DEFETIONS OF ES ND FRES INTRODUTION : When a structure is loaded, its stressed elements deform. In a truss, bars in tension elongate and bars in compression shorten. eams bend. s this deformation occur, the structure changes shape and points on the structure displace. lthough these deflections are normally small, as a part of the total design the engineer must verify that these deflections are within the limits specified by the governing design code to ensure that the structure is serviceable. arge deflections cause cracking of non structural elements such as plaster ceiling, tile walls or brittle pipes. Since the magnitude of deflections is also a measure of a member s stiffness, limiting deflections also ensures that ecessive vibrations of building floors. In this chapter we consider several methods of computing deflections and slopes at points along the ais of beams and frames. These methods are based on the differential equation of the elastic curve of a beam. This equation relates curvature at a point along beam s longitudinal ais to the bending moment at that point and the properties of the cross section and the material. If the elastic curve seems difficult to establish, it is suggested that the moment diagram for the beam or frame be drawn first. positive moment tends to bend a beam concave upward. ikewise a negative moment bend the beam concave downward.
Therefore if the moment diagram is known, it will be easy to construct the elastic curve. In particular, there must be an inflection point at the point where the curve changes from concave down to concave up, since this is a zero moment. θ θ θ θ SHETIY oment Diagrams Elastic urves Inflection point Inflection point
DOUE INTEGRTION ETHOD The double integration method is a procedure to establish the equations for slope and deflection at points along the elastic curve of a loaded beam. The equations are derived by integrating the differential equation of the elastic curve twice. The method assumes that all deformations are produced by moment. GEOETRY OF URVES The slope of the curve at point dy d tanθ If the angles are small, the slope can be written dy d θ y Elastic curve From the geometry of triangular segment O ρ. d θ ds O d ine tangent at Dividing each side by ds dθ ψ ρ ds dθ/ds represents the change in slope per unit length of distance along the curve, is called curvature and denoted by symbol Ψ. Since slopes are small in actual beams dsd we can epress the curvature, ρ dθ dθ ψ d ρ ds θ dθ ine tangent at Differentiating both sides of the second equation d θ d y d d
Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings. ψ dθ ρ d d y d Eample : For the cantilever beam in figure establish the equations for slope and deflection by the double integration method. lso determine the magnitude of the slope Θ and deflection at the tip of the cantilever. is constant. y - d d dy d y y ( ) ( ) + + + + + 6 oundary conditions when when......... y... dy d... then... then...... dy θ + d y + 6 θ
Eample : Use the double integration method, establish the equations for slope and deflection for the uniformly loaded beam in figure. Evaluate the deflection at mid-span and the slope at support. is constant. y q/ d q -dy d dy q q d y q q d dy q q + d 4 6 4 q q y + + 4 oundary... onditions... y... 4 q q... y... 4 dy q q q θ d 4 6 4 4 q q q y 4 4 4 +... q 4 Deflection at mid-span substitute / and the slope at substitute 4 5q y 84 q θ 4
OENT-RE ETHOD There are two moment-area theorems. These theorems were developed by Otto ohr and later stated formally by harles E. Greene in 87. These theorems provide a semi-graphical technique for determining the slope of the elastic curve and its deflection due to bending. They are particularly advantageous when used to solve problems involving beams especially those subjected to serious of concentrated loadings or having segments with different moments of inertia. The first theorem is used to calculate a change in slope between two points on the elastic curve. The second theorem is used to compute the vertical distance (called a tangential deviation) between a point on the elastic curve and a line tangent to the elastic curve at a second point. These quantities are illustrated in the following figure. Elastic curve θ t θ θ Tangent at Derivation of the moment-area theorems : The relation between slope and moment and bending stiffness is dθ d dθ d It states that the change in slope of the tangents on either side of the element d is equal to the area under the / diagram. Integrating from point on the elastic curve to point we have, θ d
The ordinates of the moment curve must be divided by the bending stiffness to produce / curve. The last equation forms the basis for the first moment-area theorem. The change in slope between any two points on a continuous elastic curve is equal to the area under the / curve between these points. The tangential deviation dt can be epressed by dt dθ. / diag. Substitute dθ in the equation, dt ( d). Elastic curves d dθ d dt dθ Tangent at To evaluate t we must sum all increments of dt by integrating the contribution of all the infinitesimal segment between points and. t dt. d Tangent at t θ Remembering that the quantity d/ represents an infinitesimal area under the / diagram and that is the distance from the area to point, we can interpret the integral as the moment about point of the area under the / diagram between points and
This result constitutes the second moment-area theorem. The tangential deviation at a point on a continuous elastic curve from a tangent drawn to the elastic curve at a second point is equal to the moment about of the area under the / diagram between the two points. Eample : ompute the slope and deflection at the tip of the cantilever beam. is constant. y t v Tangent at is always horizontal. θ θ + θ θ θ θ The slope at mid-point : / DIGR θ θ 8 t + v 8 inus sign indicates that the tangent lies above elastic curve
Eample : Determine the deflection at points and of the beam shown below. Values for the moment of inertia of each segment are I 8. 6 mm 4, and I 4. 6 mm 4. Take E Ga, 4 m 5 N.m y inspection the moment diagram for the beam is a rectangle. If we construct the / diagram relative to I. The couple causes the beam to deflect concave up. We are required to find the vertical displacements at and. These displacements can be related directly to the deviations between the tangents. 5 / diagram 5 tan t 5. *4. 9 6 *4*.5m t t t 75 5 *4*5 + 5.96m 9.6mm **.5 Since both answers are positive, they indicate that points and lie above the tangent at.
Eample : Determine the slope at point of the beam. is constant. 8 kn 4 m θ θ θ θ t θ / DIGR ' 4/ θ / D t θ θ 4 4 θ * t *6 *4 + * ** 6 4 6 θ... θ 8 6 t D Horizontal tangent 4 5 t D * * 5 t ' t 6 8 ocation of maimum deflection occurs at the point where the slope of the beam is zero. So, the change in slope must be Θ, between the support and the point of the ma. deflection. let the base of the triangle be ' t ' 4 t 6 4 5 4 ma D θ 5 4. 47m 9.86
Eample : Determine the deflection at point of the beam. E Ga, I5-6 m 4 6 kn/m ' t 6 8 t ' t ' t t 8 m 8 m - 9/ t ' t 9 9 t *8 8 + *8 + *8* 64 9 48 t *8 8* 64 48 768 * 768. knm 6 *. kn / m.5 * 4mm 6.4m. 4 *8
ONJUGTE E ETHOD The method was first presented by Otto ohr in 86. This method relies only the principles of statics and hence its application will be more familiar. The basis for the method comes from the similarity between both dq d d d q Q or... q d d which relate a beam s internal shear and moment to its applied loading, and d θ d d y d which relate the slope and deflection of its elastic curve to the internal moment divided by. Or integrating Q θ ( q) d d y ( q) d d These equations indicate that the shear in a beam can be obtained by integrating the load once and the moment by integrating the load twice. Since the curvature of a beam is proportional to the bending moment, slope and deflection of the beam can be obtained by successively integrating the moment. We want to replace the integration indicated in the equations by drawing the shear and bending moment diagrams. To do this we will consider a beam having the same length as the real beam, bot referred to here as the conjugate beam. It is loaded with the / diagram of the real beam. Shear and moment diagrams of the conjugate beam represent one and two integrations, respectively of the / diagram of the real beam. We thus conclude that shear and moment diagrams of the conjugate beam represent the slope and deflection of the real beam. Now we can therefore state two theorems related to conjugate beam, namely
Theorem : The slope at a point in the real beam is equal to the shear at the corresponding point in the conjugate beam. Theorem : The displacement at a point in the real beam is equal to the moment at the corresponding point in the conjugate beam. Since each of the previous equations requires integration it is important that the proper boundary conditions be used when they are applied. RE E ONJUGTE E θ pin Q pin θ roller Q roller θ fied Q free θ free Q fied θ Internal pin Q hinge θ Internal roller Q hinge θ hinge Q Internal roller
RE E ONJUGTE E If we treat positive values of the / diagram applied to the beam as a distributed load acting upward and negative values of / as a downward load, positive shear denotes a negative slope and a negative shear a positive slope. Further negative values of moment indicate a downward deflection and positive values of moment an upward deflection. s a rule, neglecting aial force, statically determinate real beams have statically indeterminate conjugate beams; and statically indeterminate real beams, as in the last figure above become unstable conjugate beams. lthough this occurs, the / loading will provide the necessary equilibrium to hold the conjugate beam stable.
Eample : Determine the slope and deflection of point of the steel beam shown in figure. The reactions are given in the figure. Take E Ga, I 6 mm 4 kn kn. knm. 4.6 m 4.6 m 7.667 m Real eam onjugate eam / diagram is negative, so the distributed load acts downward.76 + Q.76.76 Q 6. *. Q.49 θ.49rad 6.76 Q.76 + * 7.667 784.5.68 m 784.5 6. *. 6 ositive shear indicates negative slope, and the negative moment of the beam indicates downward displacement.
Eample : Determine the maimum deflection of the steel beam shown below. Reactions are given. I 6. 6 mm 4 kn 45 45 ma 8 kn The distributed load acts upward, since Real / diagram is positive. Eternal eam reactions are shown on the free body diagram. aimum deflection of the real beam occurs at the point where the slope 9 m m of the beam is zero. This corresponds to the same point in the conjugate beam where the shear is zero. ssumong this 6 kn 8 point acts within the region <<9 m. from, we can isolate the section below, 8 Eternal reactions 7 6 * 6.7 6.7 * 6.7 * 6.7 * +.. 6. * 6. onjugate eam 6.68 8 9 45 Internal reactions m F y 45 + 6.7 m 9 m Q
Eample: Determine the displacement of the pin at and the slope of the each beam segment connected to the pin for the beam shown in the figure. E Ga I5. 6 mm 4 5 kn knm / diagram has been drawn in parts using the principle of superposition. Notice that negative regions of this diagram develop a downward Real distributed load and the positive regions have a eam distributed load that acts upward..6.6 4.6 87. 6 676.7 6.8 4.8 m 6 m.67 767. onjugate eam 59.74 76.5 m To determine Θ r the conjugate beam is sectioned just to the right of and shear force is computed F Q 76 r 76 59.74 Q + r 7.74 7.74 Q 6. *5. r Q.768 θ r y.768 rad 59.74 6
The internal moment at yields the displacement of the pin 76 59.74 + *.76 * 4.6 884.8 884.8. * 5. 6 6.7 m Q 76 59.74 The slope Θ l can be found from a section of beam just to the left of thus Q 767. 76 59.74 F l 767. 76 59.74 Q + l l 449.7 449.7 Q θ 6. *5. l Q.667 θ l y.667 rad 6
EXTERN WORK ND STRIN ENERGY ost energy methods are based on the conservation of the energy which states that work done by all the eternal forces acting on a structure, W, is transformed into internal work or strain energy U, which is developed when the structure deforms. If the material s elastic limit is not eceeded, the elastic strain energy will return the structure to its undeformed state when the loads are removed. WU Eternal work done by a force : When a force F undergoes a displacement d in the same direction as the force, the work done is, dw F. d If the total displacement is, the work becomes W F. d F F W F. d Represents the triangular area. d
Suppose that is already applied to the bar and that another force F is now applied, so the bar deflects further by an amount, the work done by (not F ) when the bar undergoes the further deflection is then F. W The work of F does not change its magnitude. Work is simply the force magnitude () times the displacement ( ) F Eternal work done by a moment: The work of a moment is defined by the product of the magnitude of the moment and the angle dθ through which it rotates. If the total angle of rotation is Θ radians, the work becomes s in the case of force, if the moment is applied to a structure gradually, from zero to, the work is then W θ dw W θ. dθ. dθ However if the moment is already applied to the structure and other loadings further distort the structure by an amount Θ, then the rotates Θ, and the work is W. θ
Strain Energy due to an aial force, E N σ E. ε N U E N E N. N E Strain Energy due to the moment dθ Tangent to right end d dθ du U d. dθ d. d The strain energy for the beam is determined by integrating this result over the beams entire length d Tangent to left end dθ
rinciple of work and energy pplication of this method is limited to only a few select problems. It will be noted that only one load may be applied to the structure and only the displacement under the load can be obtained. Eample: Using principle of work and energy, determine the deflection at the tip of the cantilever beam. d d U W ) (. ) ( ) ( +, -...?, HOEWORK
If we take a deformable structure, and apply a series of loads to it, it will cause internal loads and internal displacements d N, d N, d Internal forces nd Internal displacements N Eternal Forces Equilibrium Equations Internal Forces Eternal Displacements ompatibility Equations internal Displacements, Eternal and internal forces Eternal and internal deflections dθ
rinciple of work and energy states. i. i i i N j. dθ. j Work of eternal forces Work of internal forces Virtual work method (Unit oad ethod) Virtual work is a procedure for computing a single component of deflection at any point on a structure. The method permits the designer to include in deflection computations the influence of support settlements, temperature change, and fabrication errors. To compute a component of deflection by the method of virtual work, the designer applies a unit force to the structure at the point and in the direction of the desired displacement. This force is often called a dummy load, the displacement it will undergo is produced by other effects. These other effects include the real loads, temperature change, support settlement and so forth. In other words, the deflected shape due to the real forces is assumed to be the virtual deflected shape of the dummy structure, and the principle of work and energy is applied. pplication of unit force method to trusses Virtual forces Real displacements. N. j j N. E α. T. Due to real loads Due to temperature change Given fabrication error
Eample: Determine the horizontal and vertical displacement at point. ll members have a cross-sectional area of 5. -4 m and a modulus of elasticity of Ga..5.864.kN.75.65.kN.m.m 7.5 6.5 87.5 5kN 4.m 4. m 5 D.84 D.65 D 5kN EER (m) N (kn) Nv Nh N.Nv. N.Nh. 6 7,5 -,5 -,75 -,5-84,75 5-6,5,84,65-6,65-95, 5-87,5 -,84,65 78,875-585,98 D 5-5 D 8 SU 48,75-865,65 V c h c N. Nv. E N. Nh. E 48.75.875m 6. *.5 865.65.7m 6. *.5 The negative sign indicates that the horizontal displacement is to the left, not to the right as assumed
Eample: The truss shown below is distorted during fabrication because member is.7 mm short. What vertical deflection is introduced at point E because of this misfit..7mm 6.m 9 E 9.m D.75.65. 65.75.kN.65.75.65. V E N j..75 * (.7mm ) j 9.55 mm
Eample: Determine the vertical displacement that occurs at point as a result temperature change of + in members D and D. The coefficient of thermal epansion is.7* -6 /. D.94.94.75m.75m 5m D D.94*.7* 6 m.94mm **6.5 D.65. 65 EER D D d,94,94 Nv -,65 -,65 Nv.d -,7 -,7.75. 75.kN D,75,75 -,74 V. N..74mm j j
Eample: Determine the change in member length for member D of the truss so that, when combined with the prescribed loading, will produce a net vertical displacement of zero at point D. E Ga..kN.kN (. mm ) (5. mm ) (. mm ).m 6 6. 97 (8mm ) D.kN 4.m (8mm ) 4.m.m.m D 8.5 8.944.kN 4 6 ember ( m ) N Nv E N.Nv./E.5.75.44 D 5 4,5 5,657-8,5-6,97 -,5, -,44 6 -,4 -,9,6787. D.kN.8 D D 4,47 6-8,944,75,8 6 sum, -,795 7,744E-5
V N. Nv.. D.7744m E.7744mm.75 kn We want to shorten the member so that upward deflection at point D be.7744 mm.. N *.7744 j..44 mm j (.75) * pplication of unit force method to beams and frames Virtual forces dθ. p. E d Real displacements q. d θ Due to real loading q. p d Integral of the product of moment functions is needed. repare an integral table to simplify the calculations
l d F f ) ( ) ( k i const k F const i f...... : ) (.... : ) ( i k X k i linear k F linear i f..... : ) (... : ) ( i k X k i k F i i f 6..... : ) ( : ) ( i k X k Second degree curve horizontal tangent at left end k i k F i f 4..... : ) ( : ) ( i X
Eample: Determine the rotation at and to an applied moment on the beam as shown. q. p p d * θ * θ q p. p. q q d d 6 i. k i. k 6.. q 6 Eample: What is the vertical deflection of the free end of the cantilever beam. / / -/ p - q p. q / 5 d i.(k + k ) + 6 6 48
Eample: Determine both the vertical and the horizontal deflection at for the frame shown. E Ga I 6 mm 4 5 kn - - 4 5 m m D m p - q 4 - q 5 - v 6 p. q d * ( + 8) + 6 5 * k.( i * 4 + i ) +. i. k..58 m p. h q 5 5 d i. k..5. m
Deflection of structures consisting of fleural members and aially loaded members Eample: Find the horizontal displacement at. E Ga I5. 6 mm 4 5 mm able 4 m kn/m 48 8 kn 4 kn m 6 m + 4 4 - + 6 p + +.5. 6 *5. 6. knm q E. 6 *5. 6. kn.5 The internal work for the given structure consists of two parts : the work due to bending of the frame and the work due to aial deformation of the cable. h c 6 p. + q d 6 + N p. N q. E.7 m 7 mm * * + 6 * 48 * 6 4 * 6 * + 8 *.5 * 4 E
kn/m D What are the relative rotation and the vertical displacement at 4 m m m 6 kn m 4 m m D What are the rotation and vertical displacement at D 6 kn D m 4 m m t t D 8/
XWE S REIRO THEORE et us consider the beam in the figure. ecause of the load F the beam deflects an amount δ F at point and an amount δ F at point. F Where δ and δ are the deflections at points and due to a unit load at point δ ij deflection at point i due to a unit load at j δf δ F F δf δ F F F W W i place of deflection j place of unit load Now we w ll formulate an epression for the work due to F and F pply the forces F and F simultaneously the resulting work can be written W ( F + ) F ( F ( δ F + δ F ) + F ( δ F + δ )) ( ( ) ) δ F + δ + δ F F + δ F F δ F + δf δ F + δ F
If we apply F first the amount of work performed is F W F. δ F δ F δ F Net we apply F to the beam on which F is already acting. The additional work resulting from the application of F F F W F ( δ F ) + F ( δ F ) δ F δ F ½ factor is absent on the first term because F remains constant at its full value during the entire displacement. The total work due to F and F. W ( δ ) F + δ F + δf F In a linear system, the work performed by two forces is independent of the order in which the forces are applied. Hence the two works must be equal. ( δ + δ ) ( δ F + F F + δ ) F ( ) δ + δ δ ( δ δ ) δ + δ δ ( δ ) F + δ F + δf F This relationship is known as awell s reciprocal theorem
δ δ δ δ δ θ θ δ θ θ θ θ onsider two different loading on a linear elastic structure. The virtual work done by the forces of the first system acting through the displacements of the second system is equal to the virtual work done by the forces of the second system acting through the corresponding displacements of the first system..
STIGINO S FIRST THEORE δ F + δf F F The work done by the forces on the deflections of the beam is δ F + δ F Taking the derivative of work with respect to F, one obtain W W F ( δ F + δ F F + δ ) F δ F + δ F Since the strain energy U stored in a deformed structure is equal to the work performed by the eternal loads U i i The partial derivative of the strain energy in a structure with respect to one of the eternal loads acting on the structure is equal to the displacement at that force in the direction of the force. This relation is known as astigliano s theorem F F U U R R R R This theorem can be used to analyze the redundant structures
F F U R R This method is called as the method of least work. Redundant must have a value that will make the strain energy in the structure a minimum. The known values of the displacements at certain points can be used for additional equations to the equations of equilibrium. The strain energy for an aially loaded member The strain energy for a truss U N U E N E To find a joint displacement of a truss astigliano s theorem can be used U N i N i δ i E δ I : joint displacement in the direction of the eternal force i N : aial force in member E : odulus of elasticity : ross sectional area To take the partial derivative, force i must be replaced by a variable force
Eample: Determine the vertical displacement of joint of the truss shown in the figure. The cross sectional area of each member is 4 mm and E Ga 4 kn vertical force is applied to do truss at joint, since this is where the vertical displacement is to be determined. m Since does not actually eist as a real load on the truss. We require p in the table. ember N Ν/ N () (m) N. Ν/. 4 m 4 m.667+ -.8+.5.677 -.8.5 8 5.67 -.4 -.8-.5 -.8 -.5 5.4.67 kn.m 4 kn,8-.5,8+.5 δ v c N N E.67 E.67 6 4* ** 6 4 kn,667+.m.mm.5-.5.5+.5
Eample: Determine the vertical displacement of joint of the truss shown in the figure. ssume that 5mm, E Ga. F E The kn force at is replaced with a variable force at joint m kn kn m m D m -(8.856+.47).+..+. F -(.+.) 9.49-.47 6.667+.667 E -(9.49+.94) 6.667+.667 D (.+.) (6.667+.667) ember N N/ N () N. N/. D.+. 6.667+.667 6.667+.667..667.667 9.98 4. 4. δ v c N N E 69.56 E DE -9.49-.94 -.94-8.89 4.4. EF F F E -.-. -8.856-.47.+. 9.49-.47 -. -.47. -.47 - -8.76 4.4 4.4 9.98 56.47 9.98 69.56 6 5* ** 6 E 6 Sum 69.56.5685m 5.685mm
8.8 8.8 D F E + find the horizontal displacement of point D + + E 4.4 E F 4.4-8.8-8.8 F - - EF 4.4-8.8-8.8 DE 6 + D 6 + 6 + N. N/. N () N/ N embe r mm m E E N N h c.77.769 ** 5* 8 8 6 6 δ Sum 8
STIGINO S THEORE FOR ES ND FRES The strain energy for beam or frame member is given by Substituting this equation into the astigliano s first theorem ı U ı d U ı ı d ı l d Eternal displacement Eternal force applied to the beam or frame in the direction of the desired displacement Internal moment in the beam or frame, epressed as function of and caused by both the force and the real loads. E odulus of elasticity I oment of inertia of cross sectional area computed about the neutral ais. If the slope Θ at a point is to be determined, the partial derivative of the internal moment with respect to an eternal couple moment acting at the point must be computed. θ l d The above equations are similar to those used for the method of virtual work ecept the partial derivatives replace moments due to unit loads like the case of trusses, slightly more calculations is generally required to determine the partial derivatives and apply astigliano s theorem rather than use the method of virtual force method.
Eample : determine the displacement of point of the beam shown in figure. E Ga I5* 6 mm 4 kn/m vertical force is placed on the beam at kn/m m The moment at an arbitrary point on the beam and its derivative 6 + ( + ) ( + 6) R R m + 6 + Setting yields 6 + 6 δ astigliano s theorem d ( 6 + 6)( ) d 6 4 4 + 6 9 + 6 5 5 6 5* ** 6.5m 5mm
Eample : determine the vertical displacement of point of the beam. Take E Ga I5* 6 mm y kn 8 kn/m 4m 4m < < 4m... (4 +.5) 4... 4 < < 8m... 64 + 4 (8 +.5)....5... 4 4 4.5... 44 8 vertical force is applied at point. ater this force will be set equal to a fied value of kn. y 4+.5 4m 8 kn/m 4m 8+.5 δ ver c 4 (4 4 4 4 4 ) 46.67 6 * *5* 6 d + + 8 4 (44 8)(4.5) 576 7 + 8.4m 4.mm 4 d 46.67
Eample : determine the rotation at and horizontal displacement at point D using astigliano s first theorem kn kn 5m 5 m 8 m D member... +... [ ]... + 6 kn member... kn 5m + kn 5 m variable couple is applied at point < < 5m... + 6... [ ] 6 c 5 < < m... + 6 ( 5)... [ ] D 8 m + +6 + kn θ 8 d d + 5 ( 6 ) + ( ) 5 d 64 + 6 * 5 7 * 5 +. * 5 + * 4. ( 5) ( 5 ) + ( * 5 ) 5.8
Eample : determine the reactions for the propped cantilever beam. / / R Reaction at point is taken as redundant -R /-R U R R < < (( < 5 6... <... R R) ( R) + R... R R( )......... R d + R)( ) d + R( )( ) d Homework, Take the moment at as redundant force
R.6 4 kn/m Using astigliano s theorem determine the horizontal reaction at point E, and draw the shear force and bending moment diagrams of the given frame..6 sinθ.6 cosθ.9488 6m 6m D E R m 5 m Symmetrical structure is subjected to a symmetrical loading so the vertical reactions at and E will be the half of the eternal load -5R.6 kn/m.6 7. kn/m 44 kn 44 kn -(45.5+.9487R) -6.6+.6R Q R 44 5R N N (45.5+.9487R) Q.6R 6.6 U R R R 8.4kN Free ody diagram of left column, Write equilibrium equations to find, N, and Q at point d... R 5 d +.6 member... R... R member... R (.6 + 5)... 5R (6.6.6R) 5R.8 6. [ (.6R 6.6) 5R.8 ](.6 5) d
4.47 8.4 - + 8.4 5.76 m.4 + -.4 4.47 SHER FORE DIGR - + 8.4 8.4 OOK EH EER FRO THE INSIDE OF THE FRE. TKE THE ORIGIN T EFT END OF THE EER 44 44 6.4 OENT DIGR IS DRWN ON TENSION SIDE 9 - - + + - - 9 66.47 ENDING OENT DIGR (KN)