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UNIVERSIDAD CENTRAL DEL ECUADOR FACULTAD DE INGENIERÍA, CIENCIAS FÍSICAS Y MATEMÁTICA CARRERA DE INGENIERÍA CIVIL HORMIGÓN ARMADO 1 ING. JORGE FRAGA

DEBER No. 2 TEMA: ANÁLISIS DE SECCIÓN DE VIGAS

NOMBRE: GUAMÁN SINCHI JOFFRE JAVIER

CURSO: QUINTO

PARALELO: SEGUNDO

FECHA DE ENTREGA: LUNES, 14 DE MAYO DE 2018

ABRIL 2018 – SEPTIEMBRE 2018

EJERCICIO 1: Hallar el momento nominal de la siguiente viga. Datos: Rec: 30mm f’c: 24 MPa Es: 2x10^5 MPa fy: 420 MPa As: 2 Ф 20 + 1Ф16 (mm) Фe = 10 (mm)

mm

ÀREA: A1: A2:

𝟐𝜫 Ф² 𝟒 𝜫 Ф² 𝟒

=

=

𝟐𝜫 (20)² 𝟒

𝜫 (16)² 𝟒

= A1= 628.319 mm²

= A2= 201.06 mm²

CENTROIDE: Ῡ:

𝜮𝑨𝒊 𝑨𝒕

=

(𝟔𝟐𝟖.𝟑𝟏𝟗)(𝟓𝟎)+(𝟐𝟎𝟏.𝟎𝟔)(𝟒𝟖) 𝟖𝟐𝟗.𝟑𝟖

d = 500-49.4 = 450.6 mm

= Ῡ= 49.52 mm

ANÁLISIS 1er CASO FALLA POR TRACCIÓN fs = fy

C = a/β

Cc = Ts

C= 68.30/0.85

0.85f’c (a)(b) = As . fy

C = 80.35 mm

𝑨𝒔.𝒇𝒚

a = 𝒃.(𝟎.𝟖𝟓)(𝒇′ 𝒄) =

(𝟖𝟐𝟗.𝟑𝟖)(𝟒𝟐𝟎) 𝟐𝟓𝟎.(𝟎.𝟖𝟓)(𝟐𝟒)

a= 68.30 mm Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄) £𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

£y = fy/Es £y = 420/(2x10^5) £y = 0.0021 mm/mm

0.003(450.48 − 80.35) 80.35

£s= 0.0138 mm/mm

Comparación £𝒔 𝑽𝒔 £𝒚 0.0138 > 0.0021 (falla por tracción) si cumple

Momento Nominal 𝒂 𝒂 𝑴𝒏 = 𝑨𝒔. 𝒇𝒚. (𝒅 − ) = 𝒂. 𝒃. (𝟎. 𝟖𝟓)(𝒇′𝒄) (𝒅 − ) 𝟐 𝟐 𝑀𝑛 = (829.38). (420). (450.48 −

68.30 68.30 ) ( 1x10^ − 6) = (68.30). (250). (0.85)(24) (450.48 − ) 2 2 Mn= 145.02 kN.m

EJERCICIO #1

Cc =348.33 kN

d = 450.48

500

s

mm

0.85*f 'c

250

= 0.0138

Ts = 348.34 kN

a = 68.30

= 0.003 c = 80.35

c

f'c=24MPa fy=420MPa Es=2x105MPa Rec=30mm As=2 20+1 16 E=10mm

EJERCICIO 2: hallar el momento nominal de la siguiente viga. Datos: Rec: 25 mm f’c: 21 MPa Es: 2x10^5 MPa fy: 420 MPa As: 2 Ф 32 + 2Ф22 (mm)

500

Фe = 10 (mm)

mm ÀREA: A1: A2:

𝟐𝜫 Ф² 𝟒 𝟐𝜫 Ф² 𝟒

= =

𝟐𝜫 (32)² 𝟒 𝜫 (22)² 𝟒

= A1= 1608.5 mm² = A2= 760.27 mm²

CENTROIDE: Ῡ:

𝜮𝑨𝒊 𝑨𝒕

=

(𝟏𝟔𝟎𝟖.𝟓)(𝟓𝟏)+(𝟕𝟔𝟎.𝟐𝟕)(𝟒𝟔) 𝟐𝟑𝟔𝟖.𝟕𝟔

= Ῡ= 49.4 mm

250

ANÁLISIS 1er CASO FALLA POR TRACCIÓN fs = fy

C = a/β

Cc = Ts

C= 222.94/0.85

0.85f’c (a)(b) = As . fy

C = 262.28 mm

𝑨𝒔.𝒇𝒚

a = 𝒃.(𝟎.𝟖𝟓)(𝒇′ 𝒄) =

(𝟐𝟑𝟔𝟖.𝟕𝟔)(𝟒𝟐𝟎) 𝟐𝟓𝟎.(𝟎.𝟖𝟓)(𝟐𝟏)

a= 222.94 mm Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄) £𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

£y = fy/Es £y = 420/(2x10^5) £y = 0.0021 mm/mm

0.003(450.6 − 262.28) 262.28

£s= 0.0022 mm/mm

Comparación £𝒔 𝑽𝒔 £𝒚 0.0022 > 0.0021 (falla por tracción) si cumple

Momento Nominal 𝒂 𝒂 𝑴𝒏 = 𝑨𝒔. 𝒇𝒚. (𝒅 − ) = 𝒂. 𝒃. (𝟎. 𝟖𝟓)(𝒇′𝒄) (𝒅 − ) 𝟐 𝟐 𝑀𝑛 = (2368.76)(420) (450.6 −

222.94 222.94 ) ( 1x10^ − 6) = (222.94). (250). (0.85). (21). (450.6 − ) 2 2 Mn= 337.39 kN.m

EJERCICIO #2

d

500

0.85*f 'c a

= 0.003 c = 262.28

c

Cc= 994.87 kN.m f'c=21MPa fy=420MPa Es=2x105MPa Rec=25mm As=2 32+2 22 E=10mm

Ts = 994.88 kN.m s=

mm

250

0.0022

EJERCICIO 3: Hallar el momento nominal de la siguiente viga. Datos: Rec: 25 mm f’c: 35 MPa Es: 2x10^5 MPa fy: 420 MPa As: 4 Ф 32 + 6Ф22 (mm) Фe = 10 (mm) Фsep = 25 (mm) E’s = 3 Фe 22 (mm)

ÀREA: A1: A2: A3: A’s:

𝟒𝜫 Ф² 𝟒 𝜫 Ф² 𝟒

=

𝟓𝜫 Ф² 𝟒 𝟑𝜫 Ф² 𝟒

=

𝟒𝜫 (32)² 𝟒

𝜫 (22)² 𝟒

= =

= A2= 380.13 mm²

𝟓𝜫 (22)² 𝟒 𝟑𝜫 (22)² 𝟒

= A1= 3216.99 mm²

= A3= 1900.66 mm² = A’s= 1140.40 mm²

CENTROIDE: Ῡ:

𝜮𝑨𝒊 𝑨𝒕

=

(𝟑𝟐𝟏𝟔.𝟗𝟗)(𝟓𝟏)+(𝟑𝟖𝟎.𝟏𝟑)(𝟒𝟔)+(𝟏𝟗𝟎𝟎.𝟔𝟔)(𝟏𝟎𝟑) 𝟓𝟒𝟗𝟕.𝟕𝟗

d = 500-68.63 = 431.37 mm d’ = 25+10+11 = 46 mm

= Ῡ= 68.63 mm

ANÁLISIS 1er CASO fs = fy ^ f’s = fy

C = a/β

Cc + Cs= Ts

C= 205.05/0.8

0.85f’c (a)(b) + (A’s)(f’s)= As . fy

C = 256.31 mm

0.85f’c (a)(b) + (A’s)(f’y)= As . fy 𝒇𝒚 (𝑨𝒔−𝑨′ 𝒔)

a = 𝒃.(𝟎.𝟖𝟓)(𝒇′ 𝒄) =

𝟒𝟐𝟎 (𝟓𝟒𝟗𝟕.𝟕𝟗−𝟏𝟏𝟒𝟎.𝟒) 𝟑𝟎𝟎.(𝟎.𝟖𝟓)(𝟑𝟓)

a = 205.05 mm Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄)

£y = fy/Es £y = 420/(2x10^5)

£𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

£y = 0.0021 mm/mm

0.003(431.37 − 256.31) 256.31 £s= 0.002 mm/mm

Comparación £𝒔 𝑽𝒔 £𝒚 0.002 < 0.0021 NO CUMPLE

2do CASO F’c = Es . £’s

fs = fy ^ f’s ≠ fy

£’s =

Cc + Cs = Ts 0.85f’c (a)(b) + (A’s)(f’s)= As . fs 0.85f’c (β)(c)(b) + (A’s)(Es)( 0.85(35) (0.8)(c)(300) + (1140.4)( 2x10^5)(

)= As . fy

0.003(𝑐−46)

Despejando “c” c= 245.52 mm

𝑐

𝒄

a = β.c

£𝒄(𝒄−𝒅′) 𝒄

£𝒄(𝒄−𝒅′)

)= (5497.79) . (420)

Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄)

£y = fy/Es £y = 420/(2x10^5)

£𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

£y = 0.0021 mm/mm

0.003(431.37 − 245.52) 245.52 £s= 0.0023 mm/mm £𝒄 £′𝒔 = 𝒄 (𝒄 − 𝒅′) £′𝒔 =

£′𝑠 =

£𝒄(𝒄 − 𝒅′) 𝒄

0.003(245.52 − 46) 245.52

£’s= 0.0023 mm/mm Comparación £𝒔 𝑽𝒔 £𝒚 0.0023 > 0.0021 NO CUMPLE 3er CASO F’c = Es . £s

fs ≠ fy ^ f’s ≠ fy

£s =

Cc + Cs = Ts 0.85f’c (a)(b) + (A’s)(f’s)= As . fs 0.85f’c (β)(c)(b) + (A’s)(fy)= As . (Es)(

£𝒄(𝒅−𝒄) 𝒄

a = β.c

£𝒄(𝒄−𝒅′) 𝒄

)

0.85(35) (0.8)(c)(300) + (1140.4)(420)= A(5497.79) . (2x10^5)(

0.003(431.37−𝑐) 𝑐

)

Despejando “c” c= 254.38 mm Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄) £𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

0.003(431.37 − 254.38) 254.38

£s= 0.00209 mm/mm

£y = fy/Es £y = 420/(2x10^5) £y = 0.0021 mm/mm

£𝒄 £′𝒔 = 𝒄 (𝒄 − 𝒅′) £′𝒔 = £′𝑠 =

£𝒄(𝒄 − 𝒅′) 𝒄

0.003(254.38 − 46) 254.48

£’s= 0.0025 mm/mm

Comparación £𝒔 𝑽𝒔 £𝒚 0.00209 < 0.0021 SI CUMPLE (Zona Plástica)

£′𝒔 𝑽𝒔 £𝒚 0.0025 > 0.0021 SI CUMPLE (Zona Plástica)

Momento Nominal 𝒂

𝑴𝒏 = [𝒂. 𝒃. (𝟎. 𝟖𝟓)(𝒇′𝒄) (𝒅 − 𝟐) + A’s(fy)(d-d’)] 𝑀𝑛 = [(203.504). (300). (0.85)(35) (431.37 −

203.504 2

) + (1140.4)(420)(431.37-46)]( 1x10^-6)

Mn= 783.26 kN.m

Ejercicio 4: hallar el momento nominal de la siguiente viga. Datos: Rec: 30mm f’c: 21 MPa Es: 2x10^5 MPa fy: 420 MPa As: 4 Ф 28 Фe = 10 (mm)

ÀREA: A1:

𝟒𝜫 Ф² 𝟒

=

𝟒𝜫 (28)² 𝟒

= A1= 2463.01 mm²

ANÁLISIS 0.85f’c (Ahc)= As . fs fs = fy Cc = Ts 0.85f’c (Ahc)= As . fy Ahc = Ahc =

𝑨𝒔.𝒇𝒚 (𝟎.𝟖𝟓)(𝒇′ 𝒄)

(𝟐𝟒𝟔𝟑.𝟎𝟏).(𝟒𝟐𝟎) (𝟎.𝟖𝟓)(𝟐𝟏)

Ahc= 57953.18 mm² Á𝒓𝒆𝒂 𝒅𝒆𝒍 𝒂𝒍𝒂

Ahc VS Aala

Aala= 450*100

57953.18 > 45000

Aala = 45000 mm²

Sección tipo “T”

𝑩𝒖𝒔𝒄𝒂𝒏𝒅𝒐 "𝒂"

C = a/β

150*a= 12953.18

C= 86.35/0.85

a = 86.35 mm

C = 101.59 mm

CENTROIDE: Ῡ:

𝜮𝑨𝒊 𝑨𝒕

=

(𝟒𝟓𝟎𝟎)(𝟓𝟎)+(𝟏𝟐𝟗𝟓𝟑.𝟏𝟖)(𝟏𝟎𝟎+ 𝟓𝟕𝟗𝟓𝟑.𝟏𝟖

𝟖𝟔.𝟑𝟓 ) 𝟐

= Ῡ= 70.83 mm

Análisis de Deformación £𝒄 £𝒔 = 𝒄 (𝒅 − 𝒄) £𝒄(𝒅 − 𝒄) £𝒔 = 𝒄 £𝑠 =

0.003(496 − 101.59) 101.59

£y = fy/Es £y = 420/(2x10^5) £y = 0.0021 mm/mm

£s= 0.012 mm/mm Comparación £𝒔 𝑽𝒔 £𝒚 0.012 > 0.0021 (fluencia) si cumple Momento Nominal 𝑴𝒏 = 𝑨𝒔. 𝒇𝒚. (𝒅 − Ῡ) = 𝑨𝒉𝒄. (𝟎. 𝟖𝟓)(𝒇′𝒄)(𝒅 − Ῡ) 𝑀𝑛 = (2463.01). (420). (496 − 70.83)( 1x10^ − 6) = (57953.18). (0.85)(21)(496 − 70.83)( 1x10^ − 6) Mn= 439.83 kN.m

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