Lecture notes: Structural Analsis II Space frames I. asic concepts. The design of a building is generall accomplished b considering the structure as an assemblage of planar frames, each of which is designed as an independent two-dimensional frame. In some instances however, it is necessar to consider the building as a whole and design it as a threedimensional structure. In a most general wa of speaking the term space frame indicates three dimensional structure capable of resisting loads in different planes. Displacements in a space structure ma occur in si directions: a displacement in, and directions and a rotation about, and aes. The sign convention for displacements is shown in Fig. 1a. The arrows indicate the positive directions of the displacement components, and, using the right hand screw sstem, rotations are considered positive when acting clockwise as viewed from the origin. A total of si displacement components define the positive restraints in a fied support given in Fig. 1b. a) b) R φ M Figure 1 Sign convention: a) Displacements; b) Reactions 1. Support reactions. A three dimensional structure is eternall determinate when si eternal restraints are applied to the structure, since the reactions in these restraints ma be determined b the available si equations of static equilibrium. The si equations of statics, which ma be applied at an point of the structure and at each support, are: Σ F = 0, Σ F = 0, Σ F = 0, Σ M = 0, Σ M = 0, Σ M = 0. φ φ 1.1 Space frame supports. Space frames are connected to their foundation using four different tpes of supports: 1) A spherical movable support; 2) A spherical roller support; 3) A spherical fied support; 4) A fied support. R The spherical movable support consists of two flat parallel slabs with a ball in between (Fig. 2). This tpe of support allows rotation about all the three aes, as well as the displacements along an direction in plane. Onl the displacements along - ais are prevented. Thus, onl one vertical reaction arises in this support, along the direction of - ais. R 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 97
Lecture notes: Structural Analsis II R Figure 2 Spherical movable support The second tpe of support consists of two rockers, with a ball inserted in their sockets. The lower rocker lie on ribs (rollers) set on a flat slab as shown in Fig. 3. R R Figure 3 Spherical roller support This tpe of supports permits free rotation about aes, passing through the centre of the ball and a longitudinal displacement in a direction perpendicular to the roller aes. It prevents displacement along two directions one being perpendicular to the plane of the roller aes and the other parallel to their aes. Two reactions R and R develop at support of this tpe (Fig. 3). The spherical fied support consists of a pair of similar rockers with a ball, but no rollers, so that the upper rocker can onl rotate about an aes passing through the centre of the ball, but cannot move in an direction. A support of this tpe imposes three constraints, respectivel three reactions R, R and R ma develop. The spherical fied support is illustrated in Fig. 4. R R R Figure 4 Spherical fied support The fied support which imposes si constraints can be applied onl in case of space cantilever, as shown in Fig. 5, in other case the structure will become staticall indeterminate. 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 98
Lecture notes: Structural Analsis II F F R R R M Figure 5 Space cantilever The minimum number of constraints necessar to maintain a bod in a fied position is alwas equal to the number of equilibrium equations, si for the case of space structures. If the structure restrained with si spherical movable supports does not constitute an unchangeable sstem the supports should be modified properl, or if the modification is impossible the supports must be increased properl. The si constraints of a bod should be arranged in such a wa to prevent displacements along the three aes, and, and to prevent rotation of the structure as a free bod about an ais of the space. Although the supports impose si constraints it ma happen that one part of the structure will have redundant constraints and will become staticall indeterminate, while the other will retain one or more degrees of freedom. An eample of fault distribution of the support constraints is given in Fig. 6. a) b) D D D D A A A A A A Figure 6 Space frame with si spherical movable supports: a) Fault distribution of the supports; b) orrect distribution forming unielding combination. The displacements of joint A are full restricted, which means that the translations of the structure as a free bod in three directions are prevented. The rotations about - ais with origin at point A is restricted b the support of point, the rotation about - ais is constrained b the support at point, the rotation about - ais is prevented again b the support at point (Fig. 6a). If we consider the ais formed b joints A and all the support reactions intersect this ais, 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 99 A A
Lecture notes: Structural Analsis II therefore the rotation of the structure along the ais into consideration is allowed. It could be guess in advance that the part of the frame between points A and D is staticall indeterminate, because the etension or fleion of this part is not allowed, thus an other part will retain one degree of freedom. The rotation about ais AD, will be restricted if the spherical movable support at point D being rotated as shown in Fig. 6b. ased on this eample it can be concluded that the support reactions must form at least three couples of forces in three different non coincident and non parallel planes in order to prevent the rotation of the bod in an direction. If we consider Fig. 6b reactions A and form a couple in plane; D and A make a couple in plane; reactions A and constitute a couple of forces in plane AD, which is different and non parallel to - and - planes. Whereas, if we return to Fig. 6a reactions A and form a couple of forces in plane; A and make a couple of forces in plane AD, but there is no third couple of forces in a third different plane. 2. Internal forces In general internal forces acting at the cut section of the member, of a space frame, will consist of a normal force (), two shear forces ( and Q ), two bending moments ( and M ) and torsion moment ( ). If the internal forces at a specified point in a member are to be determined, the method of sections must be used. This requires that a cut or section be made perpendicular to the ais of the member at the point where the internal loading is to be determined. A free bod diagram of either segment of the cut member is isolated and the internal loads are then determined from the si equations of equilibrium applied to the segment. In order to find the internal forces we need to establish a sign convention to define their positive and negative values. In the space frame structures each member has its own local coordinate sstem used to define positive directions of the si internal forces. Although the choice of the local coordinate sstem is arbitrar, the sign convention to be adopted here has been used in structural analsis software programs, as well as in the engineering practice. 2.1 Sign convention for local element coordinate sstems The first ais is directed along the length of the element from first to the second node of the element, the remaining two aes lie in the plane perpendicular to the element ais with an orientation that we specif. Usuall the local coordinate sstem is right handed sstem, although it is not a strong requirement. The first ais must be along the element the definition of the remaining two aes is up to us. The can be introduced arbitrar in such a wa to simplif the interpretation of the results (Fig. 7a). j j M Q j i i Q i M a) b) c) Figure 7 Sigh convention: a) Local coordinate sstem; b) Internal forces for free bod containing first node of the element; c) Internal forces for free bod containing second node of the element. 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia
Lecture notes: Structural Analsis II 2.1 Sign convention for internal forces. The sign convention which is adopted here is illustrated in Fig 7b and 7c. The positive normal force alwas acts outward of the section cut, which responds to a pure tension. In that respect on the left hand face of the cut member the normal force acts to the right. An equal but opposite normal force acts to the right hand face of the cut member. The same rule is applicable to the torsion moment. The positive direction of torsion coincides with positive normal force. The positive direction of internal shear forces depend on the element local coordinate sstem introduced in advance. When the direction of positive normal force coincides with local ais (positive ais is alwas from first to the second joint of the member), then the positive shear forces, and Q, follow the positive - and - local ais, as shown in Fig. 7b. When the positive normal force is contrar to the local ais then the positive shear forces are contrar to the local and aes. This case is illustrated in Fig. 7c. The same rules are valid for positive bending moments, and M, as vectors. II. umerical eample In the following eample we shall construct the internal forces diagrams for the given in Fig. 8 space frame structure. The introduced global coordinate sstem is shown in the same figure. D D D A 3.5 A A 3 4 Figure 8 umerical eample The introduced local coordinate sstems of the different elements of the space frame are presented in Fig. 9. The tpical sections where the internal forces must be calculated, in order to construct the relevant diagrams, are numbered from 1 to 8 in the same figure. The tpical sections are placed at least at the beginning and at the end of each element (segment) of the frame. The internal forces diagrams, in the limits of each element, could be derived b using the corresponding reference and base diagrams. 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 101
Lecture notes: Structural Analsis II 8 D A 1 E 7 6 2 3 4 5 Figure 8 Element local ees 1. Support reactions The accepted positive directions of the support reactions are shown in Fig. 7. The equations of static equilibrium are (note that we use the global coordinate sstem: Σ F = 0 A + D = 0, Σ F = 0 A + = 0, Σ F = 0 + D 4 = 0, Σ M = 0 D 3 1.5 + = 0, Σ M = 0 D 4 + D 3.5 3.5 + 4 2 = 0, Σ M = 0 A 3 D 3 + 4 + 1.5 = 0. The magnitudes of the support reactions in the sequence of their obtaining are as follow: D = ; = 146.67; D = ; A = ; = ; A =. The support reactions with the obtained directions are illustrated in Fig. 9. 2. Internal forces In order to obtain the internal forces at a specified point, we should make section cut perpendicular to the ais of the member at this point. This section cut divides the structure in two parts. The portion of the structure removed from the part into consideration should be replaced b the internal forces. The internal forces ensure the equilibrium of the isolated part subjected to the action of eternal loads and support reactions. A free bod diagram of either segment of the cut member is isolated and the internal loads could be derived b the si equations of equilibrium applied to the segment into consideration. We shall skip the derivation of internal forces in section 1 from Fig. 8, because the can be derived without an troubles. Let us go direct to the internal forces in section 2. If we pass a section cut at point 2 the space frame will be separated as shown in Fig. 10. The positive directions of internal forces, in accordance with the introduced local coordinate sstem for the 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 102
Lecture notes: Structural Analsis II member where the section 2 belongs to, for both of the separated parts, are given in the same figure. Obviousl it is more convenient to consider the left hand part of the frame. 3.5 05 3 146.667 4 Figure 9 Support reactions M Q 05 M Q 146.667 Figure 10 Internal forces in section 2 The equilibrium equations for the left hand part of the frame can be written as follow: Σ = 0 + = 0 =, Σ Q = 0 Q = 0 Q =, Σ Q = 0 Q = 0 Q =, 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 103
Lecture notes: Structural Analsis II Σ M = 0 M = 0, Σ M = 0 M + 1.5 = 0 M =, Σ M = 0 M 3 = 0 M = 22.857. M Q 146.667 05 M Q Figure 11 Internal forces in section 3 et the internal forces at section 3 (Fig. 8) ma be obtained. oth parts of the space frame as a result of section cut passed through point 3 with the introduced internal forces are shown in Fig. 11. It is easier to consider the equilibrium of left hand part again, the corresponding equilibrium equations are ver similar to those for section 2. Σ = 0 + = 0 =, Σ Q = 0 Q + = 0 Q =, Σ Q = 0 Q + 146.667 = 0 Q = 46.667, Σ M = 0 M 1.5 = 0 M =, Σ M = 0 M = 0, Σ M = 0 M + 3 = 0 M = 22.857. What follows is the determination of the internal forces in section 4. The corresponding parts of the space frame formed after passing the section cut at point 4 are shown in Fig. 12. The relevant internal forces for both the parts are depicted in the same figure. Here will be eamined the equilibrium of the right hand part. Σ = 0 + = 0 =, Σ Q = 0 Q + = 0 Q =, Σ Q = 0 Q + = 0 Q =, Σ M = 0 M 3 = 0 M =, 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 104
Lecture notes: Structural Analsis II Σ M = 0 M 3.5 + 3.5 = 0 M = 26.667, Σ M = 0 M + 1.5 3 = 0 M = 127.14. 05 M M 146.667 Q Q Figure 12 Internal forces in section 4 05 M Q M 146.667 Q Figure 13 Internal forces in section 5 The free bod diagrams for section cut in point 5 are shown in Fig. 13. In this case the equilibrium equations will be formed easier for the right hand portion of the frame. Σ = 0 = 0 =, Σ Q = 0 Q = = 0 Q =, 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 105
Lecture notes: Structural Analsis II Σ Q = 0 Q = 0, Σ M = 0 M 1.5 + 3 = 0 M = 127.14, Σ M = 0 M 3 = 0 M =, Σ M = 0 M + 3.5 3.5 = 0 M = 26.667. M Q M 05 Q 146.667 Figure 14 Internal forces in section 6 M Q M Q 05 146.667 Figure 15 Internal forces in section 7 What is left to get the internal forces in sections 6 and 7. The corresponding free bod diagrams are given in Figs. 14 and 15. The derivation of internal forces in these sections will be skipped here because the equilibrium equations are similar to those for the previous sections 2-5. 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 106
Lecture notes: Structural Analsis II Finall, the diagrams of internal forces should be constructed. The rules for the construction of internal force diagrams in space frames are identical with those for plane frame. There are some specific moments with respect to the local aes.. The normal force and torsion moment diagrams, with their signs, can be depicted in or planes, no matter which one we choose. The shear force Q should be drawn in local plane, likewise shear force must be set in local plane for the element into consideration. The bending moment diagram should be drawn in the local plane where the bending moment acts. In that respect the moment is a vector which rotates in local plane, and must b drawn in this plane. The bending moment M acts in local plane, and the corresponding diagram must be set in this local plane. The bending moment diagrams must be drawn on the tension side with no signs, as those in the plane frames. The bending moment diagram must be depicted together with Q, and vice versa the bending moment diagram M corresponds to. In such a wa the relations =± dq / d and M =± dq / d could be easil traced. The internal force diagrams are shown in Figs. 16, 17 and 18. 127.14 138.57-7.62 + 127.14 + 22.857 22.857 26.67 M - 7.62 Figure 16 ending moment, M, and shear force,, diagrams 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 107
Lecture notes: Structural Analsis II - + 26.667 46.667 + - Q 53.333 Figure 17 ending moment,, and shear force,, diagrams Q 127.14 - - + + 127.14 - Figure 18 Torsion moment and aial force diagrams 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 108
Lecture notes: Structural Analsis II 3. Verification equilibrium of the joints A 05 D 22.857 146.667 22.857 146.667 46.667 127.143 26.667 127.143 26.667 127.143 E E 127.143 Figure 19 Equilibrium of the joints 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 109
Lecture notes: Structural Analsis II References DARKOV, A. AD V. KUZETSOV. Structural mechanics. MIR publishers, Moscow, 1969 WILLIAMS, А. Structural analsis in theor and practice. utterworth-heinemann is an imprint of Elsevier, 09 HIELER, R.. Structural analsis. Prentice-Hall, Inc., Singapore, 06 11 S. Parvanova, Universit of Architecture, ivil Engineering and Geodes - Sofia 110