CHAPTER 6 Space Trusses


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1 CHAPTER 6 Space Trusses
2 INTRODUCTION A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic tetrahedral, formed by connecting six members with four joints
3 INTRODUCTION
4 OBJECTIVES To determine the stability and determinacy of space trusses To determine member forces of space trusses using tension coefficient analysis
5 1. Simple Space Truss TYPE OF SPACE TRUSSES This truss is constructed from a tetrahedron. The truss can be enlarged by adding three members.
6 TYPE OF SPACE TRUSSES 2. Compound Space Truss This truss is constructed by combining two or more simple truss.
7 TYPE OF SPACE TRUSSES 3. Complex Space Truss Complex truss is a truss that cannot be classified as simple truss or compound truss.
8 DETERMINACY & STABILITY Due to three dimensions, there will be three equations of equilibrium for each joint. ( F x = 0; F y = 0; F z = 0) The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium. Generally, the least number of required reactions for stable and externally determinate is SIX If r < 6 Unstable If r > 6 Externally Indeterminate
9 DETERMINACY & STABILITY For internal determinacy, if m = number of members; j = number of joints; r = number of supports; therefore: If m = 3j + r Stable and Internally Determinate If m < 3j + r Unstable If m > 3j + r Internally Indeterminate Internal stability can sometimes be checked by careful inspection of the member arrangement.
10 DETERMINACY & STABILITY Internally m + r = 3j m + r > 3j m + r < 3j Externally r < 6 r = 6 r > 6 Determinate Truss Indeterminate Truss Unstable Truss Unstable Truss Determinate if Truss is Stable Indeterminate Truss
11 TYPES OF SUPPORT
12 EXAMPLE 1 Ball & Socket m = 3, j = 4, r = 9 m + r = 12 3j = 12 m + r = 3j Determinate Truss
13 EXAMPLE 2 m = 15, j = 10, r = 15 Ball & Socket m + r = 30 3j = 30 m + r = 3j Determinate Truss
14 EXAMPLE 3 m = 13, j = 8, r = 12 m + r = 25 3j = 24 m + r = 25 3j = 24 Ball & Socket Indeterminate Truss in the First Degree
15 ASSUMPTIONS FOR DESIGN The members are joined together by smooth pins (no friction cannot resist moment) All loadings and reactions are applied centrally at the joints The centroid for each members are straight and concurrent at a joint Therefore, each truss member acts as an axial force member: If the force tends to elongate Tensile (T) If the force tends to shorten Compressive (C)
16 THEOREM 1: ZERO FORCE MEMBERS If all members and external force except one member at a joint, (say, joint B) lie in the same plane, then, the force in member A is zero. Member A B z P y x The force in member A is zero
17 THEOREM 2: ZERO FORCE MEMBERS If all members at a joint has zero force except for two members, (say member A and B), and both members (A and B) do not lie in a straight line, then the force in member A and B are zero. Member B Member A F By F Bx F Ay FAx 0 Both members A and B has zero force because both members do not lie in a straight line. 0 0
18 ZERO FORCE MEMBERS If members A and B lie in a straight line, then, the forces in these members MIGHT NOT be zero. In fact, referring to the example below: F Ax = F Bx F Ay = F By Member A Member B F Ay FAx F Bx FBy
19 THEOREM 3: ZERO FORCE MEMBERS If three members at a joint do not lie in the same plane and there is no external force at that joint, then the force in the three members is zero. Member B Member A Three members connected at a joint has zero force. A plane can consists of two members, say member A and B. Thus, no force can balance the component of member C that is normal to the plane. Member C
20 Identify the members of the space truss that has zero force. Theorem 1: Joint L F 1 = 0 and F 2 = 0 Theorem 3: Joint K F 3 = F 4 = F 5 = 0 and F 1 = 0 Theorem 3: Joint M F 6 = F 7 = F 8 = 0 and F 2 = 0 Theorem 3: Joint J F 9 = F 10 = F 11 = 0 and F 5 = F 6 = 0 EXAMPLE 4
21 EXAMPLE 5 Theorem 1: Joint F Members FE, FC, FD lie in a plane, except member FG. Thus, member FG has zero force. Theorem 3: Joint F Members FE, FC, FD have zero force. Theorem 3: Joint E Members ED, EA, EH have zero force.
22 ANALYSIS METHOD: TENSION COEFFICIENT To determine member forces Based on 3D particle equilibrium L x = L cos : L y = L cos ; L z = L cos F x = F cos : F y = F cos ; F z = F cos Therefore: F x L x = F L F y L y = F L F z L z = F L
23 ANALYSIS METHOD: TENSION Tension Coefficient: t = F/L COEFFICIENT F x = t L x ; F y = t L y ; F z = t L z If t positive (tension), if t negative (compression) The actual force and length are given from: L = L x 2 + L y 2 + L z 2 F = F x 2 + F y 2 + F z 2 For 3D equilibrium: F x = 0 ; F y = 0 ; F z = 0 tl x = 0 ; tl y = 0 ; tl z = 0
24 ANALYSIS METHOD: TENSION COEFFICIENT y L, F L y, F y L z, F z x A L x, F x z The component of length, L, and force, F in the x, y, z direction
25 B y ANALYSIS METHOD: TENSION COEFFICIENT O F y F X F x A D x F x = Fcosθ x F x = F L x L = F L L x F x = t L x z E F z C Where t = F/L t = tension coefficient
26 B y ANALYSIS METHOD: TENSION COEFFICIENT O F y Y F F x A D x F y = Fcosθ y F y = F L y L = F L L y F z F y = t L y E C z
27 B y ANALYSIS METHOD: TENSION COEFFICIENT F y F z O Z F F x A D x F z = Fcosθ z F z = F L z L = F L L z F z = t L z E C z
28 EXAMPLE 6 The space truss shown in the figure has roller and socket support at joint A, B, C and D z y x i) Determine the member force for all members at joint F and G 20 kn 40 kn E 40 kn 10 kn 40 kn G ii) Determine the reaction at support C A B F C 2 m 2 m 4 m D 1 m
29 1. Start at Joint G Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) GC GE GF Force (kn) F x = 0: 4t GE + 10 = 0 F z = 0: 2t GC 40 = 0 EXAMPLE 6 Solution F y = 0: 2t GE 2t GF = 0 t GE = 2.5 kn/m t GC = 20 kn/m 2(2.5) 2t GF = 0 t GF = 2.5 kn/m
30 2. Move to Joint F Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) FC FD FE FG Force (kn) F x = 0: 4t FE = 0 F y = 0: 2t FC t FD + 2t FG = 0 F z = 0: 2t FC 2t FD 40 = 0...(ii) (i) + (ii): 3t FD 45 = 0 From (i): 2t FC (15) 5 = 0 EXAMPLE 6 Solution t FE = 0 kn/m 2t FC t FD 5 = 0... (i) t FD = 15 kn/m t FC = 5 kn/m
31 3. Calculate the Reaction at Joint C Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) CF CG Force (kn) R Cx R Cy R Cz F x = 0: R Cx = 0 kn F y = 0: 2t CF + R Cy = 02(5) + R Cy = 0 R Cy = 10 kn ( ) F z = 0: 2t CF + 2t CG + R Cz = 0 EXAMPLE 6 Solution 2(5) + 2(50) + R Cz = 0 R Cz = 50 kn ( )
32 Slotted roller constraint in a cylinder z B C EXAMPLE 7 Short link Determine the force in each member of the space truss shown. 4 m D x A E z = 4 kn Ball & socket 4 m y 2 m E 2 m
33 EXAMPLE 7 Solution 1. Start with joints where there are only 3 unknowns force/reaction. Joint B 5 unknowns Joint C 5 unknowns Joint A 7 unknowns Joint E 4 unknowns Joint D Theorem 3: Three members at a joint and no external force. Thus, all members have zero forces. t DC = t DA = t DE = 0
34 Joint E EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) EC EB ED EA Force (kn) F z = 0: 4t EB + 4t EC 4 = 0...(i) F x = 0: 2t EC + 2t EB 2t ED + 2t EA = 0... (ii) where t ED = 0 F y = 0: 4t EC 4t EB 4t ED 4t EA = 0...(iii) where t ED = 0 Solve Eq. (i), (ii) & (iii): t EC = 0 kn/m, t EA = 1 kn/m and t EB = 1 kn/m
35 Joint C EXAMPLE 7 Solution Joint C has three (3) unknowns, as t CE = t CD = 0 z R Cy C Therefore, by Theorem 3: t CB = t CA = R Cy = 0 B D x A y E z = 4 kn E
36 Joint C EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) CE CA CD CB Force (kn) 0 R Cy 0 F z = 0: 4t CA = 0 F x = 0: 4t CA + 4t CB = 0 F y = 0: R Cy = 0 t CA = 0 kn/m t CB = 0 kn/m
37 Joint B EXAMPLE 7 Solution Joint B has three (3) unknowns. z C R By B R Bx D x A y E z = 4 kn E
38 Joint B EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) BC BA BE Force (kn) R Bx R By 0 F x = 0: 4t BC 2t BE + R Bx = 0 F y = 0: 4t BE R By = 0 F z = 0: 4t BA 4t BE = 0 R Bx = 2 kn R By = 4 kn t BA = 1 kn/m
39 Joint A EXAMPLE 7 Solution z C B D R Ay x R Ax A R Az y E z = 4 kn E
40 Joint A EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) AB AC AD AE Force (kn) R Ax R Ay R Az F x = 0: 4t AC 4t AD 2t AE R Ax = 0 F y = 0: 4t AE + R Ay = 0 F z = 0: 4t AB + 4t AC + R Az = 0 R Ax = 2 kn R Ay = 4 kn R Az = 4 kn
41 Method 2 EXAMPLE 7 Solution Reactions at A, B and C can also be obtained from Method 2 R By z B R Cy C R Bx D R Ay x R Ax A R Az y E z = 4 kn E
42 Method 2 M y = 04(2) + R Bx (4) = 0 R Bx = 2 kn M z = 0 R Cy = 0 kn M x = 0 R By (4) 4(4) = 0 R By = 4 kn EXAMPLE 7 Solution F x = 0 2 R Ax = 0 R Ax = 2 kn F y = 0 R Ay 4 = 0 R Ay = 4 kn F z = 0 R Az 4 = 0 R Az = 4 kn
43 CLASS EXERCISE The space truss shown in the Figure consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Analyse the force in each of the members for the given loading. y 2 m 2 m A 3 m B O D z 7 m C x 1800 N
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