ARCH 631 Note Set 2.1 F2010abn. Statics Primer

Transcription

1 RCH 631 Note Set.1 F010abn Statics Primer Notation: a = name for acceleration = area (net = with holes, bearing = in contact, etc...) (C) = shorthand for compression d = perpendicular distance to a force from a point d = difference in the direction between an area centroid ( ) and the centroid of the composite shape ( ˆ ) d = difference in the direction between an area centroid ( ) and the centroid of the composite shape ( ŷ ) F = name for force vectors, as is,, C, T and P F = force component in the direction F = force component in the direction g = acceleration due to gravit h = name for height I = moment of inertia about the centroid I = moment of inertia with respect to an -ais I = moment of inertia with respect to a -ais L = beam span length m = name for mass M = moment due to a force or internal bending moment N = name for normal force to a surface p = pressure Q = first moment area about an ais (using distances) Q = first moment area about an ais (using distances) R = name for resultant vectors R = resultant component in the direction R = resultant component in the direction tail = start of a vector (without arrowhead) tip = direction end of a vector (with arrowhead) (T) = shorthand for tension V = internal shear force w = name for distributed load w s(elf) w(t) = name for distributed load from self weight of member W = name for force due to weight = ais direction or algebra variable = the distance in the direction from a reference ais to the centroid of a shape = ais direction or algebra variable = the distance in the direction from a reference ais to the centroid of a shape = angle, in math = angle, in math = angle, in math = coefficient of static friction = angle, in a trig equation, e. sin, that is measured between the ais and tail of a vector = summation smbol Newton s Laws of Motion Newton s laws govern the behavior of phsical bodies, whether at rest or moving: First Law. particle originall at rest, or moving in a straight line with constant velocit, will remain in this state provided the particle is not subjected to an unbalanced force. Second Law. particle of mass m acted upon b an unbalanced force eperiences an acceleration that has the same direction as the force and a magnitude that is directl proportional to the force. This is epressed mathematicall as: F ma, 1

2 RCH 631 Note Set.1 F010abn where F and a are vector (directional) quantities, and m is a scalar quantit. Third Law. The mutual forces of action and reaction between two particles are equal, opposite, and collinear. Units Units are necessar to define quantities. Standards eist to relate quantities in a convention sstem, such as the International Sstem of Units (SI) or the U.S. Customar sstem. Units Mass Length Time Force SI kg m s kg m N s bsolute English lb ft s lbft Poundal s Technical lbf s English slug ft ft s lb force Engineering English lb ft s lb force lb 3 17 ft force lb( mass ). s gravitational constant conversions (pg. vii) g 3.17ft c s (English) F=mg g c 9.81m s (SI) 1in 5. 4mm 1lb N Conversions Conversion of a quantit from a categor within a unit sstem to a more useful categor or to another unit sstem is ver common. Tables of conversion can be found in most phsics, statics and design tets. Numerical ccurac Depends on 1) accurac of data ou are given ) accurac of the calculations performed The solution CNNOT be more accurate than the less accurate of #1 and # above! DEFINITIONS: precision the number of significant digits accurac the possible error

3 RCH 631 Note Set.1 F010abn relative error Relative error measures the degree of accurac: 100 degree of measurement For engineering problems, accurac rarel is less than 0.%. accurac (% ) Math for Structures 1. Parallel lines never intersect.. Two lines are perpendicular (or normal) when the intersect at a right angle = Intersecting (or concurrent) lines cross or meet at a point. 4. If two lines cross, the opposite angles are identical: 5. If a line crosses two parallel lines, the intersection angles with the same orientation are identical: 6. If the sides of two angles are parallel and intersect in the same fashion, the angles are identical. 7. If the sides of two angles are parallel, but intersect in the opposite fashion, the angles are supplementar: + = If the sides of two angles are perpendicular and intersect in the same fashion, the angles are identical. 9. If the sides of two angles are perpendicular, but intersect in the opposite fashion, the angles are supplementar: + = If the side of two angles bisects a right angle, the angles are complimentar: + γ = If a right angle bisects a straight line, the remaining angles are complimentar: + γ = The sum of the interior angles of a triangle = For a right triangle, that has one angle of 90, the sum of the other angles = For a right triangle, the sum of the squares of the sides equals the square of the hpotenuse: C C 15. Similar triangles have identical angles in the same orientation. Their sides are related b: Case 1: Case : D C C E C 3 D C E C DE C C C C

4 RCH 631 Note Set.1 F010abn 16. For right triangles: sin oppositeside hpotenuse sin C cos tan adjacentside hpotenuse oppositeside adjacent side cos tan C C C C (SOHCHTO) 17. If an angle is greater than 180 and less than 360, sin will be less than 0. If an angle is greater than 90 and less than 70, cos will be less than 0. If an angle is greater than 90 and less than 180, tan will be less than 0. If an angle is greater than 70 and less than 360, tan will be less than LW of SINES (an triangle) sin sin 19. LW of COSINES (an triangle) sin C C C C cos 0. Surfaces or areas have dimensions of width and length and units of length squared (e. in or inches inches). 1. Solids or volumes have dimension of width, length and height or thickness and units of length cubed (e. m 3 or m m m). Force is defined as mass times acceleration. So a weight due to a mass is accelerated upon b gravit: F = mg g 9.81 m ft 3. sec 17 sec 3. Weight can be determined b volume if the unit weight or densit is known: W = V where V is in units of length 3 and is in units of force/unit volume 4. lgebra: If ab= cd then it can be rewritten: a b k c d k add a constant c d a b switch sides c d a b divide both sides b b a d divide both sides b bc c b 4

5 RCH 631 Note Set.1 F010abn 5. Cartesian Coordinate Sstem Y. (,) - coordinates 6. Solving equations with one unknown: O -origin X 1 st order polnomial: a b 0 1 b a nd order polnomial a bc ( a 1, b 0, c 1) b b 4ac a 0 0 4( 1) 1 1 two answers (radical cannot be negative) 7. Solving linear equations simultaneousl: One equation consisting onl of variables can be rearranged and then substituted into the second equation: e: add 3 to both sides to rearrange divide both sides b substitute into the other equation 3 4( 5 ) add like terms 7 5 simplif Equations can be added and factored to eliminate one variable: e: multipl both sides b and add simplif 1 put =1 in an equation for simplif

6 RCH 631 Note Set.1 F010abn 8. Derivatives of polnomials constant d 0 d d d d d d d 1 a a 3 d d 3 9. The minimum and maimum of a function can be found b setting the derivative = 0 and solving for the unknown variable. 30. Calculators (and software) process equations b an order of operations, which tpicall means the process functions like eponentials and square roots before simpler functions such as + or -. E SURE to specif with parenthesis what order ou want, or ou ll get the wrong answers. It is also important to have degrees set in our calculator for trig functions. For instance, Ecel uses for sign (like -1) first, then will process eponents and square roots, times and divide, followed b plus and minus. If ou tpe 410^ and reall mean (410)^ ou will get an answer of 400 instead of General Procedure for nalsis 1. Inputs GIVEN: Outputs FIND: Critical Path SOLUTION on graph paper. Draw simple diagram of bod/bodies & forces acting on it/them. 3. Choose a reference sstem for the forces. 4. Identif ke geometr and constraints. 5. Write the basic equations for force components. 6. Count the equations & unknowns. 7. SOLVE 8. Feel the validit of the answer. (Use common sense. Check units ) 6

7 RCH 631 Note Set.1 F010abn Eample: Two forces, &, act on a particle. What is the resultant? 1. GIVEN: Two forces on a particle and a diagram with size and orientation FIND: The resultant of the two forces SOLUTION:. Draw what ou know (the diagram, an other numbers in the problem statement that could be put on the drawing.) 3. Choose a reference sstem. What would be the easiest? Cartesian, radian? 4. Ke geometr: the location of the particle as the origin of all the forces Ke constraints: the particle is free in space 5. Write equations: sizeof sizeof sizeof sin sizeof 6. Count: Unknowns:, magnitude and direction Equations: can solve 7. Solve: graphicall or with equations sizeof resultant 8. Feel : Is the result bigger than and bigger than? Is it in the right direction? (like & ) Forces Forces are vectors, which means the have a direction, size and point or line of application. Eternal forces can be moved along the line of action b the law of transmissibilit. Internal forces are within a material or at a connection between elements. R Force sstems can be classified as concurrent, collinear, coplanar, coplanarparallel, or space. F ecause the are vector quantities, the cannot be simpl added. The must be graphicall added or analticall added b resolving forces into components using trigonometr and summed. is: between & F F = Fcos F = Fsin F = tan = F F F F F R F, R F and R R R F F F F F tan R R P 7

8 RCH 631 Note Set.1 F010abn Tpes of Forces Forces can be classified as concentrated at a point or distributed over a length or area. Uniforml distributed loads are quite common and have units of lb/ft or N/m. The total load is commonl wanted from the distribution, and can be determined based on an area calculation with the load value as the height. Equivalent force sstems are the reorganization of the loads in a sstem so there is a equivalent force put at the same location that would cause the same translation and rotation (see Moments). w W W w 0 w W W/ w w W W/ w / / /3 /3 / /6 /3 To determine a distributed load due to hdrostatic pressure, the height of the water, h, is multiplied b the material densit, (6.4 lb/ft 3 ): p = h. To determine a weight of a beam member per length, the cross section area,, is multiplied b the material densit, (e. concrete = 150 lb/ft 3 ): w s.w. =. (Care must be taken with units.) Friction Friction is a resulting force from the contact of two materials and a normal force. It can be static or kinematic. Static friction is defined as the product of the normal force, N, with the coefficient of static friction,, which is a constant dependant upon the materials in contact: F μ N Moments Moments are the tendenc of forces to cause rotation and are vector quantities with rotational direction. Most phsics tets define positive rotation as counter clockwise. With the sign convention, moments can be added. F Moments are defined as the product of the force magnitude, F, with the perpendicular distance from the point of interest to the line of action of the force, d : M = F d d Moment couples can be identified with forces of equal size in opposite direction that are parallel. The equations is M = F d where F is the size of one of the forces. 8

9 RCH 631 Note Set.1 F010abn Support Conditions Reaction forces and moments occur at supports for structural elements. The force component directions and moments are determined b the motion that is resisted, for eample no rotation will mean a reaction moment. Supports are commonl modeled as these general tpes, with the drawing smbols of triangles, circles and ground: Structural nalsis, 4 th ed., R.C. Hibbeler 9

10 RCH 631 Note Set.1 F010abn Equilibrium Equilibrium is the state when all the eternal forces acting on a rigid bod form a sstem of forces equivalent to zero. There will be no rotation or translation. The forces are referred to as balanced. R F 0 R F 0 ND 0 Equilibrium for a point alread satisfies the sum of moments equal to zero because a force acting through a point will have zero moment from a zero perpendicular distance. This is a ver useful concept to appl when summing moments for a rigid bod. If the point summed about has unknown forces acting through it, that force variable will not appear in the equilibrium equation as an unknown quantit, allowing for much easier algebra. Free od Diagrams 1. Determine the free bod of interest. (What bod is in equilibrium?). Detach the bod from the ground and all other bodies ( free it). 3. Indicate all eternal forces which include: - action on the free bod b the supports & connections - action on the free bod b other bodies M - the weigh effect (=force) of the free bod itself (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 bod not b the bod. 5. Dimensions/angles should be included for moment computations and force computations. 6. Indicate the unknown angles, distances, forces or moments, such as those reactions or constraining forces where the bod is supported or connected. The line of action of an unknown should be indicated on the FD. The sense of direction is determined b the tpe of support. (Cables 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 direction is opposite the assumed direction. DON T CHNGE THE RROWS ON YOUR FD OR SIGNS IN YOUR EQUTIONS. With the 3 equations of equilibrium, there can be no more than 3 unknowns for statics. If there are, and the structure is stable, it means that it is staticall indeterminate and other methods must be used to solve the unknowns. When it is not stable, it is improperl constrained and ma still look like it has 3 unknowns. It will prove to be unsolvable. 10

11 RCH 631 Note Set.1 F010abn Conditions for Equilibrium of a Rigid od 1. Two-force bod: Equilibrium of a bod subjected to two forces on two points requires that those forces be equal and opposite and act in the same line of action.. d F 1 F d F 1 F 1 F F () () (C). Three-force bod: Equilibrium of a bod subjected to three forces on three points requires that the line of action of the forces be concurrent (intersect) or parallel ND that the resultant equal zero. F F 3 F F 3 F 3 F C d 1 C d C F 1 F 1 F 1 () -no () (C) Geometric Properties rea is an important quantit to be calculated in order to know material quantities and to find geometric properties for beam and column cross sections. Charts are available for common mathematical relationships. Centroid For a uniform material, the geometric center of the area is the centroid or center of gravit. It can be determined with calculus. First Moment rea The product of an area with respect to a distance about an ais is called the first moment area, Q. The quantit is useful for shear stress calculations and to determine the moment of inertia. Q d Q d 11

12 RCH 631 Note Set.1 F010abn Geometric Properties of reas rea = bh = b/ = h/ 1 3 I ' 36b h rea = bh b 3 h 3 rea = r = 0 = 0 d 4 I = r 4 I = r 4 /8 I = r 4 I = r 4 rea = r d 8 = 0 = 4r 3 rea = r d 4 = 4r 3 = 4r 3 16 rea = = 0 = 0 ab I = 16ah 3 /175 I = 4a 3 h/15 I = 37ah 3 /100 I = a 3 h/80 rea = 4ah 3 = 0 = rea = = ah 3 3h 5 3a = 3h

13 RCH 631 Note Set.1 F010abn Moment of Inertia The moment of inertia is the second area moment of an area, and is found using calculus. For a composite shape, the moment of inertia can be found using the parallel ais theorem: I I d I I d The theorem states that the sum of the centroid of each composite shape about an ais (subscript aes) can be added but must be added to the second moment area of the shape b the distance between parallel aes (opposite aes direction). Internal Forces If a bod is in equilibrium, it holds that an section of that bod is in equilibrium. Two forcebodies will have internal forces that are in line with the bod (end points), while three-force bodies will see an internal force that will not be aial, in addition to an internal moment called a bending moment. n aial force that is pulling the bod from both ends is referred to as a tensile force, and a force pushing on the bod at both ends is referred to as a compressive force. Cable nalsis Cables can onl see tensile forces. If cables are straight, the are two-force bodies and the geometr of the cable determines the direction of the force. If cables drape (are funcular) b having distributed or gravit loads, the internal vertical force component changes, while the internal horizontal force component does not. Truss nalsis Truss members are assembled such that the pins connecting them are the onl location of forces (internal and eternal). This loading assumption relies on there being no bending in the members, and all truss members are then two-force bodies. P Equilibrium of the joints will onl need to satisf the force components summing to 0 and the force components summing to zero. The member forces will have direction in the geometr of the member. ssuming the unknown forces in tension is represented b drawing arrows awa from the joint. When compression forces are known, the must be drawn in to the point. Equilibrium of the section will onl be possible if the section cut is through three or less members eposing three or less unknown forces. This method relies on the sum of moment equilibrium equation. The member forces are in the direction of the members, and the line of action of those forces runs through the member location in order to find the perpendicular distance. It is helpful to find points of intersection of unknown forces to sum moments. P 13

14 RCH 631 Note Set.1 F010abn Pinned Frame. rch and Compound eam nalsis Connecting or internal pins, mean a frame is made up of multiple bodies, just like a truss. ut unlike a truss, the member will not all be two-force bodies, so there ma be three equations of equilibrium required for each member in an assembl, in addition to the three equations of equilibrium for the entire structure. The force reactions on one side of the pin are equal and opposite those to the other side, so there are onl two unknown component forces per pin. F 1 F F 1 F R 1 M R1 R R R 1 R R R 3 eam nalsis Staticall determinate beams have a limited number of support arrangements for a limit of three unknown reactions. The cantilever condition has a reaction moment. L L L simpl supported (most common) overhang cantilever The internal forces and moment are particularl important for design. The aial force (commonl equal zero) is labeled P, while the transverse force is called shear, V, and the internal moment is called bending moment, M. The sign convention for positive shear is a downward force on a left section cut (or upward force on a right section cut). V The sign convention for positive bending moment M corresponds to a downward deflection (most M common or positive curvature.) That is a counter clockwise moment on a right section cut and a clockwise moment of a left section cut. V Shear and ending Moment Diagrams Diagrams of the internal shear at ever location along the beam and of the internal bending moment are etremel useful to locate maimum quantities to design the beams for. There are two primar methods to construct them. The equilibrium method relies on section cuts over distances and writes epressions based on the variable of distance. These functions are plotted as lines or curves. The semi-graphical method relies on the calculus relationship between the load curve (or load diagram), shear curve, and bending moment curve. If the area under a curve is known, the result in the net plot is a change b the amount of the area. 14

15 RCH 631 Note Set.1 F010abn The location of the maimum bending moment corresponds to the location of zero shear. On the deflected shape of a beam, the point where the shape changes from smile up to frown is called the inflection point. The bending moment value at this point is zero. Semigraphical Method Proceedure: loa w (force/length) 1. Find all support forces. V diagram:. t free ends and at simpl supported ends, the shear will have a zero value. 3. t the left support, the shear will equal the reaction force. 4. The shear will not change in until there is another load, where the shear is reduced if the load is negative. If there is a distributed load, the change in shear is the area under the loading. 5. t the right support, the reaction is treated just like the loads of step t the free end, the shear should go to zero. M diagram: 7. t free ends and at simpl supported ends, the moment will have a zero value. 8. t the left support, the moment will equal the reaction moment (if there is one). 9. The moment will not change in until there is another load or applied moment, where the moment is reduced if the applied moment is negative. If there is a value for shear on the V diagram, the change in moment is the area under the shear diagram. For a triangle in the shear diagram, the width will equal the height w! 10. t the right support, the moment reaction is treated just like the moments of step t the free end, the moment should go to zero. height = V shear width = w V V w 15

16 RCH 631 Note Set.1 F010abn Indeterminate Structures Structures with more unknowns than equations of equilibrium are staticall indeterminate. The number of ecess equations is the degree to which the are indeterminate. Other methods must be used to generate the additional equation. These structures will usuall have three-force bodies, and possibl rigid connections which mean internal aial, shear and bending moment at the members and at the joints. ending moment and shear diagrams can be constructed. P M+ M+ M+ 16

17 RCH 631 Note Set.1 F010abn Eample 1 Determine the resultant vector analticall with the component method. Eample Yes, because the ground will stop the rotation. 17

18 RCH 631 Note Set.1 F010abn Eample 3 18

19 RCH 631 Note Set.1 F010abn Eample 4 U 1800 lb. 400 lb. special case! 19

20 RCH 631 Note Set.1 F010abn Eample 5 0

21 RCH 631 Note Set.1 F010abn Eample m 0.75 m from the SOLUTION: Determine the reactions: 10 kn 6 kn F R 0 R = 0 kn F 10kN ( kn )(3m) R 0 R = 16 kn m M ( 10kN )(.5m) (6kN )(1.5 m) M 0 MR = kn-m Draw the load diagram with the distributed load as given with the reactions. Shear Diagram: R 10 kn I w = kn/m II R R M R *31.5 kn-m Label the load areas and calculate: rea I = (- kn/m )(0.75 m) = -1.5 kn rea II = (- kn/m )(.5 m) = -4.5 kn V = 0 VC = V + rea I = kn = -1.5 kn and VC = VC + force at C = -1.5 kn -10 kn= kn V = VC + rea II = kn -4.5 kn = -16 kn and V = V + force at = -16 kn +16 kn= 0 kn V+ (kn) III -1.5 C IV 16 kn ending Moment Diagram: Label the load areas and calculate: V rea III = (-1.5 kn)(0.75 m)/ = kn-m rea IV = (-11.5 kn)(.5m) = kn-m rea V = ( kn)(.5m)/ = kn-m M = 0 MC = M + rea III = kn-m = kn-m *M+ (kn-m) M = MC + rea IV + rea V = kn-m kn-m kn-m = = kn-m and M = M + moment at = kn-m kn-m = 0 kn-m M R positive bending moment