Mekanika Bahan - Analisa Tegangan Bidang
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Transcript of Mekanika Bahan - Analisa Tegangan Bidang
![Page 1: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/1.jpg)
Analisa Tegangan Bidang(tegangan kombinasi normal
dan geser)
Kuliah 14
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Mencari tegangan-tegangan utama (tegangan maksimum dan minimum) pada batang yang menderita tegangan
normal dan tegangan geser
![Page 3: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/3.jpg)
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
![Page 4: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/4.jpg)
Arah tegangan normal yang tegak lurus dengan sx1 yaitu sy1 adalah
tegangan normal yang berbeda sudut 90o. Nilai tegangan sy1
dapat dihitung dengan memasukkan q = q + 90o.
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
![Page 5: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/5.jpg)
σx-σyτxy θ tg
θ τxy*θyxdθσx
22
02cos22sin1
ss
Untuk mencari tegangan maksimum dan minimum, maka persamaan sx1 dan sy1 diturunkan terhadap q kemudian disamakan dengan nol
![Page 6: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/6.jpg)
σx-σyτxy θ tg
θ τxy*θyxdθσy
22
02cos22sin1
ss
![Page 7: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/7.jpg)
σx-σyτxy θ tg 22
22
max/min 2σyσx
2σyσxσ xy
2
2
2
22cos
2
2sin
xyyx
yx
xyyx
xy
ss
ss
q
ss
q
Dengan memasukkan harga sin2q dan cos 2q maka akan diperoleh smax/min
![Page 8: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/8.jpg)
Untuk mencari tegangan geser maksimum turunkan persamaan x1y1 terhadap q kemudian disamakan dengan nol
xy
yx
tg
θτxy*θσx-σydθ
yxd
ss
q
22
2sin22cos11
![Page 9: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/9.jpg)
xy
yx
tg
ss
q
22
2
2
2
2cos
2
22sin
xyyx
xy
xyyx
yx
ss
q
ss
ss
q
Dengan memasukkan harga sin2q dan cos 2q maka akan diperoleh xymax
22max τxy)
2σy-σx(τ
![Page 10: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/10.jpg)
22max τxy)
2σy-σx(τ
22
max/min 2σyσx
2σyσxσ xy
22
min 2σyσx
2σyσxσ xy
22
max 2σyσx
2σyσxσ xy
2minmaxτmax
ss
![Page 11: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/11.jpg)
Contoh : sebuah titik menderita tegangan sx = 11 MPa, sy = 4 MPa dan xy = 2.8 MPa dengan arah seperti terlihat pada gambar di bawah.1. Tentukan nilai-nilai tegangan pada sudut q = 36o dan gambarkan arah-arah tegangan
tersebut.2. Tentukan dan gambarkan arah tegangan-tegangan utama (tegangan max dan min
pada titik tersebut.
![Page 12: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/12.jpg)
MPa2.463- cos72*8.2)sin722
4-11(τx1y1
MPa 11.244sin72*8.2cos722
4112
411σx1
MPa 756.3sin72*8.2cos722
4112
411σy1
(1)
![Page 13: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/13.jpg)
MPa2.463- cos72*8.2)sin722
4-11(τx1y1
MPa 11.244sin72*8.2cos722
4112
411σx1
MPa 756 . 3 sin72 *8. 2 cos722
4 112
4 11 σy1
(1)
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MPa0.000015- cos38.66*8.2)sin38.662
4-11(τx1y1
MPa 11.982sin38.66*8.2cos38.662
4112
411σmax
(2) o
σx-σyτxy θ tg
33.1966.382
8.04118.2*222
MPa018.3sin38.66*8.2cos38.662
4112
411σmin
Perhitungan tegangan normal maksimum,
minimum
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(2) Gambar arah tegangan normal maksimum,
minimum
![Page 16: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/16.jpg)
MPa4.482 )cos(-51.34*8.24))sin(-51.32
4-11(τmax
MPa5.7)sin(-51.34*8.2)cos(-51.342
4112
411σx1
(2)oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
MPa5.7)sin(-51.34*8.2)cos(-51.342
4112
411σy1
Perhitungan tegangan geser maksimum
![Page 17: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/17.jpg)
(2) Penggambaran tegangan pada saat tegangan
geser mencapai nilai maksimum
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(2)MPa482.48.2
2411 2
2
max
MPa982.11482.45.7σmax
MPa018.3482.45.7σmax
MPa482.42
018.3982.11τmax
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o
σx-σyτxy θ tg
33.1966.382
8.04118.2*222
qq oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
Perputaran sumbu X1
berbeda 45o
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o
σx-σyτxy θ tg
33.1966.382
8.04118.2*222
qq oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
Pada perhitungan smax/min, akan
dihasilkan nilai x1y1 = 0 (nol)
Pada perhitungan max, akan dihasilkan sx1 =
sy1
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o
σx-σyτxy θ tg
33.1966.382
8.04118.2*222
qq oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
Berapapun nilai q selalu akan diperoleh sx1+sy1 = smax + smin = sx + sy
![Page 22: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/22.jpg)
Perjanjian arah tegangan dan perputaran sudut q
![Page 23: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/23.jpg)
Contoh : sebuah titik menderita tegangan sx = 11 MPa, sy = 4 MPa dan xy = 2.8 MPa dengan arah seperti terlihat pada gambar di bawah.1. Tentukan dan gambarkan arah tegangan-tegangan utama (tegangan max dan min
pada titik tersebut.
![Page 24: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/24.jpg)
MPa0.000015 cos38.66*)8.2()sin38.662
4-11(τx1y1
MPa 11.982sin38.66*)8.2(cos38.662
41-12
4-11σmax
o
σx-σyτxy θ tg
33.1966.382
8.0411
)8.2(*222
MPa018.3sin38.66*)8.2(cos38.662
41-12
4-11σmin
Perhitungan tegangan normal maksimum,
minimum
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Gambar tegangan normal maksimum,
minimum
![Page 26: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/26.jpg)
MPa4.482- )cos(-51.34*)8.2(4))sin(-51.32
4-11(τmax
MPa5.7)sin(-51.34*)8.2()cos(-51.342
41-12
4-11σx1
oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
MPa5.7)sin(-51.34*)8.2()cos(-51.342
41-12
4-11σy1
Perhitungan tegangan geser maksimum
![Page 27: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/27.jpg)
Penggambaran tegangan pada saat tegangan
geser mencapai nilai maksimum
![Page 28: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/28.jpg)
Contoh : sebuah titik menderita tegangan sx = 11 MPa, sy = 4 MPa dan xy = 2.8 MPa dengan arah seperti terlihat pada gambar di bawah.1. Tentukan dan gambarkan arah tegangan-tegangan utama (tegangan max dan min
pada titik tersebut.
![Page 29: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/29.jpg)
MPa0.000015- 38.66-cos*8.238.66-)sin2
4-11(τx1y1
MPa 982.1138.66-sin*8.238.66-cos2
41-12
4-11σmax
o
σx-σyτxy θ tg
33.1966.382
8.0411)8.2(*222
MPa018.338.66-sin*8.238.66-cos2
41-12
4-11σmin
Perhitungan tegangan normal maksimum,
minimum
![Page 30: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/30.jpg)
Penggambaran tegangan normal maksimum,
minimum
![Page 31: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/31.jpg)
MPa4.482 cos51.34*8.2)sin51.342
4-11(τmax
MPa5.7sin51.34*8.2cos51.342
41-12
4-11σx1
oo
xy
yx
tg
67.2534.512
25.18.2
2411
22
ss
q
MPa5.7sin51.34*8.2cos51.342
41-12
4-11σy1
Perhitungan tegangan geser maksimum
![Page 32: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/32.jpg)
Penggambaran tegangan pada saat tegangan
geser mencapai maksimum
![Page 33: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/33.jpg)
ContohSebuah balok dengan bentang 16 meter menderita beban terpusat di tengah sebesar 100 kN. Hitung dan gambarkan arah tegangan utama pada titik 1, 2, 3, 4 dan 5 pada jarak 4 meter dari titik A (Potongan I-I).
![Page 34: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/34.jpg)
Contoh
Ix = 1/12 * 0.4*0.83 = 0.0170667 m3
S2 = S4 = 0.4*0.2*0.3 m3= 0.024 m3
S3 = 0.4*0.4*0.2 m3 = 0.032 m3
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s1 = 200*0.4/0.0170667 kN/m2 = 4687.491 kN/m2 = 4.688 Mpa (tekan)s2= ½* 4.688 Mpa = 2.344 MPa(tekan)s3 = 0 Mpas4 = 2.344 Mpa (tarik)s5 = 4.688 Mpa (tarik)
Perhitungan tegangan normal
![Page 36: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/36.jpg)
Perhitungan tegangan geser
1 = 0 Mpa 2= 50*0.024/(0.4*0.0170667) kN/m2 = 175.781 kN/m2 = 0.176 MPa3 = 50*0.032/(0.4*0.0170667) kN/m2 = 234.374 kN/m2 = 0.234 MPa 4 = 0.176 MPa5 = 0 MPa
![Page 37: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/37.jpg)
Perhitungan tegangan utama pada titik 1
Pada titik 1, sx = - 4.688 Mpa, xy = 0
![Page 38: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/38.jpg)
Perhitungan tegangan utama pada titik 1 ( DITINJAU TERHADAP s mak/min )
Pada titik 1, sx = - 4.688 Mpa, xy = 0
0θ02θ0 tg2θ
MPa688.40cos02
4.688-2
4.688-σx
MPa00cos02688.4
2688.4σy
Mencari tegangan normal max. dan min 2
2
max/min 2σyσx
2σyσxσ xy
INGAT
σx-σyτxy
θ tg22
tegangan normal akan mencapai max. / min jika :
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
![Page 39: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/39.jpg)
Pada titik 1, sx = - 4.688 Mpa, xy = 0
MPa344.2cos(90)2
-4.6882
-4.688σx1
MPa344.2cos(90)2688.4
2688.4σy1
o45θ902θ02
4.688
tg2θ
MPa2.344 )sin(90)2
-4.688(τmax
Perhitungan tegangan utama pada titik 1( DITINJAU TERHADAP mak/min )
INGATtegangan geser akan mencapai max. / min jika :
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
xy
yx
tg
ss
q
22
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
22max τxy)
2σy-σx(τ
![Page 40: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/40.jpg)
Perhitungan tegangan utama pada titik 1
![Page 41: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/41.jpg)
Pada titik 2, sx = - 2.344 Mpa, xy = - 0.176 MPa
Perhitungan tegangan utama pada titik 2
sX=2.344 MPaX
Y
sX=2.344 MPa2
xy = 0.176 MPa
xy = 0.176 MPa
yx = 0.176 MPa
yx = 0.176 MPa
![Page 42: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/42.jpg)
MPa357.2)54.8sin176.0(cos8.542
-2.3442
-2.344σmax
Pada titik 2, sx = - 2.344 Mpa, xy = - 0.176 MPa
o
σx-σyτxy θ tg
27.454.82
15.0344.2
176.0*222
MPa013.0)54.8sin176.0(cos8.542
-2.3442
-2.344σmin
Perhitungan tegangan utama pada titik 2
sX=2.344 MPaX
Y
sX=2.344 MPa2
xy = 0.176 MPa
xy = 0.176 MPa
yx = 0.176 MPa
yx = 0.176 MPa
![Page 43: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/43.jpg)
MPa172.1))46.81sin(176.0()cos(-81.462
-2.3442
-2.344σx1
o73.04θ46.182θ659.60.176-
2344.2
tg2θ
MPa172.1))46.81sin(176.0()cos(-81.462
-2.3442
-2.344σy1
MPa1.185- ))cos(-81.46*176.0(6))sin(-81.42
-2.344(τmax
Pada titik 2, sx = - 2.344 Mpa, xy = - 0.176 MPa
Perhitungan tegangan utama pada titik 2
sX=2.344 MPaX
Y
sX=2.344 MPa2
xy = 0.176 MPa
xy = 0.176 MPa
yx = 0.176 MPa
yx = 0.176 MPa
![Page 44: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/44.jpg)
Perhitungan tegangan utama pada titik 2
sX=2.344 MPaX
Y
sX=2.344 MPa2
xy = 0.176 MPa
xy = 0.176 MPa
yx = 0.176 MPa
yx = 0.176 MPa
![Page 45: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/45.jpg)
Perhitungan tegangan utama pada titik 3
Pada titik 3, sx = 0 , xy = - 0.234 MPa
![Page 46: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/46.jpg)
Perhitungan tegangan utama pada titik 3
Pada titik 3, sx = 0 , xy = - 0.234 MPa
oo
σx-σyτxy θ tg
45902
0234.0*222
MPa234.0)90sin234.0(σmax
MPa234.0)54.8sin176.0(σmin
![Page 47: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/47.jpg)
Perhitungan tegangan utama pada titik 3
Pada titik 3, sx = 0 , xy = - 0.234 MPa
o0θ02θ0.234-
20
tg2θ
0σx1 0σy1
MPa0.234- cos(0))*234.0(0τmax
![Page 48: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/48.jpg)
Perhitungan tegangan utama pada titik 3
![Page 49: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/49.jpg)
Perhitungan tegangan utama pada titik 4
Pada titik 4, sx = 2.344 , xy = - 0.176 MPa
![Page 50: Mekanika Bahan - Analisa Tegangan Bidang](https://reader034.fdocuments.net/reader034/viewer/2022042423/577c84321a28abe054b7e79f/html5/thumbnails/50.jpg)
Perhitungan tegangan utama pada titik 4
Pada titik 4, sx = 2.344 , xy = - 0.176 MPa
oo
σx-σyτxy θ tg
27.454.82
15.0344.2
176.0*222
MPa357.2)54.8sin176.0(8.54-cos2
2.3442
2.344σmax
MPa013.0)54.8sin176.0(8.54-cos2
2.3442
2.344σmin
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Perhitungan tegangan utama pada titik 4
MPa172.1)46.81sin176.0(cos81.462
2.3442
2.344σx1
o73.04θ46.182θ659.60.176-
2344.2
tg2θ
MPa172.1)46.81sin176.0(cos81.462
2.3442
2.344σy1
MPa1.185- cos81.46)*176.0()sin81.462
2.344(τmax
Pada titik 4, sx = 2.344 , xy = - 0.176 MPa
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Perhitungan tegangan utama pada titik 4
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Perhitungan tegangan utama pada titik 5
Pada titik 5, sx = 4.6884 , xy = 0
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Perhitungan tegangan utama pada titik 5
Pada titik 5, sx = 4.6884 , xy = 0
0θ02θ0 tg2θ
MPa688.40cos02
4.6882
4.688σmax
MPa00cos02688.4
2688.4σmin
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Perhitungan tegangan utama pada titik 5
Pada titik 5, sx = 4.6884 , xy = 0
MPa344.2cos(90)2
4.6882
4.688σx1
MPa344.2cos(90)2688.4
2688.4σy1
o45θ902θ02
4.688
tg2θ
MPa2.344- )sin(90)2
4.688(τmax
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Perhitungan tegangan utama pada titik 5
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Perubahan arah tegangan utama pada
titik 1 - 5
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Jawaban Soal Ujian 20111 No 1 (50%)
Pada gelagar roll-sendi bekerja muatan merata q = 25 kN/m dan gaya normal eksentris tekan P = 4650 kN. Gaya P mempunyai eksentrisitas terhadap sumbu x sebesar e. Apabila perilaku tegangan-regangan bahan seperti tergambar, maka:•Gambarkan bidang momen dan bidang normal gelagar tersebut dan tentukan letak penampang kritis. •Tentukan besarnya tegangan pada serat teratas dan terbawah penampang tersebut sebagai fungsi eksentrisitas e, dalam MPa. •Tentukan besarnya nilai eksentrisitas minimum agar gelagar tersebut aman. •Berikan evaluasi pengaruh gaya normal eksentris tersebut terhadap perilaku tegangan gelagar.
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1000 kN/m2 = 1 MPa1 N/mm2 = 1 MPa1*10-3 kN / 10-6 m2 = 1MPa → 1000 kN/m2 = 1MPa
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W = 1/6 * b * h2 = 1/6 * 400 * 6002 = 24 x 106 mm3 = 0.024 m3
A = b*h = 400*600 = 240000 mm2 = 0.24 m2
s1 = 4650/0.24 = 19375 kN/m2 = 19.375 MPa.s2 = MP/W = 4650*10-3e/0.024 =193.75e kN/m2 = 0.19375e Mpa (tarik)s3 = MP/W = 4650*10-3e/0.024 =193.75e kN/m2 = 0.19375e Mpa (tekan)s4 = Mq/W = 1250/0.024 =52083.333 kN/m2 = 52.083 Mpa (tekan)s5 = Mq/W = 1250/0.024 = 52083.333 kN/m2 = 52.083 Mpa (tarik)
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Serat atas tertekan :s1 +s4 – s2 ≤ 4519.375+52.083 – 0.19375 e ≤ 45 71.458 - 0.19375 e ≤ 45 0.19375 e ≥ 26.458e ≥ 136.557 mm
Serat atas tertekan :s2 - s1 - s4 ≤ 40.19375 e – 19.375 – 52.083 ≤ 4 0.19375 e ≤ 75.458 e ≤ 389.46 mm
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Serat bawah tertekans1 + s3 – s5 ≤ 45 19.375 + 0.19375 e – 52.085 ≤ 450.19375 e ≤ 77.71e ≤ 401.08 mm
Serat bawah tertariks5 - s1 - s3 ≤ 452.085 – 19.375 - 0.19375 e ≤ 40.19375 e ≥ 28.71e ≥ 148.18 mm
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148.18 ≤ e ≤ 389.46
e minimum = 148 mmJarak e maksimum = 300 mm
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Jawaban Soal Ujian 2011 No 2 (50%)Balok di atas tumpuan sendi dan roll dengan bentang 10 meter menderita beban merata q = 50 kN/m dan beban terpusat P = 100 kN. Balok juga menderita beban normal N = 1000 kN pada titik K. Penampang balok terlihat seperti pada potongan I-I.1. Hitung dan gambarkan tegangan utama pada posisi momen
maksimum pada posisi titik A.2. Hitung dan gambarkan tegangan utama pada posisi gaya geser
maksimum pada posisi titik A.
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VS = (100*4.5+50*10*5)/10 = 295 kNVR = (100*5.5+50*10*5)/10 = 305 kN
Posisi 5.5 m dari tumpuan S.D1 = 295 – 50*5.5 = 20 kN D2 = 20 – 100 = -80 kNM = 295*5.5-0.5*50*5.52 = 866.25 kNm
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A = 40*70-2*10*40 = 2000 cm2 = 0.2 m2
Ix = 1/12*40*703-2*1/12*10*403 = 1036666.667 cm4 = 0.010367 m4
Ix = 2*1/12*40*153+ 1/12*20*403+2*40*15*(35-7.5)2 = 1036666.667 cm4 = 0.010367 m4
SA = 40*15*(35-7.5) = 16500 cm3 = 0.0165 m3
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A = 0.2 m2 Ix = 0.010367 m4
SA = 0.0165 m3Pada posisi Momen
maksimum :ML = 866.25 kNmN = 1000 kNex = 30 cm = 0.3 mMex = 300 kNmMR = 566.25 kNmD = 80 kN
sA = 1000/0.2 + 566.25*0.2/0.010367 = 15924.08604 kN/m2 = 15.924 Mpa (tekan).
A = 80*0.0165/(0.2*0.010367) = 636.635 kN/m2 = 0.637 MPa.
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A = 0.2 m2 Ix = 0.010367 m4
SA = 0.0165 m3Pada posisi D maksimum :
ML = 0 kNmN = 1000 kNex = 30 cm = 0.3 mMex = 300 kNmD = 305 kN
sA = 1000/0.2 - 300*0.2/0.010367 = -787.595 kN/m2 = -0.788 Mpa (tarik).
A = 305*0.0165/(0.2*0.010367) = 2427.173 kN/m2 = 2.427 MPa.
A = 305/80*0.637 = 2.429 MPa.
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MPa949.15))574.4sin(637.0()cos(-4.5742
-15.9242
-15.924σx1
o287.2θ574.42θ08.015.924-
637.0*2tg2θ
MPa025.0))574.4sin(637.0()cos(-4.5742
-15.9242
-15.924σy1
0 MPa0.00003 ))cos(-4.574*637.0(4))sin(-4.572
-15.924(τx1y1
σx-σyτxy θ tg 22
Pada posisi Momen maksimum :ML = 866.25 kNmN = 1000 kNex = 30 cm = 0.3 mMex = 300 kNmMR = 566.25 kNmD = 80 kNsA = 1000/0.2 + 566.25*0.2/0.010367 = 15924.08604 kN/m2 = 15.924 Mpa (tekan).
A = 80*0.0165/(0.2*0.010367) = 636.635 kN/m2 = 0.637 MPa.
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
?max σ MPa 15,924σx MPa 0σy
MPa637,0τxy
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MPa962.7))426.85sin(637.0()cos(85.4262
-15.9242
-15.924σx1
o713.42θ426.852θ499.12(0.637)*2
924.15tg2θ
MPa962.7))426.85sin(637.0()cos(85.4262
-15.9242
-15.924σy1
MPa7.987 ))cos(85.426*637.0(6))sin(85.422
-15.924(τmax
?τmax
xy
yx
tg
ss
q
22
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
MPa 15,924σx
MPa 0σy
MPa637,0τxy
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MPa962.7))426.85sin(637.0()cos(85.4262
-15.9242
-15.924σx1
o713.42θ426.852θ499.12(0.637)*2
924.15tg2θ
MPa962.7))426.85sin(637.0()cos(85.4262
-15.9242
-15.924σy1
?τmax
xy
yx
tg
ss
q
22
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
MPa 15,924σx
MPa 0σy
MPa637,0τxy
MPa987,7)637,0()2
0- 15,924-(τ 22max
22max τxy)
2σy-σx(τ
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Gambar Tegangan utama pada posisi momen maksimum
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MPa855.2))786.80sin(429.2()cos(80.7862
0.7882
0.788σx1
MPa067.2))786.80sin(429.2()cos(80.7862
0.7882
0.788σy1
o393.40θ786.802θ165.60.788
429.2*2tg2θ
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
σx-σyτxy
θ tg22
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
?max σPada posisi geser maksimum
0 MPa0.00002 ))cos(80,786*429.2(6))sin(80,782
0-0,788(τx1y1
MPa 0,788σx
MPa 0σy
MPa429,2τxy
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MPa394.0))214.9sin(429.2()cos(-9.2142
0.7882
0.788σx1
MPa394.0))214.9sin(429.2()cos(-9.2142
0.7882
0.788σy1
o607.4θ214.92θ162.02.429*2788.0tg2θ
MPa2.461 ))cos(-9.214*429.2(4))sin(-9.212
0.788(τmax
?τmax
xy
yx
tg
ss
q
22
MPa 0,788σx
MPa 0σy
MPa428,2τxy
sin2θ*τxycos2θ2
σyσx2
σyσxσx1
cos2θ*τxy)sin2θ2
σy-σx(τx1y1
sin2θ*τxycos2θ2
σyσx2
σyσxσy1
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Gambar Tegangan utama pada posisi geser maksimum
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Terima KasihSemoga Allah yang maha
pengasih dan maha penyayang selalu melimpahkan rahmat
dan karunia-Nya kepada saudara/i sekalian sehingga
sukses menempuh ujian mata kuliah ini
Windu Partono - 2012