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Video ini membahas tentang tekanan hidrostatik, bagaimana menghitung gaya yang dihasilkan oleh tekanan ini pada berbagai permukaan (vertikal, miring, dan melengkung), serta penerapannya dalam contoh soal. Konsep-konsep utama meliputi:
- Tekanan hidrostatik meningkat secara linear dengan kedalaman.
- Gaya hidrostatik dapat dihitung sebagai integral tekanan terhadap area, atau disederhanakan menjadi area segitiga atau trapesium untuk permukaan datar.
- Lokasi gaya resultan adalah pada centroid bentuk tekanan terdistribusi.
Tekanan hidrostatik meningkat secara linear seiring dengan kedalaman. Tekanan sebagai fungsi kedalaman dapat ditulis sebagai gamma (berat jenis fluida) dikalikan dengan h (kedalaman), atau sebagai rho (densitas fluida) dikalikan dengan g (percepatan gravitasi) dikalikan dengan h. Satuan untuk densitas adalah massa per volume (misalnya, kg/m³), gravitasi adalah percepatan (misalnya, m/s²), dan tinggi adalah jarak (misalnya, meter). Tekanan memiliki satuan Pascal (Pa) untuk metrik atau psi untuk satuan Inggris. Gaya dari cairan pada permukaan yang terendam adalah integral dari tekanan dikalikan dengan area kontak dari dasar hingga puncak permukaan.
Karena tekanan tidak berubah pada sumbu horizontal dan permukaan yang dianalisis biasanya persegi panjang, representasi 3D dapat disederhanakan menjadi beban terdistribusi segitiga pada bidang 2D. Diagonal beban segitiga ini adalah garis lurus dari 0 hingga rho gh, dengan kemiringan sama dengan berat jenis. Dengan penyederhanaan 2D, satuan tekanan berubah menjadi gaya per panjang. Integral dari beban terdistribusi (p dz) dari atas ke bawah sama dengan area di bawah kurva segitiga. Lokasi beban titik ekivalen adalah centroid segitiga, sepertiga dari alas.
Tidak seperti beban pada balok di mana nilai tertinggi beban (wx) diberikan, dalam masalah tekanan hidrostatik, nilai tertinggi beban terdistribusi dapat ditemukan dengan mengetahui kedalaman (h) dan jenis cairan (rho). Dengan mengetahui kedalaman, kita tahu h, dan dengan mengetahui cairan (misalnya, air), kita tahu rho. Area segitiga adalah setengah dari alas (panjang) dikalikan dengan tinggi (rho gh).
Tujuan menghitung gaya tekanan hidrostatik adalah untuk menghitung reaksi, momen, atau gaya yang harus dihasilkan oleh suatu gerbang untuk mencegah air masuk ke reservoir yang berdekatan.
Proses menemukan beban horizontal resultan dari tekanan fluida juga berlaku untuk permukaan miring. Gaya horizontal dihitung dengan cara yang sama, dengan besarnya beban titik ekivalen sama dengan area segitiga dari beban terdistribusi segitiga (rho gh * h / 2), dan terletak sepertiga dari bawah ke atas. Berat cairan di atas permukaan miring, yang merupakan beban vertikal, juga mempengaruhi permukaan. Berat cairan adalah volume dikalikan dengan berat jenis, atau area dikalikan dengan berat jenis dikalikan dengan lebar (untuk representasi 2D). Lokasi beban titik dari beban persegi panjang berada di tengah, dan dari beban segitiga berada sepertiga dari sisi segitiga.
Jika dinding atau anggota yang diminati terendam di bawah permukaan cairan (misalnya, gerbang di bagian bawah reservoir), beban terdistribusi hanya akan menjadi bagian dari tekanan hidrostatik antara bagian bawah dan atas gerbang, yang berbentuk trapesium. Untuk menghindari perhitungan centroid trapesium, bentuk tersebut dapat dibagi menjadi persegi panjang dan segitiga. Beban titik persegi panjang terletak di tengah tinggi gerbang, dan beban titik segitiga terletak sepertiga dari tinggi gerbang dari bawah. Tekanan di bagian atas dan bawah gerbang adalah rho g z, di mana z adalah kedalaman dari permukaan cairan.
Untuk bentuk dasar reservoir lainnya (misalnya, dasar melengkung), prosesnya sama: tekanan di sisi menciptakan gaya horizontal, dan berat cairan di atas kurva menciptakan beban vertikal. Berat dibagi menjadi persegi panjang di atas kurva dan berat geometri yang tersisa. Gaya ekivalen adalah berat jenis dikalikan dengan volume, yang dalam bentuk integral adalah lebar dikalikan dengan integral area terhadap z. Lokasi beban ekivalen berada di centroid volume.
Saat air pasang surut, gerbang air pasang otomatis terbuka untuk mengalirkan rawa. Untuk kondisi air pasang tinggi tertentu, reaksi horizontal yang dikembangkan pada engsel C dan blok berhenti D dihitung. Lebar gerbang adalah 6 meter dan densitas air adalah 1000 kg/m³. Diagram benda bebas dari gerbang air pasang menunjukkan reaksi di C dan D, ditambah dua gaya tekanan hidrostatik. Tekanan hidrostatik maksimum di sebelah kiri (E) adalah rho g z, dengan kedalaman 3 meter. Tekanan hidrostatik di titik D adalah untuk kedalaman 2 meter. Gaya dari sisi kiri adalah area segitiga ungu dikalikan dengan lebar gerbang, dan gaya dari sisi kanan adalah area persegi panjang merah muda dikalikan dengan lebar gerbang. Lokasi gaya ungu berada di centroid segitiga kiri (sepertiga dari tinggi dari bawah), dan lokasi gaya merah muda berada di centroid segitiga kanan (sepertiga dari tinggi 2 meter dari permukaan bawah). Dengan menjumlahkan momen di sekitar C dan menjumlahkan gaya dalam arah x, nilai untuk reaksi di D dan C dapat ditemukan.
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0:00 hydrostatic pressure increases linearly with depth the pressure as a function of depth can be written as gamma times h and since gamma is the specific weight of the fluid we can also write the function as rho g h the units for the density rho are mass over volume which can be pounds per cubic foot or kilograms per cubic meter for example the gravity g is an acceleration which is distance over time squared for example m over second square and since the height or depth is distance for example meters we find that the pressure has units of pascals for metric or psi for english units the force from the liquid on a submerged surface would be the pressure times the area of contact but since the
0:46 pressure changes with depth the force is actually the integral from the bottom to the top of the surface of the pressure times that d area now since the pressure is not changing in the horizontal axis and the surfaces we analyze are usually rectangular meaning that the width is constant along the vertical axis a reasonable representation of a 3d problem in a 2d space would be to assume the pressure on the wall as a triangular distributed load the diagonal of this triangular load would be a straight line that goes from 0 to rho gh as a straight line with a slope equal to the specific gravity with this 2d simplification which means multiplying by the horizontal axis dimension which is a constant we would get new
1:32 units for the pressure which was one's force over area as force per length which is consistent with the units of a typical distributed load the integral from top to bottom of the distributed load lowercase p times dz would be the area under the curve of the triangle just like it normally is for any of the triangular loads we studied in the previous video link below to that specific section of that video and just like any triangular load the location of the equivalent point load with that magnitude is the centroid of the triangle one third from the base as opposed to the triangular loads on a beam for example we don't need to be given the highest value for the load density in the case of a load on a beam
2:18 we would be given a wx that reads something like 20 pounds per foot and that value referred to that highest value the area of the triangle would be one half of the base which is the length in the x axis times that 20 pound per foot height in the case of hydrostatic pressure problems by knowing the depth or height and the type of liquid we have we can find that highest value from the distributed load since the pressure is rho gh by knowing the depth we know h and by knowing the liquid for example water we know rho its density the reason for calculating these hydrostatic pressure forces is to be able to calculate the reaction moment or forces that for example a gate has to generate
3:03 to prevent the water from coming into the adjacent reservoir now the process of finding the resulting horizontal load from the fluid pressure doesn't only work for vertical surfaces in contact with a liquid reservoir if the surface in contact with the liquid is inclined the horizontal forces affecting the surface will be calculated in the exact same manner the magnitude of the equivalent point load in the horizontal direction would be equal to the area of the triangle of the triangular distributed load meaning rho gh times h divided by two and just like before it's also located one third of the way from the bottom to the top of course on an inclined surface the weight of the liquid above it which is a vertical load also affects the surface
3:50 in some cases this weight will only be a triangle or the combination of a rectangle plus a triangle or just a trapezoid the weight of the liquid would be the volume times the specific weight which for a 2d representation is the area times the specific weight times the width the location of the point load from the rectangular load is located right at the center and again just like with the pressure force the location from the triangular weight is located one third of the way from the side of the triangle variation of these two situations the vertical wall and the inclined one that begins right at the surface of the liquid is if the walls or other member of interest is submerged beyond the surface of the liquid for example if there's a gate at the
4:36 bottom of the reservoir and you can think of it as a valve that opens to drain a reservoir the distributed load affecting the gate will only be the portion of the hydrostatic pressure between the bottom and the top of the gate which means the linear distributed load looks like a trapezoid and again if you don't want to deal with the centroid of a trapezoid to find the location of the equivalent point load you can divide it into a rectangle plus a triangle the rectangle's point load located at the center of the gate's height and the triangle's point load located one third of the height of the gate from the bottom the value for the pressure at both the top and the bottom would just be rho g times z where z is the depth or distance from the surface of the liquid
5:22 with these values we could calculate the area of the rectangle as well as the area of the triangle by finding the base as the difference between the distributed pressures for any other shape of the floor of the reservoir for example a curved bottom the process would be the same the pressure on the side is creating a horizontal force and the weight of the liquid above the curve is creating a vertical load the weight is divided into one a rectangle above the curve just like we did with the flat inclined wall from before and two the weight of the remaining geometry which can be calculated using everything we learned during the centroid and center of gravity main video link below if you haven't watched that yet the equivalent force will be the
6:07 specific gravity times the volume which in terms of the integral is the width in the third dimension we don't see times the area written as the integral about z of all the d areas along the z direction this is usually written as a function of z known due to the geometry of the floor slash wall the location of the equivalent load would be at the centroid of the volume again link below to that portion of the video where we study center of gravity center of mass or centroid of the volume for a shape described by a given math function let's take a look at a simple example where we make use of what we've learned today and if you want to check out more complex problems you can check out the two minute long video examples on this topic linked in the description below when the tidewater a subsides the tide
6:55 gate automatically swings open to drain the marsh b for this specific condition of high tide shown what are the horizontal reactions developed at the hinge c and the stop block d the width of the gate is 6 meters and the water density is 1000 kilograms per meter cubed remember to try to solve this problem on your own before watching the solution in this video a free body diagram of the tide gate would show the reactions at c and d that we're looking for plus the two hydrostatic pressure forces the maximum hydrostatic pressure on the left let's call it e would be equal to rho g z with the depth c being 3 meters the hydrostatic pressure at point d would be that for a depth of 2 meters
7:41 the force coming from the left side would be equal to the area of the purple triangle times the width of the gate the force coming from the right side would be equal to the area of the pink rectangle times the width of the gate the location of the purple force would be at the centroid of the left triangle which is one third of the height from the bottom the location of the pink force would also be the centroid of the right triangle which is one third of the two meter height from the bottom surface assuming that the height of block d is negligible with a simple sum of moments about c and knowing that the gate is not rotating we can find the value for the reaction at d and then with the sum of forces in x and knowing that the gate is not translating
8:27 left or right we find the value for the reaction at c for more complex examples on hydrostatic pressure as well as the other main videos of the statics course make sure to check out the links in the description below thanks for watching
