Engineering Design Assignment

Engineering Design – Dr Frame Design Task

December 14, 2011

Kerrie Noble Engineering Design Assignment 1 14/12/2011

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Contents Executive Summary....................................................................................................................... 2

Introduction ................................................................................................................................. 2

My Frame Design ......................................................................................................................... 3

Design for Buckling ...................................................................................................................... 5

Design for Bending ....................................................................................................................... 7

Design for Welding ....................................................................................................................... 7

Conventional Stress Analysis ..................................................................................................... 8

Weld as a Line Method ............................................................................................................. 8

Design for Fastening .................................................................................................................... 9

Discussion.................................................................................................................................. 10

Dr Frame 3D; ......................................................................................................................... 10

Design for Buckling; ............................................................................................................... 10

Design for Welding;................................................................................................................ 12

Design for Fastening; .............................................................................................................. 12

Conclusion ................................................................................................................................. 12


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Executive Summary

This report outlines the Dr Frame design task which was given. The frame design was to have a 1m

clearance around an object on all sides, support a load of 45kN which was to be placed on a platform

2m above the object. My design was developed and refined on the Dr Frame 3D software which then

enabled me to investigate the frame for different design aspects. The design aspects investigated were

design for buckling, design for bending, design for welding and design for fastening.

The design for buckling and design for bending calculations were to illustrate that this frame design

had a factor of safety of between 1 and 2 for both areas. The factor of safety calculated for these areas

were 11.08 and 1.95 respectively. The reasoning behind the design for buckling factor of safety being

greater than 2 in discussed at the end of the report.

The design for welding calculations showed that a weld of 10mm thickness is required for this design.

Similarly the design for fastening calculations showed a steel bolt of 10mm in diameter was the

minimum required for the detailed fastening joint.

Some of the initial values from calculations were suggested minimum values, therefore at the end of

the report I have suggested some improvements to the design. However the design shown in the

diagrams is suitable for the task given and has been shown that failure will not occur in any of the

engineering design areas outlined in the calculations provided.

Introduction

The task given for this assignment was to use Dr Frame 3D to design a frame which allowed a load to

be supported on a flat platform 2m above an object. The frame was to have 1m clearance around the

object on all sides but the frame must not be wider than the object. The dimensions used to drive my

frame design were A = 2m, B = 5m and the frame must not be wider than 2m. The load to be

supported by the flat platform was 45kN. The diagram below was given to help derive the required

dimensions for the frame design.

The dimensions for my frame design were therefore derived as;

R-R = at least 4m


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B + 2 = at least 7m

A – 1 = 1m

L = 45kN

My Frame Design

The diagrams below illustrate the overall shape of my chosen frame design. The brief stated that

there must be at least 1m clearance around the box, to ensure this was the case I have made the

measurements of the components of the frame larger than the minimum necessary. This was to allow

for the angles used in design, I felt that if I used the minimum dimension of 1m and then included a

design with some steep slopes then the 1m clearance zone would be impinged on and my design

therefore would not have followed the brief.

Diagram 1 (above) shows an annotated front view of my frame design. The length of some key

members has been shown and the 45kN force is acting on the flat platform within the design.

Diagram 2 (above) illustrates the length of members of the frame in a side view.


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Diagram 3 (above) illustrates the dimensions of the flat platform on which the load of 45kN rests and

the length of the members supporting this platform.

Diagram 4 (below) is a scaled view of the deformation which occurs when the 45kN force is placed

on the platform at the top of the frame design.


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Diagram 5 (above) illustrates the beams with the strangest and weakest axis shear. This is the shear

force which occurs due to the load. The highest value will be used for design calculations for

welding, fastening, buckling and bending.

Diagram 6 (above) depicts the beams with the highest bending moment. The value to be used in

calculating the bending stress in the design is 5 kN.

List of standard sections used;

Rectangular Tube – 3 x 3 x 5/16

Standard Pipe – 4inch

Standard W 6 x 25

Standard Pipe – 5inch

Design for Buckling

Section – rectangular tube 3 x 3 x 5/16 (inches)

75mm x 75mm x 7.8125mm


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SY – 248.2 MPa

SS –

E – 199.9 GPa

A – 2.01 x 10-3

I – 1.49 x 10-6

The member being analysed for buckling is the member which is highlighted in diagram 5.

The member is fixed at both ends so Leff from the table = 0.65L

L = 1.1

0.65 x 1.1 = 0.715

ρ = √𝐼

𝐴 = √

1.49 𝑋 10−6

2.01 𝑋 10−3 = 0.02723

Slenderness Ratio = 𝐿𝑒𝑓𝑓

𝜌 =

0.715

0.02723 = 26.261

Checking Johnston Critereon;

𝜋√2𝐸

𝑆𝑌 = 𝜋√

2 𝑋 199.9 𝑋 109

248.2 𝑋 106 = 126.087

Since 𝐿𝑒𝑓𝑓

𝜌 < 𝜋√

2𝐸

𝑆𝑌 the member therefore acts as a Johnston column. To find the critical load the

Johnston column buckling equation is used;

𝑃𝐶𝑅

𝐴= 𝑆𝑌 −

𝑆𝑌2

4𝜋2𝐸(

𝐿𝑒𝑓𝑓

𝜌)

2

𝑃𝐶𝑅

𝐴= 248.2 𝑥 106 −

(248.2 𝑥 106)2

4 𝑥 𝜋2 𝑥 199.9 𝑥 109(

0.715

0.02723)

2

𝑃𝐶𝑅

𝐴= (248.2 𝑥 106) − 74.2093(26.261) 2

𝑃𝐶𝑅

𝐴= 248148822.3

PCR = 498779.13

Factor of Safety = 𝑃𝐶𝑅

𝑃 =

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐿𝑜𝑎𝑑

𝐿𝑜𝑎𝑑 𝑝𝑙𝑎𝑐𝑒𝑑 𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟 =

498779.13

45000 = 11.08

The reasons for this factor of safety not falling between 1 and 2 will be talked about in the discussion

section of this report.


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Design for Bending

Using the Engineer’s equation for bending;

𝜎 = 𝑀𝑦

𝐼

The bending stress placed on the member (as highlighted in diagram 6) can be determined.

M = maximum bending moment = 5kN

Y = perpendicular distance to neutral axis = 0.038m

I = 1.49 x 10-6 m4

𝜎 = (5 𝑥 103)𝑥 0.038

1.49 𝑥 10−6

= 127.52 x 106

Factor of Safety = 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 =

248.2 𝑥 106

127.52 𝑥 106 = 1.95

Design for Welding

The member I have chosen to design for welding calculations is the central bar of the diamond shape

which forms the flat platform on which the 45kN load is placed. (As shown in diagram 3 above).

The diagram included above shows the schematic of the beam which I have chosen to analyse for the

design for welding section of this task, (chosen beam is the shaded section of the schematic). The

schematic also includes some calculations which were necessary to establish the length of the weld

needed.

I have chosen to use an E43 electrode of S355 grade steel

Tensile yield stress = 250 N/mm2


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The shear stress yield value required is therefore approximately;

0.75 x 250 = 187.5 N/mm2

cos𝜃 = 0.5

1.1

𝜃 = cos−10.5

1.1

𝜃 = 62.96˚

sin 62.96˚ = 3.75

𝑥

𝑥 = 3.75

sin 62.96˚

𝑥 = 4.21 𝑐𝑚 = 42.1 𝑚𝑚

Weld length = 42.1 + 42.1 + 42.1 + 42.1 + 75 + 75 = 318.4 mm

Conventional Stress Analysis

Shear stress = 22500N

Factor of safety = 2

F = 22500 x 2 = 45000N

𝜎𝑠 = 𝐹

𝑡𝐿

187.5 = 45000

𝑡 𝑥 318.4

t = 0.754 mm

𝑠 = 𝑡

cos45

𝑠 = 0.754

0.707

𝑠 = 1.07 𝑚𝑚

Using the conventional stress analysis suggests using a weld thickness of 2 mm.

Weld as a Line Method

𝜎 = 𝐹

𝐿

𝜎 = 45 000

318.4


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𝜎 = 141.33 𝑁/𝑚𝑚2

𝑡 = 𝜎

𝜎𝑠

𝑡 = 141.33

187.5

𝑡 = 0.754 𝑁/𝑚𝑚2

𝑠 = 𝑡

cos45

𝑠 = 0.754

0.707

𝑠 = 1.07 𝑚𝑚

Using the weld as a line method also suggests using a weld thickness of 2 mm.

For this type of weld a minimum value of s = 10 is needed so since the s value is below this then a

weld of thickness 10mm will be used for the weld between these members.

Design for Fastening

The diagram below shows a schematic of my proposed bolted joint design between two members.

The two members I have chosen to design this for are the two members which create the point of the

diamond-shaped platform on which the 45kN force is place, as shown in diagram 3 at the beginning

of the report.

Beam 1 shear force = 25200 N

Beam 2 shear force = 25200 N

Total shear force acting on the fastened joint = 50400 N

Factor of Safety of 2 so total shear force acting = 1008000 N


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I have chosen to use a steel 12.9 bolt in this fastening design.

Number of bolts = 12

𝜏 = 1008000 N

𝜎𝑦 = 1200 𝑀𝑃𝑎

𝐷2

4=

𝜏

𝑁 × 𝜋 × 1200

𝐷2

4=

1008000

12 × 𝜋 × 1200

𝐷2

4= 22.28169203 𝑚𝑚

𝐷2 = 89.12676813 𝑚𝑚

𝐷 = 9.44 𝑚𝑚

𝐷 = 10 𝑚𝑚

Discussion

Dr Frame 3D; Getting used to and understanding the Dr Frame programme was my first initial

hurdle. I found it took a few attempts at using the software to feel comfortable with its use. When I

had improved my skills on Dr Frame and established how to changed dimensions, section types, build

a structure from an empty window and discovered where to find the necessary mechanical properties

for the beams in question I felt more competent and able to attempt and successfully complete the

coursework element with the aid of Dr Frame.

Design for Buckling; The coursework brief asked to demonstrate the beam that has the highest axial,

compressive load had a design factor of safety of two. As the working above shows my design factor

of safety was 11.08 which is almost six times higher than what was stated. This is due to a problem

with the design factor of safety of between one and two for bending. When the factor of safety was

between one and two for buckling this had a dramatic consequence on the beam’s performance when

considering bending. This consequence is shown in the working below.

The section being used has the following values;

Section – standard pipe – 1 inch

A = 3.1863 x 10-4

I = 3.6549 x 10-8

Strong axis shear = 25.2 kN

Maximum bending moment = 5 kNm

The member is fixed at both ends so Leff from the table = 0.65L


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L = 1.1

0.65 x 1.1 = 0.715

ρ = √𝐼

𝐴 = √

3.6549 𝑥 10−8

3.1863 𝑥 10−4 = 0.01071

Slenderness Ratio = 𝐿𝑒𝑓𝑓

𝜌 =

0.715

0.01071 =66.7593

Checking Johnston Critereon;

𝜋√2𝐸

𝑆𝑌 = 𝜋√

2 𝑋 199.9 𝑋 109

248.2 𝑋 106 = 126.087

Since 𝐿𝑒𝑓𝑓

𝜌 < 𝜋√

2𝐸

𝑆𝑌 the member therefore acts as a Johnston column. To find the critical load the

Johnston column buckling equation is used;

𝑃𝐶𝑅

𝐴= 𝑆𝑌 −

𝑆𝑌2

4𝜋2𝐸(

𝐿𝑒𝑓𝑓

𝜌)

2

𝑃𝐶𝑅

𝐴= 248.2 𝑥 106 −

(248.2 𝑥 106)2

4 𝑥 𝜋2 𝑥 199.9 𝑥 109(

0.715

0.01071)

2

𝑃𝐶𝑅

𝐴= (248.2 𝑥 106) − 74.2093(66.7593) 2

𝑃𝐶𝑅

𝐴= 247869263.9

PCR = 78978.58

Factor of Safety = 𝑃𝐶𝑅

𝑃 =

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐿𝑜𝑎𝑑

𝐿𝑜𝑎𝑑 𝑝𝑙𝑎𝑐𝑒𝑑 𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟 =

78978.58

45000 = 1.76

The factor of safety for buckling using this beam has therefore been shown to be between one and

two, however when this beam is used within the engineer’s bending formula the effect this has is

shown below.

𝜎 = 𝑀𝑦

𝐼

𝜎 = (5 𝑥 103)𝑥 0.0127

3.6549 𝑥 10−8

= 173.74 x 107

Factor of Safety = 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 =

248.2 𝑥 106

173.74 𝑥 107 = 0.1429

This value shows that the beam, although it is suitable for buckling, fails under bending conditions.

To solve this problem with the design I made bending the determining factor and chose a suitable

section to satisfy the factor of safety requirements for bending given in the design brief for the


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coursework. This therefore meant the factor of safety gained for the buckling calculations would be

greater than stated but the beam would then be suitable for both buckling and bending and failing

would not occur under any of these conditions. This resulted in a better design, which would avoid

failure and would also be suitable for the task which was given.

Design for Welding; For the design for welding section of the coursework brief I had originally

chosen to use the conventional stress analysis and weld as a line methods to calculate the minimum

weld size needed for the frame design. The beam I had chosen to look at was the middle bar in the

platform section of the design. After doing this I realised that these methods are primarily used for

fillet welds and that the design primarily involved a butt weld. As part of my discussion I have

therefore included an extra calculation to ensure that the size of weld I suggested in the above

calculations was suitable for the design I have outlined.

𝑓𝑣 = 𝑃

0.707 𝑎 𝐿𝑤

As in the calculations above;

P = 45000 N

Fv = 45000 N

Lw = 0.3184 m

𝑎 = 𝑃

0.707 𝑓𝑣 𝐿𝑤

𝑎 = 45000

0.707 𝑥 45000 𝑥 0.3184

𝑎 = 4.44 𝑚

After completion of this calculation I came to the conclusion that the weld in my design was clearly

not a butt weld due to the inappropriately large number achieved at the end of the calculation and that

my earlier calculations were therefore correct. The thickness value for the weld should therefore

equal 10mm as stated above.

Design for Fastening; I simply used the equation to find the minimum diameter of the bolts required

for my design. The answer I got through this calculation was 10mm however I feel that this was a

minimum value and should require some factor of safety to be considered within the design.

Although the diameter stated for use in the report above was 10mm I would use a 12mm bolt to allow

for a factor of safety.

Conclusion

As a design, the frame illustrated in the report is suitable for the task given. It has a clearance of at

least 1m around all sides of the obstacle, the platform, on which the 45kN load is supported, is 1m

wide and is 2m above the obstacle. The frame design therefore has fulfilled the brief given at the

beginning of the task.


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After analysis from buckling, bending, welding and fastening calculations it is clear that the frame

will not fail in any of these areas due to acceptable safety factors which have been shown in the

design. As an extra safety factor I would suggest using a bolt of 12mm diameter for the designed

fastening joint. The value which was obtained from the calculation was a minimum suggested value

and I would therefore use a larger bolt to add an extra safety factor within the fastened joint to prevent

the opportunity for failure to occur.

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Engineering Design Assignment