engineering design assignment
TRANSCRIPT
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.