S O L V E D E X A M P L E S

Transcription

1 UNIVERSITY OF RCHITECTURE, CIVIL ENGINEERING ND GEODESY TECHNICL ECHNICS DEPRTENT S O L V E D E X P L E S OF C O U R S E W O R K S ON ECHNICS PRT I (KINETICS & STTICS) ssistant Professor ngel ladensk, PhD Sofia 5/6

2 COURSE WORK : KINETICS OF THE PLNE OTION OF RIGID ODIES Problem t the instant given, a mechanism occupies position shown in Fig Velocit and acceleration of point are V 5 m/s and a 6 m/s, respectivel Clinder rolls without sliding Determine: Velocities of points, C, D, E and angular velocities of all members of the mechanism; cceleration of point and angular accelerations of members and O Fig Solution: Numeration of the mechanism s members and determination of their tpes of motion embers of the mechanism are four clinder of centre (member ), rod (member ), rod C (member ) and rod O (member 4) The tpes of their motion are found b consideration of previous instant of the motion of the mechanism, ie mechanism s position O ' ' ' C' (Fig) Fig

3 Clinder () rolls, ie it performs plane motion embers () and () also perform plane motion because the have no fied points and do not remain parallel to themselves during the motion ember (4) has an unmovable point O, ie it performs rotation about point O Determination of velocities of points, C, D, E and angular velocities of all members of the mechanism The procedure of determination of all velocities begins from the given velocit of point which is a point of three mechanism s members, namel (), () and () Clinder performs plane motion represented as a rotation about instantaneous center of ero velocit, the point at which clinder touches the ground Then, the magnitude of the angular velocit of clinder () is: V 5 ω 5 s -, P where P is the distance from point to the clinder s instantaneous center of ero velocit The sense of ω follows the sense of V r (Fig) Further, the velocit of point D is obtained as: V D ω DP 5 m/s, where DP is the distance between D and the clinder s instantaneous center of ero velocit perpendicular to DP, and its sense depends on the sense of ω fter that, the velocit of point E is found as: V E ω EP 5,m/s, V r D is where EP is the distance between E and the clinder s instantaneous center of ero velocit V r E is perpendicular to EP, and its sense follows the sense of ω Fig

4 Solution continues with determination of the angular velocit of member () performing plane motion Such motion is presented as a rotation about the instantaneous center of ero velocit of bod () Here, in order to find its position, perpendiculars to the directions of the velocities of two points are drawn These points are and C, because the directions of their velocities are known: direction of V r is horiontal, while direction of V r C is inclined at an angle of 45 with respect to the horiontal Then, intersecting the perpendiculars, the instantaneous center of ero velocit, point P, is located (Fig) Hence: V ω, P where P is the distance between and P To obtain P the isosceles right-angled triangle CP is considered: C C cos 45 P ; cos 45 C 4, 4 P cos 45 C cos 45 4,4 m; P 6 m cos 45 Finall, 5 ω,5 s -, 6 of sense following the sense of V r Further, the velocit of point C is determined using the epression: V C ω CP, where CP C 4, 4 m, because both are cathetus in the isosceles triangle CP V,54,4,6 m/s C V r C is perpendicular to CP and its sense is determined b the sense of ω (Fig) Net step is to determine the angular velocit of member () gain, the plane motion is presented as rotation and the velocit of point is used Location of the instantaneous center of ero velocit of member () is found b intersection of perpendiculars to the directions of velocities of points and (Fig) Then, the angular velocit of member () is: V ω, P where P is obtained from the right-angled triangle P, as follows: Thus, P tg 4, 8 tg tg 5 ω 6,5 s -,,8 P m where the sense depends on the sense of V r Further, velocit of point is obtained appling the epression: V ω P, where P is carried out b the Pthagoras s theorem as: ( P + P P ( P ) + ( ), , 68m ) ( ) ( ) Hence, V 6,54,68 m/s, of sense following the sense of ω Finall, V r is used for obtaining of the angular velocit of member (4): V ω4 s -, O where the distance O is carried out as: 4

5 ,5,5 sin O O sin m Check with the theorem of the velocities projections: pr Vr E pr Vr E E V cos VE cos 45 5 cos,cos ; pr Vr DE D pr Vr DE E V D cos 45 VE cos cos 45,cos,, ; pr Vr pr Vr V cos V cos 6 5 cos cos 6 5 5; pr CV r pr CV r C V cos 45 VC cos 5 cos 45,6cos,6, 6; Determination of acceleration of point and angular accelerations of and O Point is the connecting point of members () and (4) It should be noted that the magnitude and direction of acceleration of point are unknown Therefore, in order to find them, the following procedure is performed First, is considered as a point of member () Thus, the formula of the acceleration is: r r r c r r a a + a + a, (I) where a r c is the acceleration of point chosen for a pole, a r r and a r are the centripetal and rotational components, respectivel, of the acceleration of point There are three unknowns in the equation (I) the magnitude and sense of a r r and the magnitude of a r, ie the problem is insoluble Further, is eamined as a point of member (4) Then, the epression of the acceleration is: r r n r t a a + a, (II) n where a r t and equation (II) the magnitude and sense of a r are the normal and tangential components, respectivel Three unknowns are available in a r and the magnitude of again However, equating the right-hand sides of (I) and (II), it is obtained: r n r t r r a + c r r a a + a + a (III) А Thus, two unknowns take part in epression (III) the magnitudes of of (III) onto two aes make the task soluble t a r It means that the problem is insoluble r a r and t a r Then, the projections Fig 4 First, the normal and centripetal accelerations in (III) are carried out as: a n ω O m/s, а c А ω 6, m/s, where the sense of 4 n a is from to O, while the sense of c а А is from to (Fig4) 5

6 Further, the tangential and rotational accelerations are introduced perpendicular to the normal and centripetal ones, respectivel Their magnitudes are: a t α 4 O α 4, а r α 4α, where α and α 4 are the angular accelerations of the members () and (4), respectivel The senses of t a r, α, and α 4 are arbitrar chosen and shown in Fig4 Then, the projection of (III) onto -ais is written, as follows: n t c a cos a sin a a,,866 a t,5 6 69, t 59, a 785,6 m/s,,5 a t 785,6 α 4 6,87 s - t a r and α 4 have positive signs, ie their senses are correctl chosen (Fig4) Further, the projection of (III) onto -ais is written as: n t r a sin a cos a, r,5 785,6, 866 а, a 5 68,5 8,5 m/s, r a r 8,5 α 7,59 s r The magnitudes of a r and α are obtained negative meaning that their senses are incorrect and have to be changed (Fig4) Finall, in order to find the total acceleration of point, n equation (II) is used, because the components a r t and a r are perpendicular to each other Тhen: ц вр a ( a ) + ( a ) + 785,6 84,95 m/s The sense and direction of parallelogram rule (Fig5) a r are determined using Fig 5 r a r, 6

7 COURSE WORK : KINETICS OF THE RELTIVE OTION OF POINT Problem pipe is pivoted at one end to point O and starts to rotate from horiontal position according to t equation ϕ ( t ) [ ϕ rad, t s] (Fig) eanwhile, point is moving relative to the pipe in accordance with equation S( t) O ( t) t +, 5t [ S m, t s] Find the absolute velocit V r and the absolute acceleration a r of the point at instant t s Fig Solution: t Rotation of the pipe in accordance with equation ϕ ( t ) is the motion that transports point The rectilinear motion of the point according to equation S ( t) O ( t) t + is the relative motion Position of the point at instant t s First step of the solution is determination of the point s position at instant t s This is performed b substitution of t t s into the equations of motions The result is: 4 ( ) t ϕ t rad 76,4 ; S( t ) O ( t) t +,5t +,5 5 m bsolute velocit of the point at instant t s The absolute velocit of point is equal to the sum of relative and transport velocit of the point: r r r V V r + V e The relative velocit is the first order derivative with respect to the time of the relative motion equation: V ( t) S& r ( t) t+,5 ( t ) +,5 4, 5 m/s V r V r r is obtained positive meaning that its sense coincides with the sense of the relative motion (Fig) The pipe performs rotation Then, the transport velocit of the point is obtained using the epression: V e ωo, where ω is the angular velocit of the pipe, and O is the distance from the center of rotation to the point The angular velocit is the first order derivative with respect to the time of the equation of rotation: t 7

8 t ω ( t ) ϕ& ( t) ω( t ) t, s - It has a positive magnitude and its sense follows the sense of ϕ (t) (Fig) Finall, the magnitude of transport velocit of is: V,5 6,665 m/s, e The V r e s direction is perpendicular too (trajector of the point due to the transport motion is circle meaning that its direction is perpendicular to the radius) with the sense depending on the sense of ω (Fig) Fig Further, since the relative and transport velocit of the point are perpendicular to each other (Fig), the absolute velocit is obtained directl b the following epression: V V r + Ve 4,5 + 6,665 8,4m/s The sense and direction of the absolute velocit are found b the parallelogram law (Fig) bsolute acceleration of the point at instant t s The absolute acceleration of the point is equal to the sum of the relative, transport and Coriolis accelerations: r r r r a ar + ae + ac The relative motion is rectilinear Then, the relative acceleration is the second order derivative with respect to the time of the relative motion equation: a ( t) S & r ( t) a ( ) r t m/s It is obtained positive, ie its sense coincides with the sense of the point relative motion (Fig) Since the transport motion of the point is rotation, the transport acceleration has two components normal and tangential: r r n r t ae ae + ae The normal acceleration is: a n e ω O, 5 8,884 m/s, and it points to the center of rotation, here, point O (Fig) The tangential component is carried out b the epression: 8

9 a t e αo, where α is the angular acceleration of the pipe, which is the second order derivative with respect to the time of the equation of rotation: α ( t) & ϕ ( t) α( t ), 667 s -, It has positive magnitude, ie its sense follows the sense of ϕ (t) (Fig) Then: a,6675,5 m/s, t e which is perpendicular to O, and of sense determined b the sense of α The magnitude of the transport acceleration is: n t a ( a ) + ( a ) 8,884 +,5 9,489 m/s e e e Fig 4 Fig The Coriolis acceleration is caused b the interaction between the angular velocit of the pipe and the relative velocit of the point It has to be obtained b the following formula: r r r a C ω V r The magnitude of the Coriolis acceleration is: r r a ω V sin( ω; V ) C r r Here, the angle between the directions of angular and relative velocit is 9, because r ω is perpendicular to the plane O (its direction is determined b the right-hand rule the four fingers of the right hand follows the direction of rotation when the thumb points in the sense of the vector), while plane O Thus: a,4,5sin 9,4,5 m/s C V r r lies in the 9

10 ccording to the cross product, vector a r C has to be perpendicular to the plane formed b r ω and and, besides, the three vectors have to form the right-handed sstem Therefore, the direction and the sense of the Coriolis acceleration coincides with the direction of the transport tangential acceleration of the point (Fig) Finall, the magnitude of the absolute acceleration of the point is obtained b projections of its components onto aes and, which are perpendicular to each other These aes are chosen because all of the accelerations can be projected with their total magnitudes onto them a a + a, n a a a 8,884 6,884 m/s, r e t ae + ac,5+ a 5,5 m/s, a ( 6,884) + 5,5 6,8 m/s The sense and direction of a r are determined b the parallelogram rule (Fig 4) Problem semicircle-shaped bod is attached to -ais and rotates according to equation ϕ ( t) t t [ ϕ rad, t s] while a point moves on its peripher according to equation S( t) ( t) π t S [ V r r, m, t s] Find the absolute velocit V r and the absolute acceleration a r of the point at instant t s, if at the same instant the bod occupies the position shown in Fig Fig Solution: The motion that transports the point is rotation of the bod in accordance with equation ϕ ( t) t t The relative motion is the motion of the point with respect to the bod in accordance with equation S( t) ( t) π t This motion is curvilinear, because the trajector of the point is circle of radius R Position of the point at instant t s The task is to find the absolute velocit and acceleration a r of point at instant t s However, the position of the semicircle-shaped bod at the same instant has been indicated b the statement Therefore, onl the position of point relative to the bod has to be found Then, substituting t t s into the equation of the relative motion, it is obtained: S( s) π π [m] Here, the more convenient wa to represent the point position is using the central angle concept, ie: π π 8 C [rad] 6 R π bsolute velocit of the point at instant t s The absolute velocit of point is equal to the sum of relative and transport velocit: r r r V V r + V e

11 The relative velocit is the first order derivative with respect to the time of the relative motion equation: V r ( t) S & ( t) π t ( t ) π 6, 8 m/s V r V r r lies on the circle s tangent at point and its sense coincides with the sense of the relative motion (Fig) Due to the rotation of the bod, the transport velocit is obtain b the epression: V e ωn, where ω is the angular velocit of the semicircle, and N is the shortest distance between the point and the ais of rotation The angular velocit is the first order derivative with respect to the time of the equation of rotation: ω ( t) ϕ& ( t) t ω ( s) s - It has a positive magnitude meaning that the sense coincides with the sense of ϕ (t) (Fig) Further, to find N, the right-angled triangle CN is considered: N sin 6 N C sin 6 Rsin 6,866, 6 m C Finall: V,6,6 m/s e TheV r e s direction is perpendicular to N, parallel to -ais (the trajector due to the rotation is circle and the velocit s direction is perpendicular to the radius) and the sense follows the sense of ω (Fig) Fig Here, relative and transport velocit of the point have mutuall perpendicular directions Then, the magnitude of the velocit of the point is: V V r + Ve 6,8 +,6 6,8 m/s The sense and direction of the absolute velocit are determined b the parallelogram rule (Fig) bsolute acceleration of the point at instant t s The absolute acceleration of the point is equal to the sum of relative, transport and Coriolis components: r r r r aa ar + ae + ac Relative motion of the point is curvilinear and the relative acceleration has normal and tangential components: r r n r t ar ar + ar When the point trajector is a curve, the normal acceleration points to the center of such curve, and its magnitude is: n Vr 6,8 ar,5 m/s R

12 The tangential component of relative acceleration is the second order derivative with respect to the time of the relative motion equation: ( t) S & ( t) π a t ( s) 6, 8 r m/s a t r t a r r lies on the tangent to the trajector of point and b virtue of positive magnitude its sense coincides with the sense of the relative motion of the point (Fig) The magnitude of the relative acceleration of is: n t a ( a ) + ( a ),5 + 6,8 4,57 m/s r r r Fig od performs rotation and the transport acceleration is the sum of normal and tangential accelerations: r r n r t ae ae + ae The normal acceleration is given b the epression: a n e ω N,6,6 m/s, and it alwas directs to the ais of rotation (Fig) The tangential component is: a t e αn, where α is the angular acceleration of the bod and it is the second order derivative with respect to the time of the equation of rotation: α ( t) & ϕ ( t) s - α( s ) s - It is carried out with negative sign, ie its sense is opposite to the sense of ϕ (t) (Fig) a,6 5, m/s t e t The sense of a r e coincides to this one of α and its direction is perpendicular to N (Fig) The magnitude of the transport acceleration of is: ц вр a ( a ) + ( a ),6 + 5, 5,8 m/s e e e The Coriolis acceleration is: r r r a C ω V r The magnitude of the Coriolis acceleration is given b the epression: r r ac ω Vr sin( ω; Vr ), where the angle between the directions of angular and relative velocit is5 (Fig4) Fig 4

13 Finall, ir is obtained: a 6,8sin5 6,8 m/s C Vector a r C has to be perpendicular to the plane formed b r ω and V r r, and the three vectors must form the righthanded sstem Therefore, the Coriolis acceleration has direction and sense as shown in Fig5 Fig 5 To obtain the magnitude of the absolute acceleration of the point, the components have to be projected onto three mutuall perpendicular aes Here, these are the aes, and Then: a a + a + a t a a + a 5,+ 6,8,8 m/s ; e C a a sin 6 a cos6 + a, n r t r a,5,866 6,8,5+,6,85 m/s ; n t ar cos 6 ar sin 6, a a,5,5 6,8,866, m/s ; a,8 +,85 + (,) 6, m/s n e The sense and direction of a r are determined b the parallelipiped rule (Fig 6) Fig 6

14 COURSE WORK : REDUCTION OF SPTIL SET OF FORCES Problem spatial set of forces is applied to homogeneous rectangular prism, as shown in Fig Reduce the set with respect to point O and determine the reduction case to which the set can be brought; Draw on the sketch of suitable scale the force resultant, the moment resultant and the angle between them; Calculate the moment of the force F r about an ais formed b points E and H of direction from E to H Data: O m O 5 m OC 4 m F 56 kn F kn F 8kN 5 knm 6 knm Fig Solution Reduction of a set of forces about a point requires the force resultant and moment resultant about the point to be obtained Resultant force R r The force resultant is going to be obtained adding its projections onto three mutuall perpendicular aes, such as,, : r r r r R R i + R j+ R k, where: R F + F + F R F + F + F R F + F + F,, Here, F,, F are the projections of the applied forces onto aes,,, respectivel It should be noted that each one of them has to be taken with the sign depending on the sense of the force relative to the sense of the aes Resolution of F r Force F r is in general position with respect to the aes,, and Therefore, to obtain its projections along the aes, the fact that F r is collinear to vector DC has to be used Thus: F λ F ; F F ; F ν F, where λ DC, DC µ DC DC µ DC, ν DC are the direction cosines of DC The are determined b the epressions: 4

15 where DC C D λ DC ; DC C D µ DC ; DC C D ν DC, DC ( C D ) + ( C D ) + ( C D ) is the magnitude of DC : DC ( ) + ( 5) + (4 ) 7,7m Further: 5 4 λ DC,44; µ DC, 77; ν DC, ,7 7,7 7,7 - Check of the direction cosines: λ DC + µ DC + ν DC (,44) + (,77) +,5657! Finall: F,4456,76 kn; F,7756 9, 6 kn; F,565756, 68 kn Negative signs show that F and F have senses opposite to the senses of the aes and (Fig) Resolution of F r Fig Projections of force F r onto the aes are found using the direction cosines of vector ED collinear to the force F r, ie D E λ ED ; ED The magnitude of ED is: D E µ ED ; ED D E ν ED ED ED ( ) + ( ) + ( ) ( ) + (5 5) + ( 4) 5 m, D E D E D E and the direction cosines are: λ ED,6 ; µ ED ; ν ED, Check of the direction cosines: λ µ + ν,6 + + (,8)! ED + ED ED 5

16 Projections of F r are obtained, as follows: F λed F,6 9, kn; F µ ED F ; F ν ED F,8 5, 6 kn Resolution of F r Force F r is parallel to -ais, ie it has projection onto -ais onl (Fig): F ; F F 8 kn; F Finall, projections of the force resultant onto aes are carried out as: F + F F,76+ 9,+ 4, 56 kn; R + R F + F + F 9,6+ + 8, 6 kn; R F F + F,68 5,6+ 6, 8 kn The magnitude of the force resultant is obtained as: R R + R + R ( 4,56) + (,6) + 6,8,87 kn oment resultant about point O oment resultant r O will be obtained b the sum of projections onto aes,, and, ie r r r r i + j+ k O O O O Here, the projections of the moments r and r onto the aes are determined first, as follows oment r is parallel to -ais, ie it has projection onto onl In contrast to r, the moment of the couple r is in general position with respect to the aes,, (Fig) The couple lies in the plane formed b points,,c, and vector r is collinear to the normal vector Therefore, the projections of r can be obtained using the direction cosines of N r C N r C of such plane (Fig) The direction cosines of vectors ling in the plane C : r i C C C C N r C Fig are calculated using the cross product of vectors C and C, two r r j k C C C C 6

17 It should be mentioned that the cross product is C C, and not about point C, the moment r first crosses vector C, and then C (Fig) r r r r r r i j k i j k r r r C C 4 4 i + j+ 5k, which means that N r C ; N r C ; N r C 5 Further, the magnitude of N r C is calculated as: r r r r N N + N + N C C C C C C, because during the rotation 7,7 m Then, the direction cosines are: r r r N C N C N C λ C r 5,7 ; µ C r, 47 ; ν C r, 549 N 7,7 N 7,7 N 7,7 C C C - Check of the direction cosines: λ µ + ν,7 +,47 +,549! C + C C Fig 4 Finall, the projections of r onto the aes,, are obtained as: λ C,76 44,7kNm; µ C,476 6,8 knm; ν C,5496,54 knm, and their senses and directions are shown in Fig4 Then, the moment equations about aes,, are written, as follows: O O + + F D F CE 5+ 44,7+,685 5,65 F D+ F E 6,8,68+ 9,4 8,59 knm; 5,kNm; 7

18 O + + F D F D F CE F O,54+,765 9,6 9,5+ 8, 54 knm Using the right-hand rule, the positive senses of the projections of resultant moment are chosen to coincide with the positive senses of the aes (Fig5) Finall, the magnitude of O O + O + O r O is calculated as: 5, Fig 5 + 8,59 +,54 55, knm Case of reduction The case of reduction is found using the scalar product of the force resultant and moment resultant: R r r R cos O β R + R + R, O O O O where β is the angle between directions of the force resultant and moment resultant Here, both R r and r O have been obtained different than ero Therefore, if their scalar product is also different than ero, ie angleβ is different than 9, then, the given set of forces can be further reduced to a wrench However, if their dot product is equal to ero, ie the angleβ is 9, then, the set of forces can be further reduced to a single resultant force calculation, it is obtained: R r r O ( 4,56)5,+ (,6)8,59+ 6,8,54 97, 8 kn m Therefore, the conclusion is that the case of reduction is wrench Further, to compute the angle between directions of the force resultant and moment resultant the previous epression rearranged with respect to cos β is used: R O + R O + R O 97,8 97,8 cosβ,574 β 4,9 R O,8755, 765,9 The force resultant, the moment resultant and the angle between them are shown in Fig6 8

19 Fig 6 4 oment of force F r about ais EH The moment of a force about an ais is a scalar, equal to the projection onto the ais of the moment of the force about a point belonging to the ais Here, point E is chosen as a point of the ais Then: EH r r F e ED F r λeh µ EH ν EH ), EH ( D F E D F where e r EH is the unit vector of ais EH (Fig7), while λ EH, µ EH, and ν EH are the direction cosines of the ais EH The are found b the following epressions: H E H E H E λ EH ; µ EH ; ν EH, EH EH EH where EH ( H E ) + ( H E ) + ( H E ) ( ) + ( 5) + (4 4) 5, 8 E D F E m λ EH,545; µ EH, 8575 ; ν EH 5,8 5,8 5,8 - Check of the direction cosines: λ EH + µ EH + ν EH,545 + (,8575) +! Finall: 9

20 ,545,8575 EH F 5 5 4,76 9,6, 68, , 5 knm,76,8575 9,6 EH F has a negative sign, which means that the sense of ais EH (Fig7),68 EH F is opposite to the positive sense of Fig 7

21 COURSE WORK 4: REDUCTION OF SET OF COPLNR FORCES Problem 4 coplanar set of forces is applied to homogeneous figure, as shown in Fig4 Locate the position of the center of gravit, point C; ppl vertical force F r of downward sense at point C and reduce the set of forces with respect to point ; Determine the reduction case to which the set can be brought F kn F 4 kn F 6 kn 4 knm where Fig 4 Solution Center of gravit To obtain the coordinates of the center of gravit the following equations are used: i and n i C n i i i i i ; n i C n i i i i, are the coordinates of the centers of gravit of the simple figures with respect to the aes, chosen for location of the center of gravit of the figure given Here, the simple figures are three rectangle, triangle, semicircle, and, first, the absolute location of their centers of gravit and their areas have to be determined Figure () is rectangle The coordinates of the center of gravit are: ( h b C m;, 5m) (Fig4a) The area of the rectangle is: 54 m Fig 4a

22 Fig 4b Figure () is right-angled triangle and the coordinates of the center of gravit with respect to the right angle are: ( h b C,9 m;, 5m) (Fig4b) The area of the triangle is:,74,5 6,75 m Figure () is semicircle The coordinates of the center of gravit are: 4R C ( R m;, 85 m) (Fig4c) π The area of the semicircle is:,4 π R 6,8 m Fig 4c fter that, the coordinates of the centers of gravit of three figures about aes and have to be found The aes and are chosen so that the entire figure lies in the first quadrant (Fig4) Thus, all of the coordinates of the centers of gravit of the simple figures are positive Using the coordinates determined earlier and following Fig4, it is obtained:,5 m; m;,5+ 4,5,5,5 m;,9+,5,4 m;,5+ 4,5+,85 5,85 m; m, ie the locations of the centers of gravit about and are: C(,5 m; m); C (,5 m;, 4m); C (5,85 m; m) Fig 4 Finall, the coordinates of the center of gravit of the given figure about aes and are: i i i +,5 6,75,5+ 6,85,85 65,475 C,4 m; + 6,75+ 6,8,5 i i

23 i i i + 6,75,4+ 6,8 44,55 C,8 m + 6,75+ 6,8,5 i and its position is given in Fig 44 i Fig 44 Reduction of the set of forces about point fter the center of gravit has alread been located, the force F r is applied (Fig45) and the reduction of the set with respect to the point begins Fig 45 Force resultant First, the aes X and Y of origin point are introduced (Fig46) Then, the force resultant is: r r r R R X + R Y, r r r r Ru F X + F X + F X, r r r r Rv F Y + F Y + F Y r v Here, F X,, F Y are projections of the applied forces onto aes X and Y (Fig46): F X F cos 6,5 5 kn; F sin 6 Y F,866 6 kn; F X ; F Y F 4 kn; F X F 6 kn; F Y

24 Further: R R X Y F X + F 5+ 6 F F 6 4 Y Fig 46 kn; 5 kn; R R X + RY + ( 5) 54,kN Finall, the angle between X-ais and resultant force is calculated The result is: RY 5 tgα R,8 α R 67, (Fig47) R X Fig 47 oment resultant The moment resultant about point is obtained b the following epression: + F X (,5) F Y + F (5,4) F,5 4+ 5,5 6+ 4,76 6, 5 7, 6 knm, 4

25 where the positive sense of the moment is counterclockwise (Fig48) The result is obtained negative meaning that the moment resultant points toward the page (Fig49) Fig 48 Case of reduction The set of forces is coplanar, which means that it can be further reduced to the single resultant force (Fig 49) The equation of the direction of single resultant force is: XR Y YRX ; X ( 5) Y 7,6 ; 7,6 7,6 nx,45 m; ny, 9m; R 5 R Y 7,6 hr,4 m R 54, X Fig 49 5

26 COURSE WORK 5: EQUILIRIU OF ODY CTED UPON Y SPTIL SET OF FORCES Problem 5 Determine the reactions of the rectangular prism supported and loaded as shown in Fig5 Check the result obtained using the relevant equilibrium equations The weight per unit volume is γ kn/m Fig 5 Solution: Determination of the weight of the bod, the resultant force of the distributed load, the projections of the moment and the force F onto the aes of the appropriate coordinate sstem First, the Cartesian coordinate sstem of origin point is introduced Point has been chosen for origin, because it is a spherical support containing three unknown reactions Thus, if the moment equations about aes,, are written, then, the three reactions at point are eliminated The weight of the rectangular prism is: G Vγ, where V D DL 5 m is the volume Then: G 9 kn G is directed down vertical force applied at the bod s center of gravit (Fig5) Distributed load q is applied on the side DLK of the rectangular prism Then, the resultant force is: D K q 5 75 kn R q The line of action and sense of R q are determined b the line of action and sense of the distributed load The point of application of R q is the projection of the distributed load s center of gravit on the prism s side DLK (Fig5) oment is in general position with respect to the aes,, Then, in order to find its projections onto these aes, the direction cosines of vector N, collinear to r, are used (Fig5) The coordinates of the points formed N are N ( ;;) and (;5;) Then, the magnitude of N is: 6

27 N ( ) + ( ) + ( ) ( ) + ( 5) + ( ) 6,64 m N N N Further, N N 5 N λ N,4867 ; µ N, 8 ; ν N, 44 N 6,64 N 6,64 N 6,64 - Check of the direction cosines: λ N + µ N + ν N ; (,4867) + (,8) +,44 ;,9999! Finall, the projections of the moment are found as: λn, ,4 knm; µ N,87 58, 4 knm; ν N,447,6 knm Force F lies on the side KLNH of the bod Then, it has the projections onto aes and : F Fcos β ; F Fsinβ (Fig5) Considering right-angled triangle LNH, angle β is obtained as: LN 5 tgβ β 59,4 NH Then, F 6cos59,4,9 kn; F 6sin 59,4 5, 45 kn (Fig5) Fig 5 Determination of support reactions First, free bod diagram is drawn, ie all supports are detached from the bod and the support reactions are introduced instead These are: - Point is a spherical support constraining the displacements parallel to the aes of the coordinate sstem Then, three support reactions are available, namel,, where their senses are supposed (Fig5); - There is a link at point which means that onl one support reaction of line of action coinciding with the direction of the link has to be introduced (Fig5); - Point L is a clindrical support and contains two support reactions, L and parallel to the aes and (Fig5) esides, all eternal loads are applied L, of lines of action 7

28 The spatial set of forces is in equilibrium when the main vector and the main moment are both equal to ero, ie R r, r O These two vector equations can be transformed into nine scalar equations, three force and si moment equations Si of them are used for the support reactions determination, while the other three are applied for the check of the reactions obtained Further, in order to determine the support reactions, the appropriate choice of equations has to be made In the best case, the independent equations in which onl one unknown reaction takes part has to be written Here, in order to find the independent equations, the free-bod diagram is analed carefull (Fig5) It is obvious that onl reaction i Fig 5 is parallel to -ais Then, the first independent equation is: ) F ; F R ; + 5,45 75 ;, 55 kn + q is positive which means that the sense supposed is correct (Fig5) The second independent equation is: 75+,6 ) ; + Rq L5 ;,6+ 75 L 5 ; L 9, 67 kn 5 The third equation containing onl one unknown is: ) ; 5 F5+ + Rq ; 5,95+,6+ 75 ; 75+,6,95, kn 5 The sign - for means that the sense supposed is not correct and has to be changed (Fig) The solution continues with the fourth independent equation: 4) ; 5 F+ G,5 + Rq ; 5 5,45+ 9,5 5,4+ 75 ; 9, ,4 5,45,4kN 5 Further, analing Fig5 the conclusion that there is no other independent equation is made Then, the determination of the last support reactions continues with the net equations: 8

29 5) ; G,5 F + L + L ; 9,5 58,4,9+ L + 9,67 ; 9,5+ 58,4+,9 9,67 L 7,95 kn; 6) ; 5 F G,5+ R + L 5 ; q 5 5,45 9, ,4+ 7,955 ; 5,45+ 9, ,4 7,955 4,6 kn 5 Check for the reactions obtained One force and five moment equations have been used for determination of the support reactions Then, two force and one moment equation have to be written for the check of the result obtained It should be noted that an equation chosen for check has to contain as man as possible support reactions Fig 54 The force equations for the check are (Fig54): 7) F ; F + D ;,,9+ 9,67 ;,9,9! i i 8) F ; G + D ;,4 9 4,6+ 7,95 ; 4,6 4,6! The moment equation is (Fig54): 9) ;,5 F + L+ L,5+,5 ;,,4,5,9 58,4+ 9,67+ 7,95,5+ 4,6,5 ; 49,5 49,45,! ll equations are obtained equal to ero meaning that the support reactions are determined correctl! Problem 5 Determine the support reactions of the rectangular plate supported and loaded as shown in Fig5 Check the result obtained using the relevant equilibrium equations ssume that the thickness of the plate is approimatel ero Neglect the weight of the plate 9

30 F // F // D q // EH ; ; Fig 5 Solution: Determination of the resultant force of the distributed load, the projections of the moment and the support reaction D onto the aes of the appropriate coordinate sstem Solution begins with the choice of the coordinate sstem Here, the coordinate sstem is chosen so that the aes pass through as man as possible supports (Fig5) Fig 5 The distributed load q acts on the half plate To determine the resultant force volume of the prism is applied, ie R q the formula for the

31 R q D q 4,54,85 54 kn R is a force parallel to -ais of sense coinciding with the sense of q The point of application of q the projection onto the plate of the distributed load s center of gravit (Fig5) oment has projections onto aes and, and the are obtained using angle δ as: cosδ ; sinδ, where δ is : DE 4,5 tgδ,975 δ 4,5 (Fig5) E 4,8 Then, 4cos 4,5 7,5 knm; 4sin 4,5 6, 4kNm (Fig5) R q is The link at point D is inclined with respect to the aes and and the reaction in the link is in general position with respect to the same aes (Fig5) Therefore, it is more convenient to work with the projections of the force D onto the aes and These projections are determined as follows: D Dsin β ; D Dcosβ, ED 4,5 where tgβ, 8 β 6,95 EH,5 Finall: D D sin 6,95, 874D ; D D cos 6,95, 4856D (Fig5) Determination of the support reactions The plate is detached from the constraints and the support reactions are introduced instead: - There are two links at point and reactions and of lines of action coinciding with directions of the links are introduced (Fig5); - There are two links at point and two reactions are introduced, as well (Fig); - The link at point D has alread been considered and the projections of the reaction, namel D and D have been introduced (Fig5); - There is a link at point E and one reaction is introduced (Fig5) It should be noted that the senses of all reactions are supposed Further, the solution continues with analsis of Fig5 Similar to the previous eample, the independent equations are four These are: ) F ; ; + 6 ; 6 kn i,4 ) ; 4,8+ F,4 ; 4,8+,4 ; 5 kn 4,8 ) ; 4,5 + Rq,5 ; 4,5 7,5+ 54,5 ; 7,5+ 54,5 4,kN 4,5 4) ; Rq,+ D4,8 ; 6,4 54,+ D 4,8 ; 6,4+ 54,,58 D,58 kn D 67, 9 kn D,87467,9 58, 64 kn 4,5,4856 The last two equations are: 5) ; + 4,5 R + D 4,5 ; 7,5+ 4,5 54+,584,5 ; 7,5+ 54,584,5 7,kN; 4,5 6) q ; E 4,8 F,4+ D 4,8 ; E 4,8,4+ 58,644,8 ;

32 kn Fig 5 Check for the reactions obtained Two force and one moment equations is used for check (Fig54): 7) F ; + F + E D ; ,64 58,64 ; 7,64 7,64! i i 8) F ; + Rq + D ; 4,+ 7, 54+,58 ; 54 54! 9) ;,,+ D,6 ; 6,4 4,, 7,,+,58,6 ; 68,5 68,54,! Fig 54

33 COURSE WORK 6: EQUILIRIU OF ODY SUJECTED TO SET OF COPLNR FORCES Problem 6 ppl independent equilibrium equations to determine the support reactions of the construction shown in Fig6 Check the result obtained Fig 6 Solution: First step of solution is introduction of the coordinate sstem of aes and (Fig6) It should be noted that the pla smbolic role here, ie the onl give horiontal and vertical directions and positive senses esides, the origin of the coordinate sstem is chosen arbitraril Determination of the resultant force R q of the distributed load and the projections of the force F onto aes and Load q is uniforml distributed and its resultant force is: q 4,5 44,5 6kN, R q which direction and sense coincides with the direction and sense of distributed load The R s point of application is the midpoint of the distance over which the distributed load acts (Fig6) q Fig 6

34 Projections of the force F onto aes and are equal, because the angle between the force s line of action and the aes is 45, ie F F F cos 45,77 4,4 kn (Fig6) Determination of the support reactions First, the free-bod diagram is drawn where the constraints are substituted for their support reactions is a point where a link inclined at an angle of 6 with respect to the horiontal is positioned Then, a reaction of line of action following direction of the link is introduced The sense is supposed arbitraril (Fig6); Point is constrained b roller and one vertical support reaction is introduced instead (Fig6); There is a horiontal link at point C the horiontal reaction C is introduced (Fig6) Projections of reaction onto aes and The support reaction at point is in general position with respect to the aes Therefore, its projections parallel to the aes have to be found The result is: cos6, 5 ; sin 6, 866 Their point of application is, and the senses follow the sense of support reaction (Fig6) Independent equilibrium equations When a coplanar set of forces is applied to the structure, the equations used for the reactions determination are five, two force and three moment equations Here, the statement requires all equations used for obtaining of support reactions to be independent Then, the careful analsis of Fig6 reveals that the two force equations Fi and F i are not independent, because each one of them contains two unknowns It means that the moment equations have to be used oreover, these moment equations need to be about points of intersection of the lines of action of two support reactions These points are named Ritter s points Thus, intersecting the lines of action of reactions and, and C, and C, points D, E and H, respectivel, are located (Fig6) Position of points D and E - Right-angled triangle D is considered (Fig6): D tg 6 D tg6,5,7 6, 6 m; - Right-angled triangle EHD is considered (Fig6): DH DH 9,6 tg 6 EH 5, m EH tg 6,7 Support reactions determination The moment equations about Ritter s points D, E, and H are written, as follows It should be noted that the positive sense of the moment is chosen to be the counterclockwise one (Fig6): ) ;,5 + F 4,9+ F + R,5+ F + H E q ; (,866 ),5 (,5)+ 4,44,9+ 4,4+ 6, ; 4,5 7,456 ; 59,69 kn ) ; 5, F (5, 4,9) + F R (5,,5) + F + D q ; 5, 4,4,+ 4,4 6, ; 5,,99 ; 5,4 kn ) ; C 9,6 F 8,6+ R,5+ F 4,9 F 6,6 ; + q C 9, ,6+ 6,5+ 4,44,9 4,46,6 ; 9,6C 47,57 ; C 6,9 kn ll of the reactions are obtained positive meaning that the senses are correct (Fig6) 4

35 Fig 6 Check of the results obtained The two force equations are used for check First, the magnitudes of Then: 4) F ; F + F C i i,5,559,69 9,85 kn;,866,86659,69 5,69 kn (Fig64) ; 4,4 9,85+ 6,9 ; 46,4 46,4! 5) F ; F + R + q ; 4,4+ 5, ,4 ; 77,4+ 77,,! and are found: The two equations equal ero which means that the reactions obtained are correct! Fig Problem 6 ppl independent equilibrium equations to determine the support reactions of the construction shown in Fig6 Check the result obtained 5

36 Fig 6 Solution: Solution begins with introduction of the coordinate sstem of aes and (Fig6) Determination of R q, F и F Load q is linearl distributed and its resultant force is: R q q,6 8,6,4 kn; R is a force of direction and sense determined b q The point of application is the projection of the q distributed loading s center of gravit on the construction (Fig6) Projections of F onto aes and are: F cos 6,866,8 kn; F F F sin 6,5 8 Determination of the support reactions kn (Fig6) Fig 6 6

37 The following support reactions are introduced: There is a pinned support at point a horiontal and a vertical reaction of supposed senses are introduced (Fig6); The roller is situated at point a vertical support reaction is introduced (Fig6) naling Fig6, it is concluded that onl one unknown reaction is parallel to -ais Then, the first independent equation is: ) F ; R + F ; i + q +,4+,8 ; 6,58 kn The negative sign means that the sense chosen for is not correct and has to be changed (Fig6) The second independent equilibrium equation is the moment one about point, because is the point of intersection of the and s lines of action (Fig6): ) ; R,4 F,5+ 4 F,6 F 5,5 ; q,4,4 4 4,5+ 4,8,6 855 ; 4 76 ; 94 kn The positive sense of moment is the counterclockwise one The third independent equation is the moment equation about D, the point of intersection lines of action (Fig6): ) ; 4 R,4 + F,5 F,6 F,5 ; D q 4,4,4 4+ 4,5,8,6 8,5 ; 4 6 ; 4 kn and s Check for the reactions obtained Since one force and two moment equations have been used for the support reaction determination, a moment equation containing all reactions is used for check This equation is the moment one about point E (Fig 6): 4) ;,5,4 +,5 F, F 4 ; E 4,5 6,58, ,5,8, 84 ; 86 86! Fig 6 7

38 COURSE WORKS 7: EQUILIRIU OF GERER ES Problem 7 Determine the support reactions and the forces in hinges of the Gerber beam shown in Fig7 Check the result obtained Fig 7 Solution: First, the number and the tpe of the beams composing the Gerber beam are determined Here, the beams are three, namel S, S S, and S CD eam S CD is the main beam, because it will be in equilibrium detached from the structure However, beams S and S S can not be in equilibrium without the main beam Therefore, the are secondar beams Then, the beams are numbered: S is beam (), S S is beam (), and S CD is beam () Finall, the coordinate sstem of aes and is introduced (Fig7) Determination of the distributed loads resultant forces Load q is applied on beam () It is linearl distributed and its resultant force is: R q S 4 6 kn R has line of action and sense following these ones of distributed load The point of application of R is the position of the center of gravit of the triangle representing the distributed load (Fig7) The load q is uniforml distributed over two beams Then, two resultant forces have to be obtained - Resultant force for the portion of q applied on beam () is: R q S 8 56 kn R is a vertical force of downward sense with point of application at the midpoint of distance S (Fig7) - Resultant force for the portion of q acting on beam () is: R q DS 84 kn R is a vertical force of downward sense with point of application at the midpoint of distance DS (Fig7) Determination of the support reactions and the forces in hinges In order to determine the support reactions and forces in hinges, the Gerber beam is separated through the hinges In other words, the entire construction is decomposed into three beams over which the applied loads and unknown reactions and forces in hinges are introduced The number and the lines of action of the support reactions depends on the tpe of the constraints, and, as usual, their senses are arbitrar chosen (Fig7) The forces in hinges are introduced at points S and S, and the act on the two beams connected b the hinge definition, the forces in hinges represent the influence of one beam to another and the ensure the equilibrium of the beam detached from entire structure The forces in hinges have vertical and horiontal lines of action and senses supposed randoml, but opposite for the two beams connected b the hinge (Fig7) Finall, it should be noted that the solution starts from the beam of fewest number unknowns Here, this is beam () eam () It is convenient the three independent equations to be written These are: ) F i ; F S ; 8 S ; S 8 kn; ) ; R + S ; 6+ S ; S 4 kn The positive sense of the moment is chosen to be the counterclockwise one (Fig7) 8

39 ) S ; R + ; 6 + ; kn The three unknowns are obtained positive meaning that the senses chosen are correct (Fig7) - Check of the result obtained: F i ; R + S ; 6+ 4 ; 6 6! Fig 7 eam () ) F i ; S S ; 8 S ; S 8kN; ) ; S R S ; 4 56 S ; S 4 kn; S is obtained negative, ie its sense is changed for the two beams to which S is applied (Fig7) ) S ; S 4 + R ; ; 76 kn - Check: F i ; S + R + S ; ; 8 8! eam () ) F i ; S C ; 8 C ; C 8kN; ) C ; S,5 R,5+ D,5+ ; 4,5,5+ D,5+ 4 ; D, ) D ; S 4 C,5+ R + ; 4 4 C, ; C, 8 - Check: F i ; S + C R + D ; 4 +,8 +, ; 6 6! Check of the entire structure In order to perform this check, the Gerber beam is recomposed and the support reactions obtained are applied to it The equations for check are the following ones: ) F i ; F C ; 8 8! ) F i ; R + R + C R + D ,8 +, kn; kn ; 4 4! Fig 7 9

40 COURSE WORK 8: EQUILIRIU OF THREE-HINGED FRES Problem 8 Determine the support reactions and the forces in hinge of the three-hinged frame shown in Fig8 Check the result obtained Fig 8 Solution: Determination of the resultant force of distributed load The load q is uniforml distributed and its resultant force is: Rq,5 4 kn R q is a horiontal force of sense to the left The point of application of R q is the midpoint of the distance over which the load is distributed (Fig8) Determination of the support reactions First, free-bod diagram of entire structure is drawn Points and are pinned supports and the are substituted for their reactions having horiontal and vertical lines of action and senses chosen arbitraril (Fig8a) However, the number of the unknown for entire three-hinged frame is four while the number of equilibrium equations is three Then, in order to find the unknowns, the three-hinged frame is separated to its composing two parts through the hinge Further, each part is loaded b the eternal forces and support reactions acting upon it In the case when a concentrated force or moment is applied directl at the hinge, then, the force or moment is assumed to act onl on one part of the construction Here, it is chosen the moment to be applied to the frame part S (Fig8b) fter that, in order to ensure the equilibrium of each part of the frame, the forces in hinges are introduced at point S It should be noted that their lines of action are horiontal and vertical while their senses are arbitrar chosen esides, the are applied to the two parts of the frame and their senses are opposite with regard to the two parts (Fig8b) General tpe of three-hinged frame does not allow the independent equilibrium equations to be written Then, a set of two equations of two unknowns has to be found The proper choice is first equation of the set to be the moment equation for entire structure, while the second equation to be the moment equation for one of the parts The points, about which the moment equations are going to be written, are the points of intersection of two unknowns Determination of and 4

41 EntStr ) (Fig8a); 5,6 + F + F 5,6 + Rq,75 ; 5, ,6 4+ 4,75 ; 5,6,5 + The positive sense of the moment is chosen to be the counterclockwise one S ) (Fig8b); S,5+,5 F,5 ;,5+,5,5 4 ; 46,8 Solving the set of equations, the result is: 5,6,5 () + 46,8 46,8,8+ + 5,6 () 46, 8+ 46,5 6,6 54,7,44 kn; 46,8+,44 7,4 kn and senses are correct Determination of have positive values, ie their and EntStr ) (Fig8а); + R, ,6 ; q 4 + 4, ,6 ; 5,6,5 + S 4) (Fig8b); S F,5 R,75+,5+, ; q 5,5 4,75+,5+, ;,5, + Fig 8a Solving the set of equations, it is obtained: 5,6,5 () + +,,5 (4) + 5,6,5,5 5, 6,5(,5 5,6 ) +, 6,75 9,6 +, 6,5 5,75,56kN;,5 5,6,56,76 kn Check for the reactions obtained (Fig8): Fig 8b 4

42 EntStr F ; i F Rq + ; 7,4 4+,76 ; 7 7! EntStr F ; i F + ;,44 5+,56 ; 5 5 Fig 8 Determination of the forces in hinge The forces in hinge are determined b two force equation for one of the frame parts Then, the are checked b two force equation for the other part (Fig84): S F ; i F S ; 7,4 S ; S 4, 4 kn S F ; i S ;,44 S ; S, 44 kn - Check: S F ; S + i Rq ; 4,4 4+,76 ; 4 4! S F ; S i F + ;,44 5+,56 ; 5 5! Fig 84 4

43 COURSE WORK 9: EQUILIRIU OF PLNE TRUSSES Problem 9 plane truss is supported and loaded as shown in Fig9 Determine: Support reactions; Force in each member using the method of joints; Force in the members marked using the method of sections Solution: Support reactions determination Fig 9 Fig 9 Here, the independent equations for determination of the support reactions are one force equation onto -ais and two moment equations about the point of intersection of and (point ) and the point of intersection of and (pointc ), in which the positive sense of the moment is assumed to be the counterclockwise one (Fig9): 4

44 i C ) F + F kn; ) F F 9 F kN; ) + F 9+ F F kn D Check: 9 4+ F 6+ ; ; 9 9! Determination of the force in each member using the method of joints Numeration of links and joints Numeration begins from the left and goes to the right The numbers of the members are in circle, while the numbers of the joints are in quotation marks (Fig9) Zero-force members In present problem, the ero-force members are three: - links () and (4) the are connected b the unloaded joint (Fig9); - link (9) it ends in joint 4 where the other two members have one and the same direction (Fig9) Forces in the links Solution begins from the first joint, here, joint It is detached from the truss and the eternal forces and the forces in members are introduced The forces in the truss members are unknown their lines of action coincide with the directions of the links while their senses are chosen to be out of the cut, ie the forces are supposed tensile Then, the equilibrium of the joint is eamined using two force equations Thus, the unknowns are obtained, as follows - Joint А - Joint i i F ; S sin 5, + 9 ; S, 75 kn (c); F ; S 4 S cos 5, ; S 56, 5 kn(t) - Joint i F ; S sin 5, S5 S7 sin 5, ; S7,5kN (c); F ; S cos5,+ S8 S7 cos 5, ; S8,5 kn (t) i - Joint 4 i i F S 8 S ; S, 5 kn (c); F S 9 i i F S 6 S ; S 56, 6 5 kn (t); F S 6 ; S 6 kn (t)

45 - Joint 5 - Joint D i F ; S 7 sin 5, S sin 5, ; S,5 kn (t); F ; i - Joint 6 S 6 + S7 cos5,+ S cos 5,+ S ; S,75 kn (t) i F ; S S cos5,+ S4 ; S4,5 kn (t) F ; S sin 5,+ S ; i 4 4,5,8+ 4 ; 4 4! - Joint В - Check i i F ; S + S cos 5, ; S 5, 5 5 kn(t); 5 F ; S + S sin 5, ; S 4kN(c) 5 i F ; S S cos 5,+ 4 ; i 4 5,5 5,5,6+ 4 ; 4 4! F ; 4 S 5 sin 5, ; 4 5,5,8 ; 4 4! Notes: There are letters (c) or (t) behind the values obtained for each force in truss members The letters shows that the rod is compressive or tensile, respectivel When the rod is compressed, the sense of the force has to be changed! The last joint s equations of equilibrium are used for the check of the results obtained ethod of joints allows solution of the problem to begin from the last joint and to go from the right to the left Then, joint will be used for the check of the results Determination of the force in truss members marked using the method of sections The essence of the method of sections is that the truss is divided into two portions b an imaginar cut through the three members marked Then, the truss portion loaded b less number of loads is considered where the eternal forces, support reactions and the forces in the members through which the cut passes are applied Further, three independent equilibrium equations are written and the unknowns are obtained Finall, the check is made 45