Suppose your client wishes to purchase an annuity that pays $50,000 each year for 5 years, with the first payment 4 years from n
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Answer:
the angle of twist of B with respect to D is -1.15°
the angle of twist of C with respect to D is 1.15°
Explanation:
The missing diagram that is supposed to be added to this image is attached in the file below.
From the given information:
The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.
For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²
The objective are :
1) To determine the angle of twist of B with respect to D
Considering the Polar moment of Inertia at the shaft
shaft =
where ;
r = 1 in /2
r = 0.5 in
shaft =
shaft = 0.098218
Now; the angle of twist at B with respect to D is calculated by using the expression
where;
are the torques at segments CD and length at segments CD
are the torques at segments BC and length at segments BC
Also ; from the diagram; the following values where obtained:
= 2.5 in
= 0.098218
G = 11 × 10⁶ lb/in²
= -60 lb.ft
= 0 lb.ft
= 5.5 in
− 0.02 rad
To degree; we have
Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction
Thus; the angle of twist of B with respect to D is 1.15°
(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
For the angle of twist of C with respect to D; we have:
0.02 rad
To degree; we have
Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
