Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
Answer:
1790 μrad.
Explanation:
Young's modulus, E is given as 10000 ksi,
μ is given as 0.33,
Inside diameter, d = 54 in,
Thickness, t = 1 in,
Pressure, p = 794 psi = 0.794 ksi
To determine shear strain, longitudinal strain and circumferential strain will be evaluated,
Longitudinal strain, eL = (pd/4tE)(1 - 2μ)
eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)
eL = 3.64 x 10-⁴ radians
Circumferential strain , eH = (pd/4tE)(2-μ)
eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)
eH = 1.79 x 10-³ radians
The maximum shear strain is 1790 μrad.
Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
Answer:
True
Explanation:
For point in xz plane the stress tensor is given by
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane