Solucion Problema 3 Concreto PDF

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Summary

��, ���� ��̈�� ≤ ������ 0,����{��. ���� − ��. ����(�� ́��−����������)} ������ < �� ́�� ≤������ 0.��. ���� �� ́�� >������ 0,:: ����= ��, ������ ���� �� �� ́70 64 64 6 ���� = 4(5. 07)= 20����2 ����, ������ =0√ 2804200��40�� = 7,98����2 ������������ �� ������ (���� =20����2)≥(����, ������ = 7,...

Description

INGENIERIA CIVIL

𝒇¨ 𝒄 ≤ 𝟐𝟖𝟎

𝟎, 𝟖𝟓 𝒇´𝒄−𝟐𝟖𝟎

𝜷𝟏 {𝟎. 𝟖𝟓 − 𝟎. 𝟎𝟓 (

𝟕𝟎

)}

0,85

𝟐𝟖𝟎 < 𝒇´𝒄 ≤ 𝟓𝟔𝟎

𝟎. 𝟔𝟓

0.85

𝒇´𝒄 > 𝟓𝟔𝟎

0,65

:: 𝜷𝟏 = 𝟎, 𝟖𝟓 𝒉 70

𝒅𝒕 64

 𝐴𝑠 = 4(5.07) = 20.28𝑐𝑚2  𝐴𝑠, 𝑚𝑖𝑛 =

0.7√280 4200

𝑥40𝑥

𝒅 64

𝒅´ 6

= 7,98𝑐𝑚2

 𝐷𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒 (𝐴𝑠 = 20.28𝑐𝑚2 ) ≥ (𝐴𝑠, 𝑚𝑖𝑛 = 7,98𝑐𝑚2 ), 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑓𝑠=𝑓𝑦 = 4200 𝑘𝑔𝑓/𝑐𝑚2  𝑇𝑠 = 𝐴𝑠 𝑥𝑓𝑦 = 20.28(4200) 𝑇𝑠 = 85176𝑘𝑔𝑓  𝑎=

𝑇𝑠 0.85𝑥280𝑥80

 𝑐=

𝑎 𝛽1

= 4.47𝑐𝑚

𝑎 = 4,47𝑐𝑚

4,47

= 0,85 = 5,26𝑐𝑚

𝑐 = 5,26𝑐𝑚  𝜀𝑦 =    

𝑓𝑦

𝐸𝑠

=

4200 2100000

= 0.002

𝜀𝑡 = 0.0335 𝐷𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒 (𝜀𝑡 = 0.0335) ≥ 0.004 → 𝜀𝑡 = 0.0335 ≥ 0.004 𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝐴𝑠 ≤ 𝐴𝑠 𝑚á𝑥 𝐷𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒 (𝜀𝑡 = 0.0335) ≥ 0.005

Pág. 1

INGENIERIA CIVIL

 𝐹𝑖𝑛𝑎𝑙𝑚𝑒𝑛𝑡𝑒, 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒 (𝑎 = 4,47𝑐𝑚) ≤ (ℎ𝑓 = 12.5) 𝑎 𝑀𝑛 = 𝑇𝑠 (𝑑 − ) 2 4,47 ) 𝑀𝑛 = 85176(64 − 2 𝑀𝑛 = 52,61 𝑇𝑜𝑛. 𝑚

Pág. 2

INGENIERIA CIVIL

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INGENIERIA CIVIL

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