# Theory and Analysis of Structures

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1 7 Theory and nalysis of Structures J.Y. Richard iew National University of Singapore N.E. Shanmugam National University of Singapore 7. Fundamental Principles oundary Conditions oads and Reactions Principle of Superposition 7. eams Relation between oad, Shear Force, and ending Moment Shear Force and ending Moment Diagrams Fixed-End eams Continuous eams eam Deflection Curved eams 7. Trusses Method of Joints Method of Sections Compound Trusses 7. Frames Slope Deflection Method Frame nalysis Using Slope Deflection Method Moment Distribution Method Method of Consistent Deformations 7.5 Plates ending of Thin Plates oundary Conditions ending of Rectangular Plates ending of Circular Plates Strain Energy of Simple Plates Plates of Various Shapes and oundary Conditions Orthotropic Plates 7.6 Shells Stress Resultants in Shell Element Shells of Revolution Spherical Dome Conical Shells Shells of Revolution Subjected to Unsymmetrical oading Cylindrical Shells Symmetrically oaded Circular Cylindrical Shells 7.7 Influence ines. Influence ines for Shear in Simple eams Influence ines for ending Moment in Simple eams Influence ines for Trusses Qualitative Influence ines Influence ines for Continuous eams 7.8 Energy Methods Strain Energy Due to Uniaxial Stress Strain Energy in ending Strain Energy in Shear The Energy Relations in Structural nalysis Unit oad Method 7.9 Matrix Methods Flexibility Method Stiffness Method Element Stiffness Matrix Structure Stiffness Matrix oading between Nodes Semirigid End Connection 00 by CRC Press C

4 00 kn 5 m C 60 5 m 0 (a) pplied load 00 Sin kn x 00 Cos kn x y y FIGURE 7. eam in equilibrium. reactions and the applied loads, which are resolved in vertical and horizontal directions, are shown in Fig. 7.b. From geometry, it can be calculated that y x. Equation (7.) can be used to determine the magnitude of the support reactions. Taking moment about gives from which 0 y y 7. kn Equating the sum of vertical forces, SF y, to zero gives and hence we get Therefore 7. + y 6. 0 y 7. kn Equilibrium in the horizontal direction, SF x 0, gives and hence (b) Support reactions 00 kn. x y x x 00 kn There are three unknown reaction components at a fixed end, two at a hinge, and one at a roller. If, for a particular structure, the total number of unknown reaction components equal the number of equations available, the unknowns may be calculated from the equilibrium equations, and the structure is then said to be statically determinate externally. Should the number of unknowns be greater than the number of equations available, the structure is statically indeterminate externally; if less, it is unstable externally. The ability of a structure to support adequately the loads applied to it is dependent not only on the number of reaction components but also on the arrangement of those components. It is possible for a structure to have as many or more reaction components than there are equations available and yet be unstable. This condition is referred to as geometric instability. 00 by CRC Press C

6 Equation (7.) shows that the rate of change of shear at any point is equal to the intensity of load applied to the beam at that point. Therefore, the difference in shear at two cross-sections C and D is X D S - S p Ú dx D C (7.5) X c We can write this in the same way for moment as M D xd - MC S dx Ú xc (7.6) Shear Force and ending Moment Diagrams In order to plot the shear force and bending moment diagrams, it is necessary to adopt a sign convention for these responses. shear force is considered to be positive if it produces a clockwise moment about a point in the free body on which it acts. negative shear force produces a counterclockwise moment about the point. The bending moment is taken as positive if it causes compression in the upper fibers of the beam and tension in the lower fiber. In other words, a sagging moment is positive and a hogging moment is negative. The construction of these diagrams is explained with an example given in Fig. 7.. Section E of the beam is in equilibrium under the action of applied loads and internal forces acting at E, as shown in Fig There must be an internal vertical force and internal bending moment to maintain equilibrium at section E. The vertical force or the moment can be obtained as the algebraic sum of all forces or the algebraic sum of the moment of all forces that lie on either side of section E. The shear on a cross-section an infinitesimal distance to the right of point is +55, and therefore the shear diagram rises abruptly from zero to +55 at this point. In portion C, since there is no additional load, the shear remains +55 on any cross-section throughout this interval, and the diagram is a horizontal, as shown in Fig. 7.. n infinitesimal distance to the left of C the shear is +55, but an infinitesimal x 0 kn kn/m 0 kn C D E F kn 0 m kn 9 FIGURE 7. ending moment and shear force diagrams. 55 kn kn/m E C D m m 6 m V C T FIGURE 7.5 Internal forces. 00 by CRC Press C

7 distance to the right of this point the concentrated load of magnitude 0 has caused the shear to be reduced to +5. Therefore, at point C, there is an abrupt change in the shear force from +55 to +5. In the same manner, the shear force diagram for portion CD of the beam remains a rectangle. In portion DE, the shear on any cross-section a distance x from point D is S 55 0 x 5 x which indicates that the shear diagram in this portion is a straight line decreasing from an ordinate of +5 at D to + at E. The remainder of the shear force diagram can easily be verified in the same way. It should be noted that, in effect, a concentrated load is assumed to be applied at a point, and hence, at such a point the ordinate to the shear diagram changes abruptly by an amount equal to the load. In portion C, the bending moment at a cross-section a distance x from point is M 55x. Therefore, the bending moment diagram starts at zero at and increases along a straight line to an ordinate of +65 at point C. In portion CD, the bending moment at any point a distance x from C is M 55(x + ) 0x. Hence, the bending moment diagram in this portion is a straight line increasing from 65 at C to 65 at D. In portion DE, the bending moment at any point a distance x from D is M 55(x + 7) 0(X + ) x /. Hence, the bending moment diagram in this portion is a curve with an ordinate of 65 at D and at E. In an analogous manner, the remainder of the bending moment diagram can easily be constructed. ending moment and shear force diagrams for beams with simple boundary conditions and subject to some selected load cases are given in Fig Fixed-End eams When the ends of a beam are held so firmly that they are not free to rotate under the action of applied loads, the beam is known as a built-in or fixed-end beam and it is statically indeterminate. The bending moment diagram for such a beam can be considered to consist of two parts viz. the free bending moment diagram obtained by treating the beam as if the ends are simply supported and the fixing moment diagram resulting from the restraints imposed at the ends of the beam. The solution of a fixed beam is greatly simplified by considering Mohr s principles, which state that:. The area of the fixing bending moment diagram is equal to that of the free bending moment diagram.. The centers of gravity of the two diagrams lie in the same vertical line, i.e., are equidistant from a given end of the beam. The construction of the bending moment diagram for a fixed beam is explained with an example shown in Fig P Q U T is the free bending moment diagram, M s, and P Q R S is the fixing moment diagram, M i. The net bending moment diagram, M, is shaded. If s is the area of the free bending moment diagram and i the area of the fixing moment diagram, then from the first Mohr s principle we have s i and Wab ( M + M ) M M Wab + (7.7) From the second principle, equating the moment about of s and i, we have ( ) Wab M + M a + ab+ b (7.8) 00 by CRC Press C

8 ODING SHER FORCE ENDING MOMENT q o / unit length x C a b R R q o a M x q ox M max q oa q o / unit length C a b R R qo b M max q o b a + b q o / unit length a C D b c R R q o b M max q o b a + b x q o a C b R R q oa M x q ox 6a M max q oa 6 P a C b R R P M max x M x P.x M max P.a M a C b Zero shear x M max M x M a R q o b R s a R R q oa s R q oas 6 R x M max q oa s + s 6 when x a s s FIGURE 7.6 Shear force and bending moment diagrams for beams with simple boundary conditions subjected to selected loading cases. Solving Eqs. (7.7) and (7.8) for M and M we get M M Wab Wa b 00 by CRC Press C

9 ODING SHER FORCE ENDING MOMENT R a R q o b R R q oa b R M max R q oa b x q o a b / When x a b P R R / R R R P R M max P P P a R a R R R P M max Pa P a R b R R R Pb/ R R Pa/ M max Pa b P P R C D a b c R a > c R R P(b + c) R P(b +a) R M C Pa(b + c) M D Pc(b + a) P P / / / R R R R R P R M max P P P P C D E /6 / / /6 R R R R R P R M C M E P M D 5 P FIGURE 7.6 (continued). Shear force can be determined once the bending moment is known. The shear force at the ends of the beam, i.e., at and, are S S M M - M - M ending moment and shear force diagrams for fixed-end beams subjected to some typical loading cases are shown in Fig Wb Wa 00 by CRC Press C

10 ODING SHER FORCE ENDING MOMENT P P P R C D E / / / / R R R R P R M C M E P 8 M D P P P P P R C D E F /8 / / / /8 R R R R P R M C M F P M D M E P q o unit load C D E S S R R R R R q o S + R M M q os M D q o 8 + M q o unit load R C D E S S R R R R R q o S M M q os q o unit load C D S R Q R R R R q o (S + ) R q o ( + S) ( S) q o /8 M q os FIGURE 7.6 (continued). a W T b S M P Wab U R Q M FIGURE 7.7 Fixed-end beam. Continuous eams Continuous beams like fixed-end beams are statically indeterminate. ending moments in these beams are functions of the geometry, moments of inertia, and modulus of elasticity of individual members, besides the load and span. They may be determined by Clapeyron s theorem of three moments, the moment distribution method, or the slope deflection method. n example of a two-span continuous beam is solved by Clapeyron s theorem of three moments. The theorem is applied to two adjacent spans at a time, and the resulting equations in terms of unknown support moments are solved. The theorem states that 00 by CRC Press C

11 ODING SHER FORCE ENDING MOMENT C q o /unit length R R R q o / R M M M q o M C q o M C a d q o /unit length D b c e R R When r is the simple support reaction R r + M M R r + M M M M q M o [e ( e) c ( c)] b q M o [d ( d) a ( a)] b x M x W q o R R 0.5q o R 0.5q o R M M q o M x 0x J 60 9x + N + M max q o /6.6 when x 0.55 M q o /0 M q o /0 W q o C R R R q o / R M q o / M 5q o M M 96 P C / / R R R P/ R M M M P/8 M C P 8 M C a P b R R P Jb N J + a N R P Ja N J + b N R M Pab M Pa b M C M Pba M P P R C D / / / R R P R M P/9 M M P/9 M FIGURE 7.8 Shear force and bending moment diagrams for built-up beams subjected to typical loading cases. Ê x M + M + MC 6Á Ë ( ) + + x ˆ (7.9) in which M, M, and M C are the hogging moment at supports,, and C, respectively, of two adjacent spans of length and (Fig. 7.9); and are the area of bending moment diagrams produced by the vertical loads on the simple spans and C, respectively; x is the centroid of from ; and x is the distance of the centroid of from C. If the beam section is constant within a span but remains different for each of the spans Eq. (7.9) can be written as 00 by CRC Press C

12 ODING SHER FORCE ENDING MOMENT P P P C D E /6 / / /6 R R M P/ M D P/7 M R R P/ M M 9P/7 P P P C D E / / / / R R R P/ R M M D P/6 P/8 M M M 5P/6 P P P P C D E F /8 / / / /8 R R R P R M M D M E 5P/ P/ M M M P/ FIGURE 7.8 (continued). C oad M M x x M C ending moment FIGURE 7.9 Continuous beams. M I Ê M ˆ M x x + Á + 6 Ë I I + I Ê I + ˆ Á Ë I C (7.0) in which I and I are the moments of inertia of the beam sections in spans and, respectively. Example 7. The example in Fig. 7.0 shows the application of this theorem. For spans C and C È M 0 + MC( 0 + 0) + M 0 6 Î Since the support at is simply supported, M 0. Therefore, M C + M 50 (7.) Considering an imaginary span D on the right side of and applying the theorem for spans C and D 00 by CRC Press C

13 00 kn 0 kn/m C I I D 5 m 0 m 0 m Spans C and C FIGURE 7.0 Example of a continuous beam M 0 M 0 M ( ) + C D M + M 500 QM -M C C D ( ) (7.) Solving Eqs. (7.) and (7.) we get M 07. knm M C 85.7 knm Shear force at is S M - M C kn Shear force at C is S C Ê M Á Ë - M ˆ Ê M - M ˆ Á + 00 Ë C C ( ) + ( + ) kN Shear force at is The bending moment and shear force diagrams are shown in Fig eam Deflection S Ê M Á Ë - M kn There are several methods for determining beam deflections: () moment area method, () conjugate beam method, () virtual work, and () Castigliano s second theorem, among others. The elastic curve of a member is the shape the neutral axis takes when the member deflects under load. The inverse of the radius of curvature at any point of this curve is obtained as C ˆ by CRC Press C

14 M R EI (7.) in which M is the bending moment at the point and EI the flexural rigidity of the beam section. Since the deflection is small, /R is approximately taken as d y/dx, and Eq. (7.) may be rewritten as: M EI dy dx (7.) In Eq. (7.), y is the deflection of the beam at distance x measured from the origin of coordinate. The change in slope in a distance dx can be expressed as M dx/ei, and hence the slope in a beam is obtained as q M - q Ú EI dx (7.5) Equation (7.5) may be stated: the change in slope between the tangents to the elastic curve at two points is equal to the area of the M/EI diagram between the two points. Once the change in slope between tangents to the elastic curve is determined, the deflection can be obtained by integrating further the slope equation. In a distance dx the neutral axis changes in direction by an amount dq. The deflection of one point on the beam with respect to the tangent at another point due to this angle change is equal to dd x dq, where x is the distance from the point at which deflection is desired to the particular differential distance. To determine the total deflection from the tangent at one point,, to the tangent at another point,, on the beam, it is necessary to obtain a summation of the products of each dq angle (from to ) times the distance to the point where deflection is desired, or d Mx dx - d Ú EI (7.6) The deflection of a tangent to the elastic curve of a beam with respect to a tangent at another point is equal to the moment of M/EI diagram between the two points, taken about the point at which deflection is desired. Moment rea Method The moment area method is most conveniently used for determining slopes and deflections for beams in which the direction of the tangent to the elastic curve at one or more points is known, such as cantilever beams, where the tangent at the fixed end does not change in slope. The method is applied easily to beams loaded with concentrated loads, because the moment diagrams consist of straight lines. These diagrams can be broken down into single triangles and rectangles. eams supporting uniform loads or uniformly varying loads may be handled by integration. Properties of some of the shapes of M/EI diagrams that designers usually come across are given in Fig. 7.. It should be understood that the slopes and deflections obtained using the moment area theorems are with respect to tangents to the elastic curve at the points being considered. The theorems do not directly give the slope or deflection at a point in the beam compared to the horizontal axis (except in one or two special cases); they give the change in slope of the elastic curve from one point to another or the deflection of the tangent at one point with respect to the tangent at another point. There are some special cases in which beams are subjected to several concentrated loads or the combined action of concentrated and uniformly distributed loads. In such cases it is advisable to separate the concentrated loads and uniformly 00 by CRC Press C

15 Center of gravity Center of gravity for half parabola a b + a + b (a) Center of gravity w a a Center of gravity y FIGURE 7. Typical M/EI diagram. a (b) w uniform load w K O(a)( a) + a (c) y kx n y n + x x n + n + q/unit length / EI q 8EI / M EI diagram δ c q δ EI q δ 8EI q δ 8EI FIGURE 7. Deflection simply supported beam under UD. distributed loads, and the moment area method can be applied separately to each of these loads. The final responses are obtained by the principle of superposition. For example, consider a simply supported beam subjected to uniformly distributed load q, as shown in Fig. 7.. The tangent to the elastic curve at each end of the beam is inclined. The deflection, d, of the tangent at the left end from the tangent at the right end is found as ql /EI. The distance from the original chord between the supports and the tangent at the right end, d, can be computed as ql /8EI. 00 by CRC Press C

16 C Q Q C N O N O O P P P Q Q D y E N O (a) (b) FIGURE 7. ending of curved beams. The deflection of a tangent at the center from a tangent at the right end, d, is determined as ql /8EI. The difference between d and d gives the centerline deflection as (5/8) x (ql /EI). Curved eams The beam formulas derived in the previous section are based on the assumption that the member to which bending moment is applied is initially straight. Many members, however, are curved before a bending moment is applied to them. Such members are called curved beams. In the following discussion all the conditions applicable to straight-beam formulas are assumed valid, except that the beam is initially curved. et the curved beam DOE shown in Fig. 7. be subjected to the load Q. The surface in which the fibers do not change in length is called the neutral surface. The total deformations of the fibers between two normal sections, such as and, are assumed to vary proportionally with the distances of the fibers from the neutral surface. The top fibers are compressed, while those at the bottom are stretched, i.e., the plane section before bending remains plane after bending. In Fig. 7. the two lines and are two normal sections of the beam before the loads are applied. The change in the length of any fiber between these two normal sections after bending is represented by the distance along the fiber between the lines and ; the neutral surface is represented by NN, and the stretch of fiber PP is P P, etc. For convenience, it will be assumed that line is a line of symmetry and does not change direction. The total deformations of the fibers in the curved beam are proportional to the distances of the fibers from the neutral surface. However, the strains of the fibers are not proportional to these distances because the fibers are not of equal length. Within the elastic limit the stress on any fiber in the beam is proportional to the strain of the fiber, and hence the elastic stresses in the fibers of a curved beam are not proportional to the distances of the fibers from the neutral surface. The resisting moment in a curved beam, therefore, is not given by the expression si/c. Hence the neutral axis in a curved beam does not pass through the centroid of the section. The distribution of stress over the section and the relative position of the neutral axis are shown in Fig. 7.b; if the beam were straight, the stress would be zero at the centroidal axis and would vary proportionally with the distance from the centroidal axis, as indicated by the dot dash line in the figure. The stress on a normal section such as is called the circumferential stress. Sign Conventions The bending moment M is positive when it decreases the radius of curvature and negative when it increases the radius of curvature; y is positive when measured toward the convex side of the beam and negative when measured toward the concave side, that is, toward the center of curvature. With these sign conventions, s is positive when it is a tensile stress. Circumferential Stresses Figure 7. shows a free-body diagram of the portion of the body on one side of the section; the equations of equilibrium are applied to the forces acting on this portion. The equations obtained are 00 by CRC Press C

17 +M σ da y Z σ da O y da X Y Y FIGURE 7. Free-body diagram of curved beam segment. C dθ R dθ + dθ C R ρ Neutral surface ρ da y dθ P O H P O OO ds O P y FIGURE 7.5 Curvature in a curved beam. Ú SF z 0 or sda 0 SM z 0 or M ysda Ú (7.7) (7.8) Figure 7.5 represents the part of Fig. 7.a enlarged; the angle between the two sections and is dq. The bending moment causes the plane to rotate through an angle Ddq, thereby changing the angle this plane makes with the plane C from dq to (dq + Ddq); the center of curvature is changed from C to C, and the distance of the centroidal axis from the center of curvature is changed from R to r. It should be noted that y, R, and r at any section are measured from the centroidal axis and not from the neutral axis. It can be shown that the bending stress s is given by the relation M Ê s Á + ar Ë Z y ˆ R + y (7.9) 00 by CRC Press C

18 in which s is the tensile or compressive (circumferential) stress at a point at distance y from the centroidal axis of a transverse section at which the bending moment is M; R is the distance from the centroidal axis of the section to the center of curvature of the central axis of the unstressed beam; a is the area of the crosssection; and Z is a property of the cross-section, the values of which can be obtained from the expressions for various areas given in Fig Detailed information can be obtained from Seely and Smith (95). Example 7. The bent bar shown in Fig. 7.6 is subjected to a load P 780 N. Calculate the circumferential stress at and, assuming that the elastic strength of the material is not exceeded. We know from Eq. (7.9) in which a the area of rectangular section (0 80 mm ) R 0 mm y 0 y +0 P 780 N M 780 0,600 N mm. From Table 7.., for rectangular section Z - y a Ú da R+ y P M Ê ˆ s + Á + a ar Ë y Z R+y Z R È R+c - + loge h Î R- c h 0 mm Hence, c 0 mm 0 È Z - + loge Î 0-0 P P 0 mm 0 mm 0 mm mm Section FIGURE 7.6 ent bar. 00 by CRC Press C

19 h Z c c 5 c 6 7 c 8 K O + 8 K R R O + 6 K R O + 8 K R O +... c R Z + KR c O K R c O K R c O h Z c 6 K R O + 5 Kc R O + 7 Kc R O +... c R Z + R R + c log e K h R c OF b h b R Z + a h ;[b h + (R + c )(b b )] log R + e K R c O (b b c )h? c c R R Z + (b + b )h ;b + b b + (R + c h )F log e KR R c O (b b c )? c h c R R Z + h (R + c ) log R + e K R c c O hf h Z c K O c 5 c K O + R R 6 K R O + 7 c 8 8 K R O +... c R Z + KR c O K R c O K R c O FIGURE 7.7 nalytical expressions for Z. Therefore 7. Trusses 780 s Ê Á Ë s Ê Á Ë ˆ ˆ N/mm (tensile) 5 N/mm (compressive) structure that is composed of a number of members pin-connected at their ends to form a stable framework is called a truss. If all the members lie in a plane, it is a planar truss. It is generally assumed that loads and reactions are applied to the truss only at the joints. The centroidal axis of each member is straight, coincides with the line connecting the joint centers at each end of the member, and lies in a plane that also contains the lines of action of all the loads and reactions. Many truss structures are threedimensional in nature. However, in many cases, such as bridge structures and simple roof systems, the three-dimensional framework can be subdivided into planar components for analysis as planar trusses 00 by CRC Press C

20 h c R Z + c c R c R c c R b h c b R Z + ;bc bc b c Kc R K c O O R O c K c R R R b c K O K c c O R O? c K c c c c b t R b Z + R a [b log e (R + c ) + (t b )log e (R + c ) + (b t)log e (R c ) b log e (R c )] t b c c c c R t/ t/ b c c c c R t/ The value of Z for each of these three sections may be found from the expression above by making b b, c c, and c c Z + R a R b log e K R + c + (t b)log R + c e K O c R c O t/ b rea a [(t b) c + bc ] c c c c R t b c c c R In the expression for the unequal I given above make c c and b t, then t/ t/ c c c b Z + R a [t log e (R + c ) + (b t) log e (R c ) b log e (R c )] rea a tc (b t)c + bc R FIGURE 7.7 (continued). without seriously compromising the accuracy of the results. Figure 7.8 shows some typical idealized planar truss structures. There exists a relation between the number of members, m, the number of joints, j, and the reaction components, r. The expression is 00 by CRC Press C

21 Warren truss Pratt truss Howe truss Fink truss Warren truss Pratt truss owstring truss FIGURE 7.8 Typical planar trusses. m j r (7.0) which must be satisfied if it is to be statically determinate internally. r is the least number of reaction components required for external stability. If m exceeds (j r), then the excess members are called redundant members, and the truss is said to be statically indeterminate. For a statically determinate truss, member forces can be found by using the method of equilibrium. The process requires repeated use of free-body diagrams from which individual member forces are determined. The method of joints is a technique of truss analysis in which the member forces are determined by the sequential isolation of joints the unknown member forces at one joint are solved and become known for the subsequent joints. The other method is known as method of sections, in which equilibrium of a part of the truss is considered. Method of Joints n imaginary section may be completely passed around a joint in a truss. The joint has become a free body in equilibrium under the forces applied to it. The equations SH 0 and SV 0 may be applied to the joint to determine the unknown forces in members meeting there. It is evident that no more than two unknowns can be determined at a joint with these two equations. Example 7. truss shown in Fig. 7.9 is symmetrically loaded and is sufficient to solve half the truss by considering joints 5. t joint, there are two unknown forces. Summation of the vertical components of all forces at joint gives 5 F sin5 0 which in turn gives the force in members and, F 90 kn (compressive). Similarly, summation of the horizontal components gives F F cos by CRC Press C

22 5 6 5 kn 7 90 kn 90 kn 90 kn 6 m 6 m 6 m 6 m 5 kn 6 m F 5 5 kn F 5 F F F 5 F F 5 5 F F F 90 kn F 5 F kn FIGURE 7.9 Example of the method of joints, planar truss. Substituting for F gives the force in member as F 5 kn (tensile) Now, joint is cut completely, and it is found that there are two unknown forces F 5 and F. Summation of the vertical components gives Therefore Summation of the horizontal components gives and hence F cos5 F 0 F 5 kn (tensile) F sin5 F 5 0 F 5 5 kn (compressive) fter solving for joints and, one proceeds to take a section around joint at which there are now two unknown forces viz. F and F 5. Summation of the vertical components at joint gives F F 5 sin Substituting for F, one obtains F kn (compressive). Summing the horizontal components and substituting for F one gets 00 by CRC Press C

23 Therefore, F 0 F 80 kn (tensile) The next joint involving two unknowns is joint. When we consider a section around it, the summation of the vertical components at joint gives F 5 90 kn (tensile) Now, the forces in all the members on the left half of the truss are known, and by symmetry the forces in the remaining members can be determined. The forces in all the members of a truss can also be determined by using the method of sections. Method of Sections In this method, an imaginary cutting line called section is drawn through a stable and determinate truss. Thus, a section divides the truss into two separate parts. Since the entire truss is in equilibrium, any part of it must also be in equilibrium. Either of the two parts of the truss can be considered, and the three equations of equilibrium SF x 0, SF y 0, and SM 0 can be applied to solve for member forces. Example 7. above (Fig. 7.0) is once again considered. To calculate the force in members 5, F 5, section should be run to cut members 5 as shown in the figure. It is required only to consider the equilibrium of one of the two parts of the truss. In this case, the portion of the truss on the left of the section is considered. The left portion of the truss as shown in Fig. 7.0 is in equilibrium under the action of the forces viz. the external and internal forces. Considering the equilibrium of forces in the vertical direction, one can obtain 5 kn F kn FIGURE 7.0 Example of the method of sections, planar truss F 5 sin5 0 Therefore, F 5 is obtained as The negative sign indicates that the member force is compressive. The other member forces cut by the section can be obtained by considering the other equilibrium equations viz. SM 0. More sections can be taken in the same way to solve for other member forces in the truss. The most important advantage of this method is that one can obtain the required member force without solving for the other member forces. Compound Trusses F -5 5 kn compound truss is formed by interconnecting two or more simple trusses. Examples of compound trusses are shown in Fig. 7.. typical compound roof truss is shown in Fig. 7.a in which two simple trusses are interconnected by means of a single member and a common joint. The compound truss shown in Fig. 7.b is commonly used in bridge construction, and in this case, three members are used to interconnect two simple trusses at a common joint. There are three simple trusses interconnected at their common joints, as shown in Fig. 7.c. 00 by CRC Press C

24 (a) Compound roof truss (b) Compound bridge truss (c) Cantilevered construction FIGURE 7. Compound truss. The method of sections may be used to determine the member forces in the interconnecting members of compound trusses, similar to those shown in Fig. 7.a and b. However, in the case of a cantilevered truss the middle simple truss is isolated as a free-body diagram to find its reactions. These reactions are reversed and applied to the interconnecting joints of the other two simple trusses. fter the interconnecting forces between the simple trusses are found, the simple trusses are analyzed by the method of joints or the method of sections. 7. Frames Frames are statically indeterminate in general; special methods are required for their analysis. Slope deflection and moment distribution methods are two such methods commonly employed. Slope deflection is a method that takes into account the flexural displacements such as rotations and deflections and involves solutions of simultaneous equations. Moment distribution, on the other hand, involves successive cycles of computation, each cycle drawing closer to the exact answers. The method is more labor intensive but yields accuracy equivalent to that obtained from the exact methods. Slope Deflection Method This method is a special case of the stiffness method of analysis. It is a convenient method for performing hand analysis of small structures. et us consider a prismatic frame member with undeformed position along the x axis deformed into configuration p, as shown in Fig. 7.. Moments at the ends of frame members are expressed in terms of the rotations and deflections of the joints. It is assumed that the joints in a structure may rotate or deflect, but the angles between the members meeting at a joint remain unchanged. The positive axes, along with the positive member-end force components and displacement components, are shown in the figure. y V M y θ ψ P ψ θ y V M b x a FIGURE 7. Deformed configuration of a beam. 00 by CRC Press C

25 The equations for end moments may be written as M M EI q + q -y l ( ) + EI q + q -y l M F ( ) + M F (7.) in which M F and M F are fixed-end moments at supports and, respectively, due to the applied load. y is the rotation as a result of the relative displacement between member ends and given as D y + y y (7.) where D is the relative deflection of the beam ends. y and y are the vertical displacements at ends and. Fixed-end moments for some loading cases may be obtained from Fig The slope deflection equations in Eq. (7.) show that the moment at the end of a member is dependent on member properties EI, length l, and displacement quantities. The fixed-end moments reflect the transverse loading on the member. Frame nalysis Using Slope Deflection Method The slope deflection equations may be applied to statically indeterminate frames with or without side sway. frame may be subjected to side sway if the loads, member properties, and dimensions of the frame are not symmetrical about the centerline. pplication of the slope deflection method can be illustrated with the following example. Example 7. Consider the frame shown in Fig. 7. subjected to side sway D to the right of the frame. Equation (7.) can be applied to each of the members of the frame as follows: Member : M EI Ê D ˆ Áq + q - + M 6 Ë 0 F q 0, M M EI Ê + - M Ë Á ˆ + 0 D q q 0 F F M 0 F 80 kn m 6 m C 6 m EI Same for all members 9 m D FIGURE 7. Example of the slope deflection method. 00 by CRC Press C

26 Hence M M EI q -y 6 ( ) EI - ( y ) 0 q (7.) (7.) in which y D 6 Member C: M M EI ( q + q - 0)+ M 9 C C FC C EI q + q - 0 C 9 ( ) + M FC 80 6 M FC - -0 ft-kips 9 M ft-kips 9 FC Hence M EI ( q + q )- 0 9 C C (7.5) EI ( ) M C q C + q (7.6) Member CD: M EI Ê D ˆ Áq + q - + M 9 Ë 0 CD C D FCD q D 0, M M EI Ê D ˆ Áq + q - + M 9 Ë 0 DC D C FDC FCD M 0 FDC Hence M CD M EI Ê ˆ EI ÁqC - 6y qc -y 9 Ë 9 DC ( ) EI Ê ˆ EI Á qc - 6y qc -y 9 Ë 9 ( ) (7.7) (7.8) 00 by CRC Press C

27 Considering moment equilibrium at joint Substituting for M and M C, one obtains SM M + M C 0 EI 0 q + qc - 9 y 9 or ( ) q + qc - 9y EI (7.9) Considering moment equilibrium at joint C Substituting for M C and M CD we get SM C M C + M CD 0 EI qc + q - y 9 or ( ) - 0 q qc y EI (7.0) For summation of base shears equal to zero, we have or SH H + H D 0 M + M M + M CD DC Substituting for M, M, M CD, and M DC and simplifying 0 q + qc - 70y 0 (7.) Solution of Eqs. (7.9) to (7.) results in. 7 q EI. q - 69 C EI 00 by CRC Press C

28 and y 0. EI (7.) Substituting for q, q C, and y from Eq. (7.) into Eqs. (7.) to (7.8) we get Moment Distribution Method M.0 knm M 5. knm M C 5. knm M C knm M CD knm M DC 8 knm The moment distribution method involves successive cycles of computation, each cycle drawing closer to the exact answers. The calculations may be stopped after two or three cycles, giving a very good approximate analysis, or they may be carried out to whatever degree of accuracy is desired. Moment distribution remains the most important hand-calculation method for the analysis of continuous beams and frames, and it may be solely used for the analysis of small structures. Unlike the slope deflection method, this method does require the solution to simultaneous equations. The terms constantly used in moment distribution are fixed-end moments, the unbalanced moment, distributed moments, and carryover moments. When all of the joints of a structure are clamped to prevent any joint rotation, the external loads produce certain moments at the ends of the members to which they are applied. These moments are referred to as fixed-end moments. Initially the joints in a structure are considered to be clamped. When the joint is released, it rotates if the sum of the fixed-end moments at the joint is not zero. The difference between zero and the actual sum of the end moments is the unbalanced moment. The unbalanced moment causes the joint to rotate. The rotation twists the ends of the members at the joint and changes their moments. In other words, rotation of the joint is resisted by the members, and resisting moments are built up in the members as they are twisted. Rotation continues until equilibrium is reached when the resisting moments equal the unbalanced moment at which time the sum of the moments at the joint is equal to zero. The moments developed in the members resisting rotation are the distributed moments. The distributed moments in the ends of the member cause moments in the other ends, which are assumed fixed; these are the carryover moments. Sign Convention The moments at the end of a member are assumed to be positive when they tend to rotate the member clockwise about the joint. This implies that the resisting moment of the joint would be counterclockwise. ccordingly, under a gravity loading condition the fixed-end moment at the left end is assumed as counterclockwise ( ve) and at the right end as clockwise (+ve). Fixed-End Moments Fixed-end moments for several cases of loading may be found in Fig pplication of moment distribution may be explained with reference to a continuous beam example, as shown in Fig. 7.. Fixed-end moments are computed for each of the three spans. t joint the unbalanced moment is obtained and the clamp is removed. The joint rotates, thus distributing the unbalanced moment to the ends of spans and C in proportion to their distribution factors. The values of these distributed moments are carried over at one half rate to the other ends of the members. When equilibrium is reached, 00 by CRC Press C

29 90 (Uniformly distributed) EI EI C EI D FIGURE 7. Example of a continuous beam by moment distribution. joint is clamped in its new rotated position and joint C is released afterwards. Joint C rotates under its unbalanced moment until it reaches equilibrium, the rotation causing distributed moments in the C ends of members C and CD and their resulting carryover moments. Joint C is now clamped and joint is released. This procedure is repeated again and again for joints and C, the amount of unbalanced moment quickly diminishing, until the release of a joint causes negligible rotation. This process is called moment distribution. The stiffness factors and distribution factors are computed as follows: The fixed-end moments are DF DF DF DF C C CD I K Â K I 0 + I 0 I KC 0 0. Â K I 0 + I 0 I KC Â K I 0 + I 5 I KCD Â K I I 0 + I 5 MF - 50; MFC - 50; MFCD -0 M 50; M 50; M 0 F FC FDC When a clockwise couple is applied near the end of a beam, a clockwise couple of half the magnitude is set up at the far end of the beam. The ratio of the moments at the far and near ends is defined as the carryover factor, 0.5 in the case of a straight prismatic member. The carryover factor was developed for carrying over to fixed ends, but it is applicable to simply supported ends, which must have final moments of zero. It can be shown that the beam simply supported at the far end is only three fourths as stiff as the one that is fixed. If the stiffness factors for end spans that are simply supported are modified by three fourths, the simple end is initially balanced to zero and no carryovers are made to the end afterward. This simplifies the moment distribution process significantly. 00 by CRC Press C

30 EI C EI EI 0 D EI EI 0 E 0 F FIGURE 7.5 Example of a nonsway frame by moment distribution. Moment Distribution for Frames Moment distribution for frames without side sway is similar to that for continuous beams. The example shown in Fig. 7.5 illustrates the applications of moment distribution for a frame without side sway. Similarly, EI DF 0 EI EI DF M M E FC FE EI ; DFC ; MFC 00 50; M -50 FE Structural frames are usually subjected to side sway in one direction or the other, due to asymmetry of the structure and eccentricity of loading. The sway deflections affect the moments, resulting in an unbalanced moment. These moments could be obtained for the deflections computed and added to the originally distributed fixed-end moments. The sway moments are distributed to columns. Should a frame have columns all of the same length and the same stiffness, the side sway moments will be the same for each column. However, should the columns have differing lengths or stiffnesses, this will not be the case. The side sway moments should vary from column to column in proportion to their I/l values. The frame in Fig. 7.6 shows a frame subjected to sway. The process of obtaining the final moments is illustrated for this frame. The frame sways to the right, and the side sway moment can be assumed in the ratio 00 0 : 00 0 (or) : by CRC Press C

31 .5 (Uniformly distributed) C D FIGURE 7.6 Example of a sway frame by moment distribution. Final moments are obtained by adding distributed fixed-end moments and.06/.99 times the distributed assumed side sway moments. Method of Consistent Deformations This method makes use of the principle of deformation compatibility to analyze indeterminate structures. It employs equations that relate the forces acting on the structure to the deformations of the structure. These relations are formed so that the deformations are expressed in terms of the forces, and the forces become the unknowns in the analysis. et us consider the beam shown in Fig. 7.7a. The first step, in this method, is to determine the degree of indeterminacy or the number of redundants that the structure possesses. s shown in the figure, the beam has three unknown reactions, R, R C, and M. Since there are only two equations of equilibrium available for calculating the reactions, the beam is said to be indeterminate to the first degree. Restraints that can be removed without impairing the load-supporting capacity of the structure are referred to as redundants. Once the number of redundants are known, the next step is to decide which reaction is to be removed in order to form a determinate structure. ny one of the reactions may be chosen to be the redundant, provided that a stable structure remains after the removal of that reaction. For example, let us take the reaction R C as the redundant. The determinate structure obtained by removing this restraint is the cantilever beam shown in Fig. 7.7b. We denote the deflection at end C of this beam, due to P, by D CP. The first subscript indicates that the deflection is measured at C, and the second subscript indicates that 00 by CRC Press C

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