Design of a Drum Brake

My Design project demonstrates a comprehensive understanding of the mechanics of the drum brake system as well as my ability to design and draw a complex mechanical system.



Cover Image


This design is based on the guidance catalog provided by Dr. Zsolt Tiba during our Machine Element classes.


A double-shoe brake is a type of drum brake that applies pressure to the inside of a rotating drum using two shoes. The brake shoes are located outside the drum in the external configuration, which makes maintenance and replacement easier. To release the brake, a thruster release mechanism employs compressed air or hydraulic pressure. When designing a drum brake, factors such as the brake's application, the weight of the object being stopped, and the desired stopping distance must all be considered. Pre-calculating these parameters can aid in ensuring that the brake is both effective and safe.

**Making a free-hand drawing of the design on A1 paper is an effective way of communicating the design to others and serving as a reference during the construction process. however unfortunately i do not have a good file about my own construction which made me bringing the standers one available at University of Debrecen.

Task setup (Step1):

As a student i was assign with a specified parameters in which i should further investigate my design which can be seen below:

NumberBrake type Mbrake [Nm]n [1/s]

Choosing the material based on previous data which i believe will withstand the needed tasks:

Lining Material Drum material TypeCoefficient of frictionPallwoed [MPa] Qallowed [W/mm^2]Tmax [°c]
Woven (dry)Cast iron or steelUR 410.27 – 0.460.53.0150

Calculating the preliminary drum diameter (Step 2):

Dpre=17[MbrakeμminPallowed3][mm]Dpre\:=\:17\cdot \left[\sqrt[3]{\frac{Mbrake}{\mu min\cdot Pallowed}}\right]\left[mm\right]
Dpre=17[2000.270.53]=208.76[mm]Dpre\:=\:17\cdot \:\left[\sqrt[3]{\frac{200}{0.27\cdot \:0.5}}\right]\:=\:208.76\:\left[mm\right]

We are choosing the Brake Drum diameter with taking into consideration the shoe dimensions standers as 200mm.


Calculate the Normal force (step3):

Mbr=2μFnD2[Nm]Mbr\:=\:2\cdot \mu \cdot Fn\cdot \frac{D}{2}\left[Nm\right]
200=20.27Fn2001032200\:=\:2\cdot 0.27\cdot Fn\cdot \frac{200\cdot 10^{-3}}{2}
Fn=2000.27200103===>3703.7NFn\:=\:\frac{200}{0.27\cdot 200\cdot 10^3}===>3703.7\:N

Checking braking bearing stress (step4):

Fna1b2Pallowed[N]Fn\:\le a1\cdot b2\cdot Pallowed\:\left[N\right]
Fn135.12600.5=4053.6NFn\:\le 135.12\cdot 60\cdot 0.5\:=\:4053.6\:N
3703.74053.6N===>thusitssafe3703.7\:\le 4053.6\:N\:===>\:thus\:its\:safe

where, we are calculating a1 and b1 by the help of the dimensions of the shoe in the table above and by the help of the brake lining and material property table above:

a1=Dsinα2a1\:=\:D\cdot sin\frac{\alpha }{2}
a1=200sin852=135.12mma1\:=\:200\cdot \:sin\frac{85\:}{2}\:=\:135.12\:mm
b1=7025=60b1\:=\:70\:-\:2\cdot 5\:=\:60

** α = 60° - 90° angle of contact

** a1 and b1 [mm] dimensions of the shoe

Checking for peripheral velocity (Step5)

νo=qmaxμPallowed[ms]\nu o\:=\:\frac{qmax}{\mu \cdot Pallowed}\:\left[\frac{m}{s}\right]
νo=3.00.460.5=13.04ms\nu o\:=\:\frac{3.0}{0.46\cdot 0.5}\:=\:13.04\:\frac{m}{s}

Service diagram step(6):

Mbr.maxn=KbrMbr.max\cdot n\:=\:Kbr


Kbr=qallowedpia1b1[W]Kbr\:=\:\frac{qallowed}{pi}\cdot \:a1\cdot \:\:b1\:\left[W\right]
Kbr=3.0pi135.1260[W]Kbr\:=\:\frac{3.0}{pi}\cdot \:135.12\cdot \:\:60\:\left[W\right]
Mbr.max=μPalloweda1b1D[Nm]Mbr.max\:=\:\mu \cdot Pallowed\cdot a1\cdot b1\cdot D\left[Nm\right]
Mbr.max=0.270.5135.1260200103=218.9[Nm]Mbr.max\:=\:0.27\cdot 0.5\cdot 135.12\cdot 60\cdot 200\cdot 10^{-3}\:=\:218.9\left[Nm\right]
nmax=VmaxDpi[1s]nmax\:=\:\frac{Vmax}{D\cdot pi}\:\left[\frac{1}{s}\right]
nmax=40200103pi=63.66[1s]nmax\:=\:\frac{40}{200\cdot 10^{-3}\cdot pi}\:=\:63.66\:\left[\frac{1}{s}\right]

**As the normal force is limited the allowed bearing stress, the maximum brake moment is limited too.

**In the rotating drum the peripheral stress arises. Its allowed value arises at the Vmax peripheral velocity of 40 m/s^-1

Choosing and calculation of preliminary proportion of lever arms on the basis step(7):

following the equation table we could determine of preliminary proportion of lever arm

Figure signis,acis,thrIac,thr
e-h(b/a)(r/c)(b/a)(d/c)(d/r)\:=\:\frac{250}{135}\:\cdot \:\frac{9}{5}\:=\:3.33
is.thr=250135205=7.4is.thr\:=\:\frac{250}{135}\:\cdot \:\frac{20}{5}\:=\:7.4

**is,ac proportion of lever arms between the shoe and the spring

**is,thr , the travel of the pressure lug of the thruster and the thrust

**iac,thr , the maximum load exerted by the thruster arises in released condition of the brake.

where the following equations are based on engineering judgment ratio of dimensions.


Calculate the necessary actuating force (step8):

Fn=is,acFcηFn\:=\:is,ac\cdot Fc\cdot \eta
Mbr=μFnDMbr\:=\:\mu \cdot Fn\cdot D


Fac=MbrμDis,acηFac\:=\:\frac{Mbr}{\mu \cdot D\cdot is,ac\cdot \eta }
Fac=2000.32001033.30.9=1122.3[N]Fac\:=\:\frac{200}{0.3\cdot 200\cdot 10^3\cdot 3.3\cdot 0.9}\:=\:1122.3\:\left[N\right]

**Fac [N], activating force provided by eg. brake spring.

**η mechanical efficiency of the linkage, approximately 0.9 for considering the friction of pin.

Choosing thruster (step9):

The maximum load exerted by the thruster arises in released condition of the brake



Δf+=2δminis,acζ\Delta f^+\:=\:2\cdot \delta min\cdot is,ac\cdot \zeta
Δf+=[mm]\Delta f^+\:=\:2\cdot 0.8\cdot 3.3\cdot 1.1\:=\:5.803\:\left[mm\right]
Fac,max=Fac+sΔf+Fac,max\:=\:Fac+s\Delta f^+
Fac,max=1122.3+13.95.803=1202.96[N]Fac,max\:=\:1122.3+13.9\cdot 5.803\:=\:1202.96\left[N\right]

when releasing the brake, a min δ shoe to drum clearance will develop causing an additional deflection of the spring which magnitude is: f^

**δmin = 0.8 and ξ = 1.1 for considering the clearance fit of pins.

** s-Spring value is described next.

Determining the necessary spring constant (step10):

0.85Fac=FacsΔf0.85Fac\:=\:Fac-s\Delta \:f^-
s=0.15FacΔf[Nmm]s\:=\:\frac{0.15\cdot Fac}{\:\Delta \:f^-}\:\left[\frac{N}{mm}\right]
s=0.151122.312.1=13.9[Nmm]s\:=\:\frac{0.15\cdot \:1122.3}{\:12.1}\:=13.9\:\left[\frac{N}{mm}\right]

where, Reduction of the spring deflection:

Δf=2δwearis,acξ\Delta \:f^-\:=\:2\cdot \delta wear\cdot is,ac\cdot \xi
Δf=21.663.311=12.1[mm]\Delta f^-\:=\:2\cdot 1.66\cdot 3.3\cdot 11\:=\:12.1\:\left[mm\right]

Prescribe the δmin and calculate δmax and δwear (step10):

When the brake is applied, the pressure lug is traveling from the upper limit downwards and takes distance Ln until applying the brake shoes to the drum.

Maximum allowed clearance is:

δmax=0.8Ln2is,thrζ\delta max\:=\:\frac{0.8\cdot Ln}{2\cdot is,thr\cdot \zeta }
δmax=0.85027.41.1=2.45\delta \:max\:=\:\frac{0.8\cdot \:50}{2\cdot \:7.4\cdot \:1.1}\:=\:2.45


δwear=δmaxδmin\delta \:\:\:wear\:=\:\delta \:\:\:\:max\:-\:\delta \:\:\:min
δwear=2.460.8=1.66\delta \:\:\:wear\:=\:2.46\:-\:0.8\:=\:1.66

The pressure lug must not get close to the lower limit because in this position the arm linkage mechanism would be backed and the shoes could not apply to the drum and as a consequence the braking moment can not develop. To avoid this situation the following requirement must be complied:

Δl=2δis,thrζ\Delta l\:=\:2\cdot \delta \cdot is,thr\cdot \zeta
Δlmax0.8Δln\frac{\Delta \:lmax}{0.8}\:\le \:\Delta \:ln
22.457.41.10.8Δln\frac{2\cdot 2.45\cdot 7.4\cdot 1.1}{0.8}\:\le \:\:\Delta \:\:ln
39.8865039.886\:\le \:\:50