Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_100

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the reactions at the supports of the beam which is acted on by the combination of uniform and parabolic loading distributions. Problem 5/100 (a) Reactions at Supports A and B ● Calculation Process 1. Present Final Formula: To determine the reactions at the pin support and the roller support , we first replace the distributed loads with their equivalent concentrated loads. Let the beam start at . The supports are at and . The loading consists of: A uniform load of from to . A parabolic load from to with intensity increasing from to . This can be split into a uniform component of and a pure parabolic component starting at zero and reaching . The equivalent forces are: 2. Substitute Values: Region 1 (): at . Region 2 (): Uniform part: at . Parabolic part (from vertex at ): at . A B x=0x = A 1 m x= B 5 m w = 1 2 kN/mx=0x=2 m x=2 mx=5 m 2 kN/m6 kN/m2 kN/m 4 kN/m R = 1 w ⋅ uniform L ,x = 11 centroid of region 1 R = 2 w ⋅ uniform L ,x = 22 centroid of region 2 R = 3 w ⋅ 3 1 max,parabolic L ,x = 23 x + vertex L 4 3 2 0≤x≤2 mR = 1 (2 kN/m)(2 m)=4 kNx = 1 1 m 2≤x≤5 m R = 2 (2 kN/m)(3 m)=6 kNx = 2 2+ = 2 3 3.5 m x=2 mR = 3 (6− 3 1 2 kN/m)(3 m)=4 kNx = 3 2+ (3)= 4 3 4.25 m Total downward force: . Location of total resultant: . 3. Partial Operations: Apply equilibrium equations: 4. Final Calculation: Since the centroid of the total load () is exactly halfway between the supports () and (), the vertical reactions are equal by symmetry