1. MAR 2020
APPLICATIONS OF ENGINEERING
DESIGN MAKE AND TEST
BY
ILKIN SAKIN OGLU AGHAYEV,
ANIETIE SAMUEL UMANAH,
CAMERON SCOTT TRAN,
HARREE ANAND BABU S,
PETROS PAPAIOANNOU,
RENO DELANSA LO

2. ILKIN SAKIN OGLU AGHAYEV,
Advantage: Lightweight, Simple, & Minimum welding

3. ILKIN SAKIN OGLU AGHAYEV,

4. ANIETIE SAMUEL UMANAH

5. DESIGN PARTICULARS
Statically Determinate?
Lets find out
Equation => m + r = 2j
 M = 7, r = 3,j = 5
 => 7 + 3 = 2(5)
 => 10 = 10 (Statically Determinate)
Why did I choose to design this?
 Ease of Fabrication
 Managing cost of material
 Easy calculation of member forces
Length of Each member
 AB and CD= 230.94mm
 BE and CE = 282.15mm
 AE and DE = 393.40mm
 BC = 434.43mm
 Overall length of the structure =
760mm
ANIETIE SAMUEL UMANAH

6.  Ray = 245.25N,
 Rax = 0,
 Rby = 245.25N,
 Fab = 333.3N (compression)
 Fae = 172.7N (Tension)
 Fbe = 452.45N (Tension)
 Fbc = 515.06 (Compression)
 Fce = 452.45N(Tension)
 Fcd = 333.3N (compression)
 Fde = 172.7N(Tension)
Why did I choose that factor of Safety?
 Generally, Factor of safety is required to bring the structure
from the state of collapse to a usable state.
ANIETIE SAMUEL UMANAH

7. CAMERON SCOTT TRAN
1) +Warren Truss Design
2) +Simple Construction, Strong design
3) -Heavy, many members
4) -Blocks, load lift clearance

8. CAMERON SCOTT TRAN
Rudimentary Calculations:
M=2J-3 for stable truss
M= 11, J=7, so 11=14-3. Therefore truss is stable
M+R=2J
M=11, R=3 and J=7, so 11+3=14, therefore truss is statically determinate.

9. HARREE ANAND BABU S
40.8⁰
98.4⁰
65.9⁰
61.3⁰
52.8⁰
560mm
385mm
760mm
300mm

10. HARREE ANAND BABU S
Bridge Design Information Total Length of all members =
2740mm
 Height of the Bridge = 300mm
 No. of Members = 7
 Equation m + 3 = 2n
 m = 7, n = 5
 10 = 10 , Therefore it is statically
determined
Reasond for choosing this design:
 Fabrication of bridge will be easy
 Will be light and strong
 Forces acting on each member is reasonable

11. PETROS PAPAIOANNOU
Specifications:
 All Lengths are in metres
and angles in degrees.
 Horizontal member has a set
height of 0.4 m and length
of 0.44 m.
 The load is applied at a
height of 0.1 m.

12. PETROS PAPAIOANNOU
Why did I choose this design?
1. Easy to carry on with
calculations.
2. It contains a sense of
innovation:
 Trapezoid shape
 Load is not connected by
members to the supports.

13. RENO DELANSA LO
Consideration during design process
 Ease of manufacturing
 Inspired by traditional design
 Prevent buckling on certain
member
 Forces distributed in an ideal
proportion

14. FINAL DESIGN

15. FINAL DESIGN

16. FINAL DESIGN

17. 90mm
0.4908kN
Cross Section of the compression members
Cross Section of the tension members

18. 𝑦 =
𝐴𝑦
𝐴
=
261.1875
9.75
= 2.6858 𝑚𝑚
𝐼 𝑏𝑎𝑠𝑒 = 𝐴𝑦2
+ 𝐼 𝑜𝑤𝑛 = 131.2325 + 35.824 = 167.05 𝑚𝑚
𝐼 𝑁𝐴 = 𝐼 𝑏𝑎𝑠𝑒 − 𝐴 ∗ 𝑦2
= 167.05 − 9.75 ∗ 2.68582
= 96.724 𝑚𝑚4
𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝐺𝑦𝑟𝑎𝑡𝑖𝑜𝑛 𝑟 =
𝐼
𝐴
=
96.724
9.75
= 3.14 𝑚𝑚
𝑀𝑎𝑥 𝐹 = 444𝑁 𝑤𝑖𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 200𝑚𝑚 𝑎𝑛𝑑 𝐿𝑒 = 100𝑚𝑚
𝝀 =
𝐿𝑒
𝜋 ∗ 𝑟
𝜎 𝑦
𝐸
=
100
𝜋 ∗ 3.14
282
210000
= 𝟎. 𝟑𝟕
𝝈 𝒄𝒓 =
𝜎 𝑦
1 + 𝜆2
=
282
1 + 0.372
= 248 𝑁/𝑚𝑚2
𝑴𝒂𝒙 𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝑭𝒐𝒓𝒄𝒆 = 𝟒𝟒𝟒𝑵 𝑤𝑖𝑡ℎ 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝟏. 𝟑
𝑃 = (max 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 ∗ 𝑆𝐹) = 577.2𝑁
𝐴 = 9.75 𝑚𝑚2
𝜎𝑐 =
577.3
9.75
= 59.2
𝑁
𝑚𝑚2
𝑺𝑨𝑭𝑬
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝑭𝒐𝒓𝒄𝒆 = 𝟐𝟕𝟐 𝑵 𝒘𝒊𝒕𝒉 𝒔𝒂𝒇𝒆𝒕𝒚 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝟏. 𝟑
𝑃 = 353.6𝑁 𝑎𝑛𝑑 𝐴 = 5.0 𝑚𝑚2
𝜎𝑐 =
353.6
5.0
= 70.72
𝑁
𝑚𝑚2
𝑺𝑨𝑭𝑬
Scantling Area y Ay Ay2 Iown
1 9.5*0.5 4.75 5.25 24.9375 130.92 35.72
2 10*0.5 5 0.25 1.25 0.3125 0.104
9.75 26.1875 131.2325 35.824
With the density of 7580kg/m3, the group prediction of the bridge is about 350 grams.

19. MAR 2020
APPLICATIONS OF ENGINEERING
DESIGN MAKE AND TEST