Framed Structures PLANE FRAMES. Objectives:

Framed Structures 2 Objectives: ifferentiate between perfect, imperfect and redundant frames. To compute the member forces in a frame by graphical method. To compute the forces in a truss by method of joints. To compute the forces in a truss by method of sections. To compute the forces in a truss by method of tension coefficients. 2.1 PLN FRMS The plane trusses are here by termed as plane frames. In this chapter, pin jointed plane frames which are statically determinate are considered. statically determinate frame can be completely analysed by using statics. The number of unknown forces is the same as the number of equations obtained from static equilibrium. The main methods in analysing statically determinate pin jointed plane frames are (i) Graphical solution Force diagram (ii) Method of resolution at joints (iii) Method of sections and (iv) Tension coefficient method. The first three methods are used in plane frames (trusses) and the fourth method is used for analysing the space frame. truss is an assemblage of three or more members which are hinged or pinned. load applied on the truss is transmitted to all joints so that the members are in pure compression or tension. onsider a simple truss made up of three members hinged at the ends to form a triangle. load W is acting at the apex of the triangle and due to symmetry, the reactions are W/2 at each support. W W/2 FIG. 2.1 W/2 ue to the application of the load, the joint and pulls the member out and for equilibrium at joint there should be an equal and opposite force should move away from joint. In otherwords, member in tension. ue to the downward load W, the joint is pushed vertically downwards. The forces in the members and are in compression as the joint is pushed. force in the

36 asic Structural nalysis member should counteract this force and hence this member force act towards the joint. Hence, the effect of the application of the load results in pure compression or tension. W W/2 FIG. 2.2 W/2 It should be noted that the member forces marked in the above figure are the internal forces. It can be remembered as, when the force is acting towards the joint it is compressive in nature. When the force is acting away from the joint, it is tensile in nature. The compression members are called as strut while tension members are called as tie. 2.2 PRFT, IMPRFT N RUNNT PIN JOINT FRMS 2.2.1 Perfect Frame perfect frame has just sufficient number of members to prevent the frame being unstable. The simplest perfect frame is in the form of a triangle as shown in Fig. 2.2. It has three members and three joints, namely, and. If two more members are added to form another triangle, then one more joint has been added. The frame will remain as a perfect frame as long as any number of triangles are added. Thus, the frame will remain as a perfect frame. In essence, the first three joints or pins require three bars to connect them. While the remaining ( j 3) joints require 2( j 3) extra bars. If m is the total number of bars or members required to connect j joints together. m 3 = 2( j 3) m = 2 j 3 i.e., number of members to form a perfect frame = Twice the number of joints 3. 2.2.2 Imperfect Frame n imperfect frame is one where the number of members is less than (2 j 3). In Fig. 2.3 the number of joints are 4 and the number of members are four and hence

Framed Structures 37 The combined force diagram is drawn as follows: 1. Starting from the force, the known forces, viz., and working clockwise round the frame, are set down in order and to scale as ab,bc and ca. 2. onsider the apex joint ab the centre of the clock, and the letters are read clockwise around this centre. 3. Therefore, from the letter b draw a line parallel to 1 and from the letter a draw a line parallel to 1 and both intersect at the point 1. From the force diagram, the magnitude of the forces are obtained. The member force 1 = Member force 1 = 5.7 5 = 28.5 kn. Member force 1 = 2.8 5 = 14 kn. 4. onsider the joint at the left hand support reaction. Read clockwise in the frame diagram. Member 1 is inclined and in the force diagram a to 1 is downwards and hence mark the arrow correspondingly in the force diagram from 1 to it is towards right mark this direction at that joint. 50 kn 1 25 60 FIG. 2.6 60 Frame diagram 25 5. onsider the apex joint. Read clockwise 1 is inclined member. In the force diagram, b to 1 is upwards and hence mark the arrow upward for member 1 at the apex joint. Member 1 is a sloping member. From the force diagram, 1 to a is upwards. Therefore, mark the arrow upwards at the apex joint. (compression). a 5.7 cm 1 c Scale 1 cm = 5 kn b FIG. 2.7 Force diagram

38 asic Structural nalysis 50 kn 25 kn 4 m 4 m 2 3 6 m 1 FIG. 2.16 a 2 1 b Reaction at the top hinge 3 c Reaction at the roller FIG. 2.17 d Force in kn Member Strut Tie 2 36 3 36 3 43 1 86 12 41.5 1 50.0 23 50 Reaction at hinge 103 Reaction at roller 66 2.4 MTHO OF JOINTS In the method of joints, the member forces are determined using the equilibrium conditions at that particular joint. In this resolution of forces at the joint, the free body diagram at that joint is considered. The procedure is explained as follows:

Framed Structures 39 6. onsider the right hand support reaction and again read clockwise. 1 is inclined member. In the force diagram 1 to b is downwards. Therefore, mark the arrow downwards at joint. 7. To determine the member force 1, from the force diagram it is noted that the force is acting from c towards left to 1. Mark the arrow 1 in the same direction in frame diagram. The final forces are listed below. Forces in kn Member Strut Tie 1 28.5 1 28.5 1 14.0 2.3.1 Numerical Problems on Symmetrical Frame and Symmetrical Loading XMPL 2.1: etermine the forces in the members graphically. 10 kn 20 kn 10 kn F G FIG. 2.8 SOLUTION ue to Symmetry: Using the ow notations V = V = 10 + 20 + 10 2 = 20 kn 20 kn 10 kn 10 kn 2 3 4 1 5 20 kn 20 kn FIG. 2.9

40 asic Structural nalysis The combined force diagram is drawn as follows: 1. The loads,,, and are marked to scale. 2. Start with a joint of the left hand reaction. raw a line through the point a a line parallel to 1 and from the point e draw a horizontal line parallel to 1. They intersect at a point and is marked as 1. 3. Move to the next joint where 10 kn load is acting; Through b draw a line parallel to 2 and from 1 draw a line parallel to 12. These two lines intersect at the point 2. 4. fter locating point 2 in the force diagram. onsider the joint where the members 1 2,2 3, 3 and 1 meet. The point 3 is located on intersection of line e1 and a line drawn through point 2 and parallel to 2 3 of the frame diagram. 5. The point 4 is located by drawing a line through 3 and parallel to3 4 in the load diagram which intersects the line drawn from in the force diagram and parallel to 4. 6. The point 5 is marked from point 4. raw a line parallel to 4 5 of the frame diagram from point 4 and this cuts the horizontal line through e. 7. Using the force diagram, the magnitude of the forces and the directions are obtained. 8. It is to be remembered that the arrows indicate not what is being done to the member but what the member is doing at the joint at each end. Hence, if the arrow is acting towards the joint it is compression and if the arrow is acting away from the joint then it is tensile force. Force in kn Member Strut Tie 1 5 36.5 2 4 31.5 1 5 31.0 12 45 8.25 23 34 8.25 20 kn a 10 kn 10 kn 2 3 4 1 5 4 b 5,1 3 e FIG. 2.10 Frame diagram 2 c d FIG. 2.11 Force diagram

Framed Structures 41 Force in kn Member Strut Tie 1 12.5 3 12.5 12 11.8 23 11.8 1 24.0 2 18.5 3 24.0 2.3.3 Numerical xample on antilever Frames XMPL 2. 3: Use the graphical method and determine the member forces and the reaction at the supports. 4 m 50 kn 4 m 25 kn 6 m FIG. 2.15 In the force diagram of the cantilever truss, these is no need of reactions before starting of the same. 1. The load line is drawn as in the previous examples. a b, b c, start with a joint at the free end and reading clockwise, draw line from b parallel to 3 of the frame diagram and from point draw line parallel to 3 and the intersection of the above lines give point 3. 2. The point 2 is located as follows. From the point 3, draw a line parallel to 23 and from a draw a line parallel to 2. The intersection gives the point 2. 3. The point 1 is obtained as follows. line is drawn from point 3, parallel to 3 and from point 2 draw line parallel to 21 and the intersection of 1. 4. fter marking the points 1, 2 and 3 the member forces and their nature are tabulated here.

42 asic Structural nalysis 2.3.2 Numerical xample on Frame with Loads Suspended from the ottom hord of the Frame in ddition to Loads on the Top hord XMPL 2.2: Find the forces in all the members of the truss graphically 20 kn 60 60 10 kn FIG. 2.12 10 kn 1. The loads,,, and are marked to scale. 2. Start with the left support joint. Read clockwise and draw. raw a line from a parallel to 1 in the load diagram. raw another line e from the load diagram and parallel to the 1 of the frame diagram. They intersect at the point 1. 3. The point 2 is located by considering the joint adjacent to the left support. raw a line from point 1 parallel to 1 2 of the frame diagram. From the load diagram, draw a horizontal line through d and the line intersect at point 2. 4. From the point 2 draw a line parallel to 2 3 of the frame diagram and from point e draw a horizontal line and the intersection of above two lines give point 3. 5. etermine the magnitude and nature of the forces from the forces diagram and tabulate. 20 kn 1 a,c 20 1 2 3 20 2 d 10 10 FIG. 2.13 Frame diagram 3 FIG. 2.14 Force diagram b,e

Framed Structures 43 1. heck the stability and assess its determinacy of the truss. 2. If the truss is of cantilever type, the reactions need not be computed in general. If the truss is stable and determinate where one support is hinge and the other support is on rollers; compute the reactions at the supports. 3. raw the free body diagram at each joint and analyse the member forces at a joint where only two members meet. Then, consider the adjacent joint where only two unknown forces to be determined. This process is repeated till the analysis of all joints are completed. 4. The results are tabulated along with magnitude of member forces and the nature of forces. The forces are tensile if they are pulling (acting away) the joint. The forces are compressive in nature if they are pushing (acting towards) the joint. NUMRIL XMPL XMPL 2.4: nalyse the truss shown in Fig. 2.18 by method of joints. (May 2010, RV) 20 1 70 kn 2 3 4 6 5 H 4 V 4 V 6 FIG. 2.18 SOLUTION The reactions at the supports are found out by summing up the forces in horizontal and vertical directions and also by taking moments of applied forces about the hinge support. H = 0; 20 H 4 = 0 V = 0; H 4 = 20 kn V 4 +V 6 = 70 kn M 4 = 0; 20 3 + 70 3 6V 6 = 0 V 6 = 45 kn V 4 = 25 kn

44 asic Structural nalysis n m < 2n j 3 4 < (2 4) 3 4 < 5 n j = 4 n m = 4 FIG. 2.3 2.2.3 Redundant Frame redundant frame is one where the number of member or members are more than (2 j 3). In Fig. 2.4, the number of joints are n m > (2n j 3) 6 > (2 4) 3 6 > 5 n j = 4 n m = 6 FIG. 2.4 i.e., redundant frame is having more member/members necessary to produce stability. 2.3 GRPHIL SOLUTION-FOR IGRMS onsider the perfect frame in Fig. 2.5. The forces include the applied load and the reactions at P and Q. 50 kn 1 P FIG. 2.5 ue to symmetry the reactions are 25 kn at joint P and Q respectively. In graphical method the loads and reactions are read clockwise. They are represented by capital letters written on either side of the force, commonly known as OW S Notation. They are denoted with letters,, and the space inside the member is denoted by numbers. Note that the letters,, are marked in the middle length of the members and not at the joints. The load at the apex 50 kn is denoted as load. The reaction at the right support is denoted as load. The reaction at the left support 25 kn is denoted as load. The member force in the horizontal member is denoted as force 1. Q

Framed Structures 45 Joint 4 F 14 H = 0; F 45 20 = 0 F 45 = 20 kn 4 20 F 45 V = 0; F 14 + 25 = 0 25 kn F 14 = 25 kn Joint 1 20 kn 1 V = 0; 45 F 12 25 F 15 sin45 = 0 F 15 F 15 = 35.4 kn 25 kn H = 0; 20 F 12 + F 15 cos45 = 0 20 F 12 + 35.4cos45 = 0 F 12 = 45.03 kn Joint 2 70 kn 45.03 F 23 = 0 F 23 = 45.03 kn 45.03 2 F 23 V = 0; F 25 = 70.0 kn F 25 Joint 5 35.4 70 kn 45 45 20 5 F 53 V = 0; F 53 cos45 + 35.4sin45 70 = 0 F 53 = 63.61 kn

Framed Structures 47 M = 0; 2(1.5)+4(4.5) 6V = 0 V = 3.5 kn V = 2.5 kn onsider Joint F V = 0 F sin60 = 2.5 F = 2.89 kn 60 F H = 0 F = F cos60 = 2.89cos60 2.5 kn = 1.45 kn Joint F V = 0 F sin60 = 3.5 F = 4.04 kn 60 H = 0 F cos60 F = 0 F F = 4.04cos60 = 2.02 kn 3.5 kn Joint F F V = 0 F sin60 + F sin60 = 0 60 60 1.45 kn 2.02 kn H = 0 F = F 2.02 + F cos60 1.45 F cos60 = 0 2.02 F cos60 1.45 F cos60 = 0 0.57 = 2F cos60 F = 0.57; F = 0.57

48 asic Structural nalysis Joint F F θ 25 kn V = 0 H = 0 F sinθ = 25 F = 25 sinθ = 25 = 41.67 kn 0.6 F + F cosθ = 0 F + 41.67 0.8 = 0 F = 33.34 kn Joint 50 kn 33.34 H = 0 F + 33.34 = 0 F = 33.34 kn F F V = 0 F 50 = 0 F = 50 kn Joint 50 kn 41.67 cosθ = 4/5 = 0.8 sinθ = 3/5 = 0.6 F θ 1 sinθ 1 = 4/5 = 0.8 θ θ cosθ 1 = 3/5 = 0.6 F H = 0 41.67cosθ + F cosθ F sinθ 1 = 0 41.67(0.8)+F (0.8) 0.8 F = 0 F F = 41.67 (2.1)

50 asic Structural nalysis Joint F 4 kn 60 60 H = 0; V = 0; F F cos60 F cos60 = 0 F F cos60 = 4.04cos60 4 F sin60 + 4.04sin60 = 0 F = 0.57 kn F F = 4.04 kn Substituting F in the above equation F = 1.73 kn 2 kn 1.73 4 kn 2.89 0.57 0.57 4.04 1.45 2.02 FIG. 2.23 XMPL 2.7: etermine the magnitude and nature of forces in all the number of the pin jointed plane truss shown in Fig. 2.24 by method of joints. (VTU, June 08) 25 kn 50 kn 25 kn 4 m 4 m 6 m FIG. 2.24

52 asic Structural nalysis XMPL 2. 9: etermine the forces in members and tabulate neatly. Use method of joints. (VTU, ec. 06) 8 kn H 8 kn F 8 kn G 4 m 4 kn θ 4 kn 2 m FIG. 2.28 SOLUTION ue to Symmetry: V = V = 2(4)+3(8) = 16 kn 2 Joint 4 kn θ 16 kn F F F V = 0 H = 0 4 F F sinθ + 16 = 0 12 0.555 F F = 0 F F = 21.62 kn F F F cosθ = 0 F = 21.62 0.832 = 18.0 kn Joint H = 0; F = 18.0 kn 18.0 F

Framed Structures 53 V = 0; 50 41.67sinθ + F cosθ 1 + F sinθ = 0 50 41.67(0.6)+0.6F + 0.6F = 0 0.6F + 0.6F = 75 F + F = 125 (2.2) Solving qns. (1) and (2); F = 83.3 kn, F = 41.67 kn Joint F θ 2 83.3 tanθ 2 = 8/6 sinθ 2 = 0.8 cosθ 2 = 0.6 V = 0; F = 83.3cosθ 2 = 50 kn 25 kn 50 kn 25 kn 33.3 33.3 41.67 50 41.67 50 8.33 FIG. 2.25

58 asic Structural nalysis Joint 120 30 kn θ 1 F F V = 0 H = 0 F sinθ 1 30 = 0 F = 30 = 42.43 kn 0.707 120 + F + 42.43cosθ 1 = 0 F = 42.43 0.707 + 120 = 30 kn +120 90 kn 60 kn 30 kn 30 kn 134.17 134.17 G 30 42.43 H 1.25 m 2.50 m 120 90 F FIG. 2.30 XMPL 2.11: etermine the forces in all members of all ollman truss by method of joints. (VTU, July 2005) 20 kn 20 kn 20 kn 10 m 10 m 10 m 10 m θ 1 θ 2 θ 3 10 m F G FIG. 2.31 H

Framed Structures 59 SOLUTION ue to Symmetry From geometry Joint,, F F = 20 kn, F G = 20 kn, F H = 20 kn V = V = 3(20) 2 = 30 kn tanθ 1 = 10 tanθ 2 = 10/20 tanθ 3 = 10/10 30 sinθ 1 = 0.316 sinθ 2 = 0.447 sinθ 3 = 0.707 cosθ 1 = 0.949 cosθ 2 = 0.895 cosθ 3 = 0.707 20 kn F F Joint F F F θ 3 F 20 kn θ 1 F F H = 0; V = 0; F F cosθ 3 + F F cosθ 1 = 0 0.707F F + 0.949F F = 0 20 + F F sinθ 3 + F F sinθ 1 = 0 0.707F F + 0.316F F = 20 on solving the above equations Joint G F F = 21.22 kn F F F F F = 22.37 15.81 kn F G 20 kn F G H = 0; F G cosθ 2 + F G cosθ 2 = 0 F G = F G θ 2 G θ 2 V = 0; 20 + F G sinθ 2 + F G sinθ 2 = 0 2 F G sinθ 2 = 20 2 F G (0.447)=20 F G = 22.37 kn

60 asic Structural nalysis Joint 60 H = 0 F = 0 60 kn 50 kn 78.1 kn 60.0 kn 78.1 kn 60.0 kn FIG. 2.33 XMPL 2.13: Find the forces in the members of the truss in Fig. 2.34. Tabulate your results neatly. (VTU, July 2004) 10 kn 20 kn G 10 kn 30 60 4 m 10 kn H 4 m F 4 m FIG. 2.34

Framed Structures 65 S.No Member Force (kn) Nature 1 42.20 ompressive 2 42.20 ompressive 3 G 25.77 ompressive 4 G 29.76 ompressive 5 F 25.77 Tensile 6 F 25.77 Tensile 7 51.55 Tensile 8 20.00 ompressive 9 29.24 Tensile 10 F 0.00 11 FG 0.00 XMPL 2.14: nalyse the thetruss shown in infig. the 2.35 Fig. by 2.35method by method of joints. of joints(vtu, checkug. the forces 2000) in the members, & by the method of sections. (VTU, ug. 2000) 10 kn F 20 kn 4 m FIG. 2.35 Joint 10 kn F F H = 0 V = 0 10 F F = 0 F = 0 F F F = 10 kn

66 64 asic Structural nalysis Joint 42.2 20 kn 60 30 30 60 F F H = 0 V = 0 20cos60 F cos30 F cos60 + 42.2cos30 = 0 F cos30 + F cos60 = 46.55 0.866 F + 0.5 F = 46.55 20sin60 F sin30 + F sin60 + 42.2sin30 = 0 0.5 F + 0.866 F = 3.78 F = 42.20 kn F = 20.00 kn Joint G F G 30 60 F GF G 30 F G = 29.76 H = 0 V = 0 F G cos30 F G cos30 + F GF cos60 = 0 0.866 F G 0.866 F G + 0.5 F GF = 0 0.866 F G + 0.5 F GF = 0.866 29.76 0.866 F G + 0.5 F GF = 25.77 F G sin30 + F GF sin60 + 29.76sin30 = 0 0.5 F G + 0.866 F GF = 14.88 F G = 25.77 kn, F GF = 0 Joint 10 kn 60 30 30 V = 0; 10sin60 + 42.20sin30 F sin60 + 25.77sin30 = 0 F = 29.24 kn 42.20 60 60 25.77 F

Framed Structures 67 Resolving vertically, F F sinθ 1 = 40 F F = 40 = 50 kn 0.8 Resolving all the forces in the horizontal direction F = F F cosθ 1 = 50 0.6 = 30 kn Joint F The truss is supported on rollers at the joint F. Only one reaction will be acting perpendicular to the base of the roller. Hence, F F GF θ θ 1 F 110 kn F F = 50 kn H = 0; V = 0; F GF cosθ 50cosθ 1 = 0 F GF = 50 0.6/0.832 F GF = 36.06 kn 110 36.06sinθ 50sinθ 1 + F F = 0 F F = 36.06(0.555)+50(0.8) 110 = 50 kn The ve sign indicates that we have to change the nature of the force in F F and hence F F is compressive. Joint 29.95 kn F F H Resolving all the forces meeting at the joint in the vertical direction and as no forces are acting in the vertical direction; F H = 0. Resolving the forces at the joint along the horizontal direction F = 29.95 kn

70 asic Structural nalysis Joint H F H 36 kn H F H, H and G are on the same line. Hence, resolving the forces meeting at the joint H, HG Resolving all the forces along the line, F HG = 36 kn Joint 29.95 kn F F G Resolving all the forces in the vertical direction: F G = 0 Resolving all the forces in the horizontal direction: F = 29.95 kn Joint G F G 36 kn G F GF Resolving all the forces meeting at the joint G, along the line HGF and perpendicular that line; F G = 0; F GF = 36 kn

Framed Structures 71 The computed forces are shown in Fig. 2.40. 50 kn 30 kn 30 kn 30 kn 30 kn 40 kn 36 kn H 36 kn 50 kn 50 kn 4 m G 36 kn F FIG. 2.40 2.5 MTHO OF STIONS statically determinate frame can be completely analysed by static methods. The number of unknowns is the same as the number of equations obtained from static equilibrium conditions. In method of sections, we isolate a portion of a frame by a section. This section causes the internal force appear to act as external forces in the isolated portion. The unknown forces are evaluated by using equations of equilibrium. The three equations of equilibrium are used to evaluate the unknowns. Thus, the section should cut only three members. (which include the one whose force is to be determined) and take moments about a point through which the lines of action of other two members intersect. The work is made simple by judicious of choice of such sections. This method is not desirable if used to determine all the forces in the members of a frame. However, it is readily useful to determine the forces in selected members of the truss. 2.5.1 pplication of Method of Sections onsider a truss shown in Fig. 2.41. It is required to calculate the force in the member GF of the bottom boom. X 4 m 8 m G 8 m F 8 m X FIG. 2.41

Framed Structures 73 The reaction at is 66.7 kn acting upwards and hence it is taken as negative. The applied load at G is 50 kn and is acting downwards and therefore it is positive. The vertical component due to F G is downwards, i.e., F G sin45. This is negative. Summing up all the above forces in the vertical direction and equating upward forces to downward forces. F G sin45 + 50 = 66.7 F G = 23.62 kn. This is compressive as the force is acting towards the joint. 8. It must be remembered that the arrows must be considered in respect to the nearest point of that portion of the frame which remains after the cut has been made. 2.5.2 Numerical Problems XMPL 2.17: etermine the nature and magnitude of forces in members,i and HI of the truss shown in Fig. 2.45 by using method of sections. 3 kn 6 kn 6 kn X 3 kn 60 6 m G 6 m H 6 m I 6 m 60 F V 6 kn 12 kn X 6 kn V F FIG. 2.43 SOLUTION V = 0 V +V F = 42 M = 0 3(3)+6(9)+6(15)+3(21)+6(6)+12(12)+6(18) 24V F = 0 V F = 21 kn V = 21 kn

Framed Structures 75 XMPL 2.18: etermine the forces in the members, F and FG by method of sections. 30 kn 60 kn 30 kn 3.6 4.8 m F H 4.8 4.8 m G FIG. 2.45 4.8 4.8 SOLUTION ue to symmetry, the reactions 30 + 60 + 30 V = V = = 60 kn 2 To determine the forces in, F and FG cut a section X-X as shown in Fig. 2.48 onsider left part of the truss and analyse the equilibrium. X θ 30 kn 4.8 4.8 θ 60 kn 3.6 F θ θ 1 FIG. 2.46 X To determine the force in, take moment about F where other two members of the cut section F and FG meet. 60 4.8 = F 3.6 F = 80 kn

Framed Structures 77 We know that θ = 36 52 and θ 1 = 14 2 θ 2 = FG = θ + θ 1 = 50 54 sinθ 2 = 6 F FG (4.66) 30(4.8)+60(9.6)=0 = 6sinθ 2 = 4.66 m F FG = 92.7 kn (Tensile) XMPL 2.19: etermine the forces in, G and FG by method of sections. 2.5 kn 2.5 kn 5 kn 2.6 m 2.6 m 30 F G H V FIG. 2.49 SOLUTION The reactions at, i.e., V and H and the reaction at ; i.e., V are found using equilibrium equations. V = 0; V +V = 2.5sin60 + 5sin60 + 2.5sin60 V +V = 8.66 H = 0; 2.5cos60 + 5cos60 + 2.5cos60 H = 0 H = 5 kn M = 0; 5(2.6)+2.5(5.2) 9V = 0 V = 2.88 kn V = 5.78 kn

Framed Structures 79 XMPL 2.20: Find the forces in the members, F and FG. Use method of sections. (VTU, ug. 2004) 10 kn 20 kn 10 kn 30 60 60 4 m H 10 kn 4 m F G 4 m FIG. 2.51 SOLUTION The length of panel, are found using geometry. cos30 = 6/ = 6/cos30 = 6.9 = = /2 = 3.46 m The reactions V, H and V are determined using the equilibrium equations. H = 0; V = 0; M = 0; 10cos60 + 20cos60 + 10cos60 H = 0 H = 20 kn V +V = 10sin60 + 20sin60 + 10sin60 + 10 V +V = 44.64 kn 20(3.46)+10(6.92)+10(4) 12V = 0 V = 14.87 KN To determine the forces in the members, and F cut a section through these members and for convenience consider the right part of the truss.

Framed Structures 81 M = 0 F FG G = 0 F FG = 0 XMPL 2.21: etermine the forces in the members F, F, of the truss shown in Fig. 2.54 F 50 kn FIG. 2.54 SOLUTION The reactions are found out by using equilibrium equations. Taking moment about ; V +V = 50 50 3 9V = 0 V = 16.7 kn; V = 33.3 kn To determine the forces in F, F and, cut a section through these members and consider the right-hand portion of the truss. F 3 = 16.7 6 F = 33.4 kn To determine the force F F ; take moment about ; where other two members of the cut truss, F F and F meet.

82 asic Structural nalysis F F F F F F 6 m θ 16.7 kn V FIG. 2.55 tanθ = 3/6 i.e. θ = 26 33 54. In sinθ = = 3sin26 33 54. = 1.34 m. F 1.34 = 16.7 3 F = 37.4 kn To determine the force in F, take moment about where other two members of the cut section, viz. F F and F meet. xtend Fdownwards to such that is perpendicular to the projected line of F. onsidering the ; sinθ = 3 = 3sinθ = 3 sin26 33 54. = 1.34 m M = 0; F F = 0 F F = 0

84 asic Structural nalysis F F F θ 15 kn onsider Joint F F F F FG F 59.8 t joint F, there is no vertical force acting. Hence, resolving the forces in the vertical direction F F = 0. Resolving the forces in the horizontal direction, onsider Joint H = 0; F FG 59.8 = 0 F FG = 59.8 kn F G F F = 61.73 t joint ; resolving the forces perpendicular to the line, F G = 0 as the forces F and F cannot give components in the perpendicular direction. Resolving the forces along the line of forces. F the force F is obtained as F = 61.73 kn. Joint G F G F G G 59.8 Resolving the forces in the vertical direction at joint G; F G = 0 and resolving the forces in the horizontal direction F G = 59.8 kn.

86 88 asic Structural nalysis XMPL 2.23: truss of 10 m span is loaded as shown in Fig. 2.58 find the forces in the members of the truss by using method of sections. 25 kn 30 kn 60 60 60 30 F 10 m G FIG. 2.58 SOLUTION In G V = 0 M = 0 V +V = 25 + 30 sin30 = 10 = 10sin30 V +V = 55 kn = 5m cos30 = 10 25 5cos60 + 30 6.25 10V = 0 = 8.66 m V = 25 kn V = 30 kn sin30 = G G = sin30 = 4.33sin30 = 2.165 m = = 4.3 F = 5cos60 G = 10 4.33cos60 G = 6.25 m

88 asic Structural nalysis 25 kn 30 kn 60 5 m ' FIG. 2.60 sin60 = = 5sin60 = 4.33 M = 0 30 10 25 7.5 F 4.33 = 0 F = 26 kn To determine the force in, take moment about. 30 2.5 F 4.33 = 0 F = 17.32 kn Joint F V = 0 F sin60 + 30 = 0 60 F F = 34.64 kn

Framed Structures 89 Joint F V = 0 F 30 25 = F sin30 F = 50 kn 25 kn 25 kn 35 34.6 kn 26 kn 30 kn 50 26 kn 60 60 60 17.32 kn 10 m 43.3 kn FIG. 2.61 XMPL 2.24: Figure 2.62 shows a pinjointed truss supported by a hinge at and roller at G. etermine the force in each of the five members meeting at joint. F 4 m 45º L 45º K 4 m 180 kn 125 kn J 6 @ 4 m I H G FIG. 2.62

Framed Structures 91 93 To determine the force in J, cut a section through, J and KJ. X F J 233.3 L 180 kn 45 K θ 125 kn X J FIG. 2.63 tanθ = 4/8 sinθ = 0.4472 In the cut truss, two cut members, viz. and KJ are horizontal. They do not give any vertical component of the force. Hence to maintain equilibrium in the vertical direction, the force F J which is indirect at an angle θ should give a vertical component to balance the effect due to the vertical reaction at and the applied downward loads. Hence resolving vertically. F J sinθ + 233.3 180 125 = 0. F J = 160.33 kn Using method of joints at joint (as shown in Fig. 2.64) F θ 45 106.33 329.9 F K 180 FIG. 2.64 H = 0; 329.99 cos45 F + 160.33 cosθ F K cos45 = 0 F + 0.707 F K = 376.74

92 asic Structural nalysis V = 0; 329.99 sin45 180 + F K sin45 163.33 sinθ = 0 F K = 26.07 kn F = 358.31 kn XMPL 2.25: etermine the forces in the members, G and GF of the truss shown in Fig. 2.65 by method of sections. VTU, ec. 2006 10 kn 60 20 kn 60 10 kn 30 G 60 3 @ = 9 m F 60 FIG. 2.65 SOLUTION The reactions are found using equilibrium equations H = 0; H 10cos60 20cos60 10cos60 = 0 H = 20 kn V = 0; V +V 10sin60 20sin60 10sin60 = 0 V +V = 40sin60 V +V = 34.64 kn M = 0; 20(2.6)+10(5.2) 9V = 0 V = 11.55 kn V = 23.09 kn To determine the forces in F, F G and F GF cut a section X X as shown in Fig. 2.66. The force F GF is obtained by taking moment about. 11.55 4.5 = F GF 4.5tan30 To determine the force in G, take moment about

Framed Structures 93 X 4.56 tan 30 11.55 30 60 G Z 4.5 m H FIG. 2.66 F G = 0 F GF X To determine the force in, take moment about G. F G = 11.55 3 F (3sin30)=11.55 3 F = 23.1 kn 2.6 MTHO OF TNSION OFFIINTS The method of tension coefficients was developed by Southwell (1920) and is applicable for plane and space frames. ontemporily this was developed by Muller reslau independently. T ( x,y,z ) z ( x,y,z ) y x FIG. 2.67

94 asic Structural nalysis Let be any member of a truss having length L and the tensile force be T. The force T can be expressed as T = t L where t is a tension coefficient, L is the length of member. The component of T in the direction of x axis is t (x x ) and similarly the components in the direction of y axis is t (y y ) and in z direction it is t (z = z) respectively. If at the joint, let the components of the external loads be X,Y and Z acting in x,y and z directions respective then for equilibrium of the joint where the members, and Q meet t (x x )+t (x x )+ +t Q (x Q x )+X = 0 t (y y )+t (y y )+ +t Q (y Q y )+Y = 0 t (z z )+t (z z )+ +t Q (z Q z )+Z = 0 t each joint, a similar set of three equations are formed for each joint and the tension coefficients are determined. The force in the member is obtained by multiplying this tension coefficient with the corresponding length of the members. This is known as method of tension coefficients. The procedure is as follows: 1. Mark the positive directions of x,y and z axis. 2. ssume all the members are in tension, i.e., the force is moving away from the joint. 3. Write down the equilibrium equations at each joint. 4. Solve the equations for unknown tension coefficients. 5. alculate T = L t. 2.6.1 Numerical Problems XMPL 2. 26: The Fig. 2.68 shows a Warren type cantilever truss along with the imposed loads. etermine the forces in all the members using the method of joints / tension coefficients. (nna-univ, ec. 2008) 3 kn 60 3 kn 60 60 2 kn 2 kn FIG. 2.68

Framed Structures 95 Joint F F 60 2 kn V = 0 H = 0 F sin60 2 = 0 F = 2 sin60 = 2.31 F F cos60 = 0 F = 2.31cos60 F = 1.16 kn Joint F 3 60 60 F F = 2.31 V = 0 H = 0 3 2.31sin60 + F sin60 = 0 F = 5.77 kn F + 5.77cos 60 + F cos 60 = 0 F = 4.04 kn Joint F F F 60 60 2 kn 5.77 1.16 V = 0 H = 0 F sin60 5.77sin60 2 = 0 F = 8.08 kn F F 8.08cos60 5.77cos60 1.16 = 0 F F = 8.08 kn Joint 3 kn V = 0 4.04 60 60 H = 0 F 8.08 F F 3 8.08sin60 + F F sin60 = 0 F F = 11.54 kn 4.04 + 8.08cos60 + 11.54cos60 F = 0 F = 13.85 kn

96 asic Structural nalysis 3 kn 3 kn 13.85 4.04 11.54 8.08 5.77 2.31 F 8.08 1.16 2 kn 2 kn FIG. 2.69 Using tension coefficient method 3 kn 3 kn (16,6.93) (12,6.93) (4,6.93) F(16,0) (8,0) 2 kn x (0,0) 2 kn FIG. 2.70 t Joint H = 0, t (x x )+t (x x )=0 t (4 0)+t (8 0)=0 4t + 8t = 0 V = 0, t (y y )+t (y y ) 2 = 0 6.93t + 0 = 2 Joint t = 0.2886 t = 0.1443 H = 0; t (x x )+t (x x )+t (x x )=0 0.2886(0 4)+t (8 4)+t (12 4)=0 4 t + 8 t = 1.1544

Framed Structures 97 V = 0; t (y y )+t (y y )+t (y y ) 3 = 0 0.2886(0 6.93)+t (0 6.93)+t (6.93 6.93)=3 Joint t = 0.7215 t =+0.5051 H = 0; t (x x )+t (x x )+t F (x F x )+t (x x )=0 0.7215(4 8)+t (12 8)+t F (16 8) 0.1443(0 8)=0 4t + 8t F = 4.04 V = 0; t (y y )+t (y y )+t F (y F y )+t (y y ) 2 = 0 t (6.93 0)+t (6.93 0)+t F (0 0)+t (0 0)=2 6.93 t + 6.93 t = 2.00 6.93 t + 6.93 0.7215 = 2.00 Solving the above two equations: Joint t = 1.0101 t F = 1.0100 H = 0, t (x x )+t F (x F x )+t (x x )+t (x x )=0 t (16 12)+t F (16 12)+t (8 12)+t (4 12)=0 4t + 4t F 4t 8t = 0 4t + 4t F = 4(1.0101)+8 0.5051 t +t F = 2.0203 V = 0, t (y y )+t F (y F y )+t (y y )+t (y y ) 3 = 0 t (6.93 6.93)+t F (0 6.93)+t (0 6.93)+t (6.93 6.93) 3 = 0 6.93t F 6.93(1.0101)=3 t F = 1.443 t = 3.463 T = 3.463 4 = 13.85 T F = 1.443 8 = 11.54 T = 0.5051 8 = 4.04 T =+1.0101 8 =+8.08 T = 0.2886 8 = 2.31 T = 0.7215 8 = 5.77 T = 0.1443 8 = 1.15 T F = 1.0101 8 = 8.08

98 asic Structural nalysis 3 kn 3 kn 13.85 4.04 11.54 8.08 5.77 2.31 F 8.08 1.15 2 kn 2 kn FIG. 2.71 XMPL 2. 27: nalyse the space truss shown in Fig. 2.72 using tension coefficient method. (nna Univ, 2009). y 200 kn 50 kn 7 m 1 m 1 m 8. x 4.5 m z FIG. 2.72 Joint x y z 0 0 0 8.3 0 0 3.0 7 1 3 0 4.5

Framed Structures 99 Resolving the forces at joint in x direction t (x x )+t (x x )+t (x x )+50 = 0 t (0 3)+t (3 3)+t (8.3 3)+50 = 0 3t + 5.3t = 50 (1) Resolving the forces at joint in y direction t (y y )+t (y y )+t (y y ) 200 = 0 t (0 7)+t (0 7)+t (0 7)=200 7t 7t 7t = 200 (2) Resolving the forces at the joint in z direction t (z z )+t (z z )+t (z z )=0 t (0 1)+t (4.5 1)+t (0 1)=0 t + 3.5t t = 0 (3) 3 5.3 0 t 7 7.0 7.0 t = 50 200 1 1.0 +3.5 t 0 t = 8.166 t = 14.056 t = 6.349 L = (3 0) 2 +(7 0) 2 +(1 0) 2 = 7.68 m. L = (3 8.3) 2 +(7 0) 2 +(1 0) 2 = 4.68 m. L = (3 3) 2 +(7 0) 2 +(1 4.5) 2 = 7.8. T = t L = 62.71 T = t L = 65.78 T = t L = 49.71 (ompression) (ompression) (ompression) XMPL 2. 28: Figure 2.73 shows an elevation and plan of tripod having legs of unequal lengths resting without slipping on a sloping plane, if the pinjointed apex carries a vertical load of 100 kn. alculate the force in each leg.

100 asic Structural nalysis 2 m 2.7 m 1.35 m 1. 0.7 m 1. y 1. x 1. FIG. 2.73 Joint x y z 0 0 0 2.7 +1.3 2 0 1.3 3.3 1.35 +1.3 1.6 Resolving the forces at the joint is x direction t (x x )+t (x x )+t (x x )=0 t ( 2.7 0)+t (0 0)+t (1.35 0)=0 2.7t + 1.35t = 0

Framed Structures 101 Resolving the forces at the joint in y direction t (y y )+t (y y )+t (y y )=0 t (1.3 0)+t ( 1.3 0)+t (1.30 0)=0 Resolving the forces at the joint in z direction 1.3t 1.3t + 1.3t = 0 t (z z )+t (z z )+t (z z )=0 t ( 2 0)+t ( 3.3 0)+t (2.6 0) 100 = 0 2 t 3.3 t + 2.6 t = 100 2.7 0 1.35 t +1.3 1.3 1.3 t = 0 0 2 3.3 +2.6 t 100 t = 14.925; t = 44.776; t = 29.851 L = 2.7 2 + 1.3 2 + 2 2 = 3.602 m L = 0 2 + 1.3 2 + 3.3 2 = 3.547 m L = 1.35 2 + 1.3 2 + 2.6 2 = 3.205 m T = t L = 53.76 (ompressive) T = t L = 158.82 (ompressive) T = t L = 95.67 (ompressive) XMPL 2. 29: Figure 2.74 shows plan and elevation of a symmetrical statically determinate space truss supporting a vertical load of 75 kn at. etermine the forces in the members and F.,F z, levation x 75 kn

102 asic Structural nalysis x y F FIG. 2.74 Joint x y z +6 0 +3 +6 +3 0 +6 3 0 0 0 0 +3 3 +3 F +3 +3 +3 Resolving all the forces meeting at joint in z direction as t (z z )+t (z z )+t (z z )+t F (z F z ) 75 = 0 t (3 0)+t (0 0)+t (0 0)+t F (3 0)=75 3t + 3t F = 75 t +t F = 25 s the structure and loading are symmetrical t = t F and hence t = 12.5

Framed Structures 103 Resolving all the forces in the directions of y axis at the joint t (y y )+t (y y )+t F (y F y )+t (y y )=0 t ( 3 + 3)+t (0 + 3)+t F (3 + 3)+t (0 3)=0 3t + 6t F 3 12.5 = 0 3t + 6t F = 37.5 Resolving all the forces in the direction of z-axis t (z z )+t (z z )+t F (z F z )+t (z z )=0 t (0 3)+t (3 3)+t F (3 3)+t (0 3)=0 3t 3t = 0 t = t Resolving all the forces in the direction of x-axis t (x x )+t (x x )+t F (x F x )+t (x x )=0 t (6 3)+t (6 3)+t F (3 3)+t (0 3)=0 3t + 3t 3t = 0 3( 12.5)+3t 3 12.5 = 0 3t = 75 t = 25 Substituting in the equation: 3 25 + 6 t F = 37.5 t F = 6.25 L = (x x ) 2 +(y y ) 2 +(Z Z ) 2 = 3 2 + 3 2 + 3 2 = 5.2m L F = (x F x ) 2 +(y F y ) 2 +(Z F Z ) 2 = (3 3) 2 +(3 + 3) 2 +(3 3) 2 = 6m T = t L = 12.5 5.2 = 65 kn (Tensile) T F = t F L F = 6.25 6 = 37.5 kn (ompressive)

104 asic Structural nalysis XMPL 2.30: space frame shown in figure is supported at,, and in a horizontal plane through ball joints. The member F is horizontal and is at a height of above the base. The loads at the joints and F shown in Fig. 2.75 act in a horizontal plane. Find the forces in all of the members of the frame. (nna Univ, 2004) 10 F y 15 kn 20 x 2 m 2 m FIG. 2.75 Joint x y z 0 6 0 0 0 0 7 0 0 7 6 0 2 3 3 F 5 3 3 Resolving all the forces at joint in x direction t (x x )+t (x x )+t F (x F x )+10 = 0 t (0 2)+t (0 2)+t F (5 2)+10 = 0 Resolving all the forces in y direction at joint t (y y )+t (y y )+t F (y F y )+15 = 0 t (6 3)+t (0 3)+t F (3 3)+15 = 0 3t 3t = 15 Resolving all the forces in z direction at joint t (z z )+t (z z )+t F (z F z )+0 = 0 t (0 3)+t (0 3)+t F (3 3)=0 3t 3t = 0

Framed Structures 105 Solving the equations 2 2 +3 t 3 3 0 t = 3 3 0 t F 10 15 0 t = 2.5, t =+2.5, t F = 3.33 L = (0 2) 2 +(6 3) 2 +(0 3) 2 = 4.69 m L = (0 2) 2 +(0 3) 2 +(0 3) 2 = 4.69 m L F = (5 2) 2 +(3 3) 2 +(3 3) 2 = 3m T = t L = 2.5 4.69 = 11.725 (ompressive) T = t L =+2.5 4.69 =+11.725 (Tensile) T F = t F L F = 3.33 3.00 = 9.99 (ompressive) Resolving all the forces in x direction at joint F t F (x x F )+t F (x x F )+t F (x x F )=0 3.33(2 5)+t F (7 5)+t F (7 5)=0 2t F + 2t F = 10 t F +t F = 5 Resolving all the forces in y direction at the joint F t F (y y F )+t F (y y F )+t F (y y F )+20 = 0 3.33(3 3)+t F (6 3)+t F (0 3)=0 3t F 3t F = 0 Solving the above t F = 2.5, t F = 2.5 L F = (7 5) 2 +(6 3) 2 +(0 3) 2 = 4.69 m L F = (7 5) 2 +(0 3) 2 +(0 3) 2 = 4.69 m T F = t F L F = 11.73 kn T F = t F L F = 11.73 kn

106 asic Structural nalysis RVIW QUSTIONS Remembrance: 2.1 efine plane and space truss? 2.2 What are the different types of analysis of trusses? 2.3 List the assumptions made in truss analysis? 2.4 xplain the steps involved in method of joints? 2.5 xplain the steps involved in method of sections? 2.6 xplain the method of tension coefficients? 2.7 What type of analysis is used in determining the forces of a space truss? 2.8 For what kind of trusses can be analysed by method of joints and method of sections? 2.9 efine Tension coefficient? 2.10 Which method is preferable to find out the forces in a few members of a truss? 2.11 What is the primary function of a truss? 2.12 What is the minimum numbers of elements to make a simple truss? Understanding: 2.1 istinguish between a simple truss, compound truss and complex truss? 2.2 What kind of stresses developed if the loads are not applied at the joints? 2.3 What are the limitations of method of joints? 2.4 Identify the truss members having zero forces joints?

Framed Structures 107 10 kn 10 kn 40 kn 80 kn 40 kn 10 kn 4 m 4 m F F F F I 2 m H 2 m G 2 m 40 kn P=20 kn F G 4 m H 4 m 5 m 100 kn

108 asic Structural nalysis XRIS PROLMS 2.1 Find the forces in all the members of the truss shown in figure. (nna Univ, pril, 2008) 4 kn 1.5 m θ 2 m 2 m 6 kn (ns = 2.5 kn, = 7.5, =+6, =+6.0, =+6.0) 2.2 etermine the forces in the members of the truss as shown in figure by method of joints. (VTU, ec. 2008) 20 kn 4 m 4 m 30 kn (ns =+30 kn, =+30, =+30, = 12.5, = 37.5) 2.3 etermine the nature and magnitude of forces induced in all the members of the frame loaded as shown in figure by method of joints. (VTU, Feb. 2002)

Framed Structures 109 20 kn 10 kn 5 kn 60 60 (ns = 22.75, = 11.16, = 11.9, = 11.38, = 10.95, = 0.43, = 0.43) 2.4 etermine the forces in all the members of the following truss by method of joints. (VTU, Feb. 2003) 2 kn 2 kn 45 F (ns = 0.95 kn, = 0.658, = 1.88, =+1.33, F = 1.3 F =+2.67 kn, F = 1.33,F =+1.88, = 0) 2.5 etermine the forces in all the members by method of joints. 4 m 8 m G 8 m F 8 m 50 kn 100 kn (ns = 94.3, = 133.36, = 166.6, = 117.8, G = 23.55, G = 66.7, GF = 149.93, F = 83.3, G =+94.3, F = 23.55, F = 117.8 kn)

110 asic Structural nalysis 2.6 plane truss of 6 m span is subjected to a point load of 30 kn as shown in figure. etermine the forces in, and F by method of sections. (MSRIT-2009) 2 2 F 2 m 2 m 30 kn (ns = 7.46 kn, F = 20, = 18.86 kn) 2.7 etermine the forces by using method of joints. 50 kn 30 kn 60º 30º 6 m 60º (ns = 57.66 kn, = 28.83, = 40.4, =+20.2, =+9.97) 2.8 etermine the forces by method of joints. F 50 kn 4 4 m

Framed Structures 111 (ns = +18.75, = 15.63, F = 15.63, F = 18.75, = 0, = 50, = 15.63, F = 46.88) 2.9 etermine the magnitude and nature of forces in all the members of the plane truss shown in figure. (nna Univ. ec 2008) 30 20 kn 40 kn 60 4 m 60 4 m (ns = 11.55, = 23.1, = 46.18, = 23.1, = +23.1, = 0, = 0) 2.10 Find the forces in all members of the pinjointed plane truss shown in figure. Use method of joints. (VTU, July 2007) 10 kn 20 kn 10 kn 30 30 F G 2.11 Find the forces in all members of the truss shown in figure by method of joints. (VTU, ec 2007) 20 kn 60 60 10 kn (ns = 20.21, = 14.43, = 14.43, = 10.11, = 7.22, = 8.66, = 14.43 kn)

Framed Structures 113 2.14 Find the forces in members RS, ST and TQby method of sections. R 10 kn 20 kn T S P 60 60 60 30 T 6.92 m 10 Q (ns RS = 1990 kn, TQ= 9.96 kn, ST =+20 kn) 2.15 plane truss is subjected to point loads as shown in figure. Find the forces in the member H and by method of sections. (MSRIT-2009) 40 30 kn H 2.7 2.7 3.6 3.6 3.6 (ns H = 5.5 kn; = 44.4 kn) 2.16 Figure shows a pin jointed frame which is hinged to the foundation at and is resting on rollers at. etermine the magnitude and nature of force in.

Framed Structures 115 2.18 Figure shows a plan of a tripod, the feet, and being in the same horizontal plane and apex being 3.75 m above the plane. Horizontal loads of 100 kn and 150 kn are applied at in the directions shown. Find the forces in the members assuming that the joints are pin-joints. (nna Univ, June 2009). 5.4 m 3.6 m x 15 10 kn y (ns = 6.4 kn, = 19.1, = 9.62)