Material science

1. MATERIAL SCIENCE
COMPETITION DAY
Eng. Kareem. H. Mokhtar

2. Find the diameter of an 18 m long steel
wire that will stretch no more than .009
m when a load of 387 kg is hung at the
end of the wire. The Young’s Modulus for
steel wire is 200x109 N/m2 .
2 points

3. ■ Y=
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
delta 𝐿
𝐿𝑜
=
3800∗18
𝐴∗0.009
■ A= 3.8 x 10 -5 m2 = (Pi/4) D2
■ Dia = 0.007 m
Answer

4. Human bone has a Young’s Modulus of
14.5 x108 N/m2. Under compression, it
can withstand a stress of about 160 x 106
N/m2(stress) before breaking. Assuming
a human adult thigh bone (femur) is
about 0.5 m long, how much change in
length the bone withstand before
breaking.
2 Points

5. Answer
■ Y=
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
𝐿𝑜
=
160∗106
𝑑𝑒𝑙𝑡𝑎 𝐿
0.5
■ Delta L = 0.055 m

6. As shown in the binary phase diagram,
an alloy with a composition C0 = 48% is
cooled down from liquid to form solid. At
the temperature T, Cs =26%, CL = 72%,
the fraction
of the solid phase is
3 points

7. Answer
■ Solid fraction =
72−48
72−26
= 0.52 = 52%

8. A cylindrical rod has a diameter and
length of 10mm is to be deformed using a
tensile load of 27500N. It must NOT
experience either plastic deformation or
length change of more than 0.02 mm,
which one of these material will you
choose
4 points

9. Answer
■ Stress=
𝐹
𝐴
=
27500
𝑝𝑖
4
0.01 2
= 3.5 x 108 = 350 Mpa
■ First criteria: do not enter plastic region  only steel and titanium alloy
■ Second criteria:
■ For steel : Y=207x109 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
3.5 𝑥 108
𝑑𝑒𝑙𝑡𝑎 𝐿
0.01
 delta L = 0.0169 mm
■ For Ti: Y=107x109 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
3.5 𝑥 108
𝑑𝑒𝑙𝑡𝑎 𝐿
0.01
 delta L = 0.0327 mm

10. A cylindrical alloy is stressed in
compression by submerging it under
water, if its original and final diameters
are 20 and 19.75 mm respectively and its
original and final length are 7.5cm and 7
cm. compressibility of this alloy is 9x 10-
10 m2/N, find the depth at which the
alloy is submerged
3 Points

11. Answer
■ B=
1
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦
= 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑔 𝑥 ℎ
𝑑𝑒𝑙𝑡𝑎 𝑉
𝑉𝑜
=
1000𝑥 9,8𝑥 ℎ
delta V/ 𝑉𝑜
 h= 10558 m
■ Vo= 2.36x 10-5 m3
■ Vf= 2.14x 10-5 m3
■ deltaV= 2.2 x 10-6 m3

12. If we have Cu-20% Ni, Cu, and Ni and
we want to produce 100Kg of alloy
that has 40% Cu-Ni, find 2 different
ways to produce the alloy and pick
the best one knowing that
1 Kg of Ni = 25$
1 Kg of Cu= 20$
1 Kg of Cu-20%Ni= 23$
5 Points

13. Answer
■ To produce 100Kg of alloy Cu-60% Ni
■ First method (using Cu and Ni)
■ We need 40Kg Cu and 60Kg Ni
■ Second method (using Ni and Cu-20%Ni)
– Total amount of Ni = amount of Ni in (Cu-20%Ni) + amount of Ni
■ 60 = 0.2 x + (100-x)
■ X = 50KG
■ So we need 50Kg of alloy Cu-20% Ni and 50 kg of pure Ni
■ To pick one calculate the cost Cu and Ni have lower cost

14. ■ If Pb-Sn alloy that has a composition of 65% Pb has 2 phases alpha and Liquid,
■ Given that the amount of alpha phase is equal to the amount of liquid phase
■ What is the temperature at which the alloy is found.
■ 3 Points

15. ■ You need to produce a Cu-Ni alloy with the following
properties:
• yield strength of 20,000 psi or higher
• tensile strength not less than 60,000 psi
• % elongation of 20% or higher
Mention the available composition “range” for the alloy,
and choose the best option, and mention why?
5 marks