**Transcription for Educational Website**---**Problem Statement:**Determine the shear force at point C if P = 3.3 kN/m.**Diagram Explanation:**The diagram shows a simply supported beam with triangular distributed loading. The structure includes:- **Supports:** - Support A (left support) and support B (right support) are represented as rollers, indicating that the beam is simply supported.- **Load Distribution:** - The triangular distributed load acts over the entire length of the beam, starting from zero at the left end (near support A) and increasing linearly to a maximum value at the right end (near support B). - The maximum intensity of the distributed load is labeled as P = 3.3 kN/m at the right end.- **Dimensions:** - The beam is divided into equal sections of 1.5 meters each: - From A to C: 1.5 meters - From C to an unspecified point to the right: 1.5 meters - Total length from A to B: 4.5 meters- **Important Points:** - Point C is the specific point of interest where the shear force needs to be determined. It is located at 1.5 meters from the left support (A).This setup requires the calculation of shear force resulting from the triangular distributed load, taking into account the equilibrium of the beam and applying principles of static equilibrium.

To find: The shear force at point C. Given: The value of force is P=3.3 kN. Free body diagram:Calculation: The clockwise moment taken as positive and anticlockwise moment takes as negative. Take the moment about a point B. RA×3-12×P×4.5×1.5+12×P×1.5×0.5=0RA×3-12×3.3×4.5×1.5+12×3.3×1.5×0.5=0RA=3.3 kN Balance the force in vertical direction. RA+RB=12×P×4.5+12×P×1.53.3+RB=12×3.3×4.5+12×3.3×1.5RB=6.6 kNThe downward force consider as negative and upward force consider as positive. The weight at point C can be obtained as, 3.34.5=w32.2 kNm=w The shear force at point C from left can be obtained as, VC=-12×w×3+RA=-12×2.2×3+3.3=0 kN Thus, the shear force at point C is 0 kN.