Answer:
a) C= 1/120
b) P(X>=5) = 0.333
Explanation:
The attached file contains the explanation for the answers
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.
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
The tension in the rope at the lowest point is 270 N
Explanation:
Given;
weight of the ball, W = 150 N
length of the rope, r = 4 m
velocity of the ball, v = 5.6 m/s
When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.
T = W + F
Centripetal force, F = mv²/r
where;
m is the mass of the ball
m = W/g
m = 150 / 9.8 = 15.306 kg
Centripetal force, F = mv²/r
F = (15.306 x 5.6²)/4
F = 120 N
T = W + F
T = 150 + 120
T = 270 N
Therefore, the tension in the rope at the lowest point is 270 N
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m