Basic knowledge of mechanics

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(1) The concept of force: Force is a mutual mechanical interaction between objects; this interaction causes a change in the motion state of the objects or leads to their deformation. The effect of a force on an object is determined by three factors: the magnitude of the force, its direction, and the point where it acts. Changing any one of these factors will result in a change in the effect of that force on the object. Force is a vector, which can be represented by a line segment with an arrow, as shown in Figure 3-1. The length of the segment AB represents the magnitude of the force in proportion, while the orientation of the segment and the direction indicated by the arrow represent the direction of the force. The starting point A or the ending point B of the segment indicates the point where the force acts. A set of forces acting on an object is called a force system. If two force systems produce the same effect on the same object, then these two force systems are called equivalent force systems. If the effect of a single force R on an object is the same as the effect of a system of forces on that same object, then force R is called the resultant force of that system of forces, and each force in the system is referred to as a component force of the resultant force R. Finding the resultant force of a known system of forces is called the combination of forces ； On the contrary, determining the component forces from the resultant force is called the decomposition of forces. Forces acting on an object often appear in the following two forms. ① Concentrated force: If the area over which a force acts is very small, it can be approximated as acting at a single point. Such a force is called a concentrated force. As shown in Figure 3-2(a), both gravity G and tension T can be considered as concentrated forces. The unit of concentration is “Newton” (N) or “Kilonewton” (KN). ② distributed load: A force that is spread continuously over a large area or large volume is called a distributed load. If the load is distributed evenly in terms of magnitude, it is called a uniformly distributed load. For example, the weight of a uniform rod with a constant cross-section is a uniformly distributed load, as shown in Figure (b). The magnitude of the uniformly distributed load is expressed by the load intensity q, which represents the amount of load per unit length; its unit is Newtons per meter (N/m). The point of application of the resultant force of a uniformly distributed load is at the midpoint of the loaded section (the dashed line in the figure represents the resultant force Q). Its direction is the same as that of the load intensity q, and its magnitude is equal to the product of the load intensity q and the length l of the loaded section, that is, Q = ql. (2) Concepts of rigid body and equilibrium: Any object will deform to some extent under the action of forces, but in typical engineering problems, the deformation is extremely small and has little impact on the study of an object’s equilibrium, so it can be ignored. An object that does not deform under the action of force is called a rigid body. Of course, rigid bodies actually do not exist; they are merely idealized models of real objects. If an object is at rest or moving in a straight line at a constant speed relative to a reference frame, then the object is said to be in equilibrium. If an object is in equilibrium under the action of a force system, then such a force system is called a balanced force system. (3) Basic properties of force Through long-term practice, humanity has not only established the concept of force but also summarized its various properties, namely the axioms of statics. Axiom 1: Axiom of two-force equilibrium. The sufficient and necessary condition for two forces acting on a rigid body to keep it in equilibrium is that these two forces have equal magnitudes, opposite directions, and act along the same line, as shown in Figure 3-3. An object that is in equilibrium under the action of two forces is called a two-force member. According to the axiom of two-force equilibrium, the two forces acting on such a member must pass through the line connecting the points where these forces act; they must also be equal in magnitude and opposite in direction, regardless of the shape of the member. The axiom of two-force equilibrium applies only to rigid bodies, not to deformable bodies. Just like a rope, it cannot remain in equilibrium when subjected to a pair of equal, opposite, and collinear forces at its ends. Axiom 2: Axiom of equilibrium force systems. Adding or subtracting any equilibrium force system from a force system does not change the effect of the original force system on the rigid body. From axiom 1 and axiom 2, an important consequence can be derived: a force acting at a point on a rigid body can be moved along its line of action within the rigid body without altering the effect of that force on the rigid body. This consequence is known as the principle of transmissibility of force. As shown in Figure 3-4, applying a horizontal force F to push the cart at point A, or applying an equal horizontal force F to pull the cart at point B, can produce the same effect. It can be seen that the three elements of force can be extended to: the magnitude of the force, its direction, and the point of application. Axiom 3: The parallelogram law of forces. Two forces acting at a point on an object have their resultant force also acting at that point. The magnitude and direction of the resultant force are determined by the diagonal of the parallelogram formed by these two forces as adjacent sides, as shown in Figure 3-5. The resultant force R of the two known forces F1 and F2 acting at point A is expressed using vector equations as: R = F1 + F2. Using the parallelogram law of forces, a force can also be broken down into two perpendicular components; this type of decomposition is known as the orthogonal decomposition of forces. Axiom 4: The axiom of action and reaction. The action force and reaction force between two objects are always equal in magnitude, opposite in direction, and lie on the same line; they act on two different objects respectively. It should be noted that the action and reaction forces are distinct from a pair of forces in the axiom of two-force equilibrium. The action force and reaction force act on two different objects, whereas the pair of forces in the axiom of force equilibrium act on the same object. (4) Projection of a force ① Projection of a force on coordinate axes: The projection of a force F on the coordinate axes is defined as the distance between the feet of the perpendiculars drawn from the two ends of the force F to the selected coordinate axes x and y. As shown in Figure 3-6, ab is referred to as the projection of the force F on the x-axis, and is denoted by Fx ； a′b′ is called the projection of the force F on the y-axis, and is denoted by Fy. Fx = Fcosα, Fy = Fsinα. The projections of a force on the coordinate axes are scalar quantities, and their signs are determined as follows: if the direction from the starting point a (a′) to the ending point b (b′) is consistent with the direction of the x (y) axis, then the projection of the force on that axis is positive; otherwise, it is negative. ② The theorem of the projection of the resultant force states that the projection of the resultant force on a coordinate axis is equal to the algebraic sum of the projections of its individual components on the same coordinate axis. This relationship is known as the theorem of the projection of the resultant force; that is, Rx = F1x + F2x + … + Fnx = ∑Fx, and Ry = F1y + F2y + … + Fny = ∑Fy. In the above equations, “∑” is an abbreviation that denotes “algebraic sum”. (5) Resultant force of a planar concurrent force system: If the lines of action of all the forces acting on an object lie in the same plane, such a force system is called a planar force system. In a planar force system, if the lines of action of all forces converge at a single point, such a system is called a planar concurrent force system. When there is a system of forces F1, F2, …, Fn acting at a point in a plane, the projection theorem for resultant forces states that the projections of the resultant force R on the coordinate axes are given by: Rx = F1x + F2x + … + Fnx = ∑Fx, and Ry = F1y + F2y + … + Fny = ∑Fy. By using these two projections of the resultant force on the x and y axes, it is possible to determine both the magnitude and direction of the resultant force R. Magnitude of the resultant force R: Direction of the resultant force R: In the formula, α is the acute angle between the resultant force R and the x-axis; the specific direction of the resultant force R is determined by the signs of ∑Fx and ∑Fy. 3.1.2 Torque and Couple (1) Torque ① Concept of torque As shown in Figure 3-7, when using a wrench to turn a nut, the force F causes the wrench and the nut to rotate around point 0. It is known from experience that the degree to which the nut rotates depends not only on the magnitude of the force F but also on the perpendicular distance d from point 0 to the line of action of force F. Point 0 is called the center of moment (i.e., the center of rotation of the object), and the perpendicular distance d from point 0 to the line of action of force F is called the lever arm. The product of the magnitude of force F and the lever arm d is called torque, denoted as M0(F) = ±Fd (3-1). Torque is used to describe the rotational effect of a force on an object. In a plane, when a force causes an object to rotate, there are two different directions of rotation. To distinguish between these two directions, the sign of torque is defined as follows: when a force causes the object to rotate counterclockwise, the torque is positive; otherwise, it is negative. The unit of torque depends on the units of force and the lever arm; the commonly used units are Newtons·meter (N.m) or kilonewtons·meter (KN.m). From the formula for calculating torque, it can be seen that if the line of action of the force passes through the center of moment, the lever arm is zero; therefore, the torque is also zero. In this case, the force cannot cause the object to rotate around the center of moment. ② The theorem of resultant moment: Suppose a planar force system is composed of forces F1, F2, …, Fn, with a resultant force R. According to the definition of the resultant force, its effect on an object is equal to the sum of the effects of each individual force in the system on that object; therefore, the rotational effect of these forces on the object is also equal to the sum of the rotational effects of each individual force in the system. The rotational effect of a force on an object is measured by torque; therefore, the torque exerted by the resultant force on a point in a plane is equal to the algebraic sum of the torques exerted by each component force of the force system on that point. This principle is known as the theorem of total moment, and it is expressed as: Mo(R) = Mo(F1) + Mo(F2) + … + Mo(Fn) = ∑Mo(F). When calculating moments, if the lever arm can be easily determined, equation (3-1) can be used for direct calculation ； If it is difficult to determine the lever arm, the force can be decomposed orthogonally into two forces (decomposed in the direction that most facilitates finding the lever arm), and the torque can then be calculated using the theorem of resultant torque. (2) Couple ① Concept of a couple: As shown in Figure 3-8, a force system composed of two forces of equal magnitude, opposite directions, and parallel but not coincident lines of action is called a couple, which is commonly represented by (F, F′). The plane in which the two forces of a couple lie is called the plane of action of the couple, while the perpendicular distance d between the lines of action of the two forces is known as the couple arm. A couple exerts only a rotational effect on an object. The rotational effect of a couple on an object can be measured by the couple moment. The product of the magnitude of one of the forces in a couple and the distance between their lines of action is defined as the torque, denoted as M(F, F′) or simply as M. That is, M = M(F, F′) = ±Fd. The sign in this expression is determined as follows: when the couple causes an object to rotate counterclockwise, the torque is positive; otherwise, it is negative. The unit of couple moment is the same as that of torque, which is Newton-meters (N.m) or kilonewton-meters (KN.m). The rotational effect of a couple on an object is determined by three factors: the magnitude of the couple moment, the direction of the couple, and the plane in which the couple acts; these are known as the three elements of a couple. The representation of a couple is shown in Figure 3-9. ② Properties of a couple: A couple has no resultant force; a couple can only be balanced by another couple. The projection of a couple force on any coordinate axis is zero; therefore, a couple does not cause any translational effect on an object, only a rotational effect. The torque exerted by a couple on any point within its plane of action is constant, always equal to the magnitude of the couple moment, and independent of the position of the center of torque. ③ Equivalent conditions for couples: Any couple with the same three elements is an equivalent couple, and they can be substituted for one another. From the equivalent conditions of the aforementioned couple, it can be seen that the couple force can move freely within its plane of action without altering its effect on the object. b While maintaining the magnitude and direction of the couple moment, it is possible to change both the magnitude of the force in the couple and the length of its lever arm, without altering its effect on the object. ④ Composition of planar couple systems: Several couples acting on the same object form a couple system. A system of couple forces acting in the same plane is called a planar couple system. The result of combining a system of planar couples is a single resultant couple, whose moment is equal to the algebraic sum of the moments of the individual couples in the system. That is, M = m1 + m2 + … + mn = ∑m. 3.1.3 Force Analysis (1) Constraints and Reaction Forces In daily life and in engineering practice, it is common to see that an object is unable to move in certain directions due to the constraints imposed by other objects around it. Such an object whose movement is restricted is called a constrained body, while the other objects that impose these restrictions on the constrained body are known as its constraints. For example, a hanging light bulb cannot move downward due to the restriction imposed by the rope; thus, the light bulb is a non-free body, and the rope represents the constraint on the light bulb. Since the constraints limit the motion of the non-free body in certain directions, they exert a force on that non-free body. The force exerted by a constraint on a non-free body is called a constraint reaction force. The forces acting on an object can generally be divided into two categories: one type is the force that causes the object to move or to have a tendency to move; such forces are called driving forces. Examples include gravity, water pressure, wind pressure, spring force, and certain loads acting on the object. Driving forces are usually known ； The other type is the constraint reaction that restricts the motion of an object; such constraint reactions are usually unknown. Since the constraint reaction acts to restrict the motion of the constrained body, its direction is always opposite to the direction of motion of that constrained body, which is the basic principle for determining the direction of the constraint reaction. In practical engineering, objects come in a wide variety of shapes, and the ways in which they come into contact with one another are also diverse. As a result, there are many different types of constraints, but some of them share common characteristics and can be grouped together. The following introduces several common types of constraints in engineering. ① Soft-body constraints: Constraints made up of soft ropes, tape, chains, etc., are called soft-body constraints. The characteristic of a flexible constraint is that it can only withstand tensile forces, not compressive forces; therefore, the direction of the reaction force exerted by such a constraint is always along the center line of the flexible body and away from the non-free body, with its point of action located at the junction between the flexible body and the non-free body. As shown in Figure 3-10, when a crane is used to lift a heavy object, the restraint exerted by the steel cables on the object is a flexible-body restraint, and the object is subjected to the pulling forces T1 and T2 from the steel cables. ② Smooth surface constraint A constraint composed of completely smooth surfaces is called a smooth surface constraint. In practical engineering applications, perfectly smooth surfaces do not exist. However, when the friction between two objects is very small, or when this friction has little impact on the problem being studied, it can be ignored, and the contact surface between the two objects can be considered perfectly smooth. Components such as guide rails and cylinders can thus be regarded as having smooth-surface constraints. A smooth surface constraint can only restrict the motion of a non-free body in the direction of the normal to the contact surface; therefore, the direction of the reaction force exerted by such a constraint is along the normal at the point of contact, passing through that point and pointing toward the non-free body. This force is commonly denoted by N, as shown in Figure 3-11. ③ Fixed hinge constraint: The typical configuration of a hinge joint is shown in Figure 3-12(a). Objects A and B have identical circular holes, and they are connected to each other by a cylindrical pin; this pin prevents relative movement between the two objects but does not restrict their relative rotation. This type of constraint is known as a hinge constraint. When one of the objects connected by a hinge is fixed, it is referred to as a fixed hinge, and its simplified representation is shown in Figures 3-12(b) or (c). For hinge constraints, the line of action of the constraint reaction must pass through the center of the hinge, but its direction is unknown; it is usually represented by a pair of orthogonal components, Nx and Ny, as shown in Figures 3-12(b) or (c). However, for the hinge supports on two-force members, the line of action of the restraining reaction forces lies along the line connecting the centers of the two hinges, as shown in Figure 3-13. ④ A movable hinge constraint, as shown in Figure 3-14(a), involves installing several rollers beneath the hinge support, which are then placed on a supporting surface; this is known as a movable hinge. This type of support can only restrict the movement of an object perpendicular to the supporting surface; it cannot restrict movement along the tangent direction of the supporting surface or rotation around the pin. Therefore, the restraining reaction force of the movable hinge support passes through the center of the hinge and is perpendicular to the supporting surface; Figure 3-14(b) shows a schematic diagram of the movable hinge support. ○5 Fixed-end constraints As shown in Figure 3-15(a), a part of an object that is fixedly embedded within another object to form a constraint is called a fixed-end constraint. Examples include workpieces clamped in a chuck, utility poles inserted into the ground, and balconies in buildings. The fixed-end constraint not only limits the movement of an object but also restricts its rotation; therefore, its restraining force is typically represented by two orthogonal components, Nx and Ny, along with a restraining torque, as shown in Figure 3-15(b). (2) Force diagram: In order to clearly show the forces acting on an object, the object in question is separated from the constraints imposed by surrounding objects, and a simplified diagram of it is drawn. All the forces acting on this object—including applied forces and reaction forces from the constraints—are shown on this diagram. The object after being separated from its constraints is called a free body, and the diagram that shows all the forces acting on such a free body is known as a force diagram. The steps to draw a force diagram are as follows: ① Select a separate body – Choose the object of study based on the given conditions, and draw a simplified diagram of that object as a separate body. It should be noted that the shape and orientation of the separated entity drawn must be the same as those of the original object; they cannot be changed arbitrarily. ② Draw the driving forces. The driving forces are generally known, and it is sufficient to draw them on the object under study based on those known values. ③ Draw the constraint reactions. An object under study is often subject to multiple constraints; to ensure that no constraint reactions are missed, it is necessary first to determine how many constraints exist and what type each constraint is. Then, corresponding constraint reactions should be drawn based on the type of constraints, without drawing them arbitrarily. The following examples illustrate how to draw force diagrams. Example 3.1 A small ball with a weight of G is held by a rope and placed on a smooth inclined plane, as shown in Figure 3-16(a). Try to draw a force diagram of the small ball. Solution: ① Choose the small ball as the object of study and draw its separated parts. ② Draw the primary forces. The primary force acting on the small ball is gravity G, with its point of application at the ball’s center O, acting vertically downward. ③ Draw the constraint reaction forces. The sphere is subject to two types of constraints: one is a flexible constraint provided by a rope, with the constraint force acting at the point of connection between the rope and the sphere, C; this force is directed along the direction of the rope and away from the sphere. The other is a smooth surface constraint formed by an inclined plane; the constraint reaction force acts at the point of contact B between the inclined plane and the ball, in a direction along the normal to the inclined plane and the ball (that is, perpendicular to the inclined plane and pointing toward the center O of the ball). The force diagram of the small ball is shown in Figure 3-16(b). 3.1.4 Equilibrium of Planar Force Systems (1) The theorem of force translation: As shown in Figure 3-17(a), suppose a force F acts at point A on an object. Choosing any point O on the object, to translate force F to point O, two forces F′ and F″ of equal magnitude but opposite direction are applied at point O; their magnitudes are equal to that of F, and their lines of action are parallel to that of F, as shown in Figure 3-17(b). Among the three forces F, F′, and F″, F and F″ form a couple, known as an additional couple. Its moment of force is equal to the moment of force exerted by F about point O, namely M = Mo(F) = Fd. The remaining force F′, having the same magnitude and direction as F, can be considered as the result of translating force F; as shown in Figure 3-17(c), the combined effect of force F′ and the couple (F, F″) is identical to the effect of the original force F on the object. It can be seen that the force acting on an object can be translated to any point within the object; however, after this translation, a couple must be added, whose moment is equal to the moment of the original force about the new point of application. This is the theorem of force translation. The theorem of translation of force applies only to rigid bodies. (2) Simplification of a planar arbitrary force system: In a planar force system, if the lines of action of the forces are distributed arbitrarily, such a system is called a planar arbitrary force system. Suppose a plane arbitrary force system F1, F2, …, Fn acts on an object, as shown in Figure 3-18(a). Any point O within the plane of this force system is selected, and it is called the center of reduction. According to the theorem of force translation, by translating each force in the force system to point O, a planar concurrent force system (F1′, F2′, …, Fn′) and a planar couple system (M1, M2, …, Mn) are obtained, as shown in Figure 3-18(b). A planar system of concurrent forces (F1′, F2′, …, Fn′) can be combined into a resultant force R′. R′ is known as the resultant vector of a planar force system, and its magnitude can be calculated using the following formula: Since translating the forces does not change their projections on the coordinate axes, the resultant vector can be determined based on the projections of the forces before translation. Therefore, the above formula can be written as follows: A planar system of couples (M1, M2, …, Mn) can be combined into a resultant couple. The moment of this resultant couple, denoted as Mo, is known as the moment of inertia of a planar force system, and its magnitude can be calculated using the following formula: Mo = M1 + M2 + … + Mn. The magnitude of each couple moment is equal to the moment exerted by the corresponding force in the original force system about the center of simplicity; that is, M1 = Mo(F1), M2 = Mo(F2), Mn = Mo(Fn). Thus, the formula for calculating the moment of inertia can be written as Mo = Mo(F1) + Mo(F2) + … + Mo(Fn) = ∑Mo(F). This formula indicates that the moment of inertia is equal to the algebraic sum of the moments exerted by the various forces in the original force system about the center of simplicity. (3) Equilibrium of a planar arbitrary force system: As shown in the above analysis, a planar arbitrary force system can be simplified to a resultant vector R′ and a moment of inertia Mo. If both the resultant force and the moment are zero, it means that the force system does not cause any movement or rotation of the object; the object is in a state of equilibrium. Therefore, the condition for equilibrium of a planar force system is that both the simplified resultant force and the moment be zero. That is: R′=0, Mo=∑Mo(F)=0. From this, it follows that the equilibrium equations for any force system in a plane are ∑Fx=0, ∑Fy=0, and ∑Mo(F)=0. The meaning of these equilibrium equations is that when a force system in a plane is in equilibrium, the algebraic sum of the projections of all forces in that system on any two chosen rectangular coordinate axes is zero each ； At the same time, the algebraic sum of the moments of all forces in the force system about any point in the plane is also equal to zero. A planar arbitrary force system has three independent equilibrium equations, and these equations can be used to determine three unknowns. Example 3.2: A beam with two supports and an overhang is shown in Figure 3-19(a). A uniformly distributed load of q = 8 kN/m acts along the entire length of the beam. In addition, there is a concentrated force F = 8 kN acting between the two supports, as well as a couple of magnitude m = 2 kN·m; the value of a is 1 m. Determine the reaction forces at the supports A and B. Solution: ○1 Select beams AB as the objects of study, and draw their force diagram as shown in Figure 3-19(b). ○2 Establish a rectangular coordinate system as shown in Figure 3-19(b). ○3 equilibrium equations to solve for the unknown forces. ∑Fx=0, NAx=0, ∑MA(F)=0. NB×2a – m – Fa – q×3a×(0.5a)=0 kN. ∑Fy=0, NAy + NB – q×3a – F=0; therefore NAy = q×3a + F – NB = 8×3×1 + 8 – 11 = 21 kN. From this example, the general steps for solving problems using equilibrium equations can be derived: ① Select the object of study. The selected object of study should be one that is subjected to both known and unknown forces. ② Draw the force diagram of the object under study. ③ Establish a rectangular coordinate system. The coordinate system established should clearly reflect the geometric relationship among the various forces, and the coordinate axes should be perpendicular to the unknown forces as much as possible. ④List the equilibrium equations and solve for the unknown forces. The centroid should be placed as close as possible at the intersection of the unknown forces to facilitate problem-solving. If the value of a certain unknown force turns out to be negative, it indicates that the actual direction of that force is opposite to the assumed direction. In this case, there is no need to modify the direction of that force in the force diagram; it is sufficient to explain it in the answer. If the value of this force is used later, the negative sign should be included as well. (4) Equilibrium of a planar concurrent force system: When the resultant force R of a planar concurrent force system is equal to zero, this force system does not cause any change in the object’s state of motion; in other words, the force system is in equilibrium. Therefore, the equilibrium condition for a planar system of concurrent forces is that the resultant force is equal to zero. It follows that the equilibrium equations for a planar system of concurrent forces are ∑Fx=0 and ∑Fy=0; that is, when such a system is in equilibrium, the algebraic sum of the projections of all forces on any two chosen coordinate axes is zero. A planar concurrent force system has two independent equilibrium equations, and these equations can be used to determine two unknowns. Example 3.3 A cylindrical container with a weight of P = 15 KN is placed on idlers A and B, as shown in Figure 3-20 (a). Determine the restraining reaction forces of the idler rollers on the container. Solution: ① Taking the container as the object of study, its force diagram is shown in Figure 3-20 (b); it is clearly a planar concurrent force system. ② Establish a Cartesian coordinate system xoy, as shown in Figure 3-20(b). ③Set up equilibrium equations to solve for the unknown forces. ∑Fx = 0; NA cos60° + NB cos60° = 0 → NA = NB. ∑Fy = 0; NA sin60° + NB sin60° – P = 0 → NA = NB = P/(2sin60°) = 8.66 KN. (5) Equilibrium of a planar couple system: The result of combining planar couples is a resultant couple, whose moment M = ∑m. When the resultant couple moment is zero, the object is in a state of equilibrium; therefore, the condition for equilibrium of a system of planar couples is that the algebraic sum of the couple moments of all the individual couples is zero. That is, the expression ∑m＝0 is called the equilibrium equation for a system of planar couples. Example 3.4: A four-axis drilling machine is used to drill four holes in a workpiece, as shown in Figure 3-21(a). The cutting torque exerted by each drill bit on the workpiece is 10 N·m. The two bolts A and B, which hold the workpiece in place, are in smooth contact with the workpiece, and the distance between A and B is 0.2 m. Find the forces acting on the two bolts. Solution: ① Choose the workpiece as the object of study, as shown in Figure 3-21 (b). ②The external forces acting on the workpiece are four couples; since couples can only be balanced by other couples, the reaction forces exerted by the two bolts on the workpiece must form a couple that balances these four external couples. According to the equilibrium conditions for a system of planar couples: ∑m=0, NA(AB) – 4m = 0, hence NA = NB = 200 N. 3.2 Overview of the load-bearing capacity of components In practical engineering applications, various mechanical and structural elements are widely used. The parts that make up these machines and the components of engineering structures are collectively referred to as components. In the operating state, various components are subjected to loads, and the magnitude of these loads can be determined based on static analysis. However, under external loads, how to ensure that the components can function properly remains an issue that requires further resolution. The load that a component can bear is limited; if the load is too high, it may suffer damage or experience excessive deformation. For example, when a crane is lifting heavy objects, if the object is too heavy or the rope is too thin, the rope will break and cause an accident. Both component failure and excessive deformation are unacceptable in engineering. When a component is operating safely, its capacity to bear loads is referred to as its load-bearing capacity. To ensure that components can function safely and properly, it is necessary for them to have sufficient capacity to resist failure; this capacity to resist failure is known as strength. At the same time, components are sometimes required to have sufficient resistance to deformation, and this ability to resist deformation is known as stiffness. Furthermore, some components may lose their original equilibrium state when subjected to loads; for example, an elongated rod under high pressure might suddenly change from a straight equilibrium state to a bent or broken one, thereby losing its functional capacity and causing serious accidents. Therefore, such components are also required to have the ability to maintain their original equilibrium state while in operation; this ability to maintain the original equilibrium state is known as stability. To ensure the safe and reliable operation of components under load, they must possess sufficient strength, stiffness, and stability. Generally speaking, these requirements can be met by selecting high-quality materials or larger cross-sectional dimensions for the components ； However, this leads to material waste and a bulky structure. Obviously, there is a conflict between safety and economy, as well as between safety and weight. The task of determining the load-bearing capacity of a component is to provide the necessary theoretical foundations and calculation methods for selecting appropriate materials, as well as determining suitable cross-sectional shapes and geometric dimensions for that component, all while ensuring both safety and cost-effectiveness. In mechanical and engineering structures, the geometric shapes of components are diverse, but bars are the most common and fundamental type of component. A member is a component whose length dimension is much larger than its dimensions in the other two directions. A large number of engineering components can be simplified to bars. Such as drive shafts in machines, and beams and columns in engineering structures. The line connecting the centroids of the various cross-sections of a member is called the axis, while the cross-sections perpendicular to the axis are known as transverse sections. The loading conditions of components during operation vary, and the deformations that occur as a result of these loads also differ. For members, the basic forms of deformation that occur under load include axial tension and compression, shear and compression, torsion, and bending. 3.3 Axial Tension and Compression 3.3.1 Concepts of Axial Tension and Compression In practical engineering applications, there are many members that undergo axial tension or compression deformation, such as the bolts in the bolted connection structure shown in Figure 3-22, and the AB and BC members in the bracket shown in Figure 3-23(a). Although these members have different shapes and various loading and connection methods, axially stretched or compressed members can all be simplified to the analytical model shown in Figure 3-24. The stress characteristics of axially stretched or compressed members are as follows: the two forces acting on the member are equal in magnitude, opposite in direction, and their lines of action coincide with the axis of the member ； The deformation characteristic of axially stretched or compressed members is that the member elongates or shortens in the direction of its axis. This form of deformation is called axial tension (Figure 3-24a) or axial compression (Figure 3-24b). 3.3.2 Internal forces in the cross-section under axial tension and compression (1) Concept of internal forces: When studying the load-bearing capacity of a member, the loads acting on the entire member along with the reaction forces from constraints are collectively referred to as external forces. There are forces acting between one part of an object and another part, or between particles, which maintain the connection between the various parts of a component as well as the shape of the bar. When a component deforms under the action of external forces, the interaction forces between its various internal parts also change. These interaction forces that arise within the component as a result of external forces are known as internal forces. The internal force increases as the external force increases, but there is a limit to the increase in internal force; if this limit is exceeded, the component will fail. (2) Section method – Axial force: Since internal forces are the forces of interaction between adjacent parts of a body, the section method can be used to represent these internal forces. As shown in Figure 3-25(a), to determine the internal forces at a certain cross-section m-m of the member, it is possible to imagine cutting the member along this cross-section using a plane. One of the resulting parts (such as the left half) can be taken as the object of study, while the other part (such as the right half) is discarded, as shown in Figure 3-25(b). The effect of the removed part on the remaining part is then represented by internal forces, with their resultant force denoted as N. Since the entire rod was originally in equilibrium, any part of it after being cut still remains in equilibrium. The method of dividing a member into sections using an imaginary cut, based on the equilibrium equation N – F = 0 which yields N = F, in order to determine the internal forces, is called the section method. It is a general method for finding internal forces. When a rod is subjected to axial tension or compression, since the line of action of the external force F coincides with the axis of the rod, the line of action of the internal force N must also lie along the axis of the rod. This type of internal force is known as axial force, and it is commonly denoted by the symbol N. Through analysis, another method for calculating axial force can be derived: the axial force at a given section is equal to the algebraic sum of all external forces acting on one side of that section; external forces acting away from the section are taken as positive values, while those acting toward the section are taken as negative values. That is: (3) Axial force diagram. When a member is subjected to more than two axial external forces, the axial forces in different sections of the member will vary. To visually represent the variation of axial force along the axis of a member, coordinates parallel to the member’s axis are used to indicate the position of each cross-section, while coordinates perpendicular to the member’s axis are used to represent the value of the axial force. The graph resulting from this representation of the variation of axial force along the member’s axis is called an axial force diagram. *It is customary to plot positive axial forces on the upper side of the horizontal axis, and negative axial forces on the lower side. Example 3.5 As shown in Figure 3-26(a), the member is subjected to forces F1 = 10 kN, F2 = 20 kN, F3 = 5 kN, and F4 = 15 kN. Determine the axial force diagram of the member. Solution: ○1) Internal force analysis – Axial force N1 at section 1-1: N1 = -F1 = -10 kN. Axial force N2 at section 2-2 ； N2 = F2 - F1 = 10 kN. The axial force at section 3-3: N3 = F4 = 15 kN. ○2 Draw the axial force diagram: as shown in Figure 3-26(b). .3.3.3 Strength under axial tension and compression (1) The concept of stress: When the material is the same, the criterion for determining whether a member will fail is not the magnitude of the internal forces, but rather the degree of concentration of those forces. The intensity of internal forces is the internal force per unit cross-sectional area, commonly referred to as stress. Stress indicates the degree of force acting at a point on a cross-section; when the stress reaches a certain level, the member fails. Stress is a vector that can generally be decomposed into a component σ perpendicular to the cross-section and a component τ tangent to the cross-section. The component perpendicular to the cross-section, σ, is called normal stress, while the component tangent to the cross-section, τ, is called shear stress. In China’s legal units of measurement, the unit symbol for stress is Pa, which is named “Pascal” or abbreviated as “Pa”; 1 Pa = 1 N/m2. In practical engineering applications, this unit is too small; therefore, MPa (megapascals) and GPa (gigapascals) are commonly used. 1 Mpa = 1 N/mm2 = 106 Pa, and 1 Gpa = 109 Pa. (2) Stress on the cross-section: Analysis shows that during axial tension and compression, the magnitude of the stress at each point on the cross-section is equal; its direction is consistent with the axial force and perpendicular to the cross-section, hence it is called normal stress, as shown in Figure 3-27. Its calculation formula is given by, where σ represents the normal stress on the cross-section ； N — Axial force on the cross-section ； A — Cross-sectional area. The positive or negative sign of normal stress corresponds to axial force; that is, tensile stress is positive while compressive stress is negative. (3) Allowable stress and strength criteria ○1 Allowable stress: There is a limit to the stress that a material can withstand, and this limit varies depending on the type of material. The stress at which a member loses its ability to function properly is called the ultimate stress, denoted by σ0. To ensure the safe and reliable operation of components under external forces, taking into account factors such as the difficulty in accurately estimating the loads on these components, the approximations in calculation methods, and the unevenness of actual materials, a component is in a dangerous state when the stress within it approaches the ultimate stress level. To this end, sufficient strength reserve must be left for the components during operation. The ultimate stress is divided by a coefficient greater than 1 to determine the maximum stress that can be applied safely during operation; this stress is known as the allowable stress of the material, and it is usually represented by a symbol. In the formula, σ0 represents the ultimate stress of the material, while n is the safety factor. The strength condition states that, in order to ensure sufficient strength in axially loaded members, the maximum normal stress σmax (referred to as the working stress) within the member must not exceed the allowable stress of the material under tension (or compression); that is, σmax ≤ … This formula is known as the strength condition for tension (or compression) and serves as the basis for calculating the strength of such members. Based on the strength criteria, it is possible to address three issues: strength verification, design of cross-sectional dimensions, and determination of load-bearing capacity. Example 3.6: A tripod is formed by two rods, AB and BC, connected together by hinges, as shown in Figure 3-28. The cross-sectional areas of these two rods are A1 = 100 mm2 and A2 = 250 mm2 respectively. The material used for these rods is Q235, with an allowable stress of 120 MPa. Let the load acting on node B be F = 20 kN; ignoring the self-weight of the rods, verify the strength of the two rods. Solution: ○1 Calculate the axial forces. Bars AB and BC are two-force members that undergo axial tension or compression deformation. By using the section method to cut the two rods apart, their forces are as shown in Figure 3-28(b). From the equilibrium equation –NBCsin60. –When F = 0, the value of NBC is negative in kN, indicating that the BC bar undergoes compressive deformation. From -NBCcos600 - NAB = 0, we obtain NAB = -NBCcos600 = -(-23.09) × 0.5 = 11.55 (kN). Since NAB is positive, it indicates that member AB undergoes tensile deformation. (2) Strength check: The normal stress in rod AB (MPa) < Therefore, the strength of rod AB is sufficient. The normal stress of BC bar (MPa) < Therefore, the strength of BC bar is sufficient. 3.3.4 Deformation under axial tension and compression (1) Deformation and strain Tests show that when subjected to axial tension, the rod elongates in the longitudinal direction while its transverse dimensions decrease ； Under axial compression, the rod shortens in the longitudinal direction while its transverse dimensions increase, as shown in Figure 3-29. The deformation of a member in the axial direction is called longitudinal deformation, while the deformation perpendicular to the axial direction is called transverse deformation. ○1 Absolute deformation: The total amount of elongation or shortening of a member is called absolute deformation. Let the original length of the straight rod be l, and its transverse dimension be b. When an axial force is applied, the length of the rod becomes l1 and its transverse dimension becomes b1. The absolute longitudinal deformation of the rod is △l = l1 – l, while the absolute transverse deformation is △b = b1 – b. ○2 Relative deformation: Absolute deformation only indicates the magnitude of the deformation of the rod, but it does not reflect the degree of that deformation. The degree of deformation of a member is measured by the amount of deformation per unit length of the member. The deformation of a member per unit length is called relative deformation, also known as linear strain. The linear strains corresponding to the above two types of absolute deformation are longitudinal linear strain and transverse linear strain. (2) Hooke’s law: A rod deforms under the action of a load, and there is a certain relationship between this deformation and the load. Experiments show that when the normal stress in a rod subjected to axial tension or compression does not exceed a certain limit, its absolute axial deformation Δl is proportional to the axial force N and the length of the rod l, and inversely proportional to the cross-sectional area A of the rod. By introducing the material-specific proportionality constant E, the above equation is known as Hooke’s law. The above equation can be rewritten as , that is, or . This form is another expression of Hooke’s law. Thus, Hooke’s law can be simplified as follows: as long as the stress does not exceed a certain limit, stress is proportional to strain. The aforementioned stress limit is called the proportional limit σP. The proportional limits of various materials vary and can be determined experimentally. The proportionality constant E is known as the elastic modulus of the material. With all other conditions remaining constant, the greater the elastic modulus E, the smaller the absolute deformation Δl of the rod. This shows that the value of E indicates the material’s ability to resist elastic deformation under tension or compression, and it serves as a measure of the material’s stiffness. Since strain ε is a dimensionless quantity, the unit of elastic modulus E is the same as that of stress σ, commonly expressed in GPa (gigapascals). Its value varies depending on the material and can be determined experimentally. A list of the elastic moduli of commonly used materials in engineering is provided in Table 3-1 for reference. Table 3-1: E values of common materials Material Name E / (GPa×10²) Material Name E / (GPa×10²) Low-carbon steel Alloy steel Gray cast iron 2 ~ 2.2 1.9 ~ 2.1 1.15 ~ 1.6 Copper and its alloys Rubber 0.74 ~ 1.30 0.00008 3.3.6 Mechanical properties of materials under tension and compression When discussing tensile (or compressive) strength and deformation earlier, the mechanical properties of materials such as the elastic modulus E and the proportional limit σP were covered. The mechanical properties of a material refer to the characteristics related to strength and deformation that the material exhibits under external forces. The mechanical properties of materials must be determined through testing. To study the mechanical properties of materials, tests under static load (where the load increases slowly and steadily) are usually conducted. This section mainly introduces the axial tension and compression tests of low-carbon steel and cast iron under normal temperature and static loads. (1) Tensile test of low-carbon steel: Certain properties of a material are related to the size and shape of the specimen; in order to enable comparison of test results for different materials, the material should be processed into standard specimens. During the test, the specimen is clamped in the fixture of the testing machine, and loading is applied slowly; an automatic plotter then draws the curve showing the relationship between the load F and the elongation Δl within the gauge length, as shown in Figure 3-30(a). This curve is known as the tensile diagram or F-Δl curve. Both the vertical and horizontal axes of the F-Δl curve are related to the dimensions of the specimen. To eliminate the influence of these dimensions, the vertical axis value is divided by the cross-sectional area of the specimen, and the horizontal axis value is divided by the gauge length; this yields the stress-strain relationship curve, also known as the stress-strain diagram or σ-ε curve, as shown in Figure 3-30(b). As can be seen from the graph, the entire stretching process can be roughly divided into four stages. ①Elastic stage: In the initial stage of stretching, the relationship between σ and ε is linear, represented by the line oa; this indicates that within this stage σ is proportional to ε, i.e., σ∝ε. The slope of the line; therefore, the elastic modulus of the material is equal to the slope of that line. As can be seen from the σ-ε curve, the stress corresponding to the highest point A of the line OA is the maximum stress at which stress is proportional to strain; this value is known as the material’s proportional limit σP. The σP of Q235A steel is 200 MPa. Beyond the proportional limit σP, from point a to point a/, the relationship between σ and ε is no longer linear, but the deformation remains elastic; that is, the deformation will completely disappear once the tensile force is removed. The stress corresponding to point a is the maximum limit at which elastic deformation occurs; this value is known as the elastic limit, denoted by σe. Since points a and a/ are very close to each other, in practice no strict distinction is made between the elastic limit and the proportional limit. ②In the yield stage, beyond point b, a small serrated segment bc appears on the σ-ε curve that is nearly horizontal; this indicates that the strain increases rapidly, while the stress fluctuates within a very narrow range, essentially not increasing at all, as if the material has lost its ability to resist deformation. This phenomenon, in which the stress remains essentially constant while the strain increases significantly, resulting in noticeable plastic deformation, is known as material yield or flow. Graphically, the process corresponding to b c is called the yield stage. The lowest stress σs corresponding to the yield stage is called the yield limit (yield point). The σs of Q235A steel is 235 MPa. When the stress reaches the yield limit, the material will undergo significant plastic deformation. Since the plastic deformation of parts affects the proper operation of machinery, the yield limit σs is an important indicator for measuring material strength. ③Strengthening stage: After the yield stage, the curve begins to rise gradually starting from point c; this indicates that in order for the specimen to continue deforming, stress must be increased, and the material regains its resistance capacity. This phenomenon is known as strengthening. The process from point c to point d is referred to as the strengthening stage of the material. The stress corresponding to the peak value d in the strengthening stage is called the strength limit, denoted by σb. The ultimate strength σb of Q235A steel is 400 MPa. The strength limit σb is the maximum stress value that the material can withstand before fracturing; therefore, it is another important indicator for measuring material strength. ④Local necking stage: Before reaching the strength limit, the deformation of the specimen is uniform. Beyond point d, within a certain local area of the specimen, the longitudinal deformation increases significantly while the cross-sectional area shrinks sharply, resulting in necking, as shown in Figure 3-31. Once necking occurs in the specimen, it will be quickly pulled apart. The aforementioned stretching process of low-carbon steel goes through four stages: elasticity, yield, strengthening, and local necking. There are three characteristic points, with the corresponding stresses being the proportional limit, yield limit, and strength limit, in that order. ○5 Measurement of material plasticity: After the specimen is pulled apart, the elastic deformation disappears, but the plastic deformation remains. In engineering, the plastic deformation remaining in a specimen after fracture is used to represent the plastic properties of a material. Two common plasticity indicators are elongation and reduction of area. The elongation rate is the percentage of relative elongation of the specimen after it breaks; it is denoted by δ. In the formula, l represents the original length of the gauge length of the specimen ； l1 is the length of the gauge length after the specimen breaks. The larger the δ value, the better the plasticity of the material. Low-carbon steel has a elongation rate of 20% to 30%, indicating good plasticity. In engineering, materials with an elongation rate δ ≥ 5% are often referred to as plastic materials. b: Reduction of cross-sectional area – the percentage decrease in the cross-sectional area of the specimen after it breaks, denoted by Ψ. In this formula, A represents the original cross-sectional area of the specimen ； A1 is the minimum cross-sectional area at the fracture site after the specimen breaks. (2) Tensile test of cast iron: Cast iron is a brittle material widely used in engineering, and its stress-strain curve during tensile testing is a slightly curved line, as shown in Figure 3-32. There are no obvious straight sections in the graph, but at low stresses, the σ-ε curve is similar to a straight line, indicating that Hooke’s law can be approximately applied when the stress is not high. When drawn, cast iron exhibits no yield or necking; it breaks suddenly under relatively low tensile stress. The fracture surface is flat and perpendicular to the axis, with very little deformation at the time of fracture, with the strain typically ranging from 0.4% to 0.5%. The maximum stress at which cast iron breaks, that is, its ultimate tensile strength, is the only indicator for measuring the strength of cast iron. (3) Compression test of materials: Compression specimens for metallic materials are generally made in the form of short cylinders to prevent them from bending under compression during testing. The length of a cylinder is generally 1.5 to 3 times its diameter. ○1 Compression test of low-carbon steel: The σ-ε curve for low-carbon steel during compression is shown in Figure 3-33. Compared with the σ-ε curve for tension, which is indicated by the dashed line in the figure, the two curves are essentially identical before yield. This indicates that the elastic modulus E, proportional limit, and yield limit of low-carbon steel under compression are essentially the same as those under tension. After the yield stage, the specimen undergoes significant plastic deformation; it becomes flatter as more pressure is applied, and its cross-sectional area increases continuously. The specimen first takes on a bulged shape and eventually turns into a disc-shaped form; as a result, it is not possible to determine the strength limit under compression. ○2 Compression test of cast iron: The σ-ε curve for cast iron during compression is shown in Figure 3-34, and it is similar to its σ-ε curve during tension (dashed line). The entire curve has no straight segments; there is no yield limit, only a strength limit. The difference is that the ultimate compressive strength of cast iron is much higher than its ultimate tensile strength (about 3 to 4 times higher). Therefore, brittle materials are suitable for use as compression members. Furthermore, its rupture port forms an angle of about 450–500 degrees with the axis. In summary, the main difference in mechanical properties between plastic materials and brittle materials is as follows: ○1 Plastic materials exhibit significant plastic deformation upon failure, and some show yield behavior before fracturing ； Fragile materials, on the other hand, suddenly fracture when deformed only slightly, without exhibiting any yield phenomenon. ○2 The proportional limit, yield limit, and elastic modulus of plastic materials during tension are the same as those during compression. Since plastic materials generally do not allow reaching the yield limit, they exhibit the same strength and stiffness under tension and compression ； Fragile materials, on the other hand, are different: their strength and stiffness under compression are greater than those under tension, and their compressive strength is much higher than their tensile strength. (4) Stress concentration: In a straight rod with a constant cross-section that is subjected to axial tension or compression, the stress is evenly distributed across the cross-section. However, for members with sharp changes in cross-sectional dimensions, experimental and theoretical analyses have shown that at the points where the cross-section of the member changes suddenly, the stress is no longer evenly distributed. This phenomenon of localized increase in stress caused by a sudden change in cross-section is known as stress concentration. As shown in Figure 3-35, holes, grooves, and cuts in the member result in stress concentration; away from these areas, the stress decreases rapidly and approaches its average value. The more drastic the change in cross-section, the greater the stress concentration, and the higher the maximum stress that occurs in local areas. The ratio of the maximum stress to the average stress in the local area where there is a cross-sectional mutation is called the stress concentration factor, which is usually denoted by α. In other words, the stress concentration factor α indicates the degree of stress concentration; the larger α, the more severe the stress concentration. To reduce the degree of stress concentration, the transition at points where the cross-section changes should be as smooth as possible. To this end, sharp-edged grooves and holes should be avoided as much as possible on the members, and the shoulder sections of the circular shafts should be transitioned with rounded corners. 3.4 Shearing and Extrusion 3.4.1 Shearing (1) Concept of Shearing As shown in Figure 3-36, when using a shear machine to cut steel plates, the upper and lower blades of the shear machine apply two forces F to the steel plate, causing relative displacement at various sections between the lines of action of these forces, until the plate is finally cut apart. This type of deformation, in which the cross-section shifts relative to each other, is called shear deformation. The plane that results from this relative displacement is called the shear plane; it is always located between two opposite external forces and is parallel to the line of action of those forces. The characteristic of the force acting on shear deformation is that the external forces have equal magnitudes, opposite directions, and their lines of action are very close to each other. The characteristic of deformation is the relative displacement of various sections between the lines of action of the two forces. Common fasteners used in machinery, such as rivets, pins, bolts, and keys, are all components that endure shear forces. (2) Shear stress and shear strength conditions Figure 3-37 shows two steel plates connected by bolts. The plates are subjected to an external force F; as a result, the bolts on either side are also subjected to this force F. The bolts undergo shear deformation, and internal forces are generated at all sections between the lines of action of these two external forces. Suppose the bolt is cut into two sections along the cross-section m–m, and any one of these sections is chosen as the object of study. Since the external force F is parallel to the cross-section, the internal force Q on the cross-section must also be parallel to it. This internal force that is parallel to the cross-section is called shear force. The magnitude of the shear force is given by the equilibrium condition Q = F. When a member is subjected to shear, the shear force per unit area on the shear plane is called the shear stress τ. The distribution of shear stress on the shear plane is relatively complex. In practical engineering applications, for the sake of simplifying calculations, it is generally assumed that this stress is uniformly distributed across the shear plane. The formula for calculating it is as follows: where τ represents shear stress, in MPa ； Q — Shear force on the shear plane, N ； A — Shear area, mm2. To ensure the safety and reliability of components during operation, the practical strength condition for shear is given by the formula, where is the allowable shear stress of the material; this value is determined through testing and can be found in relevant manuals. 3.4.2 Compression (1) Concept of compression: Connectors such as rivets, bolts, pins, and keys, in addition to bearing shear forces, also exert pressure on each other at the contact surfaces between the connector and the components it connects; this phenomenon is known as compression. As shown in the bolt connection in Figure 3-37, the left side of the hole in the upper steel plate presses against the left side of the upper part of the bolt, while the right side of the hole in the lower steel plate presses against the right side of the lower part of the bolt. (2) Compression stress and compression strength conditions: The surface on a component where compression deformation occurs is called the compression surface; it is the contact surface between two objects and is generally perpendicular to the direction of the external force. The pressure acting on the extrusion surface is called the extrusion force, denoted by the symbol Fjy. The squeezing force on a unit extrusion surface is called the extrusion stress. It is denoted by the symbol бjy. The distribution of extrusion stress is also relatively complex. In engineering, it is approximately assumed that the extrusion stress is uniformly distributed over the extrusion surface; therefore, the extrusion stress can be calculated using the following formula, where Ajy represents the area of the extrusion surface. When the contact surface is flat, the contact area is equal to the area of the compression surface. For fasteners such as rivets and bolts, whose compression surface is cylindrical, the distribution of compression stress is quite complex. To simplify the calculations, the projection dt of the contacting cylindrical surface onto the diameter plane is usually used as the compression area, as shown in Figure 3-38(d). The strength condition for compression is given by the formula, where σc is the allowable compression stress of the material; this value is determined through experiments and can be found in relevant handbooks. Example 3.7 As shown in Figure 3-37, two steel plates with a thickness of δ = 8 mm are connected by a bolt, and they are subjected to a load F = 10 KN. Allowable shear stress of bolt = 60 MPa ； The allowable crushing stress for steel plates and bolts = 180 MPa. Find the diameter d of the bolt. Solution: Under the action of the load F, the bolt undergoes shear and compression deformation. ① The bolt diameter is calculated based on the shear strength condition. As shown in Figure 3-37(b), the bolt is cut along the section m-n. Using the section method, the shear force is determined to be Q = F = 10 KN. The area of the shear plane can be found using the shear strength condition; taking d = 15 mm. ② The bolt diameter is calculated based on the compression strength condition: the compressive force is Fjy = F, and the area of the compression surface is Ajy = dδ. The compression strength condition also provides a value in millimeters. To ensure that the bolt can function safely, it is necessary to meet both the shear strength and compression strength conditions; therefore, the bolt diameter should be 15 mm. 3.5 Torsion of Circular Shafts 3.5.1 Concept of Torsion: In a plane perpendicular to the axis of a rod, when a pair of couple forces of equal magnitude but opposite directions acts on it, the various cross-sections of the rod rotate relative to each other around the axis; this type of deformation is known as torsional deformation, as shown in Figure 3-39. In practical engineering, there are many members that are subjected to torsion; for example, taps used for threading, mixing shafts, and drive shafts are all members that experience torsion. In engineering, members that undergo torsional deformation mostly have circular or annular cross-sections; therefore, only the torsion problem of solid circular shafts with constant cross-section is discussed here. 3.5.2 Calculation of external torque The external force acting on the shaft during torsion is a torque. However, in practical engineering applications, what is usually given are the power transmitted by the shaft and its rotational speed; the relationship between them is expressed by the formula, where M represents the external torque acting on the shaft, in N.m ； P — Power transmitted by the shaft, KW ； n — rotational speed of the axis, r/min. As can be seen from the formula, at a constant power, the external torque is inversely proportional to the rotational speed. Therefore, on the same machine, the diameter of the high-speed shaft is smaller than that of the low-speed shaft. 3.5.3 Torsional Internal Forces: An imaginary section m-n is used to divide the circular shaft into two sections; the left section is taken as the subject of study, as shown in Figure 3-40. Since the entire circular shaft is in equilibrium, the left section after cutting must also be in equilibrium. Furthermore, since the acting surface of the external couple is perpendicular to the axis, the internal force in the m-n section must also be a couple whose acting surface is perpendicular to the axis. The moment of this internal couple is called torque, denoted by the symbol Mn. Using the section method, it can be deduced that the torque at a certain section is equal to the algebraic sum of all external couple moments on either side of that section. The positive or negative value of the external torque is determined by the right-hand rule: the four fingers of the right hand are used to indicate the direction of the torque, and if the thumb points away from that cross-section, the torque is considered positive; otherwise, it is negative. When a member is subjected to two or more external couples, the torque at different sections varies. To visually represent how torque changes with the section location, a graph showing the relationship between torque and section position is often drawn, known as a torque diagram. The horizontal axis represents the position of each cross-section, while the vertical axis represents the torque value at that corresponding cross-section. Positive torque values are plotted above the horizontal axis, while negative values are plotted below it. Example 3.8 As shown in Figure 3-41, the rotational speed of the drive shaft is n = 200 r/min. The input power to wheel A is PA = 10 KW, while the output powers of wheels B and C are PB = 6 KW and PC = 4 KW respectively. Draw the torque diagram for this shaft. Solution: ① Calculate the external torque in N.m. ② Compute the torque at each section separately and draw a torque diagram. For section BA: Mn1 = -MB = -287 N.m. For section AC: Mn2 = MC = 191 N.m. Draw the torque diagram on a suitable scale. As can be seen from the graph, the maximum torque occurs in section BA; that is, Mnmax = |Mn1| = 287 N·m. 3.5.4 Strength of a circular shaft under torsion (1) Stress in a circular shaft under torsion According to theory and experiments, only shear stress exists on the cross-section of a shaft under torsion. The direction of the shear stress at each point is perpendicular to the radius at that point, and the magnitude of the shear stress varies proportionally to the distance from that point to the center of the shaft. The shear stress is zero at the center of the shaft, while it is greatest at the surface of the shaft. The distribution of shear stress is shown in Figure 3-42. When a circular shaft is twisted, the maximum shear stress in its cross-section is given by the formula, where τmax represents the maximum shear stress in the cross-section, in MPa ； Mn —— Torque on the cross-section, in N.mm. Wn —— Torsional section modulus, a geometric quantity that depends only on the shape and dimensions of the cross-section, in mm3. The formula for calculating the torsional section modulus of a solid circular shaft is , while the formula for a hollow circular shaft is . Here, α is the ratio of the outer diameter to the inner diameter of the hollow circular shaft, that is . (2) Strength condition for torsion of circular shafts: In the case of torsion of a circular shaft, the section where the maximum shear stress occurs is called the critical section. To ensure the safe operation of a circular shaft under torsion, it is necessary to limit the maximum shear stress within the shaft to not exceed the material’s allowable shear stress. Therefore, the strength condition for torsion of a circular shaft is given by the formula, where Mnmax is the torque on the critical section, in N.mm ； Wn——torsional section modulus, mm3 ； —— Allowable shear stress of the material, MPa. 3.5.5 Stiffness of a circular shaft under torsion (1) Deformation of a circular shaft under torsion: As shown in Figure 3-39, the deformation of a circular shaft under torsion is commonly expressed by the angle through which the cross-sections at both ends of the shaft rotate relative to each other; this angle is known as the twist angle. In engineering, the twist angle per unit length θ is commonly used to indicate the degree of torsional deformation; that is, in the formula, θ represents the twist angle per unit length, inº/m ； Mn —— Torque on the cross-section, N.mm ； G —— the shear elastic modulus of a material, which is a measure of the material’s ability to resist shear deformation. The G value for ordinary steel is 80×103 MPa ； Ip——polar moment of inertia, which depends only on the shape and dimensions of the cross-section, in mm4. For a solid circular shaft, the formula for calculating the polar moment of inertia is as follows; for a hollow circular shaft, the formula for calculating the polar moment of inertia is as follows. (2) Stiffness conditions for torsion of circular shafts: In the case of shaft components, excessive torsional deformation can affect the precision of mechanical transmission or cause vibration. Therefore, to ensure the proper operation of the shaft, in addition to meeting the strength requirements, torsional deformation must also be limited. Typically, it is required that the maximum torsional angle per unit length of the shaft, θmax, does not exceed the allowable torsional angle per unit length, i.e., where —— is the allowable torsional angle per unit length, inº/m. The value can be found in relevant manuals, depending on the operating conditions of the shaft and the precision requirements of the machine. Example 3.9: If the drive shaft in Example 3.8 is a solid steel circular shaft, with an allowable shear stress of 40 MPa, an allowable torsional angle per unit length of 1º/m, and a shear elastic modulus G of 80 GPa, and if the diameter of the shaft is D = 40 mm, determine whether the strength and stiffness of this drive shaft are sufficient. Solution: A torque diagram was prepared in Example 3.8, and from this diagram it can be seen that Mnmax = 287 N•m. ①The strength of the drive shaft is calculated as τmax = Mnmax/Wn = Mnmax/(πd3/16) = 287×103 ÷ (3.14×403÷16) = 22.9 MPa, which is less than 40 MPa; therefore, the strength of the drive shaft is sufficient. ②The stiffness of the drive shaft = 180°÷π×10³×Mnmax÷(G×πd⁴/32) = 180°÷3.14×10³×287×10³÷(80×10³×π×404÷32) = 0.82°/m, which is less than 1°/m; therefore, the stiffness of the drive shaft is sufficient. 3.6 Bending of Straight Beams 3.6.1 Concept of Bending Deformation Bending deformation is very common in practical engineering applications. In the plane passing through the axis of the rod, when subjected to a couple or an external force perpendicular to the axis, the axis of the rod changes from a straight line to a curve. This type of deformation is called bending deformation. For example, the crossbeam of a bridge crane is subjected to its own weight as well as the weight of the load being lifted ； Horizontal containers are subjected to the weight of their own structure and the weight of the materials inside them ； Tall tower structures will undergo bending deformation due to wind loads and other factors. In engineering, a member that undergoes primarily bending deformation or only bending deformation is called a beam. (1) Plane bending: Beams used in engineering often have a symmetry axis in their cross-section. The plane formed by the axis of symmetry of the cross-section and the axis of the beam is called the longitudinal symmetry plane, as shown in Figure 3-43. If the external forces acting on the beam (including couples) are all within the longitudinal symmetry plane and perpendicular to the beam’s axis, then the beam’s axis becomes a planar curve within the longitudinal plane; this type of bending is known as planar bending. This section focuses only on plane bending problems. (2) Types of beams In practical engineering, beams are classified into three basic forms based on their support conditions, as shown in Figure 3-44. ○1 Simple beam: One end of the beam is supported by a fixed hinge, while the other end is supported by a movable hinge. ○2 Exposed beam: Its supports are the same as those of a simply supported beam, but one end (or both ends) of the beam extends beyond the supports. ○3 Cantilever beam: One end of the beam is fixed, while the other end is free. The distance between the two supports of a beam is called the span. 3.6.2 Internal forces in the bending of a straight beam (1) Shear force and bending moment: When a beam is subjected to external forces, internal forces are generated within it. As shown in Figure 3-45, the beam is hypothetically divided into two sections, and the left section is taken for study. Since the entire beam is in equilibrium, the left section after cutting must also be in equilibrium. Since the external force is perpendicular to the axis of the beam, there must be an internal force Q acting within section m-m to balance this external force; this internal force Q is known as shear force. At the same time, since external forces also exert a torque on the cross-section within the longitudinal symmetry plane, there must be an internal couple with its plane of action lying in the longitudinal symmetry plane acting on the cross-section m-m to balance this external torque. The moment of this internal couple is called bending moment, denoted by the symbol M. Typically, the span of beams is relatively large; in such cases, bending moment has a greater impact on the strength of the beam, while shear force has a lesser impact. Therefore, this section discusses only the effect of bending moment on beams. Using the section method, it can be determined that the bending moment at a given section is equal to the algebraic sum of the moments exerted by all external forces (including external couples) on the center of that section. The sign of the torque exerted by an external force on the center of the cross-section is defined as follows: assuming the cross-section is fixed, if a certain external force causes the beam to bend upward, the torque exerted by that force is positive; conversely, if it causes the beam to bend downward, the torque is negative. Example 3.10: For the simply supported beam shown in Figure 3-46, subjected to a load P = 300 N, determine the bending moments at sections 1-1 and 2-2. Solution: ○1 To find the reaction forces at the supports, using the equations of static equilibrium: ∑MA(F) = 0 → -200P + 600RB = 0; thus RB = 100 N. ∑Fy = 0 → RA + RB – P = 0; hence RA = P – RB = 300 – 100 = 200 N. ○2 To determine the moments: M1 = 100 × RA = 100 × 200 = 20,000 (N·mm); M2 = 300 × RB = 300 × 100 = 30,000 (N·mm). (2) Bending moment diagram: The bending moment on a beam’s cross-section generally varies depending on its position along the beam. If x is used to represent the position of a cross-section along the beam’s axis, then the bending moment at each cross-section can be expressed as a function of x, i.e., M = M(x), which is the bending moment equation. A graph that shows how the bending moment M varies with the position in the cross-section is called a bending moment diagram. The horizontal axis represents the position of each cross-section, while the vertical axis represents the bending moment value at that cross-section. Positive torque values are plotted above the horizontal axis, while negative values are plotted below it. Example 3.11: Attempt to draw the bending moment diagram of the simply supported beam shown in Figure 3-47(a). Solution: ○1 To find the reaction forces at the supports, these can be determined using the equations of static equilibrium. ○The bending moment equations: for the AC segment, the equation is (0≤x1≤a); for the CB segment, the equation is (a≤x2≤l). To draw the bending moment diagram, it can be seen from these equations that the bending moment in both the AC and CB segments is a linear function of x. Therefore, the bending moment diagrams are straight lines, and all that is needed is to determine the endpoints of these lines in order to draw them. The resulting bending moment diagram is shown in Figure 3-47(b). 3.6.3 Strength of straight beams under bending (1) Bending normal stress: The bending deformation of a beam in which both shear force and bending moment are present in its cross-section is known as flexural bending. Bending deformation in a beam cross-section caused only by bending moment is called pure bending. This section discusses only the stresses under pure bending, but the stress calculation formulas are still applicable to the stresses generated by bending moments in shear bending. As shown in Figure 3-48, it is a beam with a rectangular cross-section; consider that the beam is composed of numerous longitudinal fibers. Experimental observations show that the fibers on the side where the beam protrudes stretch, while the fibers on the side where the beam is recessed shorten. Since deformation is continuous, there must be a layer of fibers in the middle that neither stretches nor shortens; this layer of fibers is called the neutral layer. The line where the neutral layer intersects the cross-section is called the neutral axis; this axis passes through the center of the cross-section and is perpendicular to the longitudinal symmetry plane in which the external force acts. Based on experimental observations and theoretical analysis, it can be concluded that when a beam is in pure bending, there are only normal stresses; the stress on one side of the neutral axis is tensile while it is compressive on the other side. Moreover, the stress at any point on the cross-section is proportional to the distance of that point from the neutral axis, as shown in Figure 3-49. The formula for calculating the maximum normal stress in the cross-section is as follows: where σmax represents the maximum normal stress in the cross-section, in MPa ； M — Bending moment in the cross-section, N.mm ； WZ — the bending moment of inertia of the cross-section about the neutral axis Z; it is a geometric quantity that depends on the shape and size of the cross-section, with units of mm3. The bending moment of inertia for common cross-sections is shown in Table 3-2; the bending moment of inertia for various steel sections commonly used in engineering can be found in relevant handbooks. Table 3-2 Bending section modulus WZ for common cross-sections: rectangular section, circular section, annular section; large-diameter equipment or pipes. WZ (2) Calculation of normal stress strength in beams: For beams with uniform cross-sections, the bending section modulus is the same across all sections, and the section with the maximum bending moment is the critical section. Therefore, to ensure the safe operation of the beam, the maximum stress in the critical sections should be limited to not exceed the allowable bending stress {σ} of the material. That is, as shown in Figure 3-50(a) of Example 3.12, the length of the handle is l = 30 cm, its diameter is d = 2.5 cm, the applied force is P = 200 N, and the allowable stress of the material is 100 MPa. Test the handle strength. Solution: ○1 To find the maximum bending moment, the handle is simplified as a cantilever beam subjected to a concentrated force. Its bending moment diagram is drawn using the method for drawing bending moment diagrams, as shown in Figure 3-50(b). As can be seen from the graph, the maximum bending moment is Mmax = Pl = 200 × 300 = 6×104 (N·mm). ○2 Verification of the handle’s strength: The flexural section modulus of the handle’s cross-section is WZ = 0.1d3 = 0.1 × 253 = 1.56×103 (mm3). The maximum bending stress at the fixed end of the handle is (MPa)