Answer:
σ =5.39Mpa
Explanation:
step one:
The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces
Flexural strength test Flexural strength is calculated using the equation:
σ = FL/ (bd^2 )----------1
Where
σ = Flexural strength of concrete in Mpa
F= Failure load (in N).
L= Effective span of the beam
b= Breadth of the beam
step two:
Given data
F=40.45 kN= 40450N
b=0.15m
d=0.15m
L=0.45m
step three:
substituting into the expression we have
σ = 40450*0.45/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.0225 )
σ =18202.5/0.003375
σ =5393333.3
σ =5393333.3/1000000
σ =5.39Mpa
Therefore the flexure strength of the concrete is 5.39Mpa
Answer:
vacancy formation energy of Ni is 1.400 eV
Explanation:
given data
number of vacancies in Ni = 4.7 x
atomic weight = 58.69 g/mol
density = 8.8 g/cm³
solution
we get here N that is
N =
...........1
N = 
N =
and here no of vacancy will be
Nv =
.................2
put here value

take ln both side
Qv = 1.400 eV
so vacancy formation energy of Ni is 1.400 eV
Answer:
W = 112 lb
Explanation:
Given:
- δb = 0.025 in
- E = 29000 ksi (A-36)
- Area A_de = 0.002 in^2
Find:
Compute Weight W attached at C
Solution:
- Use proportion to determine δd:
δd/5 = δb/3
δd = (5/3) * 0.025
δd = 0.0417 in
- Compute εde i.e strain in DE:
εde = δd / Lde
εde = 0.0417 / 3*12
εde = 0.00116
- Compute stress in DE, σde:
σde = E*εde
σde = 29000*0.00116
σde = 33.56 ksi
- Compute the Force F_de:
F_de = σde *A_de
F_de = 33.56*0.002
F_de = 0.0672 kips
- Equilibrium conditions apply:
(M)_a = 0
3*W - 5*F_de = 0
W = (5/3)*F_de
W = (5/3)* 0.0672 = 112 lb