The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation. For A992 steel, G = 11 × 103 ksi. (1) Determine the angle of twist of B with respect to D.(2) Determine the angle of twist of C with respect to D. Answer unit: degree or radians, two decimal places

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

a 1 inch nut

Explanation: