Answer:
minimum length of a surface crack is 15.043 mm
Explanation:
given data
strain fracture toughness K = 78 MPa
tensile stress = 345 MPa
Y = 1.04
to find out
minimum length of a surface crack
solution
we find here length of critical interior flaw from formula that is
α =
....................1
put here value we get
α = 
α = 15.043 mm
so minimum length of a surface crack is 15.043 mm
Answer:
center left-turn lane
Explanation:
A <em>center left turn lane</em> will be marked as described. The arrows, if present, generally indicate that left turns are permitted from the lane with these markings.
__
If the double yellow lines are solid, they are considered to be a "barrier" and are not to be crossed.
Answer: r = 0.8081; s = -0.07071
Explanation:
A = (150i + 270j) mm
B = (300i - 450j) mm
C = (-100i - 250j) mm
R = rA + sB + C = 0i + 0j
R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j
R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j
Equating the i and j components;
150r + 300s - 100 = 0
270r - 450s - 250 = 0
150r + 300s = 100
270r - 450s = 250
solving simultaneously,
r = 0.8081 and s = -0.07071
QED!
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :


J = 207.6 
So the shear stress at point A is :



Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
H is the answer
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