Internal Expanding shoe brake.pdf

Internal expanding shoebreak
(drumbrake)
* Twin Leading Shoebrake
By: Kalola Meet M. and Jaimin Joshi

This system was used popularly in front wheels because
itis very eﬀective at braking in forward direction.
Twin leading shoe drum brake system use two single
piston wheel cylinders –one near the topof the backing
plate and one near thebottom.
Eachwheel cylinder activates one of the brake shoes.

It is called twin leading shoe brake because both the shoes are
arranged in a leading shoe conﬁguration inthe forward direction.
When applied in the reverse direction, the breaking force is very
less and hence it is usually accompanied by one of the other
types of brakes on rear wheels to be used as parkingbrakes.
This type is well suited for motorcycles they are driven mostly
in the forward direction and rarely inreverse.

Assumptions
(i)The intensity of normal pressure between the friction
lining and the brake drum at any point is proportional to its
vertical distance from thepivot.
(ii) The brake drum and the shoe arerigid.
(iii)The centrifugal force acting on theshoe is
negligible.
(iv) The coefﬁcient of friction isconstant.

calculation

Brake Drum
Assuming diameter to be 250mm.
P = 1500N

SHOE
θ1= 30o
θ2 =1200

• For curvedbeams,
• The maximum stress occurs
either at the inner fibreor at
the outer fibre.
• Stress at outerfibre:
• Stress atinner fibre:
Ro = radius of outer fibre (mm)
Ri = radius of inner fibre (mm)
R = radius of centroidal axis (mm)
RN = radius of neutral axis (mm)
hi = distance of inner fibre from neutral axis
(mm)
ho = distance of outer fibre from neutral axis
(mm)
Mb = bending moment with respect to
centroidal axis (N-mm)
A = area of the cross-section (mm2) The
eccentricity ebetween centroidal and
neutral axes is givenby,

Spring

1
2
3
• D =5d
• K =1.31
• τ = (1.31x 8 x 1500 x 5x d) /( 3.14 xd3
)
= 25031.87/d2

Taking diﬀerent values ofd
1) d= 7mm
τ = 510.68N/mm2
Using table, Sut = 1260N/mm2
τ =0.4Sut
= 500N/mm2
τ >τper
2)d =8mm
τ =391.12
Using table, Sut = 1220N/mm2
τ =0.4Sut
= 488N/mm2
τ <τper
Thus d = 8mm isoptimum
D= 5d = 40mm

• k = 1500 /6
= 250N/mm
• 250 = (81370x 84
) /(8 x 403
xN)
• N= 2.6 or 3
4
5

Total no. of coils = N +2
Nt = 5
Solid Length =Nt * d
= 40mm
6
Free Length = Solid length + δ
= 46 mm
Pitch = 46/6
Pitch = 9.2 mm

Pin
•

Friction LiningMaterial

Friction LiningMaterial
μ = 0.4
Maximum intensity of pressure= 2.06 N/mm2
Taking w = b = 20mm
thickness = 3mm
θ1= 30o
θ2 =1200

Braking Torque
• We can use the following equation tocompare the required braking
torque with generated brakingtorque.
Generated Braking Torque = 351.74Nm Required
braking torque = 250Nm

Mn = 552183.75 Nmm
Mf = 306429.68 Nmm
P = (Mn – Mf )/C = 245754.07 /187.5
= 1310.68N
Checking forforce

MODEL in Solidworks

SHOE

Assembly

Assembly Drawing

Detailed Drawings

FrictionLining

Backplate

Drum

Washer

Nut

Construction
• Drum brakes are a brake system with brake drums (rotor)
which rotate with the wheels. Inside each drum are brake shoes
fitted with brake linings (friction material). Pistons (pressure
mechanism) press against the drums from the inside to
generate braking force, thus making is possible to decelerate
and stop the vehicle.

References
• Friction lining material -
http://www.stillchampion.com/aft-132.html
• Book - Elements of machine design by V.B.Bhandari
• http://www.stillchampion.com/aft-132.html
• Spring - https://steelpurchase.com/1045-s45c-c45-
steel/

Internal Expanding shoe brake.pdf

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