Structure Design-I ( Analysis of truss by method of joint.)

(Lecture 4) Analysis of truss by method of joint.

1. Lecture 4 Analysis of truss by Method of joint Structure I

2. Introduction A frame having hinge/pin joints is called truss. A truss is an assembly of individual linear members arranged in a combination of triangles to form a rigid frame which cannot be deformed by application of external forces. The primary principle being that triangular configuration is the most stable configuration to resist coplanar forces. Members are so arranged that loads and reactions act at joints only.

3. Some typical truss

4. Relation between number of member and joints in a truss The relation between members and joint is: m = 2j – 3 …….( i ) where, m = number of members in a truss j = number of joints in a truss If number of members in a truss are equal to the value of ‘m’ given by eqn ( i ), it is known as a “Perfect truss”. Perfect truss is a stable truss and can be analyzed by equation of statics. If the number of member are less than 2j – 3 (m < 2j- 3), it is an unstable truss known as deficient or imperfect truss and will collapse under external loads. On the other hand if number of members provided are more than required i.e. m > 2j – 3, such a truss is known as Redundant truss and cannot be analyzed by equation of statics.

5. Analysis of Perfect truss Method of analysis- i ) Analytical method a ) Method of joints b) Method of sections ii) Graphical method

6. Sign conventions

7. Method of Joints The concept of this method is that if the truss as a whole is in equilibrium then each joint is also in equilibrium and should satisfy equations: Σ H = 0 ; Σ V = 0

8. Steps for analysis by Method of Joint Draw the proportional diagram of truss showing loads and probable directions of reactions at each support. Calculate the reactions in the same manner as done for beams. Visually indentify “zero force member” and mark the same: Zero force member: At a joint if two or more members meet at 90⁰. Then the member opposite to which No force/reaction/member exists will have a zero force and the same is marked on member as shown in figure-

9. Contd. Choose a joint which has not more than two members having unknown forces. Separate the joint from truss and draw its FBD. Assume the direction of forces in unknown members. Repeat step 4 above till the forces are calculated in each member. Tabulate the values and sense of force (compression/tension) in each member of truss in “Force table” at the end of solution.

10. Numerical Calculate the magnitude and nature of forces in members of the pin jointed truss shown in figure by method of joints.

11. Solution As in figure, angles are not given, so first of all, we’ll have to find the angles, Refer to free body diagram- by sine rule, 4/sin Ѳ = 5/sin90⁰ = 3/sin α or, Sin Ѳ = 4sin90⁰/5 Ѳ = sin -1 0.80 Ѳ = 53⁰ (Approx) and sin α = 3Sin 90⁰/5 α = sin -1 0.60 α = 36⁰ (Approx) Check: 3cos Ѳ + 4cos α = 5 or, 3cos 53⁰ + 4cos36⁰ = 5 3x0.60 + 4x0.80 = 5 5 = 5 hence, Ѳ & α are correct.

12. Contd. Reactions: Σ M A = 0; R B x5 = 10x3cos 53⁰ R B = 3.60kN Σ M B = 0; R A x5 = 10x4cos 36⁰ R A = 6.40kN Check: 6.40 + 3.60 = 10 hence, reactions are OK. Members forces, Joint A (see in figure) Σ V = 0; -F CA Sin53⁰ + 6.40 = 0 F CA = 8kN (C) Σ H = 0; +F AB – F CA Cos53 = 0 F AB = 8x0.6 = 4.8kN (T)

13. Contd. Joint ‘B’ Σ H = 0; - 4.8 + F CB Cos36 = 0 F CB = 4.8/Cos36 = 6kN Joint ‘C’ Let us check equilibrium of joint ‘C’ Σ V = 0; -10+8Cos36+6Cos53 = 0 0 = 0 hence, Joint C is in Equilibrium

14. Force table S.No . Member Force ( kN ) 1. AB 4.8 (T) 2. AC 8.0 (C) 3. BC 6.0 (C)

15. Numerical 2. Analyze the truss shown in figure by method of joints.

16. Solution Reactions: As the truss is symmetrically loaded, so the both reactions will be equal, R A = R B = (30x2 + 60x3)/2 = 120kN

17. Contd. Joint A: Σ V = 0; 120-30-F AB sin30⁰ = 0 F AB = 180kN (C) Σ H = 0; - F AB cos 30⁰+F AF = 0 F AF = 155.88kN (T)

18. Contd. Joint B: Σ H = 0; 180cos30⁰- F BC Cos30⁰- F BF Sin30⁰ = 0 155.80 – 0.87F BC – 0.50F BF = 0 ……..( i ) Σ V = 0; -60+180sin30⁰ + F BF cos30⁰- F BC sin30⁰ = 0 30+0.87F BF - 0.50F BC = 0 …….(ii) Multiplying eqn ( i ) with 0.50 and eqn (ii) by 0.87 and solving the two eqns - 77.90 – 0.44F BC – 0.25F BF = 0 26.10 – 0.44F BC – 0.76F BF = 0 - + + ------------------------------------------ 51.80 - 1.01F BF = 0 F BF = 51.29kN Substituting in eqn ( i ), 155.80 -0.87F BC – 0.50x51.29 = 0 F BC = 149.61kN

19. Contd. Joint ‘F’ Σ V = 0; F FC Sin 60⁰-51.29Sin 60⁰ = 0 F FC = 51.29kN Σ H = 0; -155.80+51.29Cos 60⁰+51.29Cos 60⁰ +F FG = 0 F FG = 104.59kN

20. Contd. Joint ‘E’ Joint ‘E’ is indicated to joint A solved earlier, so F ED = 180kN and F EC = 155.88kN Joint ‘D’ Joint D is indicated to joint B, hence, F DC = F BC = 149.61kN F DG = F BF = 51.29kN

21. Contd. Joint ‘G’ Σ V = 0; F FC = F GC = 51.29kN Σ H = 0; -51.29Cos 60⁰-104.59-51.29Cos 60⁰+155.88 = 0 0 = 0 Hence, solution is correct.

22. Exercise Analyze the truss shown in figure by method of joints.