EXP NO – 1
NAME –UID –
Branch: Automobile Engineering
Section:
Semester:3
Date of performance: 15-09-21
Subject Name: TOM lab
Subject Code: AEP-216
Aim: To draw the diagrammatical representation (velocity diagram
and acceleration diagram) of Four Bar Mechanism and Slider Crank
Mechanism.
THEORY : Slider-crank mechanism, arrangement of mechanical parts
designed to convert straight-line motion to rotary motion, as in a
reciprocating piston engine, or to convert rotary motion to straightline motion, as in a reciprocating piston pump. Four-bar mechanism
can be used for many mechanical purposes, including to convert
rotational motion to reciprocating motion (e.g., pump jack) convert
reciprocating motion to rotational motion (e.g., bicycle examples)
constrain motion (e.g., knee joint and suspension examples)
Velocity of a point on a link by Relative Velocity Method
The relative velocity method is based upon the relative velocity of
the various points of the link. Consider two points A and B on a link.
Let the absolute velocity of the point A i.e.VA is known in magnitude
and direction and the absolute velocity of the point B i.e. VB is
known in direction only. Then the velocity of B may be determined
by drawing the velocity diagram.
Calculation and Results
The crank of a slider crank mechanism rotates clockwise at a
constant speed of 300 r.p.m. The crank is 150 mm and the
connecting rod is 600 mm long. Determine : 1. Linear velocity and
acceleration of the midpoint of the connecting rod, and 2. angular
velocity and angular acceleration of the connecting rod, at a crank
angle of 45° from inner dead centre position.
Given :
N_{ BO }=300 r.p.m . \text { or } \omega_{ BO }=2 \pi \times 300 /
60=31.42 rad / sNBO=300r.p.m. or ωBO=2π×300/60=31.42rad/s ; OB = 150 mm
= 0.15 m ; B A = 600 mm = 0.6 m
We know that linear velocity of B with respect to O or velocity of B,
v_{ BO }=v_{ B }=\omega_{ BO } \times O B=31.42 \times 0.15=4.713 m / svBO
=vB=ωBO×OB=31.42×0.15=4.713m/s
…(Perpendicular to
BO)
1. Linear velocity of the midpoint of the connecting rod
First of all draw the space diagram, to some suitable scale; as shown in Fig. (a).
Now the velocity diagram, as shown in Fig.(b), is drawn as discussed below:
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent
the velocity of B with respect to O or simply velocity of B i.e. v_{ BO } \text { or
} v_{ B }vBO or vB, such that
\text { vector } o b=v_{ BO }=v_{ B }=4.713 m / s vector ob=vBO=vB=4.713m/s
2. From point b, draw vector ba perpendicular to BA to represent the velocity
of A with respect to B i.e. v_{ AB }vAB , and from point o draw vector oa
parallel to the motion of A (which is along AO) to represent the velocity of A
i.e. v_{ A }vA. The vectors ba and oa intersect at a.
By measurement, we find that velocity of A with respect to B,
v_{ AB }=\text { vector } b a=3.4 m / svAB= vector ba=3.4m/s
and \text { Velocity of } A, v_{ A }=\text { vector } o a=4 m / s Velocity of A,vA
= vector oa=4m/s
3. In order to find the velocity of the midpoint D of the connecting rod A B,
divide the vector ba at d in the same ratio as D divides A B, in the space
diagram. In other words,
bd / ba = BD/BA
Note: Since D is the midpoint of A B, therefore d is also midpoint of vector ba.
4. Join od. Now the vector od represents the velocity of the midpoint D of the
connecting rod i.e. v_{ D }vD.
By measurement, we find that
v_{ D }=\text { vector } o d=4.1 m / svD= vector od=4.1m/s
Acceleration of the midpoint of the connecting rod
We know that the radial component of the acceleration of B with respect to O
or the acceleration of B,
a_{ BO }^{r}=a_{ B }=\frac{v_{ BO }^{2}}{O B}=\frac{(4.713)^{2}}{0.15}=148.1 m
/ s ^{2}aBOr=aB=OBvBO2=0.15(4.713)2=148.1m/s2
and the radial component of the acceleraiton of A with respect to B,
a_{ AB }^{r}=\frac{v_{ AB }^{2}}{B A}=\frac{(3.4)^{2}}{0.6}=19.3 m / s ^{2}aABr
=BAvAB2=0.6(3.4)2=19.3m/s2
Now the acceleration diagram, as shown in Fig. (c) is drawn as discussed
below:
1. Draw vector o^{\prime} b^{\prime}o′b′ parallel to BO, to some suitable scale,
to represent the radial component of the acceleration of B with respect to O or
simply acceleration of B i.e. a_{ BO }^{r} \text { or } a_{ B }aBOr or aB,such that
\text { vector } o^{\prime} b^{\prime}=a_{ BO }^{r}=a_{ B }=148.1 m / s
^{2} vector o′b′=aBOr=aB=148.1m/s2
Note: Since the crank OB rotates at a constant speed, therefore there will be
no tangential component of the acceleration of B with respect to O.
2. The acceleration of A with respect to B has the following two components:
(a) The radial component of the acceleration of A with respect to B i.e. a_{ AB
}^{r}aABr, and
(b) The tangential component of the acceleration of A with respect to B i.e. a_{
AB }^{t}aABt. These two components are mutually perpendicular.
Therefore from point b^{\prime}b′ , draw vector b^{\prime}b′x parallel to A B
to represent a_{ AB }^{r}=19.3 m / s ^{2}aABr=19.3m/s2 and from point x draw
vector xa^{\prime}a′ perpendicular to vector b^{\prime}b′ x whose magnitude
is yet unknown.
3. Now from o^{\prime}o′, draw vector o^{\prime} a^{\prime}o′a′ parallel to
the path of motion of A (which is along AO) to represent the acceleration of A
i.e. a_{ A }aA . The vectors x a^{\prime}xa′ and o^{\prime}
a^{\prime}o′a′ intersect at a^{\prime}a′. Join a^{\prime} b^{\prime}a′b′.
4. In order to find the acceleration of the midpoint D of the connecting rod A B,
divide the vector a^{\prime} b^{\prime}a′b′ at d^{\prime}d′ in the same ratio
as D divides A B. In other words
b^{\prime} d^{\prime} / b^{\prime} a^{\prime}=B D / B Ab′d′/b′a′=BD/BA
Note: Since D is the midpoint of A B, therefore d^{\prime}d′ is also midpoint of
vector b^{\prime} a^{\prime}b′a′.
5. Join o^{\prime} d^{\prime}o′d′. The vector o^{\prime}
d^{\prime}o′d′ represents the acceleration of midpoint D of the connecting
rodi.e. a_{ D }aD.
By measurement, we find that
a_{ D }=\text { vector } o^{\prime} d^{\prime}=117 m / s ^{2}aD
= vector o′d′=117m/s2
2. Angular velocity of the connecting rod
We know that angular velocity of the connecting rod A B,
\omega_{ AB }=\frac{v_{ AB }}{B A}=\frac{3.4}{0.6}=5.67 rad / s ^{2} \text {
(Anticlockwise about B)}ωAB=BAvAB=0.63.4
=5.67rad/s2 (Anticlockwise about B)
Angular acceleration of the connecting rod
From the acceleration diagram, we find that
a_{ AB }^{t}=103 m / s ^{2}aABt=103m/s2
…(By measurement)
We know that angular acceleration of the connecting rod A B,
\alpha_{ AB }=\frac{a_{ AB }^{t}}{B A}=\frac{103}{0.6}=171.67 rad / s ^{2}(\text
{ Clockwise about B) }αAB=BAaABt=0.6103=171.67rad/s2( Clockwise about B)
GRAPH
Question: PQRS is a four bar chain with link PS fixed. The lengths of
the links are PQ= 62.5 mm ; QR = 175 mm ; RS = 112.5 mm ; and PS =
200 mm. The crank PQ rotates at 10 rad/s clockwise. Draw the
velocity and acceleration diagram when angle QPS = 60° and Q and R
lie on the same side of PS. Find the angular velocity and angular
acceleration of links QR and RS.
Given :
\omega_{ QP }=10 rad / sωQP=10rad/s ; PQ = 62.5 mm = 0.0625 m ; QR = 175
mm = 0.175 m ; RS = 112.5 mm = 0.1125 m ; PS = 200 mm = 0.2 m
We know that velocity of Q with respect to P or velocity of Q,
v_{ QP }=v_{ Q }=\omega_{ QP } \times P Q=10 \times 0.0625=0.625 m / svQP
=vQ=ωQP×PQ=10×0.0625=0.625m/s
…(Perpendicular to PQ)
Angular velocity of links QR and RS
First of all, draw the space diagram of a four bar chain, to some suitable scale, as shown in
Fig.(a). Now the velocity diagram as shown in Fig. (b), is drawn as discussed below:
1. Since P and S are fixed points, therefore these points lie at one place in velocity diagram.
Draw vector pq perpendicular to PQ, to some suitable scale, to represent the velocity of Q with
respect to P or velocity of Q i.e. v_{ QP } \text { or } v_{ Q }vQP or vQ such that
\text { vector } p q=v_{ QP }=v_{ Q }=0.625 m / s vector pq=vQP=vQ=0.625m/s
2. From point q, draw vector qr perpendicular to QR to represent the velocity of R with respect
to Q (i.e. v_{ RQ }vRQ) and from point s, draw vector sr perpendicular to SR to represent the
velocity of R with respect to S or velocity of R (i.e. v_{ RS } \text { or } v_{ R }vRS or vR). The
vectors qr and sr intersect at r. By measurement, we find that
v_{ RQ }=\text { vector } q r=0.333 m / s \text { , and } v_{ RS }=v_{ R }=\text {
vector } s r=0.426 m / svRQ= vector qr=0.333m/s , and vRS=vR
= vector sr=0.426m/s
We know that angular velocity of link QR,
\omega_{ QR }=\frac{v_{ RQ }}{R Q}=\frac{0.333}{0.175}=1.9 rad / s \text {
(Anticlockwise) }ωQR=RQvRQ=0.1750.333=1.9rad/s (Anticlockwise)
and angular velocity of link RS,
\omega_{ RS }=\frac{v_{ RS }}{S R}=\frac{0.426}{0.1125}=3.78 rad / s \text {
(Clockwise) }ωRS=SRvRS=0.11250.426=3.78rad/s (Clockwise)
Angular acceleration of links QR and RS
Since the angular acceleration of the crank PQ is not given, therefore there will be no tangential
component of the acceleration of Q with respect to P.
We know that radial component of the acceleration of Q with respect to P (or the acceleration
of Q),
a_{ QP }^{r}=a_{ QP }=a_{ Q }=\frac{v_{ QP }^{2}}{P
Q}=\frac{(0.625)^{2}}{0.0625}=6.25 m / s ^{2}aQPr=aQP=aQ=PQvQP2=0.0625(0.625)2
=6.25m/s2
Radial component of the acceleration of R with respect to Q,
a_{ RQ }^{r}=\frac{v_{ RQ }^{2}}{Q R}=\frac{(0.333)^{2}}{0.175}=0.634 m / s
^{2}aRQr=QRvRQ2=0.175(0.333)2=0.634m/s2
and radial component of the acceleration of R with respect to S (or the acceleration of R),
a_{ RS }^{r}=a_{ RS }=a_{ R }=\frac{v_{ RS }^{2}}{S
R}=\frac{(0.426)^{2}}{0.1125}=1.613 m / s ^{2}aRSr=aRS=aR=SRvRS2=0.1125(0.426)2
=1.613m/s2
The acceleration diagram, as shown in Fig. (c) is drawn as follows :
1. Since P and S are fixed points, therefore these points lie at one place in the acceleration
diagram. Draw vector p^{\prime} q^{\prime}p′q′ parallel to PQ, to some suitable scale, to
represent the radial component of acceleration of Q with respect to P or acceleration of Q
i.e a_{ QP }^{r} \text { or } a_{ Q }aQPr or aQ such that
\text { vector } p^{\prime} q^{\prime}=a_{ OP }^{r}=a_{ Q }=6.25 m / s
^{2} vector p′q′=aOPr=aQ=6.25m/s2
2. From point q^{\prime}q′, draw vector q^{\prime} xq′x parallel to QR to represent the
radial component of acceleration of R with respect to Q i.e. a_{ RQ }^{r}aRQr such that
\text { vector } q^{\prime} x=a_{ RQ }^{r}=0.634 m / s ^{2} vector q′x=aRQr
=0.634m/s2
3. From point x, draw vector x r^{\prime}xr′ perpendicular to QR to represent the tangential
component of acceleration of R with respect to Q i.e a_{ RQ }^{t}aRQt whose magnitude is not
yet known.
4. Now from point S^{\prime}S′, draw vector S^{\prime}S′y parallel to SR to represent the
radial component of the acceleration of R with respect to S i.e. a_{ RS }^{ r }aRSr such that
\text { vector } s^{\prime} y=a_{ RS }^{r}=1.613 m / s ^{2} vector s′y=aRSr
=1.613m/s2
5. From point y, draw vector y r^{\prime}yr′ perpendicular to SR to represent the tangential
component of acceleration of R with respect to S i.e. a_{ RS }^{t}aRSt.
6. The vectors x r^{\prime} \text { and } y r^{\prime}xr′ and yr′ intersect
at r^{\prime}r′. Join p^{\prime} r \text { and } q^{\prime} r^{\prime}p′r and q′r′. By
measurement, we find that
a_{ RQ }^{t}=\text { vector } x r^{\prime}=4.1 m / s ^{2} \text { and } a_{ RS
}^{t}=\text { vector } y r^{\prime}=5.3 m / s ^{2}aRQt
= vector xr′=4.1m/s2 and aRSt= vector yr′=5.3m/s2
We know that angular acceleration of link QR,
\alpha_{ QR }=\frac{a_{ RQ }^{t}}{ QR }=\frac{4.1}{0.175}=23.43 rad / s ^{2}
\text { (Anticlockwise) }αQR=QRaRQt=0.1754.1=23.43rad/s2 (Anticlockwise)
and angular acceleration of link RS,
\alpha_{ RS }=\frac{a_{ RS }^{t}}{S R}=\frac{5.3}{0.1125}=47.1 rad / s ^{2}(\text {
Anticlockwise })αRS=SRaRSt=0.11255.3=47.1rad/s2( Anticlockwise )