Framed Structures PLANE FRAMES. Objectives:
|
|
- Tamsyn Cross
- 5 years ago
- Views:
Transcription
1 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
2 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 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 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 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
3 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 = = 28.5 kn. Member force 1 = = 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 FIG Frame diagram 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
4 38 asic Structural nalysis 50 kn 25 kn 4 m 4 m m 1 FIG a 2 1 b Reaction at the top hinge 3 c Reaction at the roller FIG d Force in kn Member Strut Tie Reaction at hinge 103 Reaction at roller 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:
5 Framed Structures 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 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 = = 20 kn 20 kn 10 kn 10 kn kn 20 kn FIG. 2.9
6 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 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 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 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 kn a 10 kn 10 kn b 5,1 3 e FIG Frame diagram 2 c d FIG Force diagram
7 Framed Structures 41 Force in kn Member Strut Tie 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 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 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 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 fter marking the points 1, 2 and 3 the member forces and their nature are tabulated here.
8 42 asic Structural nalysis 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 kn FIG 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 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 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 etermine the magnitude and nature of the forces from the forces diagram and tabulate. 20 kn 1 a,c d FIG Frame diagram 3 FIG Force diagram b,e
9 Framed Structures 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 by method of joints. (May 2010, RV) kn H 4 V 4 V 6 FIG 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; V 6 = 0 V 6 = 45 kn V 4 = 25 kn
10 44 asic Structural nalysis n m < 2n j 3 4 < (2 4) 3 4 < 5 n j = 4 n m = 4 FIG 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 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
11 Framed Structures 45 Joint 4 F 14 H = 0; F = 0 F 45 = 20 kn 4 20 F 45 V = 0; F = 0 25 kn F 14 = 25 kn Joint 1 20 kn 1 V = 0; 45 F F 15 sin45 = 0 F 15 F 15 = 35.4 kn 25 kn H = 0; 20 F 12 + F 15 cos45 = 0 20 F cos45 = 0 F 12 = kn Joint 2 70 kn F 23 = 0 F 23 = kn F 23 V = 0; F 25 = 70.0 kn F 25 Joint kn F 53 V = 0; F 53 cos sin45 70 = 0 F 53 = kn
12 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 = 2.89cos 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 = kn 2.02 kn H = 0 F = F F cos F cos60 = F cos F cos60 = = 2F cos60 F = 0.57; F = 0.57
13 48 asic Structural nalysis Joint F F θ 25 kn V = 0 H = 0 F sinθ = 25 F = 25 sinθ = 25 = kn 0.6 F + F cosθ = 0 F = 0 F = kn Joint 50 kn H = 0 F = 0 F = kn F F V = 0 F 50 = 0 F = 50 kn Joint 50 kn 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 = cosθ + F cosθ F sinθ 1 = (0.8)+F (0.8) 0.8 F = 0 F F = (2.1)
14 50 asic Structural nalysis Joint F 4 kn H = 0; V = 0; F F cos60 F cos60 = 0 F F cos60 = 4.04cos60 4 F sin sin60 = 0 F = 0.57 kn F F = 4.04 kn Substituting F in the above equation F = 1.73 kn 2 kn kn FIG XMPL 2.7: etermine the magnitude and nature of forces in all the number of the pin jointed plane truss shown in Fig by method of joints. (VTU, June 08) 25 kn 50 kn 25 kn 4 m 4 m 6 m FIG. 2.24
15 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 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 = F F = 0 F F = kn F F F cosθ = 0 F = = 18.0 kn Joint H = 0; F = 18.0 kn 18.0 F
16 Framed Structures 53 V = 0; sinθ + F cosθ 1 + F sinθ = (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 = kn Joint F θ 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 FIG. 2.25
17 58 asic Structural nalysis Joint kn θ 1 F F V = 0 H = 0 F sinθ 1 30 = 0 F = 30 = kn F cosθ 1 = 0 F = = 30 kn kn 60 kn 30 kn 30 kn G H 1.25 m 2.50 m F FIG 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 H
18 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 = sinθ 2 = sinθ 3 = cosθ 1 = cosθ 2 = cosθ 3 = 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 = F F F F = F F sinθ 3 + F F sinθ 1 = F F F F = 20 on solving the above equations Joint G F F = kn F F F F F = 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 = kn
19 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 XMPL 2.13: Find the forces in the members of the truss in Fig Tabulate your results neatly. (VTU, July 2004) 10 kn 20 kn G 10 kn m 10 kn H 4 m F 4 m FIG. 2.34
20 Framed Structures 65 S.No Member Force (kn) Nature ompressive ompressive 3 G ompressive 4 G ompressive 5 F Tensile 6 F Tensile Tensile ompressive Tensile 10 F 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 Joint 10 kn F F H = 0 V = 0 10 F F = 0 F = 0 F F F = 10 kn
21 66 64 asic Structural nalysis Joint kn F F H = 0 V = 0 20cos60 F cos30 F cos cos30 = 0 F cos30 + F cos60 = F F = sin60 F sin30 + F sin sin30 = F F = 3.78 F = kn F = kn Joint G F G F GF G 30 F G = H = 0 V = 0 F G cos30 F G cos30 + F GF cos60 = F G F G F GF = F G F GF = F G F GF = F G sin30 + F GF sin sin30 = F G F GF = F G = kn, F GF = 0 Joint 10 kn V = 0; 10sin sin30 F sin sin30 = 0 F = kn F
22 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 = = 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 = /0.832 F GF = kn sinθ 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 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 = kn
23 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 kn F F G Resolving all the forces in the vertical direction: F G = 0 Resolving all the forces in the horizontal direction: F = 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
24 Framed Structures 71 The computed forces are shown in Fig 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 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 pplication of Method of Sections onsider a truss shown in Fig 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
25 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 sin = 66.7 F G = 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 Numerical Problems XMPL 2.17: etermine the nature and magnitude of forces in members,i and HI of the truss shown in Fig 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 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
26 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 m F H m G FIG SOLUTION ue to symmetry, the reactions V = V = = 60 kn 2 To determine the forces in, F and FG cut a section X-X as shown in Fig onsider left part of the truss and analyse the equilibrium. X θ 30 kn θ 60 kn 3.6 F θ θ 1 FIG X To determine the force in, take moment about F where other two members of the cut section F and FG meet = F 3.6 F = 80 kn
27 Framed Structures 77 We know that θ = and θ 1 = 14 2 θ 2 = FG = θ + θ 1 = 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 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 + 5sin sin60 V +V = 8.66 H = 0; 2.5cos60 + 5cos cos60 H = 0 H = 5 kn M = 0; 5(2.6)+2.5(5.2) 9V = 0 V = 2.88 kn V = 5.78 kn
28 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 m H 10 kn 4 m F G 4 m FIG 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; 10cos cos cos60 H = 0 H = 20 kn V +V = 10sin sin sin V +V = kn 20(3.46)+10(6.92)+10(4) 12V = 0 V = 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.
29 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 F 50 kn FIG SOLUTION The reactions are found out by using equilibrium equations. Taking moment about ; V +V = V = 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 = 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.
30 82 asic Structural nalysis F F F F F F 6 m θ 16.7 kn V FIG tanθ = 3/6 i.e. θ = In sinθ = = 3sin = 1.34 m. F 1.34 = 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 sin = 1.34 m M = 0; F F = 0 F F = 0
31 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 = 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 = 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.
32 86 88 asic Structural nalysis XMPL 2.23: truss of 10 m span is loaded as shown in Fig find the forces in the members of the truss by using method of sections. 25 kn 30 kn F 10 m G FIG SOLUTION In G V = 0 M = 0 V +V = sin30 = 10 = 10sin30 V +V = 55 kn = 5m cos30 = cos V = 0 = 8.66 m V = 25 kn V = 30 kn sin30 = G G = sin30 = 4.33sin30 = m = = 4.3 F = 5cos60 G = cos60 G = 6.25 m
33 88 asic Structural nalysis 25 kn 30 kn 60 5 m ' FIG sin60 = = 5sin60 = 4.33 M = F 4.33 = 0 F = 26 kn To determine the force in, take moment about F 4.33 = 0 F = kn Joint F V = 0 F sin = 0 60 F F = kn
34 Framed Structures 89 Joint F V = 0 F = F sin30 F = 50 kn 25 kn 25 kn kn 26 kn 30 kn kn kn 10 m 43.3 kn FIG 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 4 m I H G FIG. 2.62
35 Framed Structures To determine the force in J, cut a section through, J and KJ. X F J L 180 kn 45 K θ 125 kn X J FIG tanθ = 4/8 sinθ = 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 = 0. F J = kn Using method of joints at joint (as shown in Fig. 2.64) F θ F K 180 FIG H = 0; cos45 F cosθ F K cos45 = 0 F F K =
36 92 asic Structural nalysis V = 0; sin F K sin sinθ = 0 F K = kn F = kn XMPL 2.25: etermine the forces in the members, G and GF of the truss shown in Fig by method of sections. VTU, ec kn kn kn 30 G 60 = 9 m F 60 FIG 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 = kn M = 0; 20(2.6)+10(5.2) 9V = 0 V = kn V = kn To determine the forces in F, F G and F GF cut a section X X as shown in Fig The force F GF is obtained by taking moment about = F GF 4.5tan30 To determine the force in G, take moment about
37 Framed Structures 93 X 4.56 tan G Z 4.5 m H FIG F G = 0 F GF X To determine the force in, take moment about G. F G = F (3sin30)= 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
38 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 Numerical Problems XMPL 2. 26: The Fig 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 kn 2 kn FIG. 2.68
39 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 F F = 2.31 V = 0 H = sin60 + F sin60 = 0 F = 5.77 kn F cos 60 + F cos 60 = 0 F = 4.04 kn Joint F F F kn V = 0 H = 0 F sin sin60 2 = 0 F = 8.08 kn F F 8.08cos cos = 0 F F = 8.08 kn Joint 3 kn V = H = 0 F 8.08 F F sin60 + F F sin60 = 0 F F = kn cos cos60 F = 0 F = kn
40 96 asic Structural nalysis 3 kn 3 kn F kn 2 kn FIG 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 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 = t + 0 = 2 Joint t = t = H = 0; t (x x )+t (x x )+t (x x )= (0 4)+t (8 4)+t (12 4)=0 4 t + 8 t =
41 Framed Structures 97 V = 0; t (y y )+t (y y )+t (y y ) 3 = (0 6.93)+t (0 6.93)+t ( )=3 Joint t = t = H = 0; t (x x )+t (x x )+t F (x F x )+t (x x )= (4 8)+t (12 8)+t F (16 8) (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)= t t = t = 2.00 Solving the above two equations: Joint t = t F = 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) t +t F = V = 0, t (y y )+t F (y F y )+t (y y )+t (y y ) 3 = 0 t ( )+t F (0 6.93)+t (0 6.93)+t ( ) 3 = t F 6.93(1.0101)=3 t F = t = T = = T F = = T = = 4.04 T = =+8.08 T = = 2.31 T = = 5.77 T = = 1.15 T F = = 8.08
42 98 asic Structural nalysis 3 kn 3 kn F kn 2 kn FIG XMPL 2. 27: nalyse the space truss shown in Fig 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 Joint x y z
43 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) t t = t 0 t = t = t = 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 = T = t L = T = t L = (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.
44 100 asic Structural nalysis 2 m 2.7 m 1.35 m m 1. y 1. x 1. FIG Joint x y z 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 t = 0
45 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 t = t t = t 100 t = ; t = ; t = L = = m L = = m L = = m T = t L = (ompressive) T = t L = (ompressive) T = t L = (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
46 102 asic Structural nalysis x y F FIG Joint x y z F 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
47 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 = 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 = 0 3t = 75 t = 25 Substituting in the equation: t F = 37.5 t F = 6.25 L = (x x ) 2 +(y y ) 2 +(Z Z ) 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 = = 65 kn (Tensile) T F = t F L F = = 37.5 kn (ompressive)
48 104 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 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 Joint x y z F 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
49 Framed Structures 105 Solving the equations t t = t F 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 = = (ompressive) T = t L = = (Tensile) T F = t F L F = = 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 = (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 = kn T F = t F L F = kn
50 106 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?
51 Framed Structures 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
52 108 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)
53 Framed Structures kn 10 kn 5 kn (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, = , = 166.6, = 117.8, G = 23.55, G = 66.7, GF = , F = 83.3, G =+94.3, F = 23.55, F = kn)
54 110 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, = kn) 2.7 etermine the forces by using method of joints. 50 kn 30 kn 60º 30º 6 m 60º (ns = kn, = 28.83, = 40.4, =+20.2, =+9.97) 2.8 etermine the forces by method of joints. F 50 kn 4 4 m
55 Framed Structures 111 (ns = , = 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) 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 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 kn (ns = 20.21, = 14.43, = 14.43, = 10.11, = 7.22, = 8.66, = kn)
56 Framed Structures Find the forces in members RS, ST and TQby method of sections. R 10 kn 20 kn T S P 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) kn H (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.
57 Framed Structures 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 kn y (ns = 6.4 kn, = 19.1, = 9.62)
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
More informationPin-Jointed Frame Structures (Frameworks)
Pin-Jointed rame Structures (rameworks) 1 Pin Jointed rame Structures (rameworks) A pin-jointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless
More informationTYPES OF STRUCUTRES. HD in Civil Engineering Page 1-1
E2027 Structural nalysis I TYPES OF STRUUTRES H in ivil Engineering Page 1-1 E2027 Structural nalysis I SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented:
More informationTruss Analysis Method of Joints. Steven Vukazich San Jose State University
Truss nalysis Method of Joints Steven Vukazich San Jose State University General Procedure for the nalysis of Simple Trusses using the Method of Joints 1. raw a Free Body iagram (FB) of the entire truss
More information7 STATICALLY DETERMINATE PLANE TRUSSES
7 STATICALLY DETERMINATE PLANE TRUSSES OBJECTIVES: This chapter starts with the definition of a truss and briefly explains various types of plane truss. The determinancy and stability of a truss also will
More informationContinuing Education Course #207 What Every Engineer Should Know About Structures Part B Statics Applications
1 of 6 Continuing Education Course #207 What Every Engineer Should Know About Structures Part B Statics Applications 1. As a practical matter, determining design loads on structural members involves several
More informationCalculating Truss Forces Unit 2 Lesson 2.1 Statics
alculating Truss Forces alculating Truss Forces Principles of Engineering 22 Forces ompression body being squeezed Tension body being stretched Truss truss is composed of slender members joined together
More informationStructural Analysis. Method of Joints. Method of Pins
Structural nalysis / Method of Pins I m reading a book about an1gravity. It s impossible to put down. Pop Quiz What is today s date? 2 1 Trusses Trusses are common means of transferring loads from their
More informationQUESTION BANK ENGINEERS ACADEMY. Hinge E F A D. Theory of Structures Determinacy Indeterminacy 1
Theory of Structures eterminacy Indeterminacy 1 QUSTION NK 1. The static indeterminacy of the structure shown below (a) (b) 6 (c) 9 (d) 12 2. etermine the degree of freedom of the following frame (a) 1
More informationThe bending moment diagrams for each span due to applied uniformly distributed and concentrated load are shown in Fig.12.4b.
From inspection, it is assumed that the support moments at is zero and support moment at, 15 kn.m (negative because it causes compression at bottom at ) needs to be evaluated. pplying three- Hence, only
More informationDeterminate portal frame
eterminate portal frame onsider the frame shown in the figure below with the aim of calculating the bending moment diagram (M), shear force diagram (SF), and axial force diagram (F). P H y R x x R y L
More informationIf the number of unknown reaction components are equal to the number of equations, the structure is known as statically determinate.
1 of 6 EQUILIBRIUM OF A RIGID BODY AND ANALYSIS OF ETRUCTURAS II 9.1 reactions in supports and joints of a two-dimensional structure and statically indeterminate reactions: Statically indeterminate structures
More information0.3 m. 0.4 m. 0.3 m. A 2 kn. 0.4 m. tan γ = 7. (BC = kn) γ = Fx = BC cos θ + AC cos γ =0
Problem 6.4 etermine the aial forces in the members of the truss. kn 0.3 m 0.4 m 0.6 m 1. m Solution: irst, solve for the support reactions at and, and then use the method of joints to solve for the forces
More informationIntroduction. 1.1 Introduction. 1.2 Trigonometrical definitions
Introduction 1.1 Introduction Stress analysis is an important part of engineering science, as failure of most engineering components is usually due to stress. The component under a stress investigation
More informationSTATICALLY INDETERMINATE STRUCTURES
STATICALLY INDETERMINATE STRUCTURES INTRODUCTION Generally the trusses are supported on (i) a hinged support and (ii) a roller support. The reaction components of a hinged support are two (in horizontal
More informationCH. 5 TRUSSES BASIC PRINCIPLES TRUSS ANALYSIS. Typical depth-to-span ratios range from 1:10 to 1:20. First: determine loads in various members
CH. 5 TRUSSES BASIC PRINCIPLES Typical depth-to-span ratios range from 1:10 to 1:20 - Flat trusses require less overall depth than pitched trusses Spans: 40-200 Spacing: 10 to 40 on center - Residential
More informationENGINEERING MECHANICS SOLUTIONS UNIT-I
LONG QUESTIONS ENGINEERING MECHANICS SOLUTIONS UNIT-I 1. A roller shown in Figure 1 is mass 150 Kg. What force P is necessary to start the roller over the block A? =90+25 =115 = 90+25.377 = 115.377 = 360-(115+115.377)
More informationMethod of Consistent Deformation
Method of onsistent eformation Structural nalysis y R.. Hibbeler Theory of Structures-II M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering and Technology, Wah antt FRMES Method
More informationFr h mg rh h. h 2( m)( m) ( (0.800 kg)(9.80 m/s )
5. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity
More informationMEE224: Engineering Mechanics Lecture 4
Lecture 4: Structural Analysis Part 1: Trusses So far we have only analysed forces and moments on a single rigid body, i.e. bars. Remember that a structure is a formed by and this lecture will investigate
More information6.5 Cables: Concentrated Loads
6.5 ables: oncentrated Loads 6.5 ables: oncentrated Loads Procedures and Strategies, page 1 of 3 Procedures and Strategies for Solving Problems Involving ables With oncentrated Loads 1. Pass sections through
More informationCHAPTER 4 Stress Transformation
CHAPTER 4 Stress Transformation ANALYSIS OF STRESS For this topic, the stresses to be considered are not on the perpendicular and parallel planes only but also on other inclined planes. A P a a b b P z
More informationCalculating Truss Forces. Method of Joints
Calculating Truss Forces Method of Joints Forces Compression body being squeezed Tension body being stretched Truss truss is composed of slender members joined together at their end points. They are usually
More informationVector Mechanics: Statics
PDHOnline Course G492 (4 PDH) Vector Mechanics: Statics Mark A. Strain, P.E. 2014 PDH Online PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax: 703-988-0088 www.pdhonline.org www.pdhcenter.com
More informationSTATICS--AN INVESTIGATION OF FORCES
STTIS--N INVESTIGTION O ORES Two areas of study to investigate forces. Statics where the forces acting on a material are balanced so that the material is either stationary or in uniform motion. or fluid
More information8 ft. 5 k 5 k 5 k 5 k. F y = 36 ksi F t = 21 ksi F c = 10 ksi. 6 ft. 10 k 10 k
E 331, Fall 2004 HW 2.0 Solution 1 / 7 We need to make several decisions when designing a truss. First, we need to determine the truss shape. Then we need to determine the height of the truss and the member
More informationSTRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS
1 UNIT I STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define: Stress When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The
More informationMECHANICS OF STRUCTURES SCI 1105 COURSE MATERIAL UNIT - I
MECHANICS OF STRUCTURES SCI 1105 COURSE MATERIAL UNIT - I Engineering Mechanics Branch of science which deals with the behavior of a body with the state of rest or motion, subjected to the action of forces.
More informationStatics deal with the condition of equilibrium of bodies acted upon by forces.
Mechanics It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics
More informationHong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering Structural Mechanics. Chapter 1 PRINCIPLES OF STATICS
PRINCIPLES OF STTICS Statics is the study of how forces act and react on rigid bodies which are at rest or not in motion. This study is the basis for the engineering principles, which guide the design
More informationPoint Equilibrium & Truss Analysis
oint Equilibrium & Truss nalsis Notation: b = number of members in a truss () = shorthand for compression F = name for force vectors, as is X, T, and F = name of a truss force between joints named and,
More informationCHAPTER 5 Statically Determinate Plane Trusses
CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS TYPES OF ROOF TRUSS ROOF TRUSS SETUP ROOF TRUSS SETUP OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse
More informationCHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS
CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS 1 TYPES OF ROOF TRUSS ROOF TRUSS SETUP 2 ROOF TRUSS SETUP OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse
More informationThis lesson is an important one since it will deal with forces acting in conjunction with one another, against one another, and the resultant of a
1 This lesson is an important one since it will deal with forces acting in conjunction with one another, against one another, and the resultant of a number of forces acting through a common point (known
More information8.1 Internal Forces in Structural Members
8.1 Internal Forces in Structural Members 8.1 Internal Forces in Structural Members xample 1, page 1 of 4 1. etermine the normal force, shear force, and moment at sections passing through a) and b). 4
More informationEngineering Mechanics Department of Mechanical Engineering Dr. G. Saravana Kumar Indian Institute of Technology, Guwahati
Engineering Mechanics Department of Mechanical Engineering Dr. G. Saravana Kumar Indian Institute of Technology, Guwahati Module 3 Lecture 6 Internal Forces Today, we will see analysis of structures part
More information2008 FXA THREE FORCES IN EQUILIBRIUM 1. Candidates should be able to : TRIANGLE OF FORCES RULE
THREE ORCES IN EQUILIBRIUM 1 Candidates should be able to : TRIANGLE O ORCES RULE Draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object. State that
More informationSECTION 6.3: VECTORS IN THE PLANE
(Section 6.3: Vectors in the Plane) 6.18 SECTION 6.3: VECTORS IN THE PLANE Assume a, b, c, and d are real numbers. PART A: INTRO A scalar has magnitude but not direction. We think of real numbers as scalars,
More informationME Statics. Structures. Chapter 4
ME 108 - Statics Structures Chapter 4 Outline Applications Simple truss Method of joints Method of section Germany Tacoma Narrows Bridge http://video.google.com/videoplay?docid=-323172185412005564&q=bruce+lee&pl=true
More informationDeflections. Deflections. Deflections. Deflections. Deflections. Deflections. dx dm V. dx EI. dx EI dx M. dv w
CIVL 311 - Conjugate eam 1/5 Conjugate beam method The development of the conjugate beam method has been atributed to several strucutral engineers. any credit Heinrich üller-reslau (1851-195) with the
More informationBTECH MECHANICAL PRINCIPLES AND APPLICATIONS. Level 3 Unit 5
BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5 FORCES AS VECTORS Vectors have a magnitude (amount) and a direction. Forces are vectors FORCES AS VECTORS (2 FORCES) Forces F1 and F2 are in
More informationMEM202 Engineering Mechanics - Statics MEM
E Engineering echanics - Statics E hapter 6 Equilibrium of Rigid odies k j i k j i R z z r r r r r r r r z z E Engineering echanics - Statics Equilibrium of Rigid odies E Pin Support N w N/m 5 N m 6 m
More informationthree Equilibrium 1 and planar trusses ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SPRING 2015 lecture ARCH 614
ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SPRING 2015 lecture three equilibrium and planar trusses Equilibrium 1 Equilibrium balanced steady resultant of forces
More informationSupport Idealizations
IVL 3121 nalysis of Statically Determinant Structures 1/12 nalysis of Statically Determinate Structures nalysis of Statically Determinate Structures The most common type of structure an engineer will analyze
More informationMethod of Sections for Truss Analysis
RH 331 Note Set 5.2 F2013abn Method of Sections for Truss nalysis Notation: () = shorthand for compression = name for load or axial force vector (T) = shorthand for tension Joint onfigurations (special
More information40 N cos 35 =49N. T 1 = 28 T y
16 (a) Analyzing vertical forces where string 1 and string 2 meet, we find T 1 = 40 N cos 35 =49N (b) Looking at the horizontal forces at that point leads to T 2 = T 1 sin 35 = (49 N) sin 35 =28N (c) We
More informationReg. No. : Question Paper Code : B.Arch. DEGREE EXAMINATION, APRIL/MAY Second Semester AR 6201 MECHANICS OF STRUCTURES I
WK 4 Reg. No. : Question Paper Code : 71387 B.Arch. DEGREE EXAMINATION, APRIL/MAY 2017. Second Semester AR 6201 MECHANICS OF STRUCTURES I (Regulations 2013) Time : Three hours Maximum : 100 marks Answer
More informationFree-Body Diagrams. Introduction
Free-Body Diagrams Introduction A Free-Body Diagram is a basic two or three-dimensional representation of an object used to show all present forces and moments. The purpose of the diagram is to deconstruct
More informationDetermine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
More informationIntroduction Aim To introduce the basic concepts in mechanics, especially as they apply to civil engineering.
Mechanics Introduction im To introduce the basic concepts in mechanics, especially as they apply to civil engineering. Mechanics Mathematical models describing the effects of forces and motion on objects
More informationTUTORIAL SHEET 1. magnitude of P and the values of ø and θ. Ans: ø =74 0 and θ= 53 0
TUTORIAL SHEET 1 1. The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. Knowing that the tension in the cable is 1349N, determine the moment
More informationUNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.
UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude
More informationPlane Trusses Trusses
TRUSSES Plane Trusses Trusses- It is a system of uniform bars or members (of various circular section, angle section, channel section etc.) joined together at their ends by riveting or welding and constructed
More informationWhere, m = slope of line = constant c = Intercept on y axis = effort required to start the machine
(ISO/IEC - 700-005 Certified) Model Answer: Summer 07 Code: 70 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
More informationTheory of structure I 2006/2013. Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES
Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES Introduction A structure refers to a system of connected parts used to support a load. Important examples related to civil engineering include buildings,
More informationEngineering Mechanics: Statics in SI Units, 12e
Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body
More informationBeams. Beams are structural members that offer resistance to bending due to applied load
Beams Beams are structural members that offer resistance to bending due to applied load 1 Beams Long prismatic members Non-prismatic sections also possible Each cross-section dimension Length of member
More informationChapter 2 A Mathematical Toolbox
Chapter 2 Mathematical Toolbox Vectors and Scalars 1) Scalars have only a magnitude (numerical value) Denoted by a symbol, a 2) Vectors have a magnitude and direction Denoted by a bold symbol (), or symbol
More informationPhysics 101 Lecture 5 Newton`s Laws
Physics 101 Lecture 5 Newton`s Laws Dr. Ali ÖVGÜN EMU Physics Department The Laws of Motion q Newton s first law q Force q Mass q Newton s second law q Newton s third law qfrictional forces q Examples
More informationSubject : Engineering Mechanics Subject Code : 1704 Page No: 1 / 6 ----------------------------- Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word
More informationME 201 Engineering Mechanics: Statics
ME 201 Engineering Mechanics: Statics Unit 7.1 Simple Trusses Method of Joints Zero Force Members Simple Truss structure composed of slender members joined together at their end points Planar Truss Simple
More informationAnnouncements. Equilibrium of a Rigid Body
Announcements Equilibrium of a Rigid Body Today s Objectives Identify support reactions Draw a free body diagram Class Activities Applications Support reactions Free body diagrams Examples Engr221 Chapter
More informationBE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS)
BE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS) All questions carry equal marks(10 marks) Q.1 (a) Write the SI units of following quantities and also mention whether it is scalar or vector: (i)
More informationPhysics 8 Wednesday, October 28, 2015
Physics 8 Wednesday, October 8, 015 HW7 (due this Friday will be quite easy in comparison with HW6, to make up for your having a lot to read this week. For today, you read Chapter 3 (analyzes cables, trusses,
More informationModule 6. Approximate Methods for Indeterminate Structural Analysis. Version 2 CE IIT, Kharagpur
Module 6 Approximate Methods for Indeterminate Structural Analysis Lesson 35 Indeterminate Trusses and Industrial rames Instructional Objectives: After reading this chapter the student will be able to
More informationEquilibrium Equilibrium and Trusses Trusses
Equilibrium and Trusses ENGR 221 February 17, 2003 Lecture Goals 6-4 Equilibrium in Three Dimensions 7-1 Introduction to Trusses 7-2Plane Trusses 7-3 Space Trusses 7-4 Frames and Machines Equilibrium Problem
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS TUTORIAL 2 - BEAMS CONTENT Be able to determine the forces acting
More informationMechanics Lab & Problem Sheet 2: Structures ( )
Mechanics Lab & Problem Sheet 2: Structures (2009-2010) Name (please print): P/1 Tutor (please print): Lab group Notes In the session you will be divided into small groups (about five per group). For the
More informationStatics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3
Statics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3 2 3 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the
More informationHATZIC SECONDARY SCHOOL
HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT STATIC EQUILIBRIUM MULTIPLE CHOICE / 33 OPEN ENDED / 80 TOTAL / 113 NAME: 1. State the condition for translational equilibrium. A. ΣF = 0 B. ΣF
More informationSIGN CONVENTION OF STRESS RESULTANTS
SIGN CONVENTION OF STRESS RESULTANTS A quick guide to understanding the sign conventions used in the ush Me ull Me models National HE STEM rogramme INTRODUCTION Representing stress resultants graphically
More informationChapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
More informationEngineering Mechanics
2019 MPROVEMENT Mechanical Engineering Engineering Mechanics Answer Key of Objective & Conventional Questions 1 System of forces, Centoriod, MOI 1. (c) 2. (b) 3. (a) 4. (c) 5. (b) 6. (c) 7. (b) 8. (b)
More informationSRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK
SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple
More informationASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /13
ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER 2 2012/13 COURSE NAME: ENGINEERING MECHANICS - STATICS CODE: ENG 2008 GROUP: AD ENG II DATE: May 2013 TIME: DURATION: 2 HOURS INSTRUCTIONS: 1. This
More informationCHAPTER 5 ANALYSIS OF STRUCTURES. Expected Outcome:
CHAPTER ANALYSIS O STRUCTURES Expected Outcome: Able to analyze the equilibrium of structures made of several connected parts, using the concept of the equilibrium of a particle or of a rigid body, in
More informationENGINEERING MECHANICS STATIC
Trusses Simple trusses The basic element of a truss is the triangle, three bars joined by pins at their ends, fig. a below, constitutes a rigid frame. The term rigid is used to mean noncollapsible and
More informationThere are three main types of structure - mass, framed and shells.
STRUCTURES There are three main types of structure - mass, framed and shells. Mass structures perform due to their own weight. An example would be a dam. Frame structures resist loads due to the arrangement
More informationShear Force V: Positive shear tends to rotate the segment clockwise.
INTERNL FORCES IN EM efore a structural element can be designed, it is necessary to determine the internal forces that act within the element. The internal forces for a beam section will consist of a shear
More informationUNIT-IV SLOPE DEFLECTION METHOD
UNITIV SOPE EETION ETHO ontinuous beams and rigid frames (with and without sway) Symmetry and antisymmetry Simplification for hinged end Support displacements Introduction: This method was first proposed
More informationCourse Name : Physics I Course # PHY 107. Lecture-5 : Newton s laws - Part Two
Course Name : Physics I Course # PHY 107 Lecture-5 : Newton s laws - Part Two Abu Mohammad Khan Department of Mathematics and Physics North South University https://abukhan.weebly.com Copyright: It is
More informationSTATICS. Bodies. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Design of a support
4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies 2010 The McGraw-Hill Companies,
More informationOUT ON A LIMB AND HUNG OUT TO DRY :
27-Nov-12 17:03 1 of 2 h p://edugen.wileyplus.com/edugen/courses/crs1404/pc/c10/c2hlchbhcmq3mjewyzewxzeuegzvcm0.enc?course=crs1404&id=ref CHAPTER 10 OUT ON A LIMB AND HUNG OUT TO DRY : A LOOK AT INTERNAL
More informationCalculate the force F needed to produce a horizontal component of 300 N on the sledge (1)
1. A heavy sledge is pulled across snowfields. The diagram shows the direction of the force F exerted on the sledge. Once the sledge is moving, the average horizontal force needed to keep it moving at
More informationChapter 2 Basis for Indeterminate Structures
Chapter - Basis for the Analysis of Indeterminate Structures.1 Introduction... 3.1.1 Background... 3.1. Basis of Structural Analysis... 4. Small Displacements... 6..1 Introduction... 6.. Derivation...
More informationEquilibrium of a Particle
ME 108 - Statics Equilibrium of a Particle Chapter 3 Applications For a spool of given weight, what are the forces in cables AB and AC? Applications For a given weight of the lights, what are the forces
More informationUNIT I ENERGY PRINCIPLES
UNIT I ENERGY PRINCIPLES Strain energy and strain energy density- strain energy in traction, shear in flexure and torsion- Castigliano s theorem Principle of virtual work application of energy theorems
More informationChapter 2: Deflections of Structures
Chapter 2: Deflections of Structures Fig. 4.1. (Fig. 2.1.) ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 1 (2.1) (4.1) (2.2) Fig.4.2 Fig.2.2 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 2
More informationPrecalculus Summer Assignment 2015
Precalculus Summer Assignment 2015 The following packet contains topics and definitions that you will be required to know in order to succeed in CP Pre-calculus this year. You are advised to be familiar
More informationUnit M1.4 (All About) Trusses
Unit M1.4 (ll bout) Trusses Readings: DL 1.9 16.001/002 -- Unified Engineering Department of eronautics and stronautics Massachusetts Institute of Technology LERNING OBJETIVES FOR UNIT M1.4 Through participation
More informationThe Great Indian Hornbill
The Great Indian Hornbill Our mascot The Great Indian Hornbill is on the verge of extinction and features in International Union for Conservation of Nature (IUCN) red list of endangered species. Now these
More informationJurong Junior College 2014 J1 H1 Physics (8866) Tutorial 3: Forces (Solutions)
Jurong Junior College 2014 J1 H1 Physics (8866) Tutorial 3: Forces (Solutions) Take g = 9.81 m s -2, P atm = 1.0 x 10 5 Pa unless otherwise stated Learning Outcomes (a) Sub-Topic recall and apply Hooke
More informationENGI 1313 Mechanics I
ENGI 1313 Mechanics I Lecture 25: Equilibrium of a Rigid Body Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland spkenny@engr.mun.ca
More informationSTRESS. Bar. ! Stress. ! Average Normal Stress in an Axially Loaded. ! Average Shear Stress. ! Allowable Stress. ! Design of Simple Connections
STRESS! Stress Evisdom! verage Normal Stress in an xially Loaded ar! verage Shear Stress! llowable Stress! Design of Simple onnections 1 Equilibrium of a Deformable ody ody Force w F R x w(s). D s y Support
More informationChapter 7 FORCES IN BEAMS AND CABLES
hapter 7 FORES IN BEAMS AN ABLES onsider a straight two-force member AB subjected at A and B to equal and opposite forces F and -F directed along AB. utting the member AB at and drawing the free-body B
More informationCE 201 Statics. 2 Physical Sciences. Rigid-Body Deformable-Body Fluid Mechanics Mechanics Mechanics
CE 201 Statics 2 Physical Sciences Branch of physical sciences 16 concerned with the state of Mechanics rest motion of bodies that are subjected to the action of forces Rigid-Body Deformable-Body Fluid
More informationFigure Two. Then the two vector equations of equilibrium are equivalent to three scalar equations:
2004- v 10/16 2. The resultant external torque (the vector sum of all external torques) acting on the body must be zero about any origin. These conditions can be written as equations: F = 0 = 0 where the
More informationInternal Internal Forces Forces
Internal Forces ENGR 221 March 19, 2003 Lecture Goals Internal Force in Structures Shear Forces Bending Moment Shear and Bending moment Diagrams Internal Forces and Bending The bending moment, M. Moment
More informationMechanics 2. Revision Notes
Mechanics 2 Revision Notes October 2016 2 M2 OCTOER 2016 SD Mechanics 2 1 Kinematics 3 Constant acceleration in a vertical plane... 3 Variable acceleration... 5 Using vectors... 6 2 Centres of mass 7 Centre
More information8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method
The basis for the method comes from the similarity of eqn.1 &. to eqn 8. & 8. To show this similarity, we can write these eqn as shown dv dx w d θ M dx d M w dx d v M dx Here the shear V compares with
More information