• Author / Uploaded
• Ivan J. Jurado Vilca

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1.- DIMENSIONAMIENTO DE LA PANTALLA:

𝜙 𝐾𝐴 = tan2 (45° − ) 2 𝐾𝐴 = tan2 (45° −

32 ) = 0.307 2

𝑠 ℎ𝑝 2 𝑀𝑢 = 1.7𝐾𝑎. 𝜌 . ℎ𝑝 + . 𝐾𝑎. ( ) 𝑐 2 3

4.50 2 𝑀𝑢 = 1.7𝑥(0.307)(1.9)(4.50/6) + 1 ( ) 𝑥(0.307) 2 3

𝑀𝑢 = 18.17 𝑡 − 𝑚 … … … … 1 ADEMAS: 𝑀𝑢 = ∅𝑏𝑑2 𝑓𝑐. 𝑤(10.59𝑤) … … … … … … . .2 𝑤=

𝑝 + 𝑦 (0.004𝑥4200) = = 0.096 𝑓𝑐 175

𝐸𝑁𝑇𝑂𝑁𝐶𝐸𝑆: 𝐷𝐸 1 𝑌 2 18.17𝑋105 = 0.9𝑥100𝑥𝑑2 𝑥175𝑥0.096(1 − 59𝑥0.096) 𝑑 = 35.69𝑐𝑚

𝑇 =𝑑+𝑟+

∅𝑎𝑐𝑒𝑟𝑜 2

𝑇 = 35.69 + 4 +

1.59 = 40.49𝑐𝑚 2

Usar: 𝑇2 = 45𝑐𝑚 𝑑 = 40.21

2.- VERIFICACIÓN POR CORTE: 𝑉𝑑𝑢 = 1.7𝑉𝑑 1 𝑠 𝑉𝑑𝑢 = 1.7( ) 𝐾𝑎. 𝜌 . (ℎ𝑝 − 𝑑)2 + 1.7 . 𝐾𝑎. (ℎ𝑝 − 𝑑) 2 𝑐 1 𝑉𝑑𝑢 = 1.7 ( ) 𝑥1.9.0.307 . (4.50 − 0.4021)2 + 1.7𝑥(1).0.307. (4.50 − 0.4021) 2 𝑉𝑑𝑢 = 10.45 𝑡𝑜𝑛

𝑉𝑐 = ∅0.53√𝑓´𝑐 𝑏𝑥𝑑 𝑉𝑐 = 0.85𝑥0.53√175 10𝑥1𝑥0.4021 𝑉𝑐 = 23.96 𝑡𝑜𝑛 𝑉𝑢 = 10.45 < 𝑉𝑎𝑑𝑚 = 23.96 (𝑜𝑘)

DIMENSIONAMIENTO DE LA ZAPATA: 𝐻𝑧 = 𝑇2 + 50𝑐𝑚 𝐻𝑧 = 50𝑐𝑚 𝐻𝑣 ≥ 𝑓𝑠𝐷 𝐻𝑎

1 𝑠 𝐵1 ≥ ( 𝐾𝑎. 𝑌ℎ2 + 𝑥𝐻𝑎𝑥𝐾𝑎)𝐹𝑠𝐷)/𝑓𝐻𝑌𝑚 + 𝑠/𝑐 𝑥𝐹 2 𝑐 1 0.307.1.9𝑥52 + (1)𝑥5𝑥0.307𝑥1.5 2 𝐵1 ≥ 0.39𝑥5𝑥2 + 1𝑥039 𝐵1 ≥ 3.086 + 𝑇 𝐵1 = 3.10𝑚 Calculo de B2 𝑀𝑟 ≥ 𝐹𝑠𝑉 𝑀𝑎 𝑠 𝐻 𝑀𝑎 = 𝐾𝑎𝑌ℎ3 + 𝑥𝐻𝑥𝐾𝑎 ( ) 𝑐 2 𝐵 𝑀𝑟 = 𝑌𝑚𝑥𝐵𝑥𝐻 ( ) 2 Entonces B2=0.001 (pequeño) 𝑃𝑒𝑟𝑜 𝐵2 𝑚𝑖𝑛𝑖𝑚𝑜 = 𝐻𝑧 𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝐵2 = 0.50

pi p1 p2 p3

peso(ton) brazo(m) 0.45x4.50x2.4 0.50x3x2.40 2.65x1.9x4.5 31.84

momento 0.725 1.8 2.275

𝐹𝑠𝐷 =

3.52 7.78 51.58 62.85

𝐻𝑟 𝐻𝑎

1 𝑠 𝐻𝑎 = 𝐾𝑎𝑝ℎ2 + 𝐻𝑥𝐾𝑎 2 𝑐 1 𝐻𝑎 = (0.307)𝑥1.9𝑥5𝑥5 + 1𝑥5𝑥. 307 2 𝐻𝑟 = 31.84𝑥0.39 + 1𝑥3.10𝑥0.39 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 ∶ 𝐹𝑠𝐷 = 1.53 > 1.50 ( 𝑜𝑘 ) Entonces 𝑀𝑎 = 15.99 𝑀𝑟 69.21 = = 4.33 > 1.75( 𝑜𝑘 ) 𝑀𝑎 15.99

PRESIONES SOBRE EL TERRENO 𝑋𝑜 = 𝑒=

𝑀𝑟 − 𝑀𝑎 69.21 − 15.99 = 𝑃 31.84 + 3.10

𝐵 3.60 − 𝑋𝑜 = − 1.52 = 0.28 2 2

𝐵 3.60 = = 0.60 > 𝑒 = 0.28(𝑜𝑘) 6 6 𝑙𝑢𝑒𝑔𝑜 𝑞1 =

𝑝 6𝑒 (1 + ) 𝐵 𝐵

𝑞1 = 14.24 𝑞2 = 5.18 𝑞1 < 𝑞𝑎𝑑𝑚. (𝑜𝑘)𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑒