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420, , Strength, Ratio 1/:, , of, , Materials, W, , We know that due to a central point loacd, the case of a, beam of span ,, , D, , simply supported, , H2, , Slope at, , the, , ()/2, , support8, , 16E, , and, , Fig. 8.22, , WI'3, , Deflection at the centre, , 48 E, W2, , Hence, , p16E, , Since the angle is very small YA= 6pX AD, , Also, , 3, , deflection at the centre, Yo, 48 EI, , Equation (i) and (i), we get., w2, , w3, , 16 EI, , 48 EI, , or,, , -), , 3, , 5, , or,, , 73(Ans.)D, , 8.7. MACAULAY'S METHOD, -, , In, is, , Macaulay's method a single equation is formed for, constructed in sucha way that, the constants, , all, , loadings on a beam, the equa, , of integration apply, , beam. This method is also, to all, called method of, portions, singularity, This is a convenient method, functions., for, determining the deflection of a beam subjected, loads or in general, discontinuous loads., to, This method is, as, explained follows:, Fig. 8.23 shows abeam of span I, simply, and W, at distances a and b, from the end A. Letsupported at A and B and carrying the, point lo, R and R, be the reactions at A, B respev, and, W, , W2, YD, , C, , b, , X, , Rp, , X,, Fig. 8.23, , -Consider a section X,X, belween A and, This, , M, =RA*x, , C, , distant x from A., Oiment is given, The bending moment, , expression (for the bending moment), , and x = a., , holds good for, all values of x, be, , 7.r=t

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Chapter: 8, Consider a, , section, , X2 X, between C and D, , :, , and distant, , moment, , bending, , from end A. The, , x, , 421, , of Beams, , Deflection, , isgiven by:, , ..ii), , M, = RA *- W, r - a), This expression holds good for all values of x between x, -Consider, , a, , section, , X, X, between D, , and B, , and, , a, , =, , =, , r, , b., , distantx from A. The bending moment, , is, , given, , by, , .i), , M,=RA X - W, (x-a)-W,r-b), This expression holds good for all values of x between x = b and x =l., , At any section, in general, the bending moment is given by:, , M,=El, , =R^xx|-W a, , a)|- W2 (r, , -, , -, , equation it may be noted that as the magnitude of x, additional expressions appear, the law of loading changes,, In the above, , For value of x between:, (i) x=0 and x a only the first, =, , termn, , .(8.23), , b), goes, , on, , increasing, , so that, , of the above equation should be considered., , should be considered., only the first wo terms of the above equation, (ii) x a and x b, should be considered., (iit) x b and x l . . all the terms of the above equation, as follows:, Integration eqn. (8.23), we get the general expression for slope, =, , =, , =, , =, , . . ., , Slope equation (8.24), , E-R+G|-Wro,b, 2, , 2, dx, The following points are worth noting, , (), , The constant of integration, , (i), , The, , quantity (x, , -, , C,, , should be written after the first, , a) should be integrated, , ( -a), , term, , and not as, , as, , of the above, -, , constant, , C is valid for all value of, , Integrating eqn. (8.24), Ely=, , we, , ax, , 2, , Similarly the quantity (r- b) should be integrated as a whole i.e. as, (in) The, , equation., , -b, , O, 2, , r., , get the deflection equation,, , C, RX+Gx+, 6, , Wx, , -, , W,r-by ...Deflection equation (8.25), , a, , 6, , 6, , lt may be further noted that:, , ()(r-a)* has been integrated, , to, , and (r - b¥ has been integrated to, , -b, , 3, , 3, , () Constant C, is written after C, x. The constant C, is valid for all values of x., (n), , and, , If the end, , Example:, x, , =, , l, y, , constants, conditions are known, the, , If the beam is, , simply supported the, , C,, , and, , C, can, , deflection is, , be evaluated., , zero at, , =0, , A and B, i.e. when, , r, , =0, , we get, C, 0, 0 and y =0 in deflection equation,, the constant, can be calculated., Putting r = l and y 0 in the deflection equation, When the constants C, and C, are known the slope and deflection at any section at any section, Can be determined., , Futting x, , =, , =, , =, , C

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422, , Strength of Materials, W, , Example 8.10. A beam AB of, , X, , length I simply supported at the ends, carries a point load Wat a distance a, , from the left end. Find:, (i) The deflection, , under the, , R, =, , Wb/l, , RR, , load., (ii), , The maximum deflection., , Solution. Fig. 8.24 shows a, Fig. 8.24, , beam AB of span I carrying a point load, Wat C., , An IC engine intake manifold., Let., , AC = a, CB = b, and a > b, , To find reactions taking moments about A, we get, Rg xl= Wxa, , Rg, Also., , Wa, , R+R=W, , RW-, , =w|1, , -w--wG:T, Wb, , Wa, , (l-a=ED, , RA, and R, Hence., ulay3, The bending moment at any section XX at a distance x from the end A, following Macau, method, is given by:, , M, =EId4 -Wb, , W (r -a)

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Chapter : 8, , Deflection of Beams, , 423, , Integrating for slope, we get, , CW(r-aWr- a, E-W, dr, , ..(ii), , 2, , Integrating again for deflection. we get, , Ely-, , +Cx+C|C|-Wr-, , ...(iii), , 6, , At A, the deflection is zero, i.e. when x = 0, y =0, , C=0, At B, the deflection is zero, i.e. when x =l, y = 0, , Wb.+C1, , W( -, , I6, , 6, , Wb, , -a)_Wb, , Cl=, , 6, , a), , 6, , Wbl, , b ) or, 6, , C-, , or, , 5, Hence the slope and deflection at any section are given by, E, , Wox, 21, dx, , (: 1-a =b), , 6, , 6, , -, , b'), , -2)-Wx-a), , .Slope equation, , 2, , 61, , OX- W2_hhrW(x-a, 6, 6l, 6, Deflection under the load,Yc:, , Ely, , Ely=, = Wo, , ), , To find y, , putting x =a in the deflection equation, we get, E ly c, Wba3, , 61, But,, , ...Deflection equation, , 6, , V2- b)a =-, , 2-b?-a, , l=a+ bb, , Elyc, , b + 2ab -b-a') -"(2, ab)=wDa(2+, 61, 61, =, , Wab2, 3EI, , Wab, 31, , (Ans.), , (i) The maximum deflection, Ymax, The maximum deflection will occur, on the larger segment AC, Moreover the slope is zero at, ne point of maximun deflection. Therefore, equating the slope at a section in AC to zero we get,, , Whx_Wb(b), 2/, 6/, , 3, , P-b, , or, , or, , +2ab, , The maximum deflection can be obtained by putting this value of x in dellection equation., , Thus,, , Elymus, , Wh-b, , 3, , Wb- b

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424, , Strength of Materials, , 23, , -b3V2, , Wb(2-b2 3/2, , 9/31, [Puttingl = (a + b),we get, , Wb-b/2, Ymax, , 93 ElIl, , Example 8.11. A beam with a, , the, , (Ans.), , span, , Ymar, , kNat 3 metres tr, of 4.5 metres carries a point load of30, and E, , =, , 4.5, , m, , 10 mt, left support. ffor the section, Iyy 54.97 x, (i) The deflection under the load, (i) The position and amount of marimum deflection., =, , Solution. Refer, , Wb(a+2ab3/2, Ans., 9/3 Ell, , Fig. 8.25. The span of the beam,, , l, , =, , 200 GN/m, find:, , 30 kN, a, , =, , 3, , +b-1.5 m, , m, , c, V., , mar, , = 2.45 m - l=4.5 m, , Fig. 8.25, , Point load,, , W, , 30 kN;, , a, , =, , 3m; b= 1.5, , m, , Moment of inertia,, , 1 54.97x 10- m, E 200 GN/m, ), , The deflection under the load, y,:, , Wab, 3Ell, , 30x1000x3x1.5, 3x 200 x 10°x 54.97 x 106x 4.5, = -0.00409 m= - 4.09 mm, Downward deflection under the load = 4.09 mm, , (Ans.), , ii) The position (x) and amount of maximum delflection (y,), , 4.5-1.53 2.45m from thelef end. (Ans.), , Position,, , 3, , Maximum deflection,, , Wb(-b2)3/2, Ymax, , 9/3 EN, 30 x 1000 x, , 1.5x(4.5- 1.5, , 9/3 x 200x 10°x 54.97 x 10x 4.5, Hence, dowmward deflection = 4.456 mm, , (Ans.), , x, , 10' mm, , =, , -, , 4.456 m

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Deflection of Beams, , Chapter:8, Example, , girder of uniform section, 14, , 8.12. A steel, , concentrated loads, ends. It carries, Calculate:, ends respectively, , wo, , of 90 kN and 60 kN at two points, , is, , long, , metres, , 3, , metres, , 425, its, from the, , simply supported, and . 5 metres, , at, , ofthe girder at the points under the two loads., , ()The deflection, , (i) The maxinmum deflection., 64 x 101 mt, I, Take:, Solution. Span of the steel girder,, , and E, , 210, , =, , 10 kN/m*., , x, , I= 14.0 m, I= 64 x 10m, E= 210 x 10 kN/m2, , Moment of inertia,, Young's modulus, , Let, , R, , and, , R, be the reactions at the support A, , and B, , 60 kN, , 90 kN, , -3 m, , -4.5 m, , 6.5 m-, , E, , respectively., , Rp 60 kN, , RA 90kN, Fig. 8.26, , Taking, , A,, , moments about, , R, , 14, , x, , R, , we, , get, 3, , =, , 90, , =, , 60 kN, , x, , +, , 60, , 150, , R+R= 90+ 60, , Also,, , R, , =, , 150-60, , 840, , 9.5, , x, , =, , 90 kN, , Consider any section XX at a distance, , x, , from the end A,, , following Macaulay's method,, , thee, , bending moment is given by:, , M Eldx =90x-90(x 3)- 60(x, , ), , 9.5), , -, , Integrating, we get, El =, , 4 5 + C-45 (x - 3, , -30 (x - 9.5), , .ii), , dx, Integrating again, we get, , Ely =15 r+Cx+ C,I- 15 (x -3), , 10, , -, , (r-9.5)3, , 2i), , x = 0, y =0, , When,, , C, =0;, x= 14 m, y = 0, , When,, , 0= 15, , x, , (14) +C, x 14 - 15 (14 -3), , 41160, C, , ., , +, , 14, , C,, , -, , 19965, , -, , 911.25, , =, , (14 -9.5), , 10, , -, , 14, , C,, , +, , 20283.75, , = - 1448.84, , is given by:, Hence the deflection at any section, 1448.84 x1-15 (rEly= 15, , x, , -, , 3)-, , 10, , (r-9.5)3, , ...Deflection equation., , (i) y and y,:, Deflectioin at C. ye, , Putting x =3, , m, , Ely= 15, , in the deflection, , x3, , 1448.84, , x, , equation,, 3, , =, , 405, , we, , get, , - 4346.52, , =, , -3941.52

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426, , Strength of Materials, 3941.52, Yc, , 3941.52, , = -0.00293 m, , =, , 210x 10°x 64 x 10, , El, = -2.93 mm, , or, Downward deflection of C, , 2.93 mm (Ans.), Deflection at D, y, : Putting x = 9.5 mm in the deflection equation, we get, , Ely,= 15 x 9.5- 1448.84 x 9.5- 15 (9.5 -3), = 12860.6- 13764 -4119.4 = - 5022.8, , 5022.8, D, , EI, , 5022.8, 210x 10°x 64 x 10= -0.00373 m or - 3.73 mm, , Downward deflection of, , D=3.73 mm (Ans.), , ii) Maximum deflection, ynar, Let us assume that the deflection will be maximum, slope at he section to zero, we get, , dy, El=, 45- 1448.84, dx, or,, , 45 -, , or,, , 45, , 1448.84 45 (, , -, , 1448.84, , at, , section betwen C and D., , a, , 45(x3, , Equating the, , =0, , - 6r +9) = 0, , 45+270x - 405 =0, , 270r= + 1853.84, , or,, , X = 6.87 m, , Putting this value of x in the deflection equation, we get, , Ely, , 15, , x, , 6.87- 1448.84, , 4863.6 -, , Downward deflection,, , =, , mar, A, 8.13., beam, AB, Example, shown in Fig 8.27. Detemine:, , -, , -, , 3), , =, , 5959.3, , 5959.3, , EI, , 210x 10x 64 x 104, , max, =, , 9953.5, , 6.87 15 (6.87, 869.4 -5959.3, x, , 0.0043 m or-4.43, 4.43 mm (Ans.), , mm, , of 4 metres span is simply supported at the ends and is loaded s, , (i), , Deflection at C,, , (ii), , Maximum deflection, and, , (ii) Slope at the end A., Given: E, , =, , 200, , x, , 10° kN/n?,, , andI, , =, , 20, , x, , 10m, , 20 kN, l 0 kN/m, , DY, -, , Im-, , Co00000M0, -Im-, , RA 20 kN, , B, , 2 m-, , R20 kN, Fig. 8.27

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Deflection of Beams, , 8, , Chapter:, , Solution. Span ofthe beam, / = 4m, E = 200 x 10 kN/m°, I =20 x 10, To calculate reaction at B taking moments about A, we get, R, , X4, , 427, , m*, , 20 x 1+ 10x 2, , R= 20 kN, , R+R= 20 + 10 x 2 40, , Also, , R20 kN, Using Macaulay's method consider any section XX at a distance x from the end A; the bending, moment at the section XX,, , M,= El, Integrating. we get, , 20x|-20(r 1)--10(x2 2, dx, -, , El10x+G|-10r -1 ) | - a - 2, , ...(ii), , dx, , Integrating agian, we get, , 1-a-2, , s +Ga, , Ely-, , .ii), , X= 0, y =0, , When,, , C2 =0;, X= 4m, y=0, , When,, , 0 -3x + 4C-4-1-4-2, 12, 213.33 +4 C, 90 -6.67, C =-29.16, Hence, the, , slope, , and deflection, , El=10x, , equations are, , - 29.16 |-10(x - 1), , - 5/3(x - 2), , ...Slope equation, , dx, 10, Ely =x, - 29.16x|, , and,, (i), , - D°|-, , -2, , ..Deflection equation, , Deflection at C, yc, , Putting x, , 2m in the deflection, , equation,, , we, , get, , 10, , Elycx 2-29.16 x2-(2-1, 26.6758.32 -3.33 =-34.98, 34.98, El, , Hence,, , 34.98, , x, , 200 x 10°x 20 x 10-6, , 10, , mm, , =, , -8.74 mm, , yc8.74 mm (downward) (Ans.), , (ii), , Maximum deflection, y,r, The maximum deflection will be very, the section between D and C., the, For maximum deflection equating, , near, , to the, , slope at, , mid-point C. Let, , the section to, , El=10x-29.16- 10(r 1) =0, dx, , or,, Or,, , 10x-29.16- 10 (r-2x+, , 10x 29.16-, , 10x+20x, , 1), , =, , 0, , 10, , =, , 0, , us assume, , zero, we, , get, , that it, , occurs, , in

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428, , Strength of Materials, , 39.161.958m, , Or,, , Putting the value of x, , 20, in the deflection, 10, , EI Ymax, , equation, we get, , x(1.958), 3, , -29.16 x 1.958 (1.958 - 13, , 25.02 -57.09 -2.93, 35, , 200 x 10°x 20 x 10, max8.75 mm (downward) (Ans.), , i.e., , (ii) Slope, , Putting, , 35, , F, , max, , =-35, , x, , =, , at the, , EI, , x, , 10', , mm, , =, , -8.75 mm, , end A, 0,, slope equation., we get, , 0 in the, , El dx -29.16, e ,=, , _29.16, dr, , 29.1, , EI, , = -0.00729 radian, , 200 x 10x 20 x 108 -0.417, , = -0.00729, , T, , Hence, , e,=0.417 (Ans.), , Example 8.14. A beam AB of span 8 metres is simply, supported at the ends. It carries a unifonly distributed of 30, kN/m over its entire length and a concentrated, load of 60 kN, at 3 metres from the, support A. Determine the maximum deflection in the beam and the location where the deflection ocCurs., , Take:, , E, , =, , =, , Solution., , Length, , 200, , x, , 80, , x, , 10 kN/,, 10 m, , of span of the beanm., I= 8 m, , Moment of inertia., , I= 80, , Young's modulus., , E, , To calculate reaction at B, , x, , 10 m, , 200 x 10 kN/ m, , taking, , moments, , about A., , we, , get, , Shock-absorber and axle-shaft of an a, , Rx 8 =60 x 3 + 30 x 8 x 8/2 = 1140, R = 142.5 kN, , Also., , R+R = 60+30 x 8 300 kN, R, , = 300- 142.5 = 157.5 kN, 60 kN, , 30 kN/m, , R, , 5 m, , m, , 5KN, , 8 m, Fig. 8.28, , Rg142.5 kN

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Chapter: 8: Deflection of Beams, , 429, , Using Macaulay's method consider any section XX at a distance x from the end A; the bending, moment at, , the section, , XX,, , M,=E, , 302, , =157.5, , 60(*, , 2, , 3), , ..), , On successively integrating the above equation, we have, , dyI57.5.5+C-30(x-3, dx, , .ii), , 2, , Ely=37.5", -SA+Cx+C-10x-3), , .ii), , 6, r= 0, y = 0, , When., , C =0, X= 8m, y = 0, , When,, , 8, , 5x8 +8C 10(8-3, 4, 6, 13440 5120 +8C,-1250 7070 +8C, , 0=, , 157.5 x, , C -883.75, , Hence, the slope and deflection equations are, , E 1 5 32-5x -883.75-30(r 3), dx, , Slope equation, , -, , 157.53, , E157.5_SX-883.75x-10(x, -3), Ely, 6, =, , .Deflection equation, , i) The maximum deflection and its location:, For maximum deflection, equating the slope at the section to zero, we get, , 157.5*5- 883.75 - 30(x -3) =0, , E, By, , trial and error,, , 2, , dx, x, , =, , 3.92, , satisfies it., , m, , . Deflection is maximum at a distance of 3.92 metres from A (Ans.), To get maximum deflection (y, , putting this value of x in the deflection equation, we have, , 157.5, , Elymax, , x 3.923, 6, , 5x, , 3.92-883.75, x 3.92, 4, , 10x, , (3.92, , -, , 3, , = 1581.2 - 295.16 - 3464.3 -7.78 = -2186, , 2186, , 2186, Ymax, , Hence,, , x, , EI, , (Ans.), ymax1.366mm (downward), , Example 8.15. A beam AB of span, , 10 kN/m, , om is simply supported at the ends A and B, , and is loaded as shown in Fig., , 8.29., , If E, , =, , 200 x 10 kN/m? and I = 120 x 10 m', , determine:, i) Deflection at the mid-span., , ii) Maximum deflection., (ii), , 10 mm = - 1.366mm, , 200 x 10°x 80 x 10, , Slope at the end A., , 2 m-, , C, , D, , -2 m - -, , 8 mFig. 8.29, , 4 m-

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430, , Strength of Materials, , bSolution. Length of span of the beam, , 8, , =, , m, , Young's modulus,, , E = 200 x 10° kN/m, , Moment of inertia,, I= 120 x 10 m, lo get reaction at B taking moments about A, we get, , RX8 =10x2>, R, , Also,, , =7.5 kN, , R+R= 10 x 2 =20 kN, , R, , =, , 20-7.5, , =12.5 kN, , In order that the, for the bending moment at any section may he, general, In the form suitable for application of Macaulay's method the loading on the beam isbea expre, , expression, , arange, , shown in Fig. 8.30., , 2m-, , -2m-, , 4 m-, , 10 kN/m, DB, , C, , B, , B, , 10 kN/m upward load, , from D toB, , RA= 12.5 kN, , Rg7.5kN, , 8 mFig. 8.30, , Using Macaulay's method consider any section XX at a distance x, , moment at the section, , On, , XX,, , M,-El=, dx, , =, , 125x10G-2, , from the end A; the bend, , 10(x- 4), b, , successively integrating the above equation, we get, E _12.5, 12.5, , 10(x-2, , 10(x-4), , dx, , arn, When,, , When,, , 125 6, , Cx+ C10(-210(x24, 4), 24, , 24, , 01-X=0, y =0, C =0, x= 8m, y = 0, , 0 2.5x83 +8C-10(8 6, , 2)4, , 24, , 1066.67, C =-79.16, , 10(8 - 4), , 24, , +8C, -540 + 106.67 =8C, +633.34, , =, , Hence, slope and deflection equations are, El 1 2 . 5 2, , 79.1610(r- 23, , dx, and,, , ), For, , 10(x - 4), , ...Slopeepu, , 6, , Ely = 12.5x3, 79., , -, , 4), 16x10-2)10(r24, , 6, Deflection at the mid span, yp :, , deflection, , at the mid, , span putting, , x, , =, , 4m in, , deflection equation,, , we, , 7 equane, , Deflection, get

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Deflection, , Chapter:8:, 12.5 x 43, , ElyD, , 6, =, , of Beams, , 431, , 10(4 2, , 79.16 x 4, , 24, , 133.33 -316.64, , -, , 6.67, , 189.98, 200 x 10°x 120 x 10, , -189.98, , =, , x, , 10, , mm, , -7.195 mm, , =, , Hence. deflection at the mid span = 7.915 mm (downward) (Ans.), , i) Maximum deflection, y,., max, , Position of maximum deflection, Let us assume that the deflection will be maximum between C and D. Equating the slope to, zero, we have, , El dy, , -79.16 - 10(x-2 =0, , dx, or,, , 6, , =, 6.25x-79.16 - 10(r-2, 6, , 0, , Solving the above equation by trial and eror, we get x 3.75 m, X= 3.75 m in the deflection equation, we have, Putting, =, , Elymax, , 12.5 x 3.75, , -, , -79.16x, , 3.75103.75, , 24, , 6, = 109.86-296.85-3.91 = -190.9, 190.9, , 2), , x 10 mm = -7.954 mm, , 200 x 10°x 120 x 10, Hence, maximum deflection = 7.954 mm (downward) (Ans.), , (in) Slope at the end A, 0, For slope at A, putting x = 0 in the slope equation, we have, -79.16 or El9, = -79.16, dr, , 79.16, , =, , 200 x 10°x 120 x 10, -0.00329 x, , 180, , -, , 0.00329radian, , -0.1880, , -0.188° (Ans.), Hence,, Example 8.16. A horizontal beam ofuniform section and length l rests on supports at its ends., carries a uniformly distributed load ofwper unit runforadistance afrom the right end. Calculate, ne value ofa for which the maximum deflection will occur at the left end f uniformly distributed load., , Solution. Fig. 8.31 shows the beam with the loading To find reaction at A taking moments, , about B, we get, w/unit run, , -(1-a)-, , C, RA(wa/2/M2/-a), , RAwa(2), Fig. 8.31

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432, , Strength of Materials, RAXI=, Wa, , RA2, Also, , R +R= wa, Rg, , wa, , wa, , W (21 - a), , 2, , 2/, , Using Macaulay's method consider any section, XX at a distance x from the end A; the bending moment, , Another, , view of shock-absorber and, ade, , of an automobile., , at the section,, , M, =E4, "x, 2/, dr, , x, , =, , -, , d, , -, , af, , On successively integrating the above equation, we have, , E 4=x+C, 4/, wa, , Ely, , and,, , When,, , -, , +Cr+C2, , 12, , X=0, y, , x, , (, , - a)P, x-d -a)*, , =0, , C, = 0, , When,, , X=1, y = 0, wa, 12/, , =, , +C -, , 4, , Cl, , Or,, , Wa, , Was2, , 24, , 12, , C, , wa, , wa, , 24, , 24, , (2-a'), , (212-a?), , Hence, the slope and deflection equations are:, F, , y, , 2, , wa'x, , dx, , Wa, , (2-a)-, , 4, , 241, , -(-a), , EEly wa"_ Wa(22-a)x-r-(1-, , Slope equaio, , wasx, , 12/, , 241, , a)l".Deflection equation, , For the condition that the maximum deflection should occur at C, we have, , When,, , =(l- a). = 0, dx, , E 4Ywa 2, dx, , 4, , wa (212 a?) = 0, 24/, , wa ( a)wa (22- a), =0, 4, 241, -, , Or,, , Wa, or,, , 24, , 16( - a - (2/- a, , =0

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Chapter: 8 Deflection of Beams, wa-|6- 2la + a)- (2, Of., , 433, , a*1= 0, , 241, , 6F-12/a +6a-2P + a =0, 7 a- 12/ a + 4 P = (0, , or., or., , 12ty144/-4x7x412/ t144/-112, 14, , 14, , 12 t 32r, , 12/ +, , 14, a =0.4531, , i.e, , 5.6510453, , 14, , (neglecting +ve sign), , Putting this value ofa in the value of C, we get, , w(0.4551)22-(0.4531)"], 24, , C, , = -0.00855 w, , (2-0.4532) = - 0.015345 w3B, , Puting this value of C in deflection equation, we get, , w(0.453/)., ((-0453/)3-, , Elymax, , 0.015345wl°(, , - 0.453/), , 121, = 0.0171 wlx (0.5471) - 0.015345 w, , w, , x 0.5471, , (0.0171 x 0.5473 -0.015345 x 0.547), , wl (0.0027987 -0.0083937), = - 0.005595 wA or ymas, , 0.005595 w!4, EI, , Hence, maximum deflection, , 0.005595w, , 8.17. A simply supported, loadas shown in Fig. 8.32. Find:, , Example, , (i) The slope at end A., ii) The deflection at the, , (downward), , EI, beam carries the, , (Ans.), , triangularly distributed symmetrical, , centre., , -, , I, , -, , Rgwl4, , RAwl4, Fig. 8.32, , Solution. The support reactions atA and B,, total load on the beam AB., , Consider, , a, , section XX, , from, at a distance x, , Load intensity at, , /2, , the end A., , 2wx, , W, , XX =-, , by symmetry, are equal and each is one-half of the, , X:

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Chapter, 16w, , 8: Deflectlon of Beams, , 435, , w, , 120, , 1920, 120 E, , Hence, deflection at thee centre = W, , (downward) (Ans.), , 120 EI, , Eample S.I8. A beam om long is subjected to a 450 kNm clockwise couple as shown in the, , Fig. S.33. Ufthe unifom flexural rigidly (El) ofthe beam is 8 x 10 kNm?, determine:, )The deflection at the point of application of the couple., ), , The nmaximum deflection., X, , 450 kNmn, , C, 4 m, RA, , 2 m-, , 75 kN, , X, , R75 kN, , 6 m-, , Fig. 8.33, Solution. Span of the beam AB = óm, El = 8 x 10 kNm2, , Flexural rigidity,, , To find reaction at B taking moments about A, we get, RX 6=450, , and R, , 75 kNT, , R, , =, , 75 kN, , Consider any section XX at a distance x from the end A. The bending moment at the section,, following Macaulay's method, is given by:, , -75x+450 = -75x| + 450(r 4, , M,= El, , ..), , On successively differentiating the above equation, we get, , E, , 752, +C+450, , dx, , (r, , -, , Ely =-75x + Cx+ Ca+ 225(r - 4, , and,, , 6, X= 0, y = 0, , When,, When,, , .(ii), , C=0, , 75x 6, , +6C+225(66, , 6C,, , or,, , the, , slope, , 2700 + 900, , =, , C, , 0, , and deflection equations, Ey, , and,, , ..(), , X=6 m, y =0, , 0= -, , Hence,, , 4}, , 300, , are:, , 75, +300, , dx, , 4, , + 450(r -, , 4), , 2, , _Ely =, , + 300.x + 225(x - 4)*, 6, , Slope equation, , .Deflection equation

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438 aStrength of Materials, Length of the beanm = 6m, , Young's modulus,, E = 200 x 10°kN/m2, Moment of inertia., , l= 120 x 10- m, To find the support reactions taking moments about B, we get, Rx4+30 x 2 = 60x 2;, , Also,, , R, R+R, , =, , =, , 15 kN, =, , 90 kN, , =, , 75 kN, , 30+ 60, , R=90-, , 15, , Using Macaulay's method, consider any section XX at a distance x from the endA; the bend., ending, moment at the section XX,, , M,==EI, M,, Eld4y.= -30x|+75 (x 2)| - 60(x - 4), Integrating, we get, , El=-15x + C+37.5 (x - 2 - 30(x - 4), dx, , Integrating again, we get, , When,, , Ely=-5r+Cx+C,I+ 12.5 (r-2I- 10 (x-4), x=2m, y =0, 0=-5 x23 + Cx 2+ C,, , 2C, +C =40, When,, , x = 6m, y = 0, , 0=-5 x6, , +, , 6C, +C2 +, , 12.5, , (6-2)- 10 (6-4, , : - 1080 + 6C, + C,+800 8 0, , 6C +C 360, Solving (iv) and (v), we get, , C, , =, , 80 and, , C,, , =, , -120, , Equation for deflection will now be, , Ely=-5x+ 80x 1201+ 12.5 (r -2) I- 10 (t -4)3, , ..Deflectiom, equa, , x =0, y= YA, Ely=- 120, , ., , -, , of,, , A, , 120, El, , 120, , 200x 10°x 120 x 10-, , = -0.005 m or -5 mm, , Hence,, At, , Y5 mm (downward), X = 4m, y = Yc deflection at the point C., , Ely=-5 x4' + 80 x 4, = -320 + 320 -, , 20, Yc, , F, , EI, , 120 +, , 120 + 12.5 (4 2, 100 = -20, , 20, , x 10" mm = -0.83 mm, , 200x 10°x 120 x 10, , Y=0.83 mm (downward, In this case maximum deflection occurs at thefree end where the load 30 kN is applied; na, , Hence,, , deflection = 5mm (downward), , Ans.)