design proposal

ENG1002 Design project Sem2 2013 - Client Brief Version 2.1 (9/9/2013) 1

© University of Southern Queensland

Client Brief Version 2.1

1. Project Outline (page 3 correction in blue)

Midas Gold Pty Ltd seeks submissions from suitably qualified companies for the design an underground

passenger lift to service their new mine.

The lift is required to operate in a vertical shaft to 60m below ground level. The lift is required to transport

300 miners at the change of shifts (150 up, 150 down) within a period of 15 minutes. The steel cage is to be

suspended by multiple steel ropes from a winch drum positioned 10m above ground level. The winch is to

be used to raise and lower the cage. Figure 1 shows the dimensional detail of the shaft, lift and winch

system.

Figure 1: Proposed lift equipment layout

Any company tendering a design must clearly specify the final design parameters listed in bold for each of

the design sections below. Each design section requires a technical analysis which must be summarised in

the final design proposal.

Ground

level

Passenger cage

W x W x 2.2m

Winch drum

Radius R

10m

60m

motion

Cage guides

and braking

surfaces

Brakes on cage

W

2.2m

W

NOT TO SCALE = 0.25 m

2 ENG1002 – Introduction to Engineering and Spatial Science Applications

1.1 Design Sections

The project has been divided into four distinct sections to ensure clarity in the requirements and the

expected outcomes.

1. the passenger cage sizing (cage dimension W, number of passengers, travel time, top speed of lift, magnitude of acceleration and deceleration, quantities of steel mesh required for the

cage)

2. the winch (model number, input power requirement, torque delivered at drum)

3. the steel rope selection (rope type, total length of rope required, number of ropes)

4. the budget and costs of all components of the system

All students please refer to the IMPORTANT NOTES on the last page of this document.

1.2 Design Goals

The design goals for the project are to:

G1. maximise the rate at which workers can be transported

G2. do not exceed the budget ($40,000) for the project

G3. offer a minimum cost solution that meets the requirements

2. Specification of Requirements

2.1 Requirements

The following requirements must be met:

R1. The lift must be capable of moving 300 passengers (150 up, 150 down) the full travel of 60m within a 15 minute period.

R2. The maximum acceleration experienced by the passengers is not to exceed more than 3g’s, that is – 1g due to gravity plus 2g’s due to the movement of the lift. (1g = 9.81m/s

2 ).

R3. The minimum acceleration experienced by the passengers is not to be less than 0.5g, that is - 1g due to gravity plus -0.5g due to the movement of the lift.

2.2 Scope

The technical analysis and design work required for this project only requires the selection of

components from those provided and the specification of values for the parameters listed for each Design Section. In particular, aspects of the project that are outside the scope of the design include:

 all construction costs of the lift and lift shaft

 the design and materials for the cage guides, winch drum and counter weight

 electricity supply and the control system

 all detailed components such as cable clamps etc

 all components or materials not specified in versions of this brief

ENG1002 Design project Sem2 2013 - Client Brief Version 2.1 (9/9/2013) 3

© University of Southern Queensland

2.3 Constraints

The following constraints apply:

 A maximum budget of ($40,000) has been set for the cost of materials for the lift.

 The maximum design speed of the lift is not to exceed 5m/s.

 You are to assume that each passenger (on average) takes 0.5 seconds to enter or to exit the lift.

 The minimum area required for each passenger in the lift is 0.25m 2 .

 The winch efficiency is 70%.

 The steel rope breaking stress is 700 MPa. (pressure on cross section of rope)

 A minimum safety factor of 8 must be used for the design of the steel ropes.

 The minimum number of steel ropes to be used for the design is 3.

 The maximum number of steel ropes that can be fitted to the lift is 6.

 The minimum length of steel rope that must remain wound on the drum is 3 full wraps around the drum.

 The radius of the winch drum is 0.25 m.

2.4 Assumptions

The following simplifying assumptions have been made:

 the passenger cage has a square floor (W x W)

 the steel mesh forming the passenger cage provides the structure of the cage

 the cost of electricity to run motors is ignored

 the mass of the cables is to be ignored

 The Torque (Moment) delivered at the winch drum is related to the output power of the winch by:

(Moment) Torque = Power * ω

(Moment) Torque (N.m) = Power (W) / ω (rad/s)

where ω (omega) is the rotational speed of the drum in rad/s

Assume ω is constant during the motion of the lift, calculated from your chosen top speed of

the lift.

Notes to students:

You are expected to consider the forces on the lift cage to include the weight of the

lift cage, the weight of the passengers and the force applied via the cables generated

by the Torque of the winch. The largest magnitude of force will be applied when the

lift accelerates upward at the bottom of the shaft.

4 ENG1002 – Introduction to Engineering and Spatial Science Applications

3.0 Technical Information

Technical and cost information covering the components of the project is provided in this section.

Table 1: Technical Information related to the Passenger Cage

Quantity variable value or equation unit

Cage height h 2.2 m

Cage wall thickness w 0.02 m

Cage material – steel mesh 20% of surface area is solid

Density of steel ρ 7830 kg/m 3

Mass of a passenger (maximum) m 100 kg

Cost of steel mesh Cm 200 $/m 2

Table 2: Winch type, power rating and cost

Winch Type Input Power (kW) Cost Cw ($)

W50 50 12,250

W80 80 17,150

W100 100 24,000

W150 150 33,600

W200 200 47,000

Table 3: Steel rope type, diameter and cost (only Rope R-2 now to be considered)

Steel Rope Type Diameter (mm) Cost Cr ($/m)

R-1 13 4

R-2 16 6

R-3 22 12

R-4 29 20

R-5 35 29

ENG1002 Design project Sem2 2013 - Client Brief Version 2.1 (9/9/2013) 5

© University of Southern Queensland

Important note to students

The sections listed above are to be used to subdivide the analysis and design process and

identify the sections you are to use for your Technical Analysis, Presentation and Design

Proposal assessments, as detailed in the requirements of each assessment.

IMPORTANT: This is a closed design problem where all information required to

complete the technical analysis, calculations and evaluation of possible solutions will be

available in the Client Brief, your text books or other provided assignment material. The

problem presented is a simplified version of a real design problem, so the fine details of the

components of the proposed system are ignored.

If you find yourself seeking information beyond that provided in the Client Brief,

your text books or other assignment material then you are probably over thinking the

problem. The three assessments using this problem are able to be completed using just the

engineering fundamentals you are studying, supported by other course material and tools

like the spreadsheet. There is no need to research commercial equipment.

For the Technical Analysis assessment all students must complete a technical analysis

and prepare a short technical report on Design Section 1 (only) of the project. Your

memorandum to a (pretend) colleague is to request a technical analysis and short

technical report on either section 2 or 3.

For the Presentation assessment each student will select a design section of the project

(not section 1) on which to complete a technical analysis and prepare a short oral

presentation. [This can be the same as the section identified in your memo.] You are to

present a summarised technical analysis of that section of the design and how it

depends-on / influences any other section of the design. The presentation is to be

prepared and delivered as if to other colleagues in your company who are working with you

on the larger project.

For the Design Proposal assessment students are expected to complete the technical

analysis for the whole project, model the design on a spreadsheet, evaluate some

alternatives within the design and select a specific design solution to recommend in their

report. The recommendation must clearly specify all of the parameters listed in the design

sections in bold, as they define each section of the design.

Students should note there is more than one correct answer to this problem, as several

possible solutions will meet the requirements of the design.

Furthermore - a technical analysis of a single design section ALONE is unlikely to

identify a set of design parameters that results in the final project design, as the

sections are somewhat dependent on each other. Hence when you complete a technical

analysis on a single section of the design you are not looking for a specific ‘answer’ to

that section.

Your analysis should show the relationships between the quantities within a section

and possibly with those in other sections of the design, to help you understand the

problem. This analysis may enable you to eliminate some of the possible choices of

equipment on offer (when it is evident it cannot do the job) or you may be able to reduce the

range of values for some variables over which you expect they will to contribute to finding

a viable solution.

------------------------------------------------------------------------------------------------

To : Mark Senott , (Design Engineer)

From : Daeej Ali , (Project Engineer)

Date : 12 Sep. 2013

Subject : Design a lift

------------------------------------------------------------------------------------------------

Mark

We want a design for an underground passenger lift to service their

new mine. It’s supposed to take 300 passengers (150 up , 150 down).

I need from you the design for the winch part with the calculations

and what power we need for that. It is necessary to provide me with

the whole information about it ,model number, input power

requirement, torque delivered at drum. The winch efficiency is 70%.

I have completed the design for the cage sizing and I want you to see

it that you might have a better idea about what we want exactly.In

addition I want you to get which steel ropes should we use with and

also with the information … the length, the number, the calculations

and everything that’s about it. That should be designed with a safety

factor = 8 and The maximum number of steel ropes that can be fitted

to the lift is 6. The minimum length of steel rope that must remain

wound on the drum is 3 full wraps around the drum.

Our budget to design this project is limited with an about $40000 so

you should but that in your eye while getting what we need. We will

see how much will it cost to do it with the final results that we will get

later. I need all this to be provided and totally completed in the 20 th

of

Sep. so try to finifh it ASAP.

Regards

Ali

Mobile : 0468717944

Copy : John Terry , (Chief Engineer)

Technical Analysis (Lift Design)

By Daeej Ali

September 2013

#Introduction

This report is for designing an an underground passenger lift to service their

new mine by Midas Gold Pty Ltd seeks submissions.

The lift is required to operate in a vertical shaft to 60m below ground level. The

lift is required to transport 300 miners at the change of shifts (150 up, 150

down) within a period of 15 minutes. The steel cage is to be suspended by

multiple steel ropes from a winch drum positioned 10m above ground level. The

winch is to be used to raise and lower the cage. Figure 1 shows the dimensional

detail of the shaft, lift and winch system.

#Cage Size

To get the cage size we need to calculate a few things that would help to design

it to be suitable. We need to have the mass and the weight of the miners. Also

the cage floor area and the other parts of the cage area that is the steel mesh.

Mass of the miners = Mass of a passenger (maximum) * number of passengers

Mass of the miners =100*150 =15000 Kg

Weight of the miners = mass of the miners * g

Weight of the miners = 15000 * 9.81 = 147.15 KN

Then we can get the area of the cage floor by the equation (w*w)

W*W = 0.25*150 = 37.5 m^2

W^2 = 37.5

W = √37.5 = 6.124 m

Cage Area = 2w^2 + (4*2.2*w)

Cage Area =2w^2 + 8.8 * w m^2

Cage Area = (2*6.124^2) + (8.8*6.124) = 128.898 m^2

Cage Volume = Cage Area * Thickness

Cage Volume = 128.898 * 0.02 = 2.578 m^3

ρ (Density of the cage) = Mass of the cage / Volume of the cage

Mass of the cage = ρ (Density of the cage) * V (Volume of the cage)

Mass of the cage = 7830 * 2.578 = 20185.74 Kg

Weight of the cage = mass of the cage * g

Weight of the cage = 20185.74 * 9.81 = 198.022 KM

T = m (g + a) ,,, when it’s up

Tmax = m (g+2g)

Tmax = m * 3g

M = 15000 (mass of the miners) + 20185.74 (mass of the cage) = 35185.74 Kg

Tmax = 35185.74 * 3 * 9.81 = 1035.516 KN

T = m (g - a) ,,, when it’s down

Tmin = m (g – 0.5 g)

Tmin = m * 0.5 g

Tmin = 35185.74 * 0.5 * 9.82 = 172.586 KN

#SAFETY FACTOR = 8

Design for 1035.516 * 8 = 8284.128 KN

δ = F / A ,,,, F = δ * A

F=700*10^6 * π/4 * 0.061^2

F= 140.743 KN

-The force that can be carried by one cable.

The Winch

Efficiency = P(out) / P(in) X 100 %

70/100 = P(out) / P(in)

τ (Torque) = P(out) / w

τ = T(max) * r

τ = (1035516.328 * 8 ) * 0.25

τ = 2071032.656 Nm

P(out) = τ x w

P(out) = 2071032.656 x 20 = 41.421KW

P(in) = P(out) x 100 / 70

P(in) = 41.421x 100 / 70 = 59.173 KW

The cost of the Winch

our power input is about 60 KW , so we can find in the table that

we can use W80 which costs $17,150

The cost of the steel ropes

#Number of ropes need = 6

R-2 cost is $6 / m . We need to use 10 meters for one rope. And the

minimum length of the steel rope that must remain wound on the

drum is 3 full wraps around the drum. Also we can remember that the

radius of the drum is 0.25 m so we can figure out this.

10 + 2 π r * 3

10 + 2 π * 0.25 * 3 = 14.712 m (one rope)

And for 6 ropes

14.712 X 6 = 88.272 m

So the cost = 88.272 x 6 = $529.632

The cost of the steel cage

Total surface area of the mesh = w^2 + (2.2 x w) x 4

Total surface area of the mesh = w^2 + 8.8 x w

Total surface area of the mesh = (6.124)^2+(8.8 x 6.124)=91.395 m^2

Then from the table above we can see that it costs $200 for 1 m^2

So for the cage it will cost $200 x 91.395 = $18279

# in conclusion

The goals of this design are to maximise the rate at which

workers can be transported. In addition we don’t exceed the

budget of about $40000 that we can see with this project

design we just spent less than $36000 by offering a minimum

cost solutions that meets the requirements that we need to do

this project.

Daeej Ali

U1054594

ENG 1002 Sep. 2013