2/28/2006 Statics ( F.Robilliard) 1

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1 2/28/2006 Statics (.Robilliard) 1

2 Extended Bodies: In our discussion so far, we have considered essentially only point masses, under the action of forces. We now broaden our considerations to extended bodies. These are composed of a large number of point masses, connected together, and distributed in space in some geometrical shape. Initially we will assume our extended bodies to be rigid. Statics is the study of the special case where the forces acting on the extended body are in equilibrium, and the body is stationary. This topic finds application in such areas as the design of bridges, buildings, and other static engineering structures. To analyse these structures, we divide them into a number of components, such as columns and beams. We then consider the equilibrium of each individual component. So the first thing we need to consider, is the set of conditions for an extended rigid body to be in equilibrium. We define equilibrium to mean that the body has no acceleration. We will consider only the Statics case, where its velocity is also zero. 2/28/2006 Statics (.Robilliard) 2

3 Equilibrium: By N2, If a body has zero acceleration, there must be a zero resultant force acting on it. Consider a beam acted upon by two equal, but opposite forces, of magnitude, It is clear, that here, the body is in equilibrium. It is equally clear, here, that the body is not in equilibrium, even though the resultant force acting on it is zero. To distinguish between these two cases, we must introduce the concept of torque. 2/28/2006 Statics (.Robilliard) 3

4 Torque: Torque (also called the moment of a force ) is a measure of the twisting strength of a force about a point, called a pivot. It is the rotational analogue of force. orces cause linear acceleration; torques cause angular acceleration. Say, a force,, acts, at a point Q, on an extended body. We wish to consider the rotational effect of that force, about another point P (the pivot), which is different to Q. Let the displacement PQ = r. r Q The torque, τ, of the force,, about the pivot, P, is defined by: P τ = r x (cross product!) Note: τ is a vector, whose direction is given by the cross product definition. SI Unit: Newton metre = N m 2/28/2006 Statics (.Robilliard) 4

5 P r Q θ cos θ sin θ Magnitude of τ: τ = r x thus - τ = r x = r sin θ = r ( sin θ) Resolve into components perpendicular, and parallel, to r. Magnitude of torque = (length of r) x (component of force perpendicular to r) (the (sin θ) component exerts all the torque; the ( cos θ) component exerts zero torque about the pivot P.) r sin θ P Q r θ r cos θ θ τ = r x = r sin θ = ( r sin θ) Resolve r into components perpendicular, and parallel, to. Magnitude of torque = (perpendicular distance from the pivot point onto the line of action of the force) x (magnitude of the force) (all the torque is from the (r sin θ) component of r; the (r cos θ) component contributes zero torque about the pivot, P.) 2/28/2006 Statics (.Robilliard) 5

6 Magnitude of τ: If r is perpendicular to, we get a simplified expression for torque: P r 90 0 Q or P r Q 90 0 τ = r sinθ = r sin 90 0 = r Magnitude of torque = (magnitude of the force) x (perpendicular distance from the pivot onto the line of the force) τ = r 2/28/2006 Statics (.Robilliard) 6

7 τ = r x Direction of τ: RHR for cross products gives the direction of τ. τ. Say r and are both in the plane of the top surface of a wheel, as shown. If is in +x-direction, and r is in the +zdirection, then τ will be in the +y-direction. Say force,, rotates the wheel, from rest, about its spin axis. If the wheel be grasped, so that the fingers of the right hand point in the direction in which the wheel would rotate, then the thumb points in the direction of torque vector, τ. That is, along the spin axis of the wheel. This is the direction of the angular acceleration of the wheel, caused by the torque. 2/28/2006 Statics (.Robilliard) 7

8 Couple: Two equal, but opposite forces, separated by a perpendicular distance d, is called a couple. d total torque of couple = τ = sum of the torques of the two forces about the mid point between them =.d/2 +.d/2 = d Now that we have defined torque, we are able state the general (and sufficient) conditions, for a rigid body to be in a state of equilibrium. 2/28/2006 Statics (.Robilliard) 8

9 General Conditions for the Equilibrium of a Rigid body: 1 A rigid body, upon which a number of forces act, will be in equilibrium if vector sum of all forces is zero 2. vector sum of all the torques of all forces, about any pivot point in the body, is zero 3 Σ = 0 Σ τ P = 0 or two dimensional problems - Σ x = 0 Σ y = 0 Σ τ P = 0 where x = x-components of the forces and y = y-components of the forces 2/28/2006 Statics (.Robilliard) 9

10 Centre of Gravity: This is the effective point in a body, at which its weight force acts. or symmetrical bodies, the centre of gravity, C, will be the geometric centre of the body. C C C W W W 2/28/2006 Statics (.Robilliard) 10

11 Example: A horizontal foot bridge, of weight 1000 N, and length 8 m, is supported by a pivot at each end. What will be the reactions at the end pivots, A & B, when a walker, W, of weight 700 N, is ¾ way across the bridge? A R 4m 8m 6m 1000N W 700N (2): 8 r = 8200 r = 1025 (N).(3) (3)->(1): R = r = = 675 (N) r B + Let R = normal reaction at A r = normal reaction at B Positive directions are shown. or equilibrium: Σ y = 0 = +R +r (1) Taking torques about A (anticlockwise positive): Στ Α = 0 = +r x x 6 (2) A Answer: 675 N 1025 N 1000N 700N B 2/28/2006 Statics (.Robilliard) 11

12 Example: A beam, HE, of weight 100 N and length 4 m, is hinged to a wall at end H. The beam is held horizontally, by a cable at end E, which is secured to the wall, 3 m above the hinge. ind the tension in the cable, and the reaction of the hinge on the beam. 3m Ry 4m + T θ orces acting on the beam are shown. Resolve tension, T, vertically & horizontally. Positive directions are shown. T sinθ = T.3/5 Σ x = 0 = +Rx - T.4/5....(1) H Rx 100 N T cosθ =T.4/5 50 N 83.3 N Answer: 66.7 N E Σ y = 0 = +Ry + T.3/ (2) Taking torques about H: Στ Η = 0 = +(T.3/5).4 - (100).2.. (3) (3): T = 250/3 = 83.3 N.(4) (4)->(1): Rx = 200/3 = 66.7 N (4)->(2): Ry = 50 N 2/28/2006 Statics (.Robilliard) 12

13 Steps: Notes: 1. Label on the diagram, all forces that act on the body. 2. Resolve all forces in common, mutually perpendicular directions. 3. Choose positive directions. 4. Apply the conditions for equilibrium: Σx = Σy = Στ = Solve for unknowns, and interpret solution. If the line of action of a force passes through a pivot, it has zero torque about that pivot. We can choose any point as our pivot about which to take torques. However, to simplify the solution, it is best to choose a pivot, through which the most unknown forces pass. Thus those unknowns are eliminated from the torque equation, thereby simplifying the solution. At a hinge, there are generally both horizontal and vertical components of reaction. If one body bears on another, there will generally be a normal reaction, R; if there is friction between the surfaces, there will also be a friction force, µr, along the surface. 2/28/2006 Statics (.Robilliard) 13

14 Elasticity: So far, we have assumed that extended bodies are rigid that is, their shape is not changed, when external forces act upon them. In reality, external forces do change shape, and size. The atoms of real bodies are held together by short-range atomic forces. When an external force acts on a body, the distances between atoms can be changed, thereby changing the size, and/or shape of the body. We say that real bodies are elastic. The way in which the body changes will depend on the nature of the distorting forces, and on the elastic properties of the body. The external forces are characterised by the stress they exert on the body. Stress is the external force per unit area acting on the body. Strain is a measure of the resulting deformation of the body. Stress causes strain. 2/28/2006 Statics (.Robilliard) 14

15 Elastic Modulus: If we define stress, and strain, appropriately, and limit ourselves to small deformations of the elastic body, the strain produced is found to be proportional to the stress causing it. The proportionality constant is called the elastic modulus, and depends on the composition of the body, and on the way in which it is deformed. Stress ( Stress) = k ( Strain) where Therefore k constant of k Strain Elastic Modulus proportionality called the Stress Strain Elastic Modulus There are several ways in which a body can be deformed. A particular elastic modulus is defined for each of these modes of deformation.. We will define three elastic moduli 1. Young s Modulus, Y 2. Shear Modulus, S 3. Bulk Modulus, B 2/28/2006 Statics (.Robilliard) 15

16 1.Young s Modulus, Y: This measures the elasticity of a body, when external tensile stress changes its length. Consider a long bar, of cross-section area, A, & unstretched length, L - L A L Tensile forces,, stretch the bar an additional length, L. If the cross-section area, A, of the bar is increased, the extension, L, produced by a given tensile force,, would be reduced. Thus we define the tensile stress as the force per unit cross-section area = /A. Also, for a given tensile stress, a longer bar would extend further than a shorter bar. Thus we define the tensile strain produced as the fractional change in length = ( L)/L. Thus: Youngs Modulus Y Tensile Stress Tensile Strain ( A ) L ( ) 2/28/2006 Statics (.Robilliard) 16 L Y is a constant for a given material

17 Stress-Strain Curve: Say we apply increasing stress to an elastic (say steel) bar, and measure the resulting strain produced. The following graph shows a typical plot of stress against strain: Linear Region: rom O to L, Stress is Q proportional to Strain. Young s modulus is the Stress P gradient of the stress-strain curve in the linear L region elastic limit racture Strain Elastic Limit: In the region OP, the bar extends elastically. That is, if the tensile force is reduced to zero, the bar will return to its original unstretched length, L. However, if a bar is stretched beyond P, it becomes permanently deformed. It will not return to its original length, when the stress is removed. We call P the elastic limit. racture Point: If the stress is increased beyond the elastic limit, eventually, the bar will fracture. This occurs at Q in the figure. Young s modulus is the same for compression of the bar, as for its extension. 2/28/2006 Statics (.Robilliard) 17

18 2.Shear Modulus, S: This measures the elasticity of a body to changes in its shape. x Q h P R A Say the base of a solid block is secured, while its top is skewed to the right, by a force, as illustrated. Let the top surface area of the block = A Let the height of the block = PQ = h Let the sideways displacement of the top of the block = QR = x. Shear Modulus S Shear Stress Shear Strain ( ) A ( x ) h A stress-strain curve for shear deformation, analogous to the tensile case, can be measured. S is a constant, for a given material, in the linear region of this curve. Scissors and metal shears work by applying shear forces, beyond the shear fracture point of the material being cut. 2/28/2006 Statics (.Robilliard) 18

19 Bulk Stress = Pressure: Bulk elasticity has to do with how the volume of a body changes when it is subject to compressive forces on all of its surfaces. It is useful here, to recall the quantity pressure, p. Pressure, p, on a surface is the perpendicular force acting per unit surface area. A Total perpendicular force acting on a surface p = = Total surface area over which the force acts A Pressure is bulk stress. SI units: [p] = N m -2 = Pascal = P When the external pressure on a body is increased, its volume decreases. As long as the changes are small, the fractional change in volume is proportional to the change in pressure causing it. The constant of proportionality between the increase in pressure, and the reduction in fractional volume, is the Bulk modulus, B. 2/28/2006 Statics (.Robilliard) 19

20 3.Bulk Modulus, B: Bulk modulus, B, is a measure of the elasticity of the volume of a block of material to external compressive forces. Consider a block of initial volume, V. Say an V (V- V) additional external force, is applied to each of its six surfaces, each of area, A. That is, an additional external pressure of p = ( )/A acts on all faces of the block Say the additional external forces ( ), cause the volume of the cube to shrink from a volume V to (V-dV). Bulk Modulus B volume stress volume strain ( d ) A dp = ( dv ) ( dv ) V V dp is the increase in applied pressure required to cause a decrease of dv in the volume of the block. Negative sign required, since an increase in (d positive) produces a decrease in V (dv negative). 2/28/2006 Statics (.Robilliard) 20

21 SI Units for Elastic Moduli: Some Elastic moduli: Young s modulus, Y, Shear modulus, S, and Bulk modulus, B, all have the same units = Newton per square metre = N m -2 Some Typical Values: Material Y S B (N m -2 ) (N m -2 ) (N m -2 ) aluminium copper steel Note: Solids have all three moduli. However, liquids and gases only have Bulk moduli - their Young s, and Shear moduli are zero. Note: We will need to use the idea of bulk modulus of a gas, when we study sound, later in this course. 2/28/2006 Statics (.Robilliard) 21

22 Example: A mass of 1 tonne is hung from the end of a vertical steel rod, of square cross section 1 cm x 1 cm, and unstretched length 1 metre. By how much will the rod be stretched? = 1000 g =10 4 N A = 0.01x0.01 m 2 L = 1.0 m = 1000 g =10 4 N rom previous table: Y = 20x10 10 N m- 2 take g = 10 m s -2 Y thus L = = = ( A ) L ( ) L L Y A 10 ( 4 )( 1.0) ( 10 )( ) = mm 3 m Steel bar will stretch an additional 0.5 mm. 2/28/2006 Statics (.Robilliard) 22

23 2/28/2006 Statics (.Robilliard) 23

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