Free Conveyor Design

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Belt conveyor design

This example shows the steps necessary to design an industrial belt conveyor.

Tutorial

Conveyor requirements:

Length, Lb: 20 ft (6.1 m)

Width, wb: 16 in (0.0406 m)

Slope, α: +15°(rising)

Step 1: Belt selection

Step 2: Pulley and drive design

Step 3: Frame design

Step 4: Supports design

Load: cardboard boxes

Total load weight, W: 400 lbs (181.4 kg)

Speed, s: 60 ft/min (18.3 m/min)

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Step 1: Belt selectionStep 1: Belt selection

1.A. Belt applicable operational constraints:

•Overall length.

•Width.

•Temperature range.

•Food grade.

•Chemical resistance.

•Antistatic.

•Flammability rating.

•Conveying side adhesiveness.

•Suitable for a slider bed or carrying rollers.

Our application requires an adhesive belt in order to convey the boxes on an upward slope. It will be supported by a slider bed. The adhesiveness between the load and a

belt sample is to be tested; this by placing the load on top of the belt, both on an

angled surface and increasing the angle. A rubber/polyester belt is a good candidate.

1.B. Belt required data:

Thickness, t: 0.06 in. (1.5 mm)

Belt mass, m: 0.39 lbs/ft² (2 kg/m²)

Min. pulley diameter: 1 in (25 mm)

Tensile force, k1%: 40 lbf/in (7 N/mm)

Max. tensile force, Tm: 63 lbf/in (11

N/mm)

Friction coefficient on driving pulley, fp:

0.15

Friction coefficient on slider bed, fb: 0.2

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1.C. Geometry of the driving pulley:

Pulley diameter dp: 4 in (101.6 mm) {higher than min. pulley diameter, final value to be validated in Step 2 }

Pulley width, wp: 18 in (457.2 mm)

Belt wrapping angle, θ: 180°(smaller value than actual angle)

Driven (tensioning) pulley has the same geometry for economic reason.

1.D.1 Operation forces and torque:

The tension in the belt is at its maximum level when the fully loaded conveyor is accelerating to its rated speed after a stop.

The time to accelerate, ta, is to be:

Short enough:

•To minimize overload current in the electric motor and its amplifier (consult ratings).

•To maximize conveyor output.

Long enough:

•To prevent conveyed load from slipping or tilting.

•To minimize motor horsepower.

For our application, ta = 3 sec.

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Pulley speed, ω = 12 s / (30 dp) = 6 rad/sec (π = 3.1416)

ω = 1000 s / (30 dp) = 6 rad/sec

Pulley acceleration, a = ω / ta = 2 rad/s2

Pulley weight, Wp = (0.283 lbs/in3) π dp2 wp / 4 = 64 lbs per pulley

Wp = (7833 Kg/m3) π dp2 wp / 4 = 29 kg per pulley

Pulleys mass inertia, Jp = 2 (Wp dp2 / 8) = 256 lbs-in2 (0.075 kg-m2)

Belt weight, Wb = m wb (24 Lb + π dp) / 144 = 21.3 lbs

Wb = m wb ( 2 Lb + π dp) / 1000 = 9.68 kg

Belt mass inertia, Jb = Wb dp2 / 4 = 85.4 lbs-in2 (0.025 kg-m2)

Load mass inertia, J = W dp2 / 4 = 1600 lbs-in2 (0.468 kg-m2)

Total mass inertia, Jt = Jp + Jb + J = 1941 lbs-in2 (0.568 kg-m2)

Acceleration torque, Ta = Jt a / 386 = 10.06 lbf-in

Ta = Jt a = 1.136 N-m

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Friction force between belt and slider bed,

Ff = (W + 12 m Lb wb / 144) fb cos(α) = 79.3 lbf

Ff = 9.81 (W + m Lb wb) fb cos(α) = 353 N

Friction torque, Tf = Ff dp / 2 = 158.6 lbf-in (17.9 N-m)

Lifting force, Fl = W sin(α) = 103.5 lbf (460.7 N)

Lifting torque, Tl = Fl dp / 2 = 207 lbf-in (23.4 N-m)

Total torque, Tt = Ta + Tf + Tl = 375.7 lbf-in (42.5 N-m)

Power, P = Tt ω / 6600 = 0.342 Hp

P = Tt ω = 255 Watts

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1.E Belt tension and stress

Tension (carrier side) F1 = {Tt / (dp / 2)} / { 1 – 1 / (2.718 (fp θ) ) } = 499 lbf (2221 N)

Tension (return side) F2 = F1 / (2.718 (fp θ) ) = 311.5 lbf (1386 N)

Belt stress, σb = F1 / Lb = 31.9 lbf/in (5.46 N/mm)

Belt safety factor, SFb = Tm / σb = 2.02, selected belt is adequate.

Safety factor for electric motor shall be lower than belt safety factor to prevent breakage.

Safety factors for all other conveyor parts shall be higher than belt safety factor.

Snub rollers

Snub rollers are used to:

•Increase the angle of contact between the belt and the driving pulley and improve the

torque transmitted.

•Decrease the distance between the belt carrying side and return side.

•Improve belt tracking with adjustable snub rollers.

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Step 2: Pulley and drive designStep 2: Pulley and drive design

2.A Pulley design

Pulleys with cylindrical-conical shape are to be used in order to help belt tracking and prevent runoff. The geometry is derived from experience and is sometimes supplied

by belt manufacturers. For our application, the following pulley is used for the drive

end and the idler end.

Pulley material: UNS G43400 Surface finish: Ra 60 µin (1.6 µm)

Pulley deflection under maximum load

Section inertia, Ip = π (dp4 – di

4) / 64 = 12.52 in4 (5.211 x 10-6 m4)

Deflection, y = {5 (F1 + F2) wp3} / {384 x 29000000 Ip) = 0.0002 in

y = {5 (F1 + F2) wp3} / {384 x 2.07 x 1011 Ip) x 1000 = 0.004 mm

The maximum deflection is smaller than the slope height in the conical portion of the pulley. The pulley will be able to track the belt.

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2.B.1 Shaft design

Drive shaft design is subjected to torsion stress from the motor torque and bending stress from the pulley tension, the drive sprocket, and the bearing supports

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2.B.2 Drive shaft loads

Force from gearmotor torque, Fm = 2 Tt / Ds = 188 lbf (838 N)

(sprocket pitch diameter, Ds= 3.989 in (101 mm)

Along the conveyor axis:

Fx = Fm cos(17°) = 180 lbf (801 N)

Q = (F1 + F2) / C = 47.7 lbf/in (8.35 N/mm) (distributed belt force)

R1 = ( Q C (C/2 + D) + F (A + B + C + D) ) / (B + C + D) = 606 lbf (2696 N)

(left bearing reaction)

R2 = Fx + Q C – R1 = 386 lbf (1718 N) (right bearing reaction)

M1x = -Fx A = -391 lbf-in (-44.2 N-m)

Perpendicular to conveyor axis:

Fy = Fm sin(17°) = 41 lbf (183 N)

M1y = -Fy A = -119 lbf-in (-13.5 N-m)

Bending torque at left bearing :

M1 = (M1x2 + M1y

2)0.5 = 409 lbf-in (46.2 N-m)

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2.B.3 Drive shaft design

Drive shaft geometry

Shaft material: UNS G10450

Tensile yield strength, Sy = 58700 psi (405 MPa)

Ultimate strength, Su = 97900 psi (675 MPa)

Endurance strength Se = 0.55 x 1 x 0.814 x 1 x 1 x (0.5 Su) = 21914 psi (151 MPa)

Shaft safety factor, SFd = 2.1 (greater than belt safety factor)

Minimum shaft diameter

dm = { (32 SFd / π) x ( (Tt / Sy)2 + (M1 / Se)

2 )0.5 }0.33 = 0.750 in (19.05 mm)



2.C Commercial parts

Gearmotor: integrated worm gear reducer and AC motor with electrical brake, ratio 22.5:1 ,

power 0.5 Hp (0.37 kW), maximum output torque 340 lbf-in (38.5 N-m ), service factor 1.8 .

Overload protection from motor amplifier. Electrical installation must be conformed to local

government regulations.

Bearings: ball bearing oval flanged unit, bore 25 mm, dynamic load rating 14 kN, static load

rating 7 kN, limiting speed 7000 rpm, grub screw locking, with bearing shields.

Sprockets: roller chain steel sprocket, no.40 (1/2”pitch), 25 teeth, keyway + 2 setscrews.

Roller chain: no.40 (1/2”pitch) steel roller chain. To be lubricated with oil (refer to

manufacturer for oil viscosity)


Steps 3 and 4 explain the design of the conveyor frame, supports and components

with:

•Loads and reactions.

•Stresses and deflections.

•Detailed drawings.

To access Steps 3 and 4 (14 pages), please subscribe to

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