MMAN3110 – MECHANICAL DESIGN 1
ASSIGNMENT – WELDING
31ST AUGUST 2011
Z3253384
THOMAS BEROUKAS
1
X15 Experimental Aircraft Launch Support Structure
Introduction
The B52 bomber aircraft is home to the experimental X15 rocket aircraft. This smaller size
craft will be mounted to the underside of the wing from the B52 bomber. This design requires
a support structure consisting of cantilevered arms which will be welded to an existing
connector block attached to hard points under the wing. At the end of each cantilever beam
there will be lugs welded to the selected material for a cable to be attached at each point. The
report following will select each the three cantilever beams and all welds required in
supporting the beam to the connector block or their associated lugs.
Figure 1: Top & side view of the connector block, cantilever beam layout as provided in the
design brief.
Design Brief
The report requires the selection of three beams as located in the views above as well as
designing six individual welds to satisfy the loads, which can be identified at the ends of the
cantilever beams. In particular, loads exerted onto the weld sections require in depth
calculations and iterations. A free body diagram for all beams in the x-y planes will be
constructed in turn to determine the bending moment diagrams and shear force diagrams and
other critical loadings. All beams will be selected from a provided “One Steel” catalogue
which allows for the identification of yield and ultimate stresses. Further calculations will be
made to clarify all weld specifications including forces such as bending, shear and torsion
forces. An engineering drawing will be produced based on the specifications.
2
Calculations
Beam 1
Single front universal I-beam at front of connector block to hold the nose of the aircraft
Free body diagram for beam one
𝐹𝑥 = 10,000 sin(20) = 3420 𝑁
𝐹𝑦 = 10,000 cos(20) = 9396.93 𝑁
∑ 𝐹𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = 0
𝐹𝑦 = 𝑅𝑦 = 9396.93 𝑁
Shear Force Diagram
Bending Moment Diagram
𝑀𝑎𝑡 𝐵𝑙𝑜𝑐𝑘 𝑊𝑒𝑙𝑑 = (𝐹)(𝐿) = (9396.93)(2050) = 19,263,706.5 𝑁. 𝑚𝑚
𝑀𝑎𝑡 𝐿𝑢𝑔 𝑊𝑒𝑙𝑑 = (𝐹)(𝐿) = (9396.93)(50) = 469,846.5 𝑁. 𝑚𝑚
3
Beam 2 & 3
Two rear universal i-beams to support the wings of the aircraft, both sets of calculations are identical
although the beam is mirrored across the rear of the connector block
Free body diagram for beam two and three (mirrored)
𝐹𝑦 = 15,000 𝑁
∑ 𝐹𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = 0
𝐹𝑦 = 𝑅𝑦 = 15,000 𝑁
Shear Force Diagram
Bending Moment Diagram
𝑀𝑎𝑡 𝐵𝑙𝑜𝑐𝑘 𝑊𝑒𝑙𝑑 = (𝐹)(𝐿) = (15,000)(1850) = 27,750,000 𝑁. 𝑚𝑚
𝑀𝑎𝑡 𝐿𝑢𝑔 𝑊𝑒𝑙𝑑 = (𝐹)(𝐿) = (15,000)(50) = 750,000 𝑁. 𝑚𝑚
4
Factor of Safety
Stresses allowed within the AISC Code for Weld Metal include a weld safety factor of n =
1.67. This factor of safety will be used universally for the entirety of the design.
Material for beam & Allowable Stress
"Grade 350" structural steel has been selected for the major connector block component and
therefore the materials selected will be of similar quality to the existing material. Within the
Australia Standard “AS1554.1-2000,” a weld is assumed to have the same or greater strength
then the original parent materials. The selection of Grade 350 steel from the “One Steel"
catalogue will be selected due to its ease of availability with an ultimate tensile strength of Su
= 480 MPa. The AWS Electrode number E70xx with a tensile strength of Sut = 480 MPa,
yield strength of Sy = 393 MPa and 22% elongation will be used for welding.
Beam Selection
Grade 350 steel has an ultimate strength of Su = 480 MPa and the use of an ultimate tensile
strength needs a factor of safety of at least three. Therefore
Beam 1
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)
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 360
=
= 216 𝑀𝑃𝑎
𝑛
1.67
Using largest value of maximum bending moment acting on beam in x-y plane where M = 19,263,707
N.mm, the required section modulus:
𝑍=
𝑀
𝜎𝑎𝑙𝑙𝑜𝑤
=
(19,263,706.5 N. mm)
= 89,183.82 𝑚𝑚3
(216 MPa)
𝑍 = 89.183 × 103 𝑚𝑚3
5
Beam 2 & 3
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)
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 360
=
= 216 𝑀𝑃𝑎
𝑛
1.67
Using largest value of maximum bending moment acting on beam in x-y plane where M = 19,263,707
N.mm, the required section modulus:
𝑍=
𝑀
𝜎𝑎𝑙𝑙𝑜𝑤
=
(27,750,000 N. mm)
= 128,472.22 𝑚𝑚3
(216 MPa)
𝑍 = 127.472 × 103 𝑚𝑚3
6
Beam Selections & Sections
Based on the design brief the required size of universal beams is chosen from the Hot Rolled
and Steel Structural Steel catalogue provided by “One Steel”.
Figure 4: diagram of the cross section of the universal beam from the one steel catalogue.
Using the calculated section modulus the 310 Universal Beam 40.4 have been chosen for all
beams in an attempt to minimize product costs in bulk if this practical design becomes mass
produced for a variety of aircraft.
Label / Variable
Beam Size
Weight per unit length
Flange Width (bf)
Web Thickness (tw)
Flange Thickness (tf)
Root Radius (r1)
Depth of Section (d)
Depth Between Flanges (d1)
Section Modulus x-axis (Zx)
Section Modulus y-axis (Zy)
Yield Stress at flange (σy(f))
Yield Stress at web (σy(w))
Value
310 UB 40.4
40.4
165
6.1
10.2
11.4
304
284
569×103
92.7×103
320
320
Units
kg.m-1
mm
mm
mm
mm
mm
mm
mm3
mm3
MPa
MPa
Table 1: Catalogue Information of 310 UB 40.4 universal beam.
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Weld Outline Shape
All joint welds, excluding the lugs will have welds across the top and bottom section of the
universal beams as this section is furthest from the centre of the beam, which has the highest
contribution to the moment of inertia value. The shear stress distribution acting across the
beam varies in a parabolic shape from zero at the extreme values to a maximum in the centre
of the web. The bending stress distribution changes linearly from zero in the centre to the
maximum at both extremities with a max tension or compression at each end. Shear stress
will be taken simply as the shear force over the total area of the cross-section of the weld.
Figure 5: Shear and bending stress distribution along the cross-section
Since the universal beam 1 is perpendicular to the plane of the wall there will not be any
torsion in the material. Since there is no torsion acting on the beam the shear stress
contribution is much less and bending stress is much more dominant. Although in beam 2 &
3 there will be a torsion component at the connector block welds as the beam is attached at a
45˚ in the x-plane.
Figure 7: Sketch of the weld pattern for all block welds.
As mentioned earlier the weld pattern will be created at the most distant points from the
centre. This allows for an easy accessible weld in regards to the manufacturer, which will
further reflect a lower price in construction.
Lug Design
The common lug, which will be used on all three beams, is a pre fabricated product which
will be purchased by a third party who specialized in the production of fittings and
connections. The design which can be viewed in the final CAD assembly drawing is a base
plate with the dimensions of 320x180x16mm. The beam will be welded to the rear flat face,
and the lug will be pre-casted to the front face, centralized to remove any torsion. The total
distance from the centre of the hole to the rear of the back plate is 50mm which will be used
for calculation purposes
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Weld calculations
The calculations for the forces acting on the welds per unit length are show below. The
substitution of values and subsequent calculations has been performed using pen and paper
calculations, only results and formulas will be shown. The calculations provided below are
for the final iteration, there were preliminary designs which failed, therefore highlighting the
importance of iteration in my results.
Beam 1Connector Block Weld
Fy = Vy = −9396.93 N
MA = 19,263,706.5 N.mm
The area of throat weld is taken where the initial weld length is at unity:
Au = (2)(b) = (2)(165) = 330 mm2
Unit Second Moment of Area of x-axis
𝐼𝑢 =
𝑏𝑑2
2
=
(165)(304)2
= 7,624,320 𝑚𝑚3
2
Shear force per unit length magnitude (fs):
𝑓𝑠 =
𝑉𝑦 −9396.93
=
= −28.47 𝑁. 𝑚𝑚−1
𝐴𝑢
330
Bending force per unit length magnitude (fb):
𝑓𝑏 =
𝑀𝑎 𝑦 (19,263,706.5)(152)
=
= 384.045 𝑁. 𝑚𝑚−1
𝐼𝑢
7,624,320
Vector addition of the shear force and bending force per unit length (ft):
𝑓𝑡𝑜𝑡𝑎𝑙 = √(−28.47)2 + (384.045)2 = 385.098 𝑁. 𝑚𝑚−1
The formulas for transverse loading based on the distortion energy theory due to no torsion
component in this weld:
𝑓𝑡 =
𝑆𝑢 ℎ
3 × 1.21𝑛
Rearranging the formula above for the required weld size gives:
ℎ=
3 × 1.21𝑓𝑡 𝑛 3 × 1.21 × 385.099 × 1.67
=
= 4.86 𝑚𝑚
𝑆𝑢
480
Therefore ℎ = 5 𝑚𝑚
9
Beam 2&3 Connector Block Welds
Fy = Vy = -15,000 N
MA = 27,750,000 N.mm
The area of throat weld is taken where the initial weld length is at unity:
Au = (2)(b) = (2)(165) = 330 mm2
Unit Second Moment of Area of x-axis
𝑏𝑑2
(165)(304)2
𝐼𝑢 =
=
= 7,624,320 𝑚𝑚3
2
2
Shear force per unit length magnitude (fs):
𝑓𝑠 =
𝑉𝑦 −15,000
=
= −45.45 𝑁. 𝑚𝑚−1
𝐴𝑢
330
Bending force per unit length magnitude (fb):
𝑓𝑏 =
𝑀𝑎 𝑦 (27,750,000)(152)
=
= 553.230 𝑁. 𝑚𝑚−1
𝐼𝑢
7,624,320
Torsion force per unit length magnitue (ft):
𝑓𝑡 =
𝑇𝑟
𝐽𝑤
In order to solve this equation we need to define T, Jw and r:
T = F × r = (15,000)(1272.79) = 19,091,850 N. mm*
*r has been calculated through general trigonometric calculations
𝐽𝑤 =
𝑑(3𝑏 2 + 𝑑2 )
304(3(1652 ) + (3042 ))
=
= 8,820,610.667 𝑚𝑚3
6
6
Breaking the beam into extreme points please refer to the diagram below
10
@ Point A
𝑟𝑎 = √(82.5)2 + (152)2 = 172.94 𝑚𝑚
𝑓𝑡𝑎 =
𝑇𝑟 (19,091,850)(172.94)
=
= 374.322 𝑁. 𝑚𝑚−1
𝐽𝑤
8,820,610.667
@ Point B
𝑟𝑏 = √(𝑜)2 + (152)2 = 152 𝑚𝑚
𝑓𝑡𝑏 =
𝑇𝑟 (19,091,850)(152)
=
= 328.998 𝑁. 𝑚𝑚−1
𝐽𝑤
8,820,610.667
@ Point C
𝑟𝑐 = √(82.5)2 + (152)2 = 172.94 𝑚𝑚
𝑓𝑡𝑐 =
𝑇𝑟 (19,091,850)(172.94)
=
= 374.322 𝑁. 𝑚𝑚−1
𝐽𝑤
8,820,610.667
Vector Addition for force totals at each point (ftotal):
@ Point A
𝑓𝑡𝑦 ↑= 264.68 𝑁. 𝑚𝑚−1
𝑓𝑡𝑥 →= 264.68 𝑁. 𝑚𝑚−1
𝑓𝑠 ↓= 45.45 𝑁. 𝑚𝑚−1
𝑓𝑏 ∅ = 328.998 𝑁. 𝑚𝑚−1
𝑓𝑡𝑜𝑡𝑎 = √(𝑓𝑡𝑦 − 𝑓𝑠 )2 + (𝑓𝑡𝑥 )2 + (𝑓𝑏 )2 = 475.66 𝑁. 𝑚𝑚−1
@Point C (Which conveniently is the critical point after interation)
𝑓𝑡𝑦 ↓= 264.68 𝑁. 𝑚𝑚−1
𝑓𝑡𝑥 ←= 264.68 𝑁. 𝑚𝑚−1
𝑓𝑠 ↓= 45.45 𝑁. 𝑚𝑚−1
𝑓𝑏 ∅ = 328.998 𝑁. 𝑚𝑚−1
𝑓𝑡𝑜𝑡𝑐 = √(𝑓𝑡𝑦 + 𝑓𝑠 )2 + (𝑓𝑡𝑥 )2 + (𝑓𝑏 )2 = 523.904 𝑁. 𝑚𝑚−1
Therefore due to a torsional component the formula for parallel loading based on the distortion
energy theory:
𝑓𝑡𝑜𝑡𝑐 = 𝑓𝑝 =
11
𝑆𝑤 ℎ
3√2𝑛
Rearranging the formulas above for the required weld size gives:
3√2𝑓𝑝 𝑛
3√2(523.904)(1.67)
=
= 7.73 𝑚𝑚
𝑆𝑢
480
ℎ=
Therefore ℎ = 8 𝑚𝑚
Beam 1 Lug Weld
Fy = Vy = −9396.93 N
MA = 469,846.5 N.mm
The area of throat weld is taken where the initial weld length is at unity:
Au = (2)(b) = (2)(165) = 330 mm2
Unit Second Moment of Area of x-axis
𝐼𝑢 =
𝑏𝑑2
2
=
(165)(304)2
= 7,624,320 𝑚𝑚3
2
Shear force per unit length magnitude (fs):
𝑓𝑠 =
𝑉𝑦 −9396.93
=
= −28.47 𝑁. 𝑚𝑚−1
𝐴𝑢
330
Bending force per unit length magnitude (fb):
𝑓𝑏 =
𝑀𝑎 𝑦 (469,846.5)(152)
=
= 9.367 𝑁. 𝑚𝑚−1
𝐼𝑢
7,624,320
Vector addition of the shear force and bending force per unit length (ft):
𝑓𝑡𝑜𝑡𝑎𝑙 = √(−28.47)2 + (9.367)2 = 29.97 𝑁. 𝑚𝑚−1
The formulas for transverse loading based on the distortion energy theory due to no torsion
component in this weld:
𝑓𝑡 =
𝑆𝑢 ℎ
3 × 1.21𝑛
Rearranging the formula above for the required weld size gives:
ℎ=
3 × 1.21𝑓𝑡 𝑛 3 × 1.21 × 29.97 × 1.67
=
= 0.379 𝑚𝑚
𝑆𝑢
480
Therefore ℎ = 1 𝑚𝑚
12
Beam 2&3 Lug Welds
Fy = Vy = −15,000 N
MA = 750,000 N.mm
The area of throat weld is taken where the initial weld length is at unity:
Au = (2)(b) = (2)(165) = 330 mm2
Unit Second Moment of Area of x-axis
𝑏𝑑2
(165)(304)2
𝐼𝑢 =
=
= 7,624,320 𝑚𝑚3
2
2
Shear force per unit length magnitude (fs):
𝑓𝑠 =
𝑉𝑦 −15,000
=
= −45.45 𝑁. 𝑚𝑚−1
𝐴𝑢
330
Bending force per unit length magnitude (fb):
𝑓𝑏 =
𝑀𝑎 𝑦 (750,000)(152)
=
= 14.952 𝑁. 𝑚𝑚−1
𝐼𝑢
7,624,320
Vector addition of the shear force and bending force per unit length (ft):
𝑓𝑡𝑜𝑡𝑎𝑙 = √(14.952)2 + (45.45)2 = 47.846 𝑁. 𝑚𝑚−1
The formulas for transverse loading based on the distortion energy theory due to no torsion
component in this weld:
𝑓𝑡 =
𝑆𝑢 ℎ
3 × 1.21𝑛
Rearranging the formula above for the required weld size gives:
ℎ=
3 × 1.21𝑓𝑡 𝑛 3 × 1.21 × 29.97 × 1.67
=
= 0.379 𝑚𝑚
𝑆𝑢
480
Therefore ℎ = 1 𝑚𝑚
13
Concluding Discussion
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. This was modified to the result displayed above including
two simple horizontal weld lines on the very bottom and top of each beam. This allowed for
simple calculations and weld application with minimal change in strength and design
integrity.
The force per unit length due to shear, torsion and bending was analysed at the extreme
points on the beam for possible critical stresses. There were two equal critical stresses on
each weld which did not have a torsion component disrupting the symmetric distribution of
stresses acting on the beam. Whilst the beams which consisted of a torsional component
could easily identify which location would home the critical values. This was done by vector
observation and analysis.
Although some of the weld sizes are conceptually 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.
14