Problem on Shaft based on Strength Criteria Video Lecture | Design of Machine Elements - Mechanical Engineering

Hello friends here in this video we will
see a problem of a shaft based on
strength criteria for that purpose we
have a question here I'll read what is
given here a solid shaft is transmitting
1 megawatt at 240 RPM determine the
diameter of shaft if the maximum
permissible torque exceeds main torque
by 20% take maximum allowable shear
stress as 60 mega Pascal now this is the
question which we have whatever is given
in this question first we will write
that in the form of data and then we
will start solving the problem so let us
get started in data first I will write
the type of shaft the type of shaft is
solid because it is given in the
question that it is a solid shaft next
is transmitting 1 megawatt this 1
megawatt is called as the power which is
transmitted so power denoted by P that
is 1 megawatt 1 megawatt means it is 10
raise to 6 watts at 240 rpm so this is
the speed of the shaft determine the
diameter of the shaft this is the
question so diameter I will denote it by
capital D which we have to calculate if
the maximum permissible torque exceeds
main torque by 20% now this is a
condition they have given that maximum
permissible torque I will denote it by T
max that is equal to now the condition
is maximum permissible torque which is T
max exceeds me torque
by 20% so it is exceeding the main
torque means maximum torque is first of
all it is equal to main torque then in
that I have to add 20% of the main
torque so 20% of main torque so
therefore I can write here t max is
equal to this T mean if I keep common so
T mean into bracket this will be 1 plus
20% if I write it in the form of number
it would be 0.2 so hence Tmax is equal
to 1.2 times of T mean now this relation
is what is given in this problem that
team axis 1.2 times of T mean next they
have said that take maximum allowable
shear stress as 60 mega Pascal maximum
allowable shear stress means it is tau
and the value is 60 mega Pascal or we
can say 60 Newton per millimeter square
so this is the data for the problem now
here we have to find how much will be
the diameter of solid shaft so let us
try to get the solution
in solution first I'll draw the cross
section of the shaft it is circular in
shape the diameter is capital D that is
what we have to calculate this is the
cross section of solid shaft now how to
start the problem since power and speed
is given I start from the problem of
power and from that I will get the value
of torque so I can say that since power
transmitted by solid shaft is given by
the formula of power transmitted P is
equal to 2 pi n t upon 60 this is the
formula of power now therefore I can say
that here I keep T on one side 60 goes
on the other side so it is P into 60
divided by 2 pi n here I will put the
values power P it is given in the
question it was 1 megawatt but we want
the value inward so I will keep this 10
raise to 6 volt so 10 raise to 6 into 60
divided by 2 pi into n the speed that
speed is 240 rpm
so from this if I calculate I will get
the value of torque that is equal to
which I have already calculated my
answer is thirty nine point seven nine
into 10 raised to 3 Newton meter
now this torque value which I am getting
this is nothing but the mean torque or
we can say average tour now after we
have got the average torque in the
problem they have said that maximum
torque is 20% more than the main torque
here they have mentioned that T max is
20% more of T mean so we will find T max
after this I will say that therefore T
max is 1.2 times of T mean so 1.2 into t
minus thirty nine point seven nine into
10 raised to 3 so therefore T max my
answer is that is forty-seven 0.75 into
10 raised to 3 Newton meter now whenever
we have torque in terms of Newton meter
and we want to calculate the diameter of
the shaft always convert the torque into
Newton mm so that we can get the
diameter in terms of mm so here after
I've got this answer I will say that
this is or it is 4 point 7 5 into 10
raise to 6 Newton mm for the conversion
I am multiplying by 10 raised to 3 to
replace meter by mm now this much is the
maximum torque and as we know shaft is a
rotating machine element this will
rotate so the torque which is required
for its rotation the maximum torque
value we will take for designing of the
shaft that is venable we are designing
the shaft
in that case we have to take the maximum
torque into account so after this I will
say that since by using strengths
criteria because the problem is based on
strength criteria here I am using
strength and what it says it says that T
upon J is equal to tau by R now this is
called as the strength criteria and why
it is called a strength criteria because
here we have shear stress because of
this stress we are designing the shaft
based on its strength
because stress or strength it indicates
the material condition of the shop so
now I will say that therefore here we
have T J is polar moment of inertia that
is PI by 30 to be raise to 4 is equal to
tau upon R is radius so it is diameter
by 2 now once I have written the formula
based on strength criteria here this D
and D will get cancelled out so here
instead of D raise to 4 we will be
having D cube this 2 and 32 it gets
cancelled so finally I can write this
formula in a more better manner that is
T is equal to this PI by 16 will go on
the other side so it is PI by 16 and
even this d cube PI by 16 D cube into
tau now this is another formula which we
can use in case of strength criteria
that is this is the reduced formula so
we can use it for solid shaft now your I
will go on putting the values this
torque is there it is nothing but
maximum torque as I have told whenever
we are designing the shaft we have to
consider the maximum
and then decide so maximum torque is
47.7 5 into 10 raise to 6 so I will
write the value now this 16 I will
multiply here n PI n tau PI into tau
value it is given as 60 Newton per mm
square so on one side I only have D cube
next after this when we are calculating
this at first we would be getting the
answer of d cube and when I take the
cube root of this answer then I will be
getting the deed so finally the value of
D which I have that is 150 9 point 4 4
mm or we can round off this value and it
becomes 160 mm so this is our answer so
if we have a shaft which is circular in
section and it is a solid shaft
transporting one megawatt power which is
given at 240 rpm then the diameter of
that solid shaft it should be equal to
160 mm that is we have calculated the
diameter of this shaft so when the
question was in front of us they were
telling us to determine the diameter of
this shaft and for the given conditions
which they have given that maximum
torque exceeds main torque by 20% power
was 1 megawatt 240 rpm was the speed of
the shaft and it was subjected to
maximum shear stress of 60 mega Pascal
so for all these conditions the answer
is that we should use a shaft of
diameter 160 mm which would follow or
which would suit these conditions so
with this we complete this question