Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = 
......................1
put here value and we get
critical stress = 
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Answer:
The answer is as given in the explanation.
Explanation:
The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.
- Flow is entirely laminar, there's no boundary layer release.
- Flow is streamlined, ie, it follows the geometrical path imposed by the curvature.
By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:
And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:
where P_∞ is Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.
Answer:
The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.
Explanation:
The net force over the window is calculated by multiplying the difference in pressure by the area of the window:
F = Δp*A
The pressure inside the plane is around 1 atm, hence the difference in pressure is:
Δp = 1atm - 0.35 atm = 0.65 atm
Expressing in the EE unit system:
Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2
Replacing in the force:
F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf
For the international unit system, we re-calculate the window's area and the difference in pressure:
A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2
Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2
Replacing in the force:
F = 65861 N/m^2 *0.0516 m^2 = 3398 N
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:
0.124
Explanation:
We calculate the hydraulic gradient by the formulas below.
I = (change in h)/(change in l)-----eqn 1
I = (hk-hl)/change in L ----- equation 2
At k the headloss = hk,
At L the headloss = hL
The distance of water travel is change in I
Total head at k
hk = 543+23
= 566 ft
Total head at L
hL = 461+74
= 535 ft
Change in L = 250
When we substitute these values in equation 2
566-535/250
= 0.124
The hydraulic gradient is 0.124
