Engineering Mechanics - Statics - chap05 - friction

810工程力学参考书-回复什么是810工程？力学又是什么？那么推荐的参考书有哪些呢？接下来，我将逐步回答这些问题。

810工程是中国科学技术大学为期两年的本科课程之一，也是工程力学课程的一部分。

它旨在培养学生的工程力学基础知识和能力，帮助他们理解和解决复杂的力学问题。

这门课程通常在大学工程学院的力学课程中教授，并且对于工程专业的学生具有重要的意义。

力学是研究物体的运动和力的学科。

它是工程学科中最重要的基础学科之一。

力学包括静力学、动力学和变形力学三个主要分支。

静力学研究物体处于静止或平衡状态下的力学性质，动力学研究物体在运动过程中的力学性质，而变形力学研究物体在受力作用下的形变和变形。

了解了810工程和力学的基本概念后，我们可以来看一些推荐的参考书。

以下是一些经典的力学参考书：1. "工程力学"（作者：靳伟民）：这本书是中国科学技术大学教授靳伟民创作的经典力学教材。

它详细介绍了工程力学的基本概念、原理和方法。

这本书很适合初学者，内容系统全面，易于理解。

2. "Engineering Mechanics: Statics and Dynamics"（作者：R. C.Hibbeler）：这是一本国际知名的力学教材，作者是美国工程教育界的权威之一。

该书结构严谨，内容深入浅出，既适合本科生学习，也适合作为研究生和工程师的参考书。

3. "Mechanics of Materials"（作者：Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek）：这本书主要介绍了材料的力学性质和行为，是学习变形力学的重要参考书。

它涵盖了材料力学、弹性力学、塑性力学、断裂力学等内容。

以上这些参考书都是经典之作，内容全面，且有助于理解和应用810工程力学课程的基本概念和原理。

《理论⼒学》课程教学⼤纲72《理论⼒学》课程教学⼤纲⼀、课程的基本情况课程中⽂名称:理论⼒学课程英⽂名称：Theoretical Mechanics课程代码：0701017课程类别：专业基础课课程性质：必修课总学时：72 讲课学时：72 实验学时：0课程学分：4授课对象：机械类及相近专业本科学⽣前导课程：⾼等数学⼆、教学⽬的理论⼒学为⼯程类学科的基础课程之⼀。

理论⼒学教育不仅可以培养学⽣的⼒学素质，⽽且可以加强学⽣的⼯程概念。

这对于他们向其他学科或其他⼯程领域扩展是很有利的。

是相关专业后续课程的理论和设计以及⼯程设计规范的基础。

通过理论⼒学的学习可以使学⽣了解刚体的特性及其研究⽅法；了解理论⼒学基本理论以及在⼯程实际中的应⽤；了解理论⼒学与其他相关课程的联系。

三、教学基本要求绪论0.1理论⼒学的研究对象0.2理论⼒学的发展简史0.3理论⼒学的内容和研究⽅法0.4理论⼒学的应⽤及和其他学科的关系基本要求：1.了解理论⼒学的研究对象和内容。

2.了解理论⼒学与其他课程的关系。

3.了解理论⼒学在各个不同领域的应⽤。

重点与难点:1.理论⼒学的内容和研究⽅法。

第⼀部分静⼒学第1章静⼒学公理和物体的受⼒分析1.1静⼒学公理1.2约束和约束⼒1.3物体的受⼒分析和受⼒图基本要求：1.深⼊地理解⼒、刚体、平衡和约束等重要概念，深⼊理解⼒的基本性质。

2.明确基本约束的特征,正确地对物体系统进⾏受⼒分析。

重点与难点:1.约束和物体系的受⼒分析。

2.物体系统的受⼒分析和受⼒图。

第2章平⾯汇交⼒系与平⾯⼒偶系2.1平⾯汇交⼒系合成与平衡⼏何法2.2平⾯汇交⼒系合成与平衡的解析法2.2平⾯⼒对点之矩概念及计算2.3平⾯⼒偶基本要求：1.清晰的理解汇交⼒系和⼒偶系的合成。

2.熟练解汇交⼒系的平衡问题。

3.清晰的理解⼒对点的矩，并能熟练的计算。

4.深⼊理解⼒偶和⼒偶矩的概念。

明确⼒偶的性质和⼒偶的等效条件。

重点与难点:1.汇交⼒系的平衡⽅程。

工
程
静
力
Statics)
学
(Engineering
静力学－－研究刚体上的作用力。 静力学研究的三大问题：
1. 作用在刚体上的力系的等效替换或简化
等效替换－－两个作用效果相同的力系间的相互 替换。 简化－－用简单的力系替换复杂的力系。 2. 作用在刚体上的力系的平衡条件 平衡条件－－物体处于平衡时，作用力系应满足的 条件。
约束与约束力
光滑铰链约束
用铰链连接的杆
FR

结论与讨论
关于二力构件
请判断下列结构中哪些构件是二力构件， 哪些构件则不是二力构件。

结论与讨论
关于二力构件
请判断下列结构中哪些构件是二力构件， 哪些构件则不是二力构件。
3. 加减平衡力系公理
若在作用于刚体上的已知力系上添加或减去任何 平衡力系，则对刚体的作用效应并不改变。
3. 约束和约束力
(3) 几种常见约束的约束力 (a) 柔性体（绳索、皮带、链条）
T1’ T1
T
T 2’ T2
作用线：沿柔性体 方向：离开非自由体（拉力）
(3) 几种常见约束的约束力 (b) 光滑面
R N
作用线：沿约束面的公法线
方向：指向非自由体（压力）
(c) 滚轴支承
N
N
N


约束与约束力
五．力偶及其性质
1．力偶的概念 力偶: 由两个等值、反向、不共线的（平行）力组成的力系称 为力偶，记作（F，F，）
m o F’ B d A F
力偶没有合力，不能用一个力来代替，也不能用一个力与之平 衡它是力学中的又一基本要素，其作用使物体发生转动，以力偶 矩表示。 m(F , F ) mo (F ) mo (F ) F OA F OB Fd 大小：

Part II：Mechanics of MaterialsTorsionStudy object: A long straight member subjected to a torsional loading •How to determine the stress distribution?•How to determine the angle of twist?•Statically indeterminate analysis of the member in torsion?Torsion:If a member is subjected to the action of a pair of moments which are of a common magnitude and opposite senses,and lie in planes perpendicular to the longitudinal axis,the member is said to be in torsion.Screwdriver bar Transmission shaftTorque: a moment that tends to twist a member about its longitudinal axis. Shaft: a member that deforms mainly in torsion.Assumptions ：•The cross section remains a plane,and the size and shape remain the same (a rigid plane).•The cross section rotates about the longitudinal axis through an angle.The small element on the surface is under pure shear:By observation ：Longitudinal lines : remain straight , twisted angle;the length of shaft remains unchangedRadial lines : remain straight and rotate about the center of the cross sectionCircumferential lines : remain the same and rotate about the longitudinal axis TORSIONAL DFORMATION OF A CIRCULAR SHAFT t M tMAngle of twist : the relative angular displacement between two cross-sections,()x φ The angle of twist varies linearly with x .()x φφ∆The angle of twist of the back face: ()x φThe angle of twist of the front face: ()x φφ+∆causes the element to be subjected to a shear strain.Isolate an element from the shaft After deformation Before deformationIf andx dx ∆→d φφ∆→Since and are the same for all points on the cross section at x dφdxThe shear strain within the shaft varies linearly along any radial line from zero to .max γρd constant dx φ=at a specific position x c cdx d ///max γργφ==max γργ⎪⎭⎫ ⎝⎛=cImportant points (review)☐Assumptions:for a shaft with a circular cross section subjected to a torque.•The cross-section remains a plane•The length of the shaft and its radius remain unchanged•Its radial lines remain straight and circles remain circular☐The shear strain varies linearly along any radial line.r If the material is linear-elastic, then Hooke’s law applies, G τγ=A linear variation in shear strain leads to a corresponding linear variation in shear stress . The integral depends only on the geometry of the shaft.2max max A T dA J c c ττρ==⎰Polar moment of inertia max ()()A A T dA dA c ρρτρτ==⎰⎰Each element of area located at ,is subjected to aforce of .The resultant internal torque producedby this force is dA ρ()dT dA ρτ=()dF dA τ=TORSIONAL FORMULAmax γργ⎪⎭⎫ ⎝⎛=c max τρτ⎪⎭⎫ ⎝⎛=cT J ρτ=wheremaxτT :c :J :Torsional formulaPolar moment of inertia()cc A cd d dA J 0403022)41(222ρπρρπρπρρρ⎰⎰⎰====42c J π=Note: J is a geometric property of the circular area and is always positive. The commonunit is mm 4Solid shaft Tubular shaft ()4402i c c J -=πO 2cρρd d 2d =A O 2c i 2c 0ρρdThe internal torque T develops a linear distribution of shear stress along each radial line in the plane of the cross-section,it also develops an associated shear-stress distribution along an axial plane.Why?T Jρτ=The internal torque T develops a linear distribution of shear stress along each radial line in the plane of the cross-section,it also develops an associated shear-stress distribution along an axial plane.Why?(complementary property of shear stress))(x TAbsolute Maximum Torsional Stress of a shaft -A torque diagramThis diagram is a plot of the internal torque T versus its position x along the shaft length. Sign convention: By right-hand rule, if the thumb directs outward from the shaft, then the internal torque is positive.The Procedure of Analysis to determine the shear stress for a shaft under torsion1. Internal Loading (Torque)Section the shaft perpendicular to its axis at the point where the shear stress is to be e free-body diagram and the equilibrium equations to obtain the internal torque.2. Cross Section Properties （Polar moment of inertia ）3. Shear Stress Distribution (Torsional formula)T J ρτ=42c J π=()4402i c c J -=πJTc =max τThe shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at point A and B , located at section a-a of the shaft.Example 1ABInternal Torque.The bearing force reactions on the shaft are zero, since the applied torques satisfy moment equilibrium about the shaft’s axis.The free body diagram of the left segment.mkN T T m kN m kN M x ⋅==-⋅-⋅=1250030004250 ;0∑ABmkN T ⋅=1250AB Cross Sectional Property. The polar moment of inertia for the shaft is 474 )10(97.4) 75(2mm mm J ==πShear Stress. Since point A is at ρ=c = 75 mm and point B at ρ= 15 mmGPa mm kN mm mm m kN J Tc A 89.1/89.1)10(97.4)75)(1250(247==⋅==τGPa mmmm m kN J T B 377.0)10(97.4)15)(1250(47=⋅==ρτAns.Important points (review)☐For a linear elastic homogenous material, the shear stress along any radial line of the shaft varies linearly from zero to a maximum value at the outer surface.☐The shear stress is also linearly distributed along an adjacent axial plane of the shaft due to the complementary property of shear stress.☐Torsional formula:valid for a shaft with circular cross-section and made of homogenous material with a linear-elastic behavior.Deformation of the shaftdxd γφρ=(linear elastic material)d dxρφγ=)(/)(x J x T ρτ=Gx J x T )(/)(ργ=γτG =dxGx J x T d )()(=φANGLE OF TWISTdxGx J xT d )()(=φIntegrating over the entire length L of the shaft,⎰=Ldx Gx J x T 0)()(φTORSIONJGTL =φ∑=JGTL φConstant Torque and Cross-Sectional Area⎰=L dx Gx J x T 0)()(φ⎰=Ldx E x A x P 0)()(δAEPL =δ(Axially loaded bar)TTT 2T 1T 3Sign ConventionRight-hand rule,the torque and angle will be positive,provided the thumb is directed outward from the shaftTo determine the angle of twist of one end of a shaft with respect to the other end:1. Internal Loading (Torque)•The method of section and the equation of moment equilibrium2. Angle of Twist•The polar moment of inertia J (x );•or •A consistent sign convention for the shaftGdx x J x T )(/)(⎰=φJG TL /=φ✓If the torque varies continuously along the shaft’s length, a section should be made at the arbitrary position, T (x )✓If several constant external torques exist, the internal torques in each segment between any two external torques much be determined (a torque diagram)The Procedure of AnalysisExample 2The two solid steel shafts shown are coupled together using the meshed gears. Determine the angle of twist of end A of the shaft AB when the torque T=45N·m is applied.G=80GPa.Shaft AB is free to rotate within bearing E and F,whereas shaft DC is fixed at D.Each shaft has a diameter of20mm.1. Internal Torque. F ree body diagrams of the shafts are shown in figures. Step 1:Solution300Nm 150.0m/45N m 150.0/,0m 150.00=⋅===⨯-→=∑T F F T MABx ()()mN 5.22m 075.0N 003m 075.0,0-m 075.00⋅=⨯=⨯==⨯=F T T F Mx D x D CDx ∑→2. Angle of Twist.To solve the problem, we need to calculate the rotation of gear C with respect to the fixed end D in shaft DC .rad 0269.0]N/m )10(80[)m 010.0)(2/()m 5.1)(m N 5.22(294/+=⋅+==πφJG TL DC DC Since the two gears are in mesh, of gear C causes gear B to rotate C φBφrad0134.0 )m 075.0)(rad 0269.0()m 15.0(=→=B B φφm 010.0=c GPa80=GThen we need to determine the angle of twist of end A with respect to end B of shaft AB .The rotation of end A is therefore determined by adding and since bothangles are in the same direction .B φB A /φrad0.0850 rad 0.0716rad 0134.0/+=+=+=B B A A φφφAns.m 010.0=c GPa 80=G rad0134.0 =B φHomework assignments: 5-3, 5-9, 5-27, 5-38,5-58, 5-71()()()()()[]rad 0716.0m/N 1080m 010.02/m 2m N 45294/=⋅+==πϕJG TL AB BAEquilibrium:;0=--=∑B A xT T T MThere are two unknowns, therefore this problem is indeterminate.Compatibility or the kinematic condition: two ends are fixed:/=B A φSTATICALLY INDETERMINATE TORQUE-LOADED MEMBERS=-JGL T JG L T BC B AC A ()BC ACL LL +=⎪⎭⎫⎝⎛=L L T T BC A ⎪⎭⎫ ⎝⎛=L L T T AC B TORSION；0∑==B A xT T T M-- ；0=/BA φThe Procedure of AnalysisTo determine the unknown torques in statically indeterminate shafts:•Equilibrium equationsDraw a free-body diagram of the shaft to identify all internal torques.Write the equations of moment equilibrium about the axis of the shaft.•CompatibilityExpress the compatibility condition in terms of the rotational displacements caused by the reactive torques.•Solving unknownsSolve the equilibrium and compatibility equations for the unknown reactive torques.Pay attention to the sign of the results.Example 3The shaft is made from a steel tube,which is bonded to a brass core.A torque of T=250N·m is applied at its end,plot the shear-stress distribution along a radial line of its cross-section.(G st=80GPa,G br=36GPa)Solution:Equilibrium. A free-body diagram of the shaft . The reaction has been represented by twounknown amount of torques resisted by the steel, T st and by the brass, T br .m N 250=⋅+--br st T T Compatibility. The angles of twist at fixed end A should be the same for both the steel and brass since they are bonded together .brst φφφ==Applying the load-displacement relationship , , we haveJG TL /=φbrbr br st st st J G LT J G L T =brst T T 33.33=(1)(2)m N 28.7m N 72.242⋅=⋅=⇒br st T TThe shear stress in the brass core varies from zero at its center to the maximum at the interface, using the torsional formulaMPa 63.4mm)/2)(10(mm)mm/m)(10 m)(10N 28.7()(43max =⋅==πτbr br br J c T For the steel, the minimum shear stress is at this interfaceMPa 30.10])mm 10(mm) /2)[(20(mm) mm/m)(10 m)(10N 72.242()(443min=-⋅==πτst inner st st J c T The maximum shear stress is at the outer surfaceMPa 630.20])mm 10(mm) /2)[(20(mm)mm/m)(20 m)(10N 72.242()(443max =-⋅==πτst outer st st J c Tm N 28.7mN 72.242⋅=⋅=br st T Trad )10(1286.0N/mm)10(36N/mm 63.43232-===G τγAns.Homework assignments: 5-75, 5-82, 5-85The shear stress is discontinuous at the interface because the materials have different moduli .The stiffer material (steel)carries more shear stress.However,the shear strain is continuous at the interface.Shear strain at the interface:MPa63.4)(max =br τMPa30.10)(min =st τMPa630.20)(max =st τ。