A cantilever beam of length L = 6 m carries a triangular distributed load: zero at the free end, increasing linearly to 400 N/m at the fixed end. Find support reactions.
Ay−F+By=0cap A sub y minus cap F plus cap B sub y equals 0
Substitute values ($G = 200$, $P = 500$): $$ 3F_B - 200(2) - 500(4) = 0 $$ $$ 3F_B - 400 - 2000 = 0 $$ $$ 3F_B = 2400 $$ $$ F_B = 800 , \textN $$
: Sadrži detaljno rešene primere za srednjoškolski i viši nivo. Tehnička mehanika - Statika (PDF) statika zadaci za srednju skolu fixed
Q=q⋅lq=10 kN/m⋅3 m=30 kNcap Q equals q center dot l sub q equals 10 kN/m center dot 3 m equals 30 kN
: Sadrži detaljne primere slaganja sila i analitičkog određivanja reakcija, poput zadataka sa kuglom na strmoj ravni.
(suma vertikalnih sila): Prema gore djeluje reakcija Ry. Prema dolje djeluje sila F = 1000 N. Jednadžba: Ry – 1000 N = 0 Rješenje: Ry = 1000 N (Vertikalna reakcija jednaka je sili, što je intuitivno). A cantilever beam of length L = 6
FB⋅6 m=600 N⋅2 mcap F sub cap B center dot 6 m equals 600 N center dot 2 m FB⋅6=1200cap F sub cap B center dot 6 equals 1200 FB=200 Ncap F sub cap B equals 200 N
Prvi i najvažniji korak je "oslobađanje tela od veza". Zamenite užad, podloge i oslonce odgovarajućim silama reakcije.
These problems test the decomposition of the weight force. Tehnička mehanika - Statika (PDF) Q=q⋅lq=10 kN/m⋅3 m=30
YA+32.5−30−30=0cap Y sub cap A plus 32.5 minus 30 minus 30 equals 0 YA+32.5−60=0cap Y sub cap A plus 32.5 minus 60 equals 0
A more advanced topic involves calculating reactions at supports for beams under various loads (point loads, uniformly distributed loads). For a beam in equilibrium: ΣFx = 0, ΣFy = 0, ΣM = 0. These three equations allow you to determine unknown reactions for statically determinate beams.
October 26, 2023 Subject: Analysis of typical statics problems, methodologies, and solutions for high school physics curriculum.