Answer:
A) 209.12 GPa
B) 105.41 GPa
Explanation:
We are given;
Modulus of elasticity of the metal; E_m = 67 GPa
Modulus of elasticity of the oxide; E_f = 390 GPa
Composition of oxide particles; V_f = 44% = 0.44
A) Formula for upper bound modulus of elasticity is given as;
E = E_m(1 - V_f) + (E_f × V_f)
Plugging in the relevant values gives;
E = (67(1 - 0.44)) + (390 × 0.44)
E = 209.12 GPa
B) Formula for upper bound modulus of elasticity is given as;
E = 1/[(V_f/E_f) + (1 - V_f)/E_m]
Plugging in the relevant values;
E = 1/((0.44/390) + ((1 - 0.44)/67))
E = 105.41 GPa
Answer:
load factor = 0.782
Shrink Factor = 0.833
no of truck is 62500
Explanation:
given data
soil weighs in situ condition = 2,520 lbs/CY
soil weighs in loose condition = 1,970 lb/CY
soil weighs in embanked state = 3,025 lbs/CY
average volume = 16 LCY
soil from a borrow pit = 1 million CCY
solution
first we get here Load Factor that is express as
load factor =
load factor = 0.782
and Shrink Factor will be as
Shrink Factor =
Shrink Factor = 0.833
and
no of truck will be
no of truck =
no of truck is 62500