CHAPTER I. 1. VECTORS and SCALARS

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1 Engineering Mechanics I - Statics CHPTER I 1. VECTORS and SCLRS 1.1 Introduction Mechanics is a phsical science which deals with the state of rest or motion of rigid bodies under the action of forces. It is divided into three parts: mechanics of rigid bodies, mechanics of deformable bodies, and mechanics of fluids. Thus it can be inferred that Mechanics is a phsical science which deals with the eternal effects of force on rigid bodies. Mechanics of rigid bodies is divided into two parts: Statics and Dnamics. Statics: deals with the equilibrium of rigid bodies under the action of forces. Dnamics: deals with the motion of rigid bodies caused b unbalanced force acting on them. Dnamics is further subdivided into two parts: ƒ Kinematics: dealing with geometr of motion of bodies with out reference to the forces causing the motion, and ƒ Kinetics: deals with motion of bodies in relation to the forces causing the motion. Basic Concepts: The concepts and definitions of Space, Time, Mass, Force, Particle and Rigid bod are basic to the stud of mechanics. In this course, the bodies are assumed to be rigid such that what ever load applied, the don t deform or change shape. But translation or rotation ma eist. The loads are assumed to cause onl eternal movement, not internal. In realit, the bodies ma deform. But the changes in shapes are assumed to be minimal and insignificant to affect the condition of equilibrium (stabilit) or motion of the structure under load. When we deal Statics/Mechanics of rigid bodies under equilibrium condition, we can represent the bod or sstem under a load b a particle or centerline. Thus, the general response in terms of other load of the bodies can be spotted easil. Fundamental Principles The three laws of Newton are of importance while studing mechanics: First Law: particle remains at rest or continues to move in a straight line with uniform velocit if there is no unbalanced force on it. Second Law: The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force. F = m a Third Law: The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. 1 UU, Department of Mining Engineering Instructor: Tewodros N.

2 Engineering Mechanics I - Statics The first and third laws have of great importance for Statics whereas the second one is basic for dnamics of Mechanics. nother important law for mechanics is the Law of gravitation b Newton, as it usual to compute the weight of bodies. ccordingl: F = G m 1 m 2 r 2 thus the weight of a mass m W = mg 1.2 SCLRS ND VECTORS 1.2.1Definition and properties fter generall understanding quantities as Fundamental or Derived, we shall also treat them as either Scalars or Vectors. Scalar quantities: - are phsical quantities that can be completel described (measured) b their magnitude alone. These quantities do not need a direction to point out their application (Just a value to quantif their measurabilit). The onl need the magnitude and the unit of measurement to full describe them. E.g. Time[s], Mass [Kg], rea [m2], Volume [m3], Densit [Kg/m3], Distance [m], etc. Vector quantities: - Like Scalar quantities, Vector quantities need a magnitude. But in addition, the have a direction, and sometimes point of application for their complete description. Vectors are represented b short arrows on top of the letters designating them. E.g. Force [N, Kg.m/s2], Velocit [m/s], cceleration [m/s2], Momentum [N.s, kg.m/s], etc Tpes of Vectors Generall vectors fall into the following three basic classifications: Free Vectors: are vectors whose action in space is not confined or associated with a unique line in space; hence the are free in space. E.g. Displacement, Velocit, cceleration, Couples, etc. Sliding Vectors: are vectors for which a unique line in space along the action of the quantit must be maintained. E.g. Force acting on rigid bodies. NB: From the above we can see that a force can be applied an where along its line of action on a rigid bod with out altering its eternal effect on the bod. This principle is known as Principle of Transmissibilit. 2 UU, Department of Mining Engineering Instructor: Tewodros N.

3 Engineering Mechanics I - Statics Fied Vectors: are vectors for which a unique and well-defined point of application is specified to have the same eternal effect. E.g. Force acting on non-rigid (deformable) bodies Representation of Vectors ) Graphical representation Graphicall, a vector is represented b a directed line segment headed b an arrow. The length of the line segment is equal to the magnitude of the vector to some predetermined scale and the arrow indicates the direction of the vector. Tail θ Head Length of the line equals, to some scale, the magnitude of the vector and the arrow indicates the direction of the vector NB: The direction of the vector ma be measured b an angle υ from some known reference direction. B) lgebraic (arithmetic) representation lgebraicall a vector is represented b the components of the vector along the three dimensions. E.g.: = a i + a j + az k, Where a, a and a are components of the vector aes respectivel. z along the, and z NB: The vectors i, j and k are unit vectors along the respective aes. a = cosθ = l, l = cosθ a = cosθ = m, m = cosθ az = cosθ z = n, n = cosθ z, where l, m, n are the directional cosines of the vector. Thus, 2 = a 2 + a 2 + a 2 l 2 + m 2 + n 2 = 1 z Properties of vectors Equalit of vectors: Two free vectors are said to be equal if and onl if the have the same magnitude and direction. B C 3 UU, Department of Mining Engineering Instructor: Tewodros N.

4 Engineering Mechanics I - Statics The Negative of a vector: is a vector which has equal magnitude to a given vector but opposite in direction. - Null vector: is a vector of zero magnitude. null vector has an arbitrar direction. Unit vector: is an vector whose magnitude is unit. unit vector along the direction of a certain vector, sa vector (denoted b u ) can then be found b dividing vector b its magnitude. u = Generall, an two or more vectors can be aligned in different manner. But the ma be: * Collinear-Having the same line of action. * Coplanar- Ling in the same plane. * Concurrent- Passing through a common point. 1.3 Operations with Vectors Scalar quantities are operated in the same wa as numbers are operated. But vectors are not and have the following rules: Vector ddition or Composition of Vectors Composition of vectors is the process of adding two or more vectors to get a single vector, a Resultant, which has the same eternal effect as the combined effect of individual vectors on the rigid bod the act. There are different techniques of adding vectors ) Graphical Method I. The parallelogram law The law states, if and B are two free vectors drawn on scale, the resultant (the equivalent vector) of the vectors can be found b drawing a parallelogram having sides of these vectors, and the resultant will be the diagonal starting from the tails of both vectors and ending at the heads of both vectors. B (a.) B R (b.) B 4 UU, Department of Mining Engineering Instructor: Tewodros N.

5 Engineering Mechanics I - Statics Once the parallelogram is drawn to scale, the magnitude of the resultant can be found b measuring the diagonal and converting it to magnitude b the appropriate scale. The direction of the resultant with respect to one of the vectors can be found b measuring the angle the diagonal makes with that vector. Note: s we can see in the above figure. + B = R = B +, vector addition is commutative The other diagonal of the parallelogram gives the difference of the vectors, and depending from which verte it starts, it represents either B or B - - B B - B B - Diagonal = B Diagonal = B Since the two diagonal vectors in the above figure are not equal, of course one is the negative vector of the other, vector subtraction is not commutative. i.e. B B NB. Vector subtraction is addition of the negative of one vector to the other. II. The Triangle rule The Triangle rule is a corollar to the parallelogram aiom and it is fit to be applied to more than two vectors at once. It states If the two vectors, which are drawn on scale, are placed tip (head) to tail, their resultant will be the third side of the triangle which has tail at the tail of the first vector and head at the head of the last. R B R = + B Thus the Triangle rule can be etended to more than two vectors as, If a sstem of vectors are joined head to tail, their resultant will be the vector that completes the polgon so formed, and it starts from the tail of the first vector and ends at the head of the last vector. R C B R = + B + C NB. From the Triangle rule it can easil be seen that if a sstem of vectors when joined head to tail form a closed polgon, their resultant will be a null vector. III. naltic method. 5 UU, Department of Mining Engineering Instructor: Tewodros N.

6 Engineering Mechanics I - Statics The analtic methods are the direct applications of the above postulates and theorems in which the resultant is found mathematicall instead of measuring it from the drawings as in the graphical method.. Trigonometric rules: The resultant of two vectors can be found analticall from the parallelogram rule b appling the cosine and the sine rules. Consider the following parallelogram. nd let υ be the angle between the two vectors B β D R θ α B C Consider triangle BC From cosine law, 2 R = B B 2 B cos(θ ) R = 2 + B 2 2 B cos(θ ), This is the magnitude of the RESULTNT of the two vectors, Similarl, the inclination,β, of the resultant vector from can be found b using sine law sin β B = sin θ R β = sin 1 sin θ * B ), R, which is the angle the resultant makes with vector. Decomposition of vectors: Decomposition is the process of getting the components of a given vector along some other different ais. Practicall decomposition is the reverse of composition. Consider the following vector. nd let our aim be to find the components of the vector along the n and t aes. D t C Φ Φ θ t n α θ n (a) (b) From Triangle (B), α = 180 ( θ + φ ) B From sine law then, 6 Instructor: braham ssefa

7 Engineering Mechanics I - Statics n sin φ = n = sin φ sin α sin α Similarl, t = sinθ sin α The above are general epressions to get the components of a vector along an ais. In most cases though, components are sought along perpendicular aes, i.e. α=180-(θ+φ) = 90 sin α = 1 n = sin φ = cosθ t = sinθ = cosφ B. Component method of vector addition This is the most efficient method of vector addition, especiall when the number of vectors to be added is large. In this method first the components of each vector along a convenient ais will be calculated. The sum of the components of each vector along each ais will be equal to the components of their resultant along the respective aes. Once the components of the resultant are found, the resultant can be found b parallelogram rule as discussed above. 1.4 Vector Multiplication: Dot and Cross products Multiplication of vectors b scalars Let n be a non-zero scalar and be a vector, then multipling b n gives as a vector whose magnitude is n and whose direction is in the direction of if n is positive or is in opposite direction to if n is negative. Multiplication of vectors b scalars obes the following rules: i. Scalars are distributive over vectors. n( + B) = n + nb ii. Vectors are distributive over scalars. (n + m) = n + m iii. Multiplication of vectors b scalars is associative. (nm) = n(m ) = m(n ) Multiplication of vector b a vector In mechanics there are a few phsical quantities that can be represented b a product of vectors. Eg. Work, Moment, etc There are two tpes of products of vector multiplication 7 Instructor: braham ssefa

8 Engineering Mechanics I - Statics Dot Product: Scalar Product The scalar product of two vectors and B which are θ degrees inclined from each other denoted b B cosθ.b ( dot B) will result in a scalar of magnitude.b = B cosθ i.e If the two vectors are represented analticall as = a i + a j + a z k and B = b i + b j + b z k, then.b = a b + a b + a z b z Cross Product: Vector Product The vector product of two vectors and B that are θ degrees apart denoted b B ( cross B) is a B sin θ vector of magnitude and direction perpendicular to the plane formed b the vectors and B. The sense of the resulting vector can be determined b the right-hand rule. B = B sin θ, perpendicular to the plane formed b and B i.e. If the two vectors are represented analticall as, = a i + a j + a z k and B = b i + b j + b z k then the cross product B will be the determinant of the three b three matri as, i j k r M = a a a z b b b z B = ( a b z a z b )i + ( a z b a b z ) j + ( a b a b ) k NB. Vector product is not commutative; in fact, B = B Moment of a Vector The moment of a vector V about an point O is given b: r r M o = r V Where: r is a position vector from point O to an point on the line of action of the vector. O i j k O r r r z r V V V z V Position vector r is defined as a fied vector that locates a point in space relative to another point in space. 8 Instructor: braham ssefa

9 Engineering Mechanics I - Statics CHPTER TWO 2. FORCE SYSTEMS 2.1 Introduction Def n : force can be defined as the action of one bod on another that changes/tends to changes the state of the bod acted on. force can be applied on a bod as; Contact force:-pplied b direct mechanical contact of the acting bod on the acted one (Created b push and pull). Remote action (Bod force):-pplied b remote action as in gravitational, electrical, Magnetic, etc forces. The action of a force on a bod can be divided as internal and eternal. Internal force is a force eerted b one part of a bod on another part of the same bod. Eternal force is a force eerted on a bod b some other bod. n eternal force can then be applied on a bod as: pplied force Reactive force In Engineering mechanics, onl eternal effects of forces, hence eternal forces are considered Force sstems sstem of forces can be grouped into different categories depending on their arrangement in space. Coplanar Forces:-are forces which act on the same plane. Depending on their arrangement on the plane too, coplanar forces can further be divided as: Coplanar collinear forces:-are coplanar forces acting on the same line-collinear. Coplanar parallel forces:-re forces which are on the same plane and parallel Coplanar concurrent forces:-re forces on the same plane whose lines of action intersect at a point. General coplanar forces: Non coplanar forces:-are forces which act on different planes 9 Instructor: braham ssefa

10 Engineering Mechanics I - Statics gain it can further be broken as I. Non coplanar parallel forces:-is sstem of non planar forces but which are parallel. II. Non-coplanar concurrent forces:-are non planar forces whose lines of action meet at a point. III. General Non coplanar force: Composition and Resolution of Forces Composition of forces Composition of forces is the process of combining two or more forces in to a single resultant force, which has the same eternal effect as that of the applied sstem of forces. In the previous chapter we defined force to be either sliding or fied vector depending on what tpe of bodies it acts-rigid or deformable bodies respectivel. In engineering mechanics we will be considering rigid bodies onl; hence we can treat force as sliding vector. s discussed in the previous chapter, we have two laws of adding vectors: The parallelogram rule The triangle rule The parallelogram rule Consider the following planar force sstems acting on the rigid bod. 10 Instructor: braham ssefa

11 Engineering Mechanics I - Statics F 2 F 2 F 1 R F 1 a b To appl the parallelogram rule of vector addition, the vectors should be placed in such a wa that the form a parallelogram and the shouldn t change their eternal effect. The principle of transmissibilit states that force ma be applied at an point on its line of action, without altering the resultant effects of the force eternal to the rigid bod on which it acts. Thus, b the principle of Transmissibilit we can move each force on its line to meet at with out affecting the eternal effect as shown in (b) above. Once the parallelogram is formed the resultant can be found as in the previous chapter and its line of action will pass through. The triangle rule The triangle law can also be used to find the resultant of the above forces, but it needs moving of the line of action of one of the forces. We can change the line of action of either force, but the start of the first vector should coincide with the point of intersection of the line of action of the forces. F 2 R F 1 F 2 R F 1 F 1 R F 2 a F 2 moved out of its line of action but F 1 starts at b F 1 moved out of its line of action but F 2 starts at c The first vector, F 2 doesn t start at the intersection of the line of action of the two vectors s can be seen in the (a) and (b) parts the first force in the combination starts at the intersection of the line of action of the forces. In the (c) part, however, the first force F 2 doesn t start at the intersection. lthough the resulting resultant has the same magnitude and direction as the previous ones, its line of action is different; hence its eternal effect is also different Resolution of forces 11 Instructor: braham ssefa

12 Engineering Mechanics I - Statics Resolution as defined earlier is the reverse of composition. It is the process of getting the components of a vector along different aes. t Φ θ F n F t Φθ F n F α s derived earlier F n = F F t = F sin sin sin sin φ α θ α I. Two Dimensional Force Sstems 2.2. Rectangular components Rectangular components of a force are the components of the force along the rectangular coordinate aes. Y F F s α=90, F = Fsin(90-θ)=Fcosθ F =Fsinθ θ F X lgebraicall, a force is represented b its scalar component along the coordinate aes and a unit vector along that ais. F = F i + F j Where F and F are the scalar components of F along the X and Y-aes, and i and along the and -aes respectivel. 12 j are unit vectors NB. Depending to which quadrant the vector corresponds; the scalar components can be negative Equivalent force sstems (Moments and Couples) Moment In addition to its tendenc to move a bod in the direction of its application, a force also tends to rotate the bod about an ais which doesn t intersect the line of action of the force and which is not parallel to it. This tendenc of a force to rotate a bod about a given ais is known as the moment, M, of the force. The moment of a force is also known as torque. Instructor: braham ssefa

13 Engineering Mechanics I - Statics The fig below shows a two-dimensional bod acted upon b a force F in its plane. M d F The ais of the f O orce to rotate the bod about the ais O-O normal to the plane of the bod hence the moment is proportional both to the magnitude of the force and the moment arm d, which is the perpendicular distance from the ais to the line of action of the force. Hence the magnitude of the moment is defined as: M = F d The moment will be a vector perpendicular to the plane of the bod-parallel to the ais o-o-and its sense depends on the direction in which the force tends to rotate the bod. The right hand rule can be used to identif this sense; curl our fingers in the direction of the tendenc to rotate, the thumb will point in the direction of the moment vector. For a coplanar force sstem vector representation of the moment is unnecessar as it can be represented as its tendenc to rotate that plane-clockwise or counter clockwise. Sign convention In representing moment b its tendenc to rotate, it is a good practice to assign one of the senses, clockwise or counter clockwise, a positive direction and the other negative. Here we will be treating counter clockwise moment as positive moment and clockwise moment as negative moment. Note: - One can assign the positive sense to either the clockwise or counterclockwise moments. What is important is in a given problem; he should stick to his assignment. Principle of moment: - One of the most important principles in mechanics is Varignon s theorem, or the principle of moment, which for coplanar forces states the moment of a force about an point is equal to the sum of the moments of the components of the force about the same point. Proof Let P and Q be the components of R as shown in fig below C D R Q β r P B F q p O E G α θ Let point O be an arbitrar point in the plane of the forces through which the moments are sought. 13 U, FoT, Department o O f Civil Engineering Instructor: braham ssefa

14 Engineering Mechanics I - Statics Draw the - aes b coinciding the -ais with the line joining O with. CG = GF+CF =Rsinα CF = Psinβ BE = GF = Qsinθ Rsinα = Psinβ + Qsinθ Constructing perpendicular distances (moment arms) to all the forces from O: p, Sinα = r q & Sinβ = Sinθ = O O O R r O = P p q + Q O O Rr = Pp + Qq, Which proves Varignon' s theorem In case it is easier to find the moments of the components of a force about an ais through a point, one can then easil determine the moment of the force about an ais through that same point b appling the principle of moment Couples The moment produced b two equal and opposite and non-collinear forces is known as couple. Consider the action of equal and opposite forces F and F a distance d apart. O a d -F F O These two forces can t be combined in to a single force of the same effect on the bod, as their sum in ever direction is zero. But the effect of the forces on the bod isn t zero. The combined moment of the two forces about an ais normal to their plane and passing through an point such as o in their plane is the couple, M. The magnitude of the couple M = F (a + d ) Fa = Fd It can, therefore, be concluded that the moment of a couple is independent of the moment center selected-hence a couple can be represented as free vector. NB. couple is unchanged as long as the magnitude and direction of its vector remains constant, i.e. a given couple will not be altered b changing the value of F and d as long as their product remains the same. Likewise a couple is not affected b allowing the forces act in an one of parallel planes. 14 Instructor: braham ssefa

15 Engineering Mechanics I - Statics -F d F M M M F M -2F 2F ½ d -F d F -F d 2.4. Resolution of a force into a force and a couple The effect of a force on a bod has been described in terms of the tendenc to push or pull the bod in the direction of the force and to rotate the bod about an ais which doesn t intersect the line of the force. The representation of this dual effect can be facilitated b replacing the given force b an equal and parallel force at the new point sought and a couple to compensate for the change in the moment of the forces. The resolution process can best be illustrated b the following figures. B -F B F B F d = = F M=Fd F The given force F acting at is replaced b an equal force at point B and the anti clockwise couple M=F d. The transfer process can be seen from the middle figure and it involves the following procedure: ppl two equal and apposite forces of F and F at B where F is equal in magnitude and parallel to the force acting at. The forces applied at B will cancel each other hence the will have no effect on the bod. The Forces F at and F at B form a couple; hence can be replaced b the counter clockwise couple M=F d The original force at can be replaced b an equal and parallel force at B and a corresponding couple as shown in the right figure. Note: - The transformation described above can be performed in the reverse order. i.e. force F acting at a point B and a couple M acting on the bod can be combined into a single resultant force. This is performed b moving F until its moment about B becomes equal to the moment M of the couple to be ecluded Resultants The resultant of a force sstem is the simplest force combination that can replace the original forces without altering the eternal effect of the sstem on the rigid bod to which the forces are applied. The equilibrium of a bod is the condition where the resultant if all forces that act on it is zero. When the resultant is not zero, the acceleration of the bod is described b equating the force 15 Instructor: braham ssefa

16 Engineering Mechanics I - Statics resultant to the product of the mass and the acceleration of the bod. Thus, the determination of the resultant is basic to both statics and dnamics. The most common tpe of force sstem occurs when the forces all act in a single plane (coplanar forces). The resultant can be computed b using the parallelogram rule or using analtical methods.. Parallelogram rule F 2 F 2 F 1 R F 1 a b B. naltic Method II. Three Dimensional Force Sstems z 2.6. Rectangular Components F F z k θz θ θ F i F j 16 Instructor: braham ssefa

17 Engineering Mechanics I - Statics F i, F j and F z k are rectangular components of F. Thus, F = F i + F j + F z k, F = Fcosθ, F = Fcosθ, F z = Fcosθ z F = Fcosθ i + Fcosθ j + Fcosθ z k = F (cosθ i + cosθ j + cosθ z k) F = n F, F = F 2 + F 2 + F z 2 n = unit vector in the direction of F In solving three-dimensional problems, one must usuall find the, and z scalar components of a given or unknown force. In most cases, the direction of a force is described; / b two points on the line of action of the force, or B/ b two angles which orient the line of action.. Using two points: If the coordinates of points ( 1, 1,z 1 ) and B( 2, 2,z 2 ) on the line of action of the force are known and the direction of the force is from to B, the force ma be written as; F = F n B = F.B/ B = F. ( 2 1 )i + ( 2 1 )j + (z 2 z 1 )k B. Using two angles: ( 2-1 ) 2 + ( 2-1 ) 2 + (z 2 -z 1 ) 2 z F z F Φ F θ F F F = F i + F j + F z k F = F.cosΦ, F z = F.sinΦ, F = F.cosθ = F.cosΦ.cosθ, F = F.sinθ = F.cosΦ.sinθ Rectangular components of a force F ma be written with the aid of dot or scalar product operation. If the unit vector n = αi + βj + γk and F = F(li + mj + nk), the projection of F in the n direction is given b: i/ s a scalar Fn = F. n = F(li + mj + nk). (αi + βj + γk) = F(lα + mβ + nγ) 17 Instructor: braham ssefa

18 Engineering Mechanics I - Statics ii/ s a vector Fn = (F.n) n If θ is the angle between F and n, F. n = Fncosθ, θ = cos-1((f. n)/ F ) It should be observed that the dot product relationship applies to non intersecting vectors as well as to intersecting vectors Moment and Couple In three dimensions, the determination of the perpendicular distance between a point or line and the line of action of the force can be a tedious computation. The use of a vector approach using cross-product multiplication becomes advantageous Moment z F z M Z F F r F O r z r M r M The moment M o of F about an ais through O is given b; Mo = r X F = (r i + r j + r z k) X (F i + F j + F z k) = i j k r r r z F F F z = (r.f z r z.f )i + (r z.f r.f z )j + (r.f r.f )k =M i + M j + M z k 18 Instructor: braham ssefa

19 If n is a unit vector in the λ direction, the moment M λ of F about an ais λ through O is epressed b; M λ = Mo. n = ( r X F. n) --- which is the scalar magnitude. Or M λ = (Mo. n) n = (r X F. n) n - vector epression for the moment of F about an ais λ through O Couple M -F d F B r r B O r -If the vector r joins an point B on the line of action of F to an point on the line of action of F. The combined moment (couple) of the two forces about O is; M = r X F + r B X (-F) = (r r B ) X F = r X F The moment of the couple, M = r X F. It is the same about all points. Thus, - The moment of a couple is a free vector, whereas the moment of a force about a point (which is also the moment about a defined ais through the point) is a sliding vector whose direction is along the ais through the point. - couple tends to produce a pure rotation of the bod about an ais normal to the plane of the forces which constitute the couple Resolution of a force into a couple and a force. Force-Couple Sstem F M = r X F F = B r B The force F at point is replaced b an equal force F at point B and the couple M = r X F. Couple vectors obe all of the rules, which govern vector quantities. 19 Instructor: braham ssefa

20 2.8. Resultants -n sstem of forces ma be replaced b its resultant force R and the resultant couple M. For the sstem of forces F1, F2, F3, acting on a rigid bod, the resultant force R and the resultant couple M is given b; R = F1 + F2 + F = Σ Fi M = M1 + M2 + M = Σ Mi Eg. M1 M2 O F1 F1 r 2 r 1 = O F2 F2 M1 = r1 X F1, M2 = r2 X F2 = O M R = F1 + F2. r 1, r 2 = vectors from O to an point on the line of action of F1 and F2 respectivel. In three dimensions, the magnitudes of the resultants and their components are; R = Σ F, R = Σ F, R z = Σ F z R = (ΣF ) 2 + (ΣF ) 2 + (ΣF z ) 2 M = Σ (r X F), M = Σ (r X F), M z = Σ (r X F) z M = M 2 + M 2 + M z 2 The magnitude and direction of M depends on the particular point selected. The magnitude and direction of R, however, are the same no matter which point is selected WRENCH RESULTNT When the resultant couple vector M is parallel to the resultant fore R, as shown in figure below, the resultant is said to be a wrench. common eample of a positive wrench is found with the application of screwdriver. 20 Instructor: braham ssefa

21 n general force sstem ma be represented b a wrench applied along a unique line of action as shown below. 21 Instructor: braham ssefa

22 CHPTER III 3. EQUILIBRIUM 3.1 Introduction In the previous chapter, we have seen sstems of forces. In this chapter stabilit of force sstems, named as equilibrium of a bod. Thus a bod is said to be in equilibrium when the resultant of all the forces acting on it is zero. That is, the resultant force vector R and the resultant couple vector r M are both zero. Epressed mathematicall: r R = F = 0 ; M = M = 0 Note that these are both necessar and sufficient conditions for equilibrium. 3.2 Equilibrium in Two-Dimension Mechanical sstem isolation and free bod diagram (FBD) Before considering equilibrium conditions, it is ver much essential and absolutel necessar to define unambiguousl the particular bod or mechanical sstem to be analzed and represent clearl and completel all forces which act on the bod. Modeling the action of forces in Two Dimensional nalsis The most important step in drawing the free bod diagram will be to show the eternal forces eerted on the rigid bod. On of the eternal forces will be forces eerted b contacts with supports and reactions. The different support and contact forces are shown in the figure below. diagram showing a bod/group of bodies considered in the analsis with all forces and relevant dimensions is called free bod diagram (FBD). It is after such diagram is clearl drawn that the equilibrium equations be used to determine some of the unknown forces. Therefore free bod diagram is the most important single step in the solution of problems in mechanics. Steps for the construction of free bod diagram Decide which bod or combination of bodies to be considered. The bod or combination chosen is isolated b a diagram that represents its complete eternal boundar. ll forces that act on the isolated bod b the removed contacting and attracting bodies and known forces represented in their proper positions on the diagram of the isolated bod. 22 Instructor: braham ssefa

23 Each unknown force should be represented b a vector arrow with the unknown magnitude and direction. The fore eerted on the bod to be isolated b the bod to be removed is indicated and its sense shall be opposite to the movement of the bod which would occur if the contacting or supporting member were removed. 23 Instructor: braham ssefa

24 The choice of coordinate aes should be indicated directl on the diagram and relevant dimensions should be represented Equilibrium Conditions It was stated that a bod is in equilibrium if the resultant force vector and the resultant couple vector are both zero. These requirements can be stated in the form of vector equation of equilibrium, which in two dimensions can be written as: ΣF X = 0 ΣF Y = 0 ΣMo =ΣMz = 0 Categories of Equilibrium in Two Dimensions The following categories of equilibrium conditions can be identified due to the nature of forces considered. Categories of equilibrium in two-dimension Force Sstem Free Bod diagram Independent Equations 1. Collinear F 2 F 3 Σ F X = 0 F 1 2. Concurrent F F 4 1 F 2 F 3 ΣF X = 0 ΣF Y = 0 3. Parallel F 1 F 2 F 4 F 3 ΣF X = 0 ΣM Z = 0 F 1 F 2 4. General M F 3 ΣFX = 0 ΣFY = 0 ΣMZ = 0 F 4 lternative equilibrium equations In two-dimensional bod, the maimum number of unknown variables is three. nd the three equilibrium equations are sufficient to solve the unknown variables. Thus, whatever the combination, three total equations are maimall needed. What we have seen is two forces and one moment equations. But we could have the following combinations. 24 Instructor: braham ssefa

25 One force and two moment equations: B.. ΣF X = 0 ΣM = 0 ΣM B = 0 Three-moment equation B.. C. ΣM = 0 ΣMB = 0 ΣMC = Equilibrium in Three-Dimensions Equilibrium Conditions -The necessar and sufficient conditions for complete equilibrium in three dimensions are; Σ F = 0 OR Σ F = 0, Σ F = 0 and Σ Fz = 0 and Σ M =0 OR Σ M = 0, Σ M = 0 and Σ Mz = 0 Notes; -In appling the vector form of the above equations, we first epress each of the forces in terms of the coordinate unit vectors i, j and k. - For the first equation, Σ F = 0, the vector sum will be zero onl if the coefficients of i, j and k in the epression are, respectivel, zero. These three sums when each is set equal to zero ield precisel the three scalar equations of equilibrium, Σ F = 0, Σ F = 0 and Σ Fz = 0 - For the second equation, Σ M = 0, where the moment sum ma be taken about an convenient point o, we epress the moment of each force as the cross product r X F, where r is the position vector from o to an point on the line of action of the force F. Thus, Σ M = Σ (r X F) = 0. The coefficients of i, j and k in the resulting moment equation when set equal to zero, respectivel, produce the three scalar moment equations Σ M = 0, Σ M = 0 and Σ Mz = 0. Free Bod Diagram (FBD) shall alwas be drawn before analsis of the force sstem. Usuall either pictorial view or orthogonal projects of the FBD are used. 25 Instructor: braham ssefa

26 CEng1001- Engineering Mechanics I- Statics MODELING THE CTION OF FORCES IN THREE--DIMENSIONL NLYSIS Tpe of Contact and Force Origin ction on Bod to be Isolated 1. Member in contact with smooth surface, or ball-supported member JZ I I JZ I I Force must be normal to t.he surface and directed toward the member. 2. Member in contact with rough _ surface The possibilit eists for a force F tangent to the surface (friction force) to act on the member, as well as a normal force N. 3. Roller or wheel support with lateral constraint IZ / / '-... ' 4. BaH-and-socket joint ' JZ. 1 p /... N. '- lateral force P eerted b the guide on the wheel can eist, in addition to the normal forcen. ball-and-socket joint free to!}. Fied connection (embedded or welded) 6. Thrust-bearing support fz 26 Instructor: braham ssefa

27 27 Instructor: braham ssefa

28 CHPTER IV 4. NLYSIS OF SIMPLE STRUCTURES 4.1 Introduction - n engineering structure is an connected sstem of members built to support or transfer forces and to safel withstand the loads applied to it. - In this chapter we shall analze the internal forces acting in several tpes of structures, namel, trusses, frames and simple machines. Constraints and Statical Determinac Equilibrium equations, once satisfied, are both necessar and sufficient conditions to establish the equilibrium of a bod. However the don t necessaril provide all the information that is required to determine all the unknown forces that ma act on a bod in equilibrium. If the number of unknown forces is more than the number of independent equilibrium equations, the equilibrium equations alone are not enough to determine the unknown forces, possibl reaction forces at the constraints. The adequac of the constraints to prevent possible movement of the bod depends on the number, arrangement and characteristics of the constraints. a) Complete fiit adequate constraints b) Incomplete fiit partial constraints c) Incomplete fiit partial constraints d) Ecessive fiit redundant constraints Problem Solution It is found important to develop a logical and sstematic approach in the solution of problems of mechanics, which includes the following steps: Identif clearl the quantities that are known and unknown. Make an unambiguous choice of the bod/group of bodies/ to be isolated and draw its complete FBD, labeling all eternal known and unknown forces and couples which act on it. Designate a convenient set of aes and choose moment centers with a view to simplifing the calculations. 28 Instructor: braham ssefa

29 Identif and state the applicable force and moment principles or equations which govern the equilibrium condition of problem. Match the number of independent equations with the number of unknowns in each problem. Carr out the solution and check the results. 4.2 Plane Trusses truss is a framework composed of members joined at their ends to form a rigid structure. When the members of the truss lie essentiall in a single plane, the truss is known as a plane truss. Eamples of commonl used trusses that can be analzed as plane as plane trusses are; - i/ Bridge Trusses Pratt Howe Warren K Baltimore ii/ Roof Trusses Fink Pratt Howe 29 Warren Instructor: braham ssefa

30 The basic element of a plane truss is the triangle. Structures that are built from a basic triangle in the manner described are known as simple trusses. When more members are present than are needed to prevent collapse, the truss is staticall indeterminate. staticall indeterminate truss cannot be analzed b the equations of equilibrium alone. dditional members or supports that are not necessar for maintaining the equilibrium position are called redundant. -Three bars joined b pins at their ends constitute a rigid frame. -Four or more bars pin-jointed to form a polgon of as man sides constitute a non rigid frame. - We can make the non rigid frame stable or rigid b adding diagonal bars. -The term rigid is used in the sense of non-collapsible and also in the sense that deformation of the members due to induced internal strains is negligible. ll members in a simple truss are assumed to be two-force members. The members ma be in tension (T) or in compression ( C ). Tension Fig. Two-force members Compression The weight of truss members is assumed small compared with the force it supports. If it is not, or if the small effect of the weight is to be accounted for, the weight W of the member ma be replaced b two forces, each W/2 if the member is uniform, with one force acting at each end of the member. These forces, in effect, are treated as loads eternall applied to the pin connections. ccounting for the weight of a member in this wa gives the correct result for the average tension or compression along the member but will not account for the effect of bending of the member. - When welded or riveted connections are used to join structural members, the assumption of a pin-jointed connection is usuall satisfactor if the centerlines of the members are concurrent at the joint. - We also assume in the analsis of simple trusses that all eternal forces are applied at the pin connections. This condition is satisfied in most trusses. In bridge trusses the deck is usuall laid on cross beams that are supported at the joints. Force analsis of plane trusses Two methods for the force analsis of simple trusses will be given. The eternal reactions are usuall determined b computation from the equilibrium equations applied to the truss as a whole before the force analsis of the remainder of the truss is begun. 30 Instructor: braham ssefa

31 4.2.1 Method of joints This method for finding the forces in the members of a simple truss consists of satisfing the conditions of equilibrium for the forces acting on the connecting pin of each joint. - The method deals with the equilibrium of concurrent forces, and onl two independent equilibrium equations are involved. ( Σ F = 0 and Σ F = 0 for each joint) - We begin the analsis with an joint where at least one known load eists and where not more than two unknown forces are present. Taking free bod diagram of a joint, tension will alwas be indicated b an arrow awa from the pin, and compression will alwas be indicated b an arrow toward the pin. - In some instances it is not possible to initiall assign the correct direction of one or both of the unknown forces acting on a given pin. In this event we ma make an arbitrar assignment. negative value from the computation indicates that the assumed direction is incorrect Method of sections On the analsis of plane trusses b the method of joints, we took advantage of onl two of the three equilibrium equations, since the procedures involve concurrent forces at each joint. We ma take advantage of the third or moment equation of equilibrium b selecting an entire section of the truss for the free bod in equilibrium under the action of a non-concurrent sstem of forces. This method of sections has the basic advantage that the force in almost an desired member ma be found directl from an analsis of a section, which has cut that member. Thus it is not necessar to proceed with the calculation from joint to joint until the member in question has been reached. -In choosing a section of the truss, we note that, in general, not more than three members whose forces are unknown ma be cut, since these are onl three available equilibrium relations which are independent. - It is essential to understand that in the method of sections an entire portion of the truss is considered a single bod in equilibrium. Thus, the forces in members internal to the section are not involved in the analsis of the section as a whole. - To classif the free bod and the forces acting eternall on it, the section is preferabl passed through the members and not the joints. In some cases the methods of sections and joints can be combined for an efficient solution.. The moment equations are used to great advantage in the method of sections. One should choose a moment center, either on or off the section, through which as man unknown forces as possible pass.. It is not alwas possible to assign the proper sense of an unknown force when the free-bod diagram of a section is initiall drawn. With an arbitrar assignment made, a positive answer will verif the assumed sense and a negative result will indicate that the force is in the sense opposite to that assumed. 31 Instructor: braham ssefa

32 4.3 Frames and Simple Machines structure is called a frame or machine if at least one of its individual members is a multiforce member. multiforce member is defined as one with three or more forces acting on it or one with two or more forces and one or more couples acting on it. Frames are structures which are designed to support applied loads and are usuall fied in position. - Machines are structures which contain moving parts and are designed to transmit forces or couples from input values to output values. In this article attention is focused on the equilibrium of interconnected rigid bodies which contain multi force members. The forces acting on each member of a connected sstem are found b isolating the member with a free-bod diagram and appling the established equations of equilibrium. The principle of action and reaction must be carefull observed when we represent the forces of interaction on the separate free-bod diagrams. If the frame or machine constitutes a rigid unit b itself when removed from its supports, the analsis is best begun b establishing all the forces eternal to the structure considered as a single rigid bod. We then dismember the structure and consider the equilibrium of each part separatel. The equilibrium equations for the several parts will be related through the terms involving the forces of interaction. Rigid Non-collapsible -If the structure is not a rigid unit b itself but depends on its eternal supports for rigidit, as in the figure below, then the calculation of the eternal support reactions cannot be completed until the structure is dismembered and the individual parts are analzed. Non rigid Collapsible In most cases we find that the analsis of frames and machines is facilitated b representing the forces in terms of their rectangular components. 32 Instructor: braham ssefa

33 It is not alwas possible to assign ever force or its components in the proper sense when drawing the free bod diagrams and it becomes necessar for us to make an arbitrar assignment. -In an event it is absolutel necessar that a force be consistentl represented on the diagrams for interacting bodies, which involve the force in question. For eample, for two bodies connected b the pin in the figure below the force components must be consistentl represented in opposite directions on the separate free-bod diagrams. If we choose to use vector notation in labeling the forces, then we must be careful to use a plus sign for an action and a minus sign for the corresponding reaction. -Situations occasionall arise where it is necessar to solve two or more equations simultaneousl in order to separate the unknowns. In most instances, however, we ma avoid simultaneous solutions b careful choice of the member or group of members for the free-bod diagram and b a careful choice of moment aes which will eliminate undesired terms from the equations. 33 Instructor: braham ssefa

34 CHPTER V 5. INTERNL CTIONS IN BEMS Introduction Beams are generall horizontal structural members subjected to lateral or transversal loads, i.e. forces or moments having their vectors perpendicular to the ais of the bar. For instance, the main members supporting floors of buildings are beams. 5.1 Classification of beams and Diagrammatic Conventions Classification of beams Beams or an other structures are classified into two general parts. The are either staticall determinate or indeterminate. For staticall determinate beams, the number of unknown reactions equals three and then using the three equilibrium equations, we can determine force at an part of the structure-but not for the staticall indeterminate ones. I) Staticall determinate beams are: a) Simpl supported beams (simple beams) P w N/m L span b) Cantilever beam L w N/m L c) Overhanging beam P 1 P2 d) Compound beam (one eample) P Hinge 34 Instructor: braham ssefa

35 II) Staticall indeterminate beams a) Propped beams P 1 P 2 W N/m L L P b) Fied or restrained beam W N/m L P 1 P 2 c) Continuous beam WN/m L L 1 L 2 L Diagrammatic Conventions for supports a) Ring tpe support Beam R Pin a) ctual b) Diagrammatic R Resists horizontal & vertical forces. b) Roller tpe of support Beam C Roller R Resists vertical force onl Rollers D Instructor: braham ssefa

36 c) Link tpe of support Pins B Beam Link R Resist a force onl in the direction of line B. d) Fied support Mc R c ctual Diagrammatic R c Resists horizontal & vertical forces & moment 5.2 Diagrammatic representations of internal actions in beams The method of sections can be applied to obtain the forces that eist at a section of a beam. Consider a beam with certain concentrated & distributed forces acting on it. The eternall applied forces & the reactions at the support keep the whole bod in equilibrium. (a) P 2 W 1 P 1 W 2 R B (b) P 2 W 1 M M R P v W P 1 1 W 2 R B P v R R R B 36 Instructor: braham ssefa

37 Now consider an imaginar cut - normal to the ais of the beam, which separates the beam into two segments. If the whole bod is in equilibrium, an part of it likewise is in equilibrium. t a section of a beam, vertical forces, horizontal forces, and moments are necessar to maintain the part of the beam in equilibrium. These quantities take on a special significance in beams & therefore will be discussed separatel. SHER IN BEM To maintain the segment of a beam shown, in Fig (a) above, in equilibrium there must be an internal vertical force V at the cut to satisf the equation F =0. This internal force V, acting at right angles to the ais of the beam, is called the shear or the shearing force. Positive shear: - downward internal force acting on the left side of a cut or upward force acting on the right side of the same cut corresponds to a positive shear. That is a positive shear tend to rotate an element counterclockwise and vise versa. XIL FORCE IN BEMS In addition to the shear V, a horizontal forces such as P ma be necessar at a section of a beam to satisf the condition of equilibrium in -ais i.e. F =0 If the horizontal force P acts toward the cut, it is termed a trust; if awa from the cut, it is termed as aial tension and if it is towards to the cut, it is aial compression. In referring to either of these forces the term aial force is used. R (resultant of all forces to the left of section) +v +v +v +v Beam segment +v Fig. Definition of Positive shear 37 Instructor: braham ssefa

38 BENDING MOMENT IN BEM The remaining condition of static equilibrium for a planar problem, i.e ( M z =0) can be satisfied onl b developing a couple or an internal resisting moment within the cross sectional area of the cut to counteract the moment caused b the eternal forces. This internal resisting moment tends to bend a beam in the plane of the load and is usuall referred to as bending moment. Positive B.M +M +M +M +M +M positive BM is defined as one that produces compression in the top part and tension in the lower part of a beam's crosssection. Positive Bending Negative Bending 5.3 Tpes of Loads and reactions Beams are subjected to variet of loads. In general loads on beams can be classified as concentrated or distributed. Commonl forces are acted on beams through a post, a hanger, or a bolted detail as shown in the figure ( a) below. In such arrangements the force is applied over a ver limited portion of the beam and can be idealized as Concentrated Load for the purpose of beam analsis as shown in figure (b). Most commonl forces are applied over a considerable portion of the beam. Such forces are termed Distributed Loads. Man tpes of distributed loads occur. mong these, two kinds are particularl important: the uniforml distributed loads and the uniforml varing loads. Refer to figure (a), (b) and (c). 38 Instructor: braham ssefa

39 Finall it is also important to notice that a beam can be subjected to a Concentrated Moment essentiall at a point (as shown in the figure below). (c) 5.4 Shear, ial Force, nd Bending-Moment Diagrams Shear, ial-force & Bending moment diagrams are merel the graphical visualization of the shear, aial and moment equations plotted on V-, P- & M- aes. The are usuall drawn below the loading diagram. That is to show the variation of the internal forces with respect to the horizontal distance. * Shear & moment diagrams are eceedingl important. From them a designer sees at a glance the kind of performance that is designed from a beam at ever section. Eample: - Construct the shear, aial force and bending moment diagrams for the weightless beam shown subjected to the inclined force P=5KN. a b P=5KN c 3 4 d (a) 5m a b B c d C 5m Solution: - 39 Instructor: braham ssefa

40 a 4KN 3KN a 3KN C B (b) Free Bod Diagram 2KN 2KN 4 KN.m 3KN 2KN 3 KN 2 KN (c) Section a-a 2m 3KN 10KN.m 3KN Section b-b 2KN 5-2KN (d) 3KN 2KN KN 10KN.m 3KN 2KN P=0 (e) 3 KN 2 KN 4 KN 4 KN.m 3 KN 2 KN P=0 (f) 8m 2 KN + 0 2m - _ 6m -3 KN 10 KN.M BENDING MOMENT 4 KN.M + 4 KN.M DIGRM 40-2 KN 2m SHER FORCE DIGRM XIL FORCE DIGRM Instructor: braham ssefa

41 Eample 2 Write shear & moment equations for the beam loaded as shown in the figure & Sketch the shear & moment diagrams. a 20KN/m b c 30KN R 1 =63KN a C B b R 2 =67KN 5 m 5 m 4 m c D The sections in the beam should be taken at locations where the loading conditions change of load points and are designated b the letters,b,c,&d. (a) (20) KN a /2 R 1 =63KN a M B V B 100KN (-2.5) m b (b) R 1 =63KN 2.5m 100KN -2.5 c b M B (c) M CD R 1 =63KN 10m R 2 =67KN V CD -10 c c 14-30KN 41 M CD V CD c Instructor: braham ssefa

42 30KN =3.15 SHER FORCE DIGRM -37KN 99.2 KN.m BENDING MOMENT DIGRM 120 E KN.m ELSTIC DIGRM Note:- the value of making M B maimum can be found b differentiating M B with respect to and equating the result to zero. Some conclusions: - - dm B /d=v B M B = d (M B ) = 63 2 *10 = d d (M ) = = 0 = 63 / 20 = 3.15 d B Maimum moment corresponds to the section of zero shear. The beam between & E is concave up, and between E & D it is concave down. It is not surprising that the moment diagram has positive values corresponding to the region E, while for the position ED, where the beam is concave down; the moment diagram has negative values. Sketching the shape of the beam therefore provides a check of the sign of bending moment. t point E, where the beam changes its shape from concave up to concave down, we have that is called a point of inflection, it corresponds to the section of zero bending moment. Its position ma be calculated b settling M Bc = =0 = Instructor: braham ssefa

43 5.4 Relations between Static functions and their application The relationships b/n load intensit function q() & bending moment function M() that eist at an point on a loaded beam, provides: - method of construction shear & moment diagrams without writing shear & moment equations: - method of constructing the loading diagram or the bending moment diagram from the given shear force diagram. To develop these relationships consider a beam loaded or literaril, such as shown below. q a b q() P a b d Cutting planes The enlarged view of a small longitudinal element of the beam, b/n the cutting planes a-a & b-b is shown below. ppling the condition of equilibrium of forces in the -direction & on this element, we obtain: [+ F = 0] v( ) [v( ) + dv( )] + q( )d = 0 v( ) v( ) dv( ) + q( )d = 0 q( )d = dv( ) (a) q( ) = dv ( ) (b) d M() q()d d/2 C q() M()+dM() v() V()+dv () From Equation " the slope of the shear force diagram at an point along the beam equals, in magnitude & direction, to the ordinate of the load intensit function at the same point. If we integrate equation (a) 2 v 2 dv( ) = V V = ΔV 1 v q( )d = d 2 q( )d = V2 V 1 (c) 1 The integration is simpl the area under the load intensit. 43 Instructor: braham ssefa

44 From equation (c) " The area under the load intensit, or q() diagram between coordinates 1 & 2 on the ais of the beam, is equal to the change shear force ordinates," From a moment summation about point c we have: [ Mc=0] M ( ) + dm ( ) q( )d d M ( ) v( )d = 0 2 The third term in this equation is the square of the differential that is negligible in comparison with other terms; hence the equation reduce to dm ( ) v( )d = 0 dm ( ) = v( )d (d ) v( ) = d (M / ) d v( ) = d M ( ) (e) d From equation (e) " the slope of the bending moment diagram at an location along the beam, is equal to the shear force ordinate at the same location on the span". If we integrate equation (d), we get: V ( )d = M 2 dm ( ) = M M = ΔM M 1 2 v( )d =M area under SFD M ( f ) 1 From equation (f) " the area under the shear force diagram, between coordinates 1 & 2 on the longitudinal ais, is equal to the change in bending moment ordinates at those coordinates,". Eample 1: Using the semi-graphicall technique, sketch the shear force and bending moment diagrams for the beam loaded as shown below. w=3kn/m P=12KN B C D E a) Space diagram 2m 2m 1m 2m 44 Instructor: braham ssefa

45 w=3kn/m 12KN R B =16.8KN 10.8 slope= R E =7.2KN b) Load intensit diagram c) Shear force slope= -3 diagram -6 F d) Bending moment diagram e) Deflected curve Notation: L = area under the load intensit diagram s = area under the shear force diagram V n = shear to the left of the point n. V r = shear to the right of the point n. i) Shear force diagram V =0 (free end, no concentrated load applied) V - B=V +( L ) -B =0+(2)(-3)= -6KN + - V B B B= V +R = =10.8KN V c = V + B +( L ) B-C =10.8+(2)(-3)=4.8KN V - D=V c +( L ) C-D =4.8+0=4.8KN V + D=V - D +P=4.8-12= -7.2KN V - E=V + D +( L ) DE = = -7.2KN + - V E E E=V +R = =0KN ii) Bending Moment Diagrams M =0 (Free end, no couple applied) M B =M +( s ) -B =0+1/2*(-6)(2)= -6KN 45 Instructor: braham ssefa

46 M C =M B +( S ) BC = -6+(1/2)*6*2+4.8*2 = 9.6 KN.m M D = M C + ( s ) C-D = *1=14.4 KN.m 46 Instructor: braham ssefa

47 M E = M D + ( s ) DE =14.4+2(-7.2)=0 Infection point F 1 /=6/2 1 =3 2 = M F =M B +( S ) B-F = -6+(1/2)**3+(10.8-3)* = -6+(3/2) M F = =0 2 -(10.8/1.5)+(6/1.5)= =0 7.2 ± = ± 5.99 = 2 = = 6.59m, = = 0.61m 2 2 Eample 2: Using the Semi-graphicall technique, sketch the load intensit & shear force diagrams corresponding to the bending moment diagram shown below. F 2m B 1m C 1m D 2m KN.m KN.m -4.5 KN.m E 2 nd degree -10KN.m 0.083m -10KN.m V D + V - 3 E 1 2m SFD(KN) - V E 5KN 5KN w=3kn/m -10 KN.m Load intensit diagram KN 5.75 KN Solution * sudden fall or rise in the shear force diagram indicates that, at that location a concentrated load is applied. * From a vertical fall (or rise) in the bending moment diagrams; we conclude that, a concentrated moment, is applied at that point. 47 Instructor: braham ssefa

48 Notation: - Slope of segments on shear force diagrams=d s - Slope of segments on BM diagrams = D m i) Shear force diagram: Computing the slopes in the segments first; (D m ) -B =-10-0/2= -5KN (D m ) B-C = /1=+5.25KN (D m ) C-D = /1=0.25KN (D s ) DF =0-0.25/0.0833= -3.0KN/m V D /D F =D S =(V + D-V - E)/2-2D s =V + D-V - E V E =+2D S +V D + =-5.75 KN ii) Load intensit diagram We observe the following sudden jumps; Point, a sudden drop of -5 KN Point B, a sudden rise of =10.25 KN Point C, a sudden drop of -5.25*0.25 = -5 KN Point E, a sudden rise of 5.75 KN Slope of shear diagram = load intensit Segment B = Slope = 0 Segment BC= Slope=0 Segment CD= Slope=0 Segment DE= -3KN/m. Sudden rise in BM diagram at E Eercise: - Draw moment & load diagrams corresponding to the given shear diagram. 2 nd degree m 1m 1m 2m SFD (KN) 48 Instructor: braham ssefa

49 CHPTER VI 6. CENTROIDS Introduction ctuall, concentrated forces do not eist in the eact sense, since ever eternal force applied mechanicall to a bod is distributed over a finite contact area however small. ¾ When forces are applied over a region whose dimensions are not negligible compared with other pertinent dimensions, then we must account for the actual manner in which the force is distributed b summing up the effects of the distributed force over the entire region. We carr out this process b using the procedures of mathematical integration. For this purpose we need to know the intensit of the force at an location. There are three categories into which such problems fall; i/ Line Distribution :- when a force is distributed along a line. The loading is epressed as force per unit length of line ( N ). m ii/ rea Distribution :- when a force is distributed over an area. The loading is epressed as force per unit area. ( N 2 )). m iii/ Volume Distribution :- when a force is distributed over the volume of a bod (bod force). ( N 3 ). m ¾ The bod force due to the earth s gravitational attraction (weight) is b far the most commonl encountered distributed force. The following sections of the chapter deal with the determination of the point in a bod through which the resultant gravitational forces acts and the associated geometrical properties of lines, areas and volumes Centroids of lines, reas, and Volumes of Figures and Bodies Center of Mass or Center of Gravit z G z G dw r W r z z dm 49 Instructor: braham ssefa

50 ppling the principle of moments, the moment of the resultant gravitational force W about an ais equals the sum of the moments about the same ais of the gravitational forces dw acting on all particles treated as infinitesimal elements of the bod. = W dw = W zdw z = W W = dw ppling moment principle about the -ais, the moment about the -ais of the elemental weight, dw is.dw. The sum of these moments for all elements of the bod is dw..w = dw Similarl, dw Substituting W = mg and dm dm =, = z = m m dw = gdm zdm The above equations ma be epressed in vector form; m r dm r =, r = i + j + zk (position vector for the elemental mass, dm ). r = i + j + z k m (Position vector for G) If the densit of the bod is not constant, dm = ρdv = ρdv, ρdv = ρdv = zρdv ρdv z ρdv The point (,, z) is known as the center of mass, and coincides with the center of gravit as long as the gravit field is treated as uniform and parallel. -The center of mass has a special significance in calculating the dnamic response of a bod to unbalanced forces. In most problems the calculation of the position of the center of mass ma be simplified b an intelligent choice of reference aes. In general the aes should be placed so as to simplif the equations of the boundaries as much as possible. Thus polar coordinates will be useful for bodies having circular boundaries. nother important clue ma be taken from considerations of smmetr. Whenever there eists a line or plane of smmetr in a homogenous bod, a coordinate ais or plane should be chosen to coincide with this line or plane. The center of mass will alwas lie on such a line or plane, since the moments due to smmetricall located elements will alwas cancel, and the bod can be considered composed of pairs of these elements. -The location of the center of mass is alwas facilitated b the observation of smmetr when it eists. 50 Instructor: braham ssefa

51 6.2 Centroids of Lines, reas and Volumes When speaking of an actual phsical bod, we use the term center of mass. If the densit is uniform throughout the bod, the positions of the centroid and the center of mass are identical, whereas if the densit varies, these two points will, in general, not coincide. The term centroid is used when the calculation concerns a geometrical shape onl. The calculation of centroids fall within three distinct categories, depending on whether the shape of the bod involved can be modeled as a line, an area, or a volume. 1/ Centroids of Lines z dl C z z L = dl L = dl L z = zdl L It should be noted that, in general, the centroid C will not lie on the line. If the rod lies in a single plane, such as the - plane, onl two coordinates will require calculation. 2/ Centroids of reas z d - If ρ and t are constant over the entire area, the coordinates of the center of mass of the bod also become the coordinates of the Centroid C of the surface area. C = d z z = d z = zd 51 Instructor: braham ssefa

52 The centroid C for the curved surface will in general not lie on the surface. If the area is a flat surface, sa, the - plane, onl the coordinates of C in that plane will be unknown. 3/ Centroids of Volumes -For a general bod of volume V and constant densit, the coordinates of the center of mass also become the coordinates of the centroid C of the bod. = dv, = dv, z = zdv V V V Choice of Element for Integration -With mass centers and centroids the concept of the moment principle is simple enough; the difficulties reside primaril with the choice of the differential element and with setting up the integrals. In particular there are five guidelines to be speciall observed. i/ Order of element ;- whenever possible, a first order differential element should be selected in preference to a higher order element so that onl one integration will be required to cover the entire figure. Eg. l d d d d = d.d d = l.d Selected ii/ Continuit ;- whenever possible, we choose an element which can be integrated in one continuous operation to cover the figure. Eg. l d a/ b/ d 1 The horizontal strip in fig (a), would be preferable to the vertical strip in fig (b), which, if used, would require two separate integrals because of the discontinuit in the epression for the height of the strip at = Instructor: braham ssefa

53 iii/ Discarding higher-order terms Higher-order terms ma alwas be dropped compared with the lower-order terms. d d -The vertical strip of area under the curve is given b the first order term d = d, and the 2 nd order triangular area ½*d.d is discarded. In the limit, of course, there is no error. iv/ Choice of Coordinates ;-choose the coordinate sstem which best matches the boundaries of the figure. Eg. =k 2 r θ (a) (b) -The boundaries of the area in (a) are most easil described in rectangular coordinates, whereas the boundaries of the circular sector in (b) are best suited to polar coordinates. V/ Centroidal Coordinate of Element:- When a 1 st or 2 nd order differential element is adopted, it is essential to use the coordinate of the centroid of the element for the moment arm in setting up the moment of the differential element. -The moment of da about the -ais is c.d, c where c is the -coordinate of the centroid C of the element. C c d c d c =, =, z = z d c c = dv c, = dv z, z = c dv V V V -It is essential to recognize that the subscript C serves as a reminder that the moment arms appearing in the numerators of the integral epressions for moments are alwas the coordinates of the centroids of the particular elements chosen. 53 Instructor: braham ssefa

54 6.3. Composite bodies and Figures -Consider a bod whose parts have masses m1, m2, m3 with the respective mass center coordinates 1, 2, 3 in the -direction; 2 1 G1,m1 3 G2, m2 G3,m3 G The moment principle gives; (m1 + m2 + m3). = m1.1 + m2.2 + m3.3 m. () = (m.).i (m ) i = m i. is the -coordinate of the center of mass of the whole. Similarl, = (m.), z = (m.z) m m nalogous relations hold for composite lines, areas and volumes, where the m s are replaced b L s, s and V s respectivel. It should be pointed out that if a hole or cavit is considered in one of the component parts of a composite bod or figure, the corresponding mass represented b the cavit or hole is treated as a negative quantit. 54 Instructor: braham ssefa

55 CHPTER VII 7. RE Moments of Inertia 7.1. Introduction - When forces are distributed continuousl over an area on which the act, it is often necessar to calculate the moment of these forces about some ais either in or perpendicular to the plane of the area. The intensit of the force (pressure or stress) is proportional to the distance of the force from the moment ais. - The elemental force acting on an element of area, then, is proportional to distance times differential area, and the elemental moment is proportional to distance squared times differential area. We see, therefore that the total moment involves an integral that has the form (distance) 2 d(area). This integral is known as the moment of inertia or the second moment of the area. O d O σ = k Fig. Stress distribution in cross section of a bending beam. df = σ.d = k.d dm = df. = k 2 d M = dm = k 2 d = k. 2 d The term 2 d is called second moment of area or area moment of inertia of the cross-section. It appears so frequentl in design formulas that it needs a separate treatment. In general, for an area, it will be found as follows. d.i = 2 d d O r I = 2.d d.i = 2.d I = 2.d 55 Instructor: braham ssefa

56 The second moment of area about the z-ais (pole O) is; di z = r 2.d I z = r 2.d 0r di z = r 2.d = ( ).d = 2.d + 2.d I z = ( ).d = I + I I z = I + I, - I and I are called rectangular moments of inertia. - I z is called polar moment of inertia and is the sum of rectangular moments of inertia about aes passing through the point. The choice of elements of integration is similar to that of Centroids. The parallel aes Theorem (Transfer es) - It is often necessar to get moment of inertia of an area about aes parallel to centroidal aes. So this theorem provides relationship between centroidal moments of inertia and moments of inertia about parallel aes. o o d o d c o d d fig.* o B definition: di = (d + o ) 2.d = (d 2 + 2d.o + o 2 ).d I = o 2.d + 2d. od + d 2. d = I o + d 2. I = I o + d 2., Similarl I = I o + d 2. Where ; - I o and I o are centroidal rectangular moments of inertia and ; - I and I are rectangular moments of inertia about the - and -aes. I z = I zo + d 2. Hence, the parallel ais theorem can be stated as; - The moment of inertia of an area with respect to an ais is equal to the moment of inertia about the parallel ais through the centroid of the area plus the product of the area and the square of the distance between the two aes. Two things to note are; 56 Instructor: braham ssefa

57 i/ The aes should be parallel ii/ One of the aes should be centroidal Radius of Gration - The radius is a measure of the distribution of the area from the ais in question. C O O O k K -The distance k and k are known as the radius of gration of the area about the - and -ais respectivel. k r - rectangular or polar moment of inertia ma be epressed b specifing the radius of gration and the area. I = k 2. k = I / I = k 2. Or I z =kz 2 2. k = k z = k k I / kz = I z / The parallel-ais theorem also hold for radii of gration. The transfer relation becomes; k 2 = k 2 + d 2,Where ;- k is the radius of gration about a centroidal ais parallel to the ais about which k applies and d is the distance between the two aes. The aes ma be either in the plane or normal to the plane of the area Composite reas The moment of inertia of a composite area about a particular ais is simpl the sum of the moments of inertia of its component parts about the same ais. -It is often convenient to regard a composite area as being composed of positive and negative parts. We ma then treat the moment of inertia of a negative area as a negative quantit. 57 Instructor: braham ssefa

58 -When a composite area is composed of a large number of parts, it is convenient to tabulate the results for each of the parts in terms of its area, its centroidal moment of inertia I, the distance d from its centroidal ais to the ais about which the moment of inertia of the entire section is being computed, and the product d2..for a composite area in the - plane, for eg, and with the notation of fig.*, where I is the same as Io and I is the same as Io, the tabulation would include; Part rea, d d d 2 d 2 I I 1 2. Sums d 2 d 2 I I -From the sums of the four columns, then the moments of inertia for the composite area about the - and -aes become; I = I + d 2, I = I + d 2 Note; - lthough we ma add the moments of inertia of the individual parts of a composite area about a given ais, we ma not add their radii of gration. The radius of gration for the composite area about the ais in question is given b k = I/, where I is the total moment of inertia and is the total area of the composite figure. Similarl, the radius of gration k about a polar ais through some point equals I z /, where, I z = I + I for - ais through that point Products of Inertia & Transfer of es Products of Inertia In certain problems involving unsmmetrical cross sections and in the calculation of moments of inertia about rotated aes, an epression di = d occurs, which has the integrated form ; I =..d Where ;- and are the coordinates of the element of area d = d.d. -I is called the product of inertia of the area with respect to the - ais. -Unlike moments of inertia, which are alwas positive for positive areas, the product of inertia ma be positive, negative or zero. -The product of inertia is zero whenever either one of the reference aes is an ais of smmetr. Eg Instructor: braham ssefa

59 Here we see in the fig. that the sum of the terms (-)d and (+)d due to smmetricall placed elements vanishes. Since the entire area ma be considered to be composed of pairs of such elements, it follows that the product of inertia I for the entire area is zero. Transfer of es B definition the product of inertia of the area in fig.* with respect to the -and -aes in terms of the coordinates o, o to the centroidal aes is; I = (o + d) (o + d).d = o.od + d. o.d + d. o.d + d.d. d The 1 st integral is b definition the product of inertia about the centroidal aes, which we write I. The middle two integrals are both zero since the 1 st moment of the area about its own centroid is necessaril zero. The third integral is merel d.d.. Thus, the transfer-of-ais theorem for products of inertia becomes; I = I + d.d Rotation of es: The product of inertia is useful when we need to calculate the moment of inertia of an area about inclined aes. This consideration leads directl to the important problem of determining the aes about which the moment of inertia is a maimum and a minimum. In the figure below, the moments of inertia of the area about the and aes are I' = '2 d = ( cosθ sin θ ) 2 d I' = '2 d = ( sin θ + cosθ ) 2 d, where and have been replaced b their equivalent epressions as seen from the geometr figure. Epanding and substituting the trigonometric identities and defining relations for I, I, I give us 59 Instructor: braham ssefa

60 In a similar manner the product of inertia about the inclined aes as dding the above equations gives I + I = I + I = I z. The angle that makes I and I a maimum or a minimum ma be determined b setting the derivatives of either I or I with respect to θ equal to zero. tan 2α = 2I I I The above equation gives two values for 2α which differ b. Consequentl the two solutions for α will differ b /2. One value of defines the ais of maimum moment of inertia, and the other value defines the ais of minimum moment of inertia. These two rectangular aes are known as the principal aes of inertia. Thus the maimum and minimum moments of inertia become: Mohr s Circle of Inertia: The relationships stated above can be presented graphicall b a diagram known as Mohr s Circle as shown below. For given values of I, I, I the corresponding values of I, I, I ma be determined from the diagram for an desired angle θ. The coordinates of an point C are (I, I ), and those of the corresponding point D are (I, I ). lso the angle between O and OC is 2 θ or twice the angle from the ais to the ais. 60 Instructor: braham ssefa