pg III-54
2
< N ecesita arm adura portorsión D ISEÑ O PO R CO RTE Y TO RSIÓ N = − = Ø 2 3 =0.85 7.44( 120 22 3 2 ) = 1.8 − = 2.44 − 1.8 = 0.65 ton −m 0 = Ø 1 1 = 0.66+ 0.33 1 1 ≤1.5 = 0.66+ 0.33 12 16 = 2.9> 1.5 Se tom a =1.5 0 = 64000 0.85 1.5 112 16 4200 =0.002 2 /
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Transcript of pg III-54
𝑐 < 𝑢 Necesita armadura por torsión
DISEÑO POR CORTE Y TORSIÓN 𝑀𝑡𝑠=𝑀𝑡𝑢 − 𝑀𝑡𝑐 𝑀𝑡𝑐=Ø 𝑐 𝑥2𝑦3 =0.85𝑥7.44(120𝑥223 2)
𝑀𝑡𝑐=1.8 𝑡𝑜𝑛− 𝑚 𝑀𝑡𝑠=2.44− 1.8 = 0.65 ton− m
𝐴0𝑠 = 𝑀𝑡𝑠Ø𝑅𝑥1𝑦1𝑓𝑦
𝑅= 0.66+ 0.33𝑦1𝑥1 ≤ 1.5
𝑅= 0.66+ 0.33൬1216൰= 2.9 > 1.5
Se toma 𝑅= 1.5 𝐴0𝑠 = 640000.85𝑥1.5𝑥112𝑥16𝑥4200 = 0.002 𝑐𝑚2/𝑐𝑚
POR CORTE 𝑉𝑆 = 𝑉𝑢 − 𝑉𝑐
𝑉𝑐 = Ø𝑣𝑐𝑏𝑑= 0.85ۉ
0.53ξ175ۇට1+ 10.25 ی
19𝑥120ۊ
𝑉𝑐 = 6.3 𝑡𝑜𝑛 𝑉𝑢 = 3.33𝑥0.85𝑥19𝑥120 = 6.7 𝑡𝑜𝑛 𝑉𝑆 = 6.7− 6.3 𝑉𝑆 = 0.4 ton 𝐴0𝑠 = 4000.85𝑥19𝑥4200 = 0.006 𝑐𝑚2/𝑐𝑚
Para tener estribos de 2 ramas 2𝐴0𝑠 + 𝐴𝑣𝑠 = 2ሺ0.007ሻ+ 0.006 = 0.020 𝑐𝑚2/𝑐𝑚