Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_196

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the reactions at and for the beam subjected to the couples and distributed load. At , the distributed load is increasing at the rate of per foot. The distributed load is given by the function . The load starts at at the left end () and ends at at . A clockwise couple of is applied at the left end, and a counter-clockwise couple of is applied at the right end (). Support is a pin located at and support is a roller located at . Problem 5/196 (a) Determination of Distributed Load Function and Resultant Force 1. Formula: The distributed load is . The total resultant force is and its centroidal location is . 2. Substitution: At . At . At . 3. Calculation: . . . . from the left end. 4. Result: acting at . AB x=010 lb/ft w=k + 0 k x+ 1 k x 2 2 w=200 lb/ft x=0w=500 lb/ftx=24 ft 2700 lb-ft1200 lb-ft x=28 ftAx=4 ftB x=20 ft w(x)=k + 0 k x+ 1 k x 2 2 WW= w (x)dx ∫ 0 L =xˉ W xw(x)dx∫ x=0,w=200 lb/ft⇒k = 0 200 x= 0, = dx dw 10 lb/ft/ft⇒k = 1 10 x=24 ft,w=500 lb/ft⇒200+10(24)+k (24)= 2 2 500⇒ 440+576k = 2 500 k = 2 = 576 60 lb/ft 48 5 3 w(x)=200+10x+ x 48 5 2 W= (200+ ∫ 0 24 10x+ x)dx= 48 52 [200x+5x+ 2 x] = 144 53 0 24 4800+2880+480=8160 lb xw(x)dx=∫ (200x+ ∫ 0 24 10x+ 2 x)dx= 48 5 3 [100x+ 2 x+ 3 10 3 x] = 192 5 4 0 24 57600+46080+8640=112320 lb-ft =xˉ = 8160 112320 13.765 ft W=8160 lb=xˉ13.765 ft ● Final Conclusion: The distributed load is equivalent to a single downward force of located from the left end of the beam. (b) Support R