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The Proposed Arrangement The Proposed Arrangement L1 L2 4.0m 1.0m 6 kN 10 kN A B E RHS CHS D Member DE (RHS) Mild Steel Pin (fy= 250 MPa) 5mm thick plate Tension Rod with 10 kN load through C...

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The Proposed A
angement
The Proposed A
angement
L1
L2
4.0m
1.0m
6 kN
10 kN
A
B
E
RHS
CHS
D
Member
DE (RHS)
Mild Steel Pin (fy=
250 MPa)
5mm thick
plate
Tension Rod
with 10 kN
load through
C


University of Ballarat
University of Ballarat

School of Science & Engineering

ENCOR2020 FUNDAMENTALS OF ENGINEERING (SOLID MECHANICS)


Design Exercise - Design of a Load Support System for Strength and Specific
Deformation Characteristics


The Scenario

Suppose you have been approached by the manager of a factory, who requires the design of an
unusual piece of equipment which he has devised to perform a specific function within his
operations.

The schematic diagram of the proposed system is shown overleaf. A circular hollow section
(CHS) is to span between supports A and B. The supports are essentially simple supports, except
that at A the CHS is constrained so that it is unable to twist about its longitudinal axis at this
point; i.e. it is torsionally restrained at this point. The length of this member is set by other
considerations and is shown on the diagram. This beam supports, and is rigidly welded to, a
cross-member D-E, which is proposed to be a rectangular hollow section (RHS), bent about its
weaker (y-y) axis in order to avoid possible lateral buckling. Member D-E is perpendicular to
member A-B and the location of the connection point C along member A-B has also been
predetermined and is shown on the diagram.
At ends D and E of the RHS, vertical downwards loads are applied via tension rods. The
proposed a
angement for the connection of the rods to the RHS is also shown. The magnitudes
of the loads at D and E have been determined and are shown, but their positions are flexible at
this stage. That is, lengths C-D and C-E are yet to be determined.
Apart for the need to design this load support system so that it possesses adequate strength, there
is another key requirement which must be satisfied if the system is to perform its required
function: the deflection of point D under the action of the loads shown must be approximately
zero.
The Proposed A
angement
See Appendix 1.

Requirement for this Exercise
The design requirements of this exercise are as follows:
a) Design the two main components of this system, (i.e. the CHS and the RHS) in order to
satisfy strength and stiffness requirements. For each component, this will require selection of a
suitable steel hollow section (section property tables are attached) and determination of the
equired lengths L1 and L2. For an efficient design, member sizes should not be excessive; that
is, where possible, member stresses should be close to their maximum allowable under the
design loading case shown.
) The proposed connection detail between the tension rods and the RHS at D and E was shown
on the previous page. Determine the minimum permissible pin diameter which could be used
here, based upon shear strength, and then go ahead and check whether the bearing stresses in the
connection are acceptable.
c) CHS and RHS dimensions and properties from AS XXXXXXXXXXare in Appendix 2. Please note
that the symbols in the AS may not be the same as in the textbook, for example, elastic section
modulus is Z in AS, while it is S in the textbook, respectively.
Design Data
i) Steel Yield Stresses
 The CHS and RHS members are cold-formed, Grade 350 steel (σy = 350 MPa)
 The plates and pins used to connect the tension rods to the RHS at D and E are Grade 250
mild steel (σy = 250 MPa)
ii) Allowable Stresses
 Allowable bending stress in the bending members = 0.45 σy
 Allowable torsional shear stress in the twisting member = 0.25 σy
 Allowable average shear stress in the pin at connection C = 0.3 σy
 Allowable bearing stress in the connection at C = σy
iii) Other Data
 E = 200 GPa and
 G = 80 GPa for steel


Guidelines for the Design
• As already mentioned, a key criterion in the design of this system is that the net deflection of
point D should be zero under the loading case shown. The deflections of points D and E are the
net result of three types of deformation:
o vertical deflection of C due to bending deformation of member A-B
o torsional deformation (i.e. twist) of member A-B
o bending deformation of the cantilevers C-D and C-E.
It is possible to devise a combination of values for L1, L2 and CHS and RHS sizes which will
achieve this outcome.
• A suggested approach to the design of this system is as follows:
o determine a suitable size for the CHS member A-B, based on bending strength (that
is, so that under the loading shown the maximum bending stress in the member is less
than the specified allowable bending stress, though preferably by a small margin, for
an efficient design)
o based on this CHS size, determine the resulting vertical deflection of the member at C
o assuming this member is also stressed to its maximum allowable shear stress (due to
torsion) due to this loading case, it will also be possible to determine the maximum
angular twist that can occur at C
o now, probably using trial and e
or, it will be possible to come up with a suitable
combination of L1/L2/CHS size/RHS size which will achieve δD = 0. Note well that
the RHS is to be a
anged so that it is bending about its weak axis; thus, its IYY and
zYY properties will be the important ones for determining deflection and bending
stresses (the next step)
o the final step is to check the bending stresses in the RHS. If these are found to be
excessive you have two options. Firstly, you could continue with a trial and e
or
approach until both deflection and stress requirements are satisfied. However, you
may find that the variables involved are so inter-related that this is hard to achieve.
An acceptable alternative approach would be to weld steel plates to the top and
ottom of the RHS in the vicinity of the high stresses in order to strengthen the
member. If this approach is taken, then note the following:
 a suitable plate thickness must be specified, within the range 3 – 10mm
(rounded to an even mm)
 assume the plate width = 30mm less than the width of the RHS, to allow room
for welding and the rounded corners on the RHS
• the final achieved maximum bending stress in the strengthened section must
e calculated and shown to be acceptable, based on the specified allowable
ending stress of 0.45 fy. Note, however, that the steel plates will be of a yield
stress of fy = 250 MPa, not the 350 MPa used in the CHS and RHS sections. This
will not affect your calculation of the stresses (both types of steel have the same E
value), but the allowable stress will obviously be less.
• The relationship between the torque on the CHS and the torsional shear stress involves the
polar moment of inertia Ip. This is identified as the Torsion Constant J in the ‘Dimensions and
Properties’ table for the circular hollow sections. (Note: The torsion constant is equal to the polar
moment of inertia only for shapes which cannot warp. A circular tube has a relatively high
torsion constant because it doesn’t warp under torsion. However, if a saw cut is made through
the tube wall the torsion constant is drastically reduced because the cross section can change
shape under very small torsion loads. Thus two shapes with very similar geometric properties
can have substantially different torsion constants.)


Submission
A fully detailed set of calculations is to be submitted, together with a summary of the final
design decisions; i.e. lengths and sizes of members, and diameter of pin.
Note the following:
• If a trial and e
or approach is adopted, all trials and their outcomes must be shown
• A trial and e
or approach may be presented in tabular form if desired. However, a
complete set of fully detailed manual computations must be shown for the final adopted
design.
Assessment will be based upon:
• Technical co
ectness;
• Neatness of presentation;
• The extent to which the calculations are logically and clearly set out, with clear but
concise explanations, neat diagrams, etc.

Due

Thursday 31th May 2012

    Design Exercise - Design of a Load Support System for Strength and Specific Deformation Characteristics
    The Scenario
    Requirement for this Exercise
    Design Data
    Guidelines for the Design
    Submission
    Due
Answered Same Day Dec 20, 2021

Solution

Robert answered on Dec 20 2021
125 Votes
Microsoft Word - Solution
Let’s assume a suitable RHS and CHS have been chosen and their following properties are known to us:
(i) Properties of CHS:-
ICHS = Second moment of area (mm4)
ZpCHS = Plastic Section Modulus (mm3)
JCHS = Torsion Constant (Polar moment of inertia) (mm4)
ZtCHS = Torsion Modulus (mm3)
(ii) Properties of RHS:-
Iyy = Second moment of area about minor axis (mm4)
Zyy = Plastic Section Modulus about minor axis (mm3)
(iii) Common Properties:-
E = 200 * 103 N/mm2.
G = 80 * 103 N/mm2.
Now, we have a design requirement which says that vertical deflection of the point D should approx.
0.Let’s find out the vertical deflection of the point D using virtual work method.
The proposed a
angement is known to us. Clearly this structural system has Degree of Static
Indeterminacy equal to 0. Now let’s first see the F.B.D. of RHS DE.


(6*L1 ‐ 10*L2) kN.m 
10 kN 
6 kN 
16 kN 
Free Body Diagram of DE 
10*L2 


E Bending Moment Diagram of DE 
6*L1 
Now, In order to find the vertical deflection of point D due to the loading, apply a unit virtual load
downwards at D. Here we are using virtual work method to find the deflection.
12.8 kN 


(6*L1 – 10*L2) kN.m 
(6*L1 – 10*L2) kN.m 
B 16 kN 
3.2 kN 
Free Body Diagram of AB 
12.8 kN.m 
A  B C 
Bending Moment Diagram of AB
(6*L1 – 10*L2) kN.m 
A  C 

Torsion Moment Diagram of AB
Due to this unit virtual load the bending moment and torsion diagrams will be as below:-
Free Body Diagram of DE 
E

L1  




L1  
L1  
B 1 
0.2  
Free Body Diagram of AB 
0.8  

L1 

E Bending Moment Diagram of DE  
0.8 
A ...
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