Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
=3.335 *10^-4
ε(avg) =150 *10^-6
orientation of γmax
θ = 31.71 or -58.29
To determine the direction of γmax
= 1.67 *10^-4
Answer:
specialist focus in one very specific thing, generalist focus and many things, they are like a jack of all trades
Answer:
Explanation:
given data
Load P = 35 kN
Width of bar W = 50.8 mm
Breadth of bar B = 25 mm
Ratio of crack length to width α = a/W = 0.2
solution
we get here KI for a rectangular bar that is express as
................................1
here Y is the geometrical function
so
Y =
Y =
Y =
Y = 0.9878
so put here value in equation 1
= 5210.45 × 10³
= 5.21 MPa