Point Equilibrium & Truss Analysis

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Brian Taylor
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1 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, e. F = free bod diagram F = force component in the direction, as is T F = force component in the direction, as is T n = number of joints in a truss N = normal force (perpendicular to something) R = name for resultant vectors R = resultant component in the direction R = resultant component in the direction T = name for a tension force (T) = shorthand for tension = ais direction, or horizontal dimension = ais direction, or vertical dimension = coefficient of static friction = angle, in a trig equation, e. sin, that is measured between the ais and tail of a vector = summation smbol EQUILIRIUM is the state where the resultant of the forces on a particle or a rigid bod is zero. There will be no rotation or translation. The forces are referred to as balanced. e: 2 forces of same size, opposite direction X e: 4 forces, polgon rule shows that it closes X nalticall, for a point: R F ( or h) 0 R F ( or v) 0 (scalar addition) NEWTON S FIRST LW: If the resultant force acting on a particle is zero, the particle will remain at rest (if originall at rest) or will move with constant speed in a straight line (if originall in motion). R F 0 R F 0 1

2 It is SOLUTELY NEESSRY to consider all the forces acting on a bod (applied directl and indirectl) using a FREE OY IGRM. Omission of a force would ruin the conditions for equilibrium. FREE OY IGRM (aka F): Sketch of a significant isolated particle of a bod or structure showing all the forces acting on it. Forces can be from eternall applied forces weight of the rigid bod reaction forces or constraints forces developed within a section member ollinear Force Sstem ll forces act along the same line. Onl one equilibrium equation is needed: F 0 Equivalentl: R F 0 and R F 0 oncurrent Force Sstem ( inline) ll forces act through the same point. Onl two equilibrium equations are needed: R F 0 and R F 0 How to solve when there are more than three forces on a free bod: 1. Resolve all forces into and components using known and unknown forces and angles. (Tables are helpful.) 2. etermine if an unknown forces are related to other forces and write an equation. 3. Write the two equilibrium equations (in and ). 4. Solve the equations simultaneousl when there are the same number of equations as unknown quantities. (see math handout) ommon problems have unknowns of: 1) magnitude of force 2) direction of force FREE OY IGRM STES FOR OINT; 1. etermine the point of interest. (What point is in equilibrium?) 2. etach the point from and all other bodies ( free it). 2

3 3. Indicate all eternal forces which include: - action on the point b the supports & connections - action on the point b other bodies - the weigh effect (=force) of an bod attached to the point (force due to gravit) 4. ll forces should be clearl marked with magnitudes and direction. The sense of forces should be those acting on the point not from the point. 5. imensions/angles should be included for force component computations. 6. Indicate the unknown forces, such as those reactions or constraining forces where the bod is supported or connected. Force Reactions can be categorized b the tpe of connections or supports. force reaction is a force with known line of action, or a force of unknown direction. The line of action of the force is directl related to the motion that is prevented. prevents motion: up and down prevents motion: vertical & horizontal The line of action should be indicated on the F. The sense of direction is determined b the tpe of support. (ables are in tension, etc ) If the sense isn t obvious, assume a sense. When the reaction value comes out positive, the assumption was correct. When the reaction value comes out negative, the assumption was opposite the actual sense. ON T HNGE THE RROWS ON YOUR F OR SIGNS IN YOUR EQUTIONS. With the 2 equations of equilibrium for a point, there can be no more than 2 unknowns. Friction There will be a force of resistance to movement developed at the contact face between objects when one is made to slide against the other. This is known as static friction and is determined from the reactive force, N, which is normal to the surface, and a coefficient of friction,, which is based on the materials in contact. F μ N If the static friction force is eceeded b the pushing force, there will still be friction, but it is called kinetic friction, and it is smaller than static friction, so it is moving. The friction resistance is independent of the amount of contact area. 3

Materials range Metal on ice 0.03-0.05 Metal on metal 0.-0.60 Metal on wood 0.20-0.60 Metal on stone 0.30-0.70 Wood on wood 0.30-0.70 Steel on steel 0.75 Stone on stone 0.40-0.

70 LE STRUTURES: ables with Several oncentrated Loads or Fied Geometr In order to completel constrain cables, the number of unknown support reactions will be more than the available

The tension in the cable IS NOT the same everwhere, but the horizontal component in a cable segment WILL E.

4 Materials range Metal on ice Metal on metal Metal on wood Metal on stone Wood on wood Steel on steel 0.75 Stone on stone Rubber on concrete luminum on aluminum LE STRUTURES: ables with Several oncentrated Loads or Fied Geometr In order to completel constrain cables, the number of unknown support reactions will be more than the available number of equilibrium equations. We can solve because we have additional equations from geometr due to the slope of the cable. The tension in the cable IS NOT the same everwhere, but the horizontal component in a cable segment WILL E. 2 m 4 m 6 m T T 45 kn 45 kn Truss Structures truss is made up of straight two-force members connected at its ends. The triangular arrangement produces stable geometr. Loads on a truss are applied at the joints onl. Joints are pin-tpe connections (resist translation, not rotation). Forces of action and reaction on a joint must be equal and opposite. Members in TENSION are being pulled. Members in OMRESSION are being squeezed. Eternal forces act on the joints. 4

5 Truss configuration: Three members form a rigid assembl with 3 (three) connections. To add members and still have a rigid assembl, 2 (two) more must be added with one connection between. For rigidit: b = 2n 3, where b is number of members and n is number of joints Method of Joints The method takes advantage of the conditions of equilibrium at each joint. 1. etermine support reaction forces. 2. raw a F of each member N each joint 3. Identif geometr of angled members 4. Identif zero force members and other special (eas to solve) cases 5. Each pin is in equilibrium ( F 0 and F 0 for a concurrent force sstem) Tools available: Tip-to-tail method for 3 joint forces must close nalticall, there will be at most 2 unknowns with 2 equilibrium equations. 6. Total equations = 2n = b+3 (one force per member + 3 support reactions) dvantages: an find ever member force isadvantages: Lots of equations, eas to lose track of forces found. F E 5

6 Joint onfigurations (special cases to recognize for faster solutions) ase 1) Two odies onnected or (0) (0) F has to be equal (=) to F ase 2) Three odies onnected with Two odies in Line or or even (0) (0) (0) F and F have to be equal, and F has to be 0 (zero). ase 3) Three odies onnected and a Force 2 odies aligned & 1 od and a Force are ligned Four odies onnected - 2 odies ligned and the Other 2 odies ligned E E F has to equal F, and [F has to equal ] or [F has to equal F E ] 6

7 Graphical nalsis The method utilizes what we know about force triangles and plotting force magnitudes to scale. 1. raw an accurate form diagram of the truss at a convenient scale with the loads and support reaction forces. 2. etermine the support reaction forces. 3. Working clockwise and from left to right, appl interval notation to the diagram, assigning capital letters to the spaces between eternal forces and numbers to internal spaces. 4. onstruct a load line to a convenient scale of length to force b using the interval notation and working clockwise around the truss from the upper left plotting the lengths of the vertical and horizontal loads. 5. Starting at a left joint where we know there are fewer than three forces, we draw reference lines in the direction of the unknown members so that the intersect. Label the intersection with the number of the internal space. 6. Go to the net joint (clockwise and left to right) with two unknown forces and repeat for all joints. The diagram should close. 7. Measure the line segments and appl interval notation to determine their sense: roceeding clockwise around the joint, follow the notation. The direction toward the joint is compressive. The direction awa from the joint is tensile. Eample 1 (pg 72 & 77) Using the method of joints, determine all member forces

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9 Eample 2 Using the method of joint, determine all member forces. SOLUTION: Find the joints with 2 (or less unknowns) for F s : and H, while looking for an special cases like E, which has crossed members and forces. FE = FEF and FE = 500 (tension). (heck off members found:,,,,, E, E, EF, G, F, FG, GH, FH) LS 0 LS 500 LS 200 LS LS Let s use first (but H is just as acceptable). raw the point, adding the known force, and draw the unknown member forces awa from the point, assuming tension (shown as dashed). Find the geometr of member from the rise of 10 ft and the run of ft. The hpotenuse will be = 18.03: F F F F F 0 F =(-412.5)*18.03/10 = () and substituting the (negative) value of F into the F, F = (T) (heck off members found:,,,,, E, E, EF, G, F, FG, GH, FH) Review which joints have 2 (or less) unknowns: and H. Let s use. ecause we know F is in compression we draw the force into the point. We need the geometr of member with a rise of 5 (-10) and a run of with a 2 2 hpotenuse of 5 =.81: F F 0 F = () F F F 0 (substituting the negative value of F ) F = (T) (heck off members found:,,,,, E, E, EF, G, F, EF, FG, GH, FH) Review which joints have 2 (or less) unknowns: and H. Let s use. oth F and F are tensile, so we can draw them awa. The geometr of E is familiar with the rise the same as the run for an angle of 45 : F F cos 45 FE 0 F F sin 45 0 F = () and substituting the (negative) value of F into the F, FE = (T) = FEF (! from above) (heck off members found:,,,,, E, E, EF, G, F, FG, GH, FH) Review which joints have 2 (or less) unknowns: and H. Let s use. raw F and F as compressive forces. nd the geometr has been found due to smmetr, with the angle of FF a negative 45 : F cos 45 FF cos( 45) FG F sin FF sin( 45) FG Solve simultaneousl because there isn t an eas wa to find one unknown equal to a value multiplied b the other unknown: F F 0.949F 0 F F G F 0.316F 0 F G LS LS add: F F 0.633FG 0 FG = () and substituting, FF = () (heck off members found:,,,,, E, E, EF, G, F, FG, GH, FH) FE LS 10 FE 500 LS F F F LS FE 0 LS 79.5 LS LS LS 45 F FG FF FEF F

10 Eample 2 (continued) Review which joints have 2 (or less) unknowns: G, F and H. Let s use F (because H reall looks like mirrored). raw FF as compressive and FEF in tension. The angle from for FF is negative 45 : F cos( 45) FFH 0 FFH = (T) F 27.6 sin( 45) 200 FFG 0 FFG = (T) (heck off members found: LS,,,,, E, E, EF, G, F, FG, GH, FH) 0 LS 500 LS 200 LS LS Review which joints have 2 (or less) unknowns; which are G and H. Let s use G and pretend that we have onl found FGF (and not FG) in order to show a set of equations that use substitution with the algebra. The geometr has been found due to smmetr: F FG FGH 0 rearranging: FG FGH FGH F FG FGH Substituting: F ( FGH ) FGH Simplifing 0.277FGH FGH = () and FG = () (which validates the earlier answer found of () with respect to rounding errors in all fractions and trig functions) (heck off members found:,,,,, E, E, EF, G, F, FG, GH, FH) (Tpicall, the last joint left will verif that the joint is in equilibrium with values found, but in this eercise the last joint was used to show the algebraic method of substitution.) 27.6 LS FFG LS 200 LS FG FFH LS FGH 10 10

ARCH 614 Note Set 2 S2011abn. Forces and Vectors

ARCH 614 Note Set 2 S2011abn. Forces and Vectors orces and Vectors Notation: = name for force vectors, as is A, B, C, T and P = force component in the direction = force component in the direction h = cable sag height L = span length = name for resultant