Chapter 5 Equilibrium of a Rigid Bod Objectives Develop the equations of equilibrium for a rigid bod Concept of the free-bod diagram for a rigid bod Solve rigid-bod equilibrium problems using the equations of equilibrium Chapter 5 Outline Conditions for Rigid Equilibrium Free-Bod Diagrams Equations of Equilibrium Two and Three-Force Members Free Bod Diagrams Equations of Equilibrium Constraints and Statical Determinac 43
5.1 Conditions for Rigid-Bod Equilibrium The equilibrium of a bod is epressed as F R M R O F 0 M O 0 Consider summing moments about some other point, such as point A, we require MA r FR MR O 0 44
5. Free Bod Diagrams Support Reactions If a support prevents the translation of a bod in a given direction, then a force is developed on the bod in that direction. If rotation is prevented, a couple moment is eerted on the bod. 45
5. Free Bod Diagrams 46
5. Free Bod Diagrams 47
5. Free Bod Diagram Weight and Center of Gravit Each particle has a specified weight Sstem can be represented b a single resultant force, known as weight W of the bod Location of the force application is known as the center of gravit 48
Eample 5.1 Draw the free-bod diagram of the uniform beam. The beam has a mass of 100kg. Solution Free-Bod Diagram Support at A is a fied wall Two forces acting on the beam at A denoted as A, A, with moment M A Unknown magnitudes of these vectors For uniform beam, Weight, W = 100(9.81) = 981N acting through beam s center of gravit, 3m from A 49
5.4 Two- and Three-Force Members Two-Force Members When forces are applied at onl two points on a member, the member is called a two-force member Onl force magnitude must be determined Three-Force Members When subjected to three forces, the forces are concurrent or parallel 50
5.5 3D Free-Bod Diagrams 51
5.5 Free-Bod Diagrams 5
5.7 Constraints for a Rigid Bod Redundant Constraints More support than needed for equilibrium Staticall indeterminate: more unknown loadings than equations of equilibrium 53
5.7 Constraints for a Rigid Bod Improper Constraints Instabilit caused b the improper constraining b the supports When all reactive forces are concurrent at this point, the bod is improperl constrained 54
Chapter 6 Structural Analsis Objectives Determine the forces in the members of a truss using the method of joints and the method of sections Analze forces acting on the members of frames and machines composed of pin-connected members Outline Simple Trusses The Method of Joints Zero-Force Members The Method of Sections Space Trusses Frames and Machines 55
6.1 Simple Truss A truss composed of slender members joined together at their end points Planar Trusses The analsis of the forces developed in the truss members is D Similar to roof truss, the bridge truss loading is also coplanar Assumptions for Design All loadings are applied at the joint - Weight of the members neglected The members are joined together b smooth pins - Assume connections provided the center lines of the joining members are concurrent 56
6.1 Simple Truss Simple Truss Form of a truss must be rigid to prevent collapse The simplest form that is rigid or stable is a triangle Method of Joints For truss, we need to know the force in each members Forces in the members are internal forces For eternal force members, equations of equilibrium can be applied Force sstem acting at each joint is coplanar and concurrent F = 0 and F = 0 must be satisfied for equilibrium 57
Eample 6.1 Determine the force in each member of the truss and indicate whether the members are in tension or compression. Solution unknown member forces at joint B 1 unknown reaction force at joint C unknown member forces and unknown reaction forces at point A For Joint B, F 0; 500N F F BC F cos 45 BC sin 45 0; N F BA N 0 F 0 F BA BC 707.1N ( C) 500N( T ) 58
Solution For Joint C, F F F C CA 0; 707.1cos 45 0; 707.1sin 45 N N 0 0 C F CA 500N( T ) 500N For Joint A, F 500N F 500N A 0; 0; A 0 0 A A 500N 500N 59
6.3 Zero-Force Members Method of joints is simplified using zero-force members Zero-force members is supports with no loading In general, when 3 members form a truss joint, the 3 rd member is a zero-force member provided no eternal force or support reaction is applied to the joint 60
Eample 6.4 Using the method of joints, determine all the zero-force members of the roof truss. Assume all joints are pin connected. Solution For Joint G, F 0 F 0 GC GC is a zero-force member. For Joint D, F 0 F 0 DF 61
Solution For Joint F, F 90, F FC For Joint B, 0 F 0 FC cos 0 FBH kn F HC satisf F = 0 and therefore HC is not a zero-force member. 6
6.4 Method of Sections Used to determine the loadings within a bod If a bod is in equilibrium, an part of the bod is in equilibrium To find forces within members, an imaginar section is used to cut each member into and epose each internal force as eternal Consider the truss and section a-a as shown Member forces are equal and opposite to those acting on the other part Newton s Law 63
Eample 6.5 Determine the force in members GE, GC, and BC of the truss. Indicate whether the members are in tension or compression. Solution Draw FBD of the entire truss F M A 0; 0; 400N A 0 400N 100N(8m) 400N(3m) D A (1m) 0 D 900N F 0; A 100N 900N 0 A 300N 64
Solution Draw FBD for the section portion ) ( 500 0 5 3 300 0; ) ( 800 0 ) (3 ) (8 300 0; ) ( 800 0 ) (3 ) (3 400 ) (4 300 0; T N F F N F C N F m F m N M T N F m F m N m N M GC GC GE GE C BC BC G 65
Eample 6.5 Reaction force A A D 400N 300N 900N D 1500N 100N 900N E 800N 900N 400N 1500N C 500N 800N 900N 100N 100N A 500N B 0N G 800N 400N 800N 800N 800N 500N 500N 300N 0N 66
6.6 Frames Composed of pin-connected multi-force members Frames are stationar Appl equations of equilibrium to each member to determine the unknown forces Eample 6.9 For the frame, draw the free-bod diagram of (a) each member, (b) the pin at B and (c) the two members connected together. Solution Part (a) BA and BC are not two-force AB is subjected to the resultant forces from the pins 67
A 6kN A 1kN B 0 B 4kN C 4kN M 3kN m 68 A
Chapter 7 Internal Force Objectives Method of sections for determining the internal loadings in a member Develop procedure b formulating equations that describe the internal shear and moment throughout a member Analze the forces and stud the geometr of cables supporting a load Outline Internal Forces Developed in Structural Members Shear and Moment Equations and Diagrams Relations between Distributed Load, Shear and Moment Cables 69
7.1 Internal Forces in Structural Members The design of an structural or mechanical member requires the material to be used to be able to resist the loading acting on the member These internal loadings can be determined b the method of sections Force component N, acting normal to the beam at the cut session V acting tangent to the session are normal or aial force and the shear force Couple moment M is referred as the bending moment 70
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Eample 7.3 Determine the internal force, shear force and the bending moment acting at point B of the two-member frame. Solution Support Reactions FBD of each member Member AC M A = 0; -400kN(4m) + (3/5)F DC (8m)= 0 F DC = 333.3kN + F = 0; -A + (4/5)(333.3kN) = 0 A = 66.7kN + F = 0; A 400kN + 3/5(333.3kN) = 0 A = 00kN 7
Solution Support Reactions Member AB + F = 0; N B 66.7kN = 0 N B = 66.7kN + F = 0; 00kN 00kN V B = 0 V B = 0 M B = 0; M B 00kN(4m) 00kN(m) = 0 M B = 400kN.m 73
7. Shear and Moment Equations Beams structural members designed to support loadings perpendicular to their aes A simpl supported beam is pinned at one end and roller supported at the other A cantilevered beam is fied at one end and free at the other 74
7.3 Relations between Distributed Load, Shear and Moment Distributed Load Consider beam AD subjected to an arbitrar load w = w() and a series of concentrated forces and moments Distributed load assumed positive when loading acts downwards 75
7.3 Relations between Distributed Load, Shear and Moment Distributed Load Distributed loading has been replaced b a resultant force F = w() that acts at a fractional distance k ( ) from the right end, where 0 < k <1 ) ( ) ( 0 ) ( ) ( 0; ) ( 0 ) ( ) ( 0; k w V M M M k w M V M w V V V w V F 76
7.3 Relations between Distributed Load, Shear and Moment Distributed Load Slope of the shear diagram Slope of shear diagram dv d w() dm V d Negative of distributed load intensit Shear moment diagram Change in shear M BC Vd Area under shear diagram Change in moment V BC w( ) d Area under shear diagram 77
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Eample 7.9 Draw the shear and moment diagrams for the overhang beam. The support reactions are shown. Shear Diagram Shear of kn at end A of the beam is at = 0. Positive jump of 10 kn at = 4 m due to the force. Moment Diagram M 4 8kN m M M 0 4 0 80
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7.4 Cables Cable Subjected to Concentrated Loads For a cable of negligible weight, it will subject to constant tensile force Known: h, L 1, L, L 3 and loads P 1 and P Form equations of equilibrium Use Pthagorean Theorem to relate the three segmental lengths 8
Eample 7.11 Determine the tension in each segment of the cable. FBD for the entire cable. F 0; A E 0 M E 0; A (18 m) 4 kn(15 m) 15 kn(10 m) 3 kn( m) 0 A 1kN F 0; 1kN 4kN 15kN 3kN E 0 E 10kN Consider leftmost section which cuts cable BC since sag C = 1m. M C 0; A (1 m) 1 kn(8 m) 4 kn(5 m) 0 A E 6.33kN F 0; T cos 6.33kN 0 BC BC F 0; 1kN 4kN T sin 0 BC BC BC 51.6, T 10.kN BC 83
7.4 Cables Cable Subjected to a Distributed Load Consider weightless cable subjected to a load w = w() For FBD of the cable having length Since the tensile force changes continuousl, it is denoted on the FBD b T Distributed load is represented b second integration, 1 F H w( ) d d 84
Cable T cos ( T T )cos( ) 0 T sin ww ( T T )sin( ) 0 1 w( ) T cos T sin 0 1 Tcos d [ T cos ( T T ) cos ] 0 ( T cos ) 0 d 1 dt ( sin ) [ T sin w ( T T )sin( )] 0 w 0 d 1 d w Tcos Tsin 0 tan d Tcos F T sin H constant wd T sin 1 tan d d ( wd) d Tcos F 85 H
Solution Note w() = w o 1 F Perform two integrations Boundar Conditions at = 0 H wod d 1 w o C 1 C FH 0, 0, d / d 0 Therefore, wo C1 C 0 Curve becomes F Boundar Condition at = L/ h For constant, F Tension, T = F H /cosθ H wo L 8h 4h and L H Slope at point B d d L/ w L 1 o tanma ma tan F H Therefore T ma F H cos( ma ) Using triangular relationship T ma 4F H w o L 86
Solution For a differential segment of cable length ds, Determine total length b integrating, Integrating ields, d L h ds L / 0 8 1 L h h L L h L 4 sinh 4 4 1 1 d d d d d ds 1 87
7.4 Cables Cable Subjected to its Own Weight When weight of the cable is considered, the loading function becomes a function of the arc length s rather than length FBD of a segment of the cable is shown 88
7.4 Cables Cable Subjected to its Own Weight T cos ( T T )cos( ) 0 T sin ws ( T T )sin( ) 0 1 ws( s) T cos T sin 0 1 Tcos constant FH s 1 d ( T sin ) w 0 s ds d sin 1 tan wds d cos F H Appl equilibrium equations to the force sstem d 1 T cos FH T sin w( s) ds w( s) ds d F Replace d/d b ds/d for direct integration H d ds ds d d d d 1 Therefore ds d 1/ ds w( s) H 1 1 w( s ds 1 1 ds F 1/ ) FH 89
Eample 7.13 Determine the deflection curve, the length, and the maimum tension in the uniform cable. The cable weights w o = 5N/m. Solution For smmetr, origin located at the center of the cable. Deflection curve epressed as = f() ds 1/ 1/ F w H o 1 1/ F w s C 1 ds Substitute u Perform second integration or 1 / F w s C F w H o H sinh F w H o o 1 1 u C sinh 1 1 F H du ( wo / FH ) ds w s C o 1 ds 1/ H o 1 C 90
Solution Evaluate constants d/d = 0 at s = 0, then C 1 = 0 s=0 at =0, then C =0 solve for s ds w F d d o H 1 1 1 C w s F d d o H F w w F s H o o H sinh 1 1 1 sinh C C w s F w F o H o H F w d d H o sinh 3 cosh C F w w F H o o H Boundar Condition = 0 at = 0 For deflection curve, This equations defines a catenar curve. o H w F C 3 1 cosh F w w F H o o H 91
Solution Boundar Condition = h at = L/ F H wo h cosh 1 wo FH F H 50N Since w o = 5N/m, h = 6m and L = 0m, 6m cosh 1 5N / m FH For deflection curve, 9.19 cosh 0.109 1m F H 45. 9N = 10m, for half length of the cable 45.9 5 N / m sinh 10m 1.1m 5 N / m 45.9N Hence 4.m Maimum tension occurs when is maimum at s = 1.1m d d T ma s1.1m ma 5 N / m 1.1m tanma 1.3, ma 5.8 45.9N FH 45.9N 75.9N cos cos 5.8 9
Chapter 8 Friction Objectives Introduce the concept of dr friction To present specific applications of frictional force analsis on wedges, screws, belts, and bearings To investigate the concept of rolling resistance Chapter Outline Characteristics of Dr Friction Problems Involving Dr Friction Wedges, Screws, Flat Belts, Collar Bearings, Pivot Bearings, and Disks, Journal Bearings Rolling Resistance 93
8.1 Characteristics of Dr Friction Theor of Dr Friction: Impending Motion Constant of proportionalit μ s is known as the coefficient of static friction Angle ϕ s that R s makes with N is called the angle of static friction tan Tpical Values of μ s Contact Materials s 1 Fs N tan 1 sn N tan 1 s Coefficient of Static Friction μ s Metal on ice Wood on wood Leather on wood Leather on metal Aluminum on aluminum 0.03 0.05 0.30 0.70 0.0 0.50 0.30 0.60 1.10 1.70 94
Chapter 9 Center of Gravit and Centroid Objectives Concept of the center of gravit, center of mass, and the centroid Determine the location of the center of gravit and centroid for a sstem of discrete particles and a bod of arbitrar shape 9.1 Center of Gravit and Center of Mass Mass Center ~ m ~ m ~ zm ;, z m m m Consider a particle having weight of dw ~ dw ; dw ~ dw ; z dw ~ zdw dw 95
Eample 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc. Solution For differential length of the element dl d dl d d 1d d Since = and then d/d = dl 1d The centroid is located at dl 1 1 4 1d 4 1 d L 0 0 0.6063 0.410m 1 1 dl 4 1 d 4 1 d 1.479 L dl 0 0 1 4 1 L d 0 0.8484 1 dl 4 1d 1.479 0 L 0.574m 96
9. Composite Bodies Eample 9.10 Locate the centroid of the plate area. Solution Composite Parts Plate divided into 3 segments. Area of small rectangle considered negative. Moment Arm Location of the centroid for each piece is determined and indicated in the diagram. Summations ~ A 4 A 11.5 ~ A 14 A 11.5 0.348 mm 1.mm 97
9.5 Fluid Pressure Magnitude of depends on the specific weight or mass densit of the fluid and the depth z of the point from the fluid surface p z gz Valid for incompressible fluids Flat Plate of Constant Width As pressure varies linearl with depth, the distribution of pressure over the plate s surface is represented b a trapezoidal volume having an intensit of w1 bp1 brz1 at depth z 1 and at depth z Magnitude of the resultant force F R = volume of this loading diagram Curved Plate of Constant Width w bp brz 98
Eample 9.14 Determine the magnitude and location of the resultant hdrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; w = 1000kg/m 3. Solution The water pressures at depth A and B are A B gz w gz w A B (1000 kg (1000 kg / m / m 3 3 )(9.81m / s )(9.81m / s )(m) 19.6kPa )(5m) 49.05kPa For intensities of the load at A and B, w w A B b b A B (1.5m)(19.6kPa) 9.43kN / m (1.5m)(49.05kPa) 73.58kN / m FR 1 (3)(9.4 73.6) 154.5 N This force acts through the centroid of the area, 1 (9.43) 73.58 h (3) 1. 9m 3 9.43 73.58 measured upwards from B 99
Chapter 10 Moments of Inertia Objectives Method for determining the moment of inertia for an area Introduce product of inertia and show determine the maimum and minimum moments of inertia for an area Outline Definitions of Moments of Inertia for Areas Parallel-Ais Theorem for an Area Radius of Gration of an Area Moments of Inertia for Composite Areas Product of Inertia for an Area Moments of Inertia for an Area about Inclined Aes Mohr s Circle for Moments of Inertia Mass Moment of Inertia 100
10.1 Definition of Moments of Inertia for Areas Centroid for an area is determined b the first moment of an area about an ais Second moment of an area is referred as the moment of inertia Moment of Inertia moments of inertia of the differential plane area da di da di da A I da I da A Formulate the second moment of da about z ais dj O r da where r is perpendicular from the pole (z ais) to the element da Polar moment of inertia for entire area, 101 J r da I I z A
10. Parallel Ais Theorem for an Area Determine the moment of inertia of area about a corresponding parallel. For moment of inertia of da about ais di For entire area I A A ' ' d ' d da d da da For polar moment of inertia A ' da d ' da da 0; 0 I I Ad and I I Ad z J J Ad C A da 10
10.3 Radius of Gration of an Area Radius of gration of a planar area has units of length and is a quantit used in the design of columns in structural mechanics For radii of gration I I Jz k k kz A A A 103
Eample 10.1 Determine the moment of inertia for the rectangular area with respect to (a) the centroidal ais, (b) the ais b passing through the base of the rectangular, and (c) the pole or z ais perpendicular to the - plane and passing through the centroid C. Solution I A ' da h/ h/ ' ( bd') h/ h/ ' d 1 1 bh 3 B appling parallel ais theorem, 1 3 h 1 3 I b I Ad bh bh bh 1 3 For polar moment of inertia about point C, 1 3 1 I' hb and JC I I' bh( h b ) 104 1 1
10.5 Product of Inertia for an Area Moment of inertia for an area is different for ever ais Product of inertia for an element of area da located at a point (, ) is defined as di I da da A Parallel Ais Theorem For the product of inertia of da with respect to the and aes ' d ' d da di A For the entire area, di A ' A ' d ' d ' da d A Forth integral represent the total area A, I I ' ' Ad da ' da d d A da ' d d A da 105
10.6 Moments of Inertia for an Area about Inclined Aes r r, r T u v cos sin u Ar, sin cos v r A u cos sin r, v sin cos I da, I da, I da Consider moment of inertia matri I I Iuu Iuv I= I I I Iuv I vv T Then I A IA and IA I A Principal Moments of Inertia Or find the eigenvalue of I or I matri T v 0 r θ u 106
Eample 10.8 Determine the principal moments of inertia for the beam s cross-sectional area with respect to an ais passing through the centroid. Moment and product of inertia of the cross-sectional area, I 9 4 9 4 9 4 mm I 5.6010 mm I 3.0010.90 10 mm z Solution.9 3 I 3 5.6 ( -.9)( -5.6)+9 =0 8.57.4 0 eigenvalue of ( I) 0.96 or 7.54 0.96.9 3 1.94 3 3 eigenvector I, 0, 3 0.96.9 3 1.94 1.94 7.54.9 3 4.6 3 7.54 5.6 4 3 3 0, 3 1.94 4.64 107
10.7 Mohr s Circle for Moments of Inertia The circle constructed is known as a Mohr s circle with radius R I I and center at (a, 0) where I a / I I 108
10.7 Mohr s Circle for Moments of Inertia Determine I, I and I Establish the, aes for the area, with the origin located at point P of interest and determine I, I and I Principal of Moments of Inertia Points where the circle intersects the abscissa give the values of the principle moments of inertia I min and I ma Product of inertia will be zero at these points Principle Aes This angle represent twice the angle from the ais to the area in question to the ais of maimum moment of inertia I ma The ais for the minimum moment of inertia I min is perpendicular to the 109 ais for I ma
10.8 Mass Moment of Inertia Mass moment of inertia is defined as the integral of the second moment about an ais of all the elements of mass dm which compose the bod For bod s moment of inertia about the z ais, I m r dm The ais that is generall chosen for analsis, passes through the bod s mass center G When ρ being a constant, I r V dv For moment of inertia of bod about the z ais, I r dm d ' ' dm m m ' ' dm d ' dm d m m Parallel Ais Theorem For moment of inertia about the z ais, I = I G + md m dm Radius of Gration I mk or k I m 110