SIMPLE STRESS & STRAIN

JISHNU V ENGINEER BHELPROPERTIES OF SURFACES(CENTRE OF GRAVITY & MOMENT OF INERTIA)

2.1 CENTRE OF GRAVITYCentre of gravity(C.G) of a body is defined as the point through which the entire weight of the body acts. A body has only one centre of gravity for all its positions.Centroid is defined as the point at which the total area of a plane figure is assumed to be concentrated. It is the CG of a plane figureCG of a rectangle is the point where the diagonals meetCG of a triangle is the point where the medians meetCG of a circle is at its centre

2.1 CENTRE OF GRAVITYConsider a plane figure of total area A composed of a number of small areas a1, a2, a3,...., anLet x1= Distance of CG of a1 from axis of reference OYx2= Distance of CG of a2 from axis of reference OYx3= Distance of CG of a1 from axis of reference OY;;xn= Distance of CG of an from axis of reference OY

2.1 CENTRE OF GRAVITYThe moments of all areas about axis OY= a1x1 + a2x2 + a3x3+ ....... + anxn

Let G be the centre of gravity of the total area about the axis OY whose distance from OY is Xc

2.2 CENTRE OF GRAVITY OF PLANE FIGURES BY METHOD OF MOMENTSMoment of total area about OY= AXcSum of moments of small areas about axis OY must be equal to the moment of total area about OY.Hence a1x1 + a2x2 + a3x3+ ....... + anxn= AXcHence Xc= (a1x1 + a2x2 + a3x3+ ....... + anxn)/A

Similarly

2.2 CENTRE OF GRAVITY OF PLANE FIGURES BY METHOD OF INTEGRATIONIf the areas are large in number (that is i tends to infinity), then the summations in the above equations can be replaced by integration. And when the number of split up areas are large, the size of component areas will be small, hence a can be replaced by dA in the above equation.HenceXc=xdA/dA Yc=ydA/dA

2.3 IMPORTANT NOTES TO CALCULATE CENTRE OF GRAVITYAxes about which moments of areas are taken, is known as axis of reference. Axis of reference of plane figures is generally taken as the lowest line of the figure for determining Axis of reference of plane figures is generally taken as the lowest line of the figure for determining c and left line of the figure for calculation XcIf the given section is symmetrical about X-X or Y-Y, CG will lie in the line of symmetry

2.2 AREA MOMENT OF INERTIA

Consider an area A as shown. Let dA be an elemental area of the area A with coordinates x and y. The term xdA is called the moment of inertia of area A about y axis and term ydA is called the moment of inertia of area A about x-axis.

Ixx= ydA & Iyy= xdA

When dA is very small, mathematically ydA= ydAxdA= xdA

2.4 AREA MOMENT OF INERTIAIxx= ydA & Iyy= xdAIf r is the distance of an area dA (which is a part of area A) from an axis AB, then the sum of terms rdA (ie rdA) to cover the entire area is called the moment of inertia of the area A about axis AB or second moment of area of area A about axis AB.Moment of inertia is a fourth dimensional term as it is obtained by multiplying area by distance squared. Hence the SI unit is m^4

2.5 RADIUS OF GYRATIONMathematical term defined by the expression, k= (I/A) is called radius of gyration.I=AkHence the radius of gyration can be considered a that distance at which the complete area is squeezed and kept as a strip of negligible width such that there is no change in the moment of inertia

2.6 THEOREMS OF MOMENT OF INERTIA1) Perpendicular axis theorem:-The area moment of inertia about an axis perpendicular to its plane at any point is equal to the sum of moment of inertia about two mutually perpendicular axes through the point O and lying in the same plane of area.Polar moment of inertia is defined as the product of area and square of distance between CG from the axis of reference perpendicular to the area.

2.6 THEOREMS OF MOMENT OF INERTIAProof:- Consider an infinitesimal elemental area dA with co-ordinates (x,y).

Ixx=ydA;Iyy=xdA; Izz=rdAr= x + yHence r x dA= x dA + y x dAHence Izz= Ixx + Iyy

2.6 THEOREMS OF MOMENT OF INERTIAParallel axis theorem:- Moment of inertia about any axis in the plane of an area is equal to sum of moment of inertia about a parallel centroidal axis and the product of area and square of the distance between the two parallel axesReferring to the figure given below, the theorem means IAB= IGG + Ayc

2.6 THEOREMS OF MOMENT OF INERTIAIAB Moment of inertia about axis ABIGG Moment of inertia about centroidal axis GG parallel to ABA The area of plane figure given. Yc The distance between the axis AB and the parallel centroidal axis GGProof:- Consider an elemental strip dA whose CG is at a distance y from centroidal axis G-G

2.6 THEOREMS OF MOMENT OF INERTIAIGG= y dAIAB= (y + yc) dAIAB= (y + yc + 2yyc) dAy is a variable and yc is a constant; hence y dA= IGG and yc dA= Ayc.y dA/ A=distance of centroid from the axis of reference= 0 as GG is passing through the centroid. Hence IAB= IGG + Ayc

2.7 DETERMINATION OF MOMENT OF INERTIA1) Moment of inertia of a rectangular section about the centroidal axis in the plane of sectionConsider a rectangular section of length b and depth d. Let X-X be the horizontal axis passing through the centroid of the area.

2.7 DETERMINATION OF MOMENT OF INERTIAConsider a rectangular elemental strip whose CG is at a distance y from the horizontal centroidal axis.Area of the strip dA= b x dyArea moment of inertia of strip about the centroidal axis,

Ixx =bd/12 Similarly Iyy =db/12

2.7 DETERMINATION OF MOMENT OF INERTIA2) Moment of inertia of a rectangular section about an axis passing through the base of the rectangle Consider a rectangular section of length b and depth d. Let AB be the horizontal axis passing through the base of the rectangle.

IAB= IGG + A x Yc= bd/12 + bd x (d/2)IAB= bd/3

2.7 DETERMINATION OF MOMENT OF INERTIA3) Moment of inertia of a hollow rectangular section about the centroidal axis in the plane of section

Moment of inertia of main section about X-X axis= b2d2/12Moment of inertia of the cut out section about X-X axis= b1d1/12

2.7 DETERMINATION OF MOMENT OF INERTIAMoment of inertia of the hollow rectangular section about X-X axis= (b2d2/12- b1d1/12)Similarly Moment of inertia of the hollow rectangular section about Y-Y axis= (d2b2/12- d1b1/12)Moment of inertia of the hollow rectangular section about any axis= ( MI of outer rectangular section about the axis - MI of cut-out rectangular section about the axis )

2.7 DETERMINATION OF MOMENT OF INERTIA4) Moment of inertia of a circular section passing through the centre and lying in the plane of the figureConsider an elementary circular ring of radius r and thickness dr. Area of the circular ring =2rdrThe moment of inertia about an axis passing through the centre O of the circle and perpendicular to the plane of area,

Izz= R^4/2= D^4/32According to perpendicular axis theorem, Ixx=Iyy= Izz= D^4/64

2.7 DETERMINATION OF MOMENT OF INERTIA5) Moment of inertia of a hollow circular section

Moment of inertia of outer circle about X-X axis= D^4/64Moment of inertia of the cut-out circle= d^4/64Moment of inertia of the hollow circular section about X-X axis Ixx= x (D^4 d^4) /64Similarly Iyy= x (D^4 d^4) /64

2.7 DETERMINATION OF MOMENT OF INERTIA6) Moment of inertia of a semicircular areaINN= Moment of inertia of the semicircular lamina about an axis passing through the centre of the semicircle=1/2 x Moment of inertia of a circular lamina about an axis passing through the centre and lying in the plane of the figure= D^4/128

2.7 DETERMINATION OF MOMENT OF INERTIAMoment of inertia of the semicircle about an axis passing through the CG of the semicircle= Inn + A x Yc

Ixx= D^4/128 + D/8 x (2D/3)= 0.11R^4

2.7 DETERMINATION OF MOMENT OF INERTIA7) Moment of inertia of a triangular section about its baseConsider an elemental strip DE of the triangle at a distance y from the vertex opposite to the base of the triangle. From the above figure , DE/b= (d-y)/dDE=b (1- y/d)

Area of the elemental strip dA= DE x dy = b (1-y/d) x dyMoment of inertia of the area about N-N,

INN =bd/12

2.7 DETERMINATION OF MOMENT OF INERTIA8) Moment of inertia of a triangle about the centroidal axis parallel to the base of the triangleINN =bd/12=Igg + Ayc

bd/12 - bd/2 x d/9= IggIgg= bd/36

2.8 MASS MOMENT OF INERTIAConsider a body of mass M lying in the XY plane. Let X= distance of CG of the body from OY axisY= distance of CG of the body from OX axis

Moment of mass M about OY axis= Mx

2.8 MASS MOMENT OF INERTIASecond moment of mass M about OY axis= MxSecond Moment of mass is known as mass moment of inertiaMass moment of inertia(IM) about an axis is hence defined as the product of mass of a body and the square of its perpendicular distance from the axis of reference.

2.8 MASS MOMENT OF INERTIASuppose the body is split up into small masses m1, m2, m3,....., mn. Let the distance of CGs of masses be at distances r1, r2, r3,.....rn from an axis of reference. Then mass moment of inertia about that axis is given byIM= mi x riIf the small masses are large in number, then the summation in above equation can be replaced by integral,Im= rdmPhysical meaning of mass moment of inertia:- It is the resistance of a rotating body against the change in angular velocity

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT CENTROIDAL AXIS PARALLEL TO THE BASE OF THE PLATEConsider a rectangular plate of width b, depth d and thickness t composed of a material of density .

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT CENTROIDAL AXIS PARALLEL TO THE BASE OF THE PLATEMass of the plate= x b x d x tConsider an elementary rectangular strip of width b, depth dy and thickness t at a distance y from the centroidal axis X-X . The area of the strip is, dA= b x dyMass of the strip, dm= Volume of the strip x density= Thickness x Area of the strip x density=t x b x dy x Mass moment of inertia of the strip = y dm= yx (t x b x dy x )

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT CENTROIDAL AXIS PARALLEL TO THE BASE OF THE PLATEMass moment of inertia of the entire mass about XX axis= y dm = yx (t x b x dy x )

=bt x d/12= x t x b d/12= Density x thickness x Moment of inertia of the rectangular section about the centroidal axis parallel to the base

bt x d= Mass of the rectangular plate= M

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT CENTROIDAL AXIS PARALLEL TO THE BASE OF THE PLATE

Hence moment of inertia of a rectangular plate about the centroidal axis parallel to its base Imxx= Md/12Moment of inertia of a rectangular plate about the centroidal axis perpendicular to its base, Imyy= Mb/12Moment of inertia of a hollow rectangular plate of outer section dimensions B, D and inner section dimensions b, d is given by the equation, Imxx= 1/12 (MD - md); where M is the mass of outer section and m is the mass of cut-out section.

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT AN AXIS PASSING THROUGH ITS BASEConsider a rectangular plate of width b, depth d and thickness t composed of a material of density .Mass of the plate= x b x d x tConsider an elementary rectangular strip of width b, depth dy and thickness t at a distance y from the base AB . The area of the strip is, dA= b x dy

Mass of the strip, dm= Volume of the strip x density= Thickness x Area of the strip x density=t x b x dy x

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT AN AXIS PASSING THROUGH ITS BASEMass moment of inertia of the strip = y dm= yx (t x b x dy x )Mass moment of inertia of the entire mass about XX axis= y dm = yx (t x b x dy x )

=bt x d/3= x t x b d/3= Density x thickness x Moment of inertia of the rectangular section about the centroidal axis parallel to the base

2.9 MASS MOMENT OF INERTIA OF A RECTANGULAR PLATE ABOUT AN AXIS PASSING THROUGH ITS BASEbt x d= Mass of the rectangular plate= MHence moment of inertia of a rectangular plate about a horizontal axis passing through to its base Imxx= Md/3Moment of inertia of a rectangular plate about the centroidal axis perpendicular to its base and passing through the vertical side, Imyy= Mb/3

2.10 MASS MOMENT OF INERTIA OF A CIRCULAR PLATEConsider a circular plate of radius R, thickness t, with O as its centre. Consider an elementary circular ring of radius r and width dr as shown.

Area of the circular ring, dA= (2r)drVolume of the circular ring, dV=(2r)dr x tMass of the circular ring, dm=(2r)dr x t x

2.10 MASS MOMENT OF INERTIA OF A CIRCULAR PLATEMass moment of inertia of the circular ring about axis perpendicular to the plane of the figure, Imzz= rdm=

Imzz= R^4t /2Imzz=Imxx=1/2 x Imzz= R^4t /4Imxx= D^4t /64Mass of the circular disc, M= Area of the disc x thickness of the disc x density= D/4 x t x Imxx= Md/16

2.11 MASS MOMENT OF INERTIA OF A HOLLOW CIRCULAR CYLINDERLetR2 Outer radius of the cylinderR1 Inner radius of the cylinderL Length of cylinderM Mass of cylinderDensity of the material of cylinderM= x (R2 - R1) x L x

2.11 MASS MOMENT OF INERTIA OF A HOLLOW CIRCULAR CYLINDERArea of the circular ring, dA= (2r)drVolume of the circular ring, dV=(2r)dr x LMass of the circular ring, dm=(2r)dr x L x Mass moment of inertia of the circular ring about axis perpendicular to the plane of the figure, Imzz= rdm

2.11 MASS MOMENT OF INERTIA OF A HOLLOW CIRCULAR CYLINDER=L x (R2^4 R1^4)/2= x (R2 - R1) L x (R2 + R1)/2=M x(R2 + R1)/2 According to perpendicular axis theorem, Imxx=Imyy=Imzz/2= M x (R2 + R1)/4

2.12 MASS MOMENT OF INERTIA OF A UNIFORM RODConsider a uniform thin rod AB of length L. Let m be the mass per unit length of the rod and M be the mass of the rod. Hence M=mL

Consider a strip of length dx at a distance x from the Y-Y axis of reference.

2.12 MASS MOMENT OF INERTIA OF A UNIFORM RODMass of the strip dm=mdxMass moment of inertia about axis YY,Imyy= xdm

=mL/3=ML/3Hence Imyy= ML/3

2.13 PRODUCT OF AREAConsider a small elemental area dA of a body of area A. Moment of area dA about X-axis= y.dAMoment of y.dA about Y-axis=x.(ydA)xydA is known as the product of inertia(Ixy) of area dA about x-axis and y-axis. xydA is known as the product of inertia of the area A about x-axis and y-axis.Ixy= xydAHence product of area is obtained if an elemental area is multiplied by the product of its co-ordinates and is integrated over an area.

2.13 PRODUCT OF AREAThe product of inertia can also be written mathematically as Ixy= (xiyiAi)= x1y1A1 + x2y2A2 + .......+xnynAnWhere (xi,yi) are the co-ordinates of the CG of area Ai.Notes: Product of inertia can be positive (if both x and y have the same sign) or negative (if both x and y have different signs) If the area is symmetrical w.r.t one or both the axis, then the product of inertia will be zero The product of inertia w.r.t the centroidal axis will be always zero.

2.14 PRINCIPAL AXESPrincipal axes are the axes about which the product of inertia is zero.Consider a body of area A. Consider a small area dA. The product of inertia of the total area A w.r.t x and y axes is given by

2.14 PRINCIPAL AXESLet now the axes be rotated by 90 in CCW direction keeping the total area in the same position. Let X1 and Y1 be the new axes. The co-ordinates of the area dA w.r.t new axes are x and y.Hence the product of inertia w.r.t new axes=(xydA);Ixy=xydALet now the axes be rotated by 90 in CCW direction keeping the total area in the same position. Let X1 and Y1 be the new axes. The co-ordinates of the area dA w.r.t new axes are x and y.Hence the product of inertia w.r.t new axes=(xydA);

2.14 PRINCIPAL AXESComparing the two configurations, x= y and y=-xHence (xydA)= -(xydA).That means, the product of area changes its sign from positive to negative while rotating the axes through an angle of 90 in CCW direction. Hence it is also possible that on rotating the axes through a certain angle, the product of inertia becomes zero. The new axes about which product of inertia is zero is called principal axes.