Strength of MaterialsHandout No.13 Deflection of Beams Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:
[email protected] www.mediafire.com/haniazizameen Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 13.1 Itroduction Under the action of applied forces the axis of a beam deflects from its initial position. Basic differential equations for the deflection of beams will be developed in this chapter. 13 .2 Equation of the Elastic Curve When a beam is loaded perpendicular to its axis ,as shown in Fig.(13-1) it defects from its initial position. Fig(13-1) The deflection of the beam due to the applied loads should not be excessive as it is always recommended that the max. deflection of the beam should not exceed 1/360 of its length. In order to derive the equation of the elastic curve consider any two points A and B on the curve as shown in Fig(13-2). Fig(13-2) For convenience the arc AB whose length is ds is enlarged as shown in Fig(13-3). Let the length ds subtend an angle d , then ds = d 1 d d dx or (13-1) ds dx ds Assuming that ds tends to zero , dy we have = tan dx Fig(13-3) Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 2 d2y dx 2 sec 2 d dx 1 tan 2 d dx 1 dy dx d dx . (13-2) also, as ds tends to zero , (ds)2 = (dx)2 + (dy)2 or ds dx 2 1 dy dx 2 . (13-3) substituting for Eq(13-1) , yields 1 d ds from Eq(13-2) and for from Eq(13-3) into dx dx d2y dx 2 1 Since the slope dy 2 dx 3 (12-4) 2 dy dy of the elastic curve is small , is still small dx dx and therefore it is neglected, in case of beam deflected under a load , hence , 1 d2y .(13-5) dx 2 EI We established the equation M = in article (8-3) , substituting for 2 (1/R) from Eq.(13-5) gives d 2y M = EI .. (13-6) dx 2 It is known as the equation of the elastic or deflection curve The equation d2y EI = M must be expressed in the form dx 2 d2y M M dM E = , the function then being integrated since F = dx I I dx 2 3 dF d y and q = , from Article ( 7.3 , Eq.(8-4)) , it follows that EI =F dx dx 3 d4y and EI =q dx 4 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 13.3 Methods of Computing the Deflection The two methods that will be employed in computing the deflection of a loaded beam are the double integration method and the moment area method. 13.3.1 Double Integration Method d2y From equation M = EI . If the bending moment can be expressed dx 2 as a function of x , successive integration will give expressions for dy EI and EIy . The constants of integration may be determined from dx the end fixing conditions. dy M ( x ) dx C1 EI dx EIy = M( x ) dx dy C1x C2 Where y is the deflection of the beam , and C1 & C2 are constant of integration. 13.3.1.1 (a) Standard Case of Beam Deflections Cantilever Beam with Concentrated Load let y be the deflection at any distance x from the fixed end M(x) = Ra . x+M = Wx +WL M(x) = W(L x) d2y EI = W(L x) dx 2 x2 dy = W(Lx ) + C1 EI 2 dx x2 x3 EIy = W (L ) + C1x + C2 2 6 The Boundary condition dy When x =0 , =0 C1= 0 dx Lx 2 x 3 EIy = w ( ) + C2 2 6 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen when x = 0 , y = 0 C2 = 0 2 3 Lx x ) EIy = W ( 2 6 The max. slope and deflection occur at the free end, when x = L dy WL2 i.e. dx max 2EI WL3 ymax = 3EI Cantilever Beam with Uniform Distributed Load M(x) = Ra x + M+q x (x/2) x2 L2 q = q L.x + q 2 2 2 d y L x EI 2 q (L x) 2 dx d2y EI 2 (q/2) (L2 2Lx + x2) dx dy x2 x3 2 EI = (q/2) (L x 2L ) + C1 dx 2 3 x2 x3 x4 ) + C1x + C2 EIy = (q/2) (L2 L 2 3 12 The Boundary condition dy When x = 0 , 0 , so that C1 = 0 dx 2 x3 x4 2 x EIy = (q/2) ( L L ) + C2 2 3 12 When x = 0 , y = 0 , so that , C2 = 0 The max. slope and deflection occur at the free end . Where x = L dy q L3 i.e. dx max 6 EI ymax = q L4 8 EI (b) Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen (c) Cantilever Beam with Moment at its End M(x) = M d2y EI =M dx 2 dy EI = Mx + C1 dx x2 + C1x + C2 EIy = M 2 The Boundary condition When x=0 , (dy/dx) = 0 , so that C1 = 0 x2 EIy = M + C2 2 When x = 0 , y =0 C2 = 0 the max. slope and deflection occur at the free end , where x = L dy dx ML EI max ML2 ymax = 2 EI (d) Cantilever Beam with Distributed Load 0 q L Ra = o 2 MH 0 Fy qoL L +M=0 2 3 q L2 q o L M= o 2 2 2 q L = o 3 qx x M(x) = Ra .x M 2 3 q o L2 q o x 3 q L M(x) = o x 3 6L 2 2 2 d y qoL qoL qo x 3 EI = x 3 6L 2 dx 2 Ra L + Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen dy q o L 2 q o L2 q ox 4 EI = x x + C1 3 24L dx 4 q L x 3 q o L2 x 2 q o x 5 EIy = o C1x + C2 4 3 3 2 24L 5 The Boundary condition dy At x = 0 , =0 C1 = 0 dx q o L 3 q o L2 2 qo x EIy = x x 5 + C2 6 12 120L at x = 0 , y = 0 C2 =0 q o L 3 q o L2 2 qo x x x5 6 12 120L The max. Slope and deflection occur at the free end where x = L EIy = dy dx ymax. 1 qoL 2 L EI 4 q o L2 qo 4 L L 3 24L q o L4 6 qo 5 L 120L max 1 qo L 3 L = EI 12 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen e) Simply Supported Beam with Central Concentrated Load Taking the original at the central L x ) + Wx M(x) = Ra . ( 2 WL W = x Wx 2 2 2 W L d 2y x EI 2 = 2 2 dx dy EI dx EIy = W Lx 2 2 x2 2 C1 W Lx 2 x 3 + C1x + C2 2 4 6 The Boundary condition dy AT x = 0 , =0 C1 = 0 dx W L3 L When x = ,y=0 C2 = 2 24 2 2 3 3 W Lx x L EIy = 2 4 6 24 The max. slope occurs at the ends, where x = L 2 dy WL2 dx max 16EI The max. deflection occurs at the center , where x = 0 y max WL3 48EI Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen f ) Simply Supported Beam with Uniform Distributed Load Fy 0 R a R b qL M A 0 R a L qL.(L / 2) Ra= qL /2 , R b= qL/2 0 x<L M(x) = Ra x q x (x/2) M(x)= (qL / 2) (qx2 / 2) EI (d2y/dx2) = (qLx/2) (qx2/2) EI (dy/dx) = (qLx2/4) (qx3/6)+C1 EIy = (qLx3/12) (qx4/24)+C1 x +C2 The Boundary conditions C2 =0 At x=0 , y=0 At x = L , y =0 C1 = ( qL3/24) (or we can apply other boundary condition is dy/dx = 0 , at x=L/2 ) EIy = ( qL/24) x4 + (qL/12)x3 (qL3/24) The max. deflection occurs at x = L/2 ymax = (( 5 q L4) / (384EI) ) 13.3.1.2 Macaulay s Method In this method we take whole the beam as the interval of x , and it is convential to use pointed brackets for terms such as < x a > , ( which is neglected in case of negative or zero ) to find M (x) Case (a) Concentrated Load 0 x<3a M(x) = Ra x P1 <x a > P2 < x 2a > a > P2 < x 2a > EI (d2y/dx2) = M(x) EI (d2y/dx2) = Ra . x P1 < x Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen EI (dy/dx) = Ra . x2/2 (P1 /2) < x EIy = Ra .( x3/6 ) (P1 /6) < x Case ( b ) Applied Moment a >3 a >2 (P2 /2) < x 2a >2 +C1 (P2 /6) < x 2a >3 +C1 x + C2 M(x) = (M/L ) . x M<x a >0 , This term < x a >0 = 1 EI (d2y/dx2) = M(x) EI (d2y/dx2) = (Mx/L) M < x a>0 EI (dy/dx) = (M/L) (x2/2) M < x a >1 +C1 EIy = (M/6) x3 (M/2) < x a >2 + C1 x +C2 Case ( c ) Distribution Load If the beam shown in Fig (13-4a) carries a uniform distributed load q per unit length over the whole span the M(x) = Ra. x W <x a> (q x2/2) If the distributed load only covers the part DB (Fig.(13-4b)) M(x)=Ra.x W < x a > q < x b > <x b > / 2 M(x)=Ra.x W < x a > (q/2) < x b >2 - a- -bFig(13-4) Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen If the distributed load does not continue to the end of the beam remote from the origin as shown in Fig(13-5) Fig(13-5) It must be continue to the end and a compensating load added under neath as shown in Fig.(13-6) Fig(13-6) M(x)=Ra x W < x a > q < x b > < x b > /2 + q < x c > < x c >/2 M(x) = Ra x W < x a > (q/2) < x b >2 + (q/2) <x c >2 In general , any distributed load , when started , must continue to the end remote from the origin , so that the load system shown in Fig.(13-7a) must be converted to that shown in Fig(13-7b) before Macaulay s Method can be applied . Fig(13-7) Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 13.3.1.3 Examples The following examples explain the differences ideas of the deflection problems . Example (13-1) Fig.(13-8) show a cantilever beam . Find the ymax for the cantilever beam Fig(13-8) Solution Fy 0 R a P M B 0 R a .a M M(x)= Ra x M P<x M P.a a> EI (d2y/dx2) = P x Pa P<x a> (i) (ii) EI ( dy / dx) = P x2/2 P a x (P/2) < x a >2 + C1 EIy = (P/6) x3 P.a (x2/2) (P/6) < x a >3 +C1 x +C2 Boundary conditions y = 0 at x = 0 , put in eq(ii) neglected 0 = (P/6) < a > + C2 C2 = 0 C1 0 (dy/dx) = 0 at x=0 , put into Eq( i ) y = (1/EI) ( (Px3/6) (Pa/2) x2 ymax at x = L (P/6) < x a >3 ) ymax = (1/EI) *( (PL3/6) (P a L2 /2) (P/6) < L a >3) = (PL3/6) = (PL3/6) (PaL2/2) (P/6) [(L a)(L2 2aL+a2)] (PaL2/2) (P/6) [L3 2aL+a2L aL2+2a2L a3] Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen = Pa2L( 1 6 1 )+ 3 Pa3/6 = Pa2L ( 1 2 ) 6 + (Pa3/6) ymax = (1/EI) (( Pa2/6) (3L a)) Example (13-2) Fig.(13-9) shows the simply supported beam 5 m long, subjected to the uniform distributed load at 3m from the left side as shown in the figure Find the deflection at the right end . Fig(13-9) Solution M(x) = 650 x 400 (x2/2) + (400/2) < x 3 >2 + 950 < x 4 > EI (dy/dx) = 650 (x2/2) 400 (x3/6) + (400/60) < x 3 >3 +(950/2) < x 4 > 2 + C1 EIy = 650 (x3/6) 400 (x4/24) + (400/24) < x 3 >4 +(950/6) < x 4 > 3 + C1x+C2 Boundary condition C2 = 0 y = 0 at x = 0 y = 0 at x = 4 C1= 671 y = (1/EI) * [(325x3/3) (50x4/4) + (50/3) < x +(475/3) <x 4 >3 671x ] at right hand x = 5 y = (1/EI) * 195 3>4 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-3) Fig.(13-10) shows a simply supported beam subjected to the load indicated in the figure. Find the midspan deflection , select origin at midspan position of the elastic curve . Fig(13-10) Solution Vo=(wL/4) (wL/4) = 0 Mo = (wL/4)(L/2) (wL/2)(1/2) [(L/2 ).(2/3)] = (1/24) wL2 EI (d2y/dx2) = (wL2/24) (wx3/3L) EI(dy/dx) = (wL2/24)x (w/12L)x4 + C1 EIy = (wL2/48) x2 (w/60L)x5 + C1 x +C2 Boundary Condition y = 0 at x = 0 C2 = 0 at x=0 C1=0 dy/dx = 0 y= (1/EI) [ ( wL2/48)x2 ymax at x = L/2 (w/60L) x5] ymax = (9 w L4 )/ (1920 EI) Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-4) Fig.(13-11) shows a beam subjected to a concentrated load (P) . Find ymax . Fig(13-11) Solution M(x) = Ra . x P < x a > EI(d2y/dx2) Ra x P < x a > EI ( dy/dx) = Ra (x2/2) (P/2) < x a >2 + C1 EIy = (P.b/L)(x3/6) (P/6) < x a>3 + C1x + C2 The boundary condition x=0 at y=0 x=L at y = 0 C1 C2 = 0 0 = (P.b/L) (L3/6) (P/6) < L a>3 +C1L ( Pb / L)( L3 / 6) (P / 6)( L a )( L2 L 2aL a 2 ) .. (i) at maximum deflection , the slope is horizontal i.e. dy/dx = 0 0 = Ra (x2/2) P <x a>2 + C1 Ra (x2/2) (P/2) ( x2 2ax + a2) +C1 (P.b/4)x2 (P/2) x2 + (P/2) 2 a x (P/2)a2 +C1 = 0 we obtain the value of x L2 a2 3 sub into Eq(i) , the max. deflection is P.a.( L2 a 2 ) 3 / 2 y max 9 3 LEI Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-5) Fig.(13-12) shows a wooden flag-post 6 m high , is 50 mm square for the upper 3m and 100mm square for the lower 3 m . Find the deflection of the top due to a horizontal pull of 40 N at that point applied in a direction parallel to one edge of the section E = 10 GPa . Fig(13-12) Solution The total defection at the top is made up of: (1) The deflection at B (y1). (2) The slope at B, multiplied by the distance BC (y2). (3) The further deflection due to bending of BC (y3). These deflection are shown in Fig(13-12c). Let the second moments of area of parts AB and BC be I1 & I2 receptively. Then y=y1+y2+y3 40 * 33 120 * 32 40 * 32 120 * 3 40 * 33 *3 3EI1 2EI1 3EI1 EI1 3EI 2 360 * 12 7 10 * 109 0.14 1 0.054 0.09936m Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-6) Fig.(13-13) shows the two beams AB and CD. are of the same material and have the same cross-section. The support at B is at the same level as the fixed end A. Find the reactions at B and D if the beam CD arrives on its whole length a uniformly distributed load of 1 kN/m. Fig(13-13) Solution Let reaction at B be R and the force in the spacer at D be P, as shown below Then the downward deflection at B due P must equal the upward deflection due to R, since the point B is level with A, P * 43 P * 4 2 R * 53 *1 i.e. 3EI 2EI 3EI 125 P R (i) 88 The deflection at D must be the same for the upper beam as it is for the lower beam, since the spacer is assumed rigid. For the deflection at D on the lower beam, it will be convenient to move the force R to D and introduce an anticlockwise moment R*1 to compensate. 1 * 4 4 P * 43 P R R * 4 2 i.e. 8EI 3EI 3EI 2EI from which 16P 11R=12 ( ii ) from Eq(i) & Eq(ii) , P=1.454 kN R=1.023 kN Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-7) Fig(13-14) shows the overhung crankpin of a locomotive can be considered as a cantilever of length L , and the distributed load applied to the pin by the hydrodynamic lubricating film can be assumed to be of the form k(Lx-x2) per unit length x is the distance from the built in end and k is a constant. Find the expression for the deflection at the free end of the pin. Fig(13-14) Solution dx d3 y EI 3 dx EI d4y 4 k Lx k Lx 2 2 x2 x3 3 C1 kL3 6 C2 When x=L, S.F.=0, so that C1= d2y EI 2 dx Lx 3 k 6 x4 12 L3 x 6 kL4 When x=L, B.M.=0, so that C2= 12 dy Lx 4 x 5 L3 x 2 L4 x EI k dx 24 60 12 12 dy 0 , so that C3=0 dx Lx 5 x 6 L3 x 3 L4 x 2 EIy k 120 360 36 24 When x = 0 , y = 0, so that C4 = 0 When x=0, C3 C4 Therefore the deflection at free end k L6 L6 L6 L6 7kL6 EI 120 360 36 24 360EI Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-8) Fig.(13-15) shows a cantilever of circular section tapers uniformly from a diameter D at the free end to 2D at the fixed end. It carries a single concentrated load at the free end. Find the diameter of a cantilever of uniform diameter, which would have the same end deflection. Prove any formula used for calculating the deflection of the tapered cantilever. Fig(13-15) Solution The simplest expression for the second moment of area of a typical section will be obtained by taking the point O as the origin for x. x Then diameter of section = D L 4 x I D4 4 64 L 2 d y M Wx L E 2 k x 3 Lx 4 4 I dx x D4 4 64 L 64WL4 where k D4 dy x 2 Lx 3 E k C1 dx 2 3 dy k 0 , so that C1= dx 12L2 Ey = k [ ( x 1 /2 ( Lx 2 )/6+ x /(12L2) ] +C2 When x = 2 L , y=0 C2 = 3k/(8L) When x = L , y= (k/E) [ (1/(2L)) (1/(6L2))+(1/(12L2) (3/(8L))] y= (64WL3) / (24E D4) when x=2L, Deflection at free end of cantilever of uniform diameter = 64WL3 24E D 4 64WL3 3E d 4 WL3 3E 64 d 4 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 4 d= D 8 = 1.682 D Example(13-9) Fig(13-16) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy . Fig(13-16) Solution R1+R2 = 12 4*5 R2 + 4*2 +4 =0 R2 = 8 kN R1 = 4 kN M(x) = R1x 2 (x2/2) 4 <x 2> + (2/2) < x 2>2+R2<x 4> (2/2)<x 4>2 M(x)= 4x x2 <x 2> + <x 2>2+ R2 <x 4> <x 4>2 EI(dy/dx) =4x2/2 x3 /3 (4/2) <x 2>2 + (1/3)<x 2>3 + (8/2)<x 4>2 (1/3) < x 4 >3 +C1 EIy =2x3/3 (1/3)(x4 /4) (2/3) <x 2>3 + (1/3)(1/4)<x 2>4 + (4/3)<x 4> (1/3) < x 4 >4 +C1x+C2 EIy = (2/3)x3 x4/12 (2/3) <x 2>3 + (1/12)<x 2>4+(4/3)<x 4> (1/3) <x 4>4 +C1x+C2 The boundary conditions y = 0 at x = 0 C2 0 and y = 0 at x = 4 0 = (2/3)(4)3 (44/12) (2/3)<4 2>3 + (1/12) < x 2 >4 + C1*4 C1 = 4.33 EIy = (2/3)(2)3 (1/3)(24/4) (2/3)<2 2>3+(1/12)<2 2>4+(3/4)<2 4 > (1/3) <2 4 >4 + ( 4.33)*2 EIy = 4.66 kN.m3 Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-10) Fig(13-17 ) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy . Fig(13-17) Solution R1+R2=600 4R1 600*1.5=0 R1=225 N R2=375 N 4R2+600*2.5=0 R2=375 N R1=225 N M(x) =R1x (300/2) <x 1.5>2 + (300/2)<x 3.5>2 d2y EI 2 =255x 150<x 1.5>2+150<x 3.5>2 d x dy 225 2 150 150 x x 1.5 3 x 3.5 3 C1 EI = dx 2 3 3 112.5 3 50 50 x x 1.5 4 x 3.5 4 C1x C 2 EIy= 3 4 4 y = 0 at x = 0 C2 = 0 y = 0 at x = 4 m 112.5 3 50 50 0 ( 4) 4 1.5 4 4 3.5 4 4C1 3 4 4 C1= 478.125 112.5 3 50 50 ( 2) 4 1.5 4 2 3.5 4 ( 478.125) * 2 EIy = 3 4 4 3 EIy = 657 N.m Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-11) Fig(13-18) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy . Fig(13-18) Solution R1+ R2 = 1200 + 400 = 1600 4R1 1200*2.5+400=0 R1=650 N R2=950 N M(x) =R1.x 400(x2/2)+R2 <x 4> + (400/2) < x 3 >2 d2y EI 2 =650x 200x2+950<x 4>+200<x 3>2 d x dy 650 2 200 3 950 200 x x x 4 2 x 3 3 C1 EI = dx 2 3 2 3 325 3 66.7 4 475 66.7 x x x 4 4 x 3 4 C1x C 2 EIy = 3 4 3 4 y= 0 at x = 0 C2 = 0 y = 0 at x = 4 325 3 66.7 4 475 66.7 0 ( 4) ( 4) 4 4 4 4 3 4 4C1 3 4 3 4 C1= 670.3 325 3 66.7 4 475 66.7 (5) (5) 5 4 4 5 3 4 5 * ( 670.3) EIy = 3 4 3 4 3 EIy = 193.42 N.m Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen Example(13-12) Fig(13-19) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy . Fig(13-19) Solution R1+R2=1600 6R1 800*5 800=0 R1=800 N R2=800 N M(x) =R1.x 400(x2/2)+(400/2) <x 2>2+(400/2)<x 4>2 d2y EI 2 =800x 200x2 + 200 <x 2>2 + 200 <x 4> 2 d x x2 x 3 200 dy 200 EI = 800 200 x 2 3 x 4 3 C1 dx 2 3 3 3 400 3 66.7 4 475 66.7 EIy = x x x 2 4 x 4 4 C1x C 2 3 4 3 4 y = 0 at x = 0 C2 = 0 y = 0 at x = 6 400 3 66.7 4 66.7 66.7 0 (6) ( 6) 6 2 4 6 4 6C1 3 4 4 4 C1= 1865.2 400 3 66.7 4 66.7 66.7 EIy = (3) (3) 3 2 4 3 4 4 1865.2 * 3 3 4 4 4 3 = 3.3296 N.m Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen 13-3-2 Moment Area Method The alternative method of computing the deflection of beams is the moment area method. The deflection computed by the moment area method can be considerably simplified especially if the deflection is required at a specified point on the beam. Fig(13-20) Consider any two points C & D on the elastic cure of the beam, as shown in Fig(13-20b) which is loaded arbitrarily as shown in Fig(13-20b) if tangents are taken at points A & B, they will intersect at an angle d . The arc length of CD = ds The deflection of any point on the beam is so small. Hence the length ds = dx ds = d 1 d d ds dx 1 M i.e. hence (from article 8-3 ) EI Strength of materials Handout No.13- Deflection of Beams- Dr. Hani Aziz Ameen M dx -7) EI Form the above derivation it can be concluded :d First theorem curve is The deviation dt = x d B B tB/ A A dt A xd -8) Substituting Eq(13-7) into Eq(13-8) gives B M tB/ A x dx A EI tB/ A where (M dx)= shaded area (Fig(13-20c)) tB/A (M diagram / EI) multiplied by the horizontal distance from the center of gravity of the whole shaded area to the vertical through point B. Second theorem equal to 1 EI 1 XB x Mdx EI XA 1 area AB .x B EI where x B is the horizontal distance from the center of gravity of the whole shaded area to the vertical through point B. (see Fig(13-20c)). tB/ A Note: The deviations is very often different from the deflection.