Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_95

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the force and moment reactions at the support A of the cantilever beam subjected to the load distribution shown. The beam is of length , and the load intensity is given by , where at . Problem 5/95 (a) Vertical reaction force at support A 1. Present Final Formula: The vertical reaction at the support must balance the total resultant force of the distributed load. From the diagram, the total length is and the intensity is . Given , we have , so . 2. Substitute Values: 3. Partial Operations: Perform the integration: 4. Final Calculation: ● Final Conclusion: The vertical reaction force at support A is acting upwards. 2bw= k xw=w 0 x=b R A R = A R= w (x)dx ∫ 0 L L=2bw(x)=k x w(b)=w 0 w = 0 k bk= b w 0 R = A x dx ∫ 0 2b b w 0 1/2 R = A x = b w 0 [ 3 2 3/2 ] 0 2b (2b)−0 b w 0 ( 3 2 3/2 ) R = A ⋅ b w 0 ⋅ 3 2 2 b 2b R = A w b≈ 3 4 2 0 1.886w b 0 R = A w b 3 4 2 0 (b) Moment reaction at support A 1. Present Final Formula: The moment reaction at A must balance the moment produced by the distributed load about point A (located at ). 2. Substitute Values: 3. Partial Operations: Evaluate the integrals: Integral 1: Integral 2: Combine the results: 4. Final Calculation: ● Final Conclusion: The reaction moment at support A is acting counter-clockwise. ✨ Final Answer Summary (a) (Upwards) M A x=2b M = A (2b− ∫ 0 2b x)w(x)dx M = A (2b− ∫ 0 2b x) x dx= ( b w 0 1/2 ) 2b xdx− xd b w 0 ( ∫ 0 2b 1/2 ∫ 0 2b 3/2 2b x = [ 3 2 3/2 ] 0 2b 2b ⋅2 b =