PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 5.12 The bent rod in the figure is supported at A by a journal bearing, at D by a ball-and-socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned on the shaft. In other words, no couple moments are required at this support. CHAPTER 06 STRUCTURAL ANALYSIS CHAPTER OBJECTIVES ▪ ▪ To show how to determine the forces in the members of a truss using the method of joints and the method of sections. To analyze the forces acting on the members of frames and machines composed of pinconnected members. 6.1 SIMPLE TRUSS A truss is a structure composed of slender members joined together at their end points. PUP Civil Engineering Department #WeLearnAsOne . . Assumptions for Design • All loadings are applied at the joints. • The members are joined together by smooth pins. • Because of these two assumptions, each truss member will act as a two-force member, and therefore the force acting at each end of the member will be directed along the axis of the member. If the force tends to elongate the member, it is a tensile force (T); whereas if it tends to shorten the member, it is a compressive force (C). 6.2 THE METHOD OF JOINTS • This method is based on the fact that if the entire truss is in equilibrium, then each of its joints is also in equilibrium. PUP Civil Engineering Department #WeLearnAsOne • . . If the free-body diagram of each joint is drawn, the force equilibrium equations can then be used to obtain the member forces acting on each joint. Procedure for Analysis • Draw the free-body diagram of a joint having at least one known force and at most two unknown forces. • Assume force to be either in tension or compression. • Orient the x and y axes such that the forces on the free-body diagram can be easily resolved into their x and y components and then apply the two force equilibrium equations ƩFx = 0 and ƩFy = 0. Solve for the two unknown member forces and verify their correct sense. • Using the calculated results, continue to analyze each of the other joints. Be sure to choose a joint having at most two unknowns and at least one known force. SAMPLE PROBLEM | 6.1 Determine the force in each member of the truss shown. Indicate whether the members are in tension or compression. 6.3 ZERO FORCE MEMBERS • Truss analysis using the method of joints is greatly simplified if we can first identify those members which support no loading. These zero-force members are used to increase the stability of the truss during construction and to provide added support if the loading is changed. PUP Civil Engineering Department #WeLearnAsOne • • • . . The zero-force members of a truss can generally be found by inspection of each of the joints. If only two non-collinear members form a truss joint and no external load or support reaction is applied to the joint, the two members must be zero-force members. If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction has a component that acts along this member. SAMPLE PROBLEM | 6.2 Using the method of joints, determine all the zero-force members of the Fink roof truss shown. Assume all joints are pin connected. PUP Civil Engineering Department #WeLearnAsOne . . 6.4 THE METHOD OF SECTIONS • • When we need to find the force in only a few members of a truss, we can analyze the truss using the method of sections. It is based on the principle that if the truss is in equilibrium then any segment of the truss is also in equilibrium. PUP Civil Engineering Department #WeLearnAsOne . . Procedure for Analysis Free-Body Diagram • Make a decision on how to “cut” or section the truss through the members where forces are to be determined. • Before isolating the appropriate section, it may first be necessary to determine the truss’s support reactions. If this is done then the three equilibrium equations will be available to solve for member forces at the section. • Draw the free-body diagram of that segment of the sectioned truss which has the least number of forces acting on it. • Use one of the two methods described above for establishing the sense of the unknown member forces. Equations of Equilibrium • Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces, so that the third unknown force can be determined directly from the moment equation. • If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these unknowns to determine directly the third unknown force. SAMPLE PROBLEM | 6.3 Determine the force in members GE, GC, and BC of the truss shown. Indicate whether the members are in tension or compression. SAMPLE PROBLEM | 6.4 Determine the force in member CF of the truss shown. Indicate whether the member is in tension or compression. Assume each member is pin connected. PUP Civil Engineering Department #WeLearnAsOne . . 6.6 FRAMES AND MACHINES • • Frames and machines are two types of structures which are often composed of pinconnected multiforce members, i.e., members that are subjected to more than two forces. Frames are used to support loads, whereas machines contain moving parts and are designed to transmit and alter the effect of forces. In order to determine the forces acting at the joints and supports of a frame or machine, the structure must be disassembled and the free-body diagrams of its parts must be drawn. SAMPLE PROBLEM | 6.5 For the frame shown, draw the free-body diagram of (a) each member, (b) the pins at B and A, and (c) the two members connected together. Procedure for Analysis Free-Body Diagram • Draw the free-body diagram of the entire frame or machine, a portion of it, or each of its members. The choice should be made so that it leads to the most direct solution of the problem. • Identify the two-force members. Remember that regardless of their shape, they have equal but opposite collinear forces acting at their ends. • When the free-body diagram of a group of members of a frame or machine is drawn, the forces between the connected parts of this group are internal forces and are not shown on the free-body diagram of the group. Equations of Equilibrium • Count the number of unknowns and compare it to the total number of equilibrium equations that are available. In two dimensions, there are three equilibrium equations that can be written for each member. • Sum moments about a point that lies at the intersection of the lines of action of as many of the unknown forces as possible. • If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reverse of that shown on the free-body diagram. SAMPLE PROBLEM | 6.6 Determine the horizontal and vertical components of force which the pin at C exerts on member BC of the frame. PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 6.7 The compound beam shown is pin connected at B. Determine the components of reaction at its supports. Neglect its weight and thickness. SAMPLE PROBLEM | 6.8 The two planks are connected together by cable BC and a smooth spacer DE. Determine the reactions at the smooth supports A and F, and also find the force developed in the cable and spacer. SAMPLE PROBLEM | 6.9 The 75-kg man in the figure attempts to lift the 40-kg uniform beam off the roller support at B. Determine the tension developed in the cable attached to B and the normal reaction of the man on the beam when this is about to occur. PUP Civil Engineering Department #WeLearnAsOne . . CHAPTER 08 FRICTION CHAPTER OBJECTIVES ▪ ▪ To introduce the concept of dry friction and show how to analyze the equilibrium of rigid bodies subjected to this force. To present specific applications of frictional force analysis on wedges, screws, belts, and bearings. 8.1 CHARACTERISTICS OF DRY FRICTION • • • Friction is a force that resists the movement of two contacting surfaces that slide relative to one another. This force always acts tangent to the surface at the points of contact and is directed so as to oppose the possible or existing motion between the surfaces. We will study the effects of dry friction, which is sometimes called Coulomb friction since its characteristics were studied extensively by the French physicist CharlesAugustin de Coulomb in 1781. Dry friction occurs between the contacting surfaces of bodies when there is no lubricating fluid. Theory of Dry Friction • Equilibrium. PUP Civil Engineering Department #WeLearnAsOne . . • Impending Motion. In cases where the surfaces of contact are rather “slippery,” the frictional force F may not be great enough to balance P, and consequently the block will tend to slip. • Motion. If the magnitude of P acting on the block is increased so that it becomes slightly greater than Fs, the frictional force at the contacting surface will drop to a smaller value Fk, called the kinetic frictional force. • The above effects regarding friction can be summarized by referring to the graph below which shows the variation of the frictional force F versus the applied load P. Characteristics of Dry Friction As a result of experiments that pertain to the foregoing discussion, we can state the following rules which apply to bodies subjected to dry friction. PUP Civil Engineering Department #WeLearnAsOne • • • • • . . The frictional force acts tangent to the contacting surfaces in a direction opposed to the motion or tendency for motion of one surface relative to another. The maximum static frictional force Fs that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to severely deform or crush the contacting surfaces of the bodies. The maximum static frictional force is generally greater than the kinetic frictional force for any two surfaces of contact. However, if one of the bodies is moving with a very low velocity over the surface of another, Fk becomes approximately equal to Fs, i.e., s ≈ k. When slipping at the surface of contact is about to occur, the maximum static frictional force is proportional to the normal force, such that F s = sN. When slipping at the surface of contact is occurring, the kinetic frictional force is proportional to the normal force, such that F k = kN. 8.2 PROBLEMS INVOLVING DRY FRICTION Types of Friction Problems • No Apparent Impending Motion. • Impending Motion at All Points of Contact. • Impending Motion at Some Points of Contact. SAMPLE PROBLEM | 8.1 The uniform crate shown has a mass of 20 kg. If a force P = 80 N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is s = 0.3. SAMPLE PROBLEM | 8.2 It is observed that when the bed of the dump truck is raised to an angle of θ = 25° the vending machines will begin to slide off the bed. Determine the static coefficient of friction between a vending machine and the surface of the truckbed. PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 8.3 The uniform 10-kg ladder rests against the smooth wall at B, and the end A rests on the rough horizontal plane for which the coefficient of static friction is s = 0.3. Determine the angle of inclination θ of the ladder and the normal reaction at B if the ladder is on the verge of slipping. SAMPLE PROBLEM | 8.4 Beam AB is subjected to a uniform load of 200 N/m and is supported at B by post BC. If the coefficients of static friction at B and C are B = 0.2 and C = 0.5, determine the force P needed to pull the post out from under the beam. Neglect the weight of the members and the thickness of the beam. 8.3 WEDGES • A wedge is a simple machine that is often used to transform an applied force into much larger forces, directed at approximately right angles to the applied force. Wedges also can be used to slightly move or adjust heavy loads. SAMPLE PROBLEM | 8.5 The uniform stone has a mass of 500 kg and is held in the horizontal position using a wedge at B. If the coefficient of static friction is s = 0.3 at the surfaces of contact, determine the minimum force P needed to remove the wedge. Assume that the stone does not slip at A. PUP Civil Engineering Department #WeLearnAsOne . . 8.4 FRICTIONAL FORCES ON FLAT BELTS • Whenever belt drives or band brakes are designed, it is necessary to determine the frictional forces developed between the belt and its contacting surface. Frictional Analysis Since dθ is of infinitesimal size, sin(dθ/2) = dθ/2 and cos(dθ/2) = 1. Also, the product of the two infinitesimals dT and dθ/2 may be neglected when compared to infinitesimals of the first order. As a result, these two equations become Eliminating dN yields PUP Civil Engineering Department #WeLearnAsOne . . Integrating this equation between all the points of contact that the belt makes with the drum, and noting that T = T1 at θ = 0 and T = T2 at θ = b, yields Solving for T2, we obtain SAMPLE PROBLEM | 8.6 The maximum tension that can be developed in the cord is 500 N. If the pulley at A is free to rotate and the coefficient of static friction at the fixed drums B and C is s = 0.25, determine the largest mass of the cylinder that can be lifted by the cord. CHAPTER 09 CENTER OF GRAVITY AND CENTROID CHAPTER OBJECTIVES ▪ To discuss the concept of the center of gravity, center of mass, and the centroid. PUP Civil Engineering Department #WeLearnAsOne ▪ ▪ . . To show how to determine the location of the center of gravity and centroid for a body of arbitrary shape and one composed of composite parts. To use the theorems of Pappus and Guldinus for finding the surface area and volume for a body having axial symmetry. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY • Knowing the resultant or total weight of a body and its location is important when considering the effect this force produces on the body. The point of location is called the center of gravity, and in this section we will show how to find it for an irregularly shaped body. We will then extend this method to show how to find the body’s center of mass, and its geometric center or centroid. Center of Gravity • A body is composed of an infinite number of particles of differential size, and so if the body is located within a gravitational field, then each of these particles will have a weight dW. • These weights will form a parallel force system, and the resultant of this system is the total weight of the body, which passes through a single point called the center of gravity, G. • Consider the rod shown, • Consider the plate shown, PUP Civil Engineering Department #WeLearnAsOne • . . Consider the three-dimensional body shown, Center of Mass of a Body • In order to study the dynamic response or accelerated motion of a body, it becomes important to locate the body’s center of mass Cm. This location can be determined by substituting dW = g dm. • Provided g is constant, it cancels out, and so PUP Civil Engineering Department #WeLearnAsOne . . Centroid of a Volume • If the body is made from a homogeneous material, then its density ρ (rho) will be constant. Therefore, a differential element of volume dV has a mass dm = ρ dV. • Substituting and canceling out ρ, we obtain formulas that locate the centroid C or geometric center of the body Centroid of an Area • If an area lies in the x–y plane and is bounded by the curve y = f(x), then its centroid will be in this plane and can be determined from the integrals PUP Civil Engineering Department #WeLearnAsOne • . . These integrals can be evaluated by performing a single integration if we use a rectangular strip for the differential area element. Centroid of a Line • If a line segment (or rod) lies within the x–y plane and it can be described by a thin curve y = f(x), then its centroid is determined from • The length of the differential element is given by the Pythagorean theorem, dL = √(dx) 2 + (dy)2, which can also be written in the form SAMPLE PROBLEM | 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc as shown. SAMPLE PROBLEM | 9.2 Locate the centroid of the circular wire segment shown. PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 9.3 Determine the distance y measured from the x axis to the centroid of the area of the triangle shown. SAMPLE PROBLEM | 9.4 Locate the centroid for the area of a quarter circle shown. PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 9.5 Locate the centroid of the area shown. SAMPLE PROBLEM | 9.6 Locate the y centroid for the paraboloid of revolution as shown. SAMPLE PROBLEM | 9.7 Determine the location of the center of mass of the cylinder shown if its density varies directly with the distance from its base, i.e., ρ = 200z kg/m3. PUP Civil Engineering Department #WeLearnAsOne . . 9.2 COMPOSITE BODIES • • • • • A composite body consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular, semicircular, etc. Such a body can often be sectioned or divided into its composite parts and, provided the weight and location of the center of gravity of each of these parts are known, we can then eliminate the need for integration to determine the center of gravity for the entire body. Rather than account for an infinite number of differential weights, we have instead a finite number of weights. When the body has a constant density or specific weight, the center of gravity coincides with the centroid of the body. The centroid for composite lines, areas, and volumes can be found using relations analogous to the previous equations; however, the W’s are replaced by L’s, A’s, and V’s, respectively. PUP Civil Engineering Department #WeLearnAsOne • . . Centroids for common shapes of lines, areas, shells, and volumes that often make up a composite body are given in the following tables. (or look at the inside back cover of the book) Geometric Properties of Line and Area Elements PUP Civil Engineering Department #WeLearnAsOne Center of Gravity and Mass Moment of Inertia of Homogenous Solids . . PUP Civil Engineering Department #WeLearnAsOne SAMPLE PROBLEM | 9.8 Locate the centroid of the wire shown. SAMPLE PROBLEM | 9.9 Locate the centroid of the plate area shown. . . PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 9.10 Locate the center of mass of the assembly shown. The conical frustum has a density of ρc = 8 Mg/m3, and the hemisphere has a density of ρh = 4 Mg/m3. There is a 25-mm-radius cylindrical hole in the center of the frustum. 9.3 THEOREMS OF PAPPUS AND GULDINUS • • The two theorems of Pappus and Guldinus are used to find the surface area and volume of any body of revolution. They were first developed by Pappus of Alexandria during the fourth century A.D. and then restated at a later time by the Swiss mathematician Paul Guldin or Guldinus (1577– 1643). First Theorem : Surface Area • If we revolve a plane curve about an axis that does not intersect the curve we will generate a surface area of revolution. • Therefore the first theorem of Pappus and Guldinus states that the area of a surface of revolution equals the product of the length of the generating curve and the distance traveled by the centroid of the curve in generating the surface area. PUP Civil Engineering Department #WeLearnAsOne . . Second Theorem : Volume • A volume can be generated by revolving a plane area about an axis that does not intersect the area. • Therefore the second theorem of Pappus and Guldinus states that the volume of a body of revolution equals the product of the generating area and the distance traveled by the centroid of the area in generating the volume. SAMPLE PROBLEM | 9.11 Show that the surface area of a sphere is A = 4R2 and its volume is V = (4/3)R3. CHAPTER 10 MOMENTS OF INERTIA CHAPTER OBJECTIVES ▪ To develop a method for determining the moment of inertia for an area. PUP Civil Engineering Department #WeLearnAsOne ▪ ▪ . . To introduce the product of inertia and show how to determine the maximum and minimum moments of inertia for an area. To discuss the mass moment of inertia. 10.1 DEFINITION OF MOMENTS OF INERTIA FOR AREAS Whenever a distributed load acts perpendicular to an area and its intensity varies linearly, the calculation of the moment of the loading about an axis will involve an integral of the form ∫ y 2dA • The integral ∫ y2dA is sometimes referred to as the “second moment” of the area about an axis (the x axis), but more often it is called the moment of inertia of the area. Moment of Inertia • By definition, the moments of inertia of a differential area dA about the x and y axes are dIx = y2 dA and dIy = x2 dA, respectively. • For the entire area A the moments of inertia are determined by integration; i.e., • • We can also formulate this quantity for dA about the “pole” O or z axis. This is referred to as the polar moment of inertia. It is defined as dJO = r2 dA, where r is the perpendicular distance from the pole (z axis) to the element dA. For the entire area the polar moment of inertia is PUP Civil Engineering Department #WeLearnAsOne • • • . . This relation between JO and Ix, Iy is possible since r2 = x2 + y2. From the above formulations it is seen that Ix, Iy, and JO will always be positive since they involve the product of distance squared and area. Furthermore, the units for moment of inertia involve length raised to the fourth power, e.g., m4, mm4, or ft4, in4. 10.2 PARALLEL-AXIS THEOREM FOR AN AREA • The parallel-axis theorem can be used to find the moment of inertia of an area about any axis that is parallel to an axis passing through the centroid and about which the moment of inertia is known. • The form of each of the three equations states that the moment of inertia for an area about an axis is equal to its moment of inertia about a parallel axis passing through the area’s centroid plus the product of the area and the square of the perpendicular distance between the axes. 10.3 RADIUS OF GYRATION OF AN AREA • • The radius of gyration of an area about an axis has units of length and is a quantity that is often used for the design of columns in structural mechanics. Provided the areas and moments of inertia are known, the radii of gyration are determined from the formulas PUP Civil Engineering Department #WeLearnAsOne • . . The form of these equations is easily remembered since it is similar to that for finding the moment of inertia for a differential area about an axis. For example, Ix = kx2A; whereas for a differential area, dIx = y2 dA. SAMPLE PROBLEM | 10.1 Determine the moment of inertia for a rectangular area, of base b and height h, with respect to (a) the centroidal x’ axis, (b) the axis x b passing through the base of the rectangle, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C. SAMPLE PROBLEM | 10.2 Determine the moment of inertia for the shaded area shown about the x axis. SAMPLE PROBLEM | 10.3 Determine the moment of inertia with respect to the centroidal x axis for a circular area of radius r. 10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS • • A composite area consists of a series of connected “simpler” parts or shapes, such as rectangles, triangles, and circles. Provided the moment of inertia of each of these parts is known or can be determined about a common axis, then the moment of inertia for the composite area about this axis equals the algebraic sum of the moments of inertia of all its parts. PUP Civil Engineering Department #WeLearnAsOne Geometric Properties of Line and Area Elements . . PUP Civil Engineering Department #WeLearnAsOne . . SAMPLE PROBLEM | 10.4 Determine the moment of inertia of the area shown about the x axis. SAMPLE PROBLEM | 10.5 Determine the moments of inertia for the cross-sectional area of the member shown about the x and y centroidal axes. SAMPLE PROBLEM | 10.6 Determine the centroidal moment of inertia I x’ and Iy’ of the shaded area shown.
