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Pengaruh Geser thd balok Beton

Pengaruh Geser thd balok Beton. 2ton. 2ton. Gaya geser. 3. 1. 2ton. 2ton. 2ton. 2ton. 2ton. 4. 2. 2ton. 2ton. Gaya geser. 2ton. 4ton. 2ton. Bid “ D “ ( “Vu “). Gaya geser. Bidang ' D '. Gaya geser. 30 ton. 20 ton. 10 ton. 1/3L. 2/3 L. 20 ton. 10 ton. 10 ton. 20 ton.

uriel-whitfield
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Pengaruh Geser thd balok Beton

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  1. Pengaruh Geser thd balok Beton

  2. 2ton 2ton Gaya geser 3 1 2ton 2ton 2ton 2ton 2ton 4 2

  3. 2ton 2ton Gaya geser 2ton 4ton 2ton Bid “ D “ ( “Vu “)

  4. Gaya geser Bidang ' D '

  5. Gaya geser

  6. 30 ton 20 ton 10 ton 1/3L 2/3 L 20 ton 10 ton 10 ton 20 ton Bid " Vu " 20 ton 20 ton 10 ton 10 ton Gaya geser

  7. Gaya geser Ra Rb

  8. Gaya geser Ra Rb

  9. Gaya geser Ra Rb

  10. Gaya geser Ra Rb

  11. Gaya geser Ra Rb

  12. Gaya geser Ra Rb

  13. 120 kg Gaya geser 2,5 m 2,5 m 120 kg 60 kg 60kg

  14. 120 kg Gaya geser 72 kg 48 kg 3m 2m 120 kg 72 kg 48 kg Bidang D

  15. 120 kg Gaya geser 4m 1m 90 kg 120 kg 30 kg Bidang D

  16. Gaya geser q= 30kg / m L = 5 m 75 ton 75 ton

  17. 2m 2,5 m Gaya geser “Vu” max jarak 0m --- ½ m dr tepi 120 kg 90 kg 75 kg 2 / 2,5 x 75 kg = 60 kg 75 kg Bidang ' D ' Gaya geser 0,5 m 2,5 m

  18. Gaya geser 2m Bid “Vu” max jarak 0m --- ½ m dr tepi 12000 kg 9000 kg 3000 kg 7500kg 2 / 2,5 x 7500 kg = 6000 kg Bidang ' D ' 7500 kg Rata rata = (7500+6000)/2=6750 kg

  19. ‘D’ maxyg terjadi dari tepi jarak ‘ h m’=0 m h m=0 m P’=16154 kg/lebar balok T q=3873 kg/m /lebar balok T 7.5 m 3.75 m 7.5 m 7.5 / 7.5 x 16154 = 16154kg 0 kg ½ x 7.5 x 3873 = 14524 kg Bidang ' D ' 14524kg D tot 0 m dr tepi = 16154kg + 14524 kg = 30678 kg

  20. ‘D’ maxyg terjadi dari tepi jarak ‘ 1m’ 1m q=3873 kg per 1 balok T P’=16154 kg per 1 balok T 7.5 m 3.75 m 1m 6.5 m 6.5 / 7.5 x 16154 =14000kg 16154 kg ½ x 7.5 x 3873 = 14524 kg 2.75 / 3,75 x 14524 kg = 10651 kg Bidang ' D ' D tot 1m dr tepi = 14000+ 10651 = 24651kg

  21. ‘D’ maxyg terjadi dari tepi jarak 2m P’=16154 kg per 1 balok T 2m q=3873 kg per 1 balok T 7.5 m 3.75 m 5.5 m 2m 5.5 / 7.5 x 16154 =11846 kg 16154 kg 1.75/3.75x14524=6778kg 14524 kg 10651 kg Bidang ' D ' 14524 kg D tot 2m dr tepi = 11846 + 6778 = 18624 kg

  22. ‘D’ maxyg terjadi dari tepi jarak 3m P’=16154 kg per 1 balok T q=3873 kg per 1 balok T 3m 7.5 m 3.75 m 4.5 m 3m 4.5 / 7.5 x 16154 = 9692 kg 16154 kg 0.75/3.75 x 14524 = 2905 kg 14524 kg 10651 kg 6778kg 14524 kg D tot 3m dr tepi = 9692 + 2905 = 12597 kg Bidang ' D '

  23. Kuat geser Beton ( Vn ) 600 mm ( d ) 28 600 mm ( d ) 300 mm ( bw ) s d d 28 fy = 414 N / mm² fc’ = 27,5 N / mm² 300mm Tahanan geser max yg bisa ditahan beton c = fc’ / 6 N / mm²=  27,5 / 6 =0.87N/ mm² . Kuat geser max yg bisa ditahan beton ( V c )=(  c ) x bw x d Vc =0.87 x 600 x 300 =156600 N As=luas pot tulangan begel s=jarak begel /sengkang d = tinggi balok fy =kekuatan baja per mm² Kuat geser max yg bisa ditahan 2 tul. baja  8 mm tegak( satu.begel) Vb =As x fy = 2 x ¼  8² x 414 = 41598,72 Newton Jumlah begel selebar bid geser = nb =d/s batang Kuat geser max yg bisa ditahan tul. baja ( begel/sengkang ) ,selebar bidang geser ( V s ) = Vb x nb = As x fy x d/s s = V s / (As x fy x d)

  24. 2.7 2 200mm C = 200x 27,5=3500 N (Kekuatan batas beton luas 200 mm ) 2 13

  25. Kuat Geser beton animasi

  26. Kuat Geser beton animasi

  27. Kuat Geser beton d D( Vu ) animasi D bw Gaya geser terfaktor yg terjadi pd pot tertentu (Vu) = D Newton

  28. Kuat Geser beton animasi

  29. Kuat Geser beton animasi

  30. Kuat Geser beton bw D D( Vu ) Luas pot Geser = bw x d d animasi

  31. Kuat Geser Tulangan geser animasi Kuat geser 1 tulangan geser tegak = Asx fy As = ¼  x  ²xfy As =luas pot =1/4 x x  ²  =diameter tul (mm)

  32. Kuat Geser Tulangan geser Tul tegak Tul tegak Tul tegak Tul tegak Jumlah tul tegak setinggi d = d/s buah Kekuatan 1 baja tegak = As x fy ( newton) Untuk d/s buah, bisa menaha geser= As x fy x d/s N Sumbangan kuat geser tul tegak pada luas pot geser ,bila dipasang setiap jarak s mm . Vs=. = As x fy x d / ( s ) Newton

  33. bw (  c )= fc’ / 6 N / mm² Kuat Geser beton D D( Vu ) d bw d Luas bid geser = bw x d mm² Vc =  fc’ / 6 x bw x d N

  34. d D d D( Vu ) s s bw Luas pot Geser = bw x d Jumlah tul tegak setinggi d = d/s buah Kekuatan 1 baja tegak = As x fy ( newton) Untuk d/s buah, bisa menaha geser= As x fy x d/s N Sumbangan kuat geser tul tegak pada luas pot geser ,bila dipasang setiap jarak s mm . Vs=. = As x fy x d / ( s ) Newton Kekuatan 1 baja tegak = As x fy (newton) As = ¼  x  ²

  35. Kuat Geser Tulangan geser Sumbangan kuat geser tul begel pada luas pot geser s Kekuatan 1 baja tegak = As x fy N Untuk nb buah begel, bisa menahan geser= 2x nb x As x fy N =2xnb x (1/4x  x D²)x fy N Jumlah tul begel per potongan nb = d/s buah nb =jml tul begel per pot geser d= tinggi balok dlm mm S=jarak pemasangan begel dlm mm D= Diameter tulangan begel dlm mm Fy=tegangan ijin baja dlm Newton / mm² N =satuan newton

  36. Kuat Geser beton bw D D( Vu ) Luas pot Geser = bw x d d animasi Gaya geser terfaktor yg terjadi pd pot tertentu (Vu) = D Newton Total Penahan Geser:  Vn =  ( Vc + Vs )  = 0,8 . Vu harus <  Vn Oke

  37. Gaya geser yg terjadi (Vu ) Kuat geser ( Vn ) Gaya Geser max jarak 0.5m ---1½ m dr tepi 28 150mm ( s = 150 mm ). P = 12000 kg 500 mm ( d ) q= 3000kg / m 500 mm ( d ) fy= 414 N / mm² fc’ = 27,5 N / mm² L = 5 m 300 mm ( bw ) 1,5 m 300 mm ( bw ) Tahanan geser max yg bisa ditahan beton ( c )= fc’ / 6 N / mm²=  27,5 / 6 =0.87N/ mm² Kuat geser max beton ( V c )=(  c ) x bw x d Vc =0.87 x 500 x 300 =130500 N 8400 kg 12000 kg 3600 kg Bid ‘D’pd muatan terpusat Gaya geser = 8400 kg 1 / 2,5 x 750kg = 3000 kg 7500kg Kuat geser max yg bisa ditahan 2 tul. baja ( sengkang ) =As x fy = 2 x ¼ 8² x 414 = 41598,72 Newton Jumlah Sengkang selebar bid geser = d/s batang Kuat geser max yg bisa ditahan tul. baja ( sengkang ) ,selebar bidang geser ( V s ) =As x fy x d /s = 41598.72 x 500/150=138662,40 Newton 1,0 m 6000 kg 7500kg 2,5 m Bid ‘D’ pd muatan terbagi rata Rata rata Gaya geser=6000+3000) / 2 = 4500 kg Total Gaya geser =8400 kg + 4500 kg = 12900 kg = 129000 newton ( 1kg = 10 newton ) Vu = 129000 Newton  Vn =  ( Vc + Vs ) = 0,8 .( 130500 +138662,40 ) = 215329,9 Newton Vu <  VnOke

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