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Outline of solution to Final 1999

Problem No.1

The equation for the gel/space ratio X, is given. However, the volume of entrained air, Vair, is not known. You therefore need to use the following equation as well.

From the above equation, we find: Vair = 0.08361 + 0.05042 w/c
Substituting into the equation for X, and set X = 0.72, = 0.8
w/c is found to be 0.394 or approximately, 0.4.

Problem No.2

NOTE: By mistake, I put the draft version of the exam question on the web. In the final version, we change the steel volume from 3% to 0.3%. The calculation, is, however, very similar.

Cement with continuous fiber reinforcement can be treated as a parallel system.
Stress in cement (assumed uncracked)
= (composite stress/composite modulus)x(cement modulus)
Composite modulus = 0.003x210 + 0.997x20 = 20.57 GPa
Stress in cement = 6/20.57*20 = 5.8338 MPa > 5 MPa
Cement has therefore already cracked
All the loading is hence carried by the fiber.
Area of fiber = 25(0.1)2
Area of composite = (99.7/0.3) x area of steel
Load carried by the specimen = 6 x (99.7/0.3) x 25(0.1)2

Problem No.3

This problem has been discussed in the tutorial on steel.


Problem No.4
(a)
(i)
In this case, sec(a/W) is very close to 1.0




Kc = 50 ( 0.0005)1/2 = 1.982 x 10-6 Nm-1.5

(ii)
Maximum allowable crack size to avoid brittle fracture

aT = (1/) (Kc/y)2 = 0.125 mm


(b)
Application of Paris law gives the following:

1/ai C 1/af = A24Nf = B Nf
af = 2 ai when Nf =10,000
1/ai = 20,000 B
When af = 4 ai , B Nf = . ai , implying Nf = 15,000
Additional number of cycles to failure = 5,000

(c)
Paris law only applies to high cycle fatigue. In low cycle fatigue (i.e., a low number of cycles to failure), a significant portion of the material has gone into yielding. Linear elastic fracture mechanics then no longer holds.




Problem No.5 (Note: is set to be 5000 GPa.s in the final version of the paper)
(i)
For applied stress of 100 MPa, the spring will NEVER yield. The system is therefore exactly the same as the Kelvin-Voigt system discussed in class.

Initially, at time = 0, all applied stress is carried by the dashpot,
slope of strain vs time = 100/ = 100/5,000,000 = 2 x 10-5/s


For the Kelvin-Voigt model, final strain rate = 0
Schematic plot of strain vs time is the same as that for the parallel model given
in the notes.
(ii) For applied stress = 150 MPa, initial slope = 150/ = 3 x 10-5/s
Since 150 MPa is larger than the yield strength of the spring (120 MPa), the spring will eventually go into yielding when the system starts to deform. After the spring yields, it carries a constant stress of 120 MPa. The remaining 30 MPa acts on the dashpot. The final slope of the strain vs time curve is then
30/ = 6 x 10-6/s
Schematic plot of behavior:


Strain



Time