BAB IVMOMEN INERSIA LUAS BENDA PENOPANG

(MOMEN LUAS KEDUA)

Momen laus ada dua yaitu momen luas kedua ( second moment of area ), dan momen luas tk satu. Momen luas kedua adalah sifat suatu bentuk benda yang dapat digunakan untuk memprediksi ketahanan suatu papan terhadap lenturan dan simpangan. Simpangan suatu balok yang dibebani tidak hanya tergantung bebansaja, melainkan tergantung geometri dari penampang melintang. Oleh karena itu balok dengan dengan momen area dari momen inersia besar, seperti balok bentuk I, sering digunakan dalam konstruksi balok yang disusun berlawanan dengan balok yang lain dalam luas yang sama. Hal ini sama saja dengan momen polar dari momen inersia yang mempunyai karakter kemampuan obyek tersebut untuk menahan tosi atau momen puntir (lihat Gambar 4.1).

Gambar 4.1. Konstruksi yang banyak menggunakan penopang dengan konstruksi I, karena momen luas keduanya besar.

Second moment of area tidaklah sama dengan momen inersia , yang digunakan untuk menghitung percepatan sudut. Banyak enginer yang merujuk pada momen luas kedua sebagai momen inersia dan menggunakan simbol yang sama yaitu Iatau J untuk keduanya hal ini yang membingungkan. Padahal dilihat dari unitnya berbeda momen luas kedua mempunyai unit panjang pangkat empat, sedangkan momen inersia mempunyai unit massa kali panjang kuadrat.

Terkadang suatu struktur penopang menerima puntiran, kopel puntir atau momen puntiran. Puntiran tersebut menimbulkan tegangan geseran yang disebut sebagai tegangan geser puntir. (lihat Gambar 4.1).

Gambar 4.1. Batang yang mengalami puntiran (torsion)

Besarnya tegangan yang diakibatkan oleh momen puntir/torsi pada penampang batang dituliskan dengan formula sebagai brerikut (lihat BAB II dan III). t = T . r / Ip (4.1)dimana : t = tegangan geser torsi, T = besaran momen torsi, r = Jari-jari batang terputir,Jp = Ip = momen inersia polar penampang tergeser (momen inersia luas), dengan Ip = πd 4/32 untuk lingkaran pejal, Ip = π/32(d2

4 - di4) untuk lingkaran berlubang. Pada

persamaan 4.1 besarnya momen inersia harus diketahui. Karena bentuknya tidak selalu dengan penampang lingkaran momen inersia harus dihitung berdasarkan rumus dasar:

(4.2a)

Ix = momen luas kedua terhadap sumbu x, y = jarak tegak lurus terhadap sumbu x terhadap elemen dA, dA = elemen unit luas.

4.1. Menghitung Momen Inersia

Momen luas kedua suatu penampang adalah salah satu parameter geometri yang sangat penting dalam analisis struktur. Untuk penampang yang beraturan, seperti persegi, lingkaran dsb, telah dibuatkan tabel, namun bagaimana cara memperolehnya yaitu berbesis dari momen inersia luar permukaan yaitu momen inersia terhadap sumbu x (pers 4.2a). Kalau untuk sumbu y, tinggal menukar sumbu y menjadi x.Dari formula dasar itulah kita bisa menurunkan formula momen inersia untuk bentuk geometri apapun.

Hal tersebut di atas dapat digunakan sendirinya manakala bentuk penampangnya simetris misal terhadap sumbu x. Apabila sulit diperoleh maka dapat dicari momen luas kedua dari hasil kedua sumbu x- dan y-axis dan hasil dari perkaliannya adalah momen lusa kedua, Ixy, yang dapat digunakan.

(4.2b)

Iy = momen luas kedua terhadap sumbu y, x = jarak tegak lurus dari sumbu y terhadap elemen dA, dA = an elemental area.

Bentuk Persegi

Salah satu bentuk balok penopang bangunan atau beban berpenampang persegi empat berukuran , b lebar, h tinggi. Sumbu x sebagai pusat puntiran terletak pada sumbu netral atau garis berat (tengah-tengah) . Berdasarkan formula dasar , maka kita harus meninjau sebuah elemen kecil . Elemen ini mempunyai ukuran dan . Sehingga bisa kita tuliskan luas elemen . Jika kita kumpulkan semua elemen

yang mempunyai nilai yang sama dengan b tetap dan variabelnya adalah y

maka elemen , kini menjadi , sehingga (pers.4.2)

b bernilai konstan untuk setiap nilai , kita keluarkan saja b dari kurungan cacing tersebut, Sekarang, tinggal menentukan batas atas dan batas bawah dari dy. Berdasarkan gambar di atas, maka batas bawahnya adalah - h/2 dan batas atas adalah h/2. Sehingga

Bagaimana Dengan Momen Inersia Terhadap Bukan Sumbu Netral?

Misalnya, pada gambar di atas, kita mau menentukan tapi sumbu x-x tidak pada garis berat, melainkan seperti pada gambar di bawah. Dengan cara yang sama dengan menggati batas integrasi dari 0 – h, maka diperoleh:

Coba kita geser lebih jauh lagi ke atas. Lihat gambar di bawah.

Bila batas bawah = , dan batas atas = terhadap sumbu x = 0, maka, dapat digunakan formula yang sudah diketahui yaitu bh3 /12 atau bh3 /3, dengan cara:

1. Gunakan bentuk benda dari segitiga atau segi empat yang sudah diketahui2. jumlahkan bentuk bangunan tersebut dengan prinsip penjumlah momen luas dari

persamaan momen yang terintegrasi.

Dari contoh di atas tidak lain adalah luas persegi, sementara yo + h/2 adalah jarak titik berat ke sumbu momen inersia!.dari luas persegi empat yang berpusat pada garis atau sumbu x di tengah-tengah yaitu + h/2 dengan – h/2.

Secara umum bisa dituliskan:

dimana, adalah momen inersia terhadap sumbu x tertentu, adalah momen inersia terhadap sumbu netral (garis berat), adalah luas bangun/penampang, adalah jarak dari titik berat ke sumbu momen inersia yang dicari.

Dalam beberapa buku penggunaan Ix dan Ixx sama tergantung pengguna. Ada yang menggunakan dengan notasi sumbu x-axis digunakan Ix, kadang kala diambil penggal sumbu x-x' sehingga digunakan Ixx, demikian juga Iy dan Iyy.

4.2. Produk moment luas (area)

Produk momen luas antara sumbu x dan y adalah Ixy didefinisikan:

dA = an elemental area, x = the perpendicular distance to the element dA from the axis y, y = the perpendicular distance to the element dA from the axis x

Produk momen area sangat signifikan untuk mencari stress pembengkokan pada penampang melintang asymmetric, yang dapat memberikan nilai negatif dan positif.Sistem koordinat yang menghasilkan momen area nol sebagai rujukan dan sumbu utama, dengan pendekatan maximum dan minimum. Pusat perlakuan lenturan dari sumbu utama dan simetris dengan satuan (mm4, in4 dsb.)

Teorima axis (sumbu) sejajar.

The parallel axis theorem can be used to determine the moment of an object about any axis, given the moment of inertia (second moment of area) of the object about the parallel axis through the object's center of mass (or centroid) and the perpendicular distance between the axes.

Ix = the second moment of area with respect to the x-axis IxCG = the second moment of area with respect to an axis parallel to x and passing

through the centroid of the shape (coincides with the neutral axis) A = area of the shape d = the distance between the x-axis and the centroidal axis

Axis rotation

The following formulae can be used to calculate moments of the section in a co-ordinate system rotated relative to the original co-ordinate system:

φ = the angle of rotation(anticlockwise sense):

x * = xcosφ + ysinφy * = − xsinφ + ycosφ

Ix, Iy and Ixy = the second moments and the product moment of area in the original coordinate system

Ix*, Iy

* and Ixy* = the second moments and the product moment of area in the rotated

coordinate system.

The value of the angle φ, which will give a product moment of zero, is equal to:

This angle is the angle between the axes of the original coordinate system and the principal axes of the cross section.

Stress in a beam

The general form of the classic bending formula for a beam in co-ordinate system having origin located at the neutral axis of the beam is (Pilkey 2002, p. 17):

σ is the normal stress in the beam due to bending x = the perpendicular distance to the centroidal y-axis y = the perpendicular distance to the centroidal x-axis My = the bending moment about the y-axis Mx = the bending moment about the x-axis Ix = the second moment of area about x-axis Iy = the second moment of area about y-axis Ixy = the product moment of area

If the coordinate system is chosen to give a product moment of area equal to zero, the formula simplifies to:

If additionally the beam is only subjected to bending about one axis, the formula simplifies further:

Circular cross and Hollow cylindrical cross section

D = diameter, r = radius. DO = outside diameter, DI = inside diameter, rO = outside radius, rI

= inside radius

This equation is useful in calculating the required strength of masts. Taking the area moment of inertia calculated from the previous formula, and entering it into Euler's formula gives the maximum force that a mast can theoretically withstand.

E is [Young's modulus|Young (elastic) modulus of material], I is the second moment of area of examined object,l is the length of panel

Composite cross section

When it is easier to compute the moment for an item as a combination of pieces, the second moment of area is calculated by applying the parallel axis theorem to each piece and adding the terms:

y = distance from x-axis, x = distance from y-axis, A = surface area of part, Ilocal is the second moment of area for that part of the composite, in the appropriate direction (i.e. Ix or Iy respectively).

"I-beam" cross section

I-beam I-beam diagram, moment by subtraction

b = width (x-dimension), h = height (y-dimension), tw = width of central webbing, h1 = inside distance between flanges (usually referred to as hw, the height of the web)

The I-beam can be analyzed as either three pieces added together or as a large piece with two pieces removed from it. Either of these methods will require use of the formula for composite cross section. This section only covers doubly symmetric I-beams, meaning the shape has two planes of symmetry. This formula uses the method of a block with two pieces removed. (While this may not be the easiest way to do this calculation, it is instructive in demonstrating how to subtract moments). Since the I-beam is symmetrical with respect to the y-axis the Ix has no component for the centroid of the blocks removed being offset above or below the x axis.

When computing Iy it is necessary to allow for the fact that the pieces being removed are offset from the Y axis, this results in the Ax2 term.

A = Area contained within the middle of one of the 'C' shapes of created by two flanges and

the webbing on one side of the cross section = , x = distance of the centroid of the

area contained in the 'C' shape from the y-axis of the beam =

Doing the same calculation by combining three pieces, the center webbing plus identical contributions for the top and bottom piece:

I-beam diagram, moment by addition

Since the centroids of all three pieces are on the y-axis Iy can be computed just by adding the moments together.

However, this time the law for composition with offsets must be used for Ix because the centroids of the top and bottom are offset from the centroid of the whole I-beam.

A = Area of the top or bottom piece= y = offset of the centroid of the top or bottom piece from the centroid of the whole

I-beam=

Any cross section defined as polygon

The second moments of area for any cross section defined as a simple polygon on XY plane can be computed in a generic way by summing contributions from each segment of a polygon.

For each segment defined by two consecutive points of the polygon, consider a triangle with two corners at these points and third corner at the origin of the coordinates. Integration by the area of that triangle and summing by the polygon segments yields:

ai = xiyi + 1 − xi + 1yi is twice the (signed) area of the elementary triangle, index i passes over all n points in the polygon, which is considered closed, i.e. point

n+1 is point 1

These formulae imply that points defining the polygon are ordered in anticlockwise manner; for clockwisely defined polygons it will give negative values. See polygon area for calculating area and centroid of the section using similar formulae.

The following is list of area moments of inertia. The area moment of inertia or second moment of area has a unit of dimension length4, and should not be confused with the mass moment of inertia. Each is with respect to a horizontal axis through the centroid of the given shape, unless otherwise specified.

Description Figure Area moment of inertia Comment Reference

a filled circular area of radius r

[1]

an annulus of inner radius r1 and outer

radius r2

For thin tubes, this is approximately equal

to: or π times the cube of the

average radius times the thickness.

a filled circular sector of angle θ in radians and radius r with respect to

an axis through the centroid of the sector and the

centre of the circle

a filled semicircle with radius r with

respect to a horizontal line

passing through the centroid of the

area

[2]

a filled semicircle as above but with respect to an axis collinear with the

base

This is a consequence of the parallel axis

theorem and the fact that the distance

between these two

axes is

[2]

a filled semicircle as above but with

respect to a vertical axis through the

centroid

[2]

a filled quarter circle with radius r entirely in the 1st quadrant of the

Cartesian coordinate system

[3]

a filled quarter circle as above but with respect to a

horizontal or vertical axis through the

centroid

This is a consequence of the parallel axis

theorem and the fact that the distance

between these two

axes is

[3]

a filled ellipse whose radius

along the x-axis is a and whose radius along the y-axis is

b

a filled rectangular area with a base width of b and

height h

[4]

a filled rectangular area as above but with respect to an axis collinear with

the base

This is a trivial result from the parallel axis

theorem

[4]

a filled triangular area with a base width of b and height h with

respect to an axis through the

centroid

[5]

a filled triangular area as above but with respect to an axis collinear with

the base

This is a consequence of the parallel axis

theorem

[5]

a filled regular hexagon with a side length of a

The result is valid for both a horizontal and

a vertical axis through the centroid.

The following is a list of moments of inertia. Mass moments of inertia have units of dimension mass × length2. It is the rotational analogue to mass. It should not be confused with the second moment of area (area moment of inertia), which is used in bending calculations. The following moments of inertia assume constant density throughout the object.

NOTE: The axis of rotation is taken to be through the center of mass, unless otherwise specified.

Description Figure Moment(s) of inertia Comment

Thin cylindrical shell with open ends, of radius r and mass m

This expression assumes the shell thickness is negligible. It is a special case of the next object for r1=r2.

Also, a point mass (m) at the end of a rod of

length r has this same moment of inertia.

Thick-walled cylindrical tube with open ends, of inner radius r1, outer radius r2, length h and mass m

[1]

or when defining the normalized thickness tn = t/r and letting r = r2,

then

With a density of ρ and the same geometry

Solid cylinder of radius r, height h and mass m

This is a special case of the previous object for r1=0.

Thin, solid disk of radius r and mass m

This is a special case of the previous object for h=0.

Thin circular hoop of radius r and mass m

This is a special case of a torus for b=0. (See below.)

Solid sphere of radius r and mass m

A sphere can be taken to be made up of a stack of infinitesimal thin, solid discs, where the radius differs from 0 to r.

Hollow sphere of radius r and mass m

Similar to the solid sphere, only this time considering a stack of infinitesimal thin, circular hoops.

Oblate Spheroid of major a, minor b and mass m

—

Right circular cone with radius r, height h and mass m

—

Solid cuboid of height h, width w, and depth d, and mass m

For a similarly oriented cube with

sides of length s, .

Thin rectangular plane of height h and of width w and mass m

—

Thin rectangular plane of height h and of width w and mass m(Axis of rotation at the end of the plate)

—

Rod of length L and mass m

This expression assumes that the rod is an infinitely thin (but rigid) wire. This is a special case of the previous object for w=L and h=d=0.

Rod of length L and mass m(Axis of rotation at the end of the rod)

This expression assumes that the rod is an infinitely thin (but rigid) wire.

Torus of tube radius a, cross-sectional radius b and mass m.

About a diameter:

About the vertical axis: —

Plane polygon with

vertices , ,

, ..., and mass m uniformly distributed on its interior, rotating about an axis perpendicular to the plane and passing through the origin.

—