P2 2013 1 - Caldeiron Prova e Resolução

d) Cálculo Swx E( Fc ) Swc = ∫ Ṅ M dx \[ \int \] Fc ) Swc = \[ - \] 1, 3, 5, 8, 1,5 = \[ - \] 1, 5 \times 1,5 = \[ - \] 1, 5, 1,5 \times 4, 8625 3 \ 3 \ 3 \[ \times \] 3 - 1, 3, \ 3, \X \ 0,9 EFE ) Swc = \ 10, 9125 D Swc = \ 10, 9125 Swc = 1, 091 \times 10^{-3} m (7) Swc = 1, 091 \times 10^{-3} m (8) --- (illegible) \( \frac{a}{s} \) \[ \begin{array}{c} 3 \ 4 \end{array} \] (4) \( \frac{b}{k} = \frac{V}{8} \) \left( il \ illegible \right) Slado de carregamento \( \eq \) \ 1 \[ T c \ Y \] \[ \c jn \] \( \frac{Z}{g} \) El momento de apoyo ΣFxo = ⇒ Hq = 0 M^{qo} = \frac{a}{s} - \frac{8}{5} + \left( 1 \right) ΣFy = \frac{-1}{3} \[ V0 = \right. M^{c,qx} ⇒ \left( \frac{1, J , 3 , 0}{3} \right) M_{o} = \frac{z}{5} m (8) (4) D- Diagrama \begin{array}{c} \ D \ E \end{array} Mo_{a} = -4m/M_{D} = 0 da) Calculs Swx EFE ) Swx = \[ - \] 1 \ 24, 5\ 1, 5\ 1, 3 \[ - \] \[ \frac{-1,3}{3} \ \, \ \[ \times \] \ \overline{\ , 0 \times 3 (4) EF\, M_{b} = 14,55 \ g Sta\,s = 1,15 \times 10^{-3} \, mod (7) \text{Ans.} = 1 .145 \times ;,10^{3} \text{nad} (12)