Answer:
Check the explanation
Explanation:
Given
1) CU traixial compression test
2) Devatoric stress at failure = бd = 50 kN/
3) Confining pressure at failure = бd = 48 kN/
4) Pore pressure at failure = u = 18 kN/
5) Unconfined compression stress = q = 20 kN/
6) Undrained cohesion = q/2 = 20 kN/
To find:
1) Effective and total stress strength failure envelope
Kindly check the attached image below .
Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m
Answer:
% reduction in area==PR=0.734=73.4%
% elongation=EL=0.42=42%
Explanation:
given do=12.8 mm
df=6.60
Lf=72.4 mm
Lo=50.8 mm
% reduction in area=((*(do/2)^2)-(
*(df/2)^2)))/
*(do/2)^2
substitute values
% reduction in area=73.4%
% elongation=EL=((Lf-Lo)/Lo))*100
% elongation=((72.4-50. 8)/50.8)*100=42%
Answer:
the life (N) of the specimen is 46400 cycles
Explanation:
given data
ultimate strength Su = 1600 MPa
stress amplitude σa = 900 MPa
to find out
life (N) of the specimen
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 1600
Se = 800 Mpa
and we know
Se for steel is 700 Mpa for Su ≥ 1400 Mpa
so we take endurance limit Se is = 700 Mpa
and strength of friction f = 0.77 for 232 ksi
because for Se 0.5 Su at cycle = (1600 × 0.145 ksi ) = 232
so here coefficient value (a) will be
a =
a =
a = 2168.3 Mpa
so
coefficient value (b) will be
a = -log
b = -log
b = -0.0818
so no of cycle N is
N =
put here value
N =
N = 46400
the life (N) of the specimen is 46400 cycles