fc tepat pd tendon (bagian bawah balok ) = - {( 2129 x 10^ 3 N ) / (286 X 10^ 3 mm ^2 )
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fc tepat pd tendon (bagian bawah balok ) = - {( 2129 x 10^ 3 N ) / (286 X 10^ 3 mm ^2 ) - {(2129 x 10^ 3 N x 340 mm x 340mm ) / ( 3, 39 x 10^10mm ^4 ) } + [{ 1/ 8 x 14,6 N / mm x (19.8 x 10^ 3mm ) ^2 x 340 mm } / { 3,39 x 10^ 10 mm ^ 4 }] + 3,39 N / mm ^ 2 (beban akibat balok (Wg)) = - 7,44 N / mm^ 2 - 7,25 N / mm^ 2 + 7 N / mm^ 2 + 3,39 N / mm ^ 2 = -4,3 N / mm ^ 2 = -4,3 Mpa CR = 11 x 0,87 x 0,65 x 4,3 Mpa = 26,75 Mpa Kehilangan prategang akibat Susut ( SH ) SH = ( USH ) ( SSF ) ( PSH ) USH = 27 - 3 x ( 4,415 ksi ) = 13,76 ksi > 12 ksi OKE SSF = 0, 86 V / S =3 SSF = 0,86 ----------------------Tabel 6 PSH umur 30 hari = ( AUS ) 1095 - ( AUS ) 30 = 1 - 0,42 = 0,58----------------------Tabel 7 48
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fc tepat pd tendon (bagian bawah balok ) = - {( 2129 x 10^ 3 N ) / (286 X 10^ 3 mm ^2 ) - {(2129 x 10^ 3 N x 340 mm x 340mm ) / ( 3, 39 x 10^10mm ^4 ) } + [{ 1/ 8 x 14,6 N / mm x (19.8 x 10^ 3mm ) ^2 x 340 mm } / { 3,39 x 10^ 10 mm ^ 4 }] - PowerPoint PPT Presentation
Transcript of fc tepat pd tendon (bagian bawah balok ) = - {( 2129 x 10^ 3 N ) / (286 X 10^ 3 mm ^2 )
fc tepat pd tendon (bagian bawah balok )
= - {( 2129 x 10^ 3 N ) / (286 X 10^ 3 mm ^2 )
- {(2129 x 10^ 3 N x 340 mm x 340mm ) / ( 3, 39 x 10^10mm ^4 ) }
+ [{ 1/ 8 x 14,6 N / mm x (19.8 x 10^ 3mm ) ^2 x 340 mm } / { 3,39 x 10^ 10 mm ^ 4 }]
+ 3,39 N / mm ^ 2 (beban akibat balok (Wg))
= - 7,44 N / mm^ 2 - 7,25 N / mm^ 2 + 7 N / mm^ 2 + 3,39 N / mm ^ 2
= -4,3 N / mm ^ 2 = -4,3 Mpa
CR = 11 x 0,87 x 0,65 x 4,3 Mpa = 26,75 Mpa
Kehilangan prategang akibat Susut ( SH )
SH = ( USH ) ( SSF ) ( PSH )
USH = 27 - 3 x ( 4,415 ksi ) = 13,76 ksi > 12 ksi OKE
SSF = 0, 86 V / S =3 SSF = 0,86 ----------------------Tabel 6
PSH umur 30 hari = ( AUS ) 1095 - ( AUS ) 30
= 1 - 0,42 = 0,58----------------------Tabel 7
SH = 13,76 ksi x 0,86 x 0,58 = 6,86 ksi= 47 , 3 Mpa
48
