Simple Band Brake Video Lecture | Design of Machine Elements - Mechanical Engineering

[Music]
assalamu alaikum
this is a facade a a welcome to my
channel have you checked my other
channel if no go and check it out so
what we are going to learn today it is a
type of break called simple band break
this is a diagram of the simple band
break this is a band which is connected
to the liver this is shaft on which a
wheel is mounted and band is wrapped
around the wheel lead and shafts are
connected by mean of a key this is the
contact angle of Bend and wheel denoted
by theta T 1 and T 2 are the tensions in
the band as the pulleys rotating
clockwise so this side is tied side and
this is slack side it means t1 is
greater than t2 unreal t1 is for tight
side and t2 is for loose side tension B
is distance between the pivot point O
and capital B point small L is the
length of the lever B is the external
force to apply the brake he is a pole
force when we apply P the position of
the river will be like this and Bend
will grab the wheel and offer frictional
resistance to its rotation in this brake
the braking force is equal to t1 minus
t2 tension in tight side - tension in
slack side and braking torque is equal
to the braking force into radius of the
pulley or wheel or whatever you have in
your case
ratio between tensions can be found by
this relation in it mu is coefficient of
friction between bands and Wheels
material and theta is lap angle or
contact angle maximum tension can be
found by this relation team X or T 1 is
equal to tensile stress into area of
band w's width of bed
t is thickness of band now it's time to
solve a problem
[Music]
you
read and gather data from the problem
statement
[Music]
you
this end of the band is fast on to the
fixed end of the lever which is pivot
point oh this is lever arm and distance
between o and capital V is 125
millimeter lever length is 720
millimeter problem says that the
bisecting line of contact angle is
perpendicular to the lever
this is power in a part we have to find
the pull force and in be part we have to
find the band width while thickness and
tensile stress is given due to drums
rotation this side would be tie and this
side would be slack first thing first
the direction of pull force when we pull
the lever upward the band will grab the
drum and it would be not possible in
case of pulling the lever downward so
the apply force would be upward first
find out tensions
in question theta is in degree but in
elation it would be used in radium
we have tensions ratio now apply power
formula from it we will find TB which is
braking torque power and RPMs are given
you
we also know that for simple band break
TB is equal to t1 minus t2 into our TB
and our are known
so we have T 1 minus T 2 is equal to
triple five seven Newton we have
calculated t1 over t2 earlier so t1 is
equal to three point two five T to
substitute this value in above relation
and we have t to use this value to find
t1 next taking moment about OPV point
pull force multiplied by distance now 42
force as t2 is slag side so its
direction would be in opposite of the
direction of the drums rotation pulse
moment is in anti-clockwise and moment
due to t2 is in clockwise direction let
me draw the system again
you
with tea to force the distance would be
OD now find OD in this triangle
45 is equal to oay / OD
and from figure is half of the OB or 125
millimeter
you
now we have od moment due to t1 is 0
because it is passing through the pivot
point Oh
by simplifying we have pull value next
we have to find the width of the band
first apply a stress formula from here
we will find area which is width into
thickness
you
and maximum tension is equal to t1
so solving it we have value of be
[Music]