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Concreto Armado 205/04/2025

DIMENSIONAMENTO À FLEXÃO SIMPLES Exemplo: Determinar a armad...

DIMENSIONAMENTO À FLEXÃO SIMPLES Exemplo: Determinar a armadura longitudinal para uma viga de concreto armado, com os seguintes dados: Msd_{\text{sd}} = 25 kNm; b = 15 cm; h = 40 cm; fck_{\text{ck}} = 30 MPa; CA-50; c = 3,0 cm; Da NBR 6118 (ABNT, 2023), item 17.2.2 e Figura 8.2

αc=0,85\alpha_c = 0,85 p/ fck_{\text{ck}} ≤ 50MPa αc=0,85[1(fck50)200]\alpha_c = 0,85\left[1 - \frac{(f_{\text{ck}}-50)}{200}\right] p/ fck_{\text{ck}}> 50MPa

ηc=1\eta_c = 1 p/ fck_{\text{ck}} ≤ 40MPa ηc=(40/fck)1/3\eta_c = (40/f_{\text{ck}})^{1/3} p/ fck_{\text{ck}}> 40MPa

λ=0,8\lambda = 0,8 p/ fck_{\text{ck}} ≤ 50MPa λ=0,8fck50400\lambda = 0,8 - \frac{f_{\text{ck}}-50}{400} p/ fck_{\text{ck}}> 50MPa

DIMENSIONAMENTO À FLEXÃO SIMPLES
Exemplo: Determinar a armadura longitudinal para uma viga de concreto armado, com os
seguintes dados: M$_{\text{sd}}$ = 25 kNm; b = 15 cm; h = 40 cm; f$_{\text{ck}}$ = 30 MPa; CA-50; c = 3,0 cm;
Da NBR 6118 (ABNT, 2023), item 17.2.2 e Figura 8.2

$\alpha_c = 0,85$ 					p/ f$_{\text{ck}}$ ≤ 50MPa
$\alpha_c = 0,85\left[1 - \frac{(f_{\text{ck}}-50)}{200}\right]$ 		p/ f$_{\text{ck}}$> 50MPa

$\eta_c = 1$ 					p/ f$_{\text{ck}}$ ≤ 40MPa
$\eta_c = (40/f_{\text{ck}})^{1/3}$ 			p/ f$_{\text{ck}}$> 40MPa

$\lambda = 0,8$ 					p/ f$_{\text{ck}}$ ≤ 50MPa
$\lambda = 0,8 - \frac{f_{\text{ck}}-50}{400}$ 		p/ f$_{\text{ck}}$> 50MPa
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