for UDL Equivalent Load Position:** - The equivalent load acts at the midpoint of the $5$ m distributed load, so at $2.5$ m from the left support. 8. **Summary:** - Point loads are treated as concentrated forces. - UDLs are replaced by a single equivalent point load for analysis. This explanation covers the types of loading and their representation on beams with free body diagrams and equivalent loads.