Vibrating Screen Support Design
Introduction
A sugar refining mill requires the design of a vibrating screen supported by a welded
universal beam column which is subjected to acceleration due to an inertial force. The loads
exerted on the column must be determined as well as material and section properties. In
particular, loadings exerted onto the weld outline sections require great emphasis with
iterations where necessary. A free body diagram for each bracket in the x-y & x-z planes will
be constructed to determine the maximum bending moment and other crucial loadings.
Material for the beam must be chosen from a catalogue which conversely leads to the
determination of yield stress and ultimate stress. An engineering drawing will be produced
based on the specifications.
Design Brief
Figure 1: Side view of the screen unit layout as provided in the design brief.
An elevated vibrating screen unit has the purpose of grading sugar into commercial sizes. The
screen will be modeled as a simply supported beam with the centroid of the weight acting as
a point force. This can be treated as a static problem as the screen supported by two pairs of
brackets that are welded to a column of undisclosed length which serve as vibration isolation
mountings and as support reactions.
Figure 2: Possible bracket detail. The primary assembly of the welding analysis.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Bracket B is perfectly perpendicular to the column which will be modeled as a simple
cantilever beam with an induced bending moment due to the screen’s loading. Bracket A is
similar but in which is offset which will require an extra set of calculations compared to
Bracket B. The brackets are the primary focus for this analysis as the welding calculations are
performed after the static analysis from the screen to the bracket.
Figure 3: A sketch of the plan. The dimensions provided serve as necessary parameters for
this analysis.
However this problem is not entirely static as an acceleration of 0.15g induced by an inertial
force which can move in either direction will be present, adding another set of force
components to make the analysis more difficult. The universal column is provided however
the designer is to select an appropriate beam cross-section as the brackets based on a number
of criteria. An engineering drawing will be provided to formalise the details.
Calculations
Factors of Safety
Bending is one of the significant loads present on the bracket. Stresses Permitted by the
AISC Code for Weld Metal the weld factor of safetyi is determined to be n = 1.67 (or n = 5/3
if a rational expression is desired).
With readily determinable knowledge about the loading on the beams the factor of safety will
also be taken to be n = 1.67. This will be the universal factor of safety in this design.
Free-Body Diagrams and maximum bending moments
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Calculations will be shown below with the free body diagrams of the screen which exerts
loads onto the brackets as reaction forces. Those same reaction forces will permit the
calculation of the maximum bending moment. Since the screen undergoes an acceleration of
0.15g in either direction it will be treated as an inertial or ‘shaking’ force.
The length of each bracket taken without offset will be 442.5 mm based on the dimensions
set in the problem definition. Note that due to the offset the length of bracket A will not be
the same as bracket B.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
3
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of screen
Page: 4
Calculations
Results
Free body diagram configuration 1
Free body diagram configuration 2
The screen is symmetric, need to analyse one plane only. When analysing the
screen as a 2D problem, only consider two of four supports and half the
weight.
Inertial force:
∑ 𝐹𝑥 = 𝑚𝑎
Finertial = ±10,290 N
RH = Finertial = ±0.15mg = ±10,290 N
Note that the inertial force can be in either direction.
∑ 𝐹𝑦 = 0
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
4
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of screen
Page: 5
Calculations
𝑅𝑎 + 𝑅𝑏 =
𝑚𝑔
4
∑ 𝑀𝐴 = 0
𝑅𝑏 𝐿 =
𝑅𝑉 = 𝑅𝑎 = 𝑅𝑏 =
𝑚𝑔 𝐿
2 2
𝑚𝑔 (7000)(9.8)
=
= 17,150 𝑁
4
4
Results
Vertical reaction
force at each end:
RV = 17,150 N
Horizontal reaction
force due to inertial
force
RH = ±10,290 N
Shear force diagram
Mmax = 23,152.50
N.m
Bending moment diagram
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
5
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of bracket A in x-y plane
Page: 6
Calculations
Results
Angle between offset length and length of beam:
𝜃 = tan−1 (
300
) = 34.14°
442.5
LA = √(300)2 + (442.5)2 = 534.609 mm
∑ 𝐹𝑥 = 0
∑ 𝐹𝑦 = 0
𝑅𝑉 = 17150𝑁
(One vertical reaction and one applied point force)
∑ 𝑀𝑂 = 0
Mxy_max = RV × LA = (−17150)(534.609) = −9168535.803 N.mm
= −9168.536 N.m
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Shear force and bending moment diagrams of bracket A in x-y
plane
Page: 7
Calculations
Results
Maximum bending
moment of Bracket
A in x-y plane:
MA_xy
= −9168.536 N.m
Shear force diagram
Bending moment
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
7
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of bracket A in x-z plane
Page: 8
Calculations
Results
Using the same length LA as from previous calculations
∑ 𝐹𝑥 = 0
∑ 𝐹𝑧 = 0
𝑅𝐻 = 10290𝑁
(One horizontal reaction and one applied point force)
The inertial force can be applied in either direction of equal magnitude.
Consider one case only:
∑ 𝑀𝑂 = 0
Mxz_max = RH × LA
= (−10290)(534.609)
= −5501121.482 N.mm
= −5501.121 N.m
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
8
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Shear force and bending moment diagrams of bracket A in x-z
plane
Page: 9
Calculations
Results
Maximum bending
moment of Bracket
A in x-z plane:
MA_xz =
−5501.121 N.m
Shear force diagram
Bending moment
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
9
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of bracket B in x-y plane
Page: 10
Calculations
Results
∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0
𝑅𝑉 = 17150𝑁
(One vertical reaction and one applied point force)
∑ 𝑀𝑂 = 0
Mxy_max = RV × L = (−17150)(442.5) = −7588875 N.mm = −7588.875 N.m
Maximum bending
moment of Bracket
B in x-y plane:
MB_xy =
Shear force diagram
−7588.875 N.m
Bending moment
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
10
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Free body diagram of bracket B in x-z plane
Page: 11
Calculations
Results
∑ 𝐹𝑥 = 0
∑ 𝐹𝑧 = 0
𝑅𝐻 = 10290𝑁
(One horizontal reaction and one applied point force)
∑ 𝑀𝑂 = 0
Mxz_max = RH × L = (−10290)(442.5) = −4553325 N.mm = −4553.325 N.m
Maximum bending
moment of Bracket
B in x-z plane:
MB_xz =
−4553.325 N.m
Shear force diagram
Bending moment diagram
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Material for beam
Grade AS/NZS 3679.1-350 steel is a structural steel. Yield strength of this particular steel
may vary from 340 to 360 MPa for our application depending on the thickness of up to and
not including 40mm.
Allowable Stress
Based on the AS1554.1-2000 Australian Standard, the weld is assumed to have the same
strength of the parent steel. Grade 350 steel is chosen since it is available in the Onesteel
catalogue with an ultimate tensile strength of Su = 480 MPa.
The AWS Electrode number E70xx with a tensile strength of Sut = 482 MPa, yield strength of
Sy = 393 MPa and 22% elongation will be used for welding. Calculations will be shown
below:
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: Allowable stresses and required section modulus
Page: 13
Calculations
Results
Grade 350 steel has an ultimate strength of Su = 480 MPa
Allowable normal stress:
2𝑆𝑢 2(480)
=
= 192 𝑀𝑃𝑎
𝑛
3(1.67)
𝜎𝑛𝑜𝑟𝑚𝑎𝑙 =
Allowable shear stress:
𝜏𝑎𝑙𝑙𝑜𝑤 =
(480)
𝑆𝑢
=
= 96 𝑀𝑃𝑎
3𝑛 3(1.67)
AS1554.1-2000
Australian Standard
Allowable normal
stress:
σnormal = 192 MPa
Definition of section modulus:
𝑍=
𝐼
𝑐
Allowable bending stress:
𝜎𝑎𝑙𝑙𝑜𝑤 =
𝑀𝑐 𝑀
=
𝐼
𝑍
Allowable stress based on factor of safety of yield stress:
𝜎𝑎𝑙𝑙𝑜𝑤 =
𝑆𝑦 393
=
= 235.3 𝑀𝑃𝑎
𝑛
1.67
Using largest value of maximum bending moment acting on beam in x-y
plane where M = 9,168,535.803 N.mm, the required section modulus:
𝑍=
𝑀
𝜎𝑎𝑙𝑙𝑜𝑤
=
(9168535.803 N. mm)
= 38960 𝑚𝑚3
(235.8 MPa)
Allowable shear
stress:
τall = 96 MPa
Limiting stress based
on yield stress:
σallow = 235.3 MPa
Required section
modulus:
Z = 38.960 × 10-6 m3
𝑍 = 38.960 × 10−6 𝑚3
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Column and beam sections
Based on the design brief the required size of columns is chosen from the Hot Rolled and
Steel Structural Steel Products Third Edition catalogue provided by onesteel market mills.
Figure 4: Diagram showing diagram of the cross section of the universal column.
Data
Variable
Value
Size of Columns
310 UC 118
Flange Width
bf
307
Web Thickness
tw
11.9
Root Radius
r1
16.5
Depth of Section
d
315
Section Modulus
Zx
1.76×10-3
(x-axis)
Section Modulus
Zy
588×10-6
(y-axis)
Table 1: Specifications of the 310 UC 118 universal column.
Units
mm
mm
mm
mm
m3
m3
The brackets are to be welded to the 310 UC 118 universal columns and a flat plate 20 mm
thick by 200 mm square is required for the foot mounting. As a result, brackets are restricted
to a width between 200 mm to 307 mm.
Based on the required section modulus and a compromise between weight, moment of inertia,
section modulus and yield stresses, the 530 UB 82 universal beams are chosen.
Data
Size of Beams
Weight per unit
length
Flange Width
Web Thickness
Flange Thickness
Root Radius
Depth of Section
Variable
bf
tw
tf
r1
d
Value
530 UB 82
82
kg.m-1
209
9.6
13.2
14
528
mm
mm
mm
mm
mm
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
Units
14
Depth Between
d1
502
Flanges
Section Modulus
Zx
1.81×10-3
(x-axis)
Section Modulus
Zy
193×10-6
(y-axis)
Yield Stress at flange
σy(f)
300
Yield Stress at web
σy(w)
320
Table 2: Specifications of the 530 UC 82 universal beam.
mm
m3
m3
MPa
MPa
Weld Outline Shape
The welds will be mostly done across the top and bottom section as the sections furthest from
the centre of the beam contribute to the moment of inertia. The shear stress distribution across
the beam varies in a parabolic fashion from zero at the edges to maximum at the centre. The
bending stress distribution varies linearly from zero at the centre to the maximum at both
edges at maximum tension or compression at either end. Shear stress will be taken simply as
the shear force over the total area of the cross-section of the weld.
Type A –
Figure 5: Diagram showing the shear and bending stress distribution along the cross-section
of the Type A beam in one direction.
Since the Type A bracket length is not directly perpendicular to the plane of the wall there
will be torsion present. Torsion acting on the beam in addition to the shear force contributes
to the shear stress acting on the beam in addition to the bending stress.
Type B –
Figure 6: Diagram showing the shear and bending stress distribution along the cross-section
of the Type B beam in one direction.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Since there is no torsion acting on the beam the shear stress contribution is much less and
bending stress is much more dominant.
Figure 7: Sketch of the proposed weld pattern for the Type A and type B beams.
The weld pattern is applied furthest from the centre. The ease of the weld geometry justifies
this choice in addition to the fact the contribution to the second moment of area increases
significantly with distance from the centroid of the cross-section. This weld pattern is readily
available from tables of torsional and bending properties of fillet weldsii.
Category
Weld
Throat Area
(A)
2
1.414hd
3
1.414hd
Unit Second
Moment of
Area
(Iu)
𝑑3
6
Unit Second Polar
Moment of Area
(Ju)
𝑑(3𝑏 2 + 𝑑2 )
6
𝑏𝑑 2
2
Weld calculations
The calculations for the forces acting on the welds per unit length are show below. The
substitution of values and subsequent calculations will be performed using an external
program such as Microsoft Excel or MATLAB and only results and formulas with relatively
simple expressions will be shown. The calculations provided below are for the initial iteration
even will result in the same value after several iterations.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Since there are horizontal and vertical forces acting on each bracket different formulas will be
used, with bending stress as an example, which is given byiii:
𝑓𝑏 = −
𝑀𝑥 𝑦 𝑀𝑦 𝑥
+
𝐼𝑥
𝐼𝑦
This formula for unsymmetrical bending must be used to superimpose the resultant moments
acting on the beam.
Figure 8: Extreme points on the I-beam.
Due to the loadings present in the horizontal and vertical directions due to the weight and the
inertial force respectively as well as torsion, the stress distribution may not be symmetrical
across a chosen axis.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 18
Calculations
Results
The sign for the inertial force does not change the final result.
Fx = Vx = 10290 N
Fy = Vy = −17150 N
MA_x = −9168.536 N.m
MA_y = ±5501.121 N.m
MB_x = −7588.875 N.m
MB_y = ±4553.325 N.m
r = 300 mm
T = Fyr = (17150)(300) = 5145000 = 5145 N.m
The area of throat weld is taken where the initial weld length is at unity:
Au_x = 1.414hd = 746.592 mm2 = 746.592×10-6 m2
Iu_x =
Au_y = 1.414hb = 295.526 mm2 = 295.526×10-6 m2
29.1329 × 10-3 m3
Unit Second Moment of Area of x-axis
𝐼𝑢_𝑥 =
𝑏𝑑2
2
−3
= 29132928 𝑚𝑚3 = 29.1329 × 10
𝑚3
Iu_y =
24.5330 × 10-3 m3
Unit Second moment of Area of y-axis
𝐼𝑢_𝑦 =
𝑑3
= 24532992 𝑚𝑚3 = 24.5330 × 10−3 𝑚3
6
Unit Polar Second Moment of Area
Ju =
36.0648 × 10-3 m3
𝑑(3𝑏 2 + 𝑑2 )
𝐽𝑢 =
= 36064776 𝑚𝑚3 = 36.0648 × 10−3 𝑚3
6
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
18
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 19
Calculations
Results
Shear force per unit length on x-axis (fs_x), y-axis (fs_y) and magnitude (fs):
𝑓𝑠_𝑥 =
𝑉𝑥
10290
=−
= 13.7826 𝑁. 𝑚𝑚−1
𝐴𝑢_𝑥
746.592
𝑓𝑠_𝑦 =
𝑉𝑦
17150
=
= −58.0321 𝑁. 𝑚𝑚−1
𝐴𝑢_𝑦 295.526
2 + 𝑓 2 = 59.6464 𝑁. 𝑚𝑚 −1
𝑓𝑠 = √𝑓𝑠_𝑥
𝑠_𝑦
Bending force per unit length on brackets A (fb_a) and B (fb_b):
Note that x and y indicates the extreme half of the flange width and depth of
section respectively. Subscripts ba_a indicates the bending force per unit
length of Bracket A and point A:
𝑓𝑏𝑎_𝑎 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= −59.6520 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
For points B, C, D, E and F:
𝑓𝑏𝑎_𝑏 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= −83.0845 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑎_𝑐 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= −106.5169 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑎_𝑑 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= 106.5169 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑎_𝑒 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= 83.0845 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑎_𝑓 = −
𝑀𝑎𝑥 𝑦 𝑀𝑎𝑦 𝑥
+
= 59.6520 𝑁. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
19
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 20
Calculations
Results
Similarly bracket B at points A, B, C, D, E & F:
𝑓𝑏𝑏_𝑎 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= −49.3745 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑏_𝑏 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= −68.7697 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑏_𝑐 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= −88.1649 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑏_𝑑 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= 88.1649 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑏_𝑒 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= 68.7697 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
𝑓𝑏𝑏_𝑓 = −
𝑀𝑏𝑥 𝑦 𝑀𝑏𝑦 𝑥
+
= 49.3745 N. 𝑚𝑚−1
𝐼𝑢𝑥
𝐼𝑢𝑦
Shear force per unit length due to torsion only applies to bracket A due to the
offset of each end. The torsion force per unit length x-direction at point A on
the negative x-axis can be computed as:
𝑓𝑗_𝑎𝑥 =
𝑇𝑥−
= −14.9080 N. 𝑚𝑚−1
𝐽𝑢
At point B it only has a vertical component hence only the y-direction is
considered.
𝑓𝑗_𝑦 =
𝑇𝑦
= 37.6622 N. 𝑚𝑚−1
𝐽𝑢
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
20
J.Lam
MECH3110
𝑓𝑗_𝑐𝑥 =
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 21
Calculations
Results
𝑇𝑥
= 14.9080 N. 𝑚𝑚−1
𝐽𝑢
Shear force per unit length due to torsion in x-direction at point C. Point C is
located on the positive x-axis.
Magnitude of the shear stress due to torsion at point A and point C (square of
a number cancels out negative values) is calculated as:
2
2
𝑓𝑗_𝑎 = 𝑓𝑗_𝑐 = √𝑓𝑗_𝑎𝑥
+ 𝑓𝑗_𝑦
= 40.5054 N. 𝑚𝑚−1
The magnitudes of the torsion force acting per unit length on each point can
be tabulated as below:
Point
A
B
C
D
E
F
Torsion Force Per Unit
Length (N/mm)
40.505447
37.662233
40.505447
40.505447
37.662233
40.505447
The angle between the two torsion force per unit length components can be
computed using the following expression, at point A as an example:
𝑓𝑗_𝑦
𝜃𝐴 = tan−1 (
) = −68.4047°
𝑓𝑗_𝑎𝑥
𝜃𝐶 = tan−1 (
𝑓𝑗_𝑦
) = 68.4047°
𝑓𝑗_𝑐𝑥
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
21
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 22
Calculations
Results
Once again the angle between the two torsion force per unit length
components can be tabulated as below:
Point
A
B
C
D
E
F
Angle (degrees)
-68.404689
90
68.404689
68.404689
-90
-68.404689
A rough sketch helping visualize the magnitudes of the force per unit length
vectors at the three points.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
22
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 23
Calculations
Results
And thus the forces per unit length at point A, B and C can be computed via
vector summation at each point for bracket A by summing shear stresses due
to torsion and shear in the horizontal direction, shear stresses due to torsion
and shear in the vertical direction and finally bending stress.
2
2
2
𝑓𝑎 = √(𝑓𝑗_𝑎𝑥 + 𝑓𝑠_𝑥 ) + (𝑓𝑗𝑦 + 𝑓𝑠_𝑦 ) + 𝑓𝑏𝑎_𝑎
= 116.3569 N. 𝑚𝑚−1
2
2
2
𝑓𝑏 = √(𝑓𝑠_𝑥 ) + (𝑓𝑗𝑦 + 𝑓𝑠_𝑦 ) + 𝑓𝑏𝑎_𝑏
= 127.4770 N. 𝑚𝑚−1
2
𝑓𝑐 = √(𝑓𝑗_𝑐𝑥 + 𝑓𝑠_𝑥 ) + (𝑓𝑗𝑦 + 𝑓𝑠_𝑦 ) + 𝑓𝑏𝑎_𝑐
= 143.1940 N. 𝑚𝑚−1
Critical Stress on
Bracket A:
The data can then be tabulated for the vector summation of forces per unit
length for bracket A as:
Point C at 143.1940
N/mm
2
Point
A
B
C
D
E
F
2
Magnitude of force per unit length
(N/mm)
116.3569
127.4770
143.1940
112.1781
86.6483
63.0441
The critical stress is the point of highest stress. From this the point of highest
stress is at point C with a value of 143.1940 N/mm
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
23
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 24
Calculations
Results
Similarly the vector summation can be applied to bracket B. However torsion
is not present on bracket B and only shear and bending force per unit length
applies and the formula can be applied as:
2
2
2
𝑓𝑎 = √(𝑓𝑠_𝑥 ) + (𝑓𝑠_𝑦 ) + 𝑓𝑏𝑏_𝑎
= 169.6278 N. 𝑚𝑚−1
The data can then be tabulated for the vector summation of forces per unit
length for bracket B as:
Point
Magnitude of force per unit length
(N/mm)
77.4308
91.0327
106.4459
106.4459
91.0327
77.4308
A
B
C
D
E
F
Critical Stress on
Bracket B:
Points C and D at
106.4459 N/mm
The critical stress is the point of highest stress. From this the point of highest
stress is shared with point C and D with a value of 106.4459 N/mm. This
shows the stress distribution is symmetric about the diagonal.
For both bracket A and bracket B, the critical stresses do not exceed the
maximum stress calculated earlier.
The formulas for parallel or transverse loading based on the distortion energy
theory are given respectively asiv:
𝑓𝑝 =
𝑓𝑡 =
𝑆𝑤 ℎ
3√2𝑛
𝑆𝑢 ℎ
3 × 1.21𝑛
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
24
J.Lam
MECH3110
Title of project: Vibrating Screen Support – Welding Assignment
2010
Section of design: General welding calculations
Page: 25
Calculations
Results
Part of the loading is in parallel hence the parallel loading formula is used.
Rearranging the formulas above for the required weld size gives:
ℎ=
ℎ=
3√2𝑓𝑝 𝑛
𝑆𝑢
3 × 1.21𝑓𝑡 𝑛
𝑆𝑢
Using fp = 143.1940 N/mm for Bracket A and fp = 106.4459 N/mm for
Bracket B respectively will yield the following values.
Required weld sizes:
Use 3 mm for
Bracket A
Use 2 mm for
Bracket B
Bracket A
h = 2.1137 mm
According to AS1554.1-2000, use 3 mm weld size.
Bracket B
h = 1.5712 mm
According to AS1554.1-2000, use 2 mm weld size.
The weld sizes are appropriately small enough to use. Further iterations are
not necessary.
Although not shown here, further iterations using MATLAB show that the
required weld sizes for brackets A and B will always converge to values less
than 2 mm for each iteration using the previous average of the two weld sizes
from both brackets.
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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Discussion
For the purposes of safe design, a worst case scenario was taken, where both the weight force
and inertial force acting simultaneously are taken into account.
Free body diagrams were designed for both x-y & x-z due to the weight force and an inertial
force. The 530 UB 82 was chosen for being the beam with the smallest flange width that
could accommodate the flat plates while not letting the size and self-weight of a beam
become a contributing factor and taking up unnecessary space where the vibrating screen unit
is the be placed.
The initial weld pattern involved welding around the entire flange width however the
disadvantages far outweigh the advantages including the difficult calculations involved and
the difficulty of the actual welding and then was modified to be two simple horizontal weld
lines on the very bottom and top of each beam simplifying the calculations and weld
application itself with almost no change in strength.
The force per unit length due to shear, torsion and bending was analysed at the extreme
points on the beam for possible critical stresses. Due to the complexity of the loading, the
unsymmetrical bending stress formula had to be implemented and also multiple critical
stresses were found. There were two equal critical stresses on bracket B which did not have a
torsion component disrupting the symmetric distribution of stresses acting on the bracket.
Although the weld size conceptually is too small, multiple iterations have shown the weld
size converges to small values even while modifying variables such as factor of safety and
ultimate strength.
The rapid computational capability of MATLAB allows results to be computed immediately
while changing desired variables. The tables below show how the weld size changes with
certain variables:
Ultimate Strength (MPa)
Weld size for bracket A (mm) Weld size for bracket B (mm)
250
4.0582
3.0168
300
3.3819
2.5140
350
2.8987
2.1548
400
2.5364
1.8855
450
2.2546
1.6760
500
2.0291
1.5084
Table showing how weld sizes for each bracket A and B changes with different ultimate
strengths while retaining other variables as constant.
Factor of Safety
1
2
3
4
5
6
Weld size for bracket A (mm)
1.2657
2.5313
3.7970
5.0627
6.3283
7.5940
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
Weld size for bracket B (mm)
0.9409
1.8817
2.8226
3.7634
4.7043
5.6451
26
7
8.8597
6.5860
8
10.1253
7.5269
Table showing how weld sizes for each bracket A and B changes with different factors of
safety while retaining other variables as constant.
Iteration Number
Average weld size
Weld size for bracket Weld size for bracket
(mm)
A (mm)
B (mm)
1
1
2.1137
1.5712
2
1.8424
1.8778
1.3863
3
1.6321
1.9103
1.4088
4
1.6595
1.9055
1.4054
5
1.6554
1.9062
1.4059
6
1.6560
1.9061
1.4058
7
1.6560
1.9061
1.4058
8
1.6560
1.9061
1.4058
9
1.6560
1.9061
1.4058
10
1.6560
1.9061
1.4058
Table showing how the required weld size changes with the weld size from each previous
evaluation. Only several iterations are required to converge to the same values regardless of
the number of extra iterations.
Reaching the optimal weld size in the mid-range of available weld-sizes specified by the
AS1554.1-2000 Australian Standard can be achieved by increasing the factor of safety,
however increasing the factor of safety is only necessary in situations with greater uncertainty,
complexity or danger.
Conclusion
A free body diagram for each bracket in the x-y & x-z planes has been constructed to
determine the maximum bending moment and other crucial loadings. The yield stress and
ultimate stress has been determined from a catalogue as well as Grade AS/NZS 3679.1-350
material for the beam and a universal beam designation of 530 UB 82. Based on the
calculated combination of force per unit length due to shear, bending as well as torsion acting
on the extreme points on each bracket, critical stresses have been found located at one corner
on bracket A and two corners on bracket B. Although the weld size may seem small, costs
will be lower and the weld size can be easily modified by changing variables such as factor of
safety can be increased if necessary. Based on the specifications, an engineering drawing is
provided in the appendix.
References
R. Budynas, J. K. Nisbett, Shigley’s Mechanical Engineering Design, Eighth Edition,
McGraw-Hill, 2008, Table 9-4, page 472
i
Jason Lam – z3252911 – MECH3110 – Welded Joint Assignment
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R. Budynas, J. K. Nisbett, Shigley’s Mechanical Engineering Design, Eighth Edition,
McGraw-Hill, 2008, Tables 9-1 and 9-2, pages 466 and 470
iii
R. C. Hibbeler, Mechanics of Materials, Seventh Edition, Pearson, 2008, Equations 6-17,
page 323
iv
S.Kanapathipillai, WeldingTwo.pdf, Lecture Notes, 2010
ii
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