LECTURE 2 Geometrical Properties of Rod Cross Sections (Part 2) 1 Moments of Inertia Transformation with Parallel Transfer of Axes.

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1 V. DEMENKO MECHNCS OF MTERLS 05 LECTURE Geometrial Properties of Rod Cross Setions (Part ) Moments of nertia Transformation with Parallel Transfer of xes. Parallel-xes Theorems S Given: a b = S = 0. z z Fig. z where 0 and z are entral axes i.e. and z are axes parallel to the and z axes. The distane between z and z axes is a and the distane between and axes is b. Determine: the moments of inertia with respet to z and axes. B definition n Fig. it is seen that = z = z =. () z d d zd z = z b = a. () Substituting z and from expressions () into formula () we find ( ) (3) = z b d= z d b zd+ b d

2 f z and V. DEMENKO MECHNCS OF MTERLS 05 z signifiantl simplified z ( ) () = a d= d a d+ a d ( )( ). (5) = a z bd= zd a zd b d+ ab d axes are entral then S = S = 0 and obtained expressions are = + b z z = z + a parallel-axes theorem. (6) = + ab. z z The moment of inertia of setion with respet to an axis in its plane is equal to the moment of inertia with respet to parallel entroidal axis plus the produt of the area and the square of the distane between two axes. The produt of inertia of a setion with respet to an pair of axes in its plane is equal to the produt of inertia with respet to parallel entroidal axes plus the produt of a setion area and the oordinates of the entroid with respet to the pair of axes. t follows from the first two formulas of (6) that in the famil of parallel axes the moment of inertia with respet to the entral axis is a minimum. While determining the produt of inertia b formulas (6) it is neessar to take into aount the signs of values a and b. The are entroid oordinates O in z0 orthogonal sstem. Example Determine axial moments of inertia and produt of inertia for the right triangle relative to entral axes whih are parallel to triangle legs. Given: h b 3 = bh / z 3 = hb / Fig. z =+ bh / (the are found b integration) Determine: z z. Let us use the results of previous example and parallel axes transfer formulae.

3 V. DEMENKO MECHNCS OF MTERLS 05 3 n this ase the transfer from an arbitrar z axes to entral z axes is neessar to perform: h = 3 b z = z 3 h b z =. z fter substitutions and simplifiations we get Fig. 3 in result: Note: sign of the produt 3 3 bh hb bh = z = 36 z = seleted orthogonal sstem of oordinates. in Fig. z depends on orientation of the triangle relative to Example Calulate the produt of inertia Fig. x of the Z-setion shown The setion has width b height h and thikness t. To obtain the produt of inertia with respet to the x axes through the entroid we divide the area into three parts and use the parallel-axis theorem. The parts are as follows: () a retangle of width b t and thikness t in the upper flange () a similar retangle in the lower flange and (3) a web retangle with height h and thikness t. The produt of inertia of the web retangle with respet to the x axes is zero (from smmetr).

4 V. DEMENKO MECHNCS OF MTERLS 05 The produt of inertia ( x ) of the upper flange retangle (with respet to the x axes) is determined b using the parallel-axis theorem: in whih ( x) = x + d d x is the produt of inertia of the retangle with respet to its own entroid is the area of the retangle d is the x oordinate of the entroid of the retangle and d is the oordinate of the entroid of the retangle. Thus h t b x = 0 = ( b tt ) d = d = ; and the produt of inertia of the upper flange retangle is h t b bt ( x) = x + dd = 0 + tb ( t) ( h t)( b t) =. The produt of inertia of the lower flange retangle is the same. Therefore the produt of inertia of the entire Z-setion is twie ( x ) or bt x = ( h t)( b t). Note: This produt of inertia is positive beause the flanges lie in the first and third quadrants. Example 3 Determine entroidal axial moments of inertia of a paraboli semisegment The paraboli semisegment OB shown in Fig. 5 has base b and height h. Using the parallel-axis theorem determine the moments of inertia entroidal axes x and. x and with respet to the We an use the parallel-axis theorem (rather than integration) to find the entroidal moments of inertia beause we alread know the area the entroidal oordinates x and and the moments of inertia x and with respet to the x and axes. These quantities ma be obtained b integration (see axial moment of inertia of a paraboli semisegment). The are repeated here:

5 V. DEMENKO MECHNCS OF MTERLS 05 5 = bh x 3b 3 = h 8 = 5 3 6bh x = 05 3 = hb. 5 To obtain the moment of inertia with respet to the x axis we write the parallel-axis theorem as follows: Fig. 5 x x = +. Therefore the moment of inertia x is ( ) x = x bh bh h bh = = n a similar manner we obtain the moment of inertia with respet to the axis: ( ) = x hb bh b hb = = Example Determine the moment of inertia x with respet to the horizontal axis in Fig. 6. x through the entroid C of the beam ross setion shown The position of the entroid C was determined previousl and equals to =.8in. Note: t will be lear from beam theor that axis x is the neutral axis for bending of this beam and therefore the moment of inertia x must be determined in order to alulate the stresses and defletions of this beam. We will determine the moment of inertia x with Fig. 6 respet to axis x b appling the parallel-axis theorem to eah individual part of the omposite

6 6 V. DEMENKO MECHNCS OF MTERLS 05 area. The area is divided naturall into three parts: () the over plate () the wideflange setion and (3) the hannel setion. The following areas and entroidal distanes were obtained previousl: = = 3 = 3.0in. 0.8in. 8.8in. ; 3 = 9.85in. = 0 = 9.88in. =.80in. The moments of inertia of the three parts with respet to horizontal axes through their own entroids C C and C 3 are as follows: 3 bh 3 = = ( 6.0in. )( 0.5in. ) = 0.063in. ; = 70in. ; 3 = 3.9in. Now we an use the parallel-axis theorem to alulate the moments of inertia about axis x for eah of the three parts of the omposite area: x ( ) ( ) = + + = 0.063in. + (3.0in. ).8in. = 38in. ; x x = + = 70in. + (0.8in. )(.80in.) = 0in. ; ( ) = = 3.9in. + (8.8in. )(8.08in.) = 580in.. The sum of these individual moments of inertia gives the moment of inertia of the entire ross-setional area about its entroidal axis x : x x x x = + + = 00in.. zo. Moments of nertia Change in Coordinate xes Rotating Let us onsider a ross setion of a rod. Relate it to a sstem of oordinates solate an element d from the area with oordinates z. Let us onsider that ross setion s axial moments of inertia z and produt of inertia (Fig. 7): z are given

7 B definition V. DEMENKO MECHNCS OF MTERLS 05 7 Fig. 7 t is required to determine z z i.e. the moments of inertia with respet to axes z rotated through an angle α in relation to sstem z ( α > 0 i.e. ounter lokwise rotation is hosen as positive). t should be observed here that the point O is not the setion entroid. Using Fig.7 we find: z = zosα sin α = osα + zsinα. (7) Then = z d z = d z = zd. (8) B similar wa ( osα sinα) os α = z d= z d sinαosα zd+ sin α d. (9) z ( osα sinα) os α = + z d= d + z + sinαosα zd+ sin α z d (0) ( osα sinα)( osα sinα) = z + z d = = os α zd sin α zd sinαosα d+ sinαosα z d = ( ) = os α sin α xd + sinαosα. () z d d

8 8 V. DEMENKO MECHNCS OF MTERLS 05 Using moment of inertia definition and the formula sinαosα = sinα we ma write that os α sin α = osα and = z + z z = z + z + os α sinα sin α os α sinα sin α z z = os sin. z α + α () We note that funtions (9 0 ) are periodi with a period π. The axial moments of inertia are positive. The an be minimum or maximum but simultaneousl at the same angle rotation axes. t is evident that α p. The produt of inertia hanges its sign in 3 Sum of xial Moments of nertia + z = os sin sin sin α + z α z α + α + z z z ( ) ( ) + os α + sinα = os α + sin α + sin α + os α. (3) Thus the sum of the axial moments of inertia with respet to two mutuall perpendiular axes depends on the angle of rotation and remains onstant when the axes are rotated. and Note that + z = ρ () ( ) + z d= ρ d (5) where ρ is the distane from the origin to the element of area. Thus z ρ where ρ is the familiar polar moment of inertia. + = (6)

9 V. DEMENKO MECHNCS OF MTERLS 05 9 Prinipal xes. Prinipal Central xes. Prinipal Moments of nertia Eah of the quantities and z hanges with the axis α rotation angle but their sum remains unhanged. Consequentl there exists an angle α p at whih one of moments of inertia attains its maximum value while the other assumes a minimum value. Differentiating the expression for derivative to zero we find (9) with respet to α and equating the d osαpsinαp zosαp Z osαpsinαp 0 dα = + = ( z )sinαp = zosαp (7) tanα = p z z. For this value of the angle minimum. α p one of axial moments is maximum and the other one is f tanα p > 0 then axes should be rotated ounter lokwise. For the same angle α p the produt of inertia vanishes: z z z = sinα sin 0 p + αp. (8) xes with respet to whih the produt of inertia is zero and the axial moments takes extremal values are alled prinipal axes. f in addition the are entral suh axes alled prinipal entral axes. The axial moments of inertia with respet to prinipal axes are alled prinipal moments of inertia. Prinipal moments of inertia are determined b using the following formulas: z + z z z max = ± + min z + z z + z ( ) ( )

10 0 V. DEMENKO MECHNCS OF MTERLS 05 z + ( ) = ± z + z (9) max min The upper sign orresponds to the maximal moment of inertia and the lower one to the minimal moment. f setion has an axis of smmetr this axis is b all means the prinipal one. t means that setional parts ling on different sides of the axis and produts of inertia are equal but have opposite signs. Consequentl z = 0 and and z axes are prinipal Fig. 8 z = z = 0. i= Example 5 Determine the orientations of the prinipal entroidal axes and the magnitudes of the prinipal entroidal moments of inertia for the rosssetional area of the Z-setion shown in Fig. 9. Given: Use the following numerial data: height h = 00 mm width b = 90 mm and thikness t = 5 mm. Let us use the x axes as the referene axes through the entroid C. The moments and produt of inertia with respet to these axes an be obtained b dividing the area into three retangles and using the parallel-axis theorems. The results of suh alulations are as follows: 6 = mm x 6 = mm 6 = mm. x Substituting these values into the equation for the angle θ p Eq. (7) we get tan θ x p = = x θ p = 38. and8..

11 Thus the two values of θ p are V. DEMENKO MECHNCS OF MTERLS 05 θ = 9. and09.. p Using these values of θ p in the transformation equation for x x + x x = + osθ sin p x θp (0) we find 6 = mm and x are obtained if we substitute into equations: 6 =. 0 mm respetivel. The same values x x + x U = max = + + x + x V = min = + x x () Thus the prinipal moments of inertia and the angles to the orresponding prinipal axes are: 6 = mm U 6 =. 0 mm V θ = 9. ; p θ = 09.. Fig. 9 The prinipal axes are shown in Fig. 9 as the U V axes. Example 6 Determine the orientations of the prinipal entroidal axes and the magnitudes of the prinipal entroidal moments of inertia for the rosssetional area shown in Fig. 0. Use the following numerial data (see table). p Fig. 0

12 Parts of the omposite area h i m im V. DEMENKO MECHNCS OF MTERLS 05 b i m Geometrial properties x i m i m m 0 m x i i The oordinates of angular setion entroid C are known from assortment ( x0 = 0 =.3 0 m). The oordinates of the entroid C are determined beforehand and equals to: x = m = m. Note: the first element (-beam) was hosen as original in this alulation. Let us use the x axes as the referene axes through the entroid C. The moments and produt of inertia with respet to these axes an be obtained using the parallel-axis theorems. The results of suh alulations are as follows. = + () x x x x = x + = = = + = = x x m m x 8 8 ( ) = + = m = + (3) = + a = = = + = = a m m 8 8 ( ) = + = m

13 V. DEMENKO MECHNCS OF MTERLS 05 3 = + () x x x ( ) 8 x = x + a = = x x = + a. n last equation the produt of inertia z m is unknown. Let s determine it using Fig.. Note that 3 z 3 axes are reall prinipal axes of the setion enertia sine 3 is the axis of its simmetr. The values of prinipal entral moments of inertia of the angle are speified in assortment. Let us selet these values.: 8 = max = = min = 39 0 z 3 m ; m. Fig. Note: f in assortment onl one of these two values is speified the seond one ma be alulated using the fat that the sum of two axial moments of inertia is unhanged in axes rotation: t is also evident that 0. z 3 3 max min z + = +. (5) ppling rotational formula for produt of inertia we find 3 z3 os max = α min + sinα= 0+ sin( 90 z ) = z 3 3 Consequentl 9 39 = 0 ( ) = m. ( ) = + = x m.

14 fter substitutions the result is V. DEMENKO MECHNCS OF MTERLS 05 x 8 8 ( ) = = m. Substituting these values into the equation for the angle θ p we get x 80.3 tg θ p 0.97 p ' p θ o = = = = θ o = The prinipal moments of inertia are x x + x = U max = ± + = ± min V ( ) x m U = max = m V = min = m. Cheking the results: а) max x min > > > > > > 6. 0 ; b) max min x + = = ( ) = ; z ) UV = x osθ sin p + θp = = ( 0.673) 0 = ( ) 0 m 0.

15 V. DEMENKO MECHNCS OF MTERLS Example of home problem Solution. Use the following numerial data (see Table) and draw the setion in sale (Fig. ). Fig.

16 6 Parts of the omposite area GOST GOST V. DEMENKO MECHNCS OF MTERLS 05 h m i m i b i m xi m Geometrial properties i m x i m i max m i min m i Table The oordinates of two C and C entroids for the parts are known from assortments ( x0 = 0 =.3 0 m).. Calulation of the entroidal oordinates for omposite area. xes x are seleted as referene axes in this stud (see Fig. ). The following formulae are used 0 m x = S / = Sx / where S = S + S ; S = S + S. x x x S x and = + = = m. S are zero due to entral harater of x axes for -beam. b S = ( + ( b 0)) = 3. 0 ( + ( )) = = x 3 =+.0 0 m. h 3 S = ( ( + x0)) = 3. 0 ( ( )) =.9 0 m. x S S x 3 = =.9 0 m. 3 = =+.0 0 m. =.9 0 /58. 0 = 0.077m= 7.75m. =+.0 0 /58. 0 = m=+ 3.65m. Results: the oordinates of the C entroid are equal to: The are shown on Fig.. x = m = m.

17 V. DEMENKO MECHNCS OF MTERLS Calulation of entral moments of inertia relative to entral x axes. Let us denote the x axes as the entroidal axes of the omposite area. The moments and produt of inertia with respet to these axes an be obtained using the parallel-axis theorems. The results of suh alulations are as follows. = + x x x x = x + = = m x x = + = + = m x 8 8 ( ) = + = m. = = + a = = m = + a = + = m 8 8 ( ) = + = m.. Calulation of the produt of inertia relative to x axes. For the first part of the setion: = + x x x ( ) 8 x = x + a = = m. For seond part the similar approah is used: The value of x x x = + a. should be determined beforehand using transformation equations for produt of inertia and taking into aount that in rotation of axes the sum of axial moments of inertia is unhanged i.e. + = +. The axes rotating x max min proedure is shown in Fig.. The x 3 3 axes are seleted as referene axes in this rotation to x axes. Due to rosssetion smmetr relative to x 3 axis the angle of rotation is θ = 5 (lokwise Fig. p rotation). n our ase general view of transformation equation for produt of inertia x x = sinθ + will be rewritten as x osθ

18 8 fter substituting x V. DEMENKO MECHNCS OF MTERLS 05 x = + osθ. 3 3 sinθ 3 3 x p x p = sin( 90 ) + 0os( 90 ) = 55 0 m m. n our designations this produt will be denoted as Consequentl = x ( 6.585)(3.085) m x = + =. Total result after substitutions is x 8 8 = ( ) 0 = m. 5. Rotating entral x axes to entral prinipal position at θ p angle. Substituting the values of entral moments and produt of inertia into the equation for the angle θ p we get ( 80.3) tg θ x p 0.97 p ' p θ o = = = = θ o = x Note that this angle is lokwise due to used sign onvention. t is shown in resultant piture shown below (Fig. 3). t is important to note that in an rotation of axes to prinipal position larger of two axial moments of inertia ( = m ) beomes the largest (maximum) and smaller one ( = 538m ) beomes the minimum in value. 6. Calulation of prinipal entral moments of inertia for omposite area. The prinipal moments of inertia are determined using the formula x + x = max = ± ( ) UV + = ± min U 8 = max = m V m x 8 = min = 6. 0 m. Note both values must be positive! 7. Cheking the results: (а) Cheking the orrespondene: max > > x > min (in the ase or > > > (in the ase n our ase max x min > > > x > ) x > ).

19 V. DEMENKO MECHNCS OF MTERLS 05 9 (b) Cheking the onstan of the sum of axial moments of inertia in rotating the axes: + = + max min x = ( ) =. () Calulating the evidentl zero entral prinipal produt of inertia of the setion: z UV = x osθ sin p + θp = = ( 80.3) ( 0.673) 0 = ( ) 0 m 0.

20 0 V. DEMENKO MECHNCS OF MTERLS 05 Fig. 3

LECTURE 14 Strength of a Bar in Transverse Bending. 1 Introduction. As we have seen, only normal stresses occur at cross sections of a rod in pure

V. DEMENKO MECHNCS OF MTERLS 015 1 LECTURE 14 Strength of a Bar in Transverse Bending 1 ntroduction s we have seen, onl normal stresses occur at cross sections of a rod in pure bending. The corresponding