November 18 2024

>> YOUR LINK HERE: ___ http://youtube.com/watch?v=n0kaUty99Ss

Se da bara A - B cu lungimea AB = l , in punctul A este o incastrare • Pe bara A – B actioneaza o forta distribuita q = constant. Sa se calculeze rotirea sectiunii transversale din punctul B , fiB = ? • Diagrama de eforturi interioare T si Miz • Se traseaza direct • Regiunea I , adica tronsonul AB cu x avand originea in punctul B si variaza de la 0 ; l • Eforturile interioare T cu sensul pozitiv in jos si Miz cu sensul pozitiv in sensul trigonometric , aici se ia din partea dreapta a barei • In punctul B al barei se pune un moment concentrat fictiv in sensul trigonometric MizB = 0kNm • Miz x = MizB – qx^2/2 ; dMizx /MizB = 1 ; Lt = suma de integrale din M^2izxi*dxi/2*EiIzi ; • fiB = DLt /dMizB = suma de integrale din 2 Mizxi * dMizx /dMizB* dxi /2EiIzi = = suma de integrale din Mizxi * dMizx /dMizB* dxi /EiIzi ; • fiB = DLt /dMizB = integrala de la 0 ;l din MizB – qx^2/2 inmultit cu 1*dx/EIz ; • fiB =1/EIz* integrala de la 0 ;l din MizB – qx^2/2 *dx ; • MizB = 0kNm se inlocuieste • fiB =1/EIz* integrala de la 0 ;l din – qx^2/2 *dx • fiB =1/EIz inmultit cu -q/2*x^3/3 de la 0;l • fiB =1/EIz *-q/2*l^3/3 = - ql^3/6Eiz este rotirea sectiunii transversale din punctul 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