UNIVERSITY OF ARCHITECTURE, CIVIL ENGINEERING AND GEODESY
TECHNICAL MECHANICS DEPARTMENT
S O L V E D E X A M P L E S
OF
C O U R S E W O R K S
ON
MECHANICS – PART I
(KINEMATICS & STATICS)
Assistant Professor Angel Mladensky, PhD
Sofia 2015/2016
2
COURSE WORK 1: KINEMATICS OF THE PLANE MOTION OF RIGID BODIES
Problem 1.1
At the instant given, a mechanism occupies position shown in Fig.1.1.1. Velocity and acceleration of
point A are 15=AV m/s and 36=Aa m/s2, respectively. Cylinder rolls without sliding.
Determine:
1. Velocities of points B , C , D , E and angular velocities of all members of the mechanism;
2. Acceleration of point B and angular accelerations of members AB and OB .
Fig. 1.1.1
Solution:
1. Numeration of the mechanism’s members and determination of their types of motion
Members of the mechanism are four – cylinder of centre A (member 1), rod AB (member 2), rod AC
(member 3) and rod OB (member 4). The types of their motion are found by consideration of previous instant
of the motion of the mechanism, i.e. mechanism’s position '''' CABO (Fig.1.1.2).
Fig. 1.1.2
3
Cylinder (1) rolls, i.e. it performs plane motion. Members (2) and (3) also perform plane motion because
they have no fixed points and do not remain parallel to themselves during the motion. Member (4) has an
unmovable point O, i.e. it performs rotation about point O.
2. Determination of velocities of points B , C , D , E and angular velocities of all members of the
mechanism
The procedure of determination of all velocities begins from the given velocity of point A which is a
point of three mechanism’s members, namely (1), (2) and (3).
Cylinder performs plane motion represented as a rotation about instantaneous center of zero velocity,
the point at which cylinder touches the ground. Then, the magnitude of the angular velocity of cylinder (1) is:
151
15
1
1 ===AP
VAω s-1
,
where 1AP is the distance from point A to the cylinder’s instantaneous center of zero velocity. The sense of 1ω
follows the sense of AVr
(Fig.1.1.3).
Further, the velocity of point D is obtained as:
302.15. 11 === DPVD ω m/s,
where 1DP is the distance between D and the cylinder’s instantaneous center of zero velocity. DVr
is
perpendicular to 1DP , and its sense depends on the sense of 1ω .
After that, the velocity of point E is found as:
21,212.15. 11 === EPVE ω m/s,
where 1EP is the distance between E and the cylinder’s instantaneous center of zero velocity. EVr
is
perpendicular to 1EP , and its sense follows the sense of 1ω .
Fig. 1.1.3
4
Solution continues with determination of the angular velocity of member (2) performing plane motion.
Such motion is presented as a rotation about the instantaneous center of zero velocity of body (2). Here, in
order to find its position, perpendiculars to the directions of the velocities of two points are drawn. These points
are A and C , because the directions of their velocities are known: direction of AVr
is horizontal, while direction
of CVr
is inclined at an angle of 450 with respect to the horizontal. Then, intersecting the perpendiculars, the
instantaneous center of zero velocity, point 2P , is located (Fig.1.1.3). Hence:
2
2AP
VA=ω ,
where 2AP is the distance between A and 2P . To obtain 2AP the isosceles right-angled triangle 2ACP is
considered:
0
2
45cos=AP
AC ⇒
0245cos
ACAP = ; 045cos
3=
AC ⇒ 24,4
45cos
30==AC m; 6
45cos
24,402 ==AP m.
Finally,
5,26
152 ==ω s
-1,
of sense following the sense of AVr
.
Further, the velocity of point C is determined using the expression:
22 .CPVC ω= ,
where 24,42 == ACCP m, because both are cathetus in the isosceles triangle 2ACP .
6,1024,4.5,2 ==CV m/s.
CVr
is perpendicular to 2CP and its sense is determined by the sense of 2ω (Fig.1.1.3).
Next step is to determine the angular velocity of member (3). Again, the plane motion is presented as
rotation and the velocity of point A is used. Location of the instantaneous center of zero velocity of member (3)
is found by intersection of perpendiculars to the directions of velocities of points A and B (Fig.1.1.3). Then,
the angular velocity of member (3) is:
3
3AP
VA=ω ,
where 3AP is obtained from the right-angled triangle 3ABP , as follows:
03 30tgAB
AP= ⇒ 308,230.430. 00
3 === tgtgABAP m.
Thus,
5,6308,2
153 ==ω s
-1,
where the sense depends on the sense of AVr
.
Further, velocity of point B is obtained applying the expression:
33 .BPVB ω= ,
where 3BP is carried out by the Pythagoras’s theorem as:
2
3
22
3 )()()( APABBP += ⇒ 618,44308,2)()( 2222
33 =+=+= ABAPBP m.
Hence,
30618,4.5,6 ==BV m/s,
of sense following the sense of 3ω .
Finally, BVr
is used for obtaining of the angular velocity of member (4):
OB
VB=4ω 103
30== s
-1,
where the distance OB is carried out as:
5
030sin5,1=
OB ⇒ 3
30sin
5,10==OB m.
• Check with the theorem of the velocities projections:
pr. AAE Vr
= pr. EAE Vr
⇒ 00 45cos0cos EA VV = ⇒ 00 45cos.21,210cos.15 = ⇒ 1515 = ;
pr. DDE Vr
= pr. EDE Vr
⇒ 00 0cos45cos ED VV = ⇒ 00 0cos.21,2145cos.30 = ⇒ 21,2121,21 = ;
pr. AABVr
= pr. BABVr
⇒ 00 60cos0cos BA VV = ⇒ 00 60cos.300cos.15 = ⇒ 1515 = ;
pr. AAC Vr
= pr. CAC Vr⇒ 00 0cos45cos CA VV = ⇒ 00 0cos.6,1045cos.15 = ⇒ 6,106,10 = ;
3. Determination of acceleration of point B and angular accelerations of AB and OB
Point B is the connecting point of members (3) and (4). It should be noted that the magnitude and
direction of acceleration of point B are unknown. Therefore, in order to find them, the following procedure is
performed. First, B is considered as a point of member (3). Thus, the formula of the acceleration is:
r
BA
c
BAAB aaaarrrr
++= , (I)
where Aar
is the acceleration of point A chosen for a pole, c
BAar
and r
BAar
are the centripetal and rotational
components, respectively, of the acceleration of point B . There are three unknowns in the equation (I) – the
magnitude and sense of Bar
and the magnitude of r
BAar
, i.e. the problem is insoluble.
Further, B is examined as a point of member (4). Then, the expression of the acceleration is:
t
B
n
BB aaarrr
+= , (II)
where n
Bar
and t
Bar
are the normal and tangential components, respectively. Three unknowns are available in
equation (II) – the magnitude and sense of Bar
and the magnitude of t
Bar
. It means that the problem is insoluble
again. However, equating the right-hand sides of (I) and (II), it is obtained:
t
B
n
B aarr
+ r
BA
c
BAА aaarrr
++= . (III)
Thus, two unknowns take part in expression (III) – the magnitudes of r
BAar
and t
Bar
. Then, the projections
of (III) onto two axes make the task soluble.
Fig. 1.1.4
First, the normal and centripetal accelerations in (III) are carried out as:
3003.10. 22
4 === OBa n
B ω m/s2, 1694.5,6. 22
3 === BAа c
BА ω m/s2,
where the sense of n
Ba is from B to O , while the sense of c
BАа is from B to A (Fig.1.1.4).
6
Further, the tangential and rotational accelerations are introduced perpendicular to the normal and
centripetal ones, respectively. Their magnitudes are:
44 3. αα == OBa t
B , 33 4. αα == BAа r
BA ,
where 3α and 4α are the angular accelerations of the members (3) and (4), respectively. The senses of r
BAar
,
t
Bar
, 3α , and 4α are arbitrary chosen and shown in Fig.1.1.4.
Then, the projection of (III) onto x-axis is written, as follows:
c
BAA
t
B
n
B aaaa −=− 00 30sin30cos ,
169365,0.866,0.300 −=− t
Ba ,
62,7855,0
1693681,259=
+−=t
Ba m/s2,
87,2613
62,785
34 ===
t
Baα s
-2.
t
Bar
and 4α have positive signs, i.e. their senses are correctly chosen (Fig.1.1.4).
Further, the projection of (III) onto y-axis is written as:
r
BA
t
B
n
B aaa =−− 00 30cos30sin ,
r
BAа=−− 866,0.62,7855,0.300 ,
35,83035,680150 −=−−=r
BAa m/s2,
59,2074
35,830
43 −=
−==
r
BAaα s
-2.
The magnitudes of r
BAar
and 3α are obtained negative meaning
that their senses are incorrect and have to be changed. (Fig.1.1.4).
Finally, in order to find the total acceleration of point B ,
equation (II) is used, because the components n
Bar
and t
Bar
are
perpendicular to each other. Тhen:
95,84062,785300)()( 2222 =+=+= вр
B
ц
BB aaa m/s2.
The sense and direction of Bar
are determined using
parallelogram rule (Fig.1.1.5).
Fig. 1.1.5
7
COURSE WORK 2: KINEMATICS OF THE RELATIVE MOTION OF A POINT
Problem 2.1
A pipe is pivoted at one end to point O and starts to rotate from horizontal position according to
equation 3
)(2t
t =ϕ [ −ϕ rad, −t s] (Fig.2.1.1). Meanwhile, point M is moving relative to the pipe in accordance
with equation tttOMtS 5,0)()( 2 +== [ −S m, −t s]. Find the absolute velocity Vr
and the absolute acceleration
ar
of the point M at instant 21 =t s.
Fig. 2.1.1
Solution:
Rotation of the pipe in accordance with equation 3
)(2t
t =ϕ is the motion that transports point M . The
rectilinear motion of the point according to equation 2
)()( 2 tttOMtS +== is the relative motion.
1. Position of the point at instant 21 =t s
First step of the solution is determination of the point M ’s position at instant 21 =t s. This is performed
by substitution of 21 == tt s into the equations of motions. The result is:
3
4
3)(
2
11 ==
ttϕ rad 04,76= ;
52.5,025,0)()( 2
1
2
111 =+=+== tttOMtS m.
2. Absolute velocity of the point at instant 21 =t s
The absolute velocity of point M is equal to the sum of relative and transport velocity of the point:
er VVVrrr
+= .
The relative velocity is the first order derivative with respect to the time of the relative motion equation:
5,02)()( +== ttStVr& ⇒ 5,45,02.2)( 1 =+=tVr m/s.
rVr
is obtained positive meaning that its sense coincides with the sense of the relative motion (Fig.2.1.2).
The pipe performs rotation. Then, the transport velocity of the point is obtained using the expression:
OMVe .ω= ,
where ω is the angular velocity of the pipe, and OM is the distance from the center of rotation to the point.
The angular velocity is the first order derivative with respect to the time of the equation of rotation:
8
3
2)()(
ttt == ϕω & ⇒ 333,1
3
2.2
3
2)( 1
1 ===t
tω s-1
.
It has a positive magnitude and its sense follows the sense of )(tϕ (Fig.2.1.2).
Finally, the magnitude of transport velocity of M is:
665,65.333,1 ==eV m/s,
The eVr
’s direction is perpendicular toOM (trajectory 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.2.1.2).
Fig. 2.1.2
Further, since the relative and transport velocity of the point are perpendicular to each other (Fig.2.1.2),
the absolute velocity is obtained directly by the following expression:
04,8665,65,4 2222 =+=+= er VVV m/s.
The sense and direction of the absolute velocity are found by the parallelogram law (Fig.2.1.2).
3. Absolute acceleration of the point at instant 21 =t s
The absolute acceleration of the point M is equal to the sum of the relative, transport and Coriolis
accelerations:
Cer aaaarrrr
++= .
The relative motion is rectilinear. Then, the relative acceleration is the second order derivative with
respect to the time of the relative motion equation:
2)()( == tStar&& ⇒ 2)( 1 =tar m/s
2.
It is obtained positive, i.e. its sense coincides with the sense of the point relative motion (Fig.2.1.3).
Since the transport motion of the point is rotation, the transport acceleration has two components –
normal and tangential:
t
e
n
ee aaarrr
+= .
The normal acceleration is:
884,85.333,1. 22 === OMa n
e ω m/s2,
and it points to the center of rotation, here, point O (Fig.2.1.3).
The tangential component is carried out by the expression:
9
OMa t
e .α= ,
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:
3
2)()( == tt ϕα && ⇒ 667,0
3
2)( 1 ==tα s
-2,
It has positive magnitude, i.e. its sense follows the sense of )(tϕ (Fig.2.1.3). Then:
335,35.667,0 ==t
ea m/s2,
which is perpendicular to OM , and of sense determined by the sense of α .
The magnitude of the transport acceleration is:
489,9335,3884,8)()( 2222 =+=+= t
e
n
ee aaa m/s2.
Fig. 2.1.4
Fig. 2.1.3
The Coriolis acceleration is caused by the interaction between the angular velocity of the pipe and the
relative velocity of the point. It has to be obtained by the following formula:
rC Varrr
×= ω2 .
The magnitude of the Coriolis acceleration is:
);sin(..2 rrC VVarr
ωω= .
Here, the angle between the directions of angular and relative velocity is 090 , because ωr
is
perpendicular to the plane xOy (its direction is determined by 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 rVr
lies in the
plane xOy .
Thus:
121.5,4.333,1.290sin.5,4.33,1.2 0 ===Ca m/s2.
10
According to the cross product, vector Car
has to be perpendicular to the plane formed by ωr
and rVr
,
and, besides, the three vectors have to form the right-handed system. Therefore, the direction and the sense of
the Coriolis acceleration coincides with the direction of the transport tangential acceleration of the point
(Fig.2.1.3).
Finally, the magnitude of the absolute acceleration of the point is obtained by projections of its
components onto axes 1x and 1y , which are perpendicular to each other. These axes are chosen because all of
the accelerations can be projected with their total magnitudes onto them.
2
1
2
1 yx aaa += ,
884,6884,821 −=−=−= n
erx aaa m/s2,
335,1512335,31 =+=+= c
t
ey aaa m/s2,
81,16335,15)884,6( 22 =+−=a m/s2.
The sense and direction of ar
are determined by the parallelogram rule (Fig. 2.1.4).
--------------------------------------------------------------------------------------------------------------------------------------
Problem 2.2
A semicircle-shaped body is attached to z-axis and rotates according to equation 23)( ttt −=ϕ [ −ϕ rad,
−t s] while a point M moves on its periphery according to equation 2.)()( ttAMtS π== [ −S m, −t s]. Find
the absolute velocity Vr
and the absolute acceleration ar
of the point at instant 11 =t s, if at the same instant the
body occupies the position shown in Fig.2.2.1.
Fig. 2.2.1
Solution:
The motion that transports the point is rotation of the body in accordance with equation 23)( ttt −=ϕ .
The relative motion is the motion of the point with respect to the body in accordance with equation 2.)()( ttAMtS π== . This motion is curvilinear, because the trajectory of the point is circle of radius R.
1. Position of the point at instant 11 =t s
The task is to find the absolute velocity and acceleration ar
of point M at instant 11 =t s. However, the
position of the semicircle-shaped body at the same instant has been indicated by the statement. Therefore, only
the position of point M relative to the body has to be found. Then, substituting 11 == tt s into the equation of
the relative motion, it is obtained:
ππ === 21.)1( sSAM [m].
Here, the more convenient way to represent the point position is using the central angle concept, i.e.:
3
π==∠
R
AMACM [rad] 0
0
60180
.3
==π
π.
2. Absolute velocity of the point at instant 11 =t s
The absolute velocity of point M is equal to the sum of relative and transport velocity:
er VVVrrr
+=
11
The relative velocity is the first order derivative with respect to the time of the relative motion equation:
ttStVr .2)()( π== & ⇒ 28,61.2)( 1 == πtVr m/s.
rVr
lies on the circle’s tangent at point M and its sense coincides with the sense of the relative motion
(Fig.2.2.2).
Due to the rotation of the body, the transport velocity is obtain by the expression:
MNVe .ω= ,
where ω is the angular velocity of the semicircle, and MN is the shortest distance between the point and the
axis of rotation.
The angular velocity is the first order derivative with respect to the time of the equation of rotation:
ttt 23)()( −== ϕω & ⇒ 11.23)1( =−=sω s-1
.
It has a positive magnitude meaning that the sense coincides with the sense of )(tϕ (Fig.2.2.2).
Further, to find MN , the right-angled triangle MCN is considered:
CM
MN=060sin ⇒ 6,2866,0.360sin.60sin. 00 ==== RCMMN m.
Finally:
6,26,2.1 ==eV m/s.
The eVr
’s direction is perpendicular to MN , parallel to x-axis (the trajectory due to the rotation is circle
and the velocity’s direction is perpendicular to the radius) and the sense follows the sense of ω (Fig.2.2.2).
Fig. 2.2.2
Here, relative and transport velocity of the point have mutually perpendicular directions. Then, the
magnitude of the velocity of the point is:
8,66,228,6 2222 =+=+= er VVV m/s.
The sense and direction of the absolute velocity are determined by the parallelogram rule (Fig.2.2.2).
3. Absolute acceleration of the point at instant 11 =t s
The absolute acceleration of the point M is equal to the sum of relative, transport and Coriolis
components:
Cera aaaarrrr
++= .
Relative motion of the point is curvilinear and the relative acceleration has normal and tangential
components:
t
r
n
rr aaarrr
+= .
When the point trajectory is a curve, the normal acceleration points to the center of such curve, and its
magnitude is:
15,133
28,6 22
===R
Va rn
r m/s2.
12
The tangential component of relative acceleration is the second order derivative with respect to the time
of the relative motion equation:
π2)()( == tSta t
r&& ⇒ 28,6)1( =sa t
r m/s2.
t
rar
lies on the tangent to the trajectory of point M and by virtue of positive magnitude its sense
coincides with the sense of the relative motion of the point (Fig.2.2.3).
The magnitude of the relative acceleration of M is:
57,1428,615,13)()( 2222 =+=+= t
r
n
rr aaa m/s2.
Fig. 2.2.3
Body performs rotation and the transport acceleration is the sum of normal and tangential accelerations:
t
e
n
ee aaarrr
+= .
The normal acceleration is given by the expression:
6,26,2.1. 22 === MNa n
e ω m/s2,
and it always directs to the axis of rotation (Fig.2.2.3).
The tangential component is:
MNa t
e .α= ,
where α is the angular acceleration of the body and it is the second order derivative with respect to the time of
the equation of rotation:
2)()( −== tt ϕα && s-2
⇒ 1(α s 2) −= s-2
.
It is carried out with negative sign, i.e. its sense is opposite to the sense of )(tϕ (Fig.2.2.3).
2,56,2.2 ==t
ea m/s2.
The sense of t
ear
coincides to this one of α and its direction is perpendicular to MN (Fig.2.2.3).
The magnitude of the transport acceleration of M is:
81,52,56,2)()( 2222 =+=+= вр
e
ц
ee aaa m/s2.
The Coriolis acceleration is:
rC Varrr
×= ω2 .
The magnitude of the Coriolis acceleration is given by the
expression:
);sin(..2 rrC VVarr
ωω= ,
where the angle between the directions of angular and relative
velocity is 0150 (Fig.2.2.4).
Fig. 2.2.4
13
Finally, ir is obtained:
28,6150sin.28,6.1.2 0 ==Ca m/s2.
Vector Car
has to be perpendicular
to the plane formed by ωr
and rVr
, and
the three vectors must form the right-
handed system. Therefore, the Coriolis
acceleration has direction and sense as
shown in Fig.2.2.5. Fig. 2.2.5
To obtain the magnitude of the absolute acceleration of the point, the components have to be projected
onto three mutually perpendicular axes. Here, these are the axes x , y and z . Then:
222
zyx aaaa ++= .
08,128,62,5 =+−=+−= C
t
ex aaa m/s2;
n
e
t
r
n
ry aaaa +−= 00 60cos60sin ,
85,106,25,0.28,6866,0.15,13 =+−=ya m/s2;
00 60sin60cos t
r
n
rz aaa −−= ,
01,12866,0.28,65,0.15,13 −=−−=za m/s2;
22,16)01,12(85,1008,1 222 =−++=a m/s2
The sense and direction of ar
are determined by
the parallelipiped rule (Fig. 2.2.6).
Fig. 2.2.6
14
COURSE WORK 3: REDUCTION OF A SPATIAL SET OF FORCES
Problem 3.1
A spatial set of forces is applied to homogeneous rectangular prism, as shown in Fig.3.1.1.
1. Reduce the set with respect to point O and determine the reduction case to which the set can be brought;
2. Draw on the sketch of suitable scale the force resultant, the moment resultant and the angle between them;
3. Calculate the moment of the force 1Fr
about an axis formed by points E and H of direction from E to H .
Fig. 3.1.1
Data:
3=OA m
5=OB m
4=OC m
561 =F kN
322 =F kN
283 =F kN
251 =M kNm
622 =M kNm
Solution
Reduction of a set of forces about a point requires the force resultant and moment resultant about the
point to be obtained.
1. Resultant force Rr
The force resultant is going to be obtained adding its projections onto three mutually perpendicular axes,
such as x , y , z :
kRjRiRR zyx
rrrr++= ,
where:
xxxx FFFR 321 ++= ,
yyyy FFFR 321 ++= ,
zzzz FFFR 321 ++= .
Here, xF1 ,.…, zF3 are the projections of the applied forces onto axes x , y , z , respectively. 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 axes.
• Resolution of 1Fr
Force 1Fr
is in general position with respect to the axes x , y , and z . Therefore, to obtain its projections
along the axes, the fact that 1Fr
is collinear to vector DC has to be used. Thus:
11 FF DCx λ= ; 11 FF DCy µ= ; 11 FF DCz ν= ,
where DCλ , DCµ , DCν are the direction cosines of DC . They are determined by the expressions:
15
DC
xx DC
DC
−=λ ;
DC
yy DC
DC
−=µ ;
DC
zz DC
DC
−=ν ,
where 222 )()()( DCDCDC zzyyxxDC −+−+−= is the magnitude of DC :
071,7)04()50()30( 222 =−+−+−=DC m.
Further:
4243,0071,7
30−=
−=DCλ ; 7071,0
071,7
50−=
−=DCµ ; 5657,0
071,7
04=
−=DCν .
- Check of the direction cosines:
1222 =++ DCDCDC νµλ ⇒ 15657,0)7071,0()4243,0( 222 =+−+− ⇒ 11 = !
Finally:
76,2356.4243,01 −=−=xF kN; 6,3956.7071,01 −=−=yF kN; 68,3156.5657,01 ==zF kN.
Negative signs show that xF1 and yF1 have senses opposite to the senses of the axes x and y (Fig.3.1.2).
Fig. 3.1.2
• Resolution of 2Fr
Projections of force 2Fr
onto the axes are found using the direction cosines of vector ED collinear to the
force 2Fr
, i.e.
ED
xx EDED
−=λ ;
ED
yy EDED
−=µ ;
ED
zz EDED
−=ν .
The magnitude of ED is:
5)40()55()03()()()( 222222 =−+−+−=−+−+−= EDEDED zzyyxxED m,
and the direction cosines are:
6,05
03=
−=EDλ ; 0
5
55=
−=EDµ ; 8,0
5
40−=
−=EDν .
- Check of the direction cosines:
1222 =++ EDEDED νµλ ⇒ 1)8,0(06,0 222 =−++ ⇒ 11= !
16
Projections of 2Fr
are obtained, as follows:
2,1932.6,022 === FF EDx λ kN; 022 == FF EDy µ ; 6,2532.8,022 −=−== FF EDz ν kN.
• Resolution of 3Fr
Force 3Fr
is parallel to y -axis, i.e. it has projection onto y -axis only (Fig.3.1.2):
03 =xF ; 2833 == FF y kN; 03 =zF .
Finally, projections of the force resultant onto axes are carried out as:
xxxx FFFR 321 ++−= 56,402,1976,23 −=++−= kN;
yyyy FFFR 321 ++−= 6,112806,39 −=++−= kN;
zzzz FFFR 321 +−= 08,606,2568,31 =+−= kN.
The magnitude of the force resultant is obtained as:
87,1308,6)6,11()56,4( 222222 =+−+−=++= zyx RRRR kN.
2. Moment resultant about point O
Moment resultant OMr
will be obtained by the sum of projections onto axes x , y , and z , i.e.
kMjMiMM OzOyOxO
rrrr++= .
Here, the projections of the moments 1Mr
and 2Mr
onto the axes are determined first, as follows.
Moment 1Mr
is parallel to x -axis, i.e. it has projection onto x only. In contrast to 1Mr
, the moment of the
couple 2Mr
is in general position with respect to the axes x , y , z (Fig.3.1.3). The couple lies in the plane
formed by points A , B ,C , and vector 2Mr
is collinear to the normal vector ABCNr
of such plane (Fig.3.1.3).
Therefore, the projections of 2Mr
can be obtained using the direction cosines of ABCNr
.
Fig. 3.1.3
The direction cosines of ABCNr
are calculated using the cross product of vectors CA and CB , two
vectors lying in the plane ABC :
CBCBCB
CACACA
zzyyxx
zzyyxx
kji
CBCA
−−−
−−−=×
rrr
.
17
It should be mentioned that the cross product is CBCA× , and not CACB × , because during the rotation
about point C , the moment 2Mr
first crosses vector CA , and then CB (Fig.3.1.3).
kji
kjikji
CBCArrr
rrrrrr
151220
450
403
400500
400003 ++=
−
−=
−−−
−−−=× ,
which means that
20=x
ABCNr
; 12=y
ABCNr
; 15=z
ABCNr
.
Further, the magnitude of ABCNr
is calculated as:
73,27151220 222222
=++=++=z
ABCy
ABCx
ABCABC NNNNrrrr
m2.
Then, the direction cosines are:
7212,073,27
20===
ABC
xABC
ABC
N
Nr
r
λ ; 4327,073,27
12===
ABC
yABC
ABC
N
N
r
r
µ ; 5409,073,27
15===
ABC
zABC
ABC
N
Nr
r
ν .
- Check of the direction cosines:
1222 =++ ABCABCABC νµλ ⇒ 15409,04327,07212,0 222 =++ ⇒ 11= !
Fig. 3.1.4
Finally, the projections of 2Mr
onto the axes x , y , z are obtained as:
71,4462.7212,022 === MM ABCx λ kNm;
83,2662.4327,022 === MM ABCy µ kNm;
54,3362.5409,022 === MM ABCz ν kNm,
and their senses and directions are shown in Fig.3.1.4.
Then, the moment equations about axes x , y , z are written, as follows:
11,505.6,255.68,3171,4425.. 2121 =−++−=−++−= CEFADFMMM zzxxOx kNm;
59,84.2,193.68,3183,26.. 212 =+−=+−= BEFBDFMM xzyOy kNm;
18
OAFCEFBDFADFMM yxyxzOz .... 32112 +−−+= 54,213.285.2,193.6,395.76,2354,33 =+−−+= 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 axes (Fig.3.1.5).
Fig. 3.1.5
Finally, the magnitude of OMr
is calculated as:
22,5554,2159,811,50 222222 =++=++= OzOyOxO MMMM kNm.
3. Case of reduction
The case of reduction is found using the scalar product of the force resultant and moment resultant:
Rr
. OMr
βcos.. OMR= OzzOyyOxx MRMRMR ... ++= ,
where β is the angle between directions of the force resultant and moment resultant.
Here, both Rr
and OMr
have been obtained different than zero. Therefore, if their scalar product is also
different than zero, i.e. angle β is different than 900, then, the given set of forces can be further reduced to a
wrench. However, if their dot product is equal to zero, i.e. the angle β is 900, then, the set of forces can be
further reduced to a single resultant force.
By calculation, it is obtained:
Rr
. OMr
18,19754,21.08,659,8).6,11(11,50).56,4( −=+−+−= kN2m.
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 expression rearranged with respect to βcos is used:
2574,09,765
18,197
22,55.87,13
18,197
.
...cos −=
−=
−=
++=
O
OzzOyyOxx
MR
MRMRMRβ ⇒ 092,104=β .
The force resultant, the moment resultant and the angle between them are shown in Fig.3.1.6.
19
Fig. 3.1.6
4. Moment of force 1Fr
about axis EH
The moment of a force about an axis is a scalar, equal to the projection onto the axis of the moment of
the force about a point belonging to the axis. Here, point E is chosen as a point of the axis. Then:
EH
EH
FeMr
r =1
. )( 1FEDr
×
zyx
EDEDED
EHEHEH
FFF
zzyyxx
111
−−−=
νµλ
,
where EHer
is the unit vector of axis EH (Fig.3.1.7), while EHλ , EHµ , and EHν are the direction cosines of the
axis EH . They are found by the following expressions:
EH
xx EHEH
−=λ ;
EH
yy EHEH
−=µ ;
EH
zz EHEH
−=ν ,
where 222 )()()( EHEHEH zzyyxxEH −+−+−= 831,5)44()50()03( 222 =−+−+−= m.
5145,0831,5
03=
−=EHλ ; 8575,0
831,5
50−=
−=EHµ ; 0
831,5
44=
−=EHν .
- Check of the direction cosines:
1222 =++ EHEHEH νµλ ⇒ 10)8575,0(5145,0 222 =+−+ ⇒ 11= !
Finally:
20
EH
FM1
68,316,3976,23
405503
08575,05145,0
−−
−−−
−
= 5,81
68,316,3976,23
403
08575,05145,0
−=
−−
−
−
= kNm.
EH
FM1
has a negative sign, which means that the sense of EH
FM1
is opposite to the positive sense of
axis EH (Fig.3.1.7).
Fig. 3.1.7
21
COURSE WORK 4: REDUCTION OF A SET OF COPLANAR FORCES
Problem 4.1
A coplanar set of forces is applied to homogeneous figure, as shown in Fig.4.1.1.
1. Locate the position of the center of gravity, point C;
2. Apply vertical force 2Fr
of downward sense at point C and reduce the set of forces with respect to
point A ;
3. Determine the reduction case to which the set can be brought.
Fig. 4.1.1
301 =F kN
242 =F kN
363 =F kN
42=M kNm
Solution
1. Center of gravity
To obtain the coordinates of the center of gravity the following equations are used:
∑
∑
=
==n
i
i
n
i
ii
C
A
xA
x
1
1 ;
∑
∑
=
==n
i
i
n
i
ii
C
A
yA
y
1
1 ,
where ix and iy are the coordinates of the centers of gravity of the simple figures with respect to the axes,
chosen for location of the center of gravity of the figure given. Here, the simple figures are three – rectangle,
triangle, semicircle, and, first, the absolute location of their centers of gravity and their areas have to be
determined.
Fig. 4.1.2a
Figure (1) is rectangle. The coordinates
of the center of gravity are:
22
( 11 =
hC m; 5,2
2
1 =b
m) – (Fig.4.1.2a).
The area of the rectangle is:
204.51 ==A m2.
22
Fig. 4.1.2b
Figure (2) is right-angled triangle and
the coordinates of the center of gravity with
respect to the right angle are:
9,03
( 22 =
hC m; 5,1
3
2 =b
m) – (Fig.4.1.2b).
The area of the triangle is:
075,65,4.7,2.2
12 ==A m
2.
Fig. 4.1.2c
Figure (3) is semicircle. The coordinates
of the center of gravity are:
2(3 =RC m; 85,03
4=
π
Rm) – (Fig.4.1.2c).
The area of the semicircle is:
28,62.14,3.2
1..
2
1 22
3 === RA π m2.
After that, the coordinates of the centers of gravity of three figures about axes x and y have to be
found. The axes x and y are chosen so that the entire figure lies in the first quadrant (Fig.4.1.3). Thus, all of the
coordinates of the centers of gravity of the simple figures are positive.
Fig. 4.1.3
Using the coordinates determined
earlier and following Fig.4.1.3, it is
obtained:
5,21 =x m;
21 =y m;
5,35,15,45,02 =−+=x m;
4,15,09,02 =+=y m;
85,585,05,45,03 =++=x m;
23 =y m,
i.e. the locations of the centers of
gravity about x and y are:
5,2(1C m; 2 m);
5,3(2C m; 4,1 m);
85,5(3C m; 2 m).
Finally, the coordinates of the center of gravity of the given figure about axes x and y are:
24,3205,20
475,65
28,6075,620
85,5.28,65,3.075,65,2.20
321
332211
3
1
3
1 ==+−
+−=
+−
+−==
∑
∑
=
=
AAA
xAxAxA
A
xA
x
i
i
i
ii
C m;
23
18,2205,20
055,44
28,6075,620
2.28,64,1.075,62.20
321
332211
3
1
3
1 ==+−
+−=
+−
+−==
∑
∑
=
=
AAA
yAyAyA
A
yA
y
i
i
i
ii
C m.
and its position is given in Fig. 4.1.4.
Fig. 4.1.4
2. Reduction of the set of forces about point A
After the center of gravity has already been located, the force 2Fr
is applied (Fig.4.1.5) and the reduction
of the set with respect to the point A begins.
Fig. 4.1.5
2.1 Force resultant
First, the axes X and Y of origin point A are introduced (Fig.4.1.6). Then, the force resultant is:
YX RRRrrr
+= ,
XXXu FFFR 321
rrrr++= ,
YYYv FFFR 321
rrrr++= .
Here, XF1
r, ……., YF3
v are projections of the applied forces onto axes X and Y (Fig.4.1.6):
155,0.3060cos 0
11 === FF X kN; 26866,0.3060sin 0
11 === FF Y kN;
02 =XF ; 2422 == FF Y kN; 3633 == FF X kN; 03 =YF .
24
Fig. 4.1.6
Further:
21361531 =+−=+−= FFR XX kN;
50242621 −=−−=−−= FFR YY kN;
23,54)50(21 2222 =−+=+= YX RRR kN.
Finally, the angle between X-axis and resultant force is calculated. The result is:
381,221
50−=
−==
X
YR
R
Rtgα ⇒ 022,67−=Rα (Fig.4.1.7).
Fig. 4.1.7
2.2 Moment resultant
The moment resultant about point A is obtained by the following expression:
5,3.)24,35.(2.)5,02.( 3211 FFFFMM YXA −−+−−+= =−+−+= 5,3.3676,1.242.265,1.1542 26,71− kNm,
25
where the positive sense of the moment is counterclockwise (Fig.4.1.8). The result is obtained negative
meaning that the moment resultant points toward the page (Fig.4.1.9).
Fig. 4.1.8
2.3 Case of reduction The set of forces is coplanar, which means that it can be further reduced to the single resultant force
(Fig. 4.1.9). The equation of the direction of single resultant force is:
AXY MYRXR =− ;
26,7121.)50.( −=−− YX ;
425,150
26,71=
−
−==
Y
AX
R
Mn m; 393,3
21
26,71=
−
−=
−=
X
AY
R
Mn m;
314,123,54
26,71=
−==
R
Mh A
R m.
Fig. 4.1.9
26
COURSE WORK 5: EQUILIBRIUM OF A BODY ACTED UPON BY A SPATIAL SET OF FORCES
Problem 5.1
Determine the reactions of the rectangular prism supported and loaded as shown in Fig.5.1.1. Check the
result obtained using the relevant equilibrium equations. The weight per unit volume is 3=γ kN/m3.
Fig. 5.1.1
Solution:
1. Determination of the weight of the body, the resultant force of the distributed load, the
projections of the moment M and the force F onto the axes of the appropriate coordinate system
First, the Cartesian coordinate system xyz of origin point A is introduced. Point A has been chosen for
origin, because it is a spherical support containing three unknown reactions. Thus, if the moment equations
about axes zyx ,, are written, then, the three reactions at point A are eliminated.
• The weight of the rectangular prism is:
γ.VG = ,
where 302.3.5.. === DLBDABV m3 is the volume. Then:
903.30 ==G kN.
G is directed down vertical force applied at the body’s center of gravity (Fig.5.1.2).
• Distributed load q is applied on the side BDLK of the rectangular prism. Then, the resultant force is:
7525.2.32
1..
2
1=== qBKBDRq kN.
The line of action and sense of qR are determined by the line of action and sense of the distributed load.
The point of application of qR is the projection of the distributed load’s center of gravity on the prism’s
side BDLK (Fig.5.1.2).
• Moment M is in general position with respect to the axes zyx ,, . Then, in order to find its
projections onto these axes, the direction cosines of vector BN , collinear to Mr
, are used (Fig.5.1.2). The
coordinates of the points formed BN are )2;0;3(−N and )0;5;0(B . Then, the magnitude of BN is:
27
164,6)02()50()03()()()( 222222 =−+−+−−=−+−+−= BNBNBN zzyyxxBN m.
Further,
4867,0164,6
3−=
−=
−=
BN
xx BN
BNλ ; 8111,0164,6
5−=
−=
−=
BN
yy BN
BNµ ; 3244,0164,6
2==
−=
BN
zz BN
BNν .
- Check of the direction cosines:
1222 =++ BNBNBN νµλ ; 13244,0)8111,0()4867,0( 222 =+−+− ; 19999,0 ≈ !
Finally, the projections of the moment are found as:
04,3572.4867,0. −=−== MM BNx λ kNm; 4,5872.8111,0. −=−== MM BNy µ kNm;
36,2372.3244,0. === MM BNz ν kNm.
• Force F lies on the side KLNH of the body. Then, it has the projections onto axes x and y :
βcos.FFx = ; βsin.FFy = (Fig.5.1.2) .
Considering right-angled triangle LNH , angle β is obtained as:
3
5==
NH
LNtgβ ⇒ 004,59=β .
Then,
9,3004,59cos.60 0 ==xF kN; 45,5104,59sin.60 0 ==yF kN (Fig.5.1.2).
Fig. 5.1.2
2. Determination of support reactions
First, free body diagram is drawn, i.e. all supports are detached from the body and the support reactions
are introduced instead. These are:
- Point A is a spherical support constraining the displacements parallel to the axes of the coordinate
system. Then, three support reactions are available, namely xA , yA , zA where their senses are supposed
(Fig.5.1.3);
- There is a link at point B which means that only one support reaction of line of action coinciding
with the direction of the link has to be introduced (Fig.5.1.3);
- Point L is a cylindrical support and contains two support reactions, xL and zL , of lines of action
parallel to the axes x and z (Fig.5.1.3).
Besides, all external loads are applied.
28
The spatial set of forces is in equilibrium when the main vector and the main moment are both equal to
zero, i.e. 0=Rr
, 0=OMr
. These two vector equations can be transformed into nine scalar equations, three force
and six moment equations. Six 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 only one unknown reaction takes part has to be written.
Here, in order to find the independent equations, the free-body diagram is analyzed carefully (Fig.5.1.3).
Fig. 5.1.3
It is obvious that only reaction yA is parallel to y-axis. Then, the first independent equation is:
1) 0=∑ iyF ; 0=−+ qyy RFA ; 07545,51 =−+yA ; 55,23=yA kN.
yA is positive which means that the sense supposed is correct (Fig.5.1.3).
The second independent equation is:
2) 0=∑ zM ; 05.1. =−+ xqz LRM ; 05.1.7536,23 =−+ xL ; 67,195
36,231.75=
+=xL kN.
The third equation containing only one unknown is:
3) 01=∑ zM ; 01.5.5. =++−− qzxx RMFA ; 01.7536,235.9,305. =++−− xA ;
23,115
5.9,3036,231.75−=
−+=xA kN.
The sign “-“ for xA means that the sense supposed is not correct and has to be changed (Fig.1.3).
The solution continues with the fourth independent equation:
4) 01=∑ xM ; 01.5,2.2.5. =+−+−− qxyz RMGFA ; 01.7504,355,2.902.45,515. =+−+−− zA ;
41,325
2.45,5104,351.755,2.90=
−−+=zA kN.
Further, analyzing Fig.5.1.3 the conclusion that there is no other independent equation is made. Then,
the determination of the last support reactions continues with the next equations:
29
5) 0=∑ yM ; 02.3.2.5,1. =++−−− xzxy LLFMG ; 02.67,193.2.9,304,585,1.90 =++−−− zL ;
95,713
2.67,192.9,304,585,1.90=
−++=zL kN;
6) 0=∑ xM ; 05.1.5,2.2.5. =+−+−− zxqy LMRGFB ;
05.95,7104,351.755,2.902.45,515. =+−+−−B ;
36,145
5.95,7104,351.755,2.902.45,51−=
−+−+=B kN.
3. 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 any equation chosen for check has to contain as many as possible support reactions.
Fig. 5.1.4
The force equations for the check are (Fig.5.1.4):
7) 0=∑ ixF ; 0=+− xxx DFA ; 067,199,3023,11 =+− ; 09,309,30 =− !
8) 0=∑ izF ; 0=+−− zz DBGA ; 095,7136,149041,32 =+−− ; 036,10436,104 =− !
The moment equation is (Fig.5.1.4):
9) 01=∑ yM ; 05,1.5,1.1.1.5,1.1. =+++−−−− BLLMFAA zxyxzx ;
05,1.36,145,1.95,711.67,194,581.9,305,1.41,321.23,11 =+++−−−− ;
001,0145,149135,149 ≈−=− !
All equations are obtained equal to zero meaning that the support reactions are determined correctly!
--------------------------------------------------------------------------------------------------------------------------------------
Problem 5.2
Determine the support reactions of the rectangular plate supported and loaded as shown in Fig.5.2.1.
Check the result obtained using the relevant equilibrium equations. Assume that the thickness of the plate is
approximately zero. Neglect the weight of the plate.
30
Fig. 5.2.1
ABF //1 ;
BDF //2 ;
EHq // .
Solution:
1. Determination of the resultant force of the distributed load, the projections of the moment M
and the support reaction D onto the axes of the appropriate coordinate system
Solution begins with the choice of the coordinate system. Here, the coordinate system is chosen so that
the axes pass through as many as possible supports (Fig.5.2.2).
Fig. 5.2.2
• The distributed load q acts on the half plate. To determine the resultant force qR the formula for the
volume of the prism is applied, i.e.
31
545.8,4.5,42
1..
2
1=== qBDABRq kN.
qR is a force parallel to z-axis of sense coinciding with the sense of q . The point of application of qR is
the projection onto the plate of the distributed load’s center of gravity (Fig.5.2.2).
• Moment M has projections onto axes x and y , and they are obtained using angle δ as:
δcos.MM x = ; δsin.MM y = ,
where δ is :
9375,08,4
5,4===
AE
DEtgδ ⇒ 015,43=δ (Fig.5.2.2).
Then,
5,1715,43cos.24 0 ==xM kNm; 4,1615,43sin.24 0 ==yM kNm (Fig.5.2.2).
• The link at point D is inclined with respect to the axes y and z and the reaction in the link is in
general position with respect to the same axes (Fig.5.2.2). Therefore, it is more convenient to work with the
projections of the force D onto the axes y and z. These projections are determined as follows:
βsin.DDy = ; βcos.DDz = ,
where 8,15,2
5,4===
EH
EDtgβ ⇒ 095,60=β .
Finally:
DDDy 8741,095,60sin. == ; DDDz 4856,095,60cos. == (Fig.5.2.2).
2. 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 A and reactions yA and zA of lines of action coinciding with directions
of the links are introduced (Fig.5.2.3);
- There are two links at point B and two reactions are introduced, as well (Fig.2.3);
- The link at point D has already been considered and the projections of the reaction, namely yD and
zD have been introduced (Fig.5.2.3);
- There is a link at point E and one reaction is introduced (Fig.5.2.3).
It should be noted that the senses of all reactions are supposed.
Further, the solution continues with analysis of Fig.5.2.3. Similar to the previous example, the
independent equations are four. These are:
1) 0=∑ ixF ; 02 =+− FBx ; 036 =+xB ; 36=xB kN.
2) 01=∑ zM ; 04,2.8,4. 1 =+ FAy ; 04,2.308,4. =+yA ; 15
8,4
4,2.30−=
−=yA kN.
3) 01=∑ xM ; 05,1.5,4. =+−− qxz RMA ; 05,1.545,175,4. =+−− zA ;
11,145,4
5,1.545,17=
+−=zA kN.
4) 01=∑ yM ; 08,4.2,3. =+− zqy DRM ; 08,4.2,3.544,16 =+− zD ;
58,325,4
2,3.544,16=
+−=zD kN ⇒ 09,67
4856,0
58,32==D kN ⇒ 64,5809,67.8741,0 ==yD kN.
The last two equations are:
5) 0=∑ xM ; 05,4.3.5,4. =+−+− zqzx DRBM ; 05,4.58,323.545,4.5,17 =+−+− zB ;
31,75,4
5,4.58,323.545,17=
−+=zB kN;
6) 02=∑ zM ; 08,4.4,2.8,4. 1 =+−− yDFE ; 08,4.64,584,2.308,4. =+−− E ;
32
kN.
Fig. 5.2.3
3. Check for the reactions obtained
Two force and one moment equations is used for check (Fig.5.2.4):
7) 0=∑ iyF ; 01 =−++− yy DEFA ; 064,5864,433015 =−++− ; 064,7364,73 =− !
8) 0=∑ izF ; 0=+−+ zqzz DRBA ; 058,325431,711,14 =+−+ ; 05454 =− !
9) 02=∑ yM ; 06,1.2,3.2,3. =+−− zzzy DBAM ;
06,1.58,322,3.31,72,3.11,144,16 =+−− ;
001,054,6853,68 ≈−=− !
Fig. 5.2.4
33
COURSE WORK 6: EQUILIBRIUM OF A BODY SUBJECTED TO A SET OF COPLANAR FORCES
Problem 6.1
Apply independent equilibrium equations to determine the support reactions of the construction shown
in Fig.6.1.1. Check the result obtained.
Fig. 6.1.1
Solution:
First step of solution is introduction of the coordinate system of axes x and y (Fig.6.1.2). It should be
noted that they play symbolic role here, i.e. they only give horizontal and vertical directions and positive
senses. Besides, the origin of the coordinate system is chosen arbitrarily.
1. Determination of the resultant force qR of the distributed load and the projections of the force
1F onto axes x and y
• Load q is uniformly distributed and its resultant force is:
635,4.145,4. === qRq kN,
which direction and sense coincides with the direction and sense of distributed load. The qR ’s point of
application is the midpoint of the distance over which the distributed load acts (Fig.6.1.2).
Fig. 6.1.2
34
• Projections of the force 1F onto axes x and y are equal, because the angle between the force’s line of
action and the axes is 450, i.e.
14,14707,0.2045cos. 0
111 ==== FFF yx kN (Fig.6.1.2).
2. Determination of the support reactions
First, the free-body diagram is drawn where the constraints are substituted for their support reactions.
– A is a point where a link inclined at an angle of 600 with respect to the horizontal is positioned.
Then, a reaction of line of action following direction of the link is introduced. The sense is supposed arbitrarily
(Fig.6.1.3);
– Point B is constrained by roller and one vertical support reaction B is introduced instead (Fig.6.1.3);
– There is a horizontal link at point C – the horizontal reaction C is introduced (Fig.6.1.3).
2.1 Projections of reaction A onto axes x and y
The support reaction at point A is in general position with respect to the axes. Therefore, its projections
parallel to the axes have to be found. The result is:
AAAx 5,060cos. 0 == ;
AAAy 866,060sin. 0 == .
Their point of application is A , and the senses follow the sense of support reaction A (Fig.6.1.3).
2.2 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 analysis of Fig.6.1.3 reveals that the two
force equations ∑ = 0ixF and ∑ = 0iyF are not independent, because each one of them contains two
unknowns. It means that the moment equations have to be used. Moreover, 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 A and B , A and C , B and C , points D , E and H ,
respectively, are located (Fig.6.1.3).
• Position of points D and E
- Right-angled triangle ABD∆ is considered (Fig.6.1.3):
AB
BDtg =060 ⇒ 06,6732,1.5,360. 0 === tgABBD m;
- Right-angled triangle EHD∆ is considered (Fig.6.1.3):
EH
DHtg =060 ⇒ 23,5
732,1
06,9
600===
tg
DHEH m.
2.3 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 (Fig.6.1.3):
1) 0=∑ HM ; 01.25,1.3.9,4.3.5,3. 211 =+++++−− MFRFFAA qxyxy ;
0481.3225,1.633.14,149,4.14,143).5,0(5,3).866,0( =+++++−− AA ;
456,270531,4 =A ;
69,59=A kN.
2) 0=∑ EM ; 01.)25,123,5.(3.)9,423,5.(23,5. 211 =++−−+−− MFRFFB qxy ;
0481.3298,3.633.14,1433,0.14,1423,5. =++−+−B ;
99,13223,5 =B ;
43,25=B kN.
3) 0=∑ DM ; 006,6.9,4.25,1.06,8.06,9. 112 =−++−+ xyq FFRFMC ;
006,6.14,149,4.14,1425,1.6306,8.324806,9. =−++−+C ;
57,14706,9 =C ;
29,16=C kN.
All of the reactions are obtained positive meaning that the senses are correct (Fig.6.1.3).
35
Fig. 6.1.3
3. Check of the results obtained
The two force equations are used for check. First, the magnitudes of xA and yA are found:
85,2969,59.5,05,0 === AAx kN;
69,5169,59.866,0866,0 === AAy kN (Fig.6.1.4).
Then:
4) 0=∑ ixF ; 021 =−+− CFAF xx ;
029,163285,2914,14 =−+− ;
014,4614,46 =− !
5) 0=∑ iyF ; 01 =+−+− BRAF qyy ;
043,256369,5114,14 =+−+− ;
002,012,7714,77 ≈−=+− !
The two equations equal zero which
means that the reactions obtained are correct!
Fig. 6.1.4
--------------------------------------------------------------------------------------------------------------------------------------
Problem 6.2
Apply independent equilibrium equations to determine the support reactions of the construction shown
in Fig.6.2.1. Check the result obtained.
36
Fig. 6.2.1
Solution:
Solution begins with introduction of the coordinate system of axes x and y (Fig.6.2.2).
1. Determination of qR , xF2 и yF2
• Load q is linearly distributed and its resultant force is:
4,326,3.182
16,3.
2
1=== qRq kN;
qR is a force of direction and sense determined by q . The point of application is the projection of the
distributed loading’s center of gravity on the construction (Fig.6.2.2).
• Projections of 2F onto axes x and y are:
18,31866,0.3630cos. 0
22 === FF x kN;
185,0.3630sin. 0
22 === FF y kN (Fig.6.2.2).
2. Determination of the support reactions
Fig. 6.2.2
37
The following support reactions are introduced:
– There is a pinned support at point A – a horizontal xA and a vertical reaction yA of supposed senses
are introduced (Fig.6.2.2);
– The roller is situated at point B – a vertical support reaction B is introduced (Fig.6.2.2).
Analyzing Fig.6.2.2, it is concluded that only one unknown reaction is parallel to x-axis. Then, the first
independent equation is:
1) 0=∑ ixF ; 02 =++ xqx FRA ;
018,314,32 =++xA ;
58,63−=xA kN.
The negative sign means that the sense chosen for xA is not correct and has to be changed (Fig.6.2.2).
The second independent equilibrium equation is the moment one about point A , because A is the point
of intersection of the xA and yA ’s lines of action (Fig.6.2.2):
2) 0=∑ AM ; 05,5.6,3.4.5,1.4,2. 221 =−−+−−− yxq FFBFMR ;
05.5.186,3.18,314.5,1.42244,2.4,32 =−−+−−− B ;
3764 =B ;
94=B kN.
The positive sense of moment is the counterclockwise one.
The third independent equation is the moment equation about D , the point of intersection xA and B ’s
lines of action (Fig.6.2.2):
3) 0=∑ DM ; 05,1.6,3.5,2.4,2.4. 221 =−−+−−− yxqy FFFMRA ;
05,1.186,3.18,315,2.42244,2.4,324. =−−+−−− yA ;
1364 −=yA ;
34−=yA kN.
3. 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. 6.2.3):
4) 0=∑ EM ; 04.2,1.5,2.4,2.5,1. 22 =−−+−− yxxy FFBMAA ;
04.182,1.18,315,2.94244,2.58,635,1.34 =−−+−− ;
0286286 =− !
Fig. 6.2.3
38
COURSE WORKS 7: EQUILIBRIUM OF GERBER BEAMS
Problem 7.1
Determine the support reactions and the forces in hinges of the Gerber beam shown in Fig.7.1.1. Check
the result obtained.
Fig. 7.1.1
Solution:
First, the number and the type of the beams composing the Gerber beam are determined. Here, the
beams are three, namely 1AS , 21BSS , and CDS2 . Beam CDS2 is the main beam, because it will be in
equilibrium detached from the structure. However, beams 1AS and 21BSS can not be in equilibrium without the
main beam. Therefore, they are secondary beams. Then, the beams are numbered: 1AS is beam (1), 21BSS is
beam (2), and CDS2 is beam (3). Finally, the coordinate system of axes x and y is introduced (Fig.7.1.2).
1. Determination of the distributed loads’ resultant forces
• Load 1q is applied on beam (1). It is linearly distributed and its resultant force is:
363.242
1.
2
1111 === ASqR kN.
1R has line of action and sense following these ones of distributed load. The point of application of 1R
is the position of the center of gravity of the triangle representing the distributed load (Fig.7.1.2).
• The load 2q is uniformly distributed over two beams. Then, two resultant forces have to be obtained.
- Resultant force for the portion of 2q applied on beam (2) is:
562.28. 222 === BSqR kN.
2R is a vertical force of downward sense with point of application at the midpoint of distance 2BS (Fig.7.1.2).
- Resultant force for the portion of 2q acting on beam (3) is:
1124.28. 223 === DSqR kN.
3R is a vertical force of downward sense with point of application at the midpoint of distance 2DS (Fig.7.1.2).
2. 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 type of the constraints, and, as usual, their senses are arbitrary chosen (Fig.7.1.2). The
forces in hinges are introduced at points 1S and 2S , and they act on the two beams connected by the hinge. By
definition, the forces in hinges represent the influence of one beam to another and they ensure the equilibrium
of the beam detached from entire structure. The forces in hinges have vertical and horizontal lines of action and
senses supposed randomly, but opposite for the two beams connected by the hinge (Fig.7.1.2). Finally, it should
be noted that the solution starts from the beam of fewest number unknowns. Here, this is beam (1).
2.1 Beam (1)
It is convenient the three independent equations to be written. These are:
1) ∑ = 0ixF ; 01 =− xSF ; 038 1 =− xS ; 381 =xS kN;
2) ∑ = 0AM ; 03.2. 11 =+− ySR ; 03.2.36 1 =+− yS ; 241 =yS kN.
The positive sense of the moment is chosen to be the counterclockwise one (Fig.7.1.2).
39
3) ∑ = 01SM ; 03.1.1 =+− AR ; 03.2.36 =+− A ; 12=A kN.
The three unknowns are obtained positive meaning that the senses chosen are correct (Fig.7.1.2).
- Check of the result obtained: ∑ = 0iyF ; 011 =+− ySRA ; 0243612 =+− ; 03636 =− !
Fig. 7.1.2
2.2 Beam (2)
1) ∑ = 0ixF ; 021 =− xx SS ; 038 2 =− xS ; 382 =xS kN;
2) ∑ = 0BM ; 02.1.2. 221 =−− yy SRS ; 02.1.562.24 2 =−− yS ; 42 −=yS kN;
yS2 is obtained negative, i.e. its sense is changed for the two beams to which yS2 is applied (Fig.7.1.2).
3) ∑ = 02SM ; 01.2.4. 21 =+− RBS y ; 01.562.4.24 =+− B ; 76=B kN.
- Check: ∑ = 0iyF ; 0221 =+−+− yy SRBS ; 04567624 =+−+− ; 08080 =− !
2.3 Beam (3)
1) ∑ = 0ixF ; 02 =− xx CS ; 038 =− xC ; 38=xC kN;
2) ∑ = 0CM ; 05,2.5,0.5,1. 32 =++− MDRS y ; 0425,2.5,0.1125,1.4 =++− D ; 2,3=D kN;
3) ∑ = 0DM ; 02.5,2.4. 32 =++− MRCS yy ; 0422.1125,2.4.4 =++− yC ; 8,112=yC kN.
- Check: ∑ = 0iyF ; 032 =+−+− DRCS yy ; 02,31128,1124 =+−+− ; 0116116 =− !
3. 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:
1) ∑ = 0ixF ; 01 =− xCF ; 03838 =− !
2) ∑ = 0iyF ; =+−+−+− DRCRBRA y 321 02,31128,11256763612 =+−+−+− ; 0204204 =− !
Fig. 7.1.3
40
COURSE WORK 8: EQUILIBRIUM OF THREE-HINGED FRAMES
Problem 8.1
Determine the support reactions and the forces in hinge of the three-hinged frame shown in Fig.8.1.1.
Check the result obtained.
Fig. 8.1.1
Solution:
1. Determination of the resultant force of distributed load
The load q is uniformly distributed and its resultant force is:
425,3.12 ==qR kN.
qR is a horizontal force of sense to the left. The point of application of qR is the midpoint of the
distance over which the load is distributed (Fig.8.1.2).
2. Determination of the support reactions
First, free-body diagram of entire structure is drawn. Points A and B are pinned supports and they are
substituted for their reactions having horizontal and vertical lines of action and senses chosen arbitrarily
(Fig.8.1.2a). 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 by the external forces and support reactions
acting upon it. In the case when a concentrated force or moment is applied directly at the hinge, then, the force
or moment is assumed to act only on one part of the construction. Here, it is chosen the moment to be applied to
the frame part AS (Fig.8.1.2b). After 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 horizontal and vertical
while their senses are arbitrary chosen. Besides, they are applied to the two parts of the frame and their senses
are opposite with regard to the two parts (Fig.8.1.2b).
General type 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 xA and yA
41
1) ∑ = 0..StrEnt
BM (Fig.8.1.2a);
075,1.6,5.1.1.6,5. 12 =+−++−− qxy RMFFAA ;
075,1.42426,5.251.301.6,5. =+−++−− xy AA ;
5,2016,5 =+ yx AA .
The positive sense of the moment is chosen to be the counterclockwise one.
2) ∑ = 0AS
SM (Fig.8.1.2b);
05,2.5,2.5,2. 2 =−−+− MFAA xy ;
0425,2.305,2.5,2. =−−+− xy AA ;
8,46=− yx AA .
Solving the set of equations, the result is:
5,2016,5 =+ yx AA (1)
8,46=− yx AA (2)
8,46=− yx AA ⇒ yx AA += 8,46
5,2016,58,46 =++ yy AA
7,1546,6 =yA
44,23=yA kN;
24,7044,238,46 =+=xA kN.
xA and yA have positive values, i.e. their
senses are correct.
• Determination of xB and yB
3) ∑ = 0..StrEnt
AM (Fig.8.1.2а);
06,5.1.75,0. =+++− yxq BBRM ;
06,5.1.75,0.4242 =+++− yx BB ;
5,106,5 =+ yx BB .
4) ∑ = 0BS
SM (Fig.8.1.2b);
01,3.5,3.75,1.5,2.1 =++− yxq BBRF ;
01,3.5,3.75,1.425,2.25 =++− yx BB ;
111,35,3 =+ yx BB .
Solving the set of equations, it is obtained:
5,106,5 =+ yx BB (3)
111,35,3 =+ yx BB (4)
5,106,5 =+ yx BB ⇒ yx BB 6,55,10 −=
Fig. 8.1.2a
Fig. 8.1.2b
111,3)6,55,10(5,3 =+− yy BB
111,36,1975,36 =+− yy BB
75,255,16 −=− yB
56,1=yB kN;
76,156,1.6,55,10 =−=xB kN.
• Check for the reactions obtained (Fig.8.1.3):
42
∑ = 0..StrEnt
ixF ; 02 =+−− xqx BRFA ;
076,1423024,70 =+−− ;
07272 =− !
∑ = 0..StrEnt
iyF ; 01 =+− yy BFA ;
056,12544,23 =+− ;
02525 =− .
Fig. 8.1.3
3. Determination of the forces in hinge
The forces in hinge are determined by two force equation for one of the frame parts. Then, they are
checked by two force equation for the other part (Fig.8.1.4):
∑ = 0AS
ixF ; 02 =−− xx SFA ; 03024,70 =−− xS ; 24,40=xS kN.
∑ = 0AS
iyF ; 0=− yy SA ; 044,23 =− yS ; 44,23=yS kN.
- Check:
∑ = 0BS
ixF ; 0=+− xqx BRS ; 076,14224,40 =+− ; 04242 =− !
∑ = 0BS
iyF ; 01 =+− yy BFS ; 056,12544,23 =+− ; 02525 =− !
Fig. 8.1.4
43
COURSE WORK 9: EQUILIBRIUM OF PLANE TRUSSES
Problem 9.1
A plane truss is supported and loaded as shown in Fig.9.1.1. Determine:
1. Support reactions;
2. Force in each member using the method of joints;
3. Force in the members marked using the method of sections.
Fig. 9.1.1
Solution:
1. Support reactions determination
Fig. 9.1.2
Here, the independent equations for determination of the support reactions are one force equation onto
x-axis and two moment equations about the point of intersection of xA and yA (point A ) and the point of
intersection of xA and B (pointC ), in which the positive sense of the moment is assumed to be the
counterclockwise one (Fig.9.1.2):
44
1) 0=∑ ixF ⇒ 03 =+− FAx ⇒ 042 =+− xA ⇒ 42=xA kN;
2) 0=∑ AM ⇒ 012.4.9.3. 321 =+−−− BFFF ⇒ 012.4.429.243.36 =+−−− B ⇒ 41=B kN;
3) 0=∑ CM ⇒ 04.3.9.12. 321 =−++− FFFAy ⇒ 04.423.249.3612. =−++− yA ⇒ 19=yA kN.
• Check: 0=∑ DM ⇒ 03.6.4.9. 1 =++−− BFAA xy ; 03.416.364.429.19 =++−− ; 0339339 =− !
2. Determination of the force in each member using the method of joints
2.1 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 (Fig.9.1.2).
2.2 Zero-force members
In present problem, the zero-force members are three:
- links (3) and (4) – they are connected by the unloaded joint „1” (Fig.9.1.2);
- link (9) – it ends in joint „4” where the other two members have one and the same direction (Fig.9.1.2).
2.3 Forces in the links
Solution begins from the first joint, here, joint A . It is detached from the truss and the external 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, i.e. the forces are
supposed tensile. Then, the equilibrium of the joint is examined using two force equations. Thus, the unknowns
are obtained, as follows.
- Joint „А”
0=∑ iyF ; 01913,53sin 0
2 =+S ; 75,232 −=S kN (c);
0=∑ ixF ; 013,53cos42 0
21 =−− SS ; 25,561 =S kN(t).
- Joint „3”
0=∑ iyF ; 013,53sin13,53sin 752 =−− SSS ;
25,217 −=S kN (c);
0=∑ ixF ; 013,53cos13,53cos 782 =−+ SSS ;
5,18 −=S kN (t).
- Joint „2”
0=∑ ixF ⇒ 016 =− SS ; 25,566 =S kN (t);
0=∑ iyF ⇒ 0365 =−S ; 365 =S kN (t).
- Joint „4”
0=∑ ixF ⇒ 0108 =−− SS ; 5,110 −=S kN (c);
0=∑ iyF ⇒ 09 =S .
45
- Joint „5”
0=∑ iyF ; 013,53sin13,53sin 117 =− SS ;
25,2111 =S kN (t);
0=∑ ixF ;
013,53cos13,53cos 121176 =+++− SSSS ;
75,3012 =S kN (t).
- Joint „D”
0=∑ ixF ; 013,53cos 141110 =+− SSS ;
25,1114 =S kN (t).
0=∑ iyF ; 013,53sin24 1311 =+−− SS ;
0418,0.25,2124 =+−− ; 04141 =− !
- Joint „6”
0=∑ ixF ; 013,53cos 0
1512 =+− SS ; 25,5115 =S kN(t);
0=∑ ixF ; 013,53sin 0
1513 =+ SS ; 4113 −=S kN(c).
- Joint „В” - Check
0=∑ ixF ; 04213,53cos1514 =+−− SS ;
0426,0.25,5125,11 =+−− ; 04242 =− !
0=∑ iyF ; 013,53sin41 15 =− S ;
08,0.25,5141 =− ; 04141 =− !
Notes:
1. 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, respectively. When the rod is compressed, the sense of the force
has to be changed!
2. The last joint’s equations of equilibrium are used for the check of the results obtained.
3. Method of joints allows solution of the problem to begin from the last joint and to go from the right to
the left. Then, joint A will be used for the check of the results.
3. 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 by an imaginary cut
through the three members marked. Then, the truss portion loaded by less number of loads is considered where
the external 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. Finally, the check
is made.
46
3.1 Forces in members 6, 7, and 8
1) 0=∑ iyF ; 013,53sin3619 0
7 =−− S ; 25,217 −=S kN (c);
2) 0"5" =∑M ; 04.3.366.19 8 =−+− S ; 5,18 −=S kN (c);
3) 0"3" =∑M ; 04.3.194.42 6 =+−− S ; 25,566 =S kN (t);
• Check
4) 0=∑ ixF ; 013,53cos42 0
768 =−+−− SSS ; 06,0.25,2125,565,142 =−+−− ; 025,5625,56 =− !
3.2 Forces in members 12, 13, and 14
1) 0=∑ iyF ; 04113 =+S ; 4113 −=S kN (c);
2) 0=∑ DM ; 04.3.41 12 =− S ; 75,3012 =S kN (t);
3) 0"6" =∑M ; 03.414.424.14 =+−S ; 25,1114 =S kN (t);
• Check: 0=∑ ixF ; 0421412 =+−− SS ; 04225,1175,30 =+−− ; 04242 =− !
